<ul><li><p>CONTENTS: CASE STUDIES</p><p>CASE STUDY 1 Midsouth Chamber of Commerce (A): The Role of the Operating Manager inInformation Systems</p><p>CASE STUDY I-1 IMT Custom Machine Company, Inc.: Selection of an Information TechnologyPlatform</p><p>CASE STUDY I-2 VoIP2.biz, Inc.: Deciding on the Next Steps for a VoIP Supplier</p><p>CASE STUDY I-3 The VoIP Adoption at Butler University</p><p>CASE STUDY I-4 Supporting Mobile Health Clinics: The Childrens Health Fund of New York City</p><p>CASE STUDY I-5 Data Governance at InsuraCorp</p><p>CASE STUDY I-6 H.H. Greggs Appliances, Inc.: Deciding on a New Information TechnologyPlatform</p><p>CASE STUDY I-7 Midsouth Chamber of Commerce (B): Cleaning Up an Information SystemsDebacle</p><p>CASE STUDY II-1 Vendor-Managed Inventory at NIBCO</p><p>CASE STUDY II-2 Real-Time Business Intelligence at Continental Airlines</p><p>CASE STUDY II-3 Norfolk Southern Railway:The Business Intelligence Journey</p><p>CASE STUDY II-4 Mining Data to Increase State Tax Revenues in California</p><p>CASE STUDY II-5 The Cliptomania Web Store: An E-Tailing Start-up Survival Story</p><p>CASE STUDY II-6 Rock Island Chocolate Company, Inc.: Building a Social Networking Strategy</p></li><li><p>CASE STUDY III-1 Managing a Systems Development Project at Consumer and IndustrialProducts, Inc.</p><p>CASE STUDY III-2 A Make-or-Buy Decision at Baxter Manufacturing Company</p><p>CASE STUDY III-3 ERP Purchase Decision at Benton Manufacturing Company, Inc.</p><p>CASE STUDY III-4 The Kuali Financial System: An Open-Source Project</p><p>CASE STUDY III-5 NIBCOs Big Bang: An SAP Implementation</p><p>CASE STUDY III-6 BAT Taiwan: Implementing SAP for a Strategic Transition</p><p>CASE STUDY III-7 A Troubled Project at Modern Materials, Inc.</p><p>CASE STUDY III-8 Purchasing and Implementing a Student Management System at JeffersonCounty School System</p><p>CASE STUDY IV-1 The Clarion School for Boys, Inc. Milwaukee Division: Making InformationSystems Investments</p><p>CASE STUDY IV-2 FastTrack IT Integration for the Sallie Mae Merger</p><p>CASE STUDY IV-3 IT Infrastructure Outsourcing at Schaeffer (A):The Outsourcing Decision</p><p>CASE STUDY IV-4 IT Infrastructure Outsourcing at Schaeffer (B): Managing the Contract</p><p>CASE STUDY IV-5 Systems Support for a New Baxter Manufacturing Company Plant in Mexico</p><p>CASE STUDY IV-6 The Challenges of Local System Design for Multinationals: The MaxFli SalesForce Automation System at BAT</p><p>CASE STUDY IV-7 Meridian Hospital Systems, Inc.: Deciding Which IT Company to Join</p><p>CASE STUDY IV-8 Mary Morrisons Ethical Issue</p></li><li><p>Managing Information Technology</p><p>Prentice HallBoston Columbus Indianapolis New York San Francisco Upper Saddle River</p><p>Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal TorontoDelhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo</p><p>Seventh Edition</p><p>Carol V. Brown</p><p>Howe School of Technology Management, Stevens Institute of Technology</p><p>Daniel W. DeHayes</p><p>Kelley School of Business, Indiana University</p><p>Jeffrey A. Hoffer</p><p>School of Business Administration, The University of Dayton</p><p>E. Wainright Martin</p><p>Kelley School of Business, Indiana University</p><p>William C. Perkins</p><p>Kelley School of Business, Indiana University</p></li><li><p>Editorial Director: Sally YaganEditor in Chief: Eric SvendsenAVP/Executive Editor: Bob HoranEditorial Project Manager: Mary Kate MurrayEditorial Assistant: Jason CalcanoDirector of Marketing: Patrice Lumumba JonesSenior Marketing Manager: Anne FahlgrenProduction Manager: Debbie RyanArt Director: Jayne ConteCover Designer: Bruce KenselaarPhoto Researcher: Karen SanatarManager, Rights and Permissions: Hessa AlbaderCover Art: FotoliaMedia Editor: Denise VaughnMedia Project Manager: Lisa RinaldiFull-Service Project Management: Integra Software Services Pvt. Ltd.Printer/Binder: Edwards BrothersCover Printer: Leghigh-Phoenex ColorText Font: 10/12, Times</p><p>Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbookappear on appropriate page within the text.</p><p>Microsoft and Windows are registered trademarks of the Microsoft Corporation in the U.S.A. and othercountries. Screen shots and icons reprinted with permission from the Microsoft Corporation. This book is notsponsored or endorsed by or affiliated with the Microsoft Corporation.</p><p>Copyright 2012, 2009, 2005, 2002, 1999 Pearson Education, Inc., publishing as Prentice Hall, One LakeStreet, Upper Saddle River, New Jersey 07458. All rights reserved. Manufactured in the United States ofAmerica. This publication is protected by Copyright, and permission should be obtained from the publisher prior toany prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, pleasesubmit a written request to Pearson Education, Inc., Permissions Department, One Lake Street, Upper Saddle River,New Jersey 07458.</p><p>Many of the designations by manufacturers and sellers to distinguish their products are claimed as trademarks. Wherethose designations appear in this book, and the publisher was aware of a trademark claim, the designations have beenprinted in initial caps or all caps.</p><p>Library of Congress Cataloging-in-Publication DataManaging information technology / Carol V. Brown . . . [et al.]. 7th ed.</p><p>p. cm.Includes bibliographical references and index.ISBN-13: 978-0-13-214632-6 (alk. paper)ISBN-10: 0-13-214632-0 (alk. paper)1. Management information systems. I. Brown, Carol V. (Carol Vanderbilt), 1945- T58.6.M3568 2012658.4'038011dc22</p><p>2010048598</p><p>10 9 8 7 6 5 4 3 2 1</p><p>ISBN 10: 0-13-214632-0ISBN 13: 978-0-13-214632-6</p></li><li><p>BRIEF CONTENTS</p><p>Chapter 1 Managing IT in a Digital World 1</p><p>PART I Information Technology 17</p><p>Chapter 2 Computer Systems 19Chapter 3 Telecommunications and Networking 60Chapter 4 The Data Resource 95</p><p>PART II Applying Information Technology 187</p><p>Chapter 5 Enterprise Systems 189Chapter 6 Managerial Support Systems 223Chapter 7 E-Business Systems 253</p><p>PART III Acquiring Information Systems 327</p><p>Chapter 8 Basic Systems Concepts and Tools 329Chapter 9 Methodologies for Custom Software Development 361Chapter 10 Methodologies for Purchased Software Packages 390Chapter 11 IT Project Management 410</p><p>PART IV The Information Management System 517</p><p>Chapter 12 Planning Information Systems Resources 519Chapter 13 Leading the Information Systems Function 536Chapter 14 Information Security 561Chapter 15 Social, Ethical, and Legal Issues 575</p><p>Glossary 668Index 691</p><p>iii</p></li><li><p>This page intentionally left blank </p></li><li><p>CONTENTS</p><p>Preface xvii</p><p>Chapter 1 Managing IT in a Digital World 1Recent Information Technology Trends 2</p><p>Computer Hardware: Faster, Cheaper, Mobile 2</p><p>Computer Software: Integrated, Downloadable, Social 2</p><p>Computer Networks: High Bandwidth, Wireless, Cloudy 4</p><p>New Ways to Compete 4</p><p>New Ways to Work 5</p><p>Managing IT in Organizations 5Managing IT Resources 5</p><p>IT Leadership Roles 7</p><p>The Topics and Organization of This Textbook 8Review Questions 9 Discussion Questions 9 Bibliography 9</p><p> CASE STUDY 1 Midsouth Chamber of Commerce (A): The Role of theOperating Manager in Information Systems 10</p><p>PART I Information Technology 17</p><p>Chapter 2 Computer Systems 19Basic Components of Computer Systems 20</p><p>Underlying Structure 20</p><p>Input/Output 20</p><p>Computer Memory 21</p><p>Arithmetic/Logical Unit 23</p><p>Computer Files 23</p><p>Control Unit 25</p><p>The Stored-Program Concept 25</p><p>Types of Computer Systems 28Microcomputers 29</p><p>Midrange Systems 30</p><p>Mainframe Computers 33</p><p>Supercomputers 34</p><p>Key Types of Software 34</p><p>Applications Software 36An Example of an Application Product 37</p><p>Personal Productivity Software 38</p><p>Support Software 41The Operating System 41</p><p>Language Translators 43</p><p>v</p></li><li><p>Third Generation Languages 43</p><p>Fourth Generation Languages 46</p><p>Markup Languages 48</p><p>Object-Oriented Programming 49</p><p>Languages for Developing Web Applications 51</p><p>Database Management Systems 52</p><p>CASE Tools 54</p><p>Communications Interface Software 54</p><p>Utility Programs 54</p><p>The Changing Nature of Software 55</p><p>The Information Technology Industry 55Review Questions 56 Discussion Questions 57 Bibliography 58</p><p>Chapter 3 Telecommunications and Networking 60The Need for Networking 61</p><p>Sharing of Technology Resources 61</p><p>Sharing of Data 61</p><p>Distributed Data Processing and Client/Server Systems 62</p><p>Enhanced Communications 62</p><p>Marketing Outreach 62</p><p>An Overview of Telecommunications and Networking 62</p><p>Key Elements of Telecommunications and Networking 63Analog and Digital Signals 63</p><p>Speed of Transmission 64</p><p>Types of Transmission Lines 65</p><p>Transmission Media 65</p><p>Topology of Networks 70</p><p>Types of Networks 72</p><p>Network Protocols 86</p><p>The Exploding Role of Telecommunications and Networking 88Online Operations 88</p><p>Connectivity 89</p><p>Electronic Data Interchange and Electronic Commerce 89</p><p>Marketing 89</p><p>The Telecommunications Industry 90Review Questions 92 Discussion Questions 92 Bibliography 93</p><p>Chapter 4 The Data Resource 95Why Manage Data? 96</p><p>Technical Aspects of Managing the Data Resource 97The Data Model and Metadata 97</p><p>Data Modeling 98</p><p>Database Programming 100</p><p>vi Contents</p></li><li><p>Managerial Issues in Managing Data 101Principles in Managing Data 101</p><p>The Data Management Process 106</p><p>Data Management Policies 110Review Questions 114 Discussion Questions 114 Bibliography 114</p><p> CASE STUDY I-1 IMT Custom Machine Company, Inc.: Selection of an Information Technology Platform 116</p><p> CASE STUDY I-2 VoIP2.biz, Inc.: Deciding on the Next Steps for a VoIP Supplier 128</p><p> CASE STUDY I-3 The VoIP Adoption at Butler University 144 CASE STUDY I-4 Supporting Mobile Health Clinics: The Childrens Health</p><p>Fund of New York City 157 CASE STUDY I-5 Data Governance at InsuraCorp 166 CASE STUDY I-6 HH Gregg: Deciding on a New Information Technology</p><p>Platform 170 CASE STUDY I-7 Midsouth Chamber of Commerce (B): Cleaning up an</p><p>Information Systems Debacle 177</p><p>PART II Applying Information Technology 187</p><p>Chapter 5 Enterprise Systems 189Application Areas 189</p><p>Critical Concepts 191Batch Processing versus Online Processing 191</p><p>Functional Information Systems 192</p><p>Vertical Integration of Systems 192</p><p>Distributed Systems 192</p><p>Client/Server Systems 193</p><p>Virtualization 194</p><p>Service-Oriented Architecture and Web Services 194</p><p>Transaction Processing Systems 196Payroll System 196</p><p>Order Entry System 196</p><p>Enterprise Resource Planning Systems 198An Example ERP System: SAP ERP 199</p><p>Data Warehousing 201</p><p>Customer Relationship Management Systems 204</p><p>Office Automation 206Videoconferencing 207</p><p>Electronic Mail 208</p><p>Groupware and Collaboration 209An Example Groupware System: Lotus Notes 210</p><p>Contents vii</p></li><li><p>Intranets and Portals 213</p><p>Factory Automation 215Engineering Systems 216</p><p>Manufacturing Administration 216</p><p>Factory Operations 217</p><p>Robotics 217</p><p>Supply Chain Management Systems 217Review Questions 219 Discussion Questions 220 Bibliography 220</p><p>Chapter 6 Managerial Support Systems 223Decision Support Systems 223</p><p>Data Mining 224</p><p>Group Support Systems 228</p><p>Geographic Information Systems 229Business Adopts Geographic Technologies 230</p><p>Whats Behind Geographic Technologies 231</p><p>Issues for Information Systems Organizations 232</p><p>Executive Information Systems/Business Intelligence Systems 234</p><p>Knowledge Management Systems 237Two Recent KMS Initiatives within a Pharmaceutical Firm 239</p><p>KMS Success 240</p><p>Artificial Intelligence 241</p><p>Expert Systems 241Obtaining an Expert System 242</p><p>Examples of Expert Systems 242</p><p>Neural Networks 244</p><p>Virtual Reality 245Review Questions 250 Discussion Questions 250 Bibliography 251</p><p>Chapter 7 E-Business Systems 253Brief History of the Internet 254</p><p>E-Business Technologies 254</p><p>Legal and Regulatory Environment 257</p><p>Strategic E-Business Opportunities (and Threats) 259</p><p>B2B Applications 260</p><p>B2C Applications 263Two Dot-Com Retailers 264</p><p>Two Traditional Catalog Retailers 266</p><p>Two Traditional Store Retailers 267</p><p>Summary: B2C Retailing 268</p><p>viii Contents</p></li><li><p>Dot-Com Intermediaries 269Summary: Successful Online Intermediary Models 273</p><p>Special Issue: What Makes a Good Web Site for Consumers 273</p><p>Special Issue: What Makes a Good B2C Social Media Platform 275</p><p>Review Questions 276 Discussion Questions 276 Bibliography 277</p><p> CASE STUDY II-1 Vendor-Managed Inventory at NIBCO 279</p><p> CASE STUDY II-2 Real-Time Business Intelligence at Continental Airlines 284</p><p> CASE STUDY II-3 Norfolk Southern Railway: The Business IntelligenceJourney 294</p><p> CASE STUDY II-4 Mining Data To Increase State Tax Revenues in California 300</p><p> CASE STUDY II-5 The CliptomaniaTM Web Store 308</p><p> CASE STUDY II-6 Rock Island Chocolate Company, Inc.: Building a SocialNetworking Strategy 321</p><p>PART III Acquiring Information Systems 327</p><p>Chapter 8 Basic Systems Concepts and Tools 329The Systems View 329</p><p>What Is a System? 330</p><p>Seven Key System Elements 330</p><p>Organizations as Systems 334</p><p>Systems Analysis and Design 335</p><p>Business Processes 336Identifying Business Processes 336</p><p>Business Process Redesign 336</p><p>Processes and Techniques to Develop Information Systems 339</p><p>The Information Systems Development Life Cycle 339</p><p>Structured Techniques for Life-Cycle Development 340</p><p>Procedural-Oriented Techniques 341</p><p>Techniques for the As-Is Model 343</p><p>Techniques for the Logical To-Be Model 344</p><p>Techniques for Documenting the Physical To-Be System 348</p><p>Object-Oriented Techniques 351</p><p>Core Object-Oriented Concepts 351</p><p>Summary of Processes and Techniques to Develop Information Systems 353</p><p>Contents ix</p></li><li><p>Information Systems Controls to Minimize Business Risks 354Types of Control Mechanisms 355</p><p>Controls in the Definition and Construction Phases 355</p><p>Controls in the Implementation Phase 357Review Questions 358 Discussion Questions 359 Bibliography 359</p><p>Chapter 9 Methodologies for Custom Software Development 361Systems Development Life Cycle Methodology 361</p><p>The SDLC Steps 362</p><p>Initiating New Systems Projects 363</p><p>Definition Phase 363</p><p>Construction Phase 365</p><p>Implementation Phase 366</p><p>The SDLC Project Team 370</p><p>Managing an SDLC Project 371</p><p>SDLC Advantages and Disadvantages 371</p><p>Prototyping Methodology 373The Prototyping Steps 373</p><p>The Prototyping Project Team 375</p><p>Managing a Prototyping Project 375</p><p>Prototyping Advantages and Disadvantages 375</p><p>Prototyping Within an SDLC Process 376</p><p>Newer Approaches 377Rapid Application Development (RAD) 377</p><p>Agile Methodologies 378</p><p>Managing Software Projects Using Outsourced Staff 381</p><p>Supporting User Application Development (UAD) 382Advantages and Disadvantages of User-Developed Applications 382</p><p>Assessing the Risks from UAD 384</p><p>Guidelines for User Developers 385Review Questions 387 Discussion Questions 387 Bibliography 388</p><p>Chapter 10 Methodologies for Purchased Software Packages 390The Make-or-Buy Decision 391</p><p>Purchasing Methodology 391The Purchasing Steps 392</p><p>Project Team for Purchasing Packages 400</p><p>Managing a Purchased System Project 401</p><p>Purchasing Advantages and Disadvantages 402</p><p>Special Case: Enterprise System Packages 403</p><p>Open Source Software 405</p><p>x Contents</p></li><li><p>New Purchasing Option: Application Service Providers (ASPs) 406</p><p>Review Questions 408 Discussion Questions 408 Bibliography 409</p><p>Chapter 11 IT Project Management 410IT Portfolio Management 411</p><p>Project Management Roles 412Project Manager 412</p><p>Project Sponsor and Champion Roles 413</p><p>Project Initiation 415</p><p>Project Planning 416Scheduling 416</p><p>Budgetin..</p></li></ul>

INTRODUCTION to FINITE ELEMENT METHODS
Carlos A. Felippa Department of Aerospace Engineering Sciences and Center for Aerospace Structures University of Colorado Boulder, Colorado 80309-0429, USA Last updated Fall 2004
Material assembled from Lecture Notes for the course Introduction to Finite Elements Methods (ASEN 5007) offered from 1986 to date at the Aerospace Engineering Sciences Department of the University of Colorado at Boulder.
Preface This textbook presents an Introduction to the computer-based simulation of linear structures by the Finite Element Method (FEM). It assembles the “converged” lecture notes of Introduction to Finite Element Methods or IFEM. This is a core graduate course offered in the Department of Aerospace Engineering Sciences of the University of Colorado at Boulder. IFEM was first taught on the Fall Semester 1986 and has been repeated every year since. It is taken by both first-year graduate students as part of their M.S. or M.E. requirements, and by senior undergraduates as technical elective. Selected material in Chapters 1 through 3 is used to teach a twoweek introduction of Matrix Structural Analysis and Finite Element concepts to junior undergraduate students who are taking their first Mechanics of Materials course. Prerequisites for the graduate-level course are multivariate calculus, linear algebra, a basic knowledge of structural mechanics at the Mechanics of Materials level, and some familiarity with programming concepts learnt in undergraduate courses. The course originally used Fortran 77 as computer implementation language. This has been gradually changed to Mathematica since 1995. The changeover is now complete. No prior knowledge of Mathematica is required because that language, unlike Fortran or similar low-level programming languages, can be picked up while “going along.” Inasmuch as Mathematica supports both symbolic and numeric computation, as well as direct use of visualization tools, the use of the language is interspersed throughout the book. Book Objectives “In science there is only physics; all the rest is stamp collecting” (Lord Kelvin). The quote reflects the values of the mid-XIX century. Even now, at the dawn of the XXIth, progress and prestige in the natural sciences favors fundamental knowledge. By contrast, engineering knowledge consists of three components:1 1.
Conceptual knowledge: understanding the framework of the physical world.
2.
Operational knowledge: methods and strategies for formulating, analyzing and solving problems, or “which buttons to push.”
3.
Integral knowledge: the synthesis of conceptual and operational knowledge for technology development.
The language that connects conceptual and operational knowledge is mathematics, and in particular the use of mathematical models. Most engineering programs in the USA correctly emphasize both conceptual and operational components. They differ, however, in how well the two are integrated. The most successful curricula are those that address the tendency to “horizontal disconnection” that bedevils engineering students suddenly exposed to a vast array of subjects. Integral knowledge is unique to the engineering profession. Synthesis ability is a personal attribute that cannot be coerced, only encouraged and cultivated, the same as the best music programs do not 1
Extracted from: B. M. Argrow, Pro-active teaching and learning in the Aerospace Engineering Sciences Curriculum 2000, internal report, University of Colorado, February 2001.
i
automatically produce Mozarts. Studies indicate no correlation between good engineers and good students.2 The best that can be done is to provide an adequate (and integrated) base of conceptual and operational knowledge to potentially good engineers. Where does the Finite Element Method (FEM) fit in this framework? FEM was developed initially, and prospered, as a computer-based simulation method for the analysis of aerospace structures. Then it found its way into both design and analysis of complex structural systems, not only in Aerospace but in Civil and Mechanical Engineering. In the late 1960s it expanded to the simulation of non-structural problems in fluids, thermomechanics and electromagnetics. This “Physical FEM” is an operational tool, which fits primarily the operational knowledge component of engineering, and draws from the mathematical models of the real world. It is the form emphasized in the first part of this book. The success of FEM as a general-purpose simulation method attracted attention in the 1970s from two quarters beyond engineering: mathematicians and software entrepreneurs. The world of FEM eventually split into applications, mathematics, and commercial software products. The former two are largely housed in the comfortable obscurity of academia. There is little cross-talk between these communities. They have different perpectives. They have separate constituencies, conferences and publication media, which slows down technology transfer. As of this writing, the three-way split seems likely to continue, as long as there is no incentive to do otherwise. This book aims to keep a presentation balance: the physical and mathematical interpretations of FEM are used eclectically, with none overshadowing the other. Key steps of the computer implementation are presented in sufficient detail so that a student can understand what goes on behind the scenes of a “black box” commercial product. The goal is that students navigating this material can eventually feel comfortable with any of the three “FEM communities” they come in contact during their professional life, whether as engineers, managers, researchers or teachers. Book Organization The book is divided into four Parts. The first three are of roughly similar length. Part I: The Direct Stiffness Method. This part comprises Chapters 1 through 11. It covers major aspects of the Direct Stiffness Method (DSM). This is the most important realization of FEM, and the one implemented in general-purpose commercial finite element codes used by practicing engineers. Following a introductory first chapter, Chapters 2-4 present the fundamental steps of the DSM as a matrix method of structural analysis. A plane truss structure is used as motivating example. This is followed by Chapters 5-10 on programming, element formulation, modeling issues, and techniques for application of boundary conditions. Chapter 11 deals with relatively advanced topics including condensation and global-local analysis. Throughout these chapters the physical interpretation is emphasized for pedagogical convenience, as unifying vision of this “horizontal” framework. Part II: Formulation of Finite Elements. This part extends from Chapters 12 through 19. It is more focused than Part I. It covers the development of elements from the more general viewpoint of the variational (energy) formulation. The presentation is inductive, always focusing on specific elements and progressing from the simplest to more complex cases. Thus Chapter 12 rederives the 2
As evaluated by conventional academic metrics, which primarily test operational knowledge. One difficulty with teaching synthesis is that good engineers and designers are highly valued in industry but rarely comfortable in academia.
ii
plane truss (bar) element from a variational formulation, while Chapter 13 presents the plane beam element. Chapter 14 introduces the plane stress problem, which serves as a testbed for the derivation of two-dimensional isoparametric elements in Chapter 15 through 18. This part concludes with an overview of requirements for convergence. Part III: Computer Implementation. Chapters 20 through 29 deal with the computer implementation of the finite element method. Experience has indicated that students profit from doing computer homework early. This begins with Chapter 5, which contains an Introduction to Mathematica, and continues with homework assignments in Parts I and II. The emphasis changes in Part III to a systematic description of components of FEM programs, and the integration of those components to do problem solving. Part IV: Structural Dynamics. This part, which starts at Chapter 30, is under preparation. It is intended as a brief introduction to the use of FEM in structural dynamics and vibration analysis, and is by nature more advanced than the other Parts. Exercises Most Chapters are followed by a list of homework exercises that pose problems of varying difficulty. Each exercise is labeled by a tag of the form [type:rating] The type is indicated by letters A, C, D or N for exercises to be answered primarily by analytical work, computer programming, descriptive narration, and numerical calculations, respectively. Some exercises involve a combination of these traits, in which case a combination of letters separated by + is used; for example A+N indicates analytical derivation followed by numerical work. For some problems heavy analytical work may be helped by the use of a computer-algebra system, in which case the type is identified as A/C. The rating is a number between 5 and 50 that estimates the degree of difficulty of an Exercise, in the following “logarithmic” scale: 5
A simple question that can be answered in seconds, or is already answered in the text if the student has read and understood the material.
10
A straightforward question that can be answered in minutes.
15
A relatively simple question that requires some thinking, and may take on the order of half to one hour to answer.
20
Either a problem of moderate difficulty, or a straightforward one requiring lengthy computations or some programming, normally taking one to six hours of work.
25
A scaled up version of the above, estimated to require six hours to one day of work.
30
A problem of moderate difficulty that normally requires on the order of one or two days of work. Arriving at the answer may involve a combination of techniques, some background or reference material, or lenghty but straightforward programming.
40
A difficult problem that may be solvable only by gifted and well prepared individual students, or a team. Difficulties may be due to the need of correct formulation, advanced mathematics, or high level programming. With the proper preparation, background and tools these problems may be solved in days or weeks, while remaining inaccessible to unprepared or average students.
50
A research problem, worthy of publication if solved. iii
Most Exercises have a rating of 15 or 20. Assigning three or four per week puts a load of roughly 5-10 hours of solution work, plus the time needed to prepare the answer material. Assignments of difficulty 25 or 30 are better handled by groups, or given in take-home exams. Assignments of difficulty beyond 30 are never assigned in the course, but listed as a challenge for an elite group. Occasionally an Exercise has two or more distinct but related parts identified as items. In that case a rating may be given for each item. For example: [A/C:15+20]. This does not mean that the exercise as a whole has a difficulty of 35, because the scale is roughly logarithmic; the numbers simply rate the expected effort per item. Selecting Course Material The number of chapters has been coordinated with the 28 lectures and two midterm exams of a typical 15-week semester course offered with two 75-minute lectures per week. The expectation is to cover one chapter per lecture. Midterm exams cover selective material in Parts I and II, whereas a final exam covers the entire course. It is recommended to make this final exam a one-week take-home to facilitate computer programming assignments. Alternatively a final term project may be considered. The experience of the writer, however, is that term projects are not useful at this level, since most first-year graduate students lack the synthesis ability that develops in subsequent years. The writer has assigned weekly homeworks by selecting exercises from the two Chapters covered in the week. Choices are often given. The rating may be used by graders to weight scores. Unlike exams, group homeworks with teams of two to four students are recommended. Teams are encouraged to consult other students, as well as the instructor and teaching assistants, to get over gaps and hurdles. This group activity also lessen schedule conflicts common to working graduate students. Feedback from course offerings as well as advances in topics such as programming languages resulted in new material being incorporated at various intervals. To keep within course coverage constraints, three courses of action were followed in revising the book. Deleted Topics. All advanced analysis material dealing with variational calculus and direct approximation methods such as Rayleigh-Ritz, Galerkin, least squares and collocation, was eliminated by 1990. The few results needed for Part II are stated therein as recipes. That material was found to be largely a waste of time for engineering students, who typically lack the mathematical background required to appreciate the meaning and use of these methods in an application-independent context.3 Furthermore, there is abundant literature that interested students may consult should they decide to further their knowledge in those topics for self-study or thesis work. Appendices. “Refresher” material on vector and matrix algebra has been placed on Appendices A through D. This is material that students are supposed to know as a prerequisite. Although most of it is covered by a vast literature, it was felt advisable to help students in collecting key results for quick reference in one place, and establishing a consistent notational system. Starred Material. Chapter-specific material that is not normally covered in class is presented in fine print sections marked with an asterisk. This material belong to two categories. One is extension to the basic topics, which suggest the way it would be covered in a more advanced FEM course. The other includes general exposition or proofs of techniques presented as recipes in class for expedience or time constraints. Starred material may be used as source for term projects or take-home exams. 3
This is a manifestation of the disconnection difficulty noted at the start of this Preface.
iv
The book organization presents flexibility to instructors in organizing the coverage for shorter courses, for example in a quarter system, as well as fitting a three-lectures-per-week format. For the latter case it is recommended to cover two Chapters per week, while maintaining weekly homework assignments. In a quarter system a more drastic condensation would be necessary; for example much of Part I may be left out if the curriculum includes a separate course in Matrix Structural Analysis, as is common in Civil and Architectural Engineering. Acknowledgements Thanks are due to students and colleagues who have provided valuable feedback on the original course Notes, and helped its gradual metamorphosis into a textbook. Two invigorating sabbaticals in 1993 and 2001 provided blocks of time to develop, reformat and integrate material. The hospitality of Dr. P˚al G. Bergan of Det Norske Veritas at Oslo, Norway and Professor Eugenio O˜nate of CIMNE/UPC at Barcelona, Spain, during those sabbaticals is gratefully acknowledged.
v
vi
Chapter Contents Section 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 23 24 25 26 27 28 29 30 31
Overview . . . . . . . The Direct Stiffness Method: Breakdown . . The Direct Stiffness Method: Assembly and Solution The Direct Stiffness Method: Miscellaneous Topics Analysis of Example Truss by a CAS . . . Constructing MOM Members . . . . Finite Element Modeling: Introduction . . Finite Element Modeling: Mesh, Loads, BCs . . Multifreedom Constraints I . . . . Multifreedom Constraints II . . . . . Superelements and Global-Local Analysis . . The Bar Element . . . . . . . The Beam Element . . . . . . The Plane Stress Problem . . . . . The Linear Triangle . . . . . . The Isoparametric Representation . . . . Isoparametric Quadrilaterals . . . . Shape Function Magic . . . . . . FEM Convergence Requirements . . . (Moved to AFEM) . . . . . . Implementation of One-Dimensional Elements . FEM Programs for Plane Trusses and Frames . . Implementation of iso-P Quadrilateral Elements . Implementation of iso-P Triangular Elements . . The Assembly Procedure . . . . . FE Model Definition . . . . . . Solving FEM Equations . . . . . (under revision) . . . . . . . (under revision) . . . . . . Stress Recovery . . . . . . . (placeholder) . . . . . . . (under preparation) . . . . . . (under preparation) . . . . . .
.
.
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1-1 2-1 3-1 4-1 5-1 6-1 7-1 8-1 9-1 10-1 11-1 12-1 13-1 14-1 15-1 16-1 17-1 18-1 19-1 20-1 21-1 22-1 23-1 24-1 23-1 24-1 25-1 26-1 27-1 28-1 29-1 30-1 31-1
Matrix Algebra: Vectors . . . . . . . . . . . . . Matrix Algebra: Matrices . . . . . . . . . . . . . Matrix Algebra: Determinants, Inverses, Eigenvalues . . . . . . . . Matrix Calculus . . . . . . . . . . . . . . . History of Matrix Structural Analysis . . . . . . . . . . .
A-1 B-1 C-1 D-1 H-1
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Appendices A B C D H
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R References
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R-1
ix
1
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Overview
1–1
1–2
Chapter 1: OVERVIEW
TABLE OF CONTENTS Page
§1.1.
§1.2. §1.3.
§1.4.
§1.5. §1.6. §1.7.
§1. §1.
Where this Material Fits §1.1.1. Computational Mechanics . . . . . . §1.1.2. Statics vs. Dynamics . . . . . . . §1.1.3. Linear vs. Nonlinear . . . . . . . . §1.1.4. Discretization methods . . . . . . . §1.1.5. FEM Variants . . . . . . . . . . What Does a Finite Element Look Like? The FEM Analysis Process §1.3.1. The Physical FEM . . . . . . . . §1.3.2. The Mathematical FEM . . . . . . . §1.3.3. Synergy of Physical and Mathematical FEM Interpretations of the Finite Element Method §1.4.1. Physical Interpretation . . . . . . . §1.4.2. Mathematical Interpretation . . . . . Keeping the Course *What is Not Covered *Historical Sketch and Bibliography §1.7.1. Who Invented Finite Elements? . . . . §1.7.2. G1: The Pioneers . . . . . . . . §1.7.3. G2: The Golden Age . . . . . . . . §1.7.4. G3: Consolidation . . . . . . . . §1.7.5. G4: Back to Basics . . . . . . . . §1.7.6. Recommended Books for Linear FEM . §1.7.7. Hasta la Vista, Fortran . . . . . . . References. . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . .
1–2
. . . . . . . . . .
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1–3 1–3 1–4 1–4 1–4 1–5 1–5 1–7 1–7 1–8 1–9 1–10 1–11 1–11 1–12 1–12 1–13 1–13 1–13 1–14 1–14 1–14 1–15 1–15 1–16 1–17
1–3
§1.1
WHERE THIS MATERIAL FITS
This book is an introduction to the analysis of linear elastic structures by the Finite Element Method (FEM). This Chapter presents an overview of where the book fits, and what finite elements are. §1.1. Where this Material Fits The field of Mechanics can be subdivided into three major areas: 0002 Theoretical Mechanics
Applied Computational
(1.1)
Theoretical mechanics deals with fundamental laws and principles of mechanics studied for their intrinsic scientific value. Applied mechanics transfers this theoretical knowledge to scientific and engineering applications, especially as regards the construction of mathematical models of physical phenomena. Computational mechanics solves specific problems by simulation through numerical methods implemented on digital computers. Remark 1.1. Paraphrasing an old joke about mathematicians, one may define a computational mechanician
as a person who searches for solutions to given problems, an applied mechanician as a person who searches for problems that fit given solutions, and a theoretical mechanician as a person who can prove the existence of problems and solutions.
§1.1.1. Computational Mechanics Several branches of computational mechanics can be distinguished according to the physical scale of the focus of attention:  Nanomechanics and micromechanics   0002 Solids and Structures   Computational Mechanics Continuum mechanics Fluids   Multiphysics   Systems
(1.2)
Nanomechanics deals with phenomena at the molecular and atomic levels of matter. As such it is closely linked to particle physics and chemistry. Micromechanics looks primarily at the crystallographic and granular levels of matter. Its main technological application is the design and fabrication of materials and microdevices. Continuum mechanics studies bodies at the macroscopic level, using continuum models in which the microstructure is homogenized by phenomenological averages. The two traditional areas of application are solid and fluid mechanics. The former includes structures which, for obvious reasons, are fabricated with solids. Computational solid mechanics takes an applied sciences approach, whereas computational structural mechanics emphasizes technological applications to the analysis and design of structures. Computational fluid mechanics deals with problems that involve the equilibrium and motion of liquid and gases. Well developed subsidiaries are hydrodynamics, aerodynamics, acoustics, atmospheric physics, shock and combustion. 1–3
1–4
Chapter 1: OVERVIEW
Multiphysics is a more recent newcomer. This area is meant to include mechanical systems that transcend the classical boundaries of solid and fluid mechanics, as in interacting fluids and structures. Phase change problems such as ice melting and metal solidification fit into this category, as do the interaction of control, mechanical and electromagnetic systems. Finally, system identifies mechanical objects, whether natural or artificial, that perform a distinguishable function. Examples of man-made systems are airplanes, buildings, bridges, engines, cars, microchips, radio telescopes, robots, roller skates and garden sprinklers. Biological systems, such as a whale, amoeba, inner ear, or pine tree are included if studied from the viewpoint of biomechanics. Ecological, astronomical and cosmological entities also form systems.1 In the progression of (1.2) the system is the most general concept. A system is studied by decomposition: its behavior is that of its components plus the interaction between components. Components are broken down into subcomponents and so on. As this hierarchical breakdown process continues, individual components become simple enough to be treated by individual disciplines, but component interactions get more complex. Consequently there is a tradeoff art in deciding where to stop.2 §1.1.2. Statics vs. Dynamics Continuum mechanics problems may be subdivided according to whether inertial effects are taken into account or not: 0007 Statics Continuum mechanics (1.3) Dynamics In dynamics actual time dependence must be explicitly considered, because the calculation of inertial (and/or damping) forces requires derivatives respect to actual time to be taken. Problems in statics may also be time dependent but with inertial forces ignored or neglected. Accordingly static problems may be classed into strictly static and quasi-static. For the former time need not be considered explicitly; any historical time-like response-ordering parameter, if one is needed, will do. In quasi-static problems such as foundation settlement, metal creep, rate-dependent plasticity or fatigue cycling, a realistic measure of time is required but inertial forces are still neglected. §1.1.3. Linear vs. Nonlinear A classification of static problems that is particularly relevant to this book is Linear Statics Nonlinear
(1.4)
Linear static analysis deals with static problems in which the response is linear in the cause-andeffect sense. For example: if the applied forces are doubled, the displacements and internal stresses also double. Problems outside this domain are classified as nonlinear. 1
Except that their function may not be clear to us. “The usual approach of science of constructing a mathematical model cannot answer the questions of why there should be a universe for the model to describe. Why does the universe go to all the bother of existing? Is the unified theory so compelling that it brings about its own existence? Or does it need a creator, and, if so, does he have any other effect on the universe? And who created him?” (Stephen Hawking).
2
Thus in breaking down a car engine for engineering analysis, say, the decomposition does not usually proceed beyond the components you can buy at a parts shop.
1–4
1–5
§1.2
WHAT DOES A FINITE ELEMENT LOOK LIKE?
§1.1.4. Discretization methods A final classification of CSM static analysis is based on the discretization method by which the continuum mathematical model is discretized in space, i.e., converted to a discrete model with a finite number of degrees of freedom:  Finite Element (FEM)    Boundary Element (BEM)    Finite Difference (FDM) Spatial discretization method Finite Volume (FVM)       Spectral Meshfree
(1.5)
In CSM linear problems finite element methods currently dominate the scene as regards space discretization.3 Boundary element methods post a strong second choice in specific application areas. For nonlinear problems the dominance of finite element methods is overwhelming. Space finite difference methods in solid and structural mechanics have virtually disappeared from practical use. This statement is not true, however, for fluid mechanics, where finite difference discretization methods are still important. Finite-volume methods, which directly address the discretization of conservation laws, are important in difficult problems of fluid mechanics, for example high-Re gas dynamics. Spectral methods are based on transforms that map space and/or time dimensions to spaces (for example, the frequency domain) where the problem is easier to solve. A recent newcomer to the scene are the meshfree methods. These combine techniques and tools of finite element methods such as variational formulation and interpolation, with finite difference features such as non-local support. §1.1.5. FEM Variants The term Finite Element Method actually identifies a broad spectrum of techniques that share common features outlined in §1.3 and §1.4. Two subclassifications that fit well applications to structural mechanics are4  Displacement 0002  Stiffness  Equilibrium FEM Formulation FEM Solution Flexibility (1.6)   Mixed Mixed (a.k.a. Combined) Hybrid Using the foregoing classification, we can state the topic of this book more precisely: the computational analysis of linear static structural problems by the Finite Element Method. Of the variants listed in (1.6), emphasis is placed on the displacement formulation and stiffness solution. This combination is called the Direct Stiffness Method or DSM. 3
There are finite element discretizations in time, but they are not so widely used as finite differences.
4
The distinction between these subclasses require advanced technical concepts, which cannot be covered in an introductory treatment such as this book.
1–5
1–6
Chapter 1: OVERVIEW
(a)
(b)
(c)
3 2
4
4
2r sin(π/n)
r d
(d)
5
1
5
j
i 2π/n
r
8
6 7
Figure 1.1. The “find π” problem treated with FEM concepts: (a) continuum object, (b) a discrete approximation by inscribed regular polygons, (c) disconnected element, (d) generic element.
§1.2. What Does a Finite Element Look Like? The subject of this book is FEM. But what is a finite element? The concept will be partly illustrated through a truly ancient problem: find the perimeter L of a circle of diameter d. Since L = π d, this is equivalent to obtaining a numerical value for π. Draw a circle of radius r and diameter d = 2r as in Figure 1.1(a). Inscribe a regular polygon of n sides, where n = 8 in Figure 1.1(b). Rename polygon sides as elements and vertices as nodes. Label nodes with integers 1, . . . 8. Extract a typical element, say that joining nodes 4–5, as shown in Figure 1.1(c). This is an instance of the generic element i– j pictured in Figure 1.1(d). The element length is L i j = 2r sin(π/n). Since all elements have the same length, the polygon perimeter is L n = n L i j , whence the approximation to π is πn = L n /d = n sin(π/n). Table 1.1. Rectification of Circle by Inscribed Polygons (“Archimedes FEM”) n 1 2 4 8 16 32 64 128 256
πn = n sin(π/n) 0.000000000000000 2.000000000000000 2.828427124746190 3.061467458920718 3.121445152258052 3.136548490545939 3.140331156954753 3.141277250932773 3.141513801144301
Extrapolated by Wynn-
Exact π to 16 places
3.414213562373096 3.141418327933211 3.141592658918053 3.141592653589786
3.141592653589793
Values of πn obtained for n = 1, 2, 4, . . . 256 are listed in the second column of Table 1.1. As can be seen the convergence to π is fairly slow. However, the sequence can be transformed by Wynn’s algorithm5 into that shown in the third column. The last value displays 15-place accuracy. Some key ideas behind the FEM can be identified in this example. The circle, viewed as a source mathematical object, is replaced by polygons. These are discrete approximations to the circle. The sides, renamed as elements, are specified by their end nodes. Elements can be separated by 5
A widely used lozenge extrapolation algorithm that speeds up the convergence of many sequences. See, e.g, [180].
1–6
1–7
§1.3
THE FEM ANALYSIS PROCESS
disconnecting nodes, a process called disassembly in the FEM. Upon disassembly a generic element can be defined, independently of the original circle, by the segment that connects two nodes i and j. The relevant element property: side length L i j , can be computed in the generic element independently of the others, a property called local support in the FEM. The target property: the polygon perimeter, is obtained by reconnecting n elements and adding up their length; the corresponding steps in the FEM being assembly and solution, respectively. There is of course nothing magic about the circle; the same technique can be be used to rectify any smooth plane curve.6 This example has been offered in the FEM literature, e.g. in [111], to aduce that finite element ideas can be traced to Egyptian mathematicians from circa 1800 B.C., as well as Archimedes’ famous studies on circle rectification by 250 B.C. But comparison with the modern FEM, as covered in following Chapters, shows this to be a stretch. The example does not illustrate the concept of degrees of freedom, conjugate quantities and local-global coordinates. It is guilty of circular reasoning: the compact formula π = limn→∞ n sin(π/n) uses the unknown π in the right hand side.7 Reasonable people would argue that a circle is a simpler object than, say, a 128-sided polygon. Despite these flaws the example is useful in one respect: showing a fielder’s choice in the replacement of one mathematical object by another. This is at the root of the simulation process described below. §1.3. The FEM Analysis Process Processes using FEM involve carrying out a sequence of steps in some way. Those sequences take two canonical configurations, depending on (i) the environment in which FEM is used and (ii) the main objective: model-based simulation of physical systems, or numerical approximation to mathematical problems. Both are reviewed below to introduce terminology used in the sequel. §1.3.1. The Physical FEM
Ideal Mathematical model
A canonical use of FEM is simulation of physical systems. This must done by using models. Therefore the process is often called model-based simulation. The process is illustrated in Figure 1.2. The centerpiece is the physical system to be modeled. Accordingly, this configuration is called the Physical FEM. The processes of idealization and discretization are carried out concurrently to produce the discrete model. The solution step is handled by an equation solver often customized to FEM, which delivers a discrete solution (or solutions).
CONTINUIFICATION
FEM
Physical system
ocassionally relevant
SOLUTION
Discrete model
IDEALIZATION & DISCRETIZATION
Discrete solution VERIFICATION
solution error
simulation error: modeling & solution error VALIDATION
Figure 1.2. The Physical FEM. The physical system (left) is the source of the simulation process. The ideal mathematical model (should one go to the trouble of constructing it) is inessential.
6
A similar limit process, however, may fail in three or more dimensions.
7
This objection is bypassed if n is advanced as a power of two, as in Table 1.1, by using the half-angle recursion 1−
1 − sin2 2α, started from 2α = π for which sin π = −1.
1–7
√
2 sin α =
1–8
Chapter 1: OVERVIEW
Figure 1.2 also shows a ideal mathematical model. This may be presented as a continuum limit or “continuification” of the discrete model. For some physical systems, notably those well modeled by continuum fields, this step is useful. For others, such as complex engineering systems, it makes no sense. Indeed Physical FEM discretizations may be constructed and adjusted without reference to mathematical models, simply from experimental measurements. The concept of error arises in the Physical FEM in two ways. These are known as verification and validation, respectively. Verification is done by replacing the discrete solution into the discrete model to get the solution error. This error is not generally important. Substitution in the ideal mathematical model in principle provides the discretization error. This step is rarely useful in complex engineering systems, however, because there is no reason to expect that the mathematical model exists, and even if it does, that it is more physically relevant than the discrete model. Validation tries to compare the discrete solution against observation by computing the simulation error, which combines modeling and solution errors. As the latter is typically unimportant, the simulation error in practice can be identified with the modeling error. One way to adjust the discrete model so that it represents the physics better is called model updating. The discrete model is given free parameters. These are determined by comparing the discrete solution against experiments, as illustrated in Figure 1.3. Inasmuch as the minimization conditions are generally nonlinear (even if the model is linear) the updating process is inherently iterative.
Physical system
Experimental database
FEM
EXPERIMENTS
Parametrized discrete model
Discrete solution
simulation error
Figure 1.3. Model updating process in the Physical FEM.
§1.3.2. The Mathematical FEM The other canonical way of using FEM focuses on the mathematics. The process steps are illustrated in Figure 1.4. The spotlight now falls on the mathematical model. This is often an ordinary or partial differential equation in space and time. A discrete finite element model is generated from a variational or weak form of the mathematical model.8 This is the discretization step. The FEM equations are solved as indicated for the Physical FEM. On the left Figure 1.4 shows an ideal physical system. This may be presented as a realization of the mathematical model. Conversely, the mathematical model is said to be an idealization of this system. E.g., if the mathematical model is the Poisson’s PDE, realizations may be heat conduction or an electrostatic charge-distribution problem. This step is inessential and may be left out. Indeed Mathematical FEM discretizations may be constructed without any reference to physics. The concept of error arises when the discrete solution is substituted in the “model” boxes. This replacement is generically called verification. As in the Physical FEM, the solution error is the 8
The distinction between strong, weak and variational forms is discussed in advanced FEM courses. In the present book such forms will be largely stated (and used) as recipes.
1–8
1–9
§1.3 Mathematical model IDEALIZATION
THE FEM ANALYSIS PROCESS
Discretization & solution error
VERIFICATION
FEM
REALIZATION
SOLUTION
Ideal physical system
Discrete model IDEALIZATION & DISCRETIZATION
Discrete solution
VERIFICATION
solution error ocassionally relevant
Figure 1.4. The Mathematical FEM. The mathematical model (top) is the source of the simulation process. Discrete model and solution follow from it. The ideal physical system (should one go to the trouble of exhibiting it) is inessential.
amount by which the discrete solution fails to satisfy the discrete equations. This error is relatively unimportant when using computers, and in particular direct linear equation solvers, for the solution step. More relevant is the discretization error, which is the amount by which the discrete solution fails to satisfy the mathematical model.9 Replacing into the ideal physical system would in principle quantify modeling errors. In the Mathematical FEM this is largely irrelevant, however, because the ideal physical system is merely that: a figment of the imagination. §1.3.3. Synergy of Physical and Mathematical FEM The foregoing canonical sequences are not exclusive but complementary. This synergy10 is one of the reasons behind the power and acceptance of the method. Historically the Physical FEM was the first one to be developed to model complex physical systems such as aircraft, as narrated in §1.7. The Mathematical FEM came later and, among other things, provided the necessary theoretical underpinnings to extend FEM beyond structural analysis. A glance at the schematics of a commercial jet aircraft makes obvious the reasons behind the Physical FEM. There is no simple differential equation that captures, at a continuum mechanics level,11 the structure, avionics, fuel, propulsion, cargo, and passengers eating dinner. There is no reason for despair, however. The time honored divide and conquer strategy, coupled with abstraction, comes to the rescue. First, separate the structure out and view the rest as masses and forces, most of which are time-varying and nondeterministic. 9
This error can be computed in several ways, the details of which are of no importance here.
10
Such interplay is not exactly a new idea: “The men of experiment are like the ant, they only collect and use; the reasoners resemble spiders, who make cobwebs out of their own substance. But the bee takes the middle course: it gathers its material from the flowers of the garden and field, but transforms and digests it by a power of its own.” (Francis Bacon).
11
Of course at the (sub)atomic level quantum mechanics works for everything, from landing gears to passengers. But it would be slightly impractical to represent the aircraft by, say, 1036 interacting particles modeled by the Schr¨odinger equations. More seriously, Truesdell and Toupin correctly note that “Newtonian mechanics, while not appropriate to the corpuscles making up a body, agrees with experience when applied to the body as a whole, except for certain phenomena of astronomical scale” [162, p. 228].
1–9
1–10
Chapter 1: OVERVIEW
Second, consider the aircraft structure as built of substructures (a part of a structure devoted to a specific function): wings, fuselage, stabilizers, engines, landing gears, and so on. Take each substructure, and continue to break it down into components: rings, ribs, spars, cover plates, actuators, etc, continuing through as many levels as necessary.
al atic hem Matmodel
FEM
ary Libr
en pon e Com ret disc del o m
t
NT ONE P COMEVEL L ent ponns Com tio a u eq
TEM SYS EL V LE e plet Comution sol
em Syst ete
r disc del Eventually those components become suffimo l sica y h P tem ciently simple in geometry and connectivity sys that they can be reasonably well described by the continuum mathematical models provided, for instance, by Mechanics of Materials or the Theory of Elasticity. At that point, stop. Figure 1.5. Combining physical and mathematical modeling through multilevel FEM. Only two levels The component level discrete equations are (system and component) are shown for simplicity; obtained from a FEM library based on the intermediate substructure levels are omitted. mathematical model. The system model is obtained by going through the reverse process: from component equations to substructure equations, and from those to the equations of the complete aircraft.
This system assembly process is governed by the classical principles of Newtonian mechanics, which provide the necessary “component glue.” The multilevel decomposition process is diagramed in Figure 1.5, in which the intermediate level is omitted for simplicity. Remark 1.2. More intermediate decompo-
sition levels are used in systems such as offshore and ship structures, which are characterized by a modular fabrication process. In that case the multilevel decomposition mimics the way the system is actually fabricated. The general technique, called superelements, is discussed in Chapter 11.
Physical System
Remark 1.3. There is no point in practice in going beyond a certain component level while considering the complete system. The reason is that the level of detail can become overwhelming without adding relevant information. Usually that point is reached when uncertainty impedes further progress. Further refinement of specific components is done by the so-called global-local analysis technique outlined in Chapter 11. This technique is an instance of multiscale analysis.
Idealized and Discrete System
member support joint
;; ;;
;; ;;
IDEALIZATION
Figure 1.6. The idealization process for a simple structure. The physical system — here a roof truss — is directly idealized by the mathematical model: a pin-jointed bar assembly. For this particular structure idealized and discrete models coalesce.
For sufficiently simple structures, passing to a discrete model is carried out in a single idealization and discretization step, as illustrated for the truss roof structure shown in Figure 1.6. Other levels are unnecessary in such cases. Of course the truss may be viewed as a substructure of the roof, and the roof as a a substructure of a building. 1–10
1–11
§1.4
INTERPRETATIONS OF THE FINITE ELEMENT METHOD
§1.4. Interpretations of the Finite Element Method Just like there are two complementary ways of using the FEM, there are two complementary interpretations for teaching it. One stresses the physical significance and is aligned with the Physical FEM. The other focuses on the mathematical context, and is aligned with the Mathematical FEM. §1.4.1. Physical Interpretation The physical interpretation focuses on the flowchart of Figure 1.2. This interpretation has been shaped by the discovery and extensive use of the method in the field of structural mechanics. This historical connection is reflected in the use of structural terms such as “stiffness matrix”, “force vector” and “degrees of freedom.” This terminology carries over to non-structural applications. The basic concept in the physical interpretation is the breakdown (≡ disassembly, tearing, partition, separation, decomposition) of a complex mechanical system into simpler, disjoint components called finite elements, or simply elements. The mechanical response of an element is characterized in terms of a finite number of degrees of freedom. These degrees of freedoms are represented as the values of the unknown functions as a set of node points. The element response is defined by algebraic equations constructed from mathematical or experimental arguments. The response of the original system is considered to be approximated by that of the discrete model constructed by connecting or assembling the collection of all elements. The breakdown-assembly concept occurs naturally when an engineer considers many artificial and natural systems. For example, it is easy and natural to visualize an engine, bridge, aircraft or skeleton as being fabricated from simpler parts. As discussed in §1.3, the underlying theme is divide and conquer. If the behavior of a system is too complex, the recipe is to divide it into more manageable subsystems. If these subsystems are still too complex the subdivision process is continued until the behavior of each subsystem is simple enough to fit a mathematical model that represents well the knowledge level the analyst is interested in. In the finite element method such “primitive pieces” are called elements. The behavior of the total system is that of the individual elements plus their interaction. A key factor in the initial acceptance of the FEM was that the element interaction can be physically interpreted and understood in terms that were eminently familiar to structural engineers. §1.4.2. Mathematical Interpretation This interpretation is closely aligned with the flowchart of Figure 1.4. The FEM is viewed as a procedure for obtaining numerical approximations to the solution of boundary value problems (BVPs) posed over a domain . This domain is replaced by the union ∪ of disjoint subdomains (e) called finite elements. In general the geometry of is only approximated by that of ∪ (e) . The unknown function (or functions) is locally approximated over each element by an interpolation formula expressed in terms of values taken by the function(s), and possibly their derivatives, at a set of node points generally located on the element boundaries. The states of the assumed unknown function(s) determined by unit node values are called shape functions. The union of shape functions “patched” over adjacent elements form a trial function basis for which the node values represent the generalized coordinates. The trial function space may be inserted into the governing equations and the unknown node values determined by the Ritz method (if the solution extremizes a variational 1–11
1–12
Chapter 1: OVERVIEW
principle) or by the Galerkin, least-squares or other weighted-residual minimization methods if the problem cannot be expressed in a standard variational form. Remark 1.4. In the mathematical interpretation the emphasis is on the concept of local (piecewise) approximation. The concept of element-by-element breakdown and assembly, while convenient in the computer implementation, is not theoretically necessary. The mathematical interpretation permits a general approach to the questions of convergence, error bounds, trial and shape function requirements, etc., which the physical approach leaves unanswered. It also facilitates the application of FEM to classes of problems that are not so readily amenable to physical visualization as structures; for example electromagnetics and thermal conduction. Remark 1.5. It is interesting to note some similarities in the development of Heaviside’s operational methods, Dirac’s delta-function calculus, and the FEM. These three methods appeared as ad-hoc computational devices created by engineers and physicists to deal with problems posed by new science and technology (electricity, quantum mechanics, and delta-wing aircraft, respectively) with little help from the mathematical establishment. Only some time after the success of the new techniques became apparent were new branches of mathematics (operational calculus, distribution theory and piecewise-approximation theory, respectively) constructed to justify that success. In the case of the finite element method, the development of a formal mathematical theory started in the late 1960s, and much of it is still in the making.
§1.5. Keeping the Course The first Part of this book, which is the subject of Chapters 2 through 11, stresses the physical interpretation in the framework of the Direct Stiffness Method (DSM) on account of its instructional advantages. Furthermore the computer implementation becomes more transparent because the sequence of computer operations can be placed in close correspondence with the DSM steps. Subsequent Chapters incorporate ingredients of the mathematical interpretation when it is felt convenient to do so. Nonetheless the exposition avoids excessive entanglement with the mathematical theory when it may obfuscate the physics. In Chapters 2 and 3 the time is frozen at about 1965, and the DSM presented as an aerospace engineer of that time would have understood it. This is not done for sentimental reasons, although that happens to be the year in which the writer began thesis work on FEM under Ray Clough. Virtually all commercial codes are now based on the DSM and the computer implementation has not essentially changed since the late 1960s.12 What has greatly improved since is “marketing sugar”: user interaction and visualization. §1.6.
*What is Not Covered
The following topics are not covered in this book: 1. 2. 3. 4. 5. 6. 12
Elements based on equilibrium, mixed and hybrid variational formulations. Flexibility and mixed solution methods of solution. Kirchhoff-based plate and shell elements. Continuum-based plate and shell elements. Variational methods in mechanics. General mathematical theory of finite elements.
With the gradual disappearance of Fortran as a “live” programming language, noted in §1.7.7, changes at the computer implementation level have recently accelerated. For example C++ “wrappers” are becoming more common.
1–12
1–13 7. 8. 9. 10. 11. 12.
§1.7 *HISTORICAL SKETCH AND BIBLIOGRAPHY Buckling and stability analysis. General nonlinear response analysis. Structural optimization. Error estimates and problem-adaptive discretizations. Non-structural and multiphysics applications of FEM. Designing and building production-level FEM software and use of special hardware (e.g. vector and parallel computers)
Topics 1–6 belong to what may be called “Advanced Linear FEM”, whereas 7–8 pertain to “Nonlinear FEM”. Topics 9–11 fall into advanced applications, whereas 12 is an interdisciplinary topic that interweaves with computer science. §1.7.
*Historical Sketch and Bibliography
This section summarizes the history of structural finite elements since 1950 to date. It functions as a hub for dispersed historical references. For exposition convenience, structural “finitelementology” may be divided into four generations that span 10 to 15 years each. There are no sharp intergenerational breaks, but noticeable change of emphasis. The following summary does not cover the conjoint evolution of Matrix Structural Analysis into the Direct Stiffness Method from 1934 through 1970. This was the subject of a separate essay [56], which is also given in Appendix H. §1.7.1. Who Invented Finite Elements? Not just one individual, as the historical sketch will make clear. But if the question is tweaked to: who created the FEM in everyday use? there is no question in the writer’s mind: M. J. (Jon) Turner at Boeing over the period 1950–1962. He formulated and perfected the Direct Stiffness Method, and forcefully got Boeing to commit resources to it while other aerospace companies were enmeshed in the Force Method. He established and formulated the first continuum based finite elements. In addition to Turner, major contributors to current practice include: B. M. Irons, inventor of isoparametric models, shape functions, the patch test and frontal solvers; R. J. Melosh, who recognized the Rayleigh-Ritz link and systematized the variational derivation of stiffness elements; and E. L. Wilson, who developed the first open source (and widely imitated) FEM software. All of these pioneers were in the aerospace industry at least during part of their careers. That is not coincidence. FEM is the confluence of three ingredients, one of which is digital computation. And only large industrial companies (as well as some government agencies) were able to afford mainframe computers during the 1950s. Who were the popularizers? Four academicians: J. H. Argyris, R. W. Clough, H. C. Martin, and O. C. Zienkiewicz are largely responsible for the “technology transfer” from the aerospace industry to a wider range of engineering applications during the 1950s and 1960s. The first three learned the method from Turner directly or indirectly. As a consultant to Boeing in the early 1950s, Argyris, a Force Method expert then at Imperial College, received reports from Turner’s group, and weaved the material into his influencial 1954 serial [4]. Clough and Martin, who were then junior professors at U.C. Berkeley and U. Washington, respectively, spent “faculty internship” summers at Turner’s group during 1952 and 1953. The result of this collaboration was a celebrated paper [164], widely considered the start of the present FEM. Clough baptized the method in 1960 [25] and went on to form at Berkeley the first research group that expanded the idea into Civil Engineering applications. Olek Zienkiewicz, originally an expert in finite difference methods who learned the trade from Southwell, was convinced in 1964 by Clough to try FEM. He went on to write the first textbook on the subject [183] and to organize another important Civil Engineering research group in the University of Wales at Swansea. §1.7.2. G1: The Pioneers The 1956 paper by Turner, Clough, Martin and Topp [164], henceforth abbreviated to TCMT, is recognized as the start of the current FEM, as used in the overwhelming majority of commercial codes. Along with Argyris’
1–13
1–14
Chapter 1: OVERVIEW
serial [4] they prototype the first generation, which spans 1950 through 1962. A panoramic picture of this period is available in two textbooks [124,134]. Przemieniecki’s text is still reprinted by Dover. The survey by Gallagher [70] was influential at the time but is now difficult to access outside libraries. The pioneers were structural engineers, schooled in classical mechanics. They followed a century of tradition in regarding structural elements as a device to transmit forces. This “element as force transducer” was the standard view in pre-computer structural analysis. It explains the use of flux assumptions to derive stiffness equations in TCMT. Element developers worked in, or interacted closely with, the aircraft industry. (As noted above, only large aerospace companies were then able to afford mainframe computers.) Accordingly they focused on thin structures built up with bars, ribs, spars, stiffeners and panels. Although the Classical Force Method dominated stress analysis during the 1950s [56], stiffness methods were kept alive by use in dynamics and vibration. It is not coincidence that Turner was an world-class expert in aeroelasticity. §1.7.3. G2: The Golden Age The next period spans the golden age of FEM: 1962–1972. This is the “variational generation.” Melosh showed [115] that conforming displacement models are a form of Rayleigh-Ritz based on the minimum potential energy principle. This influential paper marks the confluence of three lines of research: Argyris’ dual formulation of energy methods [4], the Direct Stiffness Method (DSM) of Turner [165–167], and early ideas of interelement compatibility as basis for error bounding and convergence [65,114]. G1 workers thought of finite elements as idealizations of structural components. From 1962 onward a two-step interpretation emerges: discrete elements approximate continuum models, which in turn approximate real structures. By the early 1960s FEM begins to expand into Civil Engineering through Clough’s Boeing-Berkeley connection [31,32] and had been baptized [25,27]. Reading Fraeijs de Veubeke’s famous article [66] side by side with TCMT [164] one can sense the ongoing change in perspective opened up by the variational framework. The first book devoted to FEM appears in 1967 [183]. Applications to nonstructural problems had started in 1965 [182], and were treated in some depth by Martin and Carey [111]. From 1962 onwards the displacement formulation dominates. This was given a big boost by the invention of the isoparametric formulation and related tools (numerical integration, fitted natural coordinates, shape functions, patch test) by Irons and coworkers [95–98]. Low order displacement models often exhibit disappointing performance. Thus there was a frenzy to develop higher order elements. Other variational formulations, notably hybrids [127,130], mixed [89,157] and equilibrium models [66] emerged. G2 can be viewed as closed by the monograph of Strang and Fix [149], the first book to focus on the mathematical foundations. §1.7.4. G3: Consolidation The post-Vietnam economic doldrums are mirrored during this post-1972 period. Gone is the youthful exuberance of the golden age. This is consolidation time. Substantial effort is put into improving the stock of G2 displacement elements by tools initially labeled “variational crimes” [148], but later justified. Textbooks by Hughes [94] and Bathe [9] reflect the technology of this period. Hybrid and mixed formulations record steady progress [8]. Assumed strain formulations appear [105]. A booming activity in error estimation and mesh adaptivity is fostered by better understanding of the mathematical foundations [155]. Commercial FEM codes gradually gain importance. They provide a reality check on what works in the real world and what doesn’t. By the mid-1980s there was gathering evidence that complex and high order elements were commercial flops. Exotic gadgetry interweaved amidst millions of lines of code easily breaks down in new releases. Complexity is particularly dangerous in nonlinear and dynamic analyses conducted by novice users. A trend back toward simplicity starts [106,107]. §1.7.5. G4: Back to Basics The fourth generation begins by the early 1980s. More approaches come on the scene, notably the Free Formulation [16,17], orthogonal hourglass control [61], Assumed Natural Strain methods [10,145], stress hybrid
1–14
1–15
§1.7 *HISTORICAL SKETCH AND BIBLIOGRAPHY
models in natural coordinates [125,135], as well as variants and derivatives of those approaches: ANDES [49,116], EAS [141,142] and others. Although technically diverse the G4 approaches share two common objectives: (i)
Elements must fit into DSM-based programs since that includes the vast majority of production codes, commercial or otherwise.
(ii)
Elements are kept simple but should provide answers of engineering accuracy with relatively coarse meshes. These were collectively labeled “high performance elements” in 1989 [48].
“Things are always at their best in the beginning,” said Pascal. Indeed. By now FEM looks like an aggregate of largely disconnected methods and recipes. The blame should not be placed on the method itself, but on the community split noted in the book Preface. §1.7.6. Recommended Books for Linear FEM The literature is vast: over 200 textbooks and monographs have appeared since 1967. Some recommendations for readers interested in further studies within linear FEM are offered below. Basic level (reference): Zienkiewicz and Taylor [185]. This two-volume set is a comprehensive upgrade of the previous edition [184]. Primarily an encyclopœdic reference work that gives a panoramic coverage of FEM applications, as well as a comprehensive list of references. Not a textbook or monograph. Prior editions suffered from loose mathematics, largely fixed in this one. A three-volume fifth edition has appeared recently. Basic level (textbook): Cook, Malkus and Plesha [33]. The third edition is comprehensive in scope although the coverage is more superficial than Zienkiewicz and Taylor. A fourth edition has appeared recently. Intermediate level: Hughes [94]. It requires substantial mathematical expertise on the part of the reader. Recently (2000) reprinted as Dover edition. Mathematically oriented: Strang and Fix [149]. Still the most readable mathematical treatment for engineers, although outdated in several subjects. Out of print. Best value for the $$$: Przemieniecki’s Dover edition [134], list price $15.95 (2003). A reprint of a 1966 McGraw-Hill book. Although woefully outdated in many respects (the word “finite element” does not appear except in post-1960 references), it is a valuable reference for programming simple elements. Contains a fairly detailed coverage of substructuring, a practical topic missing from the other books. Comprehensive bibliography in Matrix Structural Analysis up to 1966. Most fun (if you appreciate British “humor”): Irons and Ahmad [98]. Out of print. For buying out-of-print books through web services, check the search engine in www3.addall.com (most comprehensive; not a bookseller) as well as that of www.amazon.com. A newcomer is www.campusi.com §1.7.7. Hasta la Vista, Fortran Most FEM books that include programming samples or even complete programs use Fortran. Those face an uncertain future. Since the mid-1990s, Fortran is gradually disappearing as a programming language taught in USA engineering undergraduate programs. (It still survives in Physics and Chemistry departments because of large amounts of legacy code.) So one end of the pipeline is drying up. Low-level scientific programming is moving to C and C++, mid-level to Java, Perl and Python, high-level to Matlab, Mathematica and their free-source Linux equivalents. How attractive can a book teaching in a dead language be? To support this argument with some numbers, here is a September-2003 snapshot of ongoing open source software projects listed in http://freshmeat.net. This conveys the relative importance of various languages (a mixed bag of newcomers, going-strongs, have-beens and never-was) in the present environment.
1–15
1–16
Chapter 1: OVERVIEW
Lang Projects Perc Ada 38 0.20% Assembly 170 0.89% C 5447 28.55% Cold Fusion 10 0.05% Dylan 2 0.01% Erlang 11 0.06% Forth 15 0.08% Java 2332 12.22% Logo 2 0.01% Object Pascal 9 0.05% Other 160 0.84% Pascal 38 0.20% Pike 3 0.02% PROGRESS 2 0.01% Rexx 7 0.04% Simula 1 0.01% Tcl 356 1.87% Xbasic 1 0.01% Total Projects: 19079
Lang Projects Perc Lang Projects APL 3 0.02% ASP 25 Awk 40 0.21% Basic 15 C# 41 0.21% C++ 2443 Common Lisp 27 0.14% Delphi 49 Eiffel 20 0.10% Emacs-Lisp 33 Euler 1 0.01% Euphoria 2 Fortran 45 0.24% Haskell 28 JavaScript 236 1.24% Lisp 64 ML 26 0.14% Modula 7 Objective C 131 0.69% Ocaml 20 Other Scripting Engines 82 0.43% Perl 2752 14.42% PHP 2020 PL/SQL 58 0.30% Pliant 1 Prolog 8 0.04% Python 1171 Ruby 127 0.67% Scheme 76 Smalltalk 20 0.10% SQL 294 Unix Shell 550 2.88% Vis Basic 15 YACC 11 0.06% Zope 34
References Referenced items have been moved to Appendix R. Not yet sorted.
1–16
Perc 0.13% 0.08% 12.80% 0.26% 0.17% 0.01% 0.15% 0.34% 0.04% 0.10% 10.59% 0.01% 6.14% 0.40% 1.54% 0.08% 0.18%
1–17
Exercises
Homework Exercises for Chapter 1 Overview EXERCISE 1.1 [A:15] Work out Archimedes’ problem using a circumscribed regular polygon, with n =
1, 2, 4, . . . 256. Does the sequence converge any faster?
EXERCISE 1.2 [D:20] Select one of the following vehicles: truck, car, motorcycle, or bicycle. Draw a two
level decomposition of the structure into substructures, and of selected components of some substructures. EXERCISE 1.3 [D:30] In one of the earliest articles on the FEM, Clough [27] writes:
“When idealized as an assemblage of appropriately shaped two- and three-dimensional elements in this manner, an elastic continuum can be analyzed by standard methods of structural analysis. It should be noted that the approximation which is employed in this case is of physical nature; a modified structural system is substituted for the actual continuum. There need be no approximation in the mathematical analysis of this structural system. This feature distinguishes the finite element technique from finite difference methods, in which the exact equations of the actual physical system are solved by approximate mathematical procedures.” Discuss critically the contents of this paragraph while placing it in the context of time of writing (early 1960s). Is the last sentence accurate?
1–17
2
.
The Direct Stiffness Method I
2–1
2–2
Chapter 2: THE DIRECT STIFFNESS METHOD I
TABLE OF CONTENTS Page
§2.1. §2.2. §2.3. §2.4. §2.5. §2.6. §2.7.
§2.8.
§2. §2. §2.
Why A Plane Truss? Truss Structures Idealization Members, Joints, Forces and Displacements The Master Stiffness Equations The DSM Steps Breakdown §2.7.1. Disconnection . . . . . . . . . . §2.7.2. Localization . . . . . . . . . . §2.7.3. Computation of Member Stiffness Equations Assembly: Globalization §2.8.1. Coordinate Transformations . . . . . §2.8.2. Transformation to Global System . . . . Notes and Bibliography . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . .
2–2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . .
2–3 2–3 2–4 2–4 2–6 2–7 2–7 2–7 2–8 2–8 2–9 2–9 2–11 2–12 2–12 2–13
2–3
§2.2
TRUSS STRUCTURES
This Chapter begins the exposition of the Direct Stiffness Method (DSM) of structural analysis. The DSM is by far the most common implementation of the Finite Element Method (FEM). In particular, all major commercial FEM codes are based on the DSM. The exposition is done by following the DSM steps applied to a simple plane truss structure. The method has two major stages: breakdown, and assembly+solution. This Chapter covers primarily the breakdown stage. §2.1. Why A Plane Truss? The simplest structural finite element is the 2-node bar (also called linear spring) element, which is illustrated in Figure 2.1(a). Perhaps the most complicated finite element (at least as regards number of degrees of freedom) is the curved, three-dimensional “brick” element depicted in Figure 2.1(b). Yet the remarkable fact is that, in the DSM, the simplest and most complex elements are treated alike! To illustrate the basic steps of this democratic method, it makes educational sense to keep it simple and use a structure composed of bar elements.
(a)
(b)
Figure 2.1. From the simplest through a highly complex structural finite element: (a) 2-node bar element for trusses, (b) 64-node tricubic, “brick” element for three-dimensional solid analysis.
A simple yet nontrivial structure is the pin-jointed plane truss.1 Using a plane truss to teach the stiffness method offers two additional advantages: (a) Computations can be entirely done by hand as long as the structure contains just a few elements. This allows various steps of the solution procedure to be carefully examined and understood before passing to the computer implementation. Doing hand computations on more complex finite element systems rapidly becomes impossible. (b) The computer implementation on any programming language is relatively simple and can be assigned as preparatory computer homework before reaching Part III. §2.2. Truss Structures Plane trusses, such as the one depicted in Figure 2.2, are often used in construction, particularly for roofing of residential and commercial buildings, and in short-span bridges. Trusses, whether two or three dimensional, belong to the class of skeletal structures. These structures consist of elongated structural components called members, connected at joints. Another important subclass of skeletal structures are frame structures or frameworks, which are common in reinforced concrete construction of buildings and bridges. Skeletal structures can be analyzed by a variety of hand-oriented methods of structural analysis taught in beginning Mechanics of Materials courses: the Displacement and Force methods. They can also be analyzed by the computer-oriented FEM. That versatility makes those structures a good choice 1
A one dimensional bar assembly would be even simpler. That kind of structure would not adequately illustrate some of the DSM steps, however, notably the back-and-forth transformations from global to local coordinates.
2–3
2–4
Chapter 2: THE DIRECT STIFFNESS METHOD I
member support joint Figure 2.2. An actual plane truss structure. That shown is typical of a roof truss used in building construction.
to illustrate the transition from the hand-calculation methods taught in undergraduate courses, to the fully automated finite element analysis procedures available in commercial programs. In this and the next Chapter we will go over the basic steps of the DSM in a “hand-computer” calculation mode. This means that although the steps are done by hand, whenever there is a procedural choice we shall either adopt the way which is better suited towards the computer implementation, or explain the difference between hand and computer computations. The actual computer implementation using a high-level programming language is presented in Chapter 5. To keep hand computations manageable in detail we use just about the simplest structure that can be called a plane truss, namely the three-member truss illustrated in Figure 2.3. The idealized model of the example truss as a pin-jointed assemblage of bars is shown in Figure 2.4(a), which also gives its geometric and material properties. In this idealization truss members carry only axial loads, have no bending resistance, and are connected by frictionless pins. Figure 2.4(b) displays support conditions as well as the applied forces applied to the truss joints.
Figure 2.3. The three-member example truss.
It should be noted that as a practical structure the example truss is not particularly useful — the one depicted in Figure 2.2 is far more common in construction. But with the example truss we can go over the basic DSM steps without getting mired into too many members, joints and degrees of freedom. §2.3. Idealization Although the pin-jointed assemblage of bars (as depicted in Figure 2.4) is sometimes presented as an actual problem, it actually represents an idealization of a true truss structure. The axially-carrying members and frictionless pins of this structure are only an approximation of a real truss. For example, building and bridge trusses usually have members joined to each other through the use of gusset plates, which are attached by nails, bolts, rivets or welds. See Figure 2.2. Consequently members will carry some bending as well as direct axial loading. Experience has shown, however, that stresses and deformations calculated for the simple idealized problem will often be satisfactory for overall-design purposes; for example to select the cross section of the members. Hence the engineer turns to the pin-jointed assemblage of axial force elements and uses it to carry out the structural analysis. This replacement of true by idealized is at the core of the physical interpretation of the finite element method discussed in §1.4. 2–4
2–5
§2.4
fy3, u y3
(a)
A
1
fy1, u y1
f y3 = 1 3
√
= 10 2 √ = 200 2
(2)
(1)
x (1)
f x3 = 2
L = 10 E (2) A (2) = 50
(3)
y fx1, u x1
(b)
L = 10 E (1) A(1) = 100
(2)
y fx2, u x2 2
1
x
fy2, u y2
2
;; ;; ;;
E
fx3, u x3
;; ;; ;;
L
3
(3)
(3) (3)
MEMBERS, JOINTS, FORCES AND DISPLACEMENTS
Figure 2.4. Pin-jointed idealization of example truss: (a) geometric and elastic properties, (b) support conditions and applied loads.
§2.4. Members, Joints, Forces and Displacements The idealization of the example truss, pictured in Figure 2.4, has three joints, which are labeled 1, 2 and 3, and three members, which are labeled (1), (2) and (3). These members connect joints 1–2, 2–3, and 1–3, respectively. The member lengths are denoted by L (1) , L (2) and L (3) , their elastic moduli by E (1) , E (2) and E (3) , and their cross-sectional areas by A(1) , A(2) and A(3) . Note that an element number supercript is enclosed in parenthesis to avoid confusion with exponents. Both E and A are assumed to be constant along each member. Members are generically identified by index e (because of their close relation to finite elements, see below). This index is placed as supercript of member properties. For example, the cross-section area of a generic member is Ae . The member superscript is not enclosed in parentheses in this case because no confusion with exponents can arise. But the area of member 3 is written A(3) and not A3 . Joints are generically identified by indices such as i, j or n. In the general FEM, the name “joint” and “member” is replaced by node and element, respectively. The dual nomenclature is used in the initial Chapters to stress the physical interpretation of the FEM. The geometry of the structure is referred to a common Cartesian coordinate system {x, y}, which is called the global coordinate system. Other names for it in the literature are structure coordinate system and overall coordinate system. The key ingredients of the stiffness method of analysis are the forces and displacements at the joints. In a idealized pin-jointed truss, externally applied forces as well as reactions can act only at the joints. All member axial forces can be characterized by the x and y components of these forces, denoted by f x and f y , respectively. The components at joint i will be identified as f xi and f yi , respectively. The set of all joint forces can be arranged as a 6-component column vector called f. The other key ingredient is the displacement field. Classical structural mechanics tells us that the displacements of the truss are completely defined by the displacements of the joints. This statement is a particular case of the more general finite element theory. The x and y displacement components will be denoted by u x and u y , respectively. The values of u x and u y at joint i will be called u xi and u yi . Like joint forces, they are arranged into a 6-component vector called u. Here are the two vectors 2–5
Chapter 2: THE DIRECT STIFFNESS METHOD I
of nodal forces and nodal displacements, shown side by side:     f x1 u x1  f y1   u y1      f   u  f =  x2  , u =  x2  .  f y2   u y2      f x3 u x3 f y3 u y3
2–6
(2.1)
In the DSM these six displacements are the primary unknowns. They are also called the degrees of freedom or state variables of the system.2 How about the displacement boundary conditions, popularly called support conditions? This data will tell us which components of f and u are true unknowns and which ones are known a priori. In pre-computer structural analysis such information was used immediately by the analyst to discard unnecessary variables and thus reduce the amount of hand-carried bookkeeping. The computer oriented philosophy is radically different: boundary conditions can wait until the last moment. This may seem strange, but on the computer the sheer volume of data may not be so important as the efficiency with which the data is organized, accessed and processed. The strategy “save the boundary conditions for last” will be followed here also for the hand computations. Remark 2.1. Often column vectors such as (2.1) will be displayed in row form to save space, with a transpose
symbol at the end. For example, f = [ f x1 f y1 f x2 f y2 f x3 f y3 ]T and u = [ u x1 u y1 u x2 u y2 u x3 u y3 ]T .
§2.5. The Master Stiffness Equations The master stiffness equations relate the joint forces f of the complete structure to the joint displacements u of the complete structure before specification of support conditions. Because the assumed behavior of the truss is linear, these equations must be linear relations that connect the components of the two vectors. Furthermore it will be assumed that if all displacements vanish, so do the forces.3 If both assumptions hold the relation must be homogeneous and expressable in component form as      K x1x1 K x1y1 K x1x2 K x1y2 K x1x3 K x1y3 u x1 f x1  f y1   K y1x1 K y1y1 K y1x2 K y1y2 K y1x3 K y1y3   u y1        f x2   K x2x1 K x2y1 K x2x2 K x2y2 K x2x3 K x2y3   u x2  (2.2) =  .   f y2   K y2x1 K y2y1 K y2x2 K y2y2 K y2x3 K y2y3   u y2       f x3 K x3x1 K x3y1 K x3x2 K x3y2 K x3x3 K x3y3 u x3 f y3 K y3x1 K y3y1 K y3x2 K y3y2 K y3x3 K y3y3 u y3 In matrix notation: f = K u.
(2.3)
2
Primary unknowns is the correct mathematical term whereas degrees of freedom has a mechanics flavor: “any of a limited number of ways in which a body may move or in which a dynamic system may change” (Merrian-Webster). The term state variables is used more often in nonlinear analysis, material sciences and statistics.
3
This assumption implies that the so-called initial strain effects, also known as prestress or initial stress effects, are neglected. Such effects are produced by actions such as temperature changes or lack-of-fit fabrication, and are studied in Chapter 29.
2–6
2–7
§2.7
BREAKDOWN
Here K is the master stiffness matrix, also called global stiffness matrix, assembled stiffness matrix, or overall stiffness matrix. It is a 6 × 6 square matrix that happens to be symmetric, although this attribute has not been emphasized in the written-out form (2.2). The entries of the stiffness matrix are often called stiffness coefficients and have a physical interpretation discussed below. The qualifiers (“master”, “global”, “assembled” and “overall”) convey the impression that there is another level of stiffness equations lurking underneath. And indeed there is a member level or element level, into which we plunge in the Breakdown section. Remark 2.2. Interpretation of Stiffness Coefficients. The following interpretation of the entries of K is valuable
for visualization and checking. Choose a displacement vector u such that all components are zero except the i th one, which is one. Then f is simply the i th column of K. For instance if in (2.3) we choose u x2 as unit displacement, u = [0
0
1
0
0
0 ]T ,
f = [ K x1x2 K y1x2 K x2x2 K y2x2 K x3x2 K y3x2 ]T .
(2.4)
Thus K y1x2 , say, represents the y-force at joint 1 that would arise on prescribing a unit x-displacement at joint 2, while all other displacements vanish. In structural mechanics this property is called interpretation of stiffness coefficients as displacement influence coefficients. It extends unchanged to the general finite element method.
§2.6. The DSM Steps
 The DSM steps, major and minor, are sum Disconnection marized in Figure 2.5 for the convenience of Breakdown Localization the reader. The two major processing steps  are Breakdown, followed by Assembly & (Sections 2.7 & 2.8) Member (Element) Formation Solution. A postprocessing substep may follow, Globalization although this is not part of the DSM proper. Merge The first 3 DSM substeps are: (1) disconnection, Assembly & Application of BCs Solution (2) localization, and (3) computation of member Solution stiffness equations. Collectively these form (Sections 3.2-3.4) Recovery of Derived Quantities the breakdown. The first two are marked as conceptual in Figure 2.5 because they are not actually programmed as such. These subsets are post-processing conceptual processing steps steps steps implicitly carried out through the user-provided problem definition. Processing begins at the Figure 2.5. The Direct Stiffness Method steps. member-stiffness-equation forming substep.
  
§2.7. Breakdown §2.7.1. Disconnection To carry out the first breakdown step we proceed to disconnect or disassemble the structure into its components, namely the three truss members. This task is illustrated in Figure 2.6. To each member e = 1, 2, 3 assign a Cartesian system {x¯ e , y¯ e }. Axis x¯ e is aligned along the axis of the eth member. Actually x¯ e runs along the member longitudinal axis; it is shown offset in that Figure for clarity. By convention the positive direction of x¯ e runs from joint i to joint j, where i < j. The angle formed by x¯ e and x is the orientation angle ϕ e . The axes origin is arbitrary and may be placed at the member midpoint or at one of the end joints for convenience. 2–7
2–8
Chapter 2: THE DIRECT STIFFNESS METHOD I
_ _ fyi , uyi
(a) _ _ fxi , uxi
Equivalent spring stiffness
_
x (2)
(3) y
j
k s = EA / L
x
_
i
_ (3)
y
y
x
3 _ (3)
_ _ fyj , uyj _ _ fxj , uxj
_
(b)
(2)
i
j
_
y (2)
_
y(1)
−F
_
1 x
x(1)
(1)
F L
2
Figure 2.6. Breakdown of example truss into individual members (1), (2) and (3), and selection of local coordinate systems.
d
Figure 2.7. Generic truss member referred to its local coordinate system {x, ¯ y¯ }: (a) idealization as bar element, (b) interpretation as equivalent spring.
Systems {x¯ e , y¯ e } are called local coordinate systems or member-attached coordinate systems. In the general finite element method they also receive the name element coordinate systems. §2.7.2. Localization Next we drop the member identifier e so that we are effectively dealing with a generic truss member, as illustrated in Figure 2.7(a). The local coordinate system is {x, ¯ y¯ }. The two end joints are i and j. As shown in that figure, a generic truss member has four joint force components and four joint displacement components (the member degrees of freedom). The member properties are length L, elastic modulus E and cross-section area A. §2.7.3. Computation of Member Stiffness Equations The force and displacement components of the generic truss member shown in Figure 2.7(a) are linked by the member stiffness relations ¯f = K u, ¯ which written out in full is  ¯   K¯ xi xi f xi K¯ yi xi  f¯yi    ¯ =  K¯ fx j x j xi ¯ fyj K¯
y j xi
K¯ xi yi K¯ yi yi K¯ x j yi K¯ y j yi
K¯ xi x j K¯ yi x j K¯ x j x j K¯ y j x j
(2.5)  K¯ xi y j  u¯ xi  K¯ yi y j  u¯ yi   .   ¯ u ¯xj Kx jyj u¯ y j K¯ y j y j
(2.6)
Vectors ¯f and u¯ are called the member joint forces and member joint displacements, respectively, ¯ is the member stiffness matrix or local stiffness matrix. When these relations are interpreted whereas K from the standpoint of the general FEM, “member” is replaced by “element” and “joint” by ”node.” ¯ in terms of L, E and A. The most There are several ways to construct the stiffness matrix K straightforward technique relies on the Mechanics of Materials approach covered in undergraduate 2–8
2–9
§2.8
ASSEMBLY: GLOBALIZATION
courses. Think of the truss member in Figure 2.7(a) as a linear spring of equivalent stiffness ks , an interpretation illustrated in Figure 2.7(b). If the member properties are uniform along its length, Mechanics of Materials bar theory tells us that4 EA , L
ks =
(2.7)
Consequently the force-displacement equation is EA d, L
F = ks d =
(2.8)
where F is the internal axial force and d the relative axial displacement, which physically is the bar elongation. The axial force and elongation can be immediately expressed in terms of the joint forces and displacements as F = f¯x j = − f¯xi , d = u¯ x j − u¯ xi , (2.9) which express force equilibrium5 and kinematic compatibility, respectively. Combining (2.8) and (2.9) we obtain the matrix relation6   ¯  1 f xi ¯yi  E A 0 f   ¯f =   = ¯ −1 fx j L 0 f¯y j Hence
−1 0 1 0
0 0 0 0

1 E A  0 ¯ = K  −1 L 0
0 0 0 0
  0 u¯ xi 0   u¯ yi  ¯ u, ¯  =K 0 u¯ x j 0 u¯ y j
 −1 0 0 0 . 1 0 0 0
(2.10)
(2.11)
This is the truss stiffness matrix in local coordinates. Two other methods for obtaining the local force-displacement relation (2.8) are covered in Exercises 2.6 and 2.7. §2.8. Assembly: Globalization The first substep in the assembly & solution major step, as shown in Figure 2.5, is globalization. This operation is done member by member. It refers the member stiffness equations to the global system {x, y} so it can be merged into the master stiffness. Before entering into details we must establish relations that connect joint displacements and forces in the global and local coordinate systems. These are given in terms of transformation matrices. 4
See for example, Chapter 2 of [12].
5
Equations F = f¯x j = − f¯xi follow by considering the free body diagram (FBD) of each joint. For example, take joint i as a FBD. Equilibrium along x requires −F − f¯xi = 0 whence F = − f¯xi . Doing the same on joint j yields F = f¯x j .
6
The matrix derivation of (2.10) is the subject of Exercise 2.3.
2–9
2–10
Chapter 2: THE DIRECT STIFFNESS METHOD I
(a) Displacement transformation y¯
x¯
u yi u¯ xi
u¯ yi
i
u¯ y j u y j
j ϕ
(b) Force transformation
u¯ x j
f¯y j
fy j f¯x j
ux j
j fyi
y
u xi
x
ϕ
f¯xi
f¯yi
i
fx j
fxi
Figure 2.8. The transformation of node displacement and force components from the local system {x, ¯ y¯ } to the global system {x, y}.
§2.8.1. Coordinate Transformations The necessary transformations are easily obtained by inspection of Figure 2.8. For the displacements u¯ xi = u xi c + u yi s,
u¯ yi = −u xi s + u yi c,
u¯ x j = u x j c + u y j s,
u¯ y j = −u x j s + u y j c,
.
(2.12)
where c = cos ϕ, s = sin ϕ and ϕ is the angle formed by x¯ and x, measured positive counterclockwise from x. The matrix form of this relation is      u¯ xi c s 0 0 u xi  u¯ yi   −s c 0 0   u yi  (2.13)  = .  0 0 c s u¯ x j ux j 0 0 −s c u¯ y j uyj The 4 × 4 matrix that appears above is called a displacement transformation matrix and is denoted7 by T. The node forces transform as f xi = f¯xi c − f¯yi s, etc., which in matrix form become     f¯   c −s 0 0 f xi xi f¯yi   f yi   s c 0 0   . (2.14) =   0 0 c −s  f¯x j  fx j 0 0 s c fyj f¯y j The 4 × 4 matrix that appears above is called a force transformation matrix. A comparison of (2.13) and (2.14) reveals that the force transformation matrix is the transpose TT of the displacement transformation matrix T. This relation is not accidental and can be proved to hold generally.8 Remark 2.3. Note that in (2.13) the local system (barred) quantities appear on the left-hand side, whereas in (2.14) they show up on the right-hand side. The expressions (2.13) and and (2.14) are discrete counterparts of what are called covariant and contravariant transformations, respectively, in continuum mechanics. The counterpart of the transposition relation is the adjointness property. 7
This matrix will be called Td when its association with displacements is to be emphasized, as in Exercise 2.5.
8
A simple proof that relies on the invariance of external work is given in Exercise 2.5. However this invariance was only checked by explicit computation for a truss member in Exercise 2.4. The general proof relies on the Principle of Virtual Work, which is discussed later.
2–10
2–11
§2.8
ASSEMBLY: GLOBALIZATION
Remark 2.4. For this particular structural element T is square and orthogonal, that is, TT = T−1 . But this
property does not extend to more general elements. Furthermore in the general case T is not even a square matrix, and does not possess an ordinary inverse. However the congruential transformation relations (2.15)–(2.17) do hold generally.
§2.8.2. Transformation to Global System From now on we reintroduce the member (element) index, e. The member stiffness equations in global coordinates will be written f e = K e ue . (2.15) The compact form of (2.13) and (2.14) for the eth member is e fe = (Te )T ¯f .
u¯ e = Te ue ,
(2.16)
e e Inserting these matrix expressions into ¯f = K u¯ e and comparing with (2.15) we find that the member ¯ e in the local system stiffness in the global system {x, y} can be computed from the member stiffness K {x, ¯ y¯ } through the congruential transformation
¯ e Te . Ke = (Te )T K
(2.17)
Carrying out the matrix multiplications in closed form we get 
c2 E A  sc Ke =  2 −c Le −sc e
e
sc s2 −sc −s 2
−c2 −sc c2 sc
 −sc −s 2  , sc s2
(2.18)
in which c = cos ϕ e , s = sin ϕ e , with e superscripts of c and s suppressed to reduce clutter. If the angle is zero we recover (2.10), as may be expected. Ke is called a member stiffness matrix in global coordinates. The proof of (2.17) and verification of (2.18) is left as Exercise 2.8. The globalized member stiffness matrices for the example truss can now be easily obtained by inserting appropriate values into (2.18). For member (1), with end joints 1–2, angle ϕ = 0◦ and the member properties given in Figure 2.4(a) we get  f (1)   x1 1 (1)  f y1   0    f (1)  = 10  −1 x2 (1) 0 f y2
0 0 0 0
   u (1) x1 −1 0  u (1) 0 0 y1  .   (1)  1 0 u x2  0 0 u (1)
(2.19)
y2
For member (2), with end joints 2–3, and angle ϕ = 90◦ :  f (2)   x2 0 (2)  f y2  0    f (2)  = 5  0 x3 (2) 0 f y3
   u (2) x2 0 0 0  u (2) 1 0 −1   y2  .   (2)  0 0 0 u x3  −1 0 1 u (2) y3 2–11
(2.20)
Chapter 2: THE DIRECT STIFFNESS METHOD I
2–12
Finally, for member (3), with end joints 1–3, and angle ϕ = 45◦ :  f (3)   x1 0.5 (3)  f y1   0.5    f (3)  = 20  −0.5 x3 (3) −0.5 f y3
   u (3) x1 0.5 −0.5 −0.5  u (3) 0.5 −0.5 −0.5   y1  .   (3)  −0.5 0.5 0.5 u x3  −0.5 0.5 0.5 u (3) y3
(2.21)
In the following Chapter we will complete the main DSM steps by putting the truss back together through the merge step, and solving for the unknown forces and displacements. Notes and Bibliography The Direct Stiffness Method has been the dominant FEM version since the mid-1960s, and is the procedure followed by all major commercial codes in current use. DSM was invented and developed at Boeing in the early 1950s, through the leadership of Jon Turner [170–173], and had defeated its main competitor, the Force Method, by 1970 [57]. All applications-oriented FEM books cover the DSM, although the procedural steps are sometimes not clearly identified. In particular, the textbooks recommended in §1.7.6 offer adequate expositions. Trusses, also called bar assemblies, are usually the first structures treated in Mechanics of Materials books written for undergraduate courses. Two widely used books at this level are [12] and [133]. Steps in the derivation of stiffness matrices for truss elements are well covered in a number of early treatment of finite element books, of which Chapter 5 of Przemieniecki [136] is a good example. References Referenced items have been moved to Appendix R.
2–12
2–13
Exercises
Homework Exercises for Chapter 2 The Direct Stiffness Method I EXERCISE 2.1 [D:10] Explain why arbitrarily oriented mechanical loads on an idealized pin-jointed truss
structure must be applied at the joints. [Hint: idealized truss members have no bending resistance.] How about actual trusses: can they take loads applied between joints? EXERCISE 2.2 [A:15] Show that the sum of the entries of each row of the master stiffness matrix K of any
plane truss, before application of any support conditions, must be zero. [Hint: apply translational rigid body motions at nodes.] Does the property hold also for the columns of that matrix? EXERCISE 2.3 [A:15] Using matrix algebra derive (2.10) from (2.8) and (2.9). Note: Place all equations in
matrix form first. Deriving the final form with scalar algebra and rewriting it in matrix form gets no credit. EXERCISE 2.4 [A:15] By direct multiplication verify that for the truss member of Figure 2.7(a), ¯f u¯ = F d. T
Intepret this result physically. (Hint: what is a force times displacement in the direction of the force?) EXERCISE 2.5 [A:20] The transformation equations between the 1-DOF spring and the 4-DOF generic truss
member may be written in compact matrix form as ¯ d = Td u,
¯f = F T f ,
(E2.1)
where Td is 1 × 4 and T f is 4 × 1. Starting from the identity ¯f u¯ = F d proven in the previous exercise, and using compact matrix notation, show that T f = TdT . Or in words: the displacement transformation matrix and the force transformation matrix are the transpose of each other. (This can be extended to general systems) T
EXERCISE 2.6 [A:20] Derive the equivalent spring formula F = (E A/L) d of (2.8) by the Theory of Elasticity
relations e = d u( ¯ x)/d ¯ x¯ (strain-displacement equation), σ = Ee (Hooke’s law) and F = Aσ (axial force definition). Here e is the axial strain (independent of x) ¯ and σ the axial stress (also independent of x). ¯ Finally, u( ¯ x) ¯ denotes the axial displacement of the cross section at a distance x¯ from node i, which is linearly interpolated as x¯ x¯ u( ¯ x) ¯ = u¯ xi 1 − (E2.2) + u¯ x j L L
Justify that (E2.2) is correct since the bar differential equilibrium equation: d[A(dσ/d x)]/d ¯ x¯ = 0, is verified for all x¯ if A is constant along the bar. EXERCISE 2.7 [A:15] Derive the equivalent spring formula F = (E A/L) d of (2.8) by the principle of
Minimum Potential Energy (MPE). In Mechanics of Materials it is shown that the total potential energy of the axially loaded bar is
=
L
A σ e d x¯ − Fd,
1 2
(E2.3)
0
where symbols have the same meaning as the previous Exercise. Use the displacement interpolation (E2.2), the strain-displacement equation e = d u/d ¯ x¯ and Hooke’s law σ = Ee to express as a function (d) of the relative displacement d only. Then apply MPE by requiring that ∂ /∂d = 0. ¯ e u¯ e = ¯fe , (2.15) and (2.17). (Hint: premultiply both sides of EXERCISE 2.8 [A:20] Derive (2.17) from K
¯ e u¯ e = ¯fe by an appropriate matrix). Then check by hand that using that formula you get (2.18). Falk’s scheme K is recommended for the multiplications.9 EXERCISE 2.9 [D:5] Why are disconnection and localization labeled as “conceptual steps” in Figure 2.5? 9
This scheme is useful to do matrix multiplication by hand. It is explained in §B.3.2 of Appendix B.
2–13
3
.
The Direct Stiffness Method II
3–1
3–2
Chapter 3: THE DIRECT STIFFNESS METHOD II
TABLE OF CONTENTS Page
§3.1. §3.2.
§3.3.
§3.4.
§3.5.
§3.6.
§3. §3. §3.
Introduction Assembly: Merge §3.2.1. Governing Rules . . . . . . . . . . . . §3.2.2. Hand Assembly by Augmentation and Merge . . . Solution §3.3.1. Applying Displacement BCs by Reduction . . . . §3.3.2. Solving for Displacements . . . . . . . . . PostProcessing §3.4.1. Recovery of Reaction Forces . . . . . . . . §3.4.2. Recovery of Internal Forces and Stresses . . . . . *Computer Oriented Assembly and Solution §3.5.1. *Assembly by Freedom Pointers . . . . . . . §3.5.2. *Applying DBC by Modification . . . . . . . Prescribed Nonzero Displacements §3.6.1. Application of Nonzero-DBCs by Reduction . . . §3.6.2. *Application of Nonzero-DBCs by Modification . . §3.6.3. *Matrix Forms of Nonzero-DBC Application Methods Notes and Bibliography . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . .
3–2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . .
3–3 3–3 3–3 3–4 3–6 3–6 3–7 3–7 3–7 3–8 3–8 3–9 3–9 3–10 3–10 3–11 3–12 3–13 3–13 3–14
3–3
§3.2
ASSEMBLY: MERGE
§3.1. Introduction Chapter 2 went through the initial stage of the DSM. The three breakdown steps: disconnection, localization and formation of member stiffness formation take us down all the way to the generic truss element: the highest level of fragmentation. This is followed by the assembly process. Assembly involves merging the stiffness equations of each member into the global stiffness equations. For this to make sense, the member equations must be referred to a common coordinate system, which for a plane truss is the global Cartesian system {x, y}. This is done through the globalization process covered in §2.8. On the computer the formation, globalization and merge steps are done concurrently, member by member. After all members are processed we have the free-free master stiffness equations. Next comes the solution. This process embodies two substeps: application of boundary conditions and solution for the unknown joint displacements. To apply the BCs, the free-free master stiffness equations are modified by taking into account which components of the joint displacements and forces are given and which are unknown. The modified equations are submitted to a linear equation solver, which returns the unknown joint (node) displacements. As discussed under Notes and Bibliography, on some FEM implementations — especially programs written in the 1960s and 1970s — one or more of the foregoing operations are done concurrently. The solution step completes the DSM proper. Postprocessing steps may follow, in which derived quantities such as internal forces and stresses are recovered from the displacement solution. 3
§3.2. Assembly: Merge §3.2.1. Governing Rules The key operation of the assembly process is the “placement” of the contribution of each member to the master stiffness equations. The process is technically called merge of individual members. The merge operation can be physically interpreted as reconnecting that member in the process of fabricating the complete structure. For a truss structure, reconnection means inserting the pins back into the joints. See Figure 3.1. Merge is mathematically governed by two rules of structural mechanics:
(3)
(2)
(1)
2
y 1 x
Figure 3.1. The disconnected example truss prior to merge. All member stiffness equations are in the global system. Reconnecting the truss means putting the pins back into the joints.
1.
Compatibility of displacements: The displacement of all members meeting at a joint are the same.
2.
Force equilibrium: The sum of forces exerted by all members that meet at a joint balances the external force applied to that joint.
3–3
(3.1)
3–4
Chapter 3: THE DIRECT STIFFNESS METHOD II
(a)
f3
3
− f(3) 3
−
(b)
3 f(3) 3
f(2) 3
(3)
f(2) 3
(2)
Figure 3.2. The force equilibrium of joint 3 of the example truss, depicted as a free body diagram in (a). Here f3 is the known external joint force applied on the joint. Joint forces (3) f(2) 3 and f3 are applied by the joint on the members, as illustrated in (b). Consequently (3) the forces applied by the members on the joint are −f(2) 3 and −f3 . These forces would act in the directions shown if both members (2) and (3) were in tension. The free-body (3) (2) (3) equilibrium statement is f3 − f(2) 3 − f3 = 0 or f3 = f3 + f3 . This translates into the (2) (3) (2) (3) two component equations: f x3 = f x3 + f x3 and f y3 = f y3 + f y3 , of (3.2).
The first rule is physically obvious: reconnected joints must move as one entity. The second one can be visualized by considering a joint as a free body, although care is required in the interpretation of joint forces and their signs. Notational conventions to this effect are explained in Figure 3.2 for joint 3 of the example truss, at which members (2) and (3) meet. Application of the foregoing rules at this particular joint gives Rule 1:
(3) u (2) x3 = u x3 ,
(3) u (2) y3 = u y3 .
(2) (3) (1) (2) (3) f y3 = f y3 + f y3 = f y3 + f y3 + f y3 . (3.2) (1) (2) (3) (1) (2) (3) to f x3 + f x3 and of and f y3 to f y3 + f y3 , respectively, changes nothing The addition of f x3 because member (1) is not connected to joint 3. We are just adding zeros. But this augmentation enables us to write the key matrix relation:
Rule 2:
(2) (3) (1) (2) (3) f x3 = f x3 + f x3 = f x3 + f x3 + f x3 ,
f = f(1) + f(2) + f(3) .
(3.3)
§3.2.2. Hand Assembly by Augmentation and Merge To directly visualize how the two rules (3.1) translate to merging logic, we first augment the member stiffness relations by adding zero rows and columns as appropriate to complete the force and displacement vectors. For member (1):
        
(1)  f x1  10 (1)  f y1   0 (1)   f x2   −10 (1)  =    0 f y2   (1)  0 f x3 0 (1) f y3
0 0 0 0 0 0
−10 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 0 3–4
 u (1) x1 0  (1)  0   u y1    (1)  0   u x2   . 0   u (1)  y2    0  u (1)  x3 0 u (1) y3 

(3.4)
3–5
§3.2
For member (2):
        
(2)  f x1  0 (2)  f y1   0 (2)   f x2  0 =   (2)  0 f y2   (2)  0 f x3 0 (2) f y3
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 5 0 −5
0 0 0 0 0 0
ASSEMBLY: MERGE
 (2)   u x1 0  (2)  0   u y1    (2)  0   u x2   . −5   u (2)  y2   0  u (2)  x3 5 u (2) y3
(3.5)
For member (3):         
(3)  f x1  10 (3)  f y1   10 (3)   f x2   0 (3)  =    0 f y2   (3)  −10 f x3 −10 (3) f y3
10 10 0 0 −10 −10
0 0 0 0 0 0 0 0 0 0 0 0
−10 −10 0 0 10 10
 (3)   u x1 −10  (3)  −10   u y1    (3)  0   u x2   . 0   u (3)  y2   10  u (3)  x3 10 u (3) y3
(3.6)
According to the first rule, we can drop the member identifier in the displacement vectors that appear in the foregoing matrix equations. Hence the reconnected member equations are  (1)  f x1    10 0 −10 0 0 0 u x1 (1)  f y1  0 0 0 0 0   u y1    0     f (1)    x2   −10 0 10 0 0 0   u x2  (3.7) ,   (1)  =  0 0 0 0 0   u y2   f y2   0  (1)     u x3 0 0 0 0 0 0 f  x3 u y3 0 0 0 0 0 0 (1) f y3  (2)  f x1    0 0 0 0 0 0 u x1 (2)  f y1    0 0 0 0 0 0   u y1      f (2)    x2   0 0 0 0 0 0   u x2  (3.8) ,   (2)  =   f y2   0 0 0 5 0 −5   u y2   (2)     u x3 0 0 0 0 0 0 f  x3 u y3 0 0 0 −5 0 5 (2) f y3  (3)  f x1    10 10 0 0 −10 −10 u x1 (3)  f y1  10 0 0 −10 −10   u y1    10     f (3)   0 0 0 0 0   u x2   x2   0 (3.9) .   (3)  =  0 0 0 0 0   u y2   f y2   0  (3)     u x3 −10 −10 0 0 10 10 f  x3 u y3 −10 −10 0 0 10 10 (3) f y3 These three equations can be represented in direct matrix notation as f(1) = K(1) u,
f(2) = K(2) u, 3–5
f(3) = K(3) u.
(3.10)
3–6
Chapter 3: THE DIRECT STIFFNESS METHOD II
According to the second rule, expressed in matrix form as (3.3), we have f = f(1) + f(2) + f(3) = K(1) + K(2) + K(3) u = K u,
(3.11)
so all we have to do is add the three stiffness matrices that appear above, and we arrive at the master stiffness equations:       
  f x1 20 f y1   10   f x2   −10 = f y2   0   f x3 −10 f y3 −10
10 10 0 0 −10 −10
−10 0 10 0 0 0
0 0 0 5 0 −5
−10 −10 0 0 10 10
  −10 u x1 −10   u y1    0   u x2   . −5   u y2    u x3 10 u y3 15
(3.12)
Using this technique member merging becomes simply matrix addition. This explanation of the assembly process is conceptually the easiest to follow and understand. It is virtually foolproof for hand computations. However, this is not the way the process is carried out on the computer because it would be enormously wasteful of storage for large systems. A computer-oriented procedure is discussed in §3.5. §3.3. Solution Having formed the master stiffness equations we can proceed to the solution phase. To prepare the equations for a linear solver we need to separate known and unknown components of f and u. In this Section a technique suitable for hand computation is described. §3.3.1. Applying Displacement BCs by Reduction If one attempts to solve the system (3.12) numerically for the displacements, surprise! The solution “blows up” because the coefficient matrix (the master stiffness matrix) is singular. The mathematical interpretation of this behavior is that rows and columns of K are linear combinations of each other (see Remark 3.4 below). The physical interpretation of singularity is that there are unsuppressed rigid body motions: the truss still “floats” in the {x, y} plane. To eliminate rigid body motions and render the system nonsingular we must apply the physical support conditions as displacement boundary conditions. From Figure 2.4(b) we observe that the support conditions for the example truss are u x1 = u y1 = u y2 = 0,
(3.13)
whereas the known applied forces are f x2 = 0,
f x3 = 2,
f y3 = 1.
(3.14)
When solving the overall stiffness equations by hand, the simplest way to account for support conditions is to remove equations associated with known joint displacements from the master system. To apply (3.13) we have to remove equations 1, 2 and 4. This can be systematically 3–6
3–7
§3.4
POSTPROCESSING
accomplished by deleting or “striking out” rows and columns number 1, 2 and 4 from K and the corresponding components from f and u. The reduced three-equation system is
10 0 0
0 10 10
0 10 15
u x2 u x3 u y3
=
f x2 f x3 f y3
0 = 2 . 1
(3.15)
Equation (3.15) is called the reduced master stiffness system. The coefficient matrix of this system is no longer singular. Remark 3.1. In mathematical terms, the free-free master stiffness matrix K in (3.12) has order N = 6, rank
r = 3 and a rank deficiency of d = N − r = 6 − 3 = 3 (these concepts are summarized in Appendix C.) The dimension of the null space of K is d = 3. This space is spanned by three independent rigid body motions: the two rigid translations along x and y and the rigid rotation about z. Remark 3.2. Conditions (3.13) represent the simplest type of support conditions, namely specified zero displacements. More general constraint forms, such as prescribed nonzero displacements and multifreedom constraints, are handled as described in §3.5 and Chapters 8–9.
§3.3.2. Solving for Displacements Solving the reduced system by hand (for example, via Gauss elimination) yields
u x2 u x3 u y3
=
0 0.4 . −0.2
(3.16)
This is called a partial displacement solution (also reduced displacement solution) because it excludes known displacement components. This solution vector is expanded to six components by including the three specified values (3.13) in the appropiate slots:    u x1 0  u y1   0      u   0  u =  x2  =  .  u y2   0      0.4 u x3 −0.2 u y3 
(3.17)
This is called the complete displacement solution, or simply the displacement solution. §3.4. PostProcessing The last processing step of the DSM is the solution for joint displacements. But often the analyst needs information on other mechanical quantities; for example the reaction forces at the supports, or the internal member forces. Such quantities are said to be derived because they are recovered from the displacement solution. The recovery of derived quantities is part of the so-called postprocessing steps of the DSM. Two such steps are described below. 3–7
3–8
Chapter 3: THE DIRECT STIFFNESS METHOD II
§3.4.1. Recovery of Reaction Forces Premultiplying the complete displacement solution (3.17) by K we get      20 10 −10 0 −10 −10 0 −2 10 0 0 −10 −10   0   −2   10      0 10 0 0 0 0   0  −10 f = Ku =   =  0 0 5 0 −5   0   1   0      0.4 2 −10 −10 0 0 10 10 1 −10 −10 0 −5 10 15 −0.2
(3.18)
This vector recovers the known applied forces (3.14) as can be expected. Furthermore we get three reaction forces: f x1 = f y1 = −2 and f y2 = 1, which are associated with the support conditions (3.13). It is easy to check that the complete force system is in self equilibrium for the free-free structure; this is the topic of Exercise 3.1. §3.4.2. Recovery of Internal Forces and Stresses Often the structural engineer is not so much interested in displacements as in internal forces and stresses. These are in fact the most important quantities for preliminary structural design. In pinjointed trusses the only internal forces are the axial member forces. For the example truss these forces, denoted by F (1) , F (2) and F (3) , are depicted in Figure 3.3. The average axial stress σ e is obtained on dividing F e by the cross-sectional area of the member. 3 The axial force F e in member e can be obtained as follows. Extract the displacements of member e from the complete displacement solution u to form ue . Then F (3) recover local joint displacements from u¯ e = Te ue . F (2) Compute the member elongation d e (relative displacement) and recover the axial force from the equivalent spring constitutive relation: e
F (1)
1
2
e
E A e Fe = d . (3.19) d e = u¯ ex j − u¯ exi , Le Note that u¯ eyi and u¯ ey j are not needed in computing d e .
Figure 3.3. The internal forces in the example truss are the axial forces F (1) , F (2) and F (3) in the members. Directions shown pertain to tension.
Example 3.1. Recover F (2) in example truss. Member (2) goes from node 2 to node 3 and ϕ (2) = 90◦ .
Extract the global displacements of the member from (3.17): u(2) = [ u x2 u y2 u x3 u y3 ]T = [ 0 0 0.4 0.2 ]T . Convert to local displacements using u¯ (2) = T(2) u(2) :






u¯ x2 cos 90◦ sin 90◦ 0 0 u x2 0 ◦ ◦ 0 0   u y2   −1  u¯ y2   − sin 90 cos 90 =  u¯  =  0 0 cos 90◦ sin 90◦   u x3   0 x3 u¯ y3 0 0 − sin 90◦ cos 90◦ u y3 0 Then d (2) = u¯ x3 − u¯ x2 = −0.2 − 0 = −0.2, and F (2)
1 0 0 0
0 0 0 −1




0 0 0 0 0   0  = . 1   0.4   −0.2  0 −0.2 −0.4 (3.20) = (50/10) × −0.2 = −1 (compression).
Remark 3.3. An alternative interpretation of (3.19) is to regard ee = d e /L e as the (average) member axial
strain, σ e = E e ee as (average) axial stress, and F e = Ae σ e as the axial force. This is more in tune with the Theory of Elasticity viewpoint discussed in Exercise 2.6.
3–8
3–9
§3.5 *COMPUTER ORIENTED ASSEMBLY AND SOLUTION
§3.5.
*Computer Oriented Assembly and Solution
§3.5.1. *Assembly by Freedom Pointers The practical computer implementation of the DSM assembly process departs significantly from the “augment and add” technique described in §3.1.4. There are two major differences: (I)
Member stiffness matrices are not expanded. Their entries are directly merged into those of K through the use of a “freedom pointer array” called the Element Freedom Table or EFT.
(II) The master stiffness matrix K is stored using a special format that takes advantage of symmetry and sparseness. Difference (II) is a more advanced topic that is deferred to the last part of the book. For simplicity we shall assume here that K is stored as a full square matrix, and study only (I). For the example truss the freedom-pointer technique expresses the entries of K as the sum K pq =
3
K iej
for i = 1, . . . 4, j = 1, . . . 4, p = EFTe (i), q = EFTe ( j).
(3.21)
e=1
denote the entries of the 4 × 4 globalized member stiffness matrices in (2.19) through (2.21). Entries Here K pq that do not get any contributions from the right hand side remain zero. EFTe denotes the Element Freedom Table for member e. For the example truss these tables are K iej
EFT(1) = {1, 2, 3, 4},
EFT(2) = {3, 4, 5, 6},
EFT(3) = {1, 2, 5, 6}.
(3.22)
Physically these tables map local freedom indices to global ones. For example, freedom number 3 of member (2) is u x3 , which is number 5 in the master equations; consequently EFT(2) (3) = 5. Note that (3.21) involves three nested loops: over e (outermost), over i, and over j. The ordering of the last two is irrelevant. Advantage may be taken of the symmetry of Ke and K to roughly halve the number of additions. Exercise 3.5 follows the scheme (3.21) by hand. The assembly process for general structures using this technique is studied in Chapter 25. §3.5.2. *Applying DBC by Modification In §3.3.1 the support conditions (3.13) were applied by reducing (3.12) to (3.15). Reduction is convenient for hand computations because it cuts down on the number of equations to solve. But it has a serious flaw for computer implementation: the equations must be rearranged. It was previously noted that on the computer the number of equations is not the only important consideration. Rearrangement can be as or more expensive than solving the equations, particularly if the coefficient matrix is stored in sparse form or on secondary storage. To apply support conditions without rearranging the equations we clear (set to zero) rows and columns corresponding to prescribed zero displacements as well as the corresponding force components, and place ones on the diagonal to maintain non-singularity. The resulting system is called the modified set of master stiffness equations. For the example truss this approach yields
1 0 0 0 0 1 0 0  0 0 10 0  0 0 0 1  0 0
0 0
0 0
0 0
0 0 0 0 10 10


 
0 u x1 0 0   u y1   0      0    u x2  =  0  ,  0  u y2   0      u x3 10 2 u y3 15 1
(3.23)
in which rows and columns for equations 1, 2 and 4 have been cleared. Solving this modified system produces the complete displacement solution (3.17) directly.
3–9
3–10
Chapter 3: THE DIRECT STIFFNESS METHOD II
Remark 3.4. In a “smart” stiffness equation solver the modified system need not be explicitly constructed by
storing zeros and ones. It is sufficient to mark the equations that correspond to displacement BCs. The solver is then programmed to skip those equations. However, if one is using a standard solver from, say, a library of scientific routines or a commercial program such as Matlab or Mathematica, such intelligence cannot be expected, and the modified system must be set up explicitly. fy3 = 1
§3.6. Prescribed Nonzero Displacements
3
The support conditions considered in the example truss resulted in the specification of zero displacement components; for example u y2 = 0. There are cases, however, where the known value is nonzero. This happens, for example, in the study of settlement of foundations of ground structures such as buildings and bridges, and in the analysis of motion-driven machinery components.
u y2 = +0.4 going up
u y1 = −0.5 going down
1
;; ;;
2
;; ;;
Mathematically these are called non-homogenous boundary conditions. The treatment of this generalization of the FEM equations is studied in the following subsections.
fx3 = 2
ux1 = 0 no horizontal motion
Figure 3.4. The example truss with prescribed nonzero vertical displacements at joints 1 and 2.
§3.6.1. Application of Nonzero-DBCs by Reduction We describe first a matrix reduction technique, analogous to that used in §3.3.1, which is suitable for hand computations. Recall the master stiffness equations (3.12) for the example truss: 
20  10   −10   0  −10 −10
10 −10 10 0 0 10 0 0 −10 0 −10 0
0 0 0 5 0 −5
−10 −10 0 0 10 10
   −10 u x1 −10   u y1      0   u x2    = −5   u y2      10 u x3 15 u y3
 f x1 f y1   f x2   f y2   f x3 f y3
(3.24)
Suppose that the applied forces are again (3.14) but the prescribed displacements change to u x1 = 0,
u y1 = −0.5,
u y2 = 0.4
(3.25)
This means that joint 1 goes down vertically whereas joint 2 goes up vertically, as depicted in Figure 3.4. Inserting the known data into (3.24) we get 
20  10   −10   0  −10 −10
10 10 0 0 −10 −10
−10 0 0 0 10 0 0 5 0 0 0 −5
−10 −10 0 0 10 10 3–10
   −10 0 −10   −0.5      0   u x2    = −5   0.4      10 u x3 15 u y3
 f x1 f y1   0   f y2   2 1
(3.26)
3–11
§3.6 PRESCRIBED NONZERO DISPLACEMENTS
The first, second and fourth rows of (3.26) are removed, leaving only   0
 −0.5  0 −10 0 10 0 0 0    u  −10 −10 0 0 10 10  x2  = 2  0.4  1 −10 −10 0 −5 10 15   u x3 u y3
(3.27)
Columns 1, 2 and 4 are removed by transferring all known terms from the left to the right hand side:
0 (−10) × 0 + 0 × (−0.5) + 0 × 0.4 0 10 0 0 u x2 = −3 . u x3 = 2 − (−10) × 0 + (−10) × (−0.5) + 0 × 0.4 0 10 10 u y3 1 (−10) × 0 + (−10) × (−0.5) + (−5) × 0.4 −2 0 10 15 (3.28) These are the reduced stiffness equations. Note that its coefficient matrix of (3.28) is exactly the same as in the reduced system (3.15) for prescribed zero displacements. The right hand side, however, is different. It consists of the applied joint forces modified by the effect of known nonzero displacements. These are called the modified node forces or effective node forces. Solving the reduced system yields
0 u x2 (3.29) u x3 = −0.5 . u y3 0.2 Filling the missing entries with the known values (3.25) yields the complete displacement solution (listed as row vector to save space): u = [0
−0.5
0
0.4
−0.5
0.2 ]T .
(3.30)
Taking the solution (3.30) and going through the postprocessing steps discussed in §2.3, we can find that reaction forces and internal member forces do not change. This is a consequence of the fact that the example truss is statically determinate. The force systems (internal and external) in such structures are insensitive to movements such as foundation settlements. §3.6.2. *Application of Nonzero-DBCs by Modification The computer-oriented modifification approach follows the same idea outlined in §3.5.2. As there, the main objective is to avoid rearranging the master stiffness equations. To understand the process it is useful to think of being done in two stages. First equations 1, 2 and 4 are modified so that they become trivial equations, as illustrated for the example truss and the dispolacement boundary conditions (3.25):

1  0  −10   0  −10 −10
0 0 1 0 0 10 0 0 −10 0 −10 0




0 0 0 u x1 0 0 0 0   u x2   −0.5      0 0 0   u x2  =  0       1 0 0   u y2   0.4  u x3 0 10 10 2 u y3 −5 10 15 1
(3.31)
The solution of this system recovers (3.26) by construction (for example, the fourth equation is simply 1×u y2 = 0.4). In the next stage, columns 1, 2 and 4 of the coefficient matrix are cleared by transferring all known terms
3–11
3–12
Chapter 3: THE DIRECT STIFFNESS METHOD II
to the right hand side, following the same procedure explained in (3.29). We thus arrive at
1 0 0 0 0 1 0 0  0 0 10 0  0 0 0 1  0 0
0 0
0 0
0 0
0 0 0 0 10 10




0 u x1 0 0   u x2   −0.5      0    u x2  =  0       0   u y2   0.4  u x3 10 −3 u y3 15 −2
(3.32)
As before, these are called the modified master stiffness equations. Observe that the equations retain the original number and order. Solving this system yields the complete displacement solution (3.30). Note that if all prescribed displacements are zero, forces on the right hand side are not modified, and one would get (3.23) as may be expected. Remark 3.5. The modification is not actually programmed as discussed above. First the applied forces in the right-hand side are modified for the effect of nonzero prescribed displacements, and the prescribed displacements stored in the reaction-force slots. This is called the force modification step. Second, rows and columns of the stiffness matrix are cleared as appropriate and ones stored in the diagonal positions. This is called the stiffness modification step. It is essential that the procedural steps be executed in the indicated order, because stiffness terms must be used to modify forces before they are zeroed out.
§3.6.3. *Matrix Forms of Nonzero-DBC Application Methods The reduction and modification techniques for applying DBCs can be presented in compact matrix form. First, the free-free master stiffness equations Ku = f are partitioned as follows:
K11 K21
K12 K22
000e
u1 u2
000e
=
000e
f1 . f2
(3.33)
In this matrix equation, subvectors u2 and f1 collect displacement and force components, respectively, that are known, given or prescribed. Subvectors u1 and f2 collect force and displacement components, respectively, that are unknown. Forces in f2 represent reactions on supports; consequently f2 is called the reaction vector. On transferring the known terms to the right hand side the first matrix equation becomes K11 u1 = f1 − K12 u2 .
(3.34)
This is the reduced master equation system. If the support B.C.s are homogeneous (that is, all prescribed displacements are zero), u2 = 0, and we do not need to change the right-hand side: K11 u1 = f1 .
(3.35)
Examples that illustrate (3.34) and (3.35) are (3.28) and (2.23), respectively. The computer-oriented modification technique retains the same joint displacement vector as in (3.34) through the following rearrangement:
000e 000e 000e K11 0 u1 f − K12 u2 = 1 . (3.36) u2 u2 0 I This modified system is simply the reduced equation (3.35) augmented by the trivial equation Iu2 = u2 . This system is often denoted as 000f Ku = ˆf. (3.37) Solving (3.37) yields the complete displacement solution including the specified displacements u2 .
3–12
3–13
§3.
References
For the computer implementation it is important to note that the partitioned form (3.33) is only used to permit the use of compact matrix notation. In actual programming the equations are not explicitly rearranged: they retain their original numbers. For instance, in the example truss
u1 =
u x1 u y1 u y2
≡
DOF #1 DOF #2 DOF #4
,
u2 =
u x2 u x3 u y3
≡
DOF #3 DOF #5 DOF #6
.
(3.38)
The example shows that u1 and u2 are generally interspersed throughout u. Thus, matrix operations such as K12 u2 involve indirect (pointer) addressing so as to avoid explicit array rearrangement. Notes and Bibliography The coverage of the assembly and solution steps of the DSM, along with globalization and application of BCs, is not uniform across the wide spectrum of FEM books. Authors have introduced “quirks” although the overall concepts are not affected. The most common variations arise in two contexts: (1)
Some treatments apply support conditions during merge, explicitly eliminating known displacement freedoms as the elements are processed and merged into K. The output of the assembly process is what is called here a reduced stiffness matrix.1
(2)
In the frontal solution method of Irons [98,99], assembly and solution are done concurrently. More precisely, as elements are formed and merged, displacement boundary conditions are applied, and Gauss elimination and reduction of the right hand side starts once the assembler senses (by tracking an “element wavefront”) that no more elements contribute to a certain node.
Both variants appeared in FEM programs written during the 1960s and 1970s. They were motivated by computer resource limitations of the time: memory was scarce and computing time expensive.2 On the negative side, interweaving leads to unmodular programming (which easily becomes “spaghetti code” in lowlevel languages such as Fortran). Since a frontal solver has to access the element library, which is typically the largest component of a general-purpose FEM program, it has to know how to pass and receive information about each element. A minor change deep down the element library can propagate and break the solver. Squeezing storage and CPU savings on present computers is of less significance. Modularity, which simplifies scripting in higher order languages such as Matlab is desirable because it increases “plug-in” operational flexibility, allows the use of built-in solvers, and reduces the chance for errors. These changes reflect economic reality: human time is nowadays far more expensive than computer time. A side benefit of modular assembly-solution separation is that often the master stiffness must be used in a different way than just solving Ku = f; for example in dynamics, vibration or stability analysis. Or as input to a model reduction process. In those cases the solution stage can wait. Both the hand-oriented and computer-oriented application of boundary conditions have been presented here, although the latter is still considered an advanced topic. While hand computations become unfeasible beyond fairly trivial models, they are important from a instructional standpoint. The augment-and-add procedure for hand assembly of the master stiffness matrix is due to H. Martin [112]. References Referenced items have been moved to Appendix R. 1
For the example truss, the coefficient matrix in (3.15) is a reduced stiffness whereas that in (3.23) is a modified one.
2
As an illustration, the first computer used by the writer, the “classical mainframe” IBM 7094, had a magnetic-core memory of 32,768 36-bit words (≈ 0.2 MB), and was as fast as an IBM PC of the mid 1980s. One mainframe had to serve the whole Berkeley campus, and Ph.D. students were allocated 2 CPU hours per semester. Getting a moderately complex FE model through involved heavy use of slower secondary storage such as disk or tape in batch jobs.
3–13
3–14
Chapter 3: THE DIRECT STIFFNESS METHOD II
Homework Exercises for Chapter 3 The Direct Stiffness Method II EXERCISE 3.1 [A:15] Draw a free body diagram of the nodal forces (3.18) acting on the free-free truss structure, and verify that this force system satisfies translational and rotational (moment) equilibrium. EXERCISE 3.2 [A:15] Using the method presented in §3.4.2 compute the axial forces in the three members √ of the example truss. Partial answer: F (3) = 2 2. EXERCISE 3.3 [A:20] Describe an alternative method that recovers the axial member forces of the example truss from consideration of joint equilibrium, without going through the computation of member deformations. Can this method be extended to arbitrary trusses? EXERCISE 3.4 [A:20] Suppose that the third support condition in (3.13) is u x2 = 0 instead of u y2 = 0. Rederive the reduced system (3.15) for this case. Verify that this system cannot be solved for the joint displacements u y2 , u x3 and u y3 because the reduced stiffness matrix is singular.3 Offer a physical interpretation of this failure. EXERCISE 3.5 [N:20] Construct by hand the free-free master stiffness matrix of (3.12) using the freedom-
pointer technique (3.21). Note: start from K initialized to the null matrix, then cycle over e = 1, 2, 3.
f y2 = 0 u y2
E, A(1)
(1)
H
;;
f x2 = P
u x2
2
(2)
y
1
S/2
S
;; 3
x S/2
E, A(2)
Figure E3.1. Truss structure for Exercise 3.7.
EXERCISE 3.6 [N:25] Consider the two-member arch-truss structure shown in Figure E3.1. Take span
S = 8, height H = 3, elastic modulus E = 1000, cross section areas A(1) = 2 and A(2) = 4, and horizontal crown force P = f x2 = 12. Using the DSM carry out the following steps: (a)
Assemble the master stiffness equations. Any method: augment-and-add, or the more advanced “freedom pointer” technique explained in §3.5.1, is acceptable.
(b)
Apply the displacement BCs and solve the reduced system for the crown displacements u x2 and u y2 . Partial result: u x2 = 9/512 = 0.01758.
(c)
Recover the node forces at all joints including reactions. Verify that overall force equilibrium (x forces, y forces, and moments about any point) is satisfied.
(d)
Recover the axial forces in the two members. Result should be F (1) = −F (2) = 15/2.
3
A matrix is singular if its determinant is zero; cf. §C.2 of Appendix C for a “refresher” in that topic.
3–14
3–15
Exercises
EXERCISE 3.7 [N:20] Resolve items (a) through (c) — omitting (d) — of the problem of Exercise 3.7 if the
vertical right support “sinks” so that the displacement u y3 is now prescribed to be −0.5. Everything else is the same. Use the matrix reduction scheme to apply the displacement BCs.
3–15
5
.
Analysis of Example Truss by a CAS
5–1
5–2
Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
TABLE OF CONTENTS Page
§5.1.
§5.2.
§5.3. §5.4. §5.5. §5.6. §5.7.
§5. §5. §5.
Computer Algebra Systems §5.1.1. Why Mathematica? . . . . . §5.1.2. Programming Style and Prerequisites §5.1.3. Class Demo Scripts . . . . . The Element Stiffness Module §5.2.1. Module Description . . . . . . §5.2.2. Programming Remarks . . . . §5.2.3. Case Sensitivity . . . . . . . §5.2.4. Testing the Member Stiffness Module Merging a Member into the Master Stiffness Assembling the Master Stiffness Modifying the Master System Recovering Internal Forces Putting the Pieces Together §5.7.1. The Driver Program . . . . . . §5.7.2. Is All of This Worthwhile? . . . Notes and Bibliography . . . . . . . . . . . . . References . . . . . . . . . . . . Exercises . . . . . . . . . . . . .
5–2
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5–3 5–3 5–3 5–4 5–4 5–4 5–7 5–7 5–7 5–7 5–9 5–9 5–11 5–12 5–12 5–13 5–14 5–15 5–16
5–3
§5.1 COMPUTER ALGEBRA SYSTEMS
§5.1. Computer Algebra Systems Computer algebra systems, known by the acronym CAS, are programs designed to perform symbolic and numeric manipulations following the rules of mathematics.1 The development of such programs began in the mid 1960s. The first comprehensive system — the “granddaddy” of them all, called Macsyma (an acronym for Project Mac Symbolic Manipulator) — was developed using the programming language Lisp at MIT’s famous Artificial Intelligence Laboratory over the period 1967 to 1980. The number and quality of symbolic-manipulation programs has expanded dramatically since the availability of graphical workstations and personal computers has encouraged interactive and experimental programming. As of this writing the leading general-purpose contenders are Maple and Mathematica.2 In addition there are a dozen or so more specialized programs, some of which are available free or at very reasonable cost. §5.1.1. Why Mathematica? In the present book Mathematica will be used for Chapters and Exercises that develop symbolic and numerical computation for matrix structural analysis and FEM implementations. Mathematica is a commercial product developed by Wolfram Research, web site: http://www.wolfram.com. The version used to construct the code fragments presented in this Chapter is 4.1, which was commercially released in 2001. (The latest version, released summer 2003, is 5.0.) The main advantages of Mathematica for technical computing are: 1.
Availability on a wide range of platforms that range from PCs and Macs through Unix workstations. Its competitor Maple is primarily used on Unix systems.
2.
Up-to-date user interface with above average graphics. On all machines Mathematica offers a graphics user interface called the Notebook front-end. This is mandatory for serious work.
3.
A powerful programming language.
4.
Good documentation and abundance of application books at all levels.
One common disadvantage of CAS, and Mathematica is not exception, is computational inefficiency in numerical calculations compared with a low-level implementation in, for instance, C or Fortran. The relative penalty can reach several orders of magnitude. For instructional use, however, the penalty is acceptable when compared to human efficiency. This means the ability to get FEM programs up and running in very short time, with capabilities for symbolic manipulation and graphics as a bonus. §5.1.2. Programming Style and Prerequisites The following material assumes that you are a moderately experienced user of Mathematica, or are willing to learn to be one. See Notes and Bibliography for a brief discussion of tutorial and reference materials. Practice with the program until you reach the level of writing functions, modules and scripts with relative ease. With the Notebook interface and a good primer it takes only a few hours. 1
Some vendors call that kind of activity “doing mathematics by computer.” It is more appropriate to regard such programs as enabling tools that help humans with complicated and error-prone manipulations. As of now, only humans can do mathematics.
2
Another commonly used program for engineering computations: Matlab, does only numerical computations although an interface to Maple can be purchased as a toolbox.
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Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
5–4
When approaching that level you may notice that functions in Mathematica display many aspects similar to C.3 You can exploit this similarity if you are proficient in that language. But Mathematica functions do have some unique aspects, such as matching arguments by pattern, and the fact that internal variables are global unless otherwise made local.4 Although function arguments can be modified, in practice this should be avoided because it may be difficult to trace side effects. The programming style enforced here outlaws output arguments and a function can only return its name. But since the name can be a list of arbitrary objects the restriction is not serious.5 Our objective is to develop a symbolic program written in Mathematica that solves the example plane truss as well as some symbolic versions thereof. The program will rely heavily on the development and use of functions implemented using the Module construct of Mathematica. Thus the style will be one of procedural programming.6 The program will not be particularly modular (in the computer science sense) because Mathematica is not suitable for that programming style.7 The code presented in Sections 5.2–5.7 uses a few language constructs that may be deemed as advanced, and these are briefly noted in the text so that appropriate reference to the Mathematica reference manual can be made. §5.1.3. Class Demo Scripts The cell scripts shown in Figures 5.1 and 5.2 will be used to illustrate the organization of a Notebook file and the “look and feel” of some basic Mathematica commands. These scripts will be demonstrated in class from a laptop. §5.2. The Element Stiffness Module As our first FEM code segment, the top box of Figure 5.3 shows a module that evaluates and returns the 4 × 4 stiffness matrix of a plane truss member (two-node bar) in global coordinates. The text in that box of that figure is supposed to be placed on a Notebook cell. Executing the cell, by clicking on it and hitting an appropriate key ( on a Mac), gives the output shown in the bottom box. The contents of the figure is described in further detail below. §5.2.1. Module Description The stiffness module is called ElemStiff2DTwoNodeBar. Such descriptive names are permitted by the language. This reduces the need for detailed comments.
3
Simple functions can be implemented in Mathematica directly, for instance DotProduct[x ,y ]:=x.y; more complicated functions are handled by the Module construct. These things are called rules by computer scientists.
4
In Mathematica everything is a function, including programming constructs. Example: in C for is a loop-opening keyword; in Mathematica For is a function that runs a loop according to its arguments.
5
Such restrictions on arguments and function returns are closer in spirit to C than Fortran although you can of course modify C-function arguments using pointers — exceedingly dangerous but often unavoidable.
6
The name Module should not be taken too seriously: it is far away from the concept of modules in Ada, Modula, Oberon or Fortran 90. But such precise levels of interface control are rarely needed in symbolic languages.
7
And indeed none of the CAS packages in popular use is designed for strong modularity because of historical and interactivity constraints.
5–4
5–5
§5.2 THE ELEMENT STIFFNESS MODULE
Integration example f[x_,Α_,Β_]:=(1+Β*x^2)/(1+Α*x+x^2); F=Integrate[f[x,-1,2],{x,0,5}]; F=Simplify[F]; Print[F]; Print[N[F]]; F=NIntegrate[f[x,-1,2],{x,0,5}]; Print['F=',F//InputForm];
10 0001 Log0001210002 13.0445 F000213.044522437723455 Figure 5.1. Example cell for class demo.
Fa=Integrate[f[z,a,b],{z,0,5}]; Fa=Simplify[Fa]; Print[Fa]; Plot3D[Fa,{a,-1,1},{b,-10,10},ViewPoint->{-1,-1,1}]; Print['Fa=',Fa//InputForm] 0005 00030004000400040004000400040004000400040004000400040004 0006a 00030004000400040004000400040004000400040004000400040004 2 0007 2 2 0007 00030003000300030003000300030003 0003000300030003 0001 0007 000300040004000400040004000400040004 00040004000400040004 000710 4 0004 a b 0004 a 4 0004 a b Log000126 0001 5 a0002 0004 0006 2 0001 00042 0001 a b Log 1 0004 00030003000300030003000300030003 4 0004 a2 0006 0006a 0006a 0006a 2 0006 Log 1 0001 00030003000300030003000300030003 00030003000300030003000300030003 0003000300030003 0004 2 0006 b Log 1 0001 00030003000300030003000300030003 00030003000300030003000300030003 0003000300030003 0001 0006 a2 b Log 1 0001 00030003000300030003000300030003 00030003000300030003000300030003 0003000300030003 0001 000300040004000400040004000400040004 00040004000400040004 000300040004000400040004000400040004 00040004000400040004 000300040004000400040004000400040004 00040004000400040004 4 0004 a2 4 0004 a2 4 0004 a2 00030004000400040004000400040004000400040004000400040004 00030004000400040004000400040004000400040004000400040004 00030004000400040004000400040004000400040004000400040004 000410 0006 0004 0006 a 0001 4 0004 a2 000410 0006 0004 0006 a 0001 4 0004 a2 000410 0006 0004 0006 a 0001 4 0004 a2 000300030003000300030003000300030003000300030003000300030003000300030003 0004 2 0006 b Log 00030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 000300030003000300030003000300030003000300030003000300030003000300030003 0001 0006 a2 b Log 00030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 000300030003000300030003000300030003000300030003000300030003000300030003 0004 2 0006 Log 00030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 00030004000400040004000400040004000400040004000400040004 000300040004000400040004000400040004 0004 0004000400040004 000300040004000400040004000400040004 0004 0004000400040004 4 0004 a2 4 0004 a2 4 0004 a2 00030004000400040004000400040004000400040004000400040004 00030004000400040004000400040004000400040004000400040004 00030004000400040004000400040004000400040004000400040004 10 0006 0001 0006 a 0001 4 0004 a2 10 0006 0001 0006 a 0001 4 0004 a2 10 0006 0001 0006 a 0001 4 0004 a2 000e 2 0006 Log 00030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 000300030003000300030003000300030003000300030003000300030003 0001 2 0006 b Log 00030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 000300030003000300030003000300030003000300030003000300030003 0004 0006 a2 b Log 00030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 000300030003000300030003000300030003000300030003000300030003 000e 000e 00030004000400040004000400040004000400040004000400040004 00030004000400040004000400040004000400040004000400040004 00030004000400040004000400040004000400040004000400040004 000e 2 2 40004a 40004a 4 0004 a2
1 0003000300030003000300030003000300030003000300030003000300030003 00030004000400040004000400040004000400030003000300030003000300030003 00040004000400040004 2 4 0004 a2
50 25 0 -25 1-50
10 5
0.5 0
0 -5
-0.5 -10 -1
Fa000210 0007 Sqrt00014 0004 a ^ 20002 0007 b 0004 a 0007 Sqrt00014 0004 a ^ 20002 0007 b 0007 Log000126 0001 5 0007 a0002 0004 I 0007 2 0001 00042 0001 a ^ 2 0007 b 0007 Log00011 0004 I 0007 a 000f Sqrt00014 0004 a ^ 200020002 0001 2 0007 I 0007 Log00011 0001 I 0007 a 000f Sqrt00014 0004 a ^ 200020002 0004 2 0007 I 0007 b 0007 Log00011 0001 I 0007 a 000f Sqrt00014 0004 a ^ 200020002 0001 I 0007 a ^ 2 0007 b 0007 Log00011 0001 I 0007 a 000f Sqrt00014 0004 a ^ 200020002 0001 2 0007 I 0007 Log0001000410 0007 I 0004 I 0007 a 0001 Sqrt00014 0004 a ^ 20002 000f Sqrt00014 0004 a ^ 200020002 0004 2 0007 I 0007 b 0007 Log0001000410 0007 I 0004 I 0007 a 0001 Sqrt00014 0004 a ^ 20002 000f Sqrt00014 0004 a ^ 200020002 0001 I 0007 a ^ 2 0007 b 0007 Log0001000410 0007 I 0004 I 0007 a 0001 Sqrt00014 0004 a ^ 20002 000f Sqrt00014 0004 a ^ 200020002 0004 2 0007 I 0007 Log000110 0007 I 0001 I 0007 a 0001 Sqrt00014 0004 a ^ 20002 000f Sqrt00014 0004 a ^ 200020002 0001 2 0007 I 0007 b 0007 Log000110 0007 I 0001 I 0007 a 0001 Sqrt00014 0004 a ^ 20002 000f Sqrt00014 0004 a ^ 200020002 0004 I 0007 a ^ 2 0007 b 0007 Log000110 0007 I 0001 I 0007 a 0001 Sqrt00014 0004 a ^ 20002 000f Sqrt00014 0004 a ^ 200020002 000f 2 0007 Sqrt00014 0004 a ^ 20002
Figure 5.2. Another example cell for class demo.
5–5
Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
5–6
ElemStiff2DTwoNodeBar[{{x1_,y1_},{x2_,y2_}},{Em_,A_}] := Module[{c,s,dx=x2-x1,dy=y2-y1,L,Ke}, L=Sqrt[dx^2+dy^2]; c=dx/L; s=dy/L; Ke=(Em*A/L)* {{ c^2, c*s,-c^2,-c*s}, { c*s, s^2,-s*c,-s^2}, {-c^2,-s*c, c^2, s*c}, {-s*c,-s^2, s*c, s^2}}; Return[Ke] ]; Ke= ElemStiff2DTwoNodeBar[{{0,0},{10,10}},{100,2*Sqrt[2]}]; Print['Numerical elem stiff matrix:']; Print[Ke//MatrixForm]; Ke= ElemStiff2DTwoNodeBar[{{0,0},{L,L}},{Em,A}]; Ke=Simplify[Ke,L>0]; Print['Symbolic elem stiff matrix:']; Print[Ke//MatrixForm]; Numerical elem stiff matrix: 10 10 000410 000410 0005 000e 0007 000e 0007 0007 000e 0007 10 10 000410 000410 000e 000e 0007 000e 0007 000e 0007 0007 000e 0007 000410 000410 10 10 000e 000e 0007 000e 0007 000e 0007 10 0006 000410 000410 10 Symbolic elem stiff matrix: A Em 0005 00030003000300030003000300030003 00030004000400040004000300030003000300030003 0007 2 2 L 0007 0007 0007 0007 A Em 0007 0007 00030003000300030003000300030003 00030004000400040004000300030003000300030003 0007 0007 2 2 L 0007 0007 0007 0007 A Em 0007 0007 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 0007 0007 2 2 L 0007 0007 0007 0007 A Em 0007 0004 00030003000300030003000300030003 000300030003000300030003 2 L 0006 2 00030004000400040004
A Em 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2
2 L A Em 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L A Em 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L A Em 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L
A Em 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2
2 L A Em 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L A Em 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L A Em 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L
A Em 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L 000e 000e 000e 000e 000e A Em 000e 000e 0004 00030003000300030003000300030003 00030004000400040004000300030003000300030003 000e 2 2 L 000e 000e 000e 000e 000e A Em 000e 000e 00030003000300030003000300030003 00030004000400040004000300030003000300030003 000e 2 2 L 000e 000e 000e 000e 000e A Em 000e 00030003000300030003000300030003 00030004000400040004000300030003000300030003 2 2 L
Figure 5.3. Module ElemStiff2DTwoNodeBar to form the element stiffness of a 2D 2-node truss element in global coordinates: test program and its output.
The module takes two arguments: { { x1,y1 },{ y1,y2 } }
A two-level list8 containing the {x, y} coordinates of the bar end nodes labelled as 1 and 2.9
{ Em,A }
A one-level list containing the bar elastic modulus, E and the member cross section area, A. See §5.2.3 as to why name E cannot be used.
The use of the underscore after argument item names in the declaration of the Module is a requirement for pattern-matching in Mathematica. If, as recommended, you have learned functions and modules this aspect should not come as a surprise. The module name returns the 4 × 4 member stiffness matrix internally called Ke. The logic that leads to the formation of that matrix is straightforward and need not be explained in detail. Note, however, the elegant direct declaration of the matrix Ke as a level-two list, which eliminates the fiddling around with array indices typical of standard programming languages. The format in fact closely matches the mathematical expression (3.4).
8
A level-one list is a sequence of items enclosed in curly braces. For example: { x1,y1 } is a list of two items. A level-two list is a list of level-one lists. An important example of a level-two list is a matrix.
9
These are called the local node numbers, and replace the i, j of previous Chapters. This is a common FEM programming practice.
5–6
5–7
§5.3 MERGING A MEMBER INTO THE MASTER STIFFNESS
§5.2.2. Programming Remarks The function in Figure 5.3 uses several intermediate variables with short names: dx, dy, s, c and L. It is strongly advisable to make these symbols local to avoid potential names clashes somewhere else.10 In the Module[ ..] construct this is done by listing those names in a list immediately after the opening bracket. Local variables may be initialized when they are constants or simple functions of the argument items; for example on entry to the module dx=x2-x1 initializes variable dx to be the difference of x node coordinates, namely 0003x = x2 − x1 . The use of the Return statement fulfills the same purpose as in C or Fortran 90. Mathematica guides and textbooks advise against the use of that and other C-like constructs. The writer strongly disagrees: the Return statement makes clear what the Module gives back to its invoker and is self-documenting. §5.2.3. Case Sensitivity Mathematica, like most recent computer languages, is case sensitive so that for instance E is not the same as e. This is fine. But the language designer decided that names of system-defined objects such as built-in functions and constants must begin with a capital letter. Consequently the liberal use of names beginning with a capital letter may run into clashes. For example you cannot use E because of its built-in meaning as the base of natural logarithms.11 In the code fragments presented throughout this book, identifiers beginning with upper case are used for objects such as stiffness matrices, modulus of elasticity, and cross section area, following established usage in Mechanics. When there is danger of clashing with a protected system symbol, additional lower case letters are used. For example, Em is used for the elastic modulus instead of E because the latter is a reserved symbol. §5.2.4. Testing the Member Stiffness Module Following the definition of ElemStiff2DTwoNodeBar in Figure 5.3 there are several statements that constitute the module test program that call the module and print the returned results. Two cases are tested. First, the stiffness of member (3) of the example truss, using all-numerical values. Next, some of the input arguments for the same member are given symbolic names so they stand for variables; for example the elastic module is given as Em instead of 100 as in the foregoing test. The print output of the test is shown in the lower portion of Figure 5.3. The first test returns the member stiffness matrix (3.10) as may be expected. The second test returns a symbolic form in which three symbols appear: the coordinates of end node 2, which is taken to be located at { L,L } instead of {10, 10}, A, which is the cross-section area and Em, which is the elastic modulus. Note that the returning matrix Ke is subject to a Simplify step before printing it, which is the subject of an Exercise. The ability to carry along variables is of course a fundamental capability of any CAS and the main reason for which such programs are used.
10
The “global by default” choice is the worst one, but we must live with the rules of the language.
11
In retrospect this appears to have been a highly questionable decision. System defined names should have been identified by a reserved prefix or postfix to avoid surprises, as done in Macsyma or Maple. Mathematica issues a warning message, however, if an attempt to redefine a “protected symbol” is made.
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Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
5–8
MergeElemIntoMasterStiff[Ke_,eftab_,Kin_]:=Module[ {i,j,ii,jj,K=Kin}, For [i=1, i<=4, i++, ii=eftab[[i]]; For [j=i, j<=4, j++, jj=eftab[[j]]; K[[jj,ii]]=K[[ii,jj]]+=Ke[[i,j]] ] ]; Return[K] ]; K=Table[0,{6},{6}]; Print['Initialized master stiffness matrix:']; Print[K//MatrixForm] Ke=ElemStiff2DTwoNodeBar[{{0,0},{10,10}},{100,2*Sqrt[2]}]; Print['Member stiffness matrix:']; Print[Ke//MatrixForm]; K=MergeElemIntoMasterStiff[Ke,{1,2,5,6},K]; Print['Master stiffness after member merge:']; Print[K//MatrixForm]; Initialized master stiffness matrix: 00050 0007 0007 0007 0007 0 0007 0007 0007 0007 0007 0 0007 0007 0007 0007 0007 0 0007 0007 0007 0007 0007 0 0007 0007 0007 00060
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 000e 000e 000e 0000e 000e 000e 000e 000e 0000e 000e 000e 000e 000e 0000e 000e 000e 000e 000e 0000e 000e 000e 000e 0
Member stiffness matrix: 10 10 000410 000410 0005 000e 0007 000e 0007 0007 000e 0007 10 10 000410 000410 000e 000e 0007 000e 0007 000e 0007 000e 0007 000e 0007 000410 000410 10 10 000e 0007 000e 0007 000e 0007 10 0006 000410 000410 10 Master stiffness after member merge: 10 10 0 0005 0007 0007 0007 0007 10 10 0 0007 0007 0007 0007 0007 0 0 0 0007 0007 0007 0007 0007 0 0 0 0007 0007 0007 0007 0007 000410 000410 0 0007 0007 0007 0006 000410 000410 0
0 000410 000410 000e 000e 000e 0 000410 000410 000e 000e 000e 000e 000e 0 0 0 000e 000e 000e 000e 000e 0 0 0 000e 000e 000e 000e 000e 0 10 10 000e 000e 000e 000e 0 10 10
Figure 5.4. Module MergeElemIntoMasterStiff to merge a 4 × 4 bar element stiffness into the master stiffness matrix: test program and its output.
§5.3. Merging a Member into the Master Stiffness The next fragment of Mathematica code, listed in Figure 5.4, is used in the assembly step of the DSM. Module MergeElemIntoMasterStiff receives the 4 × 4 element stiffness matrix formed by FormElemStiff2DNodeBar and “merges” it into the master stiffness matrix. The module takes three arguments: Ke
The 4 × 4 member stiffness matrix to be merged. This is a level-two list.
eftab
The column of the Element Freedom Table, defined in §3.4.1, appropriate to the member being merged; cf. (3.29). Recall that the EFT lists the global equation numbers for the four member degrees of freedom. This is a level-one list consisting of 4 integers.
Kinp
The incoming 6 × 6 master stiffness matrix. This is a level-two list.
MergeElemIntoMasterStiff returns, as module name, the updated master stiffness matrix internally called K with the member stiffness merged in. Thus we encounter here a novelty: an input-output argument. Because a formal argument cannot be modified, the situation is handled by copying the 5–8
5–9
§5.5 MODIFYING THE MASTER SYSTEM AssembleMasterStiffOfExampleTruss[]:= Module[{Ke,K=Table[0,{6},{6}]}, Ke=ElemStiff2DTwoNodeBar[{{0,0},{10,0}},{100,1}]; K= MergeElemIntoMasterStiff[Ke,{1,2,3,4},K]; Ke=ElemStiff2DTwoNodeBar[{{10,0},{10,10}},{100,1/2}]; K= MergeElemIntoMasterStiff[Ke,{3,4,5,6},K]; Ke=ElemStiff2DTwoNodeBar[{{0,0},{10,10}},{100,2*Sqrt[2]}]; K= MergeElemIntoMasterStiff[Ke,{1,2,5,6},K]; Return[K] ]; K=AssembleMasterStiffOfExampleTruss[]; Print['Master stiffness of example truss:']; Print[K//MatrixForm]; Master stiffness of example truss: 10 000410 0 000410 000410 0005 20 000e 0007 000e 0007 0007 000e 0007 10 10 0 0 000410 000410 000e 000e 0007 000e 0007 000e 0007 000e 0007 000e 0007 000410 0 10 0 0 0 000e 0007 000e 0007 000e 0007 000e 0007 000e 0007 0 0 0 5 0 00045 000e 0007 000e 0007 000e 0007 0007 000e 0007 0 10 10 000e 000e 0007 000e 0007 000410 000410 0 000e 0007 15 0006 000410 000410 0 00045 10
Figure 5.5. Module AssembleMasterStiffOfExampleTruss that forms the 6 × 6 master stiffness matrix of the example truss, test program and its output.
incoming Kin into K on entry. It is the copy which is updated and returned via the Return statement. The implementation has a strong C flavor with two nested For loops. Because the iterators are very simple, nested Do loops could have been used as well. The statements after the module provide a simple test. Before the first call to this function, the master stiffness matrix must be initialized to a zero 6 × 6 array. This is done in the first test statement using the Table function. The test member stiffness matrix is that of member (3) of the example truss, and is obtained by calling ElemStiff2DTwoNodeBar. The EFT is { 1,2,5,6 } since element freedoms 1,2,3,4 map into global freedoms 1,2,5,6. Running the test statements yields the listing given in Figure 5.4. The result is as expected. §5.4. Assembling the Master Stiffness The module AssembleMasterStiffOfExampleTruss, listed in the top box of Figure 5.5, makes use of the foregoing two modules: ElemStiff2DTwoNodeBar and MergeElemIntoMasterStiff, to form the master stiffness matrix of the example truss. The initialization of the stiffness matrix array in K to zero is done by the Table function of Mathematica, which is handy for initializing lists. The remaining statements are self explanatory. The module is similar in style to argumentless Fortran or C functions. It takes no arguments. All the example truss data is “wired in.” The output from the test program in is shown in the lower box of Figure 5.5. The output stiffness matches that in Equation (3.20), as can be expected if all fragments used so far work correctly. §5.5. Modifying the Master System Following the assembly process the master stiffness equations Ku = f must be modified to account for single-freedom displacement boundary conditions. This is done through the computer-oriented equation modification process described in §3.4.2. Module ModifiedMasterStiffForDBC carries out this process for the master stiffness matrix K, whereas ModifiedMasterForcesForDBC does this for the nodal force vector f. These two modules 5–9
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Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
ModifiedMasterStiffForDBC[pdof_,K_] := Module[ {i,j,k,nk=Length[K],np=Length[pdof],Kmod=K}, For [k=1,k<=np,k++, i=pdof[[k]]; For [j=1,j<=nk,j++, Kmod[[i,j]]=Kmod[[j,i]]=0]; Kmod[[i,i]]=1]; Return[Kmod] ]; ModifiedMasterForcesForDBC[pdof_,f_] := Module[ {i,k,np=Length[pdof],fmod=f}, For [k=1,k<=np,k++, i=pdof[[k]]; fmod[[i]]=0]; Return[fmod] ]; K=Array[Kij,{6,6}]; Print['Assembled master stiffness:']; Print[K//MatrixForm]; K=ModifiedMasterStiffForDBC[{1,2,4},K]; Print['Master stiffness modified for displacement B.C.:']; Print[K//MatrixForm]; f=Array[fi,{6}]; Print['Force vector:']; Print[f]; f=ModifiedMasterForcesForDBC[{1,2,4},f]; Print['Force vector modified for displacement B.C.:']; Print[f]; Assembled master stiffness: Kij00011, 10002 Kij00011, 20002 Kij00011, 0005 0007 0007 0007 0007 Kij00012, 10002 Kij00012, 20002 Kij00012, 0007 0007 0007 0007 0007 0007 Kij00013, 10002 Kij00013, 20002 Kij00013, 0007 0007 0007 0007 Kij00014, 10002 Kij00014, 20002 Kij00014, 0007 0007 0007 0007 0007 0007 0007 0007 Kij00015, 10002 Kij00015, 20002 Kij00015, 0006 Kij00016, 10002 Kij00016, 20002 Kij00016,
30002 30002 30002 30002 30002 30002
Kij00011, Kij00012, Kij00013, Kij00014, Kij00015, Kij00016,
40002 40002 40002 40002 40002 40002
Kij00011, Kij00012, Kij00013, Kij00014, Kij00015, Kij00016,
50002 50002 50002 50002 50002 50002
Kij00011, Kij00012, Kij00013, Kij00014, Kij00015, Kij00016,
60002 000e 000e 000e 60002 000e 000e 000e 000e 000e 60002 000e 000e 000e 000e 000e 60002 000e 000e 000e 000e 000e 000e 60002 000e 000e 000e 60002
Master stiffness modified for displacement B.C.: 0 0 0 0 00051 0 0007 000e 0007 000e 0007 000e 0007 000e 0 1 0 0 0 0 0007 000e 0007 000e 0007 000e 0007 000e 0007 000e 0 0 Kij00013, 30002 0 Kij00013, 50002 Kij00013, 60002 0007 000e 0007 000e 0007 000e 0007 000e 0007 000e 0 0 0 1 0 0 0007 000e 0007 000e 0007 000e 0007 000e 0007 0007 000e 0007 0 0 Kij00015, 30002 0 Kij00015, 50002 Kij00015, 60002 000e 000e 0007 000e 0006 0 0 Kij00016, 30002 0 Kij00016, 50002 Kij00016, 60002 Force vector: 0010fi000110002, fi000120002, fi000130002, fi000140002, fi000150002, fi0001600020011 Force vector modified for displacement B.C.: 00100, 0, fi000130002, 0, fi000150002, fi0001600020011
Figure 5.6. Modules ModifiedMasterStiffForDBC and ModifiedMasterForceForDBC that modify the master stiffness matrix and force vector of a truss to impose displacement BCs.
are listed in the top box of Figure 5.6, along with test statements. The logic of both functions, but especially that of ModifiedMasterForcesForBC, is considerably simplified by assuming that all prescribed displacements are zero, that is, the BCs are homogeneous. The more general case of nonzero prescribed values is treated in Part III of the book. Function ModifiedMasterStiffnessForDBC has two arguments: pdof
A list of the prescribed degrees of freedom identified by their global number. For the example truss this list contains three entries: {1, 2, 4}.
K
The master stiffness matrix K produced by the assembly process.
The function clears appropriate rows and columns of K, places ones on the diagonal, and returns the modified K as function value. The only slightly fancy thing in this module is the use of the Mathematica function Length to extract the number of prescribed displacement components: Length[pdof] here will return the value 3, which is the length of the list pdof. Similarly nk=Length[K] assigns 6 to nk, which is the order of matrix K. Although for the example truss these values are known a priori, the use 5–10
5–11
§5.6 RECOVERING INTERNAL FORCES IntForce2DTwoNodeBar[{{x1_,y1_},{x2_,y2_}},{Em_,A_},eftab_,u_]:= Module[ {c,s,dx=x2-x1,dy=y2-y1,L,ix,iy,jx,jy,ubar,e}, L=Sqrt[dx^2+dy^2]; c=dx/L; s=dy/L; {ix,iy,jx,jy}=eftab; ubar={c*u[[ix]]+s*u[[iy]],-s*u[[ix]]+c*u[[iy]], c*u[[jx]]+s*u[[jy]],-s*u[[jx]]+c*u[[jy]]}; e=(ubar[[3]]-ubar[[1]])/L; Return[Em*A*e] ]; p =IntForce2DTwoNodeBar[{{0,0},{10,10}},{100,2*Sqrt[2]}, {1,2,5,6},{0,0,0,0,0.4,-0.2}]; Print['Member int force (numerical):']; Print[N[p]]; p =IntForce2DTwoNodeBar[{{0,0},{L,L}},{Em,A}, {1,2,5,6},{0,0,0,0,ux3,uy3}]; Print['Member int force (symbolic):']; Print[Simplify[p]]; Member int force numerical : 2.82843 Member int force symbolic : A Em ux3 0001 uy3 0003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003000300030003 2L
Figure 5.7. Module IntForce2DTwoNodeBar for computing the internal force in a bar element.
of Length serves to illustrate a technique that is heavily used in more general code. Module ModifiedMasterForcesForDBC has similar structure and logic and need not be described in detail. It is important to note, however, that for homogeneous BCs the modules are independent of each other and may be called in any order. On the other hand, if there were nonzero prescribed displacements present the force modification must be done before the stiffness modification. This is because stiffness coefficients that are cleared in the latter are needed for modifying the force vector. The test statements are purposedly chosen to illustrate another feature of Mathematica: the use of the Array function to generate subscripted symbolic arrays of one and two dimensions. The test output is shown in the bottom box of Figure 5.6, which should be self explanatory. The force vector and its modified form are printed as row vectors to save space. §5.6. Recovering Internal Forces Mathematica provides built-in matrix operations for solving a linear system of equations and multiplying matrices by vectors. Thus we do not need to write application functions for the solution of the modified stiffness equations and for the recovery of nodal forces. Consequently, the last application functions we need are those for internal force recovery. Function IntForce2DTwoNodeBar listed in the top box of Figure 5.7 computes the internal force in an individual bar element. It is somewhat similar in argument sequence and logic to ElemStiff2DTwoNodeBar of Figure 5.3. The first two arguments are identical. Argument eftab provides the Element Freedom Table array for the element. The last argument, u, is the vector of computed node displacements. The logic of IntForce2DTwoNodeBar is straightforward and follows the method outlined in §3.2.1. Member joint displacements u¯ (e) in local coordinates {x, ¯ y¯ } are recovered in array ubar, then the longitudinal strain e = (u¯ x j − u¯ xi )/L and the internal (axial) force p = E Ae is returned as function value. As coded the function contains redundant operations because entries 2 and 4 of ubar (that is, components u¯ yi and u¯ y j ) are not actually needed to get p, but were kept to illustrate the general backtransformation of global to local displacements. Running this function with the test statements shown after the module produces the output shown in the 5–11
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IntForcesOfExampleTruss[u_]:= Module[{f=Table[0,{3}]}, f[[1]]=IntForce2DTwoNodeBar[{{0,0},{10,0}},{100,1},{1,2,3,4},u]; f[[2]]=IntForce2DTwoNodeBar[{{10,0},{10,10}},{100,1/2},{3,4,5,6},u]; f[[3]]=IntForce2DTwoNodeBar[{{0,0},{10,10}},{100,2*Sqrt[2]}, {1,2,5,6},u]; Return[f] ]; f=IntForcesOfExampleTruss[{0,0,0,0,0.4,-0.2}]; Print['Internal member forces in example truss:'];Print[N[f]]; Internal member forces in example truss: 00100., 00041., 2.828430011
Figure 5.8. Module IntForceOfExampleTruss that computes internal forces in the 3 members of the example truss.
bottom box of Figure 5.7. The first test is for member (3) of the example truss using the actual nodal displacements (3.24). It also illustrates the use of the Mathematica built in function N to produce output in floating-point form. The second test does a symbolic calculation in which several argument values are fed in variable form. The top box of Figure 5.8 lists a higher-level function, IntForceOfExampleTruss, which has a single argument: u. This is the complete 6-vector of joint displacements u. This function calls IntForce2DTwoNodeBar three times, once for each member of the example truss, and returns the three member internal forces thus computed as a 3-component list. The test statements listed after IntForcesOfExampleTruss feed the actual node displacements (3.24) to IntForcesOfExampleTruss. Running the functions with the test statements produces the output √ shown in the bottom box of Figure 5.8. The internal forces are p (1) = 0, p (2) = −1 and p (3) = 2 2 = 2.82843. §5.7. Putting the Pieces Together After all this development and testing effort documented in Figures 5.3 through 5.8 we are ready to make use of all these bits and pieces of code to analyze the example plane truss. This is actually done with the logic shown in Figure 5.9. This driver program uses the seven previously described modules ElemStiff2DTwoNodeBar MergeElemIntoMasterStiff AssembleMasterStiffOfExampleTruss ModifiedMasterStiffForDBC ModifiedMasterForcesForDBC IntForce2DTwoNodeTruss IntForcesOfExampleTruss
(5.1)
These functions must have been defined (”compiled”) at the time the driver programs described below are run. A simple way to making sure that all of them are defined is to put all these functions in the same Notebook file and to mark them as initialization cells. These cells may be executed by picking up Kernel → Initialize → Execute Initialization. (An even simpler procedure would to group them all in one cell, but that would make placing separate test statements difficult.) For a hierarchical version of (5.1), see the last CATECS slide. 5–12
5–13
§5.7 PUTTING THE PIECES TOGETHER f={0,0,0,0,2,1}; K=AssembleMasterStiffOfExampleTruss[]; Kmod=ModifiedMasterStiffForDBC[{1,2,4},K]; fmod=ModifiedMasterForcesForDBC[{1,2,4},f]; u=Simplify[Inverse[Kmod].fmod]; Print['Computed nodal displacements:']; Print[u]; f=Simplify[K.u]; Print['External node forces including reactions:']; Print[f]; p=Simplify[IntForcesOfExampleTruss[u]]; Print['Internal member forces:']; Print[p]; Computed nodal displacements: 2 1 00120, 0, 0, 0, 0003000300030003 , 0004 0003000300030003 0013 5 5 External node forces including reactions: 001000042, 00042, 0, 1, 2, 10011 Internal member forces: 00030004000400040004 00100, 00041, 2 2 0011
Figure 5.9. Driver program for numerical analysis of example truss and its output.
§5.7.1. The Driver Program The program listed in the top box of Figure 5.9 first assembles the master stiffness matrix through AssembleMasterStiffOfExampleTruss. Next, it applies the displacement boundary conditions through ModifiedMasterStiffForDBC and ModifiedMasterForcesForDBC. Note that the modified stiffness matrix is placed into Kmod rather than K to save the original form of the master stiffness for the reaction force recovery later. The complete displacement vector is obtained by the matrix calculation u=Inverse[Kmod].fmod
(5.2)
which takes advantage of two built-in Mathematica functions. Inverse returns the inverse of its matrix argument12 The dot operator signifies matrix multiply (here, matrix-vector multiply.) The enclosing Simplify function in Figure 5.9 is asked to simplify the expression of vector u in case of symbolic calculations; it is actually redundant if all computations are numerical. The remaining statements recover the node vector including reactions via the matrix-vector multiply f = K.u (recall that K contains the unmodified master stiffness matrix) and the member internal forces p through IntForcesOfExampleTruss. The program prints u, f and p as row vectors to save space. Running the program of the top box of Figure 5.9 produces the output shown in the bottom box of that figure. The results confirm the hand calculations of Chapter 3. §5.7.2. Is All of This Worthwhile? At this point you may wonder whether all of this work is worth the trouble. After all, a hand calculation (typically helped by a programable calculator) would be quicker in terms of flow time. Writing and debugging the Mathematica fragments displayed here took the writer about six hours (although about two thirds of this was spent in editing and getting the fragment listings into the Chapter.) For larger problems, however, Mathematica would certainly beat hand-plus-calculator computations, the crossover typically appearing for 10 to 20 equations. For up to about 500 equations and using floating-point arithmetic, Mathematica gives answers within minutes on a fast PC or Mac with sufficient memory but 12
This is a highly inefficient way to solve Ku = f if this system becomes large. It is done here to keep simplicity.
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Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
5–14
f={0,0,0,0,fx3,fy3}; K=AssembleMasterStiffOfExampleTruss[]; Kmod=ModifiedMasterStiffForDBC[{1,2,4},K]; fmod=ModifiedMasterForcesForDBC[{1,2,4},f]; u=Simplify[Inverse[Kmod].fmod]; Print['Computed nodal displacements:']; Print[u]; f=Simplify[K.u]; Print['External node forces including reactions:']; Print[f]; p=Simplify[IntForcesOfExampleTruss[u]]; Print['Internal member forces:']; Print[p]; Computed nodal displacements: 1 1 00120, 0, 0, 0, 0003000300030003000300030003 3 fx3 0004 2 fy3 , 0003000300030003 0004fx3 0001 fy3 0013 10 5 External node forces including reactions: 00100004fx3, 0004fx3, 0, fx3 0004 fy3, fx3, fy30011 Internal member forces: 00030004000400040004 00100, 0004fx3 0001 fy3, 2 fx30011
Figure 5.10. Driver program for symbolic analysis of example truss and its output.
eventually runs out of steam at about 1000 equations. For a range of 1000 to about 50000 equations, Matlab, using built-in sparse solvers, would be the best compromise between human and computer flow time. Beyond 50000 equations a program in a low-level language, such as C or Fortran, would be most efficient in terms of computer time.13 One distinct advantage of computer algebra systems emerges when you need to parametrize a small problem by leaving one or more problem quantities as variables. For example suppose that the applied forces on node 3 are to be left as f x3 and f y3 . You replace the last two components of array p as shown in the top box of Figure 5.10, execute the cell and shortly get the symbolic answer shown in the bottom box of that figure. This is the answer to an infinite number of numerical problems. Although one may try to undertake such studies by hand, the likelyhood of errors grows rapidly with the complexity of the system. Symbolic manipulation systems can amplify human abilities in this regard, as long as the algebra “does not explode” because of combinatorial complexity. Examples of such nontrivial calculations will appear throughout the following Chapters. Remark 5.1. The “combinatorial explosion” danger of symbolic computations should be always kept in mind.
For example, the numerical inversion of a N × N matrix is a O(N 3 ) process, whereas symbolic inversion goes as O(N !). For N = 48 the floating-point numerical inverse will be typically done in a fraction of a second. But the symbolic adjoint will have 48! = 12413915592536072670862289047373375038521486354677760000000000 terms, or O(1061 ). There may be enough electrons in this Universe to store that, but barely .. Notes and Bibliography As noted in §5.1.2 the 1350-page Mathematica Book [181] is just a reference manual. Since the contents are available online (click on Help in topbar) as part of purchase of the full system,14 buying the printed book (list: $53 but heavily discounted as used) is not necessary. In fact the latest edition as of this writing is dated 1999; updates since have gone directly to the online version. There is a nice tutorial available by Glynn and Gray [75], list: $35, dated 1999. (Theodore Gray invented the Notebook front-end that appeared in version 2.2.) It is also available on CDROM from MathWare, Ltd, P. O. Box 13
The current record for FEM structural applications is about 100 million equations, done on a massively parallel supercomputer (ASCI Red at SNL). Fluid mechanics problems with over 500 million equations have been solved.
14
The student version comes with limited help files, but is otherwise functionally complete.
5–14
5–15
§5. References
3025, Urbana, IL 61208, e-mail: [email protected] The CDROM is a hyperlinked version of the book that can be installed on the same directory as Mathematica. More advanced and comprehensive is the recent appeared Mathematica Navigator in two volumes [139,140]; list: $70 and $51 but heavily discounted on web. Beyond these, there is a large number of books at all levels that expound on the use of Mathematica for various applications ranging from pure mathematics to physics and engineering. A web search (September 2003) on www3.addall.com hit 150 book titles containing Mathematica, compared to 111 for Maple and 148 for Matlab. A google search hits 1,550,000 pages containing Mathematica, but here Matlab wins with 1,770,000 hits. Wolfram Research hosts the MathWorld web site at http://mathworld.wolfram.com, maintained by Eric W. Weisstein. It is essentially a hyperlinked, on-line version of his Encyclopœdia of Mathematics [172]. To close the topic of symbolic versus numerical computation, here is a nice summary by A. Grozin, posted on the Usenet: “Computer Algebra Systems (CASs) are programs [that] operate with formulas. Mathematica is a powerful CAS (though quite expensive). Other CASs are, e.g., Maple, REDUCE, MuPAD <..>. There are also quite powerful free CASs: Maxima and Axiom. In all of these systems, it is possible to do some numerical calculations (e.g., to evaluate the formula you have derived at some numerical values of all parameters). But it is a very bad idea to do large-scale numerical work in such systems: performance will suffer. In some special cases (e.g., numerical calculations with very high precision, impossible at the double-precision level), you can use Mathematica to do what you need, but there are other, faster ways to do such things. There are a number of programs to do numerical calculations with usual double-precision numbers. One example is Matlab; there are similar free programs, e.g., Octave, Scilab, R, .. Matlab is very good and fast in doing numerical linear algebra: if you want to solve a system of 100 linear equations whose coefficients are all numbers, use Matlab; if coefficients contain letters (symbolic quantities) and you want the solution as formulas, use Mathematica or some other CAS. Matlab can do a limited amount of formula manipulations using its “symbolic toolkit,” which is an interface to a cut-down Maple. It’s a pain to use this interface: if you want Maple, just use Maple.” As regards career advice, here is a useful suggestion by David Park, also posted on Usenet: “If you plan a technical career learn a CAS as early as possible and learn it well. It will make the technical work far easier. Of course you will still have to think. A CAS will not usually solve your problems with a few key strokes. Learning how to apply the CAS to your problems will actually force you to go through the logical steps of the problem and program general routines that you can use in your future work. In other words, a CAS documents and organizes what you have learned in a way that you can actually apply in the future. It’s not just a piece of paper with some calculations but an active and interactive document and resource.” References Referenced items have been moved to Appendix R.
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Chapter 5: ANALYSIS OF EXAMPLE TRUSS BY A CAS
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Homework Exercises for Chapter 5 Analysis of Example Truss by a CAS Before doing any of these Exercises, download the Mathematica Notebook file ExampleTruss.nb from the course web site. (Go to Chapter 5 Index and click on the link). Open this Notebook file using version 4.0 or a later one. The first eight cells contain the modules and test statements listed in the top boxes of Figures 5.3–10. The first six of these are marked as initialization cells. Before running driver programs, they should be executed by picking up Kernel → Evaluation → Execute Initialization. Verify that the output of those six cells agrees with that shown in the bottom boxes of Figures 5.3–6. Then execute the driver programs in Cells 7–8 by clicking on each cell and pressing the appropriate key: on a Mac, on a Windows PC. Compare the output with that shown in Figures 5.9–10. If the output checks out, you may proceed to the Exercises. EXERCISE 5.1 [C:10] Explain why the Simplify command in the test statements of Figure 5.3 says L>0. (One
way to figure this out is to just say Ke=Simplify[Ke] and look at the output. Related question: why does Mathematica refuse to simplify Sqrt[L^2] to L unless one specifies the sign of L in the Simplify command? EXERCISE 5.2 [C:10] Explain the logic of the For loops in the merge function MergeElemIntoMasterStiff
of Figure 5.4. What does the operator += do? EXERCISE 5.3 [C:10] Explain the reason behind the use of Length in the modules of Figure 5.6. Why not simply
set nk and np to 6 and 3, respectively? EXERCISE 5.4 [C:15] Of the seven modules listed in Figures 5.3 through 5.8, with names collected in (5.2), two can be used only for the example truss, three can be used for any plane truss, and two can be used for other structures analyzed by the DSM. Identify which ones and briefly state the reasons for your classification. EXERCISE 5.5 [C:20] Modify the modules AssembleMasterStiffOfExampleTruss, IntForcesOfExampleTruss and the driver program of Figure 5.9 to solve numerically the three-node, two-member truss of Exercise 3.7. Verify that the output reproduces the solution given for that problem. Hint: modify cells but keep a copy of the original Notebook handy in case things go wrong. EXERCISE 5.6 [C:25] Expand the logic of ModifiedMasterForcesForDBC to permit specified nonzero displacements. Specify these in a second argument called pval, which contains a list of prescribed values paired with pdof.
xynode={{0,0},{10,0},{10,10}}; elenod={{1,2},{2,3},{3,1}}; unode={{0,0},{0,0},{2/5,-1/5}}; amp=5; p={}; For [t=0,t<=1,t=t+1/5, For [e=1,e<=Length[elenod],e++, {i,j}=elenod[[e]]; xyi=xynode[[i]];ui=unode[[i]];xyj=xynode[[j]];uj=unode[[j]]; p=AppendTo[p,Graphics[Line[{xyi+amp*t*ui,xyj+amp*t*uj}]]]; ]; ]; Show[p,Axes->False,AspectRatio->Automatic];
Figure E5.1. Mystery program for Exercise 5.7.
EXERCISE 5.7 [C:20] Explain what the program of Figure E5.1 does, and the logic behind what it does. (You
may want to put it in a cell and execute it.) What modifications would be needed so it can be used for any plane struss?
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5
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Constructing MoM Members
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Chapter 5: CONSTRUCTING MOM MEMBERS
TABLE OF CONTENTS Page
§5.1. §5.2.
§5.3.
§5.4.
§5. §5. §5.
Introduction Formulation of MoM Members §5.2.1. What They Look Like . . . . . . . . §5.2.2. End Quantities, Degrees of Freedom, Joint Forces §5.2.3. Internal Quantities . . . . . . . . . . §5.2.4. Discrete Field Equations, Tonti Diagram . . . Simplex MoM Members §5.3.1. The Bar Element Revisited . . . . . . . §5.3.2. The Spar Element . . . . . . . . . . §5.3.3. The Shaft Element . . . . . . . . . . *Non-Simplex MoM Members §5.4.1. *Formulation Rules . . . . . . . . . . §5.4.2. *Example: Bar with Variable Cross Section . Notes and Bibliography . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . .
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§5.2
FORMULATION OF MOM MEMBERS
§5.1. Introduction The truss member used as example in Chapters 2–4 is an instance of a structural element. Such elements may be formulated directly using concepts and modeling techniques developed in Mechanics of Materials (MoM).1 The construction does not involve the more advanced tools that are required for the continuum finite elements that appear in Part II. This Chapter presents an overview of the technique to construct the element stiffness equations of “MoM members” using simple matrix operations. These simplified equations come in handy for a surprisingly large number of applications, particularly in skeletal structures. Focus is on simplex elements, which may be formed directly as a sequence of matrix operations. Non-simplex elements are presented as a recipe because their proper formulation requires work theorems not yet studied. The physical interpretation of the FEM is still emphasized. Consequently we continue to speak of structures built up of members (elements) connected at joints (nodes). §5.2. Formulation of MoM Members §5.2.1. What They Look Like MoM-based formulations are largely restricted to intrinsically one-dimensional members. These are structural components one of whose dimensions, called longitudinal, is significantly larger than the other two, called the tranverse dimensions. Such members are amenable to the simplified structural theories developed in MoM textbooks. This Chapter covers only straight members with geometry defined by the two end joints.2 The member cross sections are defined by the intersection of planes normal to the longitudinal dimension with the member. See Figure 5.1. Note that although the individual member will be idealized as being one-dimensional in its intrinsic or local coordinate system, it often functions as component of a two- or three-dimensional structure. This class of structural components embodies bars, beams, beam-columns, shafts and spars. Although geometrically similar, the names distinguish the main kind of internal forces the member resists and transmits: axial forces for bars, bending and shear forces for beams, axial compression and bending for beam-columns, torsion forces for shafts, and shear forces for spars. Members are connected at their end joints by displacement degrees of freedom. For truss (bar) and spar members those freedoms are translational components of the joint displacements. For other types, notably beams and shafts, nodal rotations are chosen as additional degrees of freedom. Structures fabricated with MoM members are generally three-dimensional. Their geometry is defined with respect to a global Cartesian coordinate system {x, y, z}. Two-dimensional idealizations are useful simplifications should the nature of the geometry and loading allow the reduction of the structural model to one plane of symmetry, which is chosen to be the {x, y} plane. Plane trusses and plane frameworks are examples of such simplifications. 1
Mechanics of Materials was called Strength of Materials in older texts. It covers bars, beams, shafts, arches, thin plates and shells, but only one-dimensional models are considered in introductory undergraduate courses. MoM involves ab initio phenomenological assumptions such as “plane sections remain plane” or “shear effects can be neglected in thin beams.” These came about as the byproduct of two centuries of structural engineering practice, justified by success. A similar acronym (MOM) is used in Electrical Engineering for something completely different: the Method of Moments.
2
Advanced Mechanics of Materials includes curved members. Plane arch elements are studied in Chapter 13.
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Chapter 5: CONSTRUCTING MOM MEMBERS
y¯
z
y
x
x¯
z¯
Figure 5.1. A Mechanics of Materials (MoM) member is a structural element one of whose dimensions (the longitudinal dimension) is significantly larger than the other two. Local axes {x, ¯ y¯ , z¯ } are chosen as indicated. Although the depicted member is prismatic, some applications utilize tapered or stepped members, the cross section of which varies as a function of x. ¯
In this Chapter we study generic structural members that fit the preceding class. An individual member is identified by e but this superscript will be usually suppressed in the equations below to reduce clutter. The local axes are denoted by {x, ¯ y¯ , z¯ }, with x¯ along the longitudinal direction. See Figure 5.1. The mathematical model of a MoM member is obtained by an idealization process. The model represents the member as a line segment that connects the two end joints, as depicted in Figure 5.2. §5.2.2. End Quantities, Degrees of Freedom, Joint Forces The set of mathematical variables used to link members are called end quantities or connectors. In the Direct Stiffness Method (DSM) these are joint displacements (the degrees of freedom) and the joint forces. These quantities are related by the member stiffness equations. ¯ The degrees of freedom at the end joints i and j are collected in the joint displacement vector u. This may include translations only, rotations only, or a combination of translations and rotations. ¯ Component The vector of joint forces ¯f groups components in one to one correspondence with u. pairs must be conjugate in the sense of the Principle of Virtual Work. For example if the x-translation ¯ ¯ ¯ ¯ the corresponding entry in f is the x-force ¯ f xi at i. If the rotation about at joint i: u¯ xi appears in u, ¯ ¯ ¯ the corresponding entry in f is the z-moment m¯ z j . z¯ at joint j: θz j appears in u, §5.2.3. Internal Quantities Internal quantities are mechanical actions that take place within the member. Those actions involve stresses and deformations. Accordingly two types of internal quantities appear: Internal member forces form a finite set of stress-resultant quantities collected in an array p. They are obtained by integrating stresses over each cross section, and thus are also called generalized stresses in structural mechanics. This set characterizes the forces resisted by the material. Stresses at any point in a section may be recovered if p is known. Member deformations form a finite set of quantities, chosen in one-to one correspondence with internal member forces, and collected in an array v. This set characterizes the deformations experienced by the material. They are also called generalized strains in structural theory. Strains at any point in the member can be recovered if v is known. As in the case of end quantities, internal forces and deformations are paired in one to one correspondence. For example, the axial force in a bar member must be paired either with an average 5–4
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§5.2
FORMULATION OF MOM MEMBERS
End quantities (connectors) are defined at the joints i, j
y¯
z
(e)
j
i z¯ y
x¯
Internal quantities are defined within the member and may depend on x. ¯
x
Figure 5.2. The FE mathematical idealization of a MoM member. The model is one-dimensional in x. ¯ The two end joints are the site of end quantities: joint forces and displacements, that interconnect members. The internal quantities characterize the stresses and deformations in the member.
axial deformation, or with the total elongation. Pairs that mutually correspond in the sense of the Principle of Virtual Work are called conjugate. Unlike the case of end quantities, conjugacy is not a mandatory requirement although it simplifies some expressions. §5.2.4. Discrete Field Equations, Tonti Diagram ¯ v, p and f are called the discrete field equations. There are The matrix equations that connect u, three of them. The member deformations v are linked to the joint displacements u¯ by the kinematic compatibility conditions, also called the deformation-displacement or strain-displacement equations: ¯ v = B u.
(5.1)
The internal member forces are linked to the member deformations by the constitutive equations. In the absence of initial strain effects those equations are homogeneous: p = S v.
Kinematic v = B u¯
(5.2)
Finally, the internal member forces are linked to the joint forces by the equilibrium equations. If the internal forces p are constant over the member, the relation is simply ¯f = AT p.
u¯
(5.3)
Stiffness ¯f = AT S B u = K ¯ u¯
v
Constitutive
p
Figure 5.3. Tonti diagram of the three discrete field equations (5.1)–(5.3) and the stiffness equation (5.4) for a simplex MoM member. Internal and end quantities appear inside the orange and yellow boxes, respectively.
If p is a function of x¯ the relation is of differential type. This form is studied in §5.4.
5–5
Equilibrium f¯ = AT p
p=Sv
In (5.3) the transpose of A is used for convenience.3 3
¯f
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Chapter 5: CONSTRUCTING MOM MEMBERS
These equations can be presented graphically as shown in Figure 5.3. This is a discrete variant of the so-called Tonti diagrams, which represent the governing equations as arrows linking boxes containing kinematic and static quantities. Tonti diagrams for field equations are introduced in Chapter 11. Matrices B, S and A receive the following names in the literature: A S B
Equilibrium, leverage Rigidity, material, constitutive4 Compatibility, deformation-displacement, strain-displacement
If the element is sufficiently simple, the determination of these three matrices can be carried out through MoM techniques. If the construction requires more advanced tools, however, recourse to the general methodology of finite elements and variational principles is necessary. §5.3. Simplex MoM Members Throughout this section we assume that the internal quantities are constant over the member length. Such members are called simplex elements. If so the matrices A, B and S are independent of member cross section. The derivation of the element stiffness equations reduces to a straightforward sequence of matrix multiplications. Under the constancy-along-x¯ assumption, elimination of the interior quantities p and v from (5.1)(5.3) yields the element stiffness relation ¯ u, ¯f = AT S B u¯ = K ¯
(5.4)
¯ = AT S B. K
(5.5)
whence the element stiffness matrix is
The four preceding matrix equations are illustrated in the Tonti diagram of Figure 5.3. ¯ u} ¯ are conjugate in the sense of the Principle of Virtual Work, it can be shown that If {p, v} and {f, A = B and that S is symmetric. Then ¯ = BT SB. K
(5.6)
is a symmetric matrix. Symmetry is computationally desirable for reasons outlined in Part III. ¯ must come out to be Remark 5.1. If ¯f and u¯ are conjugate (as required in §5.2.2) but p and v are not, K
symmetric even if S is unsymmetric and A 0006= B. However there are more opportunities to go wrong.
4
The name rigidity matrix for S is preferable. It is a member integrated version of the cross section constitutive equations. The latter are usually denoted by symbol R, as in §5.4.
5–6
5–7
§5.3
(a) z
y¯
−F
y
(b)
Axial rigidity EA, length L
y¯ f¯xi , u¯xi i
z¯
F
SIMPLEX MOM MEMBERS
x¯
x
EA
j f¯xj , u¯ xj
x¯
L
Figure 5.4. The prismatic bar (also called truss) member: (a) individual member shown in 3D space, (b) idealization as generic member.
§5.3.1. The Bar Element Revisited The simplest MoM element is the prismatic bar or truss member already derived in Chapter 2. See Figure 5.4. This qualifies as simplex because all internal quantities are constant. One minor difference in the derivation below is that the joint displacements and forces in the y¯ direction are omitted in the generic element since they contribute nothing to the stiffness equations. In the FEM terminology, freedoms associated with zero stiffness are called inactive. Three choices for internal deformation and force variables are considered. The results confirm that the element stiffness equations coalesce, as expected, since the external quantities are the same. Derivation Using Axial Elongation and Axial Force. The member axial elongation d is taken as deformation measure, and the member axial force F as internal force measure. Hence v and p reduce to the scalars v = d and p = F, respectively. The chain of discrete field equations is easily constructed: 0003 0003 0002 0003 0002 0002 EA f¯xi −1 u¯ xi ¯ ¯ = Bu, F= = F = AT F. d = S d, f= ¯ d = [ −1 1 ] u¯ x j 1 fx j L (5.7) Consequently 0002 0003 0002 0003 EA −1 1 −1 T T ¯ = . (5.8) K = A SB = SB B = S [ −1 1 ] 1 1 L −1 Note that A = B because F and d are conjugate. Derivation Using Mean Axial Strain and Axial Force. Instead of d we may use the mean axial strain e¯ = d/L as deformation measure whereas F is kept as internal force measure. The only change is that B becomes [ −1 1 ] /L whereas S becomes E A. Matrix A does not change. The product ¯ as in (5.8), as can be expected. Now AT is not equal to B because F and AT SB gives the same K e¯ are not conjugate, but they differ only by a factor 1/L. Derivation Using Mean Axial Strain and Axial Stress. We keep the mean axial strain e¯ = d/L as deformation measure but the mean axial stress σ¯ = F/A, (which is not conjugate to e) ¯ is taken as internal force measure. Now B = [ −1 1 ] /L, S = E and AT = A [ −1 1 ]. The product AT SB ¯ shown in (5.8). gives again the same K 5–7
5–8
Chapter 5: CONSTRUCTING MOM MEMBERS
V
(a)
(b)
y¯
x¯
z
f¯yi , u¯ yi GA s (e) i
z¯ y
x
y¯
−V
Shear rigidity GA s , length L
f¯yj , u¯ yj j x¯
L
Figure 5.5. The prismatic spar (also called shear-web) member: (a) individual member shown in 3D space, (b) idealization as generic member.
Transformation to Global Coordinates. Since u¯ yi and u¯ y j are not part of (5.7) and (5.8) the displacement transformation matrix from local to global {x, y} coordinates is 2 × 4, instead of 4 × 4 as in §3.1. On restoring the element identifier e the appropriate local-to-global transformation is  e  u xi 0002 e 0003 0002 0003  u¯ xi c s 0 0  u eyi   = Te ue , u¯ e = (5.9) = u¯ ex j 0 0 c s  u ex j  u ey j where c = cos ϕ e , s = sin ϕ e , and ϕ e is the angle from x to x, ¯ cf. Figure 3.2. The 4 × 4 globalized e e T ¯e e element stiffness matrix K = (T ) K T agrees with (3.7). §5.3.2. The Spar Element The spar or shear-web member has two joints (end nodes): i and j. This member can only resist and transmit a constant shear force V in the plane of the web, which is chosen to be the {x, ¯ y¯ } plane. See Figure 5.5. It is often used for modeling high-aspect aircraft wing structures, as illustrated in Figure 5.6. We consider here only prismatic spar members of uniform material and constant cross section, which thus qualify as simplex.
COVER PLATES SPAR
RIB The active degrees of freedom for a generic spar member of length L, as depicted in Figure 5.5(b), are u¯ yi Figure 5.6. Spar members in aircraft wing and u¯ y j . Let G be the shear modulus and As the ef(Piper Cherokee). For more impressive aircraft fective shear area. (A concept developed in Mechanics structures see CATECS slides. of Materials; for a narrow rectangular cross section, As = 5A/6.) The shear rigidity is G As . As deformation measure the mean shear strain γ = V /(G As ) is chosen. The kinematic, constitutive, and equilibrium equations are 0002 0002 0003 0003 0002 0003 ¯yi 1 f u¯ yi −1 ¯f = ¯ γ = [ −1 1 ] = Bu, V = G As γ = Sγ , = V = AT V, ¯ 1 u¯ y j fyj L (5.10)
5–8
5–9
§5.3
whence the local stiffness equations are 0002 0002 0003 ¯ 1 ¯f = f yi = AT S Bu¯ = G As ¯ −1 fyj L
−1 1
00030002
SIMPLEX MOM MEMBERS
u¯ yi u¯ y j
0003
¯ u. ¯ =K
(5.11)
Note that A 0006= B as V and γ are not work-conjugate. This difference is easily adjusted for, however; see Exercise 5.1. If the spar member is used in a two dimensional context, the displacement transformation from local to global coordinates {x, y} is 0002
u¯ eyi u¯ = u¯ ey j e
0003
0002 =

−s 0
c 0
0 −s
 u exi 0003 e  0   u yi  = Te ue , c  u ex j  u ey j
(5.12)
¯ cf. Figure 3.2. The 4 × 4 globalized where c = cos ϕ e , s = sin ϕ e , and ϕ e is the angle from x to x, e e T ¯e e spar stiffness matrix is then K = (T ) K T . More often, however, the spar member will be a component in a three-dimensional structural model, such as the aircraft wing depicted in Figure 5.6. If so the determination of the 2 × 6 transformation matrix is more involved. This is the topic of Exercise 5.5. §5.3.3. The Shaft Element The shaft, also called torque member, has two joints (end nodes): i and j. A shaft can only resist and transmit a constant torque or twisting moment T along its longitudinal axis x, ¯ as pictured in Figure 5.7(a).
(a)
Torsional rigidity GJ, length L
x¯ _ _ m xj , θxj
y¯ z
T
T
_ _ m xi , θxi y
x
z¯
m¯ xi, θ¯xi
(b) y¯ i
GJ (e)
j mxj, θ¯xj
L
Figure 5.7. The prismatic shaft (also called torque member): (a) individual member shown in 3D space, (b) idealization as generic member.
We consider here only prismatic shaft members with uniform material and constant cross section, which thus qualify as simplex. The active degrees of freedom of a generic shaft member of length L, depicted in Figure 5.7(b), are θ¯xi and θ¯x j . These are the infinitesimal end rotations about x, ¯ positive 5–9
x¯
5–10
Chapter 5: CONSTRUCTING MOM MEMBERS
according to the right-hand rule. The associated joint (node) forces are end moments denoted as m¯ xi and m¯ x j . The only internal force is the torque T , positive if acting as pictured in Figure 5.7(a). Let G be the shear modulus and G J the effective torsional rigidity.5 As deformation measure pick the relative twist angle φ = θ¯x j − θ¯xi . The kinematic, constitutive, and equilibrium equations provided by Mechanics of Materials are 0002 0002 0003 0003 0002 0003 GJ θ¯yi m ¯ −1 xi ¯ φ = [ −1 1 ] φ = Sγ , ¯f = = Bu, T = = T = BT T. (5.13) 1 m¯ x j θ¯y j L From these the local stiffness equations follow as 0002 0002 0003 1 ¯f = m¯ yi = BT S Bu¯ = G J m¯ y j L −1
−1 1
00030002
θ¯yi θ¯y j
0003
¯ u. ¯ =K
(5.14)
If the shaft is used in a two-dimensional context, the displacement transformation to global coordinates {x, y} within the framework of infinitesimal rotations, is 0002 u¯ e =
θ¯xie θ¯xej
0003
0002 =
c 0
s 0
0 c
 θxie e  0   θ yi  = Te θe , s  θxej  θ yej 0003

(5.15)
¯ Note that θe collects only where as usual c = cos ϕ e , s = sin ϕ e , and ϕ e is the angle from x to x. global node rotations components. This operation is elaborated further in Exercise 5.3. §5.4.
*Non-Simplex MoM Members
The straightforward formulation of simplex MoM elements does not immediately carry over to the case in which internal quantities p and v vary over the member; that is, depend on x. ¯ The dependence may be due to element type, varying cross section, or both. As a result, one or more of the matrices A, B and S depend on x. ¯ Such members are called non-simplex. ¯ of non-simplex The matrix multiplication rule (5.4) cannot be used to construct the element stiffness matrix K T ¯ x) ¯ would depend on x. ¯ On the other hand, K members. This can be grasped by observing that A(x) ¯ S(x)B( must be independent of x¯ because it relates the end quantities u¯ and ¯f. §5.4.1. *Formulation Rules The derivation of non-simplex MoM elements requires use of the work principles of mechanics, for example the Principle of Virtual Work or PVW. Thus, more care must be exercised in the choice of conjugate internal quantities. The following rules can be justified through the arguments presented in Part II. They are stated here as recipe, and apply only to displacement-assumed elements. Rule 1. Select internal deformations v(x) ¯ and internal forces p(x) ¯ that are conjugate in the PVW sense. Link deformations to node displacements by v(x) ¯ = B(x)u. ¯ 5
J has dimension of (length)4 . For a circular or annular cross section it reduces to the polar moment of inertia about x. ¯ The determination of J for noncircular cross sections is covered in Mechanics of Materials textbooks.
5–10
5–11
§5.4 *NON-SIMPLEX MOM MEMBERS
Rule 2. From the PVW it may be shown6 that the force equilibrium equation exists only in a differential sense: BT dp = d ¯f.
(5.16)
Here d in dp denotes differentiation with respect to x. ¯ 7 The interpretation of d ¯f is less immediate because ¯f is not a function of x. ¯ It actually means the contribution of that member slice to the building of the node force vector ¯f. See (5.18) and (5.19) below. Rule 3. The constitutive relation is p = Rv,
(5.17)
in which R, which may depend on x, ¯ must be symmetric. Note that symbol R in (5.17) replaces the S of (5.2). R pertains to a specific cross section whereas S applies to the entire member. This distinction is further elaborated in Exercise 5.9. The discrete relations supplied by the foregoing rules are displayed in the discrete Tonti diagram of Figure 5.8. Internal quantities are now eliminated starting from the differential equilibrium relation (5.16):
f¯ =
u¯
Stiffness BT R B dx¯ u ¯
L 0
Kinematic
(at each x-section)
Equilibrium
(at each x-section)
v = B u¯
v
f¯
d f¯ = BT dp p=Rv Constitutive
p
(at each x-section)
Figure 5.8. Discrete Tonti diagram of the equations for a non-simplex MoM member.
¯ d ¯f = BT dp = BT p d x¯ = BT R v d x¯ = BT R B u¯ d x¯ = BT R B d x¯ u.
(5.18)
Integrating both sides over the member length L yields ¯f =
L
d ¯f =
0
L
¯ u, ¯ BT R B d x¯ u¯ = K
(5.19)
0
because u¯ does not depend on x. ¯ Consequently the local element stiffness matrix is ¯ = K
L
BT R B d x¯
(5.20)
0
The recipe (5.20) will be justified in Part II through energy methods. It will be seen that it generalizes to arbitrary displacement-assumed finite elements in any number of space dimensions. It is used in the derivation of the stiffness equations of the plane beam element in Chapter 12. The reduction of (5.20) to (5.6) when the dependence on x¯ disappears is the subject of Exercise 5.8. §5.4.2. *Example: Bar with Variable Cross Section A two-node bar element has constant E but a continuously varying area: Ai , A j and Am at i, j and m, respectively, where m is the midpoint between end joints i and j. This can be fitted by as A(x) ¯ = Ai Ni (x) ¯ + ¯ Am Nm (x), ¯ where Ni (x) ¯ = − 12 ξ(1−ξ ), N j (x) ¯ = 12 ξ(1+ξ ) and Nm (x) ¯ = 1−ξ 2 , with ξ = 2x/L−1, A j N j (x)+ are interpolating polynomials further studied in Part II as “element shape functions.” 6
¯ = The proof follows by equating expressions of the virtual work of a slice of length d x: ¯ d ¯f .δ u¯ = dpT .δv = dpT .(B u) ¯ Since δ u¯ is arbitrary, BT dp = d ¯f. (BT dp)T .δ u.
7
The meaning of dp is simply p(x) ¯ d x. ¯ That is, the differential of internal forces as one passes from cross-section x¯ to a neighboring one x¯ + d x. ¯
T
5–11
5–12
Chapter 5: CONSTRUCTING MOM MEMBERS
As internal quantities take the strain e and the axial force p = E Ae, which are conjugate quantities. Assuming the strain e to be uniform over the element8 the MoM equations are ¯ e = Bu,
p = E A(x) ¯ e = R(x) ¯ e,
d ¯f = BT dp,
B=
1 [ −1 L
1].
(5.21)
Inserting into (5.20) and carrying out the integration yields ¯ 1 ¯ = EA K L −1
−1 , 1
with
A¯ = 16 (Ai + A j ) + 23 Am .
(5.22)
Notes and Bibliography The derivation of MoM elements using straightforward matrix algebra is typical of pre-1962 Matrix Structural Analysis (MSA). The excellent book of Pestel and Leckie [126], unfortunately out of print, epitomizes that approach. Historically this idea interweaved with Generation 1 of FEM, as outlined in §1.5.3. By 1970 simplified derivations had fallen out of favor as yokelish. But these elements do not need improvement. They still work fine: a bar or beam stiffness today is the same as 40 years ago.9 The Mechanics of Materials books by Beer-Johnston [12] and Popov [133] can be cited as being widely used in US undergraduate courses. But they are not the only ones. A September 2003 in-print book search through www3.addall.com on “Mechanics of Materials” returns 99 hits whereas one on “Strength of Materials” (the older name) compiles 112. Folding editions and paper/softback variants one gets about 60 books; by all accounts an impressive number. Spar members are discussed only in MoM books that focus on aircraft structures, since they are primarily used in modeling shear web action. On the other hand, bars, shafts and beams are standard fare. The framework presented here is a tiny part of MSA. A panoramic view, including linkage to continuum formulations from the MSA viewpoint, is presented in [51]. The source of Tonti diagrams is discussed in Chapter 11. References Referenced items have been moved to Appendix R.
8
This is characteristic of a displaceent assumed element and is justified through the method of shape functions explained in Part II.
9
The chief technical difference is the heavier use of differential equations prior to 1962, as opposed to the energy methods in vogue today. The end result for simple one-dimensional models is the same.
5–12
5–13
Exercises
Homework Exercises for Chapter 5 Constructing MoM Members EXERCISE 5.1 [A:10] Explain how to select the deformation variable v (paired to V ) of the spar member
formulated in §5.3.2, so that A = B. Draw the Tonti diagram, as in Figure 5.3 but with the actual equations, for this choice. EXERCISE 5.2 [A:15] Obtain the 4 × 4 global element stiffness matrix of a prismatic spar member in a two
dimensional Cartesian system {x, y}. Start from (5.11). Indicate where the transformation (5.12) comes from ¯ e Te in closed form. (Hint: read §3.2). Evaluate Ke = (Te )T K
EXERCISE 5.3 [A:15] Obtain the 4 × 4 global element stiffness matrix of a prismatic shaft element in a two dimensional Cartesian system {x, y}. Include only node rotation freedoms in the global displacement vector. Start from (5.14). Justify the transformation (5.15) (Hint: infinitesimal rotations transform as vectors). ¯ e Te in closed form. Evaluate Ke = (Te )T K
j (x j , y j , zj ) − uxj − x E, A
(a)
(b)
j − y i (x i , yi , z i )
− uxi
−y
L
L
− z
y
− x
i
x
z
Figure E5.1. Bar element in 3D for Exercise 5.4.
EXERCISE 5.4 [A+N:15(10+5)] A bar element moving in three dimensional space is completely defined
by the global coordinates {xi , yi , z i }, {x j , y j , z j } of its end nodes i and j, as illustrated in Figure E5.1. The 2 × 6 displacement transformation matrix T, with superscript e dropped for brevity, links u¯ e = Tue . Here u¯ e contains the two local displacements u¯ xi and u¯ x j whereas ue contains the six global displacements u xi , u yi , u zi , u x j , u y j , u z j . (a)
From vector mechanics show that T=
1 x ji L 0
y ji 0
z ji 0
0 x ji
0 y ji
0 z ji
=
cx ji c y ji cz ji 0 0 0 0 0 0 cx ji c y ji cz ji
(E5.1)
in which L is the element length, x ji = x j − xi , etc., and cx ji = x ji /L, etc., are the direction cosines of the vector going from i to j. (b)
Evaluate T for a bar going from node i at {1, 2, 3} to node j at {3, 8, 6}.
EXERCISE 5.5 [A+N:30(10+15+5)] A spar element in three dimensional space is only partially defined by the global coordinates {xi , yi , z i }, {x j , y j , z j } of its end nodes i and j, as illustrated in Figure E5.2. The problem is that axis y¯ , which defines the direction of shear force transmission, is not uniquely defined by i and j.10 Most FEM programs use the orientation node method to complete the definition. A third node k, not 10
The same ambiguity arises in beam elements in 3D space. These elements are covered in Part III.
5–13
5–14
Chapter 5: CONSTRUCTING MOM MEMBERS
Orientation node k defines plane − − as passing through {i,j,k} {x,y}
(a)
k (xk , yk , z k) G, A s u−yi i (x i , yi , z i) y z
− uyj j (x j , y j , z j) x−
a b
− y
y− − uxi
(b)
k
α m
L i
− z
h
β
j
x−
L−c
c
x Figure E5.2. Spar element in 3D for Exercise 5.5.
colinear with i and j, is provided by the user. Nodes {i, j, k} define the {x, ¯ y¯ } plane and thus z¯ . The projection of k on line i j is point m. The distance h > 0 from m to k is called h as shown in Figure E5.2(b). The 2 × 6 displacement transformation matrix T, with superscript e omitted to reduce clutter, relates u¯ e = Tue . Here u¯ e contains the local transverse displacements u¯ yi and u¯ y j whereas ue contains the global displacements u xi , u yi , u zi , u x j , u y j , u z j . (a)
Show that T=
1 xkm h 0
ykm 0
z km 0
0 xkm
0 ykm
0
z km
=
cxkm c ykm czkm 0 0 0 0 0 0 cxkm c ykm czkm
(E5.2)
in which xkm = xk − xm , etc., and cxkm = xkm / h, etc., are the direction cosines of the vector going from m to k. (Assume that the position of m is known. That computation is carried out in the next item.) (b)
Work out the formulas to compute the coordinates of point m in terms of the coordinates of {i, j, k}. Assume a, b and L are computed immediately from the input data. Using
the notation of Figure E5.2(b) p( p − a)( p − b)( p − L) with and elementary trigonometry, show that h = 2A/L, where A = p = 12 (L + a + b) (Heron’s formula), cos α = (L 2 + b2 − a 2 )/(2bL), cos β = (L 2 + a 2 − b2 )/(2a L), c = b cos α, L − c = a cos β, xm = xi (L − c)/L + x j c/L, etc.11
(c)
Evaluate T for a spar member going from node i at {1, 2, 3} to node j at {3, 8, 6}. with k at {4, 5, 6}.
EXERCISE 5.6 [A:20] Explain how thermal effects can be generally incorporated in the constitutive equation
(5.2) to produce an initial force vector for a simplex element. EXERCISE 5.7 [A:15] Draw the discrete Tonti diagram for the prismatic shaft element. The diagram should
look like in Figure 5.3 but with the actual external and internal quantities in the boxes, and matrix equations written down along the arrows. EXERCISE 5.8 [A:15] If the matrices B and R are constant over the element length L, show that expression (5.20) of the element stiffness matrix for a non-simplex member reduces to (5.6), in which S = LR.
11
An alternative and more elegant procedure, found by a student in 1999, can be sketched as follows. From Figure E5.2(b) obviously the two subtriangles imk and jkm are right-angled at m and share side km of length h. Apply Pythagoras’ theorem twice, and subtract so as to cancel out h 2 and c2 , getting a linear equation for c that can be solved directly.
5–14
5–15
Exercises
EXERCISE 5.9 [A:20] Explain in detail the quickie derivation of footnote 5. (Knowledge of the Principle of
Virtual Work is required to do this exercise.) EXERCISE 5.10 [A:25(10+5+10)] Consider a non-simplex element in which R varies with x¯ but B = A is
constant. (a)
From (5.20) prove that ¯ B, ¯ = LBT R K
¯ = 1 with R L
L
R(x) ¯ d x¯
(E5.3)
0
(b)
¯ for a tapered bar with area defined by the linear law A = Ai (1− x/L)+ Apply (E5.3) to obtain K ¯ A j x/L, ¯ where Ai and A j are the end areas at i and j, respectively. Take B = [ −1 1 ] /L.
(c)
Apply (E5.3) to verify the result (5.22) for a bar with parabolically varying cross section.
EXERCISE 5.11 [A/C+N:30(25+5)] A prismatic bar element in 3D space is referred to a global coordinate system {x, y, z}, as in Figure E5.1. The end nodes are located at {x1 , y1 , z 1 } and {x2 , y2 , z 2 }.12 The elastic modulus E and the cross
section area A are constant along the length. Denote x21 = x2 − x1 , y21 = y2 − y1 , 2 2 2 z 21 = z 2 − z 1 and L = x21 + y21 + z 21 .
Show that the element stiffness matrix in global coordinates can be compactly written13
(a)
Ke = (b)
EA T B B, L3
in which B = [ −x21
−y21
−z 21
x21
y21
z 21 ] .
(E5.4)
Compute Ke if the nodes are at {1, 2, 3} and {3, 8, 6}, with elastic modulus E = 343 and cross section area A = 1. Note: the computation can be either done by hand or with the help of a program such as the following Mathematica module, which is used in Part III:
Stiffness3DBar[ncoor_,mprop_,fprop_,opt_]:= Module[ {x1,x2,y1,y2,z1,z2,x21,y21,z21,Em,Gm,rho,alpha,A, num,L,LL,LLL,B,Ke}, {{x1,y1,z1},{x2,y2,z2}}=ncoor; {x21,y21,z21}={x2-x1,y2-y1,z2-z1}; {Em,Gm,rho,alpha}=mprop; {A}=fprop; {num}=opt; If [num,{x21,y21,z21,Em,A}=N[{x21,y21,z21,Em,A}]]; LL=x21^2+y21^2+z21^2; L=PowerExpand[Sqrt[LL]]; LLL=Simplify[LL*L]; B={{-x21,-y21,-z21,x21,y21,z21}}; Ke=(Em*A/LLL)*Transpose[B].B; 12
End nodes are labeled 1 and 2 instead of i and j to agree with the code listed below.
13
There are several ways of arriving at this result. Some are faster and more elegant than others. Here is a sketch of one of the ways. Denote by L 0 and L the lengths of the bar in the undeformed and deformed configurations, respectively. Then 2 1 2 (L
− L 20 ) = x21 (u x2 − u x1 ) + y21 (u y2 − u y1 ) + z 21 (u z2 − u z1 ) + Q ≈ Bu,
in which Q is a quadratic function of node displacements which is therefore dropped in the small-displacement linear theory. Also on account of small displacements 2 1 2 (L
− L 20 ) = 12 (L + L 0 )(L − L 0 ) ≈ L 0012L .
Hence the small axial strain is e = 0012L/L = (1/L 2 )Bue , which begins the Tonti diagram. Next is F = E A e. Finally you must show that force equilibrium at nodes requires fe = (1/L)BT F. Multiplying through gives (E5.4). A plodding way is to start from the local stiffness (5.8) and transform to global using (E5.1).
5–15
Chapter 5: CONSTRUCTING MOM MEMBERS
Return[Ke]]; ClearAll[Em,A]; Em=343; A=1; ncoor={{0,0,0},{2,6,3}}; mprop={Em,0,0,0}; fprop={A}; opt={False}; Ke=Stiffness3DBar[ncoor,mprop,fprop,opt]; Print['Stiffness of 3D Bar Element:']; Print[Ke//MatrixForm]; Print['eigs of Ke: ',Eigenvalues[Ke]]; As a check, the six eigenvalues of this particular Ke should be 98 and five zeros.
5–16
5–16
6
.
FEM Modeling: Introduction
6–1
6–2
Chapter 6: FEM MODELING: INTRODUCTION
TABLE OF CONTENTS Page
§6.1. §6.2.
FEM Terminology Idealization §6.2.1. Models . . . . . . . . . . . . §6.2.2. Mathematical Models . . . . . . . §6.2.3. Implicit vs. Explicit Modeling . . . . . §6.3. Discretization §6.3.1. Decisions . . . . . . . . . . . §6.3.2. Error Sources and Approximation . . . §6.3.3. Other Discretization Methods . . . . §6.4. The Finite Element Method §6.4.1. Interpretation . . . . . . . . . . §6.4.2. Element Attributes . . . . . . . . §6.5. Classification of Mechanical Elements §6.5.1. Primitive Structural Elements . . . . . §6.5.2. Continuum Elements . . . . . . . §6.5.3. Special Elements . . . . . . . . . §6.5.4. Macroelements . . . . . . . . . §6.5.5. Substructures . . . . . . . . . . §6.6. Assembly §6.7. Boundary Conditions §6.7.1. Essential and Natural B.C. . . . . . §6.7.2. Boundary Conditions in Structural Problems §6. Notes and Bibliography . . . . . . . . . . . . . . . §6. References . . . . . . . . . . . . . . .
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6–3 6–4 6–5 6–5 6–6 6–6 6–6 6–7 6–7 6–8 6–8 6–8 6–10 6–10 6–10 6–10 6–11 6–11 6–12 6–12 6–12 6–12 6–13 6–13
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§6.1
FEM TERMINOLOGY
Chapters 2 through 5 cover material technically known as Matrix Structural Analysis or MSA. This is a subject that historically preceded the Finite Element Method (FEM), as chronicled in Appendix H. Here we begin the coverage of the FEM proper. This is technically distinguished from MSA by the more dominant role of continuum and variational mechanics. This Chapter introduces terminology used in FEM modeling, and surveys the attributes and types of finite elements used in structural mechanics. The next Chapter gives more specific rules for defining meshes, forces and boundary conditions. §6.1. FEM Terminology The ubiquitous term “degrees of freedom,” often abbreviated to either freedom or DOF, has figured prominently in the preceding Chapters. This term, as well as “stiffness matrix” and “force vector,” originated in structural mechanics, the application for which FEM was invented. These names have carried over to non-structural applications. This “terminology overspill” is discussed next. Classical analytical mechanics is that invented by Euler and Lagrange in the XVIII century and further developed by Hamilton and Jacobi as a systematic formulation of Newtonian mechanics. Its objects of attention are models of mechanical systems ranging from material particles composed of sufficiently large number of molecules, through airplanes, to the Solar System.1 The spatial configuration of any such system is described by its degrees of freedom. These are also called generalized coordinates. The terms state variables and primary variables are also used, particularly in mathematically oriented treatments. If the number of degrees of freedom is finite, the model is called discrete, and continuous otherwise. Because FEM is a discretization method, the number of degrees of freedom of a FEM model is necessarily finite. The freedoms are collected in a column vector called u. This vector is called the DOF vector or state vector. The term nodal displacement vector for u is reserved to mechanical applications. In analytical mechanics, each degree of freedom has a corresponding “conjugate” or “dual” term, which represents a generalized force.2 In non-mechanical applications, there is a similar set of conjugate quantities, which for want of a better term are also called forces or forcing terms. They are the agents of change. These forces are collected in a column vector called f. The inner product fT u has the meaning of external energy or work. Just as in the truss problem, the relation between u and f is assumed to be of linear and homogeneous. The last assumption means that if u vanishes so does f. The relation is then expressed by the master stiffness equations: Ku = f. (6.1) K is universally called the stiffness matrix even in non-structural applications because no consensus has emerged on different names. The physical significance of the vectors u and f varies according to the application being modeled, as illustrated in Table 6.1. 1
For cosmological scales, such as galaxy clusters, the general theory of relativity is necessary. For the atomic and sub-particle world, quantum mechanics is appropriate.
2
In variational mathematics this is called a duality pairing.
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Chapter 6: FEM MODELING: INTRODUCTION
Table 6.1. Significance of u and f in Miscellaneous FEM Applications Application Problem
State (DOF) vector u represents
Conjugate vector f represents
Structures and solid mechanics Heat conduction Acoustic fluid Potential flows General flows Electrostatics Magnetostatics
Displacement Temperature Displacement potential Pressure Velocity Electric potential Magnetic potential
Mechanical force Heat flux Particle velocity Particle velocity Fluxes Charge density Magnetic intensity
If the relation between forces and displacements is linear but not homogeneous, equation (6.1) generalizes to Ku = f M + f I .
(6.2)
Here f I is the initial node force vector introduced in Chapter 29 for effects such as temperature changes, and f M is the vector of mechanical forces. The basic steps of FEM are discussed below in more generality. Although attention is focused on structural problems, most of the steps translate to other applications problems as noted above. The role of FEM in numerical simulation is schematized in Figure 6.1, which is a merged simplification of Figures 1.2 and 1.3. Although this diagram oversimplifies the way FEM is actually used, it serves to illustrate terminology. The three key simulation steps shown are: idealization, discretization and solution. Each step is a source of errors. For example, the discretization error is the discrepancy that appears when the discrete solution is substituted in the mathematical model. The reverse steps: continuification and realization, are far more difficult and ill-posed problems. The idealization and discretization steps, briefly mentioned in Chapter 1, deserve further discussion. The solution step is dealt with in more detail in Part III of this book. IDEALIZATION
DISCRETIZATION
SOLUTION
FEM Physical system
Mathematical model
REALIZATION & IDENTIFICATION
Discrete solution
Discrete model
CONTINUIFICATION
Solution error
Discretization + solution error
Modeling + discretization + solution error Figure 6.1. A simplified view of the physical simulation process, reproduced to illustrate modeling terminology.
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§6.2
IDEALIZATION
§6.2. Idealization Idealization passes from the physical system to a mathematical model. This is the most important step in engineering practice, because it cannot be “canned.” It must be done by a human. §6.2.1. Models The word “model” has the traditional meaning of a scaled copy or representation of an object. And that is precisely how most dictionaries define it. We use here the term in a more modern sense, which has become increasingly common since the advent of computers: A model is a symbolic device built to simulate and predict aspects of behavior of a system. (6.3) Note the distinction made between behavior and aspects of behavior. To predict everything, in all physical scales, you must deal with the actual system. A model abstracts aspects of interest to the modeler. The qualifier symbolic means that a model represents a system in terms of the symbols and language of another discipline. For example, engineering systems may be (and are) modeled with the symbols of mathematics and/or computer sciences.3 §6.2.2. Mathematical Models Mathematical modeling, or idealization, is a process by which an engineer or scientist passes from the actual physical system under study, to a mathematical model of the system, where the term model is understood in the sense of (6.3). The process is called idealization because the mathematical model is necessarily an abstraction of the physical reality — note the phrase aspects of behavior in (6.3). The analytical or numerical results produced by the mathematical model are physically re-interpreted only for those aspects.4 To give an example on the choices an engineer may face, suppose that the structure is a flat plate structure subjected to transverse loading. Here is a non-exhaustive list of four possible mathematical models: 1.
A very thin plate model based on Von Karman’s coupled membrane-bending theory.
2.
A thin plate model, such as the classical Kirchhoff’s plate theory.
3.
A moderately thick plate model, for example Mindlin-Reissner plate theory.
4.
A very thick plate model based on three-dimensional elasticity.
The person responsible for this kind of decision is supposed to be familiar with the advantages, disadvantages, and range of applicability of each model. Furthermore the decision may be different in static analysis than in dynamics. Why is the mathematical model an abstraction of reality? Engineering systems, particularly in Aerospace and Mechanical, tend to be highly complex. For simulation it is necessary to reduce that 3
A problem-definition input file, a digitized earthquake record, or a stress plot are examples of the latter.
4
Whereas idealization can be reasonably taught in advanced design courses, the converse process of “realization” or “identification” — see Figure 6.1 — generally requires considerable physical understanding and maturity that can only be gained through professional experience.
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Chapter 6: FEM MODELING: INTRODUCTION
complexity to manageable proportions. Mathematical modeling is an abstraction tool by which complexity can be controlled. This is achieved by “filtering out” physical details that are not relevant to the analysis process. For example, a continuum material model filters out the aggregate, crystal, molecular and atomic levels of matter. Engineers are typically interested in a few integrated quantities, such as the maximum deflection of a bridge or the fundamental periods of an airplane. Although to a physicist this is the result of the interaction of billions and billions of molecules, such details are weeded out by the modeling process. Consequently, picking a mathematical model is equivalent to choosing an information filter.
al atic hem Matmodel
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§6.2.3. Implicit vs. Explicit Modeling Figure 6.2. A reproduction of Figure 1.5 with As noted the diagram of Figure 6.1 is an oversimplisome relabeling. Illustrates implicit modeling: fication of engineering practice. The more common picking elements from an existing FEM code scenario is that pictured in Figures 1.2, 1.4 and 1.5. consents to an idealization. The latter is reproduced in Figure 6.2 for convenience. A common scenario in industry is: you have to analyze a structure or a substructure, and at your disposal is a “black box” general-purpose finite element program. Those programs offer a catalog of element types; for example, bars, beams, plates, shells, axisymmetric solids, general 3D solids, and so on. The moment you choose specific elements from the catalog you automatically accept the mathematical models on which the elements are based. This is implicit modeling. Ideally you should be fully aware of the implications of your choice. Providing such “finite element literacy” is one of the objective of this book. Unfortunately many users of commercial programs are unaware of the implied-consent aspect of implicit modeling and their legal implications.
The other extreme happens when you select a mathematical model of the physical problem with your eyes wide open and then either shop around for a finite element program that implements that model, or write the program yourself. This is explicit modeling. It requires far more technical expertise, resources, experience and maturity than implicit modeling. But for problems that fall out of the ordinary it could be the right thing to do. In practice a combination of implicit and explicit modeling is common. The physical problem to be simulated is broken down into subproblems. Those subproblems that are conventional and fit available programs may be treated with implicit modeling, whereas those that require special handling may only submit to explicit modeling. §6.3. Discretization Mathematical modeling is a simplifying step. But models of physical systems are not necessarily simple to solve. They often involve coupled partial differential equations in space and time subject to boundary and/or interface conditions. Such models have an infinite number of degrees of freedom. 6–6
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§6.3
DISCRETIZATION
§6.3.1. Decisions At this point one faces the choice of going for analytical or numerical solutions. Analytical solutions, also called “closed form solutions,” are more intellectually satisfying, particularly if they apply to a wide class of problems, so that particular instances may be obtained by substituting the values of free parameters. Unfortunately they tend to be restricted to regular geometries and simple boundary conditions. Moreover some closed-form solutions, expressed for example as inverses of integral transforms, often have to be numerically evaluated to be useful. Most problems faced by the engineer either do not yield to analytical treatment or doing so would require a disproportionate amount of effort.5 The practical way out is numerical simulation. Here is where finite element methods enter the scene. To make numerical simulations practical it is necessary to reduce the number of degrees of freedom to a finite number. The reduction is called discretization. The product of the discretization process is the discrete model. As discussed in Chapter 1, for complex engineering systems this model is the product of a multilevel decomposition. Discretization can proceed in space dimensions as well as in the time dimension. Because the present book deals only with static problems, we need not consider the time dimension and are free to concentrate on spatial discretization. §6.3.2. Error Sources and Approximation Figure 6.1 tries to convey graphically that each simulation step introduces a source of error. In engineering practice modeling errors are by far the most important. But they are difficult and expensive to evaluate, because model validation requires access to and comparison with experimental results. These may be either scarce, or unavailable in the case of a new product in the design stage. Next in order of importance is the discretization error. Even if solution errors are ignored — and usually they can — the computed solution of the discrete model is in general only an approximation in some sense to the exact solution of the mathematical model. A quantitative measurement of this discrepancy is called the discretization error. The characterization and study of this error is addressed by a branch of numerical mathematics called approximation theory. Intuitively one might suspect that the accuracy of the discrete model solution would improve as the number of degrees of freedom is increased, and that the discretization error goes to zero as that number goes to infinity. This loosely worded statement describes the convergence requirement of discrete approximations. One of the key goals of approximation theory is to make the statement as precise as it can be expected from a branch of mathematics.6 5
This statement has to be tempered in two respects. First, the wider availability and growing power of computer algebra systems, outlined in Chapter 4, has widened the realm of analytical solutions than can be obtained within a practical time frame. Second, a combination of analytical and numerical techniques is often effective to reduce the dimensionality of the problem and to facilitate parameter studies. Important examples are provided by Fourier analysis, perturbation and boundary-element methods.
6
The discretization error is often overhyped in the FEM literature, since it provides an inexhaustible source of publishable poppycock. If the mathematical model is way off reducing the discretization error buys nothing; only a more accurate answer to the wrong problem.
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§6.3.3. Other Discretization Methods It was stated in the first chapter that the most popular discretization techniques in structural mechanics are finite element methods and boundary element methods. The finite element method (FEM) is by far the most widely used. The boundary element method (BEM) has gained in popularity for special types of problems, particularly those involving infinite domains, but remains a distant second, and seems to have reached its natural limits. In non-structural application areas such as fluid mechanics and electromagnetics, the finite element method is gradually making up ground but faces stiff competition from both the classical and energybased finite difference methods. Finite difference and finite volume methods are particularly well entrenched in computational fluid dynamics spanning moderate to high Reynolds numbers. §6.4. The Finite Element Method §6.4.1. Interpretation The finite element method (FEM) is the dominant discretization technique in structural mechanics. As discussed in Chapter 1, the FEM can be interpreted from either a physical or mathematical standpoint. The treatment has so far emphasized the former. The basic concept in the physical FEM is the subdivision of the mathematical model into disjoint (non-overlapping) components of simple geometry called finite elements or elements for short. The response of each element is expressed in terms of a finite number of degrees of freedom characterized as the value of an unknown function, or functions, at a set of nodal points. The response of the mathematical model is then considered to be approximated by that of the discrete model obtained by connecting or assembling the collection of all elements. The disconnection-assembly concept occurs naturally when examining many artificial and natural systems. For example, it is easy to visualize an engine, bridge, building, airplane, or skeleton as fabricated from simpler components. Unlike finite difference models, finite elements do not overlap in space. In the mathematical interpretation of the FEM, this property goes by the name disjoint support or local support. §6.4.2. Element Attributes Just like members in the truss example, one can take finite elements of any kind one at a time. Their local properties can be developed by considering them in isolation, as individual entities. This is the key to the modular programming of element libraries. In the Direct Stiffness Method, elements are isolated by the disconnection and localization steps, which were described for the truss example in Chapter 2. The procedure involves the separation of elements from their neighbors by disconnecting the nodes, followed by referral of the element to a convenient local coordinate system.7 After that we can consider generic elements: a bar element, a beam element, and so on. From the standpoint of the computer implementation, it means that you can write one subroutine or module that constructs, by suitable parametrization, all elements of one type, instead of writing one module for each element instance. 7
Both steps are only carried out in the modeler’s mind. They are placed as part of the DSM for instructional convenience. In practice the analysis begins directly at the element level.
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§6.4
THE FINITE ELEMENT METHOD
1D
2D
2D
3D
Figure 6.3. Typical finite element geometries in one through three dimensions.
Following is a summary of the data associated with an individual finite element. This data is used in finite element programs to carry out element level calculations. Intrinsic Dimensionality. Elements can have one, two or three space dimensions.8 There are also special elements with zero dimensionality, such as lumped springs or point masses. Nodal points. Each element possesses a set of distinguishing points called nodal points or nodes for short. Nodes serve a dual purpos: definition of element geometry, and home for degrees of freedom. They are usually located at the corners or end points of elements, as illustrated in Figure 6.3. In the so-called refined or higher-order elements nodes are also placed on sides or faces, as well as perhaps the interior of the element. Geometry. The geometry of the element is defined by the placement of the nodal points. Most elements used in practice have fairly simple geometries. In one-dimension, elements are usually straight lines or curved segments. In two dimensions they are of triangular or quadrilateral shape. In three dimensions the most common shapes are tetrahedra, pentahedra (also called wedges or prisms), and hexahedra (also called cuboids or “bricks”). See Figure 6.3. Degrees of freedom. The degrees of freedom (DOF) specify the state of the element. They also function as “handles” through which adjacent elements are connected. DOFs are defined as the values (and possibly derivatives) of a primary field variable at nodal points. The actual selection depends on criteria studied at length in Part II. Here we simply note that the key factor is the way in which the primary variable appears in the mathematical model. For mechanical elements, the primary variable is the displacement field and the DOF for many (but not all) elements are the displacement components at the nodes. Nodal forces. There is always a set of nodal forces in a one-to-one correspondence with degrees of 8
In dynamic analysis, time appears as an additional dimension.
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Chapter 6: FEM MODELING: INTRODUCTION
freedom. In mechanical elements the correspondence is established through energy arguments. Constitutive properties. For a mechanical element these are relations that specify the material behavior. For example, in a linear elastic bar element it is sufficient to specify the elastic modulus E and the thermal coefficient of expansion α. Fabrication properties. For mechanical elements these are fabrication properties which have been integrated out from the element dimensionality. Examples are cross sectional properties of MoM elements such as bars, beams and shafts, as well as the thickness of a plate or shell element. This data is used by the element generation subroutines to compute element stiffness relations in the local system. §6.5. Classification of Mechanical Elements The following classification of finite elements in structural mechanics is loosely based on the “closeness” of the element with respect to the original physical structure. It is given here because it clarifies points that recur in subsequent sections, as well as providing insight into advanced modeling techniques such as hierarchical breakdown and globallocal analysis.
Physical Structural Component
Mathematical Model Name
Finite Element Discretization
bar
§6.5.1. Primitive Structural Elements beam
These resemble fabricated structural components, and are often drawn as such; see Figure 6.4. The qualifier primitive is used to distinguish them from macroelements, which is another element class described below. Primitive means that they are not decomposable into simpler elements.
tube, pipe
spar (web)
These elements are usually derived from Mechanics-of-Materials simplified theories and are better understood from a physical, rather than mathematical, standpoint. Examples are the elements discussed in Chapter 5: bars, cables, beams, shafts, spars.
shear panel (2D version of above) Figure 6.4. Examples of primitive structural elements.
§6.5.2. Continuum Elements These do not resemble fabricated structural components at all. They result from the subdivision of “blobs” of continua, or of structural components viewed as continua. Unlike structural elements, continuum elements are better understood in terms of their mathematical interpretation. Examples: plates, slices, shells, axisymmetric solids, general solids. See Figure 6.5. 6–10
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§6.5
CLASSIFICATION OF MECHANICAL ELEMENTS
Finite element idealization
Physical
Physical
plates
Finite element idealization
3D solids
Figure 6.5. Continuum element examples.
§6.5.3. Special Elements Special elements partake of the characteristics of structural and continuum elements. They are derived from a continuum mechanics standpoint but include features closely related to the physics of the problem. Examples: crack elements for fracture mechanics applications, shear panels, infinite and semi-infinite elements, contact and penalty elements, rigid-body elements. See Figure 6.6.
Infinity double node
Crack element
Infinite element
Honeycomb panel
Figure 6.6. Special element examples.
§6.5.4. Macroelements Macroelements are also called mesh units and superelements, although the latter term overlaps with substructures (defined below). These often resemble structural components, but are fabricated with simpler elements. See Figure 6.7. The main reason for introducing macroelements is to simplify preprocessing tasks. For example, it may be simpler to define a regular 2D mesh using quadrilatFigure 6.7. Macroelement examples. erals rather than triangles. The fact that the quadrilateral is really a macroelement may not be important for the majority of users. Similarly a box macroelement can save modeling times for structures that are built by such components; for example box-girder bridges. 6–11
Chapter 6: FEM MODELING: INTRODUCTION
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§6.5.5. Substructures Also called structural modules and superelements. These are sets of elements with a well defined structural function, typically obtained by cutting the complete structure into functional components. Examples: the wings and fuselage in an airplane, the deck and cables in a suspension bridge. The distinction between substructures and macroelements is not clear-cut. The main conceptual distinction is that substructures are defined “top down” as parts of a complete structure, whereas macroelements are built “bottom up” from primitive elements. The term superelement is often used in a collective sense to embrace all element groupings. This topic is further covered in Chapter 10. §6.6. Assembly The assembly procedure of the Direct Stiffness Method for a general finite element model follows rules identical in principle to those discussed for the truss example. As in that case the processs involves two basic steps: Globalization. The element equations are transformed to a common global coordinate system. Merge. The element stiffness equations are merged into the master stiffness equations by appropriate indexing and matrix-entry addition. The hand calculations for the example truss conceal the implementation complexity. The master stiffness equations in practical cases may involve thousands (or even millions) of freedoms. The use of sparse matrix techniques, and possibly peripheral storage, becomes essential. But this inevitably increases the programming complexity. The topic is elaborated upon in Part III. §6.7. Boundary Conditions A key strength of the FEM is the ease and elegance with which it handles arbitrary boundary and interface conditions. This power, however, has a down side. A big hurdle faced by FEM newcomers is the understanding and proper handling of boundary conditions. Below is a simple recipe for treating boundary conditions. The following Chapter provides specific rules and examples. §6.7.1. Essential and Natural B.C. The key thing to remember is that boundary conditions (BCs) come in two basic flavors: essential and natural. Essential BCs directly affect DOFs, and are imposed on the left-hand side vector u. Natural BCs do not directly affect DOFs and are imposed on the right-hand side vector f. The mathematical justification for this distinction requires use of concepts from variational calculus, and is consequently relegated to Part II. For the moment, the basic recipe is: 1.
If a boundary condition involves one or more degrees of freedom in a direct way, it is essential. An example is a prescribed node displacement.
2.
Otherwise it is natural.
The term “direct” is meant to exclude derivatives of the primary function, unless those derivatives also appear as degrees of freedom, such as rotations in beams and plates. 6–12
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§6.
References
§6.7.2. Boundary Conditions in Structural Problems Essential boundary conditions in mechanical problems involve displacements (but not strain-type displacement derivatives). Support conditions for a building or bridge problem furnish a particularly simple example. But there are more general boundary conditions that occur in practice. A structural engineer must be familiar with displacement B.C. of the following types. Ground or support constraints. Directly restraint the structure against rigid body motions. Symmetry conditions. To impose symmetry or antisymmetry restraints at certain points, lines or planes of structural symmetry. This allows the discretization to proceed only over part of the structure with a consequent savings in modeling effort and number of equations to be solved. Ignorable freedoms. To suppress displacements that are irrelevant to the problem.9 Even experienced users of finite element programs are sometimes baffled by this kind. An example are rotational degrees of freedom normal to smooth shell surfaces. Connection constraints. To provide connectivity to adjoining structures or substructures, or to specify relations between degrees of freedom. Many conditions of this type can be subsumed under the label multipoint constraints or multifreedom constraints. These can be notoriously difficult to handle from a numerical standpoint, and are covered in Chapters 8–9. Notes and Bibliography Most FEM textbooks do not provide a systematic treatment of modeling. This is no accident: few academic authors have experience with complex engineering systems. Good engineers are too busy (and in demand) to have time for writing books. This gap has been particularly acute since FEM came on the scene because of generational gaps: “real engineers” tend to mistrust the computer, and often for good reason. The notion of explicit versus implicit modeling, which has deep legal and professional implications, is rarely mentioned. FEM terminology is by now standard, and so is a majority of the notation. But that is not so in early publications. E.g. K is universally used10 for stiffness matrix in virtually all post-1960 books. There are a few exceptions: Przemieniecki [135] uses S. There is less unanimity on u and f for node displacement and force vectors, respectively; some books such as Zienkiewicz and Taylor [186] still use different symbols. The element classification given here attempts to systematize dispersed references. In particular, the distinction between macroelements, substructures and superelements is an ongoing source of confusion, particularly since massively parallel computation popularized the notion of “domain decomposition” in the computer science community. The all-encompassing term “superelement” emerged in Norway by 1968 as part of the implementation of the computer program SESAM; additional historical details are provided in Chapter 10. The topic of BC classification and handling is a crucial one in practice. More modeling mistakes are done in this aspect of FEM application than anywhere else. References Referenced items have been moved to Appendix R.
9 10
In classical dynamics these are called ignorable coordinates. A symbol derived from the “spring constant” k that measures the stiffness of a mechanical spring.
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FEM Modeling: Mesh, Loads and BCs
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Chapter 7: FEM MODELING: MESH, LOADS AND BCS
TABLE OF CONTENTS Page
§7.1. §7.2.
§7.3.
§7.4. §7.5.
§7.6.
§7. §7. §7.
General Recommendations Guidelines on Element Layout §7.2.1. Mesh Refinement . . . . . . . §7.2.2. Element Aspect Ratios . . . . . §7.2.3. Physical Interfaces . . . . . . §7.2.4. Preferred Shapes . . . . . . . Direct Lumping of Distributed Loads §7.3.1. Node by Node (NbN) Lumping . . §7.3.2. Element by Element (EbE) Lumping §7.3.3. *Weighted Lumping . . . . . . §7.3.4. *Energy Consistent Lumping . . Boundary Conditions Support Conditions §7.5.1. Supporting Two Dimensional Bodies §7.5.2. Supporting Three Dimensional Bodies Symmetry and Antisymmetry Conditions §7.6.1. Visualization . . . . . . . . §7.6.2. Effect of Loading Patterns . . . . Notes and Bibliography . . . . . . . . . . . . . References. . . . . . . . . . . . . Exercises . . . . . . . . . . . . .
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§7.2
GUIDELINES ON ELEMENT LAYOUT
This Chapter continues the exposition of finite element modeling principles. After some general recommendations, it gives guidelines on layout of finite element meshes, conversion of distributed loads to node forces, and handling the simplest forms of support boundary conditions. The next two Chapters deal with more complicated forms of boundary conditions called multifreedom constraints. The presentation is “recipe oriented” and illustrated by specific examples from structural mechanics. Most examples are two-dimensional. No attempt is given at rigorous justification of rules and recommendations, because that would require mathematical tools beyond the scope of this course. §7.1. General Recommendations The general rules that should guide you in the use of commercial or public FEM packages, are
•
Use the simplest type of finite element that will do the job.
•
Never, never, never mess around with complicated or special elements, unless you are absolutely sure of what you are doing.
•
Use the coarsest mesh you think will capture the dominant physical behavior of the physical system, particularly in design applications.
Three word summary: keep it simple. Initial FE models may have to be substantially revised to accommodate design changes. There is little point in using complicated models that will not survive design iterations. The time for refined models is when the design has stabilized and you have a better picture of the underlying physics, possibly reinforced by experiments or observation. §7.2. Guidelines on Element Layout The following guidelines are stated for structural applications. As noted above, they will be often illustrated for two-dimensional meshes of continuum elements for ease of visualization. §7.2.1. Mesh Refinement Use a relatively fine (coarse) discretization in regions where you expect a high (low) gradient of strains and/or stresses.1 Regions to watch out for high gradients are: •
Near entrant corners, or sharply curved edges.
•
In the vicinity of concentrated (point) loads, concentrated reactions, cracks and cutouts.
•
In the interior of structures with abrupt changes in thickness, material properties or cross sectional areas.
The examples in Figure 7.1 illustrate some of these “danger regions.” Away from such regions one can use a fairly coarse discretization within constraints imposed by the need of representing the structural geometry, loading and support conditions reasonably well. 1
Gradient is the key word. High gradient means rapid variation. A high value by itself means nothing in this context.
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Chapter 7: FEM MODELING: MESH, LOADS AND BCS
entrant corners Cutouts
Cracks Vicinity of concentrated (point) loads, and sharp contact areas weld
Load transfer (bonded joints, welds, anchors, reinforcing bars, etc.)
Abrupt thickness changes
Material interfaces
Figure 7.1. Some situations where a locally refined finite element discretization (in the red-shaded areas) is recommended.
§7.2.2. Element Aspect Ratios
Good
Bad
When discretizing two and three dimensional problems, try to avoid finite elements of high aspect ratios: elongated or “skinny” elements, such as the ones illustrated on the right of Figure 7.2. (The aspect ratio of a two- or three-dimensional element is the ratio between its largest and smallest dimension.) As a rough guideline, elements with aspect ratios exceeding 3 should be viewed with caution and those exceeding 10 with alarm. Such elements will not necessarily produce bad results — that depends on the loading and boundary conditions of the problem — but do introduce the potential for trouble.
Figure 7.2. Elements with good and bad aspect ratios.
Remark 7.1. In many “thin” structures modeled as continuous bodies the appearance of “skinny” elements is
inevitable on account of computational economy reasons. An example is provided by the three-dimensional modeling of layered composites in aerospace and mechanical engineering problems.
§7.2.3. Physical Interfaces A physical interface, resulting from example from a change in material, should also be an interelement boundary. That is, elements must not cross interfaces. See Figure 7.3. 7–4
7–5
§7.3
DIRECT LUMPING OF DISTRIBUTED LOADS
No
OK
Physical interface
Figure 7.3. llustration of the rule that elements should not cross material interfaces.
§7.2.4. Preferred Shapes In 2D FE modeling, if you have a choice between triangles and quadrilaterals with similar nodal arrangement, prefer quadrilaterals. Triangles are quite convenient for mesh generation, mesh transitions, rounding up corners, and the like. But sometimes triangles can be avoided altogether with some thought. One of the homework exercises is oriented along these lines. In 3D FE modeling, prefer strongly bricks over wedges, and wedges over tetrahedra. The latter should be used only if there is no viable alternative.2 The main problem with tetrahedra and wedges is that they can produce wrong stress results even if the displacement solution looks reasonable. §7.3. Direct Lumping of Distributed Loads In practical structural problems, distributed loads are more common than concentrated (point) loads.3 Distributed loads may be of surface or volume type. Distributed surface loads (called surface tractions in continuum mechanics) are associated with actions such as wind or water pressure, snow weight on roofs, lift in airplanes, live loads on bridges, and the like. They are measured in force per unit area. Volume loads (called body forces in continuum mechanics) are associated with own weight (gravity), inertial, centrifugal, thermal, prestress or electromagnetic effects. They are measured in force per unit volume. A derived type: line loads, result from the integration of surface loads along one transverse direction, such as a beam or plate thickness, or of volume loads along two transverse directions, such as a bar or beam area. Line loads are measured in force per unit length. Whatever their nature or source, distributed loads must be converted to consistent nodal forces for FEM analysis. These forces end up in the right-hand side of the master stiffness equations. The meaning of “consistent” can be made precise through variational arguments, by requiring that the distributed loads and the nodal forces produce the same external work. Since this requires the introduction of external work functionals, the topic is deferred to Part II. However, a simpler approach called direct load lumping, or simply load lumping, is often used by structural engineers 2
Unfortunately, many existing space-filling automatic mesh generators in three dimensions produce tetrahedral meshes. There are generators that try to produce bricks, but these often fail in geometrically complicated regions.
3
In fact, one of the objectives of a good structural design is to avoid or alleviate stress concentrations produced by concentrated forces.
7–5
7–6
Chapter 7: FEM MODELING: MESH, LOADS AND BCS Nodal force f3 at 3 is set to P, the magnitude of the crosshatched area under the load curve. This area extends halfway over adjacent element sides
;; ;; ;; ;;
Distributed load intensity (load acts downward on boundary)
f3 = P
1
2
3
Boundary
4
5
6
Finite element mesh
Figure 7.4. NbN direct lumping of distributed line load, illustrated for a 2D problem.
in lieu of the more mathematically impeccable but complicated variational approach. Two variants of this technique are described below for distributed surface loads. §7.3.1. Node by Node (NbN) Lumping The node by node (NbN) lumping method is graphically explained in Figure 7.4. This example shows a distributed surface loading acting normal to the straight boundary of a two-dimensional FE mesh. (The load is assumed to have been integrated through the thickness normal to the figure, so it is actually a line load measured as force per unit length.) The procedure is also called tributary region or contributing region method. For the example of Figure 7.4, each boundary node is assigned a tributary region around it that extends halfway to the adjacent nodes. The force contribution P of the cross-hatched area is directly assigned to node 3. This method has the advantage of not requiring the computation of centroids, as needed in the EbE technique discussed in the next subsection. For this reason it is often preferred in hand computations. It can be extended to three-dimensional meshes as well as volume loads.4 It should be avoided, however, when the applied forces vary rapidly (within element length scales) or act only over portions of the tributary regions. §7.3.2. Element by Element (EbE) Lumping In this variant the distributed loads are divided over element domains. The resultant load is assigned to the centroid of the load diagram, and apportioned to the element nodes by statics. A node force is obtained by adding the contributions from all elements meeting at that node. The procedure is illustrated in Figure 7.5, which shows details of the computation over 2–3. The total force at node 3, for instance, would be that contributed by segments 2-3 and 3-4. If applicable, EbE is more accurate than NbN lumping. In fact it agrees with consistent node lumping for simple elements that possess only corner nodes. In those cases it is not affected by sharpness of the load variation and can be even used for point loads that are not applied at the nodes. 4
The computation of tributary areas and volumes for general 2D and 3D regions can be done through the so-called Voronoi diagrams. This is an advanced topic in computational geometry (see, e.g., [41]) and thus not treated here.
7–6
7–7
§7.3 DIRECT LUMPING OF DISTRIBUTED LOADS
Force P has magnitude of crosshatched area under load curve and acts at its centroid.
;;; ;;; ;;; ;;;
1
P
2
f 2e = (b/L)P
centroid C of crosshatched area
e
f2e
Distributed load intensity (load acts downward on boundary)
a Boundary
4
5
f3e = (a/L)P
P
2
f3e
3
C
3
b
L=a+b
6
Details of element force computations
Finite element mesh
Figure 7.5. EbE direct lumping of distributed line load, illustrated for a 2D problem.
The EbE procedure is not applicable if the centroidal resultant load cannot be apportioned by statics. This happens if the element has midside faces or internal nodes in addition to corner nodes, or if it has rotational degrees of freedom. For those elements the variational-based consistent approach covered in Part II and briefly described in §7.3.3, is preferable. (a)
;;;; ;;;; ;;;; ;;;; ;; ;; ;; MAX WATER EL. 584'
SPILLWAY CREST EL. 552'
(c) y
Concrete dam
EL. 364'
x
200'
A
(b)
Water
C
B
spillway crest 188' over riverbed
(d)
Soil
Figure 7.6. Norfork Dam: (a,b) picture; (c) cross section of dam above foundation; (d) coarse FEM mesh including foundation and soil [26,30]
Example 7.1. Figure 7.6(a,b) show pictures of the Norfork Dam, a 230-ft high gravity dam discretized by triangular elements as illustrated in Figure 7.6(b). The dam has a length of 2624 ft and was constructed over the White River in Arkansas over 1941–44.5 The structure is assumed to be in plane strain. The model is a 5
As described in Notes and Bibliography, this example has historical significance, as the first realistic Civil Engineering structure modeled by the Finite Element Method.
7–7
7–8
Chapter 7: FEM MODELING: MESH, LOADS AND BCS
(b)
(a)
180 ft
9
8
x 9=−490
x8 =−350
7 x7 =−210
70 120 162
2
3 y = 84 3
p=62.4 d (lbs/sq-ft)
C
22 1
A1 y1 =180 2 y =136
d
Water
6
(c) 2 3 4
4 y = 36 4 B y =0 5 5 x5 =0
1 44 2 96 144 3 180 4
C 5
5
x6 =−70 All dimensions in ft. Plane strain model, slice 1 ft thick normal to paper
Figure 7.7. Norfork Dam example: (a) computation of wet node forces due to hydrostatic pressure; (b) NbN tributary regions; (c) EbE regions.
typical 1-ft slice of the dam and near-field soil. In the analysis of dam and marine structures, a “wet node” of a FEM discretization is one in contact with the water. The effect of hydrostatic pressure is applied to the structure through nodal forces on wet nodes. The wet nodes for a water head of 200 ft are shown in Figure 7.7(b), and are numbered 1 through 9 for convenience. The pressure in psf (lbs per sq-ft) is p = 62.4d where d is the depth in ft. Compute the hydrostatic nodal forces f x1 and f x2 using NbN and EbE. Assume that the wet face AB is vertical for simplicity. Node by Node. For node 1 go halfway to 2, a distance of (200 − 152)/2 = 24 ft. The tributary load area is a triangle extending 24 ft along y with bottom pressure 62.4 × 24 = 1497.6 psf and unit width normal to paper, giving f x1 = 12 24 × 1497.6 × 1 = 17971 lbs. For node 2 go up 24 ft and down (152 − 94)/2 = 29 ft. The tributary load area is a trapezoid extending 24 + 29 = 53 ft along y, with pressures 1497.6 psf at top and 62.4 × 77 = 4804.8 psf at bottom. This gives f x2 = 53 × (1497.6 + 4804.8)/2 × 1 = 167013 lbs. Element by Element. It is convenient to start by computing pressures at wet node levels: p1 = 0, p2 = 62.4 × 48 = 2995.2 psf, p3 = 62.4 × 106 = 6601.4. Line element 1–2, labeled as (1), receives a total load of 12 48 × 2995.2 = 72000 lbs. Since the centroid of a triangle is at 32 ft from 1 and 16 ft from 2, node 1 (1) (1) gets f x1 = 13 72000 = 24000 while node 2 gets f x2 = 23 72000 = 48000 lbs. Line element 2–3, labeled as (2), receives a total load of 58 × (2995.2 + 6601.4)/2 = 278201.4 lbs. The centroid of the 2–3 trapezoid with y extent 58 ft and loads proportional to 48 and 106 at top and bottom is located at ft from node 2 and (2) = lbs. Adding element contributions we get ft from node 3. The contribution to node 2 is therefore f x2 (1) (1) (2) f x1 = f x1 = 24000 and f x2 = f x2 + f x1 = lbs. The computations for the wet nodes 3 through 9 are left as an exercise. 2
Example 7.2. Figure 7.8 shows the mesh for the
y > 0 portion of the gravity dam of the previous example. (The node numbering is different). The specific weight for concrete of γ = 200 pcf (pounds per cubic foot). Compute the node force f y11 due to weight. For this calculation the NbN scheme is unwieldy because computation of the nodal tributary region requires construction of the Vornoi diagram. (Furthermore for constant specific weight it gives the same answer as EbE.) To apply EbE, select the elelements that contribute to 11: triangles (), (), (), (), () and ().
7–8
4
1
(1)
3
(2) (3) (4)
6
Node 7 8 9 7 9 (12) (10) 10 (9) (11) 11 11 12 (13) 13 10 12 13 (15) (19) (17) 14 (16) (20) (18) (14) 16 17 15 15 18 (29) 14 16 (21) (27) (23) (25) 19 17 (22) (28) (26) (24) 18
20
(5)
5
(7)
(6)
8
21
(8)
22
23
24
x *.00 *.00 *.00 *.00 *.00 *.00 *.00 *.00 *.00 *.00 *.00 *.00
y 152.00 152.00 152.00 94.00 94.00 94.00 94.00 40.00 40.00 40.00 40.00 40.00
Figure 7.8. Computation of weight foirce at node 11 of gravity dam example of previous figure.
7–9
§7.3 DIRECT LUMPING OF DISTRIBUTED LOADS
(a)
halfway between n−1 and n
Wn = 1 1
n−1
n
halfway between n and n+1
(b)
Wn 1
n+1
n−1
xn
xn
n
n+1
Figure 7.9. Weight functions corresponding to: (a) NbN lumping, and (b) EbE lumping.
The area of a triangle with corners at {{x1 , y1 }, {x1 , y1 }, {x1 , y1 }} is given by A = (x2 y3 − x3 y2 ) + (x3 y1 − x1 y3 ) + (x1 y2 − x2 y1 ). Applying this to the geometry of the figure one finds that the areas are A(9) = ∗, A(10) = ∗, A(11) = ∗, A(17) = ∗, A(18) = ∗, and A(19) = ∗. The weight forces on each element are A(9) = γ h∗, A(10) = γ h∗, A(11) = γ h∗, A(17) = γ h∗, A(18) = γ h∗, and A(19) = γ h∗, where h is the thickness normal to the figure (1 ft in this example). For uniform element thickness and specific weight, one third of each force goies to each element corner. Thus node 11 receives f y11 = 13 γ h(A(9) + A(10) + A(11) + A(17) + A(18) =
1 3
× 1 × 200 × 1× =
(7.1)
§7.3.3. *Weighted Lumping The NbN and EbE methods are restricted to simple elements, specifically those with corner nodes only. We outline here a general method that works for more complicated models. The mathematical justification requires energy theorems covered in Part II. Thus at this stage the technique is merely presented as a recipe. To fix the ideas consider again the 2D situation depicted in Figures 7.4 and 7.5. Denote the distributed load by q(x). We want to find the lumped force f n at an interior node n of coordinate xn . Let the adjacent nodes be n − 1 and n + 1, with coordinates xn−1 and xn+1 , respectively. Introduce a weight function Wn (x) with properties to be specified below. The lumped force is given by
0002
xn+1
fn =
Wn (x) q(x) d x
(7.2)
xn−1
For this formula to make sense, the weight function must satisfy several properties: 1.
Unit value at node n: Wn (xn ) = 1.
2.
Vanishes over any element not pertaining to n: Wn = 0 if x ≤ xn−1 or x ≥ xn+1
3.
Gives the same results as NbN or EbE for constant q over elements with corner nodes only.
Both NbN and EbE are special cases of (7.2). For NbN pick Wn (x) = 1 For EbE pick
if
1 (xn−1 2
+ xn ) ≤ x ≤ 12 (xn + xn+1 ),
 x−x  1 − xn − xn−1 n−1 Wn (x) = x − xn 1 −  xn+1 − xn 0
if xn−1 ≤ x ≤ xn , if xn ≤ x ≤ xn+1 , otherwise.
These particular weight functions are depicted in Figure 7.9.
7–9
wn (x) = 0 otherwise.
(7.3)
(7.4)
7–10
Chapter 7: FEM MODELING: MESH, LOADS AND BCS
§7.3.4. *Energy Consistent Lumping The rule (7.2) can be justified from the standpoint of the Principle of Virtual Work. Let δu n be a virtual node displacement paired to f n . Take δu n Wn (x) to be the associated displacement variation. The external virtual 0006 work of q(x) is q(x)Wn (x) δu n d x extended over the portion where Wn = 0. Equating this to f n δu n and cancelling δu n from both sides yields (7.2).
q h (per unit of x length)
1
f1 =?
f2 =?
2
W1 = − ξ(1−ξ)/2 = N1
3 f 3 =?
1
1
2
3 W2 = 1−ξ 2 = N2
1
1
2
3
W3= ξ(1+ξ)/2 = N3 2 1
If Wn is the trial displacement function actually used for the development of element equations the lumping is called energy consistent or consistent for short.
a
1
3
ξ = 1 − 2x/a x
Figure 7.10. Example of consistent load lumping.
(As will be seen later, trial functions are the union of shape functions over the patch of all elements connected to node n.) This important technique is studied in Part II of the course after energy methods are introduced. Example 7.3. Conside the mesh of 9-node quadrilaterals shown in Figure 7.10. (This is later used as a benchmark problem in Chapter 27.) The upper plate edge 1–3 is subject to a uniform normal load qh per unit length, where h is the plate thickness. The problem is to compute the node forces f 1 , f 2 and f 3 . If 1–2 and 2–3 were on two different elements, both NbN and EbE would give f 1 = f 3 = 14 qha and f 2 = 12 qha. But this lumping is wrong for an element with midside nodes. Instead, pick the weight functions Wi (i = 1, 2, 3) shown on the right of Figure 7.10.
Since there is only one element on the loaded edge, the Wi are actually the quadratic shape functions Ni for a 3-node line element, developed in later Chapters. The dimensionless variable ξ is called an isoparametric natural coordinate. Applying the rule (7.2) we get
0002
f1 = 0
a
0002
1
dx dξ = W1 (x) q h d x = W1 (ξ ) q h dξ −1
0002
1 −1
− 12 ξ(1 − ξ )q h ( 12 a) dξ = 16 qha.
(7.5)
Similarly f 2 = 23 qha and f 3 = f 1 . As a check, f 1 + f 2 + f 3 = qha, which is the total load acting on the plate edge.
§7.4. Boundary Conditions The key distinction between essential and natural boundary conditions (BC) was introduced in the previous Chapter. The distinction is explained in Part II from a variational standpoint. In this section we discuss the simplest essential boundary conditions in structural mechanics from a physical standpoint. This makes them relevant to problems with which a structural engineer is familiar. Because of the informal setting, the ensuing discussion relies heavily on examples. In structural problems formulated by the DSM, the recipe of §6.7.1 that distinguishes between essential and natural BC is: if it directly involves the nodal freedoms, such as displacements or rotations, it is essential. Otherwise it is natural. Conditions involving applied loads are natural. Essential BCs take precedence over natural BCs. The simplest essential boundary conditions are support and symmetry conditions. These appear in many practical problems. More exotic types, such as multifreedom constraints, require more advanced mathematical tools and are covered in the next two Chapters. 7–10
§7.5 (b)
y
(c)
; ; ;;;; ;;;; ; ;; B
A
SUPPORT CONDITIONS
y
;; ;;
(a)
;;;
; ;;;; ;
7–11
B
; ;;; B
x
A
A
x
Figure 7.11. Suppressing two-dimensional rigid body motions.
§7.5. Support Conditions Supports are used to restrain structures against relative rigid body motions. This is done by attaching them to Earth ground (through foundations, anchors or similar devices), or to a “ground structure” which is viewed as the external environment.6 The resulting boundary conditions are often called motion constraints. In what follows we analyze two- and three-dimensional motions separately. §7.5.1. Supporting Two Dimensional Bodies Figure 7.11 shows two-dimensional bodies that move in the plane of the paper. If a body is not restrained, an applied load will cause infinite displacements. Regardless of loading conditions, the body must be restrained against two translations along x and y, and one rotation about z. Thus the minimum number of constraints that has to be imposed in two dimensions is three. In Figure 7.11, support A provides translational restraint, whereas support B, together with A, provides rotational restraint. In finite element terminology, we say that we delete (fix, remove, preclude) all translational displacements at point A, and that we delete the translational degree of freedom directed along the normal to the AB direction at point B. This body is free to distort in any manner without the supports imposing any deformation constraints. Engineers call A and B reaction-to-ground points. This means that if the supports are conceptually removed, applied loads are automatically balanced by reactive forces at A and B, in accordance with Newton’s third law. Additional freedoms may be precluded to model greater restraint by the environment. However, Figure 7.11(a) does illustrate the minimal number of constraints. Figure 7.11(b) is a simplification of Figure 7.11(a). Here the line AB is parallel to the global y axis. We simply delete the x and y translations at point A, and the x translation at point B. If the roller support at B is modified as in 7.11(c), however, it becomes ineffective in constraining the infinitesimal rotational motion about point A because the rolling direction is normal to AB. The configuration of 7.11(c) is called a kinematic mechanism, and will result in a singular modified stiffness matrix.
6
For example, the engine of a car is attached to the vehicle frame through mounts. The car frame becomes the “ground structure,” which moves with respect to Earth ground, as Earth rotates and moves through space, etc.
7–11
7–12
Chapter 7: FEM MODELING: MESH, LOADS AND BCS
§7.5.2. Supporting Three Dimensional Bodies
; ; ;;;;; ;;;;; ; ;; y
B
;;
Figure 7.12 illustrates the extension of the freedomrestraining concept to three dimensions. The minimal number of freedoms that have to be constrained is now six and many combinations are possible. In the example of Figure 7.12, all three degrees of freedom at point A have been fixed. This prevents all rigid body translations, and leaves three rotations to be taken care of. The x displacement component at point B is deleted to prevent rotation about z, the z component is deleted at point C to prevent rotation about y, and the y component is deleted at point D to prevent rotation about x.
D
A
C
x
z
Figure 7.12. Suppressing rigid body motions in a three-dimensional body.
§7.6. Symmetry and Antisymmetry Conditions Engineers doing finite element analysis should be on the lookout for conditions of symmetry or antisymmetry. Judicious use of these conditions allows only a portion of the structure to be analyzed, with a consequent saving in data preparation and computer processing time.7 §7.6.1. Visualization Recognition of symmetry and antisymmetry conditions can be done by either visualization of the displacement field, or by imagining certain rotational ot reflection motions. Both techniques are illustrated for the two-dimensional case. A symmetry line in two-dimensional motion can be recognized by remembering the “mirror” displacement pattern shown in Figure 7.13(a). Alternatively, a 180◦ rotation of the body about the symmetry line reproduces exactly the original problem. An antisymmetry line can be recognized by the displacement pattern illustrated in Figure 7.13(b). Alternatively, a 180◦ rotation of the body about the antisymmetry line reproduces exactly the original problem except that all applied loads are reversed. Similar recognition patterns can be drawn in three dimensions to help visualization of planes of symmetry or antisymmetry. More complex regular patterns associated with sectorial symmetry (also called harmonic symmetry) as well as rotational symmetry can be treated in a similar manner, but will not be discussed here. 7
Symmetry line (a)
A' loads
A
Antisymmetry line
(b)
A'
A'
A
A'
displacement vectors
Figure 7.13. Visualizing symmetry and antisymmetry lines.
Even if symmetry or antisymmetry are not explicitly applied through boundary conditions, they provide valuable checks on the computed solution.
7–12
7–13
§7.6 SYMMETRY AND ANTISYMMETRY CONDITIONS
y (a) A
B
C
D
x
;; ;
A C
B D
;; ;; ; ;
(b)
Figure 7.14. A doubly symmetric structure under symmetric loading.
§7.6.2. Effect of Loading Patterns Although the structure may look symmetric in shape, it must be kept in mind that model reduction can be used only if the loading conditions are also symmetric or antisymmetric. Consider the plate structure shown in Figure 7.14(a). This structure is symmetrically loaded on the x-y plane. Applying the recognition patterns stated above one concludes that the structure is doubly symmetric in both geometry and loading. It is evident that no displacements in the x-direction are possible for any point on the y-axis, and that no y displacements are possible for points on the x axis. A finite element model of this structure may look like that shown in Figure 7.14(b). On the other hand if the loading is antisymmetric, as illustrated in Figure 7.15(a), the x axis becomes an antisymmetry line as none of the y = 0 points can move along the x direction. The boundary conditions to be imposed on the FE model are also different, as shown in Figure 7.15(b).
y (a)
(b) A
C
B
D
x
;; ;;;; A
B
D
C
Vertical (y) motion of C (or another node) should be constrained to eliminate the y rigid motion
Figure 7.15. A doubly symmetric structure under antisymmetric loading.
Remark 7.2. For the case shown in Figure 7.15(b) note that all rollers slide in the same direction. Thus the vertical rigid body motion along y is not precluded. To do that, one node has to be constrained in the y direction. If there are no actual physical supports, the choice is arbitrary and amounts only to an adjustment on the overall (rigid-body) vertical motion. In Figure 7.15(b) the center point C has been so chosen. But any other node could be selected as well; for example A or D. The important thing is not to overconstrain the structure by applying more than one y constraint.
7–13
7–14
Chapter 7: FEM MODELING: MESH, LOADS AND BCS
(a)
P
A
B
C
D
P
;; ;; ;;
(c)
P/2
P/2
P/2
;;; ; ; ; ;
(b)
P/2
P/2
Figure 7.16. Breaking up a point load acting on a symmetry line node.
B
C
D
2P
P
;; ;
;; ;; ;;
;; ;; ; ; ;;;;
A
; ;;;; ;; ;;
P
P/2
P/2
Figure 7.17. Breaking up a point load acting on a antisymmetry line node.
Remark 7.3. Point loads acting at nodes located on symmetry or antisymmetry lines require special care. For example, consider the doubly symmetric plate structure of Figure 7.14 under the two point loads of magnitude P, as pictured in Figure 7.16(a). If the structure is broken down into 4 quadrants as in Figure 7.16(b), P must be halved as indicated in Figure 7.16(c). The same idea applies to point loads on antisymmetry lines, but there the process is trickier, as illustrated in Figure 7.17. The load must not be applied if the node is fixed against motion, since then the node force will appear as a reaction.
Distributed loads should not be divided when the structure is broken down into pieces, since the lumping process will take care of the necessary apportionment. Notes and Bibliography FEM modeling rules in most textbooks are often diffuse, if given at all. In those that focus on the mathematical interpretation of FEM they are altogether lacking; the emphasis being on academic boundary value problems. The rule collection at the start of this Chapter attempts to place important recommendations in one place. The treatment of boundary conditions, particularly symmetry and antisymmetry, tends to be also flaky. A notable exception is [99], which is understandable since Irons worked in industry (at Rolls-Royce Aerospace Division) before moving to academia. The Norfork Dam used in Examples 7.1 and 7.2 for load calculations was the first realistic Civil Engineering structure analyzed by FEM. It greatly contributed to the acceptance of the method beyond the aerospace industry where it originated. How this seminal event came to pass is narrated by Wilson in [30], from which the following fragment is taken. Annotations are inserted in squared brackets. “On the recommendation from Dr. Roy Carlson, a consultant to the Little Rock District of the Corps of Engineers, Clough [then a Professor at UC Berkeley] submitted [in 1960] a proposal to perform a finite element analysis of Norfork Dam, a gravity dam that had a temperature induced vertical crack near the center of the section. The proposal contained a coarse mesh solution that was produced by the new program [a matrix code developed by E. L. Wilson, then a doctoral student under Clough’s supervision; the mesh is that shown in Figure 7.6(d)] and clearly indicated the ability of the new method to model structures of arbitrary geometry with different orthotropic properties within the dam
7–14
7–15
§7.
References
and foundation. The finite element proposal was accepted by the Corps over an analog computer proposal submitted by Professor Richard MacNeal of CalTech [who later directed the development of NASTRAN under a NASA contract in the late 1960s], which at that time was considered as the state-of-the-art method for solving such problems. The Norfork Dam project provided an opportunity to improve the numerical methods used within the program and to extend the finite element method to the nonlinear solution of the crack closing due to hydrostatic loading. Wilson and a new graduate student, Ian King, conducted the detailed analyses that were required by the study. The significant engineering results of the project indicated that the cracked dam was safe [since the crack would be closed as the reservoir was filled].” References Referenced items moved to Appendix R.
7–15
7–16
Chapter 7: FEM MODELING: MESH, LOADS AND BCS
Homework Exercises for Chapters 6 and 7 FEM Modeling: Mesh, Loads and BCs
P C A
;;
O
E D
B
G
F
N
;;
M
J
L
K
I
H
Figure E7.1. Inplane bent plate for Exercise 7.1.
EXERCISE 7.1 [D:10] The plate structure shown in Figure E7.1 is loaded and deforms in the plane of the paper. The applied load at D and the supports at I and N extend over a fairly narrow area. List what you think are the likely “trouble spots” that would require a locally finer finite element mesh to capture high stress gradients. Identify those spots by its letter and a reason. For example, D: vicinity of point load.
A
B
D
C
Figure E7.2. Transition meshing for Exercise 7.2.
EXERCISE 7.2 [D:15] Part of a two-dimensional FE mesh has been set up as indicated in Figure E7.2.
Region ABC D is still unmeshed. Draw a transition mesh within that region that correctly merges with the regular grids shown, uses 4-node quadrilateral elements (quadrilaterals with corner nodes only), and avoids triangles. Note: There are several (equally acceptable) solutions. EXERCISE 7.3 [A:15] A rectangular plate of constant thickness h and inplane dimensions 8a and 6a is meshed with 8 rectangular elements Figure E7.3(a). The plate specific weight is γ and acts along the −y axis direction.
(a)
Compute the node forces due to plate weight at nodes 1 through 15, using the NbN method. Obtain the node-tributary regions as sketched in Figure E7.3(b), which shows each element divided by the medians drawn as dashed lines (the tributary region of node 7 is shown in yellow). Partial answer: f y1 = −2a 2 γ h. Check that adding up all y forces at the 15 nodes one gets W = −48a 2 γ h.
(b)
Repeat the computations using the EbE method. For this, take the total weight force on each element, and assign one quarter to each corner node. (This agrees with consistent energy lumping for 4-node rectangular elements.) Do the results agree with NbN lumping?
7–16
7–17
Exercises
Plate thickness h, specific weight γ
(a)
2
1
4
3
(1)
(2)
(b) 5
(3)
1
2
4
3
5
(4)
y
6a x
(5) 11
12
9
8
7
6
(6)
(7) 13
10
6
8
7
9
10
(8) 14
15
11
12
13
14
15
8a Figure E7.3. (a) Mesh layout for Exercise 7.3. (b) shows tributary area for node 7.
EXERCISE 7.4 [A:20] Complete the computation of hydrostatic node forces on the Norfork Dam under a water head of 200 ft, initiated in Example 7.1. Check that the sum of x forces on nodes 1–5 gives 12 2002 × 62.4 × 1 lbs, and that the sum of y forces on nodes 5–9 gives −2002 × 62.4 × 1 lbs. EXERCISE 7.5 [A/N: 15] Complete the computation of weight nodal forces of the coarse mesh of the Norform dam proper, initiated in Example 7.2, reusing the data of the Figure.
Figure E7.4 depicts two instance of a pull test. In (a) a stiffer material (steel rod) is pulled out of a softer one (concrete block); in this case the load transfer diagram shows a rapid variation near the inner end of the rod. In (b) a softer material (plastic rod) is pulled out of a stiffer one (concrete rod), and the load transfer diagram is reversed.
(a)
EXERCISE 7.6 [A:10]
If the test is to be simulated by a finite element model, indicate for (a) and (b) where a finer mesh would be desirable. Explain.
;;;; ;;;; ;;;; ;;;; Load transfer
Steel
A
B
P
Concrete
(b)
Load transfer A
Plastic
B
P
Concrete
Figure E7.4. Pull tests for Exercise 7.5.
EXERCISE 7.7 [D:15] Identify the symmetry and antisymmetry lines in the two-dimensional problems illustrated in Figure E7.5. They are: (a) a circular disk under two diametrically opposite point forces (the famous “Brazilian test” for concrete); (b) the same disk under two diametrically opposite force pairs; (c) a clamped semiannulus under a force pair oriented as shown; (d) a stretched rectangular plate with a central circular hole. Finally (e) and (f) are half-planes under concentrated loads.8
Having identified those symmetry/antisymmetry lines, state whether it is possible to cut the complete structure to one half or one quarter before laying out a finite element mesh. Then draw a coarse FE mesh indicating, with rollers or fixed supports, which kind of displacement BCs you would specify on the symmetry or antisymmetry lines. Note: Do all sketches on your paper, not on the printed figures.
8
Note that (e) is the famous Flamant’s problem, which is important in the 2D design of foundations of civil structures. The analytical solution of (e) and (f) may be found, for instance, in Timoshenko-Goodier’s Theory of Elasticity, 2nd Edition, page 85ff.
7–17
7–18
Chapter 7: FEM MODELING: MESH, LOADS AND BCS
P
P
(a)
(b) P
P
P
P
;; ;; P
(c)
(d) 45
fixed
(e)
(uniform)
o
q
q
center hole
45o
P
P
(f)
P
P
infinite half plane Figure E7.5. Problems for Exercise 7.7.
EXERCISE 7.8 [D:20] You (a finite element guru) pass away and come back to the next life as an intelligent but hungry bird. Looking around, you notice a succulent big worm taking a peek at the weather. You grab one end and pull for dinner; see Figure E7.6.
After a long struggle, however, the worm wins. While hungrily looking for a smaller one your thoughts wonder to FEM and how the worm extraction process might be modeled so you can pull it out more efficiently. Then you wake up to face this homework question. Try your hand at the following “worm modeling” points. (a)
The worm is simply modeled as a string of one-dimensional (bar) elements. The “worm axial force” is of course constant from the beak B to ground level G, then decreases rapidly because of soil friction (which varies roughly as plotted in the figure above) and drops to nearly zero over D E. Sketch how a good “worm-element mesh” should look like to capture the axial force well.
(b)
On the above model, how would you represent boundary conditions, applied forces and friction forces?
(c)
Next you want a more refined anaysis of the worm that distinguishes skin and insides. What type of finite element model would be appropriate?
(d)
(Advanced) Finally, point out what need to be added to the model of (c) to include the soil as an elastic medium.
Briefly explain your decisions. Dont write equations. EXERCISE 7.9 [A/D:20] Explain from kinematics why two antisymmetry lines in 2D cannot cross at a finite
7–18
7–19
Exercises B
; ; ; ; ; ; ; ; ;;;;;;;; G
D
friction force on worm surface
E
Figure E7.6. The hungry bird.
point. As a corollary, investigate whether it is possible to have more than one antisymmetry line in a 2D elasticity problem. EXERCISE 7.10 [A/D:15] Explain from kinematics why a symmetry line and an antisymmetry line must cross at right angles. EXERCISE 7.11 [A/D:15] A 2D body has n > 1 symmetry lines passing through a point C and spanning an angle π/n from each other. This is called sectorial symmetry if n ≥ 3. Draw a picture for n = 5, say for a car wheel. Explain why C is fixed. EXERCISE 7.12 [A/D:25, 5 each] A body is in 3D space. The analogs of symmetry and antisymmetry lines are symmetry and antisymmetry planes, respectively. The former are also called mirror planes.
(a)
State the kinematic properties of symmetry and antisymmetric planes, and how they can be identified.
(b)
Two symmetry planes intersect. State the kinematic properties of the intersection line.
(c)
A symmetry plane and an antisymmetry plane planes intersect. State the kinematic properties of the intersection line. Can the angle between the planes be arbitrary?
(d)
Can two antisymmetry planes intersect?
(e)
Three symmetry planes intersect. State the kinematic properties of the intersection point.
EXERCISE 7.13 [A:25] A 2D problem is called periodic in the x direction if all fields, in particular displace-
ments, repeat upon moving over a distance a > 0: u x (x + a, y) = u x (x, y) and u y (x + a, y) = u y (x, y). Can this situation be treated by symmetry and/or antisymmetry lines? EXERCISE 7.14 [A:25] Extend the previous exercise to antiperiodicity, in which u x (x + a, y) = u x (x, y)
and u y (x + a, y) = −u y (x, y).
EXERCISE 7.15 [A:20] Prove that EbE and energy consistent lumping agree if the element shape functions are piecewise linear. EXERCISE 7.16 [A:40] If the world were spatially n-dimensional (meaning it has elliptic metric), how many
independent rigid body modes would a body have? (Prove by induction)
7–19
8
.
MultiFreedom Constraints I
8–1
8–2
Chapter 8: MULTIFREEDOM CONSTRAINTS I
TABLE OF CONTENTS Page
§8.1.
Classification of Constraint Conditions §8.1.1. MultiFreedom Constraints . . . . . §8.1.2. *MFC Matrix Forms . . . . . . . §8.2. Methods for Imposing Multifreedom Constraints §8.3. The Example Structure §8.4. The Master-Slave Method §8.4.1. A One-Constraint Example . . . . . §8.4.2. Several Homogeneous MFCs . . . . . §8.4.3. Nonhomogeneous MFCs . . . . . §8.4.4. *The General Case . . . . . . . . §8.4.5. *Retaining the Original Freedoms . . §8.4.6. Model Reduction by Kinematic Constraints §8.4.7. Assessment of the Master-Slave Method §8. Notes and Bibliography . . . . . . . . . . . . . . . §8. References . . . . . . . . . . . . . . §8. Exercises . . . . . . . . . . . . . . .
8–2
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . .
. . . . . . . . . . . . . . .
. . . . .
. . . . . . . . . . . . . . .
. . . . .
. . . . . . . . . . . . . . .
8–3 8–3 8–4 8–4 8–5 8–6 8–7 8–8 8–9 8–9 8–10 8–10 8–11 8–13 8–14 8–15
8–3
§8.1
CLASSIFICATION OF CONSTRAINT CONDITIONS
§8.1. Classification of Constraint Conditions In previous Chapters we have considered structural support conditions that are mathematically expressable as constraints on individual degrees of freedom: nodal displacement component = prescribed value.
(8.1)
These are called single-freedom constraints. Chapters 3 explains how to incorporate constraints of this form into the master stiffness equations, using hand- or computer-oriented techniques. The displacement boundary conditions studied in Chapter 7, including the modeling of symmetry and antisymmetry, lead to constraints of this form. For example: u x4 = 0,
u y9 = 0.6.
(8.2)
These are two single-freedom constraints. The first one is homogeneous while the second one is non-homogeneous. These qualifiers are defined below. §8.1.1. MultiFreedom Constraints The next step up in complexity involves multifreedom equality constraints, or multifreedom constraints for short, the last name being acronymed to MFC. These are functional equations that connect two or more displacement components: F(nodal displacement components) = prescribed value,
(8.3)
where function F vanishes if all its nodal displacement arguments do. Equation (8.3), where all displacement components are in the left-hand side, is called the canonical form of the constraint. An MFC of this form is called multipoint or multinode if it involves displacement components at different nodes. The constraint is called linear if all displacement components appear linearly on the left-hand-side, and nonlinear otherwise. The constraint is called homogeneous if, upon transfering all terms that depend on displacement components to the left-hand side, the right-hand side — the “prescribed value” in (8.3) — is zero. It is called non-homogeneous otherwise. In this and next Chapter only linear constraints will be studied. Furthermore more attention is devoted to the homogeneous case, because it arises more frequently in practice. Remark 8.1. The most general constraint class is that of inequality constraints, such as u y5 − 2u x2 ≥ 0.5.
These constraints are relatively infrequent in linear structural analysis, except in problems that involve contact conditions. They are of paramount importance, however, in other fields such as optimization and control.
8–3
8–4
Chapter 8: MULTIFREEDOM CONSTRAINTS I
Example 8.1. Here are three examples of MFCs:
u x2 = 12 u y2 ,
u x2 − 2u x4 + u x6 = 0.25,
(x5 + u x5 − x3 − u x3 )2 + (y5 + u y5 − y3 − u y3 )2 = 0.
(8.4)
The first one is linear and homogeneous. It is not a multipoint constraint because it involves the displacement components of one node: 2. The second one is multipoint because it involves three nodes: 2, 4 and 6. It is linear and non-homogeneous. The last one is multipoint, nonlinear and homogeneous. Geometrically it expresses that the distance between nodes 3 and 5 in two-dimensional motions on the {x, y} plane remains constant. This kind of constraint appears in geometrically nonlinear analysis of structures, which is a topic beyond the scope of this course. §8.1.2. *MFC Matrix Forms Matrix forms of linear MFCs are often convenient for compact notation. An individual constraint such as the second one in (8.4) can be written
0002
[1
−2
1]
u x2 u x4 u x6
0003
= 0.25.
(8.5)
(no sum on i)
(8.6)
In direct matrix notation: a¯ i u¯ i = bi ,
in which index i (i = 1, 2, . . .) identifies the constraint, a¯ i is a row vector, u¯ i collects the set of degrees of freedom that participate in the constraint, and gi is the right hand side scalar (0.25 above). The bar over a and u distinguishes (8.6) from the expanded form (8.8) discussed below. For method description and general proofs it is often convenient to expand matrix forms so that they embody all degrees of freedom. For example, if (8.5) is part of a two-dimensional finite element model with 12 freedoms: u x1 , u y1 , . . . u y6 , the left-hand side row vector may be expanded with nine zeros as follows
u  x1
 u y1    [ 0 0 1 0 0 0 −2 0 0 0 1 0 ]  u x2  = 0.25,  . 
(8.7)
. u y6
in which case the matrix notation ai u = gi
(8.8)
is used. Finally, all multifreedom constraints expressed as (8.8) may be collected into a single matrix relation: Au = g,
(8.9)
where rectangular matrix A is formed by stacking the ai ’s as rows and column vector g is formed by stacking the gi s as entries. If there are 12 degrees of freedom in u and 5 multifreedom constraints then A will be 5 × 12.
8–4
8–5
§8.3
THE EXAMPLE STRUCTURE
§8.2. Methods for Imposing Multifreedom Constraints Accounting for multifreedom constraints is done — at least conceptually — by changing the assembled master stiffness equations to produce a modified system of equations. The modification process is also called constraint application or constraint imposition. The modified system is the one submitted to the equation solver. Three methods for treating MFCs are discussed in this and the next Chapter: 1.
Master-Slave Elimination. The degrees of freedom involved in each MFC are separated into master and slave freedoms. The slave freedoms are then explicitly eliminated. The modified equations do not contain the slave freedoms.
2.
Penalty Augmentation. Also called the penalty function method. Each MFC is viewed as the presence of a fictitious elastic structural element called penalty element that enforces it approximately. This element is parametrized by a numerical weight. The exact constraint is recovered if the weight goes to infinity. The MFCs are imposed by augmenting the finite element model with the penalty elements.
3.
Lagrange Multiplier Adjunction. For each MFC an additional unknown is adjoined to the master stiffness equations. Physically this set of unknowns represent constraint forces that would enforce the constraints exactly should they be applied to the unconstrained system.
For each method the exposition tries to give first the basic flavor by working out the same example for each method. The general technique is subsequently presented in matrix form for completeness but is considered an advanced topic. Conceptually, imposing MFCs is not different from the procedure discussed in Chapter 3 for singlefreedom constraints. The master stiffness equations are assembled ignoring all constraints. Then the MFCs are imposed by appropriate modification of those equations. There are, however, two important practical differences: 1.
The modification process is not unique because there are alternative constraint imposition methods, namely those listed above. These methods offer tradeoffs in generality, programming implementation complexity, computational effort, numerical accuracy and stability.
2.
In the implementation of some of these methods — particularly penalty augmentation — constraint imposition and assembly are carried out simultaneously. In that case the framework “first assemble, then modify,” is not strictly respected in the actual implementation.
Remark 8.2. The three methods are also applicable, as can be expected, to the simpler case of a single-freedom
constraint such as (8.2). For most situations, however, the generality afforded by the penalty function and Lagrange multiplier methods are not warranted. The hand-oriented reduction process discussed in Chapters 3 and 4 is in fact a special case of the master-slave elimination method in which “there is no master.” Remark 8.3. Often both multifreedom and single-freedom constraints are prescribed. The modification
process then involves two stages: apply multifreedom constraints and apply single freedom constraints. The order in which these are carried out is implementation dependent. Most implementations do the MFCs first, either after the assembly is completed or during the assembly process. The reason is practical: single-freedom constraints are often automatically taken care of by the equation solver itself.
8–5
8–6
Chapter 8: MULTIFREEDOM CONSTRAINTS I
u1 , f1
u2 , f2
u3 , f3
u4 , f4
u5 , f5
u6 , f6
(1)
(2)
(3)
(4)
(5)
(6)
2
1
3
4
5
6
u7 , f7 x 7
Figure 8.1. A one-dimensional problem discretized with six bar finite elements. The seven nodes may move only along the x direction. Subscript x is omitted from the u’s and f ’s to reduce clutter.
§8.3. The Example Structure The one-dimensional finite element discretization shown in Figure 8.1 will be used throughout Chapters 8 and 9 to illustrate the three MFC application methods. This structure consists of six bar elements connected by seven nodes that can only displace in the x direction. Before imposing various multifreedom constraints discussed below, the master stiffness equations for this problem are assumed to be 
K 11  K 12   0   0   0  0 0
K 12 K 22 K 23 0 0 0 0
0 K 23 K 33 K 34 0 0 0
0 0 K 34 K 44 K 45 0 0
0 0 0 K 45 K 55 K 56 0
0 0 0 0 K 56 K 66 K 67
   0 u1 0   u2      0   u3      0   u4  =     0   u5      K 67 u6 K 77 u7
 f1 f2   f3   f4  ,  f5   f6 f7
(8.10)
or Ku = f.
(8.11)
The nonzero stiffness coefficients K i j in (8.10) depend on the bar rigidity properties. For example, if E e Ae /L e = 100 for each element e = 1, . . . , 6, then K 11 = K 77 = 100, K 22 = . . . = K 66 = 200, K 12 = K 23 = . . . = K 67 = −100. However, for the purposes of the following treatment the coefficients may be kept arbitrary. The component index x in the nodal displacements u and nodal forces f has been omitted for brevity. Now let us specify a multifreedom constraint that states that nodes 2 and 6 are to move by the same amount: (8.12) u2 = u6. Passing all node displacements to the right hand side gives the canonical form: u 2 − u 6 = 0.
(8.13)
Constraint conditions of this type are sometimes called rigid links because they can be mechanically interpreted as forcing node points 2 and 6 to move together as if they were tied by a rigid member.1 We now study the imposition of constraint (8.13) on the master equations (8.10) by the methods mentioned above. In this Chapter the master-slave method is treated. The other two methods: penalty augmentation and Lagrange multiplier adjunction, are discussed in the following Chapter. 1
This physical interpretation is exploited in the penalty method described in the next Chapter. In two and three dimensions rigid link constraints are more complicated.
8–6
8–7
§8.4 THE MASTER-SLAVE METHOD
§8.4. The Master-Slave Method To apply this method by hand, the MFCs are taken one at a time. For each constraint a slave degree of freedom is chosen. The freedoms remaining in that constraint are labeled master. A new set of degrees of freedom uˆ is established by removing all slave freedoms from u. This new vector contains master freedoms as well as those that do not appear in the MFCs. A matrix transformation equation that relates u to uˆ is generated. This equation is used to apply a congruential transformation to the master stiffness equations. This procedure yields a set of modified stiffness equations that ˆ Because the modified system does not contain the are expressed in terms of the new freedom set u. slave freedoms, these have been effectively eliminated. §8.4.1. A One-Constraint Example The mechanics of the process is best seen by going through an example. To impose (8.13) pick u 6 as slave and u 2 as master. Relate the original unknowns u 1 , . . . u 7 to the new set in which u 6 is missing:     u1 1 0 0 0 0 0   u1  u2   0 1 0 0 0 0       u2   u3   0 0 1 0 0 0       u  (8.14)  u4  =  0 0 0 1 0 0   3  ,      u4   u5   0 0 0 0 1 0        u5 u6 0 1 0 0 0 0 u7 u7 0 0 0 0 0 1 This is the required transformation relation. In compact form: u = Tuˆ
.
(8.15)
Replacing (8.14) into (8.11) and premultiplying by TT yields the modified system ˆ uˆ = ˆf, K
ˆf = TT f.
(8.16)
    0 u1 f1 K 67   u 2   f 2 + f 6      0   u 3   f3    =  , 0   u 4   f4      0 u5 f5 K 77 u7 f7
(8.17)
ˆ = TT K T, in which K
Carrying out the indicated matrix multiplications yields 
K 11  K 12   0   0  0 0
K 12 K 22 + K 66 K 23 0 K 56 K 67
0 K 23 K 33 K 34 0 0
0 0 K 34 K 44 K 45 0
0 K 56 0 K 45 K 55 0
Equation (8.17) is a new linear system containing 6 equations in the remaining 6 unknowns: u 1 , u 2 , u 3 , u 4 , u 5 and u 7 . Upon solving it, u 6 is recovered from the constraint (8.12). Remark 8.4. The form of modified system (8.16) can be remembered by a simple mnemonic rule: premultiply
both sides of T uˆ = u by TT K, and replace K u by f on the right hand side.
8–7
8–8
Chapter 8: MULTIFREEDOM CONSTRAINTS I
Remark 8.5. For a simple freedom constraint such as u 4 = 0 the only possible choice of slave is of course
u 4 and there is no master. The congruential transformation is then nothing more than the elimination of u 4 by striking out rows and columns from the master stiffness equations. Remark 8.6. For a simple MFC such as u 2 = u 6 , it does not matter which degree of freedom is chosen as master or unknown. Choosing u 2 as slave produces a system of equations in which now u 2 is missing:
K
11
 0  0   0 
K 12 0
0 K 33 K 34 0 K 23 0
0 K 34 K 44 K 45 0 0
0 0 K 45 K 55 K 56 0

K 12 K 23 0 K 56 K 22 + K 66 K 67



0 u1 f1 0   u 3   f3      0    u 4  =  f4  .      0   u 5   f5  K 67 u6 f2 + f6 K 77 u7 f7
(8.18)
Although (8.17) and (8.18) are algebraically equivalent, the latter would be processed faster if a skyline solver (Part III of course) is used for the modified equations.
§8.4.2. Several Homogeneous MFCs The matrix equation (8.16) in fact holds for the general case of multiple homogeneous linear constraints. Direct establishment of the transformation equation, however, is more complicated if slave freedoms in one constraint appear as masters in another. To illustrate this point, suppose that for the example system we have three homogeneous multifreedom constraints: u 2 − u 6 = 0,
u 1 + 4u 4 = 0,
2u 3 + u 4 + u 5 = 0,
(8.19)
Picking as slave freedoms u 6 , u 4 and u 3 from the first, second and third constraint, respectively, we can solve for them as u6 = u2,
u 4 = − 14 u 1 ,
u 3 = − 12 (u 4 + u 5 ) = 18 u 1 − 12 u 5 .
(8.20)
Observe that solving for u 3 from the third constraint brings u 4 to the right-hand side. But because u 4 is also a slave freedom (it was chosen as such for the second constraint) it is replaced in favor of u 1 using u 4 = − 14 u 1 . The matrix form of the transformation (8.20) is   1 u1 0  u2   1     u3       81  u4  =  − 4     u5   0    u6 0 u7 0 
0 0 1 0 0 − 12 0 0 1 0
0 1 0 0
0 0   0  u1   u2  0   u5  ,  0  u7  0 1
(8.21)
The modified system is now formed through the congruential transformation (8.16). Note that the slave freedoms selected from each constraint must be distinct; for example the choice u 6 , u 4 , u 4 would be inadmissible as long as the constraints are independent. This rule is easy to enforce when slave freedoms are chosen by hand, but can lead to implementation and numerical difficulties when it is programmed as an automated procedure, as further discussed later. 8–8
8–9
§8.4 THE MASTER-SLAVE METHOD
Remark 8.7. The three MFCs (8.19) with u 6 , u 4 and u 2 chosen as slaves and u 1 , u 2 and u 5 chosen as masters,
may be presented in the partitioned matrix form:
0002
0 0 2
0 1 4 0 1 0
00030002
u3 u4 u6
0003
0002 =
0 −1 0
1 0 0
0 0 −1
00030002
u1 u2 u5
0003 (8.22)
This may be compactly written As us + Am um = 0. Solving for the slave freedoms gives us = −A−1 s A m um . Expanding with zeros to fill out u and uˆ produces (8.21). This general matrix form is considered in §8.4.4. Note that non-singularity of As is essential for this method to work.
§8.4.3. Nonhomogeneous MFCs Extension to non-homogeneous constraints is immediate. In this case he transformation equation becomes non-homogeneous. For example suppose that (8.13) has a nonzero prescribed value: u 2 − u 6 = 0.2
(8.23)
Nonzero RHS values such as 0.2 in (8.23) may be often interpreted physically as “gaps” (thus the use of the symbol g in the matrix form). Chose u 6 again as slave: u 6 = u 2 − 0.2, and build the transformation       0 1 0 0 0 0 0   u1 u1  0   u2   0 1 0 0 0 0     u2       u3   0 0 1 0 0 0     0   u      (8.24)  u4  =  0 0 0 1 0 0   3  +  0  . u         u5   0 0 0 0 1 0   4   0     u5    u6 −0.2 0 1 0 0 0 0 u7 u7 0 0 0 0 0 0 1 In compact matrix notation, u = T uˆ + g.
(8.25)
Here the constraint gap vector g is nonzero and T is the same as before. To get the modified system applying the shortcut rule of Remark 8.4, premultiply both sides of (8.25) by TT K, replace Ku by f, and pass the data to the RHS: ˆ uˆ = ˆf, K
ˆ = TT K T, in which K
ˆf = TT (f − K g).
(8.26)
ˆ the complete displacement vector is recovered from (8.25). For the MFC Upon solving (8.26) for u, (8.23) this technique gives the system      K 11 K 12 0 0 0 0 u1 f1 0 K 56 K 67   u 2   f 2 + f 6 − 0.2K 66   K 12 K 22 + K 66 K 23      K 23 K 33 K 34 0 0   u3   f3  0  (8.27)    =  . 0   u4   f4 0 K 34 K 44 K 45  0       0 K 45 K 55 0 u5 f 5 − 0.2K 56 0 K 56 0 K 67 0 0 0 K 77 u7 f 7 − 0.2K 67 See Exercise 8.2 for multiple non-homogeneous MFCs. 8–9
8–10
Chapter 8: MULTIFREEDOM CONSTRAINTS I
§8.4.4. *The General Case For implementation in general-purpose programs the master-slave method can be described as follows. The degrees of freedoms in u are classified into three types: independent or uncommitted, masters and slaves. (The uncommitted freedoms are those that do not appear in any MFC.) Label these sets as uu , um and us , respectively, and partition the stiffness equations accordingly:
0002
Kuu T Kum T Kus
Kum Kmm T Kms
Kus Kms Kss
00030002
uu um us
0003
0002
=
fu fm fs
0003
(8.28)
The MFCs may be written in matrix form as Am um + As us = g,
(8.29)
where As is assumed square and nonsingular. If so we can solve for the slave freedoms: −1 us = −A−1 s Am um + As g = Tum + g,
Inserting into the partitioned stiffness matrix and symmetrizing
Kuu T TT Kum
Kum T TT Kmm T
uu um
=
(8.30)
fu − Kus g fm − Kms g
(8.31)
It is seen that the misleading simplicity of the handworked examples is gone. §8.4.5. *Retaining the Original Freedoms A potential disadvantage of the master-slave method in computer work is that it requires a rearrangement of the original stiffness equations because uˆ is a subset of u. The disadvantage can be annoying when sparse matrix storage schemes are used for the stiffness matrix, and becomes intolerable if secondary storage is used for that purpose. With a bit of trickery it is possible to maintain the original freedom ordering. Let us display it for the example problem under (8.13). Instead of (8.14), use the square transformation



u1 1  u2   0  u3   0     u4  =  0     u5   0    u6 0 u7 0
0 1 0 0 0 1 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0


0 u1 0   u2    0   u3    0   u4  ,  0  u5    u˜ 6 0 u7 1
where u˜ 6 is a placeholder for the slave freedom u 6 . The modified equations are

K 11 K  12  0   0   0  0 0
K 12 K 22 + K 66 K 23 0 K 56 0 K 67
0 K 23 K 33 K 34 0 0 0
0 0 K 34 K 44 K 45 0 0
0 0 0 K 45 K 55 0 0
0 K 56 0 0 0 0 0
0 0 0 0 0 0 0


(8.32)


0 u1 f1 K 67   u 2   f 2 + f 6      0    u 3   f3      0   u 4  =  f4  , 0   u 5   f5      0 u˜ 6 0 K 77 u7 f7
(8.33)
which are submitted to the equation solver. If the solver is not trained to skip zero row and columns, a one should be placed in the diagonal entry for the u˜ 6 (sixth) equation. The solver will return u˜ 6 = 0, and this placeholder value is replaced by u 2 . Note the points in common with the computer-oriented placeholder technique described in §3.4 to handle single-freedom constraints.
8–10
8–11
§8.4
THE MASTER-SLAVE METHOD
Master
Master
u1 , f1
u7 , f7 x
1
7
Figure 8.2. Model reduction of the example structure of Figure 8.1 to the end freedoms.
§8.4.6. Model Reduction by Kinematic Constraints The congruential transformation equations (8.16) and (8.26) have additional applications beyond the master-slave method. An important one is model reduction by kinematic constraints. Through this procedure the number of DOF of a static or dynamic FEM model is reduced by a significant number, typically to 1% – 10% of the original number. This is done by taking a lot of slaves and a few masters. Only the masters are left after the transformation. The reduced model is commonly used in subsequent calculations as component of a larger system, particularly during design or in parameter identification. Example 8.2. Consider the bar assembly of Figure 8.1. Assume that the only masters are the end motions u 1
and u 7 , as illustrated in Figure 8.2, and interpolate all freedoms linearly:



u1 1  u 2   5/6  u 3   4/6     u 4  =  3/6     u 5   2/6    u6 1/6 u7 0

0 1/6  2/6   u1 3/6   u , 4/6  7  5/6 1
or
ˆ u = T u.
(8.34)
ˆ uˆ = TT KTuˆ = TT f = ˆf, or in detail The reduced-model equations are K
ˆ K 11 Kˆ 17
ˆ Kˆ 17 f u1 = ˆ1 , ˆ K 77 f7 u7
(8.35)
where Kˆ 11 = Kˆ 17 = Kˆ 77 = fˆ1 =
1 (36K 11 +60K 12 +25K 22 +40K 23 +16K 33 +24K 34 +9K 44 +12K 45 +4K 55 +4K 56 +K 66 ), 36 1 (6K 12 +5K 22 +14K 23 +8K 33 +18K 34 +9K 44 +18K 45 +8K 55 +14K 56 +5K 66 +6K 67 ), 36 1 (K 22 +4K 23 +4K 33 +12K 34 +9K 44 +24K 45 +16K 55 +40K 56 +25K 66 +60K 67 +36K 77 ), 36 1 (6 f 1 +5 f 2 +4 f 3 +3 f 4 +2 f 5 + f 6 ), fˆ7 = 16 ( f 2 +2 f 3 +3 f 4 +4 f 5 +5 f 6 +6 f 7 ). 6
(8.36)
This reduces the order of the FEM model from 7 to 2. A Mathematica script to do the reduction is shown in Figure 8.3. The key feature is that the masters are picked a priori, as the freedoms to be retained in the model for further use. Remark 8.8. Model reduction can also be done by the static condensation method explained in Chapter 10. As its name indicates, condensation is restricted to static analysis. On the other hand, for such problems it is exact whereas model reduction by kinematic constraints generally introduces approximations.
8–11
8–12
Chapter 8: MULTIFREEDOM CONSTRAINTS I
(* Model Reduction Example *) ClearAll[K11,K12,K22,K23,K33,K34,K44,K45,K55,K56,K66, f1,f2,f3,f4,f5,f6]; K={{K11,K12,0,0,0,0,0},{K12,K22,K23,0,0,0,0}, {0,K23,K33,K34,0,0,0},{0,0,K34,K44,K45,0,0}, {0,0,0,K45,K55,K56,0},{0,0,0,0,K56,K66,K67}, {0,0,0,0,0,K67,K77}}; Print['K=',K//MatrixForm]; f={f1,f2,f3,f4,f5,f6,f7}; Print['f=',f]; T={{6,0},{5,1},{4,2},{3,3},{2,4},{1,5},{0,6}}/6; Print['T (transposed to save space)=',Transpose[T]//MatrixForm]; Khat=Simplify[Transpose[T].K.T]; fhat=Simplify[Transpose[T].f]; Print['Modified Stiffness:']; Print['Khat(1,1)=',Khat[[1,1]],'nKhat(1,2)=',Khat[[1,2]], 'nKhat(2,2)=',Khat[[2,2]] ]; Print['Modified Force:']; Print['fhat(1)=',fhat[[1]],' fhat(2)=',fhat[[2]] ]; K11 K12 0 0 0 0 0 0004 0001 0006 0003 0006 0003 0003 0006 0003 K12 K22 K23 0 0 0 0 0006 0006 0003 0006 0003 0006 0003 0006 0003 0006 0003 0 K23 K33 K34 0 0 0 0006 0003 0006 0003 0006 0003 0003 0006 0003 0 0 K34 K44 K45 0 0 0006 K00010003 0006 0006 0003 0006 0003 0006 0003 0003 0006 0 0 0 K45 K55 K56 0 0006 0003 0006 0003 0006 0003 0006 0003 0006 0003 0006 0003 0 0 0 0 K56 K66 K67 0006 0003 0006 0003 0006 0003 0 0 0 0 K67 K77 0005 0002 0 f00010007f1, f2, f3, f4, f5, f6, f7 1 0001 0003 0003 T transposed to save space 00010003 0003 0003 00020
5 000200020002 0002 6
2 000200020002 0002 3
1 000200020002 0002 2
1 000200020002 0002 3
1 000200020002 0002 6
1 000200020002 0002 6
1 000200020002 0002 3
1 000200020002 0002 2
2 000200020002 0002 3
5 000200020002 0002 6
00004 0006 0006 0006 0006 0006 10005
Modified Stiffness: 1 Khat 1,1 0001 0002000200020002000200020002 36 K11 0003 60 K12 0003 25 K22 0003 40 K23 0003 16 K33 0003 24 K34 0003 9 K44 0003 12 K45 0003 4 K55 0003 4 K56 0003 K66
36 1 Khat 1,2 0001 0002000200020002000200020002 6 K12 0003 5 K22 0003 14 K23 0003 8 K33 0003 18 K34 0003 9 K44 0003 18 K45 0003 8 K55 0003 14 K56 0003 5 K66 0003 6 K67
36 1 Khat 2,2 0001 0002000200020002000200020002 K22 0003 4 K23 0003 4 K33 0003 12 K34 0003 9 K44 0003 24 K45 0003 16 K55 0003 40 K56 0003 25 K66 0003 60 K67 0003 36 K77
36 Modified Force: 1 fhat 1 0001 0002000200020002 6 f1 0003 5 f2 0003 4 f3 0003 3 f4 0003 2 f5 0003 f6
6
1 fhat 2 0001 0002000200020002 f2 0003 2 f3 0003 3 f4 0003 4 f5 0003 5 f6 0003 6 f7
6
Figure 8.3. Mathematica script for the model reduction example of Figure 8.2.
§8.4.7. Assessment of the Master-Slave Method What are the good and bad points of this constraint imposition method? It enjoys the advantage of being exact (except for inevitable solution errors) and of reducing the number of unknowns. The concept is also easy to explain. The main implementation drawback is the complexity of the general case as can be seen by studying (8.28) through (8.31). The complexity is due to three factors: 1.
The equations may have to be rearranged because of the disappearance of the slave freedoms. This drawback can be alleviated, however, through the placeholder trick outlined in §8.4.5.
2.
An auxiliary linear system, namely (8.30), has to be assembled and solved to produce the transformation matrix T and vector g. 8–12
8–13 3.
§8. Notes and Bibliography
The transformation process may generate many additional matrix terms. If a sparse matrix storage scheme is used for K, the logic for allocating memory and storing these entries can be difficult and expensive.
The level of complexity depends on the generality allowed as well as on programming decisions. At one extreme, if K is stored as full matrix and slave freedom coupling in the MFCs is disallowed the logic is simple.2 At the other extreme, if arbitrary couplings are permitted and K is placed in secondary (disk) storage according to some sparse scheme, the complexity can become overwhelming. Another, more subtle, drawback of this method is that it requires decisions as to which degrees of freedom are to be treated as slaves. This can lead to implementation and numerical stability problems. Although for disjointed constraints the process can be programmmed in reliable form, in more general cases of coupled constraint equations it can lead to incorrect decisions. For example, suppose that in the example problem you have the following two MFCs: 1 u 6 2
+ 12 u 4 = u 6 ,
u 3 + 6u 6 = u 7 .
(8.37)
For numerical stability reasons it is usually better to pick as slaves the freedoms with larger coefficients. If this is done, the program would select u 6 as slave freedoms from both constraints. This leads to a contradiction because having two constraints we must eliminate two slave degrees of freedom, not just one. The resulting modified system would in fact be inconsistent. Although this defect can be easily fixed by the program logic in this case, one can imagine the complexity burden if faced with hundreds or thousands of MFCs. Serious numerical problems can arise if the MFCs are not independent. For example: 1 u 6 2
= u6,
1 u 5 3
+ 6u 6 = u 7 ,
u 2 + u 3 − u 7 = 0.
(8.38)
The last constraint is an exact linear combination of the first two. If the program blindly choses u 2 , u 3 and u 7 as slaves, the modified system is incorrect because we eliminate three equations when in fact there are only two independent constraints. Exact linear dependence, as in (8.38), can be recognized by a rank analysis of the As matrix defined in (8.29). In inexact floating-point arithmetic, however, such detection may fail.3 The complexity of slave selection is in fact equivalent to that of automatically selecting kinematic redundancies in the force method. It has led implementors of programs that use this method to require masters and slaves be prescribed in the input data, thus transfering the burden to users. The method is not generally extendible to nonlinear constraints without extensive reworking. In conclusion, the master-slave method is useful when a few simple linear constraints are imposed by hand. As a general purpose technique for finite element analysis it suffers from complexity and lack of robustness. It is worth learning this method, however, because of its great importance of the congruential transformation technique in model reduction for static and dynamic problems. 2
This is the case in model reduction, since each slave freedom appears in one and only one MFC.
3
The safest technique to identify dependencies is to do a singular-value decomposition (SVD) of As . This can be, however, prohibitively expensive if one is dealing with hundreds or thousands of constraints.
8–13
Chapter 8: MULTIFREEDOM CONSTRAINTS I
8–14
Notes and Bibliography Multifreedom constraints are treated in several of the FEM books recommended in §1.7.5, notably Zienkiewicz and Taylor [186]. The master-slave method was incorporated to treat MFCs as part of the DSM developed at Boeing during the 1950s. It is first summarily described in the DSM-overview by Turner, Martin and Weikel [168, p. 212]. The implementation differs, however, from the one described here because the relation of FEM to energy methods was not clear at the time. The master-slave method became popular through its adoption by the general-purpose NASTRAN code developed in the late 1960s [3] and early assessments of its potential [160]. The implementation unfortunately relied on user inputs to identify slave DOFs. Through this serious blunder the method gained a reputation for unreliability that persists to the present day. The important application of master-slave to model reduction, which by itself justifies teaching the method, is rarely mentioned in FEM textbooks. References Referenced items have been moved to Appendix R.
8–14
8–15
Exercises
Homework Exercises for Chapter 8 MultiFreedom Constraints I
EXERCISE 8.1 [C+N:20] The example structure of Figure 8.1 has E e Ae /L e = 100 for each element
e = 1, . . . , 6. Consequently K 11 = K 77 = 100, K 22 = . . . = K 66 = 200, K 12 = K 23 = . . . = K 67 = −100. The applied node forces are taken to be f 1 = 1, f 2 = 2, f 3 = 3, f 4 = 4, f 5 = 5, f 6 = 6 and f 7 = 7, which are easy to remember. The structure is subjected to one support condition: u 1 = 0 (a fixed left end), and to one MFC: u 2 − u 6 = 1/5. Solve this problem using the master-slave method to process the MFC, taking u 6 as slave. Upon forming the modified system (8.26) apply the left-end support u 1 = 0 using the placeholder method of §3.4. Solve the equations and verify that the displacement solution and the recovered node forces including reactions are u = [ 0 0.270 Ku = [ −27
26.5
0.275 3
0.250 4
5
0.185
−18.5
0.070
0.140 ]T
7 ]T
(E8.1)
Use Mathematica or Matlab to do the algebra is recommended. For example, the following Mathematica script solves this Exercise: (* Exercise 9.1 - Master-Slave Method *) (* MFC: u2-u6 = 1/5 - slave: u6 *) MasterStiffnessOfSixElementBar[kbar_]:=Module[ {K=Table[0,{7},{7}]}, K[[1,1]]=K[[7,7]]=kbar; For [i=2,i<=6,i++,K[[i,i]]=2*kbar]; For [i=1,i<=6,i++,K[[i,i+1]]=K[[i+1,i]]=-kbar]; Return[K]]; FixLeftEndOfSixElementBar[Khat_,fhat_]:=Module[ {Kmod=Khat,fmod=fhat}, fmod[[1]]=0; Kmod[[1,1]]=1; Kmod[[1,2]]=Kmod[[2,1]]=0; Return[{Kmod,fmod}]]; K=MasterStiffnessOfSixElementBar[100]; Print['Stiffness K=',K//MatrixForm]; f={1,2,3,4,5,6,7}; Print['Applied forces=',f]; T={{1,0,0,0,0,0},{0,1,0,0,0,0},{0,0,1,0,0,0}, {0,0,0,1,0,0},{0,0,0,0,1,0},{0,1,0,0,0,0}, {0,0,0,0,0,1}}; Print['Transformation matrix T=',T//MatrixForm]; g={0,0,0,0,0,-1/5,0}; Print['Constraint gap vector g=',g]; Khat=Simplify[Transpose[T].K.T]; fhat=Simplify[Transpose[T].(f-K.g)]; {Kmod,fmod}=FixLeftEndOfSixElementBar[Khat,fhat]; (* fix left end *) Print['Modified Stiffness upon fixing node 1:',Kmod//MatrixForm]; Print['Modified RHS upon fixing node 1:',fmod]; umod=LinearSolve[Kmod,fmod]; Print['Computed umod (lacks slave u6)=',umod]; u=T.umod+g; Print['Complete solution u=',u]; Print['Numerical u=',N[u]]; fu=K.u; Print['Recovered forces K.u with reactions=',fu]; Print['Numerical K.u=',N[fu]];
8–15
8–16
Chapter 8: MULTIFREEDOM CONSTRAINTS I
EXERCISE 8.2 [C+N:25] As in the previous Exercise but applying the following three MFCs, two of which
are non-homogeneous: u 2 − u 6 = 1/5,
u 3 + 2u 4 = −2/3,
2u 3 − u 4 + u 5 = 0.
(E8.2)
Hints. Chose u 4 , u 5 and u 6 as slaves. Much of the script shown for Exercise 8.1 can be reused. The main changes are in the formation of T and g. If you are a Mathematica wizard (or willing to be one) those can be automatically formed by saying sol=Simplify[Solve[{u2-u61/5, u3+2*u4-2/3,2*u3-u4+u50},{u4,u5,u6}]]; ums={u1,u2,u3,u4,u5,u6,u7}/.sol[[1]]; um={u1,u2,u3,u7}; T=Table[Coefficient[ums[[i]],um[[j]]],{i,1,7},{j,1,4}]; g=ums/.{u1->0,u2->0,u3->0,u4->0,u5->0,u6->0,u7->0}; Print['Transformation matrix T=',T//MatrixForm]; Print['Gap vector g=',g] If you do this, explain what it does and why it works. Otherwise form and enter T and g by hand. The numerical results (shown to 5 places) should be u = [ 0. 0.043072 −0.075033 Ku = [ −4.3072
16.118
10.268
−0.29582 −37.085
−0.14575 16.124
−0.15693
−8.1176
7. ]T .
−0.086928 ]T ,
(E8.3)
EXERCISE 8.3 [A:25] Can the MFCs be pre-processed to make sure that no slave freedom in a MFC appears
as master in another? EXERCISE 8.4 [A:25] In the general case discussed in §8.4.4, under which condition is the matrix As of
(8.29) diagonal and thus trivially invertible? EXERCISE 8.5 [A:25] Work out the general technique by which the unknowns need not be rearranged, that is, u and uˆ are the same. Use “placeholders” for the slave freedoms. (Hint: use ideas of §3.4). EXERCISE 8.6 [A/C:35] Is it possible to establish a slave selection strategy that makes As diagonal or triangular? (This requires knowledge of matrix techniques such as pivoting.) EXERCISE 8.7 [A/C:40] Work out a strategy that produces a well conditioned As by selecting new slaves as
linear combinations of finite element freedoms if necessary. (Requires background in numerical analysis and advanced programming experience in matrix algebra).
8–16
10
.
Superelements and Global-Local Analysis
10–1
Chapter 10: SUPERELEMENTS AND GLOBAL-LOCAL ANALYSIS
10–2
TABLE OF CONTENTS Page
§10.1. Superelement Concept §10.1.1. Where Does the Idea Come From? . . . . §10.1.2. Subdomains . . . . . . . . . . . §10.1.3. *Mathematical Requirements . . . . . . §10.2. Static Condensation §10.2.1. Condensation by Explicit Matrix Operations §10.2.2. Condensation by Symmetric Gauss Elimination §10.2.3. Recovery of Internal Freedoms . . . . . §10.3. Global-Local Analysis §10. Notes and Bibliography . . . . . . . . . . . . . . . . §10. References. . . . . . . . . . . . . . . . §10. Exercises . . . . . . . . . . . . . . . .
10–2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10–3 10–3 10–5 10–5 10–5 10–5 10–6 10–8 10–8 10–10 10–10 10–11
10–3
§10.1
SUPERELEMENT CONCEPT
§10.1. Superelement Concept Superelements are groupings of finite elements that, upon assembly, may be regarded as an individual element for computational purposes. These purposes may be driven by modeling or processing needs. A random assortment of elements does not necessarily make up a superelement. To be considered as such, a grouping must meet certain conditions. Informally we can say that it must form a structural component on its own. This imposes certain conditions stated mathematically in §10.1.3. Inasmuch as these conditions involve advanced concepts such as rank sufficiency, which are introduced in later Chapters, the restrictions are not dwelled upon here. As noted in Chapter 6, superelements may originate from two overlapping contexts: “bottom up” or “top down.” In a bottom up context one thinks of superelements as built from simpler elements. In a top-down context, superelements may be thought as being large pieces of a complete structure. This dual viewpoint motivates the following classification: Macroelements. These are superelements assembled with a few primitive elements. Also called mesh units when they are presented to program users as individual elements. Substructures. Complex assemblies of elements that result on breaking up a structure into distinguishable portions. When does a substructure becomes a macroelement or vice-versa? There are no precise rules. In fact the generic term superelement was coined to cover the entire spectrum, ranging from individual elements to complete structures. This universality is helped by common processing features. Both macroelements and substructures are treated exactly the same way as regards matrix processing. The basic rule is that associated with condensation of internal degrees of freedom. The technique is illustrated in the following section with a simple example. The reader should note, however, that condensation applies to any superelement, whether composed of two or a million elements.1 §10.1.1. Where Does the Idea Come From? Substructuring was invented by aerospace engineers in the early 1960s2 to carry out a first-level breakdown of complex systems such as a complete airplane, as depicted in Figure 10.1. The decomposition may continue hierarchically through additional levels as illustrated in Figure 10.2. The concept is also natural for space vehicles operating in stages, such as the Apollo short stack depicted in Figure 10.3. Three original motivating factors for substructuring can be cited. 1.
Facilitate division of labor. Substructures with different functions are done by separate design groups with specialized knowledge and experience. For instance an aircraft company may set up a fuselage group, a wing group, a landing-gear group, etc. These groups are thus protected from “hurry up and wait” constraints. More specifically: a wing design group can keep on working on refinements, improvements and experimental model verification as long as the interface information (the wing-fuselage intersection) stays sensibly unchanged.
1
Of course the computer implementation becomes totally different as one goes from macroelements to substructures, because efficient processing for large matrix systems requires exploitation of sparsity.
2
See Notes and Bibliography at the end of this Chapter.
10–3
10–4
Chapter 10: SUPERELEMENTS AND GLOBAL-LOCAL ANALYSIS
S2
S6
S4
S5
S1 S3
Figure 10.1. Complete airplane broken down into six level one substructures identified as S1 through S6 . level two substructure
level one substructure individual element
Figure 10.2. Further breakdown of wing structure. The decomposition process may continue down to the individual element level.
2.
3.
Take advantage of repetition. Often structures are built of several identical or nearly identical units. For instance, the wing substructures S2 and S3 of Figure 10.1 are mirror images on reflection about the fuselage midplane, and so are the stabilizers S4 and S5 . Even if the loading is not symmetric about that midplane, recognizing repetitions saves model preparation time. Overcome computer limitations. The computers of the 1960s operated under serious memory limitations. (For example, the first supercomputer: the Control Data 6600, had a total high-speed memory of 131072 60-bit words or 1.31 MB; that machine cost $10M in 1966 dollars.) It was difficult to fit a complex structure such as an airplane as one entity. Substructuring permitted the complete analysis to be carried out in stages through use of auxiliary storage devices such as tapes or disks.
COMMAND MODULE
SERVICE MODULE
ADAPTER LUNAR MODULE
INSTRUMENT UNIT
THIRD STAGE SIV-B
Figure 10.3. The Apollo short stack.
Of the three motivations, the first two still hold today. The third one has moved to a different plane: parallel processing, as noted in §10.1.2 below. In the late 1960s the idea was picked up and developed extensively by the offshore and shipbuilding 10–4
10–5
§10.2
STATIC CONDENSATION
industries, the products of which tend to be modular and repetitive to reduce fabrication costs. As noted above, repetition favors the use of substructuring techniques. At the other end of the superelement spectrum, the mesh units herein called macroelements appeared in the mid 1960s. They were motivated by user convenience. For example, in hand preparation of models, quadrilateral and bricks involve less human labor than triangles and tetrahedra, respectively. It was therefore natural to combine the latter to assemble the former. Going a step further one can assemble components such as “box elements” for applications such as box-girder bridges. §10.1.2. Subdomains Applied mathematicians working on solution procedures for parallel computation have developed the concept of subdomains. These are groupings of finite elements that are entirely motivated by computational considerations. They are subdivisions of the finite element model done more or less automatically by a program called domain decomposer. Although the concepts of substructures and subdomains overlap in many respects, it is better to keep the two separate. The common underlying theme is divide and conquer but the motivation is different. §10.1.3. *Mathematical Requirements A superelement is said to be rank-sufficient if its only zero-energy modes are rigid-body modes. Equivalently, the superelement does not possess spurious kinematic mechanisms. Verification of the rank-sufficient condition guarantees that the static condensation procedure described below will work properly.
§10.2. Static Condensation Degrees of freedom of a superelement are classified into two groups: Internal Freedoms. Those that are not connected to the freedoms of another superelement. Nodes whose freedoms are internal are called internal nodes. Boundary Freedoms. These are connected to at least another superelement. They usually reside at boundary nodes placed on the periphery of the superelement. See Figure 10.4. The objective is to get rid of all displacement degrees of freedom associated with internal freedoms. This elimination process is called static condensation, or simply condensation. Condensation may be presented in terms of explicit matrix operations, as shown in the next subsection. A more practical technique based on symmetric Gauss elimination is discussed later. §10.2.1. Condensation by Explicit Matrix Operations To carry out the condensation process, the assembled stiffness equations of the superelement are partitioned as follows: 00030002 0003 0002 0003 0002 ub f Kbb Kbi = b . (10.1) Kib Kii ui fi where subvectors ub and ui collect boundary and interior degrees of freedom, respectively. Take the second matrix equation: (10.2) Kib ub + Kii ui = fi , 10–5
10–6
Chapter 10: SUPERELEMENTS AND GLOBAL-LOCAL ANALYSIS
(a)
(b) i
i
b
b
b
b
b
b
b
b
b
b
Figure 10.4. Classification of superelement freedoms into boundary and internal. (a) shows the vertical stabilizer substructure S6 of Figure 10.2. (The FE mesh is pictured as two-dimensional for illustrative purposes; for an actual aircraft it will be three dimensional.) Boundary freedoms are those associated to the boundary nodes labeled b (shown in red), which are connected to the fuselage substructure. (b) shows a quadrilateral macroelement mesh-unit fabricated with 4 triangles: it has one interior and four boundary nodes.
If Kii is nonsingular we can solve for the interior freedoms: ui = Kii−1 (fi − Kib ub ),
(10.3)
Replacing into the first matrix equation of (10.1) yields the condensed stiffness equations ˜ bb ub = ˜fb . K
(10.4)
In this equation, ˜ bb = Kbb − Kbi K−1 Kib , K ii
˜fb = fb − Kbi K−1 fi , ii
(10.5)
are called the condensed stiffness matrix and force vector, respectively, of the substructure. From this point onward, the condensed superelement may be viewed, from the standpoint of further 0004bb operations, as an individual element whose element stiffness matrix and nodal force vector are K ˜ and fb , respectively. Often each superelement has its own “local” coordinate system. A transformation of (10.5) to an overall global coordinate system is necessary upon condensation. In the case of multiple levels, the transformation is done with respect to the next-level superelement coordinate system. This coordinate transformation procedure automates the processing of repeated portions. Remark 10.1. The feasibility of the condensation process (10.3)–(10.5) rests on the non-singularity of Kii . This matrix is nonsingular if the superelement is rank-sufficient in the sense stated in §10.1.3, and if fixing the boundary freedoms precludes all rigid body motions. If the former condition is verified but not the latter, the superelement is called floating. Processing floating superelements demands more advanced computational techniques, among which one may cite the use of projectors and generalized inverses [57].
10–6
10–7
§10.2
STATIC CONDENSATION
CondenseLastFreedom[K_,f_]:=Module[{pivot,c,Kc,fc, n=Length[K]}, If [n<=1,Return[{K,f}]]; Kc=Table[0,{n-1},{n-1}]; fc=Table[0,{n-1}]; pivot=K[[n,n]]; If [pivot0, Print['CondenseLastFreedom:', ' Singular Matrix']; Return[{K,f}]]; For [i=1,i<=n-1,i++, c=K[[i,n]]/pivot; fc[[i]]=f[[i]]-c*f[[n]]; For [j=1,j<=i,j++, Kc[[j,i]]=Kc[[i,j]]=K[[i,j]]-c*K[[n,j]] ]; ]; Return[Simplify[{Kc,fc}]] ]; K={{6,-2,-1,-3},{ -2,5,-2,-1},{ -1,-2,7,-4},{-3,-1,-4,8}}; f={3,6,4,0}; Print['Before condensation:',' K=',K//MatrixForm,' f=',f//MatrixForm]; {K,f}=CondenseLastFreedom[K,f];Print['Upon condensing DOF 4:', ' K=',K//MatrixForm,' f=',f//MatrixForm]; {K,f}=CondenseLastFreedom[K,f];Print['Upon condensing DOF 3:', ' K=',K//MatrixForm,' f=',f//MatrixForm];
Figure 10.5. Mathematica module to condense the last degree of freedom from a stiffness matrix and force vector. The test statements carry out the example (10.6)–(10.10).
§10.2.2. Condensation by Symmetric Gauss Elimination In the computer implementation of the the static condensation process, calculations are not carried out as outlined above. There are two major differences. The equations of the substructure are not actually rearranged, and the explicit calculation of the inverse of Kii is avoided. The procedure may be in fact coded as a variant of symmetric Gauss elimination. To convey the flavor of this technique, consider the following stiffness equations of a superelement:     3 6 −2 −1 −3 u1 5 −2 −1   u 2   6   −2 (10.6)    =  . 4 u3 −1 −2 7 −4 0 u4 −3 −1 −4 8 Suppose that the last two displacement freedoms: u 3 and u 4 , are classified as interior and are to be statically condensed out. To eliminate u 4 , perform symmetric Gauss elimination of the fourth row and column:      −2 − (−1)×(−3) −1 − (−4)×(−3) 3 − 0×(−3) 6 − (−3)×(−3) u1 8 8 8 8     (−3)×(−1) 0×(−1)  (10.7) 5 − (−1)×(−1) −2 − (−4)×(−1)  u2 =  6 − 8  ,  −2 − 8 8 8 (−3)×(−4) (−1)×(−4) (−4)×(−4) 0×(−4) u3 −1 − −2 − 7− 4− 
8
8
or

39 8  − 19 8 − 52
8
− 19 8 39 8 − 52
− 52   u 1   3  − 5   u2  =  6  .
5
8
(10.8)
2
4 u3 5 Repeat the process for the third row and column to eliminate u 3 : 00030002 0003 0002 0002 39 (−5/2)×(−5/2) − − 19 − (−5/2)×(−5/2) 3− u1 8 5 8 5 = (−5/2)×(−5/2) (−5/2)×(−5/2) 39 19 u2 − − − 6− 8
8
5
10–7
4×(−5/2) 5 4×(−5/2) 5
0003 ,
(10.9)
10–8
Chapter 10: SUPERELEMENTS AND GLOBAL-LOCAL ANALYSIS
0002
or
29 8 − 29 8
− 29 8
00030002
29 8
u1 u2
0003
0002 0003 5 = . 8
(10.10)
These are the condensed stiffness equations. Figure 10.5 shows a Mathematica program that carries out the foregoing steps. Module CondenseLastFreedom condenses the last freedom of a stiffness matrix K and a force vector f. It is invoked as { Kc,fc }=CondenseLastFreedom[K,f]. It returns the condensed stiffness Kc and force vector fc as new arrays. To do the example (10.6)–(10.10), the module is called twice, as illustrated in the test statements of Figure 10.5. Obviously this procedure is much simpler than going through the explicit matrix inverse. Another important advantage of Gauss elimination is that equation rearrangement is not required even if the condensed degrees of freedom do not appear sequentially. For example, suppose that the assembled substructure contains originally eight degrees of freedom and that the freedoms to be condensed out are numbered 1, 4, 5, 6 and 8. Then Gauss elimination is carried out over those equations only, and the condensed (3 × 3) stiffness and (3 × 1) force vector extracted from rows and columns 2, 3 and 7. An implementation of this process is considered in Exercise 10.2. Remark 10.2. The symmetric Gauss elimination procedure, as illustrated in steps (10.6)–(10.10), is primarily useful for macroelements and mesh units, since the number of stiffness equations for those typically does not exceed a few hundreds. This permits the use of full matrix storage. For substructures containing thousands or millions of degrees of freedom — such as in the airplane example — the elimination is carried out using more sophisticated sparse matrix algorithms; for example that described in [46]. Remark 10.3. The static condensation process is a matrix operation called “partial inversion” or “partial elimi-
nation” that appears in many disciplines. Here is the general form. Suppose the linear system Ax = y, where A is n × n square and x and y are n-vectors, is partitioned as
A11 A21
A12 A22
x1 x2
y = 1 . y2
(10.11)
Assuming the appropriate inverses to exist, then the following are easily verified matrix identities:
0002
A−1 −A−1 11 11 A12 −1 A21 A11 A22 − A21 A−1 11 A12
00030002
y1 x2
0003
0002 =
0003
x1 , y2
0002
−1 A11 − A12 A−1 22 A21 A12 A22 −A−1 A−1 22 A21 22
00030002
x1 y2
0003
0002 =
0003
y1 . x2
(10.12)
We say that x1 has been eliminated or “condensed out” in the left identity and x2 in the right one. In FEM applications, it is conventional to condense out the bottom vector x2 , so the right identity is relevant. If A is symmetric, to retain symmetry in (10.12) it is necessary to change the sign of one of the subvectors.
§10.2.3. Recovery of Internal Freedoms (to be added) §10.3. Global-Local Analysis As discussed in the first Chapter, complex engineering systems are often modeled in a multilevel fashion following the divide and conquer approach. The superelement technique is a practical realization of that approach. A related, but not identical, technique is multiscale analysis. The whole system is first analyzed as a global entity, discarding or passing over details deemed not to affect its overall behavior. Local details 10–8
10–9
§10.3
GLOBAL-LOCAL ANALYSIS
coarse mesh
finer meshes
Figure 10.6. Left: example panel structure for global-local analysis. Right: a FEM mesh for a one-shot analysis.
are then analyzed using the results of the global analysis as boundary conditions. The process can be continued into the analysis of further details of local models. And so on. When this procedure is restricted to two stages and applied in the context of finite element analysis, it is called global-local analysis in the FEM literature. In the global stage the behavior of the entire structure is simulated with a finite element model that necessarily ignores details such as cutouts or joints. These details do not affect the overall behavior of the structure, but may have a bearing on safety. Such details are a posteriori incorporated in a series of local analyses. The gist of the global-local approach is explained in the example illustrated in Figures 10.6 and 10.7. Although the structure is admittedly too simple to merit the application of global-local analysis, it serves to illustrate the basic ideas. Suppose one is faced with the analysis of the rectangular panel shown on the top of Figure 10.6, which contains three small holes. The bottom of that figure shows a standard (one-stage) FEM treatment using a largely regular mesh that is refined near the holes. Connecting the coarse and fine meshes usually involves using multifreedom constraints because the nodes at mesh boundaries do not match, as depicted in that figure. Figure 10.7 illustrates the global-local analysis procedure. The global analysis is done with a coarse but regular FEM mesh which ignores the effect of the holes. This is followed by local analysis of the region near the holes using refined finite element meshes. The key ingredient for the local analyses is the application of boundary conditions (BCs) on the finer mesh boundaries. These BCs may be of displacement (essential) or of force (natural) type. If the former, the applied boundary displacements are interpolated from the global mesh solution. If the latter, the internal forces or stresses obtained from the global calculation are converted to nodal forces on the fine meshes through a lumping process. The BC choice noted above gives rise to two basic variations of the global-local approach. Experience accumulated over several decades3 has shown that the stress-BC approach generally gives more reliable answers. The global-local technique can be extended to more than two levels, in which case it receives the more encompassing name multiscale analysis. Although this generalization is still largely in the realm of 3
Particularly in the aerospace industry, in which the global-local technique has been used since the early 1960s.
10–9
Chapter 10: SUPERELEMENTS AND GLOBAL-LOCAL ANALYSIS
10–10
Global analysis with a coarse mesh, ignoring holes, followed bylocal analysis of the vicinity of the holes with finer meshes:
Figure 10.7. Global-local analysis of problem of Figure 10.6.
research, it is receiving increasing attention from various science and engineering communities for complex products such as the thermomechanical analysis of microelectronic components. Notes and Bibliography Substructuring was invented by aerospace engineers in the early 1960s. Przemieniecki’s book [135] contains a fairly complete bibliography of early work. Most of this was in the form of largely inaccessible internal company or lab reports and so the actual history is difficult to trace. Macroelements appeared simultaneously in many of the early FEM codes. Quadrilateral macroelements fabricated with triangles are described in [44]. For a survey of uses of the static condensation algorithm in FEM, see [179]. The generic term superelement was coined in the late 1960s by the SESAM group at DNV Veritas [42]. The matrix form of static condensation for a complete structure is presented in [4, p. 46], as a scheme to eliminate unloaded DOF in the displacement method. It is unclear when this idea was first applied to substructures or macroelements. The first application for reduced dynamical models is by Guyan [81]. The application of domain decomposition to parallel FEM solvers has produced an enormous and highly specialized literature. Procedures for handling floating superelements using generalized inverse methods are discussed in [57]. The global-local analysis procedure described here is also primarily used in industry and as such it is rarely mentioned in academic textbooks. References Referenced items have been moved to Appendix R.
10–10
10–11
Exercises
Homework Exercises for Chapter 10 Superelements and Global-Local Analysis EXERCISE 10.1 [N:15] The free-free stiffness equations of a superelement are





88 −44 −44 0 u1 5  −44 132 −44 −44   u 2   10   −44 −44 176 −44   u  =  15  . 3 u4 0 −44 −44 220 20
(E10.1)
Eliminate u 2 and u 3 from (E10.1) by static condensation, and show (but do not solve) the condensed equation system. Use either the explicit matrix inverse formulas (10.5) or the symmetric Gauss elimination process explained in §10.2.2. Hint: regardless of method the result should be
52 −36 −36 184
u1 u4
=
15 . 30
(E10.2)
EXERCISE 10.2 [C+N:20] Generalize the program of Figure 10.5 to a module CondenseFreedom[K,f,k] that is able to condense the kth degree of freedom from K and f, which is not necessarily the last one. That is, k may range from 1 to n, where n is the number of freedoms in Ku = f. Apply that program to solve the previous Exercise.
Hint: here is a possible way of organizing the inner loop: ii=0; For [i=1,i<=n,i++, If [ik, Continue[]]; ii++; c=K[[i,k]]/pivot; fc[[ii]]=f[[i]]-c*f[[k]]; jj=0; For [j=1,j<=n,j++, If [jk,Continue[]]; jj++; Kc[[ii,jj]]=K[[i,j]]-c*K[[k,j]] ]; ]; Return[{Kc,fc}] EXERCISE 10.3 [D:15] Explain the similarities and differences between superelement analysis and global-local
FEM analysis. EXERCISE 10.4 [A:20] If the superelement stiffness K is symmetric, the static condensation process can be viewed as a special case of the master-slave transformation method discussed in Chapter 8. To prove this, take exterior freedoms ub as masters and interior freedoms ui as slaves. Assume the master-slave transformation relation def ub I 0 u= = = Tub − g. (E10.3) [ ub ] − −1 −1 ui −Kii Kib −Kii fi ˆ uˆ = ˆf coalesces with the condensed stiffness ˆ = TT KT and ˆf = TT (f − Kg), and show that K Work out K T ˆ (Take advantage of the symmetry properties Kbi equations (10.4)–(10.5) if ub ≡ u. = Kib , (Kii−1 )T = Kii−1 .) EXERCISE 10.5 [D:30] (Requires thinking) Explain the conceptual and operational differences between onestage FEM analysis and global-local analysis of a problem such as that illustrated in Figures 10.6 and 10.7. Are the answers the same? What is gained, if any, by the global-local approach over the one-stage FEM analysis? EXERCISE 10.6 [N:20] The widely used Guyan’s scheme [81] for dynamic model reduction applies the staticcondensation relation (E10.3) as master-slave transformation to both the stiffness and the mass matrix of the superelement. Use this procedure to eliminate the second and third DOF of the mass-stiffness system:

2 1 M= 0 0
1 4 1 0
0 1 4 1

0 0 , 1 2


1 −1 0 0 2 −1 0  −1 K= . 0 −1 2 −1  0 0 −1 1
10–11
(E10.4)
Chapter 10: SUPERELEMENTS AND GLOBAL-LOCAL ANALYSIS
10–12
ˆ + ωˆ 2 K)ˆ ˆ vi = 0 Compute and compare the vibration frequencies of the eigensystems (M + ωi2 K)vi = 0 and (M i before and after reduction. EXERCISE 10.7 [A:20] Two beam elements: 1–2 and 2–3, each of length L and rigidity E I are connected at node 2. The macroelement has 6 degrees of freedom: {v1 , θ1 , v2 , θ2 , v3 , θ3 }. Eliminate the two DOF of node 2 by condensation. Is the condensed stiffness the same as that of a beam element of length 2L and rigidity E I ?
(For expressions of the beam stiffness matrices, see Chapter 12.)
10–12
11
.
Variational Formulation of Bar Element
11–1
11–2
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
TABLE OF CONTENTS Page
§11.1. A New Beginning §11.2. Definition of Bar Member §11.3. Variational Formulation §11.3.1. The Total Potential Energy Functional . §11.3.2. Variation of an Admissible Function . §11.3.3. The Minimum Potential Energy Principle §11.3.4. TPE Discretization . . . . . . . §11.3.5. Bar Element Discretization . . . . . §11.3.6. Shape Functions . . . . . . . . §11.3.7. The Strain-Displacement Equation . . §11.3.8. *Trial Basis Functions . . . . . . §11.4. The Finite Element Equations §11.4.1. The Stiffness Matrix . . . . . . . §11.4.2. The Consistent Node Force Vector . . §11.5. *Accuracy Analysis §11.5.1. *Nodal Exactness and Superconvergence §11.5.2. *Fourier Patch Analysis . . . . . §11. Notes and Bibliography . . . . . . . . . . . . . . §11. References. . . . . . . . . . . . . . §11. Exercises . . . . . . . . . . . . . .
11–2
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
11–3 11–3 11–3 11–3 11–6 11–6 11–7 11–7 11–8 11–8 11–9 11–9 11–9 11–10 11–10 11–10 11–11 11–12 11–13 11–14
11–3
§11.3
VARIATIONAL FORMULATION
§11.1. A New Beginning This Chapter begins Part II of the course. This Part focuses on the construction of structural and continuum finite elements using a variational formulation based on the Total Potential Energy. Why only elements? Because the other synthesis steps of the DSM: globalization, merge, BC application and solution, remain the same as in Part I. These operations are not element dependent. Part II constructs individual elements beginning with the simplest ones and progressing to more complicated ones. The formulation of two-dimensional finite elements from a variational standpoint is discussed in Chapters 14 and following. Although the scope of that formulation is broad, exceeding structural mechanics, it is better understood by going through specific elements first. From a geometrical standpoint the simplest finite elements are one-dimensional or line elements. This means that the intrinsic dimensionality is one, although these elements may be used in one, two or three space dimensions upon transformation to global coordinates as appropriate. The simplest one-dimensional structural element is the two-node bar element, which we have already encountered in Chapters 2, 3 and 5 as the truss member. In this Chapter the bar stiffness equations are rederived using the variational formulation. For uniform properties the resulting equations are the same as those found previously using the physical or Mechanics of Materials approach. The variational method has the advantage of being readily extendible to more complicated situations, such as variable cross section or more than two nodes. §11.2. Definition of Bar Member In structural mechanics a bar is a structural component characterized by two properties: (1) One preferred dimension: the longitudinal dimension or axial dimension is much larger that the other two dimensions, which are collectively known as transverse dimensions. The intersection of a plane normal to the longitudinal dimension and the bar defines the cross sections. The longitudinal dimension defines the longitudinal axis. See Figure 11.1. (2) The bar resists an internal axial force along its longitudinal dimension. In addition to trusses, bar elements are used to model cables, chains and ropes. They are also used as fictitious elements in penalty function methods, as discussed in Chapter 9. We will consider here only straight bars, although their cross section may vary. The one-dimensional mathematical model assumes that the bar material is linearly elastic, obeying Hooke’s law, and that displacements and strains are infinitesimal. Figure 11.2 pictures the relevant quantities for a fixedfree bar. Table 11.1 collects the necessary terminology for the governing equations. Figure 11.3 displays the governing equations of the bar in a graphic format called a Tonti diagram. The formal similarity with the diagrams used in Chapter 5 to explain MoM elements should be noted, although the diagram of Figure 11.3 pertains to the continuum model rather than to the discrete one. 11–3
11–4
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
Table 11.1
Nomenclature for Mathematical Model of Axially Loaded Bar
Quantity x (.)0003 u(x) q(x) L E A EA e = du/d x = u 0003 σ = Ee = Eu 0003 p = Aσ = E A e = E Au 0003 P ∗
Meaning Longitudinal bar axis∗ d(.)/d x Axial displacement Distributed axial force, given per unit of bar length Total bar length Elastic modulus Cross section area; may vary with x Axial rigidity Infinitesimal axial strain Axial stress Internal axial force Prescribed end load
x is used in this Chapter instead of x¯ (as in Chapters 2–3) to simplify the notation.
y
cross section
;; ;;
x
z
;;;
Cross section
axial rigidity EA u(x)
q(x)
P Longitudinal axis
x
P
L
Figure 11.2. Quantities that appear in analysis of bar.
Figure 11.1. A fixed-free bar member.
§11.3. Variational Formulation To illustrate the variational formulation, the finite element equations of the bar will be derived from the Minimum Potential Energy principle. §11.3.1. The Total Potential Energy Functional In Mechanics of Materials it is shown that the internal energy density at a point of a linear-elastic material subjected to a one-dimensional state of stress σ and strain e is U = 12 σ (x)e(x), where σ is to be regarded as linked to the displacement u through Hooke’s law σ = Ee and the straindisplacement relation e = u 0003 = du/d x. This U is also called the strain energy density. Integration 11–4
11–5
§11.3 VARIATIONAL FORMULATION
Displacement BCs
Prescribed end displacements
Axial displacements
Distributed axial load
u(x)
q(x)
Kinematic
e=u'
p'+q=0 Equilibrium
Axial strains
e(x)
p = EA e
Axial force
Constitutive
p(x)
Force BCs
Prescribed end loads
Figure 11.3. Tonti diagram for the mathematical model of a bar member. The diagram displays the field equations and boundary conditions as lines connecting the boxes. Yellow (brown) boxes contain unknown (given) quantities.
over the volume of the bar gives the total internal energy 0002 U=
1 2
0002 σ e dV = V
1 2
0
L
0002 pe d x =
1 2
L
0003
0003
(E Au )u d x =
0
0002 1 2
L
u 0003 E A u 0003 d x,
(11.1)
0
in which all integrand quantities may depend on x. The external energy due to applied mechanical loads pools contributions from two sources: 1.
The distributed load q(x). This contributes a cross-section density of q(x)u(x) because q is assumed to be already integrated over the section.
2.
Any applied end load(s). For the fixed-free example of Figure 11.2 the end load P would contribute P u(L).
The second source may be folded into the first by conventionally writing any point load P acting at a cross section x = a as a contribution P δ(a) to q(x), where δ(a) denotes the one-dimensional Dirac delta function at x = a. If this is done the external energy can be concisely expressed as 0002
L
W =
qu d x.
(11.2)
0
The total potential energy of the bar is given by =U −W
(11.3)
Mathematically this is a functional, called the Total Potential Energy functional or TPE. It depends only on the axial displacement u(x). In variational calculus this is called the primary variable of the functional. When the dependence of on u needs to be emphasized we shall write [u] = U [u]−W [u], with brackets enclosing the primary variable. To display both primary and independent variables we write, for example, [u(x)] = U [u(x)] − W [u(x)]. 11–5
11–6
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
Remark 11.1. According to the rules of Variational Calculus, the Euler-Lagrange equation for is
∂ d ∂ − = −q − (E A u 0003 )0003 = 0 (11.4) ∂u d x ∂u 0003 This is the equation of equilibrium in terms of the axial displacement, usually written (E A u 0003 )0003 + q = 0, or E A u 00030003 + q = 0 if E A is constant. This equation is not explicitly used in the FEM development. It is instead replaced by δ = 0, with the variation restricted over the finite element interpolation functions.
§11.3.2. Variation of an Admissible Function The concept of admissible variation is fundamental in both variational calculus and the variationally formulated FEM. Only the primary variable(s) of a functional may be varied. For the TPE functional (11.4) this is the axial displacement u(x). Suppose that u(x) is changed to u(x) + δ u(x).1 This is illustrated in Figure 11.4, where for convenience u(x) is plotted normal to x. The functional changes from to + δ . The function δ u(x) and the scalar δ are called the variations of u(x) and , respectively. The variation δ u(x) should not be confused with the ordinary differential du(x) = u 0003 (x) d x since on taking the variation the independent variable x is frozen; that is, δx = 0.
u(x)+δu(x)
u
δu(x) u(L)
u(x)
x
u(0) = 0 Figure 11.4. Concept of admissible variation of the axial displacement function u(x). For convenience u(x) is plotted normal to the longitudinal axis. Both depicted u(x) and u(x) + δu(x) are kinematically admissible, and so is the variation δu(x).
A displacement variation δu(x) is said to be admissible when both u(x) and u(x) + δ u(x) are kinematically admissible in the sense of the Principle of Virtual Work (PVW). This agrees with the conditions stated in the classic variational calculus. A kinematically admissible axial displacement u(x) obeys two conditions: (i)
It is continuous over the bar length, that is, u(x) ∈ C0 in x ∈ [0, L].
(ii) It satisfies exactly any displacement boundary condition, such as the fixed-end specification u(0) = 0 of Figure 11.2. The variation δ u(x) depicted in Figure 11.4 is kinematically admissible because both u(x) and u(x) + δ u(x) satisfy the foregoing conditions. The physical meaning of (i)–(ii) is the subject of Exercise 11.1. §11.3.3. The Minimum Potential Energy Principle The Minimum Potential Energy (MPE) principle states that the actual displacement solution u ∗ (x) that satisfies the governing equations is that which renders stationary:2 δ = δU − δW = 0
iff u = u ∗
(11.5)
with respect to admissible variations u = u ∗ + δu of the exact displacement field u ∗ (x). 1
The symbol δ not immediately followed by a parenthesis is not a delta function but instead denotes variation with respect to the variable that follows.
2
The symbol “iff” in (11.5) is an abbreviation for “if and only if”.
11–6
11–7
§11.3
VARIATIONAL FORMULATION
Remark 11.2. Using standard techniques of variational calculus3 it can be shown that if E A > 0 the solution
u ∗ (x) of (11.5) exists, is unique, and renders [u] a minimum over the class of kinematically admissible displacements. The last attribute explains the “mininum” in the name of the principle.
§11.3.4. TPE Discretization To apply the TPE functional (11.2) to the derivation of finite element equations we replace the continuum mathematical model by a discrete one consisting of a union of bar elements. For example, Figure 11.5 illustrates the subdivision of a bar member into four two-node elements. Functionals are scalars. Therefore, corresponding to a discretization such as that shown in Figure 11.5, the TPE functional (11.4) may be decomposed into a sum of contributions of individual elements:
u 1, f1
u 2, f2
u 3, f3
u 4, f4
(1)
(2)
(3)
(4)
1
2
3
4
u 5, f5
5
u u2
u3
u4
u(x)
u5
x
u1 = 0 Figure 11.5. FEM discretization of bar member. A piecewise- linear admissible displacement trial function u(x) is drawn underneath the mesh. It is assumed that the left end is fixed; thus u 1 = 0.
= (1) + (2) + . . . + (Ne )
(11.6)
where Ne is the number of elements. The same decomposition applies to the internal and external energies, as well as to the stationarity condition (11.5): δ = δ (1) + δ (2) + . . . + δ (Ne ) = 0.
(11.7)
Using the fundamental lemma of variational calculus,4 it can be shown that (11.7) implies that for a generic element e we may write δ e = δU e − δW e = 0.
(11.8)
This variational equation is the basis for the derivation of element stiffness equations once the displacement field has been discretized over the element, as described next. Remark 11.3. In mathematics (11.8) is called a weak form. In mechanics it also states the Principle of Virtual
Work for each element: δU e = δW e , which says that the virtual work of internal and external forces on admissible displacement variations is equal if the element is in equilibrium [132].
3
See references in Notes and Bibliography at the end of Chapter.
4
See, e.g., Chapter II of Gelfand and Fomin [73].
11–7
11–8
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
§11.3.5. Bar Element Discretization Figure 11.6 depicts a generic bar element e. It has two nodes, which are labeled 1 and 2. These are called the local node numbers.5 The element is referred to its local axis x¯ = x − x1 , which measures the distance from its left end. The two degrees of freedom are u e1 and u e2 . (Bars are not necessary on these values since the directions of x¯ and x are the same.) The element length is = L e . The mathematical concept of bar finite elements is based on approximation of the axial displacement u(x) over the element. The exact displacement u ∗ is replaced by an approximate displacement u ∗ (x) ≈ u e (x)
(11.9)
over the finite element mesh. This approximate displacement, u e (x), taken over all elements e = 1, 2, . . . N e , is called the finite element trial expansion or simply trial expansion. See Figure 11.5. This FE trial expansion must belong to the class of kinematically admissible displacements defined in §11.3.2. Consequently, it must be C0 continuous over and between elements. §11.3.6. Shape Functions In a two-node bar element the only possible variation of the displacement u e that satisfies the interelement continuity requirement stated above is linear. It can be expressed by the interpolation formula 0003 e0004 u e e e e e u (x) = N1 u 1 + N2 u 2 = [ N1e N2e ] 1e = Nue . (11.10) u2 The functions N1e and N2e that multiply the node displacements u 1 and u 2 are called shape functions. These functions interpolate the internal displacement u e directly from the node values. See Figure 11.6. For the bar element, with x¯ = x − x1 measuring the distance from the left node i, the shape functions are N1e = 1 −
x¯ = 1 − ζ,
N2e =
x¯ = ζ.
5
2
= L (e)
(11.11)
¯ is a dimensionless coordinate, Here ζ = (x − x1 )/ = x/ also known as a natural coordinate. Note that d x = dζ and dζ = d x/ . The shape function N1e has the value 1 at node 1 and 0 at node 2. Conversely, shape function N2e has the value 0 at node 1 and 1 at node 2. This is a general property of shape functions. It follows from the fact that element displacement interpolations such as (11.10) are based on physical node values.
(e)
1
x- = xγ - x1 1 N (e) i
0
0 Nj(e) 1
Figure 11.6. The shape functions of the generic bar element.
Note the notational change from the labels i and j of Part I. This will facilitate transition to multidimensional elements.
11–8
11–9
§11.4
THE FINITE ELEMENT EQUATIONS
Remark 11.4. In addition to continuity, shape functions must satisfy a completeness requirement with respect
to the governing variational principle. This condition is stated and discussed in later Chapters. Suffices for now to say that the shape functions (11.11) do satisfy this requirement.
§11.3.7. The Strain-Displacement Equation The axial strain over the element is 0005 dNe du e 1 e= = (u e )0003 = dx dx
0003 0004 1 d N2e 0006 u e1 = [ −1 e u2 dx
where B=
1 [ −1
0003
ue 1 ] 1e u2
0004 = Bue ,
(11.12)
(11.13)
1]
is called the strain-displacement matrix. §11.3.8. *Trial Basis Functions (1)
Shape functions are associated with elements. A trial basis function, or simply basis function, is associated with a node. Suppose node i of a bar discretization connects elements (e1) and (e2). The trial basis function Ni is defined as
0007
1
(2)
2
(3)
(4)
4
3
u
5
N3
N3(2)
N3(3)
(thick line)
x if x ∈ element (e1) Ni (x) = (11.14) if x ∈ element (e2) Figure 11.7. Trial basis function for node 3. 0 otherwise For a piecewise linear discretizations such as the two-node bar this function has the shape of a hat. Thus it is sometimes called a hat function or chapeau function. See Figure 11.7, in which i = 3, e1 = 2, e2 = 3. The concept is important in the variational interpretation of FEM as a Rayleigh-Ritz method. Ni(e1) Ni(e2)
§11.4. The Finite Element Equations In linear FEM the discretization process for the TPE functional leads to the following algebraic form (11.15) e = U e − W e , U e = 12 (ue )T Ke ue , W e = (ue )T fe , where Ke and fe are called the element stiffness matrix and the element consistent nodal force vector, respectively. Note that in (11.15) the three energies are only function of the node displacements ue . U e and W e depend quadratically and linearly, respectively, on those displacements. Taking the variation of the discretized TPE of (11.15) with respect to the node displacements gives6 T ∂ e T e e δ e = δue = δue K u − fe = 0. (11.16) e ∂u Because the variations δue can be arbitrary, the bracketed quantity must vanish, which yields Ke ue = fe
(11.17)
These are the element stiffness equations. Hence the foregoing names given to Ke and fe are justified a posteriori. 6
The 12 factor disappears on taking the variation because U e is quadratic in the node displacements. For a review on the calculus of discrete quadratic forms, see Appendix D.
11–9
11–10
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
§11.4.1. The Stiffness Matrix For the two-node bar element, the internal energy U e is 0002 U = e
1 2
x2
0002 e E A e dx =
1 2
x1
1
e E A e dζ,
(11.18)
0
where the strain e is related to the nodal displacements through (11.12). This form is symmetrically expanded by inserting e = Bue into the second e and e = e T = (ue )T BT into the first e: 0002 U = e
1 2
1
[ u e1
0
u e2
0003 0004 1 1 −1 E A [ −1 ] 1
0003
ue 1 ] 1e u2
0004 dζ.
(11.19)
0004 0003 e0004 e T e e −1 u1 1 u Ku. dζ = 2 u e2 1
(11.20)
The nodal displacements can be moved out of the integral, giving 0002 U = e
1 2
[ u e1
u e2
1
] 0
in which
0002
1
K = e
0003
EA 2
1 −1
0002
1
E A B B dζ = T
0
0
EA 2
0003
1 −1
−1 1
0004 dζ.
(11.21)
is the element stiffness matrix. If the rigidity E A is constant over the element, 0002
1
K = EAB B e
T
0
EA dζ = 2
0003
1 −1
−1 1
0004
EA =
0003
1 −1
0004 −1 . 1
(11.22)
This is the same element stiffness matrix of the prismatic truss member derived in Chapters 2 and 5 by a Mechanics of Materials approach, but now obtained through a variational argument. §11.4.2. The Consistent Node Force Vector The consistent node force vector fe introduced in (11.15) comes from the element contribution to the external work potential W : 0002 W = e
x2
0002 q u dx =
1
T q N u dζ = ue T
0
x1
0002
1
e
0
0003
1−ζ q ζ
0004
T dζ = ue fe ,
(11.23)
dζ.
(11.24)
in which ζ = (x − x1 )/ . Consequently 0002 f = e
x2 x1
0003
1−ζ q ζ
0004
0002 dx = 0
1
0003
1−ζ q ζ
0004
If the force q is constant over the element, one obtains the same results as with the EbE load-lumping method of Chapter 7. See Exercise 11.3. 11–10
11–11 §11.5.
§11.5 *ACCURACY ANALYSIS
*Accuracy Analysis
Low order 1D elements may give surprisingly high accuracy. In particular the lowly two-node bar element can display infinite accuracy under some conditions. This phenomenon is studied in this advanced section as it provides an introduction to modified equation methods and Fourier analysis along the way. §11.5.1. *Nodal Exactness and Superconvergence Suppose that the following two conditions are satisfied: 1.
The bar properties are constant along the length (prismatic member).
2.
The distributed load q(x) is zero between nodes. The only applied loads are point forces at the nodes.
If so, a linear axial displacement u(x) as defined by (11.10) and (11.11) is the exact solution over each element since constant strain and stress satisfy, element by element, all of the governing equations listed in Figure 11.3.7 It follows that if the foregoing conditions are verified the FEM solution is exact; that is, it agrees with the analytical solution of the mathematical model.8 Adding extra elements and nodes would not change the solution. That is the reason behind the truss discretizations used in Chapters 2–3: one element per member is enough if they are prismatic and loads are applied to joints. Such models are called nodally exact. What happens if the foregoing assumptions are not met? Exactness is then generally lost, and several elements per member may be beneficial if spurious mechanisms are avoided.9 For a 1D lattice of equal-length, prismatic two-node bar elements, an interesting and more difficult result is: the solution is nodally exact for any loading if consistent node forces are used. This is proven in the subsection below. This result underlies the importance of computing node forces correctly. If conditions such as equal-length are relaxed, the solution is no longer nodally exact but convergence at the nodes is extremely rapid (faster than could be expected by standard error analysis) as long as consistent node forces are used. This phenomenon is called superconvergence in the FEM literature. §11.5.2. *Fourier Patch Analysis
q(x)
The following analysis is based on the modified differential equation (MoDE) method of Warming and Hyett [172] combined with the Fourier patch analysis approach of Park and Flaggs [121,122]. Consider a lattice of two-node prismatic bar elements of constant rigidity E A and equal length , as illustrated in Figure 11.8. The total length of the lattice is L. The system is subject to an arbitrary axial load q(x). The only requirement on q(x) is that it has a convergent Fourier series in the space direction.
j
i
xi = xj − xj xk = xj +
k EA = const
L
1 − (xj −x)/ = 1−ψ Two-element patch ijk
i
Trial basis function Nj for node j
1
1 + (x j−x)/ = 1+ψ
j
k
Figure 11.8. Superconvergence patch analysis.
From the lattice extract a patch10 of two elements connecting nodes xi , x j and xk as shown in Figure 11.8. The 7
The internal equilibrium equation p 0003 + q = E A u 00030003 + q = 0 is trivially verified because q = 0 from the second assumption, and u 00030003 = 0 because of shape function linearity.
8
In variational language: the Green function of the u 00030003 = 0 problem is included in the FEM trial space.
9
These can happen when transforming such elements for 2D and 3D trusses. See Exercise E11.7.
10
A patch is the set of all elements connected to a node; in this case j.
11–11
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
11–12
FEM patch equations at node j are
EA [ −1
2
−1 ]
ui uj uk
= fi ,
(11.25)
in which the node force f j is obtained by consistent lumping:
0002
0002
xk
fj =
q(x)N j (x) d x = xi
0002
0
1
q(x j + ψ )(1 + ψ) dψ + −1
q(x j + ψ )(1 − ψ) dψ.
(11.26)
0
Here N j (x) is the “hat” trial basis function for node j, depicted in Figure 11.8, and ψ = (x − x j )/ is a dimensionless coordinate that takes the values −1, 0 and 1 at nodes i, j and k, respectively. If q(x) is expanded in Fourier series q(x) =
M 000e
qm eiβm x ,
βm = mπ/L ,
(11.27)
m=1
(the term m = 0 requires special handling) the exact solution of the continuum equation E A u 00030003 + q = 0 is u ∗ (x) =
M 000e
u ∗m eiβm x ,
u ∗m =
m=1
qm eiβm x . E Aβm2
(11.28)
Evaluation of the consistent force using (11.26) gives fj =
M 000e
f jm ,
f jm = qm
m=1
sin2 ( 12 βm ) 1 2 2 β 4 m
eiβm x2 .
(11.29)
To construct a modified differential equation (MoDE), expand the displacement by Taylor series centered at node j. Evaluate at i and k: u i = u j − u 0003j + 2 u 00030003j /2! − 3 u 000300030003j /3! + 4 u ivj /4! + . . . and u k = u j + u 0003j + 2 u 00030003j /2 + 3 u 000300030003j /3! + 4 u ivj /4! + . . . Replace these series into (11.25) to get
000f −2E A
0010
1 00030003 2 iv 4 vi u + u + uj + .. 2! j 4! j 6!
= fj.
(11.30)
This is an ODE of infinite order. It can be reduced to an algebraic equation by assuming that the response of (11.30) to qm eiβm x is harmonic: u jm eiβm x . If so u 00030003jm = −βm2 u jm , u ivjm = βm4 u jm , etc, and the MoDE becomes
000f
0010
sin2 ( 1 βm ) β 2 2 β 4 4 1 − m + m − . . . u jm = 4E A sin2 ( 12 βm ) u jm = f jm = qm 1 22 2 eiβm x j . 2! 4! 6! β 4 m (11.31) Solving gives u jm = qm eiβm x j /(E Aβm2 ), which compared with (11.28) shows that u jm = u ∗m for any m > 0. Consequently u j = u ∗j . In other words, the MoDE (11.30) and the original ODE: E Au 00030003 + q = 0 have the same value at x = x j for any load q(x) developable as (11.27). This proves nodal exactness. In between nodes the two solutions will not agree.11 2E A βm2
The case m = 0 has to be treated separately since the foregoing expressions become 0/0. The response to a uniform q = q0 is a quadratic in x, and it is not difficult to prove nodal exactness. 11
The FEM solution varies linearly between nodes whereas the exact one is generally trigonometric.
11–12
11–13
§11.
References
Notes and Bibliography The foregoing development pertains to the simplest structural finite element: the two-node bar element. For bars this may be generalized in various directions. Refined bar elements. Adding internal nodes we can pass from linear to quadratic and cubic shape functions. These elements are rarely useful on their own right, but as accessories to 2D and 3D high order continuum elements (for example, to model edge reinforcements.) For that reason they are not considered here. The 3-node bar element is developed in exercises assigned in Chapter 16. Two- and three-dimensional truss structures. The only additional ingredients are the transformation matrices discussed in Chapters 3 and 6. Curved bar elements. These can be derived using isoparametric mapping, a device introduced later. Matrices for straight bar elements are available in any finite element book; for example Przemieniecki [135]. Tonti diagrams were introduced in the 1970s in papers now difficult to access, for example [162]. Scanned images are available, howewer, at http://www.dic.units.it/perspage/discretephysics The fundamentals of Variational Calculus may be studied in the excellent textbook [73], which is now available in an inexpensive Dover edition. The proof of the MPE principle can be found in texts on variational methods in mechanics. For example: Langhaar [103], which is the most readable “old fashioned” treatment of the energy principles of structural mechanics, with a beautiful treatment of virtual work. (Out of print but used copies may be found via the web engines cited in §1.5.2.) The elegant treatment by Lanczos [102] is recommended as reading material although it is more oriented to physics than structural mechanics. The first accuracy study of FEM discretizations using modified equation methods is by Waltz et. al. [170]; however their procedures were faulty, which led to incorrect conclusions. The first correct derivation of modified equations appeared in [172]. The topic has recently attracted interest from applied mathematicians because modified equations provide a systematic tool for backward error analysis of differential equations: the discrete solution is the exact solution of the modified problem. This is particularly important for the study of long term behavior of discrete dynamical systems, whether deterministic or chaotic. Recommended references along these lines are [79,82,155]. Nodal exactness of bar models for point node loads is a particular case of a theorem by Tong [161]. For arbitrary loads it was proven by Park and Flaggs [121,122], who followed a variant of the scheme of §11.5.2. A different technique is used in Exercise 11.8. The budding concept of superconvergence, which emerged in the late 1960s, is outlined in the book of Strang and Fix [150]. There is a monograph [171] devoted to the subject; it covers only Poisson problems but provides a comprehensive reference list until 1995. References Referenced items moved to Appendix R.
11–13
Chapter 11: VARIATIONAL FORMULATION OF BAR ELEMENT
11–14
Homework Exercises for Chapter 11 Variational Formulation of Bar Element EXERCISE 11.1 [D:10] Explain the kinematic admissibility requirements stated in §11.3.2 in terms of physics, namely ruling out the possibility of gaps or interpenetration as the bar material deforms. EXERCISE 11.2 [A/C:15] Using (11.21), derive the stiffness matrix for a tapered bar element in which the cross section area varies linearly along the element length:
A = Ai (1 − ζ ) + A j ζ,
(E11.1)
where Ai and A j are the areas at the end nodes, and ζ = x e / is the dimensionless coordinate defined in §11.3.6. Show that this yields the same answer as that of a stiffness of a constant-area bar with cross section 1 (Ai + A j ). Note: the following Mathematica script may be used to solve this exercise:12 2 ClearAll[Le,x,Em,A,Ai,Aj]; Be={{-1,1}}/Le; Ζ=x/Le; A=Ai*(1-Ζ)+Aj*Ζ; Ke=Integrate[Em*A*Transpose[Be].Be,{x,0,Le}]; Ke=Simplify[Ke]; Print['Ke for varying cross section bar: ',Ke//MatrixForm];
In this and following scripts Le stands for . EXERCISE 11.3 [A:10] Using the area variation law (E11.1), find the consistent load vector fe for a bar
of constant area A subject to a uniform axial force q = ρg A per unit length along the element. Show that this vector is the same as that obtained with the element-by-element (EbE) “lumping” method of §8.4, which simply assigns half of the total load: 12 ρg A , to each node. EXERCISE 11.4 [A/C:15] Repeat the previous calculation for the tapered bar element subject to a force q = ρg A per unit length, in which A varies according to (E11.1) whereas ρ and g are constant. Check that if Ai = A j one recovers f i = f j = 12 ρg A . Note: the following Mathematica script may be used to solve this exercise:13
ClearAll[q,A,Ai,Aj,ρ,g,Le,x]; ζ =x/Le; Ne={{1-ζ ,ζ }}; A=Ai*(1-ζ )+Aj*ζ ; q=ρ*g*A; fe=Integrate[q*Ne,{x,0,Le}]; fe=Simplify[fe]; Print['fe for uniform load q: ',fe//MatrixForm]; ClearAll[A]; Print['fe check: ',Simplify[fe/.{Ai->A,Aj->A}]//MatrixForm]; EXERCISE 11.5 [A/C:20] A tapered bar element of length , end areas Ai and A j with A interpolated as
per (E11.1), and constant density ρ, rotates on a plane at uniform angular velocity ω (rad/sec) about node i. Taking axis x along the rotating bar with origin at node i, the centrifugal axial force is q(x) = ρ Aω2 x along the length, in which x ≡ x e . Find the consistent node forces as functions of ρ, Ai , A j , ω and , and specialize the result to the prismatic bar A = Ai = A j . Partial result check: f j = 13 ρω2 A 2 for A = Ai = A j . 12
13
The ClearAll[..] at the start of the script is recommended programming practice to initialize variables and avoid “cell crosstalk.” In a Module this is done by listing the local variables after the Module keyword. The ClearAll[A] before the last statement is essential; else A would retain the previous assignation.
11–14
11–15
Exercises
EXERCISE 11.6 [A:15] (Requires knowledge of Dirac’s delta function properties.) Find the consistent load
vector fe if the bar is subjected to a concentrated axial force Q at a distance x = a from its left end. Use Equation (11.31), with q(x) = Q δ(a), in which δ(a) is the one-dimensional Dirac’s delta function at x = a. Note: the following script does it by Mathematica, but it is overkill: ClearAll[Le,q,Q,a,x]; ζ =x/Le; Ne={{1-ζ ,ζ }}; q=Q*DiracDelta[x-a]; fe=Simplify[ Integrate[q*Ne,{x,-Infinity,Infinity}] ]; Print['fe for point load Q at x=a: ',fe//MatrixForm];
EXERCISE 11.7 [C+D:20] In a learned paper, Dr. I. M. Clueless proposes “improving” the result for the
example truss by putting three extra nodes, 4, 5 and 6, at the midpoint of members 1–2, 2–3 and 1–3, respectively. His “reasoning” is that more is better. Try Dr. C.’s suggestion using the Mathematica implementation of Chapter 4 and verify that the solution “blows up” because the modified master stiffness is singular. Explain physically what happens. EXERCISE 11.8 [A:35, close to research paper level]. Prove nodal exactness of the two-node bar element for arbitrary but Taylor expandable loading without using the Fourier series approach. Hints: expand q(x) = q(x j ) + ( ψ)q 0003 (x j ) + ( ψ)2 q 00030003 (x j )/2! + . . ., where ψ = x − x j is the distance to node j, compute the consistent force f j (x) from (11.26), and differentiate the MoDE (11.30) repeatedly in x while truncating all derivatives to a maximum order n ≥ 2. Show that the original ODE: E Au 00030003 + q = 0, emerges as an identity regardless of how many derivatives are kept.
11–15
12
.
Variational Formulation of Plane Beam Element
12–1
12–2
Chapter 12: VARIATIONAL FORMULATION OF PLANE BEAM ELEMENT
TABLE OF CONTENTS Page
§12.1. Introduction §12.2. What is a Beam? §12.2.1. Terminology . . . . . . . . . . . §12.2.2. Mathematical Models . . . . . . . . . §12.2.3. Assumptions of Classical Beam Theory . . . §12.3. The Bernoulli-Euler Beam Theory §12.3.1. Element Coordinate Systems . . . . . . . §12.3.2. Kinematics . . . . . . . . . . . . §12.3.3. Loading . . . . . . . . . . . . . . §12.3.4. Support Conditions . . . . . . . . . §12.3.5. Strains, Stresses and Bending Moments . . . §12.4. Total Potential Energy Functional §12.5. Beam Finite Elements §12.5.1. Finite Element Trial Functions . . . . . . §12.5.2. Shape Functions . . . . . . . . . . . §12.6. The Finite Element Equations §12.6.1. The Stiffness Matrix of a Prismatic Beam . . §12.6.2. Consistent Nodal Force Vector for Uniform Load §12. Notes and Bibliography . . . . . . . . . . . . . . . . §12. References. . . . . . . . . . . . . . . . . §12. Exercises . . . . . . . . . . . . . . . .
12–2
. . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . .
. . . . . . .
. . . . .
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
12–3 12–3 12–3 12–3 12–4 12–4 12–4 12–5 12–5 12–5 12–5 12–6 12–8 12–8 12–8 12–9 12–9 12–10 12–11 12–11 12–12
12–3
§12.2
WHAT IS A BEAM?
§12.1. Introduction The previous Chapter introduced the TPE-based variational formulation of finite elements, which was illustrated for the bar element. This Chapter applies that technique to a more complicated one-dimensional element: the plane beam described by engineering beam theory. Mathematically, the main difference of beams with respect to bars is the increased order of continuity required for the assumed transverse-displacement functions to be admissible. Not only must these functions be continuous but they must possess continuous x first derivatives. To meet this requirement both deflections and slopes are matched at nodal points. Slopes may be viewed as rotational degrees of freedom in the small-displacement assumptions used here. §12.2. What is a Beam? Beams are the most common type of structural component, particularly in Civil and Mechanical Engineering. A beam is a bar-like structural member whose primary function is to support transverse loading and carry it to the supports. See Figure 12.1. By “bar-like” it is meant that one of the dimensions is considerably larger than the other two. This dimension is called the longitudinal dimension or beam axis. The intersection of planes normal to the longitudinal dimension with the beam member are called cross sections. A longitudinal plane is one that passes through the beam axis.
Figure 12.1. A beam is a structural member designed to resist transverse loads.
A beam resists transverse loads mainly through bending action, Bending produces compressive longitudinal stresses in one side of the beam and tensile stresses in the other. The two regions are separated by a neutral surface of zero stress. The combination of tensile and compressive stresses produces an internal bending moment. This moment is the primary mechanism that transports loads to the supports. The mechanism is illustrated in Figure 12.2.
Neutral surface
Compressive stress
Tensile stress Figure 12.2. Beam transverse loads are primarily resisted by bending action.
§12.2.1. Terminology
A general beam is a bar-like member designed to resist a combination of loading actions such as biaxial bending, transverse shears, axial stretching or compression, and possibly torsion. If the internal axial force is compressive, the beam has also to be designed to resist buckling. If the beam is subject primarily to bending and axial forces, it is called a beam-column. If it is subjected primarily to bending forces, it is called simply a beam. A beam is straight if its longitudinal axis is straight. It is prismatic if its cross section is constant. A spatial beam supports transverse loads that can act on arbitrary directions along the cross section. A plane beam resists primarily transverse loading on a preferred longitudinal plane. This Chapter considers only plane beams. 12–3
Chapter 12: VARIATIONAL FORMULATION OF PLANE BEAM ELEMENT
12–4
§12.2.2. Mathematical Models One-dimensional mathematical models of structural beams are constructed on the basis of beam theories. Because beams are actually three-dimensional bodies, all models necessarily involve some form of approximation to the underlying physics. The simplest and best known models for straight, prismatic beams are based on the Bernoulli-Euler beam theory (also called classical beam theory and engineering beam theory), and the Timoshenko beam theory. The Bernoulli-Euler theory is that taught in introductory Mechanics of Materials courses, and is the one emphasized in this Chapter. The Timoshenko beam model is presented in Chapter 13, which collects advanced material. Both models can be used to formulate beam finite elements. The Bernoulli-Euler beam theory leads to the so-called Hermitian beam elements.1 These are also known as C 1 elements for the reason explained in §12.5.1. This model neglects transverse shear deformations. Elements based on Timoshenko beam theory, also known as C 0 elements, incorporate a first order correction for transverse shear effects. This model assumes additional importance in dynamics and vibration. §12.2.3. Assumptions of Classical Beam Theory The Bernoulli-Euler or classical beam theory for plane beams rests on the following assumptions: 1.
Planar symmetry. The longitudinal axis is straight, and the cross section of the beam has a longitudinal plane of symmetry. The resultant of the transverse loads acting on each section lies on this plane.
2.
Cross section variation. The cross section is either constant or varies smoothly.
3.
Normality. Plane sections originally normal to the longitudinal axis of the beam remain plane and normal to the deformed longitudinal axis upon bending.
4.
Strain energy. The internal strain energy of the member accounts only for bending moment deformations. All other contributions, notably transverse shear and axial force, are ignored.
5.
Linearization. Transverse deflections, rotations and deformations are considered so small that the assumptions of infinitesimal deformations apply.
6.
Material model. The material is assumed to be elastic and isotropic. Heterogeneous beams fabricated with several isotropic materials, such as reinforced concrete, are not excluded.
§12.3. The Bernoulli-Euler Beam Theory §12.3.1. Element Coordinate Systems Under transverse loading one of the beam surfaces shortens while the other elongates; see Figure 12.2. Therefore a neutral surface exists between the top and the bottom that undergoes no elongation or contraction. The intersection of this surface with each cross section defines the neutral axis of that cross section.2 1
The qualifier “Hermitian” relates to the use of a interpolation formula studied by the French mathematician Hermite. The term has nothing to do with the beam model used.
2
If the beam is homogenous, the neutral axis passes through the centroid of the cross section. If the beam is fabricated of different materials — for example, a reinforced concrete beam — the neutral axes passes through the centroid of an “equivalent” cross section. This topic is covered in Mechanics of Materials textbooks; for example Popov [133].
12–4
12–5
§12.3
y, v
THE BERNOULLI-EULER BEAM THEORY
q(x)
y, v Beam cross section
x, u Neutral surface
z
Neutral axis Symmetry plane
L Figure 12.3. Terminology in Bernoulli-Euler model of plane beam.
The Cartesian axes for plane beam analysis are chosen as shown in Figure 12.3. Axis x lies along the longitudinal beam axis, at neutral axis height. Axis y lies in the symmetry plane and points upwards. Axis z is directed along the neutral axis, forming a RHS system with x and y. The origin is placed at the leftmost section. The total length (or span) of the beam member is called L. §12.3.2. Kinematics The motion under loading of a plane beam member in the x, y plane is described by the two dimensional displacement field 0003 0002 u(x, y) , (12.1) v(x, y) where u and v are the axial and transverse displacement components, respectively, of an arbitrary beam material point. The motion in the z direction, which is primarity due to Poisson’s ratio effects, is of no interest. The normality assumption of the Bernoulli-Euler model can be represented mathematically as u(x, y) = −y
∂v(x) = −yv 0004 = −yθ, ∂x
v(x, y) = v(x).
(12.2)
Note that the slope v 0004 = ∂v/∂ x = dv/d x of the deflection curve has been identified with the rotation symbol θ. This is permissible because θ represents to first order, according to the kinematic assumptions of this model, the rotation of a cross section about z positive CCW. §12.3.3. Loading The transverse force per unit length that acts on the beam in the +y direction is denoted by q(x), as illustrated in Figure 12.3. Concentrated loads and moments acting on isolated beam sections can be represented by the delta function and its derivative. For example, if a transverse point load F acts at x = a, it contributes Fδ(a) to q(x). If the concentrated moment C acts at x = b, positive CCW, it contributes Cδ 0004 (b) to q(x), where δ 0004 denotes a doublet acting at x = b. §12.3.4. Support Conditions Support conditions for beams exhibit far more variety than for bar members. Two canonical cases are often encountered in engineering practice: simple support and cantilever support. These are illustrated in Figures 12.4 and 12.5, respectively. Beams often appear as components of skeletal structures called frameworks, in which case the support conditions are of more complex type. 12–5
12–6
Chapter 12: VARIATIONAL FORMULATION OF PLANE BEAM ELEMENT
;;;;;;; ;;;;;;; ;;;;;;; ;;;;;;;
Figure 12.5. A cantilever beam is clamped at one end and free at the other. Airplane wings and stabilizers are examples of this configuration.
Figure 12.4. A simply supported beam has end supports that preclude transverse displacements but permit end rotations.
§12.3.5. Strains, Stresses and Bending Moments The Bernoulli-Euler or classical model assumes that the internal energy of beam member is entirely due to bending strains and stresses. Bending produces axial stresses σx x , which will be abbreviated to σ , and axial strains ex x , which will be abbreviated to e. The strains can be linked to the displacements by differentiating the axial displacement u(x) of (12.2): e=
d 2v ∂ 2v ∂u = −y 2 = −y 2 = −yv 00040004 = −yκ. ∂x ∂x dx
(12.3)
Here κ denotes the deformed beam axis curvature, which to first order is κ ≈ d 2 v/d x 2 = v 00040004 . The bending stress σ = σx x is linked to e through the one-dimensional Hooke’s law σ = Ee = −E y
d 2v = −E yκ, dx2
(12.4)
where E is the longitudinal elastic modulus. The most important stress resultant in classical beam theory is the bending moment M, which is defined as the cross section integral 0004
d 2v −yσ d x = E 2 M= dx A
0004
0005 Here I ≡ Izz denotes the moment of inertia A y 2 d A of the cross section with respect to the z (neutral) axis. M is considered positive if it compresses the upper portion: y > 0, of the beam cross section, as illustrated in Figure 12.6. This explains the negative sign of y in the integral (12.5). The product E I is called the bending rigidity of the beam with respect to flexure about the z axis.
y 2 d A = E I κ.
(12.5)
A y
M x
z
V Figure 12.6. Positive sign convention for M and V .
The governing equations of the Bernoulli-Euler beam model are summarized in the Tonti diagram of Figure 12.7. 12–6
12–7
§12.4
Prescribed end displacements
Displacement BCs
TOTAL POTENTIAL ENERGY FUNCTIONAL
Transverse displacements
Distributed transverse load
v(x)
q(x)
Kinematic
κ = v'
M'=q
Equilibrium
Curvature
M = EI κ
κ(x)
Bending moment
Constitutive
M(x)
Prescribed end loads
Force BCs
Figure 12.7. The Tonti diagram for the governing equations of the Bernoulli-Euler beam model.
§12.4. Total Potential Energy Functional The total potential energy of the beam is 000e=U −W
(12.6)
where as usual U and W denote the internal and external energies, respectively. As previously explained, in the Bernoulli-Euler model U includes only the bending energy: 0004 U=
1 2
0004 σe dV = V
1 2
L
0004 Mκ d x =
0
1 2
L
0004 E Iκ dx = 2
0
1 2
L
0006 00072 E I v 00040004 d x =
0
0004 1 2
L
v 00040004 E I v 00040004 d x.
0
(12.7) The external work W accounts for the applied transverse force: 0004 W =
L
qv d x.
(12.8)
0
The three functionals 000e, U and W must be regarded as depending on the transverse displacement v(x). When this dependence needs to be emphasized we write 000e[v], U [v] and W [v]. Note that 000e[v] includes up to second derivatives in v, because v 00040004 = κ appears in U . This number is called the variational index. Variational calculus tells us that since the index is 2, admissible displacements v(x) must be continuous, have continuous first derivatives (slopes or rotations), and satisfy the displacement BCs exactly. This continuity requirement can be succintly stated by saying that admissible displacements must be C 1 continuous. This condition guides the construction of beam finite elements described below. Remark 12.1. If there is an applied distributed moment m(x) per unit of beam length, the external energy
0005L
(12.8) must be augmented with a 0 m(x)θ(x) d x term. This is further elaborated in Exercises 12.4 and 12.5. Such kind of distributed loading is uncommon in practice although in framework analysis occasionally the need arises for treating a concentrated moment between nodes.
12–7
12–8
Chapter 12: VARIATIONAL FORMULATION OF PLANE BEAM ELEMENT
§12.5. Beam Finite Elements Beam finite elements are obtained by subdividing beam members longitudinally. The simplest Bernoulli-Euler plane beam element has two end nodes: 1 and 2, and four degrees of freedom (DOF). These are collected in the node displacement vector ue = [ v1 θ1 v2 θ2 ]T .
θ1
v2 v1 1
(12.9)
The element is shown in Figure 12.8, which pictures the undeformed and deformed configurations.
θ2
P'(x+u,y+v) y, v
E, I
2 x
000f
x, u
P(x,y)
Figure 12.8. The two-node Bernoulli-Euler plane beam element with four DOFs.
§12.5.1. Finite Element Trial Functions The freedoms (12.9) are used to define uniquely the variation of the transverse displacement v e (x) over the element. The C 1 continuity requirement says that both v(x) and the slope θ = v 0004 (x) = dv(x)/d x must be continuous over the entire member, and in particular between beam elements. C 1 continuity can be trivially satisfied within each element by choosing polynomial interpolation shape functions as shown below, as those are C 1 continuous. Matching nodal displacements and rotations with adjacent elements enforces the necessary interelement continuity. §12.5.2. Shape Functions
v1 = 1
The simplest shape functions that meet the C 1 continuity requirement for the nodal DOF configuration (12.9) are called the Hermitian cubic shape functions. The interpolation formula based on these functions is   v1 e e e e  θ1  v e = [ Nv1 Nθ1 Nv2 Nθ2 ]   = Ne ue . v2 θ2 (12.10) These shape functions are conveniently expressed in terms of the dimensionless “natural” coordinate 2x − 1, (12.11) ξ= 000f
Nv1e (ξ) θ1 = 1 Nθ1e (ξ) Nv2e (ξ)
v2 = 1 θ2 = 1
ξ = −1
Nθ2e (ξ)
ξ=1
Figure 12.9. Cubic shape functions of plane beam element.
where 000f is the element length. Coordinate ξ varies from ξ = −1 at node 1 (x = 0) to ξ = +1 at node 2 (x = 000f). Note that d x/dξ = 12 000f and dξ/d x = 2/000f. The shape functions in terms of ξ are e Nv1 = 14 (1 − ξ )2 (2 + ξ ),
Nθe1 = 18 000f(1 − ξ )2 (1 + ξ ),
e = 14 (1 + ξ )2 (2 − ξ ), Nv2
Nθe2 = − 18 000f(1 + ξ )2 (1 − ξ ). 12–8
(12.12)
12–9
§12.6
THE FINITE ELEMENT EQUATIONS
These four functions are depicted in Figure 12.9. The curvature κ that appears in U can be expressed in terms of the nodal displacements by differentiating twice with respect to x: d 2 v e (x) 4 d 2 v e (ξ ) 4 dNe e κ= = 2 = 2 2 u = B ue = N00040004 ue . dx2 000f dξ 2 000f dξ
(12.13)
Here B = N00040004 is the 1 × 4 curvature-displacement matrix B=
1000e ξ 6000f 000f
ξ −6 000f
3ξ − 1
000f 3ξ + 1 .
(12.14)
Remark 12.2. The 4/000f2 factor in (12.13) comes from the differentiation chain rule. If f (x) is a function of x,
and ξ = 2x/000f − 1, noting that d(2/000f)/d x = 0 one gets
0001
0 0010 0011 d f (ξ ) d(2/000f) d f (ξ ) 2 d 4 d 2 f (ξ ) d 2 f (x) = = + . (12.15) dx2 dx dξ 000f dx dξ 000f2 dξ 2
d f (ξ ) dξ 2 d f (ξ ) d f (x) = = , dx dξ d x 000f dξ
§12.6. The Finite Element Equations Insertion of (12.12) and (12.14) into the TPE functional specialized to this element, yields the quadratic form in the nodal displacements 000ee = 12 (ue )T Ke ue − (ue )T fe , where
0004
000f
K = e
0004 E I B B dx =
1
E I BT B 12 000f dξ,
(12.17)
NT q 12 000f dξ,
(12.18)
T
−1
0
(12.16)
is the element stiffness matrix and 0004 f = e
000f
0004 N q dx =
1
T
0
−1
is the consistent element node force vector. The calculation of the entries of Ke and fe for prismatic beams and uniform load q is studied next. More complex cases are treated in the Exercises. §12.6.1. The Stiffness Matrix of a Prismatic Beam If the bending rigidity E I is constant over the element it can be moved out of the ξ -integral in (12.17):  6ξ  0004 1 0004 1 000f  000f E I  3ξ − 1  000e 6ξ −6ξ dξ. (12.19) BT B dξ = Ke = 12 E I 000f  −6ξ  3ξ − 1 3ξ + 1 000f  000f 2000f −1  −1 000f 3ξ + 1 12–9
12–10
Chapter 12: VARIATIONAL FORMULATION OF PLANE BEAM ELEMENT
ClearAll[EI,l,ξ]; B={{6*ξ,(3*ξ-1)*l,-6*ξ,(3*ξ+1)*l}}/l^2; Ke=(EI*l/2)*Integrate[Ne,{ξ,-1,1}]; Ke=Simplify[Ke]; Print['Ke for prismatic beam:n',Ke//MatrixForm]; Ke for prismatic beam: 12 EI 6 EI _ 12 EI l3 l l3 6 EI 4 EI _ 6 EI l2 l l2 _ 12 EI _ 6 EI 12 EI l3 l2 l3 6 EI 2 EI _ 6 EI l2 l l2
_
ClearAll[q,l,ξ]; Ne={{2*(1-ξ)^2*(2+ξ), (1-ξ)^2*(1+ξ)*l, 2*(1+ξ)^2*(2-ξ),−(1+ξ)^2*(1-ξ)*l}}/8; fe=(q*l/2)*Integrate[Ne,{ξ,-1,1}]; fe=Simplify[fe]; Print['fe^T for uniform load q:n',fe//MatrixForm];
6 EI l2 2 EI l 6 EI l2 4 EI l
fe^T for uniform load q: lq l 2q l q _ l 2q 2 12 2 12
Figure 12.10. Using Mathematica to form Ke for a prismatic beam element.
Figure 12.11. Using Mathematica to form fe for uniform transverse load q.
Expanding and integrating over the element yields     −36ξ 2 6ξ(3ξ +1)000f 36ξ 2 6ξ(3ξ −1)000f 12 6000f −12 6000f 0004 1 2 2 2 2  EI (3ξ −1) 000f −6ξ(3ξ −1)000f (9ξ −1)000f  4000f2 −6000f 2000f2    dξ = E I  Ke = 3   12 −6000f 36ξ 2 −6ξ(3ξ +1)000f  2000f −1  000f3 symm 4000f2 symm (3ξ +1)2 000f2 (12.20) Although the foregoing integrals can be easily carried out by hand, it is equally expedient to use a CAS such as Mathematica or Maple. For example the Mathematica script listed in the top box of Figure 12.10 processes (12.20) using the Integrate function. The output, shown in the bottom box, corroborates the hand integration result. §12.6.2. Consistent Nodal Force Vector for Uniform Load If q does not depend on x it can be moved out of (12.18), giving    1 (1 − ξ )2 (2 + ξ ) 4 0004 1 0004 1 1   2   8 000f(1 − ξ ) (1 + ξ )  e T 1 1 dξ = q000f N dξ = 2 q000f f = 2 q000f    1 2  −1 −1  4 (1 + ξ ) (2 − ξ )  − 18 000f(1 + ξ )2 (1 − ξ )
 1 2 1 000f   12 . 1  2 1 − 12 000f
(12.21)
This shows that a uniform load q over the beam element maps to two transverse node loads q000f/2, as may be expected, plus two nodal moments ±q000f2 /12. The latter are called the fixed-end moments in the FEM literature. The hand result (12.21) can be verified with the Mathematica script of Figure 12.11, in which fe is printed as a row vector to save space. Example 12.1. To see the beam element in action consider the cantilever beam illustrated in Figure 12.12(a), which is subjected to two load cases. Case I: an end moment M; case (II): a transverse end force P. The beam is prismatic with constant rigidity E I , span L, and discretized with a single element. The FEM equations are constructed using the stiffness matrix (12.20) with 000f = L. The forces at end node 2 are directly set from the given loads. Applying the support conditions v1 = θ1 = 0 gives the reduced stiffness equations
E I 0012 12 L 3 −6L
−6L 4L 2
00130012
v2I θ2I
0013
=
0012
0013
0 , M
E I 0012 12 L 3 −6L
12–10
−6L 4L 2
00130012
v2I I θ2I I
0013
=
0012
0013
P . 0
(12.22)
12–11
§12.
y,v
EI constant
Load case II
M x
(b)
y,v
EI constant
q
P
;
;; ;; 1
Load case I
1
2
x
2
L L 1 = L ( 12 +α)
L
;; ;;
(a)
References
3
L 2 = L ( 12 −α)
Figure 12.12. Beam problems for Examples 12.1 and 12.2.
Solving: v2I = M L 2 /(2E I ), θ2I = M L/E I , v2I I = P L 3 /(3E I ) and θ2I I = P L 2 /(2E I ). These are the analytical values provided by BE theory. Thus a one-element idealization is sufficient for exactness. The reason is that the analytical deflection profiles v(x) are quadratic and cubic polynomials in x for cases I and II, respectively. Both are included in the span of the element shape functions. Example 12.2. The second example is a simply supported beam under uniform line load q, depicted in
Figure 12.12(b). It is prismatic with constant rigidity E I , span L, and discretized with two elements of length L 1 = L(/ + α) and L 2 = L − L 1 = L(/ − α), respectively. (Ordinarily two elements of the same length / L would be used; the scalar α ∈ (−/, /) is introduced to study the effect of unequal element sizes.) Using (12.20) and (12.21) to form the stiffness and consistent forces for both elements, assembling and applying the support conditions v1 = v3 = 0, provides the reduced stiffness equations

8L 2 1 + 2α   −24L 2 EI   (1 + 2α) 2  L 3  4L  1 + 2α 0
−24L (1 + 2α)2 192(1 + 12α 2 ) (1 − 4α 2 )3 192Lα (1 − 4α 2 )2 24L (1 − 2α)2
4L 2 1 + 2α 192Lα (1 − 4α 2 )2 16L 2 1 − 4α 2 4L 2 1 − 2α

 L(1 + 2α)2 48 θ    1 24L 1    v2  (1 − 2α)2  2  = q L    θ2  4L 2   − Lα  1 − 2α  θ3 6 0
8L 2 1 − 2α
−
    .  
L(1 − 2α)2 48
(12.23) Solving for the transverse displacement of node 2 gives v2 = q L 4 (5 − 24α 2 + 16α 4 )/(384E I ). The exact deflection profile is v(x) = q L 4 (ζ − 2ζ 3 + ζ 4 )/(24E I ) with ζ = x/L. Replacing x = L 1 = L(/ + α) yields v2exact = q L 4 (5 − 24α 2 + 16α 4 )/(384E I ), which is the same as the FEM result. The result seems prima facie surprising. First, since the analytical solution is a quartic in x we have no reason to think that a cubic element will be exact. Second, one would expect accuracy deterioration as the element sizes differ more and more with increasing α. The fact that the solution at nodes is exact for any combination of element lengths is an illustration of superconvergence, a phenomenon already discussed in §12.5. A general proof of nodal exactness is carried out in §13.7, but it does require advanced mathematical tools. Notes and Bibliography A comprehensive source of stiffness and mass matrices of plane and spatial beams is the book by Przemieniecki [136]. The derivation of stiffness matrices there is carried out using differential equilibrium equations rather than energy methods. This was in fact the common practice before 1962, as influenced by the use of transfer matrix methods on small memory computers [126]. Results for prismatic elements, however, are identical. Energy derivations were popularized by Archer [6,7], Martin [111] and Melosh [116,117]. References Referenced items have been moved to Appendix R.
12–11
Chapter 12: VARIATIONAL FORMULATION OF PLANE BEAM ELEMENT
12–12
Homework Exercises for Chapter 12 Variational Formulation of Plane Beam Element EXERCISE 12.1 [A/C:20] Use (12.17) to derive the element stiffness matrix Ke of a Hermitian beam element
of variable bending rigidity given by the inertia law I (x) = I1 (1 −
x x ) + I2 = I1 12 (1 − ξ ) + I2 12 (1 + ξ ). 000f 000f
(E12.1)
Use of Mathematica or similar CAS tool is recommended since the integrals are time consuming and error prone. Mathematica hint: write (E12.2)
EI = EI1*(1-ξ )/2 + EI2*(1+ξ )/2;
and keep EI inside the argument of Integrate. Check whether you get back (12.20) if EI=EI1=EI2. If you use Mathematica, this check can be simply done after you got and printed the tapered beam Ke, by writing ClearAll[EI]; Ke=Simplify[ Ke/.{EI1->EI,EI2->EI}]; and printing this matrix.3 EXERCISE 12.2 [A/C:20] Use (12.18) to derive the consistent node force vector fe for a Hermitian beam
element under linearly varying transverse load q defined by q(x) = q1 (1 −
x x ) + q2 = q1 12 (1 − ξ ) + q2 12 (1 + ξ ). 000f 000f
(E12.3)
Again use of a CAS is recommended, particularly since the polynomials to be integrated are quartic in ξ , and hand computations are error prone. Mathematica hint: write (E12.4)
q = q1*(1-ξ )/2 + q2*(1+ξ )/2;
and keep q inside the argument of Integrate. Check whether you get back (12.21) if q1 = q2 = q (See previous Exercise for Mathematica procedural hints). EXERCISE 12.3 [A:20] Obtain the consistent node force vector fe of a Hermitian beam element subject to
a transverse point load P at abscissa x = a where 0 ≤ a ≤ 000f. Use the Dirac’s delta function expression 0005000f q(x) = P δ(a) and the fact that for any continuous function f (x), 0 f (x) δ(a) d x = f (a) if 0 ≤ a ≤ 000f. Check the special cases a = 0 and a = 000f. EXERCISE 12.4 [A:25] Derive the consistent node force vector fe of a Hermitian beam element subject to a
linearly varying z-moment m per unit length, positive CCW, defined by the law m(x) = m 1 (1 − ξ )/2 + m 2 (1 + ξ )/2. Use the fact that the external work per unit length is m(x)θ(x) = m(x) v 0004 (x) = (ue )T (dN/d x)T m(x). For arbitrary m(x) show that this gives
0004
000f
f = e
0
∂NT m dx = ∂x
0004
1 −1
∂NT 2 1 m 000f dξ = ∂ξ 000f 2
0004
1 −1
NξT m dξ,
(E12.5)
where NξT denote the column vectors of beam shape function derivatives with respect to ξ . Can you see a shortcut that avoids the integral altogether if m is constant? EXERCISE 12.5 [A:20] Obtain the consistent node force vector fe of a Hermitian beam element subject to
a concentrated moment (“point moment”, positive CCW) C applied at x = a. Use the expression (E12.5) in which m(x) = C δ(a), where δ(a) denotes the Dirac’s delta function at x = a. Check the special cases a = 0, a = 000f and a = 000f/2. 3
ClearAll[EI] discards the previous definition (E12.2) of EI; the same effect can be achieved by writing EI=. (dot).
12–12
12–13
Exercises
EXERCISE 12.6 [A/C:25] Consider the one-dimensional Gauss integration rules.4
0004
1
. f (ξ ) dξ = 2 f (0).
(E12.6)
√ √ . f (ξ ) dξ = f (−1/ 3) + f (1/ 3).
(E12.7)
0014 5 0014 8 . 5 f (ξ ) dξ = f (− 3/5) + f (0) + f ( 3/5). 9 9 9
(E12.8)
One point : −1
0004
1
Two points:
0004
−1 1
Three points: −1
Try each rule on the monomial integrals
0004
0004
1
1
ξ dξ,
dξ, −1
0004
1 −1
ξ 2 dξ, . . .
(E12.9)
−1
until the rule fails. In this way verify that rules (E12.6), (E12.7) and (E12.8) are exact for polynomials of degree up to 1, 3 and 5, respectively. (Labor-saving hint: for odd monomial degree no computations need to be done; why?). EXERCISE 12.7 [A/C:25] Repeat the derivation of Exercise 12.1 using the two-point Gauss rule (E12.7) to
evaluate integrals in ξ . A CAS is recommended. If using Mathematica you may use a function definition to 00051 save typing. For example to evaluate −1 f (ξ ) dξ in which f (ξ ) = 6ξ 4 − 3ξ 2 + 7, by the 3-point Gauss rule (E12.8), say f[ξ ]:=6ξ ^4-3ξ ^2+7; int=Simplify[(5/9)*(f[-Sqrt[3/5]]+f[Sqrt[3/5]])+(8/9)*f[0]]; and print int. To form an element by Gauss integration define matrix functions in terms of ξ , for example Be[ξ ], or use the substitution operator /., whatever you prefer. Check whether one obtains the same answers as with analytical integration, and explain why there is agreement or disagreement. Hint for the explanation: consider the order of the ξ polynomials you are integrating over the element. EXERCISE 12.8 [A/C:25] As above but for Exercise 12.2. EXERCISE 12.9 [A/C:30] Derive the Bernoulli-Euler beam stiffness matrix (12.20) using the method of
differential equations. To do this integrate the homogeneous differential equation E I v 0004000400040004 = 0 four times over a cantilever beam clamped at node 1 over x ∈ [0, 000f] to get v(x). The process yields four constants of integration C1 through C4 , which are determined by matching the two zero-displacement BCs at node 1 and the two force BCs at node 2. This provides a 2 × 2 flexibility matrix relating forces and displacements at node j. Invert to get a deformational stiffness, and expand to 4 × 4 by letting node 1 translate and rotate. EXERCISE 12.10 [C:20] Using Mathematica, repeat Example 12.2 but using EbE lumping of the distributed force q. (It is sufficient to set the nodal moments on the RHS of (12.23) to zero.) Is v2 the same as the exact analytical solution? If not, study the ratio v2 /v2exact as function of α, and draw conclusions.
4
Gauss integration is studied further in Chapter 17.
12–13
13
.
Advanced One-Dimensional Elements
13–1
13–2
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
TABLE OF CONTENTS Page
§13.1. Introduction §13.2. Generalized Interpolation §13.2.1. Legendre Polynomials . . . . . . . . . . . §13.2.2. Generalized Stiffnesses . . . . . . . . . . . §13.2.3. Transforming to Physical Freedoms: BE Model . . . §13.2.4. Transforming to Physical Freedoms: Shear-Flexible Model §13.2.5. Hinged Plane Beam Element . . . . . . . . . §13.2.6. Timoshenko Plane Beam Element . . . . . . . . §13.2.7. Beam on Elastic Supports . . . . . . . . . . §13.3. Interpolation with Homogeneous ODE Solutions §13.3.1. Exact Winkler/BE-Beam Stiffness . . . . . . . . §13.4. Equilibrium Theorems §13.4.1. Self-Equilibrated Force System . . . . . . . . §13.4.2. Handling Applied Forces . . . . . . . . . . . §13.4.3. Flexibility Equations . . . . . . . . . . . . §13.4.4. Rigid Motion Injection . . . . . . . . . . . . §13.4.5. Applications . . . . . . . . . . . . . . . §13.5. Flexibility Based Derivations §13.5.1. Timoshenko Plane Beam-Column . . . . . . . . §13.5.2. Plane Circular Arch in Local System . . . . . . . §13.5.3. Plane Circular Arch in Global System . . . . . . . §13.6. *Accuracy Analysis §13.6.1. *Accuracy of Bernoulli-Euler Beam Element . . . . §13.6.2. *Accuracy of Timoshenko Beam Element . . . . . §13. Notes and Bibliography . . . . . . . . . . . . . . . . . . . §13. References. . . . . . . . . . . . . . . . . . . . §13. Exercises . . . . . . . . . . . . . . . . . . .
13–2
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13–3 13–3 13–3 13–4 13–4 13–5 13–5 13–6 13–7 13–8 13–9 13–13 13–13 13–14 13–15 13–16 13–17 13–17 13–18 13–19 13–22 13–24 13–24 13–26 13–27 13–28 13–29
13–3 §13.1.
§13.2 GENERALIZED INTERPOLATION
Introduction
This Chapter develops special one-dimensional elements, such as thick beams, arches and beams on elastic foundations, that require mathematical and modeling resources beyond those presented in Chapters 11–12. The techniques used are less elementary,1 and may be found in books on Advanced Mechanics of Materials, e.g. [194,195]. Readers are expected to be familiar with ordinary differential equations and energy methods. The Chapter concludes with beam accuracy analysis based on the modified equation method. All of this Chapter material would be normally bypassed in an introductory finite element course. It is primarily provided for offerings at an intermediate level, for example a first graduate FEM course in Civil Engineering. Such courses may skip most of Part I as being undergraduate material. §13.2.
Generalized Interpolation
For derivation of special and C 0 beam elements it is convenient to use a transverse-displacement cubic interpolation in which the nodal freedoms v1 , v2 , θ1 and θ2 are replaced by generalized coordinates c1 to c4 : v(ξ ) = Nc1 c1 + Nc2 c2 + Nc3 c3 + Nc4 c4 = Nc c.
(13.1)
Here Nci (ξ ) are generalized shape functions that satisfy the completeness requirement discussed in Chapter 19. Nc is a 1×4 matrix whereas c is a column 4-vector. Formula (13.1) is a generalized interpolation. It includes the Hermite interpolation (12.10–12.12) as an instance when c1 = v1 , c2 = v10005 , c3 = v2 and c4 = v20005 .
1
L3 −1
L4
0.5
6
L1
4
L2 dL2 /dξ
−0.5
0.5 −0.5
1 −1 dL1 /dξ
−1
15
dL 4 /dξ dL3 /dξ
d2L3 /dξ 2
2
−0.5
−1 0.5
1
−2
−0.5
10
d 2L 4 /dξ2
5 −5 −10 −15
0.5
1
d 2L1 /dξ 2 and d 2L 2 /dξ 2
Figure 13.1. The Legendre polynomials and their first two ξ -derivatives plotted over ξ ∈[−1, 1]. Rigid body modes (L 1 and L 2 ) in black. Deformational modes (L 3 and L 4 ) in color.
§13.2.1. Legendre Polynomials An obvious generalized interpolation is the ordinary cubic polynomial v(ξ ) = c1 + c2 ξ + c3 ξ 2 + c4 ξ 3 , but this turns out not to be particularly useful. A more seminal expression is v(ξ ) = L 1 c1 + L 2 c2 + L 3 c3 + L 4 c4 = L c,
(13.2)
where the L i are the first four Legendre polynomials2 L 1 (ξ ) = 1,
L 2 (ξ ) = ξ,
L 3 (ξ ) = 12 (3ξ 2 − 1),
L 4 (ξ ) = 12 (5ξ 3 − 3ξ ).
(13.3)
1
They do not reach, however, the capstone level of Advanced Finite Element Methods.
2
This is a FEM oriented nomenclature; L 1 through L 4 are called P0 through P3 in mathematical handbooks; e.g. Chapter 22 of Abramowitz and Stegun [1]. They are normalized as per the usual conventions: L i (1) = 1, L i (−1) = (−1)i−1 . One important application in numerical analysis is the construction of one-dimensional Gauss integration rules: the abscissas of the p-point rule are the zeros of L p+1 (ξ ).
13–3
13–4
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
Here c1 through c4 have dimension of length. Functions (13.3) and their first two ξ -derivatives are plotted in Figure 13.1. Unlike the shape functions (12.12), the L i have a clear physical meaning: Translational rigid body mode. Rotational rigid body mode. Constant-curvature deformation mode, symmetric with respect to ξ = 0. Linear-curvature deformation mode, antisymmetric with respect to ξ = 0.
L1 L2 L3 L4
These properties are also shared by the standard polynomial c1 + c2 ξ + c3 ξ 2 + c4 ξ 3 . What distinguishes the set (13.3) are the orthogonality properties
0002
0002
Q0 =
L L d x = diag [ 1 1/3 1/5 1/7 ] ,
0002
0
(L00050005 )T L00050005 d x =
0
Q3 =
Q2 =
T
(L000500050005 )T L000500050005 d x =
0
14400 diag [ 0 0 0 1 ] , 5
in which
(.)0005 ≡
48 diag [ 0 0 3 25 ] , 3
d(.) . dx
(13.4) Qn is called the covariance matrix for the n th derivative of the Legendre polynomial interpolation. The 0003 first-derivative covariance Q1 = 0 (L0005 )T L0005 d x is not diagonal, but this matrix is not used here. §13.2.2. Generalized Stiffnesses The beam stiffness matrix expressed in terms of the ci is called a generalized stiffness. Denote the beam bending and shear rigidities by R B and R S , respectively. Then Kc = KcB + KcS , where KcB comes from the bending energy and KcS from the shear energy. For the latter is its assumed that the mean shear distortion γ at a cross section is γ = ϒ 2 v 000500050005 , where ϒ is a dimensionless coefficient that depends on the mean-shear model used. Then 0002 0002
KcB =
R B (L00050005 )T L00050005 d x,
R S ϒ 2 4 (L000500050005 )T L000500050005 d x.
KcS =
0
(13.5)
0
In the case of a Bernoulli-Euler (BE) beam, the shear contribution is dropped: Kc = KcB . Furthermore if the element is prismatic, R B = E I is constant. If so KcS = Rb Q2 = E I Q2 , where Q2 is the second diagonal matrix in (13.4). With view to future use it is convenient to differentiate between symmetric and antisymmetric bending rigidities R Bs and R Ba , which are associated with the responses to modes L 3 and L 4 , respectively. Assuming R Bs and R Ba to be uniform along the element we get Kc = KcBs + KcBa ,
KcBs =
144R Bs diag [ 0 0 1 0 ] , 3
KcBa =
1200R Ba diag [ 0 0 0 1 ] , 3
(13.6)
If shear flexibility is accounted for, the contribution KcS of (13.5) is kept. Assuming R S to be constant over the element, Kc is split into 3 contributions (two bending and one shear): Kc = KcBs + KcBa + KcS ,
with
KcS = R S ϒ 2 4 Q3 =
14400R S ϒ 2 diag [ 0 0 0 1 ] .
(13.7)
§13.2.3. Transforming to Physical Freedoms: BE Model For a BE beam model, the generalized coordinates ci of (13.2) can be connected to the physical DOFs by



v1 1  θ1   0 v  = 1 1 θ1 0


−1 1 −1 c1 2/ −6/ 12/   c2  , 1 1 1   c3  c4 2/ 6/ 12/



c1 30 1  c2   −36 c  =  0 60 3 c4 6
13–4
5 −3 −5 3
30 36 0 −6


−5 v1 −3   θ1  . −5   v2  θ2 3
(13.8)
13–5
§13.2 GENERALIZED INTERPOLATION
0005 0005 In compact form: ue = G B c and c = H B ue , with H B = G−1 B . Here θ1 ≡ v1 and θ2 ≡ v2 , which reflects the fundamental “plane sections remain plane” kinematic assumption of the BE model. The physical stiffness is

Ke = HTB KcBs
12Ra 1  6Ra + KcBa H B = 3  −12Ra 6Ra
6Ra (3Ra + Rs ) 2 −6Ra (3Ra − Rs ) 2
−12Ra −6Ra 12Ra −6Ra

6Ra (3Ra − Rs ) 2  . −6Ra  2 (3Ra + Rs )
(13.9)
If Rs = Ra = E I the well known stiffness matrix (12.20) is recovered, as can be expected. The additional freedom conferred by (13.9) is exhibited later in two unconventional applications. §13.2.4. Transforming to Physical Freedoms: Shear-Flexible Model A shear flexible beam has mean shear distortion γ = ϒ 2 v 000500050005 . If ϒ is constant and v(ξ ) interpolated by (13.2), v 000500050005 = 120 c4 / 3 . Thus γ = 120ϒc4 / is constant over the element. The end rotational freedoms become θ1 = v10005 + γ and θ2 = v20005 + γ . Using = 12ϒ to simplify the algebra, the transformations (13.8) change to
       30 5 30 −5 1 −1 c1 c1 v 3 36+30 − 3  1  c2   c2  1  − 36+30 − − 6 12+10 1+ 1+ 1+ 1+  θ1    ,   =     c3 1 1 c3 60  0 −5 0 5  v2 6 12+10 6 6 3 3 c4 c4 θ2 − 1+ 1+ − 1+ 1+ (13.10) e e u = H u . Transforming K of (13.7) to physical freedoms yields In compact form, ue = G S c and c = G−1 S c S the stiffness used to construct the Timoshenko beam element in §13.2.6: 

 1 v1 0  θ1   v =  1 2 θ2 0
−1 2 1 2

0 R 0  Bs Ke = HTS Kc H S = 0 0
0 0 1 0 0 0 −1 0


0 4 −1  12R Ba + R S 2 2  2 +  −4 0 4 3 (1 + )2 2 1
2 2 −2 2

−4 2 −2 2  . (13.11) 4 −2  −2 2
y
§13.2.5. Hinged Plane Beam Element The two-node prismatic plane BE beam element depicted in Figure 13.2 has a mechanical hinge at midspan (ξ = 0). The cross sections on both sides of the hinge can rotate respect to each other. The top figure also sketches a fabrication method sometimes used in short-span pedestrian bridges. Gaps on either side of the hinged section cuts are filled with a bituminous material that permits slow relative rotations. Both the curvature κ and the bending moment M must vanish at midspan. But in a element built via cubic interpolation of v(x), κ = v 00050005 must vary linearly in both ξ and x. Consequently the mean curvature, which is controlled by the Legendre function L 2 (shown in blue on Figure 13.1) must be zero.
z
y
hinge
x
2 x
1 /2
/2
Figure 13.2. Beam element with hinge located at midspan. The top figure sketches a hinge fabrication method.
The kinematic constraint of zero mean curvature is enforced by setting the symmetric bending rigidity R Bs = 0 whereas the antisymmetric bending rigidity is the normal one: R Ba = E I . Plugging into (13.9) yields

4 3E I  2 Ke = 3  −4 2
2 2 −2 2
−4 −6 4 −2



6 2 2 3E I 3    = 3 [ 2 2 − ]  . −2  2  2 −
13–5
(13.12)
13–6
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
This matrix has rank one, as it can be expected from the last (dyadic) expression in (13.12). Ke has one nonzero eigenvalue: 6E I (4 + 2 )/ 3 , and three zero eigenvalues. The eigenvector associated with the nonzero eigenvalue pertains3 to the antisymmetric deformational mode L 3 . Matrix (13.12) can be derived by more sophisticated methods (e.g., mixed variational principles) but the present technique is the most expedient one. Example 13.1. This example deals with the prismatic
y,v (a)
EI constant
x (1) αL
2
3
; ;
1
q0
(2)
hinge L/2
L/2
L
(b)
The beam is discretized with two elements: (1) and (2), going from 1 to 2 and 2 to 3, respectively, as shown in the figure. The stiffnesses for elements (1) and (2) are those of (12.20) and (13.12), respectively, whereas (12.21) is used to build the consistent node forces for both elements.
;;
continuous beam shown in Figure 13.3(a). This has two spans with lengths αL and L, respectively, where α is a design parameter, and is subjected to uniform load q0 . The beam is free, simple supported and fixed at nodes 1, 2 and 3, respectively. There is a hinge at the center of the 2–3 span. Of interest is the design question: for which α > 0 is the tip deflection at end 1 zero?
2
1
3 α = 0.362
Figure 13.3. Beam problem for Example 13.1. (a): beam problem, (b) deflection profile when α = 0.362.
Assembling and applying the support conditions v2 = v3 = θ3 = 0, provides the reduced stiffness equations
EI L3
12/α 3 6L/α 2 6L/α 2
6L/α 2 4L 2 /α 2L 2 /α
6L/α 2 2L 2 /α 2 L (4 + 3α)/α
v1 θ1 θ2
=
q0 L 2
α Lα 2 /6 L(1 − α 2 )/6
.
(13.13)
Solving for the deflection at 1 gives v1 = q0 L 4 α(12α 2 +9α 3 −2)/(72E I ), θ1 = q0 L 3 (1−6α 2 −6α 3 )/(36E I ) and θ2 = q0 L 3 (1 − 6α 2 )/(36E I ). The equation v1 = 0 is quartic in α and has four roots, which to 8 places are α1 = −1.17137898, α2 = −0.52399776, α3 = 0, and α4 = 0.3620434. Since the latter is the only positive root, the solution is α = 0.362. The deflection profile for this value is pictured in Figure 13.3(b). §13.2.6. Timoshenko Plane Beam Element As observed in §12.2.2, the Timoshenko beam model [160] includes a first order correction for transverse shear flexibility. The key kinematic assumption changes to “plane sections remain plane but not necessarily normal to the deformed neutral surface.” This is illustrated in Figure 13.4(a) for a 2-node plane beam element. The cross section rotation θ differs from v 0005 by γ . Ignoring axial forces, the displacement field is analogous to that of the Bernoulli-Euler model (12.2) but with a shear correction: u(x, y) = −yθ,
v(x, y) = v(x),
with
θ=
∂v + γ = v0005 + γ , ∂x
γ =
V . G As
(13.14)
Here V is the transverse shear force, γ the “shear rotation” (positive CCW) averaged over the cross section, G the shear modulus and As the effective shear area.4 The product R S = G As is the shear rigidity. . To correlate def with the notation of §13.2.4, note that V = E I v 000500050005 , γ = ϒ 2 v 000500050005 = V /(G As ), so ϒ = E I /(G As 2 ) and
= 12ϒ =
12E I G A s 2
(13.15)
3
Compare the vector in the last expression in (13.12) to the last row of H B in (13.8).
4
A concept defined in Mechanics of Materials; see e.g. Chapter 10 of Popov [133] or Chapter 12 of Timoshenko and Goodier [162]. As is calculated by equating the internal shear energy 12 V γ = 12 V 2 /(G As ) to that produced by the shear stress distribution over the cross section. For a thin rectangular cross section and zero Poisson’s ratio, As = 5A/6.
13–6
13–7
§13.2 GENERALIZED INTERPOLATION
(a)
y, v θ1 γ1
v'1 =|dv/dx|1
θ2
γ
(b)
γ2
Deformed cross section
(c) γ
y M
v'2 =|dv/dx|2
v1
v2
v(x)
1
2
x, u
V Positive M, V conventions
V(+)
x
z
A positive transverse shear force V produces a CCW rotation (+γ) of the beam cross section
Figure 13.4. Two-node Timoshenko plane beam element: (a) kinematics (when developed with cubic shape functions, γ1 = γ2 = γ ); (b) M and V sign conventions; (c) concurrence of sign conventions for V and γ .
This dimensionless ratio characterizes the “shear slenderness” of the beam element. It is not an intrinsic beam property because it involves the element length. As → 0 the Timoshenko model reduces to the BE model. Replacing R Bs = R Ba = E I and R S = G As = 12E I /( 2 ) into (13.11) yields the Timoshenko beam stiffness   12 6 −12 6 EI  6 2 (4 + ) −6 2 (2 − )  Ke = 3 (13.16)  −6 12 −6  (1 + ) −12 6 2 (2 − ) −6 2 (4 + ) If = 0 this reduces to (12.20). The Mathematica module TimoshenkoBeamStiffness[Le,EI, ], listed in Figure 13.5, implements (13.16). TimoshenkoBeamStiffness[Le_,EI_,0001_]:=Module[{Ke}, Ke=EI/(Le*(1+0001))*{{ 12/Le^2, 6/Le,-12/Le^2, 6/Le }, { 6/Le , 4+0001, -6/Le , 2-0001 }, {-12/Le^2, -6/Le, 12/Le^2,-6/Le}, { 6/Le , 2-0001, -6/Le , 4+0001 }}; Return[Ke]];
Figure 13.5. Module to produce stiffness matrix for Timoshenko beam element.
The calculation of the consistent node forces for uniform transverse load is covered in Exercise 13.2. A hinged Timoshenko beam is constructed in Exercise 13.3. §13.2.7. Beam on Elastic Supports Sometimes beams, as well as other structural members, may be supported elastically along their span. Two common configurations that occur in structural engineering are: (i)
Beam resting on a continuum medium such as soil. This is the case in foundations.
(ii)
Beam supported by discrete but closely spaced flexible supports, as in the “bed of springs” pictured in Figure 13.6. This occurs in railbeds (structurally rails are beams supported by crossties) and some types of grillworks.
The Winkler foundation is a simplified elastic-support model. It is an approximation for (i) because it ignores multidimensional elasticity effects as well as friction. It is a simplification of (ii) because the discrete nature of supports is smeared out. It is nonetheless popular, particularly in foundation and railway engineering, when the presence of inherent uncertainties would not justify a more complicated model. Such uncertainties are inherent in soil mechanics.
13–7
13–8
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
The Winkler model may be viewed as a “continuification” of case (ii). Take a beam slice going from x to x + d x. The spring-reaction force acting on the beam is taken to be d f F = −k F v(x) d x. Here v(x) is the transverse deflection and k F the Winkler foundation stiffness, which has dimension of force per length-squared. Force d f F has the opposite sign of v(x), pushing up if the beam moves down and pulling down if it moves up. Beam-foundation separation effects that may occur in case (i) are ignored here because that would lead to a nonlinear contact problem.
y, v
q(x)
beam
x
;;;;;;;;; bed of springs
Figure 13.6. A beam supported by a bed of springs. Continuification of this configuration leads to the Winkler foundation model treated in this subsection.
The internal energy stored in the d x slice of Winkler springs is −/ v d f F = / k F v 2 d x. Consequently the effect of elastic supports is to modify the internal energy U Be of the beam element so that it becomes
0002 U = e
U Be
+
U Fe ,
with
U Fe
=
k F v 2 d x.
1 2
(13.17)
0
Therefore the total stiffness of the element is computed by adding the foundation stiffness to the beam stiffness. Care must be taken, however, that the same set of nodal freedoms is used. This is best handled by doing the generalized stiffness KcF first, and then using the appropriate generalized-to-physical transformation. If the transverse deflection v is interpolated with (13.2) as v = L c, the generalized Winkler foundation stiffness for constant k F is
0002
KcF = k F
L T L d x = k F Q0 ,
(13.18)
0
where Q0 is the first diagonal matrix in (13.4). This holds regardless of beam model. Now if the member resting on the foundation is modeled as a BE beam, one picks H B of (13.8) as generalized-to-physical transformation matrix to get

156 k F  22 e T K F = k F H B Q0 H B =  54 420 −13
22 4 2 13 −3 2
54 13 156 −22

−13 −3 2  , −22  4 2
(13.19)
If instead the supported member is modeled as a Timoshenko beam, one picks H S of (13.10) to get KeF = k F HTS Q0 H S

4(78+147 +70 2 ) kF  (44+77 +35 2 ) =  4(27+63 +35 2 ) 2 840(1+ ) − (26+63 +35 2 )
(44+77 +35 2 ) 2 (8+14 +7 2 ) (26+63 +35 2 ) − 2 (6+14 +7 2 )

4(27+63 +35 2 ) − (26+63 +35 2 ) (26+63 +35 2 ) − 2 (6+14 +7 2 )  4(78+147 +70 2 ) − (44+77 +35 2 )  − (44+77 +35 2 ) 2 (8+14 +7 2 ) (13.20) The module TimoshenkoWinklerStiffness[Le,kF, ] listed in Figure 13.7 implements the stiffness (13.20). To get the BE-beam Winkler stiffness (13.19), invoke with = 0. Examples of use of this module are provided in §13.3.1.
13–8
13–9
§13.3 INTERPOLATION WITH HOMOGENEOUS ODE SOLUTIONS TimoshenkoWinklerStiffness[Le_,kF_,Φ_]:=Module[{KeW}, KeW={{4*(78+147*Φ+70*Φ^2), Le*(44+77*Φ+35*Φ^2), 4*(27+63*Φ+35*Φ^2), -Le*(26+63*Φ+35*Φ^2)}, {Le*(44+77*Φ+35*Φ^2), Le^2*(8+14*Φ+7*Φ^2), Le*(26+63*Φ+35*Φ^2), -Le^2*(6+14*Φ+7*Φ^2)}, {4*(27+63*Φ+35*Φ^2), Le*(26+63*Φ+35*Φ^2), 4*(78+147*Φ+70*Φ^2), -Le*(44+77*Φ+35*Φ^2)}, {-Le*(26+63*Φ+35*Φ^2),-Le^2*(6+14*Φ+7*Φ^2), -Le*(44+77*Φ+35*Φ^2), Le^2*(8+14*Φ+7*Φ^2)}}* kF*Le/(840*(1+Φ)^2); Return[KeW]];
Figure 13.7. Stiffness matrix module for a Winkler foundation supporting a Timoshenko beam element.
§13.3.
Interpolation with Homogeneous ODE Solutions
For both BE and Timoshenko beam models, the Legendre polynomials L 1 (ξ ) through L 4 (ξ ) are exact solutions of the homogeneous, prismatic, plane beam equilibrium equation E I d 4 v/d x 4 = 0. When used as shape functions in the generalized interpolation (13.2), the resulting stiffness matrix is exact if the FEM model is loaded at the nodes, as further discussed in §13.6. The technique can be extended to more complicated onedimensional problems. It can be used to derive exact stiffness matrices if homogeneous solutions are available in closed form, and are sufficiently simple to be amenable to analytical integration. The following subsection illustrates the method for a BE beam resting on a Winkler elastic foundation. §13.3.1. Exact Winkler/BE-Beam Stiffness Consider again a prismatic, plane BE beam element resting on a Winkler foundation of stiffness k F , as pictured in Figure 13.6. The governing equilibrium equation for constant E I > 0 and k F > 0 is E I d 4 v/d x 4 + k F v = q(x). The general homogeneous solution over an element of length going from x = 0 to x = is
ζ
v(x) = e c1 sin ζ + c2 cos ζ + e
−ζ
c3 sin ζ + c4 cos ζ , with ζ = χ x/ and χ =
000e 4
kF . (13.21) 4E I
Here the ci are four integration constants to be determined from four end conditions: the nodal degrees of freedom v1 , v10005 , v2 and v20005 . These constants are treated as generalized coordinates and as before collected into vector c = [ c1 c2 c3 c4 ]T . The solution (13.21) is used as generalized interpolation with eζ sin ζ through e−ζ cos ζ as the four shape functions. Differentiating twice gives v 0005 = dv/d x and v 00050005 = d 2 v/d x 2 . The TPE functional of the element in terms of the generalized coordinates can be expressed as
0002 0014ec
=
1
E I (v 00050005 )2 + 12 k F v 2 − c0 v d x = 12 cT (KcB + KcF ) c − cT fc . 2
(13.22)
0
This defines KcB and KcF as generalized element stiffnesses due to beam bending and foundation springs, respectively, whereas fc is the generalized force associated with a transverse load q(x). The nodal freedoms are linked to generalized coordinates by





0 1 0 1 v1 c1 χ / χ / χ / −χ /  θ1     c2  v  =  c . eχ cos χ e−χ sin χ e−χ cos χ eχ sin χ 1 3 χ χ −χ −χ χ e (cos χ+ sin χ ) χ e (cos χ − sin χ ) χ e (cos χ − sin χ ) −χe (cos χ + sin χ ) θ1 c4 (13.23) In compact form this is ue = G F c. Inverting gives c = H F ue with H F = G−1 F . The physical stiffness is Ke = KeB + KeF with KeB = HTF KcB H F and KeF = HTF KcF H F . The consistent force vector is fe = HTF fc . Computation with transcendental functions by hand is unwieldy and error-prone, and at this point it is better
13–9
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
13–10
ClearAll[EI,kF,Le,χ,q0,x]; g=2-Cos[2*χ]-Cosh[2*χ]; Nf={Exp[ χ*x/Le]*Sin[χ*x/Le],Exp[ χ*x/Le]*Cos[χ*x/Le], Exp[-χ*x/Le]*Sin[χ*x/Le],Exp[-χ*x/Le]*Cos[χ*x/Le]}; Nfd=D[Nf,x]; Nfdd=D[Nfd,x]; KgF=kF*Integrate[Transpose[{Nf}].{Nf},{x,0,Le}]; KgB=EI*Integrate[Transpose[{Nfdd}].{Nfdd},{x,0,Le}]; fg= q0*Integrate[Transpose[{Nf}],{x,0,Le}]; {KgF,KgB,fg}=Simplify[{KgF,KgB,fg}]; Print['KgF=',KgF//MatrixForm]; Print['KgB=',KgB//MatrixForm]; GF=Simplify[{Nf/.x->0,Nfd/.x->0,Nf/.x->Le,Nfd/.x->Le}]; HF=Simplify[Inverse[GF]]; HFT=Transpose[HF]; Print['GF=',GF//MatrixForm]; Print['HF=',HF//MatrixForm]; facB=(EI*χ/Le^3)/(4*g^2); facF=(kF*Le)/(16*χ^3*g^2); KeB=Simplify[HFT.KgB.HF]; KeBfac=FullSimplify[KeB/facB]; Print['KeB=',facB,' * ',KeBfac//MatrixForm] KeF=Simplify[HFT.KgF.HF]; KeFfac=FullSimplify[KeF/facF]; Print['KeF=',facF,' * ',KeFfac//MatrixForm]; facf=(q0*Le)/(χ^2*g); fe=Simplify[ExpToTrig[HFT.fg]]; fefac=Simplify[fe/facf]; Print['fe=',facf,' * ',fefac];
Figure 13.8. Script to produce the exact Winkler-BE beam stiffness matrix and consistent force vector.
BEBeamWinklerExactStiffness[Le_,EI_,kF_,q0_]:=Module[{B1,B2,B3,B4,B5,B6, F1,F2,F3,F4,F5,F6,f1,f2,facB,facF,facf,KeB,KeF,fe,χ}, χ=PowerExpand[Le*((kF/(4*EI))^(1/4))]; B1 =2*χ^2*(-4*Sin[2*χ]+Sin[4*χ]+4*Sin[χ]*(Cos[χ]*Cosh[2*χ]+ 8*χ*Sin[χ]*Sinh[χ]^2)+2*(Cos[2*χ]-2)*Sinh[2*χ]+Sinh[4*χ]); B2 =2*Le*χ*(4*Cos[2*χ]-Cos[4*χ]-4*Cosh[2*χ]+ Cosh[4*χ]8*χ*Sin[2*χ]*Sinh[χ]^2+8*χ*Sin[χ]^2*Sinh[2*χ]); B3 =-(Le^2*(8*χ*Cos[2*χ]-12*Sin[2*χ]+Cosh[2*χ]*(6*Sin[2*χ]-8*χ)+3*Sin[4*χ]+ 2*(6-3*Cos[2*χ]+4*χ*Sin[2*χ])*Sinh[2*χ]-3*Sinh[4*χ])); B4 =-4*Le*χ*(χ*Cosh[3*χ]*Sin[χ]-χ*Cosh[χ]*(-2*Sin[χ]+Sin[3*χ])+(χ*(Cos[χ]+ Cos[3*χ])+Cosh[2*χ]*(-2*χ*Cos[χ]+4*Sin[χ])+2*(-5*Sin[χ]+Sin[3*χ]))*Sinh[χ]); B5 =-4*χ^2*(2*Cos[χ]*(-2+Cos[2*χ]+Cosh[2*χ])*Sinh[χ]+Sin[3*χ]* (Cosh[χ]-2*χ*Sinh[χ])+Sin[χ]*(-4*Cosh[χ]+Cosh[3*χ]+2*χ*Sinh[3*χ])); B6 =2*Le^2*(Cosh[3*χ]*(-2*χ*Cos[χ]+3*Sin[χ])+Cosh[χ]*(2*χ*Cos[3*χ]+3*(Sin[3*χ]4*Sin[χ]))+(9*Cos[χ]-3*Cos[3*χ]-6*Cos[χ]*Cosh[2*χ]+16*χ*Sin[χ])*Sinh[χ]); F1 =2*χ^2*(-32*χ*Sin[χ]^2*Sinh[χ]^2+6*(-2+Cos[2*χ])* (Sin[2*χ]+Sinh[2*χ])+6*Cosh[2*χ]*(Sin[2*χ]+Sinh[2*χ])); F2 =2*Le*χ*(4*Cos[2*χ]-Cos[4*χ]-4*Cosh[2*χ]+Cosh[4*χ]+ 8*χ*Sin[2*χ]*Sinh[χ]^2-8*χ*Sin[χ]^2*Sinh[2*χ]); F3 =Le^2*(8*χ*Cos[2*χ]+4*Sin[2*χ]-2*Cosh[2*χ]*(4*χ+Sin[2*χ])-Sin[4*χ]+ 2*(Cos[2*χ]+4*χ*Sin[2*χ]-2)*Sinh[2*χ]+Sinh[4*χ]); F4 =4*Le*χ*(χ*Cosh[3*χ]*Sin[χ]-χ*Cosh[χ]*(-2*Sin[χ]+Sin[3*χ])+(χ*Cos[χ]+ χ*Cos[3*χ]+10*Sin[χ]-2*Cosh[2*χ]*(χ*Cos[χ]+2*Sin[χ])-2*Sin[3*χ])*Sinh[χ]); F5 =-4*χ^2*(6*Cos[χ]*(-2+Cos[2*χ]+Cosh[2*χ])*Sinh[χ]+Sin[3*χ]* (3*Cosh[χ]+2*χ*Sinh[χ])+Sin[χ]*(-12*Cosh[χ]+3*Cosh[3*χ]-2*χ*Sinh[3*χ])); F6 =-2* Le^2*(-(Cosh[3*χ]*(2*χ*Cos[χ]+Sin[χ]))+Cosh[χ]*(2*χ*Cos[3*χ]+ 4*Sin[χ]-Sin[3*χ])+(Cos[3*χ]+Cos[χ]*(2*Cosh[2*χ]-3)+16*χ*Sin[χ])*Sinh[χ]); f1=2*χ*(Cosh[χ]-Cos[χ])*(Sin[χ]-Sinh[χ]); f2=-(Le*(Sin[χ]-Sinh[χ])^2); g=2-Cos[2*χ]-Cosh[2*χ]; facf=(q0*Le)/(χ^2*g); facB=(EI*χ/Le^3)/(4*g^2); facF=(kF*Le)/(16*χ^3*g^2); KeB=facB*{{B1,B2,B5,-B4},{ B2,B3,B4,B6},{B5,B4,B1,-B2},{-B4,B6,-B2,B3}}; KeF=facF*{{F1,F2,F5,-F4},{ F2,F3,F4,F6},{F5,F4,F1,-F2},{-F4,F6,-F2,F3}}; fe=facf*{f1,f2,f1,-f2}; Return[{KeB,KeF,fe}]];
Figure 13.9. Module to get the exact BE-Winkler stiffness and consistent load vector.
13–10
13–11
§13.3 INTERPOLATION WITH HOMOGENEOUS ODE SOLUTIONS
to leave that task to a CAS. The Mathematica script listed in Figure 13.8 is designed to produce KeB , KeF and fe for constant E I and k F , and uniform transverse load q(x) = q0 . The script gives






B1 B2 B5 −B4 F1 F2 F5 −F4 f1 E I χ k q B B B F F F f       F 0 3 4 6 3 4 6 , KeF = , fe = 2  2  , KeB = 3 2     3 2 B1 −B2 F1 −F2 f1 4 g 16χ g χ g symm B3 symm F3 − f2 (13.24) in which g = 2 − cos 2χ − cosh 2χ , B1 = 2χ 2 (−4 sin 2χ + sin 4χ +4 sin χ (cos χ cosh 2χ +8χ sin χ sinh2 χ )+2(cos 2χ −2) sinh 2χ + sinh 4χ ), B2 = 2 χ (4 cos 2χ − cos 4χ −4 cosh 2χ + cosh 4χ−8χ sin 2χ sinh2 χ +8χ sin2 χ sinh 2χ ), B3 = −( 2 (8χ cos 2χ−12 sin 2χ + cosh 2χ (6 sin 2χ −8χ )+3 sin 4χ + 2(6−3 cos 2χ +4χ sin 2χ) sinh 2χ −3 sinh 4χ )), B4 = −4 χ (χ cosh 3χ sin χ−χ cosh χ (−2 sin χ + sin 3χ )+(χ (cos χ + cos 3χ ) + cosh 2χ (−2χ cos χ +4 sin χ )+2(−5 sin χ+ sin 3χ )) sinh χ ), B5 = −4χ 2 (2 cos χ (−2+ cos 2χ + cosh 2χ ) sinh χ + sin 3χ (cosh χ−2χ sinh χ ) + sin χ(−4 cosh χ+ cosh 3χ +2χ sinh 3χ )) B6 = 2 2 (cosh 3χ (−2χ cos χ +3 sin χ )+ cosh χ (2χ cos 3χ +3(−4 sin χ+ sin 3χ )) +(9 cos χ −3 cos 3χ −6 cos χ cosh 2χ +16χ sin χ ) sinh χ ) F1 = 2χ 2 (−32χ sin2 χ sinh2 χ +6(−2+ cos 2χ )(sin 2χ + sinh 2χ )+6 cosh 2χ (sin 2χ + sinh 2χ)), F2 = 2 χ (4 cos 2χ − cos 4χ −4 cosh 2χ + cosh 4χ+8χ sin 2χ sinh2 χ −8χ sin2 χ sinh 2χ ), F3 = 2 (8χ cos 2χ+4 sin 2χ −2 cosh 2χ (4χ + sin 2χ )− sin 4χ +2(cos 2χ +4χ sin 2χ −2) sinh 2χ + sinh 4χ ), F4 = 4 χ (χ cosh 3χ sin χ −χ cosh χ (sin 3χ −2 sin χ )+(χ cos χ +χ cos 3χ +10 sin χ −2 cosh 2χ (χ cos χ+2 sin χ )−2 sin 3χ ) sinh χ ), F5 = −4χ 2 (6 cos χ (−2+ cos 2χ + cosh 2χ ) sinh χ + sin 3χ (3 cosh χ+2χ sinh χ ), + sin χ(−12 cosh χ +3 cosh 3χ −2χ sinh 3χ )) F6 = −2 2 (−(cosh 3χ(2χ cos χ + sin χ ))+ cosh χ (2χ cos 3χ +4 sin χ − sin 3χ ) +(cos 3χ + cos χ(2 cosh 2χ−3)+16χ sin χ ) sinh χ ), f 1 = 2χ(cosh χ − cos χ )(sin χ − sinh χ ),
f 2 = − (sin χ − sinh χ )2 .
(13.25) These expressions are used to code module BEBeamWinklerExactStiffness[Le,EI,kF,q0], which is listed in Figure 13.9. Example 13.2. A fixed-fixed BE beam rests on a Winkler foundation as shown in Figure 13.10. The beam has span 2L, and constant E I . The Winkler foundation coefficient k F is constant. As usual in foundation engineering we set (13.26) k F = E I λ4 /L 4 ,
where λ is a dimensionless rigidity to be kept as parameter.5 The beam is subjected to two load cases: (I) a central point load P at x = L, and (II) a uniform line load q0 over the right half x ≥ L. See Figure 13.10. 5
Note that λ is a true physical parameter, whereas χ is discretization dependant because it involves the element length.
13–11
13–12
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
;;;;;;;;;;; ;;;;;;;;;;; ;;;;;;;;;;; ;;;;;;;;;;; y, v
P : Load case (I) q0 : Load case (II)
EI constant
A
C
x
B
k F constant
L
L
2L
Figure 13.10. Example: fixed-fixed beam on Winkler elastic foundation.
All quantities are kept symbolic. The focus of interest is the deflection vC at midspan C (x = L). For I II and vCI I (λ) = C I I (λ)vC0 for load convenience this is rendered dimensionless by taking vCI (λ) = C I (λ)vC0 I II 3 4 cases (I) and (II)), respectively. Here vC0 = −P L /(24E I ) and vC0 = −q0 L /(48E I ) are the midspan deflections of cases (I) and (II) for λ = 0, that is, k F = 0 (no foundation). The exact deflection factors for this model are √ √ √ 13 4 6 2 cos 2λ + cosh 2λ − 2 137 8 C I (λ) = 3 =1− λ + λ .. √ √ λ 420 138600 sin 2λ + sinh 2λ √ √ √ √ 163 4 48 (cos λ/ 2− cosh λ/ 2)(sin λ/ 2− sinh λ/ 2) 20641 8 =1− λ + λ + .. C I I (λ) = 4 √ √ λ 5040 19958400 sin 2λ + sinh 2λ (13.27) Both load cases were symbolically solved with two exact elements of length L produced by the module of Figure 13.9. As can be expected, the answers reproduce the exact solutions (13.27). Using any number of those elements would match (13.27) as long as the midspan section C is at a node. Then both cases were solved with 2, 4, and 8 elements with the stiffness (13.19) produced by cubic polynomials. The results are shown in a log-log plot in Figure 13.11. Results for selected values of λ are presented in Table 13.1. As can be seen, for a “soft” foundation characterized by λ < 1, the cubic-polynomial elements gave satisfactory results and converged quickly to the exact answers, especially in load case (II). As λ grows over one, the deflections become rapidly smaller, and the polynomial FEM results exhibit higher relative errors. On the other hand, the absolute errors remain small. The conclusion is that exact elements are only worthwhile in highly rigid foundations (say λ > 5) and then only if results with small relative error are of interest. 0
0 −1
log10 CI Load Case I: Central Point Load
−2
log10 CII
−1 2 exact elements
Load Case II: Uniform Line Load Over Right Half
−2 −3
−3 −4
−4
2 poly elements 4 poly elements 8 poly elements
−5 −6 −1
−0.5
0
0.5
1
log 10 λ 1.5
2 exact elements
2 poly elements 4 poly elements 8 poly elements
−5 −6 2
log 10 λ −1
−0.5
0
0.5
1
1.5
Figure 13.11. Log-log plots of C I (λ) and C I I (λ) for Example of Figure 13.10 over range λ ∈ [0.1, 100].
13–12
2
13–13
§13.4 EQUILIBRIUM THEOREMS Table 13.1 - Results for Example of Figure 13.10 at Selected λ Values
Load case (I): Central Point Load λ exact C I Ne = 2 Ne = 4 Ne = 8 0.1 0.999997 0.999997 0.999997 0.999997 1 0.970005 0.969977 0.970003 0.970005 0.672185 0.668790 0.671893 0.672167 2 0.067652 0.049152 0.065315 0.067483 5 10 0.008485 0.003220 0.006648 0.008191 100 8.48×10−6 3.23×10−7 8.03×10−7 1.63×10−6
Load case (II): Line Load Over Right Half exact C I I Ne = 2 Ne = 4 Ne = 8 0.999997 0.999997 0.999997 0.999997 0.968661 0.969977 0.968742 0.968666 0.657708 0.668790 0.658316 0.657746 0.041321 0.049152 0.041254 0.041317 0.002394 0.003220 0.002393 0.002395 2.40×10−7 3.23×10−7 2.62×10−7 2.42×10−7
Remark 13.1. To correlate the exact stiffness and consistent forces with those obtained with polynomial shape
functions it is illuminating to expand (13.24) as power series in χ. The rationale is that as the element size 000f gets smaller, χ = 4 k F /(4E I ) goes to zero for fixed E I and k F . Mathematica gives the expansions KeB = K B0 +χ 8 K B8 +χ 12 K B12 +. . . , KeF = K F0 +χ 4 K F4 +χ 8 K F8 +. . . , fe = f0 +χ 4 f4 +. . . , (13.28) in which
KeB0
KeB12


25488 5352 23022 −5043 EI  5352 1136 2 5043 −1097 2  e = eqn (12.20) , K B8 =  23022 5043 25488 −5352  , 4365900 3 −5043 −1097 2 −5352 1136 2   528960 113504 522090 −112631 EI  113504 24384 2 112631 −24273 2  e =−  522090 112631 528960 −113504  , K F0 = eqn (13.19), 3 5959453500 −112631 −24273 2 −113504 24384 2
k F 4 e 3k F 4 e q0 q0 K B8 , KeF8 = − K , fe = [ 6 6 − ]T , fe4 = − [ 14 3 14 −3 ]T . 2E I 8E I B12 0 12 5040 (13.29) Thus as χ → 0 we recover the stiffness matrices and force vector derived with polynomial shape functions, as can be expected. Note that K B0 and K F0 decouple, which allows them to coded as separate modules. On the other hand the exact stiffnesses are coupled if χ > 0. The foregoing expansions indicate that exactness makes little difference if χ < 1. KeF4 = −
§13.4.
Equilibrium Theorems
One way to get high performance mechanical elements is to use equilibrium conditions whenever possible. These lead to flexibility methods. Taking advantage of equilibrium is fairly easy in one space dimension. It is more difficult in two and three, because it requires advanced variational methods that are beyond the scope of this book. This section surveys theorems that provide the theoretical basis for flexibility methods. These are applied to 1D element construction in §13.5. §13.4.1. Self-Equilibrated Force System First we establish a useful theorem that links displacement and force transformations. Consider a FEM discretized body such as that pictured in Figure 13.12(a). The generic potato intends to symbolize any discretized material body: an element, an element assembly or a complete structure. Partition its degrees of freedom into two types: r and s. The s freedoms (s stands for suppressed or supported) are associated with a minimal set of supports that control rigid body motions or RBMs. The r freedoms (r is for released) collect the rest. In the
13–13
13–14
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
(b)
(a)
δus Pr fr
fs
δur
fs
Pr
Ps
Ps
fr
Figure 13.12. Body to illustrate equilibrium theorems. Nodal freedoms classified into supported (s) and released (r ), each lumped to a point to simplify diagram. (a) Self equilibrated node force system. (b) Force system of (a) undergoing virtual displacements; grossly exaggerated for visibility.
figure those freedoms are shown collected at invididual points Ps and Pr for visualization convenience. Node forces, displacements and virtual displacements associated with those freedoms are partitioned accordingly. Thus 0010 0011 0010 0011 0010 0011 fs us δus f= , u= , δu = . (13.30) fr ur δur The dimension n s of fs , us and δus is 1, 3 and 6 in one-, two- and three-dimensional space, respectively. Figure 13.12(b) shows the force system {fs , fr } undergoing virtual displacements, which are exaggerated for visibility.6 Consider now the “rigid + deformational” displacement decomposition u = Gus + d, in which matrix G (of appropriate order) represents a rigid motion and d are deformational displacements. Evaluating this at the r freedoms gives δur = Gr δus + δdr , (13.31) ur = Gr us + dr , The first decomposition in (13.31), being linear in the actual displacements, is only valid only in geometrically linear analysis. That for virtual displacements is valid for a much broader class of problems. If the supported freedom motion vanishes: us = 0, then ur = dr . Thus dr represents a relative displacement of the unsupported freedoms with respect to the rigid motion Gus , and likewise for the virtual displacements. Because a relative motion is necessarily associated with deformations, the alternative name deformational displacements is justified. The external virtual work is δW = δWs + δWr = δusT fs + δurT fr . If the force system in Figure 13.12(a) is in self equilibrium and the virtual displacements are imparted by rigid motions δdr = 0 and δur = Gr δus , the virtual work must vanish: δW = δusT fs + δusT GrT fr = δusT (fs + GrT fr ) = 0. Because the δus are arbitrary, it follows that fs = −GrT fr . (13.32) fs + GrT fr = 0, These are the overall static equilibrium equations of a discrete mechanical system in self equilibrium. Sometimes it is useful to express the foregoing expressions in the complete-vector form u=
0010
us ur
0011
=
0010
0011
0010
0011
0010
I 0 δus us + , δu = Gr dr δur
0011
=
0010
0011
0010
0011
0010 0011
I 0 f δus + , f= s Gr δdr fr
=
0010
−GrT I
0011
fr . (13.33)
Relations in (13.33) are said to be reciprocal.7 Remark 13.2. The freedoms in us are virtual supports, chosen for convenience in flexibility derivations. They should not be confused with actual or physical supports. For instance Civil Engineering structures tend to have redundant physical supports, whereas aircraft or orbiting satellites have none. 6
Under virtual displacements the forces are frozen for application of the Principle of Virtual Work.
7
If the model is geometrically nonlinear, the first form in (13.33) does not hold.
13–14
13–15
§13.4 EQUILIBRIUM THEOREMS (b)
(a) Ps
qr
qs
Pr
Ps
Pr Pq
t(x)
q
(c)
δus
δur
qs Ps
(d)
−qs
qr
Ps
Pr
δuq
−q r Pr
Pq
q Figure 13.13. Processing non-self-equilibrated applied loads with flexibility methods. (a) Body under applied distributed load. (b) Substitution by resultant and self equilibration. (c) Deriving overall equilibrium conditions through the PVW. (d) Replacing the applied loads by equivalent nodal forces.
§13.4.2. Handling Applied Forces Consider now a generalization of the previous scenario. An externally applied load system of surface or body forces, not necessarily in self equilibrium, acts on the body. For example, the surface tractions pictured in Figure 13.13(a). To bring this under the framework of equilibrium analysis, a series of steps are required. First, the force system is replaced by a single resultant q, as pictured in 13.13(b).8 The point of application is Pq . Equilibrium is restored by introducing node forces qr and qs at the appropriate freedoms. The overall equilibrium condition is obtained by putting the system {q, qr , qs } through rigid-motion virtual displacements, as pictured in Figure 13.13(c). Point Pq moves through δuq , and G evaluated at Pq is Gq . The virtual work is δW = δusT (qs + GrT qr + GqT q) = 0 whence qs + GrT qr + GqT q = 0.
(13.34)
If (13.34) is sufficient to determine qs and qr , the load system of Figure 13.13(a) can be effectively replaced by the nodal forces −qs and −qr , as depicted in 13.13(d). These are called the equivalent node forces. But in general (13.34) is insufficient to fully determine qs and qr . The remaining equations to construct the equivalent forces must come from a theorem that accounts for the internal energy, as discussed in §13.4.3. Adding (13.32) and (13.34) gives the general overall equilibrium condition fs + qs + GrT (fr + qr ) + GqT q = 0,
(13.35)
which is applied in §13.4.4 to the recovery of supported freedoms. Remark 13.3. The replacement of the applied force by a resultant is not strictly necessary, as it is always possible to write out the virtual work by appropriately integrating distributed effects. The resultant is primarily useful as an instructional tool, because matrix Gq is not position dependent. Remark 13.4. Conditions (13.32) and (13.34), which were derived through the PVW, hold for general mechanical systems under mild reversibility requirements [168, §231], including geometric nonlinearities. From now on we restrict attention to systems linear in the actual displacements. 8
Although the figure shows a resultant point force, in general it may include a point moment that is not shown for simplicity. See also Remark 13.3.
13–15
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
13–16
§13.4.3. Flexibility Equations The first step in FEM equilibrium analysis is obtaining discrete flexibility equations. The stiffness equations introduced in Chapter 2 relate forces to displacements. At the element level they are fe = Ke ue . By definition, flexibility equations relate displacements to forces: ue = Fe fe , where Fe is the element flexibility matrix. So the expectation is that the flexibility can be obtained as the inverse of the stiffness: Fe = (Ke )−1 . Right? Wrong. Recall that Ke for a disconnected free-free element is singular. Its ordinary inverse does not exist. Expectations go up in smoke. The same difficulty holds for a superelement or complete structure. To get a conventional flexibility matrix9 it is necessary to remove all rigid body motions in advance. This can be done through the virtual supports introduced in §13.4.1. The support motions us are fixed, say us = 0. Flexibility equations are sought between what is left of the kinematics. Dropping the element superscript for brevity, for a linear problem one gets (13.36) Frr fr = dr . Note that ur does not appear: only the deformational or relative displacements. To recover ur it is necessary to release the supports, but if that is naively done Frr ceases to exist. This difficulty is overcome in §13.4.4. There is another key difference with stiffness methods. The DSM assembly procedure covered in Chapter 3 (and extended in Chapter 25 to general structures) does not translate into a similar technique for flexibility methods. Although it is possible to assemble flexibilities of MoM elements, the technique is neither simple nor elegant. And it becomes dauntingly complex when tried on continuum-based elements [196]. So one of the main uses of flexibility equations today is as a stepping stone on the way to derive element stiffness equations, starting from (13.36). The procedural steps are explained in §13.4.4. But how should (13.36) be derived? There are several methods but only one, based on the Total Complementary Potential Energy (TCPE) principle of variational mechanics is described here. To apply TCPE, the complementary energy 0014∗ of the body must be be expressed as a function of the nodal forces fr . For fixed supports (us = 0) and a linear system, the functional can be expressed as 0014∗ (fr ) = U ∗ (fr ) − frT dr = 12 frT Frr fr + frT br − frT dr + 0014∗0 .
(13.37)
Here U ∗ is the internal complementary energy, also called the stress energy by many authors, e.g., [197], br is a term resulting from loading actions such as as thermal effects, body or surface forces, and 0014∗0 is independent of fr . Calculation of U ∗ in 1D elements involves expressing the internal forces (axial force, shear forces, bending moments, torque, etc.) in terms of fr from statics. Application examples are given in the next section.10 The TCPE principle states that 0014∗ is stationary with respect to variations in fr when kinematic compatibility is satisfied: ∂0014∗ = Frr fr + br − dr = 0, whence dr = Frr fr + br . (13.38) ∂fr By hypothesis the deformational flexibility Frr is nonsingular. Solving for fr gives the deformational stiffness equations −1 and qr = Krr br . (13.39) fr = Krr dr − qr , with Krr = Frr The matrix Krr is the deformational stiffness matrix, whereas qr is the equivalent load vector. 9
In the FEM literature it is often called simply the flexibility. The reason is that for a long time it was believed that getting a flexibility matrix required a supported structure. With the recent advent of the free-free flexibility (see Notes and Bibliography) it becomes necessary to introduce a “deformational” or “conventional” qualifier.
10
For 2D and 3D elements the process is more delicate and demands techniques, such as hybrid variational principles, that lie beyond the scope of this material.
13–16
13–17
§13.5 FLEXIBILITY BASED DERIVATIONS
§13.4.4. Rigid Motion Injection Suppose that Frr and qr of (13.39) have been found, for example from the TPCE principle (13.38). The goal is to arrive at the free-free stiffness equations, which are partitioned in accordance with (13.30) as
0010 0011 fs fr
0010
Kss = Kr s
00110010
Ksr Krr
us ur
0011
0010
0011
q − s , qr
(13.40)
To justify the presence of Krr and qr here, set us = 0, whence ur = dr . Consequently the second equation reduces to fr = Krr dr − qr , which matches (13.39). Inserting fs and fr into (13.35) yields
0012
0013
(Kss + GrT Kr s )us + (Ksr + GrT Krr )ur + qs + GrT qr + GqT q = 0, and replacing ur = Gr us + dr ,
0012
0013
0012
0013
0012
(13.41)
0013
Kss + GrT Kr s + Ksr Gr + GrT Krr Gr us + Ksr + GrT Krr dr + qs + GrT qr + GqT q = 0.
(13.42)
Because us , dr and q can be arbitrarily varied, each bracket in (13.42) must vanish identically, giving qs = −GrT qr − GqT q,
Ksr = −GrT Krr ,
T Kr s = Ksr = −Krr Gr ,
Kss = −GrT Kr s − Ksr Gr − GrT Krr Gr = GrT Krr Gr + GrT Krr Gr − GrT Krr Gr = GrT Krr Gr . Inserting these into (13.40) yields
0010 0011 fs fr
=
0010
This can be put in the more compact form
0010 0011
0010
0011
−GrT Krr Krr
GrT Krr Gr −Krr Gr
0010
0011
0010
fs −GrT u −GrT = Krr [ −Gr I ] s − fr I ur I with T = [ −Gr I ] and Tq = [ Gq 0 ] .
0011
00110010
qr +
us ur
0010
0011
−
0010
0011
GrT qr − GqT q . qr
(13.43)
(13.44)
0011
Gq q = TT Krr T u − TT qr + TqT q, 0 (13.45)
−1 T. Alternatively (13.45) may be derived by plugging The end result is that the free-free stiffness is TT Frr dr = ur − Gr us into (13.39) and then into (13.35).
Remark 13.5. Let S be a n s × n s nonsingular matrix. The matrix R built by the prescription
R=
0010
S GS
0011
(13.46)
is called a rigid body motion matrix or simply RBM matrix. The columns of R represent nodal values of rigid motions, hence the name. The scaling provided by S may be adjusted to make R simpler. The key property is T R = 0 and thus K R = TT Krr T R = 0. Other properties are studied in [58]. §13.4.5. Applications Stiffness Equilibrium Tests. If one injects ur = Gus and qr = 0 into (13.45) the result is fr = 0 and fs = 0. That is, all node forces must vanish for arbitrary us . This test is useful at any level (element, superelement, full structure) to verify that a directly generated K (that is, a K constructed independently of overall equilibrium) is “clean” as regards rigid body modes. Element Stiffness from Flexibility. Here Frr is constructed at the supported element level, inverted to get Krr and rigid motions injected through (13.45). Applications to element construction are illustrated in §13.5. Experimental Stiffness from Flexibility. In this case Frr is obtained through experimental measurements on a supported structure or substructure.11 To insert this as a “user defined superelement” in a DSM code, it is necessary to produce a stiffness matrix. This is done again by inversion and RBM injection. 11
The classical static tests on an airplane wing are performed by applying transverse forces and torques to the wing tip with the airplane safely on the ground. These experimental influence coefficients can be used for model validation.
13–17
13–18
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
(a)
fy1
m1
(b)
E,I,G,As constant q0 x
fx1 1
fy2
fy2 m2 fx2
q0
;;
y,v
1
2
m2 f x2
2
(c)
fy2 m2 f x2
q0
+V(x) +M(x)
2
+F(x) x
Figure 13.14. Flexibility derivation of Timoshenko plane beam-column stiffness: (a) element and node forces, (b) removal of RBMs by fixing left node, (c) FBD that gives internal forces at varying x.
§13.5.
Flexibility Based Derivations
The equilibrium theorems of the foregoing section are applied to the flexibility derivation of several onedimensional elements. §13.5.1. Timoshenko Plane Beam-Column A beam-column member combines axial and bending effects. A 2-node, straight beam-column has three DOFs at each node: the axial displacement, the transverse displacement and a rotation. If the cross section is doubly symmetric, axial and bending effects are decoupled. A prismatic, plane element of this kind is shown in Figure 13.14(a). End nodes are 1–2. The bending component is modeled as a Timoshenko beam. The element is subjected to the six node forces shown, and to a uniformly distributed load q0 . To suppress rigid motions node 1 is fixed as shown in Figure 13.14(b), making the beam a cantilever. Following the notation of §13.4.1–§13.4.2,
u x1 u y1 θ1
f x1 f y1 m1
V (x) = − f y2 − q0 ( − x),
f x2 f y2 m2
0 q 0
1 0 0 , dr = ur = , fs = , fr = , q= , G(x) = 0 1 x , us = 0 0 0 (13.47) Further, Gr = G( ) and Gq = G( /2). The internal forces are the axial force F(x), the transverse shear V (x) and the bending moment M(x). These are directly obtained from statics by doing a free-body diagram at distance x from the left end as illustrated in Figure 13.14(c). With the positive convention as shown we get N (x) = − f x2 ,
u x2 u y2 θ2
M(x) = m 2 + f y2 ( − x) + 12 q0 ( − x)2 .
(13.48)
Useful check: d M/d x = V . Assuming a doubly symmetric section so that N and M are decoupled, the element TCPE functional is
0002 0014
0015
N2 M2 V2 d x − frT dr = 12 frT Frr fr + frT br − frT dr + 0014∗0 , + + 0014 = EA EI G As 0     0 0 0 E A 3   ∂ 2 0014∗ 3 (4 + ) 2  , b = q  (4 + )  . = in which Frr = r 0 12E I   0 24E I 2E I  ∂fr ∂fr 2 2 0 6E I EI 2E I ∗
1 2
Term 0014∗0 is inconsequential, since it disappears on differentiation.
13–18
(13.49)
13–19
§13.5 FLEXIBILITY BASED DERIVATIONS
ClearAll[Le,EI,GAs,Φ,q0,fx2,fy2,m2]; GAs=12*EI/(Φ*Le^2); F=fx2; V=-fy2-q0*(Le-x); M=m2+fy2*(Le-x)+(1/2)*q0*(Le-x)^2; Print['check dM/dx=V: ',Simplify[D[M,x]-V]]; Ucd=F^2/(2*EA)+M^2/(2*EI)+V^2/(2*GAs); Uc=Simplify[Integrate[Ucd,{x,0,Le}]]; Print['Uc=',Uc]; u2=D[Uc,fx2]; v2=D[Uc,fy2]; θ2=D[Uc,m2]; Frr={{ D[u2,fx2], D[u2,fy2], D[u2,m2]}, { D[v2,fx2], D[v2,fy2], D[v2,m2]}, { D[θ2,fx2], D[θ2,fy2], D[θ2,m2]}}; br={D[Uc,fx2],D[Uc,fy2],D[Uc,m2]}/.{fx2->0,fy2->0,m2->0}; Print['br=',br]; Frr=Simplify[Frr]; Print['Frr=',Frr//MatrixForm]; Krr=Simplify[Inverse[Frr]]; Print['Krr=',Krr//MatrixForm]; qr=Simplify[-Krr.br]; Print['qr=',qr]; TT={{-1,0,0},{0,-1,0},{0,-Le,-1},{1,0,0},{0,1,0},{0,0,1}}; T=Transpose[TT]; Simplify[Ke=TT.Krr.T]; Print['Ke=',Ke//MatrixForm]; GrT={{1,0,0},{0,1,0},{0,Le,1}}; Gr=Transpose[GrT]; GqT={{1,0,0},{0,1,0},{0,Le/2,1}}; Gq=Transpose[GqT]; Print['Gr=',Gr//MatrixForm,' Gq=',Gq//MatrixForm]; qv={0,q0*Le,0}; Print['qs=',Simplify[-GrT.qr-GqT.qv]];
Figure 13.15. Script to derive the stiffness matrix and consistent load vector of the prismatic, plane Timoshenko beam element of Figure 13.14 by flexibility methods.
Applying the TCPE principle yields Frr fr = br − dr . This is inverted to produce the deformational stiffness relation fr = Krr dr + qr , in which
 EA
−1 Krr = Frr
0
 0 = 
0


E I (4 + ) 6E I 2 (1 + ) (1 + ) To use (13.45) the following transformation matrices are required: −
0
 −1
0010
−GrT TT = 0
0011
 0  0 =  1  0 0

0 0 −1 0  − −1  , 0 0   1 0 0 1
1
0010
Gq Tq = 0

0  qr = q0 Krr br = q0 /2  . −q0 2 /12
 12E I − 2 6E I  3 (1 + ) (1 + )  ,
0011
0 0 = 0  0 0
0 1 /2 0 0 0

0 0 1 , 0  0 0
q=
Injecting the rigid body modes from (13.45), Ke = TT Krr T and fe = TT qr , yields
 1  0  0 E A  Ke =   −1
0 −1 0 0 0 0 0 1 0 0 0 0 0 0 e f = q0 [ 0 1/2 /12 0 0 0 0 0 0 0


0 0 0 0 0 0 0 EI 0 0 +  0 0  3 (1 + )  0   0 0 0 0 0 0 T 1/2 − /12 ] .
(13.50)
0 0 12 6 6 2 (4 + ) 0 0 −12 −6 6 2 (2 − )
0 0 0 0 0 0
0 q0 0
.
(13.51)

0 0 −12 6  −6 2 (2 − )  ,  0 0  12 −6 −6 2 (4 + )
(13.52) The bending component is the same stiffness found previously in §13.2.6; compare with (13.16). The node force vector is the same as the consistent one constructed in an Exercise. A useful verification technique is to support the beam element at end 2 and recompute Ke and fe . This should reproduce (13.52). All of the foregoing computations were carried out by the Mathematica script shown in Figure 13.15.
13–19
13–20
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
y, v E,I,G,As constant
(a) m1
fy2
fy1
fx2 x, u
|2φ|
|R|
2
fx2
1 2
m2
+M(ψ)
+V(ψ) +F(ψ)
fy2
fx1
1
(c)
(b)
fy2 fx2
ψ
2
m2
m2
Figure 13.16. Flexibility derivation of plane circular arch element: (a) element and node forces, (b) removal of RBMs by fixing left node, (c) free body diagram of varying cross section.
O
+ψ y
C +H
x 1
y
+ds
O
C
+R
−R
+ds
−H C
x
2 2 +R
+ψ 2S
2S
+ds
−H
x 2 1
−2φ
+2φ
O
+ψ
+2φ O
−R 1 2
y
+ds
2S
−2φ +H C
+ψ 2S
x
1 y
Figure 13.17. Sign conventions in derivation of circular arch element.
§13.5.2. Plane Circular Arch in Local System In this and next subsection, the flexibility method is used to construct the stiffness matrix of a curved, prismatic, plane beam-column element with circular profile, pictured in Figure 13.16(a). The local system {x, y} is defined as shown there: x is a “chord axis” that passes through end nodes 1–2, and goes from 1 to 2. Axis y is placed at +90◦ from x. No load acts between nodes. In a curved plane element of this nature, axial extension and bending are intrinsically coupled. Thus consideration of three freedoms per node is mandatory. These are the translations along x and y, and the rotation θ about z. This element can be applied to the analysis of plane arches and ring stiffeners (as in airplane fuselages and submarine pressure hulls). If the arch curvature varies along the member, it should be subdivided into sufficiently small elements over each of which the radius magnitude is sensibly constant. Care must be taken as regards sign conventions to ensure correct results when the arch convexity and node numbering changes. Various cases are pictured in Figure 13.17. The conventions are as follows: (1)
The local node numbers define a positive arclength traversal along the element midline as going from 1 to 2. The curved length (not shown in figure) and the (chord) spanlength 2S are always positive.
(2)
The arch rise H , the angular span 2φ and the arch radius R are signed quantities as illustrated in Figure 13.17. The rise is the distance from chord midpoint to arch crown C: it has the sign of its projection on y. The angular span 2φ is that subtended by the arch on moving from 1 to 2: it is positive if CCW. Finally, the radius R has the sign of φ so = 2Rφ is always positive.
Some signed geometric relations: S = R sin φ,
H = R(cos φ − 1)
(13.53)
The location of an arch section is defined by the “tilt angle” ψ measured from the circle center-to-crown symmetry line OC, positive CCW. See Figure 13.17. The differential arclength ds = R dψ always points in the positive traversal sense 1 → 2.
13–20
13–21
§13.5 FLEXIBILITY BASED DERIVATIONS
The rigid motions are removed by fixing the left end as shown in in Figure 13.16(b). The internal forces F(ψ), V (ψ) and M(ψ) at an arbitrary cross section are obtained from the FBD of Figure 13.16(c) to be F = f x2 cos ψ + f y2 sin ψ, V = f x2 sin ψ − f y2 cos ψ, M = m 2 + f x2 R(cos φ− cos ψ)+ f y2 R(sin ψ− sin φ). (13.54) For typical straight beam-column members there are only two practically useful models: Bernoulli-Euler (BE) and Timoshenko. For curved members there are many more. These range from simple corrections to BE through theory-of-elasticity-based models. The model selected here is one of intermediate complexity. It is defined by the internal complementary energy functional U∗ =
0002 0014 0
(F − M/R)2 M2 + 2E A 2E I
0015
0002
φ
0014
ds = −φ
(F − M/R)2 M2 + 2E A 2E I
0015 R dψ.
(13.55)
The assumptions enbodies in this formula are: (1) the shear energy density V 2 /(2G As ) is neglected; (2) the cross section area A and moment of inertia I are unchanged with √ respect of those of the straight member. These assumptions are reasonable if |R| > 10 r , where r = + I /A is the radius of gyration of the cross section. Further corrections are treated in Exercises. To simplify the ensuing formulas it is convenient to take EA =
4E I EI = 2 2, 001a 2 R2φ2 001a
or
001a=
2r
with r 2 =
I . A
(13.56)
This defines 001a as a dimensionless geometric parameter. Note that this is not an intrinsic measure of arch slenderness, because it involves the element length.12 The necessary calculations are carried out by the Mathematica script of Figure 13.20, which has been pared down to essentials to save space. The deformational flexibility and stiffness computed are
Frr =
F11 symm
F12 F22
F13 F23 F33
,
−1 Krr = Frr =
K 11
K 12 K 22
symm
K 13 K 23 K 33
,
in which (first expression is exact value, second a Taylor series expansion in φ) F11 = F12 = F13 = F22 = F23 =
12
3
3
2 2 2 2 2φ(2 + φ 001a ) + 2φ(1 + φ 001a ) cos 2φ − 3 sin 2φ = 15001a 2 + φ 2 (2 − 15001a 2 ) + O(φ 4 ) 3 16E I φ 60E I 3 3
sin φ φ sin φ − φ(1 + φ 2 001a 2 ) cos φ = 1 − 3001a 2 + O(φ 3 ), 4E I φ 3 12E I 3 φ
2
2 2 sin φ − φ(1 + φ 001a ) cos φ = 1 − 3001a 2 + O(φ 3 ), 2 2E I φ 6E I 3
3
2 2 2 2 2φ(2 + φ 001a ) − 2φ(1 + φ 001a ) cos 2φ − sin 2φ = 20 + 3φ 2 (5001a 2 − 2) + O(φ 4 ), 3 16E I φ 60E I 2 2
sin φ 6 + φ 2 (6001a 2 − 1) + O(φ 4 ), F33 = (1 + φ 2 001a 2 ) = (1 + φ 2 001a 2 ). 2E I φ 12E I EI (13.57)
In that respect it is similar to of the Timoshenko beam element, defined by (13.15).
13–21
13–22
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
ClearAll[φ,ψ,Ψ,Γ,R,F,M,V,Le,EA,EI]; EA=4*EI/(Ψ^2*Le^2); V=-fy2*Cos[ψ]+fx2*Sin[ψ]; F=fx2*Cos[ψ]+fy2*Sin[ψ]; M=m2+fx2*R*(Cos[ψ]-Cos[φ])+fy2*R*(Sin[ψ]+Sin[φ]); Ucd=(F-M/R)^2/(2*EA)+ M^2/(2*EI); Uc=Simplify[Integrate[Ucd*R,{ψ,-φ,φ}]]; Print['Uc=',Uc]; u2=D[Uc,fx2]; v2=D[Uc,fy2]; θ2=D[Uc,m2]; Frr=Simplify[{{ D[u2,fx2], D[u2,fy2], D[u2,m2]}, { D[v2,fx2], D[v2,fy2], D[v2,m2]}, { D[θ2,fx2], D[θ2,fy2], D[θ2,m2]}}]; Frr=FullSimplify[Frr/.{R->Le/(2*φ)}]; Print['Frr=',Frr//MatrixForm]; Krr=FullSimplify[Inverse[Frr]]; Print['Krr=',Krr//MatrixForm]; TT={{-1,0,0},{0,-1,0},{0,-2*R*Sin[φ],-1},{1,0,0},{0,1,0},{0,0, 1}}; TT=TT/.{R->Le/(2*φ)}; T=Transpose[TT]; Ke=Simplify[TT.Krr.T]; Print['Ke=',Ke//MatrixForm];
Figure 13.18. Script to produce circular arch element stiffness in local coordinates.
Introduce d1 = φ(1 + φ 2 001a 2 )(φ + sin φ cos φ) − 2 sin2 φ and d2 = φ − sin φ cos φ. Then
4E I φ 4 EI
2 2 2 2 2 4 (1 + φ 001a ) = 45001a + φ (15001a + 45001a − 1) + O(φ 4 ), K 12 = 0, 3 d1 45 3 6E I φ 4E I φ 2
K 13 = 2 φ(1 + φ 2 001a 2 ) cos φ − sin φ = 2 2 (3001a 2 − 1) + O(φ 3 ), d1 001a 3 8E I φ 12E I sin φ EI = (5 + φ 2 ) + O(φ 4 ), K 23 = − K 22 = − 3 (30 + φ 2 ) + O(φ 4 ), K 22 = 3 d2 5 3 2φ 5
EIφ 2 8φ (3 + 2φ 2 001a 2 ) − 9 + 16 cos 2φ − 7 cos 4φ − 8φ sin 2φ 2 + (1 + φ 2 001a 2 ) cos 2φ K 33 = 8 d1 d2 EI
= 180001a 2 + φ 2 (5 − 48001a 2 ) + O(φ 4 ), 2 45 001a (13.58) The constraint K 23 = −K 22 sin φ/(2φ) must be verified by any arch stiffness, regardless of the TCPE form used. The necessary transformation matrix to inject the rigid body modes K 11 =
T = [ −G I ] =
−1 0 0
0 −1 0
0 −2R sin φ −1
1 0 0 0 1 0 0 0 1
=
−1 0 0
0 −1 0
0 − sin φ/φ −1
1 0 0 0 1 0 0 0 1
. (13.59)
Applying the congruential transformation gives the free-free stiffness
 K  0 K 13 −K 11 0 −K 13 11 K 22 −K 23 0 −K 22 −K 23   0  K −K K −K K 23 K 36  13 23 33 13  Ke = TT Krr T =   −K 11 0 −K 13 K 11 0 K 13    0 −K 13
−K 22 −K 23
K 23 K 36
0 K 13
K 22 K 23
(13.60)
K 23 K 33
The only new entry not in (13.58) is K 36 = −K 33 − K 23 sin φ/φ =
EI (90001a 2 + φ 2 (12ψ 2 − 5) + O(φ 4 ) 45
As a check, if φ → 0 the entries reduce to that of the beam-column modeled with Bernoulli-Euler. For example K 11 → 4E I 001a 2 / 3 = E A/ , K 22 → 12E I / 3 , K 33 → 4E I / , K 36 → 2E I / , etc.
13–22
13–23
§13.5 FLEXIBILITY BASED DERIVATIONS
_
x
(a)
2 (x2 ,y2) 3 (x3 ,y3 ) _
y
(b)
2 3
1 (x 1 ,y1)
b 3
R 2φ
ϕ
y
(c)
1
a
2
H 2S
1
x Figure 13.19. Plane circular arch element in global coordinates: (a) geometric id (b) intrinsic geometry recovery.
Figure 13.20. Module to produce plane circular arch element stiffness in global coordinates.
§13.5.3. Plane Circular Arch in Global System To use the circular arch element in a 2D finite element program it is necessary to specify its geometry in the {x, y} plane and then to transform the stiffness (13.60) to global coordinates. The first requirement can be handled by providing the coordinates of three nodes: {xi , yi }, i = 1, 2, 3. Node 3 (see Figure) is a geometric node that serves to define the element mean curvature but has no associated freedoms. (Section to be completed).
13–23
13–24
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
§13.6.
*Accuracy Analysis
This section presents the accuracy analysis of a repeating lattice of beam elements, analogous to that done for the bar element in §12.5. The analysis uses the method of modified differential equations (MoDE) mentioned in the Notes and Bibliography of Chapter 12. It is performed using a repeated differentiations scheme similar to that used in solving Exercise 12.8. Only the case of the Bernoulli-Euler model is worked out in detail. §13.6.1. *Accuracy of Bernoulli-Euler Beam Element Consider a lattice of repeating two-node, prismatic, plane Bernoulli-Euler beam elements of rigidity E I and length , as illustrated in Figure 13.21. The system is subject to an arbitrary lateral load q(x), which is assumed infinitely differentiable in x. From the lattice extract a patch of two elements: (1) and (2), connecting nodes i– j and j–k, respectively, as shown in Figure 13.21.
q(x) j
i
xi = xj − xj xk = xj +
k EI = const
L
vj = 1
The FEM patch equations at node j are
v 
Two-element patch ijk
i
 θi  0010 0011  E I 0010 −12 −6 24 0 −12 6 0011   vj  = f j 2 2 2   3 mj 6 2 0 8 −6 2  θj 
i
vk θk
Trial functions at node j
θj = 1 (1)
j
(2)
k
Figure 13.21. Repeating beam lattice for accuracy analysis.
(13.61) Expand v(x), θ(x) and q(x) in Taylor series about x = x j , truncating at n +1 terms for v and θ, and m +1 terms for q. Using ψ = (x − x j )/ , the series are v(x) = v j + ψ v 0005j + (ψ 2 2 /2!)v 00050005j + . . . + (ψ n n /n!)v [n] j , θ(x) = [n] 0005 00050005 0005 00050005 2 2 n n 2 2 m m θ j +ψ θ j +(ψ /2!)θ j +. . .+(ψ /n!)θ j , and q(x) = q j +ψ q j +(ψ /2!)q j +. . .+(ψ /m!)q [m] j . [n] n n 13 Here v j , etc., is an abbreviation for d v(x j )/d x . Evaluate the v(x) and θ (x) series at i and k by setting ψ = ±1, and insert in (13.61). Use the q(x) series evaluated over elements (1) and (2), to compute the consistent forces f j and m j as fj = 2
00140002
0002
1
N3(1) q (1) −1
dξ
(1)
+
0015
1
N1(2) q (2) −1
dξ
(2)
, mj = 2
00140002
0002
1
N4(1) q (1) −1
dξ
(1)
+
0015
1
N2(2) q (2) −1
dξ
(2)
.
(13.62)
Here N3(1) = 14 (1 + ξ (1) )2 (2 − ξ (1) ), N4(1) = − 8 (1 + ξ (1) )2 (1 − ξ (1) ) N1(2) = 14 (1 − ξ (2) )2 (2 + ξ (2) ), and N2(2) = 8 (1 − ξ (2) )2 (1 + ξ (2) ) are the Hermitian shape functions components of the j node trial function, whereas q (1) = q(ψ (1) ), ψ (1) = − 12 (1 − ξ (1) ) and q (2) = q(ψ (2) ), ψ (2) = 12 (1 + ξ (2) ) denote the lateral loads. To show the resulting system in compact matrix form it is convenient to collect the derivatives at node j into vectors: T v j = [ v j v 0005j v 00050005j . . . v [n] j ] ,
T θ j = [ θ j θ 0005j θ 00050005j . . . θ [n] j ] ,
T q j = [ q j q 0005j q 00050005j . . . q [n] j ] .
(13.63)
The resulting differential system can be compactly written
0010
13
Svv Sθv
Svθ Sθθ
00110010
vj θj
0011
=
0010
0011
Pv qj. Pθ
(13.64)
Brackets are used instead of parentheses to avoid confusion with element superscripts. If derivatives are indexed by primes or roman numerals the brackets are omitted.
13–24
13–25
§13.6 *ACCURACY ANALYSIS
Here Svv , Svθ , Sθv and Sθθ are triangular Toeplitz matrices of order (n+1) × (n+1) whereas Pv and Pθ are generally rectangular matrices of order (n+1) × (m+1). Here is the expression of these matrices for n = 8, m = 4:

Svv
0 0  0  0  = EI 0  0  0  0 0
0 0  0  0  Sθv = E I  0  0  0 
0 0 0 0 0 0 0 0 0
−12
−12
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 −12
0 0 0 0 0 0 0

0
− − 0 − 0 1680 30 3 0 − 0 − 0  30  −12 − 3  0 − 0 30  0 −12 0 − 0   , 0 0 −12 0 −  −12 0 0 0 0    0 0 0 0 −12  0 0 0 0 0 0 0 0 0 0
0 −2 −12 0 −12 0 0 0 0 0 0 0 0 0 0 0 0 0
3
0 − 10 −2 0 0 −2 −12 0 0 −12 0 0 0 0 0 0 0 0 3
 0 0  0 0  0 0 Pv =  0 0 0 0  0 0 
3 15
0 0 0 0 0 0 0 0 0 0 0
5
− 3 10
−12
0 0 0
0
0
3
0 0 0 0 0 0
15
0 0 0 0 0
0


0 0  0  0  Pθ =  0  0  0 0 0
0 
3 15
0    3 0 −  0  10  3 0 −  0 10   , Sθθ = E I  −2 0  0    0 −2  0  −12  0 0  0 −12 0
5 560
0
 12
− 5 420
0
0 −2 0
0 0 0 0 0 0 0 0 0 0
0 
− 5 420
0
0  0  0  Svθ = E I  0  0  0 
    ,     
12
3 15
0 0 0 0 0 0 0 0
0 3 15
0 0 0 0 0 0 0
5 315
0 3 15
0 0 0 0 0 0
0 2 12 0 12 0 0 0 0 0 0 0 0 0 0 0 0 0
0 2 12 0 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 315
0 3 15
0 0 0 0 0
0 10 0 3 2 0 10 0 2 0 12 0 2 12 0 0 0 0 12 0 0 0 0 0 0 0 0 0
0 6 0 3 2 0 6 0 2 0 12 0 2 12 0 0 0 0 12 0 0 0 0 0 0 0 0 0 3

0 
5 420
3
5 420
  0   3  0 10  2 0   0 2   12 0   12 0
3 10
0 0
5 180
0 3 6
0 0
5 180
0
7 10080

0 
5 180
           
0 6 0 3 2 0 6 0 2 0 12 0 2 12 0 0 12 0 0 (13.65) 3
           
(13.66)
Elimination of θ j gives the condensed system Sv j = Pq j ,
with
0 0 0 0 0 0  0 0  S = Svv − Svθ S−1 S = E I 0 0 θθ θv 0 0  0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0
S = Svv − Svθ S−1 θθ Sθv , 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
− 5 720
0 0 0 0 0 0 0
      ,    
13–25
P = Pv − Svθ S−1 θv Pθ .
(13.67)
 0 0 0 0  0 0  P = Pv − Svθ S−1 P = 0 0 θv θ 0 0  0 0 
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
− 5 720

0  0   0    0   0   0 0 (13.68)
13–26
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
4 viii 4 iv The nontrivial differential relations14 given by S v j = P q j are E I (v iv j − v j /720) = q j − q j /720, viii 000500050005 = q iv E I v vj = q 0005j , E I v vij = q 00050005j , E I v vii j = q j , and E I v j j . The first one is a truncation of the infinite order iv MoDE. Elimination of all v j derivatives but v j yields
E I v iv j = qj,
(13.69)
exactly. That this is not a fluke can be confirmed by increasing n and while taking m = n − 4. The first 4 ˆ The last 4 rows of P, which are columns and last 4 rows of S, which are always zero, are removed to get S. ˆ also zero, are removed to get P. With m = n−4 both matrices are of order n−3 × n−3. Then Sˆ = Pˆ for any n, which leads immediately to (13.69). This was confirmed with Mathematica running n up to 24. The foregoing analysis shows that the BE cubic element is nodally exact for any smooth load over a repeating lattice if consistent load computation is used. Exercise 13.18 verifies that the property is lost if elements are not identical. Numerical experiments confirm these conclusions. The Laplace transform method only works part way: it gives a different infinite order MoDE but recovers (13.69) as n → ∞. These accuracy properties are not widely known. If all beam elements are prismatic and subjected only to point loads at nodes, overall exactness follows from a theorem by Tong [166], which is not surprising since the exact solution is contained in the FEM approximation. For general distributed loads the widespread belief is that the cubic element incurs O( 4 ) errors. The first study of this nature by Waltz et. al. [175] gave the modified differential equation (MoDE) for a uniform load q as v iv −
4 viii q v + .. = . 720 EI
(13.70)
The above terms are correct. In fact a more complete expression, obtained in this study, is
0014
0015
4 iv 6 vi 7 8 q + q − q viii + . . . 720 3024 259200 (13.71) But the conclusion that “the principal error term” is of order 4 [175, p. 1009] is incorrect. The misinterpretation is due to (13.71) being an ODE of infinite order. Truncation is fine if followed by elimination of higher derivatives. If this is done, the finite order MoDE (13.69) emerges regardless of where (13.71) is truncated; an obvious clue being the repetition of coefficients in both sides. The moral is that conclusions based on infinite order ODEs should be viewed with caution, unless corroborated by independent means. EI
4 viii 6 x 7 8 v − v + v − v xii + . . . 720 3024 259200 iv
=q−
§13.6.2. *Accuracy of Timoshenko Beam Element Following the same procedure it can be shown that the infinite order MoDE for a Timoshenko beam element repeating lattice with the stiffness (13.16) and consistent node forces is
0014 EI
v iv j
0015
2 vi 4 (1 + 5 − 5 2 ) viii 6 (20 + 7 − 70 2 + 35 3 ) x v − vj + vj + . . . + 12 j 720 60480
= qj −
4 (1 + 5 ) iv 6 (20 + 14 − 35 2 ) vi qj + qj ) + . . . , 720 60480
in which = 12E I /(G As 2 ). (13.72) This reduces to (13.71) if = 0. Elimination of higher order derivatives gives the finite order MoDE (aka FOMoDE) 2 00050005 E I 00050005 E I v iv q = qj − q . (13.73) j = qj − 12 j G As j 14
Derivatives of order 4 and higher are indicated by Roman numeral superscripts instead of primes.
13–26
13–27
§13. Notes and Bibliography
which repeats for any n > 8. This happens to be the exact governing differential equation for a statically loaded Timoshenko beam [61, p. 23]. Consequently the Timoshenko beam is nodally exact under the same conditions previously stated for the Bernoulii-Euler model. Notes and Bibliography The material in this Chapter interwines the very old and the very new. Before energy methods came to the FEM forefront by 1960 (see historical sketch in §1.7), ordinary differential equations (ODE) and flexibility methods were essential part of the toolbox repertoire of the professional structural engineer. Traces of that dominance may be found in the books by Przemieniecki [136], Pestel and Leckie [126] and the survey by Gallagher [71]. Energy derivations were popularized by Archer [6,7], Martin [111] and Melosh [116,117]. For one-dimensional elements, however, results are often identical. This can provide a valuable crosscheck. The Legendre interpolation (13.2) was introduced in [54,56] to study optimal mass-stiffness combinations for beam elements in the context of finite element templates [50]. The diagonal covariance matrices Qn given in (13.4) play a key role in model customization. The hinged element stiffness (13.12) is rederived in the Advanced FEM Lecture Notes [55] using a mixed variational principle. The separation of uncoupled rigidity effects in stiffness forms such as (13.6) and (13.7) is suggested by template theory [59]. The Timoshenko beam model was originally proposed in [160]. Timoshenko cleverly packaged the model with miscellaneous ingredients introduced earlier by Bresse and Hencky. It has become important as a tool for transient response and control simulations because its dynamic form is strictly hyperbolic.15 The Timoshenko beam element stiffness (13.16) first appeared in [170] in the guise of a spar element for use in aircraft structures; the end node freedoms of that element differring from the classical set used here. The particular form (13.16) is derived in Section 5.6 of [136] using ODEs. This beam model pertains to the class of “C 0 elements” that have been extensively studied in the FEM literature after 1968. The book of Hughes [95] provides a comprehensive treatment of such methods for beams and plates. The classical work on beams on elastic foundations is by Hetenyi [198]. Useful solutions are tabulated in Roark-Young’s handbook [140]. That the use of homogeneous solutions of governing differential equations yields nodally-exact stiffness equations was first proven generally by Tong [166] in a Galerkin context. This derivation procedure, however, was rarely used after the 1960s. Two obstacles: (1) it is largely restricted to either one dimensional elements, or to problems with special symmetries that can be modeled with ODEs;16 (2) rapidly increasing solution complexity in complicated problems discouraged hand derivations. Whereas the first limitation still holds, the increasing availability of CAS allows timely consideration of more difficult problems. The construction of the exact beam-on-Winkler-foundation element in §13.3 offers a case in point. Using Mathematica the complete derivation, checking and fully-symbolic testing took about 6 hours, whereas a hand derivation, coding and numerical testing would likely take weeks or months. The main application of exact elements appears to be a priori error estimation: how many simpler elements are needed to do the job of an exact one? The construction of stiffness matrices from flexibility information was historically one of the first techniques by which stiffness equations of MoM members were derived. The rigid body injection method of §13.4.4 largely follows Section 6.6 of [136]. The presentation of discrete-system equilibrium theorems in §13.4.2 includes a new ingredient missing from previous work: handling non-self-equilibrated loading systems. This extension removes the 40-year-old objection that flexibility methods (or more generally, schemes based on the TCPE principle) are unable to produce equivalent or consistent node forces. 15
The Bernoulli-Euler beam dynamic model is parabolic and thus exhibits an infinite transverse wave speed. Such a model is unsuitable for wave propagation problems.
16
For example, symmetrically loaded circular plates or shells of revolution.
13–27
Chapter 13: ADVANCED ONE-DIMENSIONAL ELEMENTS
13–28
The use of equilibrium methods for multidimensional finite elements was pioneered by Fraeijs de Veubeke in the 1960s and early 1970s. His obsession with solution bounding got these methods seriously stuck, however, because of difficulties in interelement connections that maintain system-level equilibrium, as well as avoidance of spurious modes. More practical extensions lead to the so-called Trefftz and equilibrium hybrid methods. These are presently the topic of active research17 but require advanced variational techniques beyond the scope of this book. Another recent advance is the discovery of the free-free flexibility as the true dual of the free-free stiffness [52,58,196]. This extension relies heavily on projection operators. As the study in §13.6 illustrates, modified equation methods in boundary value problems18 are delicate and should be used with care. Their intricacies baffled Strang and Fix who, upon doing Fourier analysis of a cubic beam element, incorrectly stated [151, p. 171] that only one of the discrete equations — that for v(x) — is consistent “and the others are completely inconsistent.” The alternative is the variational approach to error analysis. Although more robust and forgiving, predictions are often so conservative as to be of little value. References Referenced items have been moved to Appendix R.
17
Along with discontinuous Galerkin methods, a reinvention of Fraeijs de Veubeke’s weakly diffusive models.
18
They are more forgiving in initial value problems.
13–28
13–29
Exercises
Homework Exercises for Chapter 13 Advanced One-Dimensional Elements EXERCISE 13.1 [A:15] Evaluate the strain and stress fields associated with the Timoshenko beam displacement field (13.14). EXERCISE 13.2 [A/C:20] Find the consistent node forces for a Timoshenko beam element under a uniform
transverse load q. Hint: find fc for the Legendre interpolation, then premultiply by HTS . EXERCISE 13.3 [A/C:20] Construct the stiffness matrix a Timoshenko plane beam element with a hinge at the center. Hint: set R Bs = 0, R Ba = E I and R S = 12E I /( 2 ) in (13.11). EXERCISE 13.4 [A/C:15=5+10] Consider a cantilever beam of constant bending rigidity E I , constant shear rigidity G As and length L, which is modeled with one Timoshenko element. The beam is end loaded by a transverse force P. The support conditions are v1 = θ1 = 0.
(a)
Find the end displacement v2 and end rotation θ2 in terms of P, E, G, I , As and L. Compare with the analytical values P L 3 /(3E I ) + P L/(G As ) and P L 2 /(2E I ), respectively.
(b)
Why does the finite element model provides the exact answer with one element?
EXERCISE 13.5 [A:25] (Requires math ability). Discuss what happens in (13.16) if → ∞. Is the result
useful for a shear-only “spar” element? Hint: eliminate θ1 and θ2 by a master-slave MFC.
EXERCISE 13.6 [A:10] For a given number of elements N e of length = 2L/N e , relate χ and λ in Example
13.2. EXERCISE 13.7 [A/C:40] (research paper level). Derive an exact Timoshenko-beam-on-Winkler-foundation equation method.element using the differential equation method. EXERCISE 13.8 [C:30] Write Mathematica code to verify the nodal exactness conclusion of §13.6.1 using
the repeated differentiation approach. EXERCISE 13.9 [C:30] As above, but using the Laplace transform. Show that this only does half the job. EXERCISE 13.10 [A/C:35] Find the general symbolic expression of the terms in the infinite order MoDE
(13.71). EXERCISE 13.11 [A/C:40] (research paper level) Analyze nodal accuracy if the length of the beam elements in the lattice of Figure 13.21 alternates between (1 ± α) , where 0 ≤ α ≤ 12 . EXERCISE 13.12 [A/C:35] Using Mathematica, verify the results (13.72) and (13.73) in §13.6.2.
13–29
14
.
The Plane Stress Problem
14–1
14–2
Chapter 14: THE PLANE STRESS PROBLEM
TABLE OF CONTENTS Page
§14.1. Introduction §14.1.1. Plate in Plane Stress . . . . §14.1.2. Mathematical Model . . . §14.2. Plane Stress Problem Description §14.2.1. Given Problem Data . . . . §14.2.2. Problem Unknowns . . . . §14.3. Linear Elasticity Equations §14.3.1. Governing Equations . . . . §14.3.2. Boundary Conditions . . . §14.3.3. Weak Forms versus Strong Form §14.3.4. Total Potential Energy . . . §14.4. Finite Element Equations §14.4.1. Displacement Interpolation . . §14.4.2. Element Energy . . . . . §14.4.3. Element Stiffness Equations . §14. Notes and Bibliography . . . . . . . . . . . §14. References . . . . . . . . . . . §14. Exercises . . . . . . . . . . . §14. Solutions to .Exercises . . . . . . . . . .
14–2
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14–3 14–3 14–4 14–4 14–4 14–5 14–6 14–7 14–7 14–8 14–9 14–10 14–10 14–11 14–11 14–12 14–12 14–13 14–15
14–3
§14.1 INTRODUCTION
§14.1. Introduction We now pass to the variational formulation of two-dimensional continuum finite elements. The problem of plane stress will serve as the vehicle for illustrating such formulations. As narrated in §1.7.1, continuum finite elements were invented in the aircraft industry (at Boeing, early 1950s) to solve this kind of problem when it arose in the design and analysis of delta wing panels [164]. The problem is presented here within the framework of the linear theory of elasticity. §14.1.1. Plate in Plane Stress In structural mechanics, a flat thin sheet of material is called a plate.1 The distance between the plate faces is called the thickness and denoted by h. The midplane lies halfway between the two faces. z
The direction normal to the midplane is the transverse direction. Directions parallel to the midplane are called in-plane directions. The global axis z will be oriented along the transverse direction. Axes x and y are placed in the midplane, forming a right-handed Rectangular Cartesian Coordinate (RCC) system. Thus the midplane equation is z = 0. See Figure 14.1.
y x Figure 14.1. A plate structure in plane stress.
A plate loaded in its midplane is said to be in a state of plane stress, or a membrane state, if the following assumptions hold: 1.
All loads applied to the plate act in the midplane direction, and are symmetric with respect to the midplane.
2.
All support conditions are symmetric about the midplane.
3.
In-plane displacements, strains and stresses can be taken to be uniform through the thickness.
4.
The normal and shear stress components in the z direction are zero or negligible.
The last two assumptions are not necessarily consequences of the first two. For the latter to hold, the thickness h should be small, typically 10% or less, than the shortest in-plane dimension. If the plate thickness varies it should do so gradually. Finally, the plate fabrication must exhibit symmetry with respect to the midplane. To these four assumptions we add the following restriction: 5.
The plate is fabricated of the same material through the thickness. Such plates are called transversely homogeneous or (in aerospace) monocoque plates.
The last assumption excludes wall constructions of importance in aerospace, in particular composite and honeycomb sandwich plates. The development of mathematical models for such configurations requires a more complicated integration over the thickness as well as the ability to handle coupled bending and stretching effects, and will not be considered here.
1
If it is relatively thick, as in concrete pavements or Argentinian beefsteaks, the term slab is also used but not for plane stress conditions.
14–3
14–4
Chapter 14: THE PLANE STRESS PROBLEM
y Midplane
Mathematical idealization
Plate
Γ
x
Ω
Figure 14.2. Mathematical model of plate in plane stress.
Remark 14.1. Selective relaxation from assumption 4 lead to the so-called generalized plane stress state, in
which z stresses are accepted. The plane strain state is obtained if strains in the z direction are precluded. Although the construction of finite element models for those states has many common points with plane stress, we shall not consider those models here. For isotropic materials the plane stress and plane strain problems can be mapped into each other through a fictitious-property technique; see Exercise 14.1. Remark 14.2. Transverse loading on a plate produces plate bending, which is associated with a more complex configuration of internal forces and deformations. This subject is studied in more advanced courses.
§14.1.2. Mathematical Model The mathematical model of the plate in plane stress is a two-dimensional boundary value problem (BVP). The BVP is posed over a plane domain 0002 with a boundary 0003, as illustrated in Figure 14.2. In this idealization the third dimension is represented as functions of x and y that are integrated through the plate thickness. Engineers often work with internal plate forces, which result from integrating the in-plane stresses through the thickness. See Figure 14.3. §14.2. Plane Stress Problem Description §14.2.1. Given Problem Data Domain geometry. This is defined by the boundary 0003 illustrated in Figure 14.2. Thickness. Most plates used as structural components have constant thickness. If the thickness does vary, in which case h = h(x, y), it should do so gradually to maintain the plane stress state. Material data. This is defined by the constitutive equations. Here we shall assume that the plate material is linearly elastic but not necessarily isotropic. Specified Interior Forces. These are known forces that act in the interior 0002 of the plate. There are of two types. Body forces or volume forces are forces specified per unit of plate volume; for example the plate weight. Face forces act tangentially to the plate faces and are transported to the midplane. For example, the friction or drag force on an airplane skin is of this type if the skin is modeled to be in plane stress. Specified Surface Forces. These are known forces that act on the boundary 0003 of the plate. In elasticity these are called surface tractions. In actual applications it is important to know whether these forces are specified per unit of surface area or per unit length. 14–4
14–5
§14.2
PLANE STRESS PROBLEM DESCRIPTION
In-plane internal forces pyy
h x z
y
pxx
pyy pxy
y pxy
pxx
x
In-plane stresses h y
x
Positive sign convention σyy σxx σxy = σ yx
y
x In-plane displacements
In-plane strains h
h eyy
x
e xx e xy = eyx
y x
ux
uy
y
Figure 14.3. Notation for in-plane stresses, strains, displacements and internal forces for a plate in plane stress.
Displacement Boundary Conditions. These specify how the plate is supported. Points on the plate boundary may be fixed, allowed to move in one direction, or subject to multipoint constraints. In addition symmetry and antisymmetry lines may be identified as discussed in Chapter 8. If no displacement boundary conditions are imposed, the plate structure is said to be free-free. §14.2.2. Problem Unknowns The unknown fields are displacements, strains and stresses. Because of the assumed wall fabrication homogeneity the in-plane components are assumed to be uniform through the plate thickness. Thus the dependence on z disappears and all such components become functions of x and y only. Displacements. The in-plane displacement field is defined by two components: 0003 0002 u x (x, y) u(x, y) = u y (x, y)
(14.1)
The transverse displacement component u z (x, y, z) component is generally nonzero because of Poisson’s ratio effects, and depends on z. However, this displacement does not appear in the governing equations. Strains. The in-plane strain field forms a tensor defined by three independent components: ex x , e yy and ex y . To allow stating the FE equations in matrix form, these components are conventionally arranged to form a 3-component “strain vector” 0005 0004 ex x (x, y) (14.2) e(x, y) = e yy (x, y) 2ex y (x, y) 14–5
14–6
Chapter 14: THE PLANE STRESS PROBLEM
Prescribed displacements ^ u
Displacement BCs Displacements ^ u=u u on Γu Kinematic
Body forces
b
e
Ω
D σ + b = 0 Equilibrium in Ω
e=Du in Ω
Strains
Γ
σ=Ee in Ω Constitutive
Stresses
σ
Force BCs σn = t ^ or pTn = q on Γt T
^
Prescribed tractions t or forces q
Figure 14.4. The Strong Form of the plane stress equations of linear elastostatics displayed as a Tonti diagram. Yellow boxes identify prescribed fields whereas orange boxes denote unknown fields. The distinction between Strong and Weak Forms is explained in §14.3.3.
The factor of 2 in ex y shortens strain energy expressions. The shear strain components ex z and e yz vanish. The transverse normal strain ezz is generally nonzero because of Poisson’s ratio effects. This strain does not enter the governing equations as unknown, however, because the associated stress σzz is zero. This eliminates the contribution of σzz ezz to the internal energy. Stresses. The in-plane stress field forms a tensor defined by three independent components: σx x , σ yy and σx y . As in the case of strains, to allow stating the FE equations in matrix form, these components are conventionally arranged to form a 3-component “stress vector” 0005 0004 σx x (x, y) (14.3) σ(x, y) = σ yy (x, y) σx y (x, y) The remaining three stress components: σzz , σx z and σ yz , are assumed to vanish. The plate internal forces are obtained on integrating the stresses through the thickness. Under the assumption of uniform stress distribution, px x = σx x h,
p yy = σ yy h,
px y = σx y h.
(14.4)
These p’s also form a tensor. They are called membrane forces in the literature. See Figure 14.3. §14.3. Linear Elasticity Equations We shall develop plane stress finite elements in the framework of classical linear elasticity. The necessary governing equations are presented below. They are graphically represented in the Strong Form Tonti diagram of Figure 14.4.
14–6
14–7
§14.3
LINEAR ELASTICITY EQUATIONS
§14.3.1. Governing Equations The three internal fields: displacements, strains and stresses (14.1)–(14.3) are connected by three field equations: kinematic, constitutive and internal-equilibrium equations. If initial strain effects are ignored, these equations read 0004 0005 0004 00050002 0003 ex x ∂/∂ x 0 ux , 0 ∂/∂ y e yy = uy 2ex y ∂/∂ y ∂/∂ x 0005 0004 00050004 0005 0004 E 11 E 12 E 13 ex x σx x (14.5) σ yy = E 12 E 22 E 23 e yy , σx y E 13 E 23 E 33 2ex y 0002 0003 0004 σx x 0005 0002 0003 0002 0003 b 0 ∂/∂ x 0 ∂/∂ y . σ yy + x = by 0 0 ∂/∂ y ∂/∂ x σx y In these equations, bx and b y are the components of the interior body force b, E is the 3 × 3 stress-strain matrix of plane stress elastic moduli, D is the 3 × 2 symmetric-gradient operator and its transpose the 2 × 3 tensor-divergence operator.2 The compact matrix version of (14.5) is e = Du,
σ = Ee,
DT σ + b = 0,
(14.6)
in which E = ET . If the plate material is isotropic with elastic modulus E and Poisson’s ratio ν, the moduli in the constitutive matrix E reduce to E 11 = E 22 = E/(1 − ν 2 ), E 33 = 12 E/(1 + ν) = G, E 12 = ν E 11 and E 13 = E 23 = 0. See also Exercise 14.1. §14.3.2. Boundary Conditions Boundary conditions prescribed on 0003 may be of two types: displacement BC or force BC (the latter is also called stress BC or traction BC). To write down those conditions it is conceptually convenient to break up 0003 into two subsets: 0003u and 0003t , over which displacements and force or stresses, respectively, are specified. See Figure 14.5. Displacement boundary conditions are prescribed on 0003u in the form ˆ u = u.
(14.7)
Here uˆ are prescribed displacements. Often uˆ = 0. This happens in fixed portions of the boundary, as the ones illustrated in Figure 14.5. Force boundary conditions (also called stress BCs and traction BCs in the literature) are specified on 0003t . They take the form σn = ˆt. (14.8) Here ˆt are prescribed surface tractions specified as a force per unit area (that is, not integrated through the thickness), and σn is the stress vector shown in Figure 14.5. 2
The dependence on (x, y) has been omitted to reduce clutter.
14–7
14–8
Chapter 14: THE PLANE STRESS PROBLEM
n (unit t exterior normal)
;;; ; ; ;;; ; ; ;; ;;; ; ;; ; ;;; ;
+
Γu
u^ = 0 ^ Boundary displacements u are prescribed on Γu (figure depicts fixity condition)
σn
t^
Γt
σnt
^t n
^t t ^t
σ nn
Boundary tractions ^t or boundary forces q^ are prescribed on Γt
Stress BC details (decomposition of forces q^ would be similar)
Figure 14.5. Displacement and force (stress, traction) boundary conditions for the plane stress problem.
An alternative form of (14.8) uses the internal plate forces: ˆ pn = q.
(14.9)
Here pn = σn h and qˆ = ˆt h. This form is used more often than (14.8) in structural design, particularly when the plate wall construction is inhomogeneous. The components of σn in Cartesian coordinates follow from Cauchy’s stress transformation formula 0002 σn =
σx x n x + σx y n y σx y n x + σ yy n y
0003
0002 =
nx 0
0 ny
0003 0004 σx x 0005 ny σ yy , nx σx y
(14.10)
in which n x and n y denote the Cartesian components of the unit normal vector ne (also called the direction cosines of the normal). Thus (14.8) splits into two scalar conditions: tˆx = σnx and tˆy = σny . The derivation of (14.10) is the subject of Exercise 14.4. It is sometimes convenient to write the condition (14.8) in terms of normal n and tangential t directions: σnt = tˆt (14.11) σnn = tˆn , in which σnn = σnx n x + σny n y and σnt = −σnx n y + σny n x . Remark 14.3. The separation of 0003 into 0003u and 0003t is useful for conciseness in the mathematical formulation, such as the energy integrals presented below. It does not exhaust, however, all BC possibilities. Frequently at points of 0003 one specifies a displacement in one direction and a force (or stress) in the other. An example of these are roller and sliding conditions as well as lines of symmetry and antisymmetry. To cover these situations one needs either a generalization of the split, in which 0003u and 0003t are permitted to overlap, or to define another portion 0003m for “mixed” conditions. Such generalizations will not be presented here, as they become unimportant once the FE discretization is done.
14–8
14–9
§14.3
Prescribed displacements ^ u
Displacement BCs Displacements ^ u=u u on Γu
Body forces
b
δΠ= 0 in Ω
e=Du in Ω
Kinematic
LINEAR ELASTICITY EQUATIONS
Strains
σ=Ee in Ω
e
Constitutive
Γ
Ω
Equilibrium (weak)
Stresses
σ
Force BCs (weak) δΠ = 0 on Γt
Prescribed tractions t or forces q
Figure 14.6. The TPE-based Weak Form of the plane stress equations of linear elastostatics. Weak links are marked with grey lines.
§14.3.3. Weak Forms versus Strong Form We introduce now some further terminology from variational calculus. The Tonti diagram of Figure 14.4 is said to display the Strong Form of the governing equations because all relations are verified point by point. These relations, called strong links, are shown in the diagram with black lines. A Weak Form is obtained by relaxing one or more strong links. Those are replaced by weak links, which enforce relations in an average or integral sense rather than point by point. The weak links are then provided by the variational formulation chosen for the problem. Because in general many variational forms of the same problem are possible, there are many possible Weak Forms. On the other hand the Strong Form is unique. The Weak Form associated with the Total Potential Energy (TPE) variational form is illustrated in Figure 14.6. The internal equilibrium equations and stress BC become weak links, which are drawn by gray lines. These equations are given by the variational statement δ = 0, where the TPE functional is given in the next subsection. The FEM displacement formulation discussed below is based on this particular Weak Form. §14.3.4. Total Potential Energy As usual the Total Potential Energy functional for the plane stress problem is given by
= U − W. The internal energy is the elastic strain energy: 0006 1 U=2 h σT e d0002 = 0002
(14.12)
0006 1 2
h eT E e d0002. 0002
(14.13)
The derivation details are relegated to Exercise E14.5. The external energy is the sum of contributions from known interior and boundary forces: 0006 0006 T W = h u b d0002 + h uT ˆt d0003. (14.14) 0002
0003t
14–9
14–10
Chapter 14: THE PLANE STRESS PROBLEM
Note that the boundary integral over 0003 is taken only over 0003t . That is, the portion of the boundary over which tractions or forces are specified. §14.4. Finite Element Equations The necessary equations to apply the finite element method to the plane stress problem are collected here and expressed in matrix form. The domain of Figure 14.7(a) is discretized by a finite element mesh as illustrated in Figure 14.7(b). From this mesh we extract a generic element labeled e with n ≥ 3 node points. In the subsequent derivations the number n is kept arbitrary. Therefore, the formulation is applicable to arbitrary two-dimensional elements, for example those sketched in Figure 14.8.
(b)
(a)
Γ Ω (c)
Ωe
Γ Departing from previous practice in 1D elements, the element node points will be labelled 1 through n. These are called local node numbers. The Figure 14.7. Finite element discretization and element domain and boundary are denoted by extraction of generic element. 0002e and 0003 e , respectively. The element has 2n degrees of freedom. These are collected in the element node displacement vector (14.15) ue = [ u x1 u y1 u x2 . . . u xn u yn ]T . e
§14.4.1. Displacement Interpolation The displacement field ue (x, y) over the element is interpolated from the node displacements. We shall assume that the same interpolation functions are used for both displacement components.3 Thus n n 0007 0007 u x (x, y) = Nie (x, y) u xi , u y (x, y) = Nie (x, y) u yi , (14.16) i=1
i=1
where Nie (x, y) are the element shape functions. In matrix form: 0002
0003 0002 e u x (x, y) N1 u(x, y) = = u y (x, y) 0
0 N1e
N2e 0
0 N2e
.. ..
Nne 0
0003 0 ue = N ue . Nne
(14.17)
This N (with superscript e omitted to reduce clutter) is called the shape function matrix. It has dimensions 2 × 2n. For example, if the element has 4 nodes, N is 2 × 8. The interpolation condition on the element shape function Nie (x, y) states that it must take the value one at the i th node and zero at all others. This ensures that the interpolation (14.17) is correct at the nodes. Additional requirements on the shape functions are stated in later Chapters. 3
This is the so called element isotropy condition, which is studied and justified in advanced FEM courses.
14–10
14–11
§14.4
FINITE ELEMENT EQUATIONS
3
4 9
3
3
4
2
5 6
1
1
1
2
n=3
2 4
n=4
n=6
10 1
8
12 11 5
3 7 6 2
n = 12
Figure 14.8. Example plane stress finite elements, characterized by their number of nodes n.
Differentiating the finite element displacement field yields the strain-displacement relations:  ∂Ne  ∂ Nne ∂ N2e 1 0 0 . . . ∂x 0 ∂x  ∂x  e e  ∂ Nne  ∂ N2 ∂ N1 e   e 0 .. 0 e(x, y) =  0 (14.18) ∂y ∂y ∂y  u = B u .   ∂ Nne ∂ Nne ∂ N1e ∂ N1e ∂ N2e ∂ N2e . . . ∂y ∂x ∂y ∂x ∂y ∂x This B = D N is called the strain-displacement matrix. It is dimensioned 3 × 2n. For example, if the element has 6 nodes, B is 3 × 12. The stresses are given in terms of strains and displacements by σ = E e = EBue , which is assumed to hold at all points of the element. §14.4.2. Element Energy To obtain finite element stiffness equations, the variation of the TPE functional is decomposed into contributions from individual elements: δ e = δU e − δW e = 0. where
0006 U = e
1 2
and
0006 h σ e d0002 = T
0002e
e
h eT Ee d0002e
1 2
0006 W =
(14.19)
0002e
0006
e
h u b d0002 + T
0002e
e
0003e
h uT ˆt d0003 e
(14.20)
(14.21)
Note that in (14.21) 0003te has been taken equal to the complete boundary 0003 e of the element. This is a consequence of the fact that displacement boundary conditions are applied after assembly, to a free-free structure. Consequently it does not harm to assume that all boundary conditions are of stress type insofar as forming the element equations. §14.4.3. Element Stiffness Equations Inserting the relations u = Nue , e = Bue and σ = Ee into e yields the quadratic form in the nodal displacements
e = 12 ue T Ke ue − ue T fe . (14.22) 14–11
14–12
Chapter 14: THE PLANE STRESS PROBLEM
Here the element stiffness matrix is 0006 K = e
0002e
h BT EB d0002e ,
and the consistent element nodal force vector is 0006 0006 e T e f = h N b d0002 + h NT ˆt d0003 e . 0002e
0003e
(14.23)
(14.24)
In the second integral of (14.24) the matrix N is evaluated on the element boundary only. The calculation of the entries of Ke and fe for several elements of historical or practical interest is described in subsequent Chapters. Notes and Bibliography The plane stress problem is well suited for introducing continuum finite elements, from both historical and technical standpoints. Some books use the Poisson equation for this purpose, but problems such as heat conduction cannot illustrate features such as vector-mixed boundary conditions and shear effects. The first continuum structural finite elements were developed at Boeing in the early 1950s to model delta-wing skin panels [32,164]. A plane stress model was naturally chosen for the panels. The paper that gave the method its name [25] used the plane stress problem as application driver. The technical aspects of plane stress can be found in any book on elasticity. A particularly readable one is the excellent textbook by Fung [69], which is unfortunately out of print. References Referenced items have been moved to Appendix R.
14–12
14–13
Exercises
Homework Exercises for Chapter 14 The Plane Stress Problem
EXERCISE 14.1 [A+C:25] Suppose that the structural material is isotropic, with elastic modulus E and Poisson’s ratio ν. The in-plane stress-strain relations for plane stress (σzz = σx z = σ yz = 0) and plane strain (ezz = ex z = e yz = 0) as given in any textbook on elasticity, are
0004 plane stress:
plane strain:
σx x σ yy σx y
0005
0004
1 ν 0
00050004
ν 1 0
0 0 1−ν 2  1 0004 0005 σx x E(1 − ν) ν  1− σ yy = ν (1 + ν)(1 − 2ν) σx y 0 E = 1 − ν2
0005
ex x e yy , 2ex y ν 1−ν 1 0
0
0004
0  1 − 2ν 2(1 − ν)
ex x e yy 2ex y
(E14.1)
0005 .
Show that the constitutive matrix of plane strain can be formally obtained by replacing E by a fictitious modulus E ∗ and ν by a fictitious Poisson’s ratio ν ∗ in the plane stress constitutive matrix and suppressing the stars. Find the expression of E ∗ and ν ∗ in terms of E and ν. You may also chose to answer this exercise by doing the inverse process: go from plane strain to plain stress by replacing a fictitious modulus and Poisson’s ratio in the plane strain constitutive matrix. This device permits “reusing” a plane stress FEM program to do plane strain, or vice-versa, as long as the material is isotropic. Partial answer to go from plane stress to plane strain: ν ∗ = ν/(1 − ν). EXERCISE 14.2 [A:25] In the finite element formulation of near incompressible isotropic materials (as well as plasticity and viscoelasticity) it is convenient to use the so-called Lam´e constants λ and µ instead of E and ν in the constitutive equations. Both λ and µ have the physical dimension of stress and are related to E and ν by νE E λ= , µ=G= . (E14.2) (1 + ν)(1 − 2ν) 2(1 + ν)
Conversely E=
µ(3λ + 2µ) , λ+µ
ν=
λ . 2(λ + µ)
(E14.3)
Substitute (E14.3) into (E14.1) to express the two stress-strain matrices in terms of λ and µ. Then split the stress-strain matrix E of plane strain as E = E µ + Eλ (E14.4) in which Eµ and Eλ contain only µ and λ, respectively, with Eµ diagonal and E λ33 = 0. This is the Lam´e or {λ, µ} splitting of the plane strain constitutive equations, which leads to the so-called B-bar formulation of near-incompressible finite elements.4 Express Eµ and Eλ also in terms of E and ν. For the plane stress case perform a similar splitting in which where Eλ contains only λ¯ = 2λµ/(λ + 2µ) with E λ33 = 0, and Eµ is a diagonal matrix function of µ and λ¯ .5 Express Eµ and Eλ also in terms of E and ν. 4
Equation (E14.4) is sometimes referred to as the deviatoric+volumetric splitting of the stress-strain law, on account of its physical meaning in plane strain. That meaning is lost, however, for plane stress.
5
For the physical significance of λ¯ see [143, pp. 254ff].
14–13
14–14
Chapter 14: THE PLANE STRESS PROBLEM
EXERCISE 14.3 [A:20] Include thermoelastic effects in the plane stress constitutive field equations, assuming
a thermally isotropic material with coefficient of linear expansion α. Hint: start from the two-dimensional Hooke’s law including temperature: ex x =
1 (σx x − νσ yy ) + α 0011T, E
e yy =
1 (σ yy − νσx x ) + α 0011T, E
2ex y = σx y /G,
(E14.5)
in which 0011T = 0011T (x, y) and G = 12 E/(1 + ν). Solve for stresses and collect effects of 0011T in one vector of “thermal stresses.” EXERCISE 14.4 [A:15] Derive the Cauchy stress-
n(nx =dx/ds, ny =dy/ds)
ty
y
to-traction equations (14.10) using force equilibrium along x and y and the geometric relations shown in Figure E14.1. (This is the “wedge method” in Mechanics of Materials.)
σx x
x
dy ds dx
σx y = σy x
tx
σyy
Hint: tx ds = σx x dy + σx y d x, etc.
Figure E14.1. Geometry for deriving (14.10).
EXERCISE 14.5 [A:25=5+5+15] A plate is in linearly elastic plane stress. It is shown in courses in elasticity that the internal strain energy density stored per unit volume is
U = 12 (σx x ex x + σ yy e yy + σx y ex y + σ yx e yx ) = 12 (σx x ex x + σ yy e yy + 2σx y ex y ). (a)
(E14.6)
Show that (E14.6) can be written in terms of strains only as (E14.7)
U = 12 eT E e, and hence justify (14.13). (b)
Show that (E14.6) can be written in terms of stresses only as (E14.8)
U = 12 σT C σ, where C = E−1 is the elastic compliance (strain-stress) matrix. (c)
Suppose you want to write (E14.6) in terms of the extensional strains {ex x , e yy } and of the shear stress σx y = σ yx . This is known as a mixed representation. Show that
0004 U=
1 2
ex x e yy σx y
0005T 0004
A11 A12 A13
A12 A22 A23
A13 A23 A33
00050004
ex x e yy σx y
0005 ,
and explain how the entries Ai j can be calculated6 in terms of the elastic moduli E i j .
6
The process of computing A is an instance of “partial inversion” of matrix E. See Remark 11.3.
14–14
(E14.9)
15
.
The Linear Plane Stress Triangle
15–1
15–2
Section 15: THE LINEAR PLANE STRESS TRIANGLE
TABLE OF CONTENTS Page
§15.1. Introduction §15.1.1. Parametric Representation of Functions . §15.2. Triangle Geometry and Coordinate Systems §15.2.1. Triangular Coordinates . . . . . . . §15.2.2. Linear Interpolation . . . . . . . §15.2.3. Coordinate Transformations . . . . . §15.2.4. Partial Derivatives . . . . . . . . §15.2.5. *Interesting Points and Lines . . . . . §15.3. Element Derivation §15.3.1. Displacement Interpolation . . . . . §15.3.2. Strain-Displacement Equations . . . . §15.3.3. Stress-Strain Equations . . . . . . §15.3.4. The Stiffness Matrix . . . . . . . §15.3.5. The Consistent Nodal Force Vector . . §15.3.6. Element Implementation . . . . . . §15.4. *Consistency Verification §15.4.1. *Checking Continuity . . . . . . §15.4.2. *Checking Completeness . . . . . . §15.4.3. *The Tonti Diagram of the Linear Triangle §15.5. *Derivation Using Natural Strains and Stresses §15.5.4. *Natural Strains and Stresses . . . . . §15.5.5. *Covariant Node Displacements . . . §15.5.6. *The Natural Stiffness Matrix . . . . §15.6. *Elongated Triangles and Shear Locking §15.6.1. *The Inplane Bending Test . . . . . §15.6.2. *Energy Ratios . . . . . . . . . §15.6.3. *Convergence as Mesh is Refined . . . §15. Notes and Bibliography . . . . . . . . . . . . . . . §15. References . . . . . . . . . . . . . . §15. Exercises . . . . . . . . . . . . . . .
15–2
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15–3 15–3 15–3 15–4 15–5 15–5 15–6 15–7 15–7 15–7 15–8 15–8 15–8 15–9 15–10 15–10 15–11 15–11 15–12 15–12 15–12 15–13 15–14 15–14 15–14 15–15 15–16 15–16 15–17 15–18
15–3
§15.2
TRIANGLE GEOMETRY AND COORDINATE SYSTEMS
§15.1. Introduction This Chapter presents the element stiffness equations of a three-node triangle with assumed linear displacements for the plane stress problem formulated in Chapter 14. This element is called the linear triangle. It is distinguished in several respects: (1) It belongs to both the isoparametric and superparametric element families, which are covered in the next Chapter. (2) It allows closed form derivations for the stiffness matrix and consistent force vector without the need for numerical integration. (3) It cannot be improved by the addition of internal degrees of freedom. In addition the linear triangle has historical importance.1 Although not a good performer for structural stress analysis, it is still used in problems that do not require high accuracy, as well as in non-structural applications. One reason is that triangular meshes are easily generated over arbitrary domains using techniques such as Delaunay triangulation. §15.1.1. Parametric Representation of Functions The concept of parametric representation of functions is crucial in modern FEM presentations. Together with numerical integration, it has become a key tool for the systematic development of elements in two and three space dimensions.2 Without these two tools the element developer would become lost in an algebraic maze as element geometry and shape functions get more complicated. The essentials of the idea of parametric representation can be illustrated through a simple example. Consider the following alternative representations of the unit-circle function, x 2 + y 2 = 1: y=
0002
x = cos θ,
1 − x2 y = sin θ
(15.1) (15.2)
The direct representation (15.1) fits the conventional function notation, i.e., y = f (x). Given a value of x, it returns one or more y. On the other hand, the representation (15.2) is parametric: both x and y are given in terms of one parameter, the angle θ. Elimination of θ through the trigonometric identity cos2 θ + sin2 θ = 1 recovers x 2 + y 2 = 1. But there are many situations in which working with the parametric form throughout the development is more convenient. Continuum finite elements provide a striking illustration of this point.
1
The triangle was one of the two plane-stress continuum elements presented by Turner et. al. in their landmark 1956 paper [[164]. This publication is widely regarded as the start of the present FEM. Accordingly, the element is also called the Turner triangle, but the derivation was not done with assumed displacements. See Notes and Bibliography.
2
Numerical integration is not useful for the 3-node triangle, but essential in the more complicated iso-P elements covered in Chapters 16ff.
15–3
15–4
Section 15: THE LINEAR PLANE STRESS TRIANGLE
§15.2. Triangle Geometry and Coordinate Systems The geometry of the 3-node triangle shown in Figure 15.1(a) is specified by the location of its three corner nodes on the {x, y} plane. The nodes are labelled 1, 2, 3 while traversing the sides in counterclockwise fashion. The location of the corners is defined by their Cartesian coordinates: {xi , yi } for i = 1, 2, 3. The element has six degrees of freedom, defined by the six nodal displacement components { u xi , u yi }, for i = 1, 2, 3. The interpolation of the internal displacements { u x , u y } from these six values is studied in §15.3, after triangular coordinates are introduced. The area of the triangle is denoted by A and is given by 0004 0003 1 1 1 2A = det x1 x2 x3 = (x2 y3 − x3 y2 ) y1 y2 y3 + (x3 y1 − x1 y3 ) + (x1 y2 − x2 y1 ). (15.3)
(a)
The area given by (15.3) is a signed quantity. It is positive if the corners are numbered in cyclic counterclockwise order (when looking down from the +z axis), as illustrated in Figure 15.1(b). This convention is followed in the sequel.
(b)
3 (x3 ,y3 )
3
Area A > 0 2
2 (x2 ,y2) y
1
1 (x 1 ,y1) x
z up, towards you
Figure 15.1. The three-node, linear-displacement plane stress triangular element: (a) geometry; (b) area and positive boundary traversal.
§15.2.1. Triangular Coordinates Points of the triangle may also be located in terms of a parametric coordinate system: ζ1 , ζ2 , ζ3 .
(15.4)
In the literature these parameters receive an astonishing number of names, as the list given in Table 15.1 shows. In the sequel the name triangular coordinates will be used to emphasize the close association with this particular geometry. Equations ζi = constant
(15.5)
represent a set of straight lines parallel to the side opposite to the i th corner, as depicted in Figure 15.2. The equations of sides 2–3, 3–1 and 1–2 are ζ1 = 0, ζ2 = 0 and ζ3 = 0, respectively. The three corners have coordinates (1,0,0), (0,1,0) and (0,0,1). The three midpoints of the sides have coordinates ( 12 , 12 , 0), (0, 12 , 12 ) and ( 12 , 0, 12 ), the centroid has coordinates ( 13 , 13 , 13 ), and so on. The coordinates are not independent because their sum is unity: ζ1 + ζ2 + ζ3 = 1.
15–4
(15.6)
15–5
§15.2 3
TRIANGLE GEOMETRY AND COORDINATE SYSTEMS
3
ζ1 = 0
3 /
ζ2 = 0
ζ2 = 1
/ / /
1
/ / /
2
/
2
/
2
1
ζ1 = 1
ζ3 = 1
1
ζ3 = 0
Figure 15.2. Triangular coordinates.
Table 15.1 Names of element parametric coordinates Name
Applicable to
natural coordinates isoparametric coordinates shape function coordinates barycentric coordinates M¨obius coordinates triangular coordinates area (also written “areal”) coordinates
all elements isoparametric elements isoparametric elements simplices (triangles, tetrahedra, ..) triangles all triangles straight-sided triangles
Triangular coordinates normalized as per ζ1 + ζ2 + ζ3 = 1 are often qualified as “homogeneous” in the mathematical literature. Remark 15.1. In older (pre-1970) FEM publications triangular coordinates are often called area coordinates (occasionally areal coordinates). This name comes from the following interpretation: ζi = A jk /A, where A jk is the area subtended by the triangle formed by the point P and corners j and k, in which j and k are 3-cyclic permutations of i. Historically this was the way the coordinates were defined in 1960s papers. Unfortunately this interpretation does not carry over to general isoparametric triangles with curved sides and thus it is not used here.
§15.2.2. Linear Interpolation Consider a function f (x, y) that varies linearly over the triangle domain. In terms of Cartesian coordinates it may be expressed as f (x, y) = a0 + a1 x + a2 y,
(15.7)
where a0 , a1 and a2 are coefficients to be determined from three conditions. In finite element work such conditions are often the nodal values taken by f at the corners: f1, f2, f3. The expression in triangular coordinates makes direct use of those three values: 0003 0004 0003 0004 ζ1 f1 f (ζ1 , ζ2 , ζ3 ) = f 1 ζ1 + f 2 ζ2 + f 3 ζ3 = [ f 1 f 2 f 3 ] ζ2 = [ ζ1 ζ2 ζ3 ] f 2 . ζ3 f3 Formula (15.9) is called a linear interpolant for f . 15–5
(15.8)
(15.9)
15–6
Section 15: THE LINEAR PLANE STRESS TRIANGLE
§15.2.3. Coordinate Transformations Quantities that are closely linked with the element geometry are best expressed in triangular coordinates. On the other hand, quantities such as displacements, strains and stresses are often expressed in the Cartesian system {x, y}. Consequently we need transformation equations through which it is possible to pass from one coordinate system to the other. Cartesian and triangular coordinates are linked by the relation 0003 0004 0003 1 1 x = x1 y1 y
1 x2 y2
1 x3 y3
00040003
ζ1 ζ2 ζ3
0004 .
(15.10)
The first equation says that the sum of the three coordinates is one. The next two express x and y linearly as homogeneous forms in the triangular coordinates. These are obtained by applying the linear interpolant (15.9) to the Cartesian coordinates: x = x1 ζ1 + x2 ζ2 + x3 ζ3 and y = y1 ζ1 + y2 ζ2 + y3 ζ3 . Inversion of (15.10) yields 0003
0004
0003
0003 00040003 0004 1 1 2A23 x = 2A31 2A 2A y 12
00040003 0004 1 x . y (15.11) Here x jk = x j − xk , y jk = y j − yk , A is the triangle area given by (15.3) and A jk denotes the area subtended by corners j, k and the origin of the x–y system. If this origin is taken at the centroid of the triangle, A23 = A31 = A12 = A/3. ζ1 ζ2 ζ3
1 = 2A
x2 y3 − x3 y2 x3 y1 − x1 y3 x1 y2 − x2 y1
y2 − y3 y3 − y1 y1 − y2
x3 − x2 x1 − x3 x2 − x1
y23 y31 y12
x32 x13 x21
§15.2.4. Partial Derivatives From equations (15.10) and (15.11) we immediately obtain the following relations between partial derivatives: ∂x ∂y = xi , = yi , (15.12) ∂ζi ∂ζi 2A
∂ζi = y jk , ∂x
2A
∂ζi = xk j . ∂y
(15.13)
In (15.13) j and k denote the 3-cyclic permutations of i. For example, if i = 2, then j = 3 and k = 1. The derivatives of a function f (ζ1 , ζ2 , ζ3 ) with respect to x or y follow immediately from (15.13) and application of the chain rule: ∂f 1 = ∂x 2A 1 ∂f = ∂y 2A
0005 0005
∂f ∂f ∂f y23 + y31 + y12 ∂ζ1 ∂ζ2 ∂ζ3 ∂f ∂f ∂f x32 + x13 + x21 ∂ζ1 ∂ζ2 ∂ζ3 15–6
0006 0006
(15.14)
15–7
§15.3
ELEMENT DERIVATION
which in matrix form is  ∂f  1  ∂x   ∂f  = 2A ∂y
y23
y31
x32
x13
 ∂f  ∂ζ 000e 1 y12  ∂ f  x21   ∂ζ2  ∂f ∂ζ3
    .   
(15.15)
With these mathematical ingredients in place we are now in a position to handle the derivation of straight-sided triangular elements, and in particular the linear triangle. §15.2.5. *Interesting Points and Lines Some distinguished lines and points of a straight-sided triangle are briefly described here for use in other developments as well as in Exercises. The triangle medians are three lines that join the corners to the midpoints of the opposite sides, as pictured in Figure 15.3(a). The midpoint opposite corner i is labeled Mi .3 (a) 3 (b) 3 The medians 1–M , 2–M and 3–M have equations 1
2
3
ζ2 = ζ3 , ζ3 = ζ1 and ζ1 = ζ2 , respectively, in triangular coordinates. The medians intersect at the centroid C of coordinates { 13 , 13 , 13 }. Other names for the centroid are barycenter and center of gravity. If you make a real triangle out of cardboard, you can balance the triangle at this point. It can be shown that the centroid trisects the medians, that is to say, the distance from a corner to the centroid is twice the distance from the centroid to the opposite side of the triangle.
Μ1 Μ2
Η2
2
C
Η
2 Η3
Μ3
1
Η1
1
Figure 15.3. Medians and altitudes of a triangle.
The altitudes are three lines that connect each corner with their projections onto the opposing sides, as depicted in Figure 15.3(b). The projection of corner i is identified Hi , so the altitudes are 1–H1 , 2–H2 and 3–H3 . Locations Hi are called altitude feets. The altitudes intersect at the triangle orthocenter H . The lengths of those segments are the triangle heights. The triangular coordinates of Hi , H , as well as the altitude equations, are worked out in an Exercise. Another interesting point is the center OC of the circumscribed circle, or circumcircle. This is the unique circle that passes through the three corners. This point is not shown in Figure 15.3 to reduce clutter. It can be geometrically constructed by drawing the normal to each side at the midpoints; three three lines (called the perpendicular side bisectors) intersect at OC . A famous theorem by Euler asserts that the centroid, the orthocenter and the circumcircle center fall on a straight line, called the Euler line. Furthermore, C lies between OC and H , and the distance OC –H is three times the distance H –C.
§15.3. Element Derivation The simplest triangular element for plane stress (and in general, for 2D problems of variational index m = 1) is the three-node triangle with linear shape functions. The shape functions are simply the triangular coordinates. That is, Nie = ζi for i = 1, 2, 3. 3
Midpoints should not be confused with midside nodes used in the development of curved, higher order triangular elements in subsequent chapters. Midside nodes are labeled 4, 5 and 6, and are not necessarily located at side midpoints. In fact for a curved sided triangle the definition of medians and altitudes has to be changed.
15–7
Section 15: THE LINEAR PLANE STRESS TRIANGLE
15–8
§15.3.1. Displacement Interpolation For the plane stress problem we select the linear interpolation (15.9) for the displacement components u x and u y at an arbitrary point P(ζ1 , ζ2 , ζ3 ): u x = u x1 ζ1 + u x2 ζ2 + u x3 ζ3 , u y = u y1 ζ1 + u y2 ζ2 + u y3 ζ3 .
(15.16)
The interpolation is illustrated in Figure 15.4. These u x by linear uy3 P(ζ 1 ,ζ 2 ,ζ3 ) relations can be combined in a matrix form that befits the uy } interpolation u x3 expression (14.17) for an arbitrary plane stress element: 3   u x1 uy uy2 u   y1 000e
000e ux2  ux ζ1 0 ζ2 0 ζ3 0  u x2  ux e = 2   = N u . uy1 uy 0 ζ1 0 ζ2 0 ζ3  u y2    u x3 1 ux1 u y3 (15.17) Figure 15.4. Displacement interpolation over triangle. where N is the matrix of shape functions. §15.3.2. Strain-Displacement Equations The strains within the elements are obtained by differentiating the shape functions with respect to x and y. Using (15.15) and the general form (14.18) we get   u x1 0003 0004u  y23 0 y31 0 y12 0  y1  1 u  (15.18) e = D Ne ue = 0 x32 0 x13 0 x21  x2  = B ue ,  u y2  2A x y23 x13 y31 x21 y12   32 u x3 u y3 in which D denotes the symbolic strain-to-displacement differentiation operator given in (14.6), and B is the strain-displacement matrix. Note that the strains are constant over the element. This is the origin of the name constant strain triangle (CST) given it in many finite element publications. §15.3.3. Stress-Strain Equations The stress field σ is related to the strain field by the elastic constitutive equation in (14.5), which is repeated here for convenience: 0004 0003 00040003 0004 0003 E 11 E 12 E 13 ex x σx x (15.19) σ = σ yy = E 12 E 22 E 23 e yy = E e, σx y E 13 E 23 E 33 2ex y where E i j are plane stress elastic moduli. The constitutive matrix E will be assumed to be constant over the element. Because the strains are constant, so are the stresses. 15–8
15–9
§15.3
ELEMENT DERIVATION
§15.3.4. The Stiffness Matrix The element stiffness matrix is given by the general formula (14.23), which is repeated here 000f e h BT EB d000be , (15.20) K = e
where e is the triangle domain, and h is the plate thickness that appears in the plane stress problem. Since B and E are constant, they can be taken out of the integral: 000f e T h d000be (15.21) K = B EB e
If the thickness h is uniform over the element the remaining integral in (15.21) is simply h A, and we obtain the closed form y
23
h Ke = Ah BT EB = 4A
 0  y31   0  y12 0
0 x32 0 x13 0 x21

x32 y23  0003 E 11 x13   E 12 y31   E 13 x21 y12
E 12 E 22 E 23
E 13 E 23 E 33
00040003
y23 0 x32
0 x32 y23
y31 0 x13
0 x13 y31
y12 0 x21
0 x21 y12
0004 .
(15.22)
Exercise 15.1 deals with the case of a linearly varying thickness. §15.3.5. The Consistent Nodal Force Vector For simplicity we consider here only internal body forces4 defined by the vector field
000e bx b= by
(15.23)
which is specified per unit of volume. The consistent nodal force vector fe is given by the general formula (14.23) of the previous Chapter:   ζ1 0  0 ζ1  000f 000f   0 ζ h (Ne )T b d000be = h 2 (15.24) fe =  b d000be . 0 ζ  e e 2   ζ3 0 0 ζ3 The simplest case is when the body force components (15.23) as well as the thickness h are constant over the element. Then we need the integrals 000f 000f 000f e e ζ1 d = ζ2 d = ζ3 d000be = 13 A (15.25) e
4
000be
000be
For consistent force computations corresponding to distributed boundary loads over a side, see Exercise 15.4.
15–9
15–10
Section 15: THE LINEAR PLANE STRESS TRIANGLE
Trig3IsoPMembraneStiffness[encoor_,Emat_,h_]:=Module[{ x1,x2,x3,y1,y2,y3,x21,x13,x32,y12,y31,y23,A,Be,Ke}, {{x1,y1},{x2,y2},{x3,y3}}=encoor; A=Simplify[(x2*y3-x3*y2+(x3*y1-x1*y3)+(x1*y2-x2*y1))/2]; {x21,x13,x32}={x2-x1,x1-x3,x3-x2}; {y12,y31,y23}={y1-y2,y3-y1,y2-y3}; Be={{y23,0,y31,0,y12,0},{0,x32,0,x13,0,x21}, {x32,y23,x13,y31,x21,y12}}/(2*A); Ke=A*h*Transpose[Be].Emat.Be; Return[Ke]]; Figure 15.5. Implementation of linear-triangle stiffness matrix calculation as a Mathematica module.
which replaced into (15.24) gives fe =
Ah [ bx 3
by
bx
by
bx
b y ]T .
(15.26)
This agrees with the simple element-by-element force-lumping procedure, which assigns one third of the total force along the {x, y} directions: Ahbx and Ahb y , to each corner. Remark 15.2. The integrals (15.25) are particular cases of the general integration formula of monomials in triangular coordinates:
1 2A
000f
j
000be
ζ1i ζ2 ζ3k d000be =
i! j! k! , (i + j + k + 2)!
i ≥ 0, j ≥ 0, k ≥ 0.
(15.27)
which can be derived by repeated integration by parts. Here i, j, k are integer exponents. This formula only holds for triangles with straight sides, and thus is useless for higher order curved elements. Formulas (15.25) correspond to setting exponents i = 1, j = k = 0 in (15.27), and permuting {i, j, k} cyclically.
§15.3.6. Element Implementation The implementation of the linear plane stress triangle in any programming language is very simple. An implementation in the form of a Mathematica module is shown in Figure 15.5. The module needs only 8 lines of code. It is invoked as Ke=Trig3IsoPMembraneStiffness[encoor,Emat,h];
(15.28)
The arguments are encoor
Element node coordinates, arranged as a list: { { x1,y1 },{ x2,y2 },{ x3,y3 } }.
Emat
A two-dimensional list storing the 3 × 3 plane stress matrix of elastic moduli as { { E11,E12,E13 },{ E12,E22,E33 },{ E13,E23,E33 } }.
h
Plate thickness, assumed uniform over the triangle.
This module is exercised by the statements listed at the top of Figure 15.6, which form a triangle with corner coordinates { { 0,0 },{ 3,1 },{ 2,2 } }, isotropic material matrix with E 11 = E 22 = 32, E 12 = 8, E 33 = 16, others zero, and unit thickness. The results are shown at the bottom of Figure 15.6. The computation of stiffness matrix eigenvalues is always a good programming test, since 3 eigenvalues must be exactly zero and the other 3 real and positive (this is explained in Chapter 19). The last test statement draws the triangle. 15–10
15–11
§15.4 *CONSISTENCY VERIFICATION encoor={{0,0},{3,1},{2,2}}; Emat=8*{{4,1,0},{1,4,0},{0,0,2}}; Ke=Trig3IsoPMembraneStiffness[encoor,Emat,1]; Print['Ke=',Ke//MatrixForm]; Print['eigs of Ke=',Chop[Eigenvalues[N[Ke]]]//InputForm]; Show[Graphics[RGBColor[1,0,0]],Graphics[AbsoluteThickness[2]], Graphics[Polygon[encoor]],Axes->True]; 6 3 3 6 −4 2 Ke = −2 4 −2 −5 − 1 − 10
−4 2 24 − 12 − 20 10
−2 4 − 12 24 14 − 28
−2 −5 − 20 14 22 −9
−1 − 10 10 − 28 −9 38
eigs of Ke = {75.53344879465156, 32.864856509030794, 11.601694696317617, 0, 0, 0} 2 1.5 1 0.5 0.5
1
1.5
2
2.5
3
Figure 15.6. Test statements to exercise the module of Figure 15.5, and outputs.
§15.4.
*Consistency Verification
It remains to check whether the piecewise linear expansion (15.16) for the element displacements meets the completeness and continuity criteria studied in more detail in Chapter 19 for finite element trial functions. Such consistency conditions are sufficient to insure convergence towards the exact solution of the mathematical model as the mesh is refined. The variational index for the plane stress problem is m = 1. Consequently the trial functions should be 1-complete, C 0 continuous, and C 1 piecewise differentiable.
The variation of ux and uy over side 1-2 depends only on the nodal values ux1, ux2 , uy1 and uy2.
§15.4.1. *Checking Continuity Along any triangle side, the variation of u x and u y is linear and uniquely determined by the value at the nodes on that side. For example, over side 1–2 of an individual triangle, which has equation ζ3 = 0, u x = u x1 ζ1 + u x2 ζ2 + u x3 ζ3 = u x1 ζ1 + u x2 ζ2 , u y = u y1 ζ1 + u y2 ζ2 + u y3 ζ3 = u y1 ζ1 + u y2 ζ2 .
(15.29)
(e1) 1
2
(e2)
Figure 15.7. Interelement continuity check.
An identical argument holds for that side when it belongs to an adjacent triangle, such as elements (e1) and (e2) shown in Figure 15.7. Since the node values on all elements that meet at a node are the same, u x and u y match along the side, and the trial function is C 0 continuous across elements.Because the functions are continuous inside the elements, it follows that the conformity requirement is met.
15–11
15–12
Section 15: THE LINEAR PLANE STRESS TRIANGLE
§15.4.2. *Checking Completeness The completeness condition for variational order m = 1 require that the shape functions Ni = ζi be able to represent exactly any linear displacement field: u x = α0 + α1 x + α2 y,
u y = β0 + β1 x + β1 y.
(15.30)
To check this we obtain the nodal values associated with the motion (15.30): u xi = α0 + α1 xi + α2 yi and u yi = β0 + β1 xi + β2 yi for i = 1, 2, 3. Replace these in (15.17) and see if (15.30) is recovered. Here are the detailed calculations for component u x : ux =
0010 i
= α0
u xi ζi =
0010
0010
(α0 + α1 xi + α2 yi )ζi =
i
ζi + α1
i
0010
0010
i
i
(xi ζi ) + α2
0010
(α0 ζi + α1 xi ζi + α2 yi ζi )
i
(15.31)
(yi ζi ) = α0 + α1 x + α2 y.
Component u y can be similarly verified. Consequently (15.17) satisfies the completeness requirement for the plane stress problem (and in general, for any problem of variational index 1). Finally, a piecewise linear trial function is obviously C 1 piecewise differentiable and consequently has finite energy. Thus the two completeness requirements are satisfied. §15.4.3. *The Tonti Diagram of the Linear Triangle Stiffness
u
For further developments covered in more advanced courses, it is convenient to split the governing equations of the element. In the case of the linear triangle they are, omitting element superscripts:
Kinematic
Equilibrium T
f=VB σ
e=Bu e = Bu,
σ = Ee,
f = A σ = V B σ. T
T
(15.32)
Constitutive
e
Here V = h m A is the volume of the element, h m being the mean thickness. The equations (15.32) may be graphically represented with the diagram shown in Figure 15.8. This is a discrete Tonti diagram similar to those of Chapter 6. §15.5.
f
T
f=VB EBu=Ku
σ
σ=Ee Figure 15.8. Tonti matrix diagram for linear triangle.
*Derivation Using Natural Strains and Stresses
The element derivation in §15.3 uses Cartesian strains and stresses, as well as {x, y} displacements. The only intrinsic quantities are the triangle coordinates. This advanced section examines the derivation of the element stiffness matrix through natural strains, natural stresses and covariant displacements. Although the procedure does not offer obvious shortcuts over the previous derivation, it becomes important in the construction of more complicated high performance elements. It also helps reading recent literature in assumed strain elements.
15–12
(a)
3
3
ε32 = ε1 ε13= ε2
2
ε21= ε3 1
(b) τ13= τ2
τ32 = τ1 2
τ21= τ3 1
Figure 15.9. Geometry-intrinsic fields for the linear triangle: (a) natural strains 000ei , (b) natural stresses τi .
15–13
§15.5 *DERIVATION USING NATURAL STRAINS AND STRESSES
§15.5.4. *Natural Strains and Stresses Natural strains are extensional strains directed parallel to the triangle sides, as shown in Figure 15.9(a). Natural strains are denoted by 000e21 ≡ 000e3 , 000e32 ≡ 000e1 , and 000e13 ≡ 000e2 . Because they are constant over the triangle, no node value association is needed. Similarly, natural stresses are normal stresses directed parallel to the triangle sides, as shown in Figure 15.9(b). Natural stresses are denoted by τ21 ≡ τ3 , τ32 ≡ τ1 , and τ13 ≡ τ2 . Because they are constant over the triangle, no node value association is needed. The natural strains can be related to Cartesian strains by the following tensor transformation5
0003 0002=
000e1 000e2 000e3
0004

c12 =  c22 c32
0003
s12 s22 s32
s1 c1 s2 c2  s3 c3
ex x e yy 2ex y
(a)
0004
(b) 3
3
φ1
L1 = L32
= T−1 e e. L2 = L13
2
(15.33)
Here c1 = x32 /L 1 , s1 = y32 /L 1 , c2 = x13 /L 2 , s2 = y13 /L 2 , c3 = x21 /L 3 , and s3 = y21 /L 3 , are sines and cosines of the side directions with respect to {x, y}, as illustrated in Figure 15.10.
2
φ3
φ2
L3 = L21
x 1
1
Figure 15.10. Quantities appearing in natural strain and stress calculations: (a) side lengths, (b) side directions.
The inverse of this relation is
0003

0004

0003 0004 y31 y21 L 21 y12 y32 L 22 y23 y13 L 23 000e1 1 2 2 2   e= = x x L x x L x x L 000e2 = Te 0002. 31 21 1 12 32 2 23 13 3 4A2 2 2 2 000e3 (y31 x12 + x13 y21 )L 1 (y12 x23 + x21 y32 )L 2 (y23 x31 + x32 y13 )L 3 (15.34) Note that Te is constant over the triangle. From the invariance of the strain energy density σT e = τT 0002 it follows that the stresses transform as τ = Te σ and σ = T−1 e τ. That strain energy density may be expressed as ex x e yy 2ex y
U = 12 eT Ee = 12 0002T En 0002,
En = TeT ETe .
(15.35)
Here En is a stress-strain matrix that relates natural stresses to natural strains as τ = En 000e. It may be therefore called the natural constitutive matrix. §15.5.5. *Covariant Node Displacements d6
Covariant node displacements di are directed along the side directions, as shown in Figure 15.11, which defines the notation used for them. They are related to the Cartesian node displacements by
d  1
c
3
s3
0
0
 d2   c2 s2 0 0  d3   0 0 c1 s1   d=  d4  =  0 0 c3 s3    d5 d6
0 0
0 0
0 0
0 0
0 0 0 0 c2 c1
d5 d3
u  0 x1
0   u y1    0   u x2  = Td u. 0   u y2    s2 u x3 s1 u y3 (15.36)
1
d4
d1
Figure 15.11. Covariant node displacements di .
This is the “straingage rosette” transformation studied in Mechanics of Materials books.
15–13
2
d2
The inverse relation is 5
3
15–14
Section 15: THE LINEAR PLANE STRESS TRIANGLE
u 
L y 3 31  u y1   L 3 x13  u x2   = 1  0 u=  u y2  2A  0    x1
u x3 u y3
0 0
L 2 y21 L 2 x12 0 0 0 0
0 0 L 1 y12 L 1 x21 0 0
0 0 L 3 y32 L 3 x23 0 0
0 0 0 0 L 2 y23 L 2 x32


0 d1 0   d2    0    d3  = T−1 d d. 0   d4    L 1 y13 d5 L 1 x31 d6
(15.37)
The natural strains are evidently given by the relations 000e1 = (d6 − d3 )/L 1 , 000e2 = (d2 − d5 )/L 2 and 000e3 = (d4 − d1 )/L 3 . Collecting these in matrix form:
0003 0002=
000e1 000e2 000e3
0004
0003 =
d  1
0 0 −1/L 3
0 1/L 2 0
−1/L 1 0 0
0 0 1/L 3
0 −1/L 2 0
0004 d2  1/L 1   d3    = B000e d. 0  d4    0
(15.38)
d5 d6
§15.5.6. *The Natural Stiffness Matrix The natural stiffness matrix for constant h is Kn = (Ah) B000eT En B000e ,
En = TeT ETe .
(15.39)
The Cartesian stiffness matrix is K = TdT Kn Td .
(15.40)
Comparing with K = (Ah) B EB we see that T
B = Te B000e Td , §15.6.
−1 B000e = T−1 e BTd .
(15.41)
*Elongated Triangles and Shear Locking
A well known deficiency of the 3-node triangle is inability to follow rapidly varying stress fields. This is understandable since stresses within the element, for uniform material properties, are constant. But its 1D counterpart: the 2-node bar element, is nodally exact for displacements under some mild assumptions stated in Chapter 12, and correctly solves loaded-at-joints trusses with one element per member. On the other hand, the triangle can be arbitrarily way off under unhappy combinations of loads, geometry and meshing. What happens in going from 1D to 2D? New effects emerge, notably shear energy and inplane bending. These two can combine to produce shear locking: elongated triangles can become extraordinarily stiff under inplane bending because of spurious shear energy.6 The bad news for engineers is that wrong answers caused by locking are non-conservative: deflections and stresses can be so grossly underestimated that safety margins are overwhelmed. To characterize shear locking quantitatively it is convenient to use macroelements in which triangles are combined to form a 4-node rectangle. This simplifies repetition to form regular meshes. The rectangle response under in-plane bending is compared to that of a Bernoulli-Euler beam segment. It is well known that the latter is exact under constant moment. The response ratio of macroelement to beam is a good measure of triangle performance under bending. Such benchmarks are technically called higher order patch tests. Test results can be summarized by one number: the energy ratio, which gives a scalar measure of relative stiffness. 6
The deterioration can be even more pronounced for its spatial counterpart: the 4-node tetrahedron element, because shear effects are even more important in three dimensions.
15–14
15–15
; ;
§15.6 *ELONGATED TRIANGLES AND SHEAR LOCKING
M
4
y
z
3
z
b = a/γ 1
x 4
Type I: Crisscrossed 1
y
M 2
a
Thickness h/2 34
3
4
+
h
3
5
Type II: Union-Jack
1
2
2 1
b
Cross section
2
Figure 15.12. The bending test with two macroelement types.
§15.6.1. *The Inplane Bending Test The test is defined in Figure 15.12. A Bernoulli-Euler plane beam of thin rectangular cross-section of height b and thickness h is bent under applied end moments M. The beam is fabricated of isotropic material with elastic modulus E and Poisson’s ratio ν. Except for possible end effects the exact solution of the beam problem (from both the theory-of-elasticity and beam-theory standpoints) is a constant bending moment M(x) = M along the span. The associated curvature is κ = M/(E Iz ) = 12M/(Eb3 h). The exact energy taken by a beam segment of length a is Ubeam = 12 Mκa = 6M 2 a/(Eb3 h) = 241 Eb3 hκ 2 a = 241 Eb3 hθa2 /a. In the latter θa = κa is the relative rotation of two cross sections separated by a. To study the bending performance of triangles the beam is modeled with one layer of identical rectangular macroelements dimensioned a ×b and made up of triangles, as illustrated in Figure 15.12. The rectangle aspect ratio is γ = a/b. All rectangles undergo the same deformations and thus it is enough to study a individual macroelement 1-2-3-4. Two types are considered here: Crisscrossed (CC). Formed by overlaying triangles 1-2-4, 3-4-2, 2-3-1 and 4-1-2, each with thickness h/2. Using 4 triangles instead of 2 makes the macroelement geometrically symmetric. Union-Jack (UJ). Formed by placing a fifth node at the center and dividing the rectangle into 4 triangles: 1-2-5, 2-3-5, 3-4-5, 4-1-5. By construction this element is also geometrically symmetric.
−θa/2 P 4
§15.6.2. *Energy Ratios
θa/2 3 P
The assembled macroelement stiffnesses are KCC and KU+ J , of orders 8 × 8 and 10 × 10, respectively. For the b = a/γ 1 2 latter the internal node 5 is statically condensed producing P P a an 8 × 8 stiffness KU . To test performance we apply four alternating corner loads as shown in Figure 15.13. The Figure 15.13. Bending a macroelement. resultant bending moment is M = Pb. Although triangles cannot copy curvatures pointwise,7 macroelement edges can rotate since constituent triangles can expand or contract. Because of symmetries, the rotations of sides 1-2 and 3-4 are −θa /2 and θa /2, as illustrated in Figure 15.13. The corresponding corner x displacements are ±bθa /4 whereas the y displacements are zero. Assemble these into a node displacement 8-vector u M . u M = 14 bθa [ −1
0
1
0
−1
0
1
0 ]T
(15.42)
The internal energy taken by a macroelement of 8 × 8 stiffness K M under (15.42) is U M = 12 uTM K M u M , which can be expressed as a function of E, ν, a, b, h and θa .8 7
That is the reason why they can be so stiff under bending.
8
The load P could be recovered via K M u M , but this value is not needed to compute energy ratios.
15–15
Section 15: THE LINEAR PLANE STRESS TRIANGLE
15–16
ClearAll[a,b,Em,h,γ ]; b=a/γ ; Iz=h*b^3/12; Ubeam=Simplify[(1/2)*Em*Iz*θa^2/a]; Emat=Em*{{1,0,0},{0,1,0},{0,0,1/2}}; nc={{-a,-b},{a,-b},{a,b},{-a,b},{0,0}}/2; enCC={{1,2,4},{3,4,2},{2,3,1},{4,1,3}}; enUJ={{1,2,5},{2,3,5},{3,4,5},{4,1,5}}; r={0,0}; For [m=1,m<=2,m++, mtype={'CC','UJ'}[[m]]; nF={8,10}[[m]]; K=Table[0,{nF},{nF}]; f=Table[0,{nF}]; For [e=1,e<=4,e++, If [mtype'CC', enl=enCC[[e]], enl=enUJ[[e]]]; {n1,n2,n3}=enl; encoor={nc[[n1]],nc[[n2]],nc[[n3]]}; ht=h; If [mtype'CC', ht=h/2]; Ke=Trig3IsoPMembraneStiffness[encoor,Emat,ht]; eft={2*n1-1,2*n1,2*n2-1,2*n2,2*n3-1,2*n3}; For [i=1,i<=6,i++, For [j=1,j<=6,j++, ii=eft[[i]]; jj=eft[[j]]; K[[ii,jj]]+=Ke[[i,j]] ]]; ]; KM=K=Simplify[K]; If [mtype'UJ', {K,f}= Simplify[CondenseLastFreedom[K,f]]; {KM,f}=Simplify[CondenseLastFreedom[K,f]]]; Print['KM=',KM//MatrixForm]; uM={1,0,-1,0,1,0,-1,0}*θa*b/4; UM=uM.KM.uM/2; rM=Simplify[UM/Ubeam]; Print['rM=',rM]; r[[m]]=rM; ]; Plot[Evaluate[r],{γ ,0,10}];
Figure 15.14. Script to compute energy ratios for the two macroelements of Figure 15.12.
The ratio r M = U M /Ubeam is called the energy ratio. If r M > 1 the macroelement is stiffer than the beam because it take more energy to bend it to conform to the same edge rotations, and the 2D model is said to be overstiff. Results for zero Poisson’s ratio, computed by the script of Figure 15.14, are 3 rCC = 3 + γ 2 , 2
rU J =
3(1 + γ 2 )2 . 2 + 4γ 2
(15.43)
If for example γ = a/b = 10, which is an elongated rectangular shape of 10:1 aspect ratio, rCC = 153 and the crisscrossed macroelement is 153 times stiffer than the beam. For the Union-Jack configuration rU J = 10201/134 = 76.13; about twice better but still way overstiff. If γ = 1, rCC = 4.5 and rU J = 2: overstiff but not dramatically so. The effect of a nonzero Poisson’s ratio is studied in Exercise 15.10. §15.6.3. *Convergence as Mesh is Refined Note that if γ = a/b → 0, rCC → 3 and rU J → 1.5. So even if the beam of Figure 15.12 was divided into an infinite number of macroelements along x the solution will not converge. It is necessary to subdivide also along the height. If 2n (n ≥ 1) identical macroelement layers are placed along the beam height while γ is kept fixed, the energy ratio becomes r (2n) =
22n − 1 + r (1) r (1) − 1 = 1 + , 22n 22n
(15.44)
where r (1) is the ratio (15.43) for one layer. If r (1) = 1, r (2n) = 1 for all n ≥ 1, so bending exactness is maintained as expected. If n = 1 (two layers), r (2) = (3+r (1) )/4 and if n = 2 (four layers), r (4) = (7+r (1) )/8. If n → ∞, r (2n) → 1, but convergence can be slow. For example, suppose that γ = 1 (unit aspect ratio a = b) and that r (1) = rCC = 4.5. To get within 1% of the exact solution, 1 + 3.5/22n < 1.01. This is satisfed if n ≥ 5, meaning 10 layers of elements along y. If the beam span is 10 times the height, 1000 macroelements or 4000 triangles are needed for this simple problem, which is solvable exactly by one beam element. The stress accuracy of triangles is examined in Chapter 28.
15–16
15–17
§15.
References
Notes and Bibliography As a structural element, the linear triangle was first developed in the 1956 paper by Turner, Clough, Martin and Topp [164]. The target application was modeling of delta wing skin panels. Arbitrary quadrilaterals were formed by assembling triangles as macroelements. Because of its geometric flexibility, the element was soon adopted in aircraft structural analysis codes in the late 1950’s. It moved to Civil Engineering applications through the research and teaching at Berkeley of Ray Clough, who gave the method its name in [25]. The derivation method of [164] would look unfamiliar to present FEM practicioners used to the displacement method. It was based on assumed stress modes. More precisely: the element, referred to a local Cartesian system {x, y}, is put under three constant stress states: σx x , σ yy and σx y collected in array σ. Lumping the stress field to the nodes gives the node forces: f = Lσ. The strain field computed from stresses is e = E−1 σ. This is integrated to get a deformation-displacement field, to which 3 rigid-body modes are added as integration constants. Evaluating at the nodes produces e = Au, and the stiffness matrix follows on eliminating σ and e: K = LEA. For constant thickness and material properties it happens that L = V AT and so K = V AT EA happily turned out to be symmetric. (This A is the B of (15.18) times 2A.) The derivation from assumed displacements evolved later. It is not clear who worked out it first, although it is mentioned in [25,177]. The equivalence of the two forms, through energy principles, had been noted by Gallagher [70]. Early displacement derivations typically started from linear polynomials in Cartesian coordinates. For example Przemieniecki [134] begins with u x = c1 x + c2 y + c3 ,
u y = c4 x + c5 y + c6 .
(15.45)
Here the ci play the role of generalized coordinates, which have to be eventually eliminated in favor of node displacements. The same approach is used by Clough in a widely disseminated 1965 article [27]. Even for this simple element the approach is unnecessarily complicated and leads to long computations. The elegant derivation in triangular coordinates was popularized by Argyris [5]. The idea of using linear interpolation over a triangular mesh actually precedes [164] by 13 years. It appears in the Appendix of an article by Courant [37], where it is applied to a Poisson’s equation modeling St. Venant’s torsion. The idea did not influence early work in FEM, however, since as noted above the derivation in [164] was not based on displacement interpolation. The completeness check worked out in §15.4.2 is a specialization case of a general proof developed by Irons in the mid 1960s (see [98, §3.9] and references therein) for general isoparametric elements. The check works because the linear triangle is isoparametric. What are here called triangular coordinates were introduced by M¨obius in his 1827 book [117].9 They are often called barycentric coordinates on account on the interpretation discussed in [38]. Other names are listed in Table 15.1. Triangles possess many fascinating geometric properties studied even before Euclid. An exhaustive development can be found, in the form of solved exercises, in [144]. It is unclear when the monomial integration formula (15.27) was first derived. As an expression for integrands expressed in triangular coordinates it was first stated in [44]. The natural strain derivation of §15.4 is patterned after that developed for the so-called ANDES (Assumed Natural Deviatoric Strain) elements [116]. For the linear triangle it provides nothing new aside of fancy terminology. Energy ratios of the form used in §15.6 were introduced in [18] as a way to tune up the stiffness of Free-Formulation elements. References Referenced items have been moved to Appendix R. 9
He is better remembered for the “M¨obius strip” or “M¨obius band,” the first one-sided 3D surface in mathematics.
15–17
15–18
Section 15: THE LINEAR PLANE STRESS TRIANGLE
Homework Exercises for Chapter 15 The Linear Plane Stress Triangle EXERCISE 15.1 [A:15] Assume that the 3-node plane stress triangle has variable thickness defined over the
element by the linear interpolation formula h(ζ1 , ζ2 , ζ3 ) = h 1 ζ1 + h 2 ζ2 + h 3 ζ3 ,
(E15.1)
where h 1 , h 2 and h 3 are the thicknesses at the corner nodes. Show that the element stiffness matrix is still given by (15.22) but with h replaced by the mean thickness h m = (h 1 + h 2 + h 3 )/3. Hint: use (15.21) and (15.27). EXERCISE 15.2 [A:20] The exact integrals of triangle-coordinate monomials over a straight-sided triangle
are given by the formula (15.27), where A denotes the area of the triangle, and i, j and k are nonnegative integers. Tabulate the right-hand side for combinations of exponents i, j and k such that i + j + k ≤ 3, beginning with i = j = k = 0. Remember that 0! = 1. (Labor-saving hint: don’t bother repeating exponent permutations; for example i = 2, j = 1, k = 0 and i = 1, j = 2, k = 0 are permutations of the same thing. Hence one needs to tabulate only cases in which i ≥ j ≥ k). EXERCISE 15.3 [A/C:20] Compute the consistent node force vector fe for body loads over a linear triangle,
if the element thickness varies as per (E15.1), bx = 0, and b y = b y1 ζ1 + b y2 ζ2 + b y3 ζ3 . Check that for h 1 = h 2 = h 3 = h and b y1 = b y2 = b y3 = b y you recover (15.26). For area integrals use (15.27). Partial result: f y1 = (A/60)[b y1 (6h 1 + 2h 2 + 2h 3 ) + b y2 (2h 1 + 2h 2 + h 3 ) + b y3 (2h 1 + h 2 + 2h 3 )]. EXERCISE 15.4 [A/C:20]
Derive the formula for the consistent force vector fe of a linear triangle of constant thickness h, if side 1–2 (ζ3 = 0, ζ2 = 1 − ζ1 ), is subject to a linearly varying boundary force q = h ˆt such that qx = qx1 ζ1 + qx2 ζ2 = qx1 (1 − ζ2 ) + qx2 ζ2 ,
3
qy2 2
(E15.2)
q y = q y1 ζ1 + q y2 ζ2 = q y1 (1 − ζ2 ) + q y2 ζ2 .
000f
W e = (ue )T fe =
1
q x1
q x = q x1 (1−ζ 2 ) + q x2 ζ 2
x Figure E15.1. Line force on triangle side 1–2 for Exercise 15.4.
000f
1
uT q d0015 e = 0015e
qx2
y qy1
This “ line boundary force” q has dimension of force per unit of side length. Procedural Hint. Use the last term of the line integral (14.21), in which ˆt is replaced by q/ h, and show that since the contribution of sides 2-3 and 3-1 to the line integral vanish,
q y = q y1 (1−ζ 2 ) + q y2 ζ 2
uT q L 21 dζ2 ,
(E15.3)
0
where L 21 is the length of side 1–2. Replace u x (ζ2 ) = u x1 (1 − ζ2 ) + u x2 ζ2 ; likewise for u y , qx and q y , integrate and identify with the inner product shown as the second term in (E15.3). Partial result: f x1 = L 21 (2qx1 +qx2 )/6, f x3 = f y3 = 0. Note. The following Mathematica script solves this Exercise. If you decide to use it, explain the logic. ClearAll[ux1,uy1,ux2,uy2,ux3,uy3,z2,L12]; ux=ux1*(1-z2)+ux2*z2; uy=uy1*(1-z2)+uy2*z2; qx=qx1*(1-z2)+qx2*z2; qy=qy1*(1-z2)+qy2*z2; We=Simplify[L12*Integrate[qx*ux+qy*uy,{z2,0,1}]]; fe=Table[Coefficient[We,{ux1,uy1,ux2,uy2,ux3,uy3}[[i]]],{i,1,6}]; fe=Simplify[fe]; Print['fe=',fe];
15–18
15–19
Exercises
EXERCISE 15.5 [C+N:15] Compute the entries of Ke for the following plane stress triangle:
x1 = 0, y1 = 0, x2 = 3, y2 = 1, x3 = 2, y3 = 2,
0003
E=
100 25 0
25 100 0
0 0 50
0004
,
h = 1.
(E15.4)
This may be done by hand (it is a good exercise in matrix multiplication) or (more quickly) using the script of Figure 15.5. Partial result: K 11 = 18.75, K 66 = 118.75. EXERCISE 15.6 [A+C:15] Show that the sum of the rows (and columns) 1, 3 and 5 of Ke as well as the sum
of rows (and columns) 2, 4 and 6 must vanish. Check it with the foregoing script. EXERCISE 15.7 [A:20] Let point P have triangular coordinates {ζ1P , ζ2P , ζ3P }. Find the distance of P to the
three triangle sides. EXERCISE 15.8 [C+D:20] Let p(ζ1 , ζ2 , ζ3 ) represent a polynomial expression in the natural coordinates.
The integral
000f p(ζ1 , ζ2 , ζ3 ) d000b
(E15.5)
000be
over a straight-sided triangle can be computed symbolically by the following Mathematica module: IntegrateOverTriangle[expr_,tcoord_,A_,max_]:=Module [{p,i,j,k,z1,z2,z3,c,s=0}, p=Expand[expr]; {z1,z2,z3}=tcoord; For [i=0,i<=max,i++, For [j=0,j<=max,j++, For [k=0,k<=max,k++, c=Coefficient[Coefficient[Coefficient[p,z1,i],z2,j],z3,k]; s+=2*c*(i!*j!*k!)/((i+j+k+2)!); ]]]; Return[Simplify[A*s]] ]; This is referenced as int=IntegrateOverTriangle[p,{ z1,z2,z3 },A,max]. Here p is the polynomial to be integrated, z1, z2 and z3 denote the symbols used for the triangular coordinates, A is the triangle area and max the highest exponent appearing in a triangular coordinate. The module name returns the integral. For example, if p=16+5*b*z2^2+z1^3+z2*z3*(z2+z3) the call int=IntegrateOverTriangle[p,{ z1,z2,z3 },A,3] returns int=A*(97+5*b)/6. Explain how the module works. EXERCISE 15.9 [A:25] Find the triangular coordinates of the altitude feet points H1 , H2 and H3 pictured
in Figure 15.3. Once these are obtained, find the equations of the altitudes in triangular coordinates, and the coordinates of the orthocenter H . EXERCISE 15.10 [C+D:25] Explain the logic of the script listed in Figure 15.14. Then extend it to account for isotropic material with arbitrary Poisson’s ratio ν. Obtain the macroelement energy ratios as functions of γ and ν. Discuss whether the effect of a nonzero ν makes much of a difference if γ >> 1. EXERCISE 15.11 [C+D:25] To find whether shear is the guilty party in the poor performance of elongated
triangles (as alledged in §15.6) run the script of Figure 15.14 with a zero shear modulus. This can be done by setting Emat=Em*{ { 1,0,0 },{ 0,1,0 },{ 0,0,0 } } in the third line. Discuss the result. Can Em be subsequently reduced to a smaller (fictitious) value so that r ≡ 1 for all aspect ratios γ ? Is this practical?
15–19
Section 15: THE LINEAR PLANE STRESS TRIANGLE
15–20
EXERCISE 15.12 [C+D:25] Access the file Trig3PlaneStress.nb from the course Web site by clicking
on the appropriate link in Chapter 15 Index. This is a Mathematica Notebook that does plane stress FEM analysis using the 3-node linear triangle. Download the Notebook into your directory. Load into Mathematica. Execute the top 7 input cells (which are actually initialization cells) so the necessary modules are compiled. Each cell is preceded by a short comment cell which outlines the purpose of the modules it holds. Notes: (1) the plot-module cell may take a while to run through its tests; be patient; (2) to get rid of unsightly messages and silly beeps about similar names, initialize each cell twice. After you are satisfied everything works fine, run the cantilever beam problem, which is defined in the last input cell. After you get a feel of how this code operate, study the source. Prepare a hierarchical diagram of the modules,10 beginning with the main program of the last cell. Note which calls what, and briefly explain the purpose of each module. Return this diagram as answer to the homework. You do not need to talk about the actual run and results; those will be discussed in Part III. Hint: a hierarchical diagram for Trig3PlaneStress.nb begins like Main program in Cell 8 - drives the FEM analysis GenerateNodes - generates node coordinates of regular mesh GenerateTriangles - generate element node lists of regular mesh ....
10
A hierarchical diagram is a list of modules and their purposes, with indentation to show dependence, similar to the table of contents of a book. For example, if module AAAA calls BBBB and CCCC, and BBBB calld DDDD, the hierarchical diagram may look like: AAAA - purpose of AAAA BBBB - purpose of BBBB DDDD - purpose of DDDD CCCC - purpose of CCCC
15–20
16
.
The Isoparametric Representation
16–1
16–2
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
TABLE OF CONTENTS Page
§16.1. Introduction §16.2. Isoparametric Representation §16.2.1. Motivation . . . . . . . . . . . §16.2.2. Equalizing Geometry and Displacements . . §16.3. General Isoparametric Formulation §16.4. Triangular Elements §16.4.1. The Linear Triangle . . . . . . . . §16.4.2. The Quadratic Triangle . . . . . . . §16.4.3. *The Cubic Triangle . . . . . . . . §16.5. Quadrilateral Elements §16.5.1. Quadrilateral Coordinates and Iso-P Mappings §16.5.2. The Bilinear Quadrilateral . . . . . . §16.5.3. The Biquadratic Quadrilateral . . . . . §16.6. *Completeness Properties of Iso-P Elements §16.6.1. *Completeness Analysis . . . . . . . §16.6.2. Completeness Checks . . . . . . . . §16.6.3. *Completeness for Higher Variational Index §16.7. Iso-P Elements in One and Three Dimensions §16. Notes and Bibliography . . . . . . . . . . . . . . . . §16. References . . . . . . . . . . . . . . . §16. Exercises . . . . . . . . . . . . . . . . §16. Solutions to .Exercises . . . . . . . . . . . . . .
16–2
. . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . .
. . . . . . . . . .
16–3 16–3 16–3 16–4 16–5 16–5 16–6 16–6 16–6 16–6 16–6 16–7 16–7 16–8 16–8 16–9 16–10 16–11 16–11 16–11 16–12 16–15
16–3
§16.2
ISOPARAMETRIC REPRESENTATION
§16.1. Introduction The technique used in Chapter 15 for the linear triangle can be formally extended to quadrilateral elements as well as higher order triangles. But it runs quickly into technical difficulties: 1.
The construction of shape functions that satisfy consistency requirements for higher order elements with curved boundaries becomes increasingly complicated.
2.
Integrals that appear in the expressions of the element stiffness matrix and consistent nodal force vector can no longer be evaluated in simple closed form.
These difficulties can be overcome through the concepts of isoparametric elements and numerical quadrature, respectively. The combination of these two ideas transformed the field of finite element methods in the late 1960s. Together they support a good portion of what is presently used in production finite element programs. In the present Chapter the concept of isoparametric representation is introduced for two dimensional elements. This representation is illustrated on specific elements. In the next Chapter these techniques, combined with numerical integration, are applied to quadrilateral elements. §16.2. Isoparametric Representation §16.2.1. Motivation The linear triangle presented in Chapter 15 is an isoparametric element although was not originally derived as such. The two key equations are (15.10), which defines the triangle geometry, and (15.16), which defines the primary variable, in this case the displacement field. These equations are reproduced here for convenience: 0002 0003 0002 00030002 0003 1 1 1 1 ζ1 (16.1) x = x1 x2 x3 ζ2 , y y1 y2 y3 ζ3 u x = u x1 N1e + u x2 N2e + u x3 N3e = u x1 ζ1 + u x2 ζ2 + u x3 ζ3 . u y = u y1 N1e + u y2 N2e + u y3 N3e = u y1 ζ1 + u y2 ζ2 + u y3 ζ3 The interpretation of these equations is as follows. The triangular coordinates define the element geometry via (16.1). The displacement expansion (16.2) is defined by the shape functions, which are in turn expressed in terms of the triangular coordinates. For the linear triangle, shape functions and triangular coordinates coalesce. These relations are diagrammed in Figure 16.1. Evidently geometry and displacements are not treated equally. If we proceed to higher order triangular elements while keeping straight sides, only the displacement expansion is refined whereas the geometry definition remains the same. 16–3
Triangular coordinates ζ1 , ζ 2 , ζ 3
Shape functions Ni(e)
(16.2)
Geometry 1, x, y
Displacement interpolation ux , uy
Figure 16.1. Superparametric representation of triangular element.
16–4
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
Geometry 1, x, y Triangular coordinates ζ1 , ζ 2 , ζ 3
Shape functions Ni(e) Displacement interpolation ux , uy
Figure 16.2. Isoparametric representation of triangular elements.
Elements built according to the foregoing prescription are called superparametric, a term that emphasizes that unequal treatment. §16.2.2. Equalizing Geometry and Displacements On first inspection (16.2) and (16.1) do not look alike. Their inherent similarity can be displayed, however, if the second one is rewritten and adjoined to (16.1) to look as follows:       1 1 1 1 1 1 1 0003 0002 0003 0002 x2 x3  ζ1 x2 x3  N1e  x1  x   x1       (16.3) y2 y3  ζ2 =  y1 y2 y3  N2e .  y  =  y1       e u x1 u x2 u y3 ζ3 u x1 u x2 u y3 N3 ux uy u y1 u y2 u y3 u y1 u y2 u y3 This form emphasizes that geometry and displacements are given by the same parametric representation, as shown in Figure 16.2. The key idea is to use the shape functions to represent both the element geometry and the problem unknowns, which in structural mechanics are displacements. Hence the name isoparametric element (“iso” means equal), often abbreviated to iso-P element. This property may be generalized to arbitrary elements by replacing the term “triangular coordinates” by the more general one “natural coordinates.” This generalization is illustrated in Figure 16.3. Geometry 1, x, y
Natural coordinates
Shape functions Ni(e) Displacement interpolation ux , uy
Figure 16.3. Isoparametric representation of arbitrary two-dimensional elements: triangles or quadrilaterals. For 3D elements, expand the geometry list to {1, x, y, z} and the displacements to {u x , u y , u z }.
16–4
16–5
§16.4
TRIANGULAR ELEMENTS
Under this generalization, natural coordinates (triangular coordinates for triangles, quadrilateral coordinates for quadrilaterals) appear as parameters that define the shape functions. The shape functions connect the geometry with the displacements. Remark 16.1. The terms isoparametric and superparametric were introduced by Irons and coworkers at
Swansea in 1966. See Notes and Bibliography at the end of this Chapter. There are also subparametric elements whose geometry is more refined than the displacement expansion.
§16.3. General Isoparametric Formulation The generalization of (16.3) to an arbitrary two-dimensional element with n nodes is straightforward. Two set of relations, one for the element geometry and the other for the element displacements, are required. Both sets exhibit the same interpolation in terms of the shape functions. Geometric relations: 1=
n
Nie ,
i=1
x=
n
xi Nie ,
y=
i=1
n
yi Nie .
(16.4)
i=1
Displacement interpolation: ux =
n
u xi Nie ,
uy =
i=1
n
u yi Nie .
(16.5)
i=1
These two sets of equations may be combined in matrix form as   N1e     1 1 .. 1 1 N2e  x2 . . . xn    x   x1      .  . y2 . . . yn    y  =  y1   .      u x1 u x2 . . . u xn ux uy u y1 u y2 . . . u yn Ne
(16.6)
n
The first three scalar equations in (16.6) express the geometry definition, and the last two the displacement expansion. Note that additional rows may be added to this matrix expression if more variables are interpolated by the same shape functions. For example, suppose that the thickness h and a temperature field T are both interpolated from the n node values:     1 1 .. 1 1  e x2 . . . xn  N1e  x   x1  N2     y2 . . . yn    y   y1   .      (16.7)  u x  =  u x1 u x2 . . . u xn   . .        u y   u y1 u y2 . . . u yn      h h1 h2 . . . hn Nne T1 T2 . . . Tn T Note that the column of shape functions does not change. To illustrate the use of the isoparametric concept, we take a look at specific 2D isoparametric elements that are commonly used in structural and non-structural applications. These are separated into triangles and quadrilaterals because they use different natural coordinates. 16–5
16–6
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
§16.4. Triangular Elements §16.4.1. The Linear Triangle The three-noded linear triangle, studied in Chapter 15 and pictured in Figure 16.4, may be presented as an isoparametric element:     1 1 1 1 0003 0002 x2 x3  N1e  x   x1     (16.8) y2 y3  N2e .  y  =  y1     u x1 u x2 u x3 N3e ux uy u y1 u y2 u y3
3
2 1 Figure 16.4. The 3-node linear triangle.
The shape functions are simply the triangular coordinates: N1e = ζ1 ,
N2e = ζ2 ,
N3e = ζ3 .
(16.9)
The linear triangle is the only triangular element that is both superparametric and isoparametric. §16.4.2. The Quadratic Triangle The six node triangle shown in Figure 16.5 is the next complete-polynomial member of the isoparametric triangle family. The isoparametric definition is  e     N1 1 1 1 1 1 1 1  Ne  2   x   x1 x2 x3 x4 x5 x6   N3e        y  =  y1 y2 y3 y4 y5 y6   N4e       u x1 u x2 u x3 u x4 u x5 u x6  ux  N5e  uy u y1 u y2 u y3 u y4 u y5 u y6 N6e (16.10)
(a)
(b)
3
3
5
5
6
6 2
2 4
4 1
1 Figure 16.5. The 6-node quadratic triangle: (a) the superparametric version, with straight sides and midside nodes at midpoints; (b) the isoparametric version.
The shape functions are N1e = ζ1 (2ζ1 − 1), N2e = ζ2 (2ζ2 − 1), N3e = ζ3 (2ζ3 − 1), N4e = 4ζ1 ζ2 , N5e = 4ζ2 ζ3 , N6e = 4ζ3 ζ1 .
(16.11)
The element may have parabolically curved sides defined by the location of the midnodes 4, 5 and 6. The triangular coordinates for a curved triangle are no longer straight lines, but form a curvilinear system as can be observed in Figure 16.5(b). §16.4.3. *The Cubic Triangle The cubic triangle has ten nodes. This shape functions of this element are the subject of an Exercise in Chapter 18. The implementation is studied in Chapter 24.
16–6
16–7
§16.5
QUADRILATERAL ELEMENTS
§16.5. Quadrilateral Elements §16.5.1. Quadrilateral Coordinates and Iso-P Mappings Before presenting examples of quadrilateral η η elements, we must introduce the appropriate η=1 η=1 natural coordinate system for that geometry. ξ=1 The natural coordinates for a triangular ξ=−1 ξ element are the triangular coordinates ζ1 , ζ2 and ζ3 . The natural coordinates for a ξ=−1 quadrilateral element are ξ and η, which are η=−1 η=−1 illustrated in Figure 16.6 for both straight sided and curved side quadrilaterals. These are called quadrilateral coordinates. Figure 16.6. Quadrilateral coordinates.
ξ ξ=1
These coordinates vary from −1 on one side to +1 at the other, taking the value zero over the quadrilateral medians. This particular variation range (instead of taking, say, 0 to 1) was chosen by Irons and coworkers to facilitate the use of the standard Gauss integration formulas. Those formulas are discussed in the next Chapter. Remark 16.2. In some FEM derivations it is useful to visualize the quadrilateral coordinates plotted as Cartesian coordinates in the {ξ, η} plane. This is called the reference plane. All quadrilateral elements in the reference plane become a square of side 2, called the reference element, which extends over ξ ∈ [−1, 1], η ∈ [−1, 1]. The transformation between {ξ, η} and {x, y} dictated by the second and third equations of (16.4), is called the isoparametric mapping. A similar version exists for triangles. An important application of this mapping is discussed in §16.6; see Figure 16.9 there.
η
§16.5.2. The Bilinear Quadrilateral The four-node quadrilateral shown in Figure 16.7 is the simplest member of the quadrilateral family. It is defined by   1 1  x   x1     y  =  y1    u x1 ux uy u y1 
1 x2 y2 u x2 u y2
1 x3 y3 u x3 u y3
 1  e N1 x4    N2e  y4   e  .  N3 u x4 N4e u y4
(16.12)
4 η=1
ξ=1
ξ=−1
1
3
η=−1
ξ
2
Figure 16.7. The 4-node bilinear quadrilateral.
The shape functions are N1e = 14 (1 − ξ )(1 − η),
N2e = 14 (1 + ξ )(1 − η),
N3e = 14 (1 + ξ )(1 + η),
N4e = 14 (1 − ξ )(1 + η).
(16.13)
These functions vary linearly on quadrilateral coordinate lines ξ = const and η = const, but are not linear polynomials as in the case of the three-node triangle. 16–7
16–8
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
η
(a) 4
3
η=1 7 6
ξ=−1 1
ξ
3
η=1 7
4 9
8
η
(b)
6
8
ξ=1
ξ=−1
η=−1 5
1
ξ
ξ=1 η=−1 5
2
2
Figure 16.8. Two widely used higher order quadrilaterals: (a) the nine-node biquadratic quadrilateral; (b) the eight-node “serendipity” quadrilateral.
§16.5.3. The Biquadratic Quadrilateral The nine-node quadrilateral shown in Figure 16.8(a) is the next complete member of the quadrilateral family. It has eight external nodes and one internal node. It is defined by 
  1 1 x x    1     y  =  y1    u x1 ux uy u y1
1 x2 y2 u x2 u y2
1 x3 y3 u x3 u y3
1 x4 y4 u x4 u y4
1 x5 y5 u x5 u y5
1 x6 y6 u x6 u y6
1 x7 y7 u x7 u y7
1 x8 y8 u x8 u y8
  N1e  1 N2e  x9     .   y9   . .   u x9  u y9 Ne
(16.14)
9
This element is often referred to as the Lagrangian quadrilateral in the FEM literature, a term explained in the Notes and Bibliography. Its shape functions are N1e = 14 (1 − ξ )(1 − η)ξ η N2e = − 14 (1 + ξ )(1 − η)ξ η ···
N5e = − 12 (1 − ξ 2 )(1 − η)η N6e = 12 (1 + ξ )(1 − η2 )ξ ···
N9e = (1−ξ 2 )(1−η2 ) (16.15)
These functions vary quadratically along the coordinate lines ξ = const and η = const. The shape function associated with the internal node 9 is called a bubble function because of its geometric shape, as pictured in §18.4.2. Figure 16.8(a) depicts a widely used eight-node variant called the “serendipity” quadrilateral. (A name that originated from circumstances surrounding the element discovery.) The internal node is eliminated by kinematic constraints as worked out in an Exercise of Chapter 18. §16.6.
*Completeness Properties of Iso-P Elements
Some general conclusions as regards the range of applications of isoparametric elements can be obtained from a completeness analysis. More specifically, whether the general prescription (16.6) that combines (16.4) and (16.5) satisfies the completeness criterion of finite element trial expansions. This is one of the conditions for convergence to the analytical solution. The requirement is treated generally in Chapter 19, and is stated here in recipe form.
16–8
16–9
§16.6 *COMPLETENESS PROPERTIES OF ISO-P ELEMENTS
§16.6.1. *Completeness Analysis The plane stress problem has variational index m = 1. A set of shape functions is complete for this problem if they can represent exactly any linear displacement motions such as u x = α0 + α1 x + α2 y,
u y = β0 + β1 x + β2 y.
(16.16)
To carry out the check, evaluate (16.16) at the nodes u xi = α0 + α1 xi + α2 yi
u yi = β0 + β1 xi + β2 yi ,
i = 1, . . . n.
(16.17)
Insert this into the displacement expansion (16.5) to see whether the linear displacement field (16.16) is recovered. Here are the computations for the displacement component u x : ux =
n
(α0 + α1 xi + α2 yi ) Nie = α0
i=1
Nie + α1
i
xi Nie + α2
i
yi Nie = α0 + α1 x + α2 y. (16.18)
i
For the last step we have used the geometry definition relations (16.4), reproduced here for convenience:
1=
n
i=1
Nie ,
x=
n
xi Nie ,
i=1
y=
n
yi Nie .
(16.19)
i=1
A similar calculation may be made for u y . It appears that the isoparametric displacement expansion represents (16.18) for any element, and consequently meets the completeness requirement for variational order m = 1. The derivation carries without essential change to three dimensions.1 Can you detect a flaw in this conclusion? The fly in the ointment is the last replacement step of (16.18), which assumes that the geometry relations (16.19) are identically satisfied. Indeed they are for all the example elements presented in the previous sections. But if the new shape functions are constructed directly by the methods of Chapter 18, a posteriori checks of those identities are necessary. §16.6.2. Completeness Checks The first check in (16.19) is easy: the sum of shape functions must be unity. This is also called the unit sum condition. It can be easily verified by hand for simple elements. Here are two examples. Example 16.1. Check for the linear triangle: directly from the definition of triangular coordinates,
N1e + N2e + N3e = ζ1 + ζ2 + ζ3 = 1.
1
(16.20)
This derivation is due to B. M. Irons, see textbook cited in footnote 2, p. 75. The property was known since the mid 1960s and contributed substantially to the rapid acceptance of iso-P elements.
16–9
16–10
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
ξ−η plane
good mapping (compatible)
η
2
(e1) 1
(e1)
(e1)
ξ
1
1
2
η
(e2)
2
2 (e2)
2
x−y plane
bad mapping (incompatible)
(e2)
y
ξ
2
x
Figure 16.9. Good and bad isoparametric mappings of 4-node quadrilateral from the {ξ, η} reference plane onto the {x, y} physical plane.
Example 16.2. Check for the 4-node bilinear quadrilateral:
N1e + N2e + N3e + N4e = 14 (1 − ξ − η + ξ η) + 14 (1 + ξ − η − ξ η) + 14 (1 + ξ + η + ξ η) + 14 (1 − ξ + η − ξ η) = 1
(16.21)
For more complicated elements see Exercises 16.2 and 16.3. The other two checks are less obvious. For specificity consider the 4-node bilinear quadrilateral. The geometry definition equations are x=
4
xi Nie (ξ, η),
y=
i=1
4
yi Nie (ξ, η).
(16.22)
i=1
Given the corner coordinates, {xi , yi } and a point P(x, y) one can try to solve for {ξ, η}. This solution requires nontrivial work because it involves two coupled quadratics, but can be done. Reinserting into (16.22) simply gives back x and y, and nothing is gained.2 The correct question to pose is: is the correct geometry of the quadrilateral preserved by the mapping from {ξ, η} to {x, y}? In particular, are the sides straight lines? Figure 16.9 illustrate these questions. Two side-two squares: (e1) and (e2), contiguous in the {ξ, η} reference plane, are mapped to quadrilaterals (e1) and (e2) in the {x, y} physical plane through (16.22). The common side 1-2 must remain a straight line to preclude interelement gaps or interpenetration. We are therefore lead to consider geometric compatibility upon mapping. But this is equivalent to the question of interelement displacement compatibility, which is posed as item (C) in §18.1. The statement “the displacement along a side must be uniquely determined by nodal displacements on that side” translates to “the coordinates of a side must be uniquely determined by nodal coordinates on that side.” Summarizing: Unit-sum condition + interelement compatibility → completeness.
(16.23)
This subdivision of work significantly reduces the labor involved in element testing. 2
This tautology is actually a blessing, since finding explicit expressions for the natural coordinates in terms of x and y rapidly becomes impossible for higher order elements. See, for example, the complications that already arise for the bilinear quadrilateral in §23.3.
16–10
16–11
§16.
References
§16.6.3. *Completeness for Higher Variational Index The completeness conditions for variational index 2 are far more demanding because they involve quadratic motions. No simple isoparametric configurations satisfy those conditions. Consequently isoparametric formulations have limited importance in the finite element analysis of plate and shell bending.
§16.7. Iso-P Elements in One and Three Dimensions The reader should not think that the concept of isoparametric representation is confined to twodimensional elements. It applies without conceptual changes to one and three dimensions as long as the variational index remains one.3 Three-dimensional solid elements are covered in an advanced course. The use of the isoparametric formulation to construct a 3-node bar element is the subject of Exercises 16.4 through 16.6. Notes and Bibliography A detailed presentation of the isoparametric concept, with references to the original 1960 papers may be found in the textbook [98]. This matrix representation for isoparametric elements used here was introduced in [45]. The term Lagrangian element in the mathematical FEM literature identifies quadrilateral and hexahedra (brick) elements that include all polynomial terms ξ i η j (in 2D) or ξ i η j µk (in 3D) with i ≤ n, j ≤ n and k ≤ n, as part of the shape function interpolation. Such elements have (n + 1)2 nodes in 2D and (n + 1)3 nodes in 3D, and the interpolation is said to be n-bicomplete. For example, if n = 2, the biquadratic quadrilateral with (2 + 1)2 = 9 nodes is Lagrangian and 2-bicomplete. (The qualifier “Lagrangian” in this context refers to Lagrange’s interpolation formula.) References Referenced items have been moved to Appendix R
3
A limitation explained in §16.6.3.
16–11
16–12
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
Homework Exercises for Chapter 16 The Isoparametric Representation EXERCISE 16.1 [D:10] What is the physical interpretation of the shape-function unit-sum condition discussed in §16.6? Hint: the element must respond exactly in terms of displacements to rigid-body translations in the x and y directions. EXERCISE 16.2 [A:15] Check by algebra that the sum of the shape functions for the six-node quadratic
triangle (16.11) is exactly one regardless of natural coordinates values. Hint: show that the sum is expressable as 2S12 − S1 , where S1 = ζ1 + ζ2 + ζ3 . EXERCISE 16.3 [A/C:15] Complete the table of shape functions (16.23) of the nine-node biquadratic quadri-
lateral. Verify that their sum is exactly one. EXERCISE 16.4 [A:20] Consider a three-node bar element referred to the natural coordinate ξ . The two end
nodes and the midnode are identified as 1, 2 and 3, respectively. The natural coordinates of nodes 1, 2 and 3 are ξ = −1, ξ = 1 and ξ = 0, respectively. The variation of the shape functions N1 (ξ ), N2 (ξ ) and N3 (ξ ) is sketched in Figure E16.1. These functions must be quadratic polynomials in ξ : N1e (ξ ) = a0 + a1 ξ + a2 ξ 2 ,
N2e (ξ ) = b0 + b1 ξ + b2 ξ 2 ,
e
e
3 ξ=0
1 ξ=−1
1
1
3 ξ=0
2 1 ξ=1 ξ=−1
(E16.1)
N 3e (ξ)
N2 (ξ)
N1 (ξ)
1
N3e (ξ ) = c0 + c1 ξ + c2 ξ 2 .
2 ξ=1
1 ξ=−1
3 ξ=0
2 ξ=1
Figure E16.1. Isoparametric shape functions for 3-node bar element (sketch). Node 3 has been drawn at the 1–2 midpoint but it may be moved away from it, as in Exercises E16.5 and E16.6.
Determine the coefficients a0 , through c2 using the node value conditions depicted in Figure E16.1; for example N1e = 1, 0 and 0 for ξ = −1, 0 and 1 at nodes 1, 3 and 2, respectively. Proceeding this way show that N1e (ξ ) = − 12 ξ(1 − ξ ),
N2e (ξ ) = 12 ξ(1 + ξ ),
N3e (ξ ) = 1 − ξ 2 .
(E16.2)
Verify that their sum is identically one. EXERCISE 16.5
[A/C:10+10+15+5+10] A 3-node straight bar element is defined by 3 nodes: 1, 2 and 3, with axial coordinates x1 , x2 and x3 , respectively, as illustrated in Figure E16.2. The element has axial rigidity E A and length = x2 − x1 . The axial displacement is u(x). The 3 degrees of freedom are the axial node displacements u 1 , u 2 and u 3 . The isoparametric definition of the element is
0002 0003 1 x u
0002
=
1 x1 u1
1 x2 u2
1 x3 u3
00030002
N1e N2e N3e
0003
,
(E16.3)
in which Nie (ξ ) are the shape functions (E16.2) of the previous Exercise. Node 3 lies between 1 and 2 but is not necessarily at the midpoint x = 12 . For convenience define x1 = 0,
x2 = ,
16–12
x3 = ( 12 + α) ,
(E16.4)
16–13
Exercises
axial rigidity EA x, u 3 (ξ=0) x 3 = /2+α
1 (ξ= −1) x1 = 0
2 (ξ=1) x2 =
= L (e) Figure E16.2. The 3-node bar element in its local system.
where − 12 < α < 12 characterizes the location of node 3 with respect to the element center. If α = 0 node 3 is located at the midpoint between 1 and 2. See Figure E16.2. (a)
From (E16.4) and the second equation of (E16.3) get the Jacobian J = d x/dξ in terms of , α and ξ . Show that: (i) if − 14 < α < 14 then J > 0 over the whole element −1 ≤ ξ ≤ 1; (ii) if α = 0, J = /2 is constant over the element.
(b)
Obtain the 1 × 3 strain-displacement matrix B relating e = du/d x = B ue , where ue is the column 3-vector of node displacements u 1 , u 2 and u 3 . The entries of B are functions of , α and ξ . Hint: B = dN/d x = J −1 dN/dξ , where N = [ N1 N2 N3 ] and J comes from item (a).
(c)
Show that the element stiffness matrix is given by
K =
1
E A B B dx =
e
T
E A BT B J dξ.
(E16.5)
−1
0
Evaluate the rightmost integral for arbitrary α but constant E A using the 2-point Gauss quadrature rule (E13.7). Specialize the result to α = 0, for which you should get K 11 = K 22 = 7E A/(3 ), K 33 = 16E A/(3 ), K 12 = E A/(3 ) and K 13 = K 23 = −8E A/(3 ), with eigenvalues {8E A/ , 2E A/ , 0}. Note: use of a CAS is recommended for this item to save time. (d)
What is the minimum number of Gauss points needed to integrate Ke exactly if α = 0?
(e)
This item addresses the question of why Ke was computed by numerical integration in (c). Why not use exact integration? The answer is that the exact stiffness for arbitrary α is numerically useless. To see why, try the following script in Mathematica: ClearAll[EA,L,alpha,xi]; (* Define J and B={{B1,B2,B3}} here *) Ke=Simplify[Integrate[EA*Transpose[B].B*J,{xi,-1,1}, Assumptions->alpha>0&&alpha<1/4&&EA>0&&L>0]]; Print['exact Ke=',Ke//MatrixForm]; Print['exact Ke for alpha=0',Simplify[Ke/.alpha->0]//MatrixForm]; Keseries=Normal[Series[Ke,{alpha,0,2}]]; Print['Ke series about alpha=0:',Keseries//MatrixForm]; Print['Ke for alpha=0',Simplify[Keseries/.alpha->0]//MatrixForm];
At the start of this script define J and B with the results of items (a) and (b), respectively. Then run the script, the fourth line of which will trigger error messages. Comment on why the exact stiffness cannot be evaluated directly at α = 0. A Taylor series expansion about α = 0 circumvents these difficulties but the 2-point Gauss integration rule gives the correct answer without the gyrations.
16–13
16–14
Chapter 16: THE ISOPARAMETRIC REPRESENTATION
ξ=ξL
ξ=ξR
load q
x, u 1 (ξ= −1) x1 = 0
3 (ξ=0) x 3 = /2+α
2 (ξ=1) x2 =
= L (e) Figure E16.3. The 3-node bar element under a “box” axial load q.
EXERCISE 16.6 [A/C:20] Construct the consistent force vector for the 3-node bar element of the foregoing exercise, if the bar is loaded by a uniform axial force q (given per unit of x length) that extends from ξ = ξ L through ξ = ξ R , and is zero otherwise. Here −1 ≤ ξ L < ξ R ≤ 1. See Figure E16.3. Use
ξR
f = e
q NT J dξ,
(E16.6)
−ξ L
with the J = d x/dξ found in Exercise 16.5(a) and analytical integration. The answer is quite complicated and nearly hopeless by hand. Specialize the result to α = 0, ξ L = −1 and ξ R = 1.
16–14
17
.
Isoparametric Quadrilaterals
17–1
17–2
Chapter 17: ISOPARAMETRIC QUADRILATERALS
TABLE OF CONTENTS Page
§17.1. Introduction §17.2. Partial Derivative Computation §17.2.1. The Jacobian . . . . . . . §17.2.2. Shape Function Derivatives . . §17.2.3. Computing the Jacobian Matrix . §17.2.4. The Strain-Displacement Matrix §17.2.5. *A Shape Function Implementation §17.3. Numerical Integration by Gauss Rules §17.3.1. One Dimensional Rules . . . §17.3.2. Implementation of 1D Rules . . §17.3.3. Two Dimensional Rules . . . §17.3.4. Implementation of 2D Gauss Rules §17.4. The Stiffness Matrix §17.5. *Integration Variants §17.5.1. *Weighted Integration . . . . §17.5.2. *Selective Integration . . . . . §17. Notes and. Bibliography . . . . . . . . . . . §17. References. . . . . . . . . . . . §17. Exercises . . . . . . . . . . . .
17–2
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17–3 17–3 17–3 17–4 17–4 17–5 17–5 17–6 17–6 17–7 17–7 17–8 17–9 17–11 17–11 17–11 17–12 17–12 17–13
17–3
§17.2
PARTIAL DERIVATIVE COMPUTATION
§17.1. Introduction In this Chapter the isoparametric representation of element geometry and shape functions discussed in the previous Chapter is used to construct quadrilateral elements for the plane stress problem. Formulas given in Chapter 14 for the stiffness matrix and consistent load vector of general plane stress elements are of course applicable to these elements. For a practical implementation, however, we must go through the more specific steps: 1.
Construction of shape functions.
2.
Computations of shape function derivatives to form the strain-displacement matrix.
3.
Numerical integration over the element by Gauss quadrature rules.
The first topic was dealt in the previous Chapter in recipe form, and is systematically covered in the next one. Assuming the shape functions have been constructed (or readily found in the FEM literature) the second and third items are combined in an algorithm suitable for programming any isoparametric quadrilateral. The implementation of the algorithm in the form of element modules is partly explained in the Exercises of this Chapter, and covered more systematically in Chapter 23. We shall not deal with isoparametric triangles here to keep the exposition focused. Triangular coordinates, being linked by a constraint, require “special handling” techniques that would complicate and confuse the exposition. Chapter 24 discusses isoparametric triangular elements in detail. §17.2. Partial Derivative Computation Partial derivatives of shape functions with respect to the Cartesian coordinates x and y are required for the strain and stress calculations. Because shape functions are not directly functions of x and y but of the natural coordinates ξ and η, the determination of Cartesian partial derivatives is not trivial. The derivative calculation procedure is presented below for the case of an arbitrary isoparametric quadrilateral element with n nodes. §17.2.1. The Jacobian In quadrilateral element derivations we will need the Jacobian of two-dimensional transformations that connect the differentials of {x, y} to those of {ξ, η} and vice-versa. Using the chain rule:  ∂ξ ∂ξ   ∂x ∂x  0002 0003 0002 0003 0002 0003 0002 0003 0002 0003 0002 0003 dξ dx  ∂x ∂y  dx  ∂ξ ∂η  dξ T dξ −T d x =J =J , = . = ∂η ∂η  dy ∂ y ∂ y  dη dη dη dy dy ∂ξ ∂η ∂x ∂y (17.1) −1 Here J denotes the Jacobian matrix of (x, y) with respect to (ξ, η), whereas J is the Jacobian matrix of (ξ, η) with respect to (x, y):  ∂x ∂y   ∂ξ ∂η  0003 0002 ∂(ξ, η)  ∂ x ∂ x  ∂(x, y)  ∂ξ ∂ξ  J11 J12 = =  ∂ξ ∂η  J= (17.2) , J−1 = =  ∂x ∂y J21 J22 ∂(ξ, η) ∂(x, y) ∂y ∂y ∂η ∂η In FEM work J and J−1 are often called the Jacobian and inverse Jacobian, respectively; the fact that it is a matrix being understood. The scalar symbol J means the determinant of J: J = |J| = det J. 17–3
17–4
Chapter 17: ISOPARAMETRIC QUADRILATERALS
In one dimension J and J coalesce. Jacobians play a crucial role in differential geometry. For the general definition of Jacobian matrix of a differential transformation, see Appendix D. Remark 17.1. Observe that the matrices relating the differentials in (17.1) are the transposes of what we call J and J−1 . The reason is that coordinate differentials transform as covariant quantities: d x = (∂ x/∂ξ ) dξ + (∂ x/∂η) dη, etc. But Jacobians are arranged as in (17.2) because of earlier use in contravariant transformations: ∂φ/∂ x = (∂ξ/∂ x)(∂φ/∂ξ ) + (∂η/∂ x)(∂φ/∂η), as in (17.5) below.
The reader is cautioned that notations vary among application areas. As quoted in Appendix D, one author puts it this way: “When one does matrix calculus, one quickly finds that there are two kinds of people in this world: those who think the gradient is a row vector, and those who think it is a column vector.” Remark 17.2. To show that J and J−1 are in fact inverses of each other we form their product:
∂ x ∂ξ ∂ x ∂η ∂ξ ∂ x + ∂η ∂ x
∂ y ∂η ∂ x ∂ y ∂ y ∂ξ
+ ∂η ∂ x ∂ξ ∂ x 1 0 ∂x ∂x = = , (17.3) J−1 J = 0 1 ∂x ∂y ∂ x ∂ξ + ∂ x ∂η ∂ y ∂ξ + ∂ y ∂η ∂y ∂y ∂ξ ∂ y ∂η ∂ y ∂ξ ∂ y ∂η ∂ y where we have taken into account that x = x(ξ, η), y = y(ξ, η) and the fact that x and y are independent coordinates. This proof would collapse, however, if instead of {ξ, η} we had the triangular coordinates {ζ1 , ζ2 , ζ3 } because rectangular matrices have no conventional inverses. This case requires special handling and is covered in Chapter 24.
§17.2.2. Shape Function Derivatives The shape functions of a quadrilateral element are expressed in terms of the quadrilateral coordinates ξ and η introduced in §16.5.1. The derivatives with respect to x and y are given by the chain rule: ∂ Nie ∂ξ ∂ Nie ∂η ∂ Nie = + , ∂x ∂ξ ∂ x ∂η ∂ x
∂ Nie ∂ Nie ∂ξ ∂ Nie ∂η = + . ∂y ∂ξ ∂ y ∂η ∂ y
This can be put in matrix form as  ∂ N e   ∂ξ ∂η   ∂ N e   ∂x   ∂x  ∂ N e  =  ∂ξ i
∂y
 ∂Ne 
i
i
∂y
i
(17.4)
 ∂Ne  i
∂ x   ∂ξ  ∂(ξ, η)  ∂ξ  −1  ∂ξ  ∂η   ∂ Nie  = ∂(x, y)  ∂ Nie  = J  ∂ Nie  . ∂y ∂η ∂η ∂η
(17.5)
where J−1 is defined in (17.2). The computation of J is addressed in the next subsection. §17.2.3. Computing the Jacobian Matrix To compute the entries of J at any quadrilateral location we make use of the last two geometric relations in (16.4), which are repeated here for convenience: x=
n 000e
xi Nie ,
y=
i=1
n 000e
yi Nie .
(17.6)
i=1
Differentiating with respect to the quadrilateral coordinates, n 000e ∂ Nie ∂x xi = , ∂ξ ∂ξ i=1
n 000e ∂ Nie ∂y yi = , ∂ξ ∂ξ i=1
n 000e ∂ Nie ∂x xi = , ∂η ∂η i=1
17–4
n 000e ∂ Nie ∂y yi = . ∂η ∂η i=1
(17.7)
17–5
§17.2 PARTIAL DERIVATIVE COMPUTATION Quad4IsoPShapeFunDer[ncoor_,qcoor_]:= Module[ {Nf,dNx,dNy,dNξ ,dNη,i,J11,J12,J21,J22,Jdet,ξ ,η,x,y}, {ξ ,η}=qcoor; Nf={(1-ξ )*(1-η),(1+ξ )*(1-η),(1+ξ )*(1+η),(1-ξ )*(1+η)}/4; dNξ ={-(1-η), (1-η),(1+η),-(1+η)}/4; dNη= {-(1-ξ ),-(1+ξ ),(1+ξ ), (1-ξ )}/4; x=Table[ncoor[[i,1]],{i,4}]; y=Table[ncoor[[i,2]],{i,4}]; J11=dNξ .x; J12=dNξ .y; J21=dNη.x; J22=dNη.y; Jdet=Simplify[J11*J22-J12*J21]; dNx= ( J22*dNξ -J12*dNη)/Jdet; dNx=Simplify[dNx]; dNy= (-J21*dNξ +J11*dNη)/Jdet; dNy=Simplify[dNy]; Return[{Nf,dNx,dNy,Jdet}] ];
Figure 17.1. A shape function module for the 4-node bilinear quadrilateral.
because the xi and yi do not depend on ξ and η. In matrix form: 0002 J=
J11 J12 J21 J22
0003
 ∂x  ∂ξ = ∂x ∂η
 ∂Ne ∂y  1 ∂ξ   ∂ξ = PX =  ∂y  ∂ N1e ∂η ∂η
∂ N2e ∂ξ ∂ N2e ∂η
.. ..
 ∂ Nne  x1 x ∂ξ    2 e   . ∂ Nn . ∂η xn
y1  y2  .  . . yn
(17.8)
Given a quadrilateral point of coordinates ξ , η we calculate the entries of J using (17.8). The inverse Jacobian J−1 is then obtained by numerically inverting this 2 × 2 matrix. Remark 17.3. The symbolic inversion of J for arbitrary ξ , η in general leads to extremely complicated expressions unless the element has a particularly simple geometry, (for example rectangles as in Exercises 17.1–17.3). This was one of the difficulties that motivated the use of Gaussian numerical quadrature, as discussed in §17.3 below.
§17.2.4. The Strain-Displacement Matrix The strain-displacement matrix B that appears in the computation of the element stiffness matrix is given by the general expression (14.18), which is reproduced here for convenience:   ∂Ne ∂ Nne ∂ N2e 1 0 0 . . . 0   ∂x ∂x   ∂x ex x e  e e  ∂ N ∂ N ∂ N e n 1 2  e 0 .. 0 (17.9) e =  e yy  =   0 ∂y ∂y ∂ y  u = Bu .   2ex y ∂ Nne ∂ Nne ∂ N1e ∂ N1e ∂ N2e ∂ N2e . . . ∂y ∂x ∂y ∂x ∂y ∂x The nonzero entries of B are partials of the shape functions with respect to x and y. The calculation of those partials is done by computing J via (17.8), inverting and using the chain rule (17.5). §17.2.5. *A Shape Function Implementation To make the foregoing discussion more specific, Figure 17.1 shows the shape function module for the 4-node bilinear quadrilateral. This is a code fragment that returns the value of the shape functions and their {x, y} derivatives at a given point of quadrilateral coordinates {ξ, η}. The module is invoked by saying { Nf,Nfx,Nfy,Jdet }=Quad4IsoPShapeFunDer[ncoor,qcoor]
17–5
(17.10)
17–6
Chapter 17: ISOPARAMETRIC QUADRILATERALS
where the arguments are ncoor
Quadrilateral node coordinates arranged in two-dimensional list form: { { x1,y1 },{ x2,y2 },{ x3,y3 },{ x4,y4 } }.
qcoor
Quadrilateral coordinates { eta,xi } of the point.
The module returns: Nf
Value of shape functions, arranged as list { Nf1,Nf2,Nf3,Nf4 }.
Nfx
Value of x-derivatives of shape functions, arranged as list { Nfx1,Nfx2,Nfx3,Nfx4 }.
Nfy
Value of y-derivatives of shape functions, arranged as list { Nfy1,Nfy2,Nfy3,Nfy4 }.
Jdet
Jacobian determinant.
Example 17.1. Consider a 4-node bilinear quadrilateral shaped as an axis-aligned 2:1 rectangle, with 2a and a as the x and y dimensions, respectively. The node coordinate array is ncoor={ { 0,0 },{ 2*a,0 },{ 2*a,a },{ 0,a } }. The shape functions and their {x, y} derivatives are to be evaluated at the rectangle center ξ = η = 0. The appropiate call is
{ Nf,Nfx,Nfy,Jdet }=Quad4IsoPShapeFunDer[ncoor,{ 0,0 }] This returns Nf={ 1/8,1/8,3/8,3/8 }, Nfx={ -1/(8*a),1/(8*a),3/(8*a),-3/(8*a) }, Nfy={ -1/(2*a),-1/(2*a),1/(2*a),1/(2*a) } and Jdet=a^2/2. ξ = −1
§17.3. Numerical Integration by Gauss Rules The use of numerical integration is essential for practical evaluation of integrals over isoparametric element domains. The standard practice has been to use Gauss integration because such rules use a minimal number of sample points to achieve a desired level of accuracy. This economy is important for efficient element calculations, since we shall see that a matrix product is evaluated at each sample point. The fact that the location of the sample points in Gauss rules is usually given by non-rational numbers is of no concern in digital computation.
ξ=1 p=1 p=2 p=3 p=4 p=5
Figure 17.2. The first five one-dimensional Gauss rules p = 1, 2, 3, 4 depicted over the line segment ξ ∈ [−1, +1]. Sample point locations are marked with black circles. The radii of these circles are proportional to the integration weights.
§17.3.1. One Dimensional Rules The classical Gauss integration rules are defined by 000f
1 −1
F(ξ ) dξ ≈
p 000e
wi F(ξi ).
(17.11)
i=1
Here p ≥ 1 is the number of Gauss integration points (also known as sample points), wi are the integration weights, and ξi are sample-point abcissae in the interval [−1,1]. The use of the canonical interval [−1,1] is no restriction, because an integral over another range, say from a to b, can be transformed to [−1, +1] via a simple linear transformation of the independent variable, as shown in the Remark below. The first five one-dimensional Gauss rules, illustrated in Figure 17.2, are listed in Table 17.1. 17–6
17–7
§17.3
NUMERICAL INTEGRATION BY GAUSS RULES
Table 17.1 - One-Dimensional Gauss Rules with 1 through 5 Sample Points Points 1 2 3 4 5
Rule
00101
−1
00101
F(ξ ) dξ ≈ 2F(0)
√ √ F(ξ ) dξ ≈ F(−1/ 3) + F(1/ 3) −1 00101 √ √ 5 8 5 F(ξ ) dξ ≈ F(− 3/5) + F(0) + F( 3/5) 9 9 9 −1
00101
−1
F(ξ ) dξ ≈ w14 F(ξ14 ) + w24 F(ξ24 ) + w34 F(ξ34 ) + w44 F(ξ44 )
−1
F(ξ ) dξ ≈ w15 F(ξ15 ) + w25 F(ξ25 ) + w35 F(ξ35 ) + w45 F(ξ45 ) + w55 F(ξ55 )
00101
0011 0011 √ √ For the 4-point rule,√ ξ34 = −ξ24 = (3 − 2 6/5)/7, ξ = −ξ = (3 + 2 6/5)/7, 44 14 √ w14 = w44 = 12 − 16 5/6, and w24 = w34 = 12 + 16 5/6. 0011 0011 √ √ For the 5-point rule, ξ55 = −ξ15 = 13 5 + 2 10/7, ξ45 = −ξ35 = 13 5 − 2 10/7, ξ35 = 0, √ √ w15 = w55 = (322 − 13 70)/900, w25 = w45 = (322 + 13 70)/900 and w35 = 512/900.
These integrate exactly polynomials in ξ of orders up to 1, 3, 5, 7 and 9, respectively. In general a one-dimensional Gauss rule with p points integrates exactly polynomials of order up to 2 p − 1. This is called the degree of the formula. Remark 17.4. A more general integral, such as F(x) over [a, b] in which = b − a > 0, is transformed to the canonical interval [−1, 1] through the mapping x = 12 a(1 − ξ ) + 12 b(1 + ξ ) = 12 (a + b) + 12 ξ , or ξ = (2/ )(x − 12 (a + b)). The Jacobian of this mapping is J = d x/dξ = / . Thus
000f
000f
b
F(x) d x = a
000f
1
1
F(ξ ) J dξ = −1
−1
F(ξ ) 12 dξ.
(17.12)
Remark 17.5. Higher order Gauss rules are tabulated in standard manuals for numerical computation. For
example, the widely used Handbook of Mathematical Functions [1] lists (in Table 25.4) rules with up to 96 points. For p > 6 the abscissas and weights of sample points are not expressible as rational numbers or radicals, and can only be given as floating-point numbers.
§17.3.2. Implementation of 1D Rules The Mathematica module shown in Figure 17.3 returns either exact or floating-point information for the first five unidimensional Gauss rules. To get information for the i th point of the p th rule, in which 1 ≤ i ≤ p and p = 1, 2, 3, 4, 5, call the module as { xii,wi }=LineGaussRuleInfo[{ p,numer },i]
(17.13)
Logical flag numer is True to get numerical (floating-point) information, or False to get exact information. The module returns the sample point abcissa ξi in xii and the weight wi in wi. If p is not in the implemented range 1 through 5, the module returns { Null,0 }. Example 17.2. { xi,w }=LineGaussRuleInfo[{ 3,False },2] returns xi=0 and w=8/9.
17–7
17–8
Chapter 17: ISOPARAMETRIC QUADRILATERALS LineGaussRuleInfo[{rule_,numer_},point_]:= Module[ {g2={-1,1}/Sqrt[3],w3={5/9,8/9,5/9}, g3={-Sqrt[3/5],0,Sqrt[3/5]}, w4={(1/2)-Sqrt[5/6]/6, (1/2)+Sqrt[5/6]/6, (1/2)+Sqrt[5/6]/6, (1/2)-Sqrt[5/6]/6}, g4={-Sqrt[(3+2*Sqrt[6/5])/7],-Sqrt[(3-2*Sqrt[6/5])/7], Sqrt[(3-2*Sqrt[6/5])/7], Sqrt[(3+2*Sqrt[6/5])/7]}, g5={-Sqrt[5+2*Sqrt[10/7]],-Sqrt[5-2*Sqrt[10/7]],0, Sqrt[5-2*Sqrt[10/7]], Sqrt[5+2*Sqrt[10/7]]}/3, w5={322-13*Sqrt[70],322+13*Sqrt[70],512, 322+13*Sqrt[70],322-13*Sqrt[70]}/900, i=point,p=rule,info={{Null,Null},0}}, If [p1, info={0,2}]; If [p2, info={g2[[i]],1}]; If [p3, info={g3[[i]],w3[[i]]}]; If [p4, info={g4[[i]],w4[[i]]}]; If [p5, info={g5[[i]],w5[[i]]}]; If [numer, Return[N[info]], Return[Simplify[info]]]; ];
Figure 17.3. A Mathematica module that returns the first five one-dimensional Gauss rules.
§17.3.3. Two Dimensional Rules The simplest two-dimensional Gauss rules are called product rules. They are obtained by applying the one-dimensional rules to each independent variable in turn. To apply these rules we must first reduce the integrand to the canonical form: 000f
1
−1
000f
1
−1
000f F(ξ, η) dξ dη =
000f
1
1
dη −1
−1
F(ξ, η) dξ.
(17.14)
Once this is done we can process numerically each integral in turn: 000f
1 −1
000f
1
−1
000f F(ξ, η) dξ dη =
000f
1
1
dη −1
−1
F(ξ, η) dξ ≈
p1 000e p2 000e
wi w j F(ξi , η j ).
(17.15)
i=1 j=1
where p1 and p2 are the number of Gauss points in the ξ and η directions, respectively. Usually the same number p = p1 = p2 is chosen if the shape functions are taken to be the same in the ξ and η directions. This is in fact the case for all quadrilateral elements presented here. The first four two-dimensional Gauss product rules with p = p1 = p2 are illustrated in Figure 17.4. §17.3.4. Implementation of 2D Gauss Rules The Mathematica module listed in Figure 17.5 implements two-dimensional product Gauss rules having 1 through 5 points in each direction. The number of points in each direction may be the same or different. If the rule has the same number of points p in both directions the module is called in either of two ways: { { xii,etaj },wij }=QuadGaussRuleInfo[{ p, numer }, { i,j }] { { xii,etaj },wij }=QuadGaussRuleInfo[{ p, numer },k ] 17–8
(17.16)
17–9
§17.4
THE STIFFNESS MATRIX
p = 2 (2 x 2 rule)
p = 1 (1 x 1 rule)
p = 4 (4 x 4 rule)
p = 3 (3 x 3 rule)
Figure 17.4. The first four two-dimensional Gauss product rules p = 1, 2, 3, 4 depicted over a straight-sided quadrilateral region. Sample points are marked with black circles. The areas of these circles are proportional to the integration weights.
The first form is used to get information for point {i, j} of the p × p rule, in which 1 ≤ i ≤ p and 1 ≤ j ≤ p. The second form specifies that point by a “visiting counter” k that runs from 1 through p 2 ; if so {i, j} are internally extracted1 as j=Floor[(k-1)/p]+1; i=k-p*(j-1). If the integration rule has p1 points in the ξ direction and p2 points in the η direction, the module may be called also in two ways: { { xii,etaj },wij }=QuadGaussRuleInfo[{ { p1,p2 }, numer },{ i,j }] { { xii,etaj },wij }=QuadGaussRuleInfo[{ { p1,p2 }, numer },k ]
(17.17)
The meaning of the second argument is as follows. In the first form i runs from 1 to p1 and j from 1 to p2 . In the second form k runs from 1 to p1 p2 ; if so i and j are extracted by j=Floor[(k-1)/p1]+1; i=k-p1*(i-1). In all four forms, logical flag numer is set to True if numerical information is desired and to False if exact information is desired. The module returns ξi and η j in xii and etaj, respectively, and the weight product wi w j in wij. This code is used in the Exercises at the end of the chapter. If the inputs are not in range, the module returns { { Null,Null },0 }. Example 17.3. { { xi,eta },w }=QuadGaussRuleInfo[{ 3,False },{ 2,3 }] returns xi=0, eta=Sqrt[3/5]
and w=40/81. Example 17.4. { { xi,eta },w }=QuadGaussRuleInfo[{ 3,True },{ 2,3 }] returns (to 16-place precision)
xi=0., eta=0.7745966692414834 and w=0.49382716049382713. 1
Indices i and j are denoted by i1 and i2, respectively, inside the module.
17–9
17–10
Chapter 17: ISOPARAMETRIC QUADRILATERALS QuadGaussRuleInfo[{rule_,numer_},point_]:= Module[ {ξ ,η,p1,p2,i,j,w1,w2,m,info={{Null,Null},0}}, If [Length[rule]2, {p1,p2}=rule, p1=p2=rule]; If [p1<0, Return[QuadNonProductGaussRuleInfo[ {-p1,numer},point]]]; If [Length[point]2, {i,j}=point, m=point; j=Floor[(m-1)/p1]+1; i=m-p1*(j-1) ]; {ξ ,w1}= LineGaussRuleInfo[{p1,numer},i]; {η,w2}= LineGaussRuleInfo[{p2,numer},j]; info={{ξ ,η},w1*w2}; If [numer, Return[N[info]], Return[Simplify[info]]]; ];
Figure 17.5. A Mathematica module that returns two-dimensional product Gauss rules.
§17.4. The Stiffness Matrix The stiffness matrix of a general plane stress element is given by the expression (14.23), which is reproduced here: 000f e h BT EB d000fe (17.18) K = 000fe
Of the terms that appear in (17.18) the strain-displacement matrix B has been discussed previously. The thickness h, if variable, may be interpolated via the shape functions. The stress-strain matrix E is usually constant in elastic problems, but we could in principle interpolate it as appropriate should it vary over the element. To integrate (17.18) numerically by a two-dimensional product Gauss rule, we have to reduce it to the canonical form (17.14), that is 000f K =
1
000f
1
e
−1
−1
F(ξ, η) dξ dη.
(17.19)
If ξ and η are the quadrilateral coordinates, everything in (17.19) already fits this form, except the element of area d000fe . To complete the reduction we need to express d000fe in terms of the differentials dξ and dη. The desired relation is (see Remark below) d000fe = d x d y = det J dξ dη = J dξ dη. (17.20) We therefore have
∂x dη η ∂η
y
C B
∂y dη ∂η
O
F(ξ, η) = h B EB detJ. T
(17.21)
This matrix function can be numerically integrated over the domain −1 ≤ ξ ≤ +1, −1 ≤ η ≤ +1 by an appropriate Gauss product rule. 17–10
∂x dξ ∂ξ
dΩe A
ξ x ∂y dξ ∂ξ
Figure 17.6. Geometric interpretation of the Jacobian-determinant formula.
17–11
§17.5 *INTEGRATION VARIANTS
Remark 17.6. To geometrically justify the area transformation formula (17.20), consider the element of area
OACB depicted in Figure 17.6. The area of this differential parallelogram can be computed as ∂y ∂x ∂y ∂x dξ dη − dη dξ d A = O000eB × O000eA = ∂ξ ∂η ∂η ∂ξ 0012 ∂x ∂x 0012 0012 0012 0012 ∂ξ ∂η 0012 = 0012 ∂ y ∂ y 0012 dξ dη = |J| dξ dη = det J dξ dη. 0012 0012 ∂ξ ∂η
(17.22)
This formula can be extended to any number of dimensions, as shown in textbooks on differential geometry; for example [63,81,155]. §17.5.
*Integration Variants
Several deviations from the standard integration schemes described in the foregoing sections are found in the FEM literature. Two variations are described below and supplemented with motivation Exercises. §17.5.1. *Weighted Integration It is sometimes useful to form the element stiffness as a linear combination of stiffnesses produced by two different integration rules Such schemes are known as weighted integration methods. They are distinguished from the selective-integration schemes described in the next subsection in that the constitutive properties are not modified. For the 4-node bilinear element weighted integration is done by combining the stiffnesses Ke1×1 and Ke2×2 produced by 1×1 and 2×2 Gauss product rules, respectively: Keβ = (1 − β)Ke1×1 + βKe2×2 .
(17.23)
Here β is a scalar in the range [0, 1]. If β = 0 or β = 1 one recovers the element integrated by the 1×1 or 2×2 rule, respectively.2 The idea behind (17.23) is that Ke1×1 is rank-deficient and too soft whereas Ke2×2 is rank-sufficient but too stiff. A combination of too-soft and too-stiff hopefully “balances” the stiffness. An application of this idea to the mitigation of shear locking for modeling in-plane bending is the subject of Exercise E17.4. §17.5.2. *Selective Integration In the FEM literature the term selective integration is used to described a scheme for forming Ke as the sum of two or more matrices computed with different integration rules and different constitutive properties.3 We consider here the case of a two-way decomposition. Split the plane stress constitutive matrix E into two: E = EI + EII
(17.24)
This is called a stress-strain splitting. Inserting (17.24) into (17.13) the expression of the stiffness matrix becomes 000f 000f Ke =
h BT EI B d000fe + 000fe
000fe
h BT EII B d000fe = KeI + KeII .
(17.25)
2
For programming the combination (17.23) may be regarded as a 5-point integration rule with weights w1 = 4(1−β) at √ √ the sample point at ξ = η = 0 and wi = β (i = 2, 3, 4, 5) at the four sample points at ξ = ±1/ 3, η = ±1/ 3.
3
This technique is also called “selective reduced integration” to reflect the fact that one of the rules (the “reduced rule”) underintegrates the element.
17–11
Chapter 17: ISOPARAMETRIC QUADRILATERALS
17–12
If these two integrals were done through the same integration rule, the stiffness would be identical to that obtained by integrating h BT E B d000fe . The trick is to use two different rules: rule (I) for the first integral and rule (II) for the second. In practice selective integration is mostly useful for the 4-node bilinear quadrilateral. For this element rules (I) and (II) are the 1×1 and 2×2 Gauss product rules, respectively. Exercises E17.5–7 investigate stress-strain splittings (17.24) that improve the in-plane bending performance of rectangular elements. Notes and Bibliography The 4-node quadrilateral has a checkered history. It was first derived as a rectangular panel with edge reinforcements (not included here) by Argyris in his 1954 Aircraft Engineering series [4, p. 49 in the Butterworths reprint]. Argyris used bilinear displacement interpolation in Cartesian coordinates.4 After much flailing, a conforming generalization to arbitrary geometry was published in 1964 by Taig and Kerr [158] using quadrilateral-fitted coordinates called {ξ, η} but running from 0 to 1. (Reference [158] cites an 1961 English Electric Aircraft internal report as original source but [99, p. 520] remarks that the work goes back to 1957.) Bruce Irons, who was aware of Taig’s work while at Rolls Royce, changed the {ξ, η} range to [−1, 1] to fit Gauss quadrature tables. He proceeded to create the seminal isoparametric family as a far-reaching extension upon moving to Swansea [11,43,96–99]. Gauss integration is also called Gauss-Legendre quadrature. Gauss presented these rules, derived from first principles, in 1814; cf. Sec 4.11 of [77]. Legendre’s name is often adjoined because the abcissas of the 1D sample points turned out to be the zeros of Legendre polynomials. A systematic description is given in [153]. For references in multidimensional numerical integration, see Notes and Bibliography in Chapter 24. Selective and reduced integration in FEM developed in the early 1970s, and by now there is a huge literature. An excellent textbook source is [95]. References Referenced items have been moved to Appendix R.
4
This work is probably the first derivation of a continuum-based finite element by assumed displacements. As noted in §1.7.1, Argyris was aware of the ongoing work in stiffness methods at Turner’s group in Boeing, but the plane stress models presented in [170] were derived by interelement flux assumptions. Argyris used the unit displacement theorem, displacing each DOF in turn by one. The resulting displacement interpolation is what is now called a shape function.
17–12
17–13
Exercises
Homework Exercises for Chapter 17 Isoparametric Quadrilaterals The Mathematica module Quad4IsoPMembraneStiffness listed in Figure E17.1 computes the element stiffness matrix of the 4-node bilinear quadrilateral. This module is useful as a tool for the Exercises that follow. Quad4IsoPMembraneStiffness[ncoor_,Emat_,th_,options_]:= Module[{i,k,p=2,numer=False,h=th,qcoor,c,w,Nf, dNx,dNy,Jdet,Be,Ke=Table[0,{8},{8}]}, If [Length[options]2, {numer,p}=options,{numer}=options]; If [p<1||p>4, Print['p out of range']; Return[Null]]; For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}=Quad4IsoPShapeFunDer[ncoor,qcoor]; If [Length[th]4, h=th.Nf]; c=w*Jdet*h; Be={Flatten[Table[{dNx[[i]], 0},{i,4}]], Flatten[Table[{0, dNy[[i]]},{i,4}]], Flatten[Table[{dNy[[i]],dNx[[i]]},{i,4}]]}; Ke+=Simplify[c*Transpose[Be].(Emat.Be)]; ]; Return[Simplify[Ke]] ];
Figure E17.1. Mathematica module to compute the stiffness matrix of a 4-node bilinear quadrilateral in plane stress.
The module makes use of the shape function module Quad4IsoPShapeFunDer listed in Figure 17.1, and of the Gauss integration modules QuadGaussRuleInfo and (indirectly) LineGaussRuleInfo, listed in Figures 17.5 and 17.3, respectively. All modules are included in the web-posted Notebook Quad4Stiffness.nb.5 The module is invoked as Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,thick,options]
(E17.1)
The arguments are: ncoor
Quadrilateral node coordinates arranged in two-dimensional list form: { { x1,y1 },{ x2,y2 },{ x3,y3 },{ x4,y4 } }.
Emat
A two-dimensional list storing the 3 × 3 plane stress matrix of elastic moduli:
E=
E 11 E 12 E 13
E 12 E 22 E 23
E 13 E 23 E 33
(E17.2)
arranged as { { E11,E12,E33 },{ E12,E22,E23 },{ E13,E23,E33 } }. Must be symmetric. If the material is isotropic with elastic modulus E and Poisson’s ratio ν, this matrix becomes
E E= 1 − ν2 5
1 ν 0
ν 1 0
0 0 1 (1 − ν) 2
(E17.3)
This Notebook does not include scripts for doing the Exercises below, although it has some text statements at the bottom of the cell. You will need to enter the Exercise scripts yourself.
17–13
17–14
Chapter 17: ISOPARAMETRIC QUADRILATERALS
thick
The plate thickness specified either as a four-entry list: { h1,h2,h3,h4 } or as a scalar: h. The first form is used to specify an element of variable thickness, in which case the entries are the four corner thicknesses and h is interpolated bilinearly. The second form specifies uniform thickness.
options
Processing options. This list may contain two items: { numer,p } or one: { numer }. numer is a logical flag with value True or False. If True, the computations are done in floating point arithmetic. For symbolic or exact arithmetic work set numer to False.6 p specifies the Gauss product rule to have p points in each direction. p may be 1 through 4. For rank sufficiency, p must be 2 or higher. If p is 1 the element will be rank deficient by two.7 If omitted p = 2 is assumed.
The module returns Ke as an 8 × 8 symmetric matrix pertaining to the following arrangement of nodal displacements: (E17.4) ue = [ u x1 u y1 u x2 u y2 u x3 u y3 u x4 u y4 ]T .
η
y
3
4 b = a/γ
Uniform thickness h = 1 Isotropic material with elastic modulus E and Poisson's ratio ν
ξ
1
2
a
x
Figure E17.2. Element for Exercises 17.1 to 17.3.
For the following three exercises we consider the specialization of the general 4-node bilinear quadrilateral to a rectangular element dimensioned a and b in the x and y directions, respectively, as depicted in Figure E17.2. The element has uniform unit thickness h. The material is isotropic with elastic modulus E and Poisson’s ratio ν and consequently E reduces to (E17.3). The stiffness matrix of this element can be expressed in closed form.8 For convenience define γ = a/b (rectangle aspect ratio), ψ1 = (1 + ν)γ , ψ2 = (1 − 3ν)γ , ψ3 = 2 + (1 − ν)γ 2 , ψ4 = 2γ 2 + (1 − ν), ψ5 = (1 − ν)γ 2 − 4, ψ6 = (1 − ν)γ 2 − 1, ψ7 = 4γ 2 − (1 − ν) and ψ8 = γ 2 − (1 − ν). Then the stiffness matrix in closed form is

    Eh e  K = 2 24γ (1 − ν )    
4ψ3
3ψ1 4ψ4
2ψ5 3ψ2 4ψ3
−3ψ2 4ψ8 −3ψ1 4ψ4
symm
−2ψ3 −3ψ1 −4ψ6 3ψ2 4ψ3
−3ψ1 −2ψ4 −3ψ2 −2ψ7 3ψ1 4ψ4
−4ψ6 −3ψ2 −2ψ3 3ψ1 2ψ5 3ψ2 4ψ3

3ψ2 −2ψ7  3ψ1   −2ψ4  . −3ψ2   4ψ8   −3ψ1 4ψ4
(E17.5)
6
The reason for this option is speed. A symbolic or exact computation can take orders of magnitude more time than a floating-point evaluation. This becomes more pronounced as elements get more complicated.
7
The rank of an element stiffness is discussed in Chapter 19.
8
This closed form can be obtained by either exact integration, or numerical integration with a 2 × 2 or higher Gauss rule.
17–14
17–15
Exercises
EXERCISE 17.1 [C:20] Exercise the Mathematica module of Figure E17.1 with the following script:
ClearAll[Em,nu,a,b,h]; Em=48; h=1; a=4; b=2; nu=0; ncoor={{0,0},{a,0},{a,b},{0,b}}; Emat=Em/(1-nu^2)*{{1,nu,0},{nu,1,0},{0,0,(1-nu)/2}}; For [p=1, p<=4, p++, Ke= Quad4IsoPMembraneStiffness[ncoor,Emat,h,{True,p}]; Print['Gauss integration rule: ',p,' x ',p]; Print['Ke=',Chop[Ke]//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[N[Ke]]]] ]; Verify that for integration rules p=2,3,4 the stiffness matrix does not change and has three zero eigenvalues, which correspond to the three two-dimensional rigid body modes. On the other hand, for p = 1 the stiffness matrix is different and displays five zero eigenvalues, which is physically incorrect. (This phenomenon is analyzed further in Chapter 19.) Question: why does the stiffness matrix stays exactly the same for p ≥ 2? Hint: take a look at the entries of the integrand h BT EB J ;for a rectangular geometry are those polynomials in ξ and η, or rational functions? If the former, of what polynomial order in ξ and η are the entries? EXERCISE 17.2 [C:20] Check the rectangular element stiffness closed form given in (E17.5). This may be
done by hand (takes a while) or (quicker) running the script of Figure E17.3, which calls the Mathematica module of Figure E17.1.
ClearAll[Em,Ν,a,b,h,Γ]; b=a/Γ; ncoor={{0,0},{a,0},{a,b},{0,b}}; Emat=Em/(1-Ν^2)*{{1,Ν,0},{Ν,1,0},{0,0,(1-Ν)/2}}; Ke= Quad4IsoPMembraneStiffness[ncoor,Emat,h,{False,2}]; scaledKe=Simplify[Ke*(24*(1-Ν^2)*Γ/(Em*h))]; Print['Ke=',Em*h/(24*Γ*(1-Ν^2)),'*n',scaledKe//MatrixForm]; Figure E17.3. Script suggested for Exercise E17.2.
The scaling introduced in the last two lines is for matrix visualization convenience. Verify (E17.5) by printout inspection and report any typos to instructor. EXERCISE 17.3
[A/C:25=5+10+10] A Bernoulli-Euler plane beam of thin rectangular cross-section with span L, height b and thickness h (normal to the plane of the figure) is bent under end moments M as illustrated in Figure E17.4. The beam is fabricated of isotropic material with elastic modulus E and Poisson’s ratio ν. The exact solution of the beam problem (from both the theory-of-elasticity and beam-theory standpoints) is a constant bending moment M along the span. Consequently the beam deforms with uniform curvature κ = M/(E Iz ), in which Iz = 121 hb3 is the cross-section second moment of inertia about z. The beam is modeled with one layer of identical 4-node iso-P bilinear quadrilaterals through its height. These are rectangles with horizontal dimension a; in the Figure a = L/4. The aspect ratio b/a is denoted by γ . By analogy with the exact solution, all rectangles in the finite element model will undergo the same deformation. We can therefore isolate a typical element as illustrated in Figure E17.4. The exact displacement field for the beam segment referred to the {x, y} axes placed at the element center as
17–15
17–16
Chapter 17: ISOPARAMETRIC QUADRILATERALS
M
y
M z
b
Cross section
L
h
M
M
y 4 b=γa
3 x
1
2 a
Figure E17.4. Pure bending of Bernoulli-Euler plane beam of thin rectangular cross section, for Exercises 17.3–7. The beam is modeled by one layer of 4-node iso-P bilinear quadrilaterals through its height.
shown in the bottom of Figure E17.4, are u x = −κ x y,
u y = 12 κ(x 2 + νy 2 ),
(E17.6)
where κ is the deformed beam curvature M/E I . The stiffness equations of the typical rectangular element are given by the close form expression (E17.5). The purpose of this Exercise is to compare the in-plane bending response of the 4-node iso-P bilinear rectangle to that of a Bernoulli-Euler beam element (which would be exact for this configuration). The quadrilateral element will be called x-bending exact if it reproduces the beam solution for all {γ , ν}. This comparison is distributed into three items. (a)
Check that (E17.6), as a plane stress 2D elasticity solution, is in full agreement with Bernoulli-Euler beam theory. This can be done by computing the strains ex x = ∂u x /∂ x, e yy = ∂u y /∂ y and 2ex y = ∂u y /∂ x + ∂u x /∂ y. Then get the stresses σx x , σ yy and σx y through the plane stress constitutive matrix (E17.3) of an isotropic material. Verify that both σ yy and σx y vanish for any ν, and that σx x = −E κ y = −M y/Iz , which agrees with equation (13.4) in Chapter 13.
(b)
Compute the strain energy Uquad = 12 (ubeam )T Ke ubeam absorbed by the 4-node element under nodal displacements ubeam constructed by evaluating (E17.6) at the nodes 1,2,3,4. To simplify this calculation, it is convenient to decompose that vector as follows: y
x ubeam = ubeam + ubeam = 14 κab [ −1 0 1 0 −1 0 1 0 ]T
+ 18 κ(a 2 + νb2 ) [ 0 1 0 1 0 1 0 1 ]T
(E17.7)
y
Explain why Ke ubeam must vanish and consequently x x )T Ke ubeam . Uquad = 12 (ubeam
(E17.8)
This energy can be easily computed by Mathematica by using the first 4 lines of the script of the previous x is formed Exercise, except that here ncoor={ { -a,-b },{ a,-b },{ a,b },{ -a,b } }/2. If vector ubeam
17–16
17–17
Exercises
in u as a one-dimensional list, Uquad=Simplify[u.Ke.u/2]. This should come out as a function of M, E, ν, h, a and γ because κ = M/(E Iz ) = 12M/(Eha 3 γ 3 ). (c)
From Mechanics of Materials, or equation (13.7) of Chapter 13, the strain energy absorbed by the beam segment of length a under a constant bending moment M is Ubeam = 12 Mκa = M 2 a/(2E Iz ) = 6M 2 /(Eha 2 γ 3 ). Form the energy ratio r = Uquad /Ubeam and show that it is a function of the rectangle aspect ratio γ = b/a and of Poisson’s ratio ν only: r = r (γ , ν) =
1 + 2/γ 2 − ν . (2/γ 2 )(1 − ν 2 )
(E17.9)
This happens to be the ratio of the 2D model solution to the exact (beam) solution. Hence r = 1 means that we get the exact answer, that is the 2D model is x-bending exact. If r > 1 the 2D model is overstiff, and if r < 1 the 2D model is overflexible. Evidently r > 1 for all γ if 0 ≤ ν ≤ 12 . Moreover if b << a, r >> 1; for example if a = 10b and ν = 0, r ≈ 50 and the 2D model gives only about 2% of the correct solution. This phenomenon is referred to in the FEM literature as shear locking, because overstiffness is due to the bending motion triggering spurious shear energy in the element. Remedies to shear locking at the element level are studied in advanced FEM courses. Draw conclusions as to the adequacy or inadequacy of the 2D model to capture inplane bending effects, and comment on how you might improve results by modifying the discretization of Figure E17.4.9 EXERCISE 17.4
[A+C:20] A naive remedy to shear locking can be attempted with the weighted integration methodology outlined in §17.6.1. Let Ke1×1 and Ke2×2 denote the element stiffnesses produced by 1×1 and 2×2 Gauss product rules, respectively. Take (E17.10) Keβ = (1 − β)Ke1×1 + βKe2×2 where β is adjusted so that shear locking is reduced or eliminated. It is not difficult to find β if the element is rectangular and isotropic. For the definition of x-bending exact please read the previous Exercise. Inserting Keβ into the test introduced there verify that r=
β(1 + 2γ 2 − ν) . (2/γ 2 )(1 − ν 2 )
(E17.11)
Whence show that if
2/γ 2 (1 − ν 2 ) , (E17.12) 1 + 2/γ 2 − ν then r ≡ 1 for all {γ , ν} and the element is x-bending exact. A problem with this idea is that it does not make it y-bending exact because r (γ ) 0011= r (1/γ ) if γ 0011= 1. Moreover the device is not easily extended to non-rectangular geometries or non-isotropic material. β=
EXERCISE 17.5
[A+C:35] (Advanced) To understand this Exercise please begin by reading Exercise 17.3, and the concept of shear locking. The material is again assumed isotropic with elastic modules E and Poisson’s ratio ν. The 4-node rectangular element will be said to be bending exact if r = 1 for any {γ , ν} if the bending test described in Exercise 17.3 is done in both x and y directions. A bending-exact element is completely shear-lock-free. The selective integration scheme outlined in §17.6.2 is more effective than weighted integration (covered in the previous exercise) to fully eliminate shear locking. Let the integration rules (I) and (II) be the 1×1 and 9
Note that even if we make a → 0 and γ = b/a → ∞ by taking an infinite number of rectangular elements along x, the energy ratio r remains greater than one if ν > 0 since r → 1/(1 − ν 2 ). Thus the 2D model would not generally converge to the correct solution if we keep one layer through the height.
17–17
17–18
Chapter 17: ISOPARAMETRIC QUADRILATERALS
2×2 product rules, respectively. However the latter is generalized so the sample points are located at {−χ, χ}, {χ, −χ }, {χ , χ} and {−χ , χ}, with weight 1.10 Consider the stress-strain splitting
E = 1−ν 2
α β 0 β α 0 0 0 1−ν 2
where α and β are scalars. Show that if
0013
E E= 1−ν 2
1 ν 0 ν 1 0 0 0 1−ν 2
χ=
E + 1−ν 2
1−α ν−β 0 ν−β 1−α 0 0 0 0
= EI + EII ,
1 − ν2 3(1 − α)
(E17.13)
(E17.14)
the resulting element stiffness KeI + KeII is bending exact for any {α, β}. As a corollary show that that if α = ν 2 , which corresponds to the splitting
E E= 1−ν 2
1 ν 0 ν 1 0 0 0 1−ν 2
E = 1−ν 2
ν2 β 0 β ν2 0 0 0 1−ν 2
E + 1−ν 2
1−ν 2 ν−β 0 ν−β 1−ν 2 0 0 0 0
= EI + EII ,
(E17.15)
√ then χ = 1/ 3 and rule (II) becomes the standard 2×2 Gauss product rule. What are two computationally convenient settings for β? EXERCISE 17.6
[A+C:35] (Advanced) A variation on the previous exercise on selective integration to make the isotropic rectangular 4-node element bending exact. Integration rule (I) is not changed. However rule (II) has four sample points located at {0, −χ }, {χ , 0}, {0, χ} and {−χ , 0} each with weight 1.11 Show that if one selects the stress-strain splitting (E17.13) and
0013
χ=
2(1 − ν 2 ) 3(1 − α)
(E17.16)
the resulting element stiffness KeI + KeII is bending exact for any {α, β}. Discuss which choices of α reduce χ √ √ to 1/ 3 and 2/3, respectively. EXERCISE 17.7
[A+C:40] (Advanced, research paper level, requires a CAS to be tractable) Extend Exercise 17.5 to consider the case of general anisotropic material:
E=
E 11 E 12 E 13
E 12 E 22 E 23
E 13 E 23 E 33
(E17.17)
The rules for the selective integration scheme are as described in Exercise 17.5. The appropriate stress-strain splitting is
E = EI + EII =
E 11 α1 E 12 β E 13 E 12 β E 22 α2 E 23 E 13 E 23 E 33
+
E 11 (1 − α1 ) E 12 (1 − β) 0 E 12 (1 − β) E 22 (1 − α2 ) 0 0 0 0
(E17.18)
10
For a rectangular √ geometry these sample points lie on the diagonals. In the case of the standard 2-point Gauss product rule χ = 1/ 3.
11
This is called a 4-point median rule, since the four points are located on the quadrilateral medians.
17–18
17–19
Exercises
in which β is arbitrary and 1 − α1 =
3χ 2 E
1 |E| = , 2 2 3χ C11 11 (E 22 E 33 − E 23 )
1 − α2 =
3χ 2 E
1 |E| = , 2 2 3χ C22 22 (E 11 E 33 − E 13 )
2 2 2 |E| = det(E) = E 11 E 22 E 33 + 2E 12 E 13 E 23 − E 11 E 23 − E 22 E 13 − E 33 E 12 , 2 C11 = E 11 (E 22 E 33 − E 13 )/|E|,
2 C22 = E 22 (E 11 E 33 − E 13 )/|E|.
(E17.19) Show that the√resulting rectangular element is bending exact for any E and χ 0011= 0. (In practice one would select χ = 1/ 3.)
17–19
18
.
Shape Function Magic
18–1
18–2
Chapter 18: SHAPE FUNCTION MAGIC
TABLE OF CONTENTS Page
§18.1. Requirements §18.2. Direct Fabrication of Shape Functions §18.3. Triangular Element Shape Functions §18.3.1. The Three-Node Linear Triangle . . . . §18.3.2. The Six-Node Quadratic Triangle . . . §18.4. Quadrilateral Element Shape Functions §18.4.1. The Four-Node Bilinear Quadrilateral . . §18.4.2. The Nine-Node Biquadratic Quadrilateral §18.4.3. The Eight-Node “Serendipity” Quadrilateral §18.5. Does the Magic Wand Always Work? §18.5.1. Hierarchical Corrections . . . . . . §18.5.2. Transition Element Example . . . . . §18.6. *Mathematica Modules to Plot Shape Functions §18. Notes and Bibliography . . . . . . . . . . . . . . . §18. References . . . . . . . . . . . . . . . §18. Exercises . . . . . . . . . . . . . . .
18–2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18–3 18–3 18–4 18–4 18–5 18–6 18–6 18–7 18–9 18–10 18–10 18–11 18–12 18–14 18–14 18–15
18–3
§18.2
DIRECT FABRICATION OF SHAPE FUNCTIONS
§18.1. Requirements This Chapter explains, through a series of examples, how isoparametric shape functions can be directly constructed by geometric considerations. For a problem of variational index 1, the isoparametric shape function Nie associated with node i of element e must satisfy the following conditions: (A) Interpolation condition. Takes a unit value at node i, and is zero at all other nodes. (B) Local support condition. Vanishes over any element boundary (a side in 2D, a face in 3D) that does not include node i. (C) Interelement compatibility condition. Satisfies C 0 continuity between adjacent elements over any element boundary that includes node i. (D) Completeness condition. The interpolation is able to represent exactly any displacement field which is a linear polynomial in x and y; in particular, a constant value. Requirement (A) follows directly by interpolation from node values. Conditions (B), (C) and (D) are consequences of the convergence requirements discussed further in the next Chapter.1 For the moment these three conditions may be viewed as recipes. One can readily verify that all isoparametric shape function sets listed in Chapter 16 satisfy the first two conditions from construction. Direct verification of condition (C) is also straightforward for those examples. A statement equivalent to (C) is that the value of the shape function over a side (in 2D) or face (in 3D) common to two elements must uniquely depend only on its nodal values on that side or face. Completeness is a property of all element isoparametric shape functions taken together, rather than of an individual one. If the element satisfies (B) and (C), in view of the discussion in §16.6 it is sufficient to check that the sum of shape functions is identically one. §18.2. Direct Fabrication of Shape Functions Contrary to the what the title of this Chapter implies, the isoparametric shape functions listed in Chapter 16 did not come out of a magician’s hat. They can be derived systematically by a judicious inspection process. By “inspection” it is meant that the geometric visualization of shape functions plays a crucial role. The method is based on the following observation. In all examples given so far the isoparametric shape functions are given as products of fairly simple polynomial expressions in the natural coordinates. This is no accident but a direct consequence of the definition of natural coordinates. All shape functions of Chapter 16 can be expressed as the product of m factors: Nie = ci L 1 L 2 . . . L m ,
(18.1)
where L j = 0,
j = 1, . . . m.
(18.2)
are the homogeneous equation of lines or curves expressed as linear functions in the natural coordinates, and ci is a normalization coefficient. 1
Convergence means that the discrete FEM solution approaches the exact analytical solution as the mesh is refined.
18–3
18–4
Chapter 18: SHAPE FUNCTION MAGIC
(a)
(c)
(b) 3
3
ζ1 = 0
3 2 1
2
1
1
2
Figure 18.1. The three-node linear triangle: (a) element geometry; (b) equation of side opposite corner 1; (c) perspective view of the shape function N1 = ζ1 .
For two-dimensional isoparametric elements, the ingredients in (18.1) are chosen according to the following five rules. R1 Select the L j as the minimal number of lines or curves linear in the natural coordinates that cross all nodes except the i th node. (A sui generis “cross the dots” game.) Primary choices in 2D are the element sides and medians. R2 Set coefficient ci so that Nie has the value 1 at the i th node. R3 Check that Nie vanishes over all element sides that do not contain node i. R4 Check the polynomial order over each side that contains node i. If the order is n, there must be exactly n + 1 nodes on the side for compatibility to hold. R5 If local support (R3) and interelement compatibility (R4) are satisfied, check that the sum of shape functions is identically one. The examples that follow show these rules in action for two-dimensional elements. Essentially the same technique is applicable to one- and three-dimensional elements. §18.3. Triangular Element Shape Functions This section illustrates the use of (18.1) in the construction of shape functions for the linear and the quadratic triangle. The cubic triangle is dealt with in Exercise 18.1. §18.3.1. The Three-Node Linear Triangle Figure 18.1 shows the three-node linear triangle that was studied in detail in Chapter 15. The three shape functions are simply the triangular coordinates: Ni = ζi , for i = 1, 2, 3. Although this result follows directly from the linear interpolation formula of §15.2.4, it can be also quickly derived from the present methodology as follows. The equation of the triangle side opposite to node i is L j -k = ζi = 0, where j and k are the cyclic permutations of i. Here symbol L j -k denotes the left hand side of the homogeneous equation of the natural coordinate line that passes through node points j and k. See Figure 18.1(b) for i = 1, j = 2 and k = 3. Hence the obvious guess is Nie
guess
=
18–4
ci L i .
(18.3)
18–5
§18.3
(a)
(b)
3
5
6
TRIANGULAR ELEMENT SHAPE FUNCTIONS
3
2
ζ1 = 0
5
6 2
ζ1 = 1/2
4 1
1
3
5
6
4
(c)
ζ1 = 0
ζ2 = 0
2 4 1
Figure 18.2. The six-node quadratic triangle: (a) element geometry; (b) lines (in red) whose product yields N1e ; (c) lines (in red) whose product yields N4e .
This satisfies conditions (A) and (B) except the unit value at node i; this holds if ci = 1. The local support condition (B) follows from construction: the value of ζi is zero over side j–k. Interelement compatibility follows from R4: the variation of ζi along the 2 sides meeting at node i is linear and that there are two nodes on each side; cf. §15.4.2. Completeness follows since N1e + N2e + N3e = ζ1 + ζ2 + ζ3 = 1. Figure 18.1(c) depicts N1e = ζ1 , drawn normal to the element in perspective view. §18.3.2. The Six-Node Quadratic Triangle The geometry of the six-node quadratic triangle is shown in Figure 18.2(a). Inspection reveals two types of nodes: corners (1, 2 and 3) and midside nodes (4, 5 and 6). Consequently we can expect two types of associated shape functions. We select nodes 1 and 4 as representative cases. For both cases we try the product of two linear functions in the triangular coordinates because we expect the shape functions to be quadratic. These functions are illustrated in Figures 18.2(b,c) for corner node 1 and midside node 4, respectively. For corner node 1, inspection of Figure 18.2(b) suggests trying N1e
guess
=
c1 L 2-3 L 4-6 ,
(18.4)
Why is (18.4) expected to work? Clearly N1e will vanish over 2-5-3 and 4-6. This makes the function zero at nodes 2 through 6, as is obvious upon inspection of Figure 18.2(b), while being nonzero at node 1. This value can be adjusted to be unity if c1 is appropriately chosen. Clearly N1e will vanish along side 2-5-3 that does not contain 1. The equations of the lines that appear in (18.4) are L 2-3 : ζ1 = 0, L 4-6 : ζ1 − 12 = 0. (18.5) Replacing into (18.3) we get N1e = c1 ζ1 (ζ1 − 12 ),
(18.6)
To find c1 , evaluate N1e (ζ1 , ζ2 , ζ3 ) at node 1. The triangular coordinates of this node are ζ1 = 1, ζ2 = ζ3 = 0. We require that it takes a unit value there: N1e (1, 0, 0) = c1 × 1 × 12 = 1, from which c1 = 2 and finally N1e = 2ζ1 (ζ1 − 12 ) = ζ1 (2ζ1 − 1), (18.7) 18–5
18–6
Chapter 18: SHAPE FUNCTION MAGIC
3 3
6
6
1
1
5 4
5 4
2 2
N4e = 4ζ1ζ 2
e
N1 = ζ 1(2ζ1 − 1)
Figure 18.3. Perspective view of shape functions N1e and N4e for the quadratic triangle. The plot is done over a straight side triangle for programming simplicity.
as listed in §16.5.2. Figure 18.3 shows a perspective view. The other two corner shape functions follow by cyclic permutations of the corner index. For midside node 4, inspection of Figure 18.2(c) suggests trying N4e
guess
=
c4 L 2-3 L 1-3
(18.8)
Evidently (18.8) satisfies requirements (A) and (B) if c4 is appropriately normalized. The equation of sides L 2-3 and L 1-3 are ζ1 = 0 and ζ2 = 0, respectively. Therefore N4e (ζ1 , ζ2 , ζ3 ) = c4 ζ1 ζ2 . To find c4 , evaluate this function at node 4, the triangular coordinates of which are ζ1 = ζ2 = 12 , ζ3 = 0. We require that it takes a unit value there: N4e ( 12 , 12 , 0) = c4 × 12 × 12 = 1. Hence c4 = 4, which gives (18.9) N4e = 4ζ1 ζ2 as listed in §16.5.2. Figure 18.3 shows a perspective view of this shape function. The other two midside shape functions follow by cyclic permutations of the node indices. It remains to carry out the interelement continuity check. Consider node 1. The boundaries containing node 1 and common to adjacent elements are 1–2 and 1–3. Over each one the variation of N1e is quadratic in ζ1 . Therefore the polynomial order over each side is 2. Because there are three nodes on each boundary, the compatibility condition (C) of §18.1 is verified. A similar check can be carried out for midside node shape functions. Exercise 16.1 verified that the sum of the Ni is unity. Therefore the element is complete. §18.4. Quadrilateral Element Shape Functions Three quadrilateral elements, with 4, 9 and 8 nodes, respectively, which are commonly used in computational mechanics serve as examples to illustrate the construction of shape functions. Elements with more nodes, such as the bicubic quadrilateral, are not treated as they are rarely used. §18.4.1. The Four-Node Bilinear Quadrilateral The element geometry and natural coordinates are shown in Figure 18.4(a). Only one type of node (corner) and associated shape function is present. Consider node 1 as typical. Inspection of 18–6
18–7
(a)
§18.4
QUADRILATERAL ELEMENT SHAPE FUNCTIONS
η=1
η 3
4
(c)
(b)
3
4
4
ξ=1
ξ
3
1 1
1
2
2
2
Figure 18.4. The four-node bilinear quadrilateral: (a) element geometry; (b) sides (in red) that do not contain corner 1; (c) perspective view of the shape function N1e .
Figure 18.4(b) suggests trying N1e
guess
=
c1 L 2-3 L 3-4
(18.10)
This clearly vanishes over nodes 2, 3 and 4, and can be normalized to unity at node 1 by adjusting c1 . By construction it vanishes over the sides 2–3 and 3–4 that do not belong to 1. The equation of side 2-3 is ξ = 1, or ξ − 1 = 0. The equation of side 3-4 is η = 1, or η − 1 = 0. Replacing in (18.10) yields N1e (ξ, η) = c1 (ξ − 1)(η − 1) = c1 (1 − ξ )(1 − η). (18.11) To find c1 , evaluate at node 1, the natural coordinates of which are ξ = η = −1: N1e (−1, −1) = c1 × 2 × 2 = 4c1 = 1. Hence c1 =
1 4
(18.12)
and the shape function is N1e = 14 (1 − ξ )(1 − η),
(18.13)
as listed in §16.6.2. Figure 18.4(c) shows a perspective view. For the other three nodes the procedure is the same, traversing the element cyclically. It can be verified that the general expression of the shape functions for this element is Nie = 14 (1 + ξi ξ )(1 + ηi η).
(18.14)
The continuity check proceeds as follows, using N1e as example. Node 1 belongs to interelement boundaries 1–2 and 1–3. Over side 1–2, η = −1 is constant and N1e is a linear function of ξ . To see this, replace η = −1 in (18.13). Over side 1–3, ξ = −1 is constant and N1e is a linear function of η. Consequently the polynomial variation order is 1 over both sides. Because there are two nodes on each side the compatibility condition is satisfied. The sum of the shape functions is one, as shown in (16.21); thus the element is complete. 18–7
18–8
Chapter 18: SHAPE FUNCTION MAGIC
η
η=1
3 7
4
6
9
8 1
η=0 4
7
8
9
ξ
5
1
2
3
ξ=0 ξ=1
6
5 2
η=1 η=0 4
7
8
9
ξ = −1
1
η=1 3 4 ξ = −1
ξ=1
6
1 2
ξ=1
6
9
8
5
3
7
5 2
η = −1
Figure 18.5. The nine-node biquadratic quadrilateral: (a) element geometry; (b,c,d): lines (in red) whose product makes up the shape functions N1e , N5e and N9e , respectively.
4
(a)
(b)
7
4 7
8
8
3
3
9
9
1
6
1
6
5
5
2 e N1
(c)
=
1 (ξ 4
2
− 1)(η − 1)ξ η
e N5
5
(d)
1
8
9
4
9 6
4 7
8 2
= 12 (1 − ξ 2 )η(η − 1)
3
1
7
6 5
3
2
N5e = 12 (1 − ξ 2 )η(η − 1) (back view)
N9e = (1 − ξ 2 )(1 − η2 )
Figure 18.6. Perspective view of the shape functions for nodes 1, 5 and 9 of the nine-node biquadratic quadrilateral.
18–8
18–9
§18.4
η 7 4 6
8 1
η=1
η=1
3 4
QUADRILATERAL ELEMENT SHAPE FUNCTIONS
ξ
7
3
ξ + η = −1 4 ξ = −1
6
8
7
3 6
8
ξ=1
ξ=1
5 2
1
1
5
5 2
2
Figure 18.7. The eight-node serendipity quadrilateral: (a) element geometry; (b,c): lines (in red) whose product make up the shape functions N1e and N5e , respectively.
§18.4.2. The Nine-Node Biquadratic Quadrilateral The element geometry is shown in Figure 18.5(a). This element has three types of shape functions, which are associated with corner nodes, midside nodes and center node, respectively. The lines whose product is used to construct three types of shape functions are illustrated in Figure 18.5(b,c,d) for nodes 1, 5 and 9, respectively. The technique has been sufficiently illustrated in previous examples. Here we summarize the calculations for nodes 1, 5 and 9, which are taken as representatives of the three types: N1e = c1 L 2-3 L 3-4 L 5-7 L 6-8 = c1 (ξ − 1)(η − 1)ξ η. N5e = c5 L 2-3 L 1-4 L 6-8 L 3-4 = c5 (ξ − 1)(ξ + 1)η(η − 1) = c5 (1 − ξ 2 )η(1 − η).
(18.15) (18.16)
N9e = c9 L 1-2 L 2-3 L 3-4 L 4-1 = c9 (ξ − 1)(η − 1)(ξ + 1)(η + 1) = c9 (1 − ξ 2 )(1 − η2 ) (18.17) Imposing the normalization conditions we find c1 = 14 ,
c5 = − 12 ,
c9 = 1,
(18.18)
and we obtain the shape functions listed in §16.6.3. Perspective views are shown in Figure 18.6. The remaining Ni ’s are constructed through a similar procedure. Verification of the interelement continuity condition is immediate: the polynomial variation order of Nie over any side that belongs to node i is two and there are three nodes on each side. Exercise 16.2 checks that the sum of shape function is unity. Thus the element is complete. §18.4.3. The Eight-Node “Serendipity” Quadrilateral This is an eight-node quadrilateral element that results when the center node 9 of the biquadratic quadrilateral is eliminated by kinematic constraints. The geometry and node configuration is shown in Figure 18.7(a). This element was widely used in commercial codes during the 70s and 80s but it is gradually being phased out in favor of the 9-node quadrilateral. 18–9
18–10
Chapter 18: SHAPE FUNCTION MAGIC
(a)
(b)
3
4
(c)
3
(d)
3
4
3
4 6
5 1 2
1 4
7
1
2
5
2
6
1 5
2
Figure 18.8. Node configurations for which the magic recipe does not work.
The 8-node quadrilateral has two types of shape functions, which are associated with corner nodes and midside nodes. Lines whose products yields the shape functions for nodes 1 and 5 are shown in Figure 18.7(b,c). Here are the calculations for shape functions of nodes 1 and 5, which are taken again as representative cases. N1e = c1 L 2-3 L 3-4 L 5-8 = c1 (ξ − 1)(η − 1)(1 + ξ + η) = c1 (1 − ξ )(1 − η)(1 + ξ + η), (18.19) N5e = c5 L 2-3 L 3-4 L 4-1 = c5 (ξ − 1)(ξ + 1)(η − 1) = c5 (1 − ξ 2 )(1 − η).
(18.20)
Imposing the normalization conditions we find c1 = − 14 ,
c5 =
1 2
(18.21)
The other shape functions follow by appropriate permutation of nodal indices. The interelement continuity and completeness verification are similar to that carried out for the nine-node element, and are relegated to exercises. §18.5. Does the Magic Wand Always Work? The “cross the dots” recipe (18.1)-(18.2) is not foolproof. It fails for certain node configurations although it is a reasonable way to start. It runs into difficulties, for instance, in the problem posed in Exercise 18.6, which deals with the 5-node quadrilateral depicted in Figure 18.8(a). If for node 1 one tries the product of side 2–3, side 3–4, and the diagonal 2–5–4, the shape function is easily worked out to be N1e = − 18 (1 − ξ )(1 − η)(ξ + η). This satisfies conditions (A) and (B). However, it violates (C) along sides 1–2 and 4–1, because it varies quadratically over them with only two nodes per side. §18.5.1. Hierarchical Corrections A more robust technique relies on a correction approach, which employs a combination of terms such as (18.1). For example, a combination of two patterns, one with m factors and one with n factors, is (18.22) Nie = ci L c1 L c2 . . . L cm + di L d1 L d2 . . . L dn , Here two normalization coefficients: ci and di , appear. In practice trying forms such as (18.22) from scratch becomes cumbersome. The development is best done hierarchically. The first term is 18–10
18–11
§18.5 DOES THE MAGIC WAND ALWAYS WORK?
taken to be that of a lower order element, called the parent element, for which the one-shot approach works. The second term is then a corrective shape function that vanishes at the nodes of the parent element. If this is insufficient one more corrective term is added, and so on. The technique is best explained through examples. Exercise 18.6 illustrates the procedure for the element of Figure 18.8(a). The next subsection works out the element of Figure 18.8(b). §18.5.2. Transition Element Example The hierarchical correction technique is useful for transition elements, which have corner nodes but midnodes only over certain sides. Three examples are pictured in Figure 18.8(b,c,d). Shape functions that work can be derived with one, two and three hierarchical corrections, respectively. As an example, let us construct the shape function N1e for the 4-node transition triangle shown in Figure 18.8(b). Candidate lines for the recipe (18.1) are obviously the side 2–3: ζ1 = 0, and the median 3–4: ζ1 = ζ2 . Accordingly we try N1e
guess
=
c1 ζ1 (ζ1 − ζ2 ),
N1 (1, 0, 0) = 1 = c1 .
(18.23)
This function N1e = ζ1 (ζ1 − ζ2 ) satisfies conditions (A) and (B) but fails compatibility: over side 1–3 of equation ζ2 = 0, because N1e (ζ1 , 0, ζ3 ) = ζ12 . This varies quadratically but there are only 2 nodes on that side. Thus (18.23) is no good. To proceed hierarchically we start from the shape function for the 3-node linear triangle: N1e = ζ1 . This will not vanish at node 4, so apply a correction that vanishes at all nodes but 4. From knowledge of the quadratic triangle midpoint functions, that is obviously ζ1 ζ2 times a coefficient to be determined. The new guess is N1e
guess
=
ζ1 + c1 ζ1 ζ2 .
Coefficient c1 is determined by requiring that N1e vanish at 4: N1e ( 12 , 12 , 0) = c1 = −2 and the shape function is N1e = ζ1 − 2ζ1 ζ2 .
(18.24) 1 2
+ c1 41 = 0, whence (18.25)
This is easily checked to satisfy compatibility on all sides. The verification of completeness is left to Exercise 18.8. Note that since N1e = ζ1 (1 − 2ζ2 ), (18.25) can be constructed as the normalized product of lines ζ1 = 0 and ζ2 = /. The latter passes through 4 and is parallel to 1–3. As part of the opening moves in the shape function game this would be a lucky guess indeed. If one goes to a more complicated element no obvious factorization is possible.
18–11
18–12
Chapter 18: SHAPE FUNCTION MAGIC
Cell 18.1 Mathematica Module to Draw a Function over a Triangle Region
PlotTriangleShapeFunction[xytrig_,f_,Nsub_,aspect_]:=Module[ {Ni,line3D={},poly3D={},zc1,zc2,zc3,xyf1,xyf2,xyf3, xc,yc, x1,x2,x3,y1,y2,y3,z1,z2,z3,iz1,iz2,iz3,d}, {{x1,y1,z1},{x2,y2,z2},{x3,y3,z3}}=Take[xytrig,3]; xc={x1,x2,x3}; yc={y1,y2,y3}; Ni=Nsub*3; Do [ Do [iz3=Ni-iz1-iz2; If [iz3<=0, Continue[]]; d=0; If [Mod[iz1+2,3]0&&Mod[iz2-1,3]0, d= 1]; If [Mod[iz1-2,3]0&&Mod[iz2+1,3]0, d=-1]; If [d0, Continue[]]; zc1=N[{iz1+d+d,iz2-d,iz3-d}/Ni]; zc2=N[{iz1-d,iz2+d+d,iz3-d}/Ni]; zc3=N[{iz1-d,iz2-d,iz3+d+d}/Ni]; xyf1={xc.zc1,yc.zc1,f[zc1[[1]],zc1[[2]],zc1[[3]]]}; xyf2={xc.zc2,yc.zc2,f[zc2[[1]],zc2[[2]],zc2[[3]]]}; xyf3={xc.zc3,yc.zc3,f[zc3[[1]],zc3[[2]],zc3[[3]]]}; AppendTo[poly3D,Polygon[{xyf1,xyf2,xyf3}]]; AppendTo[line3D,Line[{xyf1,xyf2,xyf3,xyf1}]], {iz2,1,Ni-iz1}],{iz1,1,Ni}]; Show[ Graphics3D[RGBColor[1,0,0]],Graphics3D[poly3D], Graphics3D[Thickness[.002]],Graphics3D[line3D], Graphics3D[RGBColor[0,0,0]],Graphics3D[Thickness[.005]], Graphics3D[Line[xytrig]],PlotRange->All, BoxRatios->{1,1,aspect},Boxed->False] ]; ClearAll[f1,f4]; xyc1={0,0,0}; xyc2={3,0,0}; xyc3={Sqrt[3],3/2,0}; xytrig=N[{xyc1,xyc2,xyc3,xyc1}]; Nsub=16; f1[zeta1_,zeta2_,zeta3_]:=zeta1*(2*zeta1-1); f4[zeta1_,zeta2_,zeta3_]:=4*zeta1*zeta2; PlotTriangleShapeFunction[xytrig,f1,Nsub,1/2]; PlotTriangleShapeFunction[xytrig,f4,Nsub,1/2.5];
§18.6.
*Mathematica Modules to Plot Shape Functions
A Mathematica module called PlotTriangleShape Functions, listed in Cell 18.1, has been developed to draw perspective plots of shape functions Ni (ζ1 , ζ2 , ζ3 ) over a triangular region. The region is assumed to have straight sides to simplify the logic. The test statements that follow the module produce the shape function plots shown in Figure 18.3 for the 6-node quadratic triangle. Argument Nsub controls the plot resolution while aspect controls the x yz box aspect ratio. The remaining arguments are self explanatory. Another Mathematica module called PlotQuadrilateralShape Functions, listed in Cell 18.2, has been developed to produce perspective plots of shape functions Ni (ξ, η) over a quadrilateral region. The region is assumed to have straight sides to simplify the logic. The test statements that follow the module produce
18–12
18–13
§18.6
*MATHEMATICA MODULES TO PLOT SHAPE FUNCTIONS
Cell 18.2 Mathematica Module to Draw a Function over a Quadrilateral Region
PlotQuadrilateralShapeFunction[xyquad_,f_,Nsub_,aspect_]:=Module[ {Ne,Nev,line3D={},poly3D={},xyf1,xyf2,xyf3,i,j,n,ixi,ieta, xi,eta,x1,x2,x3,x4,y1,y2,y3,y4,z1,z2,z3,z4,xc,yc}, {{x1,y1,z1},{x2,y2,z2},{x3,y3,z3},{x4,y4,z4}}=Take[xyquad,4]; xc={x1,x2,x3,x4}; yc={y1,y2,y3,y4}; Ne[xi_,eta_]:=N[{(1-xi)*(1-eta),(1+xi)*(1-eta), (1+xi)*(1+eta),(1-xi)*(1+eta)}/4]; n=Nsub; Do [ Do [ ixi=(2*i-n-1)/n; ieta=(2*j-n-1)/n; {xi,eta}=N[{ixi-1/n,ieta-1/n}]; Nev=Ne[xi,eta]; xyf1={xc.Nev,yc.Nev,f[xi,eta]}; {xi,eta}=N[{ixi+1/n,ieta-1/n}]; Nev=Ne[xi,eta]; xyf2={xc.Nev,yc.Nev,f[xi,eta]}; {xi,eta}=N[{ixi+1/n,ieta+1/n}]; Nev=Ne[xi,eta]; xyf3={xc.Nev,yc.Nev,f[xi,eta]}; {xi,eta}=N[{ixi-1/n,ieta+1/n}]; Nev=Ne[xi,eta]; xyf4={xc.Nev,yc.Nev,f[xi,eta]}; AppendTo[poly3D,Polygon[{xyf1,xyf2,xyf3,xyf4}]]; AppendTo[line3D,Line[{xyf1,xyf2,xyf3,xyf4,xyf1}]], {i,1,Nsub}],{j,1,Nsub}]; Show[ Graphics3D[RGBColor[1,0,0]],Graphics3D[poly3D], Graphics3D[Thickness[.002]],Graphics3D[line3D], Graphics3D[RGBColor[0,0,0]],Graphics3D[Thickness[.005]], Graphics3D[Line[xyquad]], PlotRange->All, BoxRatios->{1,1,aspect},Boxed->False] ]; ClearAll[f1,f5,f9]; xyc1={0,0,0}; xyc2={3,0,0}; xyc3={3,3,0}; xyc4={0,3,0}; xyquad=N[{xyc1,xyc2,xyc3,xyc4,xyc1}]; Nsub=16; f1[xi_,eta_]:=(1/2)*(xi-1)*(eta-1)*xi*eta; f5[xi_,eta_]:=(1/2)*(1-xi^2)*eta*(eta-1); f9[xi_,eta_]:=(1-xi^2)*(1-eta^2); PlotQuadrilateralShapeFunction[xyquad,f1,Nsub,1/2]; PlotQuadrilateralShapeFunction[xyquad,f5,Nsub,1/2.5]; PlotQuadrilateralShapeFunction[xyquad,f9,Nsub,1/3];
the shape function plots shown in Figure 18.6(a,b,d) for the 9-node biquadratic quadrilateral. Argument Nsub controls the plot resolution while aspect controls the x yz box aspect ratio. The remaining arguments are self explanatory.
18–13
18–14
Chapter 18: SHAPE FUNCTION MAGIC
Notes and Bibliography The name “shape functions” for interpolation functions directly expressed in terms of physical coordinates (the node displacements in the case of isoparametric elements) was coined by Irons. The earliest published reference seems to be the paper [11]. This was presented in 1965 at the first Wright-Patterson conference, the first all-FEM meeting that strongly influenced the development of computational mechanics in Generation 2. The key connection to numerical integration was presented in [95], although it is mentioned in prior internal reports. A comprehensive exposition is given in the textbook by Irons and Ahmad [98]. The quick way of developing shape functions presented here was used in the writer’s 1966 thesis [44] for triangular elements. The qualifier “magic” arose from the timing for covering this Chapter in a Fall Semester course: the lecture falls near Halloween. References Referenced items have been moved to Appendix R.
18–14
18–15
Exercises
Homework Exercises for Chapter 18 Shape Function Magic EXERCISE 18.1 [A/C:10+10] The complete cubic triangle for plane stress has 10 nodes located as shown in Figure E18.1, with their triangular coordinates listed in parentheses.
3(0,0,1)
3 7(0,1/3,2/3)
9(2/3,0,1/3)
7
8
8(1/3,0,2/3)
6(0,2/3,1/3) 6
9 0(1/3,1/3,1/3)
0
2(0,1,0)
2 5(1/3,2/3,0) 4(2/3,1/3,0)
5
4 1
1(1,0,0)
Figure E18.1. Ten-node cubic triangle for Exercise 18.1. The left picture shows the superparametric element whereas the right one shows the isoparametric version with curved sides.
N1e
N4e
e
N0
Figure E18.2. Perspective plots of the shape functions N1e , N4e and N0e for the 10-node cubic triangle.
(a)
Construct the cubic shape functions N1e , N4e and N0e for nodes 1, 4, and 0 (the interior node is labeled as zero, not 10) using the line-product technique. [Hint: each shape function is the product of 3 and only 3 lines.] Perspective plots of those 3 functions are shown in Figure E18.2.
(b)
Construct the missing 7 shape functions by appropriate node number permutations, and verify that the sum of the 10 functions is identically one. For the unit sum check use the fact that ζ1 + ζ2 + ζ3 = 1.
EXERCISE 18.2 [A:15] Find an alternative shape function N1e for corner node 1 of the 9-node quadrilateral
of Figure 18.5(a) by using the diagonal lines 5–8 and 2–9–4 in addition to the sides 2–3 and 3–4. Show that the resulting shape function violates the compatibility condition (C) stated in §18.1. EXERCISE 18.3 [A/C:15] Complete the above exercise for all nine nodes. Add the shape functions (use a
CAS and simplify) and verify whether their sum is unity. EXERCISE 18.4 [A/C:20] Verify that the shape functions N1e and N5e of the eight-node serendipity quadri-
lateral discussed in §18.4.3 satisfy the interelement compatibility condition (C) stated in §18.1. Obtain all 8 shape functions and verify that their sum is unity. EXERCISE 18.5 [C:15] Plot the shape functions N1e and N5e of the eight-node serendipity quadrilateral studied
in §18.4.3 using the module PlotQuadrilateralShapeFunction listed in Cell 18.2.
18–15
18–16
Chapter 18: SHAPE FUNCTION MAGIC
η
N1
3
4
N5
ξ
5 1 2
Figure E18.3. Five node quadrilateral element for Exercise 18.6.
EXERCISE 18.6 [A:15]. A five node quadrilateral element has the nodal configuration shown in Figure E18.3.
Perspective views of N1e and N5e are shown in that Figure.2 Find five shape functions Nie , i = 1, 2, 3, 4, 5 that satisfy compatibility, and also verify that their sum is unity. Hint: develop N5 (ξ, η) first for the 5-node quad using the line-product method; then the corner shape functions N¯ i (ξ, η) (i = 1, 2, 3, 4) for the 4-node quad (already given in the Notes); finally combine Ni = N¯ i + α N5 , determining α so that all Ni vanish at node 5. Check that N1 + N2 + N3 + N4 + N5 = 1 identically. µ
EXERCISE 18.7 [A:15]. An eight-node “brick” finite ele-
ment for three dimensional analysis has three isoparametric natural coordinates called ξ , η and µ. These coordinates vary from −1 at one face to +1 at the opposite face, as sketched in Figure E18.4.
η
8 7
5
Construct the (trilinear) shape function for node 1 (follow the node numbering of the figure). The equations of the brick faces are:
z
4
6
ξ 3
1
1485 : ξ = −1 1265 : η = −1 1234 : µ = −1
x
2376 : ξ = +1 4378 : η = +1 5678 : µ = +1
2
y Figure E18.4. Eight-node isoparametric “brick” element for Exercise 18.7.
EXERCISE 18.8 [A:15]. Consider the 4-node transition triangular element of Figure 18.8(b). The shape
function for node 1, N1 = ζ1 − 2ζ1 ζ2 was derived in §18.5.2 by the correction method. Show that the others are N2 = ζ2 − 2ζ1 ζ2 , N3 = ζ3 and N4 = 4ζ1 ζ2 . Check that compatibility and completeness are verified. EXERCISE 18.9 [A:15]. Construct the six shape functions for the 6-node transition quadrilateral element of Figure 18.8(c). Hint: for the corner nodes, use two corrections to the shape functions of the 4-node bilinear quadrilateral. Check compatibility and completeness. Partial result: N1 = 14 (1 − ξ )(1 − η) − 14 (1 − ξ 2 )(1 − η). EXERCISE 18.10 [A:20]. Consider a 5-node transition triangle in which midnode 6 on side 1–3 is missing. Show that N1e = ζ1 − 2ζ1 ζ2 − 2ζ2 ζ3 . Can this be expressed as a line product like (18.1)?
2
Although this N1e resembles the N1e of the 4-node quadrilateral depicted in Figure 18.4, they are not the same. That in Figure E18.3 must vanish at node 5 (ξ = η = 0). On the other hand, the N1e of Figure 18.4 takes the value 14 there.
18–16
18–17
Exercises 4
3
4
Set CL1 Side 2-4 maps to this parabola; part of triangle 2-3-4 turns 'inside out'
4
3
4
2 1
2 4
3
4 2
1
Reference 2 triangular elements Set CL2 2
1
2
Figure E18.5. Mapping of reference triangles under sets (E18.1) and (E18.2). Triangles are slightly separated at the diagonal 2–4 for visualization convenience.
EXERCISE 18.11 [A:30]. The three-node linear triangle is known to be a poor performer for stress analysis. In an effort to improve it, Dr. I. M. Clueless proposes two sets of quadratic shape functions:
CL1: CL2:
N1 = ζ12 ,
N1 = ζ12 + 2ζ2 ζ3 ,
N2 = ζ22 ,
N3 = ζ32 .
N2 = ζ22 + 2ζ3 ζ1 ,
N3 = ζ32 + 2ζ1 ζ2 .
(E18.1) (E18.2)
Dr. C. writes a learned paper claiming that both sets satisfy the interpolation condition, that set CL1 will work because it is conforming and that set CL2 will work because N1 + N2 + N3 = 1. He provides no numerical examples. You get the paper for review. Show that the claims are false, and both sets are worthless. Hint: study §16.6 and Figure E18.5. EXERCISE 18.12 [A:25]. Another way of constructing shape functions for “incomplete” elements is through kinematic multifreedom constraints (MFCs) applied to a “parent” element that contains the one to be derived. Suppose that the 9-node biquadratic quadrilateral is chosen as parent, with shape functions called NiP , i = 1, . . . 9 given in §18.4.2. To construct the shape functions of the 8-node serentipity quadrilateral, the motions of node 9 are expressed in terms of the motions of the corner and midside nodes by the interpolation formulas
u x9 = α(u x1 + u x2 + u x3 + u x4 ) + β(u x5 + u x6 + u x7 + u x8 ), u y9 = α(u y1 + u y2 + u y3 + u y4 ) + β(u y5 + u y6 + u y7 + u y8 ),
(E18.3)
where α and β are scalars to be determined. (In the terminology of Chapter 9, u x9 and u y9 are slaves while boundary DOFs are masters.) Show that the shape functions of the 8-node quadrilateral are then Ni = NiP + α N9P for i = 1, . . . 4 and Ni = NiP + β N9P for i = 5, . . . 8. Furthermore, show that α and β can be determined by two conditions:
00028
Ni = 1, leads to 4α + 4β = 1.
1.
The unit sum condition:
2.
Exactness of displacement interpolation for ξ 2 and η2 leads to 2α + β = 0.
i=1
Solve these two equations for α and β, and verify that the serendipity shape functions given in §18.4.3 result. EXERCISE 18.13 [A:25] Construct the 16 shape functions of the bicubic quadrilateral.
18–17
19
.
FEM Convergence Requirements
19–1
19–2
Chapter 19: FEM CONVERGENCE REQUIREMENTS
TABLE OF CONTENTS Page
§19.1. Overview §19.2. The Variational Index §19.3. Consistency Requirements §19.3.1. Completeness . . §19.3.2. Compatibility . . §19.3.3. Matching Meshes . §19.4. Stability §19.4.1. Rank Sufficiency . §19.4.2. Jacobian Positiveness §19. Notes and Bibliography . . . . . . . §19. References . . . . . . . §19. Exercises . . . . . . . §19. Solutions to .Exercises . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . .
. . . . . . . . .
19–2
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
19–3 19–3 19–4 19–4 19–4 19–6 19–6 19–6 19–8 19–9 19–9 19–11 19–12
19–3
§19.2 THE VARIATIONAL INDEX
§19.1. Overview Chapters 12 through 18 have discussed, in piecemeal fashion, requirements for shape functions of isoparametric elements. These are motivated by convergence: as the mesh is refined, the FEM solution should approach the analytical solution of the mathematical model.1 This attribute is obviously necessary to instill confidence in FEM results from the standpoint of mathematics. This Chapter provides unified information on convergence requirements. These requirements can be grouped into three: Completeness. The elements must have enough approximation power to capture the analytical solution in the limit of a mesh refinement process. This intuitive statement is rendered more precise below. Compatibility. The shape functions should provide displacement continuity between elements. Physically these insure that no material gaps appear as the elements deform. As the mesh is refined, such gaps would multiply and may absorb or release spurious energy. Stability. The system of finite element equations must satify certain well posedness conditions that preclude nonphysical zero-energy modes in elements, as well as the absence of excessive element distortion. Completeness and compatibility are two aspects of the so-called consistency condition between the discrete and mathematical models. A finite element model that passes both completeness and continuity requirements is called consistent. This is the FEM analog of the famous Lax-Wendroff theorem,2 which says that consistency and stability imply convergence. A deeper mathematical analysis done in more advanced courses shows that completeness is necessary for convergence whereas failure of the other requirements does not necessarily precludes it. In fact there are FEM models in common use that do not satisfy compatibility. Nonetheless, the satisfaction of the three criteria guarantees convergence and may therefore be regarded as a safe choice. §19.2. The Variational Index For the mathematical statement of the completeness and continuity conditions, the variational index alluded to in previous sections plays a fundamental role. The FEM is based on the direct discretization of an energy functional 0002[u], where u (displacements for the elements considered in this book) is the primary variable, or (equivalently) the function to be varied. Let m be the highest spatial derivative order of u that appears in 0002. This m is called the variational index.
1
Of course FEM convergence does not guarantee the correctness of the mathematical model in capturing the physics. As discussed in Chapter 1, model verification against experiments is a different and far more difficult problem.
2
Proven originally for classical finite difference discretizations in fluid mechanics. More precisely, it states that a numerical scheme for the scalar conservation law, du/dt + d f /d x = 0 converges to a unique (weak) solution, if it is consistent, stable and conservative. There is no equivalent theorem for systems of conservation laws.
19–3
19–4
Chapter 19: FEM CONVERGENCE REQUIREMENTS
Example 19.1. In the bar problem discussed in Chapter 12,
0002
L
0002[u] =
00031 2
u 0004 E Au 0004 − qu
0004
d x.
(19.1)
0
The highest derivative of the displacement u(x) is u 0004 = du/d x, which is first order in the space coordinate x. Consequently m = 1. This is also the case on the plane stress problem studied in Chapter 14, because the strains are expressed in terms of first order derivatives of the displacements. Example 19.2. In the plane beam problem discussed in Chapter 13,
0002
L
00031
0004
v 00040004 E I v 00040004 − qv d x. 2
0002[v] =
(19.2)
0
The highest derivative of the transverse displacement is the curvature κ = v 00040004 = d 2 v/d x 2 , which is of second order in the space coordinate x. Consequently m = 2.
§19.3. Consistency Requirements Using the foregoing definition of variational index, we can proceed to state the two key requirements for finite element shape functions. §19.3.1. Completeness The element shape functions must represent exactly all polynomial terms of order ≤ m in the Cartesian coordinates. A set of shape functions that satisfies this condition is called m-complete. Note that this requirement applies at the element level and involves all shape functions of the element. Example 19.3. Suppose a displacement-based element is for a plane stress problem, in which m = 1. Then 1-completeness requires that the linear displacement field
u x = α0 + α1 x + α2 y,
u y = α0 + α1 x + α2 y
(19.3)
be exactly represented for any value of the α coefficients. This is done by evaluating (19.3) at the nodes to form a displacement vector ue and then checking that u = Ne ue recovers exactly (19.3). Section 16.6 presents the details of this calculation for an arbitrary isoparametric plane stress element. The analysis shows that completeness is satisfied if the sum of the shape functions is unity and the element is compatible. Example 19.4. For the plane beam problem, in which m = 2, the quadratic transverse displacement
v = α0 + α1 x + α2 x 2
(19.4)
must be exactly represented over the element. This is easily verified in for the 2-node beam element developed in Chapter 13, because the assumed transverse displacement is a complete cubic polynomial in x. A complete cubic contains the quadratic (19.4) as special case.
19–4
19–5
§19.3
(c)
(b)
(a)
i
CONSISTENCY REQUIREMENTS
i
i
bars Figure 19.1. An element patch is the set of all elements attached to a patch node, labeled i. (a) illustrates a patch of triangles; (b) a mixture of triangles and quadrilaterals; (c) a mixture of triangles, quadrilaterals, and bars.
§19.3.2. Compatibility To state this requirement succintly, it is convenient to introduce the concept of element patch, or simply patch. This is the set of all elements attached to a given node, called the patch node. The definition is illustrated in Figure 19.1, which shows three different kind of patches attached to patch node i in a plane stress problem. The patch of Figure 19.1(a) contains only one type of element: 3-node linear triangles. The patch of Figure 19.1(b) mixes two plane stress element types: 3-node linear triangles and 4-node bilinear quadrilaterals. The patch of Figure 19.1(c) combines three element types: 3-node linear triangles, 4-node bilinear quadrilaterals, and 2-node bars. We define a finite element patch trial function as the union of shape functions activated by setting a degree of freedom at the patch node to unity, while all other freedoms are zero. A patch trial function “propagates” only over the patch, and is zero beyond it. This property follows from the local-support requirement stated in §18.1: a shape function for node i should vanish on all sides or faces that do not include i. With the help of these definitions we can enunciate the compatibility requirement as follows. Patch trial functions must be C (m−1) continuous between interconnected elements, and C m piecewise differentiable inside each element. If the variational index is m = 1, the patch trial functions must be C 0 continuous between elements, and C 1 inside elements. A set of shape functions that satisfies the first requirement is called conforming. A conforming expansion that satisfies the second requirement is said to be of finite energy. Note that this condition applies at two levels: individual element, and element patch. An element endowed with conforming shape functions is said to be conforming. A conforming element that satisfies the finite energy requirement is said to be compatible.3 3
The FEM literature is a bit fuzzy as regards these terms. It seems better to leave the qualifier “conforming” to denote interelement compatibility; informally “an element that gets along with its neighbors.” The qualifier “compatible” is used in the stricter sense of conforming while possessing sufficient internal smoothness.
19–5
Chapter 19: FEM CONVERGENCE REQUIREMENTS
19–6
Figures 19.1(b,c) illustrates the fact that one needs to check the possible connection of matching elements of different types and possibly different dimensionality. §19.3.3. Matching Meshes As stated, compatibility refers to the complete finite element mesh because mesh trial functions are a combination of patch trial functions, which in turn are the union of element shape functions. This generality poses some logistical difficulties because the condition is necessarily mesh dependent. Compatibility can be checked at the element level by restricting attention to matching meshes. A matching mesh is one in which adjacent elements share sides, nodes and degrees of freedom, as in the patches shown in Figure 19.1. For a matching mesh it is sufficient to restrict consideration first to a pair of adjacent elements, and then to the side shared by these elements. Suppose that the variation of a shape function along that side is controlled by k nodal values. Then a polynomial variation of order up to k − 1 in the natural coordinate(s) can be specified uniquely over the side. This is sufficient to verify interelement compatibility for m = 1, implying C 0 continuity, if the shape functions are polynomials. This simplified criterion is the one used in previous Chapters. Specific 2D examples were given in Chapters 15 through 18. Remark 19.1. If the variational index is m = 2 and the problem is multidimensional, as in the case of plates
and shells, the check is far more involved because continuity of normal derivatives along a side is involved. This practically important scenario is examined in advanced FEM treatments. The case of non-polynomial shape functions is, on the other hand, of little practical interest.
§19.4. Stability Stability may be informally characterized as ensuring that the finite element model enjoys the same solution uniqueness properties of the analytical solution of the mathematical model. For example, if the only motions that produce zero internal energy in the mathematical model are rigid body motions, the finite element model must inherit that property. Since FEM can handle arbitrary assemblies of elements, including individual elements, this property is required to hold at the element level. In the present outline we are concerned with stability at the element level. Stability is not a property of shape functions per se but of the implementation of the element as well as its geometrical definition. It involves two subordinate requirements: rank sufficiency, and Jacobian positiveness. Of these, rank sufficiency is the most important one. §19.4.1. Rank Sufficiency The element stiffness matrix must not possess any zero-energy kinematic mode other than rigid body modes. This can be mathematically expressed as follows. Let n F be the number of element degrees of freedom, and n R be the number of independent rigid body modes. Let r denote the rank of Ke . The element is called rank sufficient if r = n F − n R and rank deficient if r < n F − n R . In the latter case, the rank deficiency is defined by d = (n F − n R ) − r 19–6
(19.5)
19–7
§19.4
STABILITY
Table 19.1 Rank-sufficient Gauss Rules for Some Plane Stress Elements Element
n
nF
nF − 3
Min n G
Recommended rule
3-node triangle 6-node triangle 10-node triangle 4-node quadrilateral 8-node quadrilateral 9-node quadrilateral 16-node quadrilateral
3 6 10 4 8 9 16
6 12 20 8 16 18 32
3 9 17 5 13 15 29
1 3 6 2 5 5 10
centroid∗ 3-point rules∗ 6-point rule∗ 2x2 3x3 3x3 4x4
∗
These triangle integration rules are introduced in §24.2.
If an isoparametric element is numerically integrated, let n G be the number of Gauss points, while n E denotes the order of the stress-strain matrix E. Two additional assumptions are made: (i)
The element shape functions satisfy completeness in the sense that the rigid body modes are exactly captured by them.
(ii) Matrix E is of full rank. Then each Gauss point adds n E to the rank of Ke , up to a maximum of n F − n R . Hence the rank of Ke will be r = min(n F − n R , n E n G ) (19.6) To attain rank sufficiency, n E n G must equal or exceed n F − n R : n E nG ≥ n F − n R
(19.7)
from which the appropriate Gauss integration rule can be selected. In the plane stress problem, n E = 3 because E is a 3 × 3 matrix of elastic moduli; see equation (14.5)2 . Also n R = 3. Consequently r = min(n F − 3, 3n G ) and 3n G ≥ n F − 3. Remark 19.2. The fact that each Gauss point adds n E n G to the rank can be proven considering the following property. Let B be a n E × n F rectangular real matrix with rank r B ≤ n E , and E an n E × n E positive-definite symmetric matrix. Then the rank of BT EB is r B . Proof: let u = 0 be a non-null n F -vector. If BT EBu = 0 then 0 = uT BT EBu = ||E1/2 Bu||. Therefore Bu = 0. Identify now B and E with the strain-displacement and stress-strain (constitutive) matrix, respectively. In the plane stress case n E = 3, n F = 2n > 3 is the number of element freedoms. Thus B has rank 3 and a fortiori BT EB must also have rank 3 since E is p.d. At each Gauss point i a contribution of wi BT EB, which has rank 3 if wi > 0, is added to Ke . By a theorem of linear algebra, the rank of Ke increases by 3 until it reaches n F − n R . Example 19.5. Consider a plane stress 6-node quadratic triangle. Then n F = 2 × 6 = 12. To attain the proper rank of 12 − n R = 12 − 3 = 9, n G ≥ 3. A 3-point Gauss rule, such as the midpoint rule defined in §24.2, makes the element rank sufficient.
19–7
19–8
Chapter 19: FEM CONVERGENCE REQUIREMENTS
3
3
4
4
1
2
1
3
3
4
2
1
3
4
2
1
2
4
1
2
Figure 19.2. Effect of displacing node 4 of the four-node bilinear quadrilateral shown on the leftmost picture, to the right.
Example 19.6. Consider a plane stress 9-node biquadratic quadrilateral. Then n F = 2 × 9 = 18. To attain
the proper rank of 18 − n R = 18 − 3 = 15, n G ≥ 5. The 2 × 2 product Gauss rule is insufficient because n G = 4. Hence a 3 × 3 rule, which yields n G = 9, is required to attain rank sufficiency.
Table 19.1 collects rank-sufficient Gauss integration rules for some widely used plane stress elements with n nodes and n F = 2n freedoms. §19.4.2. Jacobian Positiveness The geometry of the element should be such that the determinant J = det J of the Jacobian matrix defined4 in §17.2, is positive everywhere. As illustrated in Equation (17.20), J characterizes the local metric of the element natural coordinates. For a three-node triangle J is constant and in fact equal to 2A. The requirement J > 0 is equivalent to saying that corner nodes must be positioned and numbered so that a positive area A > 0 results. This is called a convexity condition. It is easily checked by a finite element program. But for 2D elements with more than 3 nodes distortions may render portions of the element metric negative. This is illustrated in Figure 19.2 for a 4-node quadrilateral in which node 4 is gradually moved to the right. The quadrilateral gradually morphs from a convex figure into a nonconvex one. The center figure is a triangle; note that the metric near node 4 is badly distorted (in fact J = 0 there) rendering the element unacceptable. This clearly contradicts the erroneous advice of some FE books, which state that quadrilaterals can be reduced to triangles as special cases, thereby rendering triangular elements unnecessary. For higher order elements proper location of corner nodes is not enough. The non-corner nodes (midside, interior, etc.) must be placed sufficiently close to their natural locations (midpoints, centroids, etc.) to avoid violent local distortions. The effect of midpoint motions in quadratic elements is illustrated in Figures 19.3 and 19.4. Figure 19.3 depicts the effect of moving midside node 5 tangentially in a 9-node quadrilateral element while keeping all other 8 nodes fixed. When the location of 5 reaches the quarter-point of side 1-2, the metric at corner 2 becomes singular in the sense that J = 0 there. Although this is 4
This definition applies to quadrilateral elements. The Jacobian determinant of an arbitrary triangular element is defined in §24.2.
19–8
19–9
§19.
3
3
7
4
6
4
6
9
9
8 2
5
8
1
2
5
1
3
5
3
7
4
6
7
4
6
9
4
6 9
9
8
8
5
2
2
3
7
1
6
9
8
1
3
7
7
4
References
8
1
5 2
1
5=2
Figure 19.3. Effect of moving midpoint 5 of a 9-node biquadratic quadrilateral tangentially toward corner 2.
disastrous in ordinary FE work, it has applications in the construction of special “crack” elements for linear fracture mechanics. Displacing midside nodes normally to the sides is comparatively more forgiving, as illustrated in Figure 19.4. This depicts a 6-node equilateral triangle in which midside nodes 4, 5 and 6 are moved inwards and outwards along the normals to the midpoint location. As shown in the lower left picture, the element may be even morphed into a “parabolic circle” (meaning that nodes 1 through 6 lie on a circle) without the metric breaking down. Notes and Bibliography The literature on the mathematics of finite element methods has grown exponentially since the monograph of Strang and Fix [149]. This is very readable but out of print. A more up-to-date exposition is the textbook by Szabo and Babuska [155]. The subjects collected in this Chapter tend to be dispersed in recent monographs and obscured by overuse of functional analysis. References Referenced items have been moved to Appendix R.
19–9
19–10
Chapter 19: FEM CONVERGENCE REQUIREMENTS
3
3
3
3
5
6 5
6
5
6
5
6
4 4
3
4
5
1
2
5 3
5
6
2
1
4
2
6
3
6
2
1 2 1
2 1
1
1
2
4 4
4
Figure 19.4. Effect of displacing midpoints 4, 5 and 6 of an equilateral 6-node triangle along the midpoint normals. Motion is inwards in first two top frames, outwards in the last four. In the lower leftmost picture nodes 1 through 6 lie on a circle.
19–10
19–11
Exercises
Homework Exercises for Chapter 19 FEM Convergence Requirements EXERCISE 19.1 [D:15] Draw a picture of a 2D non-matching mesh in which element nodes on two sides of
a boundary do not share the same locations. Discuss why enforcing compatibility becomes difficult. EXERCISE 19.2 [A:25] The isoparametric definition of the straight 3-node bar element in its local system x¯
is
0005 0006 1 x¯ v¯
0005
=
1 x¯1 u¯ 1
1 x¯2 u¯ 2
1 x¯3 u¯ 3
00060005
N1e (ξ ) N2e (ξ ) N3e (ξ )
0006
.
(E19.1)
Here ξ is the isoparametric coordinate that takes the values −1, 1 and 0 at nodes 1, 2 and 3, respectively, while N1e , N2e and N3e are the shape functions found in Exercise 16.3 and listed in (E16.2). For simplicity, take x¯1 = 0, x¯2 = L, x¯3 = 12 L + αL. Here L is the bar length and α a parameter that characterizes how far node 3 is away from the midpoint location x¯ = 12 L. Show that the minimum α’s (minimal in absolute value sense) for which J = d x/dξ ¯ vanishes at a point in the element are ±1/4 (the quarter-points). Interpret this result as a singularity by showing that the axial strain becomes infinite at a an end point. (This result has application in fracture mechanics modeling.) EXERCISE 19.3 [A:15] Consider one dimensional bar-like elements with n nodes and 1 degree of freedom per node so n F = n. The correct number of rigid body modes is 1. Each Gauss integration point adds 1 to the rank; that is N E = 1. By applying (19.7), find the minimal rank-preserving Gauss integration rules with p points in the longitudinal direction if the number of node points is n = 2, 3 or 4. EXERCISE 19.4 [A:20] Consider three dimensional solid “brick” elements with n nodes and 3 degrees of freedom per node so n F = 3n. The correct number of rigid body modes is 6. Each Gauss integration point adds 6 to the rank; that is, N E = 6. By applying (19.7), find the minimal rank-preserving Gauss integration rules with p points in each direction (that is, 1×1×1, 2×2×2, etc) if the number of node points is n = 8, 20, 27, or 64. Partial answer: for n = 27 the minimal rank preserving rule is 3 × 3 × 3. EXERCISE 19.5 [A/C:35] (Requires use of a CAS help to be tractable). Repeat Exercise 19.2 for a 9-node
plane stress element. The element is initially a perfect square, nodes 5,6,7,8 are at the midpoint of the sides 1–2, 2–3, 3–4 and 4–1, respectively, and 9 at the center of the square. Displace 5 tangentially towards 2 until the Jacobian determinant at 2 vanishes. This result is important in the construction of “singular elements” for fracture mechanics. EXERCISE 19.6 [A/C:35] Repeat Exercise 19.5 but moving node 5 along the normal to the side. Discuss the range of motion for which det J > 0 within the element. EXERCISE 19.7 [A:20] A triangular element has 3 nodes located at the midpoints of the sides. (This element is used in some non-structural applications). Develop the 3 shape functions and study whether the element satisfies compatibility and completeness.
19–11
20
.
Implementation of One-Dimensional Elements
20–1
20–2
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
TABLE OF CONTENTS Page
§20.1. The Plane Bar Element §20.1.1. Stiffness Matrix . . . . . . . §20.1.2. Stiffness Module . . . . . . . . §20.1.3. Testing the Plane Bar Module . . . §20.2. The Space Bar Element §20.2.1. Stiffness Matrix . . . . . . . . §20.2.2. Stiffness Module . . . . . . . §20.2.3. Testing the Space Bar Module . . . §20.3. The Plane Beam-Column Element §20.3.1. Stiffness Matrix . . . . . . . §20.3.2. Stiffness Module . . . . . . . . §20.3.3. Testing the Plane Beam-Column Module §20.4. *The Space Beam Element §20.4.1. *Stiffness Matrix . . . . . . . . §20.4.2. *Stiffness Module . . . . . . . §20. Notes and. Bibliography . . . . . . . . . . . . . §20. Exercises . . . . . . . . . . . . .
20–2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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20–3 20–3 20–3 20–4 20–5 20–5 20–6 20–6 20–7 20–8 20–9 20–10 20–11 20–11 20–12 20–12 20–14
20–3
§20.1
THE PLANE BAR ELEMENT
This Chapter begins Part III of the course. This Part deals with the computer implementation of the Finite Element Method for static analysis. It is organized in “bottom up” fashion. It begins with simple topics, such as programming of bar and beam elements, and gradually builds up toward more complex models and calculations. Specific examples of this Chapter illustrate the programming of one-dimensional elements: bars and beams, using Mathematica as implementation language. §20.1. The Plane Bar Element The two-node, prismatic, two-dimensional bar element was studied in Chapters 2-3 for modeling plane trusses. It is reproduced in Figure 20.1 for conveniency. It has y− two nodes and four degrees of freedom. The element y node displacements and conjugate forces are 
 u x1 u  ue =  y1  , u x2 u y2

 f x1 f  fe =  y1  . f x2 f y2
x− 2 (x 2 , y2 )
e
0006 =L
ϕ
E, A constant
1 (x 1 , y1 )
(20.1)
x Figure 20.1. Plane bar element.
The element geometry is described by the coordinates {xi , yi }, i = 1, 2 of the two end nodes. For stiffness computations, the only material and fabrication properties required are the modulus of elasticity E = E e and the cross section area A = Ae , respectively. Both are taken to be constant over the element. §20.1.1. Stiffness Matrix The element stiffness matrix in global {x, y} coordinates is given by the explicit expression derived in §3.1:  2    c sc −c2 −sc x21 x21 x21 y21 −x21 x21 −x21 y21 E A  sc s 2 −sc −s 2  E A  x21 y21 y21 y21 −x21 y21 −y21 y21  Ke =  2 = 3   . (20.2) 2 −c −sc c sc −x21 x21 −x21 y21 x21 x21 x21 y21 0006 0006 −x21 y21 −y21 y21 x21 y21 y21 y21 −sc −s 2 sc s 2 2 2 + y21 , Here c = cos ϕ = x21 /0006, s = sin ϕ = y21 /0006, in which x21 = x2 −x1 , y21 = y2 −y1 , 0006 = x21 and ϕ is the angle formed by x¯ and x, measured from x positive counterclockwise (see Figure 20.1). The second expression in (20.2) is preferable in a computer algebra system because it enhances simplification possibilities when doing symbolic work, and is the one actually implemented in the module described below. §20.1.2. Stiffness Module The computation of the stiffness matrix Ke of the two-node, prismatic plane bar element is done by Mathematica module PlaneBar2Stiffness. This is listed in Figure 20.2. The module is invoked as Ke = PlaneBar2Stiffness[ncoor, Em, A, options] (20.3) 20–3
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
20–4
PlaneBar2Stiffness[ncoor_,Em_,A_,options_]:= Module[ {x1,x2,y1,y2,x21,y21,EA,numer,L,LL,LLL,Ke}, {{x1,y1},{x2,y2}}=ncoor; {x21,y21}={x2-x1,y2-y1}; EA=Em*A; {numer}=options; LL=x21^2+y21^2; L=Sqrt[LL]; If [numer,{x21,y21,EA,LL,L}=N[{x21,y21,EA,LL,L}]]; If [!numer, L=PowerExpand[L]]; LLL=Simplify[LL*L]; Ke=(Em*A/LLL)*{{ x21*x21, x21*y21,-x21*x21,-x21*y21}, { y21*x21, y21*y21,-y21*x21,-y21*y21}, {-x21*x21,-x21*y21, x21*x21, x21*y21}, {-y21*x21,-y21*y21, y21*x21, y21*y21}}; Return[Ke]]; Figure 20.2. Mathematica stiffness module for a two-node, prismatic plane bar element.
The arguments are ncoor Em A options
Node coordinates of element arranged as { { x1,y1 },{ x2,y2 } }. Elastic modulus. Cross section area. A list of processing options. For this element is has only one entry: { numer }. This is a logical flag with the value True or False. If True the computations are carried out in floating-point arithmetic. If False symbolic processing is assumed.
The module returns the 4 × 4 element stiffness matrix as function value. ClearAll[A,Em,L]; ncoor={{0,0},{30,40}}; Em=1000; A=5; Ke= PlaneBar2Stiffness[ncoor,Em,A,{True}]; Print['Numerical Elem Stiff Matrix: ']; Print[Ke//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[N[Ke]]]]; Print['Symmetry check=',Simplify[Chop[Transpose[Ke]-Ke]]]; Numerical Elem Stiff Matrix: 36. 48. − 36. − 48. 48. 64. − 48. − 64. 0001 36. − 48. 36. 48. 0001 48. − 64. 48. 64. Eigenvalues of Ke = {200., 0, 0, 0} Symmetry check={{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}} Figure 20.3. Test of plane bar stiffness module with numerical inputs.
§20.1.3. Testing the Plane Bar Module The modules are tested by the scripts listed in Figures 20.3 and 20.4. The script shown on the top of Figure 20.3 tests a numerically defined element with end nodes located at (0, 0) and (30, 40), with E = 1000, A = 5, and numer set to True. Executing the script produces the results listed in the bottom of that figure. 20–4
20–5
§20.2 THE SPACE BAR ELEMENT ClearAll[A,Em,L]; ncoor={{0,0},{L,0}}; Ke= PlaneBar2Stiffness[ncoor,Em,A,{False}]; kfac=Em*A/L; Ke=Simplify[Ke/kfac]; Print['Symbolic Elem Stiff Matrix: ']; Print[kfac,' ',Ke//MatrixForm]; Print['Eigenvalues of Ke=',kfac,'*',Eigenvalues[Ke]]; Symbolic Elem Stiff Matrix: 1 0 00011 0 0004 0001 0006 0003 0006 0003 0006 0003 A Em 0003 0006 0003 0 0 0 00006 0006 0003 0006 0003 0006 0003 0003 L 0006 0003 00011 0 1 0 0006 0006 0003 0006 0003 0002 0 0 0 00005 A Em Eigenvalues of Ke = ∗{0, 0, 0, 2} L
Figure 20.4. Test of plane bar stiffness module with symbolic inputs.
On return from PlaneBar2Stiffness, the stiffness matrix returned in Ke is printed. Its four eigenvalues are computed and printed. As expected three eigenvalues, which correspond to the three independent rigid body motions of the element, are zero. The remaining eigenvalue is positive and equal to E A/0006. The symmetry of Ke is checked by printing (Ke )T − Ke upon simplification and chopping. The script of Figure 20.4 tests a symbolically defined bar element with end nodes located at (0, 0) and (L , 0), which is aligned with the x axis. Properties E and A are kept symbolic. Executing the script shown in the top of Figure 20.4 produces the results shown in the bottom of that figure. One thing to be noticed is the use of the stiffness scaling factor E A/0006, called kfac in the script. This is a symbolic quantity that can be extracted as factor of matrix Ke . The effect is to clean up matrix and vector output, as can be observed in the printed results. §20.2.
The Space Bar Element
To show how the previous implementation extends easily to three dimensions, this section describes the implementation of the space bar element. The two-node, prismatic, three-dimensional bar element is pictured in Figure 20.5. It has two nodes and six degrees of freedom. The element node displacements and congugate forces are arranged as
u  x1
 u y1   u z1   ue =   u x2  ,   u y2 u z2
f  x1 f  y1   f z1   fe =   f x2  .   f y2 f z2
− y
−
z
1 (x1 ,y1 ,z 1 )
2 (x 2 ,y 2 ,z 2 )
y
0006 = Le
(20.4)
z
E, A constant
x−
x Figure 20.5. The space (3D) bar element.
The element geometry is described by the coordinates {xi , yi , z i }, i = 1, 2 of the two end nodes. As in the case of the plane bar, the two properties required for the stiffness computations are the modulus of elasticity E and the cross section area A. Both are assumed to be constant over the element.
20–5
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
20–6
SpaceBar2Stiffness[ncoor_,Em_,A_,options_]:=Module[ {x1,x2,y1,y2,z1,z2,x21,y21,z21,EA,numer,L,LL,LLL,Ke}, {{x1,y1,z1},{x2,y2,z2}}=ncoor;{x21,y21,z21}={x2-x1,y2-y1,z2-z1}; EA=Em*A; {numer}=options; LL=x21^2+y21^2+z21^2; L=Sqrt[LL]; If [numer,{x21,y21,z21,EA,LL,L}=N[{x21,y21,z21,EA,LL,L}]]; If [!numer, L=PowerExpand[L]]; LLL=Simplify[LL*L]; Ke=(Em*A/LLL)* {{ x21*x21, x21*y21, x21*z21,-x21*x21,-x21*y21,-x21*z21}, { y21*x21, y21*y21, y21*z21,-y21*x21,-y21*y21,-y21*z21}, { z21*x21, z21*y21, z21*z21,-z21*x21,-z21*y21,-z21*z21}, {-x21*x21,-x21*y21,-x21*z21, x21*x21, x21*y21, x21*z21}, {-y21*x21,-y21*y21,-y21*z21, y21*x21, y21*y21, y21*z21}, {-z21*x21,-z21*y21,-z21*z21, z21*x21, z21*y21, z21*z21}}; Return[Ke]; ]; Figure 20.6. Module to form the stiffness of the space (3D) bar element.
§20.2.1. Stiffness Matrix For the space bar element, introduce the notation x21 = x2 − x1 , y21 = y2 − y1 , z 21 = z 2 − z 1 and 0006 = 2 2 2 x21 + y21 + z 21 . It can be shown1 that the element stiffness matrix in global coordinates is given by
 x x  x21 y21 x21 z 21 −x21 x21 −x21 y21 −x21 z 21 21 21 y21 y21 x21 z 21 −x21 y21 −y21 y21 −y21 z 21   x21 y21 e e  A E x z y z z 21 z 21 −x21 z 21 −y21 z 21 −z 21 z 21  21 21  21 21 . Ke =  3 x21 x21 x21 y21 x21 z 21  0006  −x21 x21 −x21 y21 −x21 z 21  −x21 y21 −x21 z 21
−y21 y21 −y21 z 21
−x21 z 21 −z 21 z 21
x21 y21 x21 z 21
y21 y21 y21 z 21
(20.5)
y21 z 21 z 21 z 21
This matrix expression in terms of coordinate differences is useful in symbolic work, because it enhances simplification possibilities. §20.2.2. Stiffness Module The computation of the stiffness matrix Ke of the two-node, prismatic space bar element, is done by Mathematica module SpaceBar2Stiffness. This is listed in Figure 20.6. The module is invoked as Ke = SpaceBar2Stiffness[ncoor, Em, A, options]
(20.6)
The arguments are ncoor Em A options
Node coordinates of element arranged as { { x1,y1,z1 },{ x2,y2,z2 } }. Elastic modulus. Cross section area. A list of processing options. For this element is has only one entry: { numer }. This is a logical flag with the value True or False. If True the computations are carried out in floating-point arithmetic. If False symbolic processing is assumed.
The module returns the 6 × 6 element stiffness matrix as function value. 1
The derivation was the subject of Exercise 6.10.
20–6
20–7
§20.3
THE PLANE BEAM-COLUMN ELEMENT
ClearAll[A,Em]; ncoor={{0,0,0},{2,3,6}}; Em=343; A=10; Ke= SpaceBar2Stiffness[ncoor,Em,A,{True}]; Print['Numerical Elem Stiff Matrix: ']; Print[Ke//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[Ke]]]; Numerical Elem Stiff Matrix: 40. 60. 120. − 40. − 60. 60. 90. 180. − 60. − 90. 120. 180. 360. − 120. − 180. − 40. − 60. − 120. 40. 60. − 60. − 90. − 180. 60. 90. − 120. − 180. − 360. 120. 180. Eigenvalues of Ke = {980., 0, 0, 0, 0, 0}
− 120. − 180. − 360. 120. 180. 360.
Figure 20.7. Testing the space bar stiffness module with numerical inputs.
ClearAll[A,Em,L]; ncoor={{0,0,0},{L,2*L,2*L}/3}; Ke= SpaceBar2Stiffness[ncoor,Em,A,{False}]; kfac=Em*A/(9*L); Ke=Simplify[Ke/kfac]; Print['Symbolic Elem Stiff Matrix: ']; Print[kfac,' ',Ke//MatrixForm]; Print['Eigenvalues of Ke=',kfac,'*',Eigenvalues[Ke]]; Symbolic Elem Stiff Matrix: 1 2 2 −1 −2 −2 2 4 4 −2 −4 −4 4 4 −2 −4 −4 A Em 2 9L −1 −2 −2 1 2 2 −2 −4 −4 2 4 4 −2 −4 −4 2 4 4 A Em Eigenvalues of Ke = ∗ {0, 0, 0, 0, 0, 18} 9L Figure 20.8. Testing the space bar stiffness module with symbolic inputs.
§20.2.3. Testing the Space Bar Module The modules are tested by the scripts listed in Figures 20.7 and 20.8. As these are similar to previous tests done on the plane bar they need not be described in detail. The script of Figure 20.7 tests a numerically defined space bar with end nodes located at (0, 0, 0) and (30, 40, 0), with E = 1000, A = 5 and numer set to True. Executing the script produces the results listed in the bottom of that Figure. The script of Figure 20.8 tests a symbolically defined bar element with end nodes located at (0, 0, 0) and (L , 2L , 2L)/3, which has length L and is not aligned with the x axis. The element properties E and A are kept symbolic. Executing the script produces the results shown in the bottom of that Figure. Note the use of a stiffness factor kfac of E A/(90006) to get cleaner printouts.
20–7
20–8
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
y− u−y1
u−x2
(a)
u−x1
θz1
θz2
u−y2 x−
e
0006 =L
ϕ
y
E, A, I constant zz
2 (x 2 , y2 )
0006
y−
2
1
x−
(b)
1 (x 1 , y1 )
x
Figure 20.9. Plane beam-column element: (a) in its local system; (b) in the global system.
§20.3. The Plane Beam-Column Element Beam-column elements model structural members that resist both axial and bending actions. This is the case in skeletal structures such as frameworks which are common in steel and reinforcedconcrete building construction. A plane beam-column element is a combination of a plane bar (such as that considered in §20.1), and a plane beam. We consider a beam-column element in its local system (x, ¯ y¯ ) as shown in Figure 20.9(a), and then in the global system (x, y) as shown in Figure 20.9(b). The six degrees of freedom and conjugate node forces of the elements are:   ¯       u¯ x1 f x1 u x1 f x1  u¯ y1   f¯y1   u y1   f y1          e  θz1   m z1   θz1   m z1  e e e ¯ u = (20.7) u¯ =  , f =  , , f =  .  u¯ x2   u¯ x2   u x2   f x2          u¯ y2 u¯ y2 u y2 f y2 θz2 m z2 θz2 m z2 The rotation angles θ and the nodal moments m are the same in the local and the global systems because they are about the z axis, which does not change in passing from local to global. The element geometry is described by the coordinates {xi , yi }, i = 1, 2 of the two end nodes. The element length is 0006 = L e . Properties involved in the stiffness calculations are: the modulus of elasticity E, the cross section area A and the moment of inertia I = Izz about the neutral axis. All properties are taken to be constant over the element. §20.3.1. Stiffness Matrix To obtain the plane beam-column stiffness in the local system we simply add the stiffness matrices derived in Chapters 12 and 13, respectively, to get  ¯e = EA K 0006
     
1
0 0
symm
0 −1 0 0 0 0 1
  0 0 0 0 0 0 12   EI  0 0  + 3  0 0 0006    0 0 0 symm 20–8
0 60006 400062
 0 0 0 0 −12 60006   0 −60006 200062   0 0 0   12 −60006 400062
(20.8)
20–9
§20.3
THE PLANE BEAM-COLUMN ELEMENT
PlaneBeamColumn2Stiffness[ncoor_,Em_,{A_,Izz_},options_]:= Module[ {x1,x2,y1,y2,x21,y21,EA,EI,numer,L,LL,LLL,Te,Kebar,Ke}, {{x1,y1},{x2,y2}}=ncoor; {x21,y21}={x2-x1,y2-y1}; EA=Em*A; EI=Em*Izz; {numer}=options; LL=Simplify[x21^2+y21^2]; L=Sqrt[LL]; If [numer,{x21,y21,EA,EI,LL,L}=N[{x21,y21,EA,EI,LL,L}]]; If [!numer, L=PowerExpand[L]]; LLL=Simplify[LL*L]; Kebar= (EA/L)*{ { 1,0,0,-1,0,0},{0,0,0,0,0,0},{0,0,0,0,0,0}, {-1,0,0, 1,0,0},{0,0,0,0,0,0},{0,0,0,0,0,0}} + (2*EI/LLL)*{ { 0,0,0,0,0,0},{0, 6, 3*L,0,-6, 3*L},{0,3*L,2*LL,0,-3*L, LL}, { 0,0,0,0,0,0},{0,-6,-3*L,0, 6,-3*L},{0,3*L, LL,0,-3*L,2*LL}}; Te={{x21,y21,0,0,0,0}/L,{-y21,x21,0,0,0,0}/L,{0,0,1,0,0,0}, {0,0,0,x21,y21,0}/L,{0,0,0,-y21,x21,0}/L,{0,0,0,0,0,1}}; Ke=Transpose[Te].Kebar.Te; Return[Ke] ]; Figure 20.10. Mathematica module to form the stiffness matrix of a two-node, prismatic plane beam-column element.
The two matrices on the right of (20.8) come from the bar stiffness (12.22) and the BernoulliEuler bending stiffness (13.20), respectively. Before adding them, rows and columns have been rearranged in accordance with the nodal freedoms (20.7). The displacement transformation matrix between local and global systems is      u¯ x1 c s 0 0 0 0 u x1  u¯ y1   −s c 0 0 0 0   u y1        θ   0 0 1 0 0 0   θz1  u¯ e =  z1  =    = T ue ,  u x2   0 0 0 c s 0   u x2       u¯ y2 u y2 0 0 0 −s c 0 θz2 θz2 0 0 0 0 0 1
(20.9)
where c = cos ϕ = (x2 − x1 )/0006, s = sin ϕ = (y2 − y1 )/0006, and ϕ is the angle between x¯ and x, measured positive-counterclockwise from x; see Figure 20.9. The stiffness matrix in the global system is obtained through the congruential transformation ¯ e T. Ke = T T K
(20.10)
An explicit expression of the entries of Ke is messy, and (unlike the bar) it is better to let the program do it. §20.3.2. Stiffness Module The computation of the stiffness matrix Ke of the two-node, prismatic plane beam-column element is done by Mathematica module PlaneBeamColumn2Stiffness. This is listed in Figure 20.10. The module is invoked as Ke = PlaneBeamColumn2Stiffness[ncoor, Em, { A,Izz }, options] The arguments are 20–9
(20.11)
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
20–10
ClearAll[L,Em,A,Izz]; ncoor={{0,0},{3,4}}; Em=100; A=125; Izz=250; Ke= PlaneBeamColumn2Stiffness[ncoor,Em,{A,Izz},{True}]; Print['Numerical Elem Stiff Matrix: ']; Print[Ke//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[Ke]]]; Numerical Elem Stiff Matrix: 2436. 48. −4800. 48. 2464. 3600. −4800. 3600. 20000. −2436. −48. 4800. −48. −2464. −3600. −4800. 3600. 10000.
−2436. −48. 4800. 2436. 48. 4800.
−48. −2464. −3600. 48. 2464. −3600.
−4800. 3600. 10000. 4800. −3600. 20000.
Eigenvalues of Ke = {34800., 10000., 5000., 0, 0, 0} Figure 20.11. Test of two-node plane beam-column element with numeric inputs.
ncoor Em A Izz options
Node coordinates of element arranged as { { x1,y1 },{ x2,y2 } }. Elastic modulus. Cross section area. Moment of inertia of cross section area respect to axis z. A list of processing options. For this element is has only one entry: { numer }. This is a logical flag with the value True or False. If True the computations are carried out in floating-point arithmetic. If False symbolic processing is assumed.
The module returns the 6 × 6 element stiffness matrix as function value. §20.3.3. Testing the Plane Beam-Column Module The beam-column stiffness are tested by the scripts shown in Figures 20.11 and 20.12. The script at the top of Figure 20.11 tests a numerically defined element of length 0006 = 5 with end nodes located at (0, 0) and (3, 4), respectively, with E = 100, A = 125 and Izz = 250. The output is shown at the bottom of that figure. The stiffness matrix returned in Ke is printed. Its six eigenvalues are computed and printed. As expected three eigenvalues, which correspond to the three independent rigid body motions of the element, are zero. The remaining three eigenvalues are positive. The script at the top of Figure 20.12 tests a plane beam-column of length L with end nodes at (0, 0) and (3L/5, 4L/5). The properties E, A and Izz are kept in symbolic form. The output is shown at the bottom of that figure. The printed matrix looks complicated because bar and beam coupling occurs when the element is not aligned with the global axes. The eigenvalues are obtained in closed symbolic form, and their simplicity provides a good check that the transformation matrix (20.9) is orthogonal. Three eigenvalues are exactly zero; one is associated with the axial (bar) stiffness and two with the flexural (beam) stiffness.
20–10
20–11
§20.4 *THE SPACE BEAM ELEMENT ClearAll[L,Em,A,Izz]; ncoor={{0,0},{3*L/5,4*L/5}}; Ke= PlaneBeamColumn2Stiffness[ncoor,Em,{A,Izz},{False}]; Print['Symbolic Elem Stiff Matrix:']; kfac=Em; Ke=Simplify[Ke/kfac]; Print[kfac,' ',Ke//MatrixForm]; Print['Eigenvalues of Ke=',kfac,'*',Eigenvalues[Ke]]; Symbolic Elem Stiff Matrix:
Em
3 ( 64 Izz +3 A L2 ) 25 L3
12 ( −12 Izz +A L2 ) 25 L3
12 ( −12 Izz +A L2 ) 25 L3
4 ( 27 Izz +4 A L2 ) 25 L3
− 245 LIzz2 −
2 − 12 ( −1225IzzL3 +A L )
−
12 ( −12 Izz +A L2 ) 25 L3
−
4 ( 27 Izz +4 A L2 ) 25 L3
24 Izz 5 L2
24 Izz 5 L2 18 Izz 5 L2 4 Izz L
18 Izz 5 L2
3 ( 64 Izz +3 A L2 ) 25 L3
−
−
18 Izz 5 L2
3 ( 64 Izz +3 A L2 ) 25 L3
−
12 ( −12 Izz +A L2 ) 25 L3
− 12 −
18 Izz 5 L2 2 Izz L
( −12 Izz +A L2 ) 25 L3
−
4 ( 27 Izz +4 A L2 ) 25 L3
3 ( 64 Izz +3 A L2 ) 25 L3
12 ( −12 Izz +A L2 ) 25 L3
12 ( −12 Izz +A L2 ) 25 L3
4 ( 27 Izz +4 A L2 ) 25 L3
− 185 LIzz2
24 Izz 5 L2
24 Izz 5 L2
18 Izz 5 L2 2 Izz L
− 185 LIzz2
24 Izz 5 L2
24 Izz 5 L2
−
−
24 Izz 5 L2
−
18 Izz 5 L2 4 Izz L
2
Eigenvalues of Ke = Em∗{0, 0, 0, 2A , 2 Izz , 6 (4 Izz +3 Izz L ) } L L L
Figure 20.12. Test of two-node plane beam-column element with symbolic inputs.
§20.4.
*The Space Beam Element
A second example in 3D is the general beam element shown in Figure 20.13. The element is prismatic and has two end nodes: 1 and 2, placed at the centroid of the end cross sections. 3 (x3 ,y3 ,z3 ) These define the local x¯ axis as directed from 1 to 2. For simplicity the cross section will be assumed to be doubly symmetric, as is the case in commercial I and x− double-T profiles. The principal moments of inertia 2 (x2 ,y2 ,z2 ) are defined by these symmetries. The local y¯ and z¯ axes are aligned with the symmetry lines of the cross y− section forming a RH system with x. ¯ Consequently the principal moments of inertia are I yy and Izz , the bars being omitted for convenience. − z The global coordinate system is {x, y, z}. To define the 1 (x ,y ,z ) 1 1 1 orientation of { y¯ , z¯ } with respect to the global system, a third orientation node 3, which must not be colinear y with 1–2, is introduced. See Figure 20.13. Axis y¯ lies in the 1–2–3 plane and z¯ is normal to 1–2–3. x z Six global DOF are defined at each node i: the 3 translations u xi , u yi , u zi and the 3 rotations θxi , θ yi , θzi .
Figure 20.13. The space (3D) beam element.
§20.4.1. *Stiffness Matrix The element global node displacements and conjugate forces are arranged as ue = [ u x1
u y1
u z1
θx1
θ y1
fe = [ f x1
f y1
f z1
m x1
m y1
θz1 m z1
u x2
u y2
f x2
f y2
u z2 f z2
θx2
θ y2
m x2
θz2 ]T , m y2
m z2 ]T .
(20.12)
The beam material is characterized by the elastic modulus E and the shear modulus G (the latter appears in the torsional stiffness). Four cross section properties are needed: the cross section area A, the moment of inertia
20–11
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
20–12
J that characterizes torsional rigidity,2 and the two principal moments of inertia I yy and Izz taken with respect to y¯ and z¯ , respectively. The length of the element is denoted by L. The Bernoulli-Euler model is used; thus the effect of tranverse shear on the beam stiffness is neglected. To simplify the following expressions, define the following “rigidity” combinations by symbols: R a = b b = E Izz /L 3 , Rz2 = E Izz /L 2 , E A/L, R t = G J/L, R by3 = E I yy /L 3 , R by2 = E I yy /L 2 , R by = E I yy /L, Rz3 Rzb = E Izz /L. Note that R a is the axial rigidity, R t the torsional rigidity, while the R b ’s are bending rigities scaled by the length in various ways. Then the 12 × 12 local stiffness matrix can be written as3

Ra 0 b 0 12Rz3   0 0  0  0  0 0   b ¯ e =  0 a 6Rz2 K  −R 0  0 −12R b  z3  0 0   0 0  0 0 b 0 6Rz2
0 0 0 0 −R a b 0 0 0 6Rz2 0 12R by3 0 −6R by2 0 0 0 Rt 0 0 0 b −6R y2 0 4R by 0 0 0 0 0 4Rzb 0 0 0 0 0 Ra b 0 0 0 −6Rz2 0 b b −12R y3 0 6R y2 0 0 0 −R t 0 0 0 −6R by2 0 2R by 0 0 0 0 0 2Rzb 0
0 0 0 0 0  b b −12Rz3 0 0 0 6Rz2  0 −12R by3 0 −6R by2 0   0 0 −R t 0 0  0 6R by2 0 2R by 0   b −6Rz2 0 0 0 2Rzb   0 0 0 0 0  b b  12Rz3 0 0 0 −6Rz2  b b 0 12R y3 0 6R y2 0   0 0 Rt 0 0   0 6R by2 0 4R by 0 b −6Rz2 0 0 0 4Rzb
(20.13)
The transformation to the global system is the subject of Exercise 20.8. §20.4.2. *Stiffness Module The computation of the stiffness matrix Ke of the two-node, prismatic plane beam-column element is done by Mathematica module SpaceBeamColumn2Stiffness. This is listed in Figure 20.14. The module is invoked as Ke = SpaceBeamColumn2Stiffness[ncoor, { Em,Gm }, { A,Izz,Iyy,Jxx }, options]
(20.14)
The arguments are ncoor Em Gm A Izz Iyy Jxx options
Node coordinates of element arranged as { { x1,y1,z1 },{ x2,y2,z2 },{ x3,y3,z3 } }, in which { x3,y3,z3 } specifies an orientation node 3 that defines the local frame. See §20.4.1. Elastic modulus. Shear modulus. Cross section area. Moment of inertia of cross section area respect to axis z¯ . Moment of inertia of cross section area respect to axis y¯ . Inertia with respect to x¯ that appears in torsional rigidity G J . A list of processing options. For this element is has only one entry: { numer }. This is a logical flag with the value True or False. If True the computations are carried out in floating-point arithmetic. If False symbolic processing is assumed.
The module returns the 12 × 12 element stiffness matrix as function value. The implementation logic and testing of this element is the subject of Exercises 20.8 and 20.9. 2
For circular and annular cross sections, J is the polar moment of inertia of the cross section wrt x. ¯ For other sections J has dimensions of (length)4 but must be calculated according to St. Venant’s theory of torsion, or approximate theories.
3
Cf. page 79 of Pzremieniecki [122]. The presentation in this book includes transverse shear effects as per Timoshenko’s beam theory. The form (20.13) results from neglecting those effects.
20–12
20–13
§20.
Notes and Bibliography
SpaceBeamColumn2Stiffness[ncoor_,{Em_,Gm_},{A_,Izz_,Iyy_,Jxx_}, options_]:= Module[ {x1,x2,y1,y2,z1,z2,x21,y21,z21,xm,ym,zm,x0,y0,z0,dx,dy,dz, EA,EIyy,EIzz,GJ,numer,ra,ry,ry2,ry3,rz,rz2,rz3,rx, L,LL,LLL,yL,txx,txy,txz,tyx,tyy,tyz,tzx,tzy,tzz,T,Kebar,Ke}, {x1,y1,z1}=ncoor[[1]]; {x2,y2,z2}=ncoor[[2]]; {x0,y0,z0}={xm,ym,zm}={x1+x2,y1+y2,z1+z2}/2; If [Length[ncoor]<=2,{x0,y0,z0}+={0,1,0}]; If [Length[ncoor]3,{x0,y0,z0}=ncoor[[3]] ]; {x21,y21,z21}={x2-x1,y2-y1,z2-z1}; {numer}=options; EA=Em*A; EIzz=Em*Izz; EIyy=Em*Iyy; GJ=Gm*Jxx; LL=Simplify[x21^2+y21^2+z21^2]; L=Sqrt[LL]; If [numer, {x21,y21,z21,EA,EIyy,EIzz,GJ,LL,L}= N[{x21,y21,z21,EA,EIyy,EIzz,GJ,LL,L}]]; If [!numer, L=PowerExpand[L]]; LLL=Simplify[LL*L]; ra=EA/L; rx=GJ/L; ry=2*EIyy/L; ry2=6*EIyy/LL; ry3=12*EIyy/LLL; rz=2*EIzz/L; rz2=6*EIzz/LL; rz3=12*EIzz/LLL; Kebar={ { ra, 0, 0, 0, 0, 0, -ra, 0, 0, 0, 0, 0}, { 0, rz3, 0, 0, 0, rz2, 0,-rz3, 0, 0, 0, rz2}, { 0, 0, ry3, 0,-ry2, 0, 0, 0,-ry3, 0,-ry2, 0}, { 0, 0, 0, rx, 0, 0, 0, 0, 0, -rx, 0, 0}, { 0, 0,-ry2, 0,2*ry, 0, 0, 0, ry2, 0, ry, 0}, { 0, rz2, 0, 0, 0,2*rz, 0,-rz2, 0, 0, 0, rz}, {-ra, 0, 0, 0, 0, 0, ra, 0, 0, 0, 0, 0}, { 0,-rz3, 0, 0, 0,-rz2, 0, rz3, 0, 0, 0,-rz2}, { 0, 0,-ry3, 0, ry2, 0, 0, 0, ry3, 0, ry2, 0}, { 0, 0, 0,-rx, 0, 0, 0, 0, 0, rx, 0, 0}, { 0, 0,-ry2, 0, ry, 0, 0, 0, ry2, 0,2*ry, 0}, { 0, rz2, 0, 0, 0, rz, 0,-rz2, 0, 0, 0,2*rz}}; {dx,dy,dz}={x0-xm,y0-ym,z0-zm};If[numer,{dx,dy,dz}=N[{dx,dy,dz}]]; tzx=dz*y21-dy*z21; tzy=dx*z21-dz*x21; tzz=dy*x21-dx*y21; zL=Sqrt[tzx^2+tzy^2+tzz^2]; If [!numer,zL=Simplify[PowerExpand[zL]]]; {tzx,tzy,tzz}={tzx,tzy,tzz}/zL; {txx,txy,txz}={x21,y21,z21}/L; tyx=tzy*txz-tzz*txy; tyy=tzz*txx-tzx*txz; tyz=tzx*txy-tzy*txx; Te={{txx,txy,txz, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {tyx,tyy,tyz, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {tzx,tzy,tzz, 0, 0, 0, 0, 0, 0, 0, 0, 0}, { 0, 0, 0, txx,txy,txz, 0, 0, 0, 0, 0, 0}, { 0, 0, 0, tyx,tyy,tyz, 0, 0, 0, 0, 0, 0}, { 0, 0, 0, tzx,tzy,tzz, 0, 0, 0, 0, 0, 0}, { 0, 0, 0, 0, 0, 0, txx,txy,txz, 0, 0, 0}, { 0, 0, 0, 0, 0, 0, tyx,tyy,tyz, 0, 0, 0}, { 0, 0, 0, 0, 0, 0, tzx,tzy,tzz, 0, 0, 0}, { 0, 0, 0, 0, 0, 0, 0, 0, 0, txx,txy,txz}, { 0, 0, 0, 0, 0, 0, 0, 0, 0, tyx,tyy,tyz}, { 0, 0, 0, 0, 0, 0, 0, 0, 0, tzx,tzy,tzz}}; Ke=Transpose[Te].Kebar.Te; Return[Ke] ];
Figure 20.14. Module to form stiffness of space (3D) beam.
Notes and Bibliography All elements implemented here are formulated in most books dealing with matrix structural analysis. Przemieniecki [122] has been recommended in Chapter 1 on account of being inexpensive. The implementation and testing procedures are not covered in any book.
20–13
Chapter 20: IMPLEMENTATION OF ONE-DIMENSIONAL ELEMENTS
20–14
Homework Exercises for Chapter 20 Implementation of One-Dimensional Elements EXERCISE 20.1 [C:15] Download the plane bar stiffness module and their testers and verify the test results reported here. Comment on whether the stiffness matrix Ke has the correct rank of 1. EXERCISE 20.2 [C:15] Download the space bar stiffness module and their testers and verify the test results reported here. Comment on whether the computed stiffness matrix Ke has the correct rank of 1. EXERCISE 20.3 [C:15] Download the plane beam-column stiffness module and their testers and verify the test results reported here. Comment on whether the computed stiffness matrix Ke has the correct rank of 3. EXERCISE 20.4 [A+C:30] Explain why the space bar element has rank 1 although it has 6 degrees of freedom and 6 rigid body modes. (According to the formula given in Chapter 19, the correct rank should be 6 − 6 = 0.) EXERCISE 20.5 [C:25] Implement the plane bar, plane beam-column and space bar stiffness element module in a lower level programming language and check them by writing a short test driver. [Do not bother about the mass modules.] Your choices are C, Fortran 77 or Fortran 90. (C++ is overkill for this kind of software). EXERCISE 20.6 [A:25] Explain why the eigenvalues of Ke of any the elements given here do not change if
the {x, y, z} global axes change. EXERCISE 20.7 [A+C:30] (Advanced) Implement a 3-node space bar element. Hint: use the results of
Exercise 16.5 and transform the local stiffness to global coordinates via a 3 × 9 transformation matrix. Test the element and verify that it has two nonzero eigenvalues. EXERCISE 20.8 [D+A:25] Explain the logic of the space beam module listed in Figure 20.14. Assume that ¯ e stored in Kebar is correct; focus instead on how the local to global transformation is built and applied. K EXERCISE 20.9 [C:25] Test the space beam element of Figure 20.14 using the scripts given in Figures E20.1 and E20.2, and report results. Comment on whether the element has the correct rank of 6. ClearAll[L,Em,Gm,A,Izz,Iyy,Jxx]; ncoor={{0,0,0},{1,8,4}}; Em=54; Gm=30; A=18; Izz=36; Iyy=72; Jxx=27; Ke= SpaceBeamColumn2Stiffness[ncoor,{Em,Gm},{A,Izz,Iyy,Jxx},{True}]; Print['Numerical Elem Stiff Matrix: ']; Print[SetPrecision[Ke,4]//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[Ke]]];
Figure E20.1. Script for numeric testing of the space beam module of Figure 20.14. ClearAll[L,Em,Gm,A,Izz,Iyy,Jxx]; ncoor={{0,0,0},{2*L,2*L,L}/3}; Ke=SpaceBeamColumn2Stiffness[ncoor,{Em,Gm},{A,Izz,Iyy,Jxx},{False}]; kfac=Em; Ke=Simplify[Ke/kfac]; Print['Numerical Elem Stiff Matrix: ']; Print[kfac,' ',Ke//MatrixForm]; Print['Eigenvalues of Ke=',kfac,'*',Eigenvalues[Ke]];
Figure E20.2. Script for symbolic testing of the space beam module of Figure 20.14.
20–14
21
.
FEM Program for Space Trusses
21–1
21–2
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
TABLE OF CONTENTS Page
§21.1. Introduction §21.2. Analysis Stages §21.3. Analysis Support Modules §21.3.1. Assembling the Master Stiffness . . . . . . §21.3.2. Defining Boundary Conditions . . . . . . §21.3.3. Modifying the Master Stiffness Equations . . . §21.3.4. Displacement Solution and Reaction Recovery . §21.3.5. Flattening and Partitioning Node-Freedom Vectors §21.3.6. Internal Force Recovery . . . . . . . . §21.3.7. The Solution Driver . . . . . . . . . . §21.4. Utility Print Modules §21.5. Utility Graphic Modules §21.5.1. Plot Module Calls . . . . . . . . . . §21.5.2. Plot View Specification . . . . . . . . . §21.6. Example 1: Bridge Plane Truss Example §21.7. Example 2: An Orbiting Truss Structure §21. Notes and Bibliography . . . . . . . . . . . . . . . . . §21. Exercises . . . . . . . . . . . . . . . . .
21–2
. . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . .
. . . .
. . . . . . . . . .
. . . . . . . . . .
21–3 21–3 21–3 21–3 21–5 21–6 21–8 21–8 21–9 21–11 21–11 21–12 21–12 21–14 21–15 21–17 21–17 21–18
21–3
§21.3
ANALYSIS SUPPORT MODULES
§21.1. Introduction This Chapter presents a complete FEM program for analysis of space trusses. Why not start with plane trusses? Three reasons. First, plane truss code already appeared in Chapter 5, although most components were tailored to the example truss of Chapters 2–3. Second, the difference between 2D and 3D implementation logic for truss analysis is insignificant. Finally, space trusses are more interesting in engineering applications, particularly for orbiting structures in aerospace. The description is done in “bottom up” fashion. That means the element level modules are presented first, followed by midlevel modules, ending with the driver program. The program includes some simple minded graphics, including animation. The graphic modules are provided in the posted Notebook but not described in detail. Data structures are explained along the way. §21.2. Analysis Stages The analysis of a structure by the Direct Stiffness Method involves three major stages: preprocessing or model definition, processing, and postprocessing. This is true for toy programs such as the one presented here, through huge commercial codes. Of course the stages here are very short. The preprocessing portion of the space truss analysis is done by a driver script, which directly sets the data structures for the problem at hand. The processing stage involves three steps: •
Assembly of the master stiffness matrix, with a subordinate element stiffness module.
•
Modification of master stiffness matrix and node force vector for displacement boundary conditions.
•
Solution of the modified equations for displacements. For the program presented here the built in Mathematica function LinearSolve is used.
Upon executing the processing steps, the displacements are available. The following postprocessing steps follow. •
Recovery of forces including reactions, done through a Ku matrix multiplication.
•
Computation of internal (axial) forces and stresses in truss members.
•
Plotting deflected shapes and member stress levels.
§21.3. Analysis Support Modules We begin by listing here modules that support processing steps. These are put into separate cells for display and testing convenience. §21.3.1. Assembling the Master Stiffness Module SpaceTrussMasterStiffness, listed in Figure 21.1, assembles the master stiffness matrix of a space truss. It uses the element stiffness formation module SpaceBar2Stiffness discussed in the previous Chapter. That module is also listed in Figure 21.1 for convenience of the reader. The assembler is invoked by K = SpaceTrussMasterStiffness[nodxyz,elenod,elemat,elefab,prcopt] 21–3
(21.1)
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
21–4
SpaceTrussMasterStiffness[nodxyz_,elenod_, elemat_,elefab_,prcopt_]:=Module[ {numele=Length[elenod],numnod=Length[nodxyz],neldof, e,eftab,ni,nj,i,j,ii,jj,ncoor,Em,A,options,Ke,K}, K=Table[0,{3*numnod},{3*numnod}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; eftab={3*ni-2,3*ni-1,3*ni,3*nj-2,3*nj-1,3*nj}; ncoor={nodxyz[[ni]],nodxyz[[nj]]}; Em=elemat[[e]]; A=elefab[[e]]; options=prcopt; Ke=SpaceBar2Stiffness[ncoor,Em,A,options]; neldof=Length[Ke]; For [i=1, i<=neldof, i++, ii=eftab[[i]]; For [j=i, j<=neldof, j++, jj=eftab[[j]]; K[[jj,ii]]=K[[ii,jj]]+=Ke[[i,j]] ]; ]; ]; Return[K]; ]; SpaceBar2Stiffness[ncoor_,Em_,A_,options_]:=Module[ {x1,x2,y1,y2,z1,z2,x21,y21,z21,EA,numer,L,LL,LLL,Ke}, {{x1,y1,z1},{x2,y2,z2}}=ncoor;{x21,y21,z21}={x2-x1,y2-y1,z2-z1}; EA=Em*A; {numer}=options; LL=x21^2+y21^2+z21^2; L=Sqrt[LL]; If [numer,{x21,y21,z21,EA,LL,L}=N[{x21,y21,z21,EA,LL,L}]]; If [!numer, L=PowerExpand[L]]; LLL=Simplify[LL*L]; Ke=(Em*A/LLL)* {{ x21*x21, x21*y21, x21*z21,-x21*x21,-x21*y21,-x21*z21}, { y21*x21, y21*y21, y21*z21,-y21*x21,-y21*y21,-y21*z21}, { z21*x21, z21*y21, z21*z21,-z21*x21,-z21*y21,-z21*z21}, {-x21*x21,-x21*y21,-x21*z21, x21*x21, x21*y21, x21*z21}, {-y21*x21,-y21*y21,-y21*z21, y21*x21, y21*y21, y21*z21}, {-z21*x21,-z21*y21,-z21*z21, z21*x21, z21*y21, z21*z21}}; Return[Ke]];
Figure 21.1. Master stiffness assembly module for a space truss. The element stiffness module SpaceBar2Stiffness, already discussed in Chapter 20, is listed for convenience.
The arguments are nodxyz
Nodal coordinates placed in a node-by-node list { { x1,y1,z1 },{ x2,y2,z2 }, . . . { xn,yn,zn } }, where n is the total number of nodes of the truss.
elenod
Element end nodes placed in an element-by-element list: { { i1,j1 }, { i2,j2 }, . . . { ie,je } }, where e is the total number of elements of the truss.
elemat
Element material properties. The only such property required for this analysis is the elastic modulus E of each element. These are put in an element-by-element list: { E1, E2, . . . Ee }, where e is the total number of elements of the truss.
elefab
Element fabrication properties. The only such property required for this analysis is the cross section area A of each element. These are put in an element-by-element list: { A1, A2, . . . Ae }, where e is the total number of elements of the truss.
prcopt
Processing option list. Only one option is required in the assembler and so prcopt is simply { numer }. Here numer is a logical flag set to { True } to tell the assembler to carry out element stiffness computations in floating-point arithmetic. Else set to { False } to keep computations in exact arithmetic or symbolic form.
The module returns the assembled stiffness, stored as a full 3n × 3n matrix, as function value. 21–4
21–5
§21.3
ANALYSIS SUPPORT MODULES
ClearAll[nodxyz,elemat,elefab,eleopt]; nodxyz={{0,0,0},{10,0,0},{10,10,0}}; elenod= {{1,2},{2,3},{1,3}}; elemat= Table[100,{3}]; elefab= {1,1/2,2*Sqrt[2]}; prcopt= {False}; K=SpaceTrussMasterStiffness[nodxyz,elenod,elemat,elefab,prcopt]; Print['Master Stiffness of Example Truss in 3D:n',K//MatrixForm]; Print['eigs of K:',Chop[Eigenvalues[N[K]]]]; Master Stiffness of Example Truss in 3D: 20 10 0 −10 0 0 −10 −10 0 10 10 0 0 0 0 −10 −10 0 0 0 0 0 0 0 0 0 0 −10 0 0 10 0 0 0 0 0 0 0 0 0 5 0 0 −5 0 0 0 0 0 0 0 0 0 0 −10 −10 0 0 0 0 10 10 0 −10 −10 0 0 −5 0 10 15 0 0 0 0 0 0 0 0 0 0 eigs of K: {45.3577, 16.7403, 7.902, 0, 0, 0, 0, 0, 0}
;;; ;;;;
Figure 21.2. Testing the space truss assembler module.
The assembler is tested by the script shown in the top cell of Figure 21.2. The script defines the example truss of Chapters 2–3 as a 3D structure with its three members placed in the {x, y} plane. See Figure 21.3. √ The axial rigidity values E A = 100, 50 and 200 2 for elements 1, 2 and 3, respectively, have to be untangled because E is placed in elemat whereas A goes to elefab.
uz3 = 0 fx3 = 2
3
; ;;;; ;
(3) (2)
y
1
(1)
z u x1= u y1 = u z1 = 0
;;; 2
x
;; ;
Details of the assembly process are discussed in Chapter 25 for more general scenarios. Here we note that the module uses the freedom-pointertable technique described in §3.4 for merging each element stiffness matrix into the master stiffness.
fy3 = 1
u y2 = u z2 = 0
Figure 21.3. The example truss in three dimensions, used as module tester.
To split E A we take E = 100 for the three elements. Thus elemat = { 100,100,100 } = Table[100,{ 3 }] whereas elefab = { 1,1/2,2*Sqrt[2] }. Running the assembler in exact arithmetic gives the 9 × 9 master stiffness shown in the bottom cell of Figure 21.2. Taking its eigenvalues gives 6 zeros, which is the expected number in three dimensions. §21.3.2. Defining Boundary Conditions The modification process described here refers to the application of displacement boundary conditions on the master stiffness equations. These are assumed to be single-freedom constraints, either homogeneous such as u x3 = 0, or nonhomogeneous such as u z6 = −0.72. Boundary condition data in FEM programs is usually specified in two levels: nodes at the first level, and freedoms assigned to that node at the second level. (The reason is that nodes are highly visible to casual users, whereas direct access to freedom numbers is difficult.) This space truss program is no exemption. This data is organized into two lists: node freedom tags and node freedom values. Their configuration is best specified through an example. 21–5
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
21–6
ModifiedMasterStiffness[nodtag_,K_] := Module[ {i,j,k,n=Length[K],pdof,np,Kmod=K}, pdof=PrescDisplacementDOFTags[nodtag]; np=Length[pdof]; For [k=1,k<=np,k++, i=pdof[[k]]; For [j=1,j<=n,j++, Kmod[[i,j]]=Kmod[[j,i]]=0]; Kmod[[i,i]]=1]; Return[Kmod]]; ModifiedNodeForces[nodtag_,nodval_,K_,f_]:= Module[ {i,j,k,n=Length[K],pdof,pval,np,d,c,fmod=f}, pdof=PrescDisplacementDOFTags[nodtag]; np=Length[pdof]; pval=PrescDisplacementDOFValues[nodtag,nodval]; c=Table[1,{n}]; For [k=1,k<=np,k++, i=pdof[[k]]; c[[i]]=0]; For [k=1,k<=np,k++, i=pdof[[k]]; d=pval[[k]]; fmod[[i]]=d; If [d0, Continue[]]; For [j=1,j<=n,j++, fmod[[j]]-=K[[i,j]]*c[[j]]*d]; ]; Return[fmod]];
Figure 21.4. Modules to modify the master stiffness matrix and node force vector to apply the displacement boundary conditions.
Consider again the example truss in 3D shown in Figure 21.3, which has 3 nodes and 9 freedoms. The node freedom tag list, internally called nodtag, is nodtag = { { 1,1,1 },{ 0,1,1 },{ 0,0,1 } }
(21.2)
Each first-level entry of this list pertains to a node. The second level is associated with freedoms: displacements in the x, y and z directions. Freedom activity is marked by tag 0 or 1. A 1-tag means that the displacement is prescribed, whereas a 0-tag indicates that the force is known. Thus { 1,1,1 } for node 1 means that the node is fixed in the three directions. The node freedom value list, internally called nodval, gives the prescribed value of the force or the displacement. For the example truss it is nodval = { { 0,0,0 },{ 0,0,0 },{ 2,1,0 } }
(21.3)
For node 1, the tag entry is { 1,1,1 } and the value entry is { 0,0,0 }. This says that u x1 = u y1 = u z1 = 0. For node 2 it says that f x2 = 0 and u y2 = u z2 = 0. For node 3 it says that f x3 = 2, f y3 = 1 and u z3 = 0. The entries of nodval can be integers, floating point numbers, or symbols, whereas those in nodtag can only be 0 or 1.1 §21.3.3. Modifying the Master Stiffness Equations The modification of the master stiffness equations K u = f for displacement BCs produces the ˆ u = ˆf. This is done by the two modules listed in Figure 21.4. These modules modified system K are not restricted to space trusses, and may in fact be used for more general problems. The stiffness modifier is invoked by Kmod = ModifiedMasterStiffness[nodtag, K] 1
(21.4)
Other tag values may be implemented in more complicated programs to mark multifreedom constraints, for example.
21–6
21–7
§21.3
ANALYSIS SUPPORT MODULES
ClearAll[K,f,v1,v2,v4]; Km=Array[K,{6,6}]; Print['Master Stiffness: ',Km//MatrixForm]; nodtag={{1,1},{0,1},{0,0}}; nodval={{v1,v2},{0,v4},{0,0}}; Kmod=ModifiedMasterStiffness[nodtag,Km]; Print['Modified Master Stiffness:',Kmod//MatrixForm]; fm=Array[f,{6}]; Print['Master Force Vector:',fm]; fmod=ModifiedNodeForces[nodtag,nodval,Km,fm]; Print['Modified Force Vector:',fmod//MatrixForm];
Master Stiffness:
K[1, 1] K[2, 1] K[3, 1] K[4, 1] K[5, 1] K[6, 1]
Modified Master Stiffness:
K[1, 2] K[1, 3] K[1, 4] K[1, 5] K[1, 6] K[2, 2] K[2, 3] K[2, 4] K[2, 5] K[2, 6] K[3, 2] K[3, 3] K[3, 4] K[3, 5] K[3, 6] K[4, 2] K[4, 3] K[4, 4] K[4, 5] K[4, 6] K[5, 2] K[5, 3] K[5, 4] K[5, 5] K[5, 6] K[6, 2] K[6, 3] K[6, 4] K[6, 5] K[6, 6] 1 0 0 0 0 0 0 1 0 0 0 0 0 0 K[3, 3] 0 K[3, 5] K[3, 6] 0 0 0 1 0 0 0 0 K[5, 3] 0 K[5, 5] K[5, 6] 0 0 K[6, 3] 0 K[6, 5] K[6, 6]
Master Force Vector: { f[1], f[2], f[3], f[4], f[5], f[6] } v1 v2 f[3] − v1 K[1, 3] − v2 K[2, 3] − v4 K[4, 3] Modified Force Vector: v4 f[5] − v1 K[1, 5] − v2 K[2, 5] − v4 K[4, 5] f[6] − v1 K[1, 6] − v2 K[2, 6] − v4 K[4, 6]
Figure 21.5. Testing the displacement boundary condition application modules.
The arguments are: nodtag
A node by node list of freedom tags, as defined in the previous subsection.
K
The master stiffness matrix K produced by the assembler module.
The modified stiffness matrix, which has the same order as K, is returned as function value. The force modifier is invoked by fmod = ModifiedNodeForces[pdof, pval, K, f]
(21.5)
The arguments are: nodtag
The node freedom tag list defined in the previous subsection.
nodval
The node freedom value list defined in the previous subsection.
K
The master stiffness matrix K produced by the assembler module (before modification). This is only used if at least one of the displacement BCs is non-homogeneous.
f
The force vector before application of the displacement BCs.
The modified force vector, which has the same order as f, is returned as function value. The modules are tested by the script listed in the top cell of Figure 21.5. It uses symbolic master stiffness equations of order 6. The test illustrates a not well known feature of Mathematica: use of Array function to generate subscripted symbolic arrays of one and two dimensions. The results, shown in the bottom cell of Figure 21.5, should be self explanatory. 21–7
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
21–8
PrescDisplacementDOFTags[nodtag_]:= Module [ {j,n,numnod=Length[nodtag],pdof={},k=0,m}, For [n=1,n<=numnod,n++, m=Length[nodtag[[n]]]; For [j=1,j<=m,j++, If [nodtag[[n,j]]>0, AppendTo[pdof,k+j]]; ]; k+=m; ]; Return[pdof]]; PrescDisplacementDOFValues[nodtag_,nodval_]:= Module [ {j,n,numnod=Length[nodtag],pval={},k=0,m}, For [n=1,n<=numnod,n++, m=Length[nodtag[[n]]]; For [j=1,j<=m,j++, If [nodtag[[n,j]]>0, AppendTo[pval,nodval[[n,j]]]]; ]; k+=m; ]; Return[pval]];
Figure 21.6. Modules to build auxiliary lists pdof and pval from node-by-node BC data.
Remark 21.1. On entry, the modification modules of Figure 21.4 build auxiliary lists pdof and pval to simplify the modification logic. pdof is a list of the prescribed degrees of freedom identified by their equation number in the master stiffness equations. For example, if freedoms 4, 7, 9 and 15 are specified, pdof = { 4,7,9,12 }. These indices are stored in ascending order. pval is a list of the prescribed displacement values listed in pdof. These lists are constructed by the modules listed in Figure 21.6. The calls are pdof = PrescribedDOFTags[nodtag] and pval = PrescribedDOFValues[nodtag,nodval]. Remark 21.2. The logic of ModifiedMasterStiffness is straightforward. Construct pdof, then clear appropriate rows and columns of K and place ones on the diagonal. Note the use of the Mathematica function Length to control loops: np=Length[pdof] sets np to the number of prescribed freedoms. Similarly n=Length[K] sets n to the order of the master stiffness matrix K, which is used to bound the row and column clearing loop. These statements may be placed in the list that declares local variables. Remark 21.3. ModifiedNodalForces has more complicated logic because it accounts for nonhomogeneous BCs. On entry it constructs pdof and pval. If nonzero values appear in pval, the original entries of f are modified as described in §4.1.2, and the end result is the effective force vector. Force vector entries corresponding to the prescribed displacement values are replaced by the latter in accordance with the prescription (4.13). If there are nonhomomogeneous BCs it is important that the stiffness matrix provided as third argument be the master stiffness before modification by ModifiedStiffnessMatrix. This is because stiffness coefficients that are cleared by ModifiedStiffnessMatrix are needed for modifying the force vector.
§21.3.4. Displacement Solution and Reaction Recovery ˆ u = ˆf, where K ˆ and ˆf are the modified stiffness and force matrices, respectively, The linear system K is solved for displacements by a built-in linear algebraic solver. In Mathematica LinearSolve is available for this purpose: u = LinearSolve[Kmod, fmod] (21.6) At this point postprocessing begins. The node forces including reactions are obtained from f = K u. This can be done simply as a matrix product: f = K.u
(21.7)
where K is the original master stiffness matrix before modification, and u the displacement vector computed by LinearSolve. 21–8
21–9
§21.3
ANALYSIS SUPPORT MODULES
FlatNodePartVector[nv_]:=Flatten[nv]; NodePartFlatVector[nfc_,v_]:= Module [ {i,k,m,n,nv={},numnod}, If [Length[nfc]0, nv=Partition[v,nfc]]; If [Length[nfc]>0, numnod=Length[nfc]; m=0; nv=Table[0,{numnod}]; For [n=1,n<=numnod,n++, k=nfc[[n]]; nv[[n]]=Table[v[[m+i]],{i,1,k}]; m+=k]]; Return[nv]];
Figure 21.7. Utility modules to flatten and node-partition node-DOF vectors.
§21.3.5. Flattening and Partitioning Node-Freedom Vectors In the analysis process one often needs displacement and force vectors in two different list formats. For example, the computed node displacement vector produced by (21.6) for the example truss in 3D is u = { 0,0,0,0,0,0,0.4,-0.2,0 } (21.8) Following the terminology of Mathematica this will be called the flat form of the displacement vector. For postprocessing purposes (especially printing and plotting) it is convenient to rearrange to the node by node form noddis = { { 0,0,0 },{ 0,0,0 },{ 0.4,-0.2,0 } }
(21.9)
This will be called the node-partitioned form of the displacement vector. Similar dual formats exist for the node force vector. In flat form this is called f and in node-partitioned form nodfor. The utility modules listed in Figure 21.7 can be used to pass from one format to the other. To flatten the node-partitioned form of a vector, say nv, say v = FlatNodePartVector[nv]
(21.10)
To node-partition a flat vector v say nv = NodePartFlatVector[nfc,v]
(21.11)
where nfc is the number of freedoms per node. For space trusses this is 3 so appropriate conversion calls are noddis = NodePartFlatVector[3,u] and nodfor = NodePartFlatVector[3,f]. Remark 21.4.
FlatNodePartVector can be directly done by the built-in function Flatten whereas NodePartFlatVector — for a fixed number of freedoms per node — can be done by Partition. The reason for the “wrappers” is to guide the conversion of Mathematica code to a lower level language such as C, where such built-in list functions are missing. Remark 21.5. The additional code in NodePartFlatVector caters to the case where the number of freedoms can vary from node to node, in which case nfc is a list (not a number) called the node freedom count, hence the abbreviation. That facility is useful in more advanced courses.
21–9
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
21–10
SpaceTrussIntForces[nodxyz_,elenod_,elemat_,elefab_, noddis_,prcopt_]:= Module[{ numnod=Length[nodxyz], numele=Length[elenod],e,ni,nj,ncoor,Em,A,options,ue,p}, p=Table[0,{numele}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; ncoor={nodxyz[[ni]],nodxyz[[nj]]}; ue=Flatten[{ noddis[[ni]],noddis[[nj]] }]; Em=elemat[[e]]; A=elefab[[e]]; options=prcopt; p[[e]]=SpaceBar2IntForce[ncoor,Em,A,ue,options] ]; Return[p]]; SpaceBar2IntForce[ncoor_,Em_,A_,ue_,options_]:= Module[ {x1,x2,y1,y2,z1,z2,x21,y21,z21,EA,numer,LL,pe}, {{x1,y1,z1},{x2,y2,z2}}=ncoor; {x21,y21,z21}={x2-x1,y2-y1,z2-z1}; EA=Em*A; {numer}=options; LL=x21^2+y21^2+z21^2; If [numer,{x21,y21,z21,EA,LL}=N[{x21,y21,z21,EA,LL}]]; pe=(EA/LL)*(x21*(ue[[4]]-ue[[1]])+y21*(ue[[5]]-ue[[2]])+ +z21*(ue[[6]]-ue[[3]])); Return[pe]]; SpaceTrussStresses[elefab_,elefor_,prcopt_]:= Module[ {numele=Length[elefab],e,elesig}, elesig=Table[0,{numele}]; For [e=1, e<=numele, e++, elesig[[e]]=elefor[[e]]/elefab[[e]] ]; Return[elesig]];
Figure 21.8. Modules to compute internal forces and stresses in a space truss.
§21.3.6. Internal Force Recovery The calculation of internal forces and stresses in a space truss involves computing axial forces in the bar elements. The modules that do those calculations given the displacement solution are listed in Figure 21.8. Module SpaceTrussIntForces computes the internal forces (axial forces) in all truss members. It is invoked by elefor = SpaceTrussIntForces[nodxyz,elenod,elemat, elefab,noddis,prcopt] (21.12) Five of the arguments: nodxyz, elenod, elemat, elefab and prcopt, are the same used in the call (21.1) to the stiffness assembler. The additional argument, noddis, contains the computed node displacements arranged in node-partitioned form noddis = { { ux1,uy1,uz1 },{ ux2,uy2,uz2 }, . . . { uxn,uyn,uzn } }
(21.13)
This form can be obtained from the computed displacement solution through the utility module (21.11). As function value SpaceTrussIntForces returns a list of element axial forces { p1, p2 . . . pe }
(21.14)
SpaceTrussIntForces makes use of SpaceBar2IntForce, which computes the internal force in an individual bar element. This is invoked as SpaceBar2IntForce p = SpaceBar2IntForces[ncoor,Em,A,ue,options] 21–10
(21.15)
21–11
§21.4 UTILITY PRINT MODULES ClearAll[nodxyz,elenod,elemat,elefab,noddis]; nodxyz={{0,0,0},{10,0,0},{10,10,0}}; elenod={{1,2},{2,3},{1,3}}; elemat= Table[100,{3}]; elefab= {1,1/2,2*Sqrt[2]}; noddis={{0,0,0}, {0,0,0}, {4/10,-2/10,0}}; prcopt={False}; elefor=SpaceTrussIntForces[nodxyz,elenod,elemat,elefab,noddis,prcopt]; Print['Int Forces of Example Truss:',elefor]; Print['Stresses:',SpaceTrussStresses[elefab,elefor,prcopt]]; Int Forces of Example Truss: { 0, −1, 2*Sqrt[2] } Stresses: { 0, −2, 1 }
Figure 21.9. Test of the internal force recovery module.
Arguments ncoor, Em, A and options are the same as in the call to SpaceBar2Stiffness described in §20.2.2. The additional argument, ue, contains the flat list of the six element node displacements in the global system arranged as { ux1,uy1,ux1,ux2,ux2,uz2 }. The recovery equation is the subject of Exercise 21.4. The last module in Figure 21.8 computes the member stresses simply by dividing the internal forces by the cross section areas. It is invoked as elesig = SpaceTrussStresses[elefab,elefor,prcopt]
(21.16)
Here elefab and prcopt are as before, and elefor contains the element forces computed by SpaceTrussIntForces. The element axial stresses are returned as function value. The statements of the top cell of Figure 21.8 exercise the internal force recovery for the example truss in 3D, requesting exact calculiations. Array p is printed in the bottom cell. The axial forces √ of 0, −1 and 2 2 agree with those determined in Chapter 3. §21.3.7. The Solution Driver It is convenient to package the sequence of operations desribed in the previous subsections, namely assembly, modification, solution, force and stress recovery, into one module called the solution driver. This is listed in Figure 21.10. It is invoked by saying { noddis,nodfor,elefor,elesig }= SpaceTrussSolution[nodxyz,elenod, elemat,elefab,nodtag,nodval,prcopt]
(21.17)
All arguments: nodxyz, elenod, elemat, elefab, nodtag, nodval and prcopt, have been described in previous subsections. The module returns four lists: noddis
Computed node displacement in node partitioned format.
nodfor
Recovered node forces including reactions in node partitioned format.
elefor
Element internal forces.
elesig
Element stresses.
Note that the listing of SpaceTrussSolution in Figure 21.10 has two commented out eigenvalue computations, one for the master stiffness K and one for the modified stiffness Kmod. Decommenting those commands comes in handy when setting up and running a new problem if errors are detected. 21–11
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
21–12
SpaceTrussSolution[nodxyz_,elenod_,elemat_,elefab_,nodtag_,nodval_, prcopt_]:= Module[{K,Kmod,f,fmod,u,noddis,nodfor,elefor,elesig}, K=SpaceTrussMasterStiffness[nodxyz,elenod,elemat,elefab,prcopt]; (* Print['eigs of K=',Chop[Eigenvalues[N[K]]]]; *) Kmod=ModifiedMasterStiffness[nodtag,K]; f=FlatNodePartVector[nodval]; fmod=ModifiedNodeForces[nodtag,nodval,K,f]; (* Print['eigs of Kmod=',Chop[Eigenvalues[N[Kmod]]]]; *) u=LinearSolve[Kmod,fmod]; u=Chop[u]; f=Chop[K.u, 10.0^(-8)]; nodfor=NodePartFlatVector[3,f]; noddis=NodePartFlatVector[3,u]; elefor=Chop[SpaceTrussIntForces[nodxyz,elenod,elemat,elefab, noddis,prcopt]]; elesig=SpaceTrussStresses[elefab,elefor,prcopt]; Return[{noddis,nodfor,elefor,elesig}]; ]; Figure 21.10. The analysis driver module.
§21.4.
Utility Print Modules
Utility print modules are used to display input and output data in tabular form. The following six modules are provided in Cell 6 of the SpaceTruss.nb notebook. To print the node coordinates in nodxyz: PrintSpaceTrussNodeCoordinates[nodxyz,title,digits]
(21.18)
To print the element nodes in elenod, element materials in elemat and element fabrications in elefab: PrintSpaceTrussElementData[elenod,elemat,elefab,title,digits]
(21.19)
To print the freedom activity data in nodtag and nodval: PrintSpaceTrussFreedomActivity[nodtag,nodval,title,digits]
(21.20)
To print the node displacements in noddis (configured in node-partitioned form): PrintSpaceTrussNodeDisplacements[noddis,title,digits]
(21.21)
To print the node forces in nodfor (configured in node-partitioned form): PrintSpaceTrussNodeForces[nodfor,title,digits]
(21.22)
To print the element internal forces in elefor and element stresses in elesig: PrintSpaceTrussElemForcesAndStresses[elefor,elesig,title,digits]
(21.23)
In all cases, title is an optional character string to be printed as a title before the table; for example 'Node coordinates of bridge truss'. To eliminate the title, specify ' (two quote marks together). The last argument of the print modules: digits, is optional. If set to { d,f } it specifies that floating point numbers are to be printed with room for at least d digits, with f digits after the decimal point. If digits is specified as a void list: { }, a preset default is used for d and f.
21–12
21–13 §21.5.
§21.5 UTILITY GRAPHIC MODULES
Utility Graphic Modules
Graphic modules that support preprocessing are placed in Cells 4 and 5 of the SpaceTruss.nb notebook. These display unlabeled elements, elements and nodes with labels, deformed shapes and element stress levels. §21.5.1. Plot Module Calls To plot elements only: PlotSpaceTrussElements[nodxyz,elenod,title,{ view,aspect,{ } }]
(21.24)
To plot element and nodes with optional labeling: PlotSpaceTrussElementsAndNodes[nodxyz,elenod,title,{ view,aspect,labels }] (21.25) To plot deformed shape of the truss under computed node displacements: PlotSpaceTrussDeformedShape[nodxyz,elenod,noddis,amplif,box,title, { view,aspect,colors }]
(21.26)
To plot element axial stress levels using a coloring scheme: PlotSpaceTrussStresses[nodxyz,elenod,elesig,sigfac,box,title, { view,aspect,{ } }]
(21.27)
In the foregoing nodxyz, elenod, noddis, noddfor elefor and elesig have been described above. The other arguments are as follows. view
A list configured as a list of two 3D vectors: { { Vhat1,Vhat2,Vhat3 },{ W1,W2,W3 } }. Vector ˆ with W with global components { W1,W2,W3 } specifies the view direction whereas vector V, global components { V1,V2,V3 }, specifies the up direction for the plot view, as discussed in §21.5.2. If a void list is provided for the argument, the default is { { 0,1,0 },{ 0,0,1 } }; this means that view direction is along the −z axis whereas the up direction is the y axis.
aspect
Vertical-to-horizontal aspect ratio of the plot as it appears on a Notebook cell. Three possibilities. If set to −1, the Mathematica default Aspect->Automatic is chosen. If set to zero, an aspect ratio is computed by the plot module as appropriate. If set to a positive number, the aspect ratio is set to that number.
labels
Only used in PlotSpaceTrussElementsAndNodes. A list with the following configuration: { { nlabels,frn,fex,fey }, { elabels,fre },{ fntfam,fntsiz,fntwgt,fntslt } }. nlabels
A logical flag. Set to True to get node labels in plot.
frn
Radius of circle drawn around each node expressed as percentage of plot size.
fex
Horizontal eccentricity in points of node label from node location.
fex
Vertical eccentricity in points of node label from node location.
elabels
A logical flag. Set to True to get element labels in plot.
fre
Radius of circle drawn around each element number expressed as percentage of plot size. from node location.
fntfam
Font family used for labels, for example 'Times'
fntsiz
Size in points of font used for labels; usually 10 through 14.
fntwgt
Font weight used for element labels. Typical settings are 'Plain' or ”Bold”. Node labels are always drawn in boldface.
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Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
Up vector, aka vertical screen direction
V = ys
Horizontal screen direction
C
e
an
en
pl
re
Sc
U = xs
y O x
W = zs View direction
z
eye point, aka camera point, observer point, view reference point
Global coordinate system (FEM) World coordinate system (graphics) Mathematica cell
Figure 21.11. Plot view in 3D as mapping of world to screen coordinates.
fntslt
Font slant used for element labels. Typical settings are 'Plain', 'Italics' or 'Slanted'. Node labels are always drawn in Plain.
amplif
The displacement amplification factor to be used by PlotSpaceTrussDeformedShape. Usually a value much larger than one (say 100 or 1000) is necessary to visualize displacements in actual structures. Becomes a list if the call is to draw several shapes (for example undeformed and deformed). For example setting amplif to { 0,100 } will draw two shapes: the undeformed configuration and one with magnification of 100. If more than one shape is to be drawn the colors specification comes in handy.
box
A list of points, specified by their {x, y, z} global coordinates, that forms a box that encloses the plot. Used in PlotSpaceTrussDeformedShape and PlotSpaceTrussStresses. The box is converted to a frame by the view projector. This is useful for various purposes, one being to do realistic animations by drawing a sequence of deformed shapes moving inside this box.
colors
Defines element colors to be used by PlotSpaceTrussDeformedShape, specified as lower case character string. Legal ones are 'black', 'red', 'blue', 'green' and ”white”. If the call is to draw several shapes (for example undeformed and deformed), this argument can be a list, such as { 'black','red' }, in which case the colors are in one to one correspondence with the amplification values in amplif. If no color is specified, black is assumed.
sigfac
A stress scaling factor for PlotSpaceTrussStresses. Normally set to 1.
Because of the confusing way Mathematica handles plots, (some features do scale with plot size while others, such as font sizes, do not), some interactive experimentation with dimension specs seems inevitable. §21.5.2. Plot View Specification Plotting 3D objects, such as the space trusses considered here, involves mapping the coordinates given in the FEM global system {x, y, z} into screen coordinates {xs , ys , z s } in which z s = 0. Objects are rendered using screen coordinates. Construction of this mapping is based on the view specification, which appears as argument of all plotting routines described in the foregoing subsection. The viewing ingredients are shown in Figure 21.11. In computer graphics, the 3D space spanned by the FEM global system {x, y, z} is called the world space for obvious reasons. The screen coordinates {xs , ys , z s } are defined by two vectors, which in computer graphics are typically identified as W and V: the view direction and up direction, respectively. The view direction is the line defined by joining the eye position at C with the origin O of {x, y, z}. The screen plane passes through O and is normal to W. Screen coordinates xs and ys are in the screen plane and
21–14
21–15
§21.6
EXAMPLE 1: BRIDGE PLANE TRUSS EXAMPLE
(a) Elastic modulus E = 1000 Cross section areas of bottom longerons: 2, top longerons: 10, battens: 3, diagonals: 1
(b) y (7)
x (13) (1) 3
4
(9)
(19)
(18) (2)
6
5
(10)
(15)
(14)
;;
1
2
(8)
10 16 10 span: 6 bays @ 10 = 60
(3)
8
7
10
(11)
(16) (20) (4)
10 (12)
(21) (17)
9
9
(5)
12
;;
10
top joints lie on parabolic profile
11
(6)
Figure 21.12. Six-bay bridge plane truss used as example problem: (a) truss structure showing supports and applied loads; (b) finite element idealization as pin-jointed truss.
going along the horizontal and vertical directions, respectively. In computer graphics the xs and ys directions are called U and V, respectively. When the plot is rendered in a Mathematica cell, U ≡ xs goes horizontally from left to right. Axes {u ≡ xs , V ≡ ys , W ≡ z s } form a RHS Cartesian coordinate system. Mappings from screen to pixel coordinates are handled by the plotting system, and need not be discussed here. In the plot modules used here, the eye point C is assumed to be at infinity.2 Thus only the view direction W, as specified by three direction numbers, is used. For example, the specification { 1,1,1 } says that W is the trisector of the {x, y, z} octant. To define the up direction V ≡ ys one has to enter a second ˆ which must not be parallel to W. Since V ˆ is not necessarily normal to W, V is “indication” vector: V, T ˆ − (V ˆ Wn )Wn , where Wn is W normalized to length one. For constructed by orthogonalization: V = V ˆ and and W are specified by view argument { { 0,1,0 },{ 2,2,1 } }, then V = [0, 1, 0] − example if V ([0, 1, 0]T [2/3, 2/3, 1/3)[2/3, 2/3, 1/3] = [0, 1, 0] − (2/3)[2/3, 2/3, 1/3] = [−4/9, 5/9, −2/9]. Note that VT W = 0.
§21.6. Example 1: Bridge Plane Truss Example This example deals with the analysis of the 6-bay bridge truss problem defined in Figure 21.12. This truss has 12 nodes and 17 elements. It is contained in the {x, y} plane and can only move in that plane. It is fixed at node 1 and on rollers at node 12. The driver is listed in Figure 21.13. Preprocessing statements appear on top, with light green background. These define the problem through specification of the following data structures.3 NodeCoordinates
Same configuration as nodxyz
2
Graphics where the eye point C is at a finite distance produce perspective plots. The Graphics3D system of Mathematica allows perpective plotting. However the plots described here use only the 2D graphics subset.
3
Note the use of longer mnemonic names in the problem driver. For example NodeCoordinates instead of nodxyz. This simplifies the preparation of problem solving assignments since driver scripts are more self-documenting. It also helps grading returned assignments.
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Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
21–16
NodeCoordinates={{0,0,0},{10,5,0},{10,0,0},{20,8,0},{20,0,0},{30,9,0}, {30,0,0},{40,8,0},{40,0,0},{50,5,0},{50,0,0},{60,0,0}}; ElemNodes={{1,3},{3,5},{5,7},{7,9},{9,11},{11,12}, {1,2},{2,4},{4,6},{6,8},{8,10},{10,12}, {2,3},{4,5},{6,7},{8,9},{10,11}, {2,5},{4,7},{7,8},{9,10}}; PrintSpaceTrussNodeCoordinates[NodeCoordinates,'Node coordinates:',{}]; numnod=Length[NodeCoordinates]; numele=Length[ElemNodes]; Em=1000; Abot=2; Atop=10; Abat=3; Adia=1; ElemMaterials= Table[Em,{numele}]; ElemFabrications={Abot,Abot,Abot,Abot,Abot,Abot,Atop,Atop,Atop,Atop, Atop,Atop,Abat,Abat,Abat,Abat,Abat,Adia,Adia,Adia,Adia}; PrintSpaceTrussElementData[ElemNodes,ElemMaterials,ElemFabrications, 'Element data:',{}]; ProcessOptions= {True}; view={{0,1,0},{0,0,1}}; labels={{True,0.06,-1.5,1.5},{True,0.12},{'Times',11,'Roman'}}; PlotSpaceTrussElementsAndNodes[NodeCoordinates,ElemNodes, 'bridge mesh',{view,-1,labels}]; NodeDOFTags= Table[{0,0,1},{numnod}]; NodeDOFValues=Table[{0,0,0},{numnod}]; NodeDOFValues[[3]]={0,-10,0}; NodeDOFValues[[5]]= {0,-10,0}; NodeDOFValues[[7]]={0,-16,0}; NodeDOFValues[[9]]={0,-10,0}; NodeDOFValues[[11]]={0,-10,0}; NodeDOFTags[[1]]= {1,1,1}; (* fixed node 1 *) NodeDOFTags[[numnod]]={0,1,1}; (* hroller @ node 12 *) PrintSpaceTrussFreedomActivity[NodeDOFTags,NodeDOFValues, 'DOF Activity:',{}]; {NodeDisplacements,NodeForces,ElemForces,ElemStresses}= SpaceTrussSolution[ NodeCoordinates,ElemNodes,ElemMaterials, ElemFabrications,NodeDOFTags,NodeDOFValues,ProcessOptions ]; PrintSpaceTrussNodeDisplacements[NodeDisplacements, 'Computed node displacements:',{}]; PrintSpaceTrussNodeForces[NodeForces, 'Node forces including reactions:',{}]; PrintSpaceTrussElemForcesAndStresses[ElemForces,ElemStresses, 'Int Forces and Stresses:',{}]; view={{0,1,0},{0,0,1}}; box={{0,-4,0},{60,-4,0},{60,10,0},{0,10,0}}; PlotSpaceTrussDeformedShape[NodeCoordinates,ElemNodes,NodeDisplacements, {0,1},box,'deformed shape (unit magnif)',{view,-1,{'black','blue'}}]; PlotSpaceTrussStresses[NodeCoordinates,ElemNodes,ElemStresses,1,box, 'axial stresses in truss members',{view,0,labels}]; Figure 21.13. Driver script for analysis of the 6-bay plane bridge truss. Preprocessing statements in light green background. Processing and postprocessing statements in light blue.
ElemNodes
Same configuration as elenod
ElemMaterials
Same configuration as elemat
ElemFabrications Same configuration as elefab. This list is built up from four repeating 21–16
21–17
§21. Notes and Bibliography
cross sectional areas: Abot, Atop, Abat and Adia, for the areas of bottom longerons, top longerons, battens and diagonals, respectively. NodeDOFTags
Same configuration as nodtag. Initialized to { 0,0,1 } for all nodes on creation so as to fix all z displacements. Then the support conditions in the {x, y} plane are specified at supported nodes. For example, NodeDOFTags[[1]] = { 1,1 } says that both node displacement components u x1 and u y1 of node 1 are specified.
NodeDOFValues
Same configuration as nodval. Initialized to { 0,0,0 } on creation for all nodes. Then the value of nonzero applied loads is set at nodes 3, 5, 7, 9 and 11. For example NodeDOFValues[[7]] = { 0,-16,0 } specifies f x7 = 0, f y7 = −16 and u z7 = 0.
The input data structures can be shown in tabular form for convenient inspection using print utility modules. Printed tables are shown on the left of Figure 21.14. Running the solution module returns computed displacements, node forces including reactions, internal forces and member stress. These are printed with utility modules. These results are shown on the right of Figure 21.14. Output plot results are collected in Figure 21.15. §21.7. Example 2: An Orbiting Truss Structure To be included in the final version of the Chapter. Notes and Bibliography The dominant philosophy in FEM implementation is to construct general purpose programs that can solve a wide range of problems. For example, static and dynamic response of arbitrary structures with linear or nonlinear behavior. This path was naturally taken once the Direct Stiffness Method became widely accepted in the mid 1960s. It is reflected in the current crop of commercial FEM programs. Their source code by now has reached into millions of lines. These codes do have a place in undergraduate engineering education, starting at the junior level. At this level students should be taught rudiments of modeling and how to use those black box programs as tools. This exposure provides also basic knowledge for senior projects that require finite element analysis. At the graduate level, however, students should understand what goes on behind the scene. But access to innards of commercial programs is precluded (and if it where, it would be difficult to follow given their complexity). The philosophy followed here is to use special purpose codes written in a high level language. These may be collectively called grey level codes. A high level language like Mathematica conceals utility operations such as matrix products, linear solvers and graphics, but permits the application code logic to be seen and studied. As a result a complete FEM program is tiny (typically a few hundreds lines), it can be built and debugged in a few hours, and may be understood as a whole by one person. On the down side of course they can do only very limited problems, but for instructional use simplicity outweights generality.
21–17
21–18
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES Node coordinates : node x0001coor 1 0.000000 2 10.000000 3 10.000000 4 20.000000 5 20.000000 6 30.000000 7 30.000000 8 40.000000 9 40.000000 10 50.000000 11 50.000000 12 60.000000 Element elem 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
y0001coor 0.000000 5.000000 0.000000 8.000000 0.000000 9.000000 0.000000 8.000000 0.000000 5.000000 0.000000 0.000000
data: nodes 00011, 30002 00013, 50002 00015, 70002 00017, 90002 00019, 110002 000111, 120002 00011, 20002 00012, 40002 00014, 60002 00016, 80002 00018, 100002 000110, 120002 00012, 30002 00014, 50002 00016, 70002 00018, 90002 000110, 110002 00012, 50002 00014, 70002 00017, 80002 00019, 100002
DOF Activity : node x0001tag y0001tag 1 1 1 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 10 0 0 11 0 0 12 0 1
modulus 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 1000.00 z0001tag 1 1 1 1 1 1 1 1 1 1 1 1
Computed node displacements : node x0001displ y0001displ 1 0.000000 0.000000 2 0.809536 00011.775600 3 0.280000 00011.792260 4 0.899001 00012.291930 5 0.560000 00012.316600 6 0.847500 00012.385940 7 0.847500 00012.421940 8 0.795999 00012.291930 9 1.135000 00012.316600 10 0.885464 00011.775600 11 1.415000 00011.792260 12 1.695000 0.000000
z0001coor 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Node forces including reactions : node x0001force y0001force z0001force 1 0.0000 28.0000 0.0000 2 0.0000 0.0000 0.0000 3 0.0000 000110.0000 0.0000 4 0.0000 0.0000 0.0000 5 0.0000 000110.0000 0.0000 6 0.0000 0.0000 0.0000 7 0.0000 000116.0000 0.0000 8 0.0000 0.0000 0.0000 9 0.0000 000110.0000 0.0000 10 0.0000 0.0000 0.0000 11 0.0000 000110.0000 0.0000 12 0.0000 28.0000 0.0000
area 2.00 2.00 2.00 2.00 2.00 2.00 10.00 10.00 10.00 10.00 10.00 10.00 3.00 3.00 3.00 3.00 3.00 1.00 1.00 1.00 1.00
x0001value 0 0 0 0 0 0 0 0 0 0 0 0
z0001displ 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
y0001value 0 0 000110 0 000110 0 000116 0 000110 0 000110 0
Int Forces and Stresses : elem axial force axial stress 1 56.0000 28.0000 2 56.0000 28.0000 3 57.5000 28.7500 4 57.5000 28.7500 5 56.0000 28.0000 6 56.0000 28.0000 7 000162.6100 00016.2610 8 000160.0300 00016.0030 9 000160.3000 00016.0300 10 000160.3000 00016.0300 11 000160.0300 00016.0030 12 000162.6100 00016.2610 13 10.0000 3.3330 14 9.2500 3.0830 15 12.0000 4.0000 16 9.2500 3.0830 17 10.0000 3.3330 18 1.6770 1.6770 19 3.2020 3.2020 20 3.2020 3.2020 21 1.6770 1.6770
z0001value 0 0 0 0 0 0 0 0 0 0 0 0
Figure 21.14. Bridge truss example: tabular printed output. On the left: node, element and freedom data. On the right: computed displacements, node forces, internal forces and member stresses. bridge mesh
bridge mesh with elem & node labels 2 1
7 1
13 3
4
9
14
19
6
10
8
20
16
8 18 2
5
3
15 7
4
9
11
10
21 5
17 11
12 6
12
deformed shape
axial stress level plot
Figure 21.15. Bridge truss example: graphic output collected in one figure.
21–18
21–19
Exercises
Homework Exercises for Chapter 21 FEM Program for Space Trusses
EXERCISE 21.1 [D:10] The logic of SpaceTrussMasterStiffness cannot be used for structures other than space trusses. Justify this assertion. EXERCISE 21.2 [D:10] The logic of ModifiedMasterStiffness and ModifiedNodeForces is not re-
stricted to space trusses, but can be used for any FEM program that stores K and f as full arrays. Justify this assertion. EXERCISE 21.3 [D:20] The logic of PrescDisplacementDOFTags and PrescDisplacementDOFValues
is not restricted to a fixed number of DOF per node. Justify this assertion. EXERCISE 21.4 [A:15] Show that the longitudinal elongation of a space bar can be computed directly from
the global displacements u x1 , u y1 , . . . u z2 from d = (x21 u x21 + y21 u y21 + z 21 u z21 )/0007,
(E21.1)
in which x21 = x2 − x1 , u x21 = u x2 − u x1 , etc, and 0007 is the bar length. Hence justify the formula used in module SpaceBar2IntForce listed in Figure 21.8 to recover the axial force p = (E A/0007)d. EXERCISE 21.5 [C:25] Analyze the structure shown in Figure E21.1. This is a pin-jointed truss model of a
200-in-high (5m) transmission tower originally proposed by Fox and Schmit in 1964 [64] as a test for early automated-synthesis codes based on FEM. It became a standard benchmark for structural optimization. The truss has 10 joints (nodes) and 25 members (elements). The truss geometry and node numbering are defined in Figure E21.1(a). Joints 1 and 2 at the top of the tower lie on the {x, z} plane. The truss (but not the loads) is symmetric about the {y, z} and {x, z} planes. Figure E21.1(b) gives the element numbers.
(a)
z 1
Coordinates: node 2 ( 37.5,0,200) node 1 (−37.5,0,200)
(b)
2
1 (1) 2 (8) (9) (5) (4) (10)
100 in
75 in
(14) joins 3-10 (20) joins 5-10
3
6
4
(13) 6 (20)
5
75 in
y
(14)
8
(21)
x
9
(23)
(15) (24)
10 100 in
200 in
4 5
(22)
7
10
(6) (7) (3) (2) (12) 3
7 (11)
(25) (18) (19)
(17)
(16)
9
200 in
Figure E21.1. 25-member space truss model of a transmission tower. (a): Geometry definition and node numbers; (b) element numbers. For member properties and loads see Tables E21.1 and E21.2.
21–19
8
21–20
Chapter 21: FEM PROGRAM FOR SPACE TRUSSES
Table E21.1
Cross section areas of transmission tower members
Element
A (sq in)
Element
A (sq in)
Element
A (sq in)
1 2 3 4 5 6 7 8 9
0.033 2.015 2.015 2.015 2.015 2.823 2.823 2.823 2.823
10 11 12 13 14 15 16 17 18
0.010 0.010 0.014 0.014 0.980 0.980 0.980 0.980 1.760
19 20 21 22 23 24 25
1.760 1.760 1.760 2.440 2.440 2.440 2.440
Table E21.2
Applied load case for transmission tower
Node
x-load (lb)
y-load (lb)
z-load (lb)
1 2 3 6
1000 0 500 500
10000 10000 0 0
−5000 −5000 0 0
Applied forces at all other nodes are zero. Own-weight loads not considered.
The members are aluminum tubes with the cross sections listed in Table E21.1.4 The modulus of elasticity is E = 107 psi for all members. The specific weight is 0.1 lb/in3 . The applied load case to be studied is given in Table E21.2. Analyze the transmission tower using the program provided in the SpaceTruss.nb Notebook (downloadable from Chapter 21 Index). Results to report: driver program cell, node displacements and element stresses. (More details on HW assignment sheet.) Note: as a quick check on model preparation, the total weight of the tower should be 555.18 lb.
4
Data taken from an optimal design reported by Venkayya, Khot and Reddy in 1968 [168].
21–20
23
.
Implementation of Iso-P Quadrilateral Elements
23–1
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
23–2
TABLE OF CONTENTS Page
§23.1. Introduction §23.2. Bilinear Quadrilateral Stiffness Matrix §23.2.1. Gauss Quadrature Rule Information . §23.2.2. Shape Function Evaluation . . . §23.2.3. Element Stiffness . . . . . . . §23.3. Test of Bilinear Quadrilateral §23.3.1. Test of Rectangular Geometry . . §23.3.2. Test of Trapezoidal Geometry . . . §23.4. *Consistent Node Forces for Body Force Field §23.5. *Recovery of Corner Stresses §23.6. *Quadrilateral Coordinates of Given Point §23. Notes and Bibliography . . . . . . . . . . . . . §23. References . . . . . . . . . . . . . §23. Exercises . . . . . . . . . . . . .
23–2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
23–3 23–3 23–3 23–4 23–5 23–7 23–7 23–8 23–10 23–11 23–12 23–12 23–13 23–14
23–3
§23.2
BILINEAR QUADRILATERAL STIFFNESS MATRIX
§23.1. Introduction This Chapter illustrates, through a specific example, the computer implementation of isoparametric quadrilateral elements for the plane stress problem. Triangles, which present some programming quirks, are covered in the next Chapter. The programming example is that of the four-node bilinear quadrilateral. It covers the computation of the element stiffness matrix, the consistent node force vector for a body forces, and stress recovery at selected points. The organization of the computations is typical of isoparametric-element modules in any number of space dimensions. §23.2. Bilinear Quadrilateral Stiffness Matrix We consider the implementation of the 4-node blinear quadrilateral for plane stress, depicted in Figure 23.1. The element stiffness matrix of simple one-dimensional elements, such as the ones discussed in Chapters 20–22, can be easily packaged in a simple module. For multidimensional elements, however, it is convenient to break up the implementation into application dependent and application independent modules, as flowcharted in Figure 23.2. The application independent modules can be “reused” in other FEM applications, for example to form thermal, fluid or electromagnetic elements.
η 4 (x 4 , y 4 )
η=1
3 (x 3 , y 3 ) ξ=1 ξ
ξ=−1 1 (x 1 , y 1 )
η=−1
2 (x 2 , y 2 )
Figure 23.1. The 4-node bilinear quadrilateral element.
For the bilinear quadrilateral stiffness computations, the separation of Figure 23.2 is done by dividing the work into three modules: Quad4IsoPMembraneStiffness. Computes the element stiffness matrix Ke of a four-node isoparametric quadrilateral element in plane stress. QuadGuassRuleInfo. Returns two-dimensional Gauss quadrature formula of product type. Quad4IsoPShapeFunDer. Evaluates the shape functions of a four-node isoparametric quadrilateraland their x/y derivatives, at a specific point. These modules are described in further detail in the following subsections, in a “bottom up” fashion. §23.2.1. Gauss Quadrature Rule Information Recall from §17.3 that Gauss quadrature rules for isoparametric quadrilateral elements have the canonical form 0002 1 0002 1 0002 10002 1 p1 p2 . 00030003 F(ξ, η) dξ dη = dη F(ξ, η) dξ = wi w j F(ξi , η j ). (23.1) −1
−1
−1
−1
i=1 j=1
Here F = hBT EB J is the matrix to be integrated, whereas p1 and p2 are the number of Gauss points in the ξ and η directions, respectively. Often, but not always, the same number p = p1 = p2 is chosen in both directions. A formula with p1 = p2 is called an isotropic integration rule because directions ξ and η are treated alike. 23–3
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
QuadGaussRuleInfo is an application independent module that implements the twodimensional product Gauss rules with 1 through 5 points in each direction. The number of points in each direction may be the same or different. Usage of this module was described in detail in §17.3.4. For the readers convenience it is listed, along with its subordinate module LineGaussRuleInfo, in Figure 23.3. This module is classified as application independent since it can reused for any quadrilateral element.
APPLICATION DEPENDENT Element Stiffness Module
Shape Function Module
Gauss Quadrature Information Module
APPLICATION INDEPENDENT
Figure 23.2. Organization of element stiffness modules into application-dependent and application-independent levels.
QuadGaussRuleInfo[{rule_,numer_},point_]:= Module[ {ξ ,η,p1,p2,i,j,w1,w2,m,info={{Null,Null},0}}, If [Length[rule]2, {p1,p2}=rule, p1=p2=rule]; If [p1<0, Return[QuadNonProductGaussRuleInfo[ {-p1,numer},point]]]; If [Length[point]2, {i,j}=point, m=point; j=Floor[(m-1)/p1]+1; i=m-p1*(j-1) ]; {ξ ,w1}= LineGaussRuleInfo[{p1,numer},i]; {η,w2}= LineGaussRuleInfo[{p2,numer},j]; info={{ξ ,η},w1*w2}; If [numer, Return[N[info]], Return[Simplify[info]]]; ]; LineGaussRuleInfo[{rule_,numer_},point_]:= Module[ {g2={-1,1}/Sqrt[3],w3={5/9,8/9,5/9}, g3={-Sqrt[3/5],0,Sqrt[3/5]}, w4={(1/2)-Sqrt[5/6]/6, (1/2)+Sqrt[5/6]/6, (1/2)+Sqrt[5/6]/6, (1/2)-Sqrt[5/6]/6}, g4={-Sqrt[(3+2*Sqrt[6/5])/7],-Sqrt[(3-2*Sqrt[6/5])/7], Sqrt[(3-2*Sqrt[6/5])/7], Sqrt[(3+2*Sqrt[6/5])/7]}, g5={-Sqrt[5+2*Sqrt[10/7]],-Sqrt[5-2*Sqrt[10/7]],0, Sqrt[5-2*Sqrt[10/7]], Sqrt[5+2*Sqrt[10/7]]}/3, w5={322-13*Sqrt[70],322+13*Sqrt[70],512, 322+13*Sqrt[70],322-13*Sqrt[70]}/900, i=point,p=rule,info={{Null,Null},0}}, If [p1, info={0,2}]; If [p2, info={g2[[i]],1}]; If [p3, info={g3[[i]],w3[[i]]}]; If [p4, info={g4[[i]],w4[[i]]}]; If [p5, info={g5[[i]],w5[[i]]}]; If [numer, Return[N[info]], Return[Simplify[info]]]; ]; Figure 23.3. Module to get Gauss-product quadrature information for a quadrilateral.
23–4
23–4
23–5
§23.2
BILINEAR QUADRILATERAL STIFFNESS MATRIX
Quad4IsoPShapeFunDer[ncoor_,qcoor_]:= Module[ {Nf,dNx,dNy,dNξ ,dNη,i,J11,J12,J21,J22,Jdet,ξ ,η,x,y}, {ξ ,η}=qcoor; Nf={(1-ξ )*(1-η),(1+ξ )*(1-η),(1+ξ )*(1+η),(1-ξ )*(1+η)}/4; dNξ ={-(1-η), (1-η),(1+η),-(1+η)}/4; dNη= {-(1-ξ ),-(1+ξ ),(1+ξ ), (1-ξ )}/4; x=Table[ncoor[[i,1]],{i,4}]; y=Table[ncoor[[i,2]],{i,4}]; J11=dNξ .x; J12=dNξ .y; J21=dNη.x; J22=dNη.y; Jdet=Simplify[J11*J22-J12*J21]; dNx= ( J22*dNξ -J12*dNη)/Jdet; dNx=Simplify[dNx]; dNy= (-J21*dNξ +J11*dNη)/Jdet; dNy=Simplify[dNy]; Return[{Nf,dNx,dNy,Jdet}] ]; Figure 23.4. Shape function module for 4-node bilinear quadrilateral.
§23.2.2. Shape Function Evaluation Quad4IsoPShapeFunDer is an application independent module that computes the shape functions Nie , i = 1, 2, 3, 4 and its x-y partial derivatives at the sample integration points. The logic, listed in Figure 23.4, is straightforward and follows closely the description of Chapter 17. The arguments of the module are the {x, y} quadrilateral corner coordinates, which are passed in ncoor, and the two quadrilateral coordinates {ξ, η}, which are passed in qcoor. The former have the same configuration as described for the element stiffness module below. The quadrilateral coordinates define the element location at which the shape functions and their derivatives are to be evaluated. For the stiffness formation these are Gauss points, but for strain and stress computations these may be other points, such as corner nodes. Quad4IsoPShapeFunDer returns the two-level list { Nf,Nx,Ny,Jdet }, in which the first three are 4-entry lists. List Nf collects the shape function values, Nx the shape function x-derivatives, Ny the shape function y-derivatives, and Jdet is the Jacobian determinant called J in Chapters 17. §23.2.3. Element Stiffness Module Quad4IsoPMembraneStiffness computes the stiffness matrix of a four-noded isoparametric quadrilateral element in plane stress. The module configuration is typical of isoparametric elements in any number of dimensions. It follows closely the procedure outlined in Chapter 17. The module logic is listed in Figure 23.5. The statements at the bottom of the module box (not shown in that Figure) test it for specific configurations. The module is invoked as Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,th,options] The arguments are: ncoor
Quadrilateral node coordinates arranged in two-dimensional list form: { { x1,y1 },{ x2,y2 },{ x3,y3 },{ x4,y4 } }. 23–5
(23.2)
23–6
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
Quad4IsoPMembraneStiffness[ncoor_,Emat_,th_,options_]:= Module[{i,k,p=2,numer=False,h=th,qcoor,c,w,Nf, dNx,dNy,Jdet,Be,Ke=Table[0,{8},{8}]}, If [Length[options]2, {numer,p}=options,{numer}=options]; If [p<1||p>4, Print['p out of range']; Return[Null]]; For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}=Quad4IsoPShapeFunDer[ncoor,qcoor]; If [Length[th]4, h=th.Nf]; c=w*Jdet*h; Be={Flatten[Table[{dNx[[i]], 0},{i,4}]], Flatten[Table[{0, dNy[[i]]},{i,4}]], Flatten[Table[{dNy[[i]],dNx[[i]]},{i,4}]]}; Ke+=Simplify[c*Transpose[Be].(Emat.Be)]; ]; Return[Simplify[Ke]] ]; Figure 23.5. Element stiffness formation module for 4-node bilinear quadrilateral.
Emat
A two-dimensional list storing the 3 × 3 plane stress matrix of elastic moduli: 0004 E=
E 11 E 12 E 13
E 12 E 22 E 23
E 13 E 23 E 33
0005 (23.3)
arranged as { { E11,E12,E33 },{ E12,E22,E23 },{ E13,E23,E33 } }. Must be symmetric. If the material is isotropic with elastic modulus E and Poisson’s ratio ν, this matrix becomes 0005 0004 1 ν 0 E E= (23.4) ν 1 0 1 − ν 2 0 0 1 (1 − ν) 2
th
The plate thickness specified either as a four-entry list: { h1,h2,h3,h4 } or as a scalar: h. The first form is used to specify an element of variable thickness, in which case the entries are the four corner thicknesses and h is interpolated bilinearly. The second form specifies uniform thickness.
options
Processing options. This list may contain two items: { numer,p } or one: { numer }. numer is a logical flag with value True or False. If True, the computations are done in floating point arithmetic. For symbolic or exact arithmetic work set numer to False.1 p specifies the Gauss product rule to have p points in each direction. p may be 1 through 4. For rank sufficiency, p must be 2 or higher. If p is 1 the element will be rank deficient by two.2 If omitted p = 2 is assumed.
1
The reason for this option is speed. A symbolic or exact computation can take orders of magnitude more time than a floating-point evaluation. This becomes more pronounced as elements get more complicated.
2
The rank of an element stiffness is discussed in Chapter 19.
23–6
23–7
§23.3 4
(a)
TEST OF BILINEAR QUADRILATERAL
3
4
(b)
a
a
3
a 1
2
2a
1
2a
2
Figure 23.6. Test quadrilateral element geometries.
The module returns Ke as an 8 × 8 symmetric matrix pertaining to the following arrangement of nodal displacements: ue = [ u x1
u y1
u x2
u y2
u x3
u y3
u x4
u y4 ]T .
(23.5)
§23.3. Test of Bilinear Quadrilateral The stiffness module is tested on the two quadrilateral geometries shown in Figure 23.6. Both elements have unit thickness and isotropic material. The left one is a rectangle of base 2a and height a. The right one is a right trapezoid with base 2a, top width a and height a. The two geometries will be used to illustrate the effect of the numerical integration rule. §23.3.1. Test of Rectangular Geometry The test statements of Figure 23.7 compute and print the stiffness of the rectangular element shown in Figure 23.6(a). This is a rectangle of base 2a and height a. The plate has unit thickness and isotropic material with E = 96 and ν = 1/3, giving the stress-strain constitutive matrix 0004 E=
108 36 0
36 108 0
0 0 36
0005 (23.6)
Using a 2 × 2 Gauss integration rule returns the stiffness matrix 
 42 18 −6 0 −21 −18 −15 0 78 0 30 −18 −39 0 −69   18   0 42 −18 −15 0 −21 18   −6   30 −18 78 0 −69 18 −39   0 e K =  0 42 18 −6 0  −21 −18 −15   0 −69 18 78 0 30   −18 −39   −15 0 −21 18 −6 0 42 −18 0 −69 18 −39 0 30 −18 78
(23.7)
Note that the rectangle dimension a does not appear in (23.7). This is a general property: the stiffness matrix of plane stress elements is independent of inplane dimension scalings. This follows from the fact that entries of the strain-displacement matrix B have dimensions 1/L, where L denotes 23–7
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
23–8
ClearAll[Em,nu,a,b,e,h,p,numer]; h=1; Em=96; nu=1/3; (* isotropic material *) Emat=Em/(1-nu^2)*{{1,nu,0},{nu,1,0},{0,0,(1-nu)/2}}; Print['Emat=',Emat//MatrixForm]; ncoor={{0,0},{2*a,0},{2*a,a},{0,a}}; (* 2:1 rectangular geometry *) p=2;(* 2 x 2 Gauss rule *)numer=False;(* exact symbolic arithmetic *) Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,h,{numer,p}]; Ke=Simplify[Chop[Ke]]; Print['Ke=',Ke//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[N[Ke]],.0000001]];
Figure 23.7. Driver for stiffness calculation of rectangular element of Figure 23.6(a) for 2 × 2 Gauss integration rule.
a characteristic inplane length. Consequently entries of BT B have dimension 1/L 2 . Integration over the element area cancels out L 2 . Using a higher order Gauss integration rule, such as 3 × 3 and 4 × 4, reproduces exactly (23.7). This is a property characteristic of the rectangular geometry, since in that case the entries of B vary linearly in ξ and η, and J is constant. Therefore the integrand h BT E BJ is at most quadratic in ξ and η, and 2 Gauss points in each direction suffice to compute the integral exactly. Using a 1 × 1 rule yields a rank-deficiency matrix, a result illustrated in detail in §23.2.2. The stiffness matrix (23.7) has the eigenvalues [ 223.64
90
78
46.3603
42
0
0
0]
(23.8)
which verifies that Ke has the correct rank of five (8 freedoms minus 3 rigid body modes). §23.3.2. Test of Trapezoidal Geometry The trapezoidal element geometry of Figure 23.6(b) is used to illustrate the effect of changing the p × p Gauss integration rule. Unlike the rectangular case, the element stiffness keeps changing as p is varied from 1 to 4. The element is rank sufficient, however, for p ≥ 2 in agreement with the analysis of Chapter 19. The computations are driven with the script shown in Figure 23.8. The value of p is changed in a loop. The flag numer is set to True to use floating-point computation for speed (see Remark 23.1). The computed entries of Ke are transformed to the nearest rational number (exact integers in this case) using the built-in function Rationalize. The strange value of E = 48×63×13×107 = 4206384, in conjunction with ν = 1/3, makes all entries of Ke exact integers when computed with the first 4 Gauss rules. This device facilitates visual comparison between the computed stiffness matrices:  1840293 1051596 −262899 −262899 −1840293 −1051596 262899 262899  3417687 −262899 1314495 −1051596 −3417687 262899 −1314495  −262899 1051596 −525798 262899 262899 −1051596 525798  1314495 −525798 1051596 262899 −1314495 525798 −1051596   −1051596 262899 262899 1840293 1051596 −262899 −262899   −3417687 262899 −1314495 1051596 3417687 −262899 1314495  262899 262899 −1051596 525798 −262899 −262899 1051596 −525798 262899 −1314495 525798 −1051596 −262899 1314495 −525798 1051596
 1051596  −262899  −262899 e K1×1 =   −1840293  −1051596 
23–8
(23.9)
23–9
§23.3
TEST OF BILINEAR QUADRILATERAL
ClearAll[Em,nu,h,a,p]; h=1; Em=48*63*13*107; nu=1/3; Emat=Em/(1-nu^2)*{{1,nu,0},{nu,1,0},{0,0,(1-nu)/2}}; ncoor={{0,0},{2*a,0},{a,a},{0,a}}; For [p=1,p<=4,p++, Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,h,{True,p}]; Ke=Rationalize[Ke,0.0000001]; Print['Ke=',Ke//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[N[Ke]],.0000001]] ]; Figure 23.8. Driver for stiffness calculation of trapezoidal element of Figure 23.6(b) for four Gauss integration rules.

Ke2×2
2062746 1092042 −485352 −303345 −1395387 −970704 −182007 182007   1092042 3761478 −303345 970704 −970704 −2730105 182007 −2002077   −485352 −303345 1274049 −485352 −182007 182007 −606690 606690   −303345 970704 −485352 1395387 182007 −2002077 606690 −364014   =  −1395387 −970704 −182007 182007 2730105 1213380 −1152711 −424683   −970704 −2730105 182007 −2002077 1213380 4792851 −424683 −60669    −182007 182007 −606690 606690 −1152711 −424683 1941408 −364014 182007 −2002077 606690 −364014 −424683 −60669 −364014 2426760

Ke3×3
2067026 1093326 −489632 −304629 −1386827 −968136 −190567 179439   1093326 3764046 −304629 968136 −968136 −2724969 179439 −2007213   −489632 −304629 1278329 −484068 −190567 179439 −598130 609258   −304629 968136 −484068 1397955 179439 −2007213 609258 −358878   =  −1386827 −968136 −190567 179439 2747225 1218516 −1169831 −429819   −968136 −2724969 179439 −2007213 1218516 4803123 −429819 −70941    −190567 179439 −598130 609258 −1169831 −429819 1958528 −358878 179439 −2007213 609258 −358878 −429819 −70941 −358878 2437032
(23.10)

Ke4×4
2067156 1093365 −489762 −304668 −1386567 −968058 −190827 179361   1093365 3764124 −304668 968058 −968058 −2724813 179361 −2007369   −489762 −304668 1278459 −484029 −190827 179361 −597870 609336   −304668 968058 −484029 1398033 179361 −2007369 609336 −358722   =  −1386567 −968058 −190827 179361 2747745 1218672 −1170351 −429975   −968058 −2724813 179361 −2007369 1218672 4803435 −429975 −71253    −190827 179361 −597870 609336 −1170351 −429975 1959048 −358722 179361 −2007369 609336 −358722 −429975 −71253 −358722 2437344
(23.11)
(23.12)
As can be seen entries change substantially in going from p = 1 to p = 2, then more slowly. The eigenvalues of these matrices are: Rule
Eigenvalues (scaled by 10−6 ) of Ke
1×1 2×2 3×3 4×4
8.77276 8.90944 8.91237 8.91246
3.68059 4.09769 4.11571 4.11627
2.26900 3.18565 3.19925 3.19966
0 2.64521 2.66438 2.66496
0 1.54678 1.56155 1.56199
0 0 0 0
0 0 0 0
0 0 0 0
(23.13)
The stiffness matrix computed by the one-point rule is rank deficient by two. The eigenvalues do not change appreciably after p = 2. Because the nonzero eigenvalues measure the internal energy 23–9
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
23–10
Quad4IsoPMembraneBodyForces[ncoor_,rho_,th_,options_,bfor_]:= Module[{i,k,p=2,numer=False,h=th, bx,by,bx1,by1,bx2,by2,bx3,by3,bx4,by4,bxc,byc,qcoor, c,w,Nf,dNx,dNy,Jdet,B,qctab,fe=Table[0,{8}]}, If [Length[options]2, {numer,p}=options, {numer}=options]; If [Length[bfor]2,{bx,by}=bfor;bx1=bx2=bx3=bx4=bx;by1=by2=by3=by4=by]; If [Length[bfor]4,{{bx1,by1},{bx2,by2},{bx3,by3},{bx4,by4}}=bfor]; If [p<1||p>4, Print['p out of range']; Return[Null]]; bxc={bx1,bx2,bx3,bx4}; byc={by1,by2,by3,by4}; For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}=Quad4IsoPShapeFunDer[ncoor,qcoor]; bx=Nf.bxc; by=Nf.byc; If [Length[th]4, h=th.Nf]; c=w*Jdet*h; bk=Flatten[Table[{Nf[[i]]*bx,Nf[[i]]*by},{i,4}]]; fe+=c*bk; ]; Return[fe] ]; Figure 23.9. Module for computation of consistent node forces from a given body force field.
taken up by the element in deformation eigenmodes, it can be seen that raising the order of the integration stiffens the element. Remark 23.1.
The formation of the trapezoidal element stiffness using floating-point computation by setting numer=True took 0.017, 0.083, 0.15 and 0.25 seconds for p = 1, 2, 3, 4, respectively, on a Mac G4/867. Changing numer=False to do exact computation increases the formation time to 0.033, 1.7, 4.4 and 44.6 seconds, respectively. (The unusually large value for p = 4 is due to the time spent in the simplification of the highly complex exact expressions produced by the Gauss quadrature rule.) This underscores the speed advantage of using floating-point arithmetic when exact symbolic and algebraic calculations are not required. §23.4.
*Consistent Node Forces for Body Force Field
The module Quad4IsoPMembraneBodyForces listed in Figure 23.9 computes the consistent force associated = {bx , b y } given over a 4-node iso-P quadrilateral in plane stress. The field is with a body force field b specified per unit of volume in componentwise form. For example if the element is subjected to a gravity acceleration field (self-weight) in the −y direction, bx = 0 and b y = −ρg, where ρ is the mass density. The arguments of the module are are exactly the same as for Quad4IsoPMembraneStiffness except for the following differences. mprop
Not used; retained as placeholder.
bfor
Body forces per unit volume. Specified as a two-item one-dimensional list: { bx,by }, or as a four-entry two-dimensional list: { bx1,by1 },{ bx2,by2 }, { bx3,by3 },{ bx4,by4 }. In the first form the body force field is taken to be constant over the element. The second form assumes body forces to vary over the element and specified by values at the four corners, from which the field is interpolated bilinearly.
23–10
23–11
§23.5
*RECOVERY OF CORNER STRESSES
Quad4IsoPMembraneStresses[ncoor_,Emat_,th_,options_,udis_]:= Module[{i,k,numer=False,qcoor,Nf, dNx,dNy,Jdet,Be,qctab,ue=udis,sige=Table[0,{4},{3}]}, qctab={{-1,-1},{1,-1},{1,1},{-1,1}}; numer=options[[1]]; If [Length[udis]4, ue=Flatten[udis]]; For [k=1, k<=Length[sige], k++, qcoor=qctab[[k]]; If [numer, qcoor=N[qcoor]]; {Nf,dNx,dNy,Jdet}=Quad4IsoPShapeFunDer[ncoor,qcoor]; Be={ Flatten[Table[{dNx[[i]], 0},{i,4}]], Flatten[Table[{0, dNy[[i]]},{i,4}]], Flatten[Table[{dNy[[i]],dNx[[i]]},{i,4}]]}; sige[[k]]=Emat.(Be.ue); ]; Return[sige] ]; Figure 23.10. Module for calculation of corner stresses.
The module returns fe as an 8 × 1 one dimensional array arranged { fx1,fy1,fx2,fy2,fx3,fy3, fx4,fy4 } to represent the vector fe = [ f x1 §23.5.
f y1
f x2
f y2
f x3
f y3
f x4
f y4 ]T .
(23.14)
*Recovery of Corner Stresses
Akthough the subject of stress recovery is treated in further detail in a later chapter, for completeness a stress computation module called Quad4IsoPMembraneStresses for the 4-node quad is listed in Figure 23.10. The arguments of the module are are exactly the same as for Quad4IsoPMembraneStiffness except for the following differences. fprop
Not used; retained as placeholder.
udis
The 8 corner displacements components. these may be specified as a 8-entry onedimensional list form: { ux1,uy1, ux2,uy2, ux3,uy3, ux4,uy4 }, or as a 4-entry two-dimensional list: { ux1,uy1 },{ ux2,uy2 },{ ux3,uy3 },{ ux4,uy4 }.
The module returns the corner stresses stored in a 4-entry, two-dimensional list: { { sigxx1,sigyy1,sigxy1 },{ sigxx2,sigyy2,sigxy2 }, { sigxx3,sigyy3,sigxy3 }, { sigxx4,sigyy4,sigxy4 } } to represent the stress array
0004
σe =
σx x1 σ yy1 σx y1
σx x2 σ yy2 σx y2
σx x3 σ yy3 σx y3
σx x4 σ yy4 σx y4
0005
(23.15)
The stresses are directly evaluated at the corner points without invoking any smoothing procedure. A more elaborated recovery scheme is presented in a later Chapter.
23–11
23–12
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
§23.6.
*Quadrilateral Coordinates of Given Point
The following inverse problem arises in some applications. Given a 4-node quadrilateral, defined by the Cartesian coordinates {xi , yi } of its corners, and an arbitrary point P(x P , y P ), find the quadrilateral coordinates ξ P , η P of P. In answering this question it is understood that the quadrilateral coordinates can be extended outside the element, as illustrated in Figure 23.11.
η=1
4
3
ξ=−1
ξ=1
P(ξ?,η?)
The governing equations are x P = x1 N1 +x2 N2 + x3 N3 + x4 N4 and y P = y1 N1 + y2 N2 + y3 N3 + y4 N4 , where N1 = 14 (1−ξ P )(1−η P ), etc. These bilinear equations are to be solved for {ξ P , η P }. Elimination of say, ξ P , leads to a quadratic equation in η P : aη2P + bη P + c = 0. It can be shown that b2 ≥ 4ac so there are two real roots: η1 and η2 . These can be back substituted to get ξ1 and ξ2 . Of the two solutions: {ξ1 , η1 } and {ξ2 , η2 } the one closest to ξ = 0, η = 0 is to be taken.
2 1
η=−1
Figure 23.11. Quadrilateral coordinates can be extended outside the element to answer the problem posed in §23.6. The six yellow-filled circles identify the four corners plus the intersections of the opposite sides. This six-point set defines the so-called complete quadrilateral, which is important in projective geometry. The evolute of the coordinate lines is a parabola.
Although seemingly straightforward, the process is prone to numerical instabilities. For example, if the quadrilateral becomes a rectangle or parallelogram, the quadratic equations degenerate to linear, and one of the roots takes off to ∞. In floating point arithmetic severe cancellation can occur in the other root. A robust numerical algorithm, which works stably for any geometry, is obtained by eliminating ξ and η in√ turn, getting = b/(b + b2 − 4ac), the minimum-modulus root of aη2P + bη P + c = 0 with the stable formula.3 ηmin P forming the other quadratic equation, and computing its minimum-modulus root the same way. In addition, x P and y P are referred to the quadrilateral center as coordinate origin. The resulting algorithm can be presented as follows. Given {x1 , y1 , . . . x4 , y4 } and {x P , y P }, compute xb = x1 − x2 + x3 − x4 ,
yb = y1 − y2 + y3 − y4 ,
xce = x1 − x2 − x3 + x4 ,
yce = y1 − y2 − y3 + y4 ,
J1 = (x3 − x4 )(y1 − y2 ) − (x1 − x2 )(y3 − y4 ), x0 =
1 4 (x 1
+ x2 + x3 + x4 ),
bξ = A − x P0 yb + y P0 xb , cη = x P0 yce − y P0 xce ,
xcx = x1 + x2 − x3 − x4 ,
y0 =
1 4 (y1
A=
1 2 ((x 3
ycx = y1 + y2 − y3 − y4 ,
− x1 )(y4 − y2 ) − (x4 − x2 )(y3 − y1 )),
J2 = (x2 − x3 )(y1 − y4 ) − (x1 − x4 )(y2 − y3 ),
+ y2 + y3 + y4 ),
x P0 = x P − x0 ,
y P0 = y P − y0 ,
bη = −A − x P0 yb + y P0 xb ,
ξP = −
2cξ
,
bξ2 − 2J1 cξ − bξ
cξ = x P0 ycx − y P0 xcx , 2cη ηP = . bη2 + 2J2 cη − bη (23.16)
One common application is to find whether P is inside the quadrilateral: if both ξ P and η P are in the range [−1, 1] the point is inside, else outside. This occurs, for example, in relating experimental data from given sensor locations4 to an existing FEM mesh. A Mathematica module that implements (23.16) is listed in Figure 23.12. 3
See Section 5.6 of [133].
4
While at Boeing in 1969 the writer had to solve a collocation problem of this nature, although in three dimensions. Pressure data measured at a wind tunnel had to be transported to an independently constructed FEM quadrilateral mesh modeling the wing skin.
23–12
23–13
§23.
References
QuadCoordinatesOfPoint[{{x1_,y1_},{x2_,y2_},{x3_,y3_}, {x4_,y4_}},{x_,y_}]:= Module[{A,J0,J1,J2, xb=x1-x2+x3-x4,yb=y1-y2+y3-y4,xcξ =x1+x2-x3-x4,ycξ =y1+y2-y3-y4, xcη=x1-x2-x3+x4,ycη=y1-y2-y3+y4,bξ ,bη,cξ ,cη, x0=(x1+x2+x3+x4)/4,y0=(y1+y2+y3+y4)/4,dx,dy,ξ ,η}, J0=(x3-x1)*(y4-y2)-(x4-x2)*(y3-y1); A=J0/2; J1=(x3-x4)*(y1-y2)-(x1-x2)*(y3-y4); J2=(x2-x3)*(y1-y4)-(x1-x4)*(y2-y3); dx=x-x0; dy=y-y0; bξ =A-dx*yb+dy*xb; bη=-A-dx*yb+dy*xb; cξ = dx*ycξ -dy*xcξ ; cη=dx*ycη-dy*xcη; ξ =2*cξ /(-Sqrt[bξ ^2-2*J1*cξ ]-bξ ); η=2*cη/( Sqrt[bη^2+2*J2*cη]-bη); Return[{ξ ,η}]]; Figure 23.12. A Mathematica module that implements the algorithm (23.16).
Notes and Bibliography For an outline of the history of the 4-node quadrilateral, see Notes and Bibliography in Chapter 17. The element is called the Taig quadrilateral in the early FEM literature, recognizing his developer [156]. This paper actually uses the exactly integrated stiffness matrix. Gauss numerical integration for quadrilaterals was advocated by Irons [95,98], who changed the range of the quadrilateral coordinates to [−1, +1] to fit tabulated Gauss rules. References Referenced items moved to Appendix R.
23–13
Chapter 23: IMPLEMENTATION OF ISO-P QUADRILATERAL ELEMENTS
23–14
Homework Exercises for Chapter 23 Implementation of Iso-P Quadrilateral Elements EXERCISE 23.1 [C:15] Figures E23.1–2 show the Mathematica implementation of the stiffness modules for the 5-node, “bilinear+bubble” iso-P quadrilateral of Figure E18.3. Module Quad5IsoPMembraneStiffness returns the 10 × 10 stiffness matrix whereas module Quad5IsoPShapeFunDer returns shape function values and Cartesian derivatives. (The Gauss quadrature module is reused.) Both modules follow the style of the 4-node quadrilateral implementation listed in Figures 23.4–5. The only differences in argument lists is that ncoor has five node coordinates: { { x1,y1 },{ x2,y2 },{ x3,y3 },{ x4,y4 },{ x5,y5 } }, and that a variable plate thickness in fprop (one of the two possible formats) is specified as { h1,h2,h3,h4,h5 }.
Quad5IsoPMembraneStiffness[ncoor_,Emat_,th_,options_]:= Module[{i,j,k,p=2,numer=False,h=th,qcoor,c,w,Nf, dNx,dNy,Jdet,B,Ke=Table[0,{10},{10}]}, If [Length[options]2, {numer,p}=options, {numer}=options]; If [p<1||p>4, Print['p out of range']; Return[Null]]; For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}=Quad5IsoPShapeFunDer[ncoor,qcoor]; If [Length[th]>0, h=th.Nf]; c=w*Jdet*h; B={ Flatten[Table[{dNx[[i]], 0},{i,5}]], Flatten[Table[{0, dNy[[i]]},{i,5}]], Flatten[Table[{dNy[[i]],dNx[[i]]},{i,5}]]}; Ke+=Simplify[c*Transpose[B].(Emat.B)]; ]; Return[Ke]; ]; Figure E23.1. Stiffness module for the 5-node “bilinear+bubble” iso-P quadrilateral.
Quad5IsoPShapeFunDer[ncoor_,qcoor_]:= Module[ {Nf,dNx,dNy,dNξ ,dNη,Nb,dNbξ ,dNbη,J11,J12,J21,J22,Jdet,ξ ,η,x,y}, {ξ ,η}=qcoor; Nb=(1-ξ ^2)*(1-η^2); (* Nb: node-5 'bubble' function *) dNbξ =2*ξ (η^2-1); dNbη=2*η*(ξ ^2-1); Nf= { ((1-ξ )*(1-η)-Nb)/4,((1+ξ )*(1-η)-Nb)/4, ((1+ξ )*(1+η)-Nb)/4,((1-ξ )*(1+η)-Nb)/4, Nb}; dNξ ={-(1-η+dNbξ )/4, (1-η-dNbξ )/4, (1+η-dNbξ )/4,-(1+η+dNbξ )/4, dNbξ }; dNη={-(1-ξ +dNbη)/4,-(1+ξ +dNbη)/4, (1+ξ -dNbη)/4, (1-ξ -dNbη)/4, dNbη}; x=Table[ncoor[[i,1]],{i,5}]; y=Table[ncoor[[i,2]],{i,5}]; J11=dNξ .x; J12=dNξ .y; J21=dNη.x; J22=dNη.y; Jdet=Simplify[J11*J22-J12*J21]; dNx= ( J22*dNξ -J12*dNη)/Jdet; dNx=Simplify[dNx]; dNy= (-J21*dNξ +J11*dNη)/Jdet; dNy=Simplify[dNy]; Return[{Nf,dNx,dNy,Jdet}] ]; Figure E23.2. The shape function module for the 5-node “bilinear+bubble” iso-P quadrilateral.
Test Quad5IsoPMembraneStiffness for the 2:1 rectangular element studied in §23.3.1, with node 5 placed at the element center. Use Gauss rules 1 × 1, 2 × 2 and 3 × 3. Take E = 96×30 = 2880 in lieu of E = 96 to
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Exercises
Quad5IsoPMembraneCondStiffness[Ke5_]:= Module[{i,j,k,n,c,Ke=Ke5,Kc=Table[0,{8},{8}]}, For [n=10,n>=9,n--, For [i=1,i<=n-1,i++, c=Ke[[i,n]]/Ke[[n,n]]; For [j=1,j<=i,j++, Ke[[j,i]]=Ke[[i,j]]=Ke[[i,j]]-c*Ke[[n,j]]; ]]]; For [i=1,i<=8,i++, For [j=1,j<=8,j++, Kc[[i,j]]=Ke[[i,j]]]]; Return[Kc] ]; Figure E23.3. A mystery module for Exercise 23.2.
get exact integer entries in Ke for all Gauss rules while keeping ν = 1/3 and h = 1. Report on which rules give rank sufficiency. Partial result: K 22 = 3380 and 3588 for the 2 × 2 and 3 × 3 rules, respectively. EXERCISE 23.2 [D:10] Module Quad5IsoPMembraneCondStiffness in Figure E23.3 is designed to receive, as only argument, the 10 × 10 stiffness Ke computed by Quad5IsoPMembraneStiffness, and returns a smaller (8 × 8) stiffness matrix. State what the function of the module is but do not describe programming details. EXERCISE 23.3 [C:20] Repeat Exercise 17.3 for the problem illustrated in Figure E17.4, but with the 5-node “bilinear+bubble” iso-P quadrilateral as the 2D element that models the plane beam. Skip item (a). Use the modules of Figures E23.1–3 to do the following. Form the 10 × 10 stiffness matrix Ke5 using Quad5IsoPMembraneStiffness with p = 2 and numer=False. Insert this Ke5 into Quad5IsoPMembraneCondStiffness, which returns a 8 × 8 stiffness Ke. Stick this Ke into equations (E17.6) and (E17.7) to get Uquad . Show that the energy ratio is
000e
000f
γ 2 (1 + ν) 2 + γ 2 (1 − ν) Uquad r= = . Ubeam (1 + γ 2 )2
(E23.1)
Compare this to the energy ratio (E17.8) for γ = 1/10 and ν = 0 to conclude that shear locking has not been eliminated, or even mitigated, by the injection of the bubble shape functions associated with the interior node.5 EXERCISE 23.4 [C:25] Implement the 9-node biquadratic element for plane stress to get its 18 × 18 stiffness matrix. Follow the style of Figures 23.3–4 or E23.1–2. (The Gauss quadrature module may be reused without change.) Test it for the 2:1 rectangular element studied in §23.3.1, with nodes 5–8 placed at the side midpoints, and node 9 at the element center. For the elastic modulus take E = 96×39×11×55×7 = 15855840 instead of E = 96, along with ν = 1/3 and h = 1, so as to get exact integer entries in Ke . Use both 2 × 2 and 3 × 3 Gauss integration rules and show that the 2 × 2 rule produces a rank deficiency of 3 in the stiffness. (If the computation with num=False takes too long on a slow PC, set num=True and Rationalize entries as in Figure 23.8.) Partial result: K 11 = 5395390 and 6474468 for the 2 × 2 and 3 × 3 rules, respectively.
5
Even the addition of an infinite number of bubble functions to the 4-node iso-P quadrilateral will not cure shear locking. This “bubble futility” has been known since the late 1960s. But memories are short. Bubbles have been recently revived by some FEM authors for other application contexts, such as multiscale modeling.
23–15
24
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Implementation of Iso-P Triangular Elements
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Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
TABLE OF CONTENTS Page
§24.1. Introduction §24.2. Gauss Quadrature for Triangles §24.2.1. Requirements for Gauss Rules . §24.2.2. Superparametric Triangles . . §24.2.3. Arbitrary Iso-P Triangles . . §24.2.4. Implementation . . . . . . §24.3. Partial Derivative Computation §24.3.1. Triangle Coordinate Partials . §24.3.2. Solving the Jacobian System . . §24.4. The Quadratic Triangle §24.4.1. Shape Function Module . . . §24.4.2. Stiffness Module . . . . . . §24.4.3. Test on Straight-Sided Triangle §24.4.4. Test on Highly Distorted Triangle §24.5. *The Cubic Triangle §24.5.1. *Shape Function Module . . §24.5.2. *Stiffness Module . . . . . §24.5.3. *Test Element . . . . . . §24. Notes and Bibliography . . . . . . . . . . . . §24. References . . . . . . . . . . . §24. Exercises . . . . . . . . . . . .
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24–3 24–3 24–3 24–4 24–5 24–6 24–6 24–7 24–8 24–9 24–9 24–11 24–12 24–12 24–14 24–14 24–16 24–17 24–18 24–19 24–20
24–3
§24.2 GAUSS QUADRATURE FOR TRIANGLES
§24.1. Introduction This Chapter continues with the computer implementation of two-dimensional finite elements. It covers the programming of isoparametric triangular elements for the plane stress problem. Triangular elements bring two sui generis implementation quirks with respect to quadrilateral elements: (1) The numerical integration rules for triangles are not product of one-dimensional Gauss rules, as in the case of quadrilaterals. They are instead specialized to the triangle geometry. (2) The computation of x-y partial derivatives and the element-of-area scaling by the Jacobian determinant must account for the fact that the triangular coordinates ζ1 , ζ2 and ζ3 do not form an independent set. We deal with these issues in the next two sections. §24.2. Gauss Quadrature for Triangles The numerical integration schemes for quadrilaterals introduced in §17.3 and implemented in §23.2 are built as “tensor products” of two one-dimensional Gauss formulas. On the other hand, Gauss rules for triangles are not derivable from one-dimensional rules, and must be constructed especially for the triangular geometry. §24.2.1. Requirements for Gauss Rules Gauss quadrature rules for triangles must possess triangular symmetry in the following sense:
If the sample point (ζ1 , ζ2 , ζ3 ) is present in a Gauss integration rule with weight w, then all other points obtainable by permuting the three triangular coordinates arbitrarily must appear in that rule, and have the same weight.
(24.1)
This constraint guarantees that the result of the quadrature process will not depend on element node numbering.1 If ζ1 , ζ2 , and ζ3 are different, (24.1) forces six equal-weight sample points to be present in the rule, because 3! = 6. If two triangular coordinates are equal, the six points coalesce to three, and (24.1) forces three equal-weight sample points to be present. Finally, if the three coordinates are equal (which can only happen for the centroid ζ1 = ζ2 = ζ3 = 1/3), the six points coalesce to one.2 Additional requirements for a Gauss rule to be numerically acceptable are:
All sample points must be inside the triangle (or on the triangle boundary) and all weights must be positive.
(24.2)
This is called a positivity condition. It insures that the element internal energy evaluated by numerical quadrature is nonnegative definite. 1
It would disconcerting to users, to say the least, to have the FEM solution depend on how nodes are numbered.
2
It follows that the number of sample points in triangle Gauss quadrature rules must be of the form 6i + 3 j + k, where i and j are nonnegative integers and k is 0 or 1. Consequently there are no rules with 2, 5 or 8 points.
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Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
(a) rule=1 degree=1
(b) rule=3 degree=2
(c) rule=−3 degree=2
0.33333
0.33333
0.33333
1.0000 0.33333
0.33333 0.33333
(d) rule=6 degree=4
(e) rule=7 degree=5
0.10995
0.22338
0.12594
0.13239
0.22338
0.13239 0.22500
0.10995
0.22338
0.10995
0.12594
0.13239
0.12594
Figure 24.1. Location of sample points (dark circles) of five Gauss quadrature rules for straight sided (superparametric) 6-node triangles. Weight written to 5 places near each sample point; sample-point circle areas are proportional to weight.
A rule is said to be of degree n if it integrates exactly all polynomials in the triangular coordinates of order n or less when the Jacobian determinant is constant, and there is at least one polynomial of order n + 1 that is not exactly integrated by the rule. Remark 24.1. The positivity requirement (24.2) is automatically satisfied in quadrilaterals by using Gauss product
rules, since the points are always inside while weights are positive. Consequently it was not necessary to call attention to it. On the other hand, for triangles there are Gauss rules with as few as 4 points that violate positivity.
§24.2.2. Superparametric Triangles We first consider superparametric straight-sided triangles geometry defined by the three corner nodes. Over such triangles the Jacobian determinant defined below is constant. The five simplest Gauss rules that satisfy the requirements (24.1) and (24.2) have 1, 3, 3, 6 and 7 points, respectively. The two rules with 3 points differ in the location of the sample points. The five rules are depicted in Figure 24.1 over 6-node straight-sided triangles; for such triangles to be superparametric the side nodes must be located at the midpoint of the sides. One point rule. The simplest Gauss rule for a triangle has one sample point located at the centroid. For a straight sided triangle, 0002 1 F(ζ1 , ζ2 , ζ3 ) d ≈ F( 13 , 13 , 13 ), (24.3) A e where A is the triangle area 0002
0003
1 1 d = 2 det x1 A= e y1
1 x2 y2
1 x3 y3
0004 =
1 2
0005
0006 (x2 y3 − x3 y2 ) + (x3 y1 − x1 y3 ) + (x1 y2 − x2 y1 ) . (24.4)
This rule is illustrated in Figure 24.1(a). It has degree 1, meaning that it integrates exactly up to linear polynomials in {ζ1 , ζ2 , ζ3 }. For example, F = 4 − ζ1 + 2ζ2 − ζ3 is exactly integrated by (24.3). 24–4
24–5
§24.2 GAUSS QUADRATURE FOR TRIANGLES (a) rule=1
(b) rule=3
(c) rule=−3
0.33333
0.33333
1.0000 0.33333
0.33333 0.33333
0.33333
(d) rule=6
0.10995
(e) rule=7
0.22338
0.13239 0.22500
0.22338 0.22338
0.12594 0.13239
0.13239
0.10995
0.12594
0.12594
0.10995
Figure 24.2. Location of sample points (dark circles) of five Gauss quadrature rules for curved sided 6-node triangles. Weight written to 5 places near each sample point; sample-point circle areas are proportional to weight.
Three Point Rules. The next two rules in order of simplicity contain three sample points: 0002 1 F(ζ1 , ζ2 , ζ3 ) d ≈ 13 F( 23 , 16 , 16 ) + 13 F( 16 , 23 , 16 ) + 13 F( 16 , 16 , 23 ). (24.5) A e 0002 1 F(ζ1 , ζ2 , ζ3 ) d ≈ 13 F( 12 , 12 , 0) + 13 F(0, 12 , 12 ) + 13 F( 12 , 0, 12 ). (24.6) A e These are depicted in Figures 24.1(b,c). Both rules are of degree 2; that is, they integrate exactly up to quadratic polynomials in the triangular coordinates. For example, the function F = 6 + ζ1 + 3ζ3 + ζ22 − ζ32 + 3ζ1 ζ3 is integrated exactly by either rule. Formula (24.6) is called the midpoint rule. Six and Seven Point Rules. There is a 4-point rule of degree 3, but it has a negative weight and so violates (24.2). There are no symmetric rules with 5 points. The next useful rules have six and seven points. There is a 6-point rule of degree 4 and a 7-point rule of degree 5, which integrate exactly up to quartic and quintic polynomials in {ζ1 , ζ2 , ζ3 }, respectively. The 7-point rule includes the centroid as sample point. The abcissas and weights are expressable as rational combinations of square roots of integers and fractions. The expressions are listed in the Mathematica implementation discussed in §24.2.4. The sample point configurations are depicted in Figure 24.1(d,e). §24.2.3. Arbitrary Iso-P Triangles If the triangle has variable metric, as in the curved sided 6-node triangle geometries shown in Figure 24.2, the foregoing formulas need adjustment because the element of area d becomes a function of position. Consider the more general case of an isoparametric element with n nodes and shape functions Ni . In §24.3 it is shown that the differential area element is given by   1 1 1 n n n 
∂ Ni
∂ Ni
∂ Ni    x x xi i i   (24.7) d = J dζ1 dζ2 dζ3 , J = 12 det  i=1 ∂ζ1 i=1 ∂ζ2 i=1 ∂ζ3   n  n n
 ∂ Ni ∂ Ni ∂ Ni  yi yi yi ∂ζ1 i=1 ∂ζ2 i=1 ∂ζ3 i=1 24–5
24–6
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS TrigGaussRuleInfo[{rule_,numer_},point_]:= Module[ {zeta,p=rule,i=point,g1,g2, info={{Null,Null,Null},0} }, If [p 1, info={{1/3,1/3,1/3},1}]; If [p 3, info={{1,1,1}/6,1/3}; info[[1,i]]=2/3]; If [p-3, info={{1,1,1}/2,1/3}; info[[1,i]]=0 ]; If [p 6, g1=(8-Sqrt[10]+Sqrt[38-44*Sqrt[2/5]])/18; g2=(8-Sqrt[10]-Sqrt[38-44*Sqrt[2/5]])/18; If [i<4, info={{g1,g1,g1},(620+Sqrt[21312553320*Sqrt[10]])/3720}; info[[1,i]]=1-2*g1]; If [i>3, info={{g2,g2,g2},(620-Sqrt[21312553320*Sqrt[10]])/3720}; info[[1,i-3]]=1-2*g2]]; If [p 7, g1=(6-Sqrt[15])/21; g2=(6+Sqrt[15])/21; If [i<4, info={{g1,g1,g1},(155-Sqrt[15])/1200}; info[[1,i]]= 1-2*g1]; If [i>3&&i<7, info={{g2,g2,g2},(155+Sqrt[15])/1200}; info[[1,i-3]]=1-2*g2]; If [i7, info={{1/3,1/3,1/3},9/40} ]]; If [numer, Return[N[info]], Return[Simplify[info]]]; ];
Figure 24.3. Module to get triangle Gauss quadrature rule information.
Here J is the Jacobian determinant, which plays the same role as J in the isoparametric quadrilaterals. If the metric is simply defined by the 3 corners, as in Figure 24.1, the geometry shape functions are N1 = ζ1 , N2 = ζ2 and N3 = ζ3 . Then the foregoing determinant reduces to that of (24.4), and J = A everywhere. But for general (curved) geometries J = J (ζ1 , ζ2 , ζ3 ), and the triangle area A cannot be factored out of the integration rules. For example the one point rule becomes 0002 F(ζ1 , ζ2 , ζ3 ) d ≈ J ( 13 , 13 , 13 ) F( 13 , 13 , 13 ). (24.8) e
whereas the midpoint rule becomes 0002 F(ζ1 , ζ2 , ζ3 ) d ≈ 13 J ( 12 , 12 , 0) F( 12 , 12 , 0) + 13 J (0, 12 , 12 ) F(0, 12 , 12 ) + 13 J ( 12 , 0, 12 ) F( 12 , 0, 12 ). e
(24.9) These can be expressed more compactly by saying that the Gauss integration rule is applied to J F. §24.2.4. Implementation The five rules pictured in Figures 24.1 and 24.2 are implemented in the module TrigGaussRuleInfo listed in Figure 24.3. The module is invoked as { { zeta1,zeta2,zeta3 },w } = TrigGaussRuleInfo[{ rule,numer },point]
(24.10)
The module has three arguments: rule, numer and i. The first two are grouped in a two-item list. Argument rule, which can be 1, 3, −3, 6 or 7, defines the integration formula as follows. Abs[rule] is the number of sample points. Of the two 3-point choices, if rule is −3 the midpoint rule is picked, else if +3 the 3-interior point rule is chosen. Logical flag numer is set to True or False to request floating-point or exact information, respectively Argument point is the index of the sample point, which may range from 1 through Abs[rule]. The module returns the list {{ζ1 , ζ2 , ζ3 }, w}, where ζ1 , ζ2 , ζ3 are the triangular coordinates of the sample point, and w is the integration weight. For example, TrigGaussRuleInfo[{ 3,False },1] returns { { 2/3,1/6,1/6 },1/3 }. If rule is not implemented, the module returns { { Null,Null,Null },0 }. 24–6
24–7
§24.3 PARTIAL DERIVATIVE COMPUTATION
§24.3. Partial Derivative Computation The calculation of Cartesian partial derivatives is illustrated in this section for the 6-node triangle shown in Figure 24.4. The results are applicable, however, to iso-P triangles with any number of nodes. The element geometry is defined by the corner coordinates {xi , yi }, with i = 1, 2, . . . 6. Corners are numbered 1,2,3 in counterclockwise sense. Side nodes are numbered 4,5,6 opposite to corners 3,1,2, respectively. Side nodes may be arbitrarily located within positive Jacobian constraints as discussed in §19.4.2. The triangular coordinates are as usual denoted by ζ1 , ζ2 and ζ3 , which satisfy ζ1 + ζ2 + ζ3 = 1. The quadratic displacement field {u x (ζ1 , ζ2 , ζ3 ), u y (ζ1 , ζ2 , ζ3 )} is defined by the 12 node displacements {u xi , u yi }, i = 1, 2, . . . , 6, as per the iso-P quadratic interpolation formula (16.10–11). That formula is repeated here for convenience: 
  1 1 x x    1     y  =  y1    ux u x1 uy u y1
1 x2 y2 u x2 u y2
1 x3 y3 u x3 u y3
1 x4 y4 u x4 u y4
1 x5 y5 u x5 u y5
3
5
6
2 4 1 Figure 24.4. The 6-node iso-P triangle.
 1 x6   y6   u x6 u y6
 N1e  N2e   e  N3   e  N4   e N5 N6e 
(24.11)
in which the shape functions and their natural derivatives are     e      ζ1 (2ζ1 −1) N1 4ζ1 −1 0 0  N2e   ζ2 (2ζ2 −1)   0   4ζ2 −1   0   ∂NT  ∂NT   e      ∂NT  N   ζ (2ζ −1)   0   0   4ζ −1  = = =  3 . NT =  3e  =  3 3 , , ,  N4   4ζ1 ζ2  ∂ζ1  4ζ2  ∂ζ2  4ζ1  ∂ζ3  0      e      4ζ2 N5 4ζ2 ζ3 0 4ζ3 4ζ3 N6e 4ζ3 ζ1 0 4ζ1 (24.12) §24.3.1. Triangle Coordinate Partials In this and following sections the superscript e of shape functions will be omitted for brevity. The bulk of the shape function logic is concerned with the computation of the partial derivatives of the shape functions (24.12) with respect to x and y at any point in the element. For this purpose consider a generic scalar function w(ζ1 , ζ2 , ζ3 ) that is quadratically interpolated over the triangle by w = w1 N1 + w2 N2 + w3 N3 + w4 N4 + w5 N5 + w6 N6 = [ w1
w2
w3
w4
w5
w6 ] NT . (24.13)
Symbol w may stand for 1, x, y, u x or u y , which are interpolated in the iso-P representation (24.11), or other element-varying quantities such as thickness, temperature, etc. Taking partials of (24.13) with respect to x and y and applying the chain rule twice yields 000f
000e ∂ Ni ∂ζ1 ∂ Ni ∂ζ2 ∂ Ni ∂ζ3 ∂w ∂ Ni = wi = wi + + , ∂x ∂x ∂ζ1 ∂ x ∂ζ2 ∂ x ∂ζ3 ∂ x 000f (24.14)
000e ∂ Ni ∂ζ1 ∂w ∂ Ni ∂ Ni ∂ζ2 ∂ Ni ∂ζ3 = wi = wi + + , ∂y ∂y ∂ζ1 ∂ y ∂ζ2 ∂ y ∂ζ3 ∂ y 24–7
24–8
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
where all sums are understood to run over i = 1, . . . 6. In matrix form:  0010 ∂ Ni  wi ∂ζ   0003 ∂w 0004 ∂ζ1 ∂ζ2 ∂ζ3  1 0010 ∂ Ni    ∂x ∂ x ∂ x ∂ x   wi ∂ζ  . ∂w = ∂ζ1 ∂ζ2 ∂ζ3  2   0010 ∂ Ni ∂y ∂y ∂y ∂y wi ∂ζ 3 Transposing both sides of (24.15) while exchanging sides yields  ∂ζ1 ∂ζ1  ∂x ∂y  00110010 0012 0010 0010  ∂ζ ∂ζ  0005 wi ∂∂ζNi wi ∂∂ζNi wi ∂∂ζNi  ∂ x2 ∂ y2  = ∂w ∂x 1 2 3   ∂ζ3 ∂ζ3 ∂x ∂y Now make w ≡ 1, x, y and stack the results row-wise:  0010 ∂ Ni 0010 ∂ Ni 0010 ∂ Ni   ∂ζ ∂ζ1   ∂1 ∂1 1 ∂ζ1 ∂ζ2 ∂ζ3 ∂x ∂y ∂y    ∂x 0010  0010 0010     ∂x ∂x ∂ζ ∂ζ ∂ N ∂ N ∂ N xi ∂ζ i xi ∂ζ i   ∂ x2 ∂ y2  =   xi ∂ζ i ∂ x ∂y  1 2 3    ∂y ∂y 0010 ∂ Ni 0010 ∂ Ni 0010 ∂ Ni ∂ζ3 ∂ζ3 yi ∂ζ yi ∂ζ yi ∂ζ ∂x ∂y ∂x ∂y 1 2 3
(24.15)
∂w 0006 . ∂y
(24.16)
  . 
(24.17)
But ∂ x/∂ x = ∂ y/∂ y = 1 and ∂1/∂ x = ∂1/∂ y = ∂ x/∂ y = ∂ y/∂ 0010 x = 0 because x and y are independent coordinates. It is shown in Remark 24.2 below that, if Ni = 1, the entries of the first row of the coefficient matrix are equal to a constant C. These entries can be scaled to unity because the first row of the right-hand side is null. Consequently we arrive at a system of linear equations of order 3 with two right-hand sides:     ∂ζ1 ∂ζ1 1 1 1 ∂y  0003 0 0 0004 0010 0010 ∂ Ni 0010 ∂ Ni  ∂ x  x i ∂ Ni   ∂ζ2 ∂ζ2  xi ∂ζ xi ∂ζ ∂ζ1 = 1 0 . (24.18)  2 3   ∂x ∂y   0010 ∂ Ni 0010 ∂ Ni 0010 ∂ Ni  0 1 ∂ζ3 ∂ζ3 yi ∂ζ yi ∂ζ yi ∂ζ 1 2 3 ∂x ∂y §24.3.2. Solving the Jacobian System By analogy with quadrilateral elements, the coefficient matrix of (24.18) will be called the Jacobian matrix and denoted by J. Its determinant scaled by one half is equal to the Jacobian J = 12 det J used in the expression of the area element introduced in §24.2.3. For compactness (24.18) is rewritten     ∂ζ1 ∂ζ1 ∂y  0003 0 0 0004 1 1 1  ∂x  ∂ζ ∂ζ  J P =  Jx1 Jx2 Jx3   ∂ x2 ∂ y2  = 1 0 . (24.19)   0 1 Jy1 Jy2 Jy3 ∂ζ3 ∂ζ3 ∂x ∂y If J = 0, solving this system gives  ∂ζ ∂ζ1  1 0004 0003 ∂y   ∂x 1 Jy23 Jx32  ∂ζ2 ∂ζ2  (24.20) = Jy31 Jx13 = P,  ∂x ∂y    2J J J y12 x21 ∂ζ3 ∂ζ3 ∂x ∂y 24–8
24–9
§24.4 THE QUADRATIC TRIANGLE
in which Jx ji = Jx j − Jxi , Jy ji = Jy j − Jyi and J = 12 det J = 12 (Jx21 Jy31 − Jy12 Jx13 ). Substituting into (24.14) we arrive at 000f
wi 000e ∂ Ni ∂w ∂ Ni ∂ Ni ∂ Ni = wi = Jy23 + Jy31 + Jy12 , ∂x ∂x 2J ∂ζ1 ∂ζ2 ∂ζ3 000f (24.21)
wi 000e ∂ Ni ∂w ∂ Ni ∂ Ni ∂ Ni Jx32 + Jx13 + Jx21 . = wi = ∂y ∂y 2J ∂ζ1 ∂ζ2 ∂ζ3 In particular, the shape function derivatives are 000f 000e ∂ Ni ∂ Ni ∂ Ni 1 ∂ Ni Jy23 + Jy31 + Jy12 , = ∂x 2J ∂ζ1 ∂ζ2 ∂ζ3 000f 000e ∂ Ni ∂ Ni ∂ Ni 1 ∂ Ni Jx32 + Jx13 + Jx21 . = ∂y 2J ∂ζ1 ∂ζ2 ∂ζ3
(24.22)
in which the natural derivatives ∂ Ni /∂ζ j can be read off (24.12). Using the 3 × 2 P matrix defined in (24.20) yields finally the compact form 0011 0012 0012 0011 ∂ Ni ∂ Ni ∂ Ni ∂ Ni ∂ Ni P. (24.23) = ∂x ∂y ∂ζ1 ∂ζ2 ∂ζ3 Remark 24.2. Here is the proof that each first row entry of the 3 × 3 matrix in (24.17) is a numerical constant, say C. Suppose the shape functions are polynomials of order n in the triangular coordinates, and let Z = ζ1 + ζ2 + ζ3 . The completeness identity is
S=
Ni = 1 = c1 Z + c2 Z 2 + . . . cn Z n ,
c1 + c2 + . . . + cn = 1.
(24.24)
where the ci are element dependent scalar coefficients. Differentiating S with respect to the ζi ’s and setting Z = 1 yields C=
∂ Ni ∂ζ1
=
∂ Ni ∂ζ2
=
∂ Ni ∂ζ3
= c1 + 2c2 Z + c3 Z 3 + . . . (n − 1)Z n−1 = c1 + 2c2 + 3c3 + . . . + cn−1 = 1 + c2 + 2c3 + . . . + (n − 2)cn ,
(24.25) which proves the assertion. For the 3-node linear triangle, S = Z and C = 1. For the 6-node quadratic triangle, S = 2Z 2 − Z and C = 3. For the 10-node cubic triangle, S = 9Z 3 /2 − 9Z 2 /2 + Z and C = 11/2. Because the first equation in (24.18) is homogeneous, the C’s can be scaled to unity.
§24.4. The Quadratic Triangle §24.4.1. Shape Function Module We specialize now the results of §24.3 to obtain the stiffness matrix of the 6-node quadratic triangle in plane stress. Taking the dot products of the natural-coordinate partials (24.12) with the node coordinates we obtain the entries of the Jacobian matrix of (24.19): Jx1 = x1 (4ζ1 −1) + 4(x4 ζ2 + x6 ζ3 ), Jx2 = x2 (4ζ2 −1) + 4(x5 ζ3 + x4 ζ1 ), Jx3 = x3 (4ζ3 −1) + 4(x6 ζ1 + x5 ζ2 ), Jy1 = y1 (4ζ1 −1) + 4(y4 ζ2 + y6 ζ3 ), Jy2 = y2 (4ζ2 −1) + 4(y5 ζ3 + y4 ζ1 ), Jy3 = y3 (4ζ3 −1) + 4(y6 ζ1 + y5 ζ2 ), (24.26)
24–9
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
24–10
Trig6IsoPShapeFunDer[ncoor_,tcoor_]:= Module[ {ζ 1,ζ 2,ζ 3,x1,x2,x3,x4,x5,x6,y1,y2,y3,y4,y5,y6, dx4,dx5,dx6,dy4,dy5,dy6,Jx21,Jx32,Jx13,Jy12,Jy23,Jy31, Nf,dNx,dNy,Jdet}, {ζ 1,ζ 2,ζ 3}=tcoor; {{x1,y1},{x2,y2},{x3,y3},{x4,y4},{x5,y5},{x6,y6}}=ncoor; dx4=x4-(x1+x2)/2; dx5=x5-(x2+x3)/2; dx6=x6-(x3+x1)/2; dy4=y4-(y1+y2)/2; dy5=y5-(y2+y3)/2; dy6=y6-(y3+y1)/2; Nf={ζ 1*(2*ζ 1-1),ζ 2*(2*ζ 2-1),ζ 3*(2*ζ 3-1),4*ζ 1*ζ 2,4*ζ 2*ζ 3,4*ζ 3*ζ 1}; Jx21= x2-x1+4*(dx4*(ζ 1-ζ 2)+(dx5-dx6)*ζ 3); Jx32= x3-x2+4*(dx5*(ζ 2-ζ 3)+(dx6-dx4)*ζ 1); Jx13= x1-x3+4*(dx6*(ζ 3-ζ 1)+(dx4-dx5)*ζ 2); Jy12= y1-y2+4*(dy4*(ζ 2-ζ 1)+(dy6-dy5)*ζ 3); Jy23= y2-y3+4*(dy5*(ζ 3-ζ 2)+(dy4-dy6)*ζ 1); Jy31= y3-y1+4*(dy6*(ζ 1-ζ 3)+(dy5-dy4)*ζ 2); Jdet = Jx21*Jy31-Jy12*Jx13; dNx= {(4*ζ 1-1)*Jy23,(4*ζ 2-1)*Jy31,(4*ζ 3-1)*Jy12,4*(ζ 2*Jy23+ζ 1*Jy31), 4*(ζ 3*Jy31+ζ 2*Jy12),4*(ζ 1*Jy12+ζ 3*Jy23)}/Jdet; dNy= {(4*ζ 1-1)*Jx32,(4*ζ 2-1)*Jx13,(4*ζ 3-1)*Jx21,4*(ζ 2*Jx32+ζ 1*Jx13), 4*(ζ 3*Jx13+ζ 2*Jx21),4*(ζ 1*Jx21+ζ 3*Jx32)}/Jdet; Return[Simplify[{Nf,dNx,dNy,Jdet}]] ];
Figure 24.5. Shape function module for 6-node quadratic triangle.
From these J can be constructed, and shape function partials ∂ Ni /∂ x and ∂ Ni /∂ y explicitly obtained from (24.22). Somewhat simpler expressions, however, result by using the following “hierarchical” side node coordinates: x4 = x4 − 12 (x1 + x2 ),
000bx5 = x5 − 12 (x2 + x3 ),
000bx6 = x6 − 12 (x3 + x1 ),
000by4 = y4 − 12 (y1 + y2 ),
000by5 = y5 − 12 (y2 + y3 ),
000by6 = y6 − 12 (y3 + y1 ).
(24.27)
Geometrically these represent the deviations from midpoint positions; thus for a superparametric element x4 = x5 = x6 = y4 = y5 = y6 = 0. The Jacobian coefficients become Jx21 = x21 + 4(000bx4 (ζ1 −ζ2 ) + (000bx5 −000bx6 )ζ3 ), Jx13 = x13 + 4(000bx6 (ζ3 −ζ1 ) + (000bx4 −000bx5 )ζ2 ), Jy23 = y23 + 4(000by5 (ζ3 −ζ2 ) + (000by4 −000by6 )ζ1 ), (Note that if all midpoint deviations vanish, Jx ji 1 det J = 12 (Jx21 Jy31 − Jy12 Jx13 ) and 2 1 P= 2J
0003
Jx32 = x32 + 4(000bx5 (ζ2 −ζ3 ) + (000bx6 −000bx4 )ζ1 ), Jy12 = y12 + 4(000by4 (ζ2 −ζ1 ) + (000by6 −000by5 )ζ3 ), Jy31 = y31 + 4(000by6 (ζ1 −ζ3 ) + (000by5 −000by4 )ζ2 ). (24.28) = x ji and Jy ji = y ji .) From this one gets J =
y23 + 4(000by5 (ζ3 −ζ2 ) + (000by4 −000by6 )ζ1 ) y31 + 4(000by6 (ζ1 −ζ3 ) + (000by5 −000by4 )ζ2 ) y12 + 4(000by4 (ζ2 −ζ1 ) + (000by6 −000by5 )ζ3 )
0004 x32 + 4(000bx5 (ζ2 −ζ3 ) + (000bx6 −000bx4 )ζ1 ) x13 + 4(000bx6 (ζ3 −ζ1 ) + (000bx4 −000bx5 )ζ2 ) . x21 + 4(000bx4 (ζ1 −ζ2 ) + (000bx5 −000bx6 )ζ3 ) (24.29)
and the Cartesian derivatives of the shape functions are 
 (4ζ1 − 1)Jy23  (4ζ2 − 1)Jy31   1  ∂NT  (4ζ3 − 1)Jy12  =  , ∂x 2J  4(ζ2 Jy23 + ζ1 Jy31 )    4(ζ3 Jy31 + ζ2 Jy12 ) 4(ζ1 Jy12 + ζ3 Jy23 )

 (4ζ1 − 1)Jx32  (4ζ2 − 1)Jx13   1  ∂NT  (4ζ3 − 1)Jx21  =  . ∂y 2J  4(ζ2 Jx32 + ζ1 Jx13 )    4(ζ3 Jx13 + ζ2 Jx21 ) 4(ζ1 Jx21 + ζ3 Jx32 ) 24–10
(24.30)
24–11
§24.4 THE QUADRATIC TRIANGLE Trig6IsoPMembraneStiffness[ncoor_,Emat_,th_,options_]:= Module[{i,k,p=3,numer=False,h=th,tcoor,w,c, Nf,dNx,dNy,Jdet,Be,Ke=Table[0,{12},{12}]}, If [Length[options]>=1, numer=options[[1]]]; If [Length[options]>=2, p= options[[2]]]; If [!MemberQ[{1,-3,3,6,7},p], Print['Illegal p']; Return[Null]]; For [k=1, k<=Abs[p], k++, {tcoor,w}= TrigGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}= Trig6IsoPShapeFunDer[ncoor,tcoor]; If [numer, {Nf,dNx,dNy,Jdet}=N[{Nf,dNx,dNy,Jdet}]]; If [Length[th]6, h=th.Nf]; c=w*Jdet*h/2; Be= {Flatten[Table[{dNx[[i]],0 },{i,6}]], Flatten[Table[{0, dNy[[i]]},{i,6}]], Flatten[Table[{dNy[[i]],dNx[[i]]},{i,6}]]}; Ke+=c*Transpose[Be].(Emat.Be); ]; If[!numer,Ke=Simplify[Ke]]; Return[Ke] ];
Figure 24.6. Stiffness matrix module for 6-node plane stress triangle.
A Mathematica shape function module Trig6IsoPShapeFunDer is shown in Figure 24.5. It receives two arguments: ncoor and tcoor. The first one is the list of {xi , yi } coordinates of the six nodes. The second is the list of three triangular coordinates { ζ1 , ζ2 , ζ3 } of the location at which the shape functions and their Cartesian derivatives are to be computed. The module returns { Nf,dNx,dNy,Jdet } as module value. Here Nf collects the shape function values, dNx and dNy the x and y shape function derivatives, respectively, and Jdet is the determinant of matrix J, equal to 2J in the notation used here. §24.4.2. Stiffness Module The numerically integrated stiffness matrix is 0002 K = e
e
hBT EB d ≈
p
wi F(ζ1i , ζ2i , ζ3i ),
where
F(ζ1 , ζ2 , ζ3 ) = hBT EB J.
(24.31)
i=1
Here p denotes the number of sample points of the Gauss rule being used, wi is the integration weight for the i th sample point, ζ1i , ζ2i , ζ3i are the sample point triangular coordinates and J = 12 det J. This data is provided by TrigGaussRuleInfo. For the 6-node triangle (24.31) is implemented in module Trig6IsoPMembraneStiffness, listed in Figure 24.6. The module is invoked as Ke = Trig6IsoPMembraneStiffness[ncoor,Emat,th,options]
(24.32)
The arguments are: ncoor
The list of node coordinates arranged as { { x1,y1 },{ x2,y2 }, ..
Emat
The plane stress elasticity matrix 0003 E=
E 11 E 12 E 13
E 12 E 22 E 23
E 13 E 23 E 33
{ x6,y6 } }.
0004 (24.33)
arranged as { { E11,E12,E33 },{ E12,E22,E23 },{ E13,E23,E33 } }. th
Plate thickness specified as either a scalar h or a six-entry list: { h1,h2,h3,h4,h5,h6 }. 24–11
24–12
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
The one-entry form specifies uniform thickness h. The six-entry form is used to specify an element of variable thickness, in which case the entries are the six node thicknesses and h is interpolated quadratically. options
Processing options list. May contain two items: { numer,rule } or one: { numer }. numer is a logical flag. If True, the computations are forced to proceed in floating-point arithmetic. For symbolic or exact arithmetic work set numer to False. rule specifies the triangle Gauss rule as described in §24.2.4; rule may be 1, 3, −3, 6 or 7. For the 6-node element the three point rules are sufficient to get the correct rank. If omitted rule = 3 is assumed.
The module returns Ke as an 12 × 12 symmetric matrix pertaining to the following arrangement of nodal displacements: ue = [ u x1 u y1 u x2 u y2 u x3 u y3 u x4 u y4 u x5 u y5 u x6 u y6 ]T .
(24.34)
§24.4.3. Test on Straight-Sided Triangle 3 (4,4)
The stiffness module of Figure 24.6 is tested on two triangle geometries, one with straight sides and constant metric and one with curved sides and highly variable metric. The straight sided triangle geometry, shown in Figure 24.7, has the corner nodes placed at (0, 0), (6, 2) and (4, 4) with side nodes 4,5,6 at the midpoints of the sides. The element has unit plate thickness. The material is isotropic with E = 288 and ν = 1/3. The element is tested with the script shown in Figure 24.8. The stiffness matrix Ke is computed by the threeinterior-point Gauss rule, specified as p=3.
5 6 2 (6,2) 4 1 (0,0) Figure 24.7. Straight sided 6-node triangle test element.
The computed Ke for integration rules with 3 or more points are identical since those rules are exact for this element if the metric is constant, which is the case here. That stiffness is 
54  27  18   0   0   9  −72   0   0   0  0 −36
27 18 0 0 9 −72 0 54 0 −18 9 36 0 72 0 216 −108 54 −36 −72 0 −18 −108 216 −36 90 0 72 9 54 −36 162 −81 0 0 36 −36 90 −81 378 0 0 0 −72 0 0 0 576 −216 72 0 72 0 0 −216 864 0 −216 144 −216 144 0 −72 0 144 −360 144 −360 −72 −288 −36 0 0 0 −36 −432 288 −144 0 0 −36 −144 288 −720
0 0 −216 144 −216 144 0 −72 576 −216 −144 0

0 0 −36 0 −36 −144  144 0 0  −360 0 0  144 0 −36   −360 −36 −144   −72 −432 288  −288 288 −720   −216 −144 0  864 0 144   0 576 −216 144 −216 864
(24.35)
The eigenvalues are: [ 1971.66 1416.75 694.82 545.72 367.7 175.23 157.68 57.54 12.899 0 0 0 ] (24.36) The 3 zero eigenvalues pertain to the three independent rigid-body modes. The 9 other ones are positive. Consequently the computed Ke has the correct rank of 9. 24–12
24–13
§24.4 THE QUADRATIC TRIANGLE ClearAll[Em,ν,a,b,e,h]; h=1; Em=288; ν=1/3; ncoor={{0,0},{6,2},{4,4},{3,1},{5,3},{2,2}}; Emat=Em/(1-ν^2)*{{1,ν,0},{ν,1,0},{0,0,(1-ν)/2}}; Print['Emat=',Emat//MatrixForm] Ke=Trig6IsoPMembraneStiffness[ncoor,Emat,h,{False,3}]; Ke=Simplify[Ke]; Print[Chop[Ke]//MatrixForm]; Print['eigs of Ke=',Chop[Eigenvalues[N[Ke]]]];
Figure 24.8. Script to form Ke of test triangle of Figure 24.7. 3 ( 0,
§24.4.4. Test on Highly Distorted Triangle
√
3/2γ)
√ 6 (−1/2, 1/ 3 )
A highly distorted test geometry places the 3 corners at the vertices √ of an equilateral triangle: {−1/2, 0}, {1/2, 0}, {0, √ 3/2} whereas√the 3 side nodes are √ located at {0, −1/(2 3)}, {1/2, 1/ 3} and {−1/2, 1/ 3}. √ The result is that the six nodes lie on a circle of radius 1/ 3, as depicted in Figure 24.9. The element has unit thickness. The material is isotropic with E = 504 and ν = 0. The stiffness Ke is evaluated for four the rank-sufficient rules: 3,-3,6,7, using the script of Figure 24.10.
√ 5 (1/2, 1/ 3 )
2 (1/2,0)
1 (−1/2,0)
√ 4 ( 0, −1/(2 3) )
Figure 24.9. Test curved sided quadratic triangle, with the 6 nodes lying on a circle.
ClearAll[Em,ν,h]; h=1; Em=7*72; ν=0; h=1; {x1,y1}={-1,0}/2; {x2,y2}={1,0}/2; {x3,y3}={0,Sqrt[3]}/2; {x4,y4}={0,-1/Sqrt[3]}/2; {x5,y5}={1/2,1/Sqrt[3]}; {x6,y6}={-1/2,1/Sqrt[3]}; ncoor= {{x1,y1},{x2,y2},{x3,y3},{x4,y4},{x5,y5},{x6,y6}}; Emat=Em/(1-ν^2)*{{1,ν,0},{ν,1,0},{0,0,(1-ν)/2}}; For [i=2,i<=5,i++, p={1,-3,3,6,7}[[i]]; Ke=Trig6IsoPMembraneStiffness[ncoor,Emat,h,{True,p}]; Ke=Chop[Simplify[Ke]]; Print['Ke=',SetPrecision[Ke,4]//MatrixForm]; Print['Eigenvalues of Ke=',Chop[Eigenvalues[N[Ke]],.0000001]] ];
Figure 24.10. Script to form Ke for the triangle of Figure 24.9, using four integration rules.
For rule=3 the result (printed to 4 places because of the SetPrecision statement in script) is  344.7  75.00  −91.80   21.00   −86.60   −24.00  −124.7   −72.00   −20.78   −36.00  −20.78 36.00
75.00 258.1 −21.00 −84.87 18.00 −90.07 96.00 0 −36.00 20.78 −132.0 −103.9
−91.80 −21.00 344.7 −75.00 −86.60 24.00 −124.7 72.00 −20.78 −36.00 −20.78 36.00
21.00 −84.87 −75.00 258.1 −18.00 −90.07 −96.00 0 132.0 −103.9 36.00 20.78
−86.60 18.00 −86.60 −18.00 214.8 0 41.57 0 −41.57 144.0 −41.57 −144.0
−24.00 −90.07 24.00 −90.07 0 388.0 0 −41.57 −24.00 −83.14 24.00 −83.14
−124.7 96.00 −124.7 −96.00 41.57 0 374.1 0 −83.14 −72.00 −83.14 72.00
24–13
−72.00 0 72.00 0 0 −41.57 0 374.1 −72.00 −166.3 72.00 −166.3
−20.78 −36.00 −20.78 132.0 −41.57 −24.00 −83.14 −72.00 374.1 0 −207.8 0
−36.00 20.78 −36.00 −103.9 144.0 −83.14 −72.00 −166.3 0 374.1 0 −41.57
−20.78 −132.0 −20.78 36.00 −41.57 24.00 −83.14 72.00 −207.8 0 374.1 0

36.00 −103.9  36.00   20.78   −144.0   −83.14   72.00  −166.3   0  −41.57   0 374.1 (24.37)
24–14
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
For rule=-3:  566.4  139.0  129.9   21.00   79.67   8.000  −364.9   −104.0   −205.5   −36.00  −205.5 −28.00
139.0 405.9 −21.00 62.93 50.00 113.2 64.00 −129.3 −36.00 −164.0 −196.0 −288.7
129.9 −21.00 566.4 −139.0 79.67 −8.000 −364.9 104.0 −205.5 28.00 −205.5 36.00
21.00 62.93 −139.0 405.9 −50.00 113.2 −64.00 −129.3 196.0 −288.7 36.00 −164.0
79.67 50.00 79.67 −50.00 325.6 0 −143.2 0 −170.9 176.0 −170.9 −176.0
8.000 113.2 −8.000 113.2 0 646.6 0 −226.3 8.000 −323.3 −8.000 −323.3
−364.9 64.00 −364.9 −64.00 −143.2 0 632.8 0 120.1 −104.0 120.1 104.0
−104.0 −129.3 104.0 −129.3 0 −226.3 0 485.0 −104.0 0 104.0 0
The stiffness for rule=6 is omitted to save space. For rule=7:  661.9  158.5  141.7   21.00   92.53   7.407  −432.1   −117.2   −190.2   −29.10  −273.8 −40.61
158.5 478.8 −21.00 76.13 49.41 125.3 50.79 −182.7 −29.10 −156.6 −208.6 −341.0
141.7 −21.00 661.9 −158.5 92.53 −7.407 −432.1 117.2 −273.8 40.61 −190.2 29.10
21.00 76.13 −158.5 478.8 −49.41 125.3 −50.79 −182.7 208.6 −341.0 29.10 −156.6
92.53 49.41 92.53 −49.41 387.3 0 −139.8 0 −216.3 175.4 −216.3 −175.4
7.407 125.3 −7.407 125.3 0 753.4 0 −207.0 7.407 −398.5 −7.407 −398.5
−432.1 50.79 −432.1 −50.79 −139.8 0 723.6 0 140.2 −117.2 140.2 117.2
−117.2 −182.7 117.2 −182.7 0 −207.0 0 562.6 −117.2 4.884 117.2 4.884

−205.5 −36.00 −205.5 196.0 −170.9 8.000 120.1 −104.0 521.9 −64.00 −60.04 0
−36.00 −164.0 28.00 −288.7 176.0 −323.3 −104.0 0 −64.00 595.8 0 180.1
−205.5 −196.0 −205.5 36.00 −170.9 −8.000 120.1 104.0 −60.04 0 521.9 64.00
−28.00 −288.7  36.00   −164.0   −176.0   −323.3   104.0  0  0  180.1   64.00 595.8 (24.38)
−190.2 −29.10 −273.8 208.6 −216.3 7.407 140.2 −117.2 602.8 −69.71 −62.78 0
−29.10 −156.6 40.61 −341.0 175.4 −398.5 −117.2 4.884 −69.71 683.3 0 207.9
−273.8 −208.6 −190.2 29.10 −216.3 −7.407 140.2 117.2 −62.78 0 602.8 69.71
−40.61 −341.0  29.10   −156.6   −175.4   −398.5   117.2  4.884   0  207.9   69.71 683.3 (24.39)

The eigenvalues of these matrices are: Rule 3 −3 6 7
Eigenvalues of Ke 702.83 665.11 553.472 553.472 481.89 429.721 429.721 1489.80 1489.80 702.833 665.108 523.866 523.866 481.890 1775.53 1775.53 896.833 768.948 533.970 533.970 495.570 1727.11 1727.11 880.958 760.719 532.750 532.750 494.987
118.391 196.429 321.181 312.123
118.391 196.429 321.181 312.123
0 0 0 0
0 0 0 0 0 0 0 0 (24.40)
Since the metric of this element is very distorted near its boundary, the stiffness matrix entries and eigenvalues change substantially as the integration formulas are advanced from 3 to 6 and 7 points. However as can be seen the matrix remains rank-sufficient. §24.5.
*The Cubic Triangle
The 10-node cubic triangle, depicted in Figure 24.11, is rarely used as such in applications because of the difficulty of combining it with other elements. Nevertheless the derivation of the element modules is instructive as this triangle has other and more productive uses as a generator of more practical elements with “drilling” rotational degrees of freedom at corners for modeling shells. Such transformations are studied in advanced FEM courses. The geometry of the triangle is defined by the coordinates of the ten nodes. A notational warning: the interior node is labeled as 0 instead of 10 (cf. Figure 24.8) to avoid confusion with notation such as x12 = x1 − x2 for coordinate differences.
24–14
24–15
§24.5 *THE CUBIC TRIANGLE
§24.5.1. *Shape Function Module The shape functions obtained in Exercise 18.1 are N1 = 12 ζ1 (3ζ1 − 1)(3ζ1 − 2), N2 = 12 ζ2 (3ζ2 − 1)(3ζ2 − 2), N3 = 12 ζ3 (3ζ3 −1)(3ζ3 −2), N4 = 92 ζ1 ζ2 (3ζ1 −1), N5 = 92 ζ1 ζ2 (3ζ2 −1), N6 = 92 ζ2 ζ3 (3ζ2 −1), N7 = 92 ζ2 ζ3 (3ζ3 −1), N8 = 92 ζ3 ζ1 (3ζ3 − 1), N9 = 92 ζ3 ζ1 (3ζ1 − 1) and N0 = 27ζ1 ζ2 ζ3 . Their natural derivatives are
 2 −2ζ +3ζ 2  1 1 9 0     0    ζ2 (6ζ1 −1)    ∂NT 9  ζ2 (3ζ2 −1)  = 2 , 0 ∂ζ1     0    ζ3 (3ζ3 −1)    ζ3 (6ζ1 −1) 6ζ2 ζ3
      ∂NT 9  = 2 ∂ζ2     

0 2 −2ζ2 +3ζ22 9
0 ζ1 (3ζ1 −1) ζ1 (6ζ2 −1) ζ3 (6ζ2 −1) ζ3 (3ζ3 −1) 0 0 6ζ1 ζ3
      ,     
      ∂NT 9  = 2 ∂ζ3     
 0 0  2 2 −2ζ 3 +3ζ3  9  0  0  . ζ2 (3ζ2 −1)  ζ2 (6ζ3 −1)   ζ1 (6ζ3 −1)   ζ1 (3ζ1 −1) 6ζ1 ζ2
(24.41)
3
As in the case of the 6-node triangle it is convenient to introduce the deviations from thirdpoints and centroid: x4 = x4 − 1 (2x1 +x2 ), x5 = x5 − 13 (x1 +2x2 ), x6 = x6 − 13 (2x2 +x3 ), 3 x7 = x7 − 13 (x2 +2x3 ), x8 = x8 − 13 (2x3 +x1 ), x9 = x9 − 1 (x3 +2x1 ), x0 = x0 − 13 (x1 +x2 +x3 ), y4 = y4 − 13 (2y1 +y2 ), 3 y5 = y5 − 13 (y1 +2y2 ), y6 = y6 − 13 (2y2 +y3 ), y7 = y7 − 1 (y2 +2y3 ), y8 = y8 − 13 (2y3 +y1 ), y9 = y9 − 13 (y3 +2y1 ), 3 and y0 = y0 − 13 (y1 + y2 + y3 ).
7
8 9
6
0
2 5
4
1
Figure 24.11. The 10-node cubic triangle.
Using (24.22) and (24.41), the shape function partial derivatives can be worked out to be


Jy23 ( 29 − 2ζ1 + 3ζ12 /2)   Jy31 ( 29 − 2ζ2 + 3ζ22 /2)   2 2   Jy12 ( 9 − 2ζ3 + 3ζ3 /2)   Jy31 ζ1 (3ζ1 − 1) + Jy23 ζ2 (6ζ1 − 1)   T  ∂N 9  Jy23 ζ2 (3ζ2 − 1) + Jy31 ζ1 (6ζ2 − 1)  , =  ∂x 4J   Jy12 ζ2 (3ζ2 − 1) + Jy31 ζ3 (6ζ2 − 1)   Jy31 ζ3 (3ζ3 − 1) + Jy12 ζ2 (6ζ3 − 1)     Jy23 ζ3 (3ζ3 − 1) + Jy12 ζ1 (6ζ3 − 1)   J ζ (3ζ − 1) + J ζ (6ζ − 1)  y12 1 1 y23 3 1 6(Jy12 ζ1 ζ2 + Jy31 ζ1 ζ3 + Jy23 ζ2 ζ3 )


Jx32 ( 29 − 2ζ1 + 3ζ12 /2)   Jx13 ( 29 − 2ζ2 + 3ζ22 /2)   2 2   Jx21 ( 9 − 2ζ3 + 3ζ3 /2)   Jx13 ζ1 (3ζ1 − 1) + Jx32 ζ2 (6ζ1 − 1)   T  ∂N 9  Jx32 ζ2 (3ζ2 − 1) + Jx13 ζ1 (6ζ2 − 1)  . =  ∂y 4J   Jx21 ζ2 (3ζ2 − 1) + Jx13 ζ3 (6ζ2 − 1)   Jx13 ζ3 (3ζ3 − 1) + Jx21 ζ2 (6ζ3 − 1)     Jx32 ζ3 (3ζ3 − 1) + Jx21 ζ1 (6ζ3 − 1)   J ζ (3ζ − 1) + J ζ (6ζ − 1)  x21 1 1 x32 3 1 6(Jx21 ζ1 ζ2 + Jx13 ζ1 ζ3 + Jx32 ζ2 ζ3 ) (24.42)
where the expressions of the Jacobian coefficients are Jx21 = x21 +
9 2
0005
0013
0014
0013
0014
000bx4 ζ1 (3ζ1 −6ζ2 −1)+ζ2 + x5 ζ2 (1−3ζ2 +6ζ1 )−ζ1 + x6 ζ3 (6ζ2 −1)
0006
+ x7 ζ3 (3ζ3 −1) + x8 ζ3 (1−3ζ3 ) + x9 ζ3 (1−6ζ1 ) + 6000bx0 ζ3 (ζ1 −ζ2 ) , Jx32 = x32 +
9 2
0005
0013
000bx4 ζ1 (1−3ζ1 ) + x5 ζ1 (1−6ζ2 ) + x6 ζ2 (3ζ2 −6ζ3 −1)+ζ3
0013
0014
0014
0006
+ x7 ζ3 (1−3ζ3 +6ζ2 )−ζ2 + x8 ζ1 (6ζ3 −1) + x9 ζ1 (3ζ1 −1) + 6000bx0 ζ1 (ζ2 −ζ3 ) , Jx13 = x13 +
9 2
0005
000bx4 (6ζ1 −1)ζ2 + x5 ζ2 (3ζ2 −1) + x6 ζ2 (1−3ζ2 ) + x7 ζ2 (1−6ζ3 )
0013
0014
0013
0014
0006
+ x8 ζ3 (3ζ3 −6ζ1 −1)+ζ1 + x9 ζ1 (1−3ζ1 +6ζ3 )−ζ3 + 6000bx0 ζ2 (ζ3 −ζ1 ) ,
24–15
(24.43)
24–16
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS Trig10IsoPShapeFunDer[ncoor_,tcoor_]:= Module[ {ζ 1,ζ 2,ζ 3,x1,x2,x3,x4,x5,x6,x7,x8,x9,x0,y1,y2,y3,y4,y5,y6,y7,y8,y9,y0, dx4,dx5,dx6,dx7,dx8,dx9,dx0,dy4,dy5,dy6,dy7,dy8,dy9,dy0, Jx21,Jx32,Jx13,Jy12,Jy23,Jy31,Nf,dNx,dNy,Jdet}, {{x1,y1},{x2,y2},{x3,y3},{x4,y4},{x5,y5},{x6,y6},{x7,y7}, {x8,y8},{x9,y9},{x0,y0}}=ncoor; {ζ 1,ζ 2,ζ 3}=tcoor; dx4=x4-(2*x1+x2)/3; dx5=x5-(x1+2*x2)/3; dx6=x6-(2*x2+x3)/3; dx7=x7-(x2+2*x3)/3; dx8=x8-(2*x3+x1)/3; dx9=x9-(x3+2*x1)/3; dy4=y4-(2*y1+y2)/3; dy5=y5-(y1+2*y2)/3; dy6=y6-(2*y2+y3)/3; dy7=y7-(y2+2*y3)/3; dy8=y8-(2*y3+y1)/3; dy9=y9-(y3+2*y1)/3; dx0=x0-(x1+x2+x3)/3; dy0=y0-(y1+y2+y3)/3; Nf={ζ 1*(3*ζ 1-1)*(3*ζ 1-2),ζ 2*(3*ζ 2-1)*(3*ζ 2-2),ζ 3*(3*ζ 3-1)*(3*ζ 3-2), 9*ζ 1*ζ 2*(3*ζ 1-1),9*ζ 1*ζ 2*(3*ζ 2-1),9*ζ 2*ζ 3*(3*ζ 2-1), 9*ζ 2*ζ 3*(3*ζ 3-1),9*ζ 3*ζ 1*(3*ζ 3-1),9*ζ 3*ζ 1*(3*ζ 1-1),54*ζ 1*ζ 2*ζ 3}/2; Jx21=x2-x1+(9/2)*(dx4*(ζ 1*(3*ζ 1-6*ζ 2-1)+ζ 2)+ dx5*(ζ 2*(1-3*ζ 2+6*ζ 1)-ζ 1)+dx6*ζ 3*(6*ζ 2-1)+dx7*ζ 3*(3*ζ 3-1)+ dx8*ζ 3*(1-3*ζ 3)+dx9*ζ 3*(1-6*ζ 1)+6*dx0*ζ 3*(ζ 1-ζ 2)); Jx32=x3-x2+(9/2)*(dx4*ζ 1*(1-3*ζ 1)+dx5*ζ 1*(1-6*ζ 2)+ dx6*(ζ 2*(3*ζ 2-6*ζ 3-1)+ζ 3)+dx7*(ζ 3*(1-3*ζ 3+6*ζ 2)-ζ 2)+ dx8*ζ 1*(6*ζ 3-1)+dx9*ζ 1*(3*ζ 1-1)+6*dx0*ζ 1*(ζ 2-ζ 3)) ; Jx13=x1-x3+(9/2)*(dx4*(6*ζ 1-1)*ζ 2+dx5*ζ 2*(3*ζ 2-1)+ dx6*ζ 2*(1-3*ζ 2)+dx7*ζ 2*(1-6*ζ 3)+dx8*(ζ 3*(3*ζ 3-6*ζ 1-1)+ζ 1)+ dx9*(ζ 1*(1-3*ζ 1+6*ζ 3)-ζ 3)+6*dx0*ζ 2*(ζ 3-ζ 1)); Jy12=y1-y2-(9/2)*(dy4*(ζ 1*(3*ζ 1-6*ζ 2-1)+ζ 2)+ dy5*(ζ 2*(1-3*ζ 2+6*ζ 1)-ζ 1)+dy6*ζ 3*(6*ζ 2-1)+dy7*ζ 3*(3*ζ 3-1)+ dy8*ζ 3*(1-3*ζ 3)+dy9*ζ 3*(1-6*ζ 1)+6*dy0*ζ 3*(ζ 1-ζ 2)); Jy23=y2-y3-(9/2)*(dy4*ζ 1*(1-3*ζ 1)+dy5*ζ 1*(1-6*ζ 2)+ dy6*(ζ 2*(3*ζ 2-6*ζ 3-1)+ζ 3)+dy7*(ζ 3*(1-3*ζ 3+6*ζ 2)-ζ 2)+ dy8*ζ 1*(6*ζ 3-1)+dy9*ζ 1*(3*ζ 1-1)+6*dy0*ζ 1*(ζ 2-ζ 3)) ; Jy31=y3-y1-(9/2)*(dy4*(6*ζ 1-1)*ζ 2+dy5*ζ 2*(3*ζ 2-1)+ dy6*ζ 2*(1-3*ζ 2)+dy7*ζ 2*(1-6*ζ 3)+dy8*(ζ 3*(3*ζ 3-6*ζ 1-1)+ζ 1)+ dy9*(ζ 1*(1-3*ζ 1+6*ζ 3)-ζ 3)+6*dy0*ζ 2*(ζ 3-ζ 1)); Jdet = Jx21*Jy31-Jy12*Jx13; dNx={Jy23*(2/9-2*ζ 1+3*ζ 1^2),Jy31*(2/9-2*ζ 2+3*ζ 2^2),Jy12*(2/9-2*ζ 3+3*ζ 3^2), Jy31*ζ 1*(3*ζ 1-1)+Jy23*ζ 2*(6*ζ 1-1),Jy23*ζ 2*(3*ζ 2-1)+Jy31*ζ 1*(6*ζ 2-1), Jy12*ζ 2*(3*ζ 2-1)+Jy31*ζ 3*(6*ζ 2-1),Jy31*ζ 3*(3*ζ 3-1)+Jy12*ζ 2*(6*ζ 3-1), Jy23*ζ 3*(3*ζ 3-1)+Jy12*ζ 1*(6*ζ 3-1),Jy12*ζ 1*(3*ζ 1-1)+Jy23*ζ 3*(6*ζ 1-1), 6*(Jy12*ζ 1*ζ 2+Jy31*ζ 1*ζ 3+Jy23*ζ 2*ζ 3)}/(2*Jdet/9); dNy={Jx32*(2/9-2*ζ 1+3*ζ 1^2),Jx13*(2/9-2*ζ 2+3*ζ 2^2),Jx21*(2/9-2*ζ 3+3*ζ 3^2), Jx13*ζ 1*(3*ζ 1-1)+Jx32*ζ 2*(6*ζ 1-1),Jx32*ζ 2*(3*ζ 2-1)+Jx13*ζ 1*(6*ζ 2-1), Jx21*ζ 2*(3*ζ 2-1)+Jx13*ζ 3*(6*ζ 2-1),Jx13*ζ 3*(3*ζ 3-1)+Jx21*ζ 2*(6*ζ 3-1), Jx32*ζ 3*(3*ζ 3-1)+Jx21*ζ 1*(6*ζ 3-1),Jx21*ζ 1*(3*ζ 1-1)+Jx32*ζ 3*(6*ζ 1-1), 6*(Jx21*ζ 1*ζ 2+Jx13*ζ 1*ζ 3+Jx32*ζ 2*ζ 3)}/(2*Jdet/9); Return[Simplify[{Nf,dNx,dNy,Jdet}]] ];
Figure 24.12. Shape function module for 10-node cubic triangle.
and Jy12 = y12 −
9 2
0005
0013
0014
000by4 ζ1 (3ζ1 −6ζ2 −1)+ζ2 + y5 (ζ2 (1−3ζ2 +6ζ1 )−ζ1 ) + y6 ζ3 (6ζ2 −1)
0006
+ y7 ζ3 (3ζ3 −1) + y8 ζ3 (1−3ζ3 ) + y9 ζ3 (1−6ζ1 ) + 6000by0 ζ3 (ζ1 −ζ2 ) , Jy23 = y23 −
9 2
0005
0013
000by4 ζ1 (1−3ζ1 ) + y5 ζ1 (1−6ζ2 ) + y6 ζ2 (3ζ2 −6ζ3 −1)+ζ3
0013
0014
0014
0006
+ y7 ζ3 (1−3ζ3 +6ζ2 )−ζ2 + y8 ζ1 (6ζ3 −1) + y9 ζ1 (3ζ1 −1) + 6000by0 ζ1 (ζ2 −ζ3 ) , Jy31 = y31 −
9 2
(24.44)
0005
000by4 (6ζ1 −1)ζ2 + y5 ζ2 (3ζ2 −1) + y6 ζ2 (1−3ζ2 ) + y7 ζ2 (1−6ζ3 )
0013
0014
0013
0014
0006
+ y8 ζ3 (3ζ3 −6ζ1 −1)+ζ1 + y9 ζ1 (1−3ζ1 +6ζ3 )−ζ3 + 6000by0 ζ2 (ζ3 −ζ1 ) . The Jacobian determinant is J = 12 (Jx21 Jy31 − Jy12 Jx13 ). The shape function module Trig10IsoPShapeFunDer that implements these expressions is listed in Figure 24.12. This has the same arguments as Trig6IsoPShapeFunDer. As there Jdet denotes 2J .
24–16
24–17
§24.5 *THE CUBIC TRIANGLE Trig10IsoPMembraneStiffness[ncoor_,Emat_,th_,options_]:= Module[{i,k,l,p=6,numer=False,h=th,tcoor,w,c, Nf,dNx,dNy,Jdet,Be,Ke=Table[0,{20},{20}]}, If [Length[options]>=1, numer=options[[1]]]; If [Length[options]>=2, p= options[[2]]]; If [!MemberQ[{1,3,-3,6,7},p], Print['Illegal p']; Return[Null]]; For [k=1, k<=Abs[p], k++, {tcoor,w}= TrigGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}= Trig10IsoPShapeFunDer[ncoor,tcoor]; If [numer, {Nf,dNx,dNy,Jdet}=N[{Nf,dNx,dNy,Jdet}]]; If [Length[th]10, h=th.Nf]; c=w*Jdet*h/2; Be= {Flatten[Table[{dNx[[i]],0 },{i,10}]], Flatten[Table[{0, dNy[[i]]},{i,10}]], Flatten[Table[{dNy[[i]],dNx[[i]]},{i,10}]]}; Ke+=c*Transpose[Be].(Emat.Be); ]; If[!numer,Ke=Simplify[Ke]]; Return[Ke] ];
Figure 24.13. Stiffness module for 10-node cubic triangle.
§24.5.2. *Stiffness Module Module Trig10IsoPMembraneStiffness, listed in Figure 24.13, implements the computation of the element stiffness matrix of the 10-node cubic plane stress triangle. The module is invoked as (24.45)
Ke = Trig10IsoPMembraneStiffness[ncoor,Emat,th,options]
this has the same arguments of the quadratic triangle module invoked in (24.32), with the following changes. The node coordinate list ncoor contains 10 entries: { { x1,y1 },{ x2,y2 },{ x3,y3 }, . . .,{ x9,y9 },{ x0,y0 } }. If the plate thickness varies, th is a list of 10 entries: { h1,h2,h3,h4,. . .,h9,h0 }. The triangle Gauss rule specified in options as { numer,rule } should be 6 or 7 to produce a rank sufficient stiffness. If omitted rule = 6 is assumed. The module returns Ke as an 20 × 20 symmetric matrix pertaining to the following arrangement of nodal displacements: (24.46) ue = [ u x1 u y1 u x2 u y2 u x3 u y3 u x4 u y4 . . . u x9 u y9 u x0 u y0 ]T . Remark 24.3. For symbolic work with this element the 7-point rule should be preferred because the exact
expressions of the abcissas and weigths at sample points are simpler than for the 6-point rule, as can be observed in Figure 24.3. This speeds up algebraic simplification. For numerical work the 6-point rule is slightly faster. §24.5.3. *Test Element
3 (4,4) 7
The stiffness module is tested on the superparametric triangle geometry shown in Figure 24.14, which has been already used for the quadratic triangle.
0
The corner nodes are placed at (0, 0), (6, 2) and (4, 4). The 6 side nodes are located at the thirdpoints of the sides and the interior node at the centroid. The plate has unit thickness. The material is isotropic with E = 1920 and ν = 0. The script of Figure 24.15 computes Ke using the 7-point Gauss rule and exact arithmetic.
6
8
2 (6,2)
9 5 4 1 (0,0)
Figure 24.14. Test 10-node triangle.
The returned stiffness matrix using either 6- or 7-point integration is the same, since for a superparametric element the integrand is quartic in the triangular coordinates. For the given combination of inputs the entries are exact integers:
24–17
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
24–18
 306  102  −42   −42   −21  21   −315   −333   171   153 e K =  −27  −9   −27   −9   99  −63   −144   180 
102 −42 −42 −21 21 −315 −333 171 153 306 42 42 −63 −105 351 369 −135 −117 42 1224 −408 −210 42 252 −180 −234 306 42 −408 1224 126 −294 108 −36 −378 450 −63 −210 126 1122 −306 −99 27 −99 27 −105 42 −294 −306 1938 27 −171 27 −171 351 252 108 −99 27 5265 −1215 −1539 −243 369 −180 −36 27 −171 −1215 6885 729 −891 −135 −234 −378 −99 27 −1539 729 5265 −1215 −117 306 450 27 −171 −243 −891 −1215 6885 −9 −1602 990 819 −171 81 81 −405 −405 −27 306 −2286 −459 1179 81 405 −405 −2025 −9 828 −468 −1611 315 81 81 81 81 −27 −180 1116 999 −2223 81 405 81 405 225 −108 36 −72 144 810 −324 810 −324 387 36 −108 −540 −684 −324 1134 −324 1134 −504 −108 36 171 −99 −4050 1620 810 −324 −828 36 −108 189 531 1620 −5670 −324 1134 0 0 0 0 0 0 −486 −486 −4860 1944 0 0 0 0 0 0 −486 −2430 1944 −6804
−27 −9 −1602 990 819 −171 81 81 −405 −405 5265 −1215 −3483 729 162 0 162 0 −972 0
−9 −27 −9 99 −63 −144 180 0 0 −27 −9 −27 225 387 −504 −828 0 0 306 828 −180 −108 36 −108 36 0 0  −2286 −468 1116 36 −108 36 −108 0 0  −459 −1611 999 −72 −540 171 189 0 0  1179 315 −2223 144 −684 −99 531 0 0 81 81 81 810 −324 −4050 1620 −486 −486   405 81 405 −324 1134 1620 −5670 −486 −2430   −405 81 81 810 −324 810 −324 −4860 1944   −2025 81 405 −324 1134 −324 1134 1944 −6804   −1215 −3483 729 162 0 162 0 −972 0 6885 1701 −4779 0 −162 0 −162 0 972   1701 5265 −1215 −810 0 162 0 −486 −486   −4779 −1215 6885 0 810 0 −162 −486 −2430   0 −810 0 5265 −1215 −1296 −486 −4860 1944   −162 0 810 −1215 6885 486 −2592 1944 −6804  0 162 0 −1296 486 5265 −1215 −972 0  −162 0 −162 −486 −2592 −1215 6885 0 972   0 −486 −486 −4860 1944 −972 0 12636 −2916 972 −486 −2430 1944 −6804 0 972 −2916 16524 (24.47)
The eigenvalues are [ 26397. 16597. 14937. 12285. 8900.7 7626.2 5417.8 4088.8 3466.8 3046.4 1751.3 1721.4 797.70 551.82 313.22 254.00 28.019 0 0 0 ]
(24.48)
The 3 zero eigenvalues pertain to the three independent rigid-body modes in two dimensions. The 17 other eigenvalues are positive. Consequently the computed Ke has the correct rank of 17.
24–18
24–19
§24. References ClearAll[Em,Ν,a,b,e,h]; Em=1920; Ν=0; h=1; ncoor={{0,0},{6,2},{4,4}}; x4=(2*x1+x2)/3; x5=(x1+2*x2)/3; y4=(2*y1+y2)/3; y5=(y1+2*y2)/3; x6=(2*x2+x3)/3; x7=(x2+2*x3)/3; y6=(2*y2+y3)/3; y7=(y2+2*y3)/3; x8=(2*x3+x1)/3; x9=(x3+2*x1)/3; y8=(2*y3+y1)/3; y9=(y3+2*y1)/3; x0=(x1+x2+x3)/3; y0=(y1+y2+y3)/3; ncoor= {{x1,y1},{x2,y2},{x3,y3},{x4,y4},{x5,y5},{x6,y6},{x7,y7}, {x8,y8},{x9,y9},{x0,y0}}; Emat=Em/(1-Ν^2)*{{1,Ν,0},{Ν,1,0},{0,0,(1-Ν)/2}}; Ke=Trig10IsoPMembraneStiffness[ncoor,Emat,h,{False,7}]; Ke=Simplify[Ke]; Print[Ke//MatrixForm]; ev=Chop[Eigenvalues[N[Ke]]]; Print['eigs of Ke=',ev];
Figure 24.15. Script for testing cubic triangle of Figure 24.14.
Notes and Bibliography The 3-node, 6-node and 10-node plane stress triangular elements are generated by complete polynomials in twodimensions. In order of historical appearance: 1.
The three-node linear triangle, also known as Constant Strain Triangle (CST) and Turner triangle, was developed as “triangular skin panel” by Turner, Clough and Martin in 1951–53 [31,32] using interelement flux assumptions, and published in 1956 [164]. It is not clear when the assumed-displacement derivation, which yields the same stiffness matrix, was done first. The displacement derivation is mentioned in passing by Clough in [25] and worked out in the theses of Melosh [114] and Wilson [177].
2.
The six-node quadratic triangle, also known as Linear Strain Triangle (LST) and Veubeke triangle, was developed by B. M. Fraeijs de Veubeke in 1962–63 [186]; published 1965 [66].
3.
The ten-node cubic triangle, also known as Quadratic Strain Triangle (QST), was developed by the writer in 1965; published 1966 [44]. Shape functions for the cubic triangle were presented there but used for plate bending instead of plane stress.
A version of the cubic triangle with freedoms migrated to corners and recombined to produce a “drilling rotation” at corners, was used in static and dynamic shell analysis in Carr’s thesis under Ray Clough [23,28]. The drillingfreedom idea was independently exploited for rectangular and quadrilateral elements, respectively, in the theses of Abu-Ghazaleh [2] and Willam [176], both under Alex Scordelis. A variant of the Willam quadrilateral, developed by Bo Almroth at Lockheed, has survived in the nonlinear shell analysis code STAGS as element 410 [137]. Numerical integration came into FEM by the mid 1960s. Five triangle integration rules were tabulated in the writer’s thesis [44, pp. 38–39]. These were gathered from three sources: two papers by Hammer and Stroud [84,85] and the 1964 Handbook of Mathematical Functions [1, §25.4]. They were adapted to FEM by converting Cartesian abscissas to triangle coordinates. The table has been reproduced in Zienkiewicz’ book since the second edition [184, Table 8.2] and, with corrections and additions, in the monograph of Strang and Fix [149, p. 184]. The monograph by Stroud [152] contains a comprehensive collection of quadrature formulas for multiple integrals. That book gathers most of the formulas known by 1970, as well as references until that year. (Only a small fraction of Stroud’s tabulated rules, however, are suitable for FEM work.) The collection has been periodically kept up to date by Cools [34–36], who also maintains a dedicated web site: http://www.cs.kuleuven.ac.be/~nines/research/ecf/ecf.html This site provides rule information in 16- and 32-digit accuracy for many geometries and dimensionalities as well as a linked “index card” of references to source publications. References Referenced items moved to Appendix R.
24–19
24–20
Chapter 24: IMPLEMENTATION OF ISO-P TRIANGULAR ELEMENTS
Homework Exercises for Chapter 24 Implementation of Iso-P Triangular Elements 3
EXERCISE 24.1 [C:20] Write an element stiffness module and a shape
function module for the 4-node “transition” iso-P triangular element with only one side node: 4, which is located between 1 and 2. See Figure E24.1. The shape functions are N1 = ζ1 − 2ζ1 ζ2 , N2 = ζ2 − 2ζ1 ζ2 , N3 = ζ3 and N4 = 4ζ1 ζ2 . Use the three interior point integration rule=3.
2
4
Test the element for the geometry of the triangle depicted in Figure 24.7, removing nodes 5 and 6, and with E = 2880, ν = 1/3 and h = 1. Report results for the stiffness matrix Ke and its 8 eigenvalues. Note: Notebook Trig6Stiffness.nb for the 6-node triangle, posted on the index of Chapter 24, may be used as “template” in support of this Exercise. Partial results: K 11 = 1980, K 18 = 1440.
1
Figure E24.1. The 4-node transition triangle for Exercise 24.1.
EXERCISE 24.2 [A/C:5+20] As in the foregoing Exercise, but now write the module for a five-node triangular
“transition” element that lacks midnode 6 opposite corner 2. Begin by deriving the five shape functions. EXERCISE 24.3 [A+C:25] Consider the superparametric straight-sided 6-node triangle where side nodes are located at the midpoints. By setting x4 = 12 (x1 + x2 ), x5 = 12 (x2 + x3 ), x3 = 12 (x3 + x1 ), y4 = 12 (y1 + y2 ), y5 = 12 (y2 + y3 ), y3 = 12 (y3 + y1 ) in (24.26), deduce that
0003
J=
1 1 1 3x1 2x1 + x2 2x1 + x3 3y1 2y1 + y2 2y1 + y3
0004
0003
ζ1 +
1 1 1 x1 + 2x2 3x2 2x2 + x3 y1 + 2y2 3y2 2y2 + y3
0004
0003
ζ2 +
1 1 1 x1 + 2x3 x2 + 2x3 3x3 y1 + 2y3 y2 + 2y3 3y3
0004
ζ3 . (E24.1)
This contradicts publications that, by mistakingly assuming that the results for the linear triangle (Chapter 15) can be extended by analogy, take 0003 0004 1 1 1 J = x1 x2 x3 (E24.2) y1 y2 y3 Show, however, that 0003 0004 y23 y32 1 2J = 2A = det J = x3 y12 + x1 y23 + x2 y31 , P= (E24.3) y31 x13 2J y x21 12 are the same for both (E24.1) and (E24.2). Thus the mistake has no effect on the computation of derivatives. EXERCISE 24.4 [A/C:15+15] Consider the superparametric straight-sided 6-node triangle where side nodes are
at the midpoints of the sides and the thickness h is constant. Using the 3-midpoint quadrature rule show that the element stiffness can be expressed in a closed form obtained in 1966 [44]:
0013
Ke = 13 Ah B1T EB1 + B2T EB2 + B3T EB3
0014
in which A is the triangle area and
0003
1 B1 = 2A
0003 1 B2 = 2A
0003 1 B3 = 2A
y32 0 y31 0 y12 0 2y23 0 2y32 0 2y23 0 0 x23 0 x13 0 x21 0 2x32 0 2x23 0 2x32 x23 y32 x13 y31 x21 y12 2x32 2y23 2x23 2y32 2x32 2y23 y23 0 y13 0 y12 0 2y31 0 2y31 0 2y13 0 0 x32 0 x31 0 x21 0 2x13 0 2x13 0 2x31 x32 y23 x31 y13 x21 y12 2x13 2y31 2x13 2y31 2x31 2y13 y23 0 y31 0 y21 0 2y21 0 2y12 0 2y12 0 0 x32 0 x13 0 x12 0 2x12 0 2x21 0 2x21 x32 y23 x13 y31 x12 y21 2x12 2y21 2x21 2y12 2x21 2y12
24–20
(E24.4)
0004 0004 (E24.5)
0004
24–21
§24. References
With this form Ke can be computed in approximately 1000 floating-point operations. Using next the 3-interior-point quadrature rule, show that the element stiffness can be expressed again as (E24.4) but with
0003
1 B1 = 6A
0003 1 B2 = 6A
0003
5y23 0 y13 0 y21 0 2y21 + 6y31 0 2y32 0 6y12 + 2y13 0 0 5x32 0 x31 0 x12 0 2x12 + 6x13 0 2x23 0 6x21 + 2x31 5x32 5y23 x31 y13 x12 y21 2x12 + 6x13 2y21 + 6y31 2x23 2y32 6x21 + 2x31 6y12 + 2y13 y32 0 5y31 0 y21 0 2y21 + 6y23 0 6y12 + 2y32 0 2y13 0 0 x23 0 5x13 0 x12 0 2x12 + 6x32 0 6x21 + 2x23 0 2x31 x23 y32 5x13 5y31 x12 y21 2x12 + 6x32 2y21 + 6y23 6x21 + 2x23 6y12 + 2y32 2x31 2y13
0004 0004 0004
y32 0 y13 0 5y12 0 2y21 0 6y31 + 2y32 0 2y13 + 6y23 0 0 x23 0 x31 0 5x21 0 2x12 0 6x13 + 2x23 0 2x31 + 6x32 x23 y32 x31 y13 5x21 5y12 2x12 2y21 6x13 + 2x23 6y31 + 2y32 2x31 + 6x32 2y13 + 6y23 (E24.6) The fact that two very different expressions yield the same Ke explains why sometimes authors rediscover the same element derived with different methods. 1 B3 = 6A
Yet another set that produces the correct stiffness is
0003 1 B1 = 6A
0003 1 B2 = 6A
0003 1 B3 = 6A
y23 0 y31 0 y21 0 2y21 0 2y12 0 2y12 0 0 x32 0 x13 0 x12 0 2x12 0 2x21 0 2x21 x32 y23 x13 y31 x12 y21 2x12 2y21 2x21 2y12 2x21 2y12 y32 0 y31 0 y12 0 2y23 0 2y32 0 2y23 0 0 x23 0 x13 0 x21 0 2x32 0 2x23 0 2x32 x23 y32 x13 y31 x21 y12 2x32 2y23 2x23 2y32 2x32 2y23 y23 0 y13 0 y12 0 2y31 0 2y31 0 2y13 0 0 x32 0 x31 0 x21 0 2x13 0 2x13 0 2x31 x32 y23 x31 y13 x21 y12 2x13 2y31 2x13 2y31 2x31 2y13
0004 0004 (E24.7)
0004
EXERCISE 24.5 [A/C:40] (Research paper level) Characterize the most general form of the Bi (i = 1, 2, 3)
matrices that produce the same stiffness matrix Ke in (E24.4). (Entails solving an algebraic Riccati equation.)
EXERCISE 24.6 [A/C:25]. Derive the 4-node transition triangle by forming the 6-node stiffness and applying
MFCs to eliminate 5 and 6. Prove that this technique only works if sides 1–3 and 2–3 are straight with nodes 5 and 6 initially at the midpoint of those sides.
24–21
25
.
The Assembly Process
25–1
25–2
Chapter 25: THE ASSEMBLY PROCESS
TABLE OF CONTENTS Page
§25.1. Introduction §25.2. Simplifications §25.3. Simplified Assemblers §25.3.1. A Plane Truss Example Structure . §25.3.2. Implementation . . . . . . §25.4. MET Assemblers §25.4.1. A Plane Stress Assembler . . . §25.4.2. Implementation . . . . . . §25.5. MET-VFC Assemblers §25.5.1. Node Freedom Arrangement . . §25.5.2. Node Freedom Signature . . . §25.5.3. The Node Freedom Allocation Table §25.5.4. The Node Freedom Map Table . §25.5.5. The Element Freedom Signature . §25.5.6. The Element Freedom Table . . §25.5.7. A Plane Trussed Frame Structure . §25.5.8. Implementation . . . . . . §25.6. *Handling MultiFreedom Constraints §25. Notes and Bibliography . . . . . . . . . . . . §25. Exercises . . . . . . . . . . . .
25–2
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25–3 25–3 25–4 25–4 25–6 25–7 25–7 25–9 25–10 25–10 25–11 25–11 25–12 25–12 25–13 25–14 25–17 25–17 25–20 25–21
25–3
§25.2
SIMPLIFICATIONS
§25.1. Introduction Chapters 20, 23 and 24 dealt with element level operations. Sandwiched between element processing and solution there is the assembly process that constructs the master stiffness equations. Assembler examples for special models were given as recipes in the complete programs of Chapters 21 and 22. In the present chapter assembly will be studied with more generality. The position of the assembler in a DSM-based code for static analysis is sketched in Figure 25.1. In most codes the assembler is implemented as a element library loop driver. This means that instead of forming all elements first and then assembling, the assembler constructs one element at a time in a loop, and immediately merges it into the master equations. This is the merge loop. Model definition data: geometry element connectivity material fabrication freedom activity
Assembler
merge loop
Element Stiffness Matrices
^
K
K
e
Modify Eqs for BCs
K
Equation Solver
Some equation solvers apply BCs and solve simultaneously Nodal displacements
ELEMENT LIBRARY To postprocessor
Figure 25.1. Role of assembler in FEM program.
Assembly is the most complicated stage of a production finite element program in terms of data flow. The complexity is not due to mathematical operations, which merely involve matrix addition, but interfacing with a large element library. Forming an element requires access to geometry, connectivity, material, and fabrication data. (In nonlinear analysis, to the state as well.) Merge requires access to freedom activity data, as well as knowledge of the sparse matrix format in which K is stored. As illustrated in Figure 25.1, the assembler sits at the crossroads of this data exchange. The coverage of the assembly process for a general FEM implementation is beyond the scope of an introductory course. Instead this Chapter takes advantage of assumptions that lead to coding simplifications. This allows the basic aspects to be covered without excessive delay. §25.2. Simplifications The assembly process is considerably simplified if the FEM implementation has these properties: •
All elements are of the same type. For example: all elements are 2-node plane bars.
•
The number and configuration of degrees of freedom at each node is the same.
•
There are no “gaps” in the node numbering sequence.
•
There are no multifreedom constraints treated by master-slave or Lagrange multiplier methods.
•
The master stiffness matrix is stored as a full symmetric matrix. 25–3
25–4
Chapter 25: THE ASSEMBLY PROCESS
(a)
(b) 1 (1,2)
E = 3000 and A = 2 for all bars 1 3
2 (3,4)
(2)
3 (5,6)
y
(1)
2
(2)
(3)
3
(4)
assembly
x
4 4
(5)
(4) 4 (7,8)
(5)
(3) 4
(1)
1 (1,2)
2 (3,4)
3 (5,6)
4 (7,8) Figure 25.2. Left: Example plane truss structure. (a): model definition; (b) disconnection → assembly process. Numbers in parentheses written after node numbers are global DOF numbers.
If the first four conditions are met the implementation is simpler because the element freedom table described below can be constructed “on the fly” from the element node numbers. The last condition simplifies the indexing to access entries of K. §25.3. Simplified Assemblers In this section all simplifying assumptions listed in §25.2 are assumed to hold. This allows us to focus on local-to-global DOF mapping. The process is illustrated on a plane truss structure. §25.3.1. A Plane Truss Example Structure The plane truss shown in Figure 25.2(a) will be used to illustrate the details of the assembly process. The structure has 5 elements, 4 nodes and 8 degrees of freedom. The disconnection-to-assembly process is pictured in Figure 25.2(b). Begin by clearing all entries of the 8 × 8 master stiffness matrix K to zero, so that we effectively start with the null matrix:   0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 2   0 0 0 0 0 0 0 0 3   0 0 0 0 0 0 0 0 4 (25.1) K=  0 0 0 0 0 0 0 0 5   0 0 0 0 0 0 0 0 6   0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 8 The numbers written after each row of K are the global DOF numbers. Element (1) joins nodes 1 and 2. The global DOFs of those nodes are 2 × 1 − 1 = 1, 2 × 1 = 1, 2 × 2 − 1 = 3 and 2 × 2 = 4. See Figure 25.2(b). Those four numbers are collected into an array called the element freedom table, or EFT for short. The element stiffness matrix K(1) is listed on 25–4
25–5
§25.3
SIMPLIFIED ASSEMBLERS
the left below with the EFT entries annotated after the matrix rows. On the right is K upon merging the element stiffness:  
1500  0  −1500 0
 0 −1500 0 0 0 0  0 1500 0 0 0 0
1500  0   −1500   0   0   0  0 0
1 2 3 4
 0 −1500 0 0 0 0 0 0 0 0 0 0 0 0  0 1500 0 0 0 0 0   0 0 0 0 0 0 0  0 0 0 0 0 0 0  0 0 0 0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 2 3 4
(25.2)
Element (2) joins nodes 2 and 3. The global DOFs of those nodes are 2 × 2 − 1 = 3, 2 × 2 = 4, 2 × 3 − 1 = 5 and 2 × 3 = 6; thus the EFT is { 3,4,5,6 }. Matrices K(2) and K upon merge are  
1500  0  −1500 0
 0 −1500 0 0 0 0  0 1500 0 0 0 0
 0 −1500 0 0 0 0 0 0 0 0 0 0 0 0  0 3000 0 −1500 0 0 0   0 0 0 0 0 0 0  0 −1500 0 1500 0 0 0   0 0 0 0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0 0 0
1500  0   −1500   0   0   0  0 0
3 4 5 6
3 4 5 6
(25.3)
Element (3) joins nodes 1 and 4. Its EFT is { 1,2,7,8 }. Matrices K(3) and K upon merge are  
 768 −576 −768 576  −576 432 576 −432    −768 576 768 −576 576 −432 −576 432
1 2 7 8
2268  −576   −1500   0   0   0  −768 576
Element (4) joins nodes 2 and 4. Its EFT is { 3,4,7,8 }. Matrices K(4)  
 0 0 0 0  0 2000 0 −2000    0 0 0 0 0 −2000 0 2000
3 4 7 8
2268  −576   −1500   0   0   0  −768 576
 1 576 −432  2  0   0   0   0   −576 7 8 432 (25.4) and K upon merge are
−576 −1500 0 0 0 −768 432 0 0 0 0 576 0 3000 0 −1500 0 0 0 0 0 0 0 0 0 −1500 0 1500 0 0 0 0 0 0 0 0 576 0 0 0 0 768 −432 0 0 0 0 −576
−576 −1500 0 0 0 −768 432 0 0 0 0 576 0 3000 0 −1500 0 0 0 0 2000 0 0 0 0 −1500 0 1500 0 0 0 0 0 0 0 0 576 0 0 0 0 768 −432 0 −2000 0 0 −576 25–5
 576 −432   0 3  −2000  4  0   0   −576 7 8 2432 (25.5)
25–6
Chapter 25: THE ASSEMBLY PROCESS
Finally, element (5) joins nodes 3 and 4. Its EFT is { 5,6,7,8 }. Matrices K(5) and K upon merge are   2268 −576 −1500 0 0 0 −768 576 0 0 0 0 576 −432   −576 432     3000 0 −1500 0 0 0  768 576 −768 −576 5  −1500 0   0 0 2000 0 0 0 −2000   576 432 −576 −432  6  0     0 −1500 0 2268 576 −768 −576  5 −768 −576 768 576 7  0   0 0 0 576 432 −576 −432  6 −576 −432 576 432 8  0   7 −768 576 0 0 −768 −576 1536 0 8 576 −432 0 −2000 −576 −432 0 2864 (25.6) Since all elements have been processed, (25.6) is the master stiffness before application of boundary conditions. PlaneTrussMasterStiffness[nodxyz_,elenod_,elemat_,elefab_, eleopt_]:=Module[{numele=Length[elenod],numnod=Length[nodxyz], e,ni,nj,eft,i,j,ii,jj,ncoor,Em,A,options,Ke,K}, K=Table[0,{2*numnod},{2*numnod}]; For [e=1, e<=numele, e++, {ni,nj}=elenod[[e]]; eft={2*ni-1,2*ni,2*nj-1,2*nj}; ncoor={nodxyz[[ni]],nodxyz[[nj]]}; Em=elemat[[e]]; A=elefab[[e]]; options=eleopt; Ke=PlaneBar2Stiffness[ncoor,Em,A,options]; For [i=1, i<=4, i++, ii=eft[[i]]; For [j=i, j<=4, j++, jj=eft[[j]]; K[[jj,ii]]=K[[ii,jj]]+=Ke[[i,j]] ]; ]; ]; Return[K] ];
Figure 25.3. Plane truss assembler module. nodxyz={{-4,3},{0,3},{4,3},{0,0}}; elenod= {{1,2},{2,3},{1,4},{2,4},{3,4}}; elemat= Table[3000,{5}]; elefab= Table[2,{5}]; eleopt= {True}; K=PlaneTrussMasterStiffness[nodxyz,elenod,elemat,elefab,eleopt]; Print['Master Stiffness of Plane Truss of Fig 25.2:']; K=Chop[K]; Print[K//MatrixForm]; Print['Eigs of K=',Chop[Eigenvalues[N[K]]]]; Master Stiffness of Plane Truss of Fig 25.2: 2268. −576. −1500. 0 0 −576. 432. 0 0 0 −1500. 0 3000. 0 −1500. 0 0 0 2000. 0 0 0 −1500. 0 2268. 0 0 0 0 576. −768. 576. 0 0 −768. 576. −432. 0 −2000. −576.
0 0 0 0 576. 432. −576. −432.
−768. 576. 0 0 −768. −576. 1536. 0
576. −432. 0 −2000. −576. −432. 0 2864.
Eigs of K={5007.22, 4743.46, 2356.84, 2228.78, 463.703, 0, 0, 0}
Figure 25.4. Plane truss assembler tester and output results.
25–6
25–7
§25.4
MET ASSEMBLERS
§25.3.2. Implementation Figure 25.3 shows a a Mathematica implementation of the assembly process just described. This assembler calls the element stiffness module PlaneBar2Stiffness of Figure 20.2. The assembler is invoked by (25.7)
K = SpaceTrussMasterStiffness[nodxyz,elenod,elemat,elefab,prcopt]
The five arguments: nodxyz, elenod, elemat, elefab, and prcopt have the same function as those described in §21.1.3 for the SpaceTrussMasterStiffness assembler. In fact, comparing the code in Figure 25.3 to that of Figure 21.1, the similarities are obvious. Running the script listed in the top of Figure 25.4 produces the output shown in the bottom of that figure. The master stiffness (25.6) is reproduced. The eigenvalue analysis verifies that the assembled stiffness has the correct rank of 8 − 3 = 5. 5(9,10)
4 (7,8)
§25.4. MET Assemblers The next complication occurs when elements of different types are to be assembled, while the number and configuration of degrees of freedom at each node remains the same. This scenario is common in simple program for analysis of specific problems: plane stress, plate bending, solids. This will be called a multiple-element-type assembler, or MET assembler for short. The only incremental change with respect to the simplest assemblers is that an array of element types, denoted here by eletyp, has to be provided.
y 4
(1)
(3)
5 in
5 (2)
5 in
trig (1)
1(1,2)
bar (3)
(4)
6 in
2(3,4)
4 (7,8)
1 in 1
2
bar (4) 1in
4 in
3(5,6)
assembly
quad (2)
3
E (plate, bars) = 10000 ksi ν (plate)= 0.25 h (plate) = 0.3 in A (bars) = 2 sq in
5(9,10)
x
1(1,2) 2(3,4)
3(5,6)
Figure 25.5. A plane stress structure used to test a MET assembler.
The assembler calls the appropriate element stiffness module according to type, and builds the local to global DOF mapping accordingly. §25.4.1. A Plane Stress Assembler This kind of assembler will be illustrated using the module PlaneStressMasterStiffness. As its name indicates, this is suitable for use in plane stress analysis programs. We make the following assumptions: 1
All nodes in the FEM mesh have 2 DOFs, namely the {u x , u y } displacements. There are no node gaps or MFCs. The master stiffness is stored as a full matrix.
2
Element types can be: 2-node bars, 3-node linear triangles, ot 4-node bilinear quadrilaterals. These elements can be freely intermixed. Element types are identified with character strings 'Bar2', 'Trig3' and 'Quad4', respectively. 25–7
25–8
Chapter 25: THE ASSEMBLY PROCESS
As noted above, the chief modification over the simplest assembler is that a different stiffness module is invoked as per type. Returned stiffness matrices are generally of different order; in our case 4 × 4, 6 × 6 and 8 × 8 for the bar, triangle and quadrilateral, respectively. But the construction of the EFT is immediate, since node n maps to global freedoms 2n − 1 and 2n. The technique is illustrated with the 4-element, 5-node plane stress structure shown in Figure 25.5. It consists of one triangle, one quadrilateral, and two bars. The assembly process goes as follows. Element (1) is a 3-node triangle with nodes 3, 5 and 4, whence the EFT is { 5,6,9,10,7,8 }. The element stiffness is 
K(1)
 500.0 0 −500.0 −600.0 0 600.0 0 1333.3 −400.0 −1333.3 400.0 0    −500.0 −400.0 2420.0 1000.0 −1920.0 −600.0   =   −600.0 −1333.3 1000.0 2053.3 −400.0 −720.0    0 400.0 −1920.0 −400.0 1920.0 0 600.0 0 −600.0 −720.0 0 720.0
5 6 9 10 7 8
(25.8)
Element (2) is a 4-node quad with nodes 2, 3, 4 and 1, whence the EFT is { 3,4,5,6,7,8,1,2 }. The element stiffness computed with a 2 × 2 Gauss rule is 
 420.27 3 −850.95  4  −164.05  5  −549.81  6 K(2)  −668.76  7  −600.15  8  412.54 1 2000.9 2 (25.9) Element (3) is a 2-node bar with nodes 1 and 4, whence the EFT is { 1,2,7,8 }. The element stiffness is   0 0 0 0 1  0 4000.0 0 −4000.0  2 (25.10) K(3) =   0 0 0 0 7 0 −4000.0 0 4000.0 8 2076.4  395.55   −735.28   −679.11 =  −331.78   −136.71  −1009.3 420.27
395.55 2703.1 −479.11 −660.61 −136.71 −1191.5 220.27 −850.95
−735.28 −479.11 2067.1 215.82 386.36 427.34 −1718.1 −164.05
−679.11 −660.61 215.82 852.12 627.34 358.30 −164.05 −549.81
−331.78 −136.71 386.36 627.34 610.92 178.13 −665.50 −668.76
−136.71 −1191.5 427.34 358.30 178.13 1433.4 −468.76 −600.15
−1009.3 220.27 −1718.1 −164.05 −665.50 −468.76 3393.0 412.54
Element (4) is a 2-node bar with nodes 1 and 2, whence the EFT is { 1,2,3,4 }. The element stiffness is   7071.1 −7071.1 −7071.1 7071.1 1  −7071.1 7071.1 7071.1 −7071.1  2 (25.11) K(4) =   −7071.1 7071.1 7071.1 −7071.1 3 7071.1 −7071.1 −7071.1 7071.1 4 Upon assembly, the master stiffness of the plane stress structure, printed with one digit after the 25–8
25–9
§25.4
MET ASSEMBLERS
PlaneStressMasterStiffness[nodxyz_,eletyp_,elenod_, elemat_,elefab_,prcopt_]:=Module[{numele=Length[elenod], numnod=Length[nodxyz],ncoor,type,e,enl,neldof, i,n,ii,jj,eftab,Emat,th,numer,Ke,K}, K=Table[0,{2*numnod},{2*numnod}]; numer=prcopt[[1]]; For [e=1,e<=numele,e++, type=eletyp[[e]]; If [!MemberQ[{'Bar2','Trig3','Quad4'},type], Print['Illegal type', ' of element e=',e,' Assembly interrupted']; Return[K]]; enl=elenod[[e]]; n=Length[enl]; eftab=Flatten[Table[{2*enl[[i]]-1,2*enl[[i]]},{i,1,n}]]; ncoor=Table[nodxyz[[enl[[i]]]],{i,n}]; If [type'Bar2', Em=elemat[[e]]; A=elefab[[e]]; Ke=PlaneBar2Stiffness[ncoor,Em,A,{numer}] ]; If [type'Trig3', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Trig3IsoPMembraneStiffness[ncoor,Emat,th,{numer}] ]; If [type'Quad4', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,th,{numer,2}] ]; neldof=Length[Ke]; For [i=1,i<=neldof,i++, ii=eftab[[i]]; For [j=i,j<=neldof,j++, jj=eftab[[j]]; K[[jj,ii]]=K[[ii,jj]]+=Ke[[i,j]] ]; ]; ]; Return[K]; ];
Figure 25.6. Implementation of MET assembler for plane stress.
decimal point, is  10464.0 −6658.5 −8080.4  −6658.5 13072.0 7491.3  −8080.4 7491.3 9147.5   7291.3 −7922.0 −6675.5   −1718.1 −164.1 −735.3  −164.1 −549.8 −679.1   −665.5 −668.8 −331.8   −468.8 −4600.2 −136.7 
0 0 0 0 −500.0 −400.0 −1920.0 −600.0 2420.0 1000.0
0 0 0  0  −600.0   −1333.3  −400.0   −720.0   1000.0 2053.3 (25.12)
[ 32883. 10517.7 5439.33 3914.7 3228.99 2253.06 2131.03 0 0 0 ]
(25.13)
0 0
0 0
7291.3 −1718.1 −164.1 −665.5 −468.8 −7922.0 −164.1 −549.8 −668.8 −4600.2 −6675.5 −735.3 −679.1 −331.8 −136.7 9774.1 −479.1 −660.6 −136.7 −1191.5 −479.1 2567.1 215.8 386.4 1027.3 −660.6 215.8 2185.5 1027.3 358.3 −136.7 386.4 1027.3 2530.9 178.1 −1191.5 1027.3 358.3 178.1 6153.4 0 0 −500.0 −400.0 −1920.0 −600.0 0 0 −600.0 −1333.3 −400.0 −720.0
The eigenvalues of K are
which displays the correct rank. §25.4.2. Implementation An implementation of the MET plane stress assembler as a Mathematica module called PlaneStressMasterStiffness is shown in Figure 25.6. The assembler is invoked by K = PlaneStressMasterStiffness[nodxyz,eletyp,elenod,elemat,elefab,prcopt] (25.14) 25–9
25–10
Chapter 25: THE ASSEMBLY PROCESS
The only additional argument is eletyp, which is a list of element types. The script listed in the top of Figure 25.7 runs module PlaneStressMasterStiffness with the inputs appropriate to the problem defined in Figure 25.5. The output shown in the bottom of the figure reproduces the master stiffness matrix (25.12) and the eigenvalues (25.13). .
Em=10000; ν=1/4; h=3/10; A=2; Emat=Em/(1-ν^2)*{{1,ν,0},{ν,1,0},{0,0,(1-ν)/2}}; nodxyz= {{0,1},{1,0},{5,0},{0,6},{5,6}}; elenod= {{3,5,4},{2,3,4,1},{1,4},{1,2}}; eletyp= {'Trig3','Quad4','Bar2','Bar2'}; elemat= {Emat,Emat,Em,Em}; elefab= {h,h,A,A}; prcopt={True}; K=PlaneStressMasterStiffness[nodxyz,eletyp,elenod,elemat,elefab,prcopt]; Print['Master Stiffness Matrix of Plane Stress Structure:']; Print[SetPrecision[Chop[K],5]//MatrixForm]; Print['Eigs of K:',Chop[Eigenvalues[N[K]]]]; Master Stiffness Matrix of Plane Stress Structure: 10464. − 6658.5 − 8080.4 7291.3 − 1718.1 − 164.05 − 665.50 − 468.76 0 0
− 6658.5 13072. 7491.3 − 7922.0 − 164.05 − 549.81 − 668.76 − 4600.2 0 0
− 8080.4 7491.3 9147.5 − 6675.5 − 735.28 − 679.11 − 331.78 − 136.71 0 0
7291.3 − 7922.0 − 6675.5 9774.1 − 479.11 − 660.61 − 136.71 − 1191.5 0 0
− 1718.1 − 164.05 − 735.28 − 479.11 2567.1 215.82 386.36 1027.3 − 500.00 − 600.00
− 164.05 − 549.81 − 679.11 − 660.61 215.82 2185.5 1027.3 358.30 − 400.00 − 1333.3
− 665.50 − 668.76 − 331.78 − 136.71 386.36 1027.3 2530.9 178.13 − 1920.0 − 400.00
− 468.76 − 4600.2 − 136.71 − 1191.5 1027.3 358.30 178.13 6153.4 − 600.00 − 720.00
0 0 0 0 − 500.00 − 400.00 − 1920.0 − 600.00 2420.0 1000.0
0 0 0 0 − 600.00 − 1333.3 − 400.00 − 720.00 1000.0 2053.3
Eigs of K: {32883., 10517.7, 5439.33, 3914.7, 3228.99, 2253.06, 2131.03, 0, 0, 0}
Figure 25.7. Script for assembling the plane stress model of Figure 25.5 and output results.
§25.5. MET-VFC Assemblers The next complication beyond multiple element types is a major one: allowing nodes to have different freedom configurations. An assembler that handles this feature will be called a MET-VFC assembler, where VFC is an acronym for Variable Freedom Configuration. A MET-VFC assembler gets closer to what is actually implemented in production FEM programs. A assembler of this kind can handle orientation nodes as well as node “gaps” automatically. Only two advanced attributes remain: allowing treatment of MFCs by Lagrange multipliers, and handling a sparse matrix storage scheme. The former can be done by an extension of the concept of element. The latter requires major programming modifications, however, and is not considered in this Chapter. The implementation of a MET-VFC assembler requires the definition of additional data structures pertaining to freedoms, at both node and element levels. These are explained in the following subsections before giving a concrete application example. 25–10
25–11
§25.5
MET-VFC ASSEMBLERS
§25.5.1. Node Freedom Arrangement A key decision made by the implementor of a FEM program for structural analysis is: what freedoms will be allocated to nodes, and how are they arranged? While many answers are possible, we focus here on the most common arrangement for a linear three-dimensional FEM program.1 At each node n three displacements {u xn , u yn , u zn } and three infinitesimal rotations {θxn , θ yn , θzn } are chosen as freedoms and ordered as follows: u xn , u yn , u zn , θxn , θ yn , θzn
(25.15)
This is a Node Freedom Arrangement or NFA. Once the decision is made on a NFA, the position of a freedom never changes.2 Thus if (25.15) is adopted, position #2 is forever associated to u yn . The length of the Node Freedom Arrangement is called the NFA length and often denoted (as a variable) by nfalen. For the common choice (25.15) of 3D structural codes, nfalen is 6. §25.5.2. Node Freedom Signature Having picked a NFA such as (25.15) does not mean that all such freedoms need to be present at a node. Take for example a model, such as a space truss, that only has translational degrees of freedom. Then only {u xn , u yn , u zn } will be used, and it becomes unnecessary to carry rotations in the master equations. There are also cases where allocations may vary from node to node. To handle such scenarios in general terms a Node Freedom Signature or NFS, is introduced. The NFS is a sequence of zero-one integers. If the jth entry is one, it means that the jth freedom in the NFA is allocated; whereas if zero that freedom is not in use. For example, suppose that the underlying NFA is (25.15) and that node 5 uses the three freedoms {u x5 , u y5 , θz5 }. Then its NFS is { 1,1,0,0,0,1 }
(25.16)
If a node has no allocated freedoms (e.g, an orientation node), or has never been defined, its NFS contains only zeros. It is convenient to decimally pack signatures into integers to save storage because allocating a full integer to hold just 0 or 1 is wasteful. The packed NFS (25.16) would be 110001. This technique requires utilities for packing and unpacking. Mathematica functions that provide those services are listed in Figure 25.8. All of them use built-in functions. A brief description follows. p = PackNodeFreedomSignature[s]
Packs the NFS list s into integer p.
s = UnpackNodeFreedomSignature[p,m] Unpacks the NFS p into a list s of length m, where m is the NFA length. (The second argument is necessary in case the NFS has leading zeros.) k = CountNodeFreedoms[p]
Returns count of ones in p.
1
This is the scheme used by most commercial codes. In a geometrically nonlinear FEM code, however, finite rotation measures must be used. Consequently the infinitesimal node rotations {θx , θ y , θz } must be replaced by something else.
2
NFA positions #7 and beyond are available for additional freedom types, such as Lagrange multipliers, temperatures, pressures, fluxes, etc. It is also possible to reserve more NFA positions for structural freedoms.
25–11
25–12
Chapter 25: THE ASSEMBLY PROCESS
PackNodeFreedomSignature[s_]:=FromDigits[s,10]; UnpackNodeFreedomSignature[p_,m_]:=IntegerDigits[p,10,m]; CountNodeFreedoms[p_]:=DigitCount[p,10,1]; Figure 25.8. Node freedom manipulation utility functions.
Example 25.1. PackNodeFreedomSignature[{ 1,1,0,0,0,1 }] returns 110001. The inverse is UnpackNodeFreedomSignature[110001,6], which returns { 1,1,0,0,0,1 }. CountNodeFreedoms[110001] returns 3.
§25.5.3. The Node Freedom Allocation Table The Node Freedom Allocation Table, or NFAT, is a node by node list of packed node freedom signatures. In Mathematica the list containing this table is internally identified as nodfat. The configuration of the NFAT for the plane stress example structure of Figure 25.5 is nodfat = { 110000,110000,110000,110000,110000 } = Table[110000,{ 5 }]. (25.17) This says that only two freedoms: {u x , u y }, are used at each of the five nodes. A zero entry in the NFAT flags a node with no allocated freedoms. This is either an orientation node, or an undefined one. The latter case happens when there is a node numbering gap, which is a common occurence when certain mesh generators are used. §25.5.4. The Node Freedom Map Table The Node Freedom Map Table, or NFMT, is an array with one entry per node. Suppose that node n has k ≥ 0 allocated freedoms. These have global equation numbers i + j, j = 1, . . . , k. This i is called the base equation index: it represent the global equation number before the first equation to which node n contributes. Obviously i = 0 if n = 1. Base equation indices for all nodes are recorded in the NFMT.3 In Mathematica code the table is internally identified as nodfmt. Figure 25.9 lists a module that constructs the NFMT given the NFAT. It is invoked as nodfmt = NodeFreedomMapTable[nodfat]
(25.18)
The only argument is the NFAT. The module builds the NFMT by simple incrementation, and returns it as function value. Example 25.2. The NFAT for the plane truss-frame structure of Figure 25.11 is nodfat = { 110001,110000,
110001,000000,110001 }. The node freedom counts are { 3,2,3,0,3 }. The call nodfmt = NodeFreedomMapTable[nodfat]
(25.19)
returns { 0,3,5,8,8 } in nodfmt. This can be easily verified visually: 0+3=3, 0+3+2=5, etc.
Figure 25.9 also lists a function TotalFreedomCount that returns the total number of freedoms in the master equations, given the NFAT as argument. 3
The “map” qualifier comes from the use of this array in computing element-to-global equation mapping.
25–12
25–13
§25.5
MET-VFC ASSEMBLERS
NodeFreedomMapTable[nodfat_]:=Module[{numnod=Length[nodfat], i,nodfmt}, nodfmt=Table[0,{numnod}]; For [i=1,i<=numnod-1,i++, nodfmt[[i+1]]=nodfmt[[i]]+ DigitCount[nodfat[[i]],10,1] ]; Return[nodfmt]]; TotalFreedomCount[nodfat_]:=Sum[DigitCount[nodfat[[i]],10,1], {i,Length[nodfat]}];
Figure 25.9. Modules to construct the NFMT from the NFAT, and to compute the total number of freedoms.
ElementFreedomTable[enl_,efs_,nodfat_,nodfmt_,m_]:=Module[ {nelnod=Length[enl],eft,i,j,k=0,ix,n,s,es,sx}, eft=Table[0,{Sum[DigitCount[efs[[i]],10,1],{i,1,nelnod}]}]; For [i=1,i<=nelnod,i++, n=enl[[i]]; s=IntegerDigits[nodfat[[n]],10,m]; ix=0; sx=Table[0,{m}]; Do [If[s[[j]]>0,sx[[j]]=++ix],{j,1,m}]; es=IntegerDigits[efs[[i]],10,m]; For [j=1,j<=m,j++, If [es[[j]]>0, k++; If [s[[j]]>0, eft[[k]]=nodfmt[[n]]+sx[[j]] ]]]; ]; Return[eft]];
Figure 25.10. Module to construct the Element Freedom Table for a MET-VFC assembler.
§25.5.5. The Element Freedom Signature The Element Freedom Signature or EFS is a data structure that shares similarities with the NFAT, but provides freedom data at the element rather than global level. More specifically, given an element, it tells to which degrees of freedom it constributes. The EFS is best described through an example. Suppose that in a general FEM program all defined structural nodes have the signature 111111, meaning the arrangement (25.15). The program includes 2-node space bar elements, which contribute only to translational degrees of freedom {u x , u y , u z }. The EFS of any such element will be efs = { 111000,111000 }
(25.20)
For a 2-node space beam element, which contributes to all six nodal freedoms, the EFS will be efs = { 111111,111111 }
(25.21)
This information, along with the NFAT and NFMT, is used to build Element Freedom Tables. §25.5.6. The Element Freedom Table The Element Freedom Table or EFT has been encountered in all previous assembler examples. It is a one dimensional list of length equal to the number of element DOF. If the ith element DOF 25–13
25–14
Chapter 25: THE ASSEMBLY PROCESS
maps to the kth global DOF, then the ith entry of EFT contains k. This allows to write the merge loop compactly. In the simpler assemblers discussed in previous sections, the EFT can be built “on the fly” simply from element node numbers. For a MET-VFC assembler that is no longer the case. To take care of the VFC feature, the construction of the EFT is best done by a separate module. A Mathematica implementation is shown in Figure 25.10. Discussion of the logic, which is not trivial, is relegated to an Exercise, and we simply describe here the interface. The module is invoked as eft = ElementFreedomTable[enl,efs,nodfat,nodfmt,m]
(25.22)
The arguments are enl
The element node list.
efs
The Element Freedom Signature (EFS) list.
nodfat
The Node Freedom Allocation Table (NFAT) described in 25.5.3
nodfmt
The Node Freedom Map Table (NFMT) described in 25.5.4
m
The NFA length, often 6.
The module returns the Element Freedom Table as function value. Examples of use of this module are provided in the plane trussed frame example below. Remark 25.1. The EFT constructed by ElementFreedomTable is guaranteed to contain only positive entries
if every freedom declared in the EFS matches a globally allocated freedom in the NFAT. But it is possible for the returned EFT to contain zero entries. This can only happen if an element freedom does not match a globally allocated one. For example suppose that a 2-node space beam element with the EFS (25.21) is placed into a program that does not accept rotational freedoms and thus has a NFS of 111000 at all structural nodes. Then six of the EFT entries, namely those pertaining to the rotational freedoms, would be zero. The occurrence of a zero entry in a EFT normally flags a logic or input data error. If detected, the assembler should print an appropriate error message and abort. The module of Figure 25.10 does not make that decision because it lacks certain information, such as the element number, that should be placed into the message.
§25.5.7. A Plane Trussed Frame Structure To illustrate the workings of a MET-VFC assembler we will follow the assembly process of the structure pictured in Figure 25.11(a). The finite element discretization, element disconnection and assembly process are illustrated in Figure 25.11(b,c,d). Although the structure is chosen to be plane to facilitate visualization, it will be considered within the context of a general purpose 3D program with the freedom arrangement (25.15) at each node. The structure is a trussed frame. It uses two element types: plane (Bernoulli-Euler) beam-columns and plane bars. Geometric, material and fabrication data are given in Figure 25.11(a). The FEM idealization has 4 nodes and 5 elements. Nodes numbers are 1, 2, 3 and 5 as shown in Figure 25.11(b). Node 4 is purposedly left out to illustrate handling of numbering gaps. There are 11 DOFs, three at nodes 1, 3 and 5, and two at node 2. They are ordered as Global DOF #: DOF: Node #:
1 u x1 1
2 u y1 1
3 θz1 1
4 u x2 2
5 u y2 2 25–14
6 u x3 3
7 u y3 3
8 θz3 3
9 u x5 5
10 u y5 5
11 θz5 5
(25.23)
25–15
§25.5 E=30000 MPa, A=0.02 m 2 , Izz =0.0004 m4 y
E=200000 MPa A=0.003 m 2
MET-VFC ASSEMBLERS
y
(b)
1 Beam-column 3 Beam-column 5
(a)
FEM idealization Bar
3m
x E=200000 MPa A=0.001 m2
4m
(c)
Bar
E=200000 MPa A=0.001 m 2
x
2 (node 4: undefined)
disassembly
4m
(d)
(2)
(1) 3(6,7,8)
1(1,2,3)
Bar
1(1,2,3)
5(9,10,11)
3(6,7,8)
5(9,10,11)
assembly (3)
(5)
(4)
2(4,5)
2(4,5)
Figure 25.11. Trussed frame structure to illustrate a MET-VFC assembler: (a) original structure showing dimensions, material and fabrication properties; (b) finite element idealization with bars and beam column elements; (c) conceptual disassembly; (d) assembly. Numbers written in parentheses after a node number in (c,d) are the global freedom (equation) numbers allocated to that node.
The NFAT and NFMT can be constructed by inspection of (25.23) to be nodfat = { 110001,110000,110001,000000,110001 } nodfmt = { 0,3,5,8,8 }
(25.24)
The element freedom data structures can be also constructed by inspection: Elem
Type
Nodes
EFS
EFT
(1) (2) (3) (4) (5)
Beam-column Beam-column Bar Bar Bar
{ 1,3 } { 3,5 } { 1,2 } { 2,3 } { 2,5 }
{ 110001,110001 } { 110001,110001 } { 110000,110000 } { 110000,110000 } { 110000,110000 }
{ 1,2,3,6,7,8 } { 6,7,8,9,10,11 } { 1,2,4,5 } { 4,5,6,7 } { 4,5,9,10 }
(25.25)
Next we list the element stiffness matrices, computed with the modules discussed in Chapter 20. The EFT entries are annotated as usual. Element (1): This is a plane beam-column element with stiffness 
K(1)
150. 0. 0. 0. 22.5 45.   0. 45. 120.  = 0. 0.  −150.  0. −22.5 −45. 0. 45. 60. 25–15
 −150. 0. 0. 0. −22.5 45.   0. −45. 60.   150. 0. 0.   0. 22.5 −45. 0. −45. 120.
1 2 3 6 7 8
(25.26)
25–16
Chapter 25: THE ASSEMBLY PROCESS
Element (2): This is a plane beam-column with element stiffness identical to the previous one   150. 0. 0. −150. 0. 0. 6 0. 22.5 45. 0. −22.5 45.  7    0. 45. 120. 0. −45. 60.  8  (25.27) K(2) =   0. 0. 150. 0. 0.  9  −150.   0. −22.5 −45. 0. 22.5 −45. 10 0. 45. 60. 0. −45. 120. 11 Element (3): This is a plane bar element with stiffness  25.6 −19.2 −25.6 19.2  −19.2 14.4 K(3) =  −25.6 19.2 25.6 19.2 −14.4 −19.2
 19.2 −14.4   −19.2 14.4
Element (4): This is a plane bar element with stiffness   0 0 0 0  0 200. 0 −200.  K(4) =   0 0 0 0 0 −200. 0 200. Element (5): This is a plane bar element with stiffness  25.6 19.2 −25.6 14.4 −19.2  19.2 K(5) =  −25.6 −19.2 25.6 −19.2 −14.4 19.2
1 2 4 5
4 5 6 7
(25.29)
 −19.2 −14.4   19.2 14.4
Upon merging the 5 elements the master stiffness matrix becomes  175.6 −19.2 0 −25.6 19.2 −150. 0 0 0 −22.5 45.  −19.2 36.9 45. 19.2 −14.4  0 45. 120. 0 0 0 −45. 60.   0 51.2 0 0 0 0  −25.6 19.2  0 0 228.8 0 −200. 0  19.2 −14.4  K =  −150. 0 0 0 0 300. 0 0  0 −22.5 −45. 0 −200. 0 245. 0   0 45. 60. 0 0 0 0 240.   0 0 0 −25.6 −19.2 −150. 0 0   0 0 0 −19.2 −14.4 0 −22.5 −45. 0 0 0 0 0 0 45. 60.
(25.28)
4 5 9 10
(25.30)
 0 0 1 0 0 2  0 0 3  −19.2 0 4  −14.4 0 5  0 0 6  −22.5 45.  7  −45. 60.  8  19.2 0 9  36.9 −45. 10 −45. 120. 11 (25.31) in which global freedom numbers are annotated for convenience. Since all elements have been processed, (25.31) is the master stiffness matrix. It eigenvalues are 0 0 0 −25.6 −19.2 −150. 0 0 175.6 19.2 0
[ 460.456 445.431 306.321 188.661 146.761 82.7415 74.181 25.4476 0 0 0 ] This has the correct rank of 11 − 3 = 8. 25–16
(25.32)
25–17
§25.6 *HANDLING MULTIFREEDOM CONSTRAINTS
Remark 25.2. For storage as a skyline matrix (Chapter 26) the template for (25.31) would look like
 175.6        K=      
−19.2 36.9
0 −25.6 45. 19.2 120. 0 51.2
19.2 −14.4 0 0 228.8
−150. 0 0 0 0 300.
 −22.5 −45. 0 −200. 0 245.
symm
45. 60. 0 0 0 0 240.
        45.   60.   0  
−25.6 −19.2 −19.2 −14.4 −150. 0 0 −22.5 0 −45. 175.6 19.2 36.9 −45. 120. (25.33)
The diagonal location pointers of (25.33) are defined by the list DLT = { 0,1,3,6,10,15,21,27,34,40,47,52 }
(25.34)
Examination of the template (25.33) reveals that some additional zero entries could be removed from the template; for example K 13 . The fact that those entries are zero is, however, fortuitous. It comes from the fact that some elements such as the beams are aligned along x, which decouples axial and bending stiffnesses.
§25.5.8. Implementation Figure 25.12 lists the Mathematica implementation of a MET-VFC assembler capable of doing the trussed frame structure of Figure 25.11. The assembler module is invoked as K = PlaneTrussedFrameMasterStiffness[nodxyz,eletyp,elenod, elemat,elefab,nodfat,prcopt]
(25.35)
The only new argument with respect to the MET assembler of §25.4.2 is nodfat
The Node Freedom Arrangement Table (NFAT).
The module returns the master stiffness matrix as function value. PlaneTrussedFrameMasterStiffness uses five other modules: the element stiffness modules PlaneBar2Stiffness and PlaneBeamColumn2Stiffness of Chapter 20, the NFMT constructor NodeFreedomMapTable of Figure 25.9, the total-DOF-counter TotalFreedomCount of Figure 25.9, and the EFT constructor ElementFreedomTable of Figure 25.10. The element fabrication list elefab has minor modifications. A plane beam-column element requires two properties: { A,Izz }, which are the cross section area and the moment of inertia about the local z axis. A bar element requires only the cross section area: A. For the example trussed frame structure elefab is { { 0.02,0.004 },{ 0.02,0.004 },0.001,0.003,0.001 } in accordance with the data in Figure 25.11. The element material property list elemat is modified on the account that the elastic moduli for the beam columns (made of reinforced concrete) and bars made of (steel) are different. Running the script listed in the top of Figure 25.13 produces the output shown in the bottom of that figure. As can be observed this agrees with (25.31) and (25.32). 25–17
25–18
Chapter 25: THE ASSEMBLY PROCESS
PlaneTrussedFrameMasterStiffness[nodxyz_,eletyp_, elenod_,elemat_,elefab_,nodfat_,prcopt_]:=Module[ {numele=Length[elenod],numnod=Length[nodxyz],numdof,nodfmt,e,enl, eftab,n,ni,nj,i,j,k,m,ncoor,Em,A,Izz,options,Ke,K}, nodfmt=NodeFreedomMapTable[nodfat]; numdof=TotalFreedomCount[nodfat]; K=Table[0,{numdof},{numdof}]; For [e=1, e<=numele, e++, enl=elenod[[e]]; If [eletyp[[e]]'Bar2', {ni,nj}=enl; ncoor={nodxyz[[ni]],nodxyz[[nj]]}; Em=elemat[[e]]; A=elefab[[e]]; options=prcopt; eftab=ElementFreedomTable[enl,{110000,110000},nodfat,nodfmt,6]; Ke=PlaneBar2Stiffness[ncoor,Em,A,options] ]; If [eletyp[[e]]'BeamCol2', {ni,nj}=enl; ncoor={nodxyz[[ni]],nodxyz[[nj]]}; eftab=ElementFreedomTable[enl,{110001,110001},nodfat,nodfmt,6]; Em=elemat[[e]]; {A,Izz}=elefab[[e]]; options=prcopt; Ke=PlaneBeamColumn2Stiffness[ncoor,Em,{A,Izz},options] ]; If [MemberQ[eftab,0], Print['Zero entry in eftab for element ',e]; Print['Assembly process aborted']; Return[K]]; neldof=Length[eftab]; For [i=1, i<=neldof, i++, ii=eftab[[i]]; For [j=i, j<=neldof, j++, jj=eftab[[j]]; K[[jj,ii]]=K[[ii,jj]]+=Ke[[i,j]] ]; ]; ]; Return[K]];
Figure 25.12. A MET-VFC assembler written for the plane trussed frame structure.
nodxyz={{-4,3},{0,0},{0,3},0,{4,3}}; eletyp= {'BeamCol2','BeamCol2','Bar2','Bar2','Bar2'}; elenod= {{1,3},{3,5},{1,2},{2,3},{2,5}}; elemat= Join[Table[30000,{2}],Table[200000,{3}]]; elefab= {{0.02,0.004},{0.02,0.004},0.001,0.003,0.001}; prcopt= {True}; nodfat={110001,110000,110001,000000,110001}; K=PlaneTrussedFrameMasterStiffness[nodxyz,eletyp, elenod,elemat,elefab,nodfat,prcopt]; K=Chop[K]; Print['Master Stiffness of Example Trussed Frame:']; Print[K//MatrixForm]; Print['Eigs of K=',Chop[Eigenvalues[N[K]]]]; Master Stiffness of Example Trussed Frame: 175.6 −19.6 0 −25.6 19.2 −150. −19.2 36.9 45. 19.2 −14.4 0 0 45. 120. 0 0 0 −25.6 19.2 0 51.2 0 0 19.2 −14.4 0 0 228.8 0 −150. 0 0 0 0 300. 0 −22.5 −45. 0 −200. 0 0 45. 60. 0 0 0 0 0 0 −25.6 −19.2 −150. 0 0 0 19.2 −14.4 0 0 0 0 0 0 0
0 −22.5 −45. 0 −200. 0 245. 0 0 −22.5 45.
0 45. 60. 0 0 0 0 240. 0 −45. 60.
0 0 0 0 0 0 −25.6 −19.2 −19.2 −14.4 −150. 0 0 −22.5 0 −45. 175.6 19.2 19.2 36.9 0 −45.
0 0 0 0 0 0 45. 60. 0 −45. 120.
Eigs of K= {460.456, 445.431, 306.321, 188.661, 146.761, 82.7415, 74.181, 25.4476, 0, 0, 0}
Figure 25.13. Script to test the assembler of Figure 25.12 with the structure of Figure 25.11.
25–18
25–19
§25.6 *HANDLING MULTIFREEDOM CONSTRAINTS
(a)
(b) FEM idealization
Same data as in Figure 25.11 plus MFC u y1 = u y5
Beam-column Beam-column Bar Bar Bar
2 (node 4:undefined)
disassembly MFC (6)
(c)
3(6,7,8)
1(1,2,3)
(d) MFC
(2)
(1)
5
3
1
1(1,2,3)
5(9,10,11)
3(6,7,8)
5(9,10,11)
assembly (3)
(5)
(4)
2(4,5)
2(4,5)
Figure 25.14. Finite element discretization, disassembly and assembly of trussed frame example structure with MFC u y1 = u y5 .
§25.6.
*Handling MultiFreedom Constraints
To see the effect of imposing an MFC through the Lagrange multiplier method on the configuration of the master stiffness matrix, suppose that the trussed frame example structure of the previous section is subjected to the constraint that nodes 1 and 5 must move vertically by the same amount. That is, u y1 = u y5
u y1 − u y5 = 0.
or
(25.36)
For assembly purposes (25.36) may be viewed as a fictitious element4 labeled as (6). See Figure 25.14. The degrees of freedom of the assembled structure increase by one to 12. They are ordered as Global DOF #: 1 DOF: u x1 Node #: 1
2 u y1 1
3 θz1 1
4 u x2 2
5 u y2 2
6 u x3 3
7 u y3 3
8 θz3 3
9 u x5 5
10 u y5 5
11 θz5 5
12 λ(6) none
(25.37)
The assembly of the first five elements proceeds as explained before, and produces the same master stiffness as (25.33), except for an extra zero row and column. Processing the MFC element (6) yields the 12 × 12
4
This device should not be confused with penalty elements, which have stiffness matrices. The effect of adding a “Lagrange multiplier element” is to append rows and columns to the master stiffness.
25–19
25–20
Chapter 25: THE ASSEMBLY PROCESS
bordered stiffness

175.6 −19.2 0 −25.6 19.2 −150. 0 0 0 0 −22.5 45. 0  −19.2 36.9 45. 19.2 −14.4  0 45. 120. 0 0 0 −45. 60. 0   −25.6 19.2 0 51.2 0 0 0 0 −25.6   19.2 −14.4 0 0 228.8 0 −200. 0 −19.2  0 0 0 0 300. 0 0 −150.  −150. K= 0 −22.5 −45. 0 −200. 0 245. 0 0   0 45. 60. 0 0 0 0 240. 0   0 0 0 −25.6 −19.2 −150. 0 0 175.6   0 0 0 −19.2 −14.4 0 −22.5 −45. 19.2  0 0 0 0 0 0 45. 60. 0 0 1. 0 0 0 0 0 0 0

0 0 0 0 0 1.  0 0 0  −19.2 0 0  −14.4 0 0  0 0 0  −22.5 45. 0 −45. 60. 0  19.2 0 0  36.9 −45. −1.   −45. 120. 0 −1. 0 0
(25.38)
in which the coefficients 1 and −1 associated with the MFC (25.36) end up in the last row and column of K. There are several ways to incorporate multiplier adjunction into an automatic assembly procedure. The most elegant ones associate the fictitious element with a special freedom signature and a reserved position in the NFA. Being of advanced nature such schemes are beyond the scope of this Chapter. Notes and Bibliography The DSM assembly process for the simplest case of §25.3 is explained in any finite element text. Few texts cover, however, the complications that arise in more general scenarios. Especially when VFCs are allowed. Assembler implementation flavors have fluctuated since the DSM became widely accepted as standard. Variants have been strongly influenced by limits on random access memory (RAM). Those limitations forced the use of “out-of-core” blocked equation solvers, meaning that heavy use was made of disk auxiliary storage to store and retrieve the master stiffness matrix in blocks. Only a limited number of blocks could reside on RAM. Thus a straightforward element-by-element assembly loop (as in the assemblers presented here) was likely to “miss the target” on the receiving end of the merge, forcing blocks to be read in, modified and saved. On large problems this swapping was likely to “trash” the system with heavy I/O, bringing processing to a crawl. One solution favored in programs of the 1965–85 period was to process all elements without assembling, saving matrices on disk. The assembler then cycled over stiffness blocks and read in the contributing elements. With smart asynchronous buffering and tuned-up direct access I/O such schemes were able to achieved reasonable efficiency. However, system dependent logic quickly becomes a maintenance nightmare. A second way out emerged by 1970: the frontal solvers referenced in Chapter 11. A frontal solver carries out assembly, BC application and solution concurrently. Element contributions are processed in a special order that traverses the FEM mesh as a “wavefront.” Application of displacement BCs and factorization can “trail the wave” once no more elements are detected as contributing to a given equation. As can be expected of trying to do too much at once, frontal solvers can be extremely sensitive to changes. One tiny alteration in the element library over which the solver sits, and the whole thing may crumble like a house of cards. The availability of large amounts of RAM (even on PCs and laptops) since the mid 90s, has had a happy consequence: interweaved assembly and solver implementations, as well as convoluted matrix blocking, are no longer needed. The assembler can be modularly separated from the solver. As a result even the most complex assembler presented here fits in one page of text. Of course it is too late for the large scale FEM codes that got caught in the limited-RAM survival game decades ago. Changing their assemblers and solvers incrementally is virtually impossible. The gurus that wrote those thousands of lines of spaghetti Fortran are long gone. The only practical way out is rewrite the whole shebang from scratch. In a commercial environment such investment-busting decisions are unlikely.
25–20
25–21
Exercises
Homework Exercises for Chapter 25 The Assembly Process EXERCISE 25.1 [D:15] Suppose you want to add a six-node plane stress quadratic triangle to the MET
assembler of §25.4. Sketch how you would modify the module of Figure 25.6 for this to happen. EXERCISE 25.2 [C:15] Exercise the module ElementFreedomTable listed in Figure 25.10 on the trussed
frame structure. Verify that it produces the last column of (25.25). EXERCISE 25.3 [D:20] Describe the logic of module ElementFreedomTable listed in Figure 25.10. Can it
return zero entries in the EFT? If yes, give a specific example of how this can happen. Hint: read the Chapter carefully. EXERCISE 25.4 [A/C:25] The trussed frame structure of Figure 25.4 is reinforced with two triangular steel
plates attached as shown in Figure EE25.1(a). The plate thickness is 1.6 mm= 0.0016 m; the material is isotropic with E = 240000 MPa and ν = 1/3. Each reinforcing plate is modeled with a single plane stress 3-node linear triangle. The triangles are numbered (6) and (7), as illustrated in Figure E25.1(c). Compute the master stiffness matrix K of the structure.5
1 Beam-column 3 Beam-column 5
(a)
FEM idealization
Plate properties: E =240000 MPa, ν=1/3 h =1.6 mm = 0.0016 m, x Other dimensions & properties same as in the trussed frame of Fig. 25.11
(c)
y
(b)
y
Bar Plate
disassembly
3(6,7,8)
1(1,2,3) (6) (3)
2 (node 4:undefined)
(d)
(2)
(1)
1(1,2,3)
5(9,10,11)
3(6,7,8)
5(9,10,11)
assembly
(7) (4)
x
Bar Plate Bar
(5) 2(4,5)
2(4,5) Figure E25.1. Plate-reinforced trussed frame structure for Exercise 25.4.
This exercise may be done through Mathematica. For this download Notebook ExampleAssemblers.nb from this Chapter index, and complete Cell 4 by writing the assembler.6 The plate element identifier is 'Trig3'. The driver script to run the assembler is also provided in Cell 4 (blue text). 5
This structure would violate the compatibility requirements stated in Chapter 19 unless the beams are allowed to deflect laterally independently of the plates. This is the fabrication actually sketched in Figure E25.1(a).
6
Cells 1–3 contain the assemblers presented in §25.3, §25.4 and §25.5, respectively, which may be used as guides. The stiffness modules for the three element types used in this structure are available in Cells 2 and 3 and may be reused for this Exercise.
25–21
25–22
Chapter 25: THE ASSEMBLY PROCESS
When reusing the assembler of Cell 3 as a guide, please do not remove the internal Print commands that show element information (red text). Those come in handy for debugging. Please keep that printout in the returned homework to help the grader. Another debugging hint: check that the master stiffness (25.31) is obtained if the plate thickness, called hplate in the driver script, is temporarily set to zero. Target:
 337.6 −19.2  −19.2 90.9  0 45.   −25.6 91.2   91.2 −14.4  K =  −312. −72.  −72. −76.5   0 45.   0 0  0 0
0 0
0 45. 120. 0 0 0 −45. 60. 0 0 0
−25.6 91.2 0 243.2 0 −192. 0 0 −25.6 −91.2 0
91.2 −14.4 0 0 804.8 0 −776. 0 −91.2 −14.4 0
−312. −72. 0 −192. 0 816. 0 0 −312. 72. 0
−72. −76.5 −45. 0 −776. 0 929. 0 72. −76.5 45.
0 45. 60. 0 0 0 0 240. 0 −45. 60.
0 0 0 −25.6 −91.2 −312. 72. 0 337.6 19.2 0
0 0 0 −91.2 −14.4 72. −76.5 −45. 19.2 90.9 −45.

0 0  0   0   0   0  45.   60.   0   −45. 120.
(E25.1)
EXERCISE 25.5 [A:30] §25.5 does not explain how to construct the NFAT from the input data. (In the trussed frame example script, nodfat was set up by inspection, which is OK only for small problems.) Explain how this table could be constructed automatically if the EFS of each element in the model is known. Note: the logic is far from trivial.
25–22
26
.
Solving FEM Equations
26–1
26–2
Chapter 26: SOLVING FEM EQUATIONS
TABLE OF CONTENTS Page
§26.1. Motivation for Sparse Solvers §26.1.1. The Curse of Fullness . . . . . . §26.1.2. The Advantages of Sparsity . . . . . §26.2. Sparse Solution of Stiffness Equations §26.2.1. Skyline Storage Format . . . . . . §26.2.2. Factorization . . . . . . . . . . §26.2.3. Solution . . . . . . . . . . . §26.2.4. Treating MFCs with Lagrange Multipliers §26.3. A SkySolver Implementation §26.3.1. Skymatrix Representation . . . . . §26.3.2. *Skymatrix Factorization . . . . . . §26.3.3. *Solving for One or Multiple RHS . . §26.3.4. *Matrix-Vector Multiply . . . . . . §26.3.5. *Printing and Mapping . . . . . . §26.3.6. *Reconstruction of SkyMatrix from Factors §26.3.7. *Miscellaneous Utilities . . . . . . §26. Exercises . . . . . . . . . . . . . . .
26–2
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26–3 26–3 26–4 26–5 26–5 26–6 26–6 26–6 26–7 26–7 26–8 26–11 26–13 26–14 26–15 26–16 26–20
26–3
§26.1
MOTIVATION FOR SPARSE SOLVERS
§26.1. Motivation for Sparse Solvers In the Direct Stiffness Method (DSM) of finite element analysis, the element stiffness matrices and consistent nodal force vectors are immediately assembled to form the master stiffness matrix and master force vector, respectively, by the process called merge. The assembly process is described in Chapter 25. For simplicity the description that follows assumes that no MultiFreedom Constraints (MFCs) are present. The end result of the assembly process are the master stiffness equations Ku = f
(26.1)
where K is the master stiffness matrix, f the vector of node forces and u the vector or node displacements. Upon imposing the displacement boundary conditions, the system (26.1) is solved for the unknown node displacements. The solution concludes the main phase of DSM computations. In practical applications the order of the stiffness system (26.1) can be quite large. Systems of order 1000 to 10000 are routinely solved in commercial software. Larger ones (say up to 100000 equations) are not uncommon and even millions of equations are being solved on suoercomputers. Presently the record is about 50 million equations on parallel computers.1 In linear FEM analysis the cost of solving this system of equations rapidly overwhelms other computational phases. Much attention has therefore given to matrix processing techniques that economize storage and solution time by taking advantage of the special structure of the stiffness matrix. The master force vector is stored as a conventional one-dimensional array of length equal to the number N of degrees of freedom. This storage arrangement presents no particular difficulties even for very large problem sizes.2 Handling the master stiffness matrix, however, presents computational difficulties. §26.1.1. The Curse of Fullness If K is stored and processed as if it were a full matrix, the storage and processing time resources rapidly becomes prohibitive as N increases. This is illustrated in Table 26.1, which summarizes the storage and factor-time requirements for orders N = 104 , 105 and 106 .3 As regards memory needs, a full square matrix stored without taking advantage of symmetry, requires storage for N 2 entries. If each entry is an 8-byte, double precision floating-point number, the required storage is 8N 2 bytes. Thus, a matrix of order N = 104 would require 8 × 108 bytes or 800 MegaBytes (MB) for storage. For large N the solution of (26.1) is dominated by the factorization of K, an operation discussed in §26.2. This operation requires approximately N 3 /6 floating point operation units. [A floating-point operation unit is conventionally defined as a (multiply,add) pair plus associated indexing and data movement operations.] Now a fast workstation can typically do 107 of these operations per second, 1
For fluid equations solved by fluid-voulume methods the number is over 100 million.
2
A force of displacement vector of, say, 1M equations uses 8MB for double precision storage. In these days of GB RAMs, that is a modest amount.
3
The factor times given in that table reflect 1998 computer technology. To update to 2003, divide times by 10 to 20.
26–3
26–4
Chapter 26: SOLVING FEM EQUATIONS
Table 26.1 Storage & Solution Time for a Fully-Stored Stiffness Matrix Matrix order N
Storage (double prec)
Factor op.units (FLOPS)
Factor time workstation (or fast PC)
Factor time supercomputer
104 105 106
800 MB 80 GB 8 TB
1012 /6 1015 /6 1018 /6
3 hrs 4 mos 300 yrs
2 min 30 hrs 3 yrs
Table 26.2 Storage & Solution Time for a Skyline Stored Stiffness Matrix √ Assuming B = N Matrix order N
Storage (double prec)
Factor op.units (FLOPS)
Factor time workstation (or fast PC)
Factor time supercomputer
104 105 106
8 MB 240 MB 8 GB
108 /2 1010 /2 1012 /2
5 sec 8 min 15 hrs
0.05 sec 5 sec 8 min
whereas a supercomputer may be able to sustain 109 or more. These times assume that the entire matrix is kept in RAM; for otherwise the elapsed time may increase by factors of 10 or more due to I/O transfer operations. The elaspsed timed estimated given in Table 26.1 illustrate that for present computer resources, orders above 104 would pose significant computational difficulties. §26.1.2. The Advantages of Sparsity Fortunately a very high percentage of the entries of the master stiffness matrix K are zero. Such matrices are call sparse. There are clever programming techniques that take advantage of sparsity that fit certain patterns. Although a comprehensive coverage of such techniques is beyond the scope of this course, we shall concentrate on a particular form of sparse scheme that is widely use in FEM codes: skyline storage. This scheme is simple to understand, manage and implement, while cutting storage and processing times by orders of magnitude as the problems get larger. The skyline storage format is a generalization of its widely used predecessor called the band storage scheme. A matrix stored in accordance with the skyline format will be called a skymatrix for short. Only symmetric skymatrices will bve considered here, since the stiffness matrices in linear FEM are symmetric. If a skymatrix of order N can be stored in S memory locations, the ratio B = S/N is called the mean bandwidth. If the entries are, as usual, 8-byte double-precision floating-point numbers, the storage requirement is 8N B bytes. The factorization of a skymatrix requires approximately 12 N B 2 √ floating-point operation units. In two-dimensional problems B is of the order of N . Under this assumption, storage requirements and estimated factorization times for N = 104 , N = 105 and 26–4
26–5
§26.2
SPARSE SOLUTION OF STIFFNESS EQUATIONS
N = 106 are reworked in Table 26.2. It is seen that by going from full to skyline storage significant reductions in computer resources have been achieved. For example, now N = 104 is easy on a workstation and trivial on a supercomputer. Even a million equations do not look far-fetched on a supercomputer as long as enough memory is available.4 In preparation for assembling K as a skymatrix one has to set up several auxiliary arrays related to nodes and elements. These auxiliary arrays are described in the previous Chapter. Knowledge of that material is useful for understanding the following description. §26.2. Sparse Solution of Stiffness Equations §26.2.1. Skyline Storage Format The skyline storage arrangement for K is best illustrated through a simple example. Consider the 6 × 6 stiffness matrix   K 11 0 K 13 0 0 K 16 K 22 0 K 24 0 0     K 33 K 34 0 0   K= (26.2)  0 K 46  K 44    K 55 K 56 symm K 66 Since the matrix is symmetric only one half, the upper triangle in the above display, need to be shown. Next we define the envelope of K as follows. From each diagonal entry move up the corresponding column until the last nonzero entry is found. The envelope separates that entry from the rest of the upper triangle. The remaining zero entries are conventionally removed:     K=  
K 11 K 22
K 13 0 K 33
K 24 K 34 K 44 K 55
symm
 K 16 0   0   K 46   K 56 K 66
(26.3)
What is left constitute the skyline profile of skyline template of the matrix. A sparse matrix that can be profitably stored in this form is called a skymatrix for brevity. Notice that the skyline profile may include zero entries. During the factorization step discussed below these zero entries will in general become nonzero, a phenomenon that receives the name fill-in. 4
On a PC or laptop with a 32-bit processor, there is a RAM “hard barrier” because of hardware addressing limitations. That is typically 2GB. As PCs migrate to 64-bit processors over the next 10 years, the barrier disappears. For example the high-end Mac G5s can presently be expanded up to 8GB RAM. The new limits are primarily dictated by power comsumption, circuitry and cooling considerations. A distributed (Beowulf) network is a practical solution to go beyond the individual machine limits.
26–5
26–6
Chapter 26: SOLVING FEM EQUATIONS
The key observation is that only the entries in the skyline template need to be stored, because fill-in in the factorization process will not occur outside the envelope. To store these entries it is convenient to use a one-dimensional skyline array: s:
[ K 11 , K 22 , K 13 , 0, K 33 , K 24 , K 34 , K 44 , K 55 , K 16 , 0, 0, K 46 , K 56 , K 66 ]
(26.4)
This array is complemented by a (N + 1) integer array p that contains addresses of diagonal locations. The array has N + 1 entries. The (i + 1)th entry of p has the location of the i th diagonal entry of K in s. For the example matrix: p:
[ 0, 1, 2, 5, 8, 9, 15 ]
(26.5)
In the previous Chapter, this array was called the Global Skyline Diagonal Location Table, or GSDLT. Equations for which the displacement component is prescribed are identified by a negative diagonal location value. For example if u 3 and u 5 are prescribed displacement components in the test example, then p : [ 0, 1, 2, −5, 8, −9, 15 ] (26.6) Remark 26.1. In Fortran it is convenient to dimension the diagonal location array as p(0:n) so that indexing begins at zero. In C this is the standard indexing.
§26.2.2. Factorization The stiffness equations (26.1) are solved by a direct method that involves two basic phases: factorization and solution. In the first stage, the skyline-stored symmetric stiffness matrix is factored as K = LDU = LDLT = UT DU,
(26.7)
where L is a unit lower triangular matrix, D is a nonsingular diagonal matrix, and U and L are the transpose of each other. The original matrix is overwritten by the entries of D−1 and U; details may be followed in the program implementation. No pivoting is used in the factorization process. This factorization is carried out by Mathematica module SymmSkyMatrixFactor, which is described later in this Chapter. §26.2.3. Solution Once K has been factored, the solution u for a given right hand side f is obtained by carrying out three stages: Forward reduction :
Lz = f,
(26.8)
Diagonal scaling :
Dy = z,
(26.9)
Back substitution :
Uu = y,
(26.10)
where y and z are intermediate vectors. These stages are carried out by Mathematica modules SymmSkyMatrixVectorSolve, which is described later. 26–6
26–7
§26.3
A SKYSOLVER IMPLEMENTATION
§26.2.4. Treating MFCs with Lagrange Multipliers In Mathematica implementations of FEM, MultiFreedom Constraints (MFCs) are treated with Lagrange multipliers. There is one multiplier for each constraint. The multipliers are placed at the end of the solution vector. Specifically, let the nummul>0 MFCs be represented in matrix form as Cu = g, where C and g are given, and let the nummul multipliers be collected in a vector λ. The multiplier-augmented master stiffness equations are u f K CT = (26.11) C 0 λ g or Ax = b. (26.12) where the symmetric matrix A, called a stiffness-bordered matrix, is of order numdof+nummul. The stiffness bordered matrix is also stored in skyline form, and the previous solution procedure applies, as long as the skyline array is properly constructed as described in the previous Chapter. The main difference with respect to the no-MFC case is that, because of the configuration (26.11), A can no longer be positive definite. In principle pivoting should be used during the factorization of A to forestall possible numerical instabilities. Pivoting can have a detrimental effect on solution efficiency because entries can move outside of the skyline template. However, by placing the λ at the end such difficulties will not be encountered if K is positive definite, and the constraints are linearly independent (that is, C has full rank). Thus pivoting is not necessary. §26.3. A SkySolver Implementation The remaining sections of this revised Chapter describe a recent implementation of the skyline solver and related routines in Mathematica. This has been based on similar Fortran codes used since 1967. §26.3.1. Skymatrix Representation In what follows the computer representation in Mathematica of a symmetric skymatrix will be generally denoted by the symbol S. Such a representation consists of a list of two numeric objects: S = { p, s }
(26.13)
Here p=GSDLT is the Global Skyline Diagonal Location Table introduced in §11.6, and s is the array of skymatrix entries, arranged as described in the previous section. This array usually consists of floating-point numbers, but it may also contain exact integers or factions, or even symbolic entries. For example, suppose that the numerical entries of the 6 × 6 skymatrix (26.10) are actually   11 13 16 22 0 24 0     33 34 0   K= (26.14)  44 46     55 56 symm 66 26–7
26–8
Chapter 26: SOLVING FEM EQUATIONS
Cell 26.1 Factorization of a Symmetric SkyMatrix SymmSkyMatrixFactor[S_,tol_]:= Module[ {p,a,fail,i,j,k,l,m,n,ii,ij,jj,jk,jmj,d,s,row,v}, row=SymmSkyMatrixRowLengths[S]; s=Max[row]; {p,a}=S; n=Length[p]-1; v=Table[0,{n}]; fail=0; Do [jj=p[[j+1]]; If [jj<0|row[[j]]0, Continue[]]; d=a[[jj]]; jmj=Abs[p[[j]]]; jk=jj-jmj; Do [i=j-jk+k; v[[k]]=0; ii=p[[i+1]]; If [ii<0, Continue[]]; m=Min[ii-Abs[p[[i]]],k]-1; ij=jmj+k; v[[k]]=a[[ij]]; v[[k]]-=Take[a,{ii-m,ii-1}].Take[v,{k-m,k-1}]; a[[ij]]=v[[k]]*a[[ii]], {k,1,jk-1}]; d-=Take[a,{jmj+1,jmj+jk-1}].Take[v,{1,jk-1}]; If [Abs[d]0, d=a[[m+j]]^2; row[[i]]+=d; row[[j]]+=d], {j,Max[1,Abs[p[[i]]]-m+1],Min[n,i]-1}], {i,1,n}]; Return[Sqrt[row]]; ];
Its Mathematica representation, using the symbols (26.13) is p= { 0,1,2,5,8,9,15 }; s= { 11,22,13,0,33,24,34,44,55,16,0,0,46,56,66 }; S= { p, s };
(26.15)
or more directly S={ { 0,1,2,5,8,9,15 },{ 11,22,13,0,33,24,34,44,55,16,0,0,46,56,66 } };
(26.16)
[The remaining sections on details of skyline processing logic, marked with a *, will not be covered in class. They are intended for a more advanced course.]
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§26.3 A SKYSOLVER IMPLEMENTATION
Cell 26.2 Factorization Test Input ClearAll[n]; n=5; SeedRandom[314159]; p=Table[0,{n+1}]; Do[p[[i+1]]=p[[i]]+ Max[1,Min[i,Round[Random[]*i]]],{i,1,n}]; a=Table[1.,{i,1,p[[n+1]]}]; Print['Mean Band=',N[p[[n+1]]/n]]; S={p,a}; Sr=SymmSkyMatrixLDUReconstruct[S]; Print['Reconstructed SkyMatrix:']; SymmSkyMatrixLowerTrianglePrint[Sr]; SymmSkyMatrixLowerTriangleMap[Sr]; Print['eigs=',Eigenvalues[SymmSkyMatrixConvertToFull[Sr]]]; x=Table[{N[i],3.,(-1)^i*N[n-i]},{i,1,n}]; Print['Assumed x=',x]; b=SymmSkyMatrixColBlockMultiply[Sr,x]; (*x=Transpose[x]; b=SymmSkyMatrixRowBlockMultiply[Sr,x];*) Print['b=Ax=',b]; Print[Timing[{F,fail}=SymmSkyMatrixFactor[Sr,10.^(-12)]]]; If [fail!=0, Print['fail=',fail]; Abort[]]; Print['F=',F]; Print['fail=',fail]; Print['Factor:']; SymmSkyMatrixLowerTrianglePrint[F]; x=SymmSkyMatrixColBlockSolve[F,b]; (*x=SymmSkyMatrixRowBlockSolve[F,b];*) Print['Computed x=',x//InputForm];
§26.3.2. *Skymatrix Factorization Module SymmSkyMatrixFactor, listed in Cell 26.1, factors a symmetric skymatrix into the product LDU where L is the transpose of U. No pivoting is used. The module is invoked as { Sf,fail } = SymmSkyMatrixFactor[S,tol] The input arguments are S
The skymatrix to be factored, stored as the two-object list { p,s }; see previous subsection.
tol
Tolerance for singularity test. The appropriate value of tol depends on the kind of skymatrix entries stored in s. If the skymatrix entries are floating-point numbers handled by default in double precision arithmetic, tol should be set to 8× or 10× the machine precision in that kind of arithmetic. The factorization aborts if, when processing the j-th row, d j ≤ tol ∗ r j , where d j is the computed j th diagonal entry of D, and r j is the Euclidean norm of the j th skymatrix row. If the skymatrix entries are exact (integers, fractions or symbols), tol should be set to zero. In this case exact singularity is detectable, and the factorization aborts only on that condition.
The outputs are: Sf
If fail is zero on exit, Sf is the computed factorization of S. It is a two-object list { p, du
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Chapter 26: SOLVING FEM EQUATIONS
Cell 26.3 Output from Program of Cells 26.1 and 26.2 Mean Band=1.6 Reconstructed SkyMatrix:
Row Row Row Row Row
1 2 3 4 5
Col 1 1.0000
Col
2
1.0000 1.0000
Col
3
Col
4
Col
5
2.0000 1.0000
1.0000 1.0000
3.0000
1 2 3 4 5 1 + 2 + 3 + + 4 + 5 + + + eigs={3.9563, 2.20906, 1., 0.661739, 0.172909} Assumed x={{1., 3., -4.}, {2., 3., 3.}, {3., 3., -2.}, {4., 3., 1.}, {5., 3., 0}} b=Ax={{1., 3., -4.}, {5., 6., 1.}, {13., 12., -1.}, {9., 6., 1.}, {22., 15., -1.}} {0.0666667 Second, {{{0, 1, 2, 4, 5, 8}, {1., 1., 1., 1., 1., 1., 1., 1.}}, 0}} F={{0, 1, 2, 4, 5, 8}, {1., 1., 1., 1., 1., 1., 1., 1.}} fail=0 Factor: Col 1 Col 2 Col 3 Col 4 Col 5 Row 1 1.0000 Row 2 1.0000 Row 3 1.0000 1.0000 Row 4 1.0000 Row 5 1.0000 1.0000 1.0000 Computed x={{1., 3., -4.}, {2., 3., 3.}, {3., 3., -2.}, {4., 3., 1.}, {5., 3., 0.}}
}, where du stores the entries of D−1 in the diagonal locations, and of U in its strict upper triangle. fail
A singularity detection indicator. A zero value indicates that no singularity was detected. If fail returns j>0, the factorization was aborted at the j-th row. In this case Sf returns the aborted factorization with ∞ stored in d j .
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§26.3 A SKYSOLVER IMPLEMENTATION
Cell 26.4 Solving for a Single RHS SymmSkyMatrixVectorSolve[S_,b_]:= Module[ {p,a,n,i,j,k,m,ii,jj,bi,x}, {p,a}=S; n=Length[p]-1; x=b; If [n!=Length[x], Print['Inconsistent matrix dimensions in', ' SymmSkyMatrixVectorSolve']; Return[Null]]; Do [ii=p[[i+1]];If [ii>=0, Continue[]]; ii=-ii; k=i-ii+Abs[p[[i]]]+1; bi=x[[i]]; If [bi0, Continue[]]; Do [jj=p[[j+1]], If [jj<0, Continue[]]; m=j-i; If [m<0, x[[j]]-=a[[ii+m]]*bi; Break[]]; ij=jj-m; If [ij>Abs[p[[j]]], x[[j]]-=a[[ij]]*bi], {j,k,n}], {i,1,n}]; Do [ii=p[[i+1]]; If [ii<0, x[[i]]=0; Continue[]]; imi=Abs[p[[i]]]; m=ii-imi-1; x[[i]]-=Take[a,{imi+1,imi+m}].Take[x,{i-m,i-1}], {i,1,n}]; Do [ii=Abs[p[[i+1]]]; x[[i]]*=a[[ii]], {i,1,n}]; Do [ii=p[[i+1]]; If [ii<0, x[[i]]=b[[i]]; Continue[]]; m=ii-Abs[p[[i]]]-1; Do [ x[[i-j]]-=a[[ii-j]]*x[[i]], {j,1,m}], {i,n,1,-1}]; Return[x] ];
A test of SymmSkyMatrixFactor on the matrix (26.14) is shown in Cells 26.2 and 26.3. The modules that print and produce skyline maps used in the test program are described later in this Chapter. §26.3.3. *Solving for One or Multiple RHS Module SymmSkyMatrixVectorSolve, listed in Cell 26.4, solves the linear system Ax = b for x, following the factorization of the symmetric skymatrix A by SymmSkyMatrixFactor. The module is invoked as x = SymmSkyMatrixVectorSolve[Sf,b] The input arguments are Sf
The factored matrix returned by SymmSkyMatrixFactor.
b
The right-hand side vector to be solved for, stored as a single-level (one dimensional) list. If the i-th entry of x is prescribed, the known value must be supplied in this vector.
The outputs are: x
The computed solution vector, stored as a single-level (one-dimensional) list. Prescribed solution components return the value of the entry in b.
Sometimes it is necessary to solve linear systems for multiple (m > 1) right hand sides. One way to do that is to call SymmSkyMatrixVectorSolve repeatedly. Alternatively, if m right hand sides are collected as columns of a rectangular matrix B, module SymmSkyMatrixColBlockSolve may be invoked as
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Chapter 26: SOLVING FEM EQUATIONS
Cell 26.5 Solving for a Block of Righ Hand Sides SymmSkyMatrixColBlockSolve[S_,b_]:= Module[ {p,a,n,nrhs,i,j,k,m,r,ii,jj,bi,x}, {p,a}=S; n=Length[p]-1; x=b; If [n!=Dimensions[x][[1]], Print['Inconsistent matrix dimensions in', ' SymmSkyMatrixBlockColSolve']; Return[Null]]; nrhs = Dimensions[x][[2]]; Do [ii=p[[i+1]];If [ii>=0, Continue[]]; ii=-ii; k=i-ii+Abs[p[[i]]]+1; Do [bi=x[[i,r]]; If [bi0, Continue[]]; Do [jj=p[[j+1]], If [jj<0, Continue[]]; m=j-i; If [m<0,x[[j,r]]-=a[[ii+m]]*bi; Break[]]; ij=jj-m; If [ij>Abs[p[[j]]], x[[j,r]]-=a[[ij]]*bi], {j,k,n}], {r,1,nrhs}], {i,1,n}]; Do [ii=p[[i+1]]; If [ii<0, Do[x[[i,r]]=0,{r,1,nrhs}];Continue[]]; imi=Abs[p[[i]]]; m=ii-imi-1; Do [ Do [ x[[i,r]]-=a[[imi+j]]*x[[i-m+j-1,r]], {j,1,m}], {r,1,nrhs}], {i,1,n}]; Do [ii=Abs[p[[i+1]]]; Do[x[[i,r]]*=a[[ii]], {r,1,nrhs}], {i,1,n}]; Do [ii=p[[i+1]]; If [ii<0, Do[x[[i,r]]=b[[i,r]],{r,1,nrhs}];Continue[]]; m=ii-Abs[p[[i]]]-1; Do [ Do [ x[[i-j,r]]-=a[[ii-j]]*x[[i,r]], {j,1,m}], {r,1,nrhs}], {i,n,1,-1}]; Return[x] ]; SymmSkyMatrixRowBlockSolve[S_,b_]:= Module[ {p,a,n,nrhs,i,j,k,m,r,ii,jj,bi,x}, {p,a}=S; n=Length[p]-1; x=b; If [n!=Dimensions[x][[2]], Print['Inconsistent matrix dimensions in', ' SymmSkyMatrixBlockRowSolve']; Return[Null]]; nrhs = Dimensions[x][[1]]; Do [ii=p[[i+1]];If [ii>=0, Continue[]]; ii=-ii; k=i-ii+Abs[p[[i]]]+1; Do [bi=x[[r,i]]; If [bi0, Continue[]]; Do [jj=p[[j+1]], If [jj<0, Continue[]]; m=j-i; If [m<0,x[[j,r]]-=a[[ii+m]]*bi; Break[]]; ij=jj-m; If [ij>Abs[p[[j]]], x[[r,j]]-=a[[ij]]*bi], {j,k,n}], {r,1,nrhs}], {i,1,n}]; Do [ii=p[[i+1]]; If [ii<0, Do[x[[r,i]]=0,{r,1,nrhs}];Continue[]]; imi=Abs[p[[i]]]; m=ii-imi-1; Do [ Do [ x[[r,i]]-=a[[imi+j]]*x[[r,i-m+j-1]], {j,1,m}], {r,1,nrhs}], {i,1,n}]; Do [ii=Abs[p[[i+1]]]; Do[x[[r,i]]*=a[[ii]], {r,1,nrhs}], {i,1,n}]; Do [ii=p[[i+1]]; If [ii<0, Do[x[[r,i]]=b[[r,i]],{r,1,nrhs}];Continue[]]; m=ii-Abs[p[[i]]]-1; Do [ Do [ x[[r,i-j]]-=a[[ii-j]]*x[[r,i]], {j,1,m}], {r,1,nrhs}], {i,n,1,-1}]; Return[x] ];
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§26.3 A SKYSOLVER IMPLEMENTATION
Cell 26.6 Multiplying Skymatrix by Individual Vector SymmSkyMatrixVectorMultiply[S_,x_]:= Module[ {p,a,n,i,j,k,m,ii,b}, {p,a}=S; n=Length[p]-1; If [n!=Length[x], Print['Inconsistent matrix dimensions in', ' SymmSkyMatrixVectorMultiply']; Return[Null]]; b=Table[a[[ Abs[p[[i+1]]] ]]*x[[i]], {i,1,n}]; Do [ii=Abs[p[[i+1]]]; m=ii-Abs[p[[i]]]-1; If [m<=0,Continue[]]; b[[i]]+=Take[a,{ii-m,ii-1}].Take[x,{i-m,i-1}]; Do [b[[i-k]]+=a[[ii-k]]*x[[i]],{k,1,m}], {i,1,n}]; Return[b] ]; (* ClearAll[n]; n=10; SeedRandom[314159]; p=Table[0,{n+1}]; Do[p[[i+1]]=p[[i]]+ Max[1,Min[i,Round[Random[]*i]]],{i,1,n}]; a=Table[1.,{i,1,p[[n+1]]}]; Print['Mean Band=',N[p[[n+1]]/n]]; S={p,a}; Sr=SymmSkyMatrixLDUReconstruct[S]; Print['Reconstructed SkyMatrix:']; SymmSkyMatrixLowerTrianglePrint[Sr]; SymmSkyMatrixLowerTriangleMap[Sr]; x=Table[1.,{i,1,n}]; b=SymmSkyMatrixVectorMultiply[Sr,x]; Print['b=Ax=',b];*)
X = SymmSkyMatrixVectorSolve[Sf,B] to provide the solution X of SX = B. This module is listed in Cell 26.5. The input arguments and function returns have the same function as those described for SymmSkyMatrixVectorSolve. The main difference is that B and X are matrices (two-dimensional lists) with the righ-hand side and solution vectors as columns. There is a similar module SymmSkyMatrixRowBlockSolve, notlisted here, which solves for multiple right hand sides stored as rows of a matrix. §26.3.4. *Matrix-Vector Multiply For various applications it is necessary to form the matrix-vector product b = Sx
(26.17)
where S is a symmetric skymatrix and x is given. This is done by module SymmSkyMatrixVectorMultiply, which is listed in Cell 26.6. Its arguments are the skymatrix S and the vector x. The function returns Sx in b. Module SymmSkyMatrixColBlockMultiply implements the multiplication by a block of vectors stored as columns of a rectangular matrix X: B = SX (26.18)
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Chapter 26: SOLVING FEM EQUATIONS
Cell 26.7 Multiplying Skymatrix by Vector Block SymmSkyMatrixColBlockMultiply[S_,x_]:= Module[ {p,a,n,nrhs,i,j,k,m,r,ii,aij,b}, {p,a}=S; n=Length[p]-1; If [n!=Dimensions[x][[1]], Print['Inconsistent matrix dimensions in', ' SymmSkyMatrixColBlockMultiply']; Return[Null]]; nrhs = Dimensions[x][[2]]; b=Table[0,{n},{nrhs}]; Do [ii=Abs[p[[i+1]]]; m=ii-Abs[p[[i]]]-1; Do [b[[i,r]]=a[[ii]]*x[[i,r]], {r,1,nrhs}]; Do [j=i-k; aij=a[[ii-k]]; If [aij0, Continue[]]; Do [b[[i,r]]+=aij*x[[j,r]]; b[[j,r]]+=aij*x[[i,r]], {r,1,nrhs}], {k,1,m}], {i,1,n}]; Return[b] ]; SymmSkyMatrixRowBlockMultiply[S_,x_]:= Module[ {p,a,n,nrhs,i,j,k,m,r,ii,aij,b}, {p,a}=S; n=Length[p]-1; If [n!=Dimensions[x][[2]], Print['Inconsistent matrix dimensions in', ' SymmSkyMatrixRowBlockMultiply']; Return[Null]]; nrhs = Dimensions[x][[1]]; b=Table[0,{nrhs},{n}]; Do [ii=Abs[p[[i+1]]]; m=ii-Abs[p[[i]]]-1; Do [b[[r,i]]=a[[ii]]*x[[r,i]], {r,1,nrhs}]; Do [j=i-k; aij=a[[ii-k]]; If [aij0, Continue[]]; Do [b[[r,i]]+=aij*x[[r,j]]; b[[r,j]]+=aij*x[[r,i]], {r,1,nrhs}], {k,1,m}], {i,1,n}]; Return[b] ];
This module is listed in Cell 26.7. Its arguments are the skymatrix S and the rectangular matrix X. The function returns SX in B. There is a similar module SymmSkyMatrixRowBlockMultiply, also listed in Cell 26.7, which postmultiplies a vector block stored as rows. §26.3.5. *Printing and Mapping Module SymmSkyMatrixUpperTrianglePrint, listed in Cell 26.8, prints a symmetric skymatrix in upper triangle form. Is is invoked as SymmSkyMatrixUpperTrianglePrint[S] where S is the skymatrix to be printed. For an example of use see Cells 262-3. The print format resembles the configuration depicted in Section 26.1. This kind of print is useful for program
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§26.3 A SKYSOLVER IMPLEMENTATION
development and debugging although of course it should not be attempted with a very large matrix.5 There is a similar module called SymmSkyMatrixLowerTrianglePrint, which displays the skymatrix entries in lower triangular form. This module is also listed in Cell 26.8. Sometimes one is not interested in the actual values of the skymatrix entries but only on how the skyline template looks like. Such displays, called maps, can be done with just one symbol per entry. Module SymmSkyMatrixUpperTriangleMap, listed in Cell 26.9, produces a map of its argument. It is invoked as SymmSkyMatrixUpperTriangleMap[S] The entries within the skyline template are displayed by symbols +, - and 0, depending on whether the value is positive, negative or zero, respectively. Entries outside the skyline template are blank. As in the case of the print module, there is module SymmSkyMatrixLowerTriangleMap which is also listed in Cell 26.9. §26.3.6. *Reconstruction of SkyMatrix from Factors In testing factorization and solving modules it is convenient to have modules that perform the “inverse process” of the factorization. More specifically, suppose that U, D (or D−1 ) are given, and the problem is to reconstruct the skymatrix that have them as factors: S = LDU,
or
S = LD−1 U
(26.19)
in which L = U . Modules SymmSkyMatrixLDUReconstruction and SymmSkyMatrixLDinvUReconstruction perform those operations. These modules are listed in Cell 26.10. Their argument is a factored form of S: U and D in the first case, and U and D−1 in the second case. T
5
Computer oriented readers may notice that the code for the printing routine is substantially more complex than those of the computational modules. This is primarily due to the inadequacies of Mathematica in handling tabular format output. The corresponding Fortran or C implementations would be simpler because those languages provide much better control over low-level display.
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Chapter 26: SOLVING FEM EQUATIONS
Cell 26.8 Skymatrix Printing SymmSkyMatrixLowerTrianglePrint[S_]:= Module[ {p,a,cycle,i,ii,ij,it,j,jj,j1,j2,jref,jbeg,jend,jt,kcmax,kc,kr,m,n,c,t}, {p,a}=S; n=Dimensions[p][[1]]-1; kcmax=5; jref=0; Label[cycle]; Print[' ']; jbeg=jref+1; jend=Min[jref+kcmax,n]; kc=jend-jref; t=Table[' ',{n-jref+1},{kc+1}]; Do [If [p[[j+1]]>0,c=' ',c='*']; t[[1,j-jref+1]]=StringJoin[c,'Col',ToString[PaddedForm[j,3]]], {j,jbeg,jend}]; it=1; Do [ii=Abs[p[[i+1]]]; m=ii-Abs[p[[i]]]-1; j1=Max[i-m,jbeg];j2=Min[i,jend]; kr=j2-j1+1; If [kr<=0, Continue[]]; If [p[[i+1]]>0,c=' ',c='*']; it++; t[[it,1]]=StringJoin[c,'Row',ToString[PaddedForm[i,3]]]; jt=j1-jbeg+2; ij=j1+ii-i; Do[t[[it,jt++]]=PaddedForm[a[[ij++]]//FortranForm,{7,4}],{j,1,kr}], {i,jbeg,n}]; Print[TableForm[Take[t,it],TableAlignments->{Right,Right}, TableDirections->{Column,Row},TableSpacing->{0,2}]]; jref=jend; If[jref0,c=' ',c='*']; t[[1,j-jref+1]]=StringJoin[c,'Col',ToString[PaddedForm[j,3]]], {j,jbeg,jend}]; it=1; Do [it++; If [p[[i+1]]>0,c=' ',c='*']; t[[it,1]]=StringJoin[c,'Row',ToString[PaddedForm[i,3]]]; j=jref; Do [j++; If [j
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§26.3 A SKYSOLVER IMPLEMENTATION Cell 26.9 Skymatrix Mapping
SymmSkyMatrixLowerTriangleMap[S_]:=Module[ {p,a,cycle,i,ii,ij,it,itop,j,jj,j1,j2,jref,jbeg,jend,jt,kcmax,kc,kr,m,n,c,t}, {p,a}=S; n=Dimensions[p][[1]]-1; kcmax=40; jref=0; Label[cycle]; Print[' ']; jbeg=jref+1; jend=Min[jref+kcmax,n]; kc=jend-jref; itop=2; If[jend>9,itop=3]; If[jend>99,itop=4]; If[jend>999,itop=5]; t=Table[' ',{n-jref+itop},{kc+1}]; it=0; If [itop>=5, it++; Do [m=Floor[j/1000]; If[m>0,t[[it,j-jref+1]]=ToString[Mod[m,10]]], {j,jbeg,jend}]]; If [itop>=4, it++; Do [m=Floor[j/100]; If[m>0,t[[it,j-jref+1]]=ToString[Mod[m,10]]], {j,jbeg,jend}]]; If [itop>=3, it++; Do [m=Floor[j/10]; If[m>0,t[[it,j-jref+1]]=ToString[Mod[m,10]]], {j,jbeg,jend}]]; it++; Do[t[[it,j-jref+1]]=ToString[Mod[j,10]],{j,jbeg,jend}]; it++; Do[If[p[[j+1]]<0,t[[it,j-jref+1]]='*'],{j,jbeg,jend}]; Do [ii=Abs[p[[i+1]]]; m=ii-Abs[p[[i]]]-1; j1=Max[i-m,jbeg];j2=Min[i,jend]; kr=j2-j1+1; If [kr<=0, Continue[]]; If [p[[i+1]]>0,c=' ',c='*']; it++; t[[it,1]]=StringJoin[ToString[PaddedForm[i,2]],c]; jt=j1-jbeg+2; ij=j1+ii-i; Do [ c=' 0'; If[a[[ij]]>0,c=' +']; If[a[[ij++]]<0,c=' -']; t[[it,jt++]]=c, {j,1,kr}], {i,jbeg,n}]; Print[TableForm[Take[t,it],TableAlignments->{Right,Right}, TableDirections->{Column,Row},TableSpacing->{0,0}]]; jref=jend; If[jref0,c=' +']; If[a[[ij++]]<0,c=' -']; t[[it,k+1]]=c, {k,1,kc}], {i,1,jend}]; Print[TableForm[Take[t,it],TableAlignments->{Right,Right}, TableDirections->{Column,Row},TableSpacing->{0,0}]]; jref=jend; If[jref
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Chapter 26: SOLVING FEM EQUATIONS
Cell 26.10 Skymatrix Reconstruction from Factors SymmSkyMatrixLDUReconstruct[S_]:= Module[ {p,ldu,a,v,n,i,ii,ij,j,jj,jk,jmj,k,m}, {p,ldu}=S; a=ldu; n=Length[p]-1; v=Table[0,{n}]; Do [jmj=Abs[p[[j]]]; jj=p[[j+1]]; If [jj<0, Continue[]]; jk=jj-jmj; v[[jk]]=ldu[[jj]]; Do [ij=jmj+k; i=j+ij-jj; ii=p[[i+1]]; If [ii<0, v[[k]]=0; Continue[]]; If [i!=j, v[[k]]=ldu[[ij]]*ldu[[ii]]]; m=Min[ii-Abs[p[[i]]],k]; a[[ij]]= v[[k]]; a[[ij]]+=Take[ldu,{ii-m+1,ii-1}].Take[v,{k-m+1,k-1}], {k,1,jk}], {j,1,n}]; Return[{p,a}]; ]; SymmSkyMatrixLDinvUReconstruct[S_]:= Module[ {p,ldu,a,v,n,i,ii,ij,j,jj,jk,jmj,k,m}, {p,ldu}=S; a=ldu; n=Length[p]-1; v=Table[0,{n}]; Do [jmj=Abs[p[[j]]]; jj=p[[j+1]]; If [jj<0, Continue[]]; jk=jj-jmj; v[[jk]]=1/ldu[[jj]]; Do [ij=jmj+k; i=j+ij-jj; ii=p[[i+1]]; If [ii<0, v[[k]]=0; Continue[]]; If [i!=j, v[[k]]=ldu[[ij]]/ldu[[ii]]]; m=Min[ii-Abs[p[[i]]],k]; a[[ij]]= v[[k]]; a[[ij]]+=Take[ldu,{ii-m+1,ii-1}].Take[v,{k-m+1,k-1}], {k,1,jk}], {j,1,n}]; Return[{p,a}]; ]; p={0,1,2,5,8,9,15}; s={11,22,13,0,33,24,34,44,55,16,0,0,46,56,66}; S={p,s}; Sr=SymmSkyMatrixLDinvUReconstruct[S]; Print[Sr//InputForm]; Print[SymmSkyMatrixFactor[Sr,0]];
§26.3.7. *Miscellaneous Utilities Finally, Cell 26.11 lists three miscellaneous modules. The most useful one is probably SymmSkyMatrixConvertToFull, which converts its skymatrix argument to a fully stored symmetric matrix. This is useful for things like a quick and dirty computation of eigenvalues: Print[Eigenvalues[SymmSkyMatrixConvertToFull[S]]]; because Mathematica built-in eigensolvers require that the matrix be supplied in full storage form.
26–18
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§26.3
A SKYSOLVER IMPLEMENTATION
Cell 26.11 Miscellaneous Skymatrix Utilities
SymmSkyMatrixConvertToFull[S_]:= Module[ {p,a,aa,n,j,jj,jmj,k}, {p,a}=S; n=Length[p]-1; aa=Table[0,{n},{n}]; Do [jmj=Abs[p[[j]]]; jj=Abs[p[[j+1]]]; aa[[j,j]]=a[[jj]]; Do [aa[[j,j-k]]=aa[[j-k,j]]=a[[jj-k]],{k,1,jj-jmj-1}], {j,1,n}]; Return[aa]; ]; SymmSkyMatrixConvertUnitUpperTriangleToFull[S_]:= Module[ {p,ldu,aa,n,j,jj,jmj,k}, {p,ldu}=S; n=Length[p]-1; aa=Table[0,{n},{n}]; Do [jmj=Abs[p[[j]]]; jj=Abs[p[[j+1]]]; aa[[j,j]]=1; Do [aa[[j-k,j]]=ldu[[jj-k]],{k,1,jj-jmj-1}], {j,1,n}]; Return[aa]; ]; SymmSkyMatrixConvertDiagonalToFull[S_]:= Module[ {p,ldu,aa,n,i,j,jj,jmj,k}, {p,ldu}=S; n=Length[p]-1; aa=Table[0,{n},{n}]; Do [jj=Abs[p[[j+1]]]; aa[[j,j]]=ldu[[jj]], {j,1,n}]; Return[aa]; ];
26–19
26–20
Chapter 26: SOLVING FEM EQUATIONS
Homework Exercises for Chapter 26 Solving FEM Equations
1
u 1, f1
u 2, f2
u 3, f3
u 4, f4
(1)
(2)
(3)
(4)
2
3
4
u 5, f5
5
Figure 26.1. Structure for Exercise 26.1
EXERCISE 26.1 [A/C:10+10+15] Consider the 4-element assembly of bar elements shown in Figure 26.1.
The only degree of freedom at each node is a translation along x. The element stiffness matrix of each element is
1 −1 (E26.1) K(e) = −1 1 (a)
Assemble the 5 × 5 master stiffness matrix K showing it as a full symmetric matrix. Hint: the diagonal entries are 1, 2, 2, 2, 1.
(b)
Show K stored as a skyline matrix using a representation like illustrated in (26.15). Hint p = { 0, 1, 3, 5, 7, 9}.
(c)
Perform the symmetric factorization K = LDLT of (26.7), where K is stored as a full matrix to simplify hand work.6 Show that the entries of D are 1, 1, 1, 1 and 0, which mean that K is singular. Why?
6
You can do this with Mathematica using the function LUDecomposition, but hand work is as quick.
26–20
27
.
A Complete Plane Stress FEM Program
27–1
27–2
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
TABLE OF CONTENTS Page
§27.1. Introduction §27.2. Analysis Stages §27.3. Model Definition §27.3.1. Benchmark Problems . . . . §27.3.2. Node Coordinates . . . . . §27.3.3. Element Type . . . . . . §27.3.4. Element Connectivity . . . . §27.3.5. Material Properties . . . . §27.3.6. Fabrication Properties . . . . §27.3.7. Node Freedom Tags . . . . §27.3.8. Node Freedom Values . . . . §27.3.9. Processing Options . . . . §27.3.10. Model Display Utility . . . . §27.3.11. Model Definition Print Utilities §27.3.12. Model Definition Script Samples §27.4. Processing §27.4.1. Processing Tasks . . . . . §27.5. Postprocessing §27.5.1. Result Print Utilities . . . . §27.5.2. Displacement Field Contour Plots §27.5.3. Stress Field Contour Plots . . . §27.5.4. Animation . . . . . . . §27.6. A Complete Problem Script Cell §27. Notes and Bibliography . . . . . . . . . . . .
27–2
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27–3 27–3 27–3 27–4 27–4 27–6 27–6 27–8 27–8 27–9 27–9 27–10 27–10 27–10 27–10 27–12 27–12 27–14 27–14 27–15 27–15 27–16 27–16 27–18
27–3
§27.3
MODEL DEFINITION
§27.1. Introduction This Chapter describes a complete finite element program for analysis of plane stress problems. Unlike the previous chapters the description is top down, i.e. starts from the main driver down to more specific modules. The program can support questions given in the take-home final exam, if they pertain to the analysis of given plane stress problems. Consequently this Chapter serves as an informal users manual. §27.2. Analysis Stages As in all DSM-based FEM programs, the analysis of plane stress problems involves three major stages: (I) preprocessing or model definition, (II) processing, and (III) postprocessing. The preprocessing portion of the plane stress analysis is done by the first part of the problem script, driver program, already encountered in Chapters 21-22. The script directly sets the problem data structures. Preprocessing tasks include: I.1 I.2
Model definition by direct setting of the data structures. Plot of the FEM mesh for verification. At the minimum this involves producing a mesh picture that shows nodes and element labels.
The processing stage involves three tasks: II.1
II.2 II.3
Assembly of the master stiffness equations Ku = f. The plane stress assembler is of multiple element type (MET). This kind of assembler was discussed in §27.4. Element types include various plane stress Iso-P models as well as bars. ˆ = ˆf. The same Application of displacement BC by a modification method that produces Ku modules described in §21.3.3 are used, since those are application problem independent. Solution of the modified equations for node displacements u. The built-in Mathematica function LinearSolve is used for this task.
Upon executing the processing steps, the nodal displacement solution is available. The postprocessing stage embodies three tasks: III.1 Recovery of node forces including reactions through built-in matrix multiplication f = Ku. III.2 Recovery of plate stresses and bar forces (if bars are present). The former are subject to interelement averaging (Chapter 28) to get nodal stresses. III.3 Print and plotting of results. §27.3. Model Definition The model-definition data may be broken down into three sets, which are listed below by order of appearance: 0002 Geometry data: node coordinates Model definition Element data: type, connectivity, material and fabrication Degree of freedom (DOF) activity data: force and displacement BCs (27.1) The element data is broken down into four subsets: type, connnectivity, material and fabrication, each of which has its own data structure. The degree of freedom data is broken into two subsets: 27–3
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
27–4
Table 27.1. Plane Stress Model Definition Data Structures Long name
Short name
Dimensions
Description
NodeCoordinates ElemTypes ElemNodes ElemMaterials ElemFabrications NodeDOFTags NodeDOFValues ProcessOptions
nodxyz elenod elenod elemat elefab nodtag nodval prcopt
numnod numele numele numele numele numnod numnod
Node coordinates in global system Element type identifiers Element node lists Element material properties Element fabrication properties Node freedom tags marking BC type Node freedom specified values Processing specifications
x 2 x x x x x
2 2
Notation: numnod: number of nodes (no numbering gaps allowed, each node has two ) displacement DOFs; numele: number of elements; variable dimension. Long names are used in user-written problem scripts. Short names are used in programmed modules. Dimensions refer to the first two level Dimensions of the Mathematica list that implements the data structure. These give a guide as to implementation in a low level language such as C.
tags and values. In addition there are miscellaneous process options, such as the symbolic versus numeric processing. These options are conveniently collected in a separate data set. Accordingly, the model-definition input to the plane stress FEM program consists of eight data structures, which are called NodeCoordinates, ElemTypes, ElemNodes, ElemMaterials, ElemFabrications, NodeDOFTags, NodeDOFValues and ProcessOptions. These are summarized in Table 27.1. The configuration of these data structures are described in the following subsections with reference to the benchmark problems and discretizations shown in Figures 27.1 and 27.2. §27.3.1. Benchmark Problems Figure 27.1(a) illustrates a rectangular steel plate in plane stress under uniform uniaxial loading in the y (vertical) direction. Note that the load q is specified as force per unit area (kips per square inch); thus q has not been integrated through the thickness h. The exact analytical solution of the problem is1 σ yy = q, σx x = σx y = 0, u y = qy/E, u x = −νu y = −νq x/E. This problem should be solved exactly by any finite element mesh as long as the model is consistent and stable. In particular, the two one-quadrilateral-element models shown in 27.1(b,c). A similar but more complicated problem is shown in Figure 27.2: a rectangular steel plate dimensioned and loaded as that of Figure 27.1(a) but now with a central circular hole. This problem is defined in Figure 27.2(a). A FEM solution is to be obtained using the two quadrilateral element models (with 4-node and 9-nodes, respectively) depicted in Figure 27.2(b,c). The main result sought is the stress concentration factor on the hole boundary, and comparison of this computed factor with the exact analytical value. 1
The displacement solution u y = qy/E and u x = −νu y assumes that the plate centerlines do not translate or rotate, a condition enfoirced in the FEM discretizations shown in Figure 27.1(b,c).
27–4
27–5
§27.3 MODEL DEFINITION
y q = 10 ksi
(a)
B
A
;;
75 kips
C
(b) 75 kips
B 1
C 3
12 in H
x
E = 10000 ksi ν = 0.25 h = 3 in
G
J 2
D
J
10 in
4 D
Model (I): 4 nodes, 8 DOFs, 1 bilinear quad
B 1
4
C 7
1 2
8
5
J 3
6
9 D
Model (II): 9 nodes, 18 DOFs, 1 biquadratic quad
Global node numbers shown
E
F
;; ;
;; ; ;; ;; ; ;;; ;;; ;;
1
(c) 25 kips 100 kips 25 kips
q
Figure 27.1. Rectangular plate under uniform uniaxial loading, and two oneelement FEM discretizations of its upper right quadrant.
§27.3.2. Node Coordinates The geometry data is specified through NodeCoordinates. This is a list of node coordinates configured as NodeCoordinates = { { x1 , y1 },{ x2 , y2 }, . . . { x N , y N } }
(27.2)
where N is the number of nodes. Coordinate values should be normally be floating point numbers;2 use the N function to insure that property if necessary. Nodes must be numbered consecutively and no gaps are permitted. Example 27.1. For Model (I) of Figure 27.1(b):
NodeCoordinates=N[{ { 0,6 },{ 0,0 },{ 5,6 },{ 5,0 } }]; Example 27.2. For Model (II) of Figure 27.1(c):
NodeCoordinates=N[{ { 0,6 },{ 0,3 },{ 0,0 },{ 5/2,6 },{ 5/2,3 },{ 5/2,0 },{ 5,6 },{ 5,3 },{ 5,0 } }]; Example 27.3. For Model (I) of Figure 27.2(a), using a bit of coordinate generation:
s={1,0.70,0.48,0.30,0.16,0.07,0.0}; xy1={0,6}; xy7={0,1}; xy8={2.5,6}; xy14={Cos[3*Pi/8],Sin[3*Pi/8]}; xy8={2.5,6}; xy21={Cos[Pi/4],Sin[Pi/4]}; xy15={5,6}; xy22={5,2}; xy28={Cos[Pi/8],Sin[Pi/8]}; xy29={5,0}; xy35={1,0}; NodeCoordinates=Table[{0,0},{35}]; Do[NodeCoordinates[[n]]=N[s[[n]] *xy1+(1-s[[n]]) *xy7], {n,1,7}]; Do[NodeCoordinates[[n]]=N[s[[n-7]] *xy8+(1-s[[n-7]]) *xy14],{n,8,14}]; Do[NodeCoordinates[[n]]=N[s[[n-14]]*xy15+(1-s[[n-14]])*xy21],{n,15,21}]; 2
Unless one is doing a symbolic or exact-arithmetic analysis. Those are rare at the level of a full FEM analysis.
27–5
27–6
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
y q = 10 ksi
A
B
C
J
12 in H
D
K
x
R = 1 in E = 10000 ksi ν = 0.25 h = 3 in G
E
F
10 in
q
; ; ; ; ; ; ;
Node 8 is exactly midway between 1 and 15
(b)
37.5 kips
75 kips
B
37.5 kips
C 15
8
1
1 7
2 9
2
3
10 4
17
11
5 12
16
8
22 18
19 6 13 20 7 26 J 14 21 27 28 35 34 33
23 24 25
;; ;; ;; ;
J
32
31
30
29
B 1
(c) 25 kips
100 kips
25 kips
C
8
15
Note: internal point of a 9-node quadrilateral is placed at intersection of the medians
2
9
1
16
3 10 22 4
11
17
2
18 12 19 6 13 24 25 20 7 26 14 27 21 28 35 34 33 31 32 5
23
30 29
;; ; ;; ; ;; ; ;; ;
(a)
K
D
Model (I): 35 nodes, 70 DOFs, 24 bilinear quads
K
D
Model (II): 35 nodes, 70 DOFs, 6 biquadratic quads
Figure 27.2. Plate with a circular hole and two FEM discretizations of its upper right quadrant. Only a few element numbers are shown to reduce clutter. Note: the meshes generated with the example scripts given later differ slightly from those shown above. Compare, for example, mesh in (b) above with that in Figure 27.3.
Do[NodeCoordinates[[n]]=N[s[[n-21]]*xy22+(1-s[[n-21]])*xy28],{n,22,28}]; Do[NodeCoordinates[[n]]=N[s[[n-28]]*xy29+(1-s[[n-28]])*xy35],{n,29,35}]; The result of this generation is that some of the interior nodes are not in the same positions as sketched in Figure 27.2(b), but this slight change hardly affects the results.
§27.3.3. Element Type The element type is a label that specifies the type of model to be used. These labels are placed into an element type list: ElemTypes = { type(1) , type(2) , . . . type Ne }
(27.3)
Here typee is the type identifier of the e-th element specified as a character string. Ne is the number of elements; no element numbering gaps are allowed. Legal type identifiers are listed in Table 27.2. Example 27.4. For Model (I) in Figure 27.2(a):
ElemTypes=Table['Quad4',{ numele }]; Example 27.5. For Model (II) in Figure 27.2(b):
ElemTypes=Table['Quad9',{ numele }]; Here numele is a variable that contains the number of elements. This can be extracted, for example, as numele=Length[ElemNodes], where ElemNodes is defined below, if ElemNodes is defined first as is often the case.
27–6
27–7
§27.3 MODEL DEFINITION Table 27.2. Element Type Identifiers Implemented in Plane Stress Program Identifier
Nodes
Model for
Description
'Bar2' 'Bar3' 'Trig3' 'Trig6' 'Trig6.-3' 'Trig10' 'Quad2' 'Quad4.1' 'Quad8' 'Quad8.2' 'Quad9' 'Quad9.2'
2 3 3 6 6 10 4 4 8 8 9 9
bar bar plate plate plate plate plate plate plate plate plate plate
2-node bar element 3-node bar element; may be curved 3-node linear triangle 6-node quadratic triangle (3-interior point Gauss rule) 6-node quadratic triangle (midpoint Gauss rule) 10-node cubic triangle (7 point Gauss rule) RS 4-node iso-P bilinear quad (2 × 2 Gauss rule) RD 4-node iso-P bilinear quad (1-point Gauss rule) RS 8-node iso-P serendipity quad (3 × 3 Gauss rule) RD 8-node iso-P serendipity quad (2 × 2 Gauss rule) RS 9-node iso-P biquadratic quad (3 × 3 Gauss rule) RD 9-node iso-P biquadratic quad (2 × 2 Gauss rule)
RS: rank sufficient, RD: Rank deficient.
§27.3.4. Element Connectivity Element connectivity information specifies how the elements are connected.3 This information is stored in ElemNodes, which is a list of element nodelists: ElemNodes = { { enl(1) }, { enl(2) }, . . . { enL Ne } }
(27.4)
Here enle denotes the lists of nodes of the element e (as given by global node numbers) and Ne is the total number of elements. Element boundaries must be traversed counterclockwise (CCW) but you can start at any corner. Numbering elements with midnodes requires more care: begin listing corners CCW, followed by midpoints also CCW (first midpoint is the one that follows first corner when traversing CCW). If element has thirdpoints, as in the case of the 10-node triangle, begin listing corners CCW, followed by thirdpoints CCW (first thirdpoint is the one that follows first corner). When elements have an interior node, as in the 9-node quadrilateral or 10-node triangle, that node goes last. Example 27.6. For Model (I) of Figure 27.1(b), which has only one 4-node quadrilateral:
ElemNodes={ { 1,2,4,3 } }; Example 27.7. For Model (II) of Figure 27.1(c), which has only one 9-node quadrilateral:
ElemNodes={ { 1,3,9,7,2,6,8,4,5 } }; Example 27.8. For Model (I) of Figure 27.2(b), numbering the elements from top to bottom and from left to right: ElemNodes=Table[{0,0,0,0},{24}]; ElemNodes[[1]]={1,2,9,8}; 3
Some FEM programs call this the “topology” data.
27–7
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
27–8
Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,2,6}]; ElemNodes[[7]]=ElemNodes[[6]]+{2,2,2,2}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,8,12}]; ElemNodes[[13]]=ElemNodes[[12]]+{2,2,2,2}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,14,18}]; ElemNodes[[19]]=ElemNodes[[18]]+{2,2,2,2}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,20,24}]; Example 27.9. For Model (II) of Figure 27.2(c), numbering the elements from top to bottom and from left to
right: ElemNodes=Table[{0,0,0,0,0,0,0,0,0},{6}]; ElemNodes[[1]]={1,3,17,15,2,10,16,8,9}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{2,2,2,2,2,2,2,2,2},{e,2,3}]; ElemNodes[[4]]=ElemNodes[[3]]+{10,10,10,10,10,10,10,10,10}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{2,2,2,2,2,2,2,2,2},{e,5,6}]; Since this particular mesh only has 6 elements, it would be indeed faster to write down the six nodelists.
§27.3.5. Material Properties Data structure ElemMaterials is a list that provides the constitutive properties of the elements: ElemMaterials = { mprop(1) , mprop(2) , . . . mprop Ne }
(27.5)
For a plate element, mprop is the stress-strain matrix of elastic moduli (also known as elasticity matrix) arranged as { { E11,E12,E33 },{ E12,E22,E23 },{ E13,E23,E33 } }. Note that although this matrix is symmetric, it must be specified as a full 3 × 3 matrix. For a bar element, mprop is simply the longitudinal elastic modulus. A common case in practice is that (i) all elements are plates, (ii) the plate material is uniform and isotropic. An isotropic elastic material is specified by the elastic modulus E and Poisson’s ratio ν. Then this list can be generated by a single Table instruction upon building the elastic ity matrix, as in the example below. Example 27.10. For all FEM discretizations in Figures 27.1 and 27.2 all elements are plates of the same
isotropic material. Suppose that values of the elastic modulus E and Poisson’s ratio ν are stored in Em and nu, respectively, which are typically declared at the beginning of the problem script. Let numele give the number of elements. Then the material data is compactly declared by saying Emat=Em/(1-nu^2)*{ { 1,nu,0 },{ nu,1,0 },{ 0,0,(1-nu)/2 } }; ElemMaterials=Table[Emat,{ numele }];
§27.3.6. Fabrication Properties Data structure ElemFabrications is a list that provides the fabrication properties of the elements: ElemFabrications = { fprop(1) , fprop(2) , . . . fprop Ne } 27–8
(27.6)
27–9
§27.3 MODEL DEFINITION
For a plate element, fprop is the thickness h of the plate, assumed constant.4 For a bar element, fprop is the cross section area. If all elements are plates with the same thickness, this list can be easily generated by a Table instruction as in the example below. Example 27.11. For all FEM discretizations in Figures 27.1 and 27.2 all elements are plates with the same thickness h, which is stored in variable th. This is typically declared at the start of the problem script. As before, numele has the number of elements. Then the fabrication data is compactly declared by saying
ElemFabrications=Table[th,{ numele }]
§27.3.7. Node Freedom Tags Data structure NodeDOFTags is a list that labels each node DOF as to whether the load or the displacement is specified. The configuration of this list is similar to that of NodeCoordinates: NodeDOFTags={ { tagx1 , tag y1 },{ tagx2 , tag y2 }, . . . { tagx N , tag y N } }
(27.7)
The tag value is 0 if the force is specified and 1 if the displacement is specified. When there are a lot of nodes, often the quickest way to specify this list is to create with a Table command that initializes it to all zeros. Then displacement BCs are inserted appropriately, as in the example below. Example 27.12. For Model (I) in Figure 27.2(a):
numnod=Length[NodeCoordinates]; NodeDOFTags=Table[{0,0},{numnod}]; (* create and initialize to zero *) Do[NodeDOFTags[[n]]={1,0},{n,1,7}]; (* vroller @ nodes 1 through 7 *) Do[NodeDOFTags[[n]]={0,1},{n,29,35}]; (* hroller @ nodes 29 through 35 *) This scheme works well because typically the number of supported nodes is small compared to the total number.
§27.3.8. Node Freedom Values Data structure NodeDOFValues is a list with the same node by node configuration as NodeDOFTags: NodeDOFValues={ { valuex1 , value y1 },{ valuex2 , value y2 }, . . . { valuex N , value y N } } (27.8) Here value is the specified value of the applied node force component if the corresponding tag is zero, and of the prescribed displacement component if the tag is one. Often most of the entries of (27.8) are zero. If so a quick way to build it is to create it with a Table command that initializes it to zero. Then nonzero values are inserted as in the example below.
4
It is possible also to specify a variable thickness by making fprop a list that contains the thicknesses at the nodes. Since the variable thickness case is comparatively rare, it will not be described here.
27–9
27–10
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
Example 27.13. For the model (I) in Figure 27.2(a) only 3 values (for the y forces on nodes 1, 8 and 15) will
be nonzero: numnod=Length[NodeCoordinates]; NodeDOFValues=Table[{0,0},{numnod}]; (* create and initialize to zero *) NodeDOFValues[[1]]=NodeDOFValues[[15]]={0,37.5}; NodeDOFValues[[8]]={0,75}; (* y nodal loads *)
§27.3.9. Processing Options Array ProcessOptions is a list of general processing options that presently contains only the numer logical flag. This is normally be set to True to specify numeric computations: ProcessOptions={ True } §27.3.10. Model Display Utility Only one graphic display utility is presently provided to show the mesh. Nodes and elements of Model (I) of Figure 27.2(a) may be plotted by saying aspect=6/5; Plot2DElementsAndNodes[NodeCoordinates, ElemNodes,aspect, 'Plate with circular hole - 4-node quad model',True,True]; Here aspect is the plot frame aspect ratio (y dimension over x dimension). The 4th argument is a plot title textstring. The last two True argument values specify that node labels and element labels, respectively, be shown. The output of the mesh plot command is shown in Figure 27.3. §27.3.11. Model Definition Print Utilities Several print utilities are provided in the plane stress program to print out model definition data in tabular form. They are invoked as follows. To print the node coordinates:
One element mesh - 4 node quad 8
1
15
1
7
2
9 2
3
8 10
3 4
6 7
12
13
17 9 14
11 4
5
16
18 10
22
15
19
5 13 11 16 20 6 14 12 17 25 21 1827 26 28 22 24 23 35 34 33 32
23 24 19 20
21 31
30
29
Figure 27.3. Mesh plot for Model (I) of Figure 27.2(a).
PrintPlaneStressNodeCoordinates[NodeCoordinates,title,digits]; To print the element types and nodes: PrintPlaneStressElementTypeNodes[ElemTypes,EleNodes,title,digits]; To print the element materials and fabrications: PrintPlaneStressElementMatFab[ElemMaterials,ElemFabrications,title,digits]; To print freedom activity data: PrintPlaneStressFreedomActivity[NodeDOFTags,NodeDOFValues,title,digits]; In all cases, title is an optional character string to be printed as a title before the table; for example 'Node coordinates'. To eliminate the title, specify ' (two quote marks together). The last argument of the print modules: digits, is optional. If set to { d,f } it specifies that floating point numbers are to be printed with room for at least d digits, with f digits after the decimal point. If digits is specified as a void list: { }, a preset default is used for d and f.
27–10
27–11
§27.3
MODEL DEFINITION
§27.3.12. Model Definition Script Samples As capstone examples, Figures 27.4 and Figures 27.5 list the preprocessing (model definition) parts of the problem scripts for Model (I) and (II), respectively, of Figure 27.1(b,c). ClearAll[Em,ν,th]; Em=10000; ν=.25; th=3; aspect=6/5; Nsub=4; Emat=Em/(1-ν^2)*{{1,ν,0},{ν,1,0},{0,0,(1-ν)/2}}; (*
Define FEM model *)
NodeCoordinates=N[{{0,6},{0,0},{5,6},{5,0}}]; PrintPlaneStressNodeCoordinates[NodeCoordinates,',{6,4}]; ElemNodes= {{1,2,4,3}}; numnod=Length[NodeCoordinates]; numele=Length[ElemNodes]; ElemTypes= Table['Quad4',{numele}]; PrintPlaneStressElementTypeNodes[ElemTypes,ElemNodes,',{}]; ElemMaterials= Table[Emat, {numele}]; ElemFabrications=Table[th, {numele}]; PrintPlaneStressElementMatFab[ElemMaterials,ElemFabrications,',{}]; NodeDOFValues=NodeDOFTags=Table[{0,0},{numnod}]; NodeDOFValues[[1]]=NodeDOFValues[[3]]={0,75}; (* nodal loads *) NodeDOFTags[[1]]={1,0}; (* vroller @ node 1 *) NodeDOFTags[[2]]={1,1}; (* fixed node 2 *) NodeDOFTags[[4]]={0,1}; (* hroller @ node 4 *) PrintPlaneStressFreedomActivity[NodeDOFTags,NodeDOFValues,',{}]; ProcessOptions={True}; Plot2DElementsAndNodes[NodeCoordinates,ElemNodes,aspect, 'One element mesh - 4-node quad',True,True];
Figure 27.4. Model definition part of Model (I) of Figure 27.1(b). ClearAll[Em,ν,th]; Em=10000; ν=.25; th=3; aspect=6/5; Nsub=4; Emat=Em/(1-ν^2)*{{1,ν,0},{ν,1,0},{0,0,(1-ν)/2}}; (*
Define FEM model *)
NodeCoordinates=N[{{0,6},{0,3},{0,0},{5/2,6},{5/2,3}, {5/2,0},{5,6},{5,3},{5,0}}]; PrintPlaneStressNodeCoordinates[NodeCoordinates,',{6,4}]; ElemNodes= {{1,3,9,7,2,6,8,4,5}}; numnod=Length[NodeCoordinates]; numele=Length[ElemNodes]; ElemTypes= Table['Quad9',{numele}]; PrintPlaneStressElementTypeNodes[ElemTypes,ElemNodes,',{}]; ElemMaterials= Table[Emat, {numele}]; ElemFabrications=Table[th, {numele}]; PrintPlaneStressElementMatFab[ElemMaterials,ElemFabrications,',{}]; NodeDOFValues=NodeDOFTags=Table[{0,0},{numnod}]; NodeDOFValues[[1]]=NodeDOFValues[[7]]={0,25}; NodeDOFValues[[4]]={0,100}; (* nodal loads *) NodeDOFTags[[1]]=NodeDOFTags[[2]]={1,0}; (* vroller @ nodes 1,2 *) NodeDOFTags[[3]]={1,1}; (* fixed node 3 *) NodeDOFTags[[6]]=NodeDOFTags[[9]]={0,1}; (* hroller @ nodes 6,9 *) PrintPlaneStressFreedomActivity[NodeDOFTags,NodeDOFValues,',{}]; ProcessOptions={True}; Plot2DElementsAndNodes[NodeCoordinates,ElemNodes,aspect, 'One element mesh - 9-node quad',True,True];
Figure 27.5. Model definition part of Model (II) of Figure 27.1(c).
27–11
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
27–12
PlaneStressSolution[nodxyz_,eletyp_,elenod_,elemat_,elefab_, nodtag_,nodval_,prcopt_]:= Module[{K,Kmod,f,fmod,u,numer=True, noddis,nodfor,nodpnc,nodsig,barele,barfor}, If [Length[prcopt]>=1, numer=prcopt[[1]]]; K=PlaneStressMasterStiffness[nodxyz,eletyp,elenod,elemat,elefab,prcopt]; If [KNull,Return[Table[Null,{6}]]]; Kmod=ModifiedMasterStiffness[nodtag,K]; f=FlatNodePartVector[nodval]; fmod=ModifiedNodeForces[nodtag,nodval,K,f]; u=LinearSolve[Kmod,fmod]; If [numer,u=Chop[u]]; f=K.u; If [numer,f=Chop[f]]; nodfor=NodePartFlatVector[2,f]; noddis=NodePartFlatVector[2,u]; {nodpnc,nodsig}=PlaneStressPlateStresses[nodxyz,eletyp,elenod,elemat, elefab,noddis,prcopt]; {barele,barfor}=PlaneStressBarForces[nodxyz,eletyp,elenod,elemat, elefab,noddis,prcopt]; ClearAll[K,Kmod]; Return[{noddis,nodfor,nodpnc,nodsig,barele,barfor}]; ];
Figure 27.6. Module to drive the analysis of the plane stress problem.
§27.4. Processing Once the model definition is complete, the plane stress analysis is carried out by calling module LinearSolutionOfPlaneStressModel listed in Figure 27.6. This module is invoked from the problem driver as {NodeDisplacements,NodeForces,NodePlateCounts,NodePlateStresses, ElemBarNumbers,ElemBarForces}= PlaneStressSolution[NodeCoordinates, ElemTypes,ElemNodes,ElemMaterials,ElemFabrications, NodeDOFTags,NodeDOFValues,ProcessOptions];
The module arguments: NodeCoordinates, ElemTypes, ElemNodes , ElemMaterials, ElemFabrications, NodeDOFTags, NodeDOFValues and ProcessOptions are the data structures described in the previous section. The module returns the following: NodeDisplacements Computed node displacements, in node-partitioned form. NodeForces
Recovered node forces including reactions, in node-partitioned form.
NodePlateCounts
For each node, number of plate elements attached to that node. A zero count means that no plate elements are attached to that node.
NodePlateStresses Averaged stresses at plate nodes with a nonzero NodePlateCounts. ElemBarNumbers
A list of bar elements if any specified, else an empty list.
ElemBarForces
A list of bar internal forces if any bar elements were specified, else an empty list.
§27.4.1. Processing Tasks PlaneStressSolution carries out the following tasks. It assembles the free-free master stiffness matrix K by calling the MET assembler PlaneStressMasterStiffness. This module is listed in Figure 27.7. A study
27–12
27–13
§27.4 PROCESSING PlaneStressMasterStiffness[nodxyz_,eletyp_,elenod_,elemat_, elefab_,prcopt_]:=Module[{numele=Length[elenod], numnod=Length[nodxyz],ncoor,type,e,enl,neldof,OKtyp,OKenl, i,j,n,ii,jj,eft,Emat,th,numer,Ke,K}, OKtyp={'Bar2','Bar3','Quad4','Quad4.1','Quad8','Quad8.2','Quad9', 'Quad9.2','Trig3','Trig6','Trig6.-3','Trig10','Trig10.6'}; OKenl= {2,3,4,4,8,8,9,9,3,6,6,10,10}; K=Table[0,{2*numnod},{2*numnod}]; numer=prcopt[[1]]; For [e=1,e<=numele,e++, type=eletyp[[e]]; If [!MemberQ[OKtyp,type], Print['Illegal type:',type, ' element e=',e,' Assembly aborted']; Return[Null] ]; enl=elenod[[e]]; n=Length[enl]; {{j}}=Position[OKtyp,type]; If [OKenl[[j]]!=n, Print ['Wrong node list length, element=',e, ' Assembly aborted']; Return[Null] ]; eft=Flatten[Table[{2*enl[[i]]-1,2*enl[[i]]},{i,1,n}]]; ncoor=Table[nodxyz[[enl[[i]]]],{i,1,n}]; If [type'Bar2', Em=elemat[[e]]; A=elefab[[e]]; Ke=PlaneBar2Stiffness[ncoor,Em,A,{numer}] ]; If [type'Bar3', Em=elemat[[e]]; A=elefab[[e]]; Ke=PlaneBar3Stiffness[ncoor,Em,A,{numer}] ]; If [type'Quad4', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,th,{numer,2}] ]; If [type'Quad4.1', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad4IsoPMembraneStiffness[ncoor,Emat,th,{numer,1}] ]; If [type'Quad8', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad8IsoPMembraneStiffness[ncoor,Emat,th,{numer,3}] ]; If [type'Quad8.2', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad8IsoPMembraneStiffness[ncoor,Emat,th,{numer,2}] ]; If [type'Quad9', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad9IsoPMembraneStiffness[ncoor,Emat,th,{numer,3}] ]; If [type'Quad9.2', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Quad9IsoPMembraneStiffness[ncoor,Emat,th,{numer,2}] ]; If [type'Trig3', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Trig3IsoPMembraneStiffness[ncoor,Emat,th,{numer}] ]; If [type'Trig6', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Trig6IsoPMembraneStiffness[ncoor,Emat,th,{numer,3}] ]; If [type'Trig6.-3', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Trig6IsoPMembraneStiffness[ncoor,Emat,th,{numer,-3}] ]; If [type'Trig10', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Trig10IsoPMembraneStiffness[ncoor,Emat,th,{numer,7}] ]; If [type'Trig10.6', Emat=elemat[[e]]; th=elefab[[e]]; Ke=Trig10IsoPMembraneStiffness[ncoor,Emat,th,{numer,6}] ]; neldof=Length[Ke]; For [i=1,i<=neldof,i++, ii=eft[[i]]; For [j=i,j<=neldof,j++, jj=eft[[j]]; K[[jj,ii]]=K[[ii,jj]]+=Ke[[i,j]] ]; ]; ]; Return[K]; ];
Figure 27.7. Plane stress assembler module.
of its code reveals that it handle the multiple element types listed in Table 27.2. The modules that compute the element stiffness matrices have been studied in previous Chapters and need not be listed here. The displacement BCs are applied by ModifiedMasterStiffness and ModifiedNodeForces. These are the same modules used in Chapter 21 for the space truss program, and thus need not be described further. The unknown node displacements u are obtained through the built in LinearSolve function, as u=LinearSolve[Kmod,fmod]. This function is of course restricted to small systems, typically less than 200 equations (it gets extremely slow for something bigger) but it has the advantages of simplicity. The node
27–13
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
27–14
PlaneStressPlateStresses[nodxyz_,eletyp_,elenod_,elemat_, elefab_,noddis_,prcopt_]:=Module[{numele=Length[elenod], numnod=Length[nodxyz],ncoor,type,e,enl,i,k,n,Emat,th, numer=True,nodpnc,nodsig}, If [Length[prcopt]>0, numer=prcopt[[1]]]; nodpnc=Table[0,{numnod}]; nodsig=Table[{0,0,0},{numnod}]; If [Length[prcopt]>=1, numer=prcopt[[1]]]; For [e=1,e<=numele,e++, type=eletyp[[e]]; enl=elenod[[e]]; k=Length[enl]; ncoor=Table[nodxyz[[enl[[i]]]],{i,1,k}]; ue=Table[{noddis[[enl[[i]],1]],noddis[[enl[[i]],2]]},{i,1,k}]; ue=Flatten[ue]; If [StringTake[type,3]'Bar', Continue[]]; If [type'Quad4', Emat=elemat[[e]]; th=elefab[[e]]; sige=Quad4IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Quad4.1', Emat=elemat[[e]]; th=elefab[[e]]; sige=Quad4IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Quad8', Emat=elemat[[e]]; th=elefab[[e]]; sige=Quad8IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Quad8.2', Emat=elemat[[e]]; th=elefab[[e]]; sige=Quad8IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Quad9', Emat=elemat[[e]]; th=elefab[[e]]; sige=Quad9IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Quad9.2', Emat=elemat[[e]]; th=elefab[[e]]; sige=Quad9IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Trig3', Emat=elemat[[e]]; th=elefab[[e]]; sige=Trig3IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Trig6', Emat=elemat[[e]]; th=elefab[[e]]; sige=Trig6IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Trig6.-3', Emat=elemat[[e]]; th=elefab[[e]]; sige=Trig6IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Trig10', Emat=elemat[[e]]; th=elefab[[e]]; sige=Trig10IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; If [type'Trig10.6', Emat=elemat[[e]]; th=elefab[[e]]; sige=Trig10IsoPMembraneStresses[ncoor,Emat,th,ue,{numer}] ]; Do [n=enl[[i]]; nodpnc[[n]]++; nodsig[[n]]+=sige[[i]], {i,1,k}]; ]; Do [k=nodpnc[[n]]; If [k>1,nodsig[[n]]/=k], {n,1,numnod}]; If [numer, nodsig=Chop[nodsig]]; Return[{nodpnc,nodsig}]; ];
Figure 27.8. Module that recovers averaged nodal stresses at plate nodes.
forces including reactions are recovered by the matrix multiplication f = K.u, where K is the unmodified master stiffness. Finally, plate stresses and bar internal forces are recovered by the modules PlaneStressPlateStresses and PlaneStressBarForces, respectively. The former is listed in Figure 27.8. This computation is actually part of the postprocessing stage. It is not described here since stress recovery is treated in more detail in a subsequent Chapter. The bar force recovery module is not described here as it has not been used in assigned problems so far.
§27.5. Postprocessing As noted above, module PlaneStressSolution carries out preprocessing tasks: recover node forces, plate stresses and bar forces. But those are not under control of the user. Here we describe result printing and plotting activities that can be specified in the problem script. 27–14
27–15
§27.5 POSTPROCESSING
§27.5.1. Result Print Utilities Several utilities are provided in the plane stress program to print solution data in tabular form. They are invoked as follows. To print computed node displacements: PrintPlaneStressNodeDisplacements[NodeDisplacements,title,digits]; To print receoverd node forces including reactions: PrintPlaneStressNodeForces[NodeForces,title,digits]; To print plate node stresses: PrintPlaneStressPlateNodeStresses[NodePlateCounts,NodePlateStresses,title,digits]; To print node displacements, node forces and node plate stresses in one table: PrintPlaneStressSolution[NodeDisplacements,NodeForces, NodePlateCounts,NodePlateStresses,title,digits]; No utility is provided to print bar forces, as problems involving plates and bars had not been assigned. In all utilities listed above, title is an optional character string to be printed as a title before the table; for example 'Node displacements for plate with a hole'. To eliminate the title, specify ' (two quote marks together). The last argument of the print modules: digits, is optional. If set to { d,f } it specifies that floating point numbers are to be printed with room for at least d digits, with f digits after the decimal point. If digits is specified as a void list: { }, a preset default is used for d and f. §27.5.2. Displacement Field Contour Plots Contour plots of displacement components u x and u y (interpolated over elements from the computed node displacements) may be produced. Displacement magnitudes are shown using a internally set color scheme based on “hue interpolation”, in which white means zero. Plots can be obtained using a script typified by ux=Table[NodeDisplacements[[n,1]],{n,numnod}]; uy=Table[NodeDisplacements[[n,2]],{n,numnod}]; {uxmax,uymax}=Abs[{Max[ux],Max[uy]}]; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes,ux, uxmax,Nsub,aspect,'Displacement component ux']; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes,uy, uymax,Nsub,aspect,'Displacement component uy']; The third argument of ContourPlotNodeFuncOver2DMesh is the function to be plotted, specified at nodes. A function maximum (in absolute value) is supplied as fourth argument. The reason for supplying this from the outside is that in many cases it is convenient to alter the actual maximum for zooming or animation purposes. As for the other arguments, Nsub is the number of element subdivisions in each direction when breaking down the element area into plot polygons. Typically Nsub is set to 4 at the start of the script and is the same for all plots of a script. Plot smoothness in terms of color grading increases with Nsub, but plot time grows as Nsub-squared. So an appropriate tradeoff is to use a high Nsub, say 8, for coarse meshes containing few elements whereas for finer meshes a value of 4 or 2 may work fine. The aspect argument specifies the ratio between the y (vertical) and x (horizontal) dimensions of the plot frame, and is usaully defined at the script start as a symbol so it is the same for all plots. The last argument is a plot title.
27–15
27–16
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM
Nodal stress sig- xx
Nodal stress sig- yy
Nodal stress sig- xy
Figure 27.9. Stress component contour plots for rectangular plate with a circular central hole, Model (I) of Figure 27.2(b).
§27.5.3. Stress Field Contour Plots Contour plots of stress components σx x , σ yy and σx y (interpolated over elements from the computed node displacements) may be produced. Stress magnitudes are shown using a internally set color scheme based on “hue interpolation”, in which white means zero. Plots can be obtained using a script typified by sxx=Table[NodePlateStresses[[n,1]],{n,numnod}]; syy=Table[NodePlateStresses[[n,2]],{n,numnod}]; sxy=Table[NodePlateStresses[[n,3]],{n,numnod}]; {sxxmax,syymax,sxymax}=Abs[{Max[sxx],Max[syy],Max[sxy]}]; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes, sxx,sxxmax,Nsub,aspect,'Nodal stress sig-xx']; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes, syy,syymax,Nsub,aspect,'Nodal stress sig-yy']; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes, sxy,sxymax,Nsub,aspect,'Nodal stress sig-xy']; For a description of ContourPlotNodeFuncOver2DMesh arguments, see the previous subsection. As an example, stress component contour plots for Model (I) of the rectangular plate with a circular central hole are shown in Figure 27.9. Remark 27.1. Sometimes it is useful to show contour plots of principal stresses, Von Mises stresses, maximum shears, etc. To do that, the appropriate function should be constructed first from the Cartesian components available in NodePlateStresses, and then the plot utility called.
§27.5.4. Animation Occasionally it is useful to animate results such as displacements or stress fields when a load changes as a function of a time-like parameter t. This can be done by performing a sequence of analyses in a For loop. Such sequence produces a series of plot frames which may be animated by doubly clicking on one of them. The speed of the animation may be controlled by pressing on the righmost buttons at the bottom of the Mathematica window. The second button from the left, if pressed, changes the display sequence to back and forth motion.
27–16
27–17
§27.6
A COMPLETE PROBLEM SCRIPT CELL
ClearAll[Em,ν,th,aspect,Nsub]; Em=10000; ν=.25; th=3; aspect=6/5; Nsub=4; Emat=Em/(1-ν^2)*{{1,ν,0},{ν,1,0},{0,0,(1-ν)/2}}; (*
Define FEM model *)
s={1,0.70,0.48,0.30,0.16,0.07,0.0}; xy1={0,6}; xy7={0,1}; xy8={2.5,6}; xy14={Cos[3*Pi/8],Sin[3*Pi/8]}; xy8={2.5,6}; xy21={Cos[Pi/4],Sin[Pi/4]}; xy15={5,6}; xy22={5,2}; xy28={Cos[Pi/8],Sin[Pi/8]}; xy29={5,0}; xy35={1,0}; NodeCoordinates=Table[{0,0},{35}]; Do[NodeCoordinates[[n]]=N[s[[n]] *xy1+(1-s[[n]]) *xy7],{n,1,7}]; Do[NodeCoordinates[[n]]=N[s[[n-7]] *xy8+(1-s[[n-7]]) *xy14],{n,8,14}]; Do[NodeCoordinates[[n]]=N[s[[n-14]]*xy15+(1-s[[n-14]])*xy21],{n,15,21}]; Do[NodeCoordinates[[n]]=N[s[[n-21]]*xy22+(1-s[[n-21]])*xy28],{n,22,28}]; Do[NodeCoordinates[[n]]=N[s[[n-28]]*xy29+(1-s[[n-28]])*xy35],{n,29,35}]; PrintPlaneStressNodeCoordinates[NodeCoordinates,',{6,4}]; ElemNodes=Table[{0,0,0,0},{24}]; ElemNodes[[1]]={1,2,9,8}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,2,6}]; ElemNodes[[7]]=ElemNodes[[6]]+{2,2,2,2}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,8,12}]; ElemNodes[[13]]=ElemNodes[[12]]+{2,2,2,2}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,14,18}]; ElemNodes[[19]]=ElemNodes[[18]]+{2,2,2,2}; Do [ElemNodes[[e]]=ElemNodes[[e-1]]+{1,1,1,1},{e,20,24}]; numnod=Length[NodeCoordinates]; numele=Length[ElemNodes]; ElemTypes= Table['Quad4',{numele}]; PrintPlaneStressElementTypeNodes[ElemTypes,ElemNodes,',{}]; ElemMaterials= Table[Emat, {numele}]; ElemFabrications=Table[th, {numele}]; (*PrintPlaneStressElementMatFab[ElemMaterials,ElemFabrications,',{}];*) NodeDOFValues=NodeDOFTags=Table[{0,0},{numnod}]; NodeDOFValues[[1]]=NodeDOFValues[[15]]={0,37.5}; NodeDOFValues[[8]]={0,75}; (* nodal loads *) Do[NodeDOFTags[[n]]={1,0},{n,1,7}]; (* vroller @ nodes 1-7 *) Do[NodeDOFTags[[n]]={0,1},{n,29,35}]; (* hroller @ node 4 *) PrintPlaneStressFreedomActivity[NodeDOFTags,NodeDOFValues,',{}]; ProcessOptions={True}; Plot2DElementsAndNodes[NodeCoordinates,ElemNodes,aspect, 'One element mesh - 4-node quad',True,True];
Figure 27.10. Preprocessing (model definition) script for problem of Figure 27.2(b).
§27.6. A Complete Problem Script Cell A complete problem script cell (which is Cell 13 in the plane stress Notebook placed in the web index of this Chapter) is listed in Figures 27.10 through 27.12. This driver cell does Model (I) of the plate with hole problem of Figure 27.2(b), which uses 4-node quadrilateral elements. The script is divided into 3 parts for convenience. Figure 27.10 lists the model definition script followed by a mesh plot command. Figure 27.11 shows the call to PlaneStressSolution analysis driver and the call to get printout of node displacements, forces and stresses. Finally, Figure 27.12 shows commands to produce contour plots of displacements (skipped) and stresses. Other driver cell examples may be studied in the PlaneStress.nb Notebook posted on the course web site. It can be observed that the processing and postprocessing scripts are largely the same. 27–17
Chapter 27: A COMPLETE PLANE STRESS FEM PROGRAM (*
27–18
Solve problem and print results *)
{NodeDisplacements,NodeForces,NodePlateCounts,NodePlateStresses, ElemBarNumbers,ElemBarForces}= PlaneStressSolution[ NodeCoordinates,ElemTypes,ElemNodes, ElemMaterials,ElemFabrications, NodeDOFTags,NodeDOFValues,ProcessOptions]; PrintPlaneStressSolution[NodeDisplacements,NodeForces,NodePlateCounts, NodePlateStresses,'Computed Solution:',{}];
Figure 27.11. Processing script and solution print for problem of Figure 27.2(b).
(*
Plot Displacement Components Distribution - skipped *)
(* ux=Table[NodeDisplacements[[n,1]],{n,numnod}]; uy=Table[NodeDisplacements[[n,2]],{n,numnod}]; {uxmax,uymax}=Abs[{Max[ux],Max[uy]}]; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes,ux, uxmax,Nsub,aspect,'Displacement component ux']; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes,uy, uymax,Nsub,aspect,'Displacement component uy']; *) (*
Plot Averaged Nodal Stresses Distribution *)
sxx=Table[NodePlateStresses[[n,1]],{n,numnod}]; syy=Table[NodePlateStresses[[n,2]],{n,numnod}]; sxy=Table[NodePlateStresses[[n,3]],{n,numnod}]; {sxxmax,syymax,sxymax}={Max[Abs[sxx]],Max[Abs[syy]],Max[Abs[sxy]]}; Print['sxxmax,syymax,sxymax=',{sxxmax,syymax,sxymax}]; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes, sxx,sxxmax,Nsub,aspect,'Nodal stress sig-xx']; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes, syy,syymax,Nsub,aspect,'Nodal stress sig-yy']; ContourPlotNodeFuncOver2DMesh[NodeCoordinates,ElemNodes, sxy,sxymax,Nsub,aspect,'Nodal stress sig-xy'];
Figure 27.12. Result plotting script for problem of Figure 27.2(b). Note that displacement contour plots have been skipped by commenting out the code.
What changes is the model definition script portion, which often benefits from ad-hoc node and element generation constructs. Notes and Bibliography Few FEM books show a complete program. More common is the display of snipets of code. These are left dangling chapter after chapter, with no attempt at coherent interfacing. This historical problem comes from early reliance on low-level languages such as Fortran. Even the simplest program may run to throusands of code lines. At 50 lines/page that becomes difficult to display snugly in a textbook while providing a running commentary. Another ages-old problem is plotting. Languages such as C or Fortran do not include plotting libraries for the simple reason that low-level universal plotting standards never existed. The situation changed when widely used high-level languages like Matlab or Mathematica appeared. The language engine provides the necessary fit to available computer hardware, concealing system dependent details. Thus plotting scripts are transportable.
27–18
28
.
Stress Recovery
28–1
28–2
Chapter 28: STRESS RECOVERY
TABLE OF CONTENTS Page
§28.1. §28.2. §28.3. §28.4. §28.5.
Introduction Calculation of Element Strains and Stresses Direct Stress Evaluation at Nodes Extrapolation from Gauss Points Interelement Averaging
28–2
28–3 28–3 28–4 28–4 28–6
28–3
§28.2
CALCULATION OF ELEMENT STRAINS AND STRESSES
§28.1. Introduction In this lecture we study the recovery of stress values for two-dimensional plane-stress elements.1 This analysis step is sometimes called postprocessing because it happens after the main processing step — the calculation of nodal displacements — is completed. Stress calculations are of interest because in structural analysis and design the stresses are often more important to the engineer than displacements. In the stiffness method of solution discussed in this course, the stresses are obtained from the computed displacements, and are thus derived quantities. The accuracy of derived quantities is generally lower than that of primary quantities (the displacements), an informal statement that may be mathematically justified in the theory of finite element methods. For example, if the accuracy level of displacements is 1% that of the stresses may be typically 10% to 20%, and even lower at boundaries. It is therefore of interest to develop techniques that enhance the accuracy of the computed stresses. The goal is to “squeeze” as much accuracy from the computed displacements while keeping the computational effort reasonable. These procedures receive the generic name stress recovery techniques in the finite element literature. In the following sections we cover the simplest stress recovery techniques that have been found most useful in practice. §28.2. Calculation of Element Strains and Stresses In elastic materials, stresses σ are directly related to strains e at each point through the elastic constitutive equations σ = Ee. It follows that the stress computation procedure begins with strain computations, and that the accuracy of stresses depends on that of strains. Strains, however, are seldom saved or printed. In the following we focus our attention on two-dimensional isoparametric elements, as the computation of strains, stresses and axial forces in bar elements is strightforward. Suppose that we have solved the master stiffness equations Ku = f,
(28.1)
for the node displacements u. To calculate strains and stresses we perform a loop over all defined elements. Let e be the element index of a specific two-dimensional isoparametric element encountered during this loop, and u(e) the vector of computed element node displacements. Recall from §15.3 that the strains at any point in the element may be related to these displacements as e = Bu(e) ,
(28.2)
where B is the strain-displacement matrix (14.18) assembled with the x and y derivatives of the element shape functions evaluated at the point where we are calculating strains. The corresponding stresses are given by σ = Ee = EBu (28.3) 1
This Chapter needs rewriting to show the use of Mathematica for stress computation. To be done in the future.
28–3
28–4
Chapter 28: STRESS RECOVERY
Table 28.1 Corner node 1 2 3 4
ξ
η
−1 −1 +1 −1 +1 +1 −1 +1
Natural Coordinates of Bilinear Quadrilateral Nodes ξ0003
η0003
√ −√3 +√ 3 +√ 3 − 3
√ −√3 −√3 +√3 + 3
Gauss node 1’ 2’ 3’ 4’
ξ
η
ξ0003
η0003
√ −1/√3 +1/√3 +1/√3 −1/ 3
√ −1/√3 −1/√3 +1/√3 +1/ 3
−1 +1 +1 −1
−1 −1 +1 +1
Gauss nodes, and coordinates ξ 0003 and η0003 are defined in §28.4 and Fig. 28.1
In the applications it is of interest to evaluate and report these stresses at the element nodal points located on the corners and possibly midpoints of the element. These are called element nodal point stresses. It is important to realize that the stresses computed at the same nodal point from adjacent elements will not generally be the same, since stresses are not required to be continuous in displacementassumed finite elements. This suggests some form of stress averaging can be used to improve the stress accuracy, and indeed this is part of the stress recovery technique further discussed in §28.5. The results from this averaging procedure are called nodal point stresses. For the moment let us see how we can proceed to compute element nodal stresses. Two approaches are followed in practice: 1.
Evaluate directly σ at the element node locations by substituting the natural coordinates of the nodal points as arguments to the shape function modules. These modules return qx and q y and direct application of (28.2)-(28.4) yields the strains and stresses at the nodes.
2.
Evaluate σ at the Gauss integration points used in the element stiffness integration rule and then extrapolate to the element node points.
Empirical evidence indicates that the second approach generally delivers better stress values for quadrilateral elements whose geometry departs substantially from the rectangular shape. This is backed up by “superconvergence” results in finite element approximation theory. For rectangular elements there is no difference. For isoparametric triangles both techniques deliver similar results (identical if the elements are straight sided with midside nodes at midpoints) and so the advantages of the second one are marginal. Both approaches are covered in the sequel. §28.3. Direct Stress Evaluation at Nodes This approach is straightforward and need not be discussed in detail. 28–4
28–5
§28.4
(a)
EXTRAPOLATION FROM GAUSS POINTS
η
(b) 3
3 4
4
4' ξ
(e)
3'
η' (e')
1' 1
ξ' 2'
1
2
2
Figure 28.1. Extrapolation from 4-node quad Gauss points: (a) 2 × 2 rule, (b) Gauss element (e0003 )
§28.4. Extrapolation from Gauss Points This will again be explained for the four-node bilinear quadrilateral. The normal Gauss integration rule for element stiffness evaluation is 2 × 2, as illustrated in Figure 28.1. The stresses are calculated at the Gauss points, which are identified as 10003 , 20003 , 30003 and 40003 in Figure 28.1. Point i 0003 is closest to node i so it is seen that Gauss point numbering essentially follows element node numbering in the counterclockwise sense. The natural coordinates of these points are listed in Table 28.1. The stresses are evaluated at these Gauss points by passing these natural coordinates to the shape function subroutine. Then each stress component is “carried” to the corner nodes 1 through 4 through a bilinear extrapolation based on the computed values at 10003 through 40003 . To understand the extrapolation procedure more clearly it is convenient to consider the region bounded by the Gauss points as an “internal element” or “Gauss element”. This interpretation is depicted in Figure 28.1(b). The Gauss element, denoted by (e0003 ), is also a four-node quadrilateral. Its quadrilateral (natural) coordinates are denoted by ξ 0003 and η0003 . These are linked to ξ and η by the simple relations √ √ √ √ ξ = ξ 0003 / 3, η = η0003 / 3, ξ 0003 = ξ 3, η0003 = η 3. (28.4) Any scalar quantity w whose values wi0003 at the Gauss element corners are known can be interpolated through the usual bilinear shape functions now expressed in terms of ξ 0003 and η0003 :  (e0003 )  N1 0003  (e )  0003 0003 0003 0003 0003 0003  N2 0003  (28.5) w(ξ , η ) = [ w1 w2 w3 w4 ]  (e )  , N 3
0003
N4(e ) where (cf. §15.6.2)
0003
N1(e ) = 14 (1 − ξ 0003 )(1 − η0003 ), 0003
N2(e ) = 14 (1 + ξ 0003 )(1 − η0003 ), 0003
N3(e ) = 14 (1 + ξ 0003 )(1 + η0003 ), 0003
N4(e ) = 14 (1 − ξ 0003 )(1 + η0003 ). 28–5
(28.6)
28–6
Chapter 28: STRESS RECOVERY
(a)
4 8
7'
3'
9 9'
6' 6
4
4' 8' 1'
1
(b)
3
7
5' 5
8 2
(c)
3'
4' 6' 6
8' 9
1
3
3'
7'
1'
2'
3
7
6 2'
2'
5'
5
2
5
1'
2
1 4
Figure 28.2. Gauss elements for higher order quadrilaterals and triangles: (a) 9-node element with 3 × 3 Gauss rule, (b) 8-node element with 3 × 3 Gauss rule, (c) 6-node element with 3-interior point rule.
√ To extrapolate w to corner 1, say, we replace its ξ 0003 and η0003 coordinates, namely ξ 0003 = η0003 = − 3, into the above formula. Doing that for the four corners we obtain √ √  0003    1 + 1 3 − 12 1 − 12 3 − 12 w1 w1 2 √ √ 1 1 1 1   −2 1 − 2 3   w20003   w2   − 2√ 1 + 2 3 √ (28.7)    = w3 − 12 1 + 12 3 − 12  w30003 1 − 12 3 √ √ w4 w40003 −1 1− 1 3 −1 1+ 1 3 2
2
2
2
Note that the sum of the coefficients in each row is one, as it should be. For stresses we apply this formula taking w to be each of the three stress components, σx x , σ yy and τx y , in turn. Extrapolation in Higher Order Elements For eight-node and nine-node isoparametric quadrilaterals the usual Gauss integration rule is 3 × 3, and the Gauss elements are nine-noded quadrilaterals that look as in Figure 28.2(a) and (b) above. For six-node triangles the usual quadrature is the 3-point rule with internal sampling points, and the Gauss element is a three-node triangle as shown in Figure 28.2(c). §28.5. Interelement Averaging The stresses computed in element-by-element fashion as discussed above, whether by direct evaluation at the nodes or by extrapolation, will generally exhibit jumps between elements. For printing and plotting purposes it is usually convenient to “smooth out” those jumps by computing averaged nodal stresses. This averaging may be done in two ways: (I)
Unweighted averaging: assign same weight to all elements that meet at a node;
(II) Weighted averaging: the weight assigned to element contributions depends on the stress component and the element geometry and possibly the element type. Several weighted average schemes have been proposed in the finite element literature, but they do require additional programming. 28–6
29
.
Thermomechanical Effects
29–1
29–2
Chapter 29: THERMOMECHANICAL EFFECTS
TABLE OF CONTENTS Page
§29.1. Thermomechanical Behavior §29.1.1. Thermomechanical Stiffness Relations §29.1.2. Globalization . . . . . . . . §29.1.3. Merge . . . . . . . . . . . §29.1.4. Solution . . . . . . . . . §29.1.5. Postprocessing . . . . . . . . §29.1.6. Worked-Out Example 1 . . . . §29.1.7. Worked-Out Example 2 . . . . . §29.2. Initial Force Effects §29.3. Pseudo Thermal Inputs §29. Notes and Bibliography . . . . . . . . . . . . . §29. References . . . . . . . . . . . . . §29. Exercises . . . . . . . . . . . . .
29–2
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
29–3 29–4 29–5 29–6 29–6 29–6 29–6 29–7 29–9 29–9 29–9 29–10 29–11
29–3
§29.1
THERMOMECHANICAL BEHAVIOR
The assumptions invoked in Chapters 2-3 for the example truss result in zero external forces under zero displacements. This is implicit in the linear-homogeneous expression of the master stiffness equation f = Ku. If u vanishes, so does f. This behavior does not apply, however, if there are initial force effects.1 If those effects are present, there can be displacements without external forces, and internal forces without displacements. A common source of initial force effects are temperature changes. Imagine that a plane truss structure is unloaded (that is, not subjected to external forces) and is held at a uniform reference temperature. External displacements are measured from this environment, which is technically called a reference state. Now suppose that the temperature of some members changes with respect to the reference temperature while the applied external forces remain zero. Because the length of members changes on account of thermal expansion or contraction, the joints will displace. If the structure is statically indeterminate those displacements will induce strains and stresses and thus internal forces. These are distinguished from mechanical effects by the qualifier “thermal.” For many structures, particularly in aerospace and mechanical engineering, such effects have to be considered in the analysis and design. There are other physical sources of initial force effects, such as moisture (hygrosteric) effects,2 member prestress, residual stresses, or lack of fit. For linear structural models all such sources may be algebraically treated in the same way as thermal effects. The treatment results in an initial force vector that has to be added to the applied mechanical forces. This subject is outlined in §29.3 from a general perspective. However, to describe the main features of the matrix analysis procedure it is sufficient to consider the case of temperature changes. In this Section we go over the analysis of a plane truss structure whose members undergo temperature changes from a reference state. It is assumed that the disconnection and localization steps of the DSM have been carried out. Therefore we begin with the derivation of the matrix stiffness equations of a generic truss member. §29.1. Thermomechanical Behavior Consider the generic plane-truss member shown in Figure 29.1. The member is prismatic and uniform. The temperature T is also uniform. To reduce clutter the member identification subscript will be omitted in the following development until the globalization and assembly steps. We introduce the concept of reference temperature Tr e f . This is conventionally chosen to be the temperature throughout the structure at which the displacements, strains and stresses are zero if no mechanical forces are applied. In structures such as buildings and bridges Tr e f is often taken to be the mean temperature during the construction period. Those zero displacements, strains and stresses, together with Tr e f , define the thermomechanical reference state for the structure. The member temperature variation from that reference state is 0002T = T −Tr e f . This may be positive or negative. If the member is disassembled or disconnected, under this variation the member length 1
Called initial stress or initial strain effects by many authors. The different names reflect what is viewed as the physical source of initial force effects at the continuum mechanics level.
2
These are important in composite materials and geomechanics.
29–3
29–4
Chapter 29: THERMOMECHANICAL EFFECTS
is free to change from L to L + dT . If the thermoelastic constitutive behavior is linear3 then dT is proportional to L and 0002T : dT = αL 0002T.
(29.1)
Here α is the coefficient of thermal expansion, which has physical units of one over temperature. This coefficient will be assumed to be uniform over the generic member. It may, however, vary from member to member. (a) fyj , uyj f yi , u yi y¯ The thermal strain is defined as eT = dT /L = α 0002T.
f xi , u xi
(29.2)
x¯
(e)
j
i
Now suppose that the member is also subject to mechanical forces, more precisely the applied axial force F shown in Figure 29.1. The member axial stress is σ = F/A. In response to this stress the length changes by d M . The mechanical strain is e M = d M /L. The total strain e = d/L = (d M + dT )/L is the sum of the mechanical and the thermal strains: e = e M + eT =
σ + α0002T E
fx j , ux j
α , 0002T
ks = E A/L
(b)
F
−F
(29.3)
L
d = d M + dT
L + d M + dT
This superposition of deformations is the basic assumption made in the thermomechanical analysis. It is physically obvious for an unconstrained member such as that depicted in Figure 29.1.
Figure 29.1. Generic truss member subjected to mechanical and thermal effects: (a) idealization as bar, (b) idealization as equivalent linear spring.
At the other extreme, suppose that the member is completely blocked against axial elongation; that is, d = 0 but 0002T 0006= 0. Then e = 0 and e M = −eT . If α > 0 and 0002T > 0 the blocked member goes in compression because σ = Ee M = −EeT = −Eα 0002T < 0. This thermal stress is further discussed in Remark 29.2.
− F = f¯xi
F = f¯xj
p = p M + pT
j
i
Figure 29.2. Equilibrium of truss member under thermomechanical forces.
3
An assumption justified if the temperature changes are small enough so that α is approximately constant through the range of interest, and no material phase change effects occur.
29–4
29–5
§29.1
THERMOMECHANICAL BEHAVIOR
§29.1.1. Thermomechanical Stiffness Relations Because e = d/L and d = u¯ x j − u¯ xi , (29.3) can be developed as u¯ x j − u¯ xi σ = + α 0002T, L E To pass to internal forces (29.4) is multiplied through by E A:
(29.4)
EA (u¯ x j − u¯ xi ) = Aσ + E A α 0002T = p M + pT = p = F. (29.5) L Here p M = Aσ denotes the mechanical axial force, and pT = E A α 0002T , which has the dimension of a force, is called (not surprisingly) the internal thermal force. The sum p = p M + pT is called the effective internal force. The last relation in (29.5), F = p = p M + pT follows from free-body member equilibrium; see Figure 29.2. Passing to matrix form:   u¯ xi EA  u¯  (29.6) F= [ −1 0 1 0 ]  yi  . u¯ yi L u¯ y j Noting that F = f¯x j = − f¯xi while f¯yi = f¯y j = 0, we can relate joint forces to joint displacements as  ¯     f xi −F  f¯yi   0    f¯  =  F  =  xj
f¯y j
0




f¯M xi −1 1 f¯M yi   0  EA  0 + E A α 0002T  =  −1 f¯M x j  1  L f¯M y j 0 0
0 0 0 0


−1 0 u¯ xi 0 0   u¯ yi  . 1 0   u¯ x j  u¯ y j 0 0
(29.7)
¯ u, ¯ or In compact matrix form this is ¯f = ¯f M + ¯fT = K ¯ u¯ = ¯f M + ¯fT . K
(29.8)
¯ is the same member stiffness matrix derived in §2.6.3. The new ingredient that appears is Here K the vector   −1 0  ¯fT = E A α 0002T  (29.9)  , 1 0 This is called the vector of thermal joint forces in local coordinates. It is an instance of an initial force vector at the element level. Remark 29.1. A useful physical interpretation of (29.8) is as follows. Suppose that the member is precluded
from joint (node) motions so that u¯ = 0. Then ¯f M + ¯fT = 0 or ¯f M = −¯fT . It follows that fT contains the negated joint forces (internal forces) that develop in a heated or cooled bar if joint motions are precluded. Because for most materials α > 0, rising the temperature of a blocked bar:0002T > 0, produces an internal compressive thermal force pT = AσT = −E AαT , in accordance with the expected physics. The quantity σT = −Eα 0002T is the thermal stress. This stress can cause buckling or cracking in severely heated structural members that are not allowed to expand or contract. This motivates the use of expansion joints in pavements, buildings and rails, and roller supports in long bridges.
29–5
Chapter 29: THERMOMECHANICAL EFFECTS
29–6
§29.1.2. Globalization At this point we restore the member superscript so that the member stiffness equations (29.7) are rewritten as ¯ e u¯ e = ¯feM + ¯feT . (29.10) K Use of the transformation rules developed in §3.1 to change displacements and forces to the global system {x, y} yields Ke ue = feM + feT , (29.11) where Te is the displacement transformation matrix (3.1), and the transformed quantities are T e e ¯ T , feM = Te T ¯fe , feT = Te T ¯feT . (29.12) Ke = T e K These globalized member equations are used to assemble the free-free master stiffness equations by a member merging process. §29.1.3. Merge The merge process is based on the same assembly rules stated in §3.1.3 with only one difference: thermal forces are added to the right hand side. The member by member merge is carried out much as described as in §3.1.4, the main difference being that the thermal force vectors feT are also merged into a master thermal force vector. Force merge can be done by augmentation-and add (for hand work) or via freedom pointers (for computer work). Illustrative examples are provided below. Upon completion of the assembly process we arrive at the free-free master stiffness equations Ku = f M + fT = f.
(29.13)
§29.1.4. Solution The master system (29.13) has formally the same configuration as the master stiffness equations (2.3). The only difference is that the effective joint force vector f contains a superposition of mechanical and thermal forces. Displacement boundary conditions can be applied by reduction or modification of these equations, simply by using effective joint forces in the descriptions of §3.2.1, §3.4.1 and §29.1. Processing the reduced or modified system by a linear equation solver yields the displacement solution u. §29.1.5. Postprocessing The postprocessing steps described in §3.4 require some modifications because the derived quantities of interest to the structural engineer are mechanical reaction forces and internal forces. Effective forces by themselves are of little use in design. Mechanical joint forces including reactions are recovered from (29.14) f M = Ku − fT To recover mechanical internal forces in member e, compute p e by the procedure outlined in §3.4.2, and subtract the thermal component: p eM = p e − E e Ae α e 0002T e .
(29.15)
This equation comes from solving (29.5) for p M . The mechanical axial stress is σ e = p eM /Ae . 29–6
29–7
§29.1 THERMOMECHANICAL BEHAVIOR
§29.1.6. Worked-Out Example 1
E = 1000, A = 12, α = 0.0005 for both (2) members; ∆T (1) = 25,Ο ∆T = −10Ο
The first worked out problem is defined in Figure 29.3. Two truss members are connected in series as shown and fixed at the ends. Properties E = 1000, A = 5 and α = 0.0005 are common to both members. The member lengths are 4 and 6. A mechanical load P = 90 acts on the roller node. The temperature of member (1) increases by 0002T (1) = 25◦ while that of member (2) drops by 0002T (2) = −10◦ . Find the stress in both members.
¯y // y
1
P = 90
3
; ; ;; ;; ; ; ;
2
(2)
(1)
(1)
L
x¯ // x
(2)
=4
L =6
Figure 29.3. Structure for worked-out Example 1.
To reduce clutter note that all y motions are suppressed so only the x freedoms are kept: u x1 = u 1 , u x2 = u 2 and u x3 = u 3 . The corresponding node forces are denoted by f x1 = f 1 , f x2 = f 2 and f x3 = f 3 . The thermal force vectors, stripped to their x¯ ≡ x components, are ¯f(1) T =
f¯(1) T1 f¯T(1) 2
= E (1)A(1)α (1) 0002T (1)
(2) −1 −150 = , f¯T = 1 150
f¯(2) T2 f¯T(2) 3
= E (2)A(2)α (2) 0002T (2)
−1 60 = . 1 −60 (29.16)
The element stiffness equations are:
1 −1 3000 −1 1
u¯ (1) 1 u¯ (1) 2
=
f¯(1) M1 (1) f¯M2
−150 + , 150
1 −1 2000 −1 1
u¯ (2) 2 u¯ (2) 3
=
f¯(2) M2 (2) f¯M3
60 + , (29.17) −60
No globalization is needed because the equations are already in the global system, and thus we can get rid of the localization marker symbols: f¯ → f , u¯ → u. Assembling by any method yields
1000
3 −3 0 −3 5 −2 0 −2 2
u1 u2 u3
=
f M1 f M2 f M3
+
−150 150 + 60 −60
=
f M1 f M2 f M3
+
−150 210 −60
.
(29.18)
The displacement boundary conditions are u 1 = u 3 = 0. The mechanical force boundary condition is f M2 = 90. On removing the first and third equations, the reduced system is 5000 u 2 = f M2 + 210 = 90 + 210 = 300, which yields u 2 = 300/5000 = +0.06. The mechanical internal forces in the members are recovered from E (1) A(1) (u 2 − u 1 ) − E (1)A(1) α (1) 0002T (1) = 3000 × 0.06 − 12000 × 0.0005 × 25 = 60, L (1) E (2) A(2) (u 3 − u 2 ) − E (2)A(2) α (2) 0002T (2) = 2000 × (−0.06) − 12000 × 0.0005 × (−10) = −72, p (2) M = L (2) (29.19) (1) (2) whence the stresses are σ = 60/12 = 5 and σ = −72/12 = −6. Member (1) is in tension and member (2) in compression. p (1) M =
§29.1.7. Worked-Out Example 2 The second example concerns the example truss of Chapters 2-3. The truss is mecanically unloaded, that is, f M x2 = f M x3 = f M y3 = 0. However the temperature of members (1) (2) and (3) changes by 0002T , −0002T and 30002T , respectively, with respect to Tr e f . The thermal expansion coefficient of all three members is assumed to be α. We will perform the analysis keeping α and 0002T as variables.
29–7
29–8
Chapter 29: THERMOMECHANICAL EFFECTS
The thermal forces for each member in global coordinates are obtained by using (29.10) and the third of (29.12):

f(1) T
f(2) T
f(3) T




0 0 0 −1 −1 1 0 0 0  0 = 100α 0002T  , 0 1 0 1 1 0 0 1 0 0     −1 0 0 −1 0 0 0 0 0   −1  = 50α 0002T  , 0 0 −1   1  0 0 1 0 0 1      1 −1 0 0 −1 −1 1 1 0 0 0 1  −1  = E (3) A(3) α (3) 0002T (3) √  = 200α 0002T  . 0 1 −1   1  1  2 0 0 0 1 1 0 1 1 0  = E (1) A(1) α (1) 0002T (1)  0 0  0 (2) (2) (2) (2)  1 = E A α 0002T  0 0
(29.20)
Merging the contribution of these 3 members gives the master thermal force vector
 −100 + 0 − 200 
 0 + 0 − 200  100 + 0 + 0 fT = α 0002T   0 − 50 + 0 
0 + 0 + 200 0 + 50 + 200
 −300 
  −200      = α 0002T  100    −50    
(29.21)
200 250
The master stiffness matrix K does not change. Consequently the master stiffness equations are
 20  10  −10   0 




10 −10 0 −10 −10 u x1 = 0 f M x1 10 0 0 −10 −10   u y1 = 0   f M y1      0 10 0 0 0   u x2  =  f M x2 = 0  + α 0002T      0 0 5 0 −5   u y2 = 0   f M y2  f M x3 = 0 u x3 −10 −10 0 0 10 10 u y3 f M y3 = 0 −10 −10 0 −5 10 15
 −300   −200   100     −50   
(29.22)
200 250
in which f M x1 , f M y1 and f M y2 are the unknown mechanical reaction forces, and the known forces and displacements have been marked. Since the prescribed displacements are zero, the reduced system is simply
10 0 0 10 0 10
0 10 15
u x2 u x3 u y3
=
0 0 0
+ α 0002T
100 200 250
= α 0002T
100 200 250
.
(29.23)
Solving (29.23) gives u x2 = u x3 = u y3 = 10α 0002T . Completing u with the prescribed zero displacements and premultiplying by K gives the complete effective force vector:
 20  10  −10 f = Ku =   0 


10 −10 0 −10 −10 0 10 0 0 −10 −10   0    0 10 0 0 0   10  α 0002T = α 0002T  0 0 5 0 −5  0    −10 −10 0 0 10 10 10 −10 −10 0 −5 10 15 10
 −300 
 −200   100     −50  .  
(29.24)
200 250
But this vector is exactly fT . Consequently f M = Ku − fT = 0.
29–8
(29.25)
29–9
§29. Notes and Bibliography
All mechanical joint forces, including reactions, vanish, and so do the internal mechanical forces. This is a consequence of the example frame being statically determinate.4 Such structures do not develop thermal stresses under any combination of temperature changes.
§29.2. Initial Force Effects As previously noted, a wide spectrum of mechanical and non-mechanical effects can be acommodated under the umbrella of the initial force concept. The stiffness equations at the local (member) level are ¯ e u¯ e = ¯feM + ¯feI = ¯fe , (29.26) K and at the global (assembled structure) level: Ku = f M + f I = f.
(29.27)
In these equations subscripts M and I identify mechanical and initial node forces, respectively. The sum of the two: ¯f at the local member level and f at the global structure level, are called effective forces. A physical interpretation of (29.26) can be obtained by considering that the structure is blocked against all motions: u = 0. Then f M = −f I , and the undeformed structure experiences mechanical forces. These translate into internal forces and stresses. Engineers also call these prestresses. Local effects that lead to initial forces at the member level are: temperature changes (studied in §4.2, in which f I ≡ fT ), moisture diffusion, residual stresses, lack of fit in fabrication, and in-member prestressing. Global effects include prescribed nonzero joint displacements (studied in §4.1) and multimember prestressing (for example, by cable pretensioning of concrete structures). As can be seen there is a wide variety of physical effects, whether natural or artificial, that lead to nonzero initial forces. The good news is that once the member equations IFEM:DSMAdd:eqn:ExampleTwoMechForces are formulated, the remaining DSM steps (globalization, merge and solution) are identical. This nice property extends to the general Finite Element Method. §29.3. Pseudo Thermal Inputs Some commercial FEM programs do not have a way to handle directly effects such as moisture, lack of fit, or prestress. But all of them can handle temperature variation inputs. Since in linear analysis all such effects can be treated as initial forces, it is possible (at least for bar elements) to model them as fictitious thermomechanical effects, by inputting phony temperature changes. The following example indicate that this is done for a bar element. Suppose that a prestress force FP is present in a bar. The total elongation is d = d M + d P where d P = FP L/(E A) is due to prestress. Equate to a thermal elongation: dT = α0002TP L and solve for 0002TP = FP /(E Aα). This is input to the program as a fictitious temperature change. If in addition there is a real temperature change 0002T one would of course specify 0002T + 0002TP . If this device is used, care should be exercised in interpreting results for internal forces and stresses given by the program. The trick is not necessary for personal or open-source codes over which you have full control. 4
For the definition of static determinacy, see any textbook on Mechanics of Materials; e.g. [12,133].
29–9
Chapter 29: THERMOMECHANICAL EFFECTS
29–10
Notes and Bibliography The additional DSM topics treated in this Chapter are covered in virtually all books on Matrix Structural Analysis, such as the often quoted one by Przemieniecki [136]. Several recent FEM books ignore these topics as too elementary. The physics of thermomechanics and the analysis of thermal stresses is covered adequatedly in textbooks such as Boley and Wiener [20], or manuals such as the widely used Roark’s [140]. ¨ sik [121] provides a comFor the separate problems of heat conduction and heat transfer, the book by Ozi¸ prehensive classic treatment. There is a vast literature on prestressed structures; search under “prestress” in http://www3.addall.com. The concepts of static determinacy and its counterpart: static indeterminacy, are important in skeletal structures such as trusses and frameworks. The pertinent design tradeoff is: insensitivity to initial force effects versus redundant safety. A discussion of this topic is beyond the scope of the book. Once going past skeletal structural systems, however, indeterminacy is the rule. References Referenced items have been moved to Appendix R.
29–10
29–11
Exercises
Homework Exercises for Chapter 29 Thermomechanical Effects EXERCISE 29.1 [N:20] Use the same data of Exercise 3.7, except that P = 0 and consequently there are no applied mechanical forces. Both members have the same dilatation coefficient α = 10−6 1/◦ F. Find the crown displacements u x2 and u y2 and the member stresses σ (1) and σ (2) if the temperature of member (1) rises by 0002T = 120 ◦ F above Tr e f , whereas member (2) stays at Tr e f .
Shortcut: the element stiffnesses and master stiffness matrix are the same as in Exercise 3.7, so if that Exercise has been previously assigned no stiffness recomputations are necessary. EXERCISE 29.2 [A:15] Consider the generic truss member of §2.4, reproduced in Figure E29.1 for conve-
nience.
_ _ fyi , uyi _ _ fxi , uxi
_ _ fyj , uyj _ _ fxj , uxj
_ y
(e)
_ x
j
i
Figure E29.1. Generic truss member.
The disconnected member was supposed to have length L, but because of lack of quality control it was fabricated with length L + δ, where δ is called the “lack of fit.” Determine the initial force vector ¯f I to be used in (29.26). Hint: find the mechanical forces that would compensate for δ and restore the desired length. EXERCISE 29.3 [A:10] Show that the lack of fit of the foregoing exercise can be viewed as equivalent to a prestress force of −(E A/L)δ. EXERCISE 29.4 [A:20] Show that prescribed nonzero displacements can, albeit somewhat artificially, be placed under the umbrella of initial force effects. Work this out for the example of §29.1.1. Hint: split node displacements into u = u H + u N , where u N (the “nonhomogeneous” part) carries the nonzero displacement values. EXERCISE 29.5 [A:40]. (Research paper level). Prove that any statically determinate truss structure is free
of thermal stresses.
29–11
30
.
Dynamics & Vibration Overview
30–1
30–2
Chapter 30: DYNAMICS & VIBRATION OVERVIEW
TABLE OF CONTENTS Page
§30.1. Introduction §30.2. Semidiscrete Equations of Motion §30.2.1. Vibrations as Equilibrium Disturbance §30.2.2. Undamped Free Vibrations . . . . §30.2.3. The Vibration Eigenproblem . . . §30.2.4. Eigensystem Properties . . . . . §30.3. Solving The Vibration Eigenproblem §30.3.1. Determinant Roots . . . . . . §30.3.2. Reduction to the Standard Eigenproblem §30.3.3. Unsymmetric Reduction . . . . . §30.3.4. Symmetry Preserving Reduction . . §30. Notes and Bibliography . . . . . . . . . . . . . §30. References . . . . . . . . . . . . . . §30. Exercises . . . . . . . . . . . . .
30–2
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30–3 30–3 30–4 30–5 30–6 30–6 30–7 30–7 30–7 30–7 30–8 30–8 30–9 30–10
30–3
§30.2
SEMIDISCRETE EQUATIONS OF MOTION
§30.1. Introduction The development in previous chapters pertain to static analysis, in which all quantities are independent of time. This kind of analysis applies also to quasi-static scenarios, where the state varies with time but does so slowly that inertial and damping effects can be ignored. For example one may imagine situations such as a roof progressively burdened by falling snow before collapse, the filling of a dam, or the construction of a tunnel. Or foundation settlements: think of the Pisa tower before leaning stopped. The quasi-static assumption is commonly used in design even for loads that vary in a faster time scale. For example, vehicles travelling over a bridge or wind effects on buildings.1 By contrast dynamic analysis is appropriate when the variation of displacements with time is so rapid that inertial effects cannot be ignored. There are numerous practical examples: earthquakes, rocket launches, vehicle crashes, explosive forming, air blasts, underground explosions, rotating machinery, airplane flutter. The structural accelerations, which are second derivatives with respect to time, must be kept in the governing equations. Damping effects, which are associated with velocities (the first temporal derivatives of displacements), may be also included. However, passive damping effects are often neglected as they tend to take energy out of a system and thus reduce the response amplitude. Dynamic analysis may be performed in the time domain or the frequency domain. The latter is restricted in scope in that it applies to linear structural models, or to linearized fluctuations about an equilibrium state. The frequency domain embodies naturally the analysis of free vibrations, which is the focus of the present Chapter. §30.2. Semidiscrete Equations of Motion The essence of structural analysis is mastering forces. In the development of FEM, this was understood by the pioneers of the first generation, as narrated in §1.7.1. With the victory of the Direct Stiffness Method (DSM) by 1970, displacements came to the foreground as primary computational variables because they scale well into complicated systems. To understand dynamic analysis, that dual role must be kept in mind. Displacements become even more important as computational variables. After all, velocities and accelerations are temporal derivatives of displacements. There is no easy way to do the job with forces only, since dynamics is about motion. On the other hand, the fundamental governing equations of structural dynamics are force balance statements. They are elaborate versions of Newtonian mechanics. This Newtonian viewpoint is illustrated in Table 30.1 for several modeling scenarios that span statics, dynamics and vibrations. For notational simplicity it is assumed that the structure has been discretized in space, for example by the FEM. The right column shows the vector form of the governing equations as force balance statements. The table defines nomenclature. When the model is time dependent, the relations shown in the right column of Table 30.1 are called semidiscrete equations of motion. The qualifier “semidiscrete” says that the time dimension has 1
The quasi-static assumption can be done during design if dynamic effects can be accounted for through appropiate safety factors. For many types of structures (e.g., buildings, bridges, offshore towers) these are specified in building codes. This saves time when dynamic effects are inherently nondeterministic, as in traffic, winds or wave effects.
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Chapter 30: DYNAMICS & VIBRATION OVERVIEW
Table 30.1. Discrete Structural Mechanics Expressed as Force Balance Statements Case Problem type I
General nonlinear dynamics
Governing force balance equations ˙ u, ¨ t) = f(u, u, u, 0002p(u, 00030004 0005 0002 00030004˙ t)0005 exter nal
internal
II
General nonlinear statics
p(u) = f(u) 0002000300040005 0002000300040005 exter nal
internal
III
Flexible structure nonlinear dynamics
˙ u, ¨ t) + pd (u, u, ˙ t) + pe (u, t) = f(u, t) p (u, u,
0002i
00030004
0005
0002 00030004 0005
0002 00030004 0005
damping
elastic
iner tial
IV Flexible structure linear dynamics Linear elastostatics
exter nal
damping
elastic
static equilibrium
damping
elastic
periodic
¨ + C u(t) ˙ + K u(t) = 0 u(t) 0002M00030004 0005 0002 00030004 0005 0002 00030004 0005 iner tial
IX Undamped free vibrations
exter nal
¨ + C u(t) ˙ + K u(t) = f p (t) u(t) 0002M00030004 0005 0002 00030004 0005 0002 00030004 0005 0002000300040005 iner tial
VIII Damped free vibrations
elastic
¨ ˙ = f(u) 0002M(u) 00030004d(t)0005 + 0002C(u) 00030004d(t)0005 + 0002K(u) 00030004d(t)0005 + p(u) 0002 00030004 0005 iner tial
VII Damped forced vibrations
damping
K u = 0002000300040005 f 0002000300040005 elastic
VI Dynamic perturbations
exter nal
¨ + C u(t) ˙ + K u(t) = f(t) u(t) 0002M00030004 0005 0002 00030004 0005 0002 00030004 0005 0002000300040005 iner tial
V
0002 00030004 0005
damping
elastic
¨ + K u(t) = 0 u(t) 0002M00030004 0005 0002 00030004 0005 iner tial
elastic
Symbol u is array of total displacement DOFs; d in case VI is a linearized perturbation of u. Symbol t denotes time. Superposed dots abbreviate time derivatives: u˙ = du/dt, u¨ = d 2 u/dt 2 , etc. The history u = u(t) is called the response of the system. This term is extendible to nonlinear statics. Initial force effects f I may be accommodated in forced cases by taking f = f I when u = 0.
not been discretized: t is still a continuous variable. This legalizes the use of time differentiation, abbreviated by superposed dots, to bring in velocities and accelerations. This table may be scanned “top down” by starting with the most general case I: nonlinear structural dynamics, branching down to more restricted but specific forms. Along the way one finds in case V an old friend: the DSM master equations K u = f for linear elastostatics, treated in previous Chapters. The last case IX: undamped free vibrations, is that treated in this and next two Chapters. Some brief comments are made as regards damped and forced vibrations. §30.2.1. Vibrations as Equilibrium Disturbance An elastic structure is placed in motion through some short-term disturbance, for example an impulse. Remove the disturbance. If wave propagation effects are ignored and the structure remains elastic, it will keep on oscillating in a combination of time-periodic patterns called vibration modes. Associated with each vibration mode is a characteristic time called vibration period. The inverse of 30–4
30–5
§30.2
SEMIDISCRETE EQUATIONS OF MOTION
a period, normalized by appropriate scaling factors, is called a vibration frequency. The structure is said to be vibrating, or more precisely undergoing free vibrations. In the absence of damping mechanisms an elastic structure will vibrate forever. The presence of even minute amounts of viscous damping, however, will cause a gradual decrease in the amplitude of the oscillations. These will eventually cease.2 If the disturbances are sufficiently small to warrant linearization, this scenario fits case VI of Table 30.1, therein labeled “dynamic perturbations.” Its main application is the investigation of dynamic stability of equilibrium configurations. If the perturbation d(t) is unbounded under some initial conditions, that equilibrium configuration3 is said to be dynamically unstable. The analysis of case VI does not belong to an introductory course because it requires advanced mathematical tools. Moreover it often involve nondeterministic (stochastic) effects. Cases VII through IX are more tractable in an introductory course. In these, fluctuations are linearized about an undeformed and unstressed state defined by u = 0. Thus d (the perturbed displacement) becomes simply u (the total displacement). Matrices M, C and K are called the mass, damping and stiffness matrices, respectively. These matrices are independent of u since they are evaluated at the undeformed state u = 0. Two scenarios are of interest in practice: 1.
Forced Vibrations. The system is subjected to a time dependent force f(t). The response u(t) ¨ + C u(t) ˙ + K u(t) = f(t) of case IV. is determined from the linear dynamics equation: M u(t) Of particular interest in resonance studies in when f(t) is periodic in time, which is case VII.
2.
Free Vibrations. The external force is zero for t > 0. The response u(t) is determined from initial conditions. If damping is viscous and light, the undamped model gives conservative answers and is much easier to handle numerically. Consequently the model of case IX is that generally adopted during design studies.
§30.2.2. Undamped Free Vibrations From the foregoing discussion it follows that case IX: undamped free vibrations is of paramount importance in design. The governing equation is ¨ + K u(t) = 0. M u(t)
(30.1)
This expresses a force balance4 in the following sense: in the absence of external loads the internal ¨ The only ingredient beyond the elastic forces K u balance the negative of the inertial forces Mu. by now familiar K is the mass matrix M. The size of these matrices will be denoted by n f , the number of degrees of freedom upon application of support conditions. Equation (30.1) is linear and homogeneous. Its general solution is a linear combination of exponentials. Under matrix definiteness conditions discussed later the exponentials can be expressed as 2
Mathematically a damped oscillation also goes on forever. Eventually, however, the motion amplitude reaches a molecular scale level at which a macroscopic idealization does not apply. At such point the oscillations in the physical structure can be considered to have ceased.
3
Usually obtained through a nonlinear static analysis. This kind of study, called dynamic stability analysis, is covered under Nonlinear Finite Element Methods.
4
Where is f = ma? To pass to internal forces change the sign of f : f int + ma = ku + ma = 0. Replace by matrices and vectors and you have (30.1).
30–5
Chapter 30: DYNAMICS & VIBRATION OVERVIEW
30–6
a combination of trigonometric functions: sines and cosines of argument ωt. A compact √ represenjωt tation of such functions is obtained by using the exponential form e , where j = −1: 0006 vi e jωi t . (30.2) u(t) = i
Here ωi is the i th circular frequency, expressed in radians per second, and vi = 0 the corresponding vibration mode shape, which is independent of t. §30.2.3. The Vibration Eigenproblem Replacing u(t) = v e jωt in (30.1) segregates the time dependence to the exponential: (−ω2 M + K) v e jωt = 0. Since e jωt is not identically zero, it can be dropped leaving the algebraic condition: (−ω2 M + K) v = 0.
(30.3)
Because v cannot be the null vector, this equation is an algebraic eigenvalue problem in ω2 . The eigenvalues λi = ωi2 are the roots of the characteristic polynomial be indexed by i: det(K − ωi2 M) = 0.
(30.4)
Dropping the index i this eigenproblem is usually written as K v = ω2 Mv.
(30.5)
If M and K satisfy some mild conditions, solutions of (30.5) are denoted by ωi and vi . This are called the vibration frequencies or eigenfrequencies, and the it vibration modes or eigenmodes, respectively. The set of all ωi is called the frequency spectrum or simply spectrum. . §30.2.4. Eigensystem Properties Both stiffness K and mass M are symmetric matrices. In addition M is nonnegative. Nothing more can be assumed in general. For example, if K incorporates Lagrangian multipliers from the treatment of a MFC, as explained in Chapter 10, it will be indefinite. If M is positive definite, the following properties hold. 1.
There are n f squared vibration frequencies ωi2 , which are roots of the characteristic polynomial (30.4). These are not necessarily distinct. A root of (30.6) that appears m times is said to have multiplicity m.5
2.
All roots ωi2 of (30.6) are real. The corresponding eigenmodes vi have real entries. 0007 2 If K is nonnegative, ωi ≥ 0 and the frequencies ωi = + ωi2 are also real and nonnegative. √ Furthermore, if K is positive definite, all ωi2 > 0 and consequently + ωi > 0.
3.
If M is nonnegative, care must be exercised; this case is discussed in an Exercise. If M is indefinite (which should never happen in structures) all of the foregoing properties are lost. 5
For example, a free-free (fully unsupported) structure has n R zero frequencies, where n R is the number of rigid body modes.
30–6
30–7
§30.3
SOLVING THE VIBRATION EIGENPROBLEM
Example 30.1. This illustrates the weird things that can happen if M is indefinite. Consider
K=

α 1
1 , 2
M=

0 1
1 , 1+β
(30.6)
where α and β vary from 1 to −1. Then M−1 K =
1 1 2α − 1 α
The eigenvalues are −2 + α + αβ ±
(1 − β) . (−1 + α + αβ)
(30.7)
α[4 − 4β + α(1 + β)2 ] . (30.8) 4α − 2 These are complex if the radicand is negative. But that is not all. If α → 0 one eigenvalue goes to ∞. If α = 0, A = M−1 K is a 2 × 2 Jordan block and one eigenvector is lost. 2 ω1,2
=
§30.3. Solving The Vibration Eigenproblem In what follows we often denote λi = ωi2 to agree more closely with the conventional notation for the algebraic eigenproblem. §30.3.1. Determinant Roots Mathematically the ωi2 are the roots of the characteristic equation (30.4). The simple minded approach is to expand the determinant to get the characteristic polynomial P(ωi2 ) and get their roots: (30.9) det(K − ωi2 M) = P(ωi2 ) = 0. This approach is deprecated by numerical analysts. It seems as welcome as anthrax. Indeed for numerical floating point computations of large systems it risks numerical overflow; moreover the roots of the characteristic polynomial can be very ill-conditioned with respect to coefficients. For small systems and using either exact or symbolic computation there is nothing wrong with this if the roots can be expressed exactly in terms of the coefficients, as in the above example. §30.3.2. Reduction to the Standard Eigenproblem The standard algebraic eigenproblem has the form Ax = λx.
(30.10)
Most library routines included in packages such as Matlab and Mathematica are designed to solve this eigenproblem. If A is symmetric the eigenvalues λi are real; moreover there exist a complete system of eigenvectors xi . If these are normalied to length one: ||xi ||2 = 1 they satisfy the orthonormality conditions 1 if i = j T , xiT Ax j = λi , (30.11) xi x j = δi j = 0 if i = j where δi j is the Kronecker delta. If the xi are collected as columns of a matrix X, the foregoing conditions can be expressed as XT X = I and XT KX = Λ = diagλi . 30–7
30–8
Chapter 30: DYNAMICS & VIBRATION OVERVIEW
§30.3.3. Unsymmetric Reduction If M is nonsingular, a simple way to reduce Kv = ω2 Mv to standard form is to premultiply both sides by M−1 whence M−1 Kv = ω2 v
⇒ Ax = λx,
with
A = M−1 K,
λ = ω2 ,
x = v.
(30.12)
The fastest way to form A is by solving MA = K for A. One nice feature of (30.12) is that the eigenvectors need not be backtransformed, as happens in symmetry-preserving methods. As in the case of the characteristic polynomial, this is deprecated by numerical analysts, also not so vehemently. Their objection is that A is not generally symmetric even if K and M are. So Ax = λx has to be submitted to an unsymmetric eigensolver. Thus risks contaminating the spectrum with complex numbers. Plus, it is slower. The writer’s experience is that (30.12) works perfectly fine for small systems. If tiny imaginary components appear, they are set to zero and life goes on. §30.3.4. Symmetry Preserving Reduction It is possible to retain symmetry by proceeding as follows. Decompose the mass matrix as M = LLT
(30.13)
This is the Cholesky decomposition, which can be carried out to completion if M is positive definite. Then (30.14) A = L−1 KL−T . The demonstraion is in one of the Exercises. The symmetric eigenproblem can be handled by standard library routines, which give back all the eigenvalues and eigenvectors. The square root of the eigenvalues give the vibration frequencies and the vibration modes are recovered from the relation Lvi = xi , which can be handled by standard library routines. Notes and Bibliography The literature on dynamics and vibrations of structures is quite large. It is sufficient to cite here titles that incorporate modern analysis methods: Clough and Penzien [25], Geradin and Rixen [62], Meirovich [95,96] and Wilson [151]. Several books in matrix methods and FEM books contain at least an introductory treatment of dynamics. Citable textbooks include Bathe [9], Cook, Malkus and Plesha [28], Hughes [78]. Despite their age, Przemieniecki [116] remains a useful source of mass matrices, and Pestel and Leckie [106] contains a catalog of transfer matrices (an early 1960 method suitable for small computers). As regards books on linear algebra matrix theory and matrix calculus see the Bibliography cited in Appendix A. The most elegant coverage is that of Strang [130]. Two comprehensive references on matrix computations in general are Golub and VanLoan [65] and Stewart [126]. The former is more up to date as regard recent literature. Bellman [14] contains more advanced material. Stewart and Sun [127] cover the sensitivity analysis of standard and generalized eigenproblems. There are comprehensive books that treat the algebraic eigenproblem. Wilkinson’s masterpiece [152] is dated in several subjects, particularly the generalized eigenproblem and the treatment of large eigenproblems. But it is still unsurpassed as the “bible” of backward error analysis. More up to date in methods is Parlett [105], which is however restricted to the symmetric eigenproblem.
30–8
30–9
§30.
References
As regards source code for matrix computations, the Handbook compilation of Algol 60 procedures by Wilkinson and Reisch [153] is elegant and still useful as template for other languages. Half of the handbook deals with eigenvalue problems. By contrast, the description of Fortran EISPACK code [60] suffers from the inherent ugliness and unreadability of Fortran IV. References Referenced items moved to Appendix R.
30–9
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Chapter 30: DYNAMICS & VIBRATION OVERVIEW
Homework Exercises for Chapter 30 - Dynamics & Vibration Overview EXERCISE 30.1 [A:15]. A 3-element model of a bar in 1D gives

2 1 M= 0 0
1 4 1 0
0 1 4 1

0 0 , 1 2


1 −1 0 0  −1 2 −1 0  K= . 0 −1 2 −1  0 0 −1 1
(E30.1)
Solve the vibration eigenproblem and show the natural frequencies and associated vibration modes. Normalize the latter so that VT MV = I (“mass normalized eigenvectors”). EXERCISE 30.2 [A:25]. In (E30.1) replace the (4,4) mass entry by 2 − α and the (4,4) stiffness entry by
1 − α/2. Using Matlab or Mathematica, solve the eigenproblem for α varing from 0to4 in 0.5 increments. Discuss what happens to the frequencies and vibration modes as α goes to 2 and beyond. Explain.
EXERCISE 30.3 [D:20]. Eigenvectors can be scaled by arbitrary nonzero factors. Discuss 4 ways in which
the eigenvectors vi of Kvi = ωi2 Mvi can be normalized, and what assumptions are necessary in each case.
30–10
31
.
Lumped and Consistent Mass Matrices
31–1
31–2
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
TABLE OF CONTENTS Page
§31.1. Introduction §31.2. Mass Matrix Construction §31.2.1. Direct Mass Lumping . . . §31.2.2. Variational Mass Lumping . . §31.2.3. Template Mass Lumping . . §31.2.4. Mass Matrix Properties . . . §31.2.5. Rank and Numerical Integration §31.3. Globalization §31.4. Mass Matrix Examples: Bars and Beams §31.4.1. The 3-Node Bar . . . . . . §31.4.2. The Bernoulli-Euler Plane Beam §31.4.3. The Plane Beam-Column . . . §31.4.4. *The Timoshenko Plane Beam . §31.4.5. Spar and Shaft Elements . . . §31.5. Mass Matrix Examples: Plane Stress §31.5.1. The Plane Stress Linear Triangle §31.5.2. Four-Node Bilinear Quadrilateral §31.6. Mass Diagonalization Methods §31.6.1. HRZ Lumping . . . . . . §31.6.2. Lobatto Lumping . . . . . §31. Notes and Bibliography . . . . . . . . . . . §31. References . . . . . . . . . . . . §31. Exercises . . . . . . . . . . .
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31–3 31–3 31–3 31–4 31–4 31–5 31–6 31–6 31–8 31–8 31–8 31–10 31–11 31–12 31–12 31–12 31–14 31–15 31–15 31–15 31–16 31–18 31–19
31–3
§31.2
MASS MATRIX CONSTRUCTION
§31.1. Introduction To do dynamic and vibration finite element analysis, you need at least a mass matrix to pair with the stiffness matrix. This Chapter provides a quick introduction to standard methods for computing this matrix. As a general rule, the construction of the master mass matrix M largely parallels of the master stiffness matrix K. Mass matrices for individual elements are formed in local coordinates, transformed to global, and merged into the master mass matrix following exactly the same techniques used for K. In practical terms, the assemblers for K and M can be made identical. This procedural uniformity is one of the great assets of the Direct Stiffness Method. A notable difference with the stiffness matrix is the possibility of using a diagonal mass matrix based on direct lumping. A master diagonal mass matrix can be stored simply as a vector. If all entries are nonnegative, it is easily inverted, since the inverse of a diagonal matrix is also diagonal. Obviously a lumped mass matrix entails significant computational advantages for calculations that involve M−1 . This is balanced by some negative aspects that are examined in some detail later. §31.2. Mass Matrix Construction The master mass matrix is built up from element contributions, and we start at that level. The construction of the mass matrix of individual elements can be carried out through several methods. These can be categorized into three groups: direct mass lumping, variational mass lumping, and template mass lumping. The last group is more general in that includes all others. Variants of the first two techniques are by now standard in the FEM literature, and implemented in all general purpose codes. Consequently this Chapter covers the most widely used methods, focusing on techniques that produce diagonally lumped and consistent mass matrices. The next Chapter covers the template approach to produce customized mass matrices. §31.2.1. Direct Mass Lumping The total mass of element e is directly apportioned to nodal freedoms, ignoring any cross coupling. The goal is to build a diagonally lumped mass matrix or DLMM, denoted here by MeL . As the simplest example, consider a 2-node prismatic bar element with length 0002, cross section area A, and mass density ρ, which can only move in the axial direction x, as depicted in Figure 31.1. The total mass of the element is M e = ρ A0002. This is divided into two equal parts and assigned to each end node to produce 0002 MeL
=
1 ρ A0002 2
1 0
0003 0 = 12 ρ A0002 I2 , 1
(31.1)
Total mass ρA0002
ρA0002
x
ρA0002
massless connector Figure 31.1. Direct mass lumping for 2-node prismatic bar element.
in which I2 denotes the 2 × 2 identity matrix. As shown in the figure, we have replaced the bar with a “dumbbell.” This process conserves the translational kinetic energy or, equivalently, the linear momentum. To show this for the bar example, take the constant x-velocity vector u˙ e = v [ 1 1 ]T . The kinetic 31–3
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Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
energy of the element is T e = 12 (u˙ e )T MeL u˙ e = 12 ρ A0002 v 2 = 12 M e v 2 . Thus the linear momentum p e = ∂ T e /∂v = M e v is preserved. When applied to simple elements that can rotate, however, the direct lumping process may not necessarily preserve angular momentum. A key motivation for direct lumping is that, as noted in §31.1, a diagonal mass matrix may offer computational and storage advantages in certain simulations, notably explicit time integration. Furthermore, direct lumping covers naturally the case where concentrated (point) masses are natural part of model building. For example, in aircraft engineering it is common to idealize nonstructural masses (fuel, cargo, engines, etc.) as concentrated at given locations.1 §31.2.2. Variational Mass Lumping A second class of mass matrix construction methods are based on a variational formulation. This is done by taking the kinetic energy as part of the governing functional. The kinetic energy of an element of mass density ρ that occupies the domain e and moves with velocity field v0006e is 0004 e 1 ρ(0006ve )T v0006e d000be . (31.2) T =2 e
Following the FEM philosophy, the element velocity field is interpolated by shape functions: v0006e = Nev u˙ e , where u˙ e are node DOF velocities and Nev a shape function matrix. Inserting into (31.3) and moving the node velocities out of the integral gives 0004 def e 1 e T T = 2 (u˙ ) ρ(Nev )T Nev d u˙ e = 12 (u˙ e )T Me u˙ e , (31.3) e
whence the element mass matrix follows as the Hessian of T e : ∂2T e M = e e = ∂ u˙ ∂ u˙
0004
e
000be
ρ (Nev )T Nev d000b.
(31.4)
If the same shape functions used in the derivation of the stiffness matrix are chosen, that is, Nev = Ne , (31.4) is called the consistent mass matrix2 or CMM. It is denoted here by MCe . For the 2-node prismatic bar element moving along x, the stiffness shape functions of Chapter 12 are Ni = 1 − (x−xi )/0002 = 1 − ζ and N j = (x−xi )/0002 = ζ . With d x = 0002 dζ , the consistent mass is easily obtained as 0004 MCe
0002
= 0
0004
1
ρ A (N ) N d x = ρ A e T
e
0
0002
1−ζ ζ
0002
0003 [1 − ζ
ζ ] 0002 dζ =
1 ρA0002 6
2 1
0003 1 . 2
(31.5)
It can be verified that this mass matrix preserves linear momentum along x. 1
Such concentrated masses in general have rotational freedoms. Rotational inertia lumping is then part of the process.
2
A better name would be stiffness-consistent. The shorter sobricket has the unfortunately implication that other choices are “inconsistent,” which is far from the truth. In fact, the consistent mass is not necessarily the best one, a topic elaborated in the next Chapter. However the name is by now ingrained in the FEM literature.
31–4
31–5
§31.2
MASS MATRIX CONSTRUCTION
§31.2.3. Template Mass Lumping A generalization of the two foregoing methods consists of expressing the mass as a linear combination of k component mass matrices: Me =
0005k i=1
µi Mie .
(31.6)
Appropriate constraints on the free parameters µi are placed to enforce matrix properties discussed in §31.2.4. Variants result according to how the component matrices Mi are chosen, and how the parameters µi are determined. The best known scheme of this nature results on taking a weighted average of the consistent and diagonally-lumped mass matrices: def
MeLC = (1 − µ)MCe + µMeL ,
(31.7)
in which µ is a free scalar parameter. This is called the LC (“lumped-consistent”) weighted mass matrix. If µ = 0 and µ = 1 this combination reduces to the consistent and lumped mass matrix, respectively. For the 2-node prismatic bar we get 0002 0003 0002 0003 0002 0003 2 1 1 0 2+µ 1−µ e 1 1 1 + µ2ρ A 0002 = 6ρ A 0002 . (31.8) M LC = (1 − µ) 6 ρ A 0002 1 2 0 1 1−µ 2+µ It is known (since the early 1970s) that the best choice with respect to minimizing low frequency dispersion is µ = /. This is proven in the next Chapter. The most general method of this class uses finite element templates to fully parametrize the element mass matrix. For the prismatic 2-node bar element one would start with the 3-parameter template 0002 0003 µ11 µ12 e M = ρA0002 , (31.9) µ12 µ22 which includes the symmetry constraint from the start. Invariance requires µ22 = µ11 , which cuts the free parameters to two. Conservation of linear momentum requires µ11 + µ12 + µ12 + µ22 = 2µ11 + 2µ12 = 1, or µ12 = / − µ11 . Taking µ = 6µ11 − 2 reduces (31.9) to (31.8). Consequently for the 2-node bar LC-weighting and templates are the same thing, because only one free parameter is left upon imposing essential constraints. This is not the case for more complicated elements. §31.2.4. Mass Matrix Properties Mass matrices must satisfy certain conditions that can be used for verification and debugging. They are: (1) matrix symmetry, (2) physical symmetries, (3) conservation and (4) positivity. Matrix Symmetry. This means (Me )T = Me , which is easy to check. For a variationally derived mass matrix this follows directly from the definition (31.4), while for a DLMM is automatic. Physical Symmetries. Element symmetries must be reflected in the mass matrix. For example, the CMM or DLMM of a prismatic bar element must be symmetric about the antidiagonal: M11 = M22 . To see this, flip the end nodes: the element remains the same and so does the mass matrix. 31–5
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
31–6
Conservation. At a minimum, total element mass must be preserved.3 This is easily verified by applying a uniform translational velocity and checking that linear momentum is conserved. Higher order conditions, such as conservation of angular momentu, are optional and not always desirable. Positivity. For any nonzero velocity field defined by the node values u˙ e = 0, (u˙ e )T Me u˙ e ≥ 0. That is, Me must be nonnegative. Unlike the previous three conditions, this constraint is nonlinear in the mass matrix entries. It can be checked in two ways: through the eigenvalues of Me , or the sequence of principal minors. The second technique is more practical if the entries of Me are symbolic. Remark 31.1. A more demanding form of the positivity constraint is to require that Me be positive definite:
(u˙ e )T Me u˙ e > 0 for any u˙ e = 0. This is more physically reassuring because one half of that quadratic form is the kinetic energy associated with the velocity field defined by u˙ e . In a continuum T can vanish only for zero velocities. But allowing T e = 0 for some nonzero u˙ e makes life easier in some situations, particularly for elements with rotational or Lagrange multiplier freedoms. The u˙ e for which T e = 0 form the null space of Me . Because of the conservation requirement, a rigid velocity field (the time derivative u˙ eR of a rigid body mode ueR ) cannot be in the mass matrix null space, since it would imply zero mass. This scenario is dual to that of the element stiffness matrix. For the latter, Ke ueR = 0, since a rigid body motion produces no strain energy. Thus ueR must be in the null space of the stiffness matrix.
§31.2.5. Rank and Numerical Integration Suppose the element has a total of n eF freedoms. A mass matrix Me is called rank sufficient or of e = n eF . Because of the positivity requirement, a rank-sufficient mass matrix full rank if its rank is r M must be positive definite. Such matrices are preferred from a numerical stability standpoint. e e e If Me has rank r M < n eF the mass is called rank deficient by d M = n eF − r M . Equivalently Me is e times singular. For a numerical matrix the rank is easily computed by taking its eigenvalues dM and looking at how many of them are zero. The null space can be extracted by functions such as NullSpace in Mathematica without the need of computing eigenvalues.
The computation of Me by the variational formulation (31.4) is often done using Gauss numerical quadrature. Each Gauss points adds n D to the rank, where n D is the row dimension of the shape function matrix Ne , up to a maximum of n eF . For most elements n D is the same as element spatial dimensionality; that is, n D = 1, 2 and 3 for 1, 2 and 3 dimensions, respectively. This property can be used to pick the minimum Gauss integration rule that makes Me positive definite. §31.3. Globalization Like their stiffness counterparts, mass matrices are often developed in a local or element frame. Should globalization be necessary before merge, a congruential transformation is applied: ¯ e Te Me = (Te )T M
(31.10)
¯ is the element mass referred to the local frame whereas Te is the local-to-global displaceHere M ment transformation matrix. Matrix Te is in principle that used for the stiffness globalization. Some procedural differences, however, must be noted. For stiffness matrices Te is often rectangular if the local stiffness has lower dimensionality. For example, the bar and spar elements formulated of e
3
We are taking about classical mechanics here; in relativistic mechanics mass and energy can be exchanged.
31–6
31–7
§31.3
GLOBALIZATION
Chapter 6 have 2 × 2 local stiffnesses. Globalization to 2D and 3D involves application of 2 × 4 and 2 × 6 transformation matrices, respectively. This works fine because the local element has zero stiffness in some directions. If the associated freedoms are explicitly kept in the local stiffness, as in Chapters 2–3, those rows and columns are zero and have no effect on the global stiffness. In contrast to stiffnesses, translational masses never vanish. One way to understand this is to think of an element moving in a translational rigid body motion u R with acceleration u¨ R . According to Newton’s second law, f R = M e u¨ R , where M e is the translational mass. This cannot be zero. The conclusion is: all translational masses must be retained in the local mass matrix. The 2-node prismatic bar moving in the {x, y} plane furnishes a simple illustration. With the freedoms arranged as ue = [ u x1 u y1 u x1 u y2 ]T , the local mass matrix constructed by consistent and diagonalized lumping are     2 0 1 0 1 0 0 0 0 2 0 1 0 1 0 0 1 ¯ Ce = 1 ρ A 0002  ¯ eL = 1 ρ A0002  M (31.11)  , M   = 2 ρ A0002I4 , 6 2 1 0 2 0 0 0 1 0 0 1 0 2 0 0 0 1 ¯ Ce respectively. Globalize via (31.10) using the transformation matrix (3.2). The result is MCe = M ¯ eL . We say that these mass matrices repeat. Verification for the DLMM is easy because and MeL = M ¯ eL Te = /ρ A0002(Te )T I4 Te = /ρ A0002(Te )T Te = /ρ A0002I4 . For the CMM, Te is orthogonal: (Te )T M however, repetition is not obvious. It is best shown by temporarily rearranging the element DOF so that instead of [ u x1 u y1 u x1 u y2 ] the x and y components are grouped: [ u x1 u x2 u y1 u y2 ]. ¯ Ce and Te accordingly yields Rearranging M 0002 0003 0002 0003 0002 0003 ˜ 0 ρ A0002 e M cI 2 1 sI 2 2 e ¯C= ˜ M , c = cos ϕ, s = sin ϕ. ˜ , T = −sI2 cI2 , with M = 6 0 M 1 2 (31.12) ¯ Carrying out the transformation in block form gives in which φ = x, x. 0002 0003 0002 0003 00030002 00030002 0003 0002 2 ˜ (cs − cs)M ˜ ˜ 0 ˜ 0 cI2 −sI2 M M cI2 sI2 (c + s 2 )M e ¯ Ce . MC = = =M = 2 2 ˜ ˜ ˜ ˜ (cs − cs)M (c + s )M sI2 cI2 −sI2 cI2 0 M 0 M (31.13) e ¯ A matrix that can be put in a block diagonal form built of identical blocks, such as MC in (31.12) ˜ are irrelevant to is called repeating block diagonal or RBD. Note that the contents and order of M the result (31.13). Hence the following generalization follows. If upon rearranging the element ¯ e is two-block RBD, and (ii) Te takes the block form shown above, the local and global DOF: (i) M matrices will coalesce. For (ii) to hold, it is sufficient that all nodal DOF be translational and be referred to the same coordinate system. The same conclusion holds for 3D; this is the subject of Exercise 31.5. This repetition rule can be summarized as follows: A RBD local mass matrix globalizes to the same matrix if all element DOFs are translational and all of them are referred to the same global system.
(31.14)
This property should be taken advantage of to skip superfluous local-to-global transformations. Frequently such operation costs more than forming the local mass matrix. 31–7
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
31–8
What if the (31.14) fails on actual computation? Then something (mass matrix or transformation) is wrong and must be fixed. As example, suppose that one tries to parrot the bar stiffness derivation process by starting with the 1D bar mass (31.1). A rectangular 2 × 4 transformation matrix Te is built by taking rows 1 and 3 of (3.2). Then the globalization (31.10) is carried out. The resulting ¯ x). The MeL is found to violate (31.14) because entries depend on the orientation angle ϕ = (x, e ¯ reason for this mistake is that M L must account for the inertia in the y direction, as in the second of (31.11), to start with. Remark 31.2. The repetition rule (31.14) can be expected to fail in the following scenarios:
1.
The element has non-translational freedoms. (Occasionally the rule may work, but that is unlikely.)
2.
The mass blocks are different in content and/or size. This occurs if different models are used in different directions. Examples are furnished by beam-column, element with curved sides or faces, and shell elements.
3.
Nodes are referred to different coordinate frames in the global system. This can happen if certain nodes are referred to special frames to facilitate the application of boundary conditions.
§31.4. Mass Matrix Examples: Bars and Beams The diagonally lumped and consistent mass matrices for the 2-node bar element were explicitly given in (31.1) and (31.5), and an optimal combination is investigated in the next Chapter. In this section the DLMM and CMM of other simple elements are worked out. More complicated ones are relegated to Exercises. The overbar of Me is omitted for brevity unless a distinction between local and global mass matrices is required. §31.4.1. The 3-Node Bar The element is prismatic with length 0002, area A, and uniform mass density ρ. Midnode 3 is at the center. The DOFs are arranged ue = [ u 1 u 2 u 3 ]T . Using the shape function in the isoparametric coordinate ξ presented in Exercise 16.2 we get the CMM
0004 1 4 −1 2 ρ A0002 MCe = ρ A (Ne )T Ne (/0002) dξ = (31.15) −1 4 2 . 30 −1 2 2 16 To produce a DLMM, the total mass of the element is divided into 3 parts: αρ A0002, αρ A0002, and (1 − 2α)ρ A0002, which are assigned to nodes 1, 2 and 3, respectively. Total mass ρA0002 x For reasons discussed later the best choice is α = /, as ρA0002 ρA0002 ρA0002 depicted in Figure 31.2. Consequently / of the total mass goes to the midpoint, and what is left to the corners, giving 3 2 1
massless connector 1 0 0 e 1 Figure 31.2. Direct mass lumping for M L = 6 ρ A0002 0 1 0 . (31.16) 3-node bar element. 0 0 4 The /:/:/ allocation happens to be Simpson’s rule for integration. This meshes in with the interpretation of diagonal mass lumping as a Lobatto integration rule, a topic discussed in §31.6. Both (31.15) and (31.16) can be used as building blocks for expanding the element to 2D or 3D space. The repetition rule (31.14) holds. 31–8
31–9
§31.4
MASS MATRIX EXAMPLES: BARS AND BEAMS
Beam2BEConsMass[Le_,Ρ_,A_,{numer_,p_}]:= Module[{i,k,Ne,NeT,Ξ,w,fac,Me=Table[0,{4},{4}]}, Ne={{2*(1-Ξ)^2*(2+Ξ), (1-Ξ)^2*(1+Ξ)*Le, 2*(1+Ξ)^2*(2-Ξ),-(1+Ξ)^2*(1-Ξ)*Le}}/8; NeT=Transpose[Ne]; fac=Ρ*A*Le/2; If [p0, Me=fac*Integrate[NeT.Ne,{Ξ,-1,1}]; Return[Me]]; For [k=1, k<=p, k++, {xi,w}= LineGaussRuleInfo[{p,numer},k]; Me+= w*fac*(NeT/.Ξ->xi).(Ne/.Ξ->xi); ]; If[!numer,Me=Simplify[Me]]; Return[Me] ]; ClearAll[Ρ,A,Le,Ξ]; mfac=(Ρ*A*Le)/420; MeC=Simplify[Beam2BEConsMass[Le,Ρ,A,{False,0}]/mfac]; For [p=1,p<=5,p++, Print['p=',p]; Me=Simplify[Beam2BEConsMass[Le,Ρ,A,{False,p}]/mfac]; Print['Me=',mfac,'*',Me//MatrixForm, ' vs exact=',mfac,'*',MeC//MatrixForm]; Print['eigs scaled ME=',Chop[Eigenvalues[N[Me/.Le->1]]]]]; uRot={-Le/2,1,Le/2,1}*Θ; TRot=Simplify[(1/2)*uRot.MeC.uRot*mfac]; TRotex=Simplify[(Le/2)*Integrate[(1/2)*Ρ*A*(Θ*Le*Ξ/2)^2,{Ξ,-1,1}]]; Print['TRot=',TRot,' should match ',TRotex]; Figure 31.3. Module to form the CMM of a prismatic 2-node Bernoulli-Euler plane beam element. Integration is done analytically if p=0, and numerically if p > 0 using a p-point Gauss rule.
§31.4.2. The Bernoulli-Euler Plane Beam The stiffness of this element was derived in Chapter 13. The 2-node plane beam element has length 0002, cross section area A and uniform mass density ρ. Only the translational inertia due to 000e0002 ˙ x) ¯ 2 d x¯ the lateral motion of the beam is considered in computing the kinetic energy T = 12 0 ρ v( of the element, whereas the rotational inertia is ignored. With the freedoms arranged as ue = [ v1 θ1 v2 θ2 ]T , use of the cubic shape functions (13.12) gives the CMM   156 220002 54 −130002 0004 1 ρ A0002  220002 400022 130002 −300022  ¯ Ce = ρ A M (/0002)(Ne )T Ne dξ = (31.17)  . 54 130002 156 −220002 420 −1 −130002 −300022 −220002 400022 in which /0002 is the Jacobian J = d x/dξ . This result may be verified using the Mathematica module Beam2BEConsMatrix listed in Figure 31.3. The arguments are self-explanatory except for p. The module computes the integral (31.17) analytically if p=0; else using a p-point 1D Gauss rule (extracted from LineGaussRuleInfo, described in Chapter 17) if 1 ≤ p ≤ 5. Beam2BEConsMatrix is run for p varying from 0 through 5 by the statements following the module. The mass matrices obtained with integration rules of 1, 2 and 3 points are       16 40002 16 −40002 86 130002 22 −50002 444 620002 156 −380002  40002 00022 40002 −00022   130002 200022 50002 −00022   620002 1100022 380002 −900022  , c , c c1   2  3  16 40002 16 −40002 22 50002 86 −130002 156 380002 444 −620002 −40002 −00022 −40002 00022 −50002 −00022 −130002 200022 −380002 −900022 −620002 1100022 (31.18) in which c1 = ρ A0002/64, c2 = ρ A0002/216 and c3 = ρ A0002/1200. The eigenvalue analysis shows that all three are singular, with rank 1, 2 and 3, respectively. The result for 4 and 5 points agrees with 31–9
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
31–10
(31.17), which has full rank. The purpose of this example is to illustrate the rank property quoted in §31.2.5: each Gauss point adds one to the rank (up to 4) since the problem is one-dimensional. The matrix (31.17) conserves linear and angular momentum; the latter property being checked by the last 3 statements of Figure 31.3. So do the reduced-integration mass matrices if p>1. To get a diagonally lumped mass matrix is trickier. ObviTotal mass ρA0002 xously the translational nodal masses must be the same as that of a bar: /ρ A0002. See Figure 31.4. But there is no consensus αρA00023 αρA00023 on rotational masses. To accommodate these variations, it is ρA0002 ρA0002 convenient to leave the latter parametrized as follows 2 1   / 0 0 0 Figure 31.4. Direct mass lumping for 22 e 0 α0002 0 0   ¯ L = ρ A0002  M , α ≥ 0. (31.19) node Bernoulli-Euler plane beam element.  0 0 / 0 0 0 0 α00022 Here α is a nonnegative parameter, typically between 0 and 1/50. The choice of α has been argued in the FEM literature over several decades, but the whole discussion is largely futile. Matching the angular momentum of the beam element gyrating about its midpoint gives α = −1/24. This violates the positivity condition stated in 31.2.4. It follows that the best possible α — as opposed to possible best — is zero. This choice gives, however, a singular mass matrix, which is undesirable in scenarios where a mass-inverse appears. Remark 31.3. This result can be readily understood physically. As shown in §32.3.2, the /ρ A0002 translational end node masses grossly overestimate (by a factor of 3 in fact) the angular momentum of the element. Hence adding any rotational lumped mass only makes things worse.
§31.4.3. The Plane Beam-Column To use the foregoing results for dynamics of a plane frame structure, such as a multistory building, the 2-node bar and plane beam elements must be combined to form a plane beam-column element with six degrees of freedom in the local system. The element is then rotated into its global position. The stiffness computation process was covered in Chapter 20. Figure 31.5, which is largely a reproduction of Figure 20.6, shows this element in its local and global configurations. In this case the global and local mass matrices are not identical because of the presence of rotational DOFs; furthermore the models in the longitudinal x¯ and lateral y¯ directions are different. Consequently the distinction between local and global masses must be carefully kept. ¯ Ce is easily obtained by augmenting (31.5) and (31.17) with The local consistent mass matrix M zeros rows and columns to fill up missing DOFs, and adding. The local-to-global transformation matrix Te is that given in Chapter 20 and reproduced here for convenience. The two matrices are     0 0 0 c ϕ sϕ 0 140 0 0 70 0 0 54 −130002  0 0 0  −sϕ cϕ 0  0 156 220002 0     2 2 ρ A0002  0 220002 40002 e 0 130002 −30002  0 1 0 0 0  0 e ¯ MC =   , T =  0 0 140 0 0  0 0 cϕ sϕ 0   0 420  70     0 54 130002 0 156 −220002 0 0 0 −sϕ cϕ 0 0 0 0 0 0 1 0 −130002 −300022 0 −220002 400022 (31.20) 31–10
31–11
§31.4 MASS MATRIX EXAMPLES: BARS AND BEAMS y−
(a) u−x1
θz2
u−x2 θz1
−
uy1
x−
(b)
2
0002
y−
u−y2 x−
2
1
y
ϕ 1
constant ρ, A, I
0002
x
Figure 31.5. The 2-node plane beam-column element: (a) referred to its local system {x, ¯ y¯ }; (b) referred to the global system {x, y}.
where cϕ = cos ϕ, sϕ = sin ϕ, and ϕ = (x, x), ¯ positive counterclockwise, see Figure 31.5(b). The globalized CMM is   148−8cϕϕ −8sϕϕ −220002sϕ 62+8cϕϕ 8sϕϕ 130002sϕ 8sϕϕ 62−8cϕϕ −130002cϕ   −8sϕϕ 148+8cϕϕ 220002cϕ   2 ρ A0002  −220002sϕ 220002cϕ 40002 −130002sϕ 130002cϕ −300022  e e T ¯ e e MC = (T ) MC T =   8sϕϕ −130002s 148−8cϕϕ −8sϕϕ 220002sϕ  420  62+8cϕϕ   8sϕϕ 62−8cϕϕ 130002cϕ −8sϕϕ 148+8cϕϕ −220002cϕ 130002sϕ −130002cϕ −300022 220002sϕ −220002cϕ 400022 (31.21) in which cϕϕ = cos 2ϕ and sϕϕ = sin 2ϕ. The global and local matrices differ for arbitrary ϕ. It ¯ Ce and MCe have the same eigenvalues, since Te is orthogonal. may be verified, however, that M §31.4.4. *The Timoshenko Plane Beam The Timoshenko plane beam model for static analysis was presented as advanced material in Chapter 13. This model is more important for dynamics and vibration than Bernoulli-Euler, and indispensable for short transient and wave propagation analysis. (As remarked in Notes and Bibliography of Chapter 13, the Bernoulli-Euler beam model has infinite phase velocity, since the equation of motion is parabolic, and thus useless for simulating wave propagation.) The Timoshenko beam incorporates two refinements over the Bernoulli-Euler model: 1.
2.
θi
In dynamics: the rotary inertia is included in the kinetic energy.
−γj
Deformed cross section
−γi
For both statics and dynamics: plane sections remain plane but not necessarily normal to the deflected midsurface. See Figure 31.6. This assumption allows the averaged shear distortion to be included in both strain and kinetic energies.
θj
−γ
y, v
v'i
v'j
vj v(x)
vi
j
i
x, u
0002 Figure 31.6. Kinematic assumptions of the Timoshenko plane beam element. (A reproduction of Figure 13.14 for the reader convenience.)
According to the second assumption, the kinetic energy of the Timoshenko beam element is given by
0004 T =
0002
1 2
000f
0010
ρ A v(x) ˙ 2 + ρ I R θ˙ (x)2 d x.
0
31–11
(31.22)
31–12
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
Here I R is the second moment of inertia to be used in the computation of the rotary inertia and θ = v 0010 + γ is the cross section rotation angle shown in Figure 31.6; γ being the section-averaged shear distortion. The element DOF are ordered ue = [ v1 θ1 v1 θ2 ]T . The lateral displacement interpolation is e e v(ξ ) = v1 Nv1 (ξ ) + v10010 Nve0010 1 (ξ ) + v2 Nv2 (ξ ) + v20010 Nve0010 2 (ξ ),
ξ=
2x − 1, 0002
(31.23)
in which the cubic interpolation functions (13.12) are used. A complication over Bernoulli-Euler is that the rotational freedoms are θ1 and θ2 but the interpolation (31.23) is in terms of v10010 = θ1 − γ and v20010 = θ2 − γ . From the analysis of §13.7 we can derive the relation
0011
v10010 v20010
0012
=
1 1+0016
0002 0016 − 0002 −0016 0002
1+ 0016 2 −0016 2
0016 0002 0016 0002


0003 v1 −0016  θ1  2 v . 0016 2 1+ 2 θ2
(31.24)
where as in (13.22) the dimensionless parameter 0016 = 12E I /(G As 00022 ) characterizes the ratio of bending and shear rigidities. The end slopes (31.24) are replaced into (31.23), the interpolation for θ obtained, and v and θ inserted into the kinetic energy (31.22). After lengthy algebra the CMM emerges as the sum of two contributions: (31.25) MCe = MCe T + MCe R , where MC T and MC R accounts for the translational and rotary inertia, respectively:
 13 MCe T =
ρA0002   (1 + 0016)2 
35
11 11 1 + 107 0016+ 13 00162 ( 210 + 120 0016+ 24 00162 )0002
9 + 103 0016+ 16 00162 70 1 1 1 13 1 + 60 0016+ 120 00162 )00022 ( 420 + 403 0016+ 24 00162 )0002 105 13 + 107 0016+ 13 00162 35
MCe R =
 ρ IR  (1 + 0016)2 0002 

1 1 1 −( 140 + 60 0016+ 120 00162 )00022 

11 11 ( 210 + 120 0016+ 241 00162 )0002  1 1 + 601 0016+ 120 00162 )00022 105
symmetric

13 1 −( 420 + 403 0016+ 24 00162 )0002
1 ( 10 − 12 0016)0002
− 65
( 152 + 16 0016 + 13 00162 )00022
(− 101 + 12 0016)0002
6 5
6 5
( 101 − 12 0016)0002
−( 301 + 16 0016 − 16 00162 )00022  (− 101 + 12 0016)0002 2 ( 15 + 16 0016 + 13 00162 )00022
symmetric
  
(31.26)
Caveat: the I in 0016 = 12E I /(G As 0002 ) is the second moment of inertia that enters in the elastic flexural elastic rigidity defined in (13.5). If the beam is homogeneous I R = I , but that is not necessarily the case if, as often happens, the beam has nonstructural attachments that contribute rotary inertia. 2
The factor of MCe R can be further transformed to facilitate parametric studies by introducing r R2 = I R /A as cross-section gyration radius and 0017 = r R /0002 as element slenderness ratio. Then the factor ρ I R /((1 + 0016)2 0002) becomes ρ A 0002 0017 2 /(1 + 0016)2 . If 0016 = 0 and 0017 = 0, MCe R vanishes and MCe T in (31.26) reduces to (31.17). Obtaining a diagonally lumped matrix can be done by the HRZ scheme explained in 31.6.1. The optimal lumped mass is derived in the next Chapter by the template method. §31.4.5. Spar and Shaft Elements The mass matrices for these 2-node elements are very similar to those of the bar, since they can be derived from linear displacement interpolation for the CMM. The only thing that changes is the matrix factor and the end DOFs. The derivation of these elements is done as Exercises.
31–12
31–13
§31.5
MASS MATRIX EXAMPLES: PLANE STRESS
Trig3IsoPMembraneConsMass[ncoor_,ρ_,h_,{numer_,p_}]:= Module[{i,k,x1,y1,x2,y2,x3,y3,A,Nfxy, tcoor,w,Me=Table[0,{6},{6}]}, For [k=1, k<=Abs[p], k++, {{x1,y1},{x2,y2},{x3,y3}}=ncoor; A=Simplify[(x2-x1)*(y3-y1)-(x1-x3)*(y1-y2)]/2; {tcoor,w}= TrigGaussRuleInfo[{p,numer},k]; Nfxy={Flatten[Table[{tcoor[[i]],0 },{i,3}]], Flatten[Table[{ 0,tcoor[[i]]},{i,3}]]}; Me+= ρ*w*A*h*Transpose[Nfxy].Nfxy; ]; If[!numer,Me=Simplify[Me]]; Return[Me] ]; Figure 31.7. CMM module for 3-node linear triangle in plane stress.
§31.5. Mass Matrix Examples: Plane Stress To illustrate the two-dimensional case, this section works out the mass matrices of two simple plane stress elements. More complicated cases are relegated to the Exercises. §31.5.1. The Plane Stress Linear Triangle The stiffness formulation of the 3-node triangle was discussed in Chapter 15. For the following derivations the plate is assumed to have constant mass density ρ, area A, uniform thickness h, and motion restricted to the {x, y} plane. The six DOFs are arranged as ue = [ u x1 u y1 u x2 u y2 u x3 u y3 ]T . The consistent mass matrix is obtained using the linear displacement interpolation (15.17). Expanding (Ne )T Ne gives a 6 × 6 matrix quadratic triangular 000e in the 2 ζ d = A/3, coordinates. This can be integrated with the formulas (15.27) exemplified by e 1 000e e ζ1 ζ2 d = A/6, etc. The result is 
 ζ1 ζ1 0 ζ1 ζ2 0 ζ1 ζ3 0 0 ζ1 ζ1 0 ζ1 ζ2 0 ζ1 ζ3  0004    ρ Ah  ζ2 ζ1 0 ζ2 ζ2 0 ζ2 ζ3 0  e MC = ρh   d = ζ2 ζ1 0 ζ2 ζ2 0 ζ2 ζ3  12 e  0   ζ3 ζ1 0 ζ3 ζ2 0 ζ3 ζ3 0 0 ζ3 ζ1 0 ζ3 ζ2 0 ζ3 ζ3

2 0  1  0  1 0
0 2 0 1 0 1
1 0 2 0 1 0
0 1 0 2 0 1
1 0 1 0 2 0
 0 1  0  (31.27) 1  0 2
This computation may be checked with the Mathematica module listed in Figure 31.7. The module is invoked as Trig3IsoPMembraneConsMass[ncoor,ρ,h,{ numer,p }] and returns matrix Me. The arguments are: ncoord passes the node coordinate list { { x1,y1 },{ x2,y2 },{ x3,y3 } }, ρ the mass density, h the plate thickness, numer is a logical flag set to True or False for numeric or symbolic computations, respectively, and p identifies the triangle integration rule as described in §24.2.1. Subordinate module TrigGaussRuleInfo is described in Chapter 24. Since the order of Me is 6, and each Gauss point adds two (the number of space dimensions) to the rank, a rule with 3 or more points is required to reach full rank, as can be verified by simple numerical experiments. The lumped mass matrix is constructed by taking the total mass of the element, which is ρ Ah, dividing it by 3 and assigning those to the corner nodes. See Figure 31.8. This process produces a 31–13
31–14
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
diagonal matrix:
ρ Ah ρ Ah diag [ 1 1 1 1 1 1 ] = I6 . 3 3 The same matrix is obtained by any other diagonalization method. MeL =
(31.28) ρAh
Total mass ρAh
Remark 31.4. If this element is used in three dimensions (for example as membrane component of a shell element), it is necessary to insert the normal-to-the-plate z mass components in either (31.27) or (31.28). According to the invariance rule (31.14) the globalization process is trivial because MCe or MeL becomes RBD on grouping the element DOFs by component. Thus the local element mass matrix repeats in the global frame.
ρAh
one third goes to each node
y x
massless wireframe
ρAh
Figure 31.8. DLMM for 3-node triangular element.
Quad4IsoPMembraneCMass[ncoor_,rho_,h_,{numer_,p_}]:= Module[{i,k,Nf,dNx,dNy,Jdet,Nfxy,qcoor,w,Me=Table[0,{8},{8}]}, For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}= Quad4IsoPShapeFunDer[ncoor,qcoor]; Nfxy={Flatten[Table[{Nf[[i]], 0},{i,4}]], Flatten[Table[{ 0,Nf[[i]]},{i,4}]]}; Me+=(rho*w*Jdet*h/2)*Transpose[Nfxy].Nfxy; ]; If[!numer,Me=Simplify[Me]]; Return[Me] ]; Figure 31.9. CMM module for 4-node bilinear quad in plane stress.
§31.5.2. Four-Node Bilinear Quadrilateral Module Quad4IsoPMembraneConsMass, listed in Figure 31.9, returns the CMM of a 4-node bilinear quadrilateral under plane stress, moving in the {x, y} plane. The plate is homogeneous with density ρ and constant thickness h. The arguments are similar to those described for the linear triangle. except that the quadrature rule pertains to quadrilateralks, and is specified as described in Chapter 23. The subordinate modules QuadGaussRuleInfo (shape functions) and Quad4IsoPShapeFunDer (Gauss quadrature information) are described in that Chapter. The integration is carried out numerically using a p× p Gauss product rule, with p specified as argument. Testing the module on a rectangular element of dimensions {a, b} returns the following CMMs for the 1×1 and 2×2 Gauss rules: 
e MC1×1
1 0 1  ρabh  0 = 32  1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0


0 4 0 2 0 1 0 4 0 2 2 0 4 0 0   ρabh  1 e  , MC2×2 0 2 0 4 =  0 72  1 0 2 0 0 1 0 2 1  2 0 1 0 0 1 0 2 0 1
31–14
1 0 2 0 4 0 2 0
0 1 0 2 0 4 0 2
2 0 1 0 2 0 4 0

0 2 0  1  . (31.29) 0  2  0 4
31–15
§31.6 MASS DIAGONALIZATION METHODS
The mass given by 1-point integration has rank 2 and 6 zero eigenvalues, and thus rank-deficient by 6. The mass given by the 2×2 rule is rank-sufficient and positive definite. Either matrix repeats e . The DLMM is obtained by assigning one fourth of on globalization. Using p > 2 returns MC2×2 the total element mass ρabh to each freedom. §31.6.
Mass Diagonalization Methods
The construction of consistent mass matrix (CMM) is fully defined by the choice of kinetic energy functional and shape functions. No procedural deviation is possible. On the other hand the construction of a diagonally lumped mass matrix (DLMM) is not a unique process, except for very simple elements in which the lumping is fully defined by conservation and symmetry considerations. A consequence of this ambiguity is that various methods have been proposed in the literature, ranging from heuristic through more scientific. This subsection gives a quick overview of the two more important methods. §31.6.1. HRZ Lumping This scheme is acronymed after the authors of [75]. It produces a DLMM given the CMM. Let M e denote the total element mass. The procedure is as follows. 1.
For each coordinate direction, select the DOFs that contribute to motion in that direction. From this set, separate translational DOF and rotational DOF subsets.
2.
Add up the CMM diagonal entries pertaining to the translational DOF subset only. Call the sum S.
3.
Apportion M e to DLMM entries of both subsets on dividing the CMM diagonal entries by S.
4.
Repeat for all coordinate directions.
Example 31.1. The see HRZ in action, consider the 3-node prismatic bar with CMM given by (31.15). Only
one direction (x) is involved and all DOFs are translational. Excluding the factor ρ A0002/30, which does not affect the results, the diagonal entries are 4, 4 and 16, which add up to S = 24. Apportion the total element mass ρ A0002 to nodes with weights 4/S = 1/6, 4/S = 1/6 and 16/S = 2/3. The result is the DLMM (31.16). Example 31.2. Next consider the 2-node Bernoulli-Euler plane beam element. Again only one direction (y) is involved but now there are translational and rotational freedoms. Excluding the factor ρ A0002/420, the diagonal entries of the CMM (31.17), are 156, 400022 , 156 and 400022 . Add the translational DOF entries: S = 156+156 = 312. Apportion the element mass ρ A0002 to the four DOFs with weights 156/312 = 1/2, 400022 /312 = 00022 /78, 156/312 = 1/2 and 400022 /312 = 00022 /78. The result is the DLMM (31.19) with α = 1/78.
The procedure is heuristic but widely used on account of three advantages: easy to explain and implement, applicable to any element as long as a CMM is available, and retaining nonnegativity. The last attribute is particularly important: it means that the DLMM is physically admissible, precluding numerical instability headaches. As a general assessment, it gives reasonable results if the element has only translational freedoms. If there are rotational freedoms the results can be poor compared to templates. §31.6.2. Lobatto Lumping A DLMM with n eF diagonal entries m i is formally equivalent to a numerical integration formula with n eF points for the element kinetic energy: Te =
0005ne
F
i=1
m i Ti ,
where
31–15
Ti = 12 u˙ i2
(31.30)
31–16
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
Table 31.1. One-Dimensional Lobatto Integration Rules Points
Abscissas ξi ∈ [−1, 1]
2 3 4
−ξ1 = 1 = ξ2 −ξ1 = 1 = ξ3 , ξ2 = 0 √ −ξ1 = 1 = ξ4 , −ξ2 = 1/ 5 = ξ3
Weights wi
Comments
w1 = w2 = 1 w1 = w3 = 13 , w2 = 43 , w1 = w4 = 16 , w2 = w3 =
5 6
Trapezoidal rule Simpson’s rule Interior points at ±0.447214
The 4-point Lobatto rule should not be confused with “Simpson’s Three-Eighths rule,” a Newton-Coates quadrature that has equidistant abscissas: −ξ1 = ξ2 = 1, −ξ2 = ξ3 = 13 , w1 = w2 = 14 , w2 = w3 = 34 . For 4 < n ≤ 10, which is rarely important in FEM work, see Table 25.6 of [1]. If the element is one-dimensional and has only translational DOF, (31.30) can be placed in correspondence with the so-called Lobatto quadrature in numerical analysis.4 . A Lobatto rule is a 1D Gaussian quadrature formula in which the endpoints of the interval ξ ∈ [−1, 1] are sample points. If the formula has p ≥ 2 abscissas, only p − 2 of those are free. Abscissas are symmetric about the origin ξ = 0 and all weights are positive. The general form is
0004
1
f (ξ ) dξ = w1 f (−1) + w p f (1) + −1
0005 p−1 i=2
wi f (ξi ).
(31.31)
The rules for p = 2, 3, 4 are collected in Table 31.1. Comparing (31.30) with (31.31) clearly indicates that if the nodes of a 1D element are placed at the Lobatto abscissas, the diagonal masses m i are simply the weights wi . This fact was first noted by Fried and Malkus [56] and further explored in [91,92]. For the type of elements noted, the equivalence works well for p = 2, 3. For p = 4 a minor difficulty arises: the interior Lobatto points are not at the thirdpoints, as can be seen in Table 31.1. If instead the element has nodes at the thirdpoints ξ = ± 13 , one must switch to the “Simpson three-eights” rule also listed in that Table, and adjust the masses accordingly. As a generalization to multiple dimensions, for conciseness we call “FEM Lobatto quadrature” one in which the connected nodes of an element are sample points of an integration rule. The equivalence with (31.30) still holds. But the method runs into some difficulties: Zero or Negative Masses. If one insists in higher order accuracy, the weights of 2D and 3D Lobatto rules are not necessarily positive. See for example the case of the 6-node triangle in the Exercises. This shortcoming can be alleviated, however, by accepting lower accuracy, or by sticking to product rules in geometries that permit them. See Exercise 31.19. Rotational Freedoms. If the element has rotational DOF, Lobatto rules do not exist. Any attempt to transform rules such as (31.31) to node rotations inevitably leads to coupling. Varying Properties. If the element is nonhomogeneous or has varying properties (for instance, a tapered bar element) the construction of Lobatto rules runs into difficulties, since the problem effectively becomes the construction of a quadrature formula with non-unity kernel. As a general assessment, the Lobatto integration technique seems advantageous only when the mass diagonalization problem happens to fit a rule already available in handbooks such as [132].
4
Also called Radau quadrature by some authors, e.g. [21]. However the handbook [1, p. 888] says that Lobatto and Radau rules are slightly different.
31–16
31–17
§31. Notes and Bibliography
Notes and Bibliography The first appearance of a mass matrix in a journal article occurs in two early-1930s papers5 by Duncan and Collar [34,35]. There it is called “inertia matrix” and denoted by [m]. The original example [34, p. 869] displays the 3 × 3 diagonal mass of a triple pendulum. In the book [55] the notation changes to A. Diagonally lumped mass matrices (DLMM) dominate pre-1963 work. Computational simplicity was not the only reason. Direct lumping gives an obvious way to account for nonstructural masses in simple discrete models of the spring-dashpot-particle variety. For example, in a multistory building “stick model” wherein each floor is treated as one DOF in lateral sway under earthquake or wind action, it is natural to take the entire mass of the floor (including furniture, isolation, etc.) and assign it to that freedom. Nondiagonal masses pop-up ocassionally in aircraft matrix analysis — e.g. wing oscillations in [55, §10.11] — as a result of measurements. As such they necessarily account for nonstructural masses due to fuel, control equipment, etc. The formulation of the consistent mass matrix (CMM) by Archer [6,7] was a major advance. All CMMs displayed in §31.3 were first derived in those papers. The underlying idea is old. In fact it follows directly from the 18th -Century Lagrange dynamic equations [84], a proven technique to produce generalized masses. If T is the kinetic energy of a discrete system and u˙ i (xi ) the velocity field defined by the nodal velocities ˙ the master (system-level) M can be generally defined as the Hessian of T with respect to nodal collected in u, velocities: 0004 ∂2T 1 ˙ ρ u˙ i u˙ i d000b, u i = u i (u), M= . (31.32) T = 2 ∂ u˙ ∂ u˙ ˙ Two key decisions remain before this could be used in FEM. This matrix is constant if T is quadratic in u. Localization: (31.32) is applied element by element, and the master M assembled by the standard DSM steps. Interpolation: the velocity field is defined by the same element shape functions as the displacement field. These had to wait until three things became well established by the early 1960s: (i) the Direct Stiffness Method, (ii) the concept of shape functions, and (iii) the FEM connection to Rayleigh-Ritz. The critical ingredient (iii) was established in Melosh’s thesis [97] under Harold Martin. The link to dynamics was closed with Archer’s contributions, and CMM became a staple of FEM. But only a loose staple. Problems persisted: (a)
Nonstructural masses are not naturally handled by CMM. In systems such as ships or aircraft, the structural mass is only a fraction (10 to 20%) of the total.
(b)
It is inefficient in some solution processes, notably explicit dynamics.6
(c)
It may not give the best results compared to other alternatives.7
(d)
For elements derived outside the assumed-displacement framework, the stiffness shape functions may be unknown or be altogether missing.
Problem (a) can be addressed by constructing “rigid mass elements” accounting for inertia (and possibly gravity or centrifugal forces) but no stiffness. Nodes of these elements must be linked to structural (elastic) 5
The brain behind these developments, which launched Matrix Structural Analysis as narrated in Appendix H, was Arthur Collar. But in the hierarchically rigid British system of the time his name, on account of less seniority, had to be last.
6
In explicit time integration, accelerations are computed at the global level as M−1 fd , where fd is the efective dynamic force. This is efficient if M is diagonal (and nonsingular). If M is the CMM, a system of equations has to be solved; moreover the stable timestep generally decreases.
7
If K results from a conforming displacement interpolation, pairing it with the CMM is a form of Rayleigh-Ritz, and thus guaranteed to provide upper bounds on natural frequencies. This is not necessarily a good thing. In practice it is observed that errors increase rapidly as one moves up the frequency spectrum. If the response is strongly influenced by intermediate and high frequencies, as in wave propagation dynamics, the CMM may give extremely bad results.
31–17
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
31–18
nodes by MFC constraints that enforce kinematic constraints. This is more of an implementation issue than a research topic, although numerical difficulties typical of rigid body dynamics may crop up. Problems (b,c,d) can be attacked by parametrization. The father of NASTRAN, Dick MacNeal, was the first to observe [87,90] that averaging the DLMM and CMM of the 2-node bar element produced better results than using either alone. This idea was further studied by Belytschko and Mullen [163] using Fourier analysis. Krieg and Key [171] had emphasized that in transient analysis the introduction of a time discretization operator brings new compensation phenomena, and consequently the time integrator and the mass matrix should be not be chosen separately. A good discussion of mass diagonalization schemes can be found in the textbook by Cook et al. [28]. The template approach addresses the problems by allowing and encouraging full customization of the mass to the problem at hand. It was first described in [47,48] for a Bernoulli-Euler plane beam using Fourier methods. It is presented in more generality in the following Chapter, where it is applied to other elements. The general concept of template as parametrized form of FEM matrices is discussed in [46]. References Referenced items have been moved to Appendix R.
31–18
31–19
Exercises
Homework Exercises for Chapter 31 Lumped and Consistent Mass Matrices EXERCISE 31.1 [A/C:15]. Derive the consistent mass for a 2-node tapered bar element of length 0002 and constant mass density ρ, moving along its axis x, if the cross section area varies as A = 12 A1 (1−ξ )+ 12 A2 (1+ξ ). Obtain also the DLMM by the HRZ scheme. Show that
MCe =
ρ0002 0011 3A1 + A2 12 A1 + A2
0012
A1 + A2 , A1 + 3A2
MeL =
0011 ρ0002 3A1 + A2 0 8(A1 + A2 )
0012
0 . A1 + 3A2
(E31.1)
EXERCISE 31.2 [A:15]. Dr. I. M. Clueless proposed a new diagonally lumped√mass scheme for the 2-node bar: placing one half of the element mass ρ A0002 at each Gauss point ξ = ±1/ 3 of a two-point rule. His argument is that this configuration conserves total mass and angular momentum, which is true. Explain in two sentences why the idea is worthless. EXERCISE 31.3 [A:25]. Extend the result (31.13) to three dimensions. The element has n e nodes and three
translational DOF per node. Arrange the element DOFs by component: Me is RBD, formed by three n e ×n e ˜ which are left arbitrary. For Te , assume nine blocks ti j I for {i, j} = 1, 2, 3, where ti j are the submatrices M, direction cosines of the local system {x, ¯ y¯ , z¯ with respect to the global system {x, y, z}, and I is the n e ×n e identity matrix. Hint: use the orthogonality properties of the ti j : ti j tik = δ jk . EXERCISE 31.4 [A:20]. Show that the local and global mass matrices of elements with only translational
DOFs repeat under these constraints: all local DOFs are referred to the same local frame, and all global DOF are referred to the same global frame. Hints: the kinetic energy must be the same in both frames, and the local-to-global transformation matrix is orthogonal. EXERCISE 31.5 [A/C:25]. Derive the CMM and DLMM for the 2-node spar and shaft elements derived in Chapter 6. Assume thet the elements are prismatic with constant properties along their length 0002. EXERCISE 31.6 [A/C:20]. Derive the consistent mass for a 2-node Bernoulli-Euler tapered plane beam element of length 0002 if the cross section area varies as A = Ai (1 − ξ )/2 + A j (1 + ξ )/2 while the mass density ρ is constant. Show that

240Ai + 72A j ρ0002  2(15Ai + 7A j )0002 e MC =  54(A + A ) 840 i j −2(7Ai + 6A j )0002

2(15Ai + 7A j )0002 54(Ai + A j ) −2(7Ai + 6A j )0002 (5Ai + 3A j )00022 2(6Ai + 7A j )0002 −3(Ai + A j )00022  . 2(6Ai + 7A j )0002 72Ai + 240A j −2(7Ai + 15A j )0002  −3(Ai + A j )00022 −2(7Ai + 15A j )0002 (3Ai + 5A j )00022
(E31.2)
EXERCISE 31.7 [A:15]. Derive the local CMM and DLMM for the 2-node spar element derived in Chapter 6. Assume that the element is prismatic with constant properties along their length 0002, and that can move in 3D space. EXERCISE 31.8 [A:15]. Derive the local CMM and DLMM for the 2-node spar shaft derived in Chapter 6.
Assume that the element is prismatic with constant properties along their length 0002, and that can move in 3D space. EXERCISE 31.9 [C:25]. Write a Mathematica script to verify the result (31.25)–(31.26). (Dont try this by
hand.) EXERCISE 31.10 [A/C:20]. Derive the consistent mass for a plane stress 3-node linear triangle with constant mass density ρ and linearly varying thickness h defined by the corner values h 1 , h 2 and h 3 .
31–19
31–20
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
EXERCISE 31.11 [A/C:20]. Derive the consistent mass for a plane stress 6-node quadratic triangle with
constant mass density ρ and uniform thickness h, moving in the {x, y} plane. The triangle has straight sides and side nodes placed at the midpoints; consequently the metric (and thus the Jacobian) is constant. Show that with the element DOFs arranged ue = [ u x1 u y1 u x2 . . . u y6 ]T , the CMM is [38, p. 35]   6 0 −1 0 −1 0 0 0 −4 0 0 0 6 0 −1 0 −1 0 0 0 −4 0 0  0  −1 0 6 0 −1 0 0 0 0 0 −4 0    0 −1  0 6 0 −1 0 0 0 0 0 −4    −1 0 −1 0 6 0 −4 0 0 0 0 0   ρh A  0 −1 0 −1 0 6 0 −4 0 0 0 0 e (E31.3) MC =   0 0 0 −4 0 32 0 16 0 16 0 180  0  0 0 0 0 0 −4 0 32 0 16 0 16     −4  0 0 0 0 0 16 0 32 0 16 0    0 −4 0 0 0 0 0 16 0 32 0 16    0 0 −4 0 0 0 16 0 16 0 32 0 0 0 0 −4 0 0 0 16 0 16 0 32 and check that all eigenvalues are positive. Hint: use the shape functions of Chapter 24 and the 6 or 7-point triangle integration rule. EXERCISE 31.12 [A:20=5+15]. Assuming the result (E31.3), use the HRZ and Lobatto integration methods to get two DLMMs for the six-node plane stress triangle. Show that HRZ gives ρh A diag[ 3 3 3 3 3 3 16 16 16 16 16 16 ], (E31.4) MeL = 57 whereas Lobatto integration, using the triangle midpoint rule (24.6), gives ρh A diag[ 0 0 0 0 0 0 1 1 1 1 1 1 ]. (E31.5) MeL = 3 EXERCISE 31.13 [A/C:30]. Derive the consistent and the HRZ-diagonalized lumped mass matrices for a plane stress 10-node cubic triangle with constant mass density ρ and uniform thickness h, moving in the {x, y} plane. The triangle has straight sides, side nodes placed at the thirdpoints and node 0 at the centroid. Consequently the metric (and thus the Jacobian) is constant. Show that with the element DOFs arranged ue = [ u x1 u y1 u x2 . . . u y0 ]T , the CMM is [38, p. 35]
 76 0  11  0   11  0   18  0  0  ρh A  0 e MC = 6720  27   0  27  0   0  0   18  0 36 0
0 76 0 11 0 11 0 18 0 0 0 27 0 27 0 0 0 18 0 36
11 0 76 0 11 0 0 0 18 0 18 0 0 0 27 0 27 0 36 0
0 11 0 76 0 11 0 0 0 18 0 18 0 0 0 27 0 27 0 36
11 0 11 0 76 0 27 0 27 0 0 0 18 0 18 0 0 0 36 0
0 11 0 11 0 76 0 27 0 27 0 0 0 18 0 18 0 0 0 36
18 0 0 0 27 0 540 0 −189 0 −135 0 −54 0 −135 0 270 0 162 0
0 18 0 0 0 27 0 540 0 −189 0 −135 0 −54 0 −135 0 270 0 162
0 0 18 0 27 0 −189 0 540 0 270 0 −135 0 −54 0 −135 0 162 0
0 0 0 18 0 27 0 −189 0 540 0 270 0 −135 0 −54 0 −135 0 162
27 0 18 0 0 0 −135 0 270 0 540 0 −189 0 −135 0 −54 0 162 0
31–20
0 27 0 18 0 0 0 −135 0 270 0 540 0 −189 0 −135 0 −54 0 162
27 0 0 0 18 0 −54 0 −135 0 −189 0 540 0 270 0 −135 0 162 0
0 27 0 0 0 18 0 −54 0 −135 0 −189 0 540 0 270 0 −135 0 162
0 0 27 0 18 0 −135 0 −54 0 −135 0 270 0 540 0 −189 0 162 0
0 0 0 27 0 18 0 −135 0 −54 0 −135 0 270 0 540 0 −189 0 162
18 0 27 0 0 0 270 0 −135 0 −54 0 −135 0 −189 0 540 0 162 0
0 18 0 27 0 0 0 270 0 −135 0 −54 0 −135 0 −189 0 540 0 162
36 0 36 0 36 0 162 0 162 0 162 0 162 0 162 0 162 0 1944 0
0 36 0  36  0 36   0 162   0 162  0  162  0  162  0 162   0 162  0 1944 (E31.6)
31–21
Exercises
Quad8IsoPMembraneConsMass[ncoor_,Ρ_,h_,{numer_,p_}]:= Module[{i,k,Nf,dNx,dNy,Jdet,Nfxy,qcoor,w,Me=Table[0,{16},{16}]}, For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}= Quad8IsoPShapeFunDer[ncoor,qcoor]; Nfxy={Flatten[Table[{Nf[[i]], 0},{i,8}]], Flatten[Table[{ 0,Nf[[i]]},{i,8}]]}; Me+=(Ρ*w*Jdet*h/2)*Transpose[Nfxy].Nfxy; ]; If[!numer,Me=Simplify[Me]]; Return[Me] ]; Figure E31.1. CMM module for 8-node serendipity quadrilateral in plane stress.
Check that all eigenvalues are positive. Show that the HRZ-diagonalized LMM is MeL =
ρh A diag[ 19 19 19 19 19 19 135 135 135 135 135 135 135 135 135 135 135 135 486 486 ]. 1353 (E31.7)
EXERCISE 31.14 [A/C:20]. Derive the consistent mass for a plane stress 4-node bilinear rectangle with
constant mass density ρ, uniform thickness h and side dimensions a and b. EXERCISE 31.15 [A:15]. Explain why, for a plane stress element moving in the {x, y} plane, each Gauss integration point adds 2 to the rank of the CMM. What would happen in three dimensions? EXERCISE 31.16 [A:25]. For any geometry, the CMM of the 4-node bilinear quadrilateral in plane stress
(assumed homogeneous and of constant thicknbess) repeats once the p × p Gauss integration rule verifies p ≥ 2. Explain why. EXERCISE 31.17 [C:15]. Using the script of Figure E31.1 derive the consistent and HRZ-diagonalized lumped mass matrices for a 8-node serendipity quadrilateral specialized to a rectangle of dimensions {a, b} that moves in the {x, y} plane. Assume constant mass density ρ and uniform thickness h. Show that
 6 0 2 0 3 0 2 0 −6 0 −8 0  0 6 0 2 0 3 0 2 0 −6 0 −8  2 0 6 0 2 0 3 0 −6 0 −6 0   0 2 0 6 0 2 0 3 0 −6 0 −6   3 0 2 0 6 0 2 0 −8 0 −6 0   0 3 0 2 0 6 0 2 0 −8 0 −6  2 0 3 0 2 0 6 0 −8 0 −8 0  ρabh  e  0 2 0 3 0 2 0 6 0 −8 0 −8 MC = 360   −6 0 −6 0 −8 0 −8 0 32 0 20 0  0 −6 0 −6 0 −8 0 −8 0 32 0 20   −8 0 −6 0 −6 0 −8 0 20 0 32 0  0 −8 0 −6 0 −6 0 −8 0 20 0 32   −8 0 −8 0 −6 0 −6 0 16 0 20 0   0 −8 0 −8 0 −6 0 −6 0 16 0 20 
−8 0 −8 0 −6 0 −6 0 16 0 20 0 32 0 −6 0 −8 0 −8 0 −6 0 20 0 16 0 20 0 −6 0 −8 0 −8 0 −6 0 20 0 16 0
0 −8 0 −8 0 −6 0 −6 0 16 0 20 0 32 0 20
−6 0 −8 0 −8 0 −6 0 20 0 16 0 20 0 32 0

0 −6  0  −8   0  −8  0  −6   0  20   0 16   0  20   0 32
(E31.8)
16 ].
(E31.9)
Which is the minimum integration rule required to get this matrix? Show that HRZ gives MeL =
ρabh diag [ 3 76
3 3 3 3
3
3
3
16
31–21
16
16
16
16
16
16
31–22
Chapter 31: LUMPED AND CONSISTENT MASS MATRICES
Quad9IsoPMembraneConsMass[ncoor_,Ρ_,h_,{numer_,p_}]:= Module[{i,k,Nf,dNx,dNy,Jdet,Nfxy,qcoor,w,Me=Table[0,{18},{18}]}, For [k=1, k<=p*p, k++, {qcoor,w}= QuadGaussRuleInfo[{p,numer},k]; {Nf,dNx,dNy,Jdet}= Quad9IsoPShapeFunDer[ncoor,qcoor]; Nfxy={Flatten[Table[{Nf[[i]], 0},{i,9}]], Flatten[Table[{ 0,Nf[[i]]},{i,9}]]}; Me+=(Ρ*w*Jdet*h/2)*Transpose[Nfxy].Nfxy; ]; If[!numer,Me=Simplify[Me]]; Return[Me] ]; Figure E31.2. CMM module for 9-node biquadratic quadrilateral in plane stress.
EXERCISE 31.18 [C:15]. Using the script of Figure E31.2 derive the consistent and HRZ-diagonalized lumped mass matrices for a 9-node biquadratic quadrilateral specialized to a rectangle of dimensions {a, b} that moves in the {x, y} plane. Assume constant mass density ρ and uniform thickness h. Show that

MeL =
0 16 0 −4 0 1 0 −4 0 8 0 −2 0 −2 0 8 0 4
−4 0 16 0 −4 0 1 0 8 0 8 0 −2 0 −2 0 4 0

0 8 0 −2 0 −2 0 8 0 4 0 −4 0 8 0 −2 0 −2 0 8 0 4 0 8 0 8 0 −2 0 −2 0 4 0  1 0 8 0 8 0 −2 0 −2 0 4  0 −2 0 8 0 8 0 −2 0 4 0  −4 0 −2 0 8 0 8 0 −2 0 4  0 −2 0 −2 0 8 0 8 0 4 0 16 0 −2 0 −2 0 8 0 8 0 4  0 64 0 4 0 −16 0 4 0 32 0  −2 0 64 0 4 0 −16 0 4 0 32   0 4 0 64 0 4 0 −16 0 32 0   −2 0 4 0 64 0 4 0 −16 0 32  0 −16 0 4 0 64 0 4 0 32 0  8 0 −16 0 4 0 64 0 4 0 32   0 4 0 −16 0 4 0 64 0 32 0  8 0 4 0 −16 0 4 0 64 0 32   0 32 0 32 0 32 0 32 0 256 0 4 0 32 0 32 0 32 0 32 0 256 (E31.10) Which is the minimum integration rule required to get this matrix? Show that HRZ gives 16  0  −4   0   1   0   −4  0  ρabh  e  8 MC = 1800   0  −2   0  −2   0   8   0  4 0
0 −4 0 16 0 −4 0 1 0 8 0 8 0 −2 0 −2 0 4
1 0 −4 0 16 0 −4 0 −2 0 8 0 8 0 −2 0 4 0
0 1 0 −4 0 16 0 −4 0 −2 0 8 0 8 0 −2 0 4
−4 0 1 0 −4 0 16 0 −2 0 −2 0 8 0 8 0 4 0
ρabh diag [ 1 1 1 1 1 36
1
1
1
4
4
4
4
4
4
4
4
16 ].
16
(E31.11)
EXERCISE 31.19 A:25] Two Lobatto formulas that may be used for the 9-node biquadratic quadrilateral
may be found listed in [132, p. 244–245]. Translated to a rectangle geometry of area A and FEM quadrilateralcoordinate notation, they are 1 A
1 A
0004
f (ξ, η) d =
16 36
+
1 36
f (ξ, η) d =
20 48
+
1 48
000be
0004 e
f (0, 0) +
0013
0013
0014
f (0, −1) + f (1, 0) + f (0, 1) + f (−1, 0)
0014
(E31.12)
f (−1, −1) + f (1, −1) + f (1, 1) + f (−1, 1) ,
f (0, 0) +
0013
4 36
6 48
0013
0014
f (0, −1) + f (1, 0) + f (0, 1) + f (−1, 0)
0014
f (−1, −1) + f (1, −1) + f (1, 1) + f (−1, 1) .
31–22
(E31.13)
31–23
Exercises
(E31.12) is a product Simpson formula, whereas (E31.13) was derived by Albrecht and Collatz in 1953. Explain (i) how these formulas can be used to set up a DLMM for a quadrilateral of arbitrary geometry, and (ii) whether (E31.12) coalesces with the HRZ result in the case of a rectangle.
31–23
A
.
Matrix Algebra: Vectors
A–1
A–2
Appendix A: MATRIX ALGEBRA: VECTORS
§A.1 MOTIVATION Matrix notation was invented1 primarily to express linear algebra relations in compact form. Compactness enhances visualization and understanding of essentials. To illustrate this point, consider the following set of m linear relations between one set of n quantities, x1 , x2 , . . . xn , and another set of m quantities, y1 , y2 , . . ., ym : a11 x1 a21 x2 ··· ai1 x1 ··· am1 x1
+ +
a12 x2 a22 x2 ··· + ai2 x2 ··· + am2 x2
+ + + +
··· ··· ··· ··· ··· ···
+ +
a1 j x j a2 j x j ··· + ai j x j ··· + am j x j
+ + + +
··· ··· ··· ··· ··· ···
+ +
a1n xn a2n xn ··· + ain xn ··· + amn xn
= = = =
y1 y2 ··· yi ··· ym
(A.1)
The subscripted a, x and y quantities that appear in this set of relations may be formally arranged as follows:      a11 a12 . . . a1 j . . . a1n x1 y1  a21 a22 . . . a2 j . . . a2n   x2   y2       . . . .   .   . .  .  . .     . . . . . . . .      (A.2)   x  =  a a . . . a . . . a  yi  ij in   j  i2  i1   .    . . . .   .  . . .     . . . . . . . . y xn am1 am2 . . . am j . . . amn m Two kinds of mathematical objects can be distinguished in (A.2). The two-dimensional array expression enclosed in brackets is a matrix, which we call A. Matrices are defined and studied in Appendix B. The one-dimensional array expressions in brackets are column vectors or simply vectors, which we call x and y, respectively. The use of boldface uppercase and lowercase letters to denote matrices and vectors, respectively, follows conventional matrix-notation rules in engineering applications. Replacing the expressions in (A.2) by these symbols we obtain the matrix form of (A.1): Ax = y
(A.3)
Putting A next to x means “matrix product of A times x”, which is a generalization of the ordinary scalar multiplication, and follows the rules explained later. Clearly (A.3) is a more compact, “short hand” form of (A.1). Another key practical advantage of the matrix notation is that it translates directly to the computer implementation of linear algebra processes in programming languages that offer array data structures. 1
By Arthur Cayley at Cambridge (UK) in 1858.
A–2
A–3
§A.2 VECTORS
§A.2 VECTORS We begin by defining a vector, a set of n numbers which we shall write in the form x  1
 x2   x=  .  . xn
(A.4)
A vector of this type is called a column vector. We shall see later that although a vector may be viewed as a special case of a matrix it deserves treatment on its own. The symbol x is the name of the vector. If n numbers are arranged in a horizontal array, as in z = [ z1
..
z2
(A.5)
zn ]
then z is called a row vector. If we use the term “vector” without a qualifier, it is understood to be a column vector such as (A.4). §A.2.1 Notational Conventions Typeset vectors will be designated by bold lowercase letters. For example: a,
b,
c,
x,
y,
z
On the other hand, handwritten or typewritten vectors, which are those written on paper or on the blackboard, are identified by putting a wiggle or bar underneath the letter. For example: a ˜ The subscripted quantities such as x1 in (A.4) are called the entries or components2 of x, while n is called the order of the vector x. Vectors of order one (n = 1) are called scalars. These are the usual quantities of analysis. For compactness one sometimes abbreviates the phrase “vector of order n” to just “n-vector.” For example, z in the example below is a 4-vector. If the components are real numbers, the vector is called a real vector. If the components are complex numbers we have a complex vector. Linear algebra embraces complex vectors as easily as it does real ones; however, we rarely need complex vectors for this exposition. Consequently real vectors will be assumed unless otherwise noted.
2
The term component is primarily used in mathematical treatments whereas entry is used more in conjunction with the computer implementation. The term element is also used in the literature but this will be avoided here as it may lead to confusion with finite elements.
A–3
A–4
Appendix A: MATRIX ALGEBRA: VECTORS

x2
x=
x1 x2
x3
x1
 x1 x =  x2  x3
x1
x2
Figure A.1. Standard visualization of 2-vectors and 3-vectors as position vectors in 2-space and 3-space, respectively. EXAMPLE A.1


4  −2  x= 3  0 is real column vector of order 4, or, briefly, a 4-vector.
EXAMPLE A.2
q = [1
1
1
1
1
1]
is a real row vector of order 6.
Occassionally we shall use the short-hand “component notation” x = [xi ]
(A.6)
for a generic vector. This comes handy when it is desirable to show the scheme of notation for the components. §A.2.2 Visualization To aid visualization a two-dimensional vector x (n = 2) can be depicted as a line segment, or arrow, directed from the chosen origin to a point on the Cartesian plane of the paper with coordinates (x1 , x2 ). See Figure A.1. In mechanics this is called a position vector in 2-space. One may resort to a similar geometrical interpretation in three-dimensional Cartesian space (n = 3); see Figure A.1. Drawing becomes a bit messier, however, although some knowledge of perspective and projective geometry helps. The interpretation extends to all Euclidean spaces of dimensions n > 3 but direct visualization is of course impaired. A–4
A–5
§A.3
VECTOR OPERATIONS
§A.2.3 Special Vectors The null vector, written 0, is the vector all of whose components are zero. The unit vector, denoted by ei , is the vector all of whose components are zero, except the i th component, which is one. After introducing matrices (Appendix B) a unit vector may be defined as the i th column of the identity matrix. The unitary vector, called e, is the vector all of whose components are unity. §A.3 VECTOR OPERATIONS Operations on vectors in two-dimensional and three-dimensional space are extensively studied in courses on Mathematical Physics. Here we summarize operations on n-component vectors that are most useful from the standpoint of the development of finite elements. §A.3.1 Transposition The transpose of a column vector x is the row vector that has the same components, and is denoted by xT : (A.7) xT = [ x 1 x 2 . . . x n ] . Similarly, the transpose of a row vector is the column vector that has the same components. Transposing a vector twice yields the original vector: (xT )T = x. EXAMPLE A.3
The transpose of a=
3 −1 6
is
b = aT = [ 3
−1
6]
and transposing b gives back a.
§A.3.2 Equality Two column vectors x and y of equal order n are said to be equal if and only if their components are equal, xi = yi , for all i = 1, . . . n. We then write x = y. Similarly for row vectors. Two vectors of different order cannot be compared for equality or inequality. A row vector cannot be directly compared to a column vector of the same order (unless their dimension is 1); one of the two has to be transposed before a comparison can be made. §A.3.3 Addition and Subtraction The simplest operation acting on two vectors is addition. The sum of two vectors of same order n, x and y, is written x + y and defined to be the vector of order n x +y  1 1 x + y  2 2 def . x+y =  .   . xn + yn
A–5
(A.8)
A–6
Appendix A: MATRIX ALGEBRA: VECTORS
y=
y1 y2

x+y=
x=
x1 x2
x1 + y1 x2 + y2
Figure A.2. In two and three-dimensional space, the vector addition operation is equivalent to the well known parallelogram composition law.
If x and y are not of the same order, the addition operation is undefined. The operation is commutative: x + y = y + x, and associative: x + (y + z) = (x + y) + z. Strictly speaking, the plus sign connecting x and y is not the same as the sign connecting xi and yi . However, since it enjoys the same analytical properties, there is no harm in using the same symbol in both cases. The geometric interpretation of the vector addition operator for two- and three-dimensional vectors (n = 2, 3) is the well known paralellogram law; see Figure A.2. For n = 1 the usual scalar addition results. For vector subtraction, replace + by − in the previous expressions. EXAMPLE A.4
The sum of a=
3 −1 6
and
b=
2 1 −4
is a + b =
5 0 2
§A.3.4 Multiplication and Division by Scalar, Span Multiplication of a vector x by a scalar c is defined by means of the relation  cx  1
cx2  def   cx =   .  . cxn
(A.9)
This operation is often called scaling of a vector. The geometrical interpretation of scaling is: the scaled vector points the same way, but its magnitude is multiplied by c. If c = 0, the result is the null vector. If c < 0 the direction of the vector is reversed. In paticular, if c = −1 the resulting operation (−1)x = −x is called reflexion about the origin or simply reflexion. A–6
A–7
§A.3
VECTOR OPERATIONS
Division of a vector by a scalar c 0006= 0 is equivalent to multiplication by 1/c. The operation is written x def ≡ x/c = (1/c)x (A.10) c The operation is not defined if c = 0. Sometimes it is not just x which is of interest but the “line” determined by x, namely the collection cx of all scalar multiples of x, including −x. We call this Span(x). Note that the span always includes the null vector. §A.3.5 Inner Product, Norm and Length The inner product of two column vectors x and y, of same order n, is a scalar function denoted by (x, y) — as well as two other forms shown below — and is defined by def
(x, y) = x y = y x = T
T
n
sc
xi yi = xi yi
(A.11)
i=1
This operation is also called dot product and interior product. If the two vectors are not of the same order, the inner product is undefined. The two other notations shown in (A.11), namely xT y and yT x, exhibit the inner product as a special case of the matrix product discussed in Appendix B. The last expression in (A.11) applies the so-called Einstein’s
summation convention, which implies sum on repeated indices (in this case, i ) and allows the operand to be dropped. The inner product is commutative: (x, y) = (y, x) but not generally associative: (x, (y, z)) 0006= ((x, y), z). If n = 1 it reduces to the usual scalar product. The scalar product is of course associative. EXAMPLE A.5
The inner product of
a=
3 −1 6
and
b=
2 1 −4
is (a, b) = aT b = 3 × 2 + (−1) × 1 + 6 × (−4) = −19.
REMARK A.1
The inner product is not the only way of “multiplying” two vectors. There are other vector products, such as the cross or outer product, which are important in many applications such as fluid mechanics and nonlinear dynamics. They are not treated here because they are not defined when going from vectors to matrices. Furthermore they are not easily generalized for vectors of dimension n ≥ 4.
The Euclidean norm or 2-norm of a real vector x is a scalar denoted by x that results by taking the inner product of the vector with itself: def
x = (x, x) =
n i=1
A–7
sc
xi2 = xi xi
(A.12)
A–8
Appendix A: MATRIX ALGEBRA: VECTORS
Because the norm is a sum of squares, it is zero only if x is the null vector. It thus provides a “meter” (mathematically, a norm) on the vector magnitude. The Euclidean length or simply length of a real vector, denoted by |x|, is the positive square root of its Euclidean norm: 000e def (A.13) |x| = + x . This definition agrees with the intuitive concept of vector magnitude in two- and three-dimensional space (n = 2, 3). For n = 1 observe that the length of a scalar x is its absolute value |x|, hence the notation |x|. The Euclidean norm is not the only vector norm used in practice, but it will be sufficient for the present course. Two important inequalities satisfied by vector norms (and not just the Euclidean norm) are the Cauchy-Schwarz inequality: |(x, y)| ≤ |x| |y|, (A.14) and the triangle inequality: |x + y| ≤ |x| + |y|.
(A.15)
EXAMPLE A.6
The Euclidean norm of x= is x = 42 + 32 = 25, and its length is |x| =
000f 0010 4 3
√ 25 = 5.
§A.3.6 Unit Vectors and Normalization A vector of length one is called a unit vector. Any non-null vector x can be scaled to unit length by dividing all components by its original length:  x /|x|  1  x2 /|x|   x/|x| =   .  .
(A.16)
xn /|x|
This particular scaling is called normalization to unit length. If x is the null vector, the normalization operation is undefined. EXAMPLE A.7
The unit-vector normalization of x=
000f 0010 4 3
is
x/|x| = x/5 =
A–8
000f
0.8 0.6
0010
A–9
§A.3
VECTOR OPERATIONS
§A.3.7 Angles, Orthonormality and Projections The angle in radians between two unit real vectors x and y, written 0006 (x, y), is the real number θ satisfying 0 ≤ θ ≤ π (or 0◦ ≤ θ ≤ 180◦ if measured in degrees). The value is defined by the cosine formula: cos θ = (x, y) (A.17) If the vectors are not of unit length, they should be normalized to such, and so the general formula for two arbitrary vectors is (x, y) x y (A.18) cos θ = ( , ) = |x| |y| |x| |y| The angle θ formed by a non-null vector with itself is zero. Note that this will always be the case for n = 1. If one of the vectors is null, the angle is undefined. The definition (A.18) agrees with that of the angle formed by oriented lines in two- and threedimensional Euclidean geometry (n = 2, 3). The generalization to n dimensions is a natural one. For some applications the acute angle φ between the line on x and the line on y (i.e., between the vector spans) is more appropriate than θ . This angle between Span(x) and Span(y) is the real number φ satisfying 0 ≤ φ ≤ π/2 and cos φ =
|(x, y)| |x| |y|
(A.19)
Two vectors x and y connected by the relation (x, y) = 0
(A.20)
are said to be orthogonal. The acute angle formed by two orthogonal vectors is π/2 radians or 90◦ . The projection of a vector y onto a vector x is the vector p that has the direction of x and is orthogonal to y − p. The condition can be stated as (p, y − p) = 0, and it is easily shown that c = (y,
where
p=c
x ) = |y| cos θ |x|
x |x|
(A.21)
(A.22)
where θ is the angle between x and y. If x and y are orthogonal, the projection of any of them onto the other vanishes. §A.3.8 Orthogonal Bases, Subspaces Let bk , k = 1, . . . m be a set of n-dimensional unit vectors which are mutually orthogonal. Then if a particular n-dimensional vector x admits the representation x=
m k=1
A–9
ck bk
(A.23)
A–10
Appendix A: MATRIX ALGEBRA: VECTORS
the coefficients ck are given by the inner products ck = (x, b ) = k
n
sc
xi bik = xi bik .
(A.24)
i=1
We also have Parseval’s equality (x, x) =
m
sc
ck2 = ck ck
(A.25)
k=1
If the representation (A.23) holds, the set bk is called an orthonormal basis for the vector x. The bk are called the base vectors. The set of all vectors x given by (A.24) forms a subspace of dimension m. The subspace is said to be spanned by the basis bk , and is called Span(bk ).* The numbers ck are called the coordinates of x with respect to that basis. If m = n, the set bk forms a complete orthonormal basis for the n-dimensional space. (The qualifier “complete” means that all n-dimensional vectors are representable in terms of such a basis.) The simplest complete orthonormal basis is of order n is the n-dimensional Cartesian basis   1 0  b1 =   ..  , 0
  0 1  b2 =   ..  , 0
···
  0 0  bn =   .. 
(A.26)
1
In this case the coordinates of x are simply its components, that is, ck ≡ xk .
* Note that if m = 1 we have the span of a single vector, which as noted before is simply a line passing through the origin of coordinates.
A–10
A–11
Exercises
Homework Exercises for Appendix A: Vectors EXERCISE A.1
Given the four-dimensional vectors



2  4  x= , 4  −8

1  −5  y= 7  5
(EA.1)
(a)
compute the Euclidean norms and lengths of x and y;
(b)
compute the inner product (x,y);
(c)
verify that the inequalities (A.15) and (A.16) hold;
(d)
normalize x and y to unit length;
(e)
compute the angles θ = 0006 (x, y) and φ given by (A.19) and (A.20);
(f)
compute the projection of y onto x and verify that the orthogonality condition (A.22) holds.
EXERCISE A.2
Given the base vectors



1 −7   b1 = 101  , 1  7

5 5   1 b2 = 10  −5  5
(a)
Check that b1 and b2 are orthonormal, i.e., orthogonal and of unit length.
(b)
Compute the coefficients c1 and c2 in the following representation:


8  −16  = c1 b1 + c2 b2 z= −2  26 (c)
(EA.2)
(EA.3)
Using the values computed in (b), verify that the coordinate-expansion formulas (A.24) and Parseval’s equality (A.25) are correct.
EXERCISE A.3
Prove that Parseval’s equality holds in general. EXERCISE A.4
What are the angles θ and φ formed by an arbitrary non-null vector x and the opposite vector −x? EXERCISE A.5
Show that (αx, βy) = αβ (x, y) where x and y are arbitrary vectors, and α and β are scalars.
A–11
(EA.4)
A–12
Appendix A: MATRIX ALGEBRA: VECTORS
Homework Exercises for Appendix A - Solutions EXERCISE A.1
(a)
x = 22 + 42 + 42 + (−8)2 = 100,
|x| = 10
y = 1 + (−5) + 7 + 5 = 100,
|y| = 10
2
2
2
2
(b) (x, y) = 2 − 20 + 28 − 40 = −30 (c) |x + y| =
|(x, y)| = 30 ≤ |x| |y| = 100 √ 32 + (−1)2 + (−11)2 + (−3)2 = 140 ≤ |x| + |y| = 10 + 10 = 20
000e

(d)


0.2 x 0.4   , = xu = 0.4  10 −0.8

0.1 y −0.5   yu = =  0.7 10 0.5
(e) cos θ = (xu , yu ) = −0.30,
θ = 107.46◦
cos φ = |(xu , yu )| = 0.30,
φ = 72.54◦
(f) c = |y| cos θ = −3     0.2 −0.6  0.4   −1.2  = p = c xu = −3  0.4   −1.2  −0.8 2.4   1.6  −3.8  y−p= 8.2  2.6 (p, y − p) = −0.96 + 4.56 − 9.84 + 6.24 = 0 EXERCISE A.2
(a)
b1 = 0.12 + (−0.7)2 + 0.12 + 0.72 = 1,
|b1 | = 1
b2 = 0.52 + 0.52 + (−0.5)2 + 0.52 = 1,
|b2 | = 1
(b1 , b2 ) = 0.05 − 0.35 − 0.05 + 0.35 = 0 (b) c1 = 30,
c2 = 10
(by inspection, or solving a system of 2 linear equations) (c) c1 = (z, b1 ) = 0.8 + 11.2 − 0.2 + 18.2 = 30 c2 = (z, b2 ) = 4.0 − 8.0 + 1.0 + 13.0 = 10 c12 + c22 = 302 + 102 = 1000 = z = 82 + (−16)2 + (−2)2 + 262 = 1000
A–12
A–13
Solutions to Exercises
EXERCISE A.3
See any linear algebra book, e.g. Strang’s. EXERCISE A.4
θ = 180◦ ,
φ = 0◦
EXERCISE A.5
(αx, βy) = αxi βyi = αβ xi yi = αβ (x, y) (summation convention used)
A–13
B
.
Matrix Algebra: Matrices
B–1
B–2
Appendix B: MATRIX ALGEBRA: MATRICES
§B.1 MATRICES §B.1.1 Concept Let us now introduce the concept of a matrix. Consider a set of scalar quantities arranged in a rectangular array containing m rows and n columns: 
a11  a21   .  .  a  i1  .  . am1
a12 a22 . . ai2 . . am2
 . . . a1n . . . a2n   .  . . .  . . . . ain   .  . . .  . . . amn
. . . a1 j . . . a2 j . . . . . . . ai j . . . . . . . am j
(B.1)
This array will be called a rectangular matrix of order m by n, or, briefly, an m × n matrix. Not every rectangular array is a matrix; to qualify as such it must obey the operational rules discussed below. The quantities ai j are called the entries or components of the matrix. Preference will be given to the latter unless one is talking about the computer implementation. As in the case of vectors, the term “matrix element” will be avoided to lessen the chance of confusion with finite elements. The two subscripts identify the row and column, respectively. Matrices are conventionally identified by bold uppercase letters such as A, B, etc. The entries of matrix A may be denoted as Ai j or ai j , according to the intended use. Occassionally we shall use the short-hand component notation (B.2) A = [ai j ]. EXAMPLE B.1
The following is a 2 × 3 numerical matrix: B=

2 4
6 9
3 1
(B.3)
This matrix has 2 rows and 3 columns. The first row is (2, 6, 3), the second row is (4, 9, 1), the first column is (2, 4), and so on.
In some contexts it is convenient or useful to display the number of rows and columns. If this is so we will write them underneath the matrix symbol. For the example matrix (B.3) we would show B
2×3
(B.4)
REMARK B.1
Matrices should not be confused with determinants. A determinant is a number associated with square matrices (m = n), defined according to the rules stated in Appendix C.
B–2
B–3
§B.1
MATRICES
§B.1.2 Real and Complex Matrices As in the case of vectors, the components of a matrix may be real or complex. If they are real numbers, the matrix is called real, and complex otherwise. For the present exposition all matrices will be real. §B.1.3 Square Matrices The case m = n is important in practical applications. Such matrices are called square matrices of order n. Matrices for which m 0004= n are called non-square (the term “rectangular” is also used in this context, but this is fuzzy because squares are special cases of rectangles). Square matrices enjoy certain properties not shared by non-square matrices, such as the symmetry and antisymmetry conditions defined below. Furthermore many operations, such as taking determinants and computing eigenvalues, are only defined for square matrices. EXAMPLE B.2
C=
12 6 8 24 2 5
3 7 11
(B.5)
is a square matrix of order 3.
Consider a square matrix A = [ai j ] of order n × n. Its n components aii form the main diagonal, which runs from top left to bottom right. The cross diagonal runs from the bottom left to upper right. The main diagonal of the example matrix (B.5) is {12, 24, 11} and the cross diagonal is {2, 24, 3}. Entries that run parallel to and above (below) the main diagonal form superdiagonals (subdiagonals). For example, {6, 7} is the first superdiagonal of the example matrix (B.5). §B.1.4 Symmetry and Antisymmetry Square matrices for which ai j = a ji are called symmetric about the main diagonal or simply symmetric. Square matrices for which ai j = −a ji are called antisymmetric or skew-symmetric. The diagonal entries of an antisymmetric matrix must be zero. EXAMPLE B.3
The following is a symmetric matrix of order 3:
S=
11 6 1
6 1 3 −1 −1 −6
The following is an antisymmetric matrix of order 4:

0  −3 W= 1 5
3 0 −7 2
B–3
−1 7 0 0
.
(B.6)

−5 −2  . 0 0
(B.7)
B–4
Appendix B: MATRIX ALGEBRA: MATRICES
§B.1.5 Are Vectors a Special Case of Matrices? Consider the 3-vector x and a 3 × 1 matrix X with the same components:
x=
x1 x2 x3
,
X=
x11 x21 x31
.
(B.8)
in which x1 = x11 , x2 = x22 and x3 = x33 . Are x and X the same thing? If so we could treat column vectors as one-column matrices and dispense with the distinction. Indeed in many contexts a column vector of order n may be treated as a matrix with a single column, i.e., as a matrix of order n × 1. Similarly, a row vector of order m may be treated as a matrix with a single row, i.e., as a matrix of order 1 × m. There are some operations, however, for which the analogy does not carry over, and one has to consider vectors as different from matrices. The dichotomy is reflected in the notational conventions of lower versus upper case. Another important distinction from a practical standpoint is discussed next. §B.1.6 Where Do Matrices Come From? Although we speak of “matrix algebra” as embodying vectors as special cases of matrices, in practice the quantities of primary interest to the structural engineer are vectors rather than matrices. For example, an engineer may be interested in displacement vectors, force vectors, vibration eigenvectors, buckling eigenvectors. In finite element analysis even stresses and strains are often arranged as vectors although they are really tensors. On the other hand, matrices are rarely the quantities of primary interest: they work silently in the background where they are normally engaged in operating on vectors. §B.1.7 Special Matrices The null matrix, written 0, is the matrix all of whose components are zero. EXAMPLE B.4
The null matrix of order 2 × 3 is

0 0
0 0
0 . 0
(B.9)
The identity matrix, written I, is a square matrix all of which entries are zero except those on the main diagonal, which are ones. EXAMPLE B.5
The identity matrix of order 4 is

1 0 I= 0 0
0 1 0 0
B–4
0 0 1 0

0 0 . 0 1
(B.10)
B–5
§B.2 ELEMENTARY MATRIX OPERATIONS
A diagonal matrix is a square matrix all of which entries are zero except for those on the main diagonal, which may be arbitrary. EXAMPLE B.6
The following matrix of order 4 is diagonal:

14  0 D= 0 0

0 0 0 −6 0 0  . 0 0 0 0 0 3
(B.11)
A short hand notation which lists only the diagonal entries is sometimes used for diagonal matrices to save writing space. This notation is illustrated for the above matrix: D = diag [ 14
−6
0
(B.12)
3 ].
An upper triangular matrix is a square matrix in which all elements underneath the main diagonal vanish. A lower triangular matrix is a square matrix in which all entries above the main diagonal vanish. EXAMPLE B.7
Here are examples of each kind:

6 0 U= 0 0
4 6 0 0
2 4 6 0

1 2 , 4 6

5  10 L= −3 −15
0 4 21 −2
0 0 6 18

0 0 . 0 7
(B.13)
§B.2 ELEMENTARY MATRIX OPERATIONS §B.2.1 Equality Two matrices A and B of same order m × n are said to be equal if and only if all of their components are equal: ai j = bi j , for all i = 1, . . . m, j = 1, . . . n. We then write A = B. If the inequality test fails the matrices are said to be unequal and we write A 0004= B. Two matrices of different order cannot be compared for equality or inequality. There is no simple test for greater-than or less-than. §B.2.2 Transposition The transpose of a matrix A is another matrix denoted by AT that has n rows and m columns AT = [a ji ]. The rows of AT are the columns of A, and the rows of A are the columns of AT . Obviously the transpose of AT is again A, that is, (AT )T = A.
B–5
(B.14)
B–6
Appendix B: MATRIX ALGEBRA: MATRICES
EXAMPLE B.8

5 A= 1
7 0
0 , 4
AT =
5 7 0
1 0 4
.
(B.15)
The transpose of a square matrix is also a square matrix. The transpose of a symmetric matrix A is equal to the original matrix, i.e., A = AT . The negated transpose of an antisymmetric matrix matrix A is equal to the original matrix, i.e. A = −AT . EXAMPLE B.9
A=
4 7 0
7 0 1 2 2 3
= AT ,
W=
0 −7 0
7 0 2
0 −2 0
= −WT
(B.16)
§B.2.3 Addition and Subtraction The simplest operation acting on two matrices is addition. The sum of two matrices of the same order, A and B, is written A + B and defined to be the matrix def
A + B = [ai j + bi j ].
(B.17)
Like vector addition, matrix addition is commutative: A+B = B+A, and associative: A+(B+C) = (A + B) + C. For n = 1 or m = 1 the operation reduces to the addition of two column or row vectors, respectively. For matrix subtraction, replace + by − in the definition (B.17). EXAMPLE B.10
The sum of A=

1 4
−3 0 2 −1
and
B=

6 7
3 −2
−3 5
is A + B =

7 11
0 0
−3 . 4
(B.18)
§B.2.4 Scalar Multiplication Multiplication of a matrix A by a scalar c is defined by means of the relation def
c A = [cai j ]
(B.19)
That is, each entry of the matrix is multiplied by c. This operation is often called scaling of a matrix. If c = 0, the result is the null matrix. Division of a matrix by a nonzero scalar c is equivalent to multiplication by (1/c). EXAMPLE B.11
If
A=

1 −3 4 2
0 , −1
B–6
3A =

3 12
−9 6
0 . −3
(B.20)
B–7
§B.3
MATRIX PRODUCTS
§B.3 MATRIX PRODUCTS §B.3.1 Matrix by Vector Before describing the general matrix product of two matrices, let us treat the particular case in which the second matrix is a column vector. This so-called matrix-vector product merits special attention because it occurs very frequently in the applications. Let A = [ai j ] be an m × n matrix, x = {x j } a column vector of order n, and y = {yi } a column vector of order m. The matrix-vector product is symbolically written y = Ax, (B.21) to mean the linear transformation def
yi =
n
sc
ai j x j = ai j x j ,
i = 1, . . . , m.
(B.22)
j=1
EXAMPLE B.12
The product of a 2 × 3 matrix and a vector of order 3 is a vector of order 2:

1 4
−3 2
0 −1
1 2 3

−5 = 5
(B.23)
This product definition is not arbitrary but emanates from the analytical and geometric properties of entities represented by matrices and vectors. For the product definition to make sense, the column dimension of the matrix A (called the premultiplicand) must equal the dimension of the vector x (called the post-multiplicand). For example, the reverse product xA does not make sense unless m = n = 1. If the row dimension m of A is one, the matrix formally reduces to a row vector (see §A.2), and the matrix-vector product reduces to the inner product defined by Equation (A.11). The result of this operation is a one-dimensional vector or scalar. We thus see that the present definition properly embodies previous cases. The associative and commutative properties of the matrix-vector product fall under the rules of the more general matrix-matrix product discussed next. §B.3.2 Matrix by Matrix We now pass to the most general matrix-by-matrix product, and consider the operations involved in computing the product C of two matrices A and B: C = AB.
(B.24)
Here A = [ai j ] is a matrix of order m × n, B = [b jk ] is a matrix of order n × p, and C = [cik ] is a matrix of order m × p. The entries of the result matrix C are defined by the formula def
cik =
n
sc
ai j b jk = ai j b jk ,
i = 1, . . . , m,
j=1
B–7
k = 1, . . . , p.
(B.25)
B–8
Appendix B: MATRIX ALGEBRA: MATRICES
We see that the (i, k)th entry of C is computed by taking the inner product of the i th row of A with the k th column of B. For this definition to work and the product be possible, the column dimension of A must be the same as the row dimension of B. Matrices that satisfy this rule are said to be product-conforming, or conforming for short. If two matrices do not conform, their product is undefined. The following mnemonic notation often helps in remembering this rule: C = A
m× p
(B.26)
B
m×n n× p
For the matrix-by-vector case treated in the preceding subsection, p = 1. Matrix A is called the pre-multiplicand and is said to premultiply B. Matrix B is called the postmultiplicand and is said to postmultiply A. This careful distinction on which matrix comes first is a consequence of the absence of commutativity: even if BA exists (it only does if m = n), it is not generally the same as AB. For hand computations, the matrix product is most conveniently organized by the so-called Falk’s   scheme: b11 · · · bik · · · b1 p  . .  . .  . . ↓ . .  · · · bnk · · · bnp b   n1 a  · · · a 11 1n . (B.27) .  .  . .  . .    .   .     ai1 → ain     · · · c   . ik  .   .  . . . . am1 · · · amn Each entry in row i of A is multiplied by the corresponding entry in column k of B (note the arrows), and the products are summed and stored in the (i, k)th entry of C. EXAMPLE B.13
To illustrate Falk’s scheme, let us form the product C = AB of the following matrices

3 A= 4
0 −1
2 , 5
B=
2 4 0
1 3 1
0 −1 −7
−5 0 4
(B.28)
The matrices are conforming because the column dimension of A and the row dimension of B are the same (3). We arrange the computations as shown below:
A=

3 4
0 −1
2 5
2 4 0 6 4
1 3 1 5 6
Here 3 × 2 + 0 × 4 + 2 × 0 = 6 and so on. B–8
0 −1 −7 −14 −34
−5 0 4 −7 0
=B (B.29) = C = AB
B–9
§B.3
MATRIX PRODUCTS
§B.3.3 Matrix Powers If A = B, the product AA is called the square of A and is denoted by A2 . Note that for this definition to make sense, A must be a square matrix. Similarly, A3 = AAA = A2 A = AA2 . Other positive-integer powers can be defined in an analogous manner. This definition does not encompass negative powers. For example, A−1 denotes the inverse of matrix A, which is studied in Appendix C. The general power Am , where m can be a real or complex scalar, can be defined with the help of the matrix spectral form and require the notion of eigensystem. A square matrix A that satisfies A = A2 is called idempotent. We shall see later that that equation characterizes the so-called projector matrices. A square matrix A whose p th power is the null matrix is called p-nilpotent. §B.3.4 Properties of Matrix Products Associativity. The associative law is verified: A(BC) = (AB)C.
(B.30)
Hence we may delete the parentheses and simply write ABC. Distributivity. The distributive law also holds: If B and C are matrices of the same order, then A (B + C) = AB + AC,
and
(B + C) A = BA + CA.
(B.31)
Commutativity. The commutativity law of scalar multiplication does not generally hold. If A and B are square matrices of the same order, then the products AB and BA are both possible but in general AB 0004= BA. If AB = BA, the matrices A and B are said to commute. One important case is when A and B are diagonal. In general A and B commute if they share the same eigensystem. EXAMPLE B.14
Matrices A=

a b
b , c
B=

a−β b
b , c−β
(B.32)
commute for any a, b, c, β. More generally, A and B = A − βI commute for any square matrix A.
Transpose of a Product. The transpose of a matrix product is equal to the product of the transposes of the operands taken in reverse order: (AB)T = BT AT .
(B.33)
The general transposition formula for an arbitrary product sequence is (ABC . . . MN)T = NT MT . . . CT BT AT . B–9
(B.34)
B–10
Appendix B: MATRIX ALGEBRA: MATRICES
Congruential Transformation. If B is a symmetric matrix of order m and A is an arbitrary m × n matrix, then (B.35) S = AT BA. is a symmetric matrix of order n. Such an operation is called a congruential transformation. It occurs very frequently in finite element analysis when changing coordinate bases because such a transformation preserves energy. Loss of Symmetry. The product of two symmetric matrices is not generally symmetric. Null Matrices may have Non-null Divisors. The matrix product AB can be zero although A 0004= 0 and B 0004= 0. Similar, it is possible that A 0004= 0, A2 0004= 0, . . . , but A p = 0. §B.4 BILINEAR AND QUADRATIC FORMS Let x and y be two column vectors of order n, and A a real square n × n matrix. Then the following triple product produces a scalar result: (B.36)
s = yT A x
1×n n×n n×1
This is called a bilinear form. Transposing both sides of (B.36) and noting that the transpose of a scalar does not change, we obtain the result s = xT AT y. (B.37) If A is symmetric and vectors x and y coalesce, i.e. AT = A,
x = y,
(B.38)
the bilinear form becomes a quadratic form (B.39)
s = xT Ax. Transposing both sides of a quadratic form reproduces the same equation. EXAMPLE B.15
The kinetic energy of a system consisting of three point masses m 1 , m 2 , m 3 is T = 12 (m 1 v12 + m 2 v22 + m 3 v32 ).
(B.40)
This can be expressed as the quadratic form (B.41)
T = 12 uT Mu
where M=
m1 0 0
0 m2 0
0 0 m3
,
B–10
u=
u1 u2 u3
.
(B.42)
B–11
Exercises
Homework Exercises for Appendix B: Matrices
EXERCISE B.1
Given the three matrices
A=
2 −1 2
4 1 0 2 3 1 5 −1 2


−2 0 , 1 2
2  1 B= 4 −3
,
C=

−3 0
1 2
2 2
(EB.1)
compute the product D = ABC by hand using Falk’s scheme. (Hint: do BC first, then premultiply that by A.) EXERCISE B.2
Given the square matrices A=

1 −4
3 , 2

3 1
0 −2
3 −1 4
−1 2 0
B=
(EB.2)
verify by direct computation that AB 0004= BA. EXERCISE B.3
Given the matrices A=
1 −1 2
0 2 0
,
B=
4 0 0
(EB.3)
(note that B is symmetric) compute S = AT BA, and verify that S is symmetric. EXERCISE B.4
Given the square matrices
A=
3 1 3
−1 0 −2
2 3 −5
,
B=
3 7 −1
−6 −14 2
−3 −7 1
(EB.4)
verify that AB = 0 although A 0004= 0 and B 0004= 0. Is BA also null? EXERCISE B.5
Given the square matrix
A=
0 a 0 0 0 0
show by direct computation that A2 0004= 0 but A3 = 0. EXERCISE B.6
Can a diagonal matrix be antisymmetric?
B–11
b c 0
(EB.5)
Appendix B: MATRIX ALGEBRA: MATRICES
B–12
EXERCISE B.7
(Tougher) Prove (B.33). (Hint: call C = (AB)T , D = BT AT , and use the matrix product definition (B.25) to show that the generic entries of C and D agree.) EXERCISE B.8
If A is an arbitrary m × n matrix, show: (a) both products AT A and AAT are possible, and (b) both products are square and symmetric. (Hint: for (b) make use of the symmetry condition S = ST and of (B.31).) EXERCISE B.9
Show that A2 only exists if and only if A is square. EXERCISE B.10
If A is square and antisymmetric, show that A2 is symmetric. (Hint: start from A = −AT and apply the results of Exercise B.8.)
B–12
B–13
Solutions to Exercises
Homework Exercises for Appendix B - Solutions
EXERCISE B.1


2 −2 0  1 B= 4 1 −3 2
2 4 1 0 A = −1 2 3 1 2 5 −1 2 EXERCISE B.2
AB =

1 −3 2 =C 2 0 2   −2 −6 0 2  1 −3  6 −12 10  = BC 9 −2000b
1 6 −36 18 23 −27 32 = ABC = D −3 3 −4

−6 3 0004= BA = −4 9
6 −10
EXERCISE B.3
S = AT BA =

23 −6
−6 8
9 −1
which is symmetric, like B. EXERCISE B.4
A=
3 −1 1 0 3 −2
3 7 −1
2 0 3 0 −5 0
However,
BA =
EXERCISE B.5
A2 = AA =
0 0 ac 0 0 0 0 0 0
−6 −14 2 0 0 0
−6 3 3 −14 7 7 2 −1 −1
−3 −7 = B 1 0 0 = AB = 0 0
0004= 0
,
A3 = AAA =
EXERCISE B.6
Only if it is the null matrix.
B–13
0 0 0
0 0 0
0 0 0
=0
B–14
Appendix B: MATRIX ALGEBRA: MATRICES
EXERCISE B.7
To avoid “indexing indigestion” let us carefully specify the dimensions of the given matrices and their transposes: AT = [a ji ] A = [ai j ], n×m
m×n
B = [b jk ],
BT = [bk j ]
n× p
p×n
Indices i, j and k run over 1 . . . m, 1 . . . n and 1 . . . p, respectively. Now call C = [cki ] = (AB)T
p×m
D = [dki ] = BT AT
p×m
From the definition of matrix product, cki =
n
ai j b jk
j=1
dki =
n
b jk ai j =
j=1
n
ai j b jk = cki
j=1
hence C = D for any A and B, and the statement is proved. EXERCISE B.8
(a) If A is m × n, AT is n × m. Next we write the two products to be investigated: AT A ,
n×m m×n
A AT
m×n n×m
In both cases the column dimension of the premultiplicand is equal to the row dimension of the postmultiplicand. Therefore both products are possible. (b) To verify symmetry we use three results. First, the symmetry test: transpose equals original; second, transposing twice gives back the original; and, finally, the transposed-product formula proved in Exercise B.7. (AT A)T = AT (AT )T = AT A (AAT )T = (AT )T AT = AAT Or, to do it more slowly, call B = AT , BT = A, C = AB, and let’s go over the first one again: CT = (AB)T = BT AT = AAT = AB = C Since C = CT , C = AAT is symmetric. Same mechanics for the second one. EXERCISE B.9
Let A be m × n. For A2 = AA to exist, the column dimension n of the premultiplicand A must equal the row dimension m of the postmultiplicand A. Hence m = n and A must be square. EXERCISE B.10
Premultiply both sides of A = −AT by A (which is always possible because A is square): A2 = AA = −AAT But from Exercise B.8 we know that AAT is symmetric. Since the negated of a symmetric matrix is symmetric, so is A2 .
B–14
C
.
Matrix Algebra: Determinants, Inverses, Eigenvalues
C–1
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
C–2
This Chapter discusses more specialized properties of matrices, such as determinants, eigenvalues and rank. These apply only to square matrices unless extension to rectangular matrices is explicitly stated. §C.1 DETERMINANTS The determinant of a square matrix A = [ai j ] is a number denoted by |A| or det(A), through which important properties such as singularity can be briefly characterized. This number is defined as the following function of the matrix elements: 0002 |A| = ± a1 j1 a2 j2 . . . an jn , (C.1) where the column indices j1 , j2 , . . . jn are taken from the set 1, 2, . . . , n with no repetitions allowed. The plus (minus) sign is taken if the permutation ( j1 j2 . . . jn ) is even (odd). EXAMPLE C.1
For a 2 × 2 matrix,
0003 0003 0003 a11 a12 0003 0003a 0003 = a11 a22 − a12 a21 . a 21
22
(C.2)
EXAMPLE C.2
For a 3 × 3 matrix,
0003 0003 0003 a11 a12 a13 0003 0003 0003 0003 a21 a22 a23 0003 = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 − a12 a21 a33 − a11 a23 a32 . 0003a a a 0003 31
32
(C.3)
33
REMARK C.1
The concept of determinant is not applicable to rectangular matrices or to vectors. Thus the notation |x| for a vector x can be reserved for its magnitude (as in Appendix A) without risk of confusion. REMARK C.2
Inasmuch as the product (C.1) contains n! terms, the calculation of |A| from the definition is impractical for general matrices whose order exceeds 3 or 4. For example, if n = 10, the product (C.1) contains 10! = 3, 628, 800 terms each involving 9 multiplications, so over 30 million floating-point operations would be required to evaluate |A| according to that definition. A more practical method based on matrix decomposition is described in Remark C.3.
§C.1.1 Some Properties of Determinants Some useful rules associated with the calculus of determinants are listed next. I.
Rows and columns can be interchanged without affecting the value of a determinant. That is |A| = |AT |.
II.
(C.4)
If two rows (or columns) are interchanged the sign of the determinant is changed. For example: 0003 0003 0003 0003 00033 0003 1 −2 0003 4 00030003 0003 0003 0003. (C.5) 0003 1 −2 0003 = − 0003 3 40003 C–2
C–3
§C.1 DETERMINANTS
III.
If a row (or column) is changed by adding to or subtracting from its elements the corresponding elements of any other row (or column) the determinant remains unaltered. For example: 0003 0003 0003 0003 0003 0003 00033 0003 00033 + 1 4 − 20003 00034 0003 4 2 0003 0003=0003 0003=0003 0003 = −10. (C.6) 0003 1 −2 0003 0003 1 −2 0003 0003 1 −2 0003
IV.
If the elements in any row (or column) have a common factor α then the determinant equals the determinant of the corresponding matrix in which α = 1, multiplied by α. For example: 0003 0003 0003 0003 00036 00033 4 00030003 8 00030003 0003 0003 (C.7) 0003 1 −2 0003 = 2 0003 1 −2 0003 = 2 × (−10) = −20.
V.
When at least one row (or column) of a matrix is a linear combination of the other rows (or columns) the determinant is zero. Conversely, if the determinant is zero, then at least one row and one column are linearly dependent on the other rows and columns, respectively. For example, consider 0003 0003 2 10003 00033 0003 0003 (C.8) 2 −1 0003 . 00031 0003 0003 2 −1 3 This determinant is zero because the first column is a linear combination of the second and third columns: column 1 = column 2 + column 3 (C.9) Similarly there is a linear dependence between the rows which is given by the relation row 1 =
VI.
7 8
row 2 +
4 5
row 3
(C.10)
The determinant of an upper triangular or lower triangular matrix is the product of the main diagonal entries. For example, 0003 0003 10003 00033 2 0003 0003 (C.11) 0003 0 2 −1 0003 = 3 × 2 × 4 = 24. 0003 0003 0 0 4 This rule is easily verified from the definition (C.1) because all terms vanish except j1 = 1, j2 = 2, . . . jn = n, which is the product of the main diagonal entries. Diagonal matrices are a particular case of this rule.
VII.
The determinant of the product of two square matrices is the product of the individual determinants: |AB| = |A| |B|. (C.12) This rule can be generalized to any number of factors. One immediate application is to matrix powers: |A2 | = |A||A| = |A|2 , and more generally |An | = |A|n for integer n.
VIII.
The determinant of the transpose of a matrix is the same as that of the original matrix: |AT | = |A|. This rule can be directly verified from the definition of determinant. C–3
(C.13)
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
C–4
REMARK C.3
Rules VI and VII are the key to the practical evaluation of determinants. Any square nonsingular matrix A (where the qualifier “nonsingular” is explained in §C.3) can be decomposed as the product of two triangular factors A = LU,
(C.14)
where L is unit lower triangular and U is upper triangular. This is called a LU triangularization, LU factorization or LU decomposition. It can be carried out in O(n 3 ) floating point operations. According to rule VII: |A| = |L| |U|.
(C.15)
But according to rule VI, |L| = 1 and |U| = u 11 u 22 . . . u nn . The last operation requires only O(n) operations. Thus the evaluation of |A| is dominated by the effort involved in computing the factorization (C.14). For n = 10, that effort is approximately 103 = 1000 floating-point operations, compared to approximately 3 × 107 from the naive application of (C.1), as noted in Remark C.1. Thus the LU-based method is roughly 30, 000 times faster for that modest matrix order, and the ratio increases exponentially for large n.
§C.1.2 Cramer’s Rule Cramer’s rule provides a recipe for solving linear algebraic equations in terms of determinants. Let the simultaneous equations be as usual denoted as Ax = y,
(C.16)
where A is a given n × n matrix, y is a given n × 1 vector, and x is the n × 1 vector of unknowns. The explicit form of (C.16) is Equation (A.1) of Appendix A, with n = m. The explicit solution for the components x1 , x2 . . ., xn of x in terms of determinants is
x1 =
0003 0003 y1 0003 0003 y2 0003 . 0003 . 0003 . 0003 yn
a12 a22 . .
a13 a23 . .
an2
an3 |A|
0003 a1n 0003 0003 a2n 0003 0003 0003 0003 0003 . . . ann .. .. . .
,
x2 =
0003 0003 a11 0003 0003 a21 0003 . 0003 . 0003 . 0003 an1
y1 y2 . .
a13 a23 . .
yn
an3 |A|
0003 a1n 0003 0003 a2n 0003 0003 0003 0003 0003 . . . ann .. .. . .
, ..
(C.17)
The rule can be remembered as follows: in the numerator of the quotient for x j , replace the j th column of A by the right-hand side y. This method of solving simultaneous equations is known as Cramer’s rule. Because the explicit computation of determinants is impractical for n > 3 as explained in Remark C.3, this rule has practical value only for n = 2 and n = 3 (it is marginal for n = 4). But such small-order systems arise often in finite element calculations at the Gauss point level; consequently implementors should be aware of this rule for such applications. EXAMPLE C.3
Solve the 3 × 3 linear system
0004
5 3 1
2 2 0
1 0 2
00050004
x1 x2 x3
C–4
0005
0004 0005 =
8 5 3
,
(C.18)
C–5
§C.2 SINGULAR MATRICES, RANK
by Cramer’s rule:
0003 00038 0003 00035 00033 x1 = 0003 00035 0003 00033 00031
2 2 0 2 2 0
0003
10003 0003 00003 6 20003 0003 = = 1, 0003 1 0003 6 00003 20003
0003 00035 0003 00033 00031 x2 = 0003 00035 0003 00033 00031
8 5 3 2 2 0
0003
0003 00035 0003 00033 00031 x3 = 0003 00035 0003 00033 00031
10003 0003 00003 6 20003 0003 = = 1, 0003 1 0003 6 00003 20003
2 2 0 2 2 0
0003
80003 0003 50003 6 30003 0003 = = 1. 0003 1 0003 6 00003 20003
(C.19)
EXAMPLE C.4
Solve the 2 × 2 linear algebraic system
by Cramer’s rule: x1 = 0003
0003 0003 0003 5 −β 0003 00030 1 + β 0003
0006
2+β −β
−β 1+β
5 + 5β
0003= , 0003 2 + β −β 0003 2 + 3β 0003 −β 1 + β 0003
00070006
x1 x2
0007
x2 = 0003
=
0006 0007 5 0
0003 0003 00032 + β 50003 0003 −β 0 0003
(C.20)
5β
0003= . 0003 2 + β −β 0003 2 + 3β 0003 −β 1 + β 0003
(C.21)
§C.1.3 Homogeneous Systems One immediate consequence of Cramer’s rule is what happens if y1 = y2 = . . . = yn = 0.
(C.22)
The linear equation systems with a null right hand side Ax = 0,
(C.23)
is called a homogeneous system. From the rule (C.17) we see that if |A| is nonzero, all solution components are zero, and consequently the only possible solution is the trivial one x = 0. The case in which |A| vanishes is discussed in the next section. §C.2 SINGULAR MATRICES, RANK If the determinant |A| of a n × n square matrix A ≡ An is zero, then the matrix is said to be singular. This means that at least one row and one column are linearly dependent on the others. If this row and column are removed, we are left with another matrix, say An−1 , to which we can apply the same criterion. If the determinant |An−1 | is zero, we can remove another row and column from it to get An−2 , and so on. Suppose that we eventually arrive at an r × r matrix Ar whose determinant is nonzero. Then matrix A is said to have rank r , and we write rank(A) = r . If the determinant of A is nonzero, then A is said to be nonsingular. The rank of a nonsingular n × n matrix is equal to n. Obviously the rank of AT is the same as that of A since it is only necessary to transpose “row” and ”column” in the definition. The notion of rank can be extended to rectangular matrices as outlined in section §C.2.4 below. That extension, however, is not important for the material covered here.
C–5
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
C–6
EXAMPLE C.5
The 3 × 3 matrix
0004 A=
3 1 2
2 2 −1
2 −1 3
0005 ,
(C.24)
has rank r = 3 because |A| = −3 = 0. EXAMPLE C.6
The matrix
0004
0005
3 2 1 (C.25) A= 1 2 −1 , 2 −1 3 already used as an example in §C.1.1 is singular because its first row and column may be expressed as linear combinations of the others through the relations (C.9) and (C.10). Removing the first row and column we are left with a 2 × 2 matrix whose determinant is 2 × 3 − (−1) × (−1) = 5 = 0. Consequently (C.25) has rank r = 2.
§C.2.1 Rank Deficiency If the square matrix A is supposed to be of rank r but in fact has a smaller rank r¯ < r , the matrix is said to be rank deficient. The number r − r¯ > 0 is called the rank deficiency. EXAMPLE C.7
Suppose that the unconstrained master stiffness matrix K of a finite element has order n, and that the element possesses b independent rigid body modes. Then the expected rank of K is r = n − b. If the actual rank is less than r , the finite element model is said to be rank-deficient. This is an undesirable property. EXAMPLE C.8
An an illustration of the foregoing rule, consider the two-node, 4-DOF plane beam element stiffness derived in Chapter 13:   12 6L −12 6L EI  4L 2 −6L 2L 2  (C.26) K= 3  12 −6L  L symm 4L 2 where E I and L are nonzero scalars. It can be verified that this 4 × 4 matrix has rank 2. The number of rigid body modes is 2, and the expected rank is r = 4 − 2 = 2. Consequently this model is rank sufficient.
§C.2.2 Rank of Matrix Sums and Products In finite element analysis matrices are often built through sum and product combinations of simpler matrices. Two important rules apply to “rank propagation” through those combinations. The rank of the product of two square matrices A and B cannot exceed the smallest rank of the multiplicand matrices. That is, if the rank of A is ra and the rank of B is rb , Rank(AB) ≤ min(ra , rb ).
(C.27)
Regarding sums: the rank of a matrix sum cannot exceed the sum of ranks of the summand matrices. That is, if the rank of A is ra and the rank of B is rb , Rank(A + B) ≤ ra + rb . C–6
(C.28)
C–7
§C.3
MATRIX INVERSION
§C.2.3 Singular Systems: Particular and Homegeneous Solutions Having introduced the notion of rank we can now discuss what happens to the linear system (C.16) when the determinant of A vanishes, meaning that its rank is less than n. If so, the linear system (C.16) has either no solution or an infinite number of solution. Cramer’s rule is of limited or no help in this situation. To discuss this case further we note that if |A| = 0 and the rank of A is r = n − d, where d ≥ 1 is the rank deficiency, then there exist d nonzero independent vectors zi , i = 1, . . . d such that Azi = 0.
(C.29)
These d vectors, suitably orthonormalized, are called null eigenvectors of A, and form a basis for its null space. Let Z denote the n × d matrix obtained by collecting the zi as columns. If y in (C.13) is in the range of A, that is, there exists an nonzero x p such that y = Ax p , its general solution is x = x p + xh = x p + Zw,
(C.30)
where w is an arbitrary d × 1 weighting vector. This statement can be easily verified by substituting this solution into Ax = y and noting that AZ vanishes. The components x p and xh are called the particular and homogeneous part, respectively, of the solution x. If y = 0 only the homogeneous part remains. If y is not in the range of A, system (C.13) does not generally have a solution in the conventional sense, although least-square solutions can usually be constructed. The reader is referred to the many textbooks in linear algebra for further details. §C.2.4 Rank of Rectangular Matrices The notion of rank can be extended to rectangular matrices, real or complex, as follows. Let A be m ×n. Its column range space R(A) is the subspace spanned by Ax where x is the set of all complex n-vectors. Mathematically: R(A) = {Ax : x ∈ C n }. The rank r of A is the dimension of R(A). The null space N (A) of A is the set of n-vectors z such that Az = 0. The dimension of N (A) is n − r . Using these definitions, the product and sum rules (C.27) and (C.28) generalize to the case of rectangular (but conforming) A and B. So does the treatment of linear equation systems Ax = y in which A is rectangular; such systems often arise in the fitting of observation and measurement data. In finite element methods, rectangular matrices appear in change of basis through congruential transformations, and in the treatment of multifreedom constraints.
§C.3 MATRIX INVERSION The inverse of a square nonsingular matrix A is represented by the symbol A−1 and is defined by the relation AA−1 = I. (C.31) The most important application of the concept of inverse is the solution of linear systems. Suppose that, in the usual notation, we have Ax = y (C.32) C–7
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
C–8
Premultiplying both sides by A−1 we get the inverse relationship x = A−1 y
(C.33)
More generally, consider the matrix equation for multiple (m) right-hand sides: A
X = Y,
n×n n×m
(C.34)
n×m
which reduces to (C.32) for m = 1. The inverse relation that gives X as function of Y is X = A−1 Y.
(C.35)
AX = I,
(C.36)
In particular, the solution of
is X = A−1 . Practical methods for computing inverses are based on directly solving this equation; see Remark C.4. §C.3.1 Explicit Computation of Inverses The explicit calculation of matrix inverses is seldom needed in large matrix computations. But ocassionally the need arises for the explicit inverse of small matrices that appear in element computations. For example, the inversion of Jacobian matrices at Gauss points, or of constitutive matrices. A general formula for elements of the inverse can be obtained by specializing Cramer’s rule to (C.36). Let B = [bi j ] = A−1 . Then A ji bi j = , (C.37) |A| in which A ji denotes the so-called adjoint of entry ai j of A. The adjoint A ji is defined as the determinant of the submatrix of order (n − 1) × (n − 1) obtained by deleting the j th row and i th column of A, multiplied by (−1)i+ j . This direct inversion procedure is useful only for small matrix orders: 2 or 3. In the examples below the inversion formulas for second and third order matrices are listed.
EXAMPLE C.9
For order n = 2: A=
0006
a11 a21
0007
a12 , a22
A−1 =
in which |A| is given by (C.2).
C–8
1 0006 a22 |A| −a21
0007
−a12 , a22
(C.38)
C–9
§C.3
MATRIX INVERSION
EXAMPLE C.10
For order n = 3:
0004 A=
where
a11 a21 a31
0003 0003 a22 0003a 0003 32 0003a b12 = − 0003 21 a 0003 31 0003a b13 = 0003 21 a b11 =
31
a12 a22 a32
a13 a23 a33
0003
0005
0004 ,
A
−1
1 = |A|
b11 b21 b31
0003 0003 0003 a12 a13 0003 0003, a32 a33 0003 0003 a13 0003 0003a b22 = 0003 11 0003, a a 0003 31 33 0003 a12 0003 0003a b23 = − 0003 11 0003, a a
a23 0003 0003, a33 0003 a23 0003 0003, a33 0003 a22 0003 0003, a32
b21 = − 0003
31
32
b12 b22 b32
b13 b23 b33
0005 ,
0003 0003 a12 0003a 0003 22 0003a b32 = − 0003 11 a 0003 21 0003a b33 = 0003 11 a
(C.39)
0003
a13 0003 0003, a23 0003 a13 0003 0003, a23 0003 a12 0003 0003, a22
b31 =
21
(C.40)
in which |A| is given by (C.3). EXAMPLE C.11
0004 A=
2 4 2 3 1 1 1 0 1
0005
0004 ,
A−1
1 =− 8
1 −2 −1
−4 0 4
2 4 −10
0005 .
(C.41)
If the order exceeds 3, the general inversion formula based on Cramer’s rule becomes rapidly useless as it displays combinatorial complexity. For numerical work it is preferable to solve the system (C.38) after A is factored. Those techniques are described in detail in linear algebra books; see also Remark C.4. §C.3.2 Some Properties of the Inverse I.
The inverse of the transpose is equal to the transpose of the inverse: (AT )−1 = (A−1 )T ,
(C.42)
(AA−1 ) = (AA−1 )T = (A−1 )T AT = I.
(C.43)
because
II.
The inverse of a symmetric matrix is also symmetric. Because of the previous rule, (AT )−1 = A−1 = (A−1 )T , hence A−1 is also symmetric.
III. The inverse of a matrix product is the reverse product of the inverses of the factors: (AB)−1 = B−1 A−1 .
(C.44)
This is easily verified by substituting both sides of (C.39) into (C.31). This property generalizes to an arbitrary number of factors. IV. For a diagonal matrix D in which all diagonal entries are nonzero, D−1 is again a diagonal matrix with entries 1/dii . The verification is straightforward. C–9
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
V.
C–10
If S is a block diagonal matrix: S   S=  
0 0 . .
0 S22 0 . .
0 0 S33 . .
0
0
0
11
.. .. .. . .
..

0 0 0 . .
   = diag [ S ] , ii  
(C.45)
Snn
then the inverse matrix is also block diagonal and is given by  S−1 −1
S
  =  
0 0 . .
0 S−1 22 0 . .
0 0 S−1 33 . .
0
0
0
11
.. .. .. . . ..
0 0 0 . .
    = diag [ S−1 ] . ii  
(C.46)
S−1 nn
VI. The inverse of an upper triangular matrix is also an upper triangular matrix. The inverse of a lower triangular matrix is also a lower triangular matrix. Both inverses can be computed in O(n 2 ) floating-point operations. REMARK C.4
The practical numerical calculation of inverses is based on triangular factorization. Given a nonsingular n × n matrix A, calculate its LU factorization A = LU, which can be obtained in O(n 3 ) operations. Then solve the linear triangular systems: UY = I, LX = Y, (C.47) and the computed inverse A−1 appears in X. One can overwrite I with Y and Y with X. The whole process can be completed in O(n 3 ) floating-point operations. For symmetric matrices the alternative decomposition A = LDLT , where L is unit lower triangular and D is diagonal, is generally preferred to save computing time and storage.
§C.4 EIGENVALUES AND EIGENVECTORS Consider the special form of the linear system (C.13) in which the right-hand side vector y is a multiple of the solution vector x: Ax = λx, (C.48) or, written in full, a11 x1 a21 x2 ··· an1 x1
+ +
a12 x2 a22 x2 ··· + an2 x2
+ + +
· · · + a1n xn · · · + a2n xn ··· ··· · · · + ann xn
= = =
λx1 λx2 ··· λxn
(C.49)
This is called the standard (or classical) algebraic eigenproblem. System (C.48) can be rearranged into the homogeneous form (A − λI) x = 0. (C.50) C–10
C–11
§C.4
EIGENVALUES AND EIGENVECTORS
A nontrivial solution of this equation is possible if and only if the coefficient matrix A − λI is singular. Such a condition can be expressed as the vanishing of the determinant 0003 0003 a12 .. a1n 0003 0003 a11 − λ 0003 0003 a22 − λ . . . a2n 0003 0003 a21 0003 = 0. 0003 |A − λI| = 0003 (C.51) . . . . 0003 . 0003 0003 . . . 0003 0003 an2 . . . ann − λ an1 When this determinant is expanded, we obtain an algebraic polynomial equation in λ of degree n: P(λ) = λn + α1 λn−1 + · · · + αn = 0.
(C.52)
This is known as the characteristic equation of the matrix A. The left-hand side is called the characteristic polynomial. We known that a polynomial of degree n has n (generally complex) roots λ1 , λ2 , . . ., λn . These n numbers are called the eigenvalues, eigenroots or characteristic values of matrix A. With each eigenvalue λi there is an associated vector xi that satisfies Axi = λi xi .
(C.53)
This xi is called an eigenvector or characteristic vector. An eigenvector is unique only up to a scale factor since if xi is an eigenvector, so is βxi where β is an arbitrary nonzero number. Eigenvectors are often normalized so that their Euclidean length is 1, or their largest component is unity. §C.4.1 Real Symmetric Matrices Real symmetric matrices are of special importance in the finite element method. In linear algebra books dealing with the algebraic eigenproblem it is shown that: (a) The n eigenvalues of a real symmetric matrix of order n are real. (b) The eigenvectors corresponding to distinct eigenvalues are orthogonal. The eigenvectors corresponding to multiple roots may be orthogonalized with respect to each other. (c) The n eigenvectors form a complete orthonormal basis for the Euclidean space E n . §C.4.2 Positivity Let A be an n × n square symmetric matrix. A is said to be positive definite (p.d.) if xT Ax > 0,
x = 0
(C.54)
A positive definite matrix has rank n. This property can be checked by computing the n eigenvalues λi of Az = λz. If all λi > 0, A is p.d. A is said to be nonnegative1 if zero equality is allowed in (C.54): xT Ax ≤ 0,
x = 0
(C.55)
A p.d. matrix is also nonnegative but the converse is not necessarily true. This property can be checked by computing the n eigenvalues λi of Az = λz. If r eigenvalues λi > 0 and n − r eigenvalues are zero, A is nonnegative with rank r . A symmetric square matrix A that has at least one negative eigenvalue is called indefinite. 1
A property called positive semi-definite by some authors.
C–11
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
C–12
§C.4.3 Normal and Orthogonal Matrices Let A be an n × n real square matrix. This matrix is called normal if AT A = AAT
(C.56)
A normal matrix is called orthogonal if AT A = AAT = I
or AT = A−1
(C.57)
All eigenvalues of an orthogonal matrix have modulus one, and the matrix has rank n. The generalization of the orthogonality property to complex matrices, for which transposition is replaced by conjugation, leads to unitary matrices. These are not required, however, for the material covered in the text. §C.5 THE SHERMAN-MORRISON AND RELATED FORMULAS The Sherman-Morrison formula gives the inverse of a matrix modified by a rank-one matrix. The Woodbury formula extends the Sherman-Morrison formula to a modification of arbitrary rank. In structural analysis these formulas are of interest for problems of structural modifications, in which a finite-element (or, in general, a discrete model) is changed by an amount expressable as a low-rank correction to the original model. §C.5.1 The Sherman-Morrison Formula Let A be a square n × n invertible matrix, whereas u and v are two n-vectors and β an arbitrary scalar. Assume that σ = 1 + β vT A−1 u = 0. Then
000e
A + βuvT
000f−1
= A−1 −
β −1 T −1 A uv A . σ
(C.58)
This is called the Sherman-Morrison formula2 when β = 1. Since any rank-one correction to A can be written as βuvT , (C.58) gives the rank-one change to its inverse. The proof is by direct multiplication as in Exercise C.5. For practical computation of the change one solves the linear systems Aa = u and Ab = v for a and b, using the known A−1 . Compute σ = 1 + βvT a. If σ = 0, the change to A−1 is the dyadic −(β/σ )abT . §C.5.2 The Woodbury Formula Let again A be a square n × n invertible matrix, whereas U and V are two n × k matrices with k ≤ n and β an arbitrary scalar. Assume that the k × k matrix Σ = Ik + βVT A−1 U, in which Ik denotes the k × k identity matrix, is invertible. Then 000e 000f−1 A + βUVT = A−1 − β A−1 UΣ−1 VT A−1 . (C.59) This is called the Woodbury formula.3 It reduces to (C.58) if k = 1, in which case Σ ≡ σ is a scalar. The proof is by direct multiplication. 2
J. Sherman and W. J. Morrison, Adjustment of an inverse matrix corresponding to changes in the elements of a given column or a given row of the original matrix, Ann. Math. Statist., 20, (1949), 621. For a history of this remarkable formula and its extensions, which are quite important in many applications such as statistics, see the paper by Henderson and Searle in SIAM Review, 23, 53–60.
3
M.A.Woodbury, Inverting modified matrices, Statist. Res. Group, Mem. Rep. No. 42, Princeton Univ., Princeton, N.J., 1950
C–12
C–13
§C.5
THE SHERMAN-MORRISON AND RELATED FORMULAS
§C.5.3 Formulas for Modified Determinants A denote the adjoint of A. Taking the determinants from both sides of A + βuvT one obtains Let 0010 |A + βuvT | = |A| + β vT 0010 A u.
(C.60)
If A is invertible, replacing 0010 A = |A| A−1 this becomes |A + βuvT | = |A| (1 + β vT A−1 u).
(C.61)
Similarly, one can show that if A is invertible, and U and V are n × k matrices, |A + βUVT | = |A| |Ik + β VT A−1 U|.
C–13
(C.62)
Appendix C: MATRIX ALGEBRA: DETERMINANTS, INVERSES, EIGENVALUES
C–14
Exercises for Appendix C: Determinants, Inverses, Eigenvalues EXERCISE C.1
If A is a square matrix of order n and c a scalar, show that det(cA) = cn det A. EXERCISE C.2
Let u and v denote real n-vectors normalized to unit length, so that uT = u = 1 and vT v = 1, and let I denote the n × n identity matrix. Show that (EC.1) det(I − uvT ) = 1 − vT u EXERCISE C.3
Let u denote a real n-vector normalized to unit length, so that uT = u = 1 and I denote the n × n identity matrix. Show that (EC.2) H = I − 2uuT is orthogonal: HT H = I, and idempotent: H2 = H. This matrix is called a elementary Hermitian, a Householder matrix, or a reflector. It is a fundamental ingredient of many linear algebra algorithms; for example the QR algorithm for finding eigenvalues. EXERCISE C.4
The trace of a n × n square matrix A, denoted trace(A) is the sum that if the entries of A are real, trace(A A) = T
n 0012 n 0012
0011n i=1
aii of its diagonal coefficients. Show
ai2j
(EC.3)
i=1 j=1
EXERCISE C.5
Prove the Sherman-Morrison formula (C.59) by direct matrix multiplication. EXERCISE C.6
Prove the Sherman-Morrison formula (C.59) for β = 1 by considering the following block bordered system
0006
A VT
U Ik
00070006 0007
0006
B I = n C 0
0007
(EC.4)
in which Ik and In denote the identy matrices of orders k and n, respectively. Solve (C.62) two ways: eliminating first B and then C, and eliminating first C and then B. Equate the results for B. EXERCISE C.7
Show that the eigenvalues of a real symmetric square matrix are real, and that the eigenvectors are real vectors. EXERCISE C.8
Let the n real eigenvalues λi of a real n × n symmetric matrix A be classified into two subsets: r eigenvalues are nonzero whereas n − r are zero. Show that A has rank r . EXERCISE C.9
Show that if A is p.d., Ax = 0 implies that x = 0. EXERCISE C.10
Show that for any real m × n matrix A, AT A exists and is nonnegative.
C–14
C–15
Exercises
EXERCISE C.11
Show that a triangular matrix is normal if and only if it is diagonal. EXERCISE C.12
Let A be a real orthogonal matrix. Show that all of its eigenvalues λi , which are generally complex, have unit modulus. EXERCISE C.13
Let A and T be real n × n matrices, with T nonsingular. Show that T−1 AT and A have the same eigenvalues. (This is called a similarity transformation in linear algebra). EXERCISE C.14
(Tough) Let A be m × n and B be n × m. Show that the nonzero eigenvalues of AB are the same as those of BA (Kahan). EXERCISE C.15
Let A be real skew-symmetric, that is, A = −AT . Show that all eigenvalues of A are purely imaginary or zero. EXERCISE C.16
Let A be real skew-symmetric, that is, A = −AT . Show that U = (I+A)−1 (I−A), called a Cayley transformation, is orthogonal. EXERCISE C.17
Let P be a real square matrix that satisfies P2 = P.
(EC.5)
Such matrices are called idempotent, and also orthogonal projectors. Show that all eigenvalues of P are either zero or one. EXERCISE C.18
The necessary and sufficient condition for two square matrices to commute is that they have the same eigenvectors. EXERCISE C.19
A matrix whose elements are equal on any line parallel to the main diagonal is called a Toeplitz matrix. (They arise in finite difference or finite element discretizations of regular one-dimensional grids.) Show that if T1 and T2 are any two Toeplitz matrices, they commute: T1 T2 = T2 T1 . Hint: do a Fourier transform to show that the eigenvectors of any Toeplitz matrix are of the form {eiωnh }; then apply the previous Exercise.
C–15
D
.
Matrix Calculus
D–1
D–2
Appendix D: MATRIX CALCULUS
In this Appendix we collect some useful formulas of matrix calculus that often appear in finite element derivations. §D.1 THE DERIVATIVES OF VECTOR FUNCTIONS Let x and y be vectors of orders n and m respectively: x 
y 1  y2 y=  . .
1
 x2   x=  .  , . xn
  , 
(D.1)
ym
where each component yi may be a function of all the x j , a fact represented by saying that y is a function of x, or y = y(x). (D.2) If n = 1, x reduces to a scalar, which we call x. If m = 1, y reduces to a scalar, which we call y. Various applications are studied in the following subsections. §D.1.1 Derivative of Vector with Respect to Vector The derivative of the vector y with respect to vector x is the n × m matrix  ∂ y1  ∂ x1  ∂ y1 ∂y def  ∂ x2 =   ∂x  .  . ∂ y1 ∂ xn
∂ y2 ∂ x1 ∂ y2 ∂ x2 . . ∂ y2 ∂ xn
··· ··· .
.
···
∂ ym ∂ x1 ∂ ym ∂ x2 . . ∂ ym ∂ xn
       
(D.3)
§D.1.2 Derivative of a Scalar with Respect to Vector If y is a scalar,
 ∂y  ∂ x1  ∂y ∂ y def  ∂x =   2 ∂x  .  . ∂y ∂ xn
    .   
(D.4)
§D.1.3 Derivative of Vector with Respect to Scalar If x is a scalar,
∂y def ∂ y1 = ∂x ∂x
∂ y2 ∂x D–2
..
∂ ym ∂x
(D.5)
D–3
§D.1 THE DERIVATIVES OF VECTOR FUNCTIONS
REMARK D.1
Many authors, notably in statistics and economics, define the derivatives as the transposes of those given above.1 This has the advantage of better agreement of matrix products with composition schemes such as the chain rule. Evidently the notation is not yet stable.
EXAMPLE D.1
Given y=
y1 , y2
x=
x1 x2 x3
(D.6)
and y1 = x12 − x2
(D.7)
y2 = x32 + 3x2 the partial derivative matrix ∂y/∂x is computed as follows:
 ∂ y1
∂ x1 ∂y  ∂ y1 =  ∂ x2 ∂x ∂ y1 ∂ x3
∂ y2 ∂ x1 ∂ y2 ∂ x2 ∂ y2 ∂ x3
  = 
2x1 −1 0
0 3 2x3
(D.8)
§D.1.4 Jacobian of a Variable Transformation In multivariate analysis, if x and y are of the same order, the determinant of the square matrix ∂x/∂y, that is 000e 000e 000e ∂x 000e J = 000e000e 000e000e (D.9) ∂y is called the Jacobian of the transformation determined by y = y(x). The inverse determinant is J
1
−1
000e 000e 000e ∂y 000e = 000e000e 000e000e . ∂x
(D.10)
One authors puts is this way: “When one does matrix calculus, one quickly finds that there are two kinds of people in this world: those who think the gradient is a row vector, and those who think it is a column vector.”
D–3
D–4
Appendix D: MATRIX CALCULUS
EXAMPLE D.2
The transformation from spherical to Cartesian coordinates is defined by x = r sin θ cos ψ,
y = r sin θ sin ψ,
z = r cos θ
(D.11)
where r > 0, 0 < θ < π and 0 ≤ ψ < 2π. To obtain the Jacobian of the transformation, let x ≡ x1 , r ≡ y1 , Then
y ≡ x2 , θ ≡ y2 ,
000e 000e 000e000e sin y2 cos y3 000e ∂x 000e 000e J = 000e000e 000e000e = 000e y1 cos y2 cos y3 ∂y 000e −y sin y sin y 1
2
z ≡ x3 ψ ≡ y3
sin y2 sin y3 y1 cos y2 sin y3 y1 sin y2 cos y3
3
(D.12)
000e
cos y2 000e 000e −y1 sin y2 000e 000e 0
(D.13)
= y12 sin y2 = r 2 sin θ.
The foregoing definitions can be used to obtain derivatives to many frequently used expressions, including quadratic and bilinear forms. EXAMPLE D.3
Consider the quadratic form y = xT Ax
(D.14)
where A is a square matrix of order n. Using the definition (D.3) one obtains ∂y = Ax + AT x ∂x
(D.15)
and if A is symmetric,
∂y = 2Ax. ∂x We can of course continue the differentiation process: ∂ ∂2 y = 2 ∂x ∂x
and if A is symmetric,
000f
∂y ∂x
(D.16)
0010 = A + AT ,
∂2 y = 2A. ∂x2
The following table collects several useful vector derivative formulas. y
∂y ∂x
Ax xT A xT x xT Ax
AT A 2x Ax + AT x
D–4
(D.17)
(D.18)
D–5
§D.2
THE CHAIN RULE FOR VECTOR FUNCTIONS
§D.2 THE CHAIN RULE FOR VECTOR FUNCTIONS Let

 x1  x2   x=  .. 

 y1  y2   y=  .  . yr
,
xn

 z1  z2   z=  .. 
and
(D.19)
zm
where z is a function of y, which is in turn a function of x. Using the definition (D.2), we can write  000f
∂z ∂x
0010T
∂z 1 ∂ x1 ∂z 2 ∂ x1
∂z 1 ∂ x2 ∂z 2 ∂ x2
.. ..
∂z 1 ∂ xn ∂z 2 ∂ xn
∂z m ∂ x1
∂z m ∂ x2
..
∂z m ∂ xn
  =  .  .

   .  . 
. .
(D.20)
Each entry of this matrix may be expanded as 0012
r 0011 ∂z i ∂ yq ∂z i = ∂ xj ∂ yq ∂ x j q=1
 0013 ∂z1 ∂ yq
Then 000f
∂z ∂x
0010T
yq ∂ x1 ∂z 2 ∂ yq ∂ yq ∂ x1
0013   = .   . 0013 ∂zm 
∂ yq ∂ yq ∂ x1
∂z 1 ∂ y1 ∂z 2 ∂ y1
  =  .  .
∂z 1 ∂ y2 ∂z 2 ∂ y2
.. ..
0013 0013
i = 1, 2, . . . , m j = 1, 2, . . . , n.
∂z 1 ∂ yq ∂ yq ∂ x2 ∂z 2 ∂ yq ∂ yq ∂ x2
.. ..
0013 ∂zm
∂ yq ∂ yq ∂ x2
∂z 1 ∂ yr ∂z 2 ∂ yr

.. ∂ y1 ∂ x1 ∂ y2 ∂ x1
    .   .
0013 0013
∂z 2 ∂ yq ∂ yq ∂ xn ∂z 2 ∂ yq ∂ yq ∂ xn
0013 ∂zm
∂ yq ∂ yq ∂ xn
∂ y1 ∂ x2 ∂ y2 ∂ x2
.. ..
∂ yr ∂ yr m .. . . . ∂z ∂ x1 ∂ x2 ∂ yr 0010 000f 0010T 000f 0010T 000f ∂y ∂z T ∂y ∂z = . = ∂y ∂x ∂x ∂y
∂z m ∂ y1
∂z m ∂ y2
(D.21)        ∂ y1 ∂ xn ∂ y2 ∂ xn
     
∂ yr ∂ xn
(D.22)
On transposing both sides, we finally obtain ∂y ∂z ∂z = , ∂x ∂x ∂y
(D.23)
which is the chain rule for vectors. If all vectors reduce to scalars, ∂ y ∂z ∂z ∂ y ∂z = = , ∂x ∂x ∂y ∂y ∂x D–5
(D.24)
D–6
Appendix D: MATRIX CALCULUS
which is the conventional chain rule of calculus. Note, however, that when we are dealing with vectors, the chain of matrices builds “toward the left.” For example, if w is a function of z, which is a function of y, which is a function of x, ∂w ∂y ∂z ∂w = . ∂x ∂x ∂y ∂z
(D.25)
On the other hand, in the ordinary chain rule one can indistictly build the product to the right or to the left because scalar multiplication is commutative. §D.3 THE DERIVATIVE OF SCALAR FUNCTIONS OF A MATRIX Let X = (xi j ) be a matrix of order (m × n) and let y = f (X),
(D.26)
be a scalar function of X. The derivative of y with respect to X, denoted by ∂y , ∂X
(D.27)
is defined as the following matrix of order (m × n): 
∂y ∂ x11 ∂y ∂ x21
∂y ∂ x12 ∂y ∂ x22
.. ..
∂y ∂ x1n ∂y ∂ x2n
∂y ∂ xm1
∂y ∂ xm2
..
∂y ∂ xmn
  ∂y = G= . ∂X   .
. .

 0014 0015 0011  ∂y ∂ y  Ei j , .  = ∂ xi j = ∂ xi j i, j . 
(D.28)
where Ei j denotes the elementary matrix* of order (m × n). This matrix G is also known as a gradient matrix. EXAMPLE D.4
Find the gradient matrix if y is the trace of a square matrix X of order n, that is y = tr(X) =
n 0011
xii .
(D.29)
i=1
Obviously all non-diagonal partials vanish whereas the diagonal partials equal one, thus G=
∂y = I, ∂X
where I denotes the identity matrix of order n. * The elementary matrix Ei j of order m × n has all zero entries except for the (i, j) entry, which is one.
D–6
(D.30)
D–7
§D.4
THE MATRIX DIFFERENTIAL
§D.3.1 Functions of a Matrix Determinant An important family of derivatives with respect to a matrix involves functions of the determinant of a matrix, for example y = |X| or y = |AX|. Suppose that we have a matrix Y = [yi j ] whose components are functions of a matrix X = [xr s ], that is yi j = f i j (xr s ), and set out to build the matrix ∂|Y| . (D.31) ∂X Using the chain rule we can write ∂|Y| ∂ yi j ∂|Y| 0011 0011 = Yi j . ∂ xr s ∂ yi j ∂ xr s i j But |Y| =
0011
(D.32)
yi j Yi j ,
(D.33)
j
where Yi j is the cofactor of the element yi j in |Y|. Since the cofactors Yi1 , Yi2 , . . . are independent of the element yi j , we have ∂|Y| = Yi j . (D.34) ∂ yi j It follows that
∂ yi j ∂|Y| 0011 0011 = Yi j . ∂ xr s ∂ x r s i j
(D.35)
There is an alternative form of this result which is ocassionally useful. Define ai j = Yi j , Then it can be shown that
A = [ai j ],
bi j =
∂ yi j , ∂ xr s
∂|Y| = tr(ABT ) = tr(BT A). ∂ xr s
B = [bi j ].
(D.36)
(D.37)
EXAMPLE D.5
If X is a nonsingular square matrix and Z = |X|X−1 its cofactor matrix, ∂|X| = ZT . ∂X
(D.38)
∂|X| = 2ZT − diag(ZT ). ∂X
(D.39)
G= If X is also symmetric, G=
D–7
D–8
Appendix D: MATRIX CALCULUS
§D.4 THE MATRIX DIFFERENTIAL For a scalar function f (x), where x is an n-vector, the ordinary differential of multivariate calculus is defined as n 0011 ∂f d xi . (D.40) df = ∂ xi i=1 In harmony with this formula, we define the differential of an m × n matrix X = [xi j ] to be 
d x11 def  d x 21 dX =   ..
d x12 d x22 . .
d xm1
d xm2
.. ..
 d x1n d x2n  . .  . 
(D.41)
. . . d xmn
This definition complies with the multiplicative and associative rules d(αX) = α dX,
d(X + Y) = dX + dY.
(D.42)
If X and Y are product-conforming matrices, it can be verified that the differential of their product is d(XY) = (dX)Y + X(dY). (D.43) which is an extension of the well known rule d(x y) = y d x + x dy for scalar functions. EXAMPLE D.6
If X = [xi j ] is a square nonsingular matrix of order n, and denote Z = |X|X−1 . Find the differential of the determinant of X: d|X| =
0011 ∂|X| i, j
∂ xi j
d xi j =
0011
Xi j d xi j = tr(|X|X−1 )T dX) = tr(ZT dX),
(D.44)
i, j
where X i j denotes the cofactor of xi j in X. EXAMPLE D.7
With the same assumptions as above, find d(X−1 ). The quickest derivation follows by differentiating both sides of the identity X−1 X = I: (D.45) d(X−1 )X + X−1 dX = 0, from which
d(X−1 ) = −X−1 dX X−1 .
000f 0010
If X reduces to the scalar x we have d
1 x
=−
D–8
dx . x2
(D.46) (D.47)
F
.
Geometric Applications of Matrices
F–1
Appendix F: GEOMETRIC APPLICATIONS OF MATRICES
F–2
In this Appendix we summarize some geometric applications of matrices. The space is 3dimensional. Indexed homogeneous Cartesian coordinates1 {x0 , x1 , x2 , x3 } are used. To pass to physical coordinates, divide x1 , x2 and x3 by x0 : {x1 /x0 , x2 /x0 , x3 /x0 . If x0 = 0 the physical coordinates are at infinity. §F.1 POINTS, PLANES Points in 3D space will de identified by X , Y , P, Q, etc. Their coordinates are put in the 4-vectors x = [ x 0 x 1 x 2 x 3 ]T ,
y = [ y0 y1 y2 y3 ]T ,
p = [ p0 p1 p2 p3 ]T , etc
(F.1)
Planes in 3D space will be identified by A, B, C, etc. The equation of plane A is written a0 x0 + a1 x1 + a2 x2 + a3 x3 = 0 or (F.2) aT x = 0, or xT a = 0 The plane A is thus defined by the 4-vector a. EXAMPLE F.1
Find the coordinates of the point where line joining points P and Q intersects plane A. Solution. Any point of P Q is R where r = λp + µq
(F.3)
If R is on plane A, then aT x = 0 so that λaT p+µaT q = 0. Absorbing a suitable multiplier into the coordinates of R we obtain its vector in the form (F.4) r = (aT q)p − (aT p)q.
§F.2 LINES Let x and y be the coordinates of points X and Y on a given line L. The 4 × 4 antisymmetric matrix L = xyT − yxT
(F.5)
is called the coordinate matrix of the line. It can be shown that L determines the line L to within a scale factor. Let A and B be two planes through L. The 4 × 4 antisymmetric matrix L∗ = abT − baT
(F.6)
is called the dual coordinate matrix of line L. It can be shown that2 L∗ L = 0.
(F.7)
1
These coordinates were independently invented in 1827 by Møbius and Feuerbach, and further developed in 1946 by E. A. Maxwell at Cambridge. See E. A. Maxwell, General Homogeneous Coordinates in Space of Three Dimensions, Cambridge University Press, 1951.
2
See E. A. Maxwell, loc. cit., page 150.
F–2
F–3
§F.2
LINES
EXAMPLE F.2
Find where line L defined by two points X and Y meets a plane A. Solution. Consider the vector p = La = (xyT − yxT )a = (yT a)x − (xT a)y
(F.8)
From the last form the point P of coordinates p must lie on the line that joins X and Y . Moreover since L is antisymmetrical, aT La = 0 so that aT p = 0. Thus P is the intersection of line L and plane A. EXAMPLE F.3
Find the plane A that joins line L to a point X . Solution.
a = L∗ x
(F.9)
where L∗ is the dual of L. The demonstration is trivial. Two immediate corollaries: (i) Line L lies in the plane A if La = 0; (ii) Line L passes through the point X if L∗ x = 0. EXAMPLE F.4
˜ 1 and L ˜ 2 , respectively. Find Consider two lines L 1 and L 2 with coordinate matrices L1 , L2 and dual matrices L the conditions for the lines to intersect. Solution. Any of the four equivalent conditions L1 L∗2 L1 = 0,
L∗1 L2 L∗1 = 0,
L2 L∗1 L2 = 0,
For the proof see E. A. Maxwell, loc. cit, page 154.
F–3
L∗2 L1 L∗2 = 0.
(F.10)
G
.
Oldies but Goodies
G–1
G–2
Appendix G: OLDIES BUT GOODIES
The following list is fairly comprehensive until about 1989. Since then many more books as well as revised editions of previous ones have appeared. Newer books usually emphasize the mathematical interpretation and thus are of limited usefulness to engineers. In fact the field by now can be safely characterized, to paraphrase T. S. Eliot, as an engineer’s wasteland. Recommendation: stick with the oldies. Unfortunately many are out of print. §G.1 MATHEMATICALLY ORIENTED A. K. Aziz (ed.), The Mathematical Foundations of the Finite Element Method with Applications to Partial Differential Equations, Academic Press, New York, 1972. G. F. Carey and J. T. Oden, Finite Elements IV: Mathematical Aspects, Prentice Hall, Englewood Cliffs, N. J., 1983. P. G. Ciarlet, The Finite Element Method for Elliptic Problems, North Holland, Amsterdam, 1978. D. F. Griffiths (ed.), The Mathematical Basis of Finite Element Methods, Clarendon Press, Oxford, UK 1984. A. R. Mitchell and R. Wait, The Finite Element Analysis in Partial Differential Equations, Wiley, New York, 1977. J. T. Oden, An Introduction to the Mathematical Theory of Finite Elements, Wiley, New York, 1976. G. Strang and G. J. Fix, An Analysis of the Finite Element Method, Prentice Hall, Englewood Cliffs, N. J., 1973. R. Wait and A. R. Mitchell, Finite Element Analysis and Applications, Wiley, Chichester, UK, 1985. E. L. Wachpress, A Rational Finite Element Basis, Wiley, New York, 1976. O. C. Zienkiewicz and K. Morgan, Finite Element and Approximations, Wiley, New York, 1983. §G.2 APPLICATIONS ORIENTED K. J. Bathe, Finite Element Procedures in Engineering Analysis, Prentice Hall, Englewood Cliffs, N. J., 1982. Second edition entitled Finite Element Analysis: From Concepts to Applications has appeared in 1996. K. J. Bathe and E. L. Wilson, Numerical Methods in Finite Element Analysis, Prentice Hall, Englewood Cliffs, N. J., 1976. E. B. Becker, G. F. Carey and J. T. Oden, Finite Elements III: Computational Aspects, Prentice Hall, Englewood Cliffs, N. J., 198 G. F. Carey and J. T. Oden, Finite Elements II: A Second Course, Prentice Hall, Englewood Cliffs, N. J., 1981. M. V. K. Chari and P. P. Silvester, Finite Elements in Electrical and Magnetic Field Problems, Wiley, Chichester, UK, 1984. T. H. Chung, The Finite Element Method in Fluid Mechanics, McGraw-Hill, New York, 1978. G–2
G–3
§G.3
SOFTWARE ORIENTED
R. D. Cook, D. S. Malkus and M. E. Plesha, Concepts and Application of Finite Element Methods, 3rd ed., Wiley, New York, 1989. A. J. Davies, The Finite Element Method, Clarendon Press, Oxford, UK, 1980. C. S. Desai, Elementary Finite Element Method, Prentice Hall, Englewood Cliffs, N. J., 1979. C. S. Desai and J. F. Abel, Introduction to the Finite Element Method, Van Nostrand, New York, 1972. G. Dhatt and G. Touzot, The Finite Element Method Displayed, Wiley, Chichester, UK, 1984. R. H. Gallagher, Finite Element Analysis, Prentice Hall, Englewood Cliffs, N. J., 1975. I. Holand and K. Bell (eds), Finite Element Methods in Stress Analysis, Tapir, Trondheim, Norway, 1969. K. H. Huebner, The Finite Element Method for Engineers, Wiley, New York, 1975. T. J. R. Hughes, The Finite Element Method: Linear Static and Dynamic Finite Element Analysis, Prentice Hall, Englewood Cliffs, N. J., 1987. B. Irons and S. Ahmad, Techniques of Finite Elements, Ellis Horwood Ltd, Chichester, UK, 1980. H. C. Martin and G. F. Carey, Introduction to Finite Element Analysis, McGraw-Hill, New York, 1973. J. T. Oden, Finite Elements of Nonlinear Continua, Wiley, New York, 1972. J. T. Oden and G. F. Carey, Finite Elements I: An Introduction, Prentice Hall, Englewood Cliffs, N. J., 1981. J. T. Oden and G. F. Carey, Finite Elements V: Special Problems in Solid Mechanics, Prentice Hall, Englewood Cliffs, N. J., 1984. J. S. Przeminiecki, Theory of Matrix Structural Analysis, McGraw-Hill, New York, 1968 (also in Dover ed). S. S. Rao, The Finite Element Method in Engineering, Pergamon Press, Oxford, 1982. J. Robinson, Integrated Theory of Finite Element Methods, Wiley, London, 1973. K. C. Rockey, H. R. Evans, D. W. Griffiths and D. A. Nethercot, The Finite Element Method: A Basic Introduction for Engineers, Collins, London, 1983. L. J. Segerlind, Applied Finite Element Analysis, Wiley, New York, 1976. P. Tong and J. N. Rosettos, Finite Element Method, MIT Press, London, 1977. O. C. Zienkiewicz, The Finite Element Method in Engineering Sciences, 3rd ed., McGraw-Hill, London, 1977. Partly superseded by Vol I of Zienkiewicz and Taylor, McGraw-Hill, 1988. Vol II appeared in 1993. G–3
G–4
Appendix G: OLDIES BUT GOODIES
§G.3 SOFTWARE ORIENTED J. E. Akin, Application and Implementation of Finite Element Methods, Academic Press, New York, 1982. E. Hinton and D. R. J. Owen, An Introduction to Finite Element Computations, Pineridge Press, Swansea, 1979. I. M. Smith, Programming the Finite Element Method, Wiley, Chichester, UK, 1982. §G.4 RECOMMENDED BOOKS FOR LINEAR FEM Basic level (reference): Zienkiewicz and Taylor (1988) Vols I (1988), II(1993). This is a comprehensive upgrade of the 1977 edition. Primarily an encyclopedic reference work that gives panoramic coverage of FEM, as well as a comprehensive list of references. Not a textbook. A fifth edition has appeared recently. Basic level (textbook): Cook, Malkus and Plesha (1989). This third edition is fairly comprehensive in scope and fairly up to date although the coverage is more superficial than Zienkiewicz and Taylor. Intermediate level: Hughes (1987). It requires substantial mathematical expertise on the part of the reader. Mathematically oriented: Strang and Fix (1973). Still the most readable mathematical treatment for engineers, although outdated in many subjects. Best value for the $$$: Przeminiecki (Dover edition, 1985, about $14). A reprint of a 1966 MacGraw-Hill book. Although woefully outdated in many respects (the word “finite element” does not appear anywhere), it is a valuable reference for programming simple elements. It contains a comprehensive bibliography up to 1966. Most fun (if you appreciate British “humor”): Irons and Ahmad (1980) For buying out-of-print books through web services, check the search engine in http://www.bookarea.com/
G–4
H
.
An Outline of MSA History
H–1
Appendix H: AN OUTLINE OF MSA HISTORY
H–2
A Historical Outline of Matrix Structural Analysis: A Play in Three Acts C. A. Felippa Department of Aerospace Engineering Sciences and Center for Aerospace Structures University of Colorado, Boulder, CO 80309-0429, USA Report CU-CAS-00-14, June 2000; submitted for publication in Computers & Structures Abstract The evolution of Matrix Structural Analysis (MSA) from 1930 through 1970 is outlined. Hightlighted are major contributions by Collar and Duncan, Argyris, and Turner, which shaped this evolution. To enliven the narrative the outline is configured as a three-act play. Act I describes the pre-WWII formative period. Act II spans a period of confusion during which matrix methods assumed bewildering complexity in response to conflicting demands and restrictions. Act III outlines the cleanup and consolidation driven by the appearance of the Direct Stiffness Method, through which MSA completed morphing into the present implementation of the Finite Element Method. Keywords: matrix structural analysis; finite elements; history; displacement method; force method; direct stiffness method; duality
§H.1 INTRODUCTION Who first wrote down a stiffness or flexibility matrix? The question was posed in a 1995 paper [1]. The educated guess was “somebody working in the aircraft industry of Britain or Germany, in the late 1920s or early 1930s.” Since then the writer has examined reports and publications of that time. These trace the origins of Matrix Structural Analysis to the aeroelasticity group of the National Physics Laboratory (NPL) at Teddington, a town that has now become a suburb of greater London. The present paper is an expansion of the historical vignettes in Section 4 of [1]. It outlines the major steps in the evolution of MSA by highlighting the fundamental contributions of four individuals: Collar, Duncan, Argyris and Turner. These contributions are lumped into three milestones: Creation. Beginning in 1930 Collar and Duncan formulated discrete aeroelasticity in matrix form. The first two journal papers on the topic appeared in 1934-35 [2,3] and the first book, couthored with Frazer, in 1938 [4]. The representation and terminology for discrete dynamical systems is essentially that used today. Unification. In a series of journal articles appearing in 1954 and 1955 [5] Argyris presented a formal unification of Force and Displacement Methods using dual energy theorems. Although practical applications of the duality proved ephemeral, this work systematized the concept of assembly of structural system equations from elemental components. FEMinization. In 1959 Turner proposed [6] the Direct Stiffness Method (DSM) as an efficient and general computer implementation of the then embryonic, and as yet unnamed, Finite Element Method. H–2
H–3
§H.2 IDEALIZATION
Physical system
BACKGROUND AND TERMINOLOGY SOLUTION
DISCRETIZATION
Discrete model
Mathematical model
Discrete solution
Solution error Discretization + solution error Modeling + discretization + solution error VERIFICATION & VALIDATION
Figure 1. Flowchart of model-based simulation (MBS) by computer.
This technique, fully explained in a follow-up article [7], naturally encompassed structural and continuum models, as well as nonlinear, stability and dynamic simulations. By 1970 DSM had brought about the demise of the Classical Force Method (CFM), and become the dominant implementation in production-level FEM programs. These milestones help dividing MSA history into three periods. To enliven and focus the exposition these will be organized as three acts of a play, properly supplemented with a “matrix overture” prologue, two interludes and a closing epilogue. Here is the program: Prologue - Victorian Artifacts: 1858–1930. Act I - Gestation and Birth: 1930–1938. Interlude I - WWII Blackout: 1938–1947. Act II - The Matrix Forest: 1947–1956. Interlude II - Questions: 1956–1959. Act III - Answers: 1959–1970. Epilogue - Revisiting the Past: 1970-date. Act I, as well as most of the Prologue, takes place in the U.K. The following events feature a more international cast. §H.2 BACKGROUND AND TERMINOLOGY Before departing for the theater, this Section offers some general background and explains historical terminology. Readers familiar with the subject should skip to Section 3. The overall schematics of model-based simulation (MBS) by computer is flowcharted in Figure 1. For mechanical systems such as structures the Finite Element Method (FEM) is the most widely used discretization and solution technique. Historically the ancestor of FEM is MSA, as illustrated in Figure 2. The morphing of the MSA from the pre-computer era — as described for example in the first book [4] — into the first programmable computers took place, in wobbly gyrations, during the transition period herein called Act II. Following a confusing interlude, the young FEM begin to settle, during the early 1960s, into the configuration shown on the right of Figure 2. Its basic components have not changed since 1970. MSA and FEM stand on three legs: mathematical models, matrix formulation of the discrete equations, H–3
H–4
Appendix H: AN OUTLINE OF MSA HISTORY
Discrete Mathematical Models
Continuum Mathematical Models
MSA
FEM
Various Matrix Formulations
Human Computers
DSM Matrix Formulation
Programmable Digital Computers
Figure 2. Morphing of the pre-computer MSA (before 1950) into the present FEM. On the left “human computer” means computations under direct human control, possibly with the help of analog devices (slide rule) or digital devices (desk calculator). The FEM configuration shown on the right settled by the mid 1960s.
and computing tools to do the numerical work. Of the three legs the latter is the one that has undergone the most dramatic changes. The “human computers” of the 1930s and 1940s morphed by stages into programmable computers of analog and digital type. The matrix formulation moved like a pendulum. It began as a simple displacement method in Act I, reached bewildering complexity in Act II and went back to conceptual simplicity in Act III. Unidimensional structural models have changed little: a 1930 beam is still the same beam. The most noticeable advance is that pre-1955 MSA, following classical Lagrangian mechanics, tended to use spatially discrete energy forms from the start. The use of space-continuum forms as basis for multidimensional element derivation was pioneered by Argyris [5], successfully applied to triangular geometries by Turner, Clough, Martin and Topp [8], and finalized by Melosh [9] and Irons [10,11] with the precise statement of compatibility and completeness requirements for FEM. Matrix formulations for MSA and FEM have been traditionally classified by the choice of primary unknows. These are those solved for by the human or digital computer to determine the system state. In the Displacement Method (DM) these are physical or generalized displacements. In the Classical Force Method (CFM) these are amplitudes of redundant force (or stress) patterns. (The qualifier “classical” is important because there are other versions of the Force Method, which select for example stress function values or Lagrange multipliers as unknowns.) There are additional methods that involve combinations of displacements, forces and/or deformations as primary unknowns, but these have no practical importance in the pre-1970 period covered here. Appropriate mathematical names for the DM are range-space method or primal method. This means that the primary unknowns are the same type as the primary variables of the governing functional. Appropriate names for the CFM are null-space method, adjoint method, or dual method. This means that the primary unknowns are of the same type of the adjoint variables of the governing functional, which in structural mechanics are forces. These names are not used in the historical outline, but are useful in placing more recent developments, as well as nonstructural FEM applications, within a general framework. The terms Stiffness Method and Flexibility Method are more diffuse names for the Displacement and Force Methods, respectively. Generally speaking these apply when stiffness and flexibility matrices, respectively, are important part of the modeling and solution process. H–4
H–5
§H.4
ACT I - GESTATION AND BIRTH: 1930-1938
§H.3 PROLOG - VICTORIAN ARTIFACTS: 1858-1930 Matrices — or “determinants” as they were initially called — were invented in 1858 by Cayley at Cambridge, although Gibbs (the co-inventor, along with Heaviside, of vector calculus) claimed priority for the German mathematician Grassmann. Matrix algebra and matrix calculus were developed primarily in the U.K. and Germany. Its original use was to provide a compact language to support investigations in mathematical topics such as the theory of invariants and the solution of algebraic and differential equations. For a history of these early developments the monograph by Muir [12] is unsurpassed. Several comprehensive treatises in matrix algebra appeared in the late 1920s and early 1930s [13–15]. Compared to vector and tensor calculus, matrices had relatively few applications in science and technology before 1930. Heisenberg’s 1925 matrix version of quantum mechanics was a notable exception, although technically it involved infinite matrices. The situation began to change with the advent of electronic desk calculators, because matrix notation provided a convenient way to organize complex calculation sequences. Aeroelasticity was a natural application because the stability analysis is naturally posed in terms of determinants of matrices that depend on a speed parameter. The non-matrix formulation of Discrete Structural Mechanics can be traced back to the 1860s. By the early 1900s the essential developments were complete. A readable historical account is given by Timoshenko [16]. Interestingly enough, the term “matrix” never appears in this book. §H.4 ACT I - GESTATION AND BIRTH: 1930-1938 In the decade of World War I aircraft technology begin moving toward monoplanes. Biplanes disappeared by 1930. This evolution meant lower drag and faster speeds but also increased disposition to flutter. In the 1920s aeroelastic research began in an international scale. Pertinent developments at the National Physical Laboratory (NPL) are well chronicled in a 1978 historical review article by Collar [17], from which the following summary is extracted. §H.4.1 The Source Papers The aeroelastic work at the Aerodynamics Division of NPL was initiated in 1925 by R. A. Frazer. He was joined in the following year by W. J. Duncan. Two years later, in August 1928, they published a monograph on flutter [18], which came to be known as “The Flutter Bible” because of its completeness. It laid out the principles on which flutter investigations have been based since. In January 1930 A. R. Collar joined Frazer and Duncan to provide more help with theoretical investigations. Aeroelastic equations were tedious and error prone to work out in long hand. Here are Collar’s own words [17, page 17] on the motivation for introducing matrices: “Frazer had studied matrices as a branch of applied mathematics under Grace at Cambridge; and he recognized that the statement of, for example, a ternary flutter problem in terms of matrices was neat and compendious. He was, however, more concerned with formal manipulation and transformation to other coordinates than with numerical results. On the other hand, Duncan and I were in search of numerical results for the vibration characteristics of airscrew blades; and we recognized that we could only advance by breaking the blade into, say, 10 segments and treating it as having 10 degrees of freedom. This approach also was more conveniently formulated in matrix terms, and readily expressed numerically. Then we found that if we put an approximate mode into one side of the equation, we calculated a better approximation on the other; and the matrix iteration procedure was born. We published our method in two papers in Phil. Mag. [2,3]; the first, dealing with conservative systems,
H–5
Appendix H: AN OUTLINE OF MSA HISTORY
H–6
in 1934 and the second, treating damped systems, in 1935. By the time this had appeared, Duncan had gone to his Chair at Hull.”
The aforementioned papers appear to be the earliest journal publications of MSA. These are amazing documents: clean and to the point. They do not feel outdated. Familiar names appear: mass, flexibility, stiffness, and dynamical matrices. The matrix symbols used are [m], [ f ], [c] and [D] = [c]−1 [m] = [ f ][m], respectively, instead of the M, F, K and D in common use today. A general inertia matrix is called [a]. As befit the focus on dynamics, the displacement method is used. Point-mass displacement degrees of freedom are collected in a vector {x} and corresponding forces in vector {P}. These are called [q] and [Q], respectively, when translated to generalized coordinates. The notation was changed in the book [4] discussed below. In particular matrices are identified in [4] by capital letters without surrounding brackets, in more agreement with the modern style; for example mass, damping and stiffness are usually denoted by A, B and C, respectively. §H.4.2 The MSA Source Book Several papers on matrices followed, but apparently the traditional publication vehicles were not viewed as suitable for description of the new methods. At that stage Collar notes [17, page 18] that “Southwell [Sir Richard Southwell, the “father” of relaxation methods] suggested that the authors of the various papers should be asked to incorporate them into a book, and this was agreed. The result was the appearance in November 1938 of “Elementary Matrices” published by Cambridge University Press [4]; it was the first book to treat matrices as a branch of applied mathematics. It has been reprinted many times, and translated into several languages, and even now after nearly 40 years, stills sells in hundreds of copies a year — mostly paperback. The interesting thing is that the authors did not regard it as particularly good; it was the book we were instructed to write, rather than the one we would have liked to write.”
The writer has copies of the 1938 and 1963 printings. No changes other than minor fixes are apparent. Unlike the source papers [2,3] the book feels dated. The first 245 pages are spent on linear algebra and ODE-solution methods that are now standard part of engineering and science curricula. The numerical methods, oriented to desk calculators, are obsolete. That leaves the modeling and application examples, which are not coherently interweaved. No wonder that the authors were not happy about the book. They had followed Southwell’s “merge” suggestion too literally. Despite these flaws its direct and indirect influence during the next two decades was significant. Being first excuses imperfections. The book focuses on dynamics of a complete airplane and integrated components such as wings, rudders or ailerons. The concept of structural element is primitive: take a shaft or a cantilever and divide it into segments. The assembled mass, stiffness or flexibility is given directly. The source of damping is usually aerodynamic. There is no static stress analysis; pre-WWII aircraft were overdesigned for strength and typically failed by aerodynamic or propulsion effects. Readers are reminded that in aeroelastic analysis stiffness matrices are generally unsymmetric, being the sum of a a symmetric elastic stiffness and an unsymmetric aerodynamic stiffness. This clean decomposition does not hold for flexibility matrices because the inverse of a sum is not the sum of inverses. The treatment of [4] includes the now called load-dependent stiffness terms, which represent another first. On reading the survey articles by Collar [17,19] one cannot help being impressed by the lack of pretension. With Duncan he had created a tool for future generations of engineers to expand and improve upon. Yet he appears almost apologetic: “I will complete the matrix story as briefly as H–6
H–7
§H.6
ACT II - THE MATRIX FOREST: 1947-1956
possible” [17, page 17]. The NPL team members shared a common interest: to troubleshoot problems by understanding the physics, and viewed numerical methods simply as helpers. §H.5 INTERLUDE I - WWII BLACKOUT: 1938-1947 Interlude I is a “silent period” taken to extend from the book [4] to the first journal publication on the matrix Force Method for aircraft [20]. Aeroelastic research continued. New demands posed by high strength materials, higher speeds, combat maneuvers, and structural damage survival increased interest in stress analysis. For the beam-like skeletal configurations of the time, the traditional flexibility-based methods such as CFM were appropriate. Flexibilities were often measured experimentally by static load tests, and fitted into the calculations. Punched-card computers and relay-calculators were increasingly used, and analog devices relied upon to solve ODEs in guidance and ballistics. Precise accounts of MSA work in aerospace are difficult to trace because of publication restrictions. The blackout was followed by a 2-3 year hiatus until those restrictions were gradually lifted, R&D groups restaffed, and journal pipelines refilled. §H.6 ACT II - THE MATRIX FOREST: 1947-1956 As Act II starts MSA work is still mainly confined to the aerospace community. But the focus has shifted from dynamics to statics, and especially stress, buckling, fracture and fatigue analysis. Turbines, supersonic flight and rocket propulsion brought forth thermomechanical effects. The Comet disasters forced attention on stress concentration and crack propagation effects due to cyclic cabin pressurization. Failsafe design gained importance. In response to these multiple demands aircraft companies staffed specialized groups: stress, aerodynamics, aeroelasticity, propulsion, avionics, and so on. A multilevel management structure with well defined territories emerged. The transition illustrated in Figure 2 starts, driven by two of the legs supporting MSA: new computing resources and new mathematical models. The matrix formulation merely reacts. §H.6.1 Computers Become Machines The first electronic commercial computer: Univac I, manufactured by a division of Remington-Rand, appeared during summer 1951. The six initial machines were delivered to US government agencies [21]. It was joined in 1952 by the Univac 1103, a scientific-computation oriented machine built by ERA, a R-R acquisition. This was the first computer with a drum memory. T. J. Watson Sr., founder of IBM, had been once quoted as saying that six electronic computers would satisfy the needs of the planet. Turning around from that prediction, IBM launched the competing 701 model in 1953. Big aircraft companies began purchasing or leasing these expensive wonders by 1954. But this did not mean immediate access for everybody. The behemoths had to be programmed in machine or assembly code by specialists, who soon formed computer centers allocating and prioritizing cycles. By 1956 structural engineers were still likely to be using their slides rules, Marchants and punched card equipment. Only after the 1957 appearance of the first high level language (Fortran I, offered on the IBM 704) were engineers and scientists able (and allowed) to write their own programs. §H.6.2 The Matrix CFM Takes Center Stage In static analysis the non-matrix version of the Classical Force Method (CFM) had enjoyed a distinguished reputation since the source contributions by Maxwell, Mohr and Castigliano. The method provides directly the internal forces, which are of paramount interest in stress-driven design. It offers considerable scope of ingenuity to experienced structural engineers through clever selection of H–7
H–8
Appendix H: AN OUTLINE OF MSA HISTORY
θB θA
MA
wB
wA
VA A
B VB
MB
Integrate governing ODEs from A to B
wB θB VB MB
Transition = matrix
wA θA VA MA
Rearrange & part-invert
Remove RBMs, rearrange & part-invert
θA θB
=
Deformational Flexibility Matrix
MA
VA MA VB MB
=
Free-free Stiffness Matrix
Remove RBMs Invert
MA =
MB
MB
wA θA wB θB
???
Expand with RBMs
Deformational Stiffness Matrix
wA θA wB θB
=
Free-free Flexibility Matrix
VA MA VB MB
(Not believed to exist; general form discovered recently; see text)
θA θB
(RBMs = rigid body modes of the element; two for the example)
Invert
Figure 3. Transition, flexibility and stiffness matrices for unidimensional linear structural elements, such as the plane beam depicted here, can be obtained by integrating the governing differential equations, analytically or numerically, over the member to relate end forces and displacements. Clever things were done with this “method of lines” approach, such as including intermediate supports or elastic foundations.
redundant force systems. It was routinely taught to Aerospace, Civil and Mechanical Engineering students. Success in hand-computation dynamics depends on “a few good modes.” Likewise, the success of CFM depends crucially on the selection of good redundant force patterns. The structures of pre-1950 aircraft were a fairly regular lattice of ribs, spars and panels, forming beam-like configurations. If the panels are ignored, the selection of appropriate redundants was well understood. Panels were modeled conservatively as inplane shear-force carriers, circumventing the difficulties of two-dimensional elasticity. With some adjustments and experimental validations, sweptback wings of high aspect ratio were eventually fitted into these models. A matrix framework was found convenient to organize the calculations. The first journal article on the matrix CFM, which focused on sweptback wing analysis, is by Levy [20], followed by publications of Rand [22], Langefors [23], Wehle and Lansing [24] and Denke[25]. The development culminates in the article series of Argyris [5] discussed in Section 6.5. §H.6.3 The Delta Wing Challenge The Displacement Method (DM) continued to be used for vibration and aeroelastic analysis, although as noted above this was often done by groups separated from stress and buckling analysis. A new modeling challenge entered in the early 1950s: delta wing structures. This rekindled interest in stiffness methods. H–8
H–9
§H.6
ACT II - THE MATRIX FOREST: 1947-1956
The traditional approach to obtain flexibility and stiffness matrices of unidimensional structural members such as bars and shafts is illustrated in Figure 3. The governing differential equations are integrated, analytically or numerically, from one end to the other. The end quantities, grouping forces and displacements, are thereby connected by a transition matrix. Using simple algebraic manipulations three more matrices shown in Figure 3 can be obtained: deformational flexibility, deformational stiffness and free-free stiffness. This well known technique has the virtue of reducing the number of unknowns since the integration process can absorb structural details that are handled in the present FEM with multiple elements. Notably absent from the scheme of Figure 3 is the free-free flexibility. This was not believed to exist since it is the inverse of the free-free stiffness, which is singular. A general closed-form expression for this matrix as a Moore-Penrose generalized stiffness inverse was not found until recently [26,27]. Modeling delta wing configurations required two-dimensional panel elements of arbitrary geometry, of which the triangular shape, illustrated in Figure 4, is the simplest and most versatile. Efforts to follow the ODE-integration approach lead to failure. (One particularly bizarre proposal, for solving exactly the wrong problem, is mentioned for fun in the label of Figure 4.) This motivated efforts to construct the stiffness matrix of the panel directly. The first attempt in this direction is by Levy [28]; this was only partly successful but was able to illuminate the advantages of the stiffness approach. The article series by Argyris [5] contains the derivation of the 8 × 8 free-free stiffness of a flat rectangular panel using bilinear displacement interpolation in Cartesian coordinates. But that geometry was obviously inadequate to model delta wings. The landmark contribution of Turner, Clough, Martin and Topp [8] finally succeeded in directly deriving the stiffness of a triangular panel. Clough [29] observes that this paper represents the delayed publication of 1952-53 work at Boeing. It is recognized as one of the two sources of present FEM implementations, the second being the DSM discussed later. Because of the larger number of unknowns compared to CFM, competitive use of the DM in stress analysis had necessarily to wait until computers become sufficiently powerful to handle hundreds of simultaneous equations.
§H.6.4 Reduction Fosters Complexity
For efficient digital computation on present computers, data organization (in terms of fast access as well as exploitation of sparseness, vectorization and parallelism) is of primary concern whereas raw problem size, up to certain computer-dependent bounds, is secondary. But for hand calculations minimal problem size is a key aspect. Most humans cannot comfortably solve by hand linear systems of more than 5 or 6 unknowns by direct elimination methods, and 5–10 times that through problemoriented “relaxation” methods. The first-generation digital computers improved speed and reliability, but were memory strapped. For example the Univac I had 1000 45-bit words and the IBM 701, 2048 36-bit words. Clearly solving a full system of 100 equations was still a major challenge. It should come as no surprise that problem reduction techniques were paramount throughout this period, and exerted noticeable influence until the early 1970s. In static analysis reduction was achieved by elaborated functional groupings of static and kinematic variables. Most schemes of the time can be H–9
H–10
Appendix H: AN OUTLINE OF MSA HISTORY
fy3
uy3 (a) 3
(b) 3
ux3
(c)
fx3 fy2
uy2 2
uy1 1
ux2
1
ux1
2
fy1 fx1
fx2
3 tn1
tn2
ts1 2 ts2 1
ts3 tn3
Figure 4. Modeling delta wing configurations required panel elements of arbitrary geometry such as the triangles depicted here. The traditional ODE-based approach of Figure 3 was tried by some researchers who (seriously) proposed finding the corner displacements in (a) produced by the concentrated corner forces in (b) on a supported triangle from the elasticity equations solved by numerical integration! Bad news: those displacements are infinite. Interior fields assumptions were inevitable, but problems persisted. A linear inplane displacement field is naturally specified by corner displacements, whereas a constant membrane force field is naturally defined by edge tractions (c). Those quantities “live” on different places. The puzzle was first solved in [8] by lumping edge tractions to node forces on the way to the free-free stiffness matrix.
understood in terms of the following classification:     primary generalized forces
   secondary     primary
generalized displacements
   secondary
0006 0006 0006 0006
applied forces fa redundant forces y condensable forces fc = 0 support reactions fs applied displacements ua redundant displacements z
(H.1)
condensable displacements uc support conditions us = 0
Here applied forces are those acting with nonzero values, that is, the ones visibly drawn as arrows by an engineer or instructor. In reduction-oriented thinking zero forces on unloaded degrees of freedom are classified as condensable because they can be removed through static condensation techniques. Similarly, nonzero applied displacements were clearly differentiated from zero-displacements arising from support conditions because the latter can be thrown out while the former must be retained. Redundant displacements, which are the counterpart of redundant forces, have been given many names, among them “kinematically indeterminate displacements” and “kinematic deficiencies.” Matrix formulation evolved so that the unknowns were the force redundants y in the CFM and the displacement redundants z in the DM. Partitioning matrices in accordance to (H.1) fostered exuberant growth culminating in the matrix forest that characterizes works of this period. To a present day FEM programmer familiar with the DSM, the complexity of the matrix forest would strike as madness. The DSM master equations can be assembled without functional labels. Boundary conditions are applied on the fly by the solver. But the computing limitations of the time must be kept in mind to see the method in the madness. H–10
H–11
§H.6
ACT II - THE MATRIX FOREST: 1947-1956
§H.6.5 Two Paths Through the Forest A series of articles published by J. H. Argyris in four issues of Aircraft Engrg. during 1954 and 1955 collectively represents the second major milestone in MSA. In 1960 the articles were collected in a book, entitled “Energy Theorems and Structural Analysis” [5]. Part I, sub-entitled General Theory, reprints the four articles, whereas Part II, which covers additional material on thermal analysis and torsion, is co-authored by Argyris and Kelsey. Both authors are listed as affiliated with the Aerospace Department of the Imperial College at London. The dual objectives of the work, stated in the Preface, are “to generalize, extend and unify the fundamental energy principles of elastic structures” and “to describe in detail practical methods of analysis of complex structures — in particular for aeronautical applications.” The first objective succeeds well, and represents a key contribution toward the development of continuum-based models. Part I carefully merges classical contributions in energy and work methods with matrix methods of discrete structural systems. The coverage is methodical, with numerous illustrative examples. The exposition of the Force Method for wing structures reaches a level of detail unequaled for the time. The Displacement Method is then introduced by duality — called “analogy” in this work: “The analogy between the developments for the flexibilities and stiffnesses .. shows clearly that parallel to the analysis of structures with forces as unknowns there must be a corresponding theory with deformations as unknowns.”
This section credits Ostenfeld [30] with being the first to draw attention to the parallel development. The duality is exhibited in a striking Form in Table II, in which both methods are presented side by side with simply an exchange of symbols and appropriate rewording. The steps are based on the following decomposition of internal deformation states g and force patterns p: p = B0 fa + B1 y,
g = A0 ua + A1 z,
(H.2)
Here the Bi and Ai denote system equilibrium and compatibility matrices, respectively. The vector symbols on the right reflect a particular choice of the force-displacement decomposition (H.1), with kinematic deficiencies taken to be the condensable displacements: z ≡ uc . This unification exerted significant influence over the next decade, particularly on the European community. An excellent textbook exposition is that of Pestel and Leckie [31]. This book covers both paths, following Argyris’ framework, in Chapters 9 and 10, using 83 pages and about 200 equations. These chapters are highly recommended to understand the organization of numeric and symbolic hand computations in vogue at that time, but it is out of print. Still in print (by Dover) is the book by Przemieniecki [32], which describes the DM and CFM paths in two Chapters: 6 and 8. The DM coverage is strongly influenced, however, by the DSM; thus duality is only superficially used. §H.6.6 Dubious Duality A key application of the duality in [5] was to introduce the DM by analogy to the then better known CFM. Although done with good intentions this approach did not anticipate the direct development of continuum-based finite elements through stiffness methods. These can be formulated from the total potential energy principle via shape functions, a technique not fully developed until the mid 1960s. The side by side presentation of Table II of [5] tried to show that CFM and DM were going through exactly the same sequence of steps. Some engineers, eventually able to write Fortran programs, concluded that the methods had similar capabilities and selecting one or the other was a matter of H–11
Appendix H: AN OUTLINE OF MSA HISTORY
H–12
taste. (Most structures groups, upholding tradition, opted for the CFM.) But the few engineers who tried implementing both noticed a big difference. And that was before the DSM, which has no dual counterpart under the decomposition (H.2), appeared. The paradox is explained in Section 4 of [1]. It is also noted there that (H.2) is not a particularly useful state decomposition. A better choice is studied in Section 2 of that paper; that one permits all known methods of Classical MSA, including the DSM, to be derived for skeletal structures as well as for a subset of continuum models. §H.7 INTERLUDE II - QUESTIONS: 1956-1959 Interlude I was a silent period dominated by the war blackout. Interlude II is more vocal: a time of questions. An array of methods, models, tools and applications is now on the table, and growing. Solid-state computers, Fortran, ICBMs, the first satellites. So many options. Stiffness or flexibility? Forces or displacements? Do transition matrix methods have a future? Is the CFM-DM duality a precursor to general-purpose programs that will simulate everything? Will engineers be allowed to write those programs? As convenient milestone this outline takes 1959, the year of the first DSM paper, as the beginning of Act III. Arguments and counter-arguments raised by the foregoing questions will linger, however, for two more decades into diminishing circles of the aerospace community. §H.8 ACT III - ANSWERS: 1959-1970 The curtain of Act III lifts in Aachen, Germany. On 6 November 1959, M. J. Turner, head of the Structural Dynamics Unit at Boeing and an expert in aeroelasticity, presented the first paper on the Direct Stiffness Method to an AGARD Structures and Materials Panel meeting [6]. (AGARD is NATO’s Advisory Group for Aeronautical Research and Development, which had sponsored workshops and lectureships since 1952. Bound proceedings or reports are called AGARDographs.) §H.8.1 A Path Outside the Forest No written record of [6] seem to exist. Nonetheless it must have produced a strong impression since published contributions to the next (1962) panel meeting kept referring to it. By 1960 the method had been applied to nonlinear problems [33] using incremental techniques. In July 1962 Turner, Martin and Weikel presented an expanded version of the 1959 paper, which appeared in an AGARDograph volume published by Pergamon in 1964 [7]. Characteristic of Turner’s style, the Introduction goes directly to the point: “In a paper presented at the 1959 meeting of the AGARD Structures and Material Panel in Aachen, the essential features of a system for numerical analysis of structures, termed the direct-stiffness method, were described. The characteristic feature of this particular version of the displacement method is the assembly procedure, whereby the stiffness matrix for a composite structure is generated by direct addition of matrices associated with the elements of the structure.”
The DSM is explained in six text lines and three equations: “For an individual element e the generalized nodal force increments {0006X e } required to maintain a set of nodal displacement increments {0006u} are given by a matrix equation {0006X e } = K e {0006u}
H–12
(H.3)
H–13
§H.8
ACT III - ANSWERS: 1959-1970
in which K e denotes the stiffness matrix of the individual element. Resultant nodal force increments acting on the complete structure are {0006X } =
0007
{0006X e } = K {0006u}
(H.4)
wherein K , the stiffness of the complete structure, is given by the summation K =
0007
Ke
(H.5)
which provides the basis for the matrix assembly procedure noted earlier.”
Knowledgeable readers will note a notational glitch. For (H.3)-(H.5) to be correct matrix equations, K e must be an element stiffness fully expanded to global (in that paper: “basic reference”) coordinates, a step that is computationally unnecessary. A more suggestive notation used in present DSM expositions is K = (L e )T K e L e , in which L e are Boolean localization matrices. Note also the use of 0006 in front of u and X and their identification as “increments.” This simplifies the extension to nonlinear analysis, as outlined in the next paragraph: “For the solution of linear problems involving small deflections of a structure at constant uniform temperature which is initially stress-free in the absence of external loads, the matrices K e are defined in terms of initial geometry and elastic properties of the materials comprising the elements; they remain unchanged throughout the analysis. Problems involving nonuniform heating of redundant structures and/or large deflections are solved in a sequence of linearized steps. Stiffness matrices are revised at the beginning of each step to account for charges in internal loads, temperatures and geometric configurations.”
Next are given some computer implementation details, including the first ever mention of user-defined elements: “Stiffness matrices are generally derived in local reference systems associated with the elements (as prescribed by a set of subroutines) and then transformed to the basic reference system. It is essential that the basic program be able to acommodate arbitrary additions to the collection of subroutines as new elements are encountered. Associated with these are a set of subroutines for generation of stress matrices S e relating matrices of stress components σ e in the local reference system of nodal displacements: ¯ (H.6) {σ e } = S e {u} The vector {u} ¯ denotes the resultant displacements relative to a local reference system which is attached to the element. .. Provision should also be made for the introduction of numerical stiffness matrices directly into the program. This permits the utilization and evaluation of new element representations which have not yet been programmed. It also provides a convenient mechanism for introducing local structural modifications into the analysis.”
The assembly rule (H.3)-(H.5) is insensitive to element type. It work the same way for a 2-node bar, or a 64-node hexahedron. To do dynamics and vibration one adds mass and damping terms. To do buckling one adds a geometric stiffness and solves the stability eigenproblem, a technique first explained in [33]. To do nonlinear analysis one modifies the stiffness in each incremental step. To apply multipoint constraints the paper [7] advocates a master-slave reduction method. Some computational aspects are missing from this paper, notably the treatment of simple displacement boundary conditions, and the use of sparse matrix assembly and solution techniques. The latter were first addressed in Wilson’s thesis work [34,35]. H–13
Appendix H: AN OUTLINE OF MSA HISTORY
H–14
§H.8.2 The Fire Spreads DSM is a paragon of elegance and simplicity. The writer is able to teach the essentials of the method in three lectures to graduate and undergraduate students alike. Through this path the old MSA and the young FEM achieved smooth confluence. The matrix formulation returned to the crispness of the source papers [2,3]. A widely referenced MSA correlation study by Gallagher [36] helped dissemination. Computers of the early 1960s were finally able to solve hundreds of equations. In an ideal world, structural engineers should have quickly razed the forest and embraced DSM. It did not happen that way. The world of aerospace structures split. DSM advanced first by word of mouth. Among the aerospace companies, only Boeing and Bell (influenced by Turner and Gallagher, respectively) had made major investments in DSM by 1965. Among academia the Civil Engineering Department at Berkeley become a DSM evangelist through Clough, who made his students — including the writer — use DSM in their thesis work. These codes were freely disseminated into the non-aerospace world since 1963. Martin established similar traditions at Washington University, and Zienkiewicz, influenced by Clough, at Swansea. The first textbook on FEM [37], which appeared in 1967, makes no mention of force methods. By then the application to non-structural field problems (thermal, fluids, electromagnetics, ..) had begun, and again the DSM scaled well into the brave new world. §H.8.3 The Final Test Legacy CFM codes continued, however, to be used at many aerospace companies. The split reminds one of Einstein’s answer when he was asked about the reaction of the old-guard school to the new physics: “we did not convince them; we outlived them.” Structural engineers hired in the 1940s and 1950s were often in managerial positions in the 1960s. They were set in their ways. How can duality fail? All that is needed are algorithms for having the computer select good redundants automatically. Substantial effort was spent in those “structural cutters” during the 1960s [32,38]. That tenacity was eventually put to a severe test. The 1965 NASA request-for-proposal to build the NASTRAN finite element system called for the simultaneous development of Displacement and Force versions [39]. Each version was supposed to have identical modeling and solution capabilities, including dynamics and buckling. Two separate contracts, to MSC and Martin, were awarded accordingly. Eventually the development of the Force version was cancelled in 1969. The following year may be taken as closing the transition depicted in Figure 2, and as marking the end of the Force Method as a serious contender for general-purpose FEM programs.
H–14
H–15
§H.10 CONCLUDING REMARKS
§H.9 EPILOGUE - REVISITING THE PAST: 1970-DATE Has MSA, now under the wider umbrella of FEM, attained a final form? This seems the case for general-purpose FEM programs, which by now are truly “1960 heritage” codes. Resurrection of the CFM for special uses, such as optimization, was the subject of a speculative technical note by the writer [40]. This was motivated by efforts of numerical analysts to develop sparse null-space methods [41–45]. That research appears to have been abandoned by 1990. Section 2 of [26] elaborates on why, barring unexpected breakthroughs, a resurrection of the CFM is unlikely. A more modest revival involves the use of non-CFM flexibility methods for multilevel analysis. The structure is partitioned into subdomains or substructures, each of which is processed by DSM; but the subdomains are connected by Lagrange multipliers that physically represent node forces. A key driving application is massively parallel processing in which subdomains are mapped on distributedmemory processors and the force-based interface subproblem solved iteratively by FETI methods [46]. Another set of applications include inverse problems such as system identification and damage detection. Pertinent references and a historical sketch may be found in a recent article [47] that presents a hybrid variational formulation for this combined approach. The true duality for structural mechanics is now known to involve displacements and stress functions, rather than displacements and forces. This was discovered by Fraeijs de Veubeke in the 1970s [48]. Although extendible beyond structures, the potential of this idea remains largely unexplored. §H.10 CONCLUDING REMARKS The patient reader who has reached this final section may have noticed that this essay is a critical overview of MSA history, rather than a recital of events. It reflects personal interpretations and opinions. There is no attempt at completeness. Only what are regarded as major milestones are covered in some detail. Furthermore there is only spotty coverage of the history of FEM itself as well as its computer implementation; this is the topic of an article under preparation for Applied Mechanics Reviews. This outline can be hopefully instructive in two respects. First, matrix methods now in disfavor may come back in response to new circumstances. An example is the resurgence of flexibility methods in massively parallel processing. A general awareness of the older literature helps. Second, the sweeping victory of DSM over the befuddling complexity of the “matrix forest” period illustrates the virtue of Occam’s proscription against multiplying entities: when in doubt chose simplicity. This dictum is relevant to the present confused state of computational mechanics. Acknowledgements The present work has been supported by the National Science Foundation under award ECS-9725504. Thanks are due to the librarians of the Royal Aeronautical Society at London for facilitating access to archival copies of pre-WWII reports and papers. Feedback suggestions from early draft reviewers will be acknowledged in the final version. References [1]
C. A. Felippa, Parametrized unification of matrix structural analysis: classical formulation and d-connected elements, Finite Elements Anal. Des., 21, pp. 45–74, 1995.
[2]
W. J. Duncan and A. R. Collar, A method for the solution of oscillations problems by matrices, Phil. Mag., Series 7, 17, pp. 865, 1934.
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[3]
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R. A. Frazer, W. J. Duncan and A. R. Collar, Elementary Matrices, and some Applications to Dynamics and Differential Equations, Cambridge Univ. Press, 1st ed. 1938, 7th (paperback) printing 1963.
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J. H. Argyris and S. Kelsey, Energy Theorems and Structural Analysis, Butterworths, London, 1960; Part I reprinted from Aircraft Engrg. 26, Oct-Nov 1954 and 27, April-May 1955.
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M. J. Turner, The direct stiffness method of structural analysis, Structural and Materials Panel Paper, AGARD Meeting, Aachen, Germany, 1959.
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M. J. Turner, H. C. Martin and R. C. Weikel, Further development and applications of the stiffness method, AGARD Structures and Materials Panel, Paris, France, July 1962, in AGARDograph 72: Matrix Methods of Structural Analysis, ed. by B. M. Fraeijs de Veubeke, Pergamon Press, Oxford, pp. 203–266, 1964.
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Appendix R: REFERENCES (IN PROGRESS)
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