The finite element method (FEM) is the dominant tool for numerical analysis in engineering, yet many engineers apply it

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*Table of contents : INTRODUCTIONMATHEMATICAL PRELIMINARIESMatrix AlgebraVectorsSecond-Order TensorsCalculusNewton`s MethodKinematics of MotionProblemsONE-DIMENSIONAL PROBLEMSThe Weak FormFinite Element ApproximationsPlugging in the Trial and Test FunctionsAlgorithm for Matrix AssemblyOne-Dimensional ElasticityProblemsLINEARIZED THEORY OF ELASTICITYCauchy`s LawPrincipal StressesEquilibrium EquationSmall-Strain TensorHooke`s LawAxisymmetric ProblemsWeak Form of the Equilibrium EquationProblemsSTEADY-STATE HEAT CONDUCTIONDerivation of the Steady-State Heat EquationFourier`s LawBoundary ConditionsWeak Form of the Steady-State Heat EquationProblemsCONTINUUM FINITE ELEMENTSThree-Node TriangleDevelopment of an Arbitrary QuadrilateralFour-Node TetrahedronEight-Node BrickElement Matrices and VectorsGauss QuadratureBending of a Cantilever BeamAnalysis of a Plate with HoleThermal Stress Analysis of a Composite CylinderProblemsSTRUCTURAL FINITE ELEMENTSSpace TrussEuler-Bernoulli BeamsMindlin-Reissner Plate TheoryDeflection of a Clamped PlateProblemsLINEAR TRANSIENT ANALYSISDerivation of the Equation of MotionSemi-Discrete Equations of MotionCentral Difference MethodTrapezoidal RuleUnsteady Heat ConductionProblemsSMALL-STRAIN PLASTICITYBasic ConceptsYield ConditionFlow and Hardening RulesDerivation of the Elastoplastic TangentFinite Element ImplementationOne-Dimensional Elastoplastic Deformation of a BarElastoplastic Analysis of a Thick-Walled CylinderProblemsTREATMENT OF GEOMETRIC NONLINEARITIESLarge-Deformation KinematicsWeak Form in the Original ConfigurationLinearization of the Weak FormSnap-Through Buckling of a Truss StructureUniaxial Tensile Test of a Rubber Dog-Bone SpecimenProblemsBIBLIOGRAPHYINDEX*

Finite Element Method Applications in Solids,

Structures, and Heat Transfer

MECHANICAL ENGINEERING A Series of Textbooks and Reference Books Founding Editor L. L. Faulkner Columbus Division, Battelle Memorial Institute

and Department of Mechanical Engineering

The Ohio State University

Columbus, Ohio

1. 2. 3. 4. 5. 6. 7. 8.

9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

Spring Designer’s Handbook, Harold Carlson Computer-Aided Graphics and Design, Daniel L. Ryan Lubrication Fundamentals, J. George Wills Solar Engineering for Domestic Buildings, William A. Himmelman Applied Engineering Mechanics: Statics and Dynamics, G. Boothroyd and C. Poli Centrifugal Pump Clinic, Igor J. Karassik Computer-Aided Kinetics for Machine Design, Daniel L. Ryan Plastics Products Design Handbook, Part A: Materials and Components; Part B: Processes and Design for Processes, edited by Edward Miller Turbomachinery: Basic Theory and Applications, Earl Logan, Jr. Vibrations of Shells and Plates, Werner Soedel Flat and Corrugated Diaphragm Design Handbook, Mario Di Giovanni Practical Stress Analysis in Engineering Design, Alexander Blake An Introduction to the Design and Behavior of Bolted Joints, John H. Bickford Optimal Engineering Design: Principles and Applications, James N. Siddall Spring Manufacturing Handbook, Harold Carlson Industrial Noise Control: Fundamentals and Applications, edited by Lewis H. Bell Gears and Their Vibration: A Basic Approach to Understanding Gear Noise, J. Derek Smith Chains for Power Transmission and Material Handling: Design and Applications Handbook, American Chain Association Corrosion and Corrosion Protection Handbook, edited by Philip A. Schweitzer Gear Drive Systems: Design and Application, Peter Lynwander Controlling In-Plant Airborne Contaminants: Systems Design and Calculations, John D. Constance CAD/CAM Systems Planning and Implementation, Charles S. Knox Probabilistic Engineering Design: Principles and Applications, James N. Siddall

24. 25. 26. 27. 28. 29. 30. 31. 32. 33.

34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52.

53. 54.

Traction Drives: Selection and Application, Frederick W. Heilich III and Eugene E. Shube Finite Element Methods: An Introduction, Ronald L. Huston and Chris E. Passerello Mechanical Fastening of Plastics: An Engineering Handbook, Brayton Lincoln, Kenneth J. Gomes, and James F. Braden Lubrication in Practice: Second Edition, edited by W. S. Robertson Principles of Automated Drafting, Daniel L. Ryan Practical Seal Design, edited by Leonard J. Martini Engineering Documentation for CAD/CAM Applications, Charles S. Knox Design Dimensioning with Computer Graphics Applications, Jerome C. Lange Mechanism Analysis: Simplified Graphical and Analytical Techniques, Lyndon O. Barton CAD/CAM Systems: Justification, Implementation, Productivity Measurement, Edward J. Preston, George W. Crawford, and Mark E. Coticchia Steam Plant Calculations Manual, V. Ganapathy Design Assurance for Engineers and Managers, John A. Burgess Heat Transfer Fluids and Systems for Process and Energy Applications, Jasbir Singh Potential Flows: Computer Graphic Solutions, Robert H. Kirchhoff Computer-Aided Graphics and Design: Second Edition, Daniel L. Ryan Electronically Controlled Proportional Valves: Selection and Application, Michael J. Tonyan, edited by Tobi Goldoftas Pressure Gauge Handbook, AMETEK, U.S. Gauge Division, edited by Philip W. Harland Fabric Filtration for Combustion Sources: Fundamentals and Basic Technology, R. P. Donovan Design of Mechanical Joints, Alexander Blake CAD/CAM Dictionary, Edward J. Preston, George W. Crawford, and Mark E. Coticchia Machinery Adhesives for Locking, Retaining, and Sealing, Girard S. Haviland Couplings and Joints: Design, Selection, and Application, Jon R. Mancuso Shaft Alignment Handbook, John Piotrowski BASIC Programs for Steam Plant Engineers: Boilers, Combustion, Fluid Flow, and Heat Transfer, V. Ganapathy Solving Mechanical Design Problems with Computer Graphics, Jerome C. Lange Plastics Gearing: Selection and Application, Clifford E. Adams Clutches and Brakes: Design and Selection, William C. Orthwein Transducers in Mechanical and Electronic Design, Harry L. Trietley Metallurgical Applications of Shock-Wave and High-Strain-Rate Phenomena, edited by Lawrence E. Murr, Karl P. Staudhammer, and Marc A. Meyers Magnesium Products Design, Robert S. Busk How to Integrate CAD/CAM Systems: Management and Technology, William D. Engelke

55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83.

Cam Design and Manufacture: Second Edition; with cam design software for the IBM PC and compatibles, disk included, Preben W. Jensen Solid-State AC Motor Controls: Selection and Application, Sylvester Campbell Fundamentals of Robotics, David D. Ardayfio Belt Selection and Application for Engineers, edited by Wallace D. Erickson Developing Three-Dimensional CAD Software with the IBM PC, C. Stan Wei Organizing Data for CIM Applications, Charles S. Knox, with contributions by Thomas C. Boos, Ross S. Culverhouse, and Paul F. Muchnicki Computer-Aided Simulation in Railway Dynamics, by Rao V. Dukkipati and Joseph R. Amyot Fiber-Reinforced Composites: Materials, Manufacturing, and Design, P. K. Mallick Photoelectric Sensors and Controls: Selection and Application, Scott M. Juds Finite Element Analysis with Personal Computers, Edward R. Champion, Jr. and J. Michael Ensminger Ultrasonics: Fundamentals, Technology, Applications: Second Edition, Revised and Expanded, Dale Ensminger Applied Finite Element Modeling: Practical Problem Solving for Engineers, Jeffrey M. Steele Measurement and Instrumentation in Engineering: Principles and Basic Laboratory Experiments, Francis S. Tse and Ivan E. Morse Centrifugal Pump Clinic: Second Edition, Revised and Expanded, Igor J. Karassik Practical Stress Analysis in Engineering Design: Second Edition, Revised and Expanded, Alexander Blake An Introduction to the Design and Behavior of Bolted Joints: Second Edition, Revised and Expanded, John H. Bickford High Vacuum Technology: A Practical Guide, Marsbed H. Hablanian Pressure Sensors: Selection and Application, Duane Tandeske Zinc Handbook: Properties, Processing, and Use in Design, Frank Porter Thermal Fatigue of Metals, Andrzej Weronski and Tadeusz Hejwowski Classical and Modern Mechanisms for Engineers and Inventors, Preben W. Jensen Handbook of Electronic Package Design, edited by Michael Pecht Shock-Wave and High-Strain-Rate Phenomena in Materials, edited by Marc A. Meyers, Lawrence E. Murr, and Karl P. Staudhammer Industrial Refrigeration: Principles, Design and Applications, P. C. Koelet Applied Combustion, Eugene L. Keating Engine Oils and Automotive Lubrication, edited by Wilfried J. Bartz Mechanism Analysis: Simplified and Graphical Techniques, Second Edition, Revised and Expanded, Lyndon O. Barton Fundamental Fluid Mechanics for the Practicing Engineer, James W. Murdock Fiber-Reinforced Composites: Materials, Manufacturing, and Design, Second Edition, Revised and Expanded, P. K. Mallick

84. 85. 86. 87. 88. 89. 90.

91. 92. 93. 94. 95. 96.

97. 98. 99. 100. 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 111. 112.

Numerical Methods for Engineering Applications, Edward R. Champion, Jr. Turbomachinery: Basic Theory and Applications, Second Edition, Revised and Expanded, Earl Logan, Jr. Vibrations of Shells and Plates: Second Edition, Revised and Expanded, Werner Soedel Steam Plant Calculations Manual: Second Edition, Revised and Expanded, V. Ganapathy Industrial Noise Control: Fundamentals and Applications, Second Edition, Revised and Expanded, Lewis H. Bell and Douglas H. Bell Finite Elements: Their Design and Performance, Richard H. MacNeal Mechanical Properties of Polymers and Composites: Second Edition, Revised and Expanded, Lawrence E. Nielsen and Robert F. Landel Mechanical Wear Prediction and Prevention, Raymond G. Bayer Mechanical Power Transmission Components, edited by David W. South and Jon R. Mancuso Handbook of Turbomachinery, edited by Earl Logan, Jr. Engineering Documentation Control Practices and Procedures, Ray E. Monahan Refractory Linings Thermomechanical Design and Applications, Charles A. Schacht Geometric Dimensioning and Tolerancing: Applications and Techniques for Use in Design, Manufacturing, and Inspection, James D. Meadows An Introduction to the Design and Behavior of Bolted Joints: Third Edition, Revised and Expanded, John H. Bickford Shaft Alignment Handbook: Second Edition, Revised and Expanded, John Piotrowski Computer-Aided Design of Polymer-Matrix Composite Structures, edited by Suong Van Hoa Friction Science and Technology, Peter J. Blau Introduction to Plastics and Composites: Mechanical Properties and Engineering Applications, Edward Miller Practical Fracture Mechanics in Design, Alexander Blake Pump Characteristics and Applications, Michael W. Volk Optical Principles and Technology for Engineers, James E. Stewart Optimizing the Shape of Mechanical Elements and Structures, A. A. Seireg and Jorge Rodriguez Kinematics and Dynamics of Machinery, Vladimír Stejskal and Michael Valásek Shaft Seals for Dynamic Applications, Les Horve Reliability-Based Mechanical Design, edited by Thomas A. Cruse Mechanical Fastening, Joining, and Assembly, James A. Speck Turbomachinery Fluid Dynamics and Heat Transfer, edited by Chunill Hah High-Vacuum Technology: A Practical Guide, Second Edition, Revised and Expanded, Marsbed H. Hablanian Geometric Dimensioning and Tolerancing: Workbook and Answerbook, James D. Meadows

113. Handbook of Materials Selection for Engineering Applications, edited by G. T. Murray 114. Handbook of Thermoplastic Piping System Design, Thomas Sixsmith and Reinhard Hanselka 115. Practical Guide to Finite Elements: A Solid Mechanics Approach, Steven M. Lepi 116. Applied Computational Fluid Dynamics, edited by Vijay K. Garg 117. Fluid Sealing Technology, Heinz K. Muller and Bernard S. Nau 118. Friction and Lubrication in Mechanical Design, A. A. Seireg 119. Influence Functions and Matrices, Yuri A. Melnikov 120. Mechanical Analysis of Electronic Packaging Systems, Stephen A. McKeown 121. Couplings and Joints: Design, Selection, and Application, Second Edition, Revised and Expanded, Jon R. Mancuso 122. Thermodynamics: Processes and Applications, Earl Logan, Jr. 123. Gear Noise and Vibration, J. Derek Smith 124. Practical Fluid Mechanics for Engineering Applications, John J. Bloomer 125. Handbook of Hydraulic Fluid Technology, edited by George E. Totten 126. Heat Exchanger Design Handbook, T. Kuppan 127. Designing for Product Sound Quality, Richard H. Lyon 128. Probability Applications in Mechanical Design, Franklin E. Fisher and Joy R. Fisher 129. Nickel Alloys, edited by Ulrich Heubner 130. Rotating Machinery Vibration: Problem Analysis and Troubleshooting, Maurice L. Adams, Jr. 131. Formulas for Dynamic Analysis, Ronald L. Huston and C. Q. Liu 132. Handbook of Machinery Dynamics, Lynn L. Faulkner and Earl Logan, Jr. 133. Rapid Prototyping Technology: Selection and Application, Kenneth G. Cooper 134. Reciprocating Machinery Dynamics: Design and Analysis, Abdulla S. Rangwala 135. Maintenance Excellence: Optimizing Equipment Life-Cycle Decisions, edited by John D. Campbell and Andrew K. S. Jardine 136. Practical Guide to Industrial Boiler Systems, Ralph L. Vandagriff 137. Lubrication Fundamentals: Second Edition, Revised and Expanded, D. M. Pirro and A. A. Wessol 138. Mechanical Life Cycle Handbook: Good Environmental Design and Manufacturing, edited by Mahendra S. Hundal 139. Micromachining of Engineering Materials, edited by Joseph McGeough 140. Control Strategies for Dynamic Systems: Design and Implementation, John H. Lumkes, Jr. 141. Practical Guide to Pressure Vessel Manufacturing, Sunil Pullarcot 142. Nondestructive Evaluation: Theory, Techniques, and Applications, edited by Peter J. Shull 143. Diesel Engine Engineering: Thermodynamics, Dynamics, Design, and Control, Andrei Makartchouk 144. Handbook of Machine Tool Analysis, Ioan D. Marinescu, Constantin Ispas, and Dan Boboc

145. Implementing Concurrent Engineering in Small Companies, Susan Carlson Skalak 146. Practical Guide to the Packaging of Electronics: Thermal and Mechanical Design and Analysis, Ali Jamnia 147. Bearing Design in Machinery: Engineering Tribology and Lubrication, Avraham Harnoy 148. Mechanical Reliability Improvement: Probability and Statistics for Experimental Testing, R. E. Little 149. Industrial Boilers and Heat Recovery Steam Generators: Design, Applications, and Calculations, V. Ganapathy 150. The CAD Guidebook: A Basic Manual for Understanding and Improving Computer-Aided Design, Stephen J. Schoonmaker 151. Industrial Noise Control and Acoustics, Randall F. Barron 152. Mechanical Properties of Engineered Materials, Wolé Soboyejo 153. Reliability Verification, Testing, and Analysis in Engineering Design, Gary S. Wasserman 154. Fundamental Mechanics of Fluids: Third Edition, I. G. Currie 155. Intermediate Heat Transfer, Kau-Fui Vincent Wong 156. HVAC Water Chillers and Cooling Towers: Fundamentals, Application, and Operation, Herbert W. Stanford III 157. Gear Noise and Vibration: Second Edition, Revised and Expanded, J. Derek Smith 158. Handbook of Turbomachinery: Second Edition, Revised and Expanded, edited by Earl Logan, Jr. and Ramendra Roy 159. Piping and Pipeline Engineering: Design, Construction, Maintenance, Integrity, and Repair, George A. Antaki 160. Turbomachinery: Design and Theory, Rama S. R. Gorla and Aijaz Ahmed Khan 161. Target Costing: Market-Driven Product Design, M. Bradford Clifton, Henry M. B. Bird, Robert E. Albano, and Wesley P. Townsend 162. Fluidized Bed Combustion, Simeon N. Oka 163. Theory of Dimensioning: An Introduction to Parameterizing Geometric Models, Vijay Srinivasan 164. Handbook of Mechanical Alloy Design, edited by George E. Totten, Lin Xie, and Kiyoshi Funatani 165. Structural Analysis of Polymeric Composite Materials, Mark E. Tuttle 166. Modeling and Simulation for Material Selection and Mechanical Design, edited by George E. Totten, Lin Xie, and Kiyoshi Funatani 167. Handbook of Pneumatic Conveying Engineering, David Mills, Mark G. Jones, and Vijay K. Agarwal 168. Clutches and Brakes: Design and Selection, Second Edition, William C. Orthwein 169. Fundamentals of Fluid Film Lubrication: Second Edition, Bernard J. Hamrock, Steven R. Schmid, and Bo O. Jacobson 170. Handbook of Lead-Free Solder Technology for Microelectronic Assemblies, edited by Karl J. Puttlitz and Kathleen A. Stalter 171. Vehicle Stability, Dean Karnopp 172. Mechanical Wear Fundamentals and Testing: Second Edition, Revised and Expanded, Raymond G. Bayer 173. Liquid Pipeline Hydraulics, E. Shashi Menon

174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. 185. 186. 187. 188. 189. 190. 191. 192. 193. 194. 195.

196. 197. 198.

Solid Fuels Combustion and Gasification, Marcio L. de Souza-Santos Mechanical Tolerance Stackup and Analysis, Bryan R. Fischer Engineering Design for Wear, Raymond G. Bayer Vibrations of Shells and Plates: Third Edition, Revised and Expanded, Werner Soedel Refractories Handbook, edited by Charles A. Schacht Practical Engineering Failure Analysis, Hani M. Tawancy, Anwar Ul-Hamid, and Nureddin M. Abbas Mechanical Alloying and Milling, C. Suryanarayana Mechanical Vibration: Analysis, Uncertainties, and Control, Second Edition, Revised and Expanded, Haym Benaroya Design of Automatic Machinery, Stephen J. Derby Practical Fracture Mechanics in Design: Second Edition, Revised and Expanded, Arun Shukla Practical Guide to Designed Experiments, Paul D. Funkenbusch Gigacycle Fatigue in Mechanical Practive, Claude Bathias and Paul C. Paris Selection of Engineering Materials and Adhesives, Lawrence W. Fisher Boundary Methods: Elements, Contours, and Nodes, Subrata Mukherjee and Yu Xie Mukherjee Rotordynamics, Agnieszka (Agnes) Muszn´yska Pump Characteristics and Applications: Second Edition, Michael W. Volk Reliability Engineering: Probability Models and Maintenance Methods, Joel A. Nachlas Industrial Heating: Principles, Techniques, Materials, Applications, and Design, Yeshvant V. Deshmukh Micro Electro Mechanical System Design, James J. Allen Probability Models in Engineering and Science, Haym Benaroya and Seon Han Damage Mechanics, George Z. Voyiadjis and Peter I. Kattan Standard Handbook of Chains: Chains for Power Transmission and Material Handling, Second Edition, American Chain Association and John L. Wright, Technical Consultant Standards for Engineering Design and Manufacturing, Wasim Ahmed Khan and Abdul Raouf S.I. Maintenance, Replacement, and Reliability: Theory and Applications, Andrew K. S. Jardine and Albert H. C. Tsang Finite Element Method: Applications in Solids, Structures, and Heat Transfer, Michael R. Gosz

Finite Element Method Applications in Solids, Structures, and Heat Transfer

Michael R. Gosz Mechanical, Materials, and Aerospace Engineering Department Illinois Institute of Technology Chicago, Illinois

Boca Raton London New York

A CRC title, part of the Taylor & Francis imprint, a member of the Taylor & Francis Group, the academic division of T&F Informa plc.

Published in 2006 by CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2006 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-10: 0-8493-3407-1 (Hardcover) International Standard Book Number-13: 978-0-8493-3407-8 (Hardcover) Library of Congress Card Number 2005053109 This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

Library of Congress Cataloging-in-Publication Data Gosz, Michael. Finite element method : applications in solids, structures, and heat transfer / by Michael Gosz. p. cm.

Includes bibliographical references and index.

ISBN 0-8493-3407-1 (alk. paper)

1. Finite element method. I. Title. TA347F5G67 2005 620'001'51825--dc22

2005053109

Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com Taylor & Francis Group is the Academic Division of Informa plc.

and the CRC Press Web site at http://www.crcpress.com

To my wonderful children Madeline and David, and to my lovely wife Mary Lee, who makes each day an exciting adventure.

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Preface

Tell me and I’ll forget. Show me, and I may not remember. Involve me, and I’ll understand. —Native American Saying Learning the ﬁnite element method can be a real mathematical challenge for both students and practicing engineers who have been in the workplace for longer than two years. The author believes, however, that anyone who has successfully completed the ﬁrst two years of an undergraduate program in engineering or the physical sciences has the potential to understand and apply the concepts of the ﬁnite element method to real-world problems. This book covers topics in linear, linear dynamic, and nonlinear ﬁnite ele ment procedures. It provides a thorough treatment written in simple language of the mathematics required to fully understand all of these ﬁnite element topics. This book is designed to be a learning aid for practicing engineers and students who wish to understand the fundamentals of the ﬁnite element method and apply it to practical problems in industry. It is not intended to be a comprehensive volume. One of the helpful features of the book is that it contains a collection of case studies. The case studies deﬁne a problem, discuss appropriate solution strategies, and warn against common pitfalls. They also emphasize how the results change when important parameters in the problem are altered, such as material properties, geometry, and mesh reﬁnement. There is no question that the ﬁnite element method is a powerful analysis tool. It is so prevalent throughout the world that most medium- and largesized manufacturing companies own or lease commercial ﬁnite element soft ware. Often practicing engineers are forced to use these codes (even though they do not have the background and/or experience) in order to get results quickly. They lack conﬁdence about the assumptions that are made, about the boundary conditions that are chosen, and whether or not the proper failure criterion is being used. This book is intended to help the practicing engineer gain conﬁdence in properly using the ﬁnite element method. The book is also intended as a textbook for advanced undergraduates and ﬁrst-year graduate students in engineering and the physical sciences. One of the most common concerns of engineers and students interested in the ﬁnite element method is whether or not they have the requisite math background to learn the material. Chapter 2 identiﬁes and reviews the basic mathematical building blocks that are absolutely necessary for understanding the ﬁnite element method. The topics include matrix algebra, fundamental

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concepts from calculus such as the Taylor series expansion and the divergence theorem, and an introduction to vectors and tensors and how they transform from one coordinate system to another. The chapter ends with a discussion of the mechanics of a continuous medium, which is especially important for understanding the concepts presented later in the book on nonlinear ﬁnite element procedures. In Chapter 3, the ﬁnite element method is introduced by considering secondorder ordinary diﬀerential equations with constant coeﬃcients. In this chap ter, the very important concepts of trial functions, test functions, and the weak form of the diﬀerential equation are introduced. The degree of con tinuity of trial and test functions is also discussed. Chapter 3 also covers the ﬁnite element assembly process, and it is shown how this process arises naturally from the weak form. The various types of boundary conditions that are encountered in the ﬁnite element method are introduced, and some one-dimensional applications are presented. The accuracy of the approximate solutions is assessed for diﬀerent element types (linear and quadratic) as the ﬁnite element mesh is reﬁned. The fundamental concepts that are needed to understand how the ﬁnite ele ment method is applied to problems in solid mechanics are presented in Chap ter 4. Here an introduction to the linearized theory of elasticity is presented. The chapter begins with a derivation of Cauchy’s law, and then proceeds with a discussion on principal stresses and a derivation of the equilibrium equations. The chapter then provides a thorough treatment of the smallstrain tensor and derives Hooke’s law for three-dimensional, plane stress, plane strain, thermoelastic, and axisymmetric formulations. The chapter concludes with a derivation of the weak form of the equilibrium equation. In Chapter 5 another important ﬁeld equation in engineering is derived — the steady-state heat equation. The chapter then goes on to discuss Fourier’s law and the boundary conditions relevant to heat conduction. The chapter ends with an example of heat conduction in a composite sphere followed by a derivation of the weak form of the steady-state heat equation. Both Chapters 4 and 5 emphasize that partial diﬀerential equations that arise in engineering and the physical sciences come from balance laws. Stated simply, a force balance gives the equilibrium equation, and an energy balance give the heat equation. A widely used class of ﬁnite elements called continuum ﬁnite elements is introduced in Chapter 6. The chapter begins with a discussion on what is meant by ﬁnite element shape functions and ends with an explanation of the isoparametric ﬁnite elements. Element types include the three-node tri angle, the arbitrary four-node quadrilateral, the four-node tetrahedron, and the eight-node brick. The computation of the ﬁnite element stiﬀness matrix for continuum elements involves evaluating integrals over the domain of the element. The method for evaluating such integrals in ﬁnite element codes is called Gauss quadrature. Several examples are given near the end of the chapter that clearly illustrate how Gauss quadrature works and its accuracy.

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The chapter concludes with three case studies. The ﬁrst case study considers the problem of a cantilever beam subjected to a concentrated end load. The case study examines the use of the four-node quadrilateral element for plane elasticity applications and discusses the numerical problem of locking that can occur under certain conditions and how to avoid it. Next, the problem of a plate with a centrally located hole subjected to a uniform distributed load is considered. Here symmetry boundary conditions are explained in detail, and a mesh convergence study is carried out to see how the approximate solution to the problem changes as the mesh is further reﬁned. The last case study considers a thermal stress problem of a copper wire surrounded by a ceramic coating. In the study, the copper wire is heated from some stress- free refer ence temperature. The temperature distribution in the composite as well as the resulting thermal stress distribution is obtained. Various types of structural ﬁnite elements are considered in Chapter 7. These include the space truss, the Euler-Bernoulli beam element, and the Mindlin-Reissner plate element. During the formulation of these elements, the engineering assumptions that place certain restrictions on the displace ment ﬁeld, called kinematic constraints, are highlighted. The chapter con tains several examples that illustrate the use of truss and beam elements and concludes with a case study of a square plate with clamped edges subjected to a uniform pressure distribution. During the case study, the performance of the Mindlin-Reissner plate element is examined. In addition, the numeri cal problem of shear locking, which can be encountered as the plate becomes very thin compared to the in-plane dimensions, is demonstrated. It is shown that the problem of shear locking can be avoided by using a technique called selective reduced integration. An introduction to ﬁnite element procedures for linear transient analysis is provided in Chapter 8. A transient analysis is nothing more than an attempt to solve a partial diﬀerential equation that depends on time. Examples of such equations are the equations of motion and the unsteady heat equation. Sub stitution of the ﬁnite element shape functions into the weak form discretizes the problem in space, yielding a system of ordinary diﬀerential equations that vary continuously in time. To obtain the complete time history for the prob lem of interest, a suitable time integration scheme must be adopted. In this chapter, two of the most popular time integration schemes are covered. These are the central diﬀerence method and the trapezoidal rule. A section of the chapter is also devoted to the topic of assessing the stability of a particular time integrator. The chapter ends with some general guidelines for time step selection. The chapter also includes two case studies. In the ﬁrst case study the problem of a one-dimensional bar subjected to an impact loading is con sidered, and the nature of stress wave propagation in the bar is studied as a function of mesh reﬁnement and time step size. In the second case study the dynamic response of a cantilever beam is obtained using the trapezoidal rule. Often in solid mechanics we encounter problems in which the relationship between strain and displacement is linear, but the stress-strain relationship

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is nonlinear. This type of problem is often referred to as a materially-only nonlinear problem. In chapter 9, due to its practical importance in solid me chanics and engineering design, the theory of small-strain, rate-independent plasticity is presented. The chapter begins with a basic treatment of onedimensional idealized stress-strain behavior and then proceeds to discuss the main ingredients that are needed in any plasticity theory: yield condition, ﬂow rule, and hardening rule. The chapter ends with the derivation of the elastoplastic tangent, which is needed for the ﬁnite element implementation of small-strain plasticity based on Newton’s method. The chapter ends by pro viding a detailed description of the general procedure for solving materially nonlinear problems. The main ideas are applied by presenting a simple ex ample of a one-dimensional elastic-plastic bar. Finally,the chapter ends with a case study of a thick-walled, elastic-plastic cylinder subjected to an internal pressure loading. Problems in solid mechanics that involve arbitrarily large translations, ro tations, and/or deformations are classiﬁed as geometrically nonlinear. In such problems, there exists a nonlinear relationship between strain and displace ment. A problem can be geometrically nonlinear even if the deformation in the solid remains small enough so that the material remains in the linearly elastic region. An introduction to the rich subject of treatment of geometri cally nonlinear problems is presented in Chapter 10. The chapter begins with a general discussion on large-strain kinematics and large-strain measures. The ﬁnite element formulation for geometrically nonlinear problems is then pre sented. The formulation begins by writing down the principle of virtual work over some ﬁxed reference conﬁguration. During the process, which is carried out through a simple change of variables, the alternative stress measures (ﬁrst Piola-Kirchhoﬀ and second Piola-Kirchhoﬀ stress tensors) naturally emerge. The concept of work conjugacy is also discussed. The chapter ends with a de scription of the total-Lagrangian procedure for solving geometrically nonlinear problems and two case studies. In the ﬁrst case study, the snap-through buck ling problem of a simple truss structure is considered, and the second study investigates the large deformation response of a neo-Hookean test specimen. This book can be used as a textbook or for self study. The material pre sented in Chapters 1–7 forms a semester-long course for advanced undergrad uates and ﬁrst-year graduate students. The course could be entitled, Intro duction to the Finite Element Method. The second part of the book, Chapters 8–10, form the subject matter of a semester-long sequel to the introductory course. The course could be called, Dynamic and Nonlinear Finite Element Procedures. The material presented in the book can also be used to teach short courses in the ﬁnite element method. Most of the material that is presented in this book is taken from my course notes and experiences teaching two fourday short courses intended for professional engineers. Each short course oﬀers a series of lectures, breaks, and hands-on problem-solving sessions. During the last two hours of each day, the students engage in a computer laboratory experience. The case studies presented throughout the book can be used for

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this purpose. The ﬁrst four-day course covers most of the material presented in Chapters 1–7. The second short course covers the material presented in Chapters 8–10. A unique feature of this book is that every ﬁgure throughout the book was generated by writing and running an individual MATLAB ∗ program. Doing this allows the three-dimensional ﬁgures to be viewed in an interactive vir tual reality environment. A virtual image gallery is available for download from the CRC Web site. Clicking on the caption of an individual ﬁgure opens up a virtual reality modeling language (VRML) ﬁle allowing students to see and explore in a three-dimensional, virtual world. The use of the VRML ﬁles requires a VRML web browser plug-in. The plug-in can be downloaded at http://www.parallelgraphics.com/products/cortona/. An instructor’s re source ﬁle containing detailed solutions to the problems presented at the end of each chapter is also available for download from the CRC Web site. Under the menu Electronic Products (located on the left side of the screen), click on Downloads & Updates. A list of books in alphabetical order with Web downloads will appear. Locate this book by a search, or scroll down to it. After clicking on the book title, a brief summary of the book will appear. Go to the bottom of this screen and click on the hyperlinked “Download” that is in a zip ﬁle.

�

∗ The

MathWorks, Inc. 3 Apple Hill Drive Natick, MA 01760-2098

xviii

Finite Element Method

Acknowledgments I thank my past and present students and my colleagues at the Illinois Institute of Technology, especially Sudhakar Nair, Kevin Cassel, and John Way, for their enthusiasm and insightful ideas that helped form my way of thinking over the years. I also thank Ted Belytschko, Brian Moran, and Jan Achenbach at Northwestern University for inspiring me to pursue a career in computational mechanics. Finally, I would like to thank my wife Mary Lee, and children Madeline and David, for putting up with me during the past year.

About the author Mike Gosz was raised in the small Wisconsin town of Kaukauna. He grad uated from Marquette Univesity with a B.S. degree (summa cum laude) in mechanical engineering. During college, he spent two years working as a manufacturing engineer at a General Motors assembly plant in Janesville, Wisconsin. He received his M.S. degree in 1989 and Ph.D. in 1993, both in theoretical and applied mechanics, from Northwestern University. He is currently an associate professor in the Mechanical, Materials, and Aerospace Engineering Department at the Illinois Institute of Technology, where he has been on the faculty since 1996. His ﬁrst faculty position (1993– 1996) was as an assistant professor in the Mechanical Engineering Department at the University of New Hampshire. Professor Gosz teaches a wide variety of undergraduate and graduate courses in the areas of solid mechanics and computational mechanics. He also regu larly teaches short courses in the ﬁnite element method to practicing engineers from companies such as Lockheed Martin, Caterpillar, BP, Motorola, Molex, just to name a few. Professor Gosz has over 30 technical publications in the areas of ﬁnite el ement methods, fracture mechanics, composite materials, and ﬂuid structure interaction. He received the IBM Faculty Development Award in 1994 from the T.J. Watson Research center in Yorktown Heights NY, and the Ralph Barnett teaching award in 1998. He is a member of ASME and the American Academy for Mechanics.

Contents

Preface

xiii

1 Introduction 2 Mathematical Preliminaries 2.1 Matrix algebra . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Multiplication of a matrix by a scalar . . . . . . . . 2.1.2 Transpose of a matrix . . . . . . . . . . . . . . . . . 2.1.3 Matrix multiplication . . . . . . . . . . . . . . . . . 2.1.4 The identity matrix . . . . . . . . . . . . . . . . . . 2.1.5 Determinant of a matrix . . . . . . . . . . . . . . . . 2.1.6 Inverse of a matrix . . . . . . . . . . . . . . . . . . . 2.1.7 Linear algebraic equations . . . . . . . . . . . . . . . 2.1.8 Integration of matrices . . . . . . . . . . . . . . . . . 2.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Vector dot product . . . . . . . . . . . . . . . . . . . 2.2.2 Vector cross product . . . . . . . . . . . . . . . . . . 2.2.3 Vector transformation . . . . . . . . . . . . . . . . . 2.3 Second-order tensors . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Tensor transformation . . . . . . . . . . . . . . . . . 2.3.2 Eigenvalues and eigenvectors of second-order tensors 2.3.3 Common operations with vectors and tensors . . . . 2.4 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Fundamental theorem of calculus . . . . . . . . . . . 2.4.2 Integration by parts . . . . . . . . . . . . . . . . . . 2.4.3 Taylor series expansion . . . . . . . . . . . . . . . . . 2.4.4 Gradient and divergence . . . . . . . . . . . . . . . . 2.4.5 The level surface . . . . . . . . . . . . . . . . . . . . 2.4.6 Divergence theorem . . . . . . . . . . . . . . . . . . 2.5 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Nonlinear equations of several variables . . . . . . . 2.6 Kinematics of motion . . . . . . . . . . . . . . . . . . . . . . 2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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xx 3 One-Dimensional Problems 3.1 The weak form . . . . . . . . . . . . . 3.2 Finite element approximations . . . . 3.2.1 The linear-u element . . . . . . 3.2.2 The quadratic-u element . . . . 3.3 Plugging in the trial and test functions 3.4 Algorithm for matrix assembly . . . . 3.5 One-dimensional elasticity . . . . . . . 3.5.1 Strain and stress calculation . . 3.5.2 Finite element results . . . . . 3.6 Problems . . . . . . . . . . . . . . . .

Finite Element Method

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4 Linearized Theory of Elasticity 4.1 Cauchy’s law . . . . . . . . . . . . . . . . . . . . . . 4.2 Principal stresses . . . . . . . . . . . . . . . . . . . . 4.3 Equilibrium equation . . . . . . . . . . . . . . . . . . 4.4 Small-strain tensor . . . . . . . . . . . . . . . . . . . 4.4.1 Relative stretch . . . . . . . . . . . . . . . . . 4.4.2 Angle change . . . . . . . . . . . . . . . . . . 4.4.3 Small-displacement gradients . . . . . . . . . 4.5 Hooke’s law . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Engineering constants . . . . . . . . . . . . . 4.5.2 Hooke’s law for plane stress and plane strain 4.5.3 Hooke’s law for thermal stress problems . . . 4.6 Axisymmetric problems . . . . . . . . . . . . . . . . 4.7 Weak form of the equilibrium equation . . . . . . . . 4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . .

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5 Steady-State Heat Conduction 5.1 Derivation of the steady-state heat equation 5.2 Fourier’s law . . . . . . . . . . . . . . . . . 5.3 Boundary conditions . . . . . . . . . . . . . 5.4 Weak form of the steady-state heat equation 5.5 Problems . . . . . . . . . . . . . . . . . . .

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6 Continuum Finite Elements 6.1 Three-node triangle . . . . . . . . . . . . . 6.1.1 The B-matrix . . . . . . . . . . . . 6.2 Development of an arbitrary quadrilateral 6.2.1 Compatibility issues . . . . . . . . 6.2.2 The bi-unit square . . . . . . . . . 6.2.3 The parent domain . . . . . . . . . 6.2.4 The B-matrix . . . . . . . . . . . . 6.2.5 Derivatives of the shape functions . 6.2.6 Area change . . . . . . . . . . . . .

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Table of Contents 6.3

xxi

Four-node tetrahedron . . . . . . . . . . . . . . 6.3.1 The B-matrix . . . . . . . . . . . . . . . 6.4 Eight-node brick . . . . . . . . . . . . . . . . . 6.4.1 The B-matrix . . . . . . . . . . . . . . . 6.4.2 Volume change . . . . . . . . . . . . . . 6.5 Element matrices and vectors . . . . . . . . . . 6.6 Gauss quadrature . . . . . . . . . . . . . . . . . 6.7 Bending of a cantilever beam . . . . . . . . . . 6.8 Analysis of a plate with hole . . . . . . . . . . . 6.9 Thermal stress analysis of a composite cylinder 6.9.1 Heat conduction analysis . . . . . . . . . 6.9.2 Thermal stress analysis . . . . . . . . . 6.9.3 Results . . . . . . . . . . . . . . . . . . 6.10 Problems . . . . . . . . . . . . . . . . . . . . .

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150 153 154 156 157 159 163 176 181 186 186 191 195 196

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8 Linear Transient Analysis 8.1 Derivation of the equation of motion . . . . . . . . . . . . 8.1.1 Weak form . . . . . . . . . . . . . . . . . . . . . . 8.2 Semi-discrete equations of motion . . . . . . . . . . . . . . 8.2.1 Properties of the mass matrix . . . . . . . . . . . . 8.2.2 Natural frequencies and normal modes . . . . . . . 8.3 Central diﬀerence method . . . . . . . . . . . . . . . . . . 8.3.1 Dynamic response of a simple pendulum . . . . . . 8.3.2 Stability of central diﬀerence method . . . . . . . . 8.3.3 Elastic wave propagation in a one-dimensional bar 8.4 Trapezoidal rule . . . . . . . . . . . . . . . . . . . . . . . .

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7 Structural Finite Elements 7.1 Space truss . . . . . . . . . . . . . . . . 7.1.1 Strain-displacement relationship 7.1.2 Element matrix . . . . . . . . . . 7.1.3 Element vector . . . . . . . . . . 7.2 Euler-Bernoulli beams . . . . . . . . . . 7.2.1 Kinematic assumptions . . . . . . 7.2.2 Finite element approximations . 7.2.3 Element matrix . . . . . . . . . . 7.2.4 Element vector . . . . . . . . . . 7.3 Mindlin-Reissner plate theory . . . . . . 7.3.1 Assumptions . . . . . . . . . . . 7.3.2 Strain-displacement relations and 7.3.3 Finite element approximations . 7.3.4 Element matrix . . . . . . . . . . 7.3.5 Element vector . . . . . . . . . . 7.4 Deﬂection of a clamped plate . . . . . . 7.5 Problems . . . . . . . . . . . . . . . . .

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xxii

8.5

8.6

Finite Element Method 8.4.1 Dynamic response of a cantilever beam . . . . . . . Unsteady heat conduction . . . . . . . . . . . . . . . . . . 8.5.1 Weak form of the unsteady heat equation . . . . . 8.5.2 Finite element approximations . . . . . . . . . . . 8.5.3 Backward diﬀerence method . . . . . . . . . . . . . 8.5.4 Stability of the backward diﬀerence method . . . . 8.5.5 Unsteady heat conduction in a composite cylinder Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .

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9 Small-Strain Plasticity 9.1 Basic concepts . . . . . . . . . . . . . . . . . . . . 9.2 Yield condition . . . . . . . . . . . . . . . . . . . . 9.2.1 Dependence on hydrostatic pressure . . . . 9.2.2 The von Mises yield condition . . . . . . . . 9.2.3 Geometry of the von Mises yield surface . . 9.2.4 Simple experiments . . . . . . . . . . . . . . 9.3 Flow and hardening rules . . . . . . . . . . . . . . 9.4 Derivation of the elastoplastic tangent . . . . . . . 9.5 Finite element implementation . . . . . . . . . . . 9.5.1 Convergence criteria . . . . . . . . . . . . . 9.5.2 Radial return stress update scheme . . . . . 9.6 One-dimensional elastoplastic deformation of a bar 9.7 Elastoplastic analysis of a thick-walled cylinder . . 9.8 Problems . . . . . . . . . . . . . . . . . . . . . . .

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10 Treatment of Geometric Nonlinearities 10.1 Large-deformation kinematics . . . . . . . . . . . . 10.1.1 The Green-Lagrange strain tensor . . . . . 10.2 Weak form in the original conﬁguration . . . . . . 10.2.1 Internal virtual work . . . . . . . . . . . . . 10.2.2 External virtual work . . . . . . . . . . . . 10.3 Linearization of the weak form . . . . . . . . . . . 10.3.1 Internal virtual work . . . . . . . . . . . . . 10.3.2 External virtual work . . . . . . . . . . . . 10.4 Snap-through buckling of a truss structure . . . . . 10.4.1 Element tangent stiﬀness matrix . . . . . . 10.4.2 Element internal force vector . . . . . . . . 10.4.3 Numerical results . . . . . . . . . . . . . . . 10.5 Uniaxial tensile test of a rubber dog-bone specimen 10.5.1 Element tangent matrix . . . . . . . . . . . 10.5.2 Element internal force vector . . . . . . . . 10.5.3 Numerical results . . . . . . . . . . . . . . . 10.6 Problems . . . . . . . . . . . . . . . . . . . . . . .

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Bibliography

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Table of Contents

xxiii

Index

397

1 Introduction

The ﬁnite element method made its debut after a series of papers was pub lished by Turner in 1959 [1]. Before 1965, the title, “Finite Element Method,” did not even exist; the method was referred to as the direct stiﬀness method (DSM). For the most part, the method was conﬁned to the structural me chanics community and the aerospace industry. An excellent article on the history of the ﬁnite element method has been published by Felippa [2]. Since the genesis of the method during the early 1960s, thousands of articles have been published on the subject. Today, the method has reached such a state of maturity that it is now thought of as a method for solving general ﬁeld problems in all areas of engineering and the physical sciences. A modern deﬁnition of the ﬁnite element method might state that it is simply a numerical technique for obtaining approximate solutions to partial diﬀerential equations. To illustrate this idea, suppose that an engineer wants to obtain the temperature distribution in a thin plate subjected to some sort of external heat source. To accomplish this task, the engineer must make a variety of assumptions, and develop an appropriate mathematical model that is both physically realistic (retains the essential physics in the problem) and is yet amenable to solution. As an example, the engineer may assume steady-state conditions (the temperature ﬁeld in the plate does not change with time), the top and bottom of the plate are perfectly insulated, the heat conduction is isotropic (the material’s resistance to heat ﬂow is the same in all directions), etc. Under such conditions, the temperature ﬁeld in the plate satisﬁes the two-dimensional Laplace’s equation. That is, ∇2 T = 0 or

∂2T ∂2T + =0 2 ∂X ∂Y 2

(1.1)

For simple geometries, a closed-form (analytical) solution to Laplace’s equa tion can be obtained; however, for complicated geometries it is usually nec essary to resort to numerical techniques. The main idea in the ﬁnite element method is to discretize or “break up” the domain of interest into a collection of points and subdomains called nodes and elements. This idea is depicted in Figure 1.1. Once the domain of interest is discretized, the ﬁeld variable of interest, temperature in this case, is interpolated throughout the area or

1

2

Finite Element Method

FIGURE 1.1 Finite element mesh. volume of the element using simple polynomial interpolation functions called shape functions. Loosely speaking, for linear ﬁeld problems the process of discretizing the domain into nodes and elements leads to a system of linear algebraic equations that can be readily solved on a digital computer. For nonlinear problems the technique is very much the same, except the equations are solved in an iterative manner, with each iteration solving a system of linear algebraic equations.

The ﬁnite element procedure The ﬁnite element procedure involves three general phases: a pre-processing phase, an analysis phase, and a post-processing phase. The three phases are illustrated in Figure 1.2. Perhaps the most time consuming and diﬃcult of the three phases is the pre-processing phase. Given a physical problem of interest, the engineer must use sound judgement to transform the real-world problem into a mathematical model that captures the essential physics and is amenable to numerical solution. After making reasonable assumptions and coming up with a suitable mathematical model, usually involving a partial differential equation along with appropriate boundary conditions, he must then decide what type of analysis to perform: linear, nonlinear, transient, steadystate, etc. He must also decide upon suitable material properties, choose the appropriate type of ﬁnite element, and then create a ﬁnite element mesh that is suﬃciently reﬁned in regions where large gradients in the ﬁeld variable are expected. Mesh generation by itself is a time-consuming process. Fortunately, with the advent of modern commercial pre-processing software, the amount of time spent in the pre-processing phase has been cut considerably and made easier than ever before. For the most part, the analysis phase of the ﬁnite element procedure is like feeding the data created in the pre-processing phase into a black box. During the analysis phase, a computer program reads the data and solves a linear system of algebraic equations, either once for linear problems, or many times (often iteratively) for nonlinear and/or dynamic problems. When the analysis phase is ﬁnished, the results of interest such as nodal displacements

Introduction

3

Pre-processing phase

Post-processing phase

¯ ≤ σy ? is σ

Analysis phase Kd = F

FIGURE 1.2 Flow chart of the ﬁnite element procedure.

and temperatures are output to a data ﬁle or ﬁles. Element quantities such as stress, strain, and heat ﬂux can also be output at the user’s request. In the ﬁnal phase of the ﬁnite element procedure, the engineer faces the diﬃ cult task of interpreting the results of the analysis phase. The post-processing phase has been made signiﬁcantly easier throughout the years through the development of user-friendly software packages with excellent graphics capa bilities. The post-processing phase still, however, requires a good working knowledge of the physics of the problem of interest. For example, in a stress analysis, the engineer may want to know if a solid body under a given set of ex ternal loads will fail. To answer this question, the engineer must decide upon an appropriate failure criterion and then carefully investigate the stress ﬁelds in the solid. The engineer typically does this by observing color contour plots. At this time the engineer must assess whether or not the nice-looking contour plots make sense. In particular, he must be able to answer questions such as: Are the boundary conditions satisﬁed? Is the ﬁnite element mesh suﬃciently reﬁned? Was the assumption of linear material behavior appropriate? Is a more complicated analysis required?

Where do we go from here? The ﬁnite element method is a rich and exciting subject. Learning the ﬁnite element method can be diﬃcult, but ultimately very satisfying if we pay at tention to the details of the method along the way. Let us now begin our

4

Finite Element Method

journey in the next chapter by studying the mathematical background that is needed to obtain an excellent understanding of the ﬁnite element method. Starting the journey in this fashion may seem slow at ﬁrst, but stay patient. You will be rewarded.

2 Mathematical Preliminaries

To climb steep hills requires a slow pace at ﬁrst. —Shakespeare

2.1

Matrix algebra

A matrix is a tool for arranging numerical values and/or functions in an organized manner. A matrix contains rows and columns. An element of a matrix is the number (or function) that occupies a speciﬁc row and column. As an example, consider the following matrix: ⎡ ⎤ 1 3 A = ⎣4 2⎦ (2.1) 5 6 In this example, and throughout the remainder of the book, we will use bold face type to denote that the quantity A is a matrix. The elements of the matrix in this example are numbers ranging from one to six. The matrix has three rows and two columns. Hence, we say that the order of the matrix is 3 × 2. Often it is required to carefully keep track of the order of matrices. Occasionally throughout the book, the order will be labeled underneath the boldface symbol, e.g., A 3×2

Keeping track of the order of matrices is especially important when performing matrix multiplication, which will be discussed below. Often a matrix has the special property of being square and/or symmetric. A square matrix is a matrix with the same number of rows as columns. A square matrix is said to be symmetric if Aij = Aji

(2.2)

where the subscripts i and j represent the row and column number, respec tively. The elements A11 , A22 , A33 , etc. are said to lie on the diagonal of the matrix. The elements that lie above and below the diagonal are said to belong to the upper and lower triangles, respectively.

5

6

2.1.1

Finite Element Method

Multiplication of a matrix by a scalar

One of the simplest operations that can be performed on a matrix is multipli cation by a scalar. A scalar is a quantity that can be characterized by a single value at every point in space. A scalar can be a function of position, but it does not have direction associated with it. Temperature is an example of a scalar quantity. Throughout the book, scalar quantities will be denoted by either lowercase alphabetic or lowercase Greek letters, unless noted otherwise. Multiplication of the matrix, A, by a scalar, α, can be deﬁned as follows. If C = αA

(2.3)

Cij = αAij

(2.4)

then the elements of C are The elements of the matrix C are obtained by multiplying each element of the matrix A by the scalar α.

2.1.2

Transpose of a matrix

The transpose of a matrix is obtained by For example, if a matrix A is given as ⎡ 1 A = ⎣5 0

then

interchanging its rows and columns. ⎤ 3 6⎦ 1

1 5 0 A = 3 6 1 T

(2.5) (2.6)

where the superscript T is used to denote the transpose. The order of the matrix A in the above example is 3 × 2, whereas the order of AT is 2 × 3.

2.1.3

Matrix multiplication

Suppose that the matrix C is deﬁned as the product of two matrices A and B, i.e., C = AB (2.7) Matrix multiplication is said to be conformable if the number of columns in the matrix A equals the number of rows in the matrix B. Let’s suppose that the order of A is 2 × 3. In order for the multiplication to be conformable, the matrix B must be of order 3 × 2. An easy way to check whether or not a matrix multiplication is conformable is to write down the order of the two matrices below each matrix. So in the example above, one could write C 3×3

A B = 3×2 2×3

(2.8)

Mathematical Preliminaries

7

Focusing on the right-hand side of equation (2.8), if the inner dimensions are equal, then the multiplication is conformable. The outer dimensions indicate the order of the resulting matrix, C, which is 3 × 3 in this case. Provided that the matrix multiplication is conformable, the elements of the matrix C are obtained as follows: Cij =

K

Aik Bkj

(2.9)

k=1

where K is the number of columns in A (or number of rows in B). In order to demonstrate a hand calculation, consider the following matrices A and B: ⎡ ⎤ 1 3 1 3 2 A = ⎣4 2⎦; B = (2.10) 4 2 1 5 6 Using equation (2.9) to calculate the product of A and B yields C11 = A11 B11 + A12 B21 = (1)(1) + (3)(4) = 13 C12 = A11 B12 + A12 B22 = (1)(3) + (3)(2) =9 etc. The result is

⎤ 13 9 5 C = ⎣ 12 16 10 ⎦ 29 27 16

(2.11)

⎡

(2.12)

To conclude this section, we emphasize that matrix multiplication, in gen eral, is not commutative. That is, AB = BA. A useful identity, however, that involves taking the transpose of the product of two matrices can be stated as follows: (2.13) (AB)T = BT AT This identity will be useful later on when it is time to substitute ﬁnite element shape functions into the weak form, discussed in Chapter 3.

2.1.4

The identity matrix

Throughout the book, the bold symbol I will be used to denote the identity matrix. The identity matrix is a square matrix whose elements are one in the diagonal positions and zero everywhere else. The components of the identity

8

Finite Element Method

matrix can be written succinctly using the Kronecker delta symbol. The Kronecker delta symbol is deﬁned as 1 : if i = j δij = 0 : if i = j So, for example, the 3 × 3 identity matrix is written as ⎡ ⎤ 1 0 0 I = ⎣0 1 0⎦ 3×3 0 0 1

2.1.5

Determinant of a matrix

Throughout the book, the determinant of a square matrix, A, will be denoted as det A. Suppose that A is a 2 × 2 matrix. The determinant of A is deﬁned as

A A

A det 2 × 2 =

11 12

A21 A22 = A11 A22 − A12 A21

(2.14)

Once we have established the deﬁnition (2.14), the determinant of higher order matrices can be deﬁned as n A det n × n = Aij Cij (2.15) j=1

Here in equation (2.15), the subscript i identiﬁes the row of the matrix. It can take on any value between 1 and n. The matrix Cij is called the cofactor matrix. The elements of the cofactor matrix are deﬁned as Cij = −1(i+j) Dij

(2.16)

where Dij is the determinant of the sub-matrix of C that is obtained by cross ing out row i and column j. As an example, let us compute the determinant of the following 3 × 3 matrix: ⎡ ⎤ 1 2 3 (2.17) A = ⎣ 2 4 −2 ⎦ 3 −2 1 Letting i = 1 in equation (2.15), we obtain det A = A11 C11 + A12 C12 + A13 C13

4 −2

− 2 2 −2 + 3 2 4

= 1

3 −2

3 1 −2 1 = −64

(2.18)

Mathematical Preliminaries

9

Undergraduate engineering students usually learn the operation carried out in (2.18) when evaluating the cross product of two vectors. In that case, the ﬁrst row of A contains three unit vectors i, j, and k. Computing the determinant of any matrix larger than 3 × 3 by hand is very time consuming, because the operation requires a large number of multiplications. Fortunately, the determinant of a matrix is rarely explicitly computed in ﬁnite element procedures. On the other hand, having the ability to perform a quick hand calculation to ﬁnd the determinant of small (2 × 2 or 3 × 3) matrices is useful for ﬁnding the inverse of these matrices required in some of the problems presented throughout the book. The inverse of a square matrix is deﬁned in the next section. If the determinant of a square matrix is found to be zero, that matrix is said to be singular.

2.1.6

Inverse of a matrix

When a matrix A is multiplied by its inverse, A−1 , the result is the identity matrix, i.e., AA−1 = A−1 A = I We emphasize that only square matrices have an inverse, and the inverse of a singular matrix does not exist. The inverse of a square matrix, A, can be found from the following formula: A−1 =

1 CT det A

(2.19)

where CT is the transpose of the cofactor matrix. As an example, we can quickly compute the inverse of a 2 × 2 matrix as follows. Suppose 2 −1 A= −1 1 It can be easily seen that det A = 1. The cofactor matrix is 1 1 C= 1 2 Notice that in this example, the cofactor matrix happens to be symmetric, i.e., C = CT , and hence 1 1 1 −1 A = 1 1 2 1 1 = 1 2 We can check our work by making sure that A−1 A = I, i.e., 2 −1 1 1 1 0 = −1 1 1 2 0 1

10

Finite Element Method

So for 2 × 2 symmetric matrices, the inverse is obtained by ﬂip-ﬂopping the diagonal terms, switching the sign of each oﬀ-diagonal term, and dividing each element by the determinant of the original 2 × 2 matrix.

2.1.7

Linear algebraic equations

Consider the following system of three equations: x1 − x2 = 10 −x1 + 2x2 − x3 = 0 −x2 + x3 = 1

(2.20)

In each equation above, the unknowns, x1 , x2 , and x3 appear linearly. This means, for example, that the unknowns are not raised to a power, and they do not appear as arguments in nonlinear functions such as sine and cosine. Solving systems of linear algebraic equations is the major computational task in every ﬁnite element analysis. Even nonlinear ﬁnite element procedures boil down to solving linear algebraic equations, albeit over and over. A method for solving systems of linear algebraic equations that is amenable to numerical computation employs matrix methods. To illustrate how the procedure works, let us write the system of equations (2.20) in the form Ax = b as follows: ⎡ ⎤⎧ ⎫ ⎧ ⎫ 1 −1 0 ⎨ x1 ⎬ ⎨ 0 ⎬ ⎣ −1 2 −1 ⎦ x2 = 0 (2.21) ⎩ ⎭ ⎩ ⎭ 1 x3 0 −1 1 If we multiply both sides of (2.21) by A−1 we obtain A−1 Ax = A−1 b =⇒ Ix = A−1 b =⇒ x = A−1 b The system of equations can be solved in the above fashion, as long as the determinant of the matrix A is not zero. If det A = 0, then A is singular and there would be inﬁnitely many vectors x that satisfy equation (2.20). Let us now proceed and attempt to solve the matrix problem. The deter minant of the matrix A is computed as

−1 −1

2 −1

−1 2

+ 1

+ 0

=0 det A = 1

0 −1

0 1

−1 1

Because det A = 0, we cannot solve for the vector x. Let us suppose, however, that the value of the variable x1 is given. This is referred to as a constraint or boundary condition. For the moment, let us take x1 = 1. Because x1 is known, the matrix problem (2.20) can be reduced to two equations with only

Mathematical Preliminaries

11

two unknowns. To obtain the reduced problem, we expand row two and row three as follows: −1(1) + 2x2 − x3 = 0 −x2 + x3 = 1 We now write (2.22) back into matrix form, obtaining 2 −1 x2 1 = −1 1 x3 1

(2.22)

(2.23)

The resulting 2 × 2 system of equations can now be solved, because the de terminant of the matrix on the left-hand side is no longer zero. Because the constraint was imposed on x1 , the reduced system of equations could also have been obtained by simply crossing out the ﬁrst row and ﬁrst column of the original matrix problem, and modifying the right-hand side (due to the fact that the prescribed value of x1 was nonzero). If the problem were changed by insisting instead that x1 = 0, the reduced system could be obtained by cross ing out the ﬁrst row and column, but leaving the right-hand side unaltered. Finally, solving the reduced system, we get 1 1 1 x2 = x3 1 2 1 2 = (2.24) 3

2.1.8

Integration of matrices

Sometimes it is convenient to store functions of one or several variables in a matrix. The deﬁnite integral of a matrix is obtained by integrating each element (function) of the matrix. The result is another matrix of the same order. For example, suppose that the matrix A is deﬁned as ⎡ 2 ⎤ 0 x3 x A = ⎣ x x4 x + 2 ⎦ 0 1 x5 and we wish to evaluate the following integral: 1 A dx

(2.25)

0

The result is obtained by integrating each of respect to x from zero to one. The result is ⎡ 1 1/3 0 A dx = ⎣ 1/2 1/5 0 0 1

the nine elements of A with ⎤ 1/4 5/2 ⎦ 1/6

12

Finite Element Method

Integration of matrices is a common operation in the derivation of ﬁnite el ement matrices, discussed in Chapter 3. Because the operation typically re quires the evaluation of many individual integrals, a symbolic math software package is a very useful tool for integrating matrices.

2.2

Vectors

A vector is a physical quantity that has both magnitude and direction. Force is an example of a vector quantity. Both magnitude and direction are needed to completely describe it. A vector u is illustrated in Figure 2.1. The X1 , X2 , X3 axes labeled in the ﬁgure form what is called a Cartesian coordinate system. There are three other vectors shown in the ﬁgure. These are labeled e1 , e2 , and e3 . They represent unit vectors (otherwise known as base vectors) pointing in the X1 , X2 , and X3 directions, respectively. It turns out that any vector u can be expressed in terms of the base vectors as follows: u = u1 e1 + u2 e2 + u3 e3

(2.26)

X2

u

u2

X1

e2 e1 e3

X3 FIGURE 2.1 Cartesian components of a vector.

u1

u3

Mathematical Preliminaries

13

The quantities u1 , u2 , and u3 in equation (2.26) are called components of u. Each component represents the the projection of u onto one of the coordinate axes. For example, the component u1 is the projection of u onto the X1 axis. Note that the components of a vector are scalar quantities; they do not have direction associated with them.

2.2.1

Vector dot product

The dot product between two vectors u and v is deﬁned as u · v = |u||v| cos θ

(2.27)

where |u| and |v| are the magnitudes of the two vectors, and cos θ is the cosine of the angle formed between them. Alternatively, the vector dot product can be written as 3 u·v = u i vi = u 1 v1 + u 2 v2 + u 3 v3 (2.28) i=1

The limit on top of the summation symbol on the right-hand side of (2.28) can be either two or three, depending on whether we are working in twodimensional or three-dimensional space. Throughout the book, we will adopt a convention called indicial notation. The use of indicial notation will allow us to write complex expressions very compactly. One of the fundamental rules of indicial notation is that whenever two subscripts are repeated in a single expression, such as in equation (2.28) above, the summation symbol is omitted, but it is still implied. Hence, us ing indicial notation, the dot product operation between two vectors can be written succinctly as (2.29) u · v = u i vi Having introduced indicial notation, the vector u can now be expressed as u = ui ei

(2.30)

The components of u are obtained by taking the dot product of u with each of the base vectors, i.e., ui = u · ei (2.31) Just to be completely clear, let us now carry out the operation spelled out in equation (2.31) above. Expanding the right-hand side, we obtain u · ei = (uj ej ) · ei = uj (ej · ei ) = uj δji = u1 δ1i + u2 δ2i + u3 δ3i = ui

(2.32)

14

Finite Element Method

There are several subtle points that should be mentioned concerning the oper ations carried out in (2.32) above. First of all, in the ﬁrst step, we were careful not to let a subscript (or index) appear more than twice in each expression. If an index appears more than twice in a single expression, then that expression becomes ambiguous. In the second step, notice that the dot product between the two base vectors ei and ej gives the Kronecker delta. This is because the dot product between distinct base vectors is zero, and the dot product between any base vector and itself is one. The ﬁnal result in (2.32) stems from the fact that the subscript i is a free index. A free index can take on any value ranging from one to three (in three-dimensional space). So notice that the equality stated in the last line holds for any value of i, due to the special property of the Kronecker delta. From now on, if we see an expression like ui δij , we will simply replace the repeated index, i, with the free index, j, and get rid of the Kronecker delta symbol, i.e., ui δij = uj .

2.2.2

Vector cross product

Another very important operation with vectors is the vector cross product. If we consider any two vectors u and v, the vector cross product u × v gives another vector that points in the direction (according to the right-hand rule) that is perpendicular to the common plane in which both vectors lie. The result is another vector, c say, whose magnitude is |c| = |u||v| sin φ

(2.33)

where φ is the angle formed between u and v. The vector c can be obtained by evaluating the following determinant

e1 e2 e3

(2.34) c =

u1 u2 u3

v1 v2 v3

where e1 , e2 , and e3 are unit base vectors. Expanding the determinant (2.34) we get c = (u2 v3 − u3 v2 ) e1 − (u1 v3 − u3 v1 )e2 + (u1 v2 − u2 v1 )e3

(2.35)

Another convenient way to express the components of c is through the use of the permutation symbol. In order to deﬁne the components of the permutation symbol, it is helpful to look at Figure 2.2. If we travel counterclockwise around the circle, we would pass the numbers (depending on the starting location) (2, 1, 3), (3, 2, 1), and (1, 3, 2). These numbers inside the parentheses are called odd permutations of the indices (i, j, k). On the other hand, if we travel clockwise, we would pass (1, 2, 3), (2, 3, 1), and (3, 1, 2). These are called even permutations of (i, j, k). Having deﬁned odd and even permutations of the

Mathematical Preliminaries

15 2

2

3

3 1

1 FIGURE 2.2 Odd and even permutations of (1, 2, 3).

indices, the permutation symbol can now be deﬁned as ⎧ ⎪ ⎨−1 : if i, j, k are odd permutations of 1, 2, 3 ijk = 0 : if any i, j, k are equal ⎪ ⎩ +1 : if i, j, k are even permutations of 1, 2, 3 So, for example, 123 = 1, 213 = −1, 113 = 0. It turns out that the compo nents of c can be expressed in terms of the permutation symbol as follows: ci = ijk uj vk

(2.36)

Expanding equation (2.36) above gives c1 = 123 u2 v3 + 132 u3 v2 = u 2 v3 − u 3 v2 c2 = 213 u1 v3 + 231 u3 v1 = −u1 v3 + u3 v1 c3 = 312 u1 v2 + 321 u2 v1 = u 1 v2 − u 2 v1 Note that the components c1 , c2 , and c3 above are precisely the same compo nents as those given in (2.35).

2.2.3

Vector transformation

Often it is necessary to obtain the components of a vector in a new coordinate system that diﬀers from the ﬁxed Cartesian coordinate system by a rotation of the coordinate axes. The idea of forming a new set of axes by rotating the ﬁxed Cartesian axes is shown in Figure 2.3. We will refer to the new set of axes as the primed set. The primed set is identiﬁed by the dashed lines in the ﬁgure. The base vectors that are associated with the primed set of axes are e1 , e2 , and e3 , or simply as ei . Now let us focus on the vector u. The projection of u onto each of the primed axes X1 , X2 , and X3 yields the components of

16

Finite Element Method X2 X1

X2

u1

u2

u

e2

e1

e2

e1

X1

FIGURE 2.3 Components of a vector in a rotated (primed) coordinate system. the vector in the primed coordinate system. These primed components can be obtained by taking the dot product of the vector u with each of the primed base vectors, i.e., (2.37) ui = u · ei We emphasize here that the vector u is the same vector, no matter what coordinate system its components are described in. We can express the vector in terms of the primed base vectors, u = ui ei

(2.38)

or, alternatively, we can express it in terms of the unprimed base vectors, u = ui ei

(2.39)

If we now substitute the expression (2.39) into equation (2.37), we obtain an expression for the primed components of the vector u in terms of the unprimed components, ui = (uj ej ) · ei = uj (ej · ei ) = (ei · ej )uj

(2.40)

The quantity ei · ej gives the projection of the primed base vectors onto the unprimed axes. If we let (2.41) aji = ei · ej then equation (2.40) can be written as ui = aji uj

(2.42)

Mathematical Preliminaries

17

The inverse of the relation (2.42) gives the unprimed components ui in terms of the primed components, i.e., ui = (uk ek ) · ei = aik uk = aij uj

2.3

(2.43)

Second-order tensors

While the physical properties of a vector (magnitude and direction) are easy to think about, the physical meaning of a second-order tensor is a little more diﬃcult to grasp. Mathematically speaking, a second-order tensor is a func tion that operates on a vector to produce another vector. This idea can be depicted by the following equation: v −→ T −→ u

(2.44)

In order to clearly deﬁne the conceptual operation above, we need to be able to express the components of the tensor T in terms of the base vectors e1 , e2 , and e3 . It turns out that any second-order tensor, T, can be expressed in terms of its components and base vectors as follows: T = Tij ei ej

(2.45)

The quantity ei ej involving the two base vectors lying side by side is called a dyadic product. Expanding (2.45) above we get T = T11 e1 e1 + T12 e1 e2 + T13 e1 e3 +T21 e2 e1 + T22 e2 e2 + T23 e2 e3 +T31 e3 e1 + T32 e3 e2 + T33 e3 e3

(2.46)

So it is now clear that a second-order tensor has nine components. The nine components can be conveniently arranged in matrix form as follows: ⎡ ⎤ T11 T12 T13 [T] = ⎣ T21 T22 T23 ⎦ (2.47) T31 T32 T33 The tensor operation (2.44) is very much like the vector dot product. The result, however, is a vector instead of a scalar. The operation can be more formally written as T · v = u (2.48)

18

Finite Element Method

By substituting the expression (2.45) into equation (2.48) we obtain u = (Tij ei ej ) · (vk ek )

(2.49)

The operation deﬁned above is an example of what is called a tensor product. In order to perform the operation, we treat the components Tij and vk as scalars. These can be moved around as follows: u = (Tij vk ei )(ej · ek )

(2.50)

Next, in equation (2.50) above, the usual vector dot product operation is carried out for the two base vectors ej and ek . Doing this we get u = Tij vk δjk ei = Tik vk ei

(2.51)

Hence, the components of the resulting vector u are ui = Tik vk = Tij vj

(2.52)

The components Tij of any second-order tensor T can be obtained as fol lows: Tij = ei · T · ej = ei · (Tkl ek el ) · ej = (ei · ek )Tkl (el · ej ) = δik Tkl δlj = Tij

(2.53)

From equation (2.53) we can see that the elements in the ﬁrst row of the matrix (2.47) are actually components of a the vector e1 · T. The second and third rows are the components of e2 · T, and e3 · T respectively. One of the fundamental second-order tensors in the study of mechanics is the stress tensor σ. When we take the tensor product e1 · σ of the stress tensor and the base vector e1 , the result is a vector having units of force per unit area. This vector acts on a plane whose unit outward normal∗ points in the direction of the X1 axis. The vector has components σ11 , σ12 , and σ13 . Physically, these components represent the normal and shear stresses acting on the plane perpendicular to the X1 axis.

Example 2.1 In continuum mechanics, the components of second-order tensors are conve niently stored in 3 × 3 matrices. The determinants of such matrices often have ∗ A unit outward normal is a unit vector passing through a point on a surface that is perpendicular to the surface and is directed out of the body.

Mathematical Preliminaries

a×b

19

c

b a

FIGURE 2.4

Volume of a parallelopiped, V = (a × b) · c.

physical signiﬁcance. As an example, suppose that a 3 × 3 matrix A is given

as

⎤ ⎡ a1 a2 a3 (2.54) A = ⎣ b1 b2 b3 ⎦ c1 c2 c3 Now let a, b, and c be vectors whose components are deﬁned by the ﬁrst, second, and third rows of A respectively, e.g., a = a1 e1 + a2 e2 + a3 e3 = ai ei b = b1 e1 + b2 e2 + b3 e3 = bi ei c = c1 e1 + c2 e2 + c3 e3 = ci ei Having deﬁned these vectors, it can be easily shown that the determinant of the matrix A can be expressed as det A = (a × b) · c = (c × a) · b = (b × c) · a

(2.55)

The operation carried out in each line of (2.55) is called a scalar triple prod uct. Referring to Figure 2.4, we can see that each triple product (and hence det A) deﬁnes the volume of a parallelopiped whose edges are deﬁned by the vectors a, b, and c.† Another important observation that we can make here is that if the vectors a, b, and c all point in the same direction, the resulting scalar triple product, and hence det A, is zero. This will have important im plications when we study eigenvalues and eigenvectors of matrices in Section 2.3.2. †a × b

gives a vector whose magnitude is the area of the base of the parallelopiped. Taking the dot product of this vector with c yields the base times the height.

20

Finite Element Method The scalar triple product can be expressed in indicial notation as (b × c) · a = (ijk bj ck ei ) · (al el ) = ijk bj ck al δil = ijk bj ck ai = ijk ai bj ck

(2.56)

As a ﬁnal remark, if Aij are the components of A, we can let ai = A1i , bj = A2j , and ck = A3k which yields an alternative expression for det A in terms of the permutation symbol, det A = ijk A1i A2j A3k

2.3.1

(2.57)

Tensor transformation

One of the deﬁnitions of a second-order tensor states that if a physical quantity transforms under a rotation of the coordinate axes according to the tensor transformation law, then that physical quantity is a second-order tensor. This deﬁnition is akin to saying if something walks like a duck, then it is a duck. We emphasize here that a tensor is a physical quantity independent of the coordinate system to which its components are referred. Because of this we can write the tensor T in terms of base vectors in the primed and unprimed coordinate systems as follows: Tij = ei · T · ej Tij = ei · T · ej

(2.58)

Using equation (2.45) along with (2.58) above, we can write the primed com ponents of the tensor T in terms of the unprimed components. Doing this we get Tij = ei · (Tkl ek el ) · ej = aki alj Tkl = api aqj Tpq

(2.59)

and vice versa, Tij = ei · (Tkl ek el ) · ej = aik ajl Tkl

(2.60)

Equation (2.59) (or equivalently (2.60)) is referred to as the tensor transfor mation law. If the components of a physical quantity transform according to one of these relations, then that quantity is a second-order tensor.

Mathematical Preliminaries

21

Example 2.2 As an example, suppose that the components of a second-order tensor in the unprimed coordinate system are given as ⎡ ⎤ 10 −2 0 T = ⎣ −2 20 0 ⎦ (2.61) 0 0 1 Now suppose we want to ﬁnd the components of this tensor in a primed coordinate system that is obtained by rotating the unprimed axes about the X3 axis by an amount θ = π/4. The components, aji , of the transformation tensor can now be calculated as follows: √ a11 = e1 · e1 = 2/2 √ a21 = e1 · e2 = 2/2 a31 = e1 · e3 = 0 √ a12 = e2 · e1 = − 2/2 √ a22 = e2 · e2 = 2/2 a32 = e2 · e3 = 0 a13 = e3 · e1 = 0 a23 = e3 · e2 = 0 a33 = e3 · e3 = 1

(2.62)

It is convenient to store the components aji in a 3 × 3 matrix Q, i.e., √ ⎤ ⎡ √ √2/2 √2/2 0 (2.63) Q = ⎣ − 2/2 2/2 0 ⎦ 0 1 0 The tensor transformation operation (2.59) can now be performed by carrying out the following matrix multiplications: T = QTQT √ √ ⎤ ⎡ √ ⎤⎡ ⎤⎡√ 10 −2 0 √2/2 √2/2 0 √2/2 −√2/2 0 = ⎣ − 2/2 2/2 0 ⎦ ⎣ −2 20 0 ⎦ ⎣ 2/2 2/2 0 ⎦ 0 0 1 0 0 1 0 1 0 ⎡ ⎤ 13 5 0 = ⎣ 5 17 0 ⎦ 0 0 1

(2.64)

Keep in mind that while the components of a second-order tensor are diﬀerent depending on what coordinate system they are referred to, the second-order tensor itself is still a physical quantity that is independent of the coordinate system to which it is referred, just like a vector.

22

2.3.2

Finite Element Method

Eigenvalues and eigenvectors of second-order tensors

Recall from the deﬁnition of a second-order tensor, T, that if we take the tensor product of T and some vector u, the result is another vector, say v. In general, both the magnitude and direction of v are not the same as those of u. Now suppose that we try to ﬁnd a unit vector n, such that when we take the tensor product of T and n, we get another vector that points in the same direction as n (but with a diﬀerent magnitude). This statement can be expressed mathematically as Tn = λn (2.65) or in component form Tij nj = λni

(2.66)

In other words, we are looking for a unit vector n such that T · n gives a vector that is proportional to n. The constant of proportionality, λ, is called an eigenvalue of T. By performing some algebraic manipulations, equation (2.66) can be written as Tij nj = λnj δij ⇒ (Tij − λδij )nj = 0

(2.67)

To see things more clearly, let us now write equation (2.67) in matrix notation. Doing this gives ⎤⎧ ⎫ ⎧ ⎫ ⎡ T11 − λ T12 T13 ⎨ n1 ⎬ ⎨ 0 ⎬ ⎣ T21 T22 − λ T23 ⎦ n2 = 0 (2.68) ⎩ ⎭ ⎩ ⎭ T31 T32 T33 − λ n3 0 Recalling our discussion in Section 2.3, it is now enlightening to view the el ements in each row of the matrix on the left-hand side of (2.68) as components of three vectors a, b, and c, e.g., a = (T11 − λ) e1 + T12 e2 + T13 e3 = ai ei b = T21 e1 + (T22 − λ) e2 + T23 e3 = bi ei c = T31 e1 + T32 e2 + (T33 − λ) e3 = ci ei

(2.69)

Using the deﬁnitions for a, b, and c above, equation (2.68) can now be recast as a·n = 0 b·n = 0 c·n = 0

(2.70)

The only way that all three of the dot products in (2.70) can be zero (for arbitrary nonzero vectors n) is if all three vectors a, b, and c all point in the

Mathematical Preliminaries

23

same direction. If this is the case, the scalar triple product (a × b) · c is zero, and hence the following determinant must also be equal to zero:

T11 − λ T12 T13

T21 T22 − λ T23

= 0 (2.71)

T31 T32 T33 − λ

Equation (2.71) can also be written in component form as |Tij − λδij | = 0

(2.72)

So it turns out that the eigenvalues of a second-order tensor T are the roots of the cubic equation obtained by expanding the determinant equation (2.71) (or equivalently (2.72)). Because the equation is a cubic polynomial, there are three eigenvalues. Once the eigenvalues are found, we can go back to equation (2.68) and solve for the eigenvectors, as demonstrated in the following example.

Example 2.3 It is desired to ﬁnd the eigenvalues and eigenvectors of a second-order tensor, T, whose components are given as ⎡ ⎤ 7 −1 2 (2.73) T = ⎣ −1 6 4 ⎦ 2 4 8 To get the eigenvalues, we must solve the following equation:

7−λ −1 2

−1 6 − λ 4

= 0

2 4 8 − λ

(2.74)

By expanding the determinant in equation (2.74) we get the following char acteristic equation: −λ3 + 21λ2 − 124λ + 176 = 0 (2.75) Solving (2.75) for the three roots λ1 , λ2 , and λ3 yields λ1 ≈ 11.36108849 λ2 ≈ 7.60076487 λ3 ≈ 2.03814664

(2.76)

To compute the eigenvectors, it is necessary to plug in the eigenvalues reported in equation (2.76) above into the following matrix equation: ⎡ ⎤⎧ ⎫ ⎧ ⎫ 7 − λ −1 2 ⎨ n1 ⎬ ⎨ 0 ⎬ ⎣ −1 6 − λ 4 ⎦ n2 = 0 (2.77) ⎩ ⎭ ⎩ ⎭ n3 2 4 8−λ 0

24

Finite Element Method

and solving for n1 , n2 , and n3 . Doing this for λ = 11.36108849 yields −4.36108849n1 − n2 + 2n3 = 0 −n1 − 5.36108849n2 + 4n3 = 0 2n1 + 4n2 − 3.36108849 = 0

(2.78)

Next, we solve the second equation in (2.77) above for n1 in terms of n2 and n3 , and plug the result back into the ﬁrst equation. Doing this we get 22.38018131n2 − 15.44435396n3 = 0 ⇒ n2 = 0.6900906541n3 n1 = 0.300362937n3

(2.79)

Finally, we make the eigenvector a unit vector by insisting that ni ni = 1. Normalizing in this way we get ⎧ ⎫ ⎧ ⎫ (1) ⎪ ⎪ ⎨ 0.23998773 ⎬ ⎨ n1 ⎬ (1) = 0.55137725 (2.80) n2 ⎪ ⎩ (1) ⎪ ⎭ ⎩ 0.79899249 ⎭ n 3

where the superscript refers to the fact that we have just solved for the com ponents of the eigenvector associated with the largest eigenvalue. Following the same procedure for the other two eigenvalues we obtain ⎧ ⎫ ⎧ ⎧ ⎫ ⎧ ⎫ ⎫ (3) ⎪ n(2) ⎪ ⎪ ⎪ n 0.89189235 ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ 0.38331962 ⎬ 1 1 (2) (3) 0.70234386 = −0.45021797 ; = (2.81) n2 n2 ⎪ ⎪ ⎩ ⎭ ⎪ ⎩ (2) ⎭ ⎩ (3) ⎪ ⎭ ⎩ −0.59981594 ⎭ 0.04279981 n n 3

2.3.3

3

Common operations with vectors and tensors

For the sake of completeness, in this section we will go over some common op erations using tensor notation that arise in solid mechanics and heat transfer. The operations begin with the tensor product and then progress to the tensor scalar products. The concepts of symmetric and skew second-order tensors will also be covered. This subsection ends with a table summarizing these common operations. Tensor product The tensor product of any two second-order tensors, A and B, is deﬁned as A · B = (Aij ei ej ) · (Bkl ek el ) = Aij Bkl δjk ei el = Aik Bkl ei el = Aik Bkj ei ej

(2.82)

Mathematical Preliminaries

25

By carefully following each step in equation (2.82) above, one will notice that the result of the tensor product operation is another tensor, C, whose components are deﬁned as (2.83) Cij = Aik Bkj The operation involves writing down, side by side, both tensors A and B in terms of the Cartesian base vectors, and then taking the dot product of the inner two base vectors. Tensor scalar product As the title suggests, the tensor scalar product between two second-order tensors produces a scalar. The operation is very common in the derivation of ﬁnite element equations. There are actually two tensor scalar products that we can deﬁne. The ﬁrst scalar product is written as α=A:B

(2.84)

β = A · ·B

(2.85)

and the second is written as In each case, the result is a scalar, but not necessarily the same. To describe the operation, the scalar product (2.84) is obtained by ﬁrst writing down, side by side, the two tensors in terms of the Cartesian base vectors, making sure that a subscript does not appear more than twice in the expression. In the second step we take the dot product of the left base vector associated with A and the left base vector associated with B. Multiplying everything together gives the desired scalar α, i.e., α=A:B = (Aij ei ej ) : (Bkl ek el ) = Aij Bkl δik δjl = Ail Bil = Aik Bik

(2.86)

The second scalar product (2.85) is very similar to the ﬁrst, except we take the dot product of the left base vector associated with A with the right base vector associated with B. The result is β = A · ·B = (Aij ei ej ) : (Bkl ek el ) = Aij Bkl δil δjk

= Aik Bki

(2.87)

As an example of the physical signiﬁcance of this operation, the tensor scalar product between the Cauchy stress tensor and the small-strain tensor gives twice the strain energy density (recoverable elastic energy per unit volume) at a point in a linearly elastic solid.

26

Finite Element Method

Symmetric and skew second-order tensors It turns out that any second-order tensor, T, can be decomposed into symmet ric and skew (antisymmetric) parts. The symmetric part, Tsym , is obtained by calculating the average of the tensor T and its transpose as follows: 1 (2.88) T + TT Tsym = 2 The skew part, Tskew , is obtained by subtracting TT from T as follows: 1 (2.89) T − TT Tskew = 2 Note that adding the symmetric and skew parts of T yields the tensor T, i.e., 1 1 T − TT Tsym + Tskew = T + TT + 2 2 1 T = T + T + T − TT 2 =T (2.90) An important operation that comes up frequently in the derivation of ﬁnite ele ment equations is the tensor scalar product between a symmetric second-order tensor, D, and another second-order tensor T, not necessarily symmetric. We can easily show that D : T = D : Tsym as follows: 1 D : T + TT 2 1 = (Dik Tik + Dik Tki ) 2 1 = (Dik Tik + Dki Tki ) 2 1 = (Dik Tik + Dik Tik ) 2 = Dik Tik =D:T

D : Tsym =

(2.91)

Note that we used the that fact that Dki = Dik in the third line of (2.91) above since D is symmetric. Note that in the fourth line we just replaced the dummy index k with i and vice versa. We end this section by listing some common tensor and vector operations in Table 2.1 below. The last row in Table 2.1 requires some further explanation. The identity can be proved as follows: (A · B) : C = (Aik Bkj ei ej ) : (Cmn em en ) = Aik Bkj Cmn δim δjn = Aik Bkj Cij

(2.92)

Mathematical Preliminaries

27

On the other hand, A : (C · BT ) = (Aij ei ej ) : (Ckl Bml ek em ) = Aij Ckl Bml δik δjm = Akm Ckl Bml = Aik Cij Bkj = Aik Bkj Cij

(2.93)

and B : (AT · C) = (Bij ei ej ) : (Alk Clm ek em ) = Bij Alk Clm δik δjm = Ali Bim Clm = Ali Bij Clj = Alk Bkj Clj = Aik Bkj Cij

(2.94)

The fact that the last lines in equations (2.92), (2.93), and (2.94) are all equal proves the identity. TABLE 2.1

Common operations with vectors and tensors. Description Symbol Result Vector dot product

a·b

Magnitude of a vector |v|

= (ai ei ) · (bk ek ) = ai bk δik = ai bi =

√

vi vi

Scalar triple product

a×b·c

= ijk ai bj ck

Trace of a tensor

tr T

=T:I = Tik δik = Tkk

Tensor transpose

TT

= Tij ej ei = Tji ei ej

Determinant

det T

= ijk T1i T2j T3k

A useful identity

A·B:C

= A : C · B T = B : AT · C

28

2.4

Finite Element Method

Calculus

The calculus is the fundamental mathematical tool that has allowed numerical techniques such as the ﬁnite element method to come into being. This section covers the essential ingredients that are needed for the development of both linear and nonlinear ﬁnite element procedures.

2.4.1

Fundamental theorem of calculus

One of the most important theorems that is presented in a ﬁrst course in calculus is the fundamental theorem of calculus. The theorem tells us that the deﬁnite integral of a function f (x) from a to b is obtained by ﬁnding the antiderivative, F (x), of the function and evaluating the diﬀerence, F (b)−F (a), i.e., b f (x) dx = F (b) − F (a) (2.95) a

In order for the theorem (2.95) to be applicable, the function F (x) must be continuous in the interval a ≤ x ≤ b. An informal proof of the theorem can be illustrated as follows. Consider the plot of some function F (x) as shown in Figure 2.5. The function f (x) rep resents the slope of the function F (x) at each point x. The quantity f (x)Δx represents the change in F (x) due to a movement Δx in the x direction. The summation of all the changes (in the limit as Δx → 0) is by deﬁnition the deﬁnite integral of f (x) from a to b. It gives the total change in F from point b a to point b. Hence a f (x) dx = F (b) − F (a).

2.4.2

Integration by parts

When deriving ﬁnite element equations, it is necessary to evaluate integrals that have the following form:

b

f (x)g(x) dx

(2.96)

a

A technique that can be used to evaluate such integrals is called integration by parts. The ﬁrst step in the procedure involves taking the derivative of the product of F (the antiderivative of f ) and the function g as follows: d [F (x)g(x)] = [F (x)g(x)] dx = f (x)g(x) + F (x)g (x)

(2.97)

Mathematical Preliminaries

29

F (x)

F (x)Δx = f (x)Δx Δx

x a

b

FIGURE 2.5 Fundamental theorem of calculus. and hence,

f (x)g(x) = [F (x)g(x)] − F (x)g (x)

(2.98)

where we have used the product rule to get the last line in (2.97). The second step is to replace f (x)g(x) in (2.96) with the right-hand side of (2.98). Doing this we get

b

b

f (x)g(x) dx = a

a

[F (x)g(x)] dx −

b

F (x)g (x) dx

(2.99)

a

Next, we apply the fundamental theorem of calculus introduced in the previous section to the ﬁrst integral on the right-hand side of (2.99) yielding a

b

b

[F (x)g(x)] dx = F (x)g(x)|a = F (b)g(b) − F (a)g(a)

(2.100)

Notice that the antiderivative of [F (x)g(x)] is simply F (x)g(x). Finally, combining (2.99) and (2.100) gives a

b

f (x)g(x) dx = −

b a

F (x)g (x) dx + {F (b)g(b) − F (a)g(a)}

(2.101)

Integration by parts is used in the development of the ﬁnite element method. Speciﬁcally, it comes into play when obtaining the weak form of the diﬀerential equation of interest. This will be discussed at great length in the next chapter.

30

2.4.3

Finite Element Method

Taylor series expansion

A key mathematical concept behind modern nonlinear ﬁnite element proce dures is the Taylor series expansion. Engineering students are usually intro duced to this concept in a ﬁrst course in calculus. Although powerful, the idea is quite simple, as will be explained below. We begin by assuming that some function, f (x), can be expressed in the form of a power series as follows: f (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + a3 (x − x0 )3 + ...

(2.102)

where a0 , a1 , a2 , etc. are unknown constants, and x0 is some point along the x axis. Here we tacitly assume that f (x0 ) exists. Once the unknown constants are found, the resulting expression is the Taylor series expansion of the function f (x) about the point x0 . The unknown constants can be found by repeatedly diﬀerentiating both sides of equation (2.102). To begin, notice that if we let x = x0 , all of the terms on the right-hand side vanish, except for a0 . Hence we conclude that a0 = f (x0 ). Diﬀerentiating both sides of equation (2.102) with respect to x, and again setting x = x0 , we get f (x0 ) = a1

(2.103)

Next, taking the second derivative of both sides of (2.102), we get f (x0 ) = 2a2

(2.104)

Repeating this process by taking the third derivative etc. gives the entire ex pansion f (x) = f (x0 ) + f |x0 (x − x0 ) +

f |x0 f |x0 (x − x0 )2 + (x − x0 )3 + ... (2.105) 2! 3!

where all derivatives in equation (2.105) are evaluated at the point x0 . For a given value of x, if the right-hand side of equation (2.105) tends toward the correct value of f (x) as the number of terms in the series is increased, the series is said to converge. If we retain only the ﬁrst two terms of the Taylor series expansion (2.105), the approximate representation for f (x) about x0 becomes f (x) = f (x0 ) + f |x0 Δx

(2.106)

The right-hand side of (2.106) is the equation of a straight line that is tangent to the curve f (x) and passes through the point f (x0 ). The second term on the right-hand side is the approximate change in the function f (x) with respect to a movement to the left or right of the point x0 . This “change” will be called Df . We note that the approximate change Df becomes exact as we get closer

Mathematical Preliminaries

31

and closer to the point x0 . The process of expanding a function in a twoterm Taylor series is called linearization. The operator D is the diﬀerential operator and follows the same rules as standard diﬀerentiation. Of practical importance is the fact that it obeys the product rule, i.e., D(f g) = (Df )(g) + (f )(Dg)

(2.107)

This property will aid in the linearization of functions that come from non linear ﬁeld problems discussed in Chapter 10. 2.4.3.1

Nonlinear functions of several variables

Now suppose that we have a function of two variables f (x, y). As in the previous section, we will express this function as a power series as follows: f (x, y) = a0 + a1 (x − x0 ) + a2 (y − y0 ) +a3 (x − x0 )2 + a4 (y − y0 )2 +a5 (x − x0 )3 + a6 (y − y0 )3 + . . .

(2.108)

If we evaluate both sides of equation (2.108) at the point (x0 , y0 ), it is clear that a0 = f (x0 , y0 ). If we take the partial derivative of both sides with respect to x and evaluate the resulting expressions at the point (x0 , y0 ) we get

∂f

a1 = (2.109) ∂x (x0 ,y0 ) Similarly, taking the partial derivative of both sides of (2.108) with respect to y we obtain

∂f

a2 = (2.110) ∂y (x0 ,y0 ) By continuing this process of taking partial derivatives of both sides of equa tion (2.108) with respect to x and y we ﬁnally obtain ∂f ∂f (x − x0 ) + (y − y0 ) f (x, y) = f (x0 , y0 ) + ∂x ∂y 2 2 ∂ f ∂ f ∂x2 ∂y 2 (x − x0 )2 + (y − y0 )2 + . . . + 2! 2!

(2.111)

In equation (2.111) above, it is understood that all partial derivatives are evaluated at the point (x0 , y0 ). In the solution of nonlinear ﬁnite element equations and nonlinear equations of several variables, it is common practice to approximate the function of several variables by the ﬁrst three terms of the Taylor series (2.111), i.e.,

∂f

∂f

Δx + Δy (2.112) f (x, y) ≈ f (x0 , y0 ) + ∂x (x0 ,y0 ) ∂y (x0 ,y0 )

32

Finite Element Method

The Taylor series approximation (2.112) above for a function of two variables is an equation for a plane that passes through point (x0 , y0 ) and is tangent to the surface z = f (x, y) at point (x0 , y0 ). Again we can write f (x, y) ≈ f (x0 , y0 ) + Df

(2.113)

∂f ∂f Δx + Δy is the approximate change in the function f (x, y) ∂x ∂y as a result of moving away from the point (x0 , y0 ). Again, the operator, D, obeys the usual rules of diﬀerentiation. where Df =

2.4.4

Gradient and divergence

In this section we deﬁne the gradient operator and explain what is meant by gradient of a scalar and gradient of a vector. We then address the concept of divergence. To begin, let us consider a scalar function f (x, y, z) of three variables x and y and z. If we use a linear Taylor series to approximate the function f in the vicinity of some point P having coordinates (x0 , y0 , z0 ), we get f (x, y, z) ≈ f (x0 , y0 , z0 ) +

∂f ∂f ∂f Δx + Δy + Δz ∂x ∂y ∂z

(2.114)

where it is understood that the partial derivatives in equation (2.114) are evaluated at (x0 , y0 , z0 ). The partial derivatives of f with respect to x, y, and z can be viewed as components of the vector ∇f , i.e., ⎧ ⎫ ∂f ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂x ⎪ ⎪ ⎨ ⎬ ∂f (2.115) ∇f = ⎪ ⎪ ∂y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂f ⎪ ⎩ ⎭ ∂z Similarly, the increments Δx = x − x0 , Δy = y − y0 , and Δz = z − z0 can also be interpreted as components of the vector Δx. The Taylor series (2.114) can now be written as f (x, y, z) ≈ f (x0 , y0 , z0 ) + ∇f · Δx

(2.116)

From our work in the previous section, we recognize that the second term on the right-hand side of (2.116) gives an approximation for the change in f due to a movement in the direction of Δx. Now suppose that n is a unit vector that points in the direction of Δx, and let be an inﬁnitesimal parameter. Then n = dx is an inﬁnitesimal vector

Mathematical Preliminaries

33

level surface

y

x tangent plane

z

P

FIGURE 2.6 Level surface and tangent plane. that points in the same direction as Δx. Taking the dot product of the two vectors ∇f and dx we get ∇f · dx = ∇f · n = Dn f

(2.117)

Here in equation (2.117), the quantity Dn f is called the directional derivative of the function f in the direction of n. The directional derivative is the rate of change of f in the direction of n. Hence, Dn f gives the change in f due to a small movement in the n direction. So we now see that the gradient of a scalar function is a very useful quantity. From ∇f we can get the rate of change of f in any direction simply by computing the directional derivative Dn f deﬁned in (2.117).

2.4.5

The level surface

Let us imagine for the moment that there is a physical scalar quantity, for ex ample temperature, associated with each point that lies on the curved surface shown in Figure 2.6. If the temperature has a constant value, T0 say, at each point on the surface, then the surface can also be represented mathematically as (2.118) T (x, y, z) = T0 Such a surface, on which the physical quantity does not change from point to point, is called a level surface. Now let n be any unit vector that lies in the tangent plane to the surface at point P . By deﬁnition of a level surface, the rate of change of the function T (x, y, z) in the direction of n is zero, i.e., Dn T = ∇T · n = 0

(2.119)

The statement (2.119) tells us that the gradient ∇T , evaluated at point P , must be perpendicular to the level surface at point P .

34

Finite Element Method

Recognizing that the level surface T (x, y, z) = T0 comes from a fully threedimensional temperature ﬁeld, T (x, y, z), we also conclude from (2.119) that the gradient acts in the direction in which that rate of change of temperature is maximum. To see this more clearly, if we march tangent to the level surface, the rate of change of temperature is zero. Because the rate of change is the temperature gradient ∇T dotted with a unit vector n, that dot product is maximum (and hence the rate of change of temperature) when n happens to be parallel to ∇T .

2.4.6

Divergence theorem

Having deﬁned the gradient of a scalar function, from now on, we can treat the gradient operator as just another vector. Letting x = X1 , y = X2 , and z = X3 , the gradient vector can be expressed as ∇=

∂ ei ∂Xi

(2.120)

Taking the dot product of ∇ and some other vector, say u, gives what is called the divergence of u, i.e., ∂ ei ) · (uj ej ) ∂Xi ∂uj ei · ej ∂Xi ∂uj δij ∂Xi ∂ui ∂Xi ui,i

∇·u = ( = = = =

(2.121)

The divergence theorem is an extremely useful tool for deriving partial dif ferential equations that arise from balance laws. It also is often used when deriving ﬁnite element equations that arise from partial diﬀerential equations. The purpose of this section is to present an informal proof of the theorem. Hopefully, this will make the interested reader feel comfortable using the the orem later on beginning in Chapter 4. To begin, consider a square domain as shown in Figure 2.7. Let us assume that the domain is of width Δx, height Δy, and thickness t (into the page). The vector q represents the ﬂux of something (e.g., heat or mass) going into and out of the square element as indicated by the arrows in the ﬁgure. We note that the ﬂux of a quantity is the amount of that quantity ﬂowing through a unit area per unit time. The components of q in this two-dimensional situation are qx and qy . For now, we will assume that no ﬂux can ﬂow into or out of the page. The quantity qx (x + Δx, y) is the ﬂux passing through the right

Mathematical Preliminaries

35 qy (y + Δy) y qx (x)

x

qx (x + Δx)

qy (y)

FIGURE 2.7 Flux going in and out of a square element.

face, and qx (x, y) is the ﬂux passing through the left face. The quantities qy (x, y) and qy (x, y + Δy) are the ﬂux components in the y direction ﬂowing into the bottom face and out the top face, respectively. For the remainder of this section, we will think of q as being heat ﬂux, the energy ﬂowing through a unit area per unit time. Having deﬁned heat ﬂux, let us now calculate the total energy per unit time, E˙ , ﬂowing out of the square element shown in Figure 2.7. To do this we need to multiply the ﬂux on each face by the cross-sectional area, and then add up the contributions from each face. Doing this yields E˙ = [qx (x + Δx, y) − qx (x, y)] tΔy + [qy (x, y + Δy) − qy (x, y)] tΔx

(2.122)

˙ be the total energy per unit volume ﬂowing out of the square. We now let W ˙ is obtained by dividing both sides of equation (2.122) by the The quantity W volume of the square element, tΔxΔy. The result, in the limit as Δx and Δy go to zero, can be expressed as ˙ = ∂qx + ∂qy W ∂x ∂y

(2.123)

˙ is called the energy density. From our work in the previous The quantity W section, the energy density can be expressed in terms of the gradient operator as follows: ˙ =∇·q W (2.124) which is the divergence of the vector q. We now see that the divergence of the vector q gives the total energy ﬂowing out of an inﬁnitesimally small region, per unit time, per unit volume. If the energy density is known at each point inside an arbitrary domain, Ω, bounded by the surface, Γ, the total energy ﬂowing out of Ω can be written

36

Finite Element Method

e+M

e

e+1

FIGURE 2.8 Flux going in and out of square domain.

as E˙ =

Ω

˙ dΩ W

=

Ω

∇ · q dΩ

(2.125)

Let us now calculate the total energy per unit time leaving an arbitrary domain Ω in a diﬀerent way. For the sake of simplicity, the domain of interest will be composed of an assemblage of square elements as shown in Figure 2.8. The square elements are arranged in M rows and N columns. Focusing on some element e labeled in the ﬁgure, let xeI and xeI+1 represent the x positions e denote the y positions of the left and right faces. Similarly let yJe and yJ+1 of the bottom and top faces. As long as no energy is created or destroyed, the energy ﬂowing out of element e through edge xeI+1 must be equal to the . In other words, the energy ﬂowing into element e + 1 through edge xe+1 I energy ﬂowing out of an element through its right edge, must be equal to the energy ﬂowing into the adjacent element through its left edge. One can also e argue that the energy ﬂowing out of element e through the top edge yJ+1 must be equal to the energy ﬂowing into element e + M through the bottom edge yJe+M . Here, M is the number of elements in a given row. Following this argument, if we perform an energy balance on the entire assemblage, on an element-by-element basis, the only contributions of energy ﬂowing into and out of the domain are indicated by the solid arrows in Figure 2.8. All of the dashed arrows cancel out of the total energy balance!

Mathematical Preliminaries

37

Let us now add up the total energy ﬂowing out of the domain, as indicated by the solid arrows. For the bottom, right, top, and left edges we get: E˙ = −

M

qy (x, y1e )tΔx +

e=1

N

qx (xeM +1 , y)tΔy

e=1

M

N

e=1

e=1

e qy (x, yN +1 )tΔx −

+

qx (xe1 , y)tΔy

(2.126)

In the limit as the element size is shrunk to zero (and the number of elements approaches inﬁnity) the energy balance statement (2.126) is equivalent to the following contour integral: ˙E = q · n dΓ Γ1 +Γ2 +Γ3 +Γ4 (2.127) + q · n dΓ Γ

where Γ1 to Γ4 in (2.127) represent the bottom, right, top, and left edges of the square domain, and Γ is the union of these four edges. In the derivation of the contour integral (2.127), we have used the following relationships between the ﬂux vector q and the unit vector n that acts perpendicular to each edge: bottom right top left

edge: edge: edge: edge:

q · n = −qy q · n = +qx q · n = +qy q · n = −qx

(2.128)

Finally, comparing equation (2.125) with equation (2.127) we arrive at the following result known as the divergence theorem: ∇ · q dV = n · q dΓ (2.129) V

Γ

Equation (2.129) above can be written in a more general form as ∇ · A dV = n · A dΓ V

(2.130)

Γ

where the quantity A can be a scalar, a vector, or a tensor. As long as the quantity A is “suﬃciently diﬀerentiable,”‡ then the volume integral of the divergence of A is equivalent to the surface integral of n · A. The divergence theorem is needed in the derivation of partial diﬀerential equations in engi neering and physics, as well as in deriving the weak forms of such equations necessary for a ﬁnite element implementation. For additional reading on vec tor calculus and the divergence theorem, see, e.g., Schey [4]. ‡ see Malvern [3] for a more detailed discussion on the continuity requirements of A for the divergence theorem to hold.

38

2.5

Finite Element Method

Newton’s method

Suppose we want to ﬁnd the value of x that satisﬁes the following equation: f (x) = 0

(2.131)

If f (x) is a nonlinear function of the variable x, ﬁnding the solution to equation (2.131) is not necessarily an easy task. A very powerful method for obtaining solutions to nonlinear equations is Newton’s method. The ﬁrst step is to approximate the function f (x) by a linear Taylor series expansion. Equation (2.131) then becomes f (x0 ) + f |x=x0 (x − x0 ) = 0

(2.132)

where x0 is called an initial guess. Letting Δx = (x − x0 ) and then solving (2.132) for Δx we get f (x0 ) (2.133) Δx = − f |x=x0 Once Δx is obtained, a (hopefully) better estimate for the solution to (2.131) is x1 = x0 + Δx. The quantity x1 can then be used as a next guess, and the entire process can be repeated. The iterative procedure can be written succinctly in two lines as follows: Δxi = −

f (xi ) f |x=xi

xi+1 = xi + Δxi

(2.134)

where i = 0, 1, 2, 3 . . . . Newton’s method for nonlinear functions of one variable can be described graphically as shown in Figure 2.9. As shown in the ﬁgure, after making the initial guess, x0 , one can check how much the function at this location, f (x0 ), deviates from zero. This diﬀerence represents the error associated with the initial guess. The error is referred to as the residual. If the residual is larger than some user-deﬁned tolerance, then the next guess, x1 , is computed using (2.134). The new residual, f (x1 ), is computed and checked against the tolerance. If it is still unacceptable, the process is repeated again. The process continues until the residual is small compared with the user-deﬁned tolerance. Newton’s method for functions of one variable is illustrated in the following example.

Example 2.4 In this example, we consider the problem of a block of mass m attached to a nonlinear spring as shown in Figure 2.10. The coordinate z shown in the

Mathematical Preliminaries

39

f (x) f (x0 )

x0

f (x2 )

x2

x1

x

f (x1 )

FIGURE 2.9 Newton’s method. ﬁgure is measured positive downward. The position z = 0 corresponds to the position of the block when the spring is unstretched (i.e., when there is no tension or compression in the spring). Now suppose that the tension in the spring, t, is related to its elongation, z, by the following equation: t = 2z 5 + 1000z 0.6

(2.135)

We desire to ﬁnd the equilibrium position of the spring when the block is subjected to the inﬂuence of gravity. To solve the problem we will take m = 1.0 kg and g = 9.81 m/s2 . To be consistent with our units, it is understood

z

FIGURE 2.10 Spring-mass example.

40

Finite Element Method

that the elongation of the spring will be in meters, and the tension in the spring will be in Newtons. Starting from some conﬁguration in which the spring is stretched, equilib rium of the block in the z direction requires that −t + mg = 0 and hence

2z 5 + 1000z 0.6 − 9.81 = 0

(2.136)

Letting f (z) = 2z 5 + 1000z 0.6 − 9.81 and f (z) = 10z 4 + 600z −0.4 we can now begin the Newton’s method iterations. A logical starting point is to pick z0 = 0 (the location of the mass when the spring is unstretched) as the initial guess. In this problem, however, the slope of the tension versus elongation curve tends to inﬁnity as z goes to zero. This guess would lead to numerical diﬃculties when the algorithm attempts to divide by zero in the ﬁrst iteration. To alleviate this problem, we will choose a small positive value z0 = 0.0001 as the initial guess and proceed. The numerical results for the ﬁrst ﬁve iterations are tabulated in Table 2.2 below.

2.5.1

Nonlinear equations of several variables

Now consider two nonlinear functions f (x, y) and g(x, y). Suppose that we wish to ﬁnd the values of x and y such that f (x, y) = 0

g(x, y) = 0

(2.137)

Here we will illustrate Newton’s method for getting approximate solutions to the system of equations (2.137) above. The ﬁrst step involves expressing the functions f (x, y) and g(x, y) as linear Taylor series expansions. This results in the following two linear equations: ∂f ∂f Δx + Δy = 0 ∂x ∂y ∂g ∂g g(x0 , y0 ) + Δx + Δy = 0 ∂x ∂y

f (x0 , y0 ) +

(2.138)

TABLE 2.2 Numerical results for spring-mass example. i zi × 10−4 (m) f (zi ) (N) 0 1 2 3 4

1.000000000 3.440267648 4.438651652 4.495393114 4.495538061

−5.828928294 −1.454829074 −0.074670893 −1.897806887 × 10−4 −1.223813939 × 10−9

f (zi ) (N/m)

Δzi (m)

23886.43023 14571.83878 13159.84655 13093.15110 13092.98224

2.440267648 × 10−4 9.983840033 × 10−5 5.674146179 × 10−6 1.449465352 × 10−8 9.347098441 × 10−14

Mathematical Preliminaries

41

Here in equation (2.138), the initial guess is the ordered pair (x0 , y0 ). It is understood that the partial derivatives in (2.138) are to be evaluated at the initial guess. The quantities Δx and Δy are deﬁned as

Δx = x − x0 Δy = y − y0

The two linearized equations (2.138) for f (x, y) and g(x, y) both represent planar surfaces. One planar surface is tangent to the surface z = f (x, y) at point (x0 , y0 ), and the other is tangent to the surface z = g(x, y) at point (x0 , y0 ). The intersection of each planar surface with the plane z = 0 forms a pair of intersecting straight lines. The point where the two straight lines intersect is an approximate solution to the system of equations (2.137). In Newton’s method, this approximate solution is subsequently used as the initial guess, and the process is repeated until the solution “converges” to the exact solution. To proceed further, let us now write the linearized system of equations (2.138) in matrix notation as follows: ⎡

∂f ⎢ ∂x ⎢ ⎣ ∂g ∂x

⎤ ∂f ∂y ⎥ ⎥ Δx = −f (x0 , y0 ) −g(x0 , y0 ) ∂g ⎦ Δy ∂y

(2.139)

The 2 × 2 matrix on the left-hand side of equation (2.140) above is called the Jacobian matrix J. The iterative incremental equations can be written as ⎡

∂f ⎢ ∂x ⎢ ⎣ ∂g ∂x

⎤ ∂f i i i ∂y ⎥ −f (xi , yi ) ⎥ Δx = −g(xi , yi ) ∂g ⎦ Δy ∂y

(2.140)

where i = 0, 1, 2, 3, . . . is the iteration counter. The entire procedure is illus trated in the following example.

42

Finite Element Method z = f (x, y) y x z z = g(x, y)

z=0 FIGURE 2.11 Newton’s method for functions of several variables.

Example 2.5 Consider the following two equations: x2 + y 2 − 1 = 0

(2.141)

x2 − y = 0

(2.142)

Both equations (2.141) and (2.142) represent a surface in three-dimensional space. These surfaces are shown in Figure 2.11. Notice that the intersection of the function z = x2 + y 2 − 1 with the plane z = 0 is a circle of radius one. The intersection of the surface z = x2 − y with the plane z = 0 is a parabola. Let us now look for (x, y) pairs in the ﬁrst quadrant that satisfy both equations (2.141) and (2.142) simultaneously. We will start with the point (1, 1) as our initial guess. The Jacobian matrix for the system of equations (2.141) and (2.142) is obtained by taking partial derivatives of the functions f (x, y) and g(x, y) with respect to x and y. The linearized system of equations is then written as 2x 2y Δx −(x2 + y 2 − 1)

(2.143)

= −(x2 − y) 2x −1 Δy The linearized representations of the functions f and g are planar surfaces, as shown in Figure 2.12. In this ﬁgure, the dark shaded surface represents the plane z = 0. The medium-shaded plane is the linearized representation of f , and the light shaded plane is the linearized representation of g. Notice that the intersection of each plane with the plane z = 0 is a line, and the intersection of the two lines is the solution to the linear system of equations (2.143).

Mathematical Preliminaries y

43 z

z = f (x, y) x

z=0

z = g(x, y)

FIGURE 2.12 Intersecting planes.

In Newton’s method, equation (2.143) is used in an iterative manner to obtain successively better estimates for the actual solution, starting with an initial guess (x0 , y0 ). During the ﬁrst iteration, the Jacobian matrix is set up and evaluated at the initial guess. The right-hand side is then set up and evaluated at the initial guess. The right-hand side is often called the residual vector. As the guesses for x and y become better and better, the right-hand side of (2.143) gets closer and closer to zero. After the right-hand side is evaluated at the initial guess, the linear system of equations is solved for Δx and Δy. Finally, the new guess is obtained by adding these increments to the previous guess. In the present example, the ﬁrst three iterations of Newton’s method are carried out in Table 2.3 below. The iterations stop when the residual is small compared with a user-deﬁned tolerance. A very attractive feature of Newton’s method from a computational point of view is that if the initial guess is not too far oﬀ from a solution to the system of nonlinear equations, and if the Jacobian matrix is formulated correctly, then the method exhibits quadratic convergence. TABLE 2.3 Newton’s method example. i J R 2.0000 2.0000 −1.0000 0 2.0000 −1.0000 0.0000 1.6667 1.3333 −0.1389 1 1.6667 −1.0000 −0.0278 1.5762 1.2381 −0.0043 2 1.5762 −1.0000 −0.0020

Δx −0.1667 −0.3333 −0.0452 −0.0476 −0.0019 −0.0010

x 0.8333 0.6667 0.7881 0.6190 0.7862 0.6180

44

Finite Element Method 1.0 0.8 0.6 |R| 0.4 0.2 0 1

3 i

2

4

5

FIGURE 2.13 Quadratic convergence.

To give a brief explanation of the what is meant by quadratic convergence, let us examine the residual vector R. The magnitude of this vector is deﬁned as |R| =

Ra Ra

(2.144)

where Ra is an element of R, and the summation convention is implied. In the present example, the magnitude of the residual vector |R| is plotted versus the iteration number as shown in Figure 2.13. Notice the rapid rate at with the magnitude of the residual vector decreases. The relative residual can be deﬁned as Ri =

|R|i+1 |R|i

(2.145)

Quadratic convergence occurs when the relative residual decays quadratically, i.e., Ri+1 ≈ (Ri )2

(2.146)

or in other words, when the new relative residual computed within iteration i + 1 is approximately equal to the square of the previous relative residual computed in iteration i.

Mathematical Preliminaries

45

p xp P XP

dx

x

dX

X

FIGURE 2.14 The undeformed and deformed conﬁgurations.

2.6

Kinematics of motion

Fundamental to the study of continuum mechanics is the notion that a con tinuous body is made up of inﬁnitely many material points. When the body is deformed by the action of external forces, for example, the motion of the material points can be described by a mathematical function that depends on the initial location of the material points and time. This concept is illustrated in Figure 2.14 below. In the ﬁgure, the position of material point P in the un deformed conﬁguration is deﬁned by the position vector XP . At any instant of time, the position of this same material point p in the deformed conﬁgu ration is deﬁned by the position vector xp . Given the general position, X, of a material point in the undeformed conﬁguration, the location of a material point, x, in the deformed conﬁguration is given by the following function or mapping: x = x(X) (2.147) Equation (2.147) can be thought of as three equations (in general nonlinear) of the three independent variables X1 , X2 , and X3 , i.e., x1 = x1 (X1 , X2 , X3 ) x2 = x2 (X1 , X2 , X3 ) x3 = x3 (X1 , X2 , X3 )

(2.148)

In other words, each component of the position vector x is a function of X1 , X2 , and X3 . Unless stated otherwise, we will assume that the mapping of

46

Finite Element Method

material points from the undeformed conﬁguration to the deformed conﬁg uration is one to one. That is, each material point is mapped to a unique location in the deformed conﬁguration (one material point cannot occupy two diﬀerent locations in the deformed conﬁguration). Let us now look at other material points in a small region surrounding point P . If this region is inﬁnitesimal, the motion of these points can be described by a two-term Taylor series expansion of equation (2.148) as follows: x1 = x1 (X1P , X2P , X3P ) ∂x1 (X1 − X1P ) + + ∂X1 x2 = x2 (X1P , X2P , X3P ) ∂x2 + (X1 − X1P ) + ∂X1 x3 = x3 (X1P , X2P , X3P ) ∂x3 (X1 − X1P ) + + ∂X1

∂x1 ∂x1 (X2 − X2P ) + (X3 − X3P ) ∂X2 ∂X3 ∂x2 ∂x2 (X2 − X2P ) + (X3 − X3P ) ∂X2 ∂X3 ∂x3 ∂x3 (X2 − X2P ) + (X3 − X3P ) (2.149) ∂X2 ∂X3

where all partial derivatives are to be evaluated at point P . The three equa tions in (2.149) can be written in matrix notation as ⎤ ⎡ ∂x1 ∂x1 ∂x1 ⎫ ⎧ ⎫ ⎧ p⎫ ⎢ ⎥⎧ 1 ∂X2 ∂X3 ⎥ ⎨ X1 − X P ⎬ ⎨ x1 ⎬ ⎨ x1 ⎬ ⎢ ∂X 1 ⎢ ∂x2 ∂x2 ∂x2 ⎥ p x2 = x (2.150) +⎢ ⎥ X2 − X2P ⎭ ⎩ ⎭ ⎩ 2p ⎭ ⎢ ∂X1 ∂X2 ∂X3 ⎥ ⎩ x3 x3 ⎣ ∂x3 ∂x3 ∂x3 ⎦ X3 − X3P ∂X1 ∂X2 ∂X3 or x − xp = J · (X − XP ) ⇒ dx = J · dX

(2.151)

where J is the 3×3 Jacobian matrix associated with the mapping between the undeformed and deformed conﬁgurations, dX = X − Xp is an inﬁnitesimal vector in the undeformed conﬁguration pointing from XP to some point X in the neighborhood of P , and dx = x − xp is an inﬁnitesimal vector pointing from xp to x in the deformed conﬁguration. In the study of kinematics of deformable bodies, the Jacobian matrix associ ated with the mapping between the undeformed and deformed conﬁgurations is known as the deformation gradient tensor F. The deformation gradient tensor contains information about the motion of material points within an inﬁnitesimal region surrounding point XP . Hence equation (2.151) can equiv alently be written as dx = F · dX (2.152)

Mathematical Preliminaries

47

In terms of its Cartesian components, the deformation gradient tensor can be written as ∂xi (2.153) Fij = ∂Xj The deformation gradient tensor is a very useful quantity, as we will see later in our study of ﬁnite element methods for large strain and deformation problems in Chapter 10. As an illustration, let ds be the length of a line segment dx in the deformed conﬁguration. The length ds can be expressed in terms of F as follows: ds2 = dx · dx = (F · dX) · (F · dX) = dX · (FT · F) · dX = dX · C · dX

(2.154)

where C is the right Cauchy-Green deformation tensor. So given any vector dX in the undeformed conﬁguration, the length of this vector in the deformed conﬁguration can be computed from equation (2.154).

2.7

Problems

Problem 2.1 Letting

⎡

⎤ 1 −1 0 2 3 4 K = ⎣ −1 2 −1 ⎦ ; B = ; 0 1 5 0 −1 1 perform the following matrix multiplications: (a) (c)

Kd T

B B

(b)

Bd

(d)

KB

⎧ ⎫ ⎨ 2⎬ d = −2 ⎩ ⎭ 1

Problem 2.2 ⎡

⎤ 1 2 0 A = ⎣2 4 2⎦; 1 4 −1 Determine the following:

Let

(a)

det A

(b)

det B

(c)

−1

(d)

B−1

A

1 −1 B= −1 3

48

Finite Element Method

Problem 2.3 Consider the following system of linear algebraic equations: d1 − d2 = R 1 −d1 + 2d2 − d3 = 0 −d2 + d3 = 1 where R1 is an unknown constant. Solve the system of equations for d2 and d3 subject to the constraint d1 = 0. After ﬁnding d2 and d3 , plug d2 back into the ﬁrst equation and solve for the constant R1 . Problem 2.4 ⎡

⎤ x 0 3x3 A = ⎣ 3 x2 + 1 x4 ⎦ 2x 6x3 − x 1

Let

1

Evaluate

−1

A dx.

Problem 2.5 The Cartesian components of a second-order tensor, T, are provided in the matrix below, ⎡ ⎤ 7 −1 −6 [Tij ] = ⎣ −1 6 −3 ⎦ −6 −5 11 With the help of Table 2.1, evaluate the following: (a)

det T

(b)

T:T

(c)

T·T

(d)

sym

(e)

tr T

(f)

Tskew

T

Problem 2.6 The Cartesian components of a symmetric second-order tensor, T, are given as ⎡ ⎤ 2 −1 0 [Tij ] = ⎣ −1 1 −1 ⎦ 0 −1 1 Determine the components of T in a primed coordinate system that is created by rotating the ﬁxed Cartesian axes through an angle of π/6 about the X3 axis.

Problem 2.7 Compute the eigenvalues and eigenvectors of the tensor deﬁned in Problem 2.6.

Mathematical Preliminaries

49

Problem 2.8 Consider the following nonlinear function: f (x) = x4 − 3x + 2 Perform two iterations of Newton’s method to obtain an estimate for the root of f (x) in the vicinity of x = 1. Problem 2.9 Consider the following system of nonlinear equations: x2 + 4y 2 = 1 4 x2 + y 2 = 1 Perform two iterations of Newton’s method to obtain an estimate for the location where the two curves intersect in the ﬁrst quadrant. Problem 2.10 Using the divergence theorem, show that the volume, V , enclosed by the surface Γ can be obtained by evaluating the following surface integral: 1 V = n · x dΓ 3 Γ where x is a position vector from the origin of a Cartesian coordinate system to a point on the surface, and n is the unit outward normal vector to the surface at that point. Problem 2.11 Consider the following two-dimensional mapping between the undeformed and deformed conﬁgurations: x1 = X12 + X2 x2 = X1 + 2X2 where x1 and x2 represent the coordinates of a point in the deformed conﬁg uration that originally occupied point (X1 , X2 ) in the original conﬁguration. Evaluate the components of the deformation gradient tensor F at the location (X1 = 0, X2 = 0). Problem 2.12 Write out the Taylor series expansions for the following functions:

50

Finite Element Method

(a)

cos θ

(b)

sin θ

(c)

eiθ

From the results obtained in parts (a), (b), and (c) show that eiθ = cos θ + i sin θ where i is the imaginary number, i.e., i2 = −1. The identity in the above equation is known as Euler’s equation.

3 One-Dimensional Problems

Everything should be made as simple as possible, but not simpler. —Albert Einstein In order to demonstrate the mathematical concepts behind the ﬁnite element method, we begin by applying the method to solve second-order, ordinary diﬀerential equations with constant coeﬃcients. Employing such a powerful numerical method to solve such simple equations is akin to using a hammer to kill a ﬂy. Nevertheless, with an understanding of the concepts presented in this chapter, the serious reader will be able to get through the rest of the book with relative ease. To begin, consider the following diﬀerential equation: au + bu + c = 0;

0≤x≤L

(3.1)

along with the boundary conditions u(0) = 0;

u (L) = 1

When students are ﬁrst introduced to diﬀerential equations, they are given the exercise of “plugging in” a function into a diﬀerential equation and checking whether or not the function is a solution to the equation. In the ﬁnite element method, such a candidate function is called a trial function. Throughout this chapter, the trial function will be denoted as u ˜(x). One of the main ideas in the ﬁnite element method is to break up the interval of interest, i.e., 0 ≤ x ≤ L in the example above, into subintervals called elements. For the sake of simplicity, it is desirable to use trial functions that are continuous piecewise. A continuous piecewise trial function is a function that does not have any “jumps,” but it can have “kinks” (locations where the slope of the trial function is not continuous). An illustration of what is meant by continuous piecewise trial functions is shown in Figure 3.1. Here, we have broken up the interval 0 ≤ x ≤ L into three elements, all of equal length. The solid line in the ﬁgure is an example of a piecewise linear trial function. The dashed line represents a piecewise quadratic trial function. The derivative of each type of trial function, u˜ , is plotted in the lower plot in Figure 3.1. Notice the jumps in the curves at the junctions between each element. These junctions are referred to as nodal locations, or simply nodes.

51

52

Finite Element Method

u ˜

linear quadratic

1

2

3

4

1

2

3

4

u ˜

FIGURE 3.1 Continuous piecewise trial functions.

When one tries to play the game of plugging in continuous piecewise trial functions into the diﬀerential equation (3.1), he encounters a problem. What is it? One can diﬀerentiate the trial function once, as shown in the lower plot of Figure 3.1, but the second derivative does not exist! The trial functions shown in Figure 3.1 are classiﬁed as C 0 continuous. This means that the functions are continuous, but their ﬁrst derivatives are piecewise continuous. There are diﬀerent classes of trial functions that are commonly used in the ﬁnite element method having varying degrees of continuity. For example, a C 1 trial function has a continuous ﬁrst derivative, but its second derivative is piecewise continuous. Trial functions that are C 1 continuous are commonly used in the formulation of beam elements discussed in Chapter 5. In order to remedy the situation of not being able to plug in the trial functions directly into the diﬀerential equation, it is necessary to derive what is called the weak form of the diﬀerential equation. The diﬀerential equation along with the boundary conditions is referred to as the strong form. The weak form is a mathematically equivalent statement to the strong form. One may like to think of it as the strong form in disguise. It turns out that the trial functions can be substituted into the weak form without any diﬃculty. After that, the game will be to ﬁnd trial functions that satisfy the weak form.

One-Dimensional Problems

3.1

53

The weak form

In the diﬀerential equation (3.1) above, because it is second order, there must be two boundary conditions attached to it. The ﬁrst boundary condition imposes a direct restriction on the function u at the left end of the interval, i.e., u(x = 0) = 0. This type of boundary condition is called a Dirichlet boundary condition, named after mathematician Lejeune Dirichlet. In the ﬁnite element literature, it is called an essential boundary condition. The boundary condition at x = L is a condition on the slope (or ﬁrst derivative) of the function u. A boundary condition that places a restriction on the ﬁrst derivative of the ﬁeld variable is called a Neumann boundary condition after mathematician John von Neumann. A Neumann condition is also called a natural boundary condition. Now that we have discussed the various types of boundary conditions that can be associated with the diﬀerential equation, we now introduce what is called a test function. A test function is a mathematical concoction with the following properties: 1. w(x) is C 0 continuous in 0 ≤ x ≤ L. 2. w(x) = 0 at all essential boundaries. In other words, a test function comes from a family of functions that are continuous on the interval of interest and are zero when evaluated at the essential boundary points. Having deﬁned the test function, the next step in the derivation of the weak form is to multiply each term in the diﬀerential equation (3.1) by w(x), i.e., au w + buw + cw = 0

(3.2)

and then to integrate both sides from 0 to L. Doing this we obtain

L

a

u w dx + b 0

L

L

uw dx + c 0

w dx = 0

(3.3)

0

Equation (3.3) is still not in a suitable form, because it contains the u term inside the ﬁrst integral on the left-hand side. Hence, we still cannot plug in C 0 trial functions into this equation. To remedy the situation, we can integrate the ﬁrst integral in equation (3.3) by parts. The goal of integrating by parts is to remove one of the derivatives on u(x) and put it on the function w(x), without changing the value of the integral. Using the techniques presented in Chapter 2, we now express the ﬁrst integral on the left-hand side of (3.3) as follows: L

a 0

L

u w dx = a

0

(u w) dx − a

L

0

u w dx

(3.4)

54

Finite Element Method

where we have used the relation (u w) = u w + u w ⇒ u w = (u w) − u w

(3.5)

From the fundamental theorem of calculus discussed in Section 2.4.1 in Chap ter 2, we see that the ﬁrst integral on the right-hand side of (3.4) has the form L f (x) dx = f (L) − f (0) 0

and hence,

L

0

(u w) dx = u (L)w(L) − u (0)w(0)

(3.6)

Because the test function w(x) is deﬁned to be zero at the essential boundary points, we see that for the present diﬀerential equation deﬁned by equation (3.1) we must have w(0) = 0. The value of the test function at all natural boundary points (at x = L in the present case) is not known and therefore the value w(x = L) is arbitrary. Equation (3.3) can now be written as

L

u w dx = b

a 0

L

L

w dx + au (L)w(L)

uw dx + c 0

(3.7)

0

Finally, from the natural boundary condition in (3.1) (u (L) = 1) we obtain the weak form of the diﬀerential equation as

L

u w dx = b

a 0

L

L

uw dx + c 0

w dx + aw(L)

(3.8)

0

While the weak form is mathematically equivalent to the diﬀerential equation (3.1), it has the advantage that only one derivative appears on the function u, and therefore continuous piecewise trial functions can now be substituted into the weak form without any diﬃculty. Notice that we can integrate the ﬁrst derivative of a continuous piecewise trial function even though it has jumps (see Figure 3.1).

3.2

Finite element approximations

In this section we will now formally deﬁne some common C 0 trial functions that can be substituted into equation (3.8).

One-Dimensional Problems

55

u ˜2 u ˜e u ˜1 x ˆ l

1

2

FIGURE 3.2 Linear-u element.

3.2.1

The linear-u element

We will begin by deﬁning piecewise linear trial functions. Let us consider the subinterval, or element, as shown in Figure 3.2. As shown in the ﬁgure, node 1 is located at the left end of the element, and node 2 is located at the right end. Over the length l of the element, the trial function varies linearly as shown in the ﬁgure. This portion of the trial function (deﬁned over a single element) ˜e . Here we use the subscript e to emphasize the fact that will be denoted as u the trial function is deﬁned over a single element. The superposed tilde is used to emphasize the fact that the trial function, in general, is approximate. The trial function has the value u1 at the left node and u2 at the right node. To facilitate the rest of our discussion, it is convenient to introduce the local coordinate x ˆ measured from the left end of the element as shown in Figure 3.2. ˆ, and We are now in position to express the trial function u ˜e as a function of x the values of the trial function u1 and u2 at the nodal points. In order to ˆ we can write accomplish this, since u ˜e is a linear function of x u ˜e (ˆ x) = c1 + c2 x ˆ

(3.9)

where c1 and c2 are constants. To determine the constants, we use the fol lowing end conditions: u ˜e (0) = u1

u ˜e (l) = u2

(3.10)

Plugging in the ﬁrst condition in (3.10) into equation (3.9) we obtain u1 = c1 . Substituting the second condition into (3.9) we get u2 = u1 + c2 l, or c2 = (u2 − u1 )/l. Plugging the expressions for c1 and c2 into equation (3.9) and simplifying we obtain u ˜e (ˆ x) = (1 −

x ˆ x ˆ )u1 + u2 l l

(3.11)

56

Finite Element Method 1 N1

N2

x ˆ 1

2

FIGURE 3.3 Shape functions for linear-u element. The two functions of x ˆ that appear in equation (3.11) are interpolation func tions called shape functions. Notice that for the linear-u element, both shape ˆ. functions are linear in x It is standard procedure in the ﬁnite element method to store the shape functions associated with a single element in a matrix called Ne . For the linear-u element we have ˆ x/l ˆ Ne = 1 − x/l x) N2 (ˆ x) (3.12) = N1 (ˆ where N1 (ˆ x) and N2 (ˆ x) are the shape functions. These shape functions are shown in Figure 3.3. Notice that the ﬁrst shape function is equal to one when evaluated at the left end of the element (node 1) and is equal to zero when evaluated at the right end (node 2). The second shape function is equal to zero at node 1 and one at node 2. All standard ﬁnite element shape functions have this property. The property can be stated succinctly as xJ ) = δIJ NI (ˆ

(3.13)

Equation (3.13) states that the shape function NI evaluated at node J, which occupies position x ˆJ , is equal to δIJ , the Kronecker delta deﬁned in Chapter 2. Here the subscripts I and J take on the values 1 to nen, where nen is the number of nodes owned by the element (two nodes in this case). The onedimensional shape functions can also be derived by choosing a local coordinate to be zero at the center of a linear-u element of length le . The domain of ˆ ≤ le /2. Letting ξ = 2ˆ x/le , the one the element then becomes −le /2 ≤ x dimensional shape functions become 1 (1 − ξ) 2 1 N2 = (1 + ξ) 2

N1 =

One-Dimensional Problems

57

Having introduced the shape function matrix Ne , it is convenient to rewrite equation (3.11) in matrix notation as follows: u1 x) N2 (ˆ x) x) = N1 (ˆ u ˜e (ˆ u2 N u = 1 ×e2 2 ×e1

(3.14)

Later on, when it comes time to plug the trial function into the weak form, x) with ˜e (ˆ it will be necessary to take the derivative of the trial function u respect to x. Since x ˆ = x − x1 , where x1 is the location of node 1, we have u ˜e =

d˜ ue dx

d˜ ue dˆ x dˆ x dx 1 1 = − u1 + u2 l l =

(3.15)

Next, we write equation (3.15) in matrix form as u1 u ˜e = −1/l 1/l u2 B u = 1 ×e2 2 ×e1

(3.16)

The matrix Be in equation (3.16) is commonly referred to as the B-matrix. The B-matrix operates on the nodal unknowns (u1 and u2 in this case), yield ing the derivative of the trial function at each point along the length of the element.

3.2.2

The quadratic-u element

Another important one-dimensional ﬁnite element is the quadratic-u element. In this element we assume that the trial function varies quadratically over the length of the element. We therefore need three nodes to uniquely specify a quadratic function over the element. The trial function is expressed as a ˆ as follows quadratic function of x ˆ + c3 x ˆ2 u ˜(ˆ x) = c1 + c2 x

(3.17)

The quadratic-u element is shown in Figure 3.4. The displacement values at each of the three nodes are labeled u1 , u2 , and u3 , respectively. Like the linear-u element, it is desirable to express the trial function in terms of shape functions and the nodal displacements. To derive the shape functions for the

58

Finite Element Method u2 u3

u1 x ˆ 2

1

3

FIGURE 3.4 Quadratic-u element. quadratic-u element, we impose the following three conditions at the nodal locations: x = −le ) = u1 u ˜e (ˆ u ˜e (ˆ x = 0) = u2 u ˜e (ˆ x = +le ) = u3

(3.18)

After substituting the conditions (3.18) into equation (3.17) we obtain u1 = c1 − c2 le + c3 le2 u2 = c1 u3 = c1 + c2 le + c3 le2

(3.19)

Solving equation (3.17) for c1 , c2 , and c3 yields 1 1 u ˜e (ξ) = − ξ(1 − ξ)u1 + (1 − ξ 2 )u2 + ξ(1 + ξ) 2 2

(3.20)

where we have used the change of variables x/le ξ = 2ˆ We can write equation (3.20) in matrix form as ⎧ ⎫ ⎨ u1 ⎬ 1 1 u2 u ˜e = − ξ(1 − ξ) 1 − ξ 2 ξ(1 + ξ) ⎩ ⎭ 2 2 u3

N u = 1 ×e3 3 ×e 1

(3.21)

One-Dimensional Problems

59 N2

1 N1

N3

ξ ξ = −1

ξ=0

ξ = +1

FIGURE 3.5 Shape functions for quadratic-u element. Notice that the variable ξ, also known as a natural or dimensionless coordi nate, ranges from −1 ≤ ξ ≤ +1. The functions that depend on the natural coordinate are the shape functions. Notice also that for the quadratic-u el ement, there are three shape functions, one associated with each node. The three shape functions are plotted in Figure 3.5. Notice again that the shape functions obey the Kronecker delta property (3.13). If we now diﬀerentiate the trial function with respect to x we obtain u ˜e =

d˜ ue dx

d˜ ue dξ dξ dx ⎧ ⎫ ⎨ u1 ⎬ 2 1 1 u2 = − + ξ −2ξ + ξ ⎩ ⎭ le 2 2 u =

3

B u = 1 ×e3 3 ×e 1

3.3

(3.22)

Plugging in the trial and test functions

For a given type of ﬁnite element, whether it be linear, quadratic, or whatever, we are now in a position to substitute the desired trial function into the weak form (3.8). To do this, it is convenient to break up the interval from 0 ≤ x ≤ L

60

Finite Element Method

into many smaller subintervals called elements. The integrals from 0 to L that appear in (3.8) can be evaluated by summing up the integrals over the individual elements. After substituting the trial function and the test function into the weak form we obtain nel nel u ˜e we dx ˆ=b u ˜e we dx ˆ a e=1

Ωe

e=1

+c

Ωe

nel e=1

Ωe

we dx ˆ + awn

(3.23)

where we = we (ˆ x) is the test function, deﬁned over an individual element, and Ωe is the domain of an individual element (either 0 ≤ x ≤ le or −le /2 ≤ x ≤ le /2). Here in equation (3.23), n is the number of nodal points along the interval 0 ≤ x ≤ L, and the quantity wn is the value of the test function evaluated at the node located at x = L. The quantity nel refers to the total number of elements along the interval 0 ≤ x ≤ L. Because the test function w(x) is arbitrary (just as long as it meets the two requirements for a test function) it is natural to assume that it varies along each element in the same manner as the trial function. That is, for given values w1 , w2 , . . ., wn of the test function at the nodal locations, the same shape functions that were used to interpolate the trial function are used to interpolate the test function nodal point data over the length of the element. This procedure of using the same interpolation functions for both the trial and test functions is consistent with a Galerkin method. Hence we take x) = Ne we w ˜e (ˆ w ˜e (ˆ x) = Be we

(3.24)

where Ne is the matrix containing the element shape functions, and Be is the element B-matrix. Again, the quantity we is a vector containing the values of the test function at the nodal points. Incidentally, both Ne and Be are 1 × 2 matrices for the linear-u element and 1 × 3 matrices for the quadratic-u element. Before we proceed further, it is useful to write down the following identities: ˜e = w ˜e u ˜e u ˜e w = (Be we )(Be ue ) = (Be we )T (Be ue ) = weT BTe Be ue

(3.25)

Similarly, we can write ˜e = w ˜e u ˜e u ˜e w = (Ne we )(Ne ue ) = (Ne we )T (Ne ue ) = weT NTe Ne ue

(3.26)

One-Dimensional Problems

61 1

1

e

2

2

4

3

FIGURE 3.6 Local and global node numbers. and ﬁnally x) = Ne we w ˜e (ˆ = (Ne we )T = weT NTe

(3.27)

When performing the manipulations (3.25) through (3.27), one must recognize that both the trial function and the test function are scalar quantities. Hence, the product of the trial function and the test function obeys the commutative law of multiplication. The identity (2.13) in Chapter 2 on page 7 was used (i.e., the transpose of the product of two matrices is equal to the product of the transpose of each matrix in reverse order). ˆ. Thus, Since weT and ue contain nodal point data, they do not depend on x these quantities can be taken outside of the integrals. We must, however, be careful to take weT out to the left and ue out to the right, so as to keep the matrix multiplications conformable. Keeping this in mind, the identities (3.25) to (3.27) can be substituted into equation (3.23) to obtain a

nel e=1

weT

Ωe

BTe Be dˆ x ue = b

nel

weT

e=1

+c

nel e=1

weT

Ωe

NTe Ne dˆ x ue

Ωe

NTe dˆ x + awn

(3.28)

The ﬁnal step in this section is to deﬁne a relationship between the vectors ue and we , which contain the values of the trial and test functions at the local nodes (nodal points owned by an individual element), and the global vectors u and w containing the values of the trial and test functions at all of the nodes within the interval 0 ≤ x ≤ L. This relationship can be accomplished by deﬁning a Boolean matrix Le (a matrix containing ones and zeros) for each element. The matrix Le contains information on how the local node numbers for a given element are related to the global node numbers. As an example, suppose that the interval 0 ≤ x ≤ L is broken up into three elements as shown in Figure 3.6. The global nodes can be numbered in any order and will be emphasized by bold text. Now suppose that each interval is a linear-u element (discussed in the previous section). The local node numbers are chosen such that local node 1 is always located at the left end of the element and local node

62

Finite Element Method

2 is always located at the right end. Focusing on the middle element in the ﬁgure, local node 1 corresponds to global node 2 and local node 2 corresponds to global node 4. In this example, the vector ue can be expressed in terms of the vector u as follows: ⎧ ⎫ ⎪ u1 ⎪ ⎪ ⎨ ⎪ ⎬ u2 u1 0 1 0 0 (3.29) = u2 0 0 0 1 ⎪ ⎪ u3 ⎪ ⎩ ⎪ ⎭ u4 or in general ue = Le u

(3.30)

Similarly, for the test function we can write

w1 w2

⎧ ⎫ ⎪ w1 ⎪ ⎪ ⎪ ⎨ ⎬ w2 0 1 0 0 = 0 0 0 1 ⎪ ⎪ w3 ⎪ ⎪ ⎩ ⎭ w4

(3.31)

Notice that the matrix Le is of order nen × n, where nen is the number of nodes owned by the element, and n is the total number of nodes in the interval 0 ≤ x ≤ L. Recognizing that weT = wT LTe and ue = Le u, we can now substitute these expressions into equation (3.28) to obtain nel T T T T Le a Be Be dˆ x−b Ne Ne dˆ x Le u w Ωe

e=1

=w

nel

T

T Le c

e=1

Ωe

Ωe

NTe

dˆ x + LTΓ NTΓ

(3.32)

The last term in equation (3.32) deserves some further explanation. Here LΓ is an nen × n Boolean matrix associated with the element that touches the natural boundary point. Recall that the natural boundary is the location where the derivative of the ﬁeld variable (u in this case) is speciﬁed. The matrix NΓ is the matrix containing the shape functions for the element that touches the natural boundary point. This shape function matrix is evaluated ˆ value associated with local node 2). at the right end of the element (at the x The interested reader will observe that equation (3.32) has the form wT (Ku − F) = 0 where K=

nel e=1

LTe

a Ωe

BTe Be

dˆ x−b

Ωe

(3.33)

NTe Ne

dˆ x Le

(3.34)

One-Dimensional Problems

63

u ˜

u3

u4

u2 1

2

u5 3

4

5

6

FIGURE 3.7 Family of trial functions for a given ﬁnite element mesh. is the global system matrix and nel T Le c F= e=1

Ωe

NTe dˆ x

+ LTΓ NTΓ

(3.35)

is the global force vector. Equations (3.34) and (3.35) deﬁne the ﬁnite element assembly process. Since the test function is arbitrary, the vector w will not in general be zero. Hence the only way that equation (3.33) can be satisﬁed is for the quantity inside the parentheses to vanish, i.e., Ku − F = 0

(3.36)

Ku = F

(3.37)

or Equation (3.36) is a consequence of the function scalar product theorem. The argument goes as follows. Consider the dot product between two vectors a and b. Suppose that we seek a vector b that satisﬁes the equation a·b=0

(3.38)

for all possible vectors a. This is a strong statement. We are seeking a vector b that satisﬁes equation (3.38) for every conceivable vector a having any magnitude and orientation. One possibility is that equation (3.38) can be satisﬁed if the vector b is chosen to be perpendicular to a. However, the only way equation (3.38) can be satisﬁed for all arbitrary vectors a is for the

64

Finite Element Method

vector b to be identically zero. Note that wT in equation (3.33) is analogous to b and Ku − F is analogous to a. We have now come to the point where we can make a very powerful state ment. When the interval 0 ≤ x ≤ L is discretized into a number of nodes and elements, the resulting ﬁnite element mesh spawns a family of trial functions. Each trial function within the family is similar in shape and satisﬁes the es sential boundary conditions but has diﬀerent values at the nodes. This idea is ˜ = 0 at global nodes illustrated in Figure 3.7, where we have speciﬁed that u 1 and 6. Which is the best trial function for the given ﬁnite element mesh? The best trial function within the family is the one that satisﬁes the equation Ku = f . Note that the trial function that satisﬁes this equation is not the ex act solution to the diﬀerential equation, but rather an approximate solution. In general, the approximate solution gets better and better by reﬁning the ﬁnite element mesh, i.e., dividing the domain of interest into more and more elements. In order to solve actual problems using the theory developed in this chapter, it is necessary to evaluate the integrals that appear in equation (3.32). To simplify matters, we ﬁrst write equation (3.37) as follows: nel nel T Le [Ke + Me ] Le u = LTe fe + LTΓ fΓ (3.39) e=1

e=1

where K=

nel

LTe [Ke + Me ] Le

(3.40)

e=1

and F=

nel

LTe fe

+ LTΓ fΓ

(3.41)

e=1

Here in equation 3.39, Ke and Me are called element matrices. Each element has associated with it element matrices that contribute to the global system matrix K. The quantities fe and fΓ are called element vectors. The assembly of all the element matrices and vectors yields the global system matrix and global vector on the right-hand side. Equations (3.40) and (3.41) represent matrix assembly and vector assembly, respectively. For the purpose of per forming the examples and problems at the end of this chapter, it is convenient to tabulate the element matrices and vectors for both the linear-u element and the quadratic-u element in Table 3.1 below. The resulting integrals can ei ther be obtained by lengthy hand calculations, or through the use of symbolic math packages such as Maple ∗ or Mathematica † . We end this section with a short discussion on the element vector fΓ . Recall that this vector arises when one or both ends of a one-dimensional domain is a

�

∗ Maplesoft, † Wolfram

�

615 Kumpf Drive, Waterloo, Ontario, Canada N2V 1K8. Research, Inc., 100 Trade Center Drive, Champaign, IL 61820-7237, USA.

One-Dimensional Problems

65

natural boundary point. In that event only the elements that touch a natural boundary point are aﬀected. If an element touches a natural boundary point, then the element vector fe for that element must be augmented by the vector fΓ . The total element vector for an element that touches a natural boundary point is fe + fΓ . The element vector fΓ is given in Table 3.2 below for both the linear-u and quadratic-u elements.

Example 3.1 In order to illustrate the machinery developed in the previous sections, let us consider the following diﬀerential equation u + 1 = 0 u(0) = 0 u (3) = 1 0≤x≤3

(3.42)

We see that the diﬀerential equation above is just a special case of equation (3.1) with a = 1, b = 0, c = 1,and L = 3. Therefore, the weak form of (3.42) TABLE 3.1

Element matrices and vectors. Matrix

Ke

a Ωe

Linear-u Quadratic-u

BTe Be

dˆ x

a 1 −1 le −1 1 " ! 7 −8 1 a −8 16 −8 3 le 1 −8 7

b Ωe

Me NTe Ne

dˆ x

−b le 2 1 12 6 " ! 4 2 −1 −b le 2 16 2 30 −1 2 4

fe

c Ωe

1 1 # c le 1 4 6 1 c le 2

TABLE 3.2

Additional element vector. aNTΓ (local node 1) 1 Linear-u a 0 # 1 Quadratic-u a 0 0 fΓ

NTe dˆ x

aNTΓ (local node 2) 0 a 1 # 0 a 0 1

66

� � �1 � 2 1

1

2

1

Finite Element Method � � �3 � 2

� � �2 � 2 1 3

4

FIGURE 3.8 Interval discretized with three linear-u elements. The global node numbers are emphasized in bold. can be written as

3

u w dx = 0

3

w dx + w(x = 3)

(3.43)

0

Let us now break up the interval 0 ≤ x ≤ 3 into three linear-u elements, each of unit length. Because b = 0, and x = 3 is a natural boundary point, the relevant element matrices are Ke , fe , and fΓ . Recall that these are given in Tables 3.1 and 3.2. Now let us suppose that the four global nodes along the interval are labeled (in bold text) sequentially from left to right as shown in Figure 3.8. The local nodes 1 and 2 are labeled at the top (left and right edges) of each element. The Boolean matrix Le for each element is given as 1 0 0 0 L1 = 0 1 0 0 0 1 0 0 L2 = 0 0 1 0 0 0 1 0 L3 = 0 0 0 1 Next, plugging in the Boolean matrices into equation (3.39) we get T L1 K1 L1 + LT2 K2 L2 + LT3 K3 L3 u = LT1 f1 + LT2 f2 + LT3 f3 + NT3 (ˆ x = 1)

(3.44)

Because the length of each element in this example is the same, and because all of the elements are linear-u elements, the element matrices and vectors are all the same, i.e., 1 −1 (3.45) K1 = K2 = K3 = −1 1 and

f1 = f2 = f3 =

1/2 1/2

(3.46)

One-Dimensional Problems

67

Notice that element 3 touches the natural boundary point at x = 3 (or equiv alently x ˆ = 1). So we have included the vector fΓ in the global right-hand-side vector. The vector fΓ in the present example is given as 0 (3.47) fΓ = 1 If we now perform the required matrix multiplications we obtain ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ ⎧ ⎫ 1 −1 0 0 0 0 0 0 0 0 0 0 ⎪ ⎪ ⎪ u1 ⎪ ⎪ ⎪ ⎪⎪ ⎨ ⎢ −1 1 0 0 ⎥ ⎢ 0 1 −1 0 ⎥ ⎢ 0 0 0 0 ⎥⎬ ⎨ u2 ⎬ ⎢ ⎥+⎢ ⎥+⎢ ⎥ ⎣ 0 0 0 0 ⎦ ⎣ 0 −1 1 0 ⎦ ⎣ 0 0 1 −1 ⎦⎪ ⎪ u3 ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎩ ⎭ ⎩ ⎭ u4 0 0 0 0 0 0 0 0 0 0 −1 1 ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ 0 ⎪ 0 ⎪ ⎪ 1/2 ⎪ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎪ ⎨ ⎪ ⎬ 1/2 1/2 0 0 = + + + (3.48) 0 ⎪ ⎪ ⎪ 1/2 ⎪ ⎪ 1/2 ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ ⎩ ⎭ ⎪ ⎩ ⎭ ⎪ ⎩ ⎭ 0 0 1/2 1 and after assembly (matrix addition), the global system of equations is ﬁnally obtained, ⎫ ⎡ ⎤⎧ ⎫ ⎧ 1 −1 0 0 ⎪ ⎪ u1 ⎪ ⎪ ⎪ ⎪ 1/2 ⎪ ⎪ ⎢ −1 2 −1 0 ⎥ ⎨ u2 ⎬ ⎨ 1 ⎬ ⎢ ⎥ (3.49) ⎣ 0 −1 2 −1 ⎦ ⎪ u3 ⎪ = ⎪ 1 ⎪ ⎪ ⎪ ⎩ ⎪ ⎭ ⎪ ⎩ ⎭ u4 3/2 0 0 −1 1 Notice that the resulting 4 × 4 left-hand side system matrix is symmetric, and it is also singular, i.e., det K = 0. When we impose the boundary condition at node 1, u1 = 0, we obtain a reduced system matrix that is nonsingular. The reduced system of equations can be written as ⎫ ⎡ ⎤⎧ ⎫ ⎧ 2 −1 0 ⎨ u2 ⎬ ⎨ 1 ⎬ ⎣ −1 2 −1 ⎦ u3 = 1 (3.50) ⎩ ⎭ ⎩ ⎭ u4 3/2 0 −1 1 and upon solution we get ⎧ ⎫ ⎧ ⎫ u1 ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ u2 7/2 = ⎪ ⎪ 6 ⎪ ⎪ u3 ⎪ ⎪ ⎪ ⎩ ⎩ ⎭ ⎪ ⎭ u4 15/2

(3.51)

The ﬁnite element solution is plotted versus the exact (analytical) solution in Figure 3.9. In the upper graph, the ﬁeld variable u is plotted versus position x. The ﬁnite element approximation is represented by the dashed line, and the exact solution is conveyed by the solid line. It turns out that the ﬁnite element solution is in perfect agreement with exact solution at the nodal points, as evidenced by the fact that the circles lie right on top of the solid

68

Finite Element Method 8 6 u 4 2 0 0

0.5

1

1.5 x

2

2.5

3

0.5

1

1.5 x

2

2.5

3

4 3 u 2 1 0 0

FIGURE 3.9 Finite element versus exact solution. curve. This is typical of one dimensional-problems, but not for all problems in general; see Hughes [5] for a more detailed treatment of this topic. On the other hand, because we have used linear-u elements, the ﬁnite element solution varies linearly between the nodal points, and hence, the ﬁnite element solution exhibits error in the regions between the nodes. In the lower plot, the derivative u is plotted versus x for both the ﬁnite element (dashed line) and exact solution (solid line). Notice that the ﬁnite element approximation for u is discontinuous at the nodal points. Because u varies linearly along each element, the ﬁnite element solution predicts that u is constant at every point within the element. Finally, it is worth mentioning that the essential boundary condition, u(0) = 0, is satisﬁed exactly by the ﬁnite element solution. The natural boundary condition is not, however. Notice the error between the ﬁnite element predic tion and the exact solution in the lower plot at x = 3. This error gets smaller and smaller as the mesh is reﬁned. For the sake of completeness, the exact solution to the diﬀerential equa tion (3.42) can be obtained by repeatedly taking the antiderivative of both sides of the equation (in other words, integrating both sides of the equation). Integrating once we obtain u = −x + C1

One-Dimensional Problems

69

where C1 is a constant of integration. The constant C1 can be obtained by enforcing the natural boundary condition at x = 3. Thus we have u (3) = 1 ⇒ 1 = −3 + C1 ,

and therefore,

C1 = 4 Having found the constant C1 , we can integrate again to obtain u=−

x2 + 4x + C2 2

where C2 is another constant. Finally, after enforcing the boundary condition, u(0) = 0, we see that C2 = 0, and the exact solution is u(x) = −

3.4

x2 + 4x 2

(3.52)

Algorithm for matrix assembly

In the example in the previous section, the ﬁnite element assembly process (the process of getting the global system matrix and right-hand-side vector from the element matrices and vectors) involved several steps. To get the global system matrix, for example, it was necessary to set up each element matrix, and then to deﬁne the Boolean matrix Le associated with each el ement. The augmented matrix for each element was obtained by pre- and post-multiplying the element matrix by the Boolean matrix. The system ma trix was ﬁnally obtained by adding up all of the augmented matrices. This procedure was presented for educational purposes only; it is never actually used in commercial ﬁnite element codes because of the large number of multi plications that are involved, which (for large matrices) is very computationally expensive. In this section we present the approach that is used in practice. This computationally eﬃcient approach can be taught like a recipe, memo rized, and then easily followed. To begin, we introduce a connectivity list. A connectivity list is a matrix containing the same number of rows as there are elements in the ﬁnite ele ment mesh. The element number is typically stored in the ﬁrst column. The remaining columns contain the global node numbers that are associated with each local node number attached to the element. In the example in the previ ous section, the ﬁnite element mesh contained three elements. Each element was a two-node linear-u element. Since the nodes were numbered sequentially from left to right, the connectivity list would look as follows: In Table 3.3, we notice that for element 1, local node 1 corresponds to global node 1, and local

70

Finite Element Method TABLE 3.3

Connectivity list. Element 1 2 1 1 2 2 2 3 3 3 4

element 3

element 2

element 1 1

-1

1

1

-1

2

1

-1

3

-1

1

2

-1

1

3

-1

1

4

1

2

2

3

3

4

-1

0

0

1

-1

0

2

1

-1 1+1 0

-1 1+1

-1

3

0

0

-1

1

4

1

2

3

4

FIGURE 3.10 Matrix assembly.

node 2 corresponds to global node 2. Similarly, for element 2, local node 1 corresponds to global node 2 and local node 2 corresponds to global node 3, etc. In this section, the connectivity list will be stored in an array (matrix) labeled ix(nen, nel), where nen is the number of nodes owned by the element, and nel is the number of elements in the mesh. For the case where there is only one unknown per node (one degree of free dom per node), the ﬁnite element matrix assembly process can be illustrated graphically in Figure 3.10. In the diagram, each element matrix is written down side by side and labeled. Focusing our attention on element 1, the recipe goes as follows. The ﬁrst and second row, and the ﬁrst and second columns are labeled 1 and 2 respectively. From the connectivity list, we see

One-Dimensional Problems element 1

71 element 2

element 3

1

1

1

2

1

3

1

2

1

3

1

4

1

1

1+1

2

1+1

3

1

4

FIGURE 3.11 Vector assembly.

that these are the global node numbers that correspond to local nodes 1 and 2 in element 1. Global node 1 is stored in ix(1, 1) and global node 2 is stored in ix(2, 1). Note that the global node number that is attached to local node i in element j is stored in ix(i, j). Because there are a total of four degrees of freedom in the ﬁnite element mesh, the global system matrix is of order 4 × 4. The rows and columns of the global system matrix are labeled consecutively from 1 to 4 as shown in the bottom portion of the ﬁgure. To describe the matrix assembly process, the number that occupies row I and column J in the 2 × 2 element matrix gets added to the global matrix in row I column J as indicated by the arrows in the ﬁgure. Starting with element 1, this process is repeated for each element until the global system matrix is completely assembled. Note that all of the numbers in the global stiﬀness matrix are initially zero, and then numerical values are added to it (element by element) until it is fully assembled. The procedure for vector assembly is the same as that for matrix assembly as depicted in Figure 3.11. The algorithm for the matrix and vector assembly process is given in Table 3.4 below. The algorithm for element matrix and vector assembly begins by deﬁning the parameter nst = nen ∗ ndf which is the number of nodes owned by the element times the number of degrees of freedom (unknowns) per node. So, for the one-dimensional linear-u element with one degree of freedom per node, nst = 2, corresponding to the number of rows in the element matrix. After deﬁning nst, each element of the global stiﬀness matrix is zeroed, and we

72

Finite Element Method TABLE 3.4

Algorithm for element matrix and vector assembly.

Algorithm : Matrix and vector assembly(K, F) nst = nen ∗ ndf

K ← 0 zero the global system

F ← 0 zero the global right-hand-side vector

for iel ← 1 to nel

k ← Ke set up the element matrix

f ← fe set up the element vector

for j ← 1 to nen loop over local node numbers

ll ← 0 initialize counter ll to zero for i ← 1 to ndf loop over degrees of freedom per node ll = ll + 1 ld(ll) = ndf ∗ (ix(j, iel) − 1) + i continue

continue

for ii ← 1 to nst

F (ld(ii)) = f (ii)

for jj ← 1 to nst

K(ld(ii), ld(jj)) = K(ld(ii), ld(jj)) + k(ii, jj)

continue

continue

continue

begin looping over the number of elements in the model. Once inside the outer element loop, the ﬁrst step is to set up the element matrix. This is typically done in practice by calling a function or subroutine for the required element type. The next step is to set up an array, which we call ld. This array stores the global equation numbers associated with each local node owned by the element that we are currently working on. So in the present example, when iel = 1, the values stored in the ld array are ld(1) = 1 and ld(2) = 2. When iel = 2, ld(1) = 2 and ld(2) = 3, etc. After the element matrix and ld arrays are set up, the algorithm proceeds to the second nested loop. This nested loop takes the numerical values stored in the element matrix and element vector and adds them to the proper locations in the global system matrix and right-hand-side vector. Once the element matrix and vector are assembled, the algorithm proceeds to the next element. The process continues until the global matrix and right-hand-side vector are

One-Dimensional Problems

73

b = ρg

L/2

FIGURE 3.12 Composite bar subjected to gravity.

fully assembled.

3.5

One-dimensional elasticity

We end this chapter by considering a more practical example. Consider a bar with a solid circular cross section as shown in Figure 3.12. Let us suppose that the bar is made of two diﬀerent materials, material 1 and material 2, labeled in the ﬁgure. The total length of the bar is 4 m, and each material section is of length 2 m. The two diﬀerent materials are assumed to be perfectly bonded at the interface located at x = 2 m. The bar is subjected to the force of gravity acting in the axial direction. The material properties, Young’s modulus, E, and mass density, ρ, for each section are taken to be those given in Table 3.5. Next, let us assume that the normal stress distribution on any cross section along the length of the bar is uniform. This means that at any given cross section, the normal stress has the same magnitude at every point on that cross section. Recall that stress has units of force per unit area. Let us also assume that all points within the bar can displace in the axial direction only. These assumptions make the problem one-dimensional. Thus far, we have been applying the ﬁnite element method to obtain approxTABLE 3.5

Mechanical properties for composite bar. Material

E (GPa) ρ (kg/m3 )

1 2

2 200

2000 7850

74

Finite Element Method

σ(x)

σ(x + Δx)

Δx x

FIGURE 3.13 Section of a one-dimensional bar. imate solutions to second-order, ordinary diﬀerential equations. The question now is: what is the governing diﬀerential equation to the problem at hand, and how do we derive it? To derive the diﬀerential equation that governs the response of the onedimensional bar described above, let us consider a small section of the bar of length Δx as shown in Figure 3.13. The left end of the section is located at position x, and the right end of the section is located at x + Δx. The normal stress acting at the right end is σ(x + Δx), and the normal stress acting at the left end is σ(x). The force of gravity generates a net force acting on the small segment of magnitude ρgΔx, where ρ is the mass density of the section material (either ρ1 or ρ2 ), g is the gravitational constant, and A is the crosssectional area. The governing diﬀerential equation that we are seeking comes from a simple force balance. Summing forces in the x direction gives A {σ(x + Δx) − σ(x)} + AρgΔx = 0

(3.53)

If we now divide both sides of equation (3.53) by AΔx we obtain σ(x + Δx) − σ(x) + ρg = 0 Δx

(3.54)

Equation (3.54) must hold for arbitrarily small Δx. Hence, in the limit as Δx goes to zero, we get lim

Δx→0

σ(x + Δx) − σ(x) + ρg = 0 Δx

(3.55)

Notice that the ﬁrst term on the left-hand side of (3.55) is the deﬁnition of the derivative, dσ/dx. Thus we can write dσ + ρg = 0 dx

(3.56)

One-Dimensional Problems

75

For linearly elastic material behavior, the normal stress, σ, is related to the extensional strain, , through the one-dimensional form of Hooke’s law, i.e., σ = E

(3.57)

where E is Young’s modulus. The extensional strain at point x is deﬁned as the change in length divided by the original length of a small section of the bar of length Δx in the limit as Δx goes to zero, i.e., = lim

Δx→0

du Δu = Δx dx

(3.58)

where u = u(x) is the axial displacement. Equation (3.58) is called a straindisplacement relationship. Plugging the strain-displacement relationship (3.58) along with the constitutive equation (3.57) into (3.56) yields Eu + ρg = 0

(3.59)

Comparing equation (3.59) above with equation (3.1) at the very beginning of this chapter, we see that equation (3.59) is just a special case of (3.1) with a = E, and c = ρg. Because both ends of the bar are clamped, no displacement can occur at either end. Thus the boundary conditions for the problem are u(0) = 0 and u(L) = 0. The ﬁnite element matrices and vectors for the present one-dimensional elasticity problem are obtained by writing down the weak form of equation (3.59) over a single element and then substituting the ﬁnite element approx imations into this expression. If we model the problem, for example, with two linear-u elements, the element matrices and vectors for elements inside material 1 and material 2 are given in Table 3.6. We can ensure continuity of displacement at the interface between the two diﬀerent materials by placing a single node at the interface where x = 2 m. Given the element matrices and vectors, the global system of equations is obtained using the ﬁnite element as sembly process described in the previous section, and enforcing the prescribed displacement boundary conditions at both ends of the bar. TABLE 3.6

Element matrices and vectors for linear-u elements inside material 1 and material 2. Region Ke fe ρ1 gle 1 E1 1 −1 material 1 1 le −1 1 2 E2 ρ2 gle 1 1 −1 material 2 1 le −1 1 2

76

3.5.1

Finite Element Method

Strain and stress calculation

After the solution to the global system of equations is obtained, the resulting nodal displacements are used to obtain the strain and stress in each element as a post-processing step. To illustrate the procedure, suppose we model the problem with linear-u elements. The nodal displacements associated with the local left and right nodes are then u1 and u2 , respectively. The strain in the element is obtained by multiplying the 1 × 2 element B-matrix by the 2 × 1 vector containing the local nodal displacements. Doing this we get = Be ue u1 = −1/le 1/le u2 u2 − u 1 = le

(3.60)

where le is the length of the element. We see here that for the case of the linearu element, the strain turns out to be constant along the length of the element. For this reason, the element is often referred to as a constant-strain element. Once the strain in the element is obtained, the stress is easily computed using Hooke’s law. We must make sure, however, to plug in the correct Young’s modulus, depending on whether the element lies in material 1 or material 2. If we model the problem instead with quadratic-u elements, for example, the strain calculation is the same, except we would use the element B-matrix deﬁned in (3.22). Letting u1 , u2 , and u3 be the displacements at local nodes 1, 2, and 3 of an individual quadratic-u element, the strain in the element is computed as B u = 1 ×e3 3 ×e 1 ⎧ ⎫ ⎨ u1 ⎬ 2 1 1 u2 = − + ξ −2ξ + ξ ⎩ ⎭ le 2 2 u 3

where ξ = 2ˆ x/le is the natural (dimensionless) coordinate deﬁned in Fig ure 3.4, and x ˆ is the local coordinate measured positive to the right starting from local node 2 located at the middle of the element. Notice that the strain along the length of a quadratic-u element varies linearly with ξ and hence ˆ. When the constitutive behavior is governed by the one-dimensional with x Hooke’s law, σ = E, the stress also varies linearly along the length of a quadratic-u element.

3.5.2

Finite element results

We end this section by presenting some ﬁnite element results for the case where the composite bar is modeled with linear-u elements. The numerical

One-Dimensional Problems

77

u/( ρE11g )

6 4 2 0 0

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

× 10−6

200 100 0

-100 0

σ/(ρ1 gL)

100 0 -100 -200 -300 -400 0

x/L FIGURE 3.14 Finite element solution (eight linear-u elements).

results for the case where the mesh is composed of four linear-u elements in each material (for a total of eight elements) are presented in Figure 3.14. In the upper plot in Figure 3.14, the displacement, u, normalized with respect to the quantity E1 /(ρ1 g) is plotted versus the dimensionless length, x/L, where L is the total length of the composite bar. As shown in the plot, the displacement varies linearly along the length of each element, and the displacement is continuous at x/L = 0.5, the interface between the two diﬀerent materials. Notice that the essential boundary conditions at x = 0 and x = L are satisﬁed exactly by the ﬁnite element solution. The strain, , along the length of the bar is plotted versus x/L in the

78

Finite Element Method

u/( ρE11g )

6 4 2 0 -2 0

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

× 10−6

200 100 0

-100 0

σ/(ρ1 gL)

100 0 -100 -200 -300 -400 0

x/L FIGURE 3.15 Finite element solution (96 linear-u elements).

middle plot in Figure 3.14. Because we are using linear-u elements, the plot reveals that the strain ﬁeld is discontinuous, and the strain is constant along the length of each individual element. The normal stress, σ, normalized with respect to ρ1 gL is plotted versus x/L in the lower plot in Figure 3.14. Like the strain ﬁeld, the stress ﬁeld is also discontinuous, and it is constant along each individual element. The magnitude of the jump (discontinuity) in stress (and/or strain) can be used as a measure of the accuracy of the numerical results. In general, as the mesh is reﬁned, the magnitude of the jump de creases and accuracy increases. The ﬁnite element results for a much ﬁner mesh composed of 96 linear-u elements are shown in Figure 3.15. In the ﬁne

One-Dimensional Problems

79

mesh, the discontinuity in strain and stress between each element is almost indistinguishable, revealing that the ﬁnite element solution will not change appreciably with further mesh reﬁnement. When this happens, the solution is said to have converged. Notice that with the ﬁne mesh, the strain and stress distributions are linear in each material section, but there is a discontinuity in the strain ﬁeld at x = L/2. This is due to the fact that the bar is made of two diﬀerent materials, which produces a kink in the displacement ﬁeld at this location. Notice that the stress ﬁeld is continuous, consistent with Newton’s third law.

3.6 Problems Problem 3.1 Consider the following diﬀerential equation: u + 2u = 0 0≤x≤1 u (0) = 3 u(1) = 0 (a)

Derive the weak form.

(b)

Derive the element matrix and vector for a linear-u element of length l = 1.

(c)

Derive the element matrix and vector for a quadratic-u element of length l = 1.

(d)

Mesh the problem with two linear-u elements, and assemble the global system of equations. Enforce the boundary condition, u3 = 0, at the right end to obtain the reduced system. Solve for the unknowns u1 and u2 .

(e)

Repeat part (d) using a mesh of one quadratic-u element.

Problem 3.2 Consider the problem of a rod with a circular solid cross section that is con nected to a rotating shaft as shown in Figure 3.16. The shaft is spinning with a constant angular velocity ω = 50 rad/s. Assume that the rod is made of steel with Young’s modulus E = 200 GPa, and mass density ρ = 7850 kg/m3 . The length of the rod is L = 1 m. Model the rod with two linear-u elements and determine: (a)

The displacement in the X direction at the end of

80

Finite Element Method

Y ω

X

FIGURE 3.16 Problem 3.2. Rod connected to a rotating shaft. the rod. (b)

The stress distribution σ(X) along the length of the rod.

Hint: For a rod rotating about a ﬁxed axis, the acceleration in the X direction at any point along the length of the rod is ω 2 X. Treat the problem as a static problem with a linearly varying body force, b = ρω 2 X, acting in the X direction. Assume that the rod can displace only in the X direction and that the rod is ﬁxed at the left end. Problem 3.3

E(X) = C1 eC2 X

X

FIGURE 3.17 Problem 3.3. Functionally graded bar subjected to a constant body force.

Consider the problem of an inhomogeneous bar with a constant circular cross section that is ﬁxed at both ends as shown in Figure 3.17. The bar of length L is subjected to a constant body force b(x) = C acting in the positive X direction. The Young’s modulus, E(X) of the bar material varies with

One-Dimensional Problems

81

TABLE 3.7

Problem 3.3. Numerical values. E1 (GPa)

E2 (GPa)

66.0

200.0

C (N/m3 ) L (m) 78.0

1.0

position along the length of the bar and is given as E(X) = C1 eC2 X C1 = E1 E2 /L C2 = ln E1 where E1 is the value of Young’s modulus at the left end of the bar, and E2 is the value at the right end. The numerical values for E1 , E2 , L, and C are given in Table 3.7. The diameter of the bar is d = 0.01 m. (a)

Determine the element matrix and vector for a linear-u element inside the domain 0 ≤ X ≤ 1 m.

(b)

Write a one-dimensional ﬁnite element program to obtain an approx imate solution to the problem.

(c)

Obtain plots of the displacement, u, versus X for several diﬀerent meshes. Reﬁne your mesh until your solution no longer changes appreciably with the number of elements.

(d)

From your most reﬁned mesh used in part (c), obtain a plot of the normal stress in the column σ(X), as a function of X.

(e)

If the ultimate compressive strength of the bar material is 10 MPa, do you expect the column to fail?

Problem 3.4 A bar with a tapered cross section is shown in Figure 3.18. The cross section

D1 D2 X FIGURE 3.18 Problem 3.4. Bar with a tapered cross section.

82

Finite Element Method

is circular with diameter D1 at the left end and D2 at the right end, and hence the cross-sectional area is a function of X, i.e., A = A(X). The bar has length L and Young’s modulus E. Discretize the bar with one linear-u element and calculate the following: (a)

The element matrix in terms of D1 , D2 , E, and L.

(b)

The element vector for the case when a constant body force b(X) = C acts in the X direction.

4 Linearized Theory of Elasticity

This chapter covers the fundamentals of the linearized theory of elasticity. The material is essential for understanding the application of the ﬁnite element method to problems in the area of solid mechanics. The chapter begins with the derivation of Cauchy’s law, which is a very useful tool for assessing the accuracy of ﬁnite element solutions. We then go on to deriving the equilibrium equation, the governing partial diﬀerential equation for problems in solid mechanics in the absence of inertia. The chapter also covers the important topics of the small strain tensor, Hooke’s law, plane stress and plane strain, and axisymmetric problems. The chapter ends with a derivation of the weak form of the equilibrium equation, which is the starting point for the ﬁnite element implementation presented in Chapter 6.

4.1

Cauchy’s law

Consider the solid body shown in Figure 4.1. The surface of the body is labeled Γ, and Ω is the interior volume (or area, depending on whether we are working in two or three dimensions). As shown in the ﬁgure, the body is loaded by a distribution of surface tractions over a portion of its boundary. A surface traction is a vector quantity that has units of force per unit area.

ti b Ω

Γt

Γu FIGURE 4.1 Solid body subjected to surface tractions and body force ﬁeld.

83

84

Finite Element Method

If we multiply a traction vector by a diﬀerential surface area, dΓ, we get a diﬀerential surface force acting in the same direction as the traction vector. The body can also be subjected to a body force distribution. A body force is a vector quantity that acts at each point throughout the volume of a body. A body force has units of force per unit volume. When a body force vector is multiplied by a diﬀerential volume element, dΩ, the result is a diﬀerential force vector acting in the same direction as the body force. We can decompose the surface, Γ, into two parts, Γu and Γt . The quantity Γu represents a portion of the boundary where the displacement components, ui , are speciﬁed. Likewise, Γt is the portion of the boundary where the com ponents, ti , of the traction vector are prescribed. We mention here that one cannot specify both the traction vector and the displacement vector at the same point. As an example, one cannot specify both t1 and u1 at the same point on Γ. If t1 is speciﬁed at a point, nothing can be said about u1 at that point. Additional information is needed about the material properties be fore the displacement ﬁeld can be fully determined for a given set of speciﬁed boundary tractions. Having deﬁned the traction vector, we now focus on the general state of stress at a point and seek a relationship between the state of stress at a point and the traction vector at that same point. This relationship is known as Cauchy’s law. To begin, consider an inﬁnitesimal tetrahedron that is cut out of a material in such a way that three of its sides have unit outward normals −e1 , −e2 , and −e3 , respectively, as shown in Figure 4.2. One of the faces (the inclined face) has area, A, and unit outward normal n. The traction vector acting on the inclined face is labeled t(n) . Now let t(e1 ) , t(e2 ) , and t(e3 ) be traction vectors acting on the planes whose unit outward normals are e1 , e2 , and e3 . As a consequence of Newton’s third law (for every action there is an equal and opposite reaction), the three traction vectors acting on the remaining (negative) faces of Cauchy’s tetrahedron are −t(e1 ) , −t(e2 ) , and −t(e3 ) . The three traction vectors, t(ei ) each have three components. We will call these components σi1 , σi2 , and σi3 . Hence, we can write t(ei ) = σij ej

(4.1)

The sum of all external force vectors acting on the faces of the tetrahedron must be equal to zero. Letting b represent the body force acting throughout the volume of the tetrahedron, the equilibrium statement can be written as 1 −t(e1 ) A1 − t(e2 ) A2 − t(e3 ) A3 + bhA + t(n) A = 0 3

(4.2)

where here we have used the fact that the volume, V , of the tetrahedron is equal to one-third its base, A, times its height, h, i.e., V =

1 Ah 3

Linearized Theory of Elasticity

85

X3

n −e1

X2

A h

−e2

X1

−e3 FIGURE 4.2 Cauchy’s tetrahedron.

We note that the height of the tetrahedron is the shortest distance from the origin to a point on the inclined face. The quantities A1 , A2 , and A3 in equation (4.2) are the areas of the faces perpendicular to e1 , e2 , and e3 , respectively. In order to proceed further, it is necessary to derive a relationship between the surface area of the inclined face and the areas, Ai , of the other three faces. To begin, let a be the length of the edge of the tetrahedron that lies on the X1 axis. Now consider the vector a e1 . The height, h, of the tetrahedron is equal to the projection of a e1 onto the n axis, i.e., h = a (e1 · n)

(4.3)

Hence, a=

h n1

The volume, V = 1/3Ah, can now be equivalently written as 1 A1 a 3 1 h = A1 3 n1

V =

(4.4)

Comparing the last line in (4.4) with the volume expression that we started with, we see that A1 = An1

86

Finite Element Method

where n1 is the component of n in the e1 direction. Using this same argument with the remaining faces, one can readily see that Ai = Ani

(4.5)

Let us now return to the balance statement (4.2). By substituting (4.5) into (4.2) we get 1 −t(e1 ) An1 − t(e2 ) An2 − t(e3 ) An3 + bhA + t(n) A = 0 3

(4.6)

The last step in the derivation of Cauchy’s law is to divide both sides of equation (4.6) by A and take the limit as h → 0. Doing this we obtain, t(n) = t(e1 ) n1 + t(e2 ) n2 + t(e3 ) n3 = t(ei ) ni = ni σij ej

(4.7)

Equation (4.7) is known as Cauchy’s law. It can be written in tensor notation as t(n) = n · σ

(4.8)

or in component form as (n)

tj

= ni σij

(4.9)

The component form of equation (4.8) comes from t(n) = n · σ = (nk ek ) · (σij ei ej ) = nk δki σij ej

= ni σij ej

(4.10)

Throughout the remainder of the book, when we write Cauchy’s law (either in tensor or component form) we will drop the superscript (n) and it will be understood implicitly that the traction vector acts on a plane whose unit outward normal is n. We will simply write t=σ·n

or

tj = ni σij

(4.11)

Cauchy’s law (4.11) reveals several important facts. First of all, recalling the deﬁnition of a second-order tensor given in Chapter 2 on page 17, it re veals that the nine components, σij , of the three traction vectors, t(ei ) , form a second-order tensor, σ. This second-order tensor is called the Cauchy stress tensor. Cauchy’s law also tells us how the components of a traction vector acting at a point on any inclined plane are related to the stress components

Linearized Theory of Elasticity

87 X2

σ22

σ23 σ21 σ12 σ32 σ33

σ11 σ31 σ13

X1

X3 FIGURE 4.3 Components of the Cauchy stress tensor. at that point to ensure equilibrium of an inﬁnitesimal volume element sur rounding that point. The nine components of the Cauchy stress tensor are shown in Figure 4.3. As shown in the ﬁgure, the stress component σ11 is the normal stress acting on the face whose unit outward normal points in the X1 direction. This face is often referred to as the positive X1 face. Similarly, the stress components σ22 and σ33 are the normal stress components acting on the positive X2 and X3 faces, respectively. The shear stress components are also shown in the ﬁgure. The stress component σ12 is the shear stress acting on the positive X1 face pointing in the X2 direction. For any particular stress component, the ﬁrst subscript refers to the face on which the stress component acts. The second subscript refers to the direction in which it acts. We end this section by making the important observation that the Cauchy stress tensor is symmetric. Recall from our study of Chapter 2 that if σij = σji , then the tensor, σ, is symmetric. One can see that the Cauchy stress tensor is symmetric by referring back to Figure 4.3 and performing individual moment balances about the X1 , X2 , and X3 axes, respectively. For example, a moment balance about the X1 axis gives σ32 ΔX1 ΔX2 ΔX3 − σ23 ΔX1 ΔX3 ΔX2 = 0 ⇒ σ32 = σ23

(4.12)

where we have used ΔX1 , ΔX2 , and ΔX3 to denote the lengths of the edges of the cube. Similarly, by summing moments about the X2 and X3 axes we get σ21 = σ12 σ31 = σ13

88

Finite Element Method t2 = 1.0 psi

X2

A

X1

FIGURE 4.4 Application of Cauchy’s law.

Because the Cauchy stress tensor is symmetric, Cauchy’s law (4.11) can also be expressed as tj = ni σij = ni σji = σji ni Changing the free index from j to i in the previous equation and using j for the repeated index, we get ti = σij nj

(4.13)

Example 4.1 To illustrate the use of Cauchy’s law and its utility in the interpretation of ﬁnite element results, consider the problem of a 10 in. × 10 in. plate with a centrally located hole as shown in Figure 4.4. The plate is loaded by a uniform traction distribution t2 = 1.0 psi along the top and bottom edges. A contour plot of the stress component σ22 is superposed on top of the plate geometry. For the given dimensions of the plate and the applied traction distribution, the maximum value of σ22 in the plate is slightly above 3.0 psi. The ratio of applied stress versus the maximum stress found in the plate is called the stress concentration factor. Here the stress concentration factor is max σ22 ≈ 3.0 t2

Linearized Theory of Elasticity

89

The location of the stress concentration is labeled A in the ﬁgure. Let us now investigate the values of the stress components σ11 and σ12 at this location. At point A, the unit outward normal to the surface is parallel to the X1 axis. Hence, n1 = 1 and n2 = 0. Because the inside of the hole is a traction-free surface, ti = 0 at all points on the inside of the circular hole. Applying Cauchy’s law (4.13) we get t1 σ11 n1 + σ12 n2 σ11 (1) + σ12 (0) σ11

=0⇒ =0⇒ =0⇒ =0

(4.14)

Similarly, because t2 also equals zero at this point we get t2 = 0 ⇒ σ21 n1 + σ22 n2 = 0 ⇒ σ21 (1) + σ22 (0) = 0 ⇒ σ21 = σ12 = 0

(4.15)

We emphasize that the stress component σ22 at point A cannot be obtained from Cauchy’s law. This stress value must be obtained from the solution of the elasticity problem.

4.2

Principal stresses

An understanding of principal stresses and principal directions is very impor tant in the analysis of solids and structures. The procedure for ﬁnding princi pal stresses and principal directions is identical to that for ﬁnding eigenvalues and eigenvectors of second-order tensors described in Chapter 2 in Section 2.3.2. Suppose that the Cauchy stress tensor is known at some point P in the solid shown in Figure 4.5. Now imagine making a variety of “cuts” through point P . Here we use the word “cut” to represent a plane, passing through P , whose unit outward normal is n as shown in the ﬁgure. Cauchy’s law tells us that if we take the dot product of the stress tensor, σ, and the unit outward normal, n, the result is a traction vector pointing in some direction. We now seek a very special direction, such that the dot product, σ · n, gives a traction vector that is parallel to n. Stated mathematically, we seek to solve the following equation: σ · n = σn (4.16)

90

Finite Element Method

P

n

FIGURE 4.5 Inclined plane passing through solid. Assuming for the moment that equation (4.16) can be solved, the scalar quan tity σ represents the magnitude of the normal stress acting on a plane having unit outward normal n. Note that no shear stress acts on this plane. The quantity σ is called a principal stress, and the unit vector n is the correspond ing principal direction. The principal stresses and directions are eigenvalues and eigenvectors of the stress tensor. Next, it is convenient to express equation (4.16) as follows: [σ − σI] · n = 0

(4.17)

where I is the second-order identity tensor. As discussed in Chapter 2, equa tion (4.17) can be satisﬁed (for nonzero vectors n) only if the following deter minate is equal to zero, |σ − σI| = 0 (4.18) Expanding equation (4.18), we get the following cubic equation: −σ 3 + I1 σ 2 + I2 σ − I3 = 0

(4.19)

I1 = tr(σ) 1 I2 = σ : σ − tr(σ)2 2 I3 = det σ

(4.20)

where

are scalar coeﬃcients called stress invariants. They are called invariants be cause no matter what coordinate system the components of the stress tensor are referred to, one will always get the same coeﬃcients in equation (4.19).

Linearized Theory of Elasticity

91

Once the three roots of equation (4.19) are found, the corresponding princi pal directions are obtained by successively plugging in each principal stress back into equation (4.17) and solving for the vector n. By convention, the principal stresses are arranged in order of algebraically largest to smallest val ues. We emphasize here that the state of stress at a point can be completely characterized by the three principal stresses along with the three principal directions.

Example 4.2 To illustrate the procedure for calculating principal stresses and principal directions, consider the following state of stress at a point: ⎡ ⎤ ⎡ ⎤ σ11 σ12 σ13 1 −10 0 ⎣ σ12 σ22 σ23 ⎦ = ⎣ −10 1 0 ⎦ MPa (4.21) σ13 σ23 σ33 0 0 0 Dropping the units for the intermediate calculations, the determinant equation to be solved is then 1 − σ −10 0 −10 1 − σ 0 = 0 0 0 0−σ Expanding the determinant above we obtain −σ 3 + 2σ 2 + 99σ = 0

(4.22)

The three roots of the characteristic equation (4.22) are found to be σ1 = 11 MPa σ2 = 0 MPa σ3 = −9 MPa

(4.23)

Notice that the three principal stresses are labeled in the order of algebraically largest to smallest values. Next, in order to calculate the principal directions, we begin by substituting the ﬁrst principal stress into equation (4.17) and solving for the components of the vector n. Doing this we get ⎡ ⎤⎧ ⎫ ⎧ ⎫ −10 −10 0 ⎨ n1 ⎬ ⎨ 0 ⎬ ⎣ −10 −10 0 ⎦ n2 = 0 (4.24) ⎩ ⎭ ⎩ ⎭ n3 0 0 −11 0 Expanding the ﬁrst row of (4.24) we see that −10n1 − 10n2 = 0 ⇒ n2 = −n1 , and n3 = 0

(4.25)

92

Finite Element Method

Notice that we would have come to the same conclusion as (4.25) if we had expanded the second row of (4.24). If we insist that the vector n be a unit vector, it follows that its magnitude must be 1, i.e., n21 + n22 + n23 = 1 Finally, by substituting (4.25) into (4.26) we get √ 2 n1 = 2√ 2 ⇒ n2 = − 2 √ √ 2 2 (1) n = e1 − e2 2 2

(4.26)

(4.27)

where the superscript (1) emphasizes that the principal direction that we just calculated is associated with the ﬁrst principal stress, i.e, n(1) is the principal direction associated with σ1 . Following the same procedure, the other two principal directions are found to be n(2) = e3 √ √ 2 2 (3) e1 + e2 n = 2 2

(4.28)

If we carefully study the three principal directions obtained in this section, we will notice that the three unit vectors n(1) , n(2) , and n(3) are mutually perpendicular, i.e., n(i) · n(j) = δij . This happens to be a property of sym metric, second-order tensors whose components are real. Such tensors have real eigenvalues and mutually perpendicular eigenvectors. In order to better illustrate the results that we have just obtained, imagine cutting out an inﬁnitesimal stress element surrounding point P in the following way. First, make two cuts that are parallel to the X1 axis. Second, make two cuts parallel to the X2 axis. Finally, make two cuts parallel to the X3 axis. Doing this we can extract a stress element that is lined up with the X1 , X2 , and X3 axes. The state of stress (4.21) can be represented graphically by drawing the normal and shear stress components acting on each of the six faces as shown in Figure 4.6 (a). Note that for the state of stress considered in the present example, the normal and shear stresses are zero on faces perpendicular to the X3 axis. Hence we can represent the state of stress by drawing a twodimensional stress element. By making cuts perpendicular to the n(1) , n(2) , and n(3) principal direc tions, we can also extract a stress element surrounding point P as shown in Figure 4.6 (b). Note that we can also represent this element in two dimen sions, because the principal direction n(2) happens to be parallel to the X3 axis. If we now let the axes X1 , X2 , and X3 line up with the principal direc tions n(1) , n(3) , and n(2) , respectively, then the stress components referred to

Linearized Theory of Elasticity this primed set of axes can be expressed ⎡ 11 0 σ = ⎣ 0 −9 0 0

93 in matrix notation as ⎤ 0 0 ⎦ MPa 0

(4.29)

If we compute the three invariants of the stress tensor from the stress compo nents in the principal orientation (4.29) we get I1 = tr(σ) = 11 − 9 + 0 = 2 1 1 (11)2 + (−9)2 − (2)2 = 99 σ : σ − tr(σ)2 = I2 = 2 2 I3 = det σ = (11)(−9)(0) = 0 which is the same as the result that we obtained using the stress components (4.21); see equation (4.22).

1 MPa

11 MPa π 4

X1

1 MPa −10 MPa

−9 MPa

(a)

(b)

FIGURE 4.6 Stress element in the aligned orientation (a) and the principal orientation (b).

4.3

Equilibrium equation

Equilibrium of a solid body requires that the net external force vector acting on the body is equal to zero. The net external force acting on the body comes from the surface traction distribution and the body force ﬁeld. The net force from the traction distribution can be obtained by “summing up” all of the diﬀerential force vectors, ti dΓ, acting on the surface. Likewise, the net force vector due to the body force distribution comes from the summation of the diﬀerential force vectors, bi dΩ. So the equilibrium statement for the body

94 can be written as

Finite Element Method

Γ

ti dΓ +

Ω

bi dΩ = 0

(4.30)

We saw in Section 4.1 that the traction vector, t, acting at a point P on Γ is related to the Cauchy stress tensor at that point through the relation t = n·σ ⇒ ti = σij nj where n = nj ej is the unit outward normal to the boundary at point P . The key to deriving a partial diﬀerential equation from a balance law is to convert all integrals in the balance statement into volume integrals. To this end, we now recognize that the ﬁrst integral in the equilibrium statement (4.30) can be recast as a volume integral by using the divergence theorem covered in Chapter 2, Section 2.4.6. The divergence theorem states that the surface integral, n · σ dΓ Γ

can be replaced by the following volume integral: ∇ · σ dΩ Ω

The vector quantity ∇ · σ is called the divergence of the stress tensor. The components of this vector can be obtained using the techniques that were introduced in Chapter 2, Section 2.3.3, i.e., ∂ ei · (σjk ej ek ) ∇·σ = ∂Xi ∂σjk δij ek = ∂Xi = σjk,j ek = σji,j ei = σij,j ei where the comma denotes partial diﬀerentiation with respect to the spatial variables, and we have used the fact that the Cauchy stress tensor is symmetric (σij = σji ). The equilibrium statement, in terms of volume integrals, can now be written as (σij, j + bi ) dΩ = 0 (4.31) Ω

Because the statement (4.31) must be true for all subdomains inside Ω, one can argue that the integrand in (4.31) must be identically zero at every point inside Ω. This fact is actually a subtle point that requires further explanation.

Linearized Theory of Elasticity

Ω2

95

f (x, y) dΓ = 0!

Ω

Γ Γ3

Γ2

Γ1

FIGURE 4.7 Illustration showing that the equilibrium statement must be true for all subdomains inside Ω.

The argument can be better understood if we take a look at Figure 4.7. Here in the ﬁgure, we are considering a square (two-dimensional) domain with boundary Γ. Let us imagine for the moment (for the sake of simplicity) that the integrand in (4.31) is a scalar function represented by the surface plot shown in the ﬁgure. Now suppose that the function is zero everywhere except inside the subdomains Ω2 and Ω3 , which are bounded by Γ2 and Γ3 labeled in the ﬁgure. In addition, suppose that the function inside Ω2 is equal in magnitude but opposite in sign to the function inside Ω3 . Obviously, the integral of such a function over the entire domain, Ω, would be zero, seemingly satisfying equation (4.31). However, the equilibrium statement must be true for all subdomains. If we cut out an arbitrary portion of the body, we require that portion to be in equilibrium. Notice for example that the integral of the function over Ω2 is not zero, violating equation (4.31). The fact that the integrand in (4.31) is actually a vector ﬁeld does not alter this argument. One can imagine a vector ﬁeld being zero everywhere inside Ω except for the subdomains Ω2 and Ω3 , in which we could deﬁne nonzero vector ﬁelds that are equal in magnitude but opposite in direction. So the governing partial diﬀerential equation (the equilibrium equation) now emerges from the balance statement (4.31) and is written as

σij,j + bi = 0 in Ω ti = t∗i on Γt ui = u∗i

on Γu

(4.32)

The partial diﬀerential equation (4.32) and boundary conditions that go along with it are referred to as the strong form. Here the superposed star indicates that the quantity (either traction or displacement component) is speciﬁed at a point on the boundary.

96

4.4

Finite Element Method

Small-strain tensor

In this section, we present the mathematical and physical arguments required to gain an understanding of the meaning of the small-strain tensor and how its components are related to the displacement components in a continuous body.

4.4.1

Relative stretch

We begin by deﬁning the relative stretch, ε(n) , at a point in the body. The relative stretch is deﬁned as the change in length divided by the original length of an inﬁnitesimal line segment that initially lies in a direction parallel to the unit vector n. Let us consider an inﬁnitesimal line segment dX(n) , with magnitude dS (n) in the original conﬁguration as shown in Figure 4.8. After deformation, this vector maps into the vector dx(n) whose magnitude is ds(n) . The relative stretch can then be written as ds(n) − dS (n) dS (n) (n) = λ −1

ε(n) =

(4.33)

where λ(n) = ds(n) /dS (n) is called the stretch of the line segment. The quantity λ(n) is also commonly referred to as the stretch ratio. Now suppose that we choose a line segment dX(1) that is initially parallel to the X1 axis. The relative stretch of this line segment is ε(1) =

ds(1) − dS (1) dS (1)

(4.34)

It turns out that the relative stretch ε(1) can be expressed in terms of the deformation gradient tensor, F, introduced at the end of Chapter 2. The X2

X2

dx(n)

dX(n)

X1 FIGURE 4.8 Relative stretch of inﬁnitesimal line segment.

X1

Linearized Theory of Elasticity

97

argument utilizes that fact that an inﬁnitesimal line segment in the original conﬁguration, dX, maps into an inﬁnitesimal line segment dx in the deformed conﬁguration through the relation dx = F · dX = dX · FT

(4.35)

Hence, the magnitude of the inﬁnitesimal line segment in the deformed con ﬁguration, ds(1) , can be expressed as ds(1) = dx(1) · dx(1) = dX(1) · (FT · F) · dX(1) = dX(1) · C · dX(1) where

C = FT · F

(4.36) (1)

is called the Green deformation tensor. The stretch, ε , can now be written as √ dX(1) · C · dX(1) − dS (1) (1) (4.37) ε = dS (1) The trick to getting equation (4.37) into a manageable form is to divide both the numerator and denominator by dS (1) and to recognize that the vector dX(1) /dS (1) is simply the unit base vector e1 . Doing this yields ε(1) = e1 · C · e1 − 1 (4.38) = C11 − 1

4.4.2

Angle change

Let us now consider two inﬁnitesimal line segments dX(1) and dX(2) that are initially parallel to the X1 and X2 axes in the undeformed conﬁguration as shown in Figure 4.9. Since these vectors are perpendicular, the angle between them is π/2. These vectors become dx(1) and dx(2) in the deformed conﬁguration. The quantity θ12 is the angle between these two vectors in the deformed conﬁguration. The engineering shear strain, γ12 , is deﬁned as the diﬀerence between the original angle, π/2, and the angle θ12 , i.e., π (4.39) γ12 = − θ12 2 It is now instructive to look at the relationship between θ12 and the in ﬁnitesimal vectors dx(1) and dx(2) . To begin, the dot product between these two vectors gives dx(1) · dx(2) = ds(1) ds(2) cos θ12

98

Finite Element Method X2

X2 dX(2) π/2

dx(2) θ12

dX(1)

dx(1) X1

X1

FIGURE 4.9 Relative stretch and angle change of two inﬁnitesimal line segments. Next, let us divide both sides of the previous equation by dS (1) and dS (2) , the magnitudes of the inﬁnitesimal line segments in the original conﬁguration. Doing this we get dx(1) dx(2) = λ1 λ2 cos θ12 (4.40) dS (1) dS (2) where λ1 and λ2 are the stretch ratios in the X1 and X2 directions, respectively. Equation (4.40) can now be written in terms of the deformation gradient tensor, F, as follows: dX(2) dX(1) T · F ·F · = λ1 λ2 cos θ12 (1) dS (2) dS

(4.41)

Recognizing that dX(1) /dS (1) = e1 , dX(2) /dS (2) = e2 , and FT · F = C, equation (4.41) becomes e1 · C · e2 = λ1 λ2 cos θ12

(4.42)

and hence, C12 (4.43) λ1 λ 2 The connection between the engineering shear strain, γ12 , and the quantity, cos θ12 , can be established by taking the sine of both sides of equation (4.39). Doing this we obtain π − θ12 sin γ12 = sin 2 π π = sin cos θ12 + cos sin θ12 2 2 = cos θ12 (4.44) cos θ12 =

where the second line in (4.44) follows from the trigonometric double-angle formula. From equation (4.44), we can express γ12 in terms of C12 and the

Linearized Theory of Elasticity

99

stretch ratios as follows: sin γ12 =

4.4.3

C12 λ1 λ2

(4.45)

Small-displacement gradients

Recall from Chapter 2 that the deformation gradient tensor, F, contains in formation about how individual points are mapped from the original conﬁg uration to the deformed conﬁguration. The components of F can be written as ∂xi ∂Xj ∂ = (ui + Xi ) ∂Xj ∂ui = + δij ∂Xj = ui,j + δij

Fij =

(4.46)

where, again, the comma denotes partial diﬀerentiation with respect to the spatial X1 − X2 − X3 coordinates. When the displacement gradients (partial derivatives of the displacement components) are small, we can neglect the products of these gradients that would otherwise be nonzero components of the Green deformation tensor. If we neglect these products, the components of the Green deformation tensor become Cij = Fki Fkj ⎡ ⎤ 2u1,1 + 1 u1,2 + u2,1 u1,3 + u3,1 u2,3 + u3,2 ⎦ ≈ ⎣ u1,2 + u2,1 2u2,2 + 1 u1,3 + u3,1 u2,3 + u3,2 2u3,3 + 1

(4.47)

Having assumed small-displacement gradients, let us now compute the relative stretch ε(1) . From equation (4.38) we can write ε(1) = C11 − 1 ≈ 2u1,1 + 1 − 1 (4.48) A Taylor series expansion of the ﬁrst term about u1,1 = 0 yields 1 1 2u1,1 + 1 ≈ 1 + (1)− 2 (2)u1,1 2 = 1 + u1,1 and hence, ε(1) ≈ u1,1

(4.49)

100

Finite Element Method

The stretches ε(2) and ε(3) can be found using the same procedure. The result is ε(2) ≈ u2,2 ε(3) ≈ u3,3 The small-strain tensor is deﬁned so that the stretch in the direction of an arbitrary unit vector n is obtained by pre- and post-multiplying the strain tensor by the unit vector n. This operation can be written as

or in matrix notation,

ε(n) = n · · n

(4.50)

ε(n) = nT n

(4.51)

Having made this deﬁnition, the components of the small-strain tensor can be written as ⎡ ⎤ 1 1 u1,1 2 (u1,2 + u2,1 ) 2 (u1,3 + u3,1 ) ⎥ ⎢ 1 u2,2 [ ij ] = ⎣ 12 (u1,2 + u2,1 ) (4.52) 2 (u2,3 + u3,2 ) ⎦ 1 1 (u + u ) (u + u ) u 1,3 3,1 1,2 2,1 3,3 2 2 Note that the diagonal components of the small-strain tensor deﬁned in (4.52) contain the relative stretches in the X1 , X2 , and X3 directions, respectively. For small-displacement gradients, the oﬀ-diagonal terms are related to the engineering shear strains γ12 , γ13 , and γ23 . To see this, let us recall the relationship (4.45) between the engineering shear strain component, γ12 , and the quantities C12 , λ1 , and λ2 , i.e., C12 λ1 λ2 C12 = √ C11 C22

sin γ12 =

(4.53)

Expressing equation (4.53) in terms of the displacement gradients we get sin γ12 =

u1,2 + u2,1 (2u1,1 + 1)(2u2,2 + 1)

(4.54)

If we view the right-hand side of (4.54) as a function of two independent variables, u1,1 and u2,2 , and expand this function as a two-term Taylor series about u1,1 = 0 and u2,2 = 0 we obtain sin γ12 = (u1,2 + u2,1 ) + (u1,2 + u2,1 )(u1,1 + u2,2 )

(4.55)

Hence, when the displacement gradients are small, we can neglect the second term on the right-hand side of (4.55). In addition, when the displacement

Linearized Theory of Elasticity

101

gradients are small, the quantity, γ12 , is much less than unity. Therefore, sin γ12 ≈ γ12 , and the expression for the engineering shear strain reduces to γ12 ≈ u1,2 + u2,1

(4.56)

Following the same argument, one can conclude that under the condition of small-displacement gradients, the engineering shear strain in the X2 − X3 and X1 − X3 planes are related to the displacement gradients as follows: γ23 ≈ u2,3 + u3,2 γ13 ≈ u1,3 + u3,1 Notice that the small-strain components 12 , 13 , and 23 in (4.52) above are equal to one-half the engineering shear strains γ12 , γ13 , and γ23 . The 1/2 factor is needed so that transforms according to the tensor transformation law given in Chapter 2, in Section 2.3.1. The components of the small-strain tensor can be written down on one line using indicial notation as follows: 1 (ui,j + uj,i ) (4.57) 2 Notice that is symmetric, and therefore we need to only keep track of six of its components. These components are typically stored in a 6 × 1 vector as ⎧ ⎫ ⎪ ⎪ 11 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 22 ⎪ ⎪ ⎨ ⎬

33 {} = (4.58) γ ⎪ ⎪ 23 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ γ13 ⎭ ⎪ ⎩ γ12

ij =

Example 4.3 A two-dimensional deformation from the original conﬁguration to the de formed conﬁguration is deﬁned as x1 = X1 + αX2 x2 = X2

(4.59)

The deformation described in (4.59) above is called simple shear. Let us now compute the relative stretch ε(n) , where (2)/2 n= (2)/2 From equation (4.59) we can compute the components of the deformation gradient tensor, 1 α F= 0 1

102

Finite Element Method

and hence,

1 α C= α α2 + 1

The stretch in the direction of n can now be obtained directly from C. We get √ ε(n) = n · C · n − 1 = 1/2 4 + 4α + 2α2 − 1 α (4.60) ≈ 2 The last line in (4.60) follows if one assumes that α is small, and expands the second line as a Taylor series about α = 0. Let us now compute the same quantity ε(n) using the small-strain tensor. Hopefully we will get the same answer. Noting that ui = xi − Xi , the smallstrain components turn out to be u1,1 1/2(u1,2 + u2,1 ) = u2,2 1/2(u1,2 + u2,1 ) 0 α/2 = α/2 0 Computing the stretch in the n direction from (4.51) we get √ 0 α/2 √ √ √2/2 ε(n) = 2/2 2/2 α/2 0 2/2 α = 2

(4.61)

which, of course, is the correct result. Note that we would not get this result if we forgot the 1/2 factor in the deﬁnition of the small shear strain components.

4.5

Hooke’s law

Much of the linearized theory of elasticity hinges on the assumptions that the material is linearly elastic and isotropic. Loosely speaking, an isotropic material is one whose mechanical properties are independent of the orientation in which the material is tested. In other words, an isotropic material does not have any preferred directions. Hooke’s law for linearly elastic materials states that there is a linear relation ship between the Cauchy stress components and the small-strain components. This relationship can be written as σij = Cijkl kl

(4.62)

Linearized Theory of Elasticity

103

where Cijkl represents the components of a fourth-order tensor called the elasticity tensor. If a material is linearly elastic and isotropic, then the components of the elasticity tensor are invariant (remain unchanged) under a rotation of the coordinate axes. An isotropic fourth-order tensor has the form Cijkl = λδij δkl + 2μδik δjl

(4.63)

where λ and μ are Lam´e’s constants. After substituting equation (4.63) into (4.62) we obtain σij = {λδij δkl + 2μδik δjl } kl = λ kk δij + 2μ ij

(4.64)

Hence, from now on, we will write the component form of Hooke’s law for isotropic materials as (4.65) σij = λ kk δij + 2μ ij The inverse of equation (4.65) can be obtained by writing the relationship in tensor notation (4.66) σ = λ tr()I + 2μ and taking the tensor scalar product of both sides with the second-order identity tensor, i.e., σ : I = λ tr()I : I + 2μ : I ⇒ σkk = λ kk δij δij + 2μ kk = 3λ kk + 2μ kk ⇒ σkk

kk = 3λ + 2μ

(4.67)

Finally, by substituting equation (4.67) into (4.65) and simplifying we obtain

ij =

4.5.1

1 λ σij − σkk δij 2μ 2μ (3λ + 2μ)

(4.68)

Engineering constants

In engineering practice, it common to use what are called engineering con stants instead of the Lam´e’s constants that appear in equation (4.65). In this section we deﬁne the engineering constants (Young’s modulus, Poisson’s ratio, and shear modulus) and present Hooke’s law in matrix notation in terms of these constants. At the end of this section, we provide relationships between the engineering constants and Lam´e’s constants. To begin, consider a state of stress at a point in a linearly elastic and isotropic material in which the only nonzero stress component is σ11 . The

104

Finite Element Method

resulting extensional strain in the X1 direction is then 11 . Under these con ditions, the normal stress σ11 and the extensional strain 11 are related by σ11 = E 11

(4.69)

where E is Young’s modulus. Similarly, if the only nonzero stress component is σ22 or σ33 , then we would have σ22 = E 22 , or σ33 = E 33 . Due to the Poisson eﬀect, the application of the stress σ11 causes the material to strain laterally in both the X2 and X3 directions. For an isotropic material, the magnitude of the lateral contraction in both the X2 and X3 directions is the same. Poisson’s ratio, ν, is deﬁned as the ratio of the lateral strain to the longitudinal strain, i.e.,

22

33 ν=− = (4.70)

11

11 Note that the strains in the X2 and X3 directions as a result of the extensional strain 11 are negative, and hence, for typical engineering materials, Poisson’s ratio is a positive constant ranging from 0 ≤ ν ≤ 0.5. The strain components 22 and 33 can now be written in terms of the applied stress, Young’s modulus, and Poisson’s ratio as follows: ν

22 = − σ11 E ν

33 = − σ11 E (4.71) Using the deﬁnitions of Poisson’s ratio and Young’s modulus, the generalized Hooke’s law can now be written as ν ν σ11 − σ22 − σ33

11 = E E E ν ν σ22

22 = − σ11 + − σ33 E E E ν ν σ33

33 = − σ11 − σ22 + E E E 1 γ23 = σ23 G 1 γ13 = σ13 G 1 γ12 = σ12 G These six equations can be written in matrix notation as: ⎧ ⎫ ⎡ ⎤⎧ ⎫ σ11 ⎪ 1/E −ν/E −ν/E 0 0 0 ⎪

11 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎥⎪ ⎪ ⎪ −ν/E 1/E −ν/E 0 0 0

⎪ ⎪ ⎪ 22 ⎥⎨ ⎪σ22 ⎪ ⎪ ⎪ ⎢ ⎪ ⎨ ⎬ ⎬ ⎢ ⎥ 1/E 0 0 0 ⎥ σ33

33 ⎢−ν/E −ν/E =⎢ (4.72) ⎥ 0 0 1/G 0 0 ⎥⎪ γ23 ⎪ σ23 ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎣ 0 0 0 0 1/G 0 ⎦⎪ γ13 ⎪ σ13 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎩ ⎭ ⎭ 0 0 0 0 0 1/G γ12 σ12

Linearized Theory of Elasticity

105

The 6 × 6 matrix in equation (4.72) is called the compliance matrix. For the sake of completeness, we can invert the matrix equation (4.72) to obtain ⎧ ⎫ ⎡ ⎤⎧ ⎫ ˜ (1 − ν) ˜ ˜ E Eν Eν 0 0 0 ⎪ ⎪ 11 ⎪ ⎪σ11 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ˜ ⎪ ⎪ 22 ⎪ ⎪ ˜ (1 − ν) ˜ Eν 0 0 0⎥ E ⎪ ⎪σ22 ⎪ ⎪ ⎢ Eν ⎪ ⎪ ⎥ ⎪ ⎪ ⎪ ⎨ ⎨ ⎬ ⎢ ˜ ⎬ ˜ ˜ (1 − ν) 0 0 0 ⎥ σ33

33 Eν Eν E ⎥ ⎢ =⎢ (4.73) γ23 ⎪ ⎪σ23 ⎪ ⎪ ⎢ ⎪ 0 0 0 G 0 0⎥ ⎪ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪γ13 ⎪ ⎪σ13 ⎪ ⎪ ⎣ ⎪ 0 0 0 0 G 0⎦⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ σ12 γ 0 0 0 0 0 G 12 where ˜= E

E (1 + ν) (1 − 2ν)

A comparison of equations (4.73) and (4.65) reveals the following relationships between Lam´e’s constants and the engineering constants Eν (1 + ν)(1 − 2ν) μ=G λ=

(4.74)

Relationship between G, E, and ν As discussed in the previous section, only two elastic constants are required to completely characterize the elastic behavior of a linearly elastic and isotropic solid. One could use Lam´e’s constants, or the engineering constants. In this section, we derive the relationship between the shear modulus and Young’s modulus and Poisson’s ratio to emphasize that only two of these constants are independent. To begin, consider a stress element on the surface of a body that is lined up with the X1 and X2 axes. Suppose that this stress element is subjected to a state of pure shear, where the faces of the element are subjected to the shear stress τ . From the generalized Hooke’s law, we can write τ = Gγ

(4.75)

where γ = 2 12 is the engineering shear strain. We know from our study of principal stresses and directions that the stress element in the principal orientation is obtained by rotating the stress element counterclockwise through an angle θ = π/4. If we now set up a primed coordinate system to be lined up with the principle directions X1 , X2 . We emphasize that the action of the shear stress τ causes no extensional strain, i.e., 11 = 22 = 0. The extensional strain 11 in the primed coordinate system is related to Young’s modulus and Poisson’s ratio by

11 =

ν τ τ − −τ = (1 + ν) E E E

(4.76)

106

Finite Element Method

Because the small-strain tensor is a second-order tensor, its components transform according to equation (2.59), the tensor transformation law. Hence the components of the strain tensor in the primed coordinate system are related to the components in the unprimed system through the relation

ij = api aqj pq

(4.77)

From equation (4.77) we can write

11 = ap1 aq1 pq = a11 a11 11 + a11 a21 12 + a21 a11 22 = a11 a21 (2 12 ) γ = 2

(4.78)

Equating equations (4.78) and (4.76) we get τ γ = (1 + ν) E 2

(4.79)

E 2 (1 + ν)

(4.80)

and hence, G=

4.5.2

Hooke’s law for plane stress and plane strain

On occasion, a three-dimensional problem can be reduced to a two-dimensional problem by making the assumption of either plane stress or plane strain. In this section, we will explain these two assumptions and express Hooke’s law for each case. To describe the assumption of plane stress, let us consider the example of a thin plate as shown in Figure 4.10. The plate lies in the X1 − X2 plane and has a constant thickness, t. The upper and lower surfaces of the plate are traction free, but tractions can act on the edges of the plate. The smallest inplane dimension of the plate is labeled W . Now consider a point, P , located on the top surface of the plate. At this location, since the top surface is traction free, the stress components σ33 , σ13 , and σ23 are all identically zero. The same is true for points that lie on the bottom surface of the plate. If the thickness of the plate is small compared to W , then it is reasonable to assume that the out-of-plane stress components do not have a chance to build up appreciably through the thickness of the plate, i.e., they remain small compared to σ11 , σ22 , and σ12 . Hence, under the assumption of plane stress we assume σ33 = σ13 = σ23 ≈ 0. As a general rule of thumb, the assumption of plane stress is justiﬁed when t/W ≤ 0.1.

Linearized Theory of Elasticity

107

X2 W

X3

t

X1

FIGURE 4.10 The assumption of plane stress. Hooke’s law for the case of plane stress can be written as ⎫ ⎧ ⎫ ⎡ ⎤⎧ 1/E −ν/E 0 ⎨ σ11 ⎬ ⎨ 11 ⎬

22 = ⎣ −ν/E 1/E 0 ⎦ σ22 ⎩ ⎭ ⎩ ⎭ γ12 σ12 0 0 1/G ν

33 = − (σ11 + σ22 ) E

(4.81)

It is important to notice that under the assumption of plane stress, the outof-plane strain component, 33 , is not in general zero. It can be calculated from the in-plane stress components σ11 and σ22 and the elastic constants. Inverting the relation (4.83) we obtain ⎫ ⎡ ⎫ ⎧ ⎤⎧ ¯ νE ¯ 0 ⎨ 11 ⎬ E ⎨ σ11 ⎬ ¯ E ¯ 0 ⎦ 22 σ22 = ⎣ ν E (4.82) ⎩ ⎭ ⎩ ⎭ σ12 γ12 0 0 G ! ¯ = E/ 1 − ν 2 . where E In engineering practice, the assumption of plane strain is commonly made when the cross-sectional geometry of the structure does not vary in one di rection, and the loading also does not vary in that same direction. Under such conditions, the analysis of a single cross section gives the behavior of the entire structure, except in the near vicinity of the ends. To explain further, let us consider the structure depicted in Figure 4.11. The geometry of the structure’s cross section does not vary in the X3 direction, and the characteristic dimension, W , labeled in the ﬁgure is large compared to the X1 − X2 in-plane dimensions of the cross section. The loading does not vary in the X3 direction. The structure shown in the ﬁgure could be viewed as a dam designed to hold back water. The hydrostatic pressure exerted by

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Finite Element Method

X2

X1 W

X3 FIGURE 4.11 The assumption of plane strain. the water on the dam does not vary appreciably in the X3 direction, although it varies linearly in the X2 direction. Now let us consider a section of the structure that is suﬃciently far from the ends outlined in the ﬁgure. Ordinarily, if the front and back surfaces of this section were free surfaces, the section would have the tendency to expand in the X3 direction due to the compressive distributed load. The constraint of the surrounding material, however, prevents this from happening, and the extensional strain, 33 , is approximately zero. The surrounding material also prevents transverse shear. Hence, for the case of plane strain, we assume

33 = 13 = 23 ≈ 0. We note that under these conditions, the normal stress, σ33 , is in general nonzero. The normal stress is needed to prevent the section from elongating or shortening in the X3 direction. Hooke’s law for the case of plane strain can be written as ⎫ ⎧ ⎫ ⎡ ⎤⎧ ˜ (1 − ν) ˜ E Eν 0 ⎨ 11 ⎬ ⎨ σ11 ⎬ ˜ ˜ (1 − ν) 0 ⎦ 22 σ22 = ⎣ Eν E ⎩ ⎭ ⎩ ⎭ σ12 γ12 0 0 G ˜ ( 11 + 22 ) σ33 = Eν

(4.83)

where again ˜= E

4.5.3

E (1 + ν) (1 − 2ν)

Hooke’s law for thermal stress problems

When a solid body made of typical engineering materials is heated, the body expands. Similarly, when the body is cooled, it contracts. For an uncon

Linearized Theory of Elasticity

109

strained body, this expansion or contraction does not cause any stress to develop inside the body. On the other hand, when the body is constrained, there are forces present trying to prevent the expansion. Under such condi tions, thermal stresses develop due to the presence of elastic strain. As an example, suppose that a stress-free bar that is clamped at both ends is sub sequently heated uniformly. Because the bar is clamped at both ends, the total strain in the bar remains zero. The thermal strain however is positive. Hence, the elastic strain is negative, and the resulting normal stress in the bar is compressive. The total strain, ij , at a point in a body can be decomposed into elastic and thermal parts. The relation can be written as

ij = eij + Tij

(4.84)

where eij is the elastic part, and Tij is the thermal part. The thermal strain at a point in the body can be written as

Tij = αij ΔT

(4.85)

where ΔT is the temperature change from some stress-free reference temper ature, and αij are coeﬃcients of thermal expansion. For isotropic materials, the amount of thermal expansion is the same in all directions, and hence for that case we have (4.86)

Tij = αΔT δij Here in equation (4.86) α is simply called the coeﬃcient of thermal expansion having units of 1/◦ C. We emphasize that stress develops in a body due to the presence of elastic strains. Hence, it should be remembered that the strain components to be used in Hooke’s law are the elastic strain components, i.e., σij = Dijkl ekl . The substitution of the strain decomposition (4.84) along with the deﬁnition of the thermal strain components (4.86) into Hooke’s law gives the following modiﬁed version of Hooke’s law for thermoelastic problems: σij = Dijkl ( kl − αδkl ΔT )

4.6

(4.87)

Axisymmetric problems

Axisymmetric problems are another important class of problems that are twodimensional simpliﬁcations of fully three-dimensional problems. In order for a problem to be considered axisymmsetric, both the geometry and the loading must possess the same axial symmetry about a ﬁxed axis. The geometry of an axisymmetric problem is obtained by revolving a two-dimensional domain

110

Finite Element Method

z r

FIGURE 4.12 Axisymmetric analysis of ﬁber pull-out problem.

about a ﬁxed axis, creating a solid of revolution. The loading is also obtained by revolving some type of load condition such as a point load or distributed load about the same ﬁxed axis. For example, a uniform surface traction dis tribution acting on the top of a circular plate could be generated by revolving a line load about an axis passing through the center of the plate. If a problem possesses such axial symmetry (both geometry and loading), then it can be solved by performing a two-dimensional axisymmetric analysis. A nice example that illustrates the ideas above is the ﬁber pull-out prob lem illustrated in Figure 4.12. An inner core (ﬁber) is embedded inside an outer cylinder (matrix). The three-dimensional geometry is shown on the left. The top of the ﬁber is subjected to a uniform traction distribution in the z direction, and the bottom of the cylinder is constrained from moving in the z direction. The traction distribution applied to the ﬁber has the tendency to ”pull out” the ﬁber from the surrounding matrix. Because this problem possesses the requisite axial symmetry (both in geometry and loading), the problem can be reduced to a two-dimensional problem shown in the center of the ﬁgure. For the sake of interest, the deformed shape of the composite cylinder is shown to the right, and a contour plot of the stress component σzz is superposed on top of the geometry. The goal of this section is to derive the relationship between strain and displacement for axisymmetric problems. To begin, it is necessary to deﬁne a cylindrical coordinate system. A cylindri cal coordinate system is formed by three mutually perpendicular coordinate curves. The coordinate curves are the intersections of coordinate surfaces, as we will discuss in the following paragraphs.

Linearized Theory of Elasticity

X3

111

X2

ξ3

ξ1 ξ2

X1

P

r θ

(a)

z

(b)

FIGURE 4.13 Coordinate curves. The intersection of a cylindrical surface of radius r and two perpendicular planar surfaces is shown in Figure 4.13. The cylinder is aligned with the X3 axis. The two planes shown in the ﬁgure can be deﬁned as X3 = z;

θ = θc

(4.88)

where z is the vertical distance from some point that lies on the X3 axis to point P labeled in the ﬁgure. From now on we will refer to the plane θ = θc as the θ-plane. The coordinate curve ξ1 is the intersection of the plane X3 = z and the θ-plane. The intersection of the cylinder and the plane X3 = z forms the coordinate curve ξ2 . Finally, the intersection of the cylinder and the θ plane forms the coordinate curve ξ3 . The point P in the ﬁgure is located at the intersection of all three coordinate curves. The location of this point can also be deﬁned by the cylindrical coordinates r, θ, and z labeled in the ﬁgure. The gradient operator in general curvilinear coordinates is deﬁned as ∇=

∂ gi ∂ξi

(4.89)

where gi are the natural unit base vectors in the curvilinear coordinate system. The natural base vectors are deﬁned as gi =

∂X ∂ξi

(4.90)

In other words, each base vector is formed by calculating the vector, dX, which represents an inﬁnitesimal change in the position vector X with respect

112

Finite Element Method

to a diﬀerential movement along the ξi curve, and dividing that vector by its magnitude. Incidentally, the base vectors gi are unit vectors, because the magnitude of dX is dξi , the arc length traveled by the tip of the vector X during the inﬁnitesimal movement along the coordinate curve. Because the base vectors gi are mutually perpendicular, the curvilinear coordinate system is called an orthogonal curvilinear coordinate system. In cylindrical coordinates, the partial derivatives with respect to the curvi linear coordinates are related to partial derivatives with respect to the param eters r, θ, and z as follows: ∂ ∂ = ∂ξ1 ∂r ∂ 1 ∂ = ∂ξ2 r ∂θ ∂ ∂ = ∂ξ3 ∂z

(4.91)

Note that the result in the second line in equation (4.91) is a result of the fact that the arc length of the diﬀerential movement dξ2 is equal to r dθ. Finally, letting er = g1 , eθ = g2 , and ez = g3 , we obtain the familiar form ∇=

∂ 1 ∂ ∂ er + eθ + ez ∂r r ∂θ ∂z

(4.92)

In general, the displacement vector, u, in cylindrical coordinates has three components, i.e., (4.93) u = ur er + uθ eθ + uz ez In axisymmetric problems, however, because both the geometry and the load ing possess axial symmetry, the displacement ﬁeld does not vary in the θ direction, and hence we can write u = ur er + uz ez

(4.94)

The main task that is left in this section is to evaluate the displacement gradient tensor using the gradient operator (4.92) and the axisymmetric form (4.94) of the displacement vector. Before doing this, however, it is convenient to investigate the derivatives of the base vectors er , eθ , and ez with respect to r, θ, and z. We will need these derivatives later in the derivation of ∇u. To begin, let us write down the base vectors er , eθ , and ez in terms of the ﬁxed Cartesian base vectors e1 , e2 , and e3 as follows: er = cos θ e1 + sin θ e2 eθ = − sin θ e1 + cos θ e2 ez = e3

(4.95)

Linearized Theory of Elasticity

113

Diﬀerentiating the ﬁrst line of (4.95) with respect to θ gives er,θ = − sin θ e1 + cos θ e2 = eθ

(4.96)

Similarly, diﬀerentiating the second line with respect to θ yields eθ,θ = − cos θ e1 − sin θ e2 = − er

(4.97)

The remaining derivatives of the base vectors are all zero. The results are summarized in Table 4.1 below. Getting the components of ∇u in cylindrical coordinates is just a lengthy exercise in the use of the chain rule once the derivatives of the base vectors are known. The operation to be carried out can be written as ∂ 1 ∂ ∂ er + eθ + ez ⊗ (ur er + uθ eθ + uz ez ) (4.98) ∇u = ∂r r ∂θ ∂z To perform the tensor operation (4.98), each term on the left side of the ⊗ symbol must operate on each term on the right side of the symbol. Each operation requires the use of the chain rule. For example, if the ﬁrst term on the left operates on the ﬁrst term on the right we get er

∂ ∂er (ur er ) = er ur,r er + er ur ∂r ∂r = ur,r er er

(4.99)

Note that the last term in (4.99) drops out because the direction of er does not change with respect to a change in r. With the help of Table 4.1, ∇u can now be written as ∇u = {ur,r er er + uθ,r er eθ + uz,r er ez } 1 1 [ur,θ eθ er + ur eθ eθ ] + [uθ,θ eθ eθ − uθ eθ er ] + r r 1 + [uz,θ eθ ez ] r + {ur,z ez er + uθ,z ez eθ + uz,z ez ez } TABLE 4.1

Derivatives of base vectors. ∂ er eθ ez ∂/∂r ∂/∂θ ∂/∂z

0 eθ 0

0 − er 0

0 0 0

(4.100)

114

Finite Element Method

For axisymmetric problems, the displacement component uθ is equal to zero, and all derivatives of the displacement components with respect to θ are also zero. Hence, for axisymmetric problems, the components of ∇u can be written in matrix form as ⎤ ⎡ ur,r 0 uz,r 1 ⎥ ⎢ (4.101) ∇u = ⎣ 0 ur 0 ⎦ r ur,z 0 uz,z Armed with the displacement gradient matrix (4.101) for axisymmetric problems, we can now compute the components of the relevant deformation tensors. For example, the components of the deformation tensor F are com puted as follows: x=X+u⇒ F = ∇u + I

(4.102)

where I is the second-order identity tensor. Using the relationship (4.102) between ∇u and F, the components of F can be written as ⎤ ⎡ 0 uz,r

ur,r + 1 1 ⎥

⎢ (4.103) F=⎣ ur + 1 0 0 ⎦ r 0 uz,z + 1 ur,z If we now assume that the displacement gradients are small and neglect products of displacement gradients such as u2r,r , then the components of Green deformation tensor, C = FT F, are approximately equal to the following: ⎤ ⎡ 0 uz,r + ur,z 2ur,r + 1 2ur ⎥ ⎢ C≈⎣ (4.104) +1 0 0 ⎦ r uz,r + ur,z 0 2uz,z + 1 Using the relationship between the components of C and the small strain tensor discussed in Section 4.4.3, the components of the small strain tensor can ﬁnally be written as ⎫ ⎧ ⎫ ⎧ ur,r ⎪ ⎪

⎪ ⎪ ⎪ ⎪ rr ⎪ ⎪ ⎪ 1 ⎨ ⎬ ⎪ ⎬ ⎨

θθ ur {} = = (4.105) r

zz ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ uz,z ⎩ ⎭ ⎪ ⎪ ⎪ ⎭ ⎩ γrz uz,r + ur,z To end this section, we note that there are four nonzero strain components for the case of axisymmetric problems. The strain components rr , θθ , and zz are the relative stretches in the radial, circumferential, and axial directions, respectively. The strain component θθ , which gives the relative stretch in the

Linearized Theory of Elasticity

115

FIGURE 4.14 Continuous piecewise surface. circumferential direction, is also known as the hoop strain. The engineering shear strain γrz gives the decrease in angle experienced by two ﬁbers that are originally perpendicular to each other and both lying in the r − z plane. For linearly elastic and axisymmetric problems, Hooke’s law becomes ⎫ ⎡ ⎫ ⎧ ⎤⎧ σrr ⎪ 2μ + λ λ λ 0 ⎪ ⎪ ⎪ ⎪ ⎪ rr ⎪ ⎪ ⎬ ⎢ ⎨ ⎥ ⎨ θθ ⎬ σθθ λ 2μ + λ λ 0 ⎥ =⎢ (4.106) σzz ⎪

zz ⎪ λ 2μ + λ 0 ⎦ ⎪ ⎪ ⎣ λ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎩ σrz

rz 0 0 0 μ where Lam´e’s constants in (4.106) above are deﬁned in equation (4.74) in terms of Young’s modulus and Poisson’s ratio.

4.7

Weak form of the equilibrium equation

Recall from our work in Section 4.3 that the equilibrium equation can be written as (4.107) σij,j + bi = 0 In a displacement-based ﬁnite element method, we seek to use continuous piecewise trial functions as candidate solutions to the partial diﬀerential equa tion (4.107). The trial function is a vector-valued function having two or three components, depending on whether we are working in two or three dimensions. At the moment, one might wonder what is meant by a continuous piecewise displacement ﬁeld in the context of two- and three-dimensional solid mechan ics problems. In order to illustrate, suppose that a series of measurements are

116

Finite Element Method

made for the displacement component, uX , at a large number of points located on the surface of a thin plate. If this surface plot turned out to be continuous piecewise, it would have the characteristics shown in Figure 4.14. Notice that the surface is continuous, i.e., there are no gaps or overlaps, but there exist creases (lines along which the slope of the surface in a given direction is dis continuous). Mathematically speaking, at a particular point lying on one of the creases, the partial derivative of the function representing the surface with respect to one or more of the spatial coordinates will be discontinuous. Sec ond partial derivatives would not exist at such points. Because the divergence of the stress (the ﬁrst term in the equilibrium equation) indirectly involves second partial derivatives of the displacement ﬁeld, we cannot directly plug in continuous piecewise trial functions into the equilibrium equation. In order to remedy this situation, we proceed like we did in Chapter 3 and derive the weak form of the equilibrium equation. In order to derive the weak form, the ﬁrst step is to introduce a vector-valued test function whose components are wi having the following properties: 1.

wi = C 0 continuous in Ω

2.

wi = 0 on Γu

(4.108)

In other words, we insist that the components of the test function be at least continuous piecewise over the domain of interest, and require wi to be zero at points on the boundary wherever ui is speciﬁed. The next step is to multiply the equilibrium equation (4.107) by the test function and then integrate the resulting expression over the domain of interest, Ω. Doing this gives σij,j wi dΩ + bi wi dΩ = 0 (4.109) Ω

Ω

Next, in order to get rid of the term involving the divergence of the stress on the left-hand side of equation (4.109) above, it is necessary to integrate the ﬁrst integral by parts. The goal is to replace the integral with an equivalent integral that does not involve ﬁrst partial derivatives of the stress ﬁeld. The integration by parts can be readily carried out with the help of the following identity: (4.110) (σij wi ),j = σij,j wi + σij wi,j where the right-hand side of equation (4.110) above is obtained using the product rule. Notice that the ﬁrst term on the right-hand side of (4.110) is the integrand of the ﬁrst integral in (4.109). Hence using equation (4.110) we can rewrite equation (4.109) as follows: (σij wi ),j dΩ − σij wi,j dΩ + bi wi dΩ = 0 (4.111) Ω

Ω

Ω

We now recognize that the integral on the left-hand side of equation (4.111) is a volume integral whose integrand is the scalar quantity, ∇ · (w · σ), the

Linearized Theory of Elasticity

117

divergence of the vector w · σ. To see this more clearly, it is helpful to go through the exercise of translating expressions from indicial notation into equivalent expressions in tensor notation and vice versa. For example, w · σ = (wi ei ) · (σkl ek el ) = wi σkl δik el = wk σkl el and hence,

(4.112)

∂ ∇ · (w · σ) = ei · (wk σkl el ) ∂Xi ∂wk σkl δil = ∂Xi ∂wk = σkl ∂Xl = σij wi,j

(4.113)

We can now employ the divergence theorem (2.129) and recast the ﬁrst integral on the left-hand side of (4.111) into the following surface integral: ∇ · (w · σ) dΩ = n · (w · σ) dΓ (4.114) Ω

Γ

Finally, noting that n · (w · σ) = (ni ei ) · (wk σkl el ) = nl wk σkl = σij nj wi = ti wi

(4.115)

and rearranging terms, equation (4.111) can be written as σij wi,j dΩ = bi wi dΩ + ti wi dΓ Ω

Ω

(4.116)

Γ

where ti are the components of the traction vector at a point on Γt . Equation (4.116) is the weak form of the equilibrium equation, or the prin ciple of virtual work. The left-hand side of this equation is commonly referred to as the internal virtual work. The right-hand side is the external virtual work. If we view the test function as having units of length, then the left-hand side of (4.116) has units of stress times volume, and the right-hand side has units of force times length, both equivalent to work. If we want to, at this stage we can plug the stress-strain law into the left-hand side of the weak form, but we will defer this exercise until after the ﬁnite element approxima tions are substituted into the weak form, which will be covered in the next chapter.

118

Finite Element Method

4.8 Problems Problem 4.1 Show that for an isotropic and linearly elastic material Hooke’s law can be written as ν

kk δij σij = 2G ij + 1 − 2ν where G is the shear modulus and ν is Poisson’s ratio Problem 4.2 Using the form of Hooke’s law obtained in Problem 4.1, show that the equilib rium equation (4.32) can be written in terms of the displacement components ui as follows: ∂ui ∂ 2 uj 1 G + + bi = 0 ∂Xj ∂Xj 1 − 2ν ∂Xj ∂Xi Problem 4.3 A solid body is subjected to all-around hydrostatic pressure of magnitude P . (a)

State the boundary condition on the surface of the body.

(b)

If the body is homogeneous, determine a stress ﬁeld inside the body that satisﬁes the equilibrium equation and the boundary conditions.

Problem 4.4 Consider the following state of stress at a point: ⎡

⎤ 10 −5 0

σ = ⎣ −5 5 0 ⎦ MPa

0 0 0

(a)

Determine the three stress invariants I1 , I2 , and I3 .

(b)

Determine the three principal stresses.

(c)

Determine the three principal directions.

(d)

Indicate on a sketch a stress element in the principal orientation.

Linearized Theory of Elasticity

119

Problem 4.5 A two-dimensional deformation from the original conﬁguration to the de formed conﬁguration is deﬁned as x1 = 3X12 + X2 x2 = X1 + X23 (a)

Determine the components of the deformation gradient tensor F.

(b)

Determine the components of the Green deformation tensor C.

(c)

Determine the relative stretch in the e1 and e2 directions at the point (X1 , X2 ) = (1, 1).

5 Steady-State Heat Conduction

I have learned through bitter experience the one supreme lesson to conserve my anger, and as heat conserved is transmitted into energy, even so our anger controlled can be transmitted into a power that can move the world. —Mahatma Gandhi The temperature distribution in solids is needed, for example, to predict ther mal stresses that can develop in constrained bodies. In this chapter, we con sider the phenomenon of steady-state heat conduction. We will restrict our attention to isotropic heat conduction, where it is assumed that a material’s resistance to heat ﬂow is the same in all directions. An energy balance is performed by calling upon the principle of conservation of energy. The principle states that the energy that ﬂows into a body per unit time minus the energy that ﬂows out per unit time is equal to the change of internal energy inside the body per unit time. When the internal energy in the body stops changing with time, the body is said to have reached steady state. Under steady-state conditions, the temperature ﬁeld no longer depends on time, but it can vary spatially throughout the body. We will begin this chapter by deriving the partial diﬀerential equation that governs steady-state heat conduction. During the process, we will deﬁne the relevant physical quantities: heat ﬂux, internal heat source, temperature gra dient, and thermal conductivity. The units associated with each of these quantities will also be discussed. After deriving the partial diﬀerential equa tion from an energy balance, we will discuss the relevant boundary conditions, such as forced convection. Finally, we will go ahead and derive the weak form which turns out to be the starting point for a temperature-based ﬁnite ele ment formulation. The chapter also includes a discussion of symmetry, which under certain circumstances can be exploited, reducing computational costs. Under conditions of small strain and deformation, it is reasonable to as sume that the heat produced by mechanical work inside a body is negligible compared to other sources. Under these conditions, we can perform an un coupled analysis, whereby the steady-state heat equation is ﬁrst solved for the temperature ﬁeld, and then the temperature ﬁeld is used as input to a separate stress analysis. The temperature ﬁeld dictates the thermal strain distribution in the solid, which in turn gives rise to thermal stresses when the solid is constrained. In other words, the thermal strain distribution loads the solid, and the resulting displacement, elastic strains, and stresses are obtained

121

122

Finite Element Method q Γq S(Xi ) Ω Γ ΓT

FIGURE 5.1 Solid body subjected to internal heat source and surface ﬂux distribution.

by performing a separate ﬁnite element analysis.

5.1

Derivation of the steady-state heat equation

In order to derive the steady-state heat equation, it is helpful to observe Figure 5.1. The ﬁgure illustrates a solid body deﬁned by the interior volume Ω and outer surface Γ. The body is subjected to a prescribed temperature distribution over a portion of the surface labeled ΓT . In addition, a heat ﬂux distribution is prescribed over the portion of the boundary labeled Γq . A heat ﬂux is a vector quantity that describes the amount of energy ﬂowing through a unit area per unit time. Hence the ﬂux vector, q, at a point on the surface has the SI units of Joules per meter squared per second, i.e., J/(m2 · s). In the sign convention adopted here, positive ﬂux points out of the body. Heat can also be generated (or dissipated) internally by means of a heat source distribution S(X1 , X2 , X3 ) inside Ω. The heat source has units of energy per unit volume per unit time, i.e., J/(m3 · s). The relevant quantities and associated units for steady-state heat conduction are summarized in Table 5.1. Note that in the table we have used the unit Watt instead of Joules per second, i.e., W = J/s. Conservation of energy (assuming steady-state conditions have been reached) states that the energy ﬂowing into the body Ω per unit time must be equal to the energy ﬂowing out of the body through Γ per unit time. The energy ﬂowing into (or out of) the body due to the heat source can be accounted for by integrating the heat source distribution over the volume Ω. The ﬂux vector at a point on the surface can be decomposed into a component that is perpendicular to the surface and one that is tangential to the surface (normal and tangential components). The tangential component of the ﬂux does not

Steady-State Heat Conduction

123

contribute to the energy balance because this energy is neither ﬂowing into or out of the body. Hence, the net ﬂow of energy out of the surface of the body is obtained by evaluating the surface integral of the normal component, qn = q · n, of the heat ﬂux vector, where n is the unit outward normal to the surface. The energy balance statement can now be written as S dΩ − n · q dΓ = 0 (5.1) Ω

Γ

Just like in our derivation of the equilibrium equation in Section 4.3, the key to massaging the balance statement (5.1) above into a partial diﬀerential equation is to convert the surface integral involving the normal component of the ﬂux vector into an equivalent volume integral. This can be accomplished by employing the divergence theorem, i.e., n · q dΓ = ∇ · q dΩ (5.2) Γ

Ω

Hence the energy balance now becomes {S − ∇ · q} dΩ = 0

(5.3)

Ω

Because the balance statement (5.3) must hold for all subdomains inside Ω, the integrand must be identically zero at all points within Ω. Hence, S−∇·q=0

(5.4)

The partial diﬀerential equation (5.4) is often referred to as the heat equa tion. The second term on the left-hand side of equation (5.4) is called the divergence of the ﬂux. This term can be written in indicial notation as fol lows: ∂ ei · (qj ej ) ∇·q = ∂Xi ∂qj δij = ∂Xi = qi,i (5.5) TABLE 5.1

Relevant quantities and units needed for steady-state heat conduction. Symbol Name SI units W k Thermal conductivity m ◦C W q Heat ﬂux vector m2 W S Heat source m3

124

Finite Element Method

and hence, the governing partial diﬀerential equation can also be written in component form as (5.6) S − qi,i = 0

5.2

Fourier’s law

Thus far, the partial diﬀerential equation that we derived in the previous section involves partial derivatives of the ﬂux vector. In this section we in troduce the constitutive relation between heat ﬂux and temperature that is known as Fourier’s law. Armed with this relation, we will be able to recast the diﬀerential equation in terms of temperature as the unknown ﬁeld variable. Fourier’s law is an experimentally determined relationship that states that the heat ﬂux at a point in a body is proportional to the temperature gradient at that point. Mathematically, Fourier’s law can be written as q = −K∇T

(5.7)

where K is a second-order tensor whose Cartesian components are the direc tionally dependent thermal conductivities of the material. The quantity K can be written in matrix notation as follows: ⎡ ⎤ k11 k12 k13 (5.8) K = ⎣ k21 k22 k23 ⎦ k31 k32 k33 The components of the thermal conductivity tensor have units of J/(m ·◦ C · s). For the sake of simplicity, throughout the remainder of this book, we will restrict our attention to materials whose resistance to heat ﬂow is the same in all directions. Such materials are called isotropic. For the case of isotropic heat conduction, the components of the thermal conductivity tensor can be written as (5.9) Kij = kδij where k is referred to as the thermal conductivity. Fourier’s law can then be written as q = −k∇T (5.10) or in component form qi = −kT,i

(5.11)

By substituting Fourier’s law (5.11) into the heat equation (5.6) we get kT,ii + S = 0 or ∇2 T + S = 0

(5.12)

Steady-State Heat Conduction

125 X2

a

b

T (X1 , X2 ) = T (−X1 , X2 )

X1

c

T (X1 , X2 ) = T (−X1 , X2 )

d

FIGURE 5.2 Exploiting symmetry. It is worth mentioning here that when the heat source is zero, the heat equa tion reduces to simply ∇2 T = 0, which is called Laplace’s equation. When the heat source is present, the equation is known as the Poisson equation.

5.3

Boundary conditions

In the present chapter, we will consider three diﬀerent types of boundary con ditions that are encountered in steady-state heat conduction analyses. First of all, the temperature, T , can be prescribed over a portion of the body. Mathematically, this condition can be stated as T = T∗

on

ΓT

(5.13)

where ΓT is the portion of the boundary over which the temperature is a pre scribed function of the spatial variables. The prescribed temperature bound ary condition is the essential boundary condition. It is a direct condition placed on the ﬁeld variable. Secondly, the normal component of the heat ﬂux, qn∗ , can be speciﬁed on Γq . This condition can be written as n · q = qn∗

on

Γq

(5.14)

Because q = −k∇T from Fourier’s law, the ﬂux condition (5.14) poses a condition on the gradient (partial derivatives) of the ﬁeld variable. The ﬂux boundary condition is the natural boundary condition.

126

Finite Element Method

FIGURE 5.3 Heat transfer problem of an inner core embedded in a sphere.

Closely related to the prescribed ﬂux condition is forced convection. In forced convection, heat transfer takes place through the surface of the body into a surrounding ﬂuid ﬂowing at a certain velocity across the body. The forced convection condition can be stated as qn∗ = h (T − T∞ )

(5.15)

where h is the ﬁlm coeﬃcient. It turns out that the ﬁnite element implemen tation of steady-state heat conduction with the forced convection condition causes no diﬃculties and still leads to a linear system of equations. Symmetry can often be exploited in steady-state heat conduction problems, reducing the computational cost. To illustrate the idea, consider the twodimensional solid shown in Figure 5.2. A two-dimensional solid in the context of heat conduction means that no heat can ﬂow in the X3 direction (into or out of the page). As shown in the ﬁgure, the geometry of the body is symmetric about both the X1 and X2 axes. Now suppose that temperature distributions are prescribed along edges a-b and c-d labeled in the ﬁgure in such a manner that both are symmetric about the X2 axis. The temperature distribution in the body would then be an even function of X1 , and hence, the heat ﬂux in the X1 direction would be zero along the line X1 = 0. It would then only be necessary to model one-half of the plate (say the right half), enforcing the zero-ﬂux condition along the left edge, similar to a traction-free edge in solid mechanics. Radiation is another way in which heat can be transferred from one body to another. The net heat ﬂux ﬂowing through the surfaces of radiating bodies is proportional to the temperature at the surface of the body raised to the fourth power. Thus, radiation gives rise to a nonlinear heat transfer problem and will not be considered here. A nice introduction to this topic can be found in Lienhard [6].

Example 5.1 To apply some of the important concepts described in this chapter, we now consider the problem of a solid spherical core embedded inside a much larger

Steady-State Heat Conduction

127 1 (1225 km/◦ C) − 225 ◦ C r

FIGURE 5.4 Inner core heated to a uniform temperature of 1000 ◦ C, while the outer surface is held at 20 ◦ C. spherical body as shown in Figure 5.3. The diameter of the inner core is 1000 m, and the diameter of the outer sphere is 5000 m. Initially, the tem perature distribution in both the inner core and outer sphere is held at a uniform temperature of 20 ◦ C. If the surface of the outer sphere is maintained at a constant temperature of 20 ◦ C and the inner core is subsequently heated to a uniform temperature of 1000 ◦ C, it is desired to obtain the steady-state temperature distribution in the outer sphere. Due to the spherical symmetry of the problem (both the geometry and loading), the temperature ﬁeld in the outer sphere will only depend on the radial coordinate r, measured positive ra dially outward from the center of the inner core. Hence T (X1 , X2 , X3 ) = T (r). Recall from the previous section that the steady-state heat equation can be written as (5.16) k∇2 T + S = 0 Because there is no heat source present, the steady-state heat equation reduces to the Laplace equation, i.e., ∇2 T = 0. To proceed with the solution, it is convenient to work in a (r, θ, and φ) spherical coordinate system. The gradient operator in spherical coordinates is given as ∇=

1 ∂ 1 ∂ ∂ er + eφ + eθ ∂r r ∂φ r sin φ ∂θ

(5.17)

where er , eφ , and eθ are unit base vectors. Because the temperature distribu tion only depends on r, the heat equation reduces to the following second-order diﬀerential equation: 1 d 2 dT r =0 (5.18) r2 dr dr The boundary conditions can be written as T (r = ri ) = 1000 ◦ C

T (r = ro ) = 20 ◦ C

(5.19)

where ri and ro are the inner and outer surfaces, respectively, of the outer core. The exact solution to equation (5.18) above can be obtained by direct

128

Finite Element Method

integration. Integrating once yields r2

dT = C1 dr

(5.20)

where C1 is an unknown constant. Integrating a second time gives −

C1 + C2 r

(5.21)

where C2 is another constant of integration. The constants C1 and C2 are found by enforcing the two boundary conditions speciﬁed in equation (5.19). Doing this yields the following temperature distribution: T (r) = C3 /r + C4

(5.22)

where C3 = 1225 km/◦ C and C4 = 225 ◦ C. A contour plot of the temperature distribution is shown in Figure 5.4.

5.4

Weak form of the steady-state heat equation

Because the heat equation involves second partial derivatives of the temper ature ﬁeld, the desire to use C 0 trial functions makes it necessary to derive the weak form of the equation as our starting point for a ﬁnite element imple mentation. In order to derive the weak form, we follow the same procedure described in Chapter 4, Section 4.7. The weak form of the heat equation is actually easier to derive than the weak form of the equilibrium equation because of the fact that temperature is a scalar quantity. To begin, we introduce a scalar-valued test function, w(X1 , X2 , X3 ). The test function is assumed to be suﬃciently smooth and to be zero at any points on the boundary where the temperature is speciﬁed, i.e., the test function is deﬁned to be zero on all essential boundaries. Next, we multiply the heat equation by the test function, and integrate the resulting expression over the domain of interest Ω. Doing this gives T,ii w dΩ + Sw dΩ = 0 (5.23) k Ω

Ω

In order to get the expression (5.23) into the desired form, the ﬁrst integral on the left-hand side needs to be integrated by parts. Noting that (T,i w),i = T,ii w + T,i w,i

(5.24)

we can rewrite equation (5.23) as (T,i w),i dΩ − k T,i w,i dΩ + Sw dΩ = 0 k Ω

Ω

Ω

(5.25)

Steady-State Heat Conduction

129

It is now instructive to write the above equation in tensor notation. Doing this yields ∇T · ∇w dΩ + Sw dΩ = 0 (5.26) k ∇ · (w∇T ) dΩ − k Ω

Ω

Ω

Note that the integrand of the ﬁrst integral on the left-hand side is the diver gence of the vector w∇T . The second integral involves the dot product of the temperature gradient vector with the gradient of the test function. Finally, the third integral involves the product of the two scalar functions S and w, i.e., the heat source and the test function. Employing the divergence theorem to recast the ﬁrst volume integral on the left-hand side of equation (5.26) as an equivalent surface integral yields ∇T · ∇w dΩ + Sw dΩ = 0 (5.27) k n · (w∇T ) dΓ − k Γ

Ω

Ω

where n is the unit outward normal to the boundary. The ﬁnal step in the derivation of the weak form involves substituting Fourier’s law, q = −k∇T , into the expression (5.27) and recognizing that n · q = qn∗ is the normal component of the heat ﬂux on the boundary Γq . The ﬁnal result is k ∇w · ∇T dΩ = wS dΩ − wqn∗ dΓ (5.28) Ω

Ω

Γq

Note that the surface integral in (5.28) above only needs to be carried out over Γq , because the test function w is zero on ΓT . The weak form of the heat equation contains only ﬁrst partial deriva tives of the temperature ﬁeld. Thus, continuous piecewise trial functions can be substituted into the weak form without any diﬃculty, leading to a temperature-based ﬁnite element method for steady-state heat conduction, as will be demonstrated in the next chapter.

5.5

Problems

Problem 5.1 Consider the steady-state heat conduction problem of a ﬁn that is insulated on its outer surface so that heat can only ﬂow in one dimension. The length of the ﬁn is L = 0.1 m, and its thermal conductivity is k = 1.0 J/(m ·◦ C · s). Suppose that the bar is subjected to the following boundary conditions: T (X = 0) = 200 ◦ C qn∗ (X = L) = h(T − T∞ )

130

Finite Element Method

where h is the ﬁlm coeﬃcient and T∞ is the remote temperature of the sur rounding ﬂuid. Letting h = 10 J/(m2 · s) and T∞ = 20 ◦ C, perform the fol lowing: (a)

Derive the governing diﬀerential equation.

(b)

Derive the weak form.

(c)

Model the problem with two linear-temperature elements to obtain an estimate for the temperature of the ﬁn at X = L.

6 Continuum Finite Elements

Although this may seem a paradox, all exact science is dominated by the idea of approximation. When a man tells you that he knows the exact truth about anything, you are safe in inferring that he is an inexact man. —Bertrand Russell

In this chapter we will explore a few of the most popular ﬁnite elements for use in continuum problems. For two-dimensional problems, we will con sider the three-node triangle and the arbitrary four-node quadrilateral. The arbitrary four-node quadrilateral falls within a class of elements called isopara metric ﬁnite elements. In the chapter we will discuss the isoparametric con cept and emphasize why it is important. We will also consider two important three-dimensional elements: the four-node tetrahedron and the eight-node brick. The development of the various element types will be performed in the context of solid mechanics. The development of all elements for heat transfer is essentially the same. The only diﬀerence is that the ﬁeld variable in solid mechanics is displacement, whereas in heat transfer it is temperature. In two- and three-dimensional solid mechanics problems, displacement is a vector ﬁeld. The displacement at each point has both magnitude and direction. On the other hand, temperature is a scalar ﬁeld. So actually, the formulation of the element types for heat transfer is somewhat easier than for solid mechanics. There is some notation that will be used in this chapter that is best de ﬁned here. We will be dealing with ﬁnite elements in two conﬁgurations: the undeformed conﬁguration, and the deformed conﬁguration. The undeformed conﬁguration of an element is its initial geometry, before any loads are applied. The deformed conﬁguration is the geometry of the element after it undergoes deformation. A position vector acting from the origin of a Cartesian coor dinate system to a point inside the undeformed element will be deﬁned as X. After the element is deformed, the same point that originally occupied position X gets transported to position x inside the deformed element. The displacement u of point X is deﬁned as the deformed position minus the orig inal position, i.e., u = x − X. This relationship between displacement and position can be written in component form as ui = xi − Xi .

131

132

Finite Element Method 3

A2

A1 (X, Y ) A3

1

d3X , d3Y

2 d2X , d2Y

d1X , d1Y

FIGURE 6.1 The three-node triangle.

6.1

Three-node triangle

The use of the three-node triangular element for plane stress elasticity prob lems was ﬁrst reported in 1956 in a paper by Turner et al. [7]. At that time, it was necessary to do everything possible to reduce three-dimensional problems into two-dimensional problems due to lack of computational resources. The three-node triangle is still popular today, primarily because of the ease with which complicated geometries can be meshed with triangles. A schematic of a three-node triangle is shown in Figure 6.1. The numbers labeled in the ﬁgure represent the local node numbers. The numbering scheme is deﬁned such that local node one is chosen to be any one of the three vertices. The remaining nodes are then numbered sequentially in a counterclockwise fashion. It is convenient to arrange the displacement components at the nodal points in a 6 × 1 vector as follows: ⎧ ⎫ ⎪ ⎪ d1X ⎪ ⎪ ⎪ ⎪ ⎪ d1Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ d2X (6.1) de = ⎪ ⎪ d2Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d3X ⎪ ⎪ ⎪ ⎭ ⎩ d3Y where daX and daY are the displacement components associated with node a in the X and Y directions, respectively. One of the main ideas in the ﬁnite element method is to interpolate the ﬁeld variable inside the element using polynomial functions. For the case of the three-node triangle, we assume that the displacement components vary linearly in X and Y . Mathematically, the ﬁnite element approximations for the displacement components can be written

Continuum Finite Elements as

u ˜X u ˜Y

133

=

e

α1 + α2 X + α3 Y β1 + β2 X + β3 Y

(6.2)

where α1 , . . . , β3 are constants. It is common practice to express equation (6.2) in terms of shape functions and the displacement components at the nodal points. To do this, we impose the following three conditions: u ˜i (X1 , Y1 ) = d1i u ˜i (X2 , Y2 ) = d2i u ˜i (X3 , Y3 ) = d3i

(6.3)

where i can be either X or Y it doesn’t make any diﬀerence which one you choose. Letting i = X and substituting the constraints (6.3) into the ﬁrst line of equation (6.2) yields ⎧ ⎫ ⎡ ⎤⎧ ⎫ 1 X1 Y1 ⎨ α1 ⎬ ⎨ d1X ⎬ d2X = ⎣ 1 X2 Y2 ⎦ α2 (6.4) ⎩ ⎭ ⎩ ⎭ d3X 1 X3 Y3 α3 The algebra is rather lengthy, but equation (6.4) can be solved for α1 , α2 , and α3 without too much trouble. The resulting expressions can be put back into equation (6.2), and after collecting terms, the ﬁnal expression can be written as (6.5) u ˜X = N1 (X, Y )d1X + N2 (X, Y )d2X + N3 (X, Y )d3X where the functions N1 , N2 , and N3 are the three shape functions. These can be written in terms of the spatial variables and nodal coordinates as follows: 1 (X32 Y − Y32 X + X2 Y3 − Y2 X3 ) 2A 1 (−X31 Y + Y31 X − X1 Y3 + Y1 X3 ) N2 (X, Y ) = 2A 1 (X21 Y − Y21 X + X1 Y2 − Y1 X2 ) N3 (X, Y ) = 2A

N1 (X, Y ) =

(6.6)

Here in equation (6.6), X1 , Y1 . . . X3 , Y3 are the X and Y coordinates of the nodes. The quantities X21 , X32 , etc., represent diﬀerences in the nodal co ordinates, i.e., X21 = X2 − X1 . The quantity A is the area of the triangular element. The area can be easily computed using the cross-product operation. Let X21 and Y21 be the components of a vector X(1) that points from local node one to local node two. Similarly, let X31 and Y31 be the components of a vector X(2) that points from local node one to local node three. The magnitude of the cross product of these two vectors yields twice the area of the triangle, i.e., 2A = X(1) × X(2) = X21 Y31 − Y21 X31

134

Finite Element Method

N1

1

3

3

2

3

N2

1

2

1

N3

2

FIGURE 6.2 Shape functions for the three-node triangle. We point out that the three shape functions are all linear in X and Y , and they obey the Kronecker delta property that was introduced in Chapter 3, i.e., (6.7) NI (XJ ) = δIJ The three shape functions are plotted in Figure 6.2. Note that each surface is planar. An alternative way of getting the shape functions for the three-node triangle is through the use of area coordinates. The area coordinates are labeled A1 , A2 , and A3 in Figure 6.1. As the point (X, Y ) moves around inside the triangle, each of these three areas changes continuously. It is left as an exercise at the end of this chapter to compute the area of the three area coordinates through the use of the cross-product operation to show that N1 (X, Y ) = A1 (X, Y )/A N2 (X, Y ) = A2 (X, Y )/A N3 (X, Y ) = A3 (X, Y )/A

(6.8)

It is easy to see that the shape functions (6.8) satisfy the Kronecker delta property and are linear in X and Y . As point (X, Y ) approaches local node a, the area coordinate Aa becomes the total area A, and hence Aa (Xa , Ya )/A = 1 ˜ can now be expressed as the product of The vector-valued trial function u a 2 × 6 matrix containing shape functions and a 6 × 1 vector containing the nodal displacements as follows: ⎧ ⎫ d1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎪ ⎪ 1Y ⎪ ⎪ ⎨ ⎬ N1 0 N2 0 N3 0 u ˜X d2X = (6.9) u ˜Y e d2Y ⎪ 0 N1 0 N2 0 N3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d3X ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ d3Y

Continuum Finite Elements

135

or ˜e u

2×1

6.1.1

N d = 2 ×e6 6 ×e 1

(6.10)

The B-matrix

Proceeding along, it is now necessary to come up with a relationship between the in-plane strain components, XX , Y Y , and γXY , deﬁned in Chapter 4, and the nodal displacement vector de , deﬁned above. Recall that the in-plane strain components are related to the displacement components as follows: XX = uX,X Y Y = uY,Y γXY = uX,Y + uY,X

(6.11)

where, again, the comma denotes partial diﬀerentiation with respect to the spatial variables. For the sake of convenience, we will store the three in-plane strain components in a 3 × 1 vector, , as follows: ⎫ ⎧ ⎨ XX ⎬ (6.12) = Y Y ⎭ ⎩ γXY Having deﬁned the strain vector, we now seek a relationship of the form ⎧ ⎫ u ˜X,X ⎨ ⎬ ˜e B d u ˜ = = 3 ×e6 6 ×e 1 (6.13) Y,Y 3×1 ⎩ ⎭ u ˜X,Y + u ˜Y,X e where Be is the B-matrix for the element. The B-matrix operates on the nodal displacement vector, giving an approximation for the strain components at a point within the element. The strain calculation is carried out as a post processing step in the ﬁnite element method after the nodal displacements are found. The components of the B-matrix can be obtained by diﬀerentiating the ˜Y as follows: displacement components u ˜X and u u ˜X,X =

3

Na,X daX

a=1

u ˜Y,Y =

3

Na,Y daY

a=1

u ˜X,X + u ˜Y,X =

3

a=1

{Na,Y daX + Na,X daY }

(6.14)

136

Finite Element Method

and then writing the resulting expression in matrix notation as ⎧ ⎫ ⎪ ⎪ d1X ⎪ ⎪ ⎪ ⎪ ⎤⎪ ⎡ ⎪ d1Y ⎪ ⎪ ⎪ ⎪ N1,X 0 N2,X 0 N3,X 0 ⎨ ⎬ d2X ⎦ ⎣ ˜e = 0 N1,Y 0 N2,Y 0 N3,Y ⎪ d2Y ⎪ ⎪ N1,Y N1,X N2,Y N2,X N3,Y N3,X ⎪ ⎪ ⎪ ⎪ ⎪ d3X ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ d3Y

(6.15)

To obtain a ﬁnal expression for the B-matrix associated with the three-node triangle, we must diﬀerentiate the shape functions. Doing this we get Y32 2A X32 = 2A Y31 = 2A X31 =− 2A Y21 =− 2A X21 = 2A

N1,X = − N1,Y N2,X N2,Y N3,X N3,Y

(6.16)

The ﬁnal result is

⎡ ⎤ −Y32 0 Y31 0 −Y21 0 1 ⎣ 0 −X31 0 X21 ⎦ 0 X32 = 3×6 2A X32 −Y32 −X31 Y31 X21 −Y21 Be

(6.17)

It turns out that the B-matrix for the three-node triangle depends only on the nodal coordinates. In other words, the elements of the B-matrix contain simply constants. As a consequence, the ﬁnite element approximation for the strain ﬁeld is constant at each point within the element. The three-node triangle is often referred to as a constant-strain triangle. An important feature of the B-matrix is that each of its rows sum to zero. For example, adding up the ﬁrst row gives Y2 − Y3 + Y3 − Y1 + Y1 − Y2 = 0 The fact that the rows of the B-matrix sum to zero makes sense from the following physical argument. Suppose that all of the nodal displacements daX and daY were chosen to have a value of 1. This type of displacement is called a rigid-body translation. If an element moves as a rigid body, no strain should develop in the element. This condition is satisﬁed if each row of Be sums to zero.

Continuum Finite Elements

137 3 4 Y X 1

2

FIGURE 6.3 Arbitrary four-node quadrilateral.

6.2

Development of a four-node arbitrary quadrilateral

Let us now consider a slightly more complicated two-dimensional element, the four-node quadrilateral. A four-node quadrilateral is shown in Figure 6.3. Like the three-node triangle, the local node numbering is chosen such that local node 1 is located at any one of the corners. The remaining nodes are then numbered sequentially in a counterclockwise fashion. There are two dis placement components associated with each node, so the nodal displacements can be stored in an 8 × 1 vector as follows: ⎧ ⎫ ⎪ d1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d1Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d2X ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ d2Y (6.18) de = ⎪ d3X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d3Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d4X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ d4Y Let us begin by interpolating the trial function with a four-term polynomial as follows: α1 + α2 X + α3 Y + α4 XY u ˜X (6.19) = u ˜Y e β1 + β2 X + β3 Y + β4 XY To derive the shape functions for the element, we proceed like we did with the three-node triangle and enforce the following conditions at the nodes: u ˜i (X1 , Y1 ) = d1i u ˜i (X2 , Y2 ) = d2i u ˜i (X3 , Y3 ) = d3i u ˜i (X4 , Y4 ) = d4i

(6.20)

138

Finite Element Method

where i can denote either X or Y . Letting i = X and enforcing the four conditions stated in (6.20) yields ⎧ ⎫ ⎡ ⎤⎧ ⎫ 1 X1 Y1 X1 Y1 ⎪ ⎪ ⎪ d1X ⎪ ⎪ ⎪ ⎪ α1 ⎪ ⎨ ⎬ ⎢ ⎥ ⎨ α2 ⎬ d2X 1 X Y X Y 2 2 2 2 ⎥ =⎢ (6.21) ⎣ 1 X3 Y3 X3 Y3 ⎦ ⎪ α3 ⎪ d3X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ d4X 1 X4 Y4 X4 Y4 α4 Except for simple element geometries like squares and rectangles, the solution of equation (6.21) is very messy. For the moment, let us assume that it can be solved in principle to obtain the following relationship between the components of the trial function and the nodal displacements: u ˜X ˜e = u u ˜Y e ⎧ ⎫ d1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d1Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d2X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ N1 0 N2 0 N3 0 N 4 0 d2Y = d3X ⎪ 0 N1 0 N 2 0 N 3 0 N 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d3Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d4X ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ d4Y N d = 2 ×e8 8 ×e 1

(6.22)

where the shape functions, N1 to N4 , are functions of X and Y , i.e., Na = Na (X, Y ). In the next section we will illustrate the shortcomings of the ﬁnite element interpolation deﬁned in equation (6.19).

6.2.1

Compatibility issues

A mesh composed of two four-node quadrilateral elements is shown in Fig ure 6.4. The original shape of each element is illustrated in Figure 6.4(a), and the initial nodal coordinates are given in Table 6.1. The nodes are displaced according to the values reported in the table. Each element is identiﬁed by a number inside a rounded box. Local node 1 for each element is chosen to be the node occupying the lower left corner. The remaining local nodes are numbered counterclockwise. The global node numbers are indicated in bold. Following the procedure outlined in the previous section, let us now obtain the shape functions for each element. Focusing on element 1, equation (6.21) becomes ⎫ ⎡ ⎧ ⎤⎧ ⎫ 1 0 0 0 ⎪ ⎪ ⎪ ⎪ d1X ⎪ ⎪ α1 ⎪ ⎪ ⎬ ⎢ ⎨ ⎥ ⎨ α2 ⎬ d2X 1 1 0 0 ⎥ =⎢ (6.23) d3X ⎪ α3 ⎪ ⎪ ⎣1 2 1 2⎦⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎪ ⎩ d4X α4 1 1 1 1

Continuum Finite Elements

139 Y

Y

4

5

� � �1 �

� � �2 �

1 1

1 2 (a)

6

X

X 3 (b)

FIGURE 6.4 Compatibility issues with the four-node quadrilateral.

where the subscripts 1 to 4 refer to local node numbers. Note that in equation (6.23) above, we are inserting the local nodal coordinates; for example, X1 refers to the X coordinate of local node one1. Solving (6.23) for α1 to α4 yields α1 = d1X α2 = −d1X + d2X α3 = −d1X − 3d3X + 2d4X α4 = d1X − d2X + d3X − d4X After plugging in these values into the ﬁrst line of equation (6.19) and collecting

TABLE 6.1

Nodal coordinates and displacements. Node X Y daX daY 1 0 0 0.0 0.0 2 1 0 -0.1 0.1 3 2 0 0.2 0.2 4 1 1 -0.3 -0.1 5 2 1 -0.2 -0.1 6 3 1 -0.2 0.1

140

Finite Element Method

terms, we obtain the following four shape functions for element 1: N1 (X, Y ) = 1 − X − Y + XY N2 (X, Y ) = X − XY N3 (X, Y ) = −Y + XY N4 (X, Y ) = 2Y − XY

(6.24)

Following the same procedure, the shape functions for element 2 are found to be N1 (X, Y ) N2 (X, Y ) N3 (X, Y ) N4 (X, Y )

= 2 − X − 2Y + XY = −1 + X + Y − XY = −2Y + XY = 3Y − XY

(6.25)

Armed with the shape functions for both elements, we can now interpolate the displacement ﬁeld within each element as follows: ˜ 1 = N1 d1 u ˜ 2 = N2 d2 u where the subscripts 1 and 2 refer to elements 1 and 2, respectively. We are now able to investigate the motion of individual points that lie inside an undeformed element. Let x1 give the position of a point in the deformed conﬁguration that originally occupied position X1 inside the undeformed el ement 1. Similarly, let x2 identify the deformed position of point X2 inside element 2. The deformed positions are related to the original positions and the nodal displacements as follows: x1 = X1 + N1 d1

x2 = X2 + N2 d2

(6.26)

The deformed positions of points that were originally located inside the undeformed elements are shown in Figure 6.4(b). This ﬁgure deserves some further discussion. The lines deﬁned by global nodes 4–5, 5–6, 1–2, and 2–3 are originally horizontal and remain straight lines after deformation. On the other hand, lines 1–4, 2–5, and 3–6, which are originally straight but slanted, are no longer straight lines in the deformed conﬁguration. This causes a severe problem along the edge 2–5. Notice that even though the deformed mesh is connected at all the nodes, there is a gap that opens up between the two elements along edge 2–5. This is called an incompatibility. This incompatible deformation would give rise to a response that is too compliant, i.e., not stiﬀ enough due to the fact that the material points between nodes 2 and 5 become disconnected.

Continuum Finite Elements

141

Y

(-1,1)

4

3

(1,1) X

(-1,-1)

1

2

(1,-1)

FIGURE 6.5 Bi-unit square mapped into arbitrary quadrilateral.

It turns out that if all of the element edges in the original conﬁguration are either horizontal or vertical (no slanted edges), the deformation (no matter what the nodal displacements are) would end up being compatible. This is true for four-node quadrilaterals that are either square or rectangular. Un fortunately, complicated geometries are diﬃcult to mesh with squares and/or rectangles. It is therefore desirable to come up with a four-node element that, when connected to other four-node elements in a mesh, results in a compat ible displacement ﬁeld. Fortunately, this problem has been overcome by the development of a class of elements called isoparametric ﬁnite elements. This topic will be discussed in the next section.

6.2.2

The bi-unit square

A square element is shown in Figure 6.5. The square is centered at the origin and has sides of length two units. As in the last example, we will interpolate the displacement components over the element using a four-term polynomial according to equation (6.19). With the nodal coordinates deﬁned in Fig ure 6.5, the equation for the constants α1 to α4 is ⎧ ⎫ ⎡ ⎤⎧ ⎫ d1X ⎪ 1 −1 −1 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ α1 ⎪ ⎨ ⎬ ⎢ ⎥ ⎨ α2 ⎬ d2X 1 1 −1 −1 ⎢ ⎥ =⎣ d3X ⎪ α3 ⎪ 1 1 1 1⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ d4X α4 1 −1 1 −1

(6.27)

142

Finite Element Method

Solving equation (6.27) yields 1 (d1X + d2X + d3X + d4X ) 4 1 α2 = (−d1X + d2X + d3X − d4X ) 4 1 α3 = (−d1X + −d2X + d3X + d4X ) 4 1 α4 = (d1X − d2X + d3X − d4X ) 4 α1 =

Finally, plugging α1 , . . . , α4 back into the ﬁrst line of equation (6.19) and collecting terms gives 1 (1 − X)(1 − Y ) 4 1 N2 = (1 + X)(1 − Y ) 4 1 N3 = (1 + X)(1 + Y ) 4 1 N4 = (1 − X)(1 + Y ) 4 N1 =

(6.28)

The shape functions deﬁned in equation (6.28) above happen to be products of one-dimensional shape functions derived in Chapter 3. For this reason, they are often referred to as bi-linear shape functions. Let us now investigate the performance of the bi-linear shape functions. To begin, let pe be an 8 × 1 vector that stores the nodal coordinates of the deformed element. Let Pe be an 8×1 vector that stores the nodal coordinates of the undeformed bi-unit square. Now let Xe be a position vector deﬁning the location of a point inside the undeformed bi-unit square, and let xe deﬁne the location of a point inside the deformed element. The vectors xe and Xe are then related through the following equations: ˜e xe = Xe + u = Xe + Ne de = Ne Pe + Ne de = Ne (Pe + de ) = Ne pe Suppose that we prescribe the nodal displacements given in Table 6.2. The bi-unit square in this deformed conﬁguration is shown in Figure 6.5. The dots show the locations of points that were originally located inside the undeformed bi-unit square. Notice that the interpolation gives rise to a deformed element whose edges remain straight. This reveals that the shape functions for the bi-unit square can be used as interpolation functions to ﬁnd the interior points

Continuum Finite Elements

143 TABLE 6.2

Nodal displacements for the bi-unit square. Node daX daY 1 2 3 4

2.0 1.9 2.3 1.5

2.6 3.1 2.9 2.5

of an arbitrary quadrilateral, given the quadrilateral’s nodal point locations. The interpolation yields a deformed element whose edges are all straight lines. Another way of saying this is that the bi-linear shape functions can be used to “map” the bi-unit square into an arbitrary quadrilateral. This mapping is commonly written as x=

4

Na xa

a=1

y=

4

Na y a ⇒

a=1

xe = Ne pe

6.2.3

The parent domain

In order to complete the development of the arbitrary four-node quadrilateral, it is necessary to deﬁne the bi-unit square in a ξ−η coordinate system as shown in Figure 6.6. The bi-unit square in the ξ−η coordinate system is called the parent domain. The shape functions for the bi-unit square can be written in terms of the dimensionless variables, ξ and η, as follows: 1 (1 − ξ)(1 − η) 4 1 N2 (ξ, η) = (1 + ξ)(1 − η) 4 1 N3 (ξ, η) = (1 + ξ)(1 + η) 4 1 N4 (ξ, η) = (1 − ξ)(1 + η) 4 N1 (ξ, η) =

(6.29)

A surface plot of the bi-linear shape function Na (ξ, η) is shown in Figure 6.7. Note that the surface, which in general is not planar, is linear along the four element edges. It takes on the value of 1 at local node a, and it is zero at the remaining nodes.

144

Finite Element Method

η

Y X x

ξ

X

FIGURE 6.6 Bi-unit square in ξ−η coordinate system.

From our previous discussion of the bi-unit square, given the nodal coor dinates, we can now construct arbitrary quadrilaterals with straight edges in the X − Y coordinate system. The deformed element in the X − Y coordinate system is called the physical domain. The mapping of the bi-unit square from the parent domain into the physical domain can be written as Xe = Ne (ξ, η)Pe where Ne is a 2 × 8 matrix containing the shape functions (6.29), and Pe is an 8 × 1 vector that contains the nodal coordinates of the element in the physical domain. Let us now assume that the undeformed element in the physical domain undergoes a deformation deﬁned by the nodal displacements. A point xe , inside the deformed element that once occupied position Xe in the physical domain, can be obtained from the following mapping: xe = Ne (ξ, η)pe where pe is an 8×1 vector that contains the nodal coordinates of the deformed element. The use of the bi-linear shape functions ensures that the edges of the deformed element remain straight lines. ˜ e is deﬁned as the deformed position xe minus the un The displacement u deformed position Xe , i.e., ˜ e = xe − Xe u = Ne (pe − Pe ) = Ne de

(6.30)

Continuum Finite Elements

145

Na (ξ, η)

η

ξ FIGURE 6.7 Surface plot of bi-linear shape function evaluated at node a.

The last line in (6.30) reveals the remarkable fact that the same bi-linear shape functions that are used to map the bi-unit square in the parent domain into an arbitrary quadrilateral in the physical domain can also be used to interpolate the displacement ﬁeld throughout an arbitrary quadrilateral. Because the bi-linear shape functions map straight lines into straight lines, the resulting deformation of an assemblage of four-node quadrilaterals will be compatible. To emphasize this concept, let us revisit the two-element mesh shown in Figure 6.4. If the bi-linear shape functions (6.29) were used instead of the shape functions (6.24) and (6.25), the resulting deformation would look like that depicted in Figure 6.8. Notice that all of the edges of the deformed elements are straight, and the resulting displacement ﬁeld is compatible, i.e., there is no longer a gap along edge 2–5. We end this section by stating that the arbitrary four-node quadrilateral belongs to a class of elements called isoparametric ﬁnite elements. An element is said to be isoparametric if the shape functions used to interpolate the displacement ﬁeld are the same (iso) as those used for mapping (parametric) the element from the parent domain to the physical domain.

6.2.4

The B-matrix

The goal of this section is to ﬁnd a relationship between the in-plane strain components stored in the vector ˜e and the vector de containing the nodal dis placements of the four-node quadrilateral. Just as we did for the three-node triangle, the strain components are obtained by diﬀerentiating the displace

146

Finite Element Method Y

Y

4

5

� � �1 �

� � �2 �

1

6

2

X

X 1

2

3

(a)

(b)

FIGURE 6.8 Compatible deformation of arbitrary quadrilaterals.

ment components as follows: ⎧ ⎫ ⎧ ⎫ u ˜X,X ⎨ ⎬ ⎨ ˜XX ⎬ ˜Y Y u ˜Y,Y = ⎩ ⎭ ⎩ ⎭ γ˜XY e u ˜X,Y + u ˜Y,X e ⎧ ⎪ ⎪ ⎨ =

4

⎪ ⎪ ⎩ 4

a=1

a=1 Na,X daX 4 a=1 Na,Y daY

(Na,Y daX + Na,X daY )

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

(6.31)

We can write equation (6.31) in matrix notation as ˜e = Be de where Be is the 3 × 8 B-matrix. Note that the B-matrix contains derivatives of the shape functions with respect to X and Y , i.e., ⎡

⎤ N1,X 0 N2,X 0 N3,X 0 N4,X 0 0 N2,Y 0 N3,Y 0 N4,Y ⎦ Be = ⎣ 0 N1,Y N1,Y N1,X N2,Y N2,X N3,Y N3,X N4,Y N4,X

(6.32)

Because the shape functions N1 to N4 are functions of the natural coordinates ξ and η, care must be exercised in taking the derivatives of these functions with respect to X and Y , as discussed the next section.

Continuum Finite Elements

6.2.5

147

Derivatives of the shape functions

When asked to diﬀerentiate the bi-linear shape functions (6.29) with respect to X and Y , the natural thing to do is to recognize that X and Y are functions of ξ and η and employ the chain rule as follows: Na,X = Na,ξ ξ,X + Na,η η,X Na,Y = Na,ξ ξ,Y + Na,η η,Y While taking the derivatives of the shape functions with respect to ξ and η poses no diﬃculty, evaluating derivatives ξ,X , ξ,Y , η,X , and η,Y can be problematic. The reason is that, even though the forward mapping from the parent domain to the physical domain is explicitly deﬁned by equation (6.30), the reverse mapping from the physical domain to the parent domain, in general, is not known in closed form. To alleviate this problem, one can instead use implicit diﬀerentiation as follows: Na,ξ = Na,X X,ξ + Na,Y Y,ξ Na,η = Na,X X,η + Na,Y Y,η

(6.33)

In equation (6.33) above, the derivatives of the spatial variables X and Y with respect to the natural coordinates can be readily evaluated given the mapping (6.30). Treating Na,X and Na,Y as unknowns and writing (6.33) in matrix form yields Na,ξ X,ξ Y,ξ Na,X = (6.34) Na,η X,η Y,η Na,Y The matrix on the right-hand side of equation (6.34) is called the Jacobian matrix and is commonly referred to as J. The physical meaning of the com ponents of J will be discussed in the next section when we discuss how an inﬁnitesimal area in the physical domain is related to the corresponding in ﬁnitesimal area in the parent domain. By inverting equation (6.34) we obtain Na,ξ Na,X = J−1 (6.35) Na,Y Na,η So in other words, the derivatives of the shape functions with respect to X and Y at some point inside the element in the physical domain are obtained by ﬁrst computing the Jacobian matrix, J, at that point, then computing its inverse, J−1 , and ﬁnally multipling J−1 by the vector containing the derivatives of the shape functions with respect to ξ and η.

6.2.6

Area change

In this section, we investigate the relationship between an inﬁnitesimal area in the physical domain and the corresponding area in the parent domain. The

148

Finite Element Method

X(2)

Y

X(1)

η X

ξ(2)

ξ

ξ(1)

FIGURE 6.9 Area change between inﬁnitesimal elements in parent and physical domains. relationship will be needed later in the formulation of the element matrices for isoparametric elements. To begin, consider an inﬁnitesimal area in the parent domain depicted in Figure 6.9. Points inside this inﬁnitesimal area in the parent domain map to points inside an inﬁnitesimal area in the physical domain through the mapping X = X(ξ, η) Y = Y (ξ, η)

(6.36)

Recalling our study of the Taylor series expansion in Chapter 2, we now expand each of the two functions in equation (6.36) as a linear Taylor series about the point (ξ0 , η0 ), which lies inside the inﬁnitesimal area in the parent domain as follows: ∂X ∂X (ξ − ξ0 ) + (η − η0 ) X ≈ X(ξ0 , η0 ) + ∂ξ ∂η ∂Y ∂Y Y ≈ Y (ξ0 , η0 ) + (ξ − ξ0 ) + (η − η0 ) (6.37) ∂ξ ∂η where the partial derivatives are understood to be evaluated at point (ξ0 , η0 ). If we restrict our attention to points X and Y that are inﬁnitesimally close to the points X(ξ0 , η0 ) and Y (ξ0 , η0 ), then the linear Taylor series representations of the functions X and Y are exact. Letting X − X(ξ0 , η0 ) = dX, Y − Y (ξ0 , η0 ) = dY , ξ − ξ0 = dξ, and η − η0 = dη, we can write equation (6.37) in matrix notation as follows: dξ dX X,ξ X,η (6.38) = dη dY Y,ξ Y,η

Continuum Finite Elements

149

It is now instructive to view dX and dY as components of an inﬁnitesi mal vector dX in the physical domain, and dξ and dη as components of an inﬁnitesimal vector dξ in the parent domain. Equation (6.38) can then be written as (6.39) dX = JT dξ where J is the Jacobian matrix, i.e., X,ξ Y,ξ J= X,η Y,η

(6.40)

One can think of equation (6.39) as the matrix equation that maps an in ﬁnitesimal line segment dξ in the parent domain into the inﬁnitesimal line segment dX in the physical domain. Now let the bottom edge of the area in the parent domain be deﬁned by the vector dξ (1) , and let the left edge of the area be deﬁned by the vector dξ (2) . The components of these two vectors in the ξ−η coordinate system are given as dξ 0 (1) (2) ; dξ = (6.41) dξ = 0 dη The magnitude of the cross product between the two vectors dξ (1) and dξ (2) yields the area, dA, of the inﬁnitesimal element in the parent domain, i.e., dA = dξ (1) × dξ (2) = dξdη The vectors dξ (1) and dξ (2) map into the vectors dX(1) and dX(2) respectively. Using the relation (6.39), we can write the components of these two vectors in the physical domain as X,ξ dξ X,η dη (1) (2) dX = ; dX = (6.42) Y,ξ dξ Y,η dη Finally, the inﬁnitesimal area, da, in the physical domain is equal to the magnitude of the cross product dX(1) × dX(2) ; hence, e2 e3

e1 da = dX(1) × dX(2) = X,ξ dξ Y,ξ dξ 0

X,η dη Y,η dη 0 = (X,ξ dξ)(Y,η dη) − (Y,ξ dξ)(X,η dη) = det J dξ dη = j dξ dη

(6.43)

Here in equation (6.43) above, j is the determinant of the Jacobian matrix, and e1 , e2 , and e3 are Cartesian base vectors. We now see that da = j dA

(6.44)

150

Finite Element Method

4 3

1 2 FIGURE 6.10 Four-node tetrahedron.

As a ﬁnal remark, the interested reader should make the connection be tween the Jacobian matrix and the deformation gradient tensor, F, deﬁned in Chapter 2. If one views the parent domain as the original conﬁguration and the physical domain as the deformed conﬁguration, then JT deﬁnes the de formation of inﬁnitesimal line segments originally lying in the parent domain.

6.3

Four-node tetrahedron

A close cousin to the three-node triangle is the four-node tetrahedron. The element is the simplest among the three-dimensional continuum elements. As shown in Figure 6.10, it has four nodes and four faces. We interpolate the ﬁeld variable throughout the volume of the element using interpolants that are linear in X, Y , and Z. Hence, for solid mechanics problems, we assume ⎫ ⎧ ⎫ ⎧ ˜X ⎬ ⎨ α1 + α2 X + α3 Y + α4 Z ⎬ ⎨u u ˜Y = β1 + β2 X + β3 Y + β4 Z ⎭ ⎩ ⎭ ⎩ u ˜Z e γ1 + γ2 X + γ3 Y + γ4 Z

(6.45)

The 12 × 1 vector containing the nodal displacements can be written as

Continuum Finite Elements

151 ⎧ ⎫ d1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d1Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎪ ⎪ 1Z ⎪ ⎪ ⎪ ⎪ d2X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d2Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ d2Z de = 12 × 1 d3X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d3Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d ⎪ ⎪ 3Z ⎪ ⎪ ⎪ ⎪ d4X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d4Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ d4Z

(6.46)

The shape functions for the element are derived by insisting that the trial function be equal to the nodal displacements when evaluated at the nodal points, i.e., u ˜X (X1 , Y1 , Z1 ) = d1X u ˜X (X2 , Y2 , Z2 ) = d2X u ˜X (X3 , Y3 , Z3 ) = d3X u ˜X (X4 , Y4 , Z4 ) = d4X

(6.47)

Inserting the constraints (6.47) into the ﬁrst line of (6.45) gives the following four equations in four unknowns: ⎡

1 X1 ⎢ 1 X2 ⎢ ⎣ 1 X3 1 X4

Y1 Y2 Y3 Y4

⎫ ⎤⎧ ⎫ ⎧ Z1 ⎪ α1 ⎪ d1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ ⎨ Z2 ⎥ ⎥ α2 = d2X ⎦ Z3 ⎪ α3 ⎪ ⎪ d3X ⎪ ⎪ ⎪ ⎭ ⎪ ⎭ ⎩ ⎪ ⎩ Z4 α4 d4X

(6.48)

It turns out that the algebra required to solve the system of equations (6.48) is rather lengthy. It is much easier to derive the shape functions for this element using volume coordinates. Consider a point, P , that lies at some location, (X, Y, Z), inside the volume of the tetrahedron. Given this location, we can deﬁne four volumes. The ﬁrst volume, V1 , is the volume of the tetrahedron having one of its vertices at point P , and the other three lying on the face deﬁned by the local nodes 2, 3, and 4. The volume coordinate, V2 , is deﬁned by the point P and local nodes 1, 3, and 4, and so on. The vertices of each volume coordinate are provided in Table 6.3. We now let a be a vector pointing from local node 1 to local node 2. Sim ilarly, let b be another vector pointing from node 1 to 3, and c be a vector pointing from node 1 to 4. These three vectors can be written in terms of

152

Finite Element Method TABLE 6.3

Volume coordinates and vertices. Volume Vertex 1 Vertex 2 Vertex 3 V1 V2 V3 V4

P P P P

2 1 1 1

Vertex 4

3 3 2 2

4 4 4 3

Cartesian base vectors as follows: a = (X2 − X1 )eX + (Y2 − Y1 )eY + (Z2 − Z1 )eZ b = (X3 − X1 )eX + (Y3 − Y1 )eY + (Z3 − Z1 )eZ c = (X4 − X1 )eX + (Y4 − Y1 )eY + (Z4 − Z1 )eZ

(6.49)

Having deﬁned the vectors a, b, and c, the volume, V , of the tetrahedron can be calculated by employing the scalar triple-product operation described in Chapter 2, Section 2.3. Doing this gives 1 [a × b] · c 6 1 = {(aX bZ − aZ bY ) cX − (aX bZ − aZ bX ) cY 6

+ (aX bY − aY bX ) cZ }

V =

(6.50)

The volume coordinates V1 to V4 can also be expressed in terms of the nodal coordinates and the spatial variables using the scalar triple-product operation carried out in (6.50) above. The expressions can be obtained by deﬁning the components of a, b, and c according to the deﬁnitions listed in Table 6.4, and then carrying out the operation (6.50). Note that in the table we have used the notation X21 = X2 − X1 , Y21 = Y2 − Y1 , etc. The shape functions for the four-node tetrahedron can be written compactly in terms of the volume coordinates as follows: Na =

Va V

(6.51)

TABLE 6.4

Deﬁnition of the components of a, b, and c for the volume coordinate calculations. Volume aX aY aZ bX bY bZ cX cY cZ V1 V2 V3 V4 V

X32 X31 X21 X21 X21

Y32 Y31 Y21 Y21 Y21

Z32 Z31 Z21 Z21 Z21

X42 X41 X41 X31 X31

Y42 Y41 Y41 Y31 Y31

Z42 Z41 Z41 Z31 Z31

X X X X

− X2 − X1 − X1 − X1 X41

Y Y Y Y

− Y2 − Y1 − Y1 − Y1 Y41

Z − Z2 Z − Z1 Z − Z1 Z − Z1 Z41

Continuum Finite Elements

153

where a is the node number, which can take on any value between 1 and 4. Note that the volume coordinates, and hence shape functions, are linear in X, Y , and Z. The shape functions also possess the Kronecker delta property, i.e., NI (XJ ) = δIJ . For example, as the point P moves inside the volume, each of the four volumes changes continuously. When the point P approaches node 1, the volume coordinate V4 approaches zero. When the point P moves to nodes 2, 3, or 4, the volume V4 also goes to zero. So in this case we have V4 (X1 , Y1 , Z1 )/V = 0, V4 (X2 , Y2 , Z2 )/V = 0, V4 (X3 , Y3 , Z3 )/V = 0, and V4 (X4 , Y4 , Z4 )/V = 1

6.3.1

The B-matrix

The B-matrix for the four-node tetrahedron can be derived by recalling the deﬁnition of the small-strain components in terms of the displacement com ponents. The 6 × 1 strain vector at a point within the element can be written as ⎫ ⎫ ⎧ ⎫ ⎧ ⎧ u ˜X,X Na,X daX ⎪ ⎪ ⎪ ⎪ ⎪ ˜XX ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ˜ u ˜ N d ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Y Y Y,Y a,Y aY ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎨ ⎬ ⎨ ⎨ ˜ZZ u ˜Z,Z Na,Z daZ ˜e = = = ˜Y,Z + u ˜Z,Y ⎪ ⎪ γ˜Y Z ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Na,Z daY + Na,Y daZ ⎪ ⎪ ⎪ ⎪u ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪u γ ˜ ˜ + u ˜ N d + N d ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ XZ X,Z Z,X a,Z aX a,X aZ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎭ ⎩ ⎩ γ˜XY e u ˜X,Y + u ˜Y,X e Na,Y daX + Na,X daY

(6.52)

˜ e = Be de yields the following 6 × 12 Writing equation (6.52) in the form σ matrix: ⎡

⎤

N1,X 0 0 N2,X 0 0 N3,X 0 0 N4,X 0 0 ⎢ 0 N1,Y 0 0 N2,Y 0 0 N3,Y 0 0 N4,Y 0 ⎥ ⎢ ⎥ 0 0 N2,Z 0 0 N3,Z 0 0 N4,Z ⎥ 0 N1,Z ⎢ 0 Be = ⎢ ⎥ ⎢ 0 N1,Z N1,Y 0 N2,Z N2,Y 0 N3,Z N3,Y 0 N4,Z N4,Y ⎥ ⎣ N1,Z 0 N1,X N2,Z 0 N2,X N3,Z 0 N3,X N4,Z 0 N4,X ⎦ N1,Y N1,X 0 N2,Y N2,X 0 N3,Y N3,X 0 N4,Y N4,X 0

The derivatives of the shape functions with respect to X, Y , and Z are given in Table 6.5. Notice that the derivatives of the shape functions are constant, and therefore the four-node tetrahedron is a constant-strain element. While in principle, complicated geometries are easy to mesh with four-node tetrahedral elements, in general, the volume must be meshed with a large number of elements to obtain good accuracy.

154

Finite Element Method

TABLE 6.5

Derivatives of the shape functions for the four-node tetrahedron. Na N1 N2 N3

6.4

∂/∂X

∂/∂Y

∂/∂Z

1 (X32 Z42 − Z32 Y42 ) 6V 1 (X32 Z42 − Z32 Y42 ) 6V 1 (X32 Z42 − Z32 Y42 ) 6V

1 (Z32 X42 − Z42 X32 ) 6V 1 (X32 Z42 − Z32 Y42 ) 6V 1 (X32 Z42 − Z32 Y42 ) 6V

1 (X32 Y42 − Y32 X42 ) 6V 1 (X32 Z42 − Z32 Y42 ) 6V 1 (X32 Z42 − Z32 Y42 ) 6V

Eight-node brick

The arbitrary isoparametric hexahedron, otherwise known as simply the eightnode brick, is a modern-day workhorse for both linear and nonlinear contin uum problems. Even though the element is considered low order, it gives good accuracy and can be used to mesh arbitrary volumes with relative ease. To begin the isoparametric formulation, we consider a tri-unit cube in a ξ − η − ζ parent domain as shown in Figure 6.11 The mapping between the parent domain and the physical domain is chosen to have the following form: X = α1 + α2 ξ + α3 η + α4 ζ + α5 ξη + α6 ξζ + α7 ηζ + α8 ξηζ Y = β1 + β2 ξ + β3 η + β4 ζ + β5 ξη + β6 ξζ + β7 ηζ + β8 ξηζ Z = γ1 + γ2 ξ + γ3 η + γ4 ζ + γ5 ξη + γ6 ξζ + γ7 ηζ + γ8 ξηζ

(6.53)

The shape functions for the eight-node brick can be derived by insisting that the values of X, Y , and Z, when evaluated at the nodes, be equal to Xa , Ya , and Za , respectively, where a denotes the local node number and can take on any value between 1 and 8. Local node 1 is chosen to occupy position (ξ = −1, η = −1, ζ = −1), and the remaining nodes are numbered according to the convection depicted in Figure 6.11. Focusing on the X components only, the constants α1 to α8 can be obtained by solving the following system of equations: ⎧ ⎫ ⎡ ⎤⎧ ⎫ X1 ⎪ 1 −1 −1 −1 1 1 1 −1 ⎪ α1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ X α 1 1 −1 −1 −1 −1 1 1 ⎪ ⎪ 2⎪ 2⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ X3 ⎪ ⎢ 1 1 1 −1 1 −1 −1 −1 ⎥ ⎪ α ⎪ ⎪ ⎪ ⎪ 3⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎨ ⎬ ⎢ ⎨ ⎥ α4 ⎬ X4 1 −1 1 −1 −1 1 −1 1 ⎢ ⎥ =⎢ (6.54) ⎥ α5 ⎪ X5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ 1 −1 −1 1 1 −1 −1 1 ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ 1 1 −1 1 −1 1 −1 −1 ⎥ ⎪ X6 ⎪ α6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎣ ⎦ X α 1 1 1 1 1 1 1 1 ⎪ ⎪ ⎪ 7⎪ 7⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ ⎪ X8 α8 1 −1 1 1 −1 −1 1 −1 If we now solve equation (6.54) for α1 to α8 and insert the resulting constants

Continuum Finite Elements

155

ζ 8

η

5 4

7

6 1 3

Y

2

ξ

Z

X FIGURE 6.11 Eight-node brick in parent and physical domains.

into the ﬁrst line of equation (6.53) and collect terms, we get the following eight shape functions: N1 (ξ, η, ζ) = 1/8(1 − ξ)(1 − η)(1 − ζ) N2 (ξ, η, ζ) = 1/8(1 + ξ)(1 − η)(1 − ζ) N3 (ξ, η, ζ) = 1/8(1 + ξ)(1 + η)(1 − ζ) N4 (ξ, η, ζ) = 1/8(1 − ξ)(1 + η)(1 − ζ) N5 (ξ, η, ζ) = 1/8(1 − ξ)(1 − η)(1 + ζ) N6 (ξ, η, ζ) = 1/8(1 + ξ)(1 − η)(1 + ζ) N7 (ξ, η, ζ) = 1/8(1 + ξ)(1 + η)(1 + ζ) N8 (ξ, η, ζ) = 1/8(1 − ξ)(1 + η)(1 + ζ)

(6.55)

The shape functions deﬁned in (6.55) above are often expressed succinctly in one line as 1 (6.56) Na = (1 ± ξ)(1 ± η)(1 ± ζ) 8 where the ± sign convention follows (6.55). The shape functions (6.55) are often called the tri-linear shape functions, because they are products of the one-dimensional shape functions derived in Chapter 3. Having derived the shape functions, the mapping between the parent and

156

Finite Element Method

physical domains can be written as X = Na (ξ, η, ζ)Xa Y = Na (ξ, η, ζ)Ya Z = Na (ξ, η, ζ)Za

(6.57)

It turns out that equation (6.57), involving the tri-linear shape functions, maps the tri-unit cube in the parent domain into an arbitrary hexahedron in the physical domain. The resulting arbitrary hexahedron in the physical domain has straight edges. The resulting element faces in the physical domain are in general not planar surfaces. Let us now use the same tri-linear shape functions deﬁned in (6.55) above to interpolate the displacement ﬁeld within an element. The displacement interpolation can be written as ⎫ ⎧ ⎫ ⎧ ˜X ⎬ ⎨ Na (ξ, η, ζ)daX ⎬ ⎨u u ˜Y = Na (ξ, η, ζ)daY (6.58) ⎭ ⎩ ⎭ ⎩ u ˜Z e Na (ξ, η, ζ)daZ It turns out that the displacement interpolation (6.58) gives rise to compatible deformation between adjacent elements.

6.4.1

The B-matrix

Recall that for three-dimensional applications, the small-strain components can be stored in a 6 × 1 vector as follows: ⎫ ⎧ ⎫ ⎧ uX,X ⎪ ⎪ ⎪ ⎪ ⎪ XX ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ uY,Y ⎪ ⎪ ⎪ ⎪ ⎪ Y Y ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ ⎨ ZZ uZ,Z = (6.59) ⎪ ⎪ γY Z ⎪ ⎪ ⎪ ⎪ uY,Z + uZ,Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ uX,Z + uZ,X ⎪ ⎪ γXZ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎩ ⎭ ⎩ γXY uX,Y + uY,X To derive the B-matrix for the eight-node brick element, we seek a matrix equation of the form ˜e B d = 6 × e24 24 ×e 1 (6.60) 6×1 where Be is the element B-matrix, and de is the vector containing the nodal displacements, i.e., ⎧ ⎫ ⎪ ⎪ d1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ d1Y ⎪ ⎪ ⎨ ⎬ (6.61) de = d1Z ⎪ ⎪ .. ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ d8Z

Continuum Finite Elements

157

The B-matrix provides the connection between the approximate strain ﬁeld within the element and the nodal displacements. Using the strain-displacement relationship (6.59), the 6 × 24 B-matrix can be expressed as B B B (6.62) Be = 6 ×e13 6 ×e23 . . . 6 ×e83 ⎡

where

Bea

⎤ 0 Na,X 0 ⎢ 0 Na,Y 0 ⎥ ⎢ ⎥ ⎢ 0 0 Na,Z ⎥ ⎥ =⎢ ⎢ 0 Na,Z Na,Y ⎥ ⎢ ⎥ ⎣ Na,Z 0 Na,X ⎦ Na,Y Na,X 0

(6.63)

is the 6 × 3 submatrix associated with node a. The derivatives of the shape functions with respect to the global X, Y , and Z coordinates are obtained by using the technique of implicit diﬀerentiation explained in Section 6.2.5. Doing this yields Na,ξ = Na,X X,ξ + Na,Y Y,ξ + Na,Z Z,ξ Na,η = Na,X X,η + Na,Y Y,η + Na,Z Z,η Na,ζ = Na,X X,ζ + Na,Y Y,ζ + Na,Z Z,ζ or in matrix notation ⎧ ⎫ ⎡ ⎫ ⎤⎧ X,ξ Y,ξ Z,ξ ⎨ Na,X ⎬ ⎨ Na,ξ ⎬ Na,η = ⎣ X,η Y,η Z,η ⎦ Na,Y ⎩ ⎭ ⎭ ⎩ Na,ζ X,ζ Y,ζ Z,ζ Na,Z

(6.64)

(6.65)

The 3 × 3 matrix deﬁned in (6.65) above is the Jacobian matrix. The deriva tives of the shape functions with respect to the global coordinates are obtained by obtaining the inverse relationship as follows: ⎫ ⎧ ⎫ ⎧ ⎨ Na,ξ ⎬ ⎨ Na,X ⎬ Na,Y = J−1 Na,η (6.66) ⎭ ⎩ ⎭ ⎩ Na,Z Na,ζ

6.4.2

Volume change

To end this section, we now illustrate how an inﬁnitesimal volume element in the physical domain is related to the corresponding inﬁnitesimal volume element in the parent domain. This relationship will be needed later for the evaluation of the element matrix, which involves evaluating a volume integral over the parent domain. We begin by writing down the mapping between the parent domain and the physical domain as follows: X = f (ξ, η, ζ)

(6.67)

158

Finite Element Method

ζ

dv

η Z

dX(1)

Y dV X ξ dξ

FIGURE 6.12 Inﬁnitesimal volume element in parent and physical domains. Equation (6.67) above simply states that the position vector, X, of a point in the physical domain, which once occupied position (ξ, η, ζ) in the parent domain, is a vector-valued function of the natural coordinates. From our work in Chapter 2, Section 2.6, we recall that given the components dξ, dη, and dζ of an inﬁnitesimal vector in the parent domain, the corresponding vector, dX, in the physical domain can be obtained as dX =

∂X ∂X ∂X dξ + dη + dζ ∂ξ ∂η ∂ζ

(6.68)

Let us now deﬁne dξ, dη, and dζ to be inﬁnitesimal vectors in the par ent domain that are mutually perpendicular and are parallel to the ξ, η, and ζ axes, respectively. To avoid clutter, only the vector dξ is labeled in Fig ure 6.12.The ﬁgure shows an inﬁnitesimal volume element, dV , in the parent domain whose edges are deﬁned by the vectors dξ, dη, and dζ. The cor responding inﬁnitesimal volume in the physical domain, dv, is also shown in the ﬁgure. Its edges are deﬁned by the three vectors dX(1) , dX(2) , and dX(3) . Only dX(1) is labeled in the ﬁgure. The three inﬁnitesimal vectors in the physical domain can be written in terms of their components and the Cartesian base vectors eX , eY , and eZ as ∂X dξ ∂ξ ∂X ∂Y ∂Z = dξ eX + dξ eY + dξ eZ ∂ξ ∂ξ ∂ξ

dX(1) =

∂X dη ∂η ∂X ∂Y ∂Z = dη e1 + dη e2 + dη e3 , ∂η ∂η ∂η

dX(2) =

Continuum Finite Elements

159

∂X dζ ∂ζ ∂X ∂Y ∂Z = dζ e1 + dζ e2 + dζ e3 ∂ζ ∂ζ ∂ζ

dX(3) =

(6.69)

From our study of Chapter 2, we recall that the volume, dv, deﬁned by the three edges, dX(1) , dX(2) , and dX(3) , can be evaluated by carrying out the following scalar triple product: (6.70) dv = dX(1) × dX(2) · dX(3) The volume, dV , of the element in the parent domain can be evaluated in the same manner, i.e., dV = {dξ × dη} · dζ = dξdηdζ

(6.71)

Finally, the scalar triple product (6.70) above (and hence dv) is equal to the following determinant: X,ξ dξ Y,ξ dξ Z,ξ dξ dv = X,η dη Y,η dη Z,η dη = j dξdηdζ (6.72) X dζ Y dζ Z dζ ,ζ

,ζ

,ζ

After expanding the determinant (6.72) and observing the result (6.71), we see that the two volumes dv and dV are related through the determinant, j, of the Jacobian matrix deﬁned in the previous section. In other words, dv = j dV

6.5

(6.73)

Element matrices and vectors

In the previous sections, we focused on deriving the shape functions and Bmatrices for a variety of two- and three-dimensional continuum elements. The process of going through these derivations is needed to gain a sound under standing of continuum element behavior. In this section we go one step further and illustrate the process of obtaining the element matrix and element vector for continuum elements. For illustration purposes, we will focus on the arbi trary four-node quadrilateral and assume the condition of either plane strain or plane stress. From our discussion in Chapter 4, the weak form of the equilibrium equation can be written over a single element as follows: σij wi,j dΩ = bi wi dΩ + ti wi dΓ (6.74) Ωe

Ωe

Γe

160

Finite Element Method

where wi are the components of the test function, σij are the Cauchy stress components, bi are the components of the body force vector, and ti are the traction components. Here in equation (6.74), Ωe refers to the interior of the element, and Γe is the surface on which the traction vector is speciﬁed. Recall that the components of the test function are chosen to be suﬃciently smooth inside the domain of interest, and zero at points on the boundary where the displacement vector is speciﬁed. For example, if the displacement component u1 is speciﬁed at some point located on the boundary, then the corresponding component of the test function, w1 , would also be zero at that point. For two-dimensional problems, it is suﬃcient to consider a family of test functions that varies only with the spatial coordinates X1 and X2 . In the following derivations, we will assume that wi = wi (X1 , X2 ), and hence, all derivatives of the test function with respect to X3 are zero. Having made the above restrictions on the test function, let us now expand the integrand on the left-hand side of equation (6.74) above. Doing this yields σij wi,j = σ11 w1,1 + σ12 w1,2 +σ21 w2,1 + σ22 w2,2

(6.75)

Next, noting that the Cauchy stress tensor is symmetric, we can combine terms on the right-hand side of equation (6.75) as follows: σij wi,j = σ11 w1,1 + σ12 (w1,2 + w2,1 ) + σ22 w2,2 = δT σ

(6.76)

where, for the sake of convenience, we have arranged the in-plane stress com ponents and the derivatives of the test function in the following 3 × 1 vectors: ⎫ ⎧ ⎫ ⎧ w1,1 ⎨ ⎬ ⎨ σ11 ⎬ w2,2 (6.77) σ = σ22 ; δ = ⎭ ⎩ ⎭ ⎩ σ12 w1,2 + w2,1 Having made these deﬁnitions, we can now write the principle of virtual work in matrix notation as δT σ wT b w T t∗ dΩ = dΩ + dΓ (6.78) 1×3 3×1 1×2 2×1 1×2 2×1 Ω Ω Γ e

e

e

where w is a 2 × 1 vector containing the components of the test function, and b and t∗ are 2 × 1 vectors containing the components of the body force vector and speciﬁed traction vector, respectively. Employing the Galerkin procedure introduced in Chapter 3, we now inter polate the test function, w , over the domain of the element using the same ﬁnite element shape functions that are used to interpolate the displacement ˜ e , then be ﬁeld. The ﬁnite element approximation for the test function, w comes ˜ e = Ne we (6.79) w

Continuum Finite Elements

161

where, for the case of the four-node arbitrary quadrilateral, Ne is the 2 × 8 matrix containing the shape functions deﬁned in equation (6.29), and we is an 8 × 1 vector containing the values of the test function at the nodes. The ﬁnite element approximation for, δ˜e , the vector containing the deriva tives of the test function, can be written as ⎧ ⎨

⎫ Na,1 wa1 ⎬ Na,2 wa2 δ˜e = ⎩ ⎭ Na,2 wa1 + Na,1 wa2

(6.80)

where the summation convention is implied (a = 1 to 4 for the case of the fournode quadrilateral). Here wai refers to the ith component of the test function evaluated at local node a. Writing equation (6.80) in matrix notation gives

⎡

N1,X 0 N2,X 0 N3,X 0 N4,X δ˜e = ⎣ 0 N1,Y 0 N2,Y 0 N3,Y 0 N1,Y N1,X N2,Y N2,X N3,Y N3,X N4,Y

⎧ ⎫ w1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ w ⎪ ⎪ 1Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎤⎪ w ⎪ ⎪ 2X ⎪ ⎪ 0 ⎨ ⎬ w 2Y ⎦ N4,Y ⎪ w3X ⎪ ⎪ N4,X ⎪ ⎪ ⎪ ⎪ w3Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ w ⎪ ⎪ 4X ⎪ ⎪ ⎩ ⎭ w4Y

= Be we

(6.81)

where the subscript X refers to the ﬁrst component (i = 1) and Y refers to the second component (i = 2). Here in the last line of equation (6.81), Be is the 3 × 8 B-matrix deﬁned in equation (6.32) containing the derivatives of the shape functions with respect to X and Y . Let us now substitute the ﬁnite element approximations (6.79) and (6.81) ˜ T = weT NTe and δ˜Te = into the weak form (6.78). Using the identities w T T we Be gives nel e=1

weT

Ωe

nel BTe σ e dΩ = weT

NTe b dΩ +

Ωe

e=1

Γe

NTe t∗ dΓ

(6.82)

Expressing the vector we in terms of the Boolean localization matrix, Le , yields wT

=w

T

nel e=1

LTe

Ωe

nel

LTe

Ωe

e=1

NeT b dΩ

BTe σ e dΩ

+ Γe

NTe t∗

dΓ

(6.83)

162

Finite Element Method

After invoking the arbitrariness of the global test function vector, w, we ﬁnally obtain nel T T Le Be σ e dΩ =

nel

LTe

Ωe

e=1

Ωe

e=1

NTe b dΩ

+ Γe

NTe t∗

dΓ

(6.84)

We now recognize that the integral on the left-hand side of equation (6.84) is an element vector, feint , that contains the internal nodal forces. Note that the internal force vector depends upon the stress ﬁeld σ e within the element, and we have not assumed anything as of yet for the constitutive (stress-strain) law. The sum of the integrals on the right-hand side is the element vector, feext , containing the external nodal forces. The fully assembled discretized equations can then be written as f int = f ext

(6.85)

When the material behavior is linearly elastic and governed by Hooke’s law, the vector σ e containing the stress components within the element can be expressed as σ e = Ce e

(6.86)

Hence, the ﬁnite element approximation for the stress ﬁeld within the element can be written as ˜ e = Ce Be de σ = Ce Be Le d

(6.87)

where Ce is the 3 × 3 elasticity matrix for the case of either plane strain or plane stress deﬁned in Chapter 4, and d is the global vector containing the nodal displacements. Substituting the last line in equation (6.86) into equation (6.84) gives nel

LTe

Ωe

e=1

=

nel e=1

LTe

Ωe

BTe Ce Be

NeT b dΩ +

Γe

dΩLe d

NTe t∗ dΓ

(6.88)

The element stiﬀness matrix, Ke , comes from the integral on the left-hand side of equation (6.88). The element external force vector comes from the integral on the right-hand side. To summarize we write BTe Ce Be dΩ (6.89) Ke = Ωe

Continuum Finite Elements feext =

163

NeT b dΩ +

Ωe

Γe

NTe t∗ dΓ

(6.90)

In practice, the integrals deﬁned in equations (6.89) and (6.90) are carried out over the area (and, if speciﬁed tractions are present, edges) of the bi-unit square in the parent domain. For example, in order to get an equivalent area integral in the parent domain, we must employ the area-change relationship derived in Section 6.2.6. A diﬀerential volume in the physical domain can be expressed as a diﬀerential volume in the parent domain as dΩe = t jdξdη

(6.91)

where in the present formulation for plane problems, t is the out-of-plane thickness. Hence, the element stiﬀness matrix can be expressed as the follow ing double integral in the parent domain: Ke = t

1

−1

1

−1

BTe Ce Be jdξdη

(6.92)

In ﬁnite element codes, the integrals over the parent domain are evaluated using a technique called Gauss quadrature. The technique gives the exact value of an integral whose integrand is a polynomial function, as explained in detail in the next section followed by some examples that demonstrate how to evaluate element matrices and vectors using Gauss quadrature.

6.6

Gauss quadrature

Gauss quadrature is often misunderstood to be a numerical integration tech nique for obtaining approximate values to integrals. While Gauss quadrature can be used to obtain approximate values of integrals, the main idea is to ﬁnd the exact values of integrals whose integrands are polynomial functions. This is accomplished using tabulated point and weight values. In this sec tion we will describe the method of Gauss quadrature for evaluating both one-dimensional and multidimensional integrals. To begin, consider the following integral whose integrand is a cubic poly nomial 1 α1 + α2 ξ + α3 ξ 2 + α4 ξ 3 dξ (6.93) −1

where α1 to α4 are arbitrary constants. The exact value, I, of the integral is 2 I = 2α1 + α3 3

(6.94)

164

Finite Element Method

We now seek to ﬁnd integration points, ξI , and weights, WI , such that nint

f (ξI )WI = I

(6.95)

I=1

where nint is the total number of integration points that are needed for the sum to give the exact value of the integral. Here in equation (6.95), the integration points, ξI , (also referred to as quadrature points) are locations within the interval −1 ≤ ξ ≤ 1. The quantities, WI , are numerical weight values associated with each integration point. Let us now try to satisfy equation (6.95) with two integration points ξ1 and ξ2 . Equation (6.95) then becomes I = f (ξ1 )W1 + f (ξ2 )W2 = (α1 + α2 ξ1 + α3 ξ12 + α4 ξ13 )W1 +(α1 + α2 ξ2 + α3 ξ22 + α4 ξ23 )W2 = (W1 + W2 )α1 + (ξ1 W1 + ξ2 W2 )α2 +(ξ12 W1 + ξ22 W2 )α3 + (ξ13 W1 + ξ23 W2 )α4

(6.96)

We now have the following two results for the same integral I: I = (W1 + W2 )α1 + (ξ1 W1 + ξ2 W2 )α2 + (ξ12 W1 + ξ22 W2 )α3

+(ξ13 W1 + ξ23 W2 )α4 2 I = 2α1 + α3 3

(6.97)

By comparing coeﬃcients in equation (6.97) above we conclude that W1 + W2 = 2 ξ1 W1 + ξ2 W2 = 0 ξ12 W1 + ξ22 W2 = 2/3 ξ13 W1 + ξ23 W2 = 0

(6.98)

We immediately see that the ﬁrst equation in (6.98) can be satisﬁed if we let W1 = W2 = 1. The second and fourth equations are then satisﬁed auto matically if we choose the quadrature point locations to be symmetric about ξ = 0, i.e., ξ1 = −ξ2 . The third line in equation (6.98) then becomes 2ξ12 = 2/3 ⇒ 1 ξ1 = ± √ 3

(6.99)

Hence, the exact value of an integral of the form (6.93) whose integrand is a cubic polynomial is √ √ I = f (−1/ 3)(1) + f (1/ 3)(1) (6.100)

Continuum Finite Elements

165

We can also obtain Gauss points and weights for higher order polynomials using the procedure described above. For example, suppose that the integrand is the following ﬁfth-order polynomial: f (ξ) = α1 + α2 ξ + α3 ξ 2 + α4 ξ 3 + α5 ξ 4 + α6 ξ 5

(6.101)

The resulting equations to be solved are then W1 + W2 + W3 = 2 ξ1 W1 + ξ2 W2 + ξ3 W3 = 0 ξ12 W1 + ξ22 W2 + ξ32 W3 = 2/3 ξ13 W1 + ξ23 W2 + ξ33 W3 = 0 ξ14 W1 + ξ24 W2 + ξ34 W3 = 2/5 ξ15 W1 + ξ25 W2 + ξ35 W3 = 0

(6.102)

Let us for the moment assume that equations (6.102) can be solved by as suming ξ2 = 0, and that the remaining integration points are symmetrically located about ξ = 0, i.e., ξ1 = −ξ3 . After making this assumption, we see that the second equation in (6.102) is satisﬁed if W1 = W3 . We are now left with the following three equations in three unknowns: 2W1 + W2 = 2 2ξ12 W1 = 2/3 2ξ14 W1 = 2/5

(6.103)

Solving the second line in (6.103) for W1 and then plugging the result into the third line gives (6.104) ξ1 = ± 3/5 Next, inserting the known value of ξ1 back into the second line yields W1 = 5/9, and hence, W2 = 8/9. The exact value of the integral is I = f (− 3/5)(5/9) + f (0)(8/9) + f ( 3/5)(5/9)

(6.105)

The procedure that we have been following for generating integration points and weights can be carried out in principle for polynomials up to any order. The integration points and weights for polynomials up to ﬁfth order are tab ulated in Table 6.6 below. For the sake of interest, the locations of these points are depicted in Figure 6.13. It turns out that n quadrature points are required to obtain the exact value of an integral whose integrand is a polynomial of order 2n−1. So, for example, for a fourth-order polynomial we have 2n − 1 = 4 or n = 5/2. Since n must be an integer, three quadrature points are required.

166

Finite Element Method ξ1 = 0 √ ξ1 = 1/ 3

√ ξ1 = −1/ 3 ξ1 = − 3/5

ξ2 = 0

ξ3 =

3/5

FIGURE 6.13 Quadrature point locations and weights. TABLE 6.6

Quadrature points and weights. ξ1 ξ2 ξ3 n 2n − 1 1

1

2

3

3

5

0 √ −1/ 3 − 3/5

– √ 1/ 3 0

w1

w2

w3

–

2

–

–

– 3/5

1

1

–

5/9

8/9

5/9

Multidimensional integrals Once the quadrature points and weights are obtained for one-dimensional in tegrals, it is straightforward to apply these rules to evaluate multidimensional integrals such as area and volume integrals. To illustrate, let us consider the following area integral:

1

1

f (ξ, η) dξ dη

I= −1

(6.106)

−1

As we recall from calculus, double integration is carried out by ﬁrst evaluat ing the inner integral with respect to ξ (keeping η ﬁxed) and then computing the outer integral with respect to η (keeping ξ ﬁxed). In other words, dou ble integration simply requires evaluating two one-dimensional integrals. To illustrate, let us carry out the double integration (6.106) using a two-point Gauss quadrature rule in both the ξ and η directions as follows:

1

I= −1

[Wξ1 f (ξ1 , η) + Wξ2 f (ξ2 , η)] dη

= Wη1 [Wξ1 f (ξ1 , η1 ) + Wξ2 f (ξ2 , η1 )] +Wη2 [Wξ1 f (ξ1 , η2 ) + Wξ2 f (ξ2 , η2 )]

Continuum Finite Elements

167 TABLE 6.7

Points and weights for 2 × 2 quadrature. I ξI ηI WI √ √ 1 −1/ 3 −1/ 3 1 √ √ 2 1/ 3 −1/ 3 1 √ √ 1/ 3 1 3 1/ 3 √ √ 1 4 −1/ 3 1/ 3

and hence I = Wη1 Wξ1 f (ξ1 , η1 ) + Wη1 Wξ2 f (ξ2 , η1 ) +Wη2 Wξ1 f (ξ1 , η2 ) + Wη2 Wξ2 f (ξ2 , η2 )

(6.107)

Employing a two-point rule in both the ξ and η directions is referred to as 2 × 2 quadrature. The operation can be written succinctly as I=

4

WI f (ξI , ηI )

I=1

where the quadrature point locations and weights are given in Table 6.7. For the sake of interest, the mapping of the quadrature points (for 2 × 2 quadrature) from the parent domain to the physical domain is shown in Figure 6.14. The X and Y locations of the quadrature points are be obtained from the isoparametric mapping (6.30), i.e., XI =

4

Na (ξI , ηI )Xa

a=1

YI =

4

Na (ξI , ηI )Ya

a=1

where Xa and Ya are the coordinates of local node a in the physical domain.

Example 6.1 To illustrate the process of evaluating one-dimensional integrals using Gauss quadrature, let us consider the following integral: X2 f (X) dX (6.108) X1

Notice that the upper and lower limits are not from −1 to 1, so the quadrature rule formulas derived above do not directly apply here. It is possible, however,

168

Finite Element Method 3 4 2 η

Y 1 X

4

3 I=2

I=4

ξ I=1

I=3

1

2

FIGURE 6.14 Quadrature point locations in parent and physical domains.

through a change of variables, to recast the integral into the proper form. To do this, we rely on the one-dimensional shape functions to map the interval from −1 to 1 into a generic interval from X1 to X2 as follows: X = N1 X 1 + N 2 X 2 1 = (1 − ξ)(X1 ) + 1/2(1 + ξ)(X2 ) 2

(6.109)

Notice that when the value of ξ is equal to −1, X is equal to X1 . When ξ is equal to 1, the value of X is X2 . Now suppose we are interested in evaluating the following integral using Gauss quadrature:

5

(1 + X 2 ) dX

(6.110)

4

We will begin by using equation (6.109) to get the integral into the proper form. The relationship between X and ξ is 1 1 (1 − ξ)(4) + (1 + ξ)(5) 2 2 9 1 = + ξ 2 2

X=

(6.111)

Continuum Finite Elements

169

Given this relationship, we can now evaluate the relationship between dX and dξ, i.e., dX 1 = ⇒ dξ 2 dX =

1 dξ 2

(6.112)

The integral (6.110) can now be written as

1

f (ξ) dξ −1

where 1 f (ξ) = 1+ 2

9 1 + ξ 2 2

2 !

Because the integrand f (ξ) is quadratic in ξ, we have 2n − 1 = 2 ⇒ n = 3/2

(6.113)

or in other words, two quadrature points are required to obtain the exact value of the integral. Carrying out the required two-point quadrature rule yields I = W1 f (ξ1 ) + W2 f (ξ2 ) 1 1 = (1)f − √ + (1)f √ 3 3 " " 2 !# 2 !# 1 9 1 1 9 1 1 1 − √ = (1) 1+ + (1) 1+ + √ 2 2 2 3 2 2 2 3 64 = 3 If we perform a quick hand calculation as a check on our work, we notice that the result above agrees with the exact value of the integral, i.e., 4

5

1+X

2

5 X 3 dX = X + 3 4 = (5 + 125/3) − (4 + 64/3) 64 = 3

170 ∗

t =

tX tY

Finite Element Method

3 4 Y 1

2

X FIGURE 6.15 Uniform traction distribution acting over an element edge.

Example 6.2 Recall that the element vector that arises from a prescribed traction (natural) boundary condition applied along an element edge is given as NTe t∗ dΓ (6.114) fe = Γe

where, for two-dimensional plane problems, Γe refers to the element edge (or edges) along which the traction is prescribed, Ne is the 2 × 8 shape function matrix, and t∗ is the 2 × 1 traction vector acting at a point on Γe . Notice that the resulting element vector is 8 × 1. To illustrate how to calculate the element vector, let us consider the situa tion illustrated in Figure 6.15. Here, a constant traction distribution, tX t∗ = tY is applied along element edge 3–4 of an arbitrary four-node quadrilateral in the undeformed conﬁguration. The nodal coordinates for the element are given in Table 6.8. Because the shape functions for the arbitrary four-node quadrilateral are given in terms of the ξ and η coordinates, it is necessary to recast the inte gral (6.114) into an equivalent integral in the parent domain. The traction distribution acts along edge 3–4, so we only need to study how edge 3–4 maps from the parent domain to the physical domain. Notice that edge 3–4 is the top edge of the bi-unit square where η = 1. In order to evaluate the integral in the parent domain, we need to know the relationship between dΓ and an inﬁnitesimal movement, dξ, along the line 3-4 in the parent domain.

Continuum Finite Elements

171

TABLE 6.8

Nodal coordinates of four-node quadrilateral. Local node Xa Ya 1 2 3 4

0.0 1.0 1.5 0.0

0.0 0.0 1.5 1.0

We begin by recalling Section 6.2.6, where we derived a relationship between an inﬁnitesimal vector d ξ in the parent domain and an inﬁnitesimal vector d X in the physical domain. This relationship is given as dX = JT dξ where J is the Jacobian matrix, i.e., X,ξ Y,ξ J = X,η Y,η Along edge 3–4 (η = 1) in the parent domain, the inﬁnitesimal movement d ξ is just a movement in the ξ direction only. Hence, dξ dξ = 0 and, as a result, the inﬁnitesimal vector in the physical domain reduces to X,ξ dX = dξ Y,ξ The magnitude of the inﬁnitesimal vector, dΓ, in the physical domain can be expressed as dΓ = t |dX| $ 2 + Y 2 dξ = t X,ξ ,ξ

(6.115)

where in the present two-dimensional formulation, t is the thickness of the element (assumed to be constant). We can now write the element vector as an integral in the parent domain as 1 $ 2 + Y 2 dξ NTe (ξ, η = 1)t∗ X,ξ (6.116) fe = t ,ξ −1

The last step in this example is to go through the algebraically lengthy task of evaluating the integral (6.116). The process is simpliﬁed if one notices that

172

Finite Element Method

the ﬁrst two shape functions N1 and N2 are identically zero along edge 3–4 where η = 1. We also note that the relationship between dΓ and dξ happens to be very simple in this example. Evaluating the derivatives X,ξ and Y,ξ we obtain 1 1 (6.117) X,ξ = (5 + η); Y,ξ = (1 + η) 8 8 and hence, dΓ = t 0.752 + 0.52 dξ l34 =t dξ 2 where l34 is the length of edge 3–4 in the physical domain, i.e., l34 = (X4 − X3 )2 + (Y4 − Y3 )2

(6.118)

(6.119)

The ﬁnal result is ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

⎫ 0 ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ 1 0 ⎬l 34 1 (1 + ξ)t fe = t dξ X ⎪2 ⎪ 2 −1 ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ 2 (1 + ξ)tY ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ (1 − ξ)tX ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎩1 ⎭ (1 − ξ)tY ⎧2 ⎫ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ l34 0 = t tX ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ tY ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ t ⎪ ⎪ X ⎪ ⎭ ⎩ ⎪ tY

(6.120)

A few closing remarks are in order here. The quantities in the element vector are called equivalent nodal loads. Equivalent nodal loads are point loads that are applied at the nodes as a result of the traction distribution acting along an edge of the element. The point loads are computed by evaluating integrals that come directly from the weak form of the governing diﬀerential equation. It is also worth mentioning that the shape functions N3 and N4 in this example, when evaluated along the line η = 1, reduce to the one-dimensional shape functions that depend only on ξ. Because N1 and N2 are zero along this line, the equivalent nodal loads could alternatively be obtained by simply carrying

Continuum Finite Elements

173

Y

31

30

(0,1)

(0,0)

(2,1)

(1,0)

22

X

26

FIGURE 6.16 Element matrix computation for arbitrary quadrilateral.

out the following integrals: f3i = t

1

−1 1

f4i = t

−1

1 (1 − ξ)ti l34 dξ 2 1 (1 + ξ)ti l34 dξ 2

where i can be either X or Y .

Example 6.3 In this example we demonstrate the process of computing the 8 × 8 element matrix for the arbitrary four-node quadrilateral. For the sake of simplicity, we will assume the condition of plane stress. The element in the physical domain is shown in Figure 6.16. The nodal coordinates and global node numbers are labeled in the ﬁgure. The connectivity list for the element is given in Table 6.9 below. We emphasize that a correct connectivity list begins with any one of the global node numbers and then continues in a counterclockwise fashion. The ﬁrst step in the process of computing the element matrix is to write down the mapping between the parent domain and the physical domain. The

TABLE 6.9

Connectivity list. Element 1 2 1

26

30

3

4

31

22

174

Finite Element Method

isoparametric mapping is written as X(ξ, η) =

4

Na (ξ, η)Xa

a=1

Y (ξ, η) =

4

Na (ξ, η)Ya

(6.121)

a=1

where Xa and Ya are the X and Y coordinates of local node a, respectively. Plugging the nodal coordinates into equation (6.121) gives X = N1 (1) + N2 (2) + N3 (0) + N4 (0) = N1 + 2N2 Y = N1 (0) + N2 (1) + N3 (1) + N4 (0) = N2 + N3

(6.122)

Recall that the shape functions for the four-node quadrilateral are deﬁned as 1 (1 − ξ) (1 − η) 4 1 N2 = (1 + ξ) (1 − η) 4 1 N3 = (1 + ξ) (1 + η) 4 1 N4 = (1 − ξ) (1 + η) 4 N1 =

(6.123)

After simplifying the algebra, the mapping (6.122) can be written succinctly as 1 (3 + ξ)(1 − η) 4 1 Y = (1 + ξ) 2

X=

(6.124)

The next step in the procedure is to ﬁrst set up the Jacobian matrix asso ciated with the mapping between the parent and physical domains, and then to compute the determinant of the Jacobian matrix. Doing this we obtain X,ξ Y,ξ J = X,η Y,η ⎤ ⎡ 1 1 (1 − η) ⎢ 2⎥ = ⎣ 41 ⎦⇒ − (3 + ξ) 0 4 1 j = (3 + ξ) 8

(6.125)

Notice that the determinant of Jacobian matrix is not constant within the element but varies linearly with the natural coordinate ξ.

Continuum Finite Elements

175

Armed with the Jacobian matrix, the matrix, Be , containing the derivatives of the shape functions with respect to the global X and Y coordinates, turns out to be ⎡ 1−ξ ⎤ −1 − ξ −1 + ξ 1+ξ ⎢ ⎢ Be = ⎣

3+ξ 0

0

2(−1 + η) 3+ξ

0

3+ξ 0

3+ξ

1−η 3+ξ

0

2+ξ+η

0

0

3+ξ

3+ξ

−1 − ξ − 2η

0

3+ξ

2(−1 + η)

1−ξ

1−η

1+ξ

2+ξ+η

−1 − ξ

−1 − ξ − 2η

−1 + ξ

3+ξ

3+ξ

3+ξ

3+ξ

3+ξ

3+ξ

3+ξ

3+ξ

⎥ ⎥ ⎦

Assuming unit thickness, the element matrix can be expressed as a double integral in the parent domain as 1 1 BTe Ce Be j dξdη (6.126) Ke = −1

−1

It turns out that the exact value of the integral on the right-hand side of equation (6.126) can be obtained using a 2 × 2 quadrature rule. Starting with the ﬁrst quadrature point, the technique involves evaluating the integrand at the quadrature point and then multiplying the result by a weight factor. The sum of the contributions from each quadrature point is the value of the integral — an 8 × 8 matrix in this case. Just to keep things as simple as possible, let’s take Young’s modulus to be E = 1 and ν = 0. The plane stress elasticity matrix is then ⎡ ⎤ E 0 0 (6.127) Ce = ⎣ 0 E 0 ⎦ 0 0 E/2 Evaluating the√matrix product BTe Ce Be j at the quadrature points (ξa ± √ 1/ 3, ηa = ±1/ 3) yields the following contributions to the element matrix: ⎡ 0.385121 −0.128374 −0.093976 −0.034398 −0.103193 0.034398 −0.187952 0.128374 ⎤ −0.128374

Ke1

−0.187952 0.128374

0.577681 0.064187 −0.239548 0.034398 −0.154789 0.029789 −0.183344

0.064187 0.073404 0.017199 0.025181 −0.017199 −0.004608 −0.064187

−0.239548 0.017199 0.132982 0.009217 0.064187 0.007982 0.042380

0.034398 0.025181 0.009217 0.027650 −0.009217 0.050362 −0.034398

−0.154789 −0.017199 0.064187 −0.009217 0.041476 −0.007982 0.049127

0.029789 −0.004608 0.007982 0.050362 −0.007982 0.142199 −0.029789

−0.183344 −0.064187 0.042380 −0.034398 0.049127 −0.029789 0.091837

0.180116 −0.023295 −0.063642 −0.086937 −0.133526 0.086937 0.017053 0.023295

−0.023295 0.350869 0.011647 −0.162227 0.014768 −0.232111 −0.003121 0.043469

−0.063642 0.011647 0.130406 0.043469 −0.031821 −0.043469 −0.034942 −0.011647

−0.086937 −0.162227 0.043469 0.130406 0.055116 0.066763 −0.011647 −0.034942

−0.133526 0.014768 −0.031821 0.055116 0.156821 −0.055116 0.008526 −0.014768

0.086937 −0.232111 −0.043469 0.066763 −0.055116 0.183237 0.011647 −0.017889

0.017053 −0.003121 −0.034942 −0.011647 0.008526 0.011647 0.009363 0.003121

0.023295 0.043469 −0.011647 −0.034942 −0.014768 −0.017889 0.003121 0.009363

0.018725 −0.006242 0.017053 −0.023295 −0.069884 0.023295 0.034106 0.006242

−0.006242 0.028088 0.003121 −0.000836 0.023295 −0.104826 −0.020174 0.077574

0.017053 0.003121 0.090058 0.011647 −0.063642 −0.011647 −0.043469 −0.003121

−0.023295 −0.000836 0.011647 0.049710 0.086937 0.003121 −0.075290 −0.051995

−0.069884 0.023295 −0.063642 0.086937 0.260811 −0.086937 −0.127285 −0.023295

0.023295 −0.104826 −0.011647 0.003121 −0.086937 0.391217 0.075290 −0.289511

0.034106 −0.020174 −0.043469 −0.075290 −0.127285 0.075290 0.136647 0.020174

0.006242 0.077574 −0.003121 −0.051995 −0.023295 −0.289511 0.020174 0.263932

0.146807 −0.034398 0.025181 −0.009217 −0.078012 0.009217 −0.093976 0.034398

−0.034398 0.101054 0.017199 −0.001235 0.081386 −0.104428 −0.064187 0.004608

0.025181 0.017199 0.013825 0.004608 0.012590 −0.004608 −0.051596 −0.017199

−0.009217 −0.001235 0.004608 0.013825 0.021807 0.039006 −0.017199 −0.051596

−0.078012 0.081386 0.012590 0.021807 0.112410 −0.021807 −0.046988 −0.081386

0.009217 −0.104428 −0.004608 0.039006 −0.021807 0.210994 0.017199 −0.145572

−0.093976 −0.064187 −0.051596 −0.017199 −0.046988 0.017199 0.192560 0.064187

0.034398 0.004608 −0.017199 −0.051596 −0.081386 −0.145572 0.064187 0.192560

⎢ −0.093976 −0.034398 =⎢ ⎣ −0.103193 0.034398 ⎡

⎢ Ke2 = ⎢ ⎣ ⎡ ⎢ Ke3 = ⎢ ⎣ ⎡ ⎢ Ke4 = ⎢ ⎣

⎥ ⎥ ⎦ ⎤ ⎥ ⎥ ⎦ ⎤ ⎥ ⎥ ⎦ ⎤ ⎥ ⎥ ⎦

176

Finite Element Method TABLE 6.10

Flow chart for computing Ke . 1. 2. 3. 4. 5. 6. 7.

Zero the element matrix Ke Set up the elasticity matrix Ce Set up the Gauss point locations and weights Loop over the quadrature points, a = 1 to nint Compute Kea = BTe (ξa , ηa )Ce Be (ξa , ηa ) j(ξa , ηa )Wa Ke = Ke + Kea Continue looping over quadrature points

Finally, adding things together yields the total contribution to the element matrix, i.e., Ke = Ke1 + Ke2 + Ke3 + Ke4 . The result is ⎡ 0.730769 ⎢ Ke = ⎢ ⎣

−0.192308 −0.115385 −0.153846 −0.384615 0.153846 −0.230769 0.192308

−0.192308 1.057692 0.096154 −0.403846 0.153846 −0.596154 −0.057692 −0.057692

−0.115385 0.096154 0.307692 0.076923 −0.057692 −0.076923 −0.134615 −0.096154

−0.153846 −0.403846 0.076923 0.326923 0.173077 0.173077 −0.096154 −0.096154

−0.384615 0.153846 −0.057692 0.173077 0.557692 −0.173077 −0.115385 −0.153846

0.153846 −0.596154 −0.076923 0.173077 −0.173077 0.826923 0.096154 −0.403846

−0.230769 −0.057692 −0.134615 −0.096154 −0.115385 0.096154 0.480769 0.057692

0.192308 −0.057692 −0.096154 −0.096154 −0.153846 −0.403846 0.057692 0.557692

⎤ ⎥ ⎥ ⎦

The ﬂow chart outlining the procedure for the element matrix computation using Gauss quadrature is given in Table 6.10.

6.7

Bending of a cantilever beam

To apply some of the concepts introduced in this chapter, let us now consider the problem of a cantilever beam loaded at the right end by a concentrated force, P = 10 kN, shown in Figure 6.17. The length of the beam is 3 m, and it has a rectangular, 20 mm × 200 mm, cross section. We will assume that the beam material is linearly elastic and isotropic with Young’s modulus E = 200 GPa. For the moment, let’s take Poisson’s ratio to be ν = 0.3. To begin, we will assume plane stress, and construct a ﬁnite element mesh using rectangular four-node elements. From our previous discussion in Sec tion 6.2, the displacement interpolation within an element then has the form α1 + α2 X + α3 Y + α4 XY u ˜X (6.128) = u ˜Y e β1 + β2 X + β3 Y + β4 XY where α1 , . . . , β4 are constants. The strain components within the element are obtained by taking the partial derivatives of the displacement components. The extensional (bending) strain in the element has the form ˜XX = α2 + α4 Y

(6.129)

Continuum Finite Elements

177 P = 10 kN

3m FIGURE 6.17 Cantilever beam.

From elementary mechanics of materials theory, the deﬂection curve for a cantilever beam loaded at the right end by a concentrated force, P , is given as 2 3 ! X P L3 X (6.130) 3 − uY = 6EI L L where L is the length of the beam, E is Young’s modulus, and I is the moment of inertia of the beam’s cross section about the neutral axis. Notice that the deﬂection curve from mechanics of materials theory is a cubic polynomial in X. The second line in (6.128) indicates that the ﬁnite element approximation for the vertical displacement within an element varies linearly with X. This suggests that a relatively large number of elements will be required in the X direction in order to achieve an accurate solution for the deﬂection of the beam. The normal stress distribution, σXX (Y ), acting on some cross section lo cated a distance X from the left end is also available from mechanics of ma terials theory. The elementary expression can be written as σXX =

−M (X)Y I

(6.131)

where M (X) is the net bending moment acting on the cross section, and Y is the distance from the beam’s neutral surface (surface on which the extensional strain is zero). Because the normal stress is proportional to Y and the material is linearly elastic, the extensional strain must also vary linearly with Y . The fact that the ﬁnite element approximation for the extensional strain varies linearly with Y suggests that only a few elements are needed in the Y direction. In the present analysis, to gain experience with how the solution changes as the mesh is reﬁned, we will consider the four diﬀerent meshes shown in Figure 6.18. The ﬁnite element solution for the end deﬂection obtained for each mesh is reported in Table 6.11. The elementary mechanics of materials prediction for the end deﬂection can be obtained by substituting X = L into

178

Finite Element Method (a)

(b)

(c)

(d)

FIGURE 6.18 2 × 5 mesh (a), 2 × 20 mesh (b), 2 × 40 mesh (c), and 4 × 40 mesh (d). equation (6.130) above. Doing this gives uY =

P L3 3EI

(10 kN)(3 m)3 (0.02 m)(0.2 m)3 9 2 3(200 × 10 N/m ) 12 = 33.75 mm =

(6.132)

Comparing the elementary solution with the deﬂection values reported in the table, the percent diﬀerence between the ﬁnite element solution for case (c) and the elementary solution is approximately 6%. Hence, as we expected, a ﬁne mesh is required to obtain good accuracy. The elementary mechanics of materials solution (6.131) for the bending stress in the beam is valid at cross sections that are suﬃciently far from clamped ends and concentrated loads. Hence, for comparison purposes, the ﬁnite element solution for the bending stress is compared to the elementary solution at a cross section located 1.5 m from the left end. The result for the 2 × 40 mesh is plotted in Figure 6.19. Here in the ﬁgure, the bending stress, σXX , is plotted versus distance, Y , from the neutral surface. The solid curve represents the solution from elementary theory, and the circular data points represent the ﬁnite element solution. The stress was output at ﬁve equallyspaced points located along the right edge of the elements contiguous to the line X = 1.5 m. Excellent agreement is obtained between the ﬁnite element TABLE 6.11

End deﬂection of beam with increasing mesh density. Mesh density 2×5 2 × 20 2 × 40 4 × 40

u ˜Y (mm) Percent error 7.5259 27.2109 31.3006 31.8934

77.7 19.4 7.3 5.5

Continuum Finite Elements

179

σXX (MPa)

150 100 50 0 -50 -100 -150 -0.1

-0.05

0

0.05

0.1

Y (m) FIGURE 6.19 Bending stress σXX along the line X = 1.5 m for the 2 × 40 mesh.

solution and elementary theory, even with only two elements through the thickness of the beam. For the sake of interest, a contour plot of the bending stress, superposed on top of the beam’s (magniﬁed) deformed geometry, is shown in Figure 6.20. Employing the 4 × 40 mesh, let us now run the problem again for two diﬀerent values of Poisson’s ratio, ν = 0.3 and ν = 0.499. In the linearized theory of elasticity discussed in Chapter 4, the Lam´e’s constant, λ, approaches inﬁnity as ν approaches 1/2. In the three-dimensional version of Hooke’s law, λ is the coeﬃcient in front of kk , the trace of the small-strain tensor. The trace of the small-strain tensor physically represents the volumetric strain deﬁned as the change in volume of an inﬁnitesimal element divided by the original volume. As Poisson’s ratio approaches 1/2, the material’s resistance to volumetric deformation goes to inﬁnity. Under such conditions, the material is said to be incompressible. An incompressible material can deform in a shear

P = 10 kN

FIGURE 6.20 Contour plot of the bending stress σXX for the 4 × 40 reﬁned mesh.

180

Finite Element Method TABLE 6.12

End deﬂection of beam for case of plane strain with ν = 0.3 and ν = 0.499. Poisson’s ratio (quadrature rule)

ν = 0.3 (2 × 2)

ν = 0.499 (2 × 2)

ν = 0.499 (1 × 1)

End deﬂection (mm)

28.83

2.76

22.92

mode, but the volume remains constant during deformation. The plane-strain results for the two values of Poisson’s ratio (ν = 0.3 and ν = 0.499) are reported in Table 6.12. Also reported in the table is the end deﬂection for the case when ν = 0.499, but only a 1 × 1 quadrature rule was used to integrate the stiﬀness matrix. For the fully integrated (2 × 2 quadrature) element, when ν = 0.499, the end deﬂection is extremely small compared to the case when ν = 0.3. This behavior is referred to as locking. For the case of plane strain when Poisson’s ratio is nearly 1/2, locking will occur no matter how much the mesh is reﬁned! The same locking behavior can occur when using the three-node triangle under conditions of plane strain, and also for the eight-node brick under nearly incompressible conditions. We note that locking does not occur in the present problem when ν = 0.499 under the condition of plane stress. When the element stiﬀness matrix is integrated using a 1 × 1 quadrature rule, the locking behavior is no longer present. The technique of underinte grating the stiﬀness matrix is called reduced integration. It is widely used to help eliminate the problem of locking in the analysis of nearly incompressible materials. The numerical problem of locking has been addressed by many researchers and can be remedied in a variety of ways. The locking of the arbi trary four-node quadrilateral can be taken care of under many circumstances by simply employing reduced integration. Unfortunately, under certain con ditions, underintegrated elements (the four-node quad being one of them) possess what are called spurious zero-energy modes. Spurious zero-energy modes are deformed conﬁgurations of the element, resembling the shape of an hourglass, in which the total energy in the element is zero. For static prob lems, this can result in a singular global stiﬀness matrix. Elements can be made more robust by using the technique of reduced integration along with what is called hourglass control or hourglass stabilization. An element with hourglass control has additional stiﬀness terms included to prevent it from going into such zero-energy modes. For further discussion on this topic see Flanagan and Belytschko [8] and Belytschko et al. [9].

Continuum Finite Elements

6.8

181

Analysis of a plate with hole

Let us now consider the problem of a 2 m × 2 m plate with a centrally located hole as shown in Figure 6.21. The top and bottom edges of the plate are subjected to a uniform traction distribution of magnitude σ0 acting in the Y direction. The radius of the hole is r = 0.1 m. In the present study, we will begin by assuming that the thickness of the plate is small compared to the width of the plate so that the condition of plane stress prevails. Unless otherwise stated, we will assume E = 200 GPa and ν = 0.3. Before we proceed, it is necessary to discuss the boundary conditions associated with the problem. Because both the geometry and the loading are symmetric about the X and Y axes, it is only necessary to model one-quarter of the plate (say the upper-right quadrant). We still however need to be careful to apply the correct prescribed displacement boundary conditions on the left and bottom edges of the quarter model. To give a mathematical explanation of the boundary conditions, it is necessary to discuss the concept of odd and even functions. A function f (X) is said to be even if f (X) = f (−X). On the other hand, the function f (X) is deﬁned to be odd if f (X) = −f (−X). Having deﬁned odd and even functions, let us now consider a point in the upper-right quadrant of the plate. If this point displaces in the positive X direction, then, due to symmetry, the mirror image of that point in the lower-right quadrant must also move the same amount in the X direction. Hence, we conclude that the displacement component uX is an even function of Y , i.e., uX (X, Y ) = uX (X, −Y ). If the Y σ0

2m X

2m FIGURE 6.21 Plate with a hole.

182

Finite Element Method

(a)

(b)

FIGURE 6.22 Coarse mesh (a) and reﬁned mesh (b).

same point in the upper-right quadrant moves in the positive Y direction, then the mirror image point in the lower-right quadrant must displace the same distance in the −Y direction. Hence uY is an odd function in Y , i.e., uY (X, Y ) = −uY (X, −Y ). Because uY is odd in Y , we conclude that this displacement component must be zero along the line Y = 0. Following the same argument, we conclude that the displacement component uX is an odd function of X, i.e., uX (X, Y ) = −uX (−X, Y ). Therefore uX must be zero along the line X = 0. These symmetry boundary conditions are enforced in practice by setting the Y degree of freedom equal to zero at nodes that lie on the X axis and setting the X degree of freedom equal to zero at nodes that lie on the Y axis. Having discussed the symmetry boundary conditions to be applied along the left and bottom edges of the plate, let us now begin our analysis by employing the two meshes composed of arbitrary four-node (plane stress) quadrilaterals shown in Figure 6.22. The coarse mesh (a) is composed of 70 elements, and the reﬁned mesh (b) has 200 elements. To enforce the boundary conditions, we simply constrain the nodes lying on the X axis from moving in the Y direction and the nodes on the Y axis from moving in the X direction. The uniform traction distribution is applied along the top edge by identifying the element edges that lie on this edge. By identifying an element edge and specifying the magnitude of the traction acting on that edge, the element vector can be calculated. Recall that the element vector is deﬁned as NTe t∗ dΓ (6.133) fe = t Γte

where Γte is the edge along which the traction distribution is applied. In commercial ﬁnite element codes, the value of this integral is stored in a library, and the element vector is set up inside the element subroutine. The

Continuum Finite Elements

183

(a) (b) FIGURE 6.23 Sparse structure of the global stiﬀness matrix (a) and banded strucutre (b).

element vector is then assembled into the global force vector following the ﬁnite element assembly process. After the boundary conditions are deﬁned, all that is left to do is to deﬁne the material properties for each element and run the analysis. Before running the analysis however, it is a good idea for computational eﬃciency to check the node numbering of the mesh. When using a direct solver, based on Gaussian elimination, the manner in which the mesh is numbered can make a huge impact on computer run time. To illustrate this idea, let us focus our attention on Figure 6.23. The ﬁgure shows the structure of the global stiﬀness matrix for the coarse mesh before renumbering (a), and the same mesh after renumbering using the Cuthill-McKee algorithm [10]. The structure of the global stiﬀness matrix is revealed by the positions in the matrix occupied by nonzero numbers. The matrix (a) to the left is said to be sparse, because it has many nonzero values occupying positions far from the diagonal. On the other hand, the matrix (b) on the right is called banded, because most of the nonzero values are clustered about the diagonal. The banded matrix requires signiﬁcantly less computer storage and, hence, a shorter run time. The number of columns away from the diagonal (either to the right or to the left) occupied by nonzero numbers is called the half bandwidth b. The half bandwidth can be calculated by the following formula: b = ndf (m + 1)

(6.134)

where ndf is the number of degrees of freedom per node, and the parameter m is the maximum diﬀerence between node numbers within an element. To get m, one needs to look at the connectivity list for each element and record the maximum diﬀerence between node numbers for each element. The global

Finite Element Method 3

3

2.5

2.5

σY Y /σ0

σY Y /σ0

184

2 1.5

2 1.5

1

1

0.5

0.5

0 0

0.5

X (m)

1

0 0

0.5

1

X (m)

FIGURE 6.24 Stress distribution σY Y /σ0 plotted versus distance X along the line Y = 0. maximum can then be extracted by comparing the element maximum values. Using a direct solver, the computer run time in seconds can be estimated using the following formula: nb2 (6.135) time ≈ 2α where n is the total number of equations in the model, and α is a parameter that deﬁnes the speed of the computer, e.g., in million ﬂoating-point opera tions per second (MFLOPS). The MFLOP rating gives an indication of the number of multiplications that can be performed per second. Modern-day PC’s are capable of achieving between 2000 and 6000 MFLOPS. As an example, suppose that a two-dimensional ﬁnite element mesh is com posed of a rectangular 500 × 500 node grid. The largest (worst) possible value of m would be 500 × 500 − 1 = 249999. Assuming a MFLOP rating of 2000, the computer run time for the worst-case scenario would be approximately 90 days! On the other hand, numbering sequentially, row by row, would give a half bandwidth of 502 − 1 = 501. The run time for that case would be only about 30 seconds. A contour plot of the stress distribution σY Y /σ0 is superposed on top of the plate geometry for the case of the course mesh back in Figure 6.21. The stress concentration factor, σYmax Y /σ0 , near the hole obtained with the course mesh is approximately 2.4. For the ﬁne mesh, the stress concentration factor turns out to be nearly 3.1. The analytical solution for the stress concentration factor has been reported by Howland [11] to be 3.14. Hence the ﬁne mesh

Continuum Finite Elements

(a)

185

(b)

FIGURE 6.25 Contour plot of σY Y /σ0 . Plane strain locked mesh with ν = 0.499 (a) and plane stress mesh with ν = 0.3.

gives excellent results. Exploring further, Figure 6.25 shows an X−Y plot of σY Y /σ0 versus X along the line Y = 0. Five output points per element were chosen to make the curves. The results for the coarse mesh are shown to the left (a) and the ﬁne-mesh results are shown to the right (b). For the coarse mesh there are six elements having an edge lying on the line Y = 0. Notice that the stress distribution along Y = 0 is piecewise continuous, but there are jumps at the nodal points. The presence of locations where the jump in stress is large suggests that mesh reﬁnement is needed in these regions. The curve obtained for the ﬁne mesh is signiﬁcantly improved, but a jump in the stress is still present at the location near the hole. At this location, a single node lies on the bottom edge, just adjacent to the element contiguous to the hole boundary. To conclude this study, just out of curiosity, let’s switch from plane stress to plane strain and run the problem again with the reﬁned mesh setting Pois son’s ratio equal to ν = 0.499. Recall that as ν approaches 1/2, a linearly elastic and isotropic material becomes incompressible. It can deform in a shear mode, but not volumetrically. A contour plot of σY Y /σ0 is shown in Figure 6.25(a). The plot exhibits a nonsensical “checkerboard” pattern and exhibits unreasonable values for the stress components. These symptoms in dicate that locking has occurred. As discussed in the previous analysis of a cantilever beam, locking can be avoided for the plane-strain, nearly incom pressible case by underintegrating the element, i.e., using 1 × 1 quadrature, and employing hourglass stabilization. The contour plot in Figure 6.25(b) is

186

Finite Element Method

z

FIGURE 6.26 Representative section of a copper core surrounded by a ceramic coating.

for the plane stress case with ν = 0.3. This plot shows the correct nature of the stress distribution.

6.9

Thermal stress analysis of a composite cylinder

To end this chapter, due to the practical importance of thermal stress prob lems, we perform a thermal stress analysis of a long composite cylinder com posed of an inner copper core and a ceramic coating. A representative section of the composite cylinder is shown in Figure 6.26. The diameter of the inner core is 0.1 m, and the thickness of the outer ceramic coating is also 0.1 m. In the analysis we will assume that the inner copper core is heated uniformly to a temperature of 200 ◦ C, and air at 20 ◦ C ﬂows over the outer surface of the composite, causing heat transfer by forced convection. In the analysis, we will take the ﬁlm coeﬃcient to be h = 10 J/(m2 ·◦ C · s), and we will assume that the inner copper core and the outer ceramic cylinder are perfectly bonded at the interface. After steady-state conditions are reached, we desire to determine the ther mal stress distribution in the composite. We will assume that the ultimate tensile strength of the ceramic coating is σu = 150 MPa. During the post processing phase of the analysis, we will address the question of whether or not we expect the coating to fail under the present circumstances. The mechanical properties for the copper and ceramic used in the analysis are reported in Table 6.13 below.

6.9.1

Heat conduction analysis

The ﬁrst step in the analysis is to determine the steady-state temperature distribution in the composite. Because both the geometry of the composite and the prescribed temperature distribution possess axial symmetry about the z axis, it is only necessary to perform a two-dimensional, axisymmetric

Continuum Finite Elements

187

TABLE 6.13

Mechanical properties for copper and ceramic. Material

E (GPa)

−6

α ( 10◦ C ) k ( m·◦JC·s )

ν

Copper

110

0.33

17

400

Ceramic

400

0.22

5

1

heat conduction analysis. In addition, because the prescribed temperature distribution does not vary in the z direction and because we have assumed that the composite cylinder is “long,” the solution for the temperature ﬁeld should not depend on z, except in the near vicinity of the ends. Therefore, it is only necessary to consider the quarter model of the representative section as shown in Figure 6.27. To explain the boundary conditions applied to the quarter model, a tem perature of T = 200 ◦ C is speciﬁed at all points inside the copper cylinder. To prevent heat from ﬂowing in the z direction, we specify a zero-ﬂux condition, i.e., qz = 0, on the top and bottom of the ceramic cylinder as shown in the ﬁgure. The forced convection condition, T = h(T − T∞ ), is applied along the right edge. Finite element implementation In the absence of an internal heat source, the weak form of the steady-state heat equation can be written over a single element as follows:

k Ωe

∇w · ∇T dΩ = −

Γe

wqn∗ dΓ

(6.136)

z

T = 200 ◦ C

qz = 0 ceramic T = h (T − T∞ )

qz = 0

r

FIGURE 6.27 Quarter model and associated boundary conditions for the heat conduction analysis.

188

Finite Element Method

where ∇w and ∇T represent the gradient of the test function and temperature ﬁeld, respectively, and qn∗ is the speciﬁed normal component of the heat ﬂux acting on Γe . The gradient operator in cylindrical coordinates can be written as ∂ 1 ∂ ∂ er + eθ + ez (6.137) ∇= ∂r r ∂θ ∂z where er , eθ , and ez are unit base vectors. In axisymmetric heat conduction problems, it is assumed that the temperature does not vary in the circumfer ential direction. The partial derivative of the temperature ﬁeld with respect to θ is then zero, and the components of the temperature gradient can be written as ⎧ ∂T ⎫ ⎪ ⎪ ⎬ ⎨ ∂r (6.138) ∇T = ⎪ ⎪ ⎩ ∂T ⎭ ∂z Let us now proceed and employ a four-node arbitrary quadrilateral ele ment for heat conduction analysis. The temperature ﬁeld over the four-node quadrilateral is interpolated using the bi-linear shape functions deﬁned in Section 6.2.3, equation (6.29). The interpolation can be written as T˜e = T˜e (r, z) = Ne Te

(6.139)

where Ne is a 1 × 4 matrix containing the bi-linear shape functions, and Te is a 4 × 1 vector containing the nodal temperatures. The ﬁnite element approximation for the temperature gradient is then ⎧ ⎫ ˜ ⎪ ⎪ ∂T ⎪ ⎪ ⎨ ⎬ ∂r (6.140) ∇T˜ = = B e Te ⎪ ⎪ ⎪ ∂T˜ ⎪ ⎩ ⎭ ∂z where Be is the 2 × 4 B-matrix containing partial derivatives of the shape functions with respect to the global r and z directions, i.e., N1,r N2,r N3,r N4,r Be (6.141) = 2×4 N1,z N2,z N3,z N4,z Using the Galerkin procedure, we now interpolate the test function over the element using the same shape functions that were employed for the tempera ture ﬁeld. Doing this yields w ˜e = Ne we ∇w ˜e = Be we

(6.142)

where we is a 4 × 1 vector containing the values of the test function at the nodes. By substituting the ﬁnite element approximations (6.140) and (6.142)

Continuum Finite Elements

189

into the weak form (6.136) we get BTe Be dΩ Te = −weT weT k Ωe

Γe

NTe qn∗ dΓ

(6.143)

Equation (6.143) above can be written succinctly as weT Ke Te = −weT fe

(6.144)

where sandwiched between weT and Te on the left-hand side is the 4 × 4 element matrix, Ke = k BTe Be dΩ (6.145) 4×4 Ωe

The element matrix for heat conduction applications is referred to as the conductance matrix. The element vector appears on the right-hand side of equation (6.144) and is given as NTe qn∗ dΓ (6.146) fe = − Γe

In the present problem, the normal component of the heat ﬂux vector, qn∗ , was speciﬁed to be zero on the top and bottom edges of the ceramic region, and hence, for elements having an edge lying on one of these boundaries, the element vector is zero. For elements with an edge lying on the line r = 0.15, the right boundary, the forced-convection condition needs to be enforced. On a forced-convection boundary, the normal component of the ﬂux is given as qn∗ = h (T − T∞ )

(6.147)

where h is the ﬁlm coeﬃcient, T is the unknown temperature at a point on the boundary, and T∞ is the temperature of the surrounding ﬂuid far away from the boundary. Recall that in the present problem we have chosen the remote temperature to be T∞ = 20 ◦ C. Substituting equations (6.145) through (6.147) into equation (6.144) and invoking the arbitrariness of the test function yields BTe Be dΩ + h NTe Ne dΓ Te = hT∞ NTe dΓ (6.148) k Ωe

Γe

Ωe

After assembling the element matrices and vectors following the usual ﬁnite element assembly procedure, the global system of equations to be solved can be written simply as KT = F (6.149) We see from equation (6.148) that the total contribution to the element matrix is composed of the conductance matrix, plus a contribution from the

190

Finite Element Method

elements that are contiguous to the forced-convection boundary. Only the elements that touch the outer surface of the ceramic get both contributions to the element matrix. In the present axisymmetric formulation, the element matrix and vector are evaluated by carrying out integrals over the ξ−η parent domain. This is accomplished by recognizing that a diﬀerential volume, dΩe , in the physical domain can be related to a diﬀerential volume in the parent domain as follows: dΩe = 2πr drdz = 2π(Na ra ) j dξdη

(6.150)

Note that the second line in equation (6.150) follows from the mapping be tween the parent domain and the physical domain, i.e., r = Na (ξ, η)ra z = Na (ξ, η)za

(6.151)

where the summation convention is implied, and ra and za are the r and z coordinates of local node a. The 4×4 element matrix can be evaluated exactly using a 2 × 2 quadrature rule. The line integrals that appear in equation (6.148) can be evaluated in the parent domain by replacing the diﬀerential surface area, dΓ, by the corre sponding area in the parent domain. For example, if local nodes 2 and 3 of an element lie on the forced-convection boundary, then dΓ would be expressed as $ 2 + z 2 dη (6.152) dΓ = 2πr r,η ,η where r,η and z,η are components of the Jacobian matrix, i.e., r z J = ,ξ ,ξ r,η z,η

(6.153)

The shape function matrix that appears inside the surface integrals would then have to be evaluated along the line ξ = 1, using a suﬃcient quadrature rule. The ﬁnite element mesh employed in the heat conduction analysis is com posed of 130 four-node axisymmetric quadrilaterals. It is shown in Fig ure 6.28(b). The mesh was designed with the subsequent thermal stress analysis in mind. In the thermal stress analysis, we expect a large discon tinuity in the stress components σzz and σθθ across the interface due to the mismatch in mechanical properties between the coper and ceramic. Hence, the mesh employed in the thermal stress analysis needs to be focused in this region. Actually, we could get away with using a much coarser mesh for the heat conduction analysis, but it is convenient to use the same mesh for both analyses. The nodal temperatures can then be directly transferred from the heat-transfer mesh to the structural mesh. The temperature distribution is needed to calculate the thermal strain distribution and resulting thermal stress distribution, as described in the next section.

Continuum Finite Elements

191

z

z ∗ =0 qn

ur = 0

uz = C

∗ = h (T − T ) qn ∞

t∗ = 0

r

uz = 0

(a)

∗ =0 qn

r

(b)

FIGURE 6.28 Finite element mesh for thermal stress analysis (a), and heat conduction anal ysis (b).

6.9.2

Thermal stress analysis

Now that the temperature distribution in the composite is known, let us proceed with the thermal stress analysis. To explain the ﬁnite element imple mentation for linear thermoelastic problems, we begin by writing down the discrete version of the weak form. Due to the absence of body forces and surface tractions in the present example, the statement can be written simply as BTe σ dΩ = 0 (6.154) Ωe

The next step is to substitute Hooke’s law for thermoelasticity problems given by equation (4.87) in Chapter 4 into equation (6.154). Hooke’s law for thermoelastic axisymmetric problems can be written as ⎤⎧ ⎡ ⎧ ⎫ ⎫ 1−ν ν ν 0 ⎪ σrr ⎪ ⎪ rr − αΔT ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎢ ν 1−ν ν ⎬ ⎨ ⎨ 0 ⎥ σθθ ⎥ θθ − αΔT ⎢ ˜ = E⎢ ν ν 1−ν 0 ⎥ ⎪ − αΔT ⎪ (6.155) ⎪ σzz ⎪ ⎦ ⎣ zz ⎪ ⎪ ⎪ 1 − 2ν ⎪ ⎭ ⎩ ⎭ ⎩ σrz γrz 0 0 0 2 where ˜= E

E (1 + ν) (1 − 2ν)

and α is the coeﬃcient of thermal expansion. The quantity ΔT is the temper ature change from a stress-free reference state. We emphasize that the strain

192

Finite Element Method

components on the right-hand side are the total strain components. Equation (6.155) above can be written compactly as σ C αΔT (6.156) = − 4×1 4×4 4×1 4×1 As discussed in Chapter 4, Section 4.6, the strain-displacement relationship for axisymmetric problems can be written as rr = ur,r 1 θθ = ur r zz = uz,z γrz = uz,r + ur,z

(6.157)

Armed with this relationship, we can now express the strain at a point within the element in terms of the element B-matrix and the nodal displacement vector. Doing this yields ˜e B d = 4 ×e8 8 ×e 1 (6.158) 4×1 The 4 × 8 B-matrix can be written in terms of the derivatives of the bi-linear shape functions with respect to r and z as follows: ⎡

0 N1,r ⎢1 ⎢ N 0 Be = ⎢ r 1,r ⎣ 0 N1,z N1,z N1,r

N2,r 0 1 N2,r 0 r 0 N2,z N2,z N2,r

N3,r 0 1 N3,r 0 r 0 N3,z N3,z N3,r

⎤ N4,r 0 1 ⎥ N4,r 0 ⎥ ⎥ r 0 N4,z ⎦ N4,z N4,r

(6.159)

Hence, the ﬁnite element approximation for the stress ﬁeld within an element can be written as ˜ e = Ce [Be de − αΔTe ] (6.160) σ Substituting equation (6.160) into equation (6.154) and rearranging terms gives T Be Ce Be dΩ de = α BTe Ce ΔTe dΩ (6.161) Ωe

Ωe

Upon assembly of the element matrices and vectors, the system of equations to be solved has the usual form Kd = F. From the right-hand side of equation (6.161) we see that the temperature ﬁeld within the element is required in order to evaluate the element vector. Given the temperature values at the nodal points, the temperature at a point inside the element is computed as follows: T˜e (ξ, η) = Na (ξ, η)Ta

(6.162)

Continuum Finite Elements

193

where Ta is the temperature at node a. Both integrals in equation (6.161) can be carried out in the parent domain as follows: 1 BTe Ce Be (Na ra )jdξdη Ke = 2π −1

fe = 2πα where

1 −1

BTe Ce ΔTe (Na ra )jdξdη

⎧ ⎫ Na Ta − Tref ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ Na Ta − Tref ΔT Na Ta − Tref ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0

(6.163)

(6.164)

is the vector containing the diﬀerence between the temperature at the quadra ture point and the stress-free reference temperature Tref . Note that here in the present axisymmetric analysis we have employed the relation dΩe = 2πrj dξdη = 2π(Na ra )j dξdη

(6.165)

Mechanical boundary conditions Before presenting the results, it is instructive to discuss the mechanical bound ary conditions to be applied to the representative section, as shown in Fig ure 6.28(a). As shown in the ﬁgure, the displacement component, uz , is constrained to be zero along the line z = 0. The plane z = 0 can be consid ered a plane of symmetry in the present problem, and hence the displacement in the z direction must be zero at all nodes lying on this plane. Due to the fact that the problem is axisymmetric, the nodes lying on the line r = 0 are constrained from moving in the r direction. Because we have assumed that the cylinder is “long,” the deformation in each representative composite section making up the cylinder must deform in an identical manner. The resulting deformation must also be compatible. For both of these conditions to be met, all points lying on the top and bottom surfaces of each section must displace the same amount in the z direction. The boundary condition, uz = C, along the top edge where C is an un known constant, is called a periodic boundary condition. If this condition were not enforced, the deformation between adjacent sections would end up being incompatible, such as that depicted in Figure 6.29. To explain how periodic boundary conditions can be implemented in a ﬁnite element code, suppose that the three elements shown in Figure 6.30 all have an edge lying on the top surface of the representative section. The top left node (5) is chosen to be the master node. Equations 9 and 10 correspond to the r and z degrees of freedom attached to this node. Now let the remaining nodes (6,7, and 8) lying on the top edge be slave nodes. The equation numbers

194

Finite Element Method

FIGURE 6.29 Incompatible deformation between adjacent cells.

associated with the r and z degrees of freedom of the slave nodes are (11,12), (13,14), and (15,16), as labeled in the ﬁgure. The next step is simply to change equation numbers 12, 14, and 16 to equation number 10. This, in eﬀect, ties the z degree of freedom associated with the slave nodes to the z degree of freedom of the master node. Once the equations are renumbered, the element matrices and vectors are assembled in the usual manner. Before the equations can be solved, however, it is necessary to zero the right-hand side of equations 12, 14, and 16 and set the pivot locations that are equal to zero on the left-hand side equal to one. The pivot positions K12 12 , K14 14 , and K16 16 will all be zero after assembly due to the fact that we removed equations 12, 14, and 16. Next, upon solution of the equation Kd = F, the z degrees of freedom associated with the slave nodes will all be zero. These degrees of freedom must subsequently be set equal to the z degree of freedom of the master node. After this displacement update, the strain in each element can calculated in the usual way, i.e., ˜e = Be de . Finally, the stress components are obtained from the modiﬁed version of Hooke’s law, ˜ e = Ce (˜e − αΔTe ). σ

z (9,10) 5

(11,12) (13,14) (15,16) 6 7 8

� � �1 � 1

� � �3 �

� � �2 � 2

3

4

r FIGURE 6.30 Implementation of periodic boundary conditions.

195

400

200

200

180

0

T (◦ C)

σθθ (MPa)

Continuum Finite Elements

-200 -400 -600 0

160 140 120 100

0.05

0.1 r (m)

0.15

80 0

0.05

0.1 r (m)

0.15

(a) (b) FIGURE 6.31 Hoop stress distribution (a), and temperature distribution (b).

6.9.3

Results

The results for the thermal stress analysis are shown in Figure 6.31(a) and Figure 6.32. The hoop stress, σθθ , is plotted versus the radial distance, r, in Figure 6.31(a). As shown in the ﬁgure, the hoop stress is constant and compressive within the inner copper region. There exists a sharp discontinuity in the hoop stress across the interface, where it suddenly attains a high tensile value of approximately 325 MPa and then gradually decreases as r approaches the outer boundary. Because the hoop stress exceeds the ultimate tensile stress of 150 MPa, we would expect the ceramic coating to fail. The failure would likely initiate in the form of radial cracks starting at the interface and then propagating outward until catastrophic failure.

(a)

(b)

FIGURE 6.32 Contour plot of the hoop stress in the ceramic coating (a), and deformed geometry when the periodic boundary condition along the top edge is not enforced (b).

196

Finite Element Method

The deformed geometry of the composite (magniﬁed 100×) is shown in Figure 6.32(a). The dashed lines represent the geometry of the undeformed composite. Just for the sake of interest, the deformed mesh without enforcing the periodic boundary condition is shown in Fig 6.32(b).

6.10

Problems

Problem 6.1 The nodal coordinates for a three-node triangle for steady-state heat conduc tion are given in Table 6.14 below. A uniform heat source, S = 30 J/(m3 · s), acts over the area of the element, and a uniform heat ﬂux distribution, qn = 10 J/(m2 · s), acts over edge 2–3. Assume that the thermal conductivity is k = 10 J/(m ·◦ C · s), and consider a unit thickness.

(a)

Derive the 3 × 3 element conductance matrix.

(b)

Derive the element vector.

TABLE 6.14

Problem 6.1. Nodal coordinates. N ode X Y 1 2 3

0.0 0.0 1.0 0.0 0.5 0.5

Problem 6.2 Consider a four-node isoparametric quadrilateral for plane-stress elasticity ap plications. The nodal coordinates, connectivity list, and nodal displacements are reported in Tables 6.15–6.17 below. Determine the stress components σXX , σY Y , and σXY at the location (ξ = 0, η = 0). Take E = 30 × 106 psi and ν = 0.3. Assume unit thickness.

Continuum Finite Elements

197 TABLE 6.15

Problem 6.2. Nodal coordinates in inches. Node X Y 20 22 26 27

1.5 0.0 0.0 1.0

1.5 1.0 0.0 0.0

TABLE 6.16

Problem 6.2. Connectivity list. Element 1 2 3 4 1

26

27

20

22

TABLE 6.17

Problem 6.2. Nodal displacements in inches. Node daX daY 20 22 26 27

0.00 −0.01 0.00 0.00

0.00 0.01 0.00 0.00

Problem 6.3 The nodal coordinates for a constant-strain triangle are given in Table 6.18 below. From a steady-state heat conduction analysis, the nodal temperatures are found to be T1 = 50 ◦ C, T2 = 70 ◦ C, and T3 = 45 ◦ C. If the stress-free reference temperature of the element is 20 ◦ C, determine the 6 × 1 equivalent nodal load vector. Assume plane strain, and take E = 200 GPa, ν = 0.3, and α = 17 × 10−6 /◦ C.

198

Finite Element Method TABLE 6.18

Problem 6.3. Nodal coordinates in meters. Node X Y 1 2 3

0.0 1.0 0.5

0.0 0.0 0.5

3 (0,1) 7

4 (1,1) (0.25,0.8) (0.8,0.7)

8 5 (0.2,0.3)

(0.7,0.25)

6 (0,0)

(1,0)

1

2

FIGURE 6.33 Problem 6.4. Patch test. Problem 6.4 Consider a mesh composed of ﬁve arbitrary quadrilateral elements for plane elasticity applications as shown in Figure 6.33. Suppose that the displacement components uX and uY are prescribed at nodes 1 through 4 according to the following equation: uX = α1 + α2 X + α3 Y

uY = β1 + β2 X + β3 Y

(6.166)

where α1 , . . . , β3 are coeﬃcients deﬁned in Table 6.19. TABLE 6.19

Problem 6.4. Coeﬃcients. α1 α2 α3 β1 0.0

0.002

0.005

0.0

β2

β3

0.003

0.001

Using a ﬁnite element program, run the problem for the case of plane strain taking E and ν to be any values that you like. Observation of the output should reveal that the displacement components at nodes 5 through 6 should

Continuum Finite Elements

199

obey the linear displacement ﬁeld deﬁned in equation (6.166), i.e., d5X = (0.0) + (.002)(0.2) + (0.005)(0.3) = 0.0019 d5Y = (0.0) + (.003)(0.2) + (0.001)(0.3) = 0.0009 etc. In addition, the strain components at each quadrature point within every element should be those obtained diﬀerentiating the displacement components in equation (6.166) with respect to X and Y , i.e., XX = uX,X = α2 = 0.002 Y Y = uY,Y = β3 = 0.001 γ12 = uX,Y + uY,X = α3 + β2 = 0.008 The ﬁnite element analysis carried out above is called a patch test. Because the arbitrary quadrilateral is capable of capturing a linear ﬁeld, passing of the test requires that the displacements at nodes 5 through 8, as well as the resulting strains at the quadrature points, be correct up to machine precision. Problem 6.5 The shape functions for the three-node triangle can be expressed in terms of the areas A1 , A2 , and A3 depicted in Figure 6.1. Compute the area coordinate A3 as follows: 1 A3 = X(1) × X 2 where X(1) = (X2 − X1 )e1 + (Y2 − Y1 )e2 , and X = (X − X1 )e1 + (Y − Y1 )e2 Show that N3 = A3 /A where N3 is the shape function deﬁned in equation (6.6). Problem 6.6 A uniform pressure of magnitude P is applied to one of the faces of an eightnode brick element. The coordinates of the local nodes are given in Table 6.20. The element faces are deﬁned such that the ﬁrst face lies on the plane ξ = −1 in the parent domain. The face deﬁnitions are given as Face Face Face Face Face Face

1: 2: 3: 4: 5: 6:

ξ = −1 ξ = +1 η = −1 η = +1 ζ = −1 ζ = +1

200

Finite Element Method TABLE 6.20

Problem 6.6. Local nodal coordinates for eight-node brick. Local node X Y Z 1 2 3 4 5 6 7 8

0.0 1.0 1.5 0.0 0.0 1.0 1.2 0.0

0.0 0.0 1.0 1.0 0.0 0.0 0.0 1.0

0.0 0.0 0.0 0.0 0.5 0.5 0.5 0.5

Determine the 24 × 1 element vector containing the equivalent nodal loads for the case when the pressure acts over the second face, i.e., where ξ = 1 in the parent domain. Problem 6.7 A three-node triangle is shown in Figure 6.34. The local node numbering and nodal coordinates (in inches) are labeled in the ﬁgure. A concentrated force of magnitude P = 1000 lb is applied at local node 2. Determine the displacement components d2X and d2Y if d1X = d1Y = d3X = d3Y = 0. Assume plane strain and take E = 30 × 106 psi and ν = 0.3. Y

(0,1)

3

P (0,0) (1,0) 1

2

X

FIGURE 6.34 Problem 6.7. Three-node triangle subjected to concentrated force.

Continuum Finite Elements

201

Problem 6.8 Compute the exact values of the following integrals using Gauss quadrature: %3 2 (a) (x + x) dx 0 %1 3 (b) (x + x5 ) dx −1 Problem 6.9 A one-dimensional, isoparametric quadratic-u element is shown in the parent domain in Figure 6.35. The shape functions in terms of the natural coordinate, ξ, are given as 1 ξ (ξ − 1) 2 N2 = 1 − ξ 2 1 N3 = ξ (ξ + 1) 2

N1 =

The nodal coordinates of the element in the physical domain are labeled in the ﬁgure. If the nodal displacements are given as d1 = 0, d2 = 1, and d3 = 2, plot the strain distribution (X) along the length of the element, where X is measured from the left end of the element. Does the result make physical sense?

1 ξ = −1

2

3

ξ=0

ξ=1

ξ

1 2 X=0 X = 0.25

3 X=1

X

FIGURE 6.35 Problem 6.9. Quadratic-u element in parent and physical domains.

7 Structural Finite Elements

The ability to simplify means to eliminate the unnecessary so that the neces sary may speak. —Hans Hofmann Most books on ﬁnite element analysis present structural ﬁnite elements be fore the continuum elements covered in the previous chapter. In the author’s opinion, it is beneﬁcial to introduce structural ﬁnite elements after contin uum elements. By doing so, the formulation of structural elements can begin by ﬁrst writing down the three-dimensional strain-displacement relationships, and then simplifying these relationships by making kinematic assumptions. The present approach requires knowledge of isoparametric ﬁnite elements dis cussed in the previous chapter. Kinematic assumptions are restrictions placed on the deformation of a solid based on sound engineering judgement. For example, in the study of axially loaded bars, it is assumed that the bar can only resist axial forces. This assumption, in turn, prompts us to assume that the member can only undergo axial extension or compression and not, for example, bending. All derivations of structural elements begin with such kinematic assumptions. Structural elements are a very powerful tool. The simplifying assumptions often enable engineers to analyze very complicated structures that otherwise could not be modeled using continuum elements due to the high computational cost. In the present chapter, we will begin our study of structural elements by considering the simplest case, the space truss. We will then go on to EulerBernoulli beams, and then to Mindlin-Reissner plates. The chapter is pre sented within the framework of small-strain kinematics.

7.1

Space truss

The main assumption in the development of the space truss is that the element can only extend or shorten in the axial direction. It is also assumed that the only nonzero stress in the element is the normal stress acting perpendicular to any given cross section. The normal stress distribution acting on each cross section is assumed to be constant. These assumptions prohibit the element

203

204

Finite Element Method 2

X

r 1

Y 1

X 2

FIGURE 7.1 Space truss in parent and physical domains.

from resisting shear and bending deformation. We begin the formulation by considering a two-node space truss as shown in Figure 7.1. The parent domain of the element is chosen to be parallel to the X axis. The length, L, and cross-sectional area, A, of the element in the parent domain correspond to the length and area of the undeformed element. To study the kinematics of the space truss, it is necessary to study the motion of points that lie on the X axis in the parent domain. These points can be mapped onto an arbitrary line in the physical domain using the following one-dimensional shape functions: 1 1− N1 (X) = 2 1 1+ N2 (X) = 2

2X L 2X L

(7.1)

Notice that while the parent domain is a one-dimensional domain deﬁned in the interval (−L/2 ≤ X ≤ L/2), the physical domain is a line in threedimensional space. Introducing the natural (dimensionless) coordinate, ξ = 2X/L, the shape functions can be expressed as 1 (1 − ξ) 2 1 N2 (ξ) = (1 + ξ) 2 N1 (ξ) =

(7.2)

The mapping of the parent domain into the physical domain can now be

Structural Finite Elements

205

written as X = N1 X1 + N2 X2 Y = N1 Y1 + N2 Y2 Z = N1 Z1 + N2 Z2

(7.3)

By writing equation (7.3) in matrix notation we get ⎧ ⎫ ⎪ ⎪ ⎪ X1 ⎪ ⎪ ⎪ ⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎪ Y1 ⎪ ⎪ ⎪ N1 0 0 N2 0 0 ⎨ ⎨X ⎬ ⎬ Z 1 Y = ⎣ 0 N1 0 0 N2 0 ⎦ ⎩ ⎭ ⎪ ⎪ X2 ⎪ 0 0 N1 0 0 N2 ⎪ Z ⎪ ⎪ ⎪ ⎪ ⎪ Y2 ⎪ ⎪ ⎪ ⎭ ⎩ Z2 Xe = Ne Pe

or

(7.4)

Here in equations (7.3) and (7.4), Xa , Ya , and Za represent the coordinates of node a in the physical domain. These coordinates (six of them) are conve niently stored in a 6 × 1 vector, Pe . For the sake of convenience, we deﬁne the local coordinate, r, to be the distance along the centerline of the element from local node 1 to some point Xe along the length of the element, as shown in Figure 7.1.

7.1.1

Strain-displacement relationship

Having deﬁned the mapping between the parent and physical domains, it is now necessary to ﬁnd a relationship between the extensional strain, rr , at each point along the length of the element and the nodal displacements. The displacement vector, u, at some location, r, along the length of the space truss is shown in Figure 7.2. The displacement vector at this location is the diﬀerence between the position vector x in the deformed conﬁguration minus the position vector X of this same point in the undeformed conﬁguration. The displacement vector can be written in terms of Cartesian base vectors as u = uX eX + uY eY + uZ eZ

(7.5)

where uX , uY , and uZ are the components of u labeled in Figure 7.2. The projection of u onto the local r axis is obtained by taking the dot product of u and the unit base vector er , i.e., ur = er · u = muX + nuY + luZ

(7.6)

206

Finite Element Method uZ u

uY 2 uX

er Z Y

1

X

X

FIGURE 7.2

Projection of the displacement vector u onto the local r axis.

where the quantities m, n, and l are the direction cosines between the local r

axis and the X, Y , and Z axes. In other words,

m = er · eX n = er · eY l = er · eZ

(7.7)

We can write equation (7.6) in matrix notation as

where

u r = AT u

(7.8)

AT = m n l

(7.9)

is a 1 × 3 matrix containing the direction cosines, and u is a 3 × 1 vector containing the displacement components uX , uY , and uZ . As discussed in Chapter 4, the extensional strain, rr , at a point along the length of the space truss is the rate of change of ur with respect to a movement in the r direction, i.e., dur (7.10) rr = dr In the present analysis of the two-node space truss, the strain-displacement relation (7.10) gives good accuracy when the axial extension of the truss is small enough to be within the linear region of the uniaxial stress-strain curve of the material, and when the rotation of the truss is “small.” A discussion of the error associated with the strain-displacement relation (7.10) as a function of rotation angle is presented in an example at the end of this section.

Structural Finite Elements

207

To proceed further, let us now introduce a 6 × 1 matrix, pe , that contains the nodal coordinates of the element in some deformed conﬁguration, i.e., ⎧ ⎫ ⎪ ⎪ x1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ y1 ⎪ ⎪ ⎨ ⎬ z1 pe = (7.11) ⎪ ⎪ x2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ y2 ⎪ ⎪ ⎩ ⎭ z2 If we subtract Pe from pe , we get, de , the vector containing the nodal dis placements. The ﬁnite element approximation for the displacement vector, ˜ e , at some point r along the element is related to de through the shape u function matrix as follows: ˜ e = Ne de (7.12) u where Ne is the 3 × 6 shape function matrix deﬁned in (7.1). The relationship (7.10) between the extensional strain and the displacement component ur can now be written in terms of the shape function matrix and the nodal displacements. Doing this yields dur ⇒ dr d T {˜rr }e = A Ne de dr dNe de = AT dr = Be de rr =

(7.13)

where Be is the 1 × 6 B-matrix yet to be determined. To complete the derivation of the strain-displacement relationship for the two-node space truss element, we need to diﬀerentiate the shape function matrix with respect to r. The technique for doing this makes use of the chain rule. The technique can be explained clearly by introducing a position vector Xe (ξ) as shown in Figure 7.3. This vector is directed from the origin to a point in the physical domain that occupies position ξ in the parent domain. Let us now introduce another position vector, Xe (ξ + Δξ), pointing from the origin to the point in the physical domain that occupies location ξ + Δξ in the parent domain. If we now look at the diﬀerence between these two vectors and divide by Δξ, in the limit as Δξ goes to zero, we get lim

Δξ→0

dXe Xe (ξ + Δξ) − Xe (ξ) = Δξ dξ

(7.14)

Note that dXe /dξ is a vector that points in the local r direction. The mag nitude of this vector, |dXe /dξ|, is the rate of change of r with respect to a

208

Finite Element Method

Xe (ξ + Δξ) Xe (ξ)

1

ξ

2 ξ + Δξ

FIGURE 7.3

The position vectors Xe (ξ) and Xe (ξ + Δξ).

movement in the ξ direction, i.e., dXe dr dξ = dξ

(7.15)

Next, invoking the chain rule, we write dNe dNe dr = ⇒ dξ dr dξ dNe dNe /dξ = dr dr/dξ

(7.16)

Referring to the last line in (7.16), the numerator on the right-hand side is easy to evaluate, since the expressions for the shape functions are given explicitly in terms of ξ in equation (7.2). The denominator, however, requires some further thought. Since dr/dξ = |dXe /dξ|, we can write dXe dNe = Pe dξ dξ

⎧ ⎫ X1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎤⎪ ⎡ Y ⎪ ⎪ 1 ⎪ ⎪ 0 0 dN2 /dξ 0 0 dN1 /dξ ⎨ ⎬ Z 1 ⎣ ⎦ 0 dN1 /dξ 0 0 dN2 /dξ 0 = ⎪ X2 ⎪ ⎪ 0 0 dN2 /dξ ⎪ 0 0 dN1 /dξ ⎪ ⎪ ⎪ Y2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Z2

Structural Finite Elements Hence,

209

⎧ ⎫ ⎪ ⎪ ⎪ X1 ⎪ ⎪ ⎪ ⎪ ⎡ ⎤⎪ ⎪ ⎪ ⎪ Y1 ⎪ −1/2 0 0 1/2 0 0 ⎨ ⎬ dXe Z1 0 1/2 0 ⎦ = ⎣ 0 −1/2 0 ⎪ X2 ⎪ dξ ⎪ 0 0 −1/2 0 0 1/2 ⎪ ⎪ ⎪ ⎪ Y2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ Z2 ⎫ ⎧ X − X1 ⎬ 1⎨ 2 Y2 − Y1 = ⎭ 2⎩ Z2 − Z1

(7.17)

The magnitude of dXe /dξ can be computed as follows: dXe dr dξ = dξ 1 = (X2 − X1 )2 + (Y2 − Y1 )2 + (Z2 − Z1 )2 2 L (7.18) = 2 where, again, L is the length of the undeformed element in the parent domain. Therefore, the derivative of the shape function matrix with respect to r can be written simply as dNe 2 dNe = (7.19) dr L dξ Referring back to equation (7.13), the extensional strain in the r direction can now be written as dNe 2 de {˜rr }e = AT L dξ = Be de (7.20) where Be is the 1 × 6 B-matrix for the two-node space truss. After diﬀerenti ating the shape function matrix with respect to ξ and performing the required matrix multiplications, the B-matrix can ﬁnally be expressed as ⎡ ⎤ −1/2 0 0 1/2 0 0 2 m n l ⎣ 0 −1/2 0 0 1/2 0 ⎦ Be = L 0 0 −1/2 0 0 1/2 1 −m −n −l m n l = (7.21) L where m = (X2 − X1 )/L n = (Y2 − Y1 )/L l = (Z2 − Z1 )/L

(7.22)

210

7.1.2

Finite Element Method

Element matrix

Having obtained the strain-displacement relationship for the two-node space truss in the previous section, we can now derive the ﬁnite element matrices and vectors for the element by starting with the weak form of the equilibrium equation discussed in Chapter 4. In this section, we will follow the practice of expressing the weak form of the equilibrium equation over a single element. Doing this, we will see that the element matrix emerges from the internal virtual work (the left-hand side of the weak form), and the element vector comes from the external virtual work (right-hand side). After deriving the el ement matrix and vector, we will be in a position to solve problems involving assemblages of space truss elements. The process of solving such problems involves following the ﬁnite element assembly procedure covered in Chapter 3, and solving systems of linear algebraic equations. Once the linear alge braic equations are solved, the stress and strain in each truss element can be calculated as a post-processing step, once the nodal displacements are known. To begin, the general expression for the principle of virtual work over a single element is written as σij wi,j dΩ = bi wi dΩ + ti wi dΓ (7.23) Ωe

Ωe

Γe

Here in equation (7.23), σij are the Cauchy stress components, wi are the components of a vector-valued test function, and bi are the components of the body force vector acting at points inside the element volume, Ωe . The quantity ti represents the traction components acting on the outer surface, Γe . It is worth mentioning here that the quantities σij , wi , bi , and ti are functions of the position vector Xe pointing from the origin to a point along the length of the truss in the physical domain. In the following discussion, for the sake of simplicity, we will ignore the body force. Let us now specialize the principle of virtual work statement for the case of the space truss element. Due to the cylindrical geometry of the space truss, the volume integrals in equation (7.23) can be replaced by line integrals. This can be done by recognizing that dΩe = Ae dr where Ae is the cross-sectional area of the element in the physical domain, and dr is a diﬀerential axial segment of the truss. In addition, because we have assumed that the only nonzero stress component is the normal stress component, σrr , the internal virtual work, δWint e , can be written as δWint e = Ae

Le /2

−Le /2

σrr wr,r dr

where Le is the length of the element in the physical domain.

(7.24)

Structural Finite Elements

211

It is now convenient to express the integral in equation (7.24) as an equiva lent integral over the parent domain through a change of variables. Using the following relationships: Le dξ 2 = Ee rr

dr = σrr

(7.25)

the integral can be written as δWint e = Ae Ee

Le 2

1

−1

rr wr,r dξ

(7.26)

where Ee is the Young’s modulus of the truss (assumed to be constant along the length of the element). We are now in a position to substitute the ﬁnite element approximations for the extensional strain and the test function into the internal work expression (7.26). Recalling equation (7.20), the ﬁnite element approximation for the extensional strain is {˜ rr }e = Be de (7.27) Similarly, if we interpolate the test function, w, along the length of the ele ment using the same linear shape functions as were used to interpolate the displacement ﬁeld, we can write wr = AT w ⇒ {w ˜r }e = AT Ne we

(7.28)

and hence, dNe we dr = Be we

{w ˜r,r }e = AT

(7.29)

Substituting the ﬁnite element approximations (7.27) and (7.29) into the in ternal work statement (7.26) gives T δWint e = we

Le 1 T Be Be dξ de Ae Ee 2 −1

(7.30)

We now recognize that the element matrix is the 4 × 4 matrix inside the curly brackets in (7.30) above, i.e., Ke =

Ae Ee Le 2

1

−1

BTe Be dξ

(7.31)

212

Finite Element Method

The integral in equation (7.31) can be carried out as follows: ⎡ ⎤ −m ⎢ ⎥ 1 ⎢ −n ⎥ ⎢ ⎥ Ae Ee Le 1 ⎢ −l ⎥ −m −n −l m n l dξ Ke = ⎢ ⎥ 2 2 Le −1 ⎢ m ⎥ ⎣ n ⎦ l and hence,

7.1.3

⎤ m2 mn ml −m2 −mn −ml ⎢ mn nl −mn −n2 −nl ⎥ n2 ⎥ ⎢ ⎢ nl l2 −ml −nl −l2 ⎥ Ke = Ae Ee ⎢ ml ⎥ 2 2 4×4 mn ml ⎥ Le ⎢ ⎥ ⎢ −m −mn2 −ml m ⎣ −mn −n −nl mn n2 nl ⎦ ml nl l2 −ml −nl −l2

(7.32)

⎡

(7.33)

Element vector

As mentioned in the previous section, the element vector is derived from the external virtual work expression. In the absence of body forces, the external virtual work for a single element can be written as ext ti wi dΓ (7.34) δWe = Γe

where ti are the components of traction vector, wi are the components of the test function, and Γe represents the outer surface of the truss. For the case of the truss element, we now replace the traction distribution around the perimeter of the truss by the net force per unit length, q, acting at some location, r, along the length of the truss. If we restrict our attention to the case where the distributed load is uniform along the length of the truss, then the external virtual work expression becomes Le ext qws dr (7.35) δWe = 0

where q is the magnitude of the distributed load, and ws is the component of the test function in the local s direction. As shown in Figure 7.4, the local s axis lies in the same plane and acts in the same direction as the distributed load, and it is perpendicular to the axial r direction. The component of the test function ws can be obtained by taking the dot product of the unit base vector es and the test function vector w, i.e., ws = es · w = mm wx + nn wy + ll wz = TT w

(7.36)

Structural Finite Elements

213 Y

q

X s

Z FIGURE 7.4 Truss element subjected to uniform distributed load q.

Here in equation (7.36), the quantities mm, nn, and ll are the direction cosines between the ﬁxed Cartesion base vectors eX , eY , and eZ and the base vector es . In other words, mm = es · eX nn = es · eY ll = es · eZ The quantity TT in equation (7.36) is the transpose of a transformation matrix that contains the direction cosines, i.e., ⎡

⎤ mm T = ⎣ nn ⎦ ll

(7.37)

Assuming that the direction and magnitude of the distributed load are both known, then the unit base vector es and hence the direction cosines can be readily calculated. The last step in the derivation of the element vector for the two-node space truss acted upon by a uniform distributed load is to substitute equation (7.36) into (7.35) and interpolate the test function w along the length of the element

214

Finite Element Method

using the linear shape functions deﬁned in (7.2). Doing this gives δWext e

Le

qTT Ne we dr r

T T = we qNe T dr

=

0

0

= weT

1

−1

qNTe T

Le dξ 2

(7.38)

where we is the vector containing the nodal values of the test function. Note that in the last line of equation (7.38) above, we have performed a change of variables, dr = (Le /2) dξ, where Le is the length of the element. We now identify the element vector as the 6 × 1 vector inside the curly brackets in (7.38), i.e., qLe 1 NTe T fe = dξ (7.39) 6×1 2 −1 6 × 3 3 × 1 After performing the required matrix multiplication in the integrand of (7.39) and carrying out the integral we obtain ⎡ ⎤ 1/2(1 − ξ) 0 0 ⎥⎡ ⎢ ⎤ 0 1/2(1 − ξ) 0 ⎥ mm 1⎢ ⎥ ⎢ qLe 0 0 1/2(1 − ξ) ⎥ ⎣ ⎢ ⎦ fe = ⎥ nn dξ 1/2(1 + ξ) 0 0 2 −1 ⎢ ⎢ ⎥ ll ⎦ ⎣ 0 1/2(1 + ξ) 0 0 0 1/2(1 + ξ) ⎡ ⎤ mm ⎢ nn ⎥ ⎢ ⎥ ⎥ qLe ⎢ ⎢ ll ⎥ = (7.40) ⎥ ⎢ 2 ⎢ mm ⎥

⎣ nn ⎦ ll As a closing remark, if a uniform body force acts over the element, the same procedure as that employed for the case of a distributed load is used to get the body force contribution to the element vector. The body force contribution is identical to the distributed load contribution, except that the quantity q is replaced by Ab, where A is the cross-sectional area of the truss, and b is the magnitude of the body force. Note that the quantity Ab has units of force per unit length. For the case of the body force, the quantities mm, nn, and ll would then refer to the direction cosines between the body force vector and the Cartesian base vectors. The total element vector, when both a traction distribution and a body force distribution act over the element, is simply the sum of the individual element vectors, i.e., fe = feq + feb .

Structural Finite Elements

215 P = 1000 lb

Y

� � �1 �

1

3

� � �2 � 2

X

FIGURE 7.5 Assemblage of space trusses.

Example 7.1 To demonstrate the use of space truss elements for structural applications, let us consider an assemblage of axially loaded members connected by pinned joints as shown in Figure 7.5. The joints, labeled 1 and 2 in the ﬁgure, are prevented from translating in all three directions. A point load (P = 1000 lb) is applied at joint 3, which is free to move in any direction. Because both members lie in the same plane, and because the load acts in this plane, the problem can be treated as two-dimensional. The goal of this exercise is to perform a hand calculation to obtain the displacement of joint 3. In addition, we seek to ﬁnd the stress and strain in each axially loaded member. To begin, we will assume that each member is linearly elastic with Young’s modulus E = 30 × 106 psi, and we will take the cross-sectional area for each member to be A = 2 in.2 . The coordinates of each joint are given in Table 7.1 below. The procedure for solving the problem involves several steps. The ﬁrst step is to write down the element matrix for each truss element and assem ble the global system matrix. Because we are treating the problem as twodimensional, the order of the element matrices will be 4 × 4, and the order of the global matrix will be 6 × 6. Because joints 1 and 2 are constrained from moving in the X and Y directions, the reduced global matrix, after enforcing the boundary conditions, will be 2 × 2. TABLE 7.1

Nodal coordinates for the truss assemblage. Joint X (in.) Y (in.) 1 2 3

0.0 30.0 14.0

0.0 -1.0 14.0

216

Finite Element Method

We will model the assemblage shown in Figure 7.5 with two linear space truss elements. Each space truss has cross-sectional area A and Young’s mod ulus E. We will denote the length of each element as L1 and L2 , respectively. The element matrix for the two-dimensional truss element can be obtained from the three-dimensional matrix (7.33) by crossing out the rows and columns that correspond to the ﬁxed degrees of freedom, i.e., row three, column three, row six, and column six. The result is ⎡

⎤ m2 mn −m2 −mn A e Ee ⎢ n2 −mn −n2 ⎥ ⎢ mn2 ⎥ Ke = ⎣ −m −mn m2 mn ⎦ Le n2 −mn −n2 mn

(7.41)

Using the global node numbering shown in Figure 7.5, the connectivity in formation for each element is summarized in Table 7.2. The local-to-global node-numbering scheme is such that for element 1, local node 1 corresponds to global node 1, and local node 2 corresponds to global node 3. For element 2, local node 1 refers to global node 3, and local node 2 is attached to global node 2. Notice that, for each element in the model, the unit vector er points in the direction from local node 1 to local node 2. The direction cosines, m and n, for each element are the projections of this unit vector onto the X and Y axes, i.e., m1 n1 m2 n2

= e1r · eX = X21 /L1 = e1r · eY = Y21 /L1 = e2r · eX = X21 /L2 = e2r · eY = Y21 /L2

(7.42)

where e1r and er2 are the unit vectors pointing in the axial direction for elements 1 and 2, respectively, and m1 , n1 , m2 , and n2 are the direction cosines. Here in equation (7.42) the notation X21 , and Y21 represents the diﬀerences X2 −X1 and Y2 − Y1 , respectively, where Xa and Ya are the X and Y coordinates of local node a attached to the element of interest.

TABLE 7.2

Connectivity list. Element Local node 1 Local node 2 � � 1 3 ��1 �� 3 2 �2 �

Structural Finite Elements

217

The element matrices can now be written as ⎡

K1 =

AE (L1 )3

K2 =

AE (L2 )3

⎤ 196 196 −196 −196 1 ⎢ ⎥ ⎢ 196 196 −196 −196 2 ⎥ ⎢ −196 −196 196 196 5 ⎥ ⎢ ⎥ ⎣ −196 −196 196 196 6 ⎦ 1 2 5 6 ⎡ ⎤ 256 −240 −256 240 5 ⎢ ⎥ ⎢ −240 225 240 −225 6 ⎥ ⎢ −256 240 256 −240 3 ⎥ ⎢ ⎥ ⎣ 240 −225 −240 225 4 ⎦ 5 6 3 4

In the element matrices above, the global equation numbers that are asso ciated with each row and column of the element matrices are labeled in the margins. The global equation numbers are obtained from the connectivity information reported in Table 7.2. For example, because there are two un knowns (degrees of freedom) at each node, there are two global equations (2n − 1 and 2n) associated with each global node number n. In other words, focusing on element one, local node one corresponds to global equations one and two, while local node two corresponds to global equations ﬁve and six. Having labeled the rows and columns of the element matrices, we can now go ahead and assemble the global matrix. Just as a reminder, the elements of the global matrix are obtained by summing the element contributions as follows:

KIJ =

nel

e Kij

(7.43)

e=1

where I and J are the global equation numbers associated with local equations i and j that belong to element e. For example, the quantity that occupies row ﬁve and column ﬁve of the global matrix is 1 2 + K11 K55 = K33 196AE 256AE = + 3 (L1 ) (L2 )3

(7.44)

When the assembly process is ﬁnished, upon substitution of the numerical values for A, E, L1 , and L2 , we get the following for the global system matrix

218

Finite Element Method

(to six digits beyond the decimal place): K = K1 + K2 ⎡

.151523 .151523 .000000 ⎢ .151523 .151523 .000000 ⎢ ⎢ .000000 .000000 .000000 =⎢ ⎢ .000000 .000000 .000000 ⎣ −.151523 −.151523 .000000 −.151523 −.151523 .000000

⎡

.000000 ⎢ .000000 ⎢ ⎢ .000000 +⎢ ⎢ .000000 ⎣ .000000 .000000

⎤

.000000 −.151523 −.151523 .000000 −.151523 −.151523 ⎥ ⎥ .000000 .000000 .000000 ⎥ ⎥ .000000 .000000 .000000 ⎥ .000000 .151523 .151523 ⎦ .000000 .151523 .151523

⎤

.000000 .000000 .000000 .000000 .000000 .000000 .000000 .000000 .000000 .000000 ⎥ ⎥ .000000 .145604 −.136504 −.145604 .136504 ⎥ ⎥ .000000 −.136504 .127972 .136504 −.127972 ⎥ ⎦ .000000 −.145604 .136504 .145604 −.136504 .000000 .136504 −.127972 −.136504 .127972

⎡

⎤

.151523 .151523 .000000 .000000 −.151523 −.151523 ⎢ .151523 .151523 .000000 .000000 −.151523 −.151523 ⎥ ⎢ ⎥ ⎢ .000000 .000000 .145604 −.136504 −.145604 .136504 ⎥ =⎢ ⎥ ⎢ .000000 .000000 −.136504 .127972 .136504 −.127972 ⎥ ⎣ −.151523 −.151523 −.145604 .136504 .297127 .015019 ⎦ −.151523 −.151523 .136504 −.127972 .015019 .279495 7

× 10

lb in.

The second step in the solution procedure is to form the right-hand-side vector, enforce the essential boundary conditions, and solve for the nodal dis placements. In the present example, since there are no body forces or external tractions present, the right-hand-side vector just contains the external point load applied at node 3. The vector also contains four unknown reactions as sociated with the ﬁxed degrees of freedom. Hence, the external force vector in the present problem is ⎧ ⎫ ⎪ R1X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ R1Y ⎪ ⎪ ⎪ ⎬ ⎨ R2X (7.45) f= ⎪ R2Y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎭ ⎩ −P The ﬁxed conditions at nodes 1 and 2 are enforced by crossing out the cor responding rows and columns of the unconstrained system. Doing this gives the following reduced system of equations to be solved: 2971270 150190 d3X 0 = (7.46) 150190 2794950 d3Y −1000 Finally, solving for the displacements at node 3 yields d3X = 1.81 × 10−5 in.;

d3Y = −3.59 × 10−4 in.

Structural Finite Elements

219

TABLE 7.3

Stress and strain results for the truss assemblage. Element � � ��1 �� �2 �

× 10−7

σ (psi)

−122

−365

−118

−354

Now that the nodal displacements are known, we can go ahead and compute the strain and stress in each element as a post-processing step. Recalling the relationship (7.20) between the extensional strain in the element and the local nodal displacement vector, we obtain the following for the strain in element 1: 1 = B1 d1

⎧ ⎫ d1X ⎪ ⎪ ⎪ ⎪ ⎨ d1Y ⎬ 1 −m1 −n1 m1 n1 = d3X ⎪ ⎪ L1 ⎪ ⎪ ⎭ ⎩ d3Y

⎫ 0 ⎪ ⎪ ⎬ √ √ √ 1 √ 0 = √ − 2/2 − 2/2 2/2 2/2 −5 1.81 × 10 ⎪ ⎪ 14 2 ⎪ ⎪ ⎭ ⎩ 3.59 × 10−4 ≈ −12.2 × 10−6

⎧ ⎪ ⎪ ⎨

(7.47)

The normal stress in element 1 is now obtained by simply multiplying the extensional strain by Young’s modulus, i.e., σ1 = E1 1 ≈ −365 psi. Note that the minus sign in front of the strain and stress indicates that element 1 is in a state of compression. The strain and stress in element 2 is found using the same procedure. The results for both elements are summarized in Table 7.3. We emphasize that the strain (and hence the stress) along the entire length of the two-node space truss element is constant. For this reason, commercial ﬁnite element packages typically report the values of stress and strain at just the centroid of the each individual space truss element.

Example 7.2 In the derivation of the strain-displacement relationship for the two-node space truss, we assumed that the extensional strain was equal to the derivative of the axial displacement, ur , with respect to r — the axial direction of the truss in its original conﬁguration. In this example, we investigate the range of validity of that assumption by examining the strain in the truss as it undergoes a general rigid-body rotation.

220

Finite Element Method

To begin, let us consider the problem of a two-node truss element as shown in Figure 7.6. The element having original length, L, lies in the X−Y plane and is parallel with the X axis in its undeformed state. The element is subjected to a stretch λ = l/L = 1.01 (7.48) where l is the ﬁnal length of the element. Keeping the stretch ﬁxed, the element is then rotated about the Z axis. We wish to determine the angle, θ, that the bar can be rotated at which the error in the strain computation using equation (7.20) reaches 10 %. After the initial stretch of the element, the extensional strain is 0 = λ−1 = 0.01. The nodal displacement vector in the deformed conﬁguration is given as ⎧ ⎫ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 0 de = (7.49) l cos θ − L ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ l sin θ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0 and the B-matrix associated with the undeformed conﬁguration is 1 −1 Be = 0 0 0 0 L L

(7.50)

The strain in the element as computed from equation (7.20) is then = Be de = λ cos θ − 1

Y θ X FIGURE 7.6 Truss element subjected to a stretch and rigid body rotation.

(7.51)

Structural Finite Elements

221

1 λ = 1.03

0.98 0.96

λ = 1.02

0.94

/0

λ = 1.01

0.92 0.9 0.88 0.86 0

0.01

0.02

θ

0.03

0.04

0.05

FIGURE 7.7 Plot of error in strain versus angle of rotation for several values of initial stretch.

A plot of the extensional strain, , in the truss obtained from equation (7.20) (normalized with respect to 0 = λ − 1) is plotted in Figure 7.7. For the case λ = 1.01, the error in the strain calculation reaches 10 % at an angle of only θ = 0.045 radians ≈ 2.6 degrees. As shown in the ﬁgure, the error decreases as the stretch increases.

7.2

Euler-Bernoulli beams

In this section, we present the formulation for a two-node Euler-Bernoulli beam element. The element is extremely useful for the analysis of beams and framed structures. In order to simplify the formulation, we will restrict our attention to two dimensions, and assume that deformation can only occur in the X −Y plane. In addition, we will assume that all loads act in the X −Y plane, and the beam’s cross section is symmetric about the Y axis so as to prevent out-of-plane deﬂection. After deriving the element matrices and vectors for the Euler-Bernoulli beam element, we will present some examples illustrating the proper use of the element at the end of this section.

222

7.2.1

Finite Element Method

Kinematic assumptions

When a beam is subjected to bending, as shown in Figure 7.8, a portion of the beam experiences positive extensional strain (tension) and a portion ex periences negative extensional strain (compression). Because the extensional strain makes a transition from tension to compression, there must exist a sur face somewhere throughout the height of the beam on which the extensional strain is zero. That surface is called the neutral surface. The intersection of the neutral surface and the cross section of the beam is called the neutral axis. In the present analysis, the coordinate axes are chosen such that the X direction points in the axial direction of the beam (to the right), and the Y direction is chosen positive downward. The Z axis then goes into the page to form a right-handed coordinate system. Before any loads are applied, it is assumed that any given cross section along the length of the beam is planar and perpendicular to the neutral surface. The main kinematic assumption in Euler-Bernoulli beam theory is that cross sections must remain planar and perpendicular to the neutral surface. This kinematic assumption is illustrated in Figure 7.9. Having made the kinematic assumptions, we can now write down the straindisplacement relationship. To this end, let us focus on point A labeled in Figure 7.9. Before deformation, point A is located some distance, Y , from the neutral surface (here the distance is negative because point A lies above the neutral surface). Because we have assumed that the beam is only subjected to transverse loads (loads parallel to the Y direction) and that cross sections remain planar and perpendicular to the neutral surface, the cross sections depicted in Figure 7.9 move downward and rotate clockwise about the Z axis. The displacement, uX , of point A in the X direction is then uX = −Y sin θ

(7.52)

where θ is the angle measured positive clockwise from the vertical undeformed

Z X

neutral surface neutral axis

Y FIGURE 7.8 Neutral surface and neutral axis of a beam subjected to bending.

Structural Finite Elements

223

A

X Y

uX

A

θ

Y

FIGURE 7.9 Kinematic assumptions. cross section to the rotated cross section. If we restrict our attention to small rotations, then sin θ ≈ θ, and we get uX = −Y θ

(7.53)

In the present two-dimensional formulation, the vertical deﬂection will be denoted as uY = v. The theory assumes that the vertical deﬂection is a function of X only, i.e., v = v(X). The extensional strain in the X direction can now be evaluated by diﬀer entiating the displacement component uX with respect to X, i.e., XX = uX,X = −Y θ,X

(7.54)

Similarly, the transverse shear strain γXY can be computed as γXY = uX,Y + uY,X = −θ + v,X = −θ + θ = 0

(7.55)

The last line in equation (7.55) follows from the fact that the slope of the deﬂection curve, v,X , is equal to the tangent of the angle θ, i.e., v,X = tan θ

(7.56)

When the angle θ is much smaller than unity, then tan θ ≈ θ and hence, the rotation, θ, depicted in Figure 7.9 is approximately equal to the slope of the deﬂection curve.

224

Finite Element Method θ1 v1

L

θ2 v2

X

FIGURE 7.10 Two-node Euler-Bernoulli beam element.

We conclude that Euler-Bernoulli beam theory does not allow for the pos sibility of transverse shear strains. In addition, we note that the extensional strain, XX , varies linearly with the distance, Y , from the neutral surface, and the extensional strain in the Y direction is zero because we have assumed that the vertical deﬂection is a function of X only. In addition to the kinematic assumptions, Euler-Bernoulli beam theory as sumes that the material is linearly elastic and that the only non-zero stress component is σXX , the normal stress in the X direction. If σXX is the only nonzero stress component, it then follows that σXX = EXX = −EY θ,X (7.57) where E is Young’s modulus. It is worth mentioning here that by assuming σXX is the only nonzero stress, we violate the kinematic assumptions described above. Because we have assumed that the vertical deﬂection, v, is a function of X only, then the extensional strain in the Y direction is zero. The normal stress in the Y direction cannot also be zero from our knowledge of Hooke’s law for the case of plane strain discussed in Chapter 4. Realizing this contradiction, we proceed anyway, as such contradictions are common in structural theories.

7.2.2

Finite element approximations

A deﬂection curve represents the vertical deﬂection of a beam as a function of distance X from the left end. In the present section we will let v(X) be the deﬂection curve. As discussed in the previous subsection, when the rotation of the beam’s cross section is small, the angle, θ, is approximately equal to the slope v,X of the deﬂection curve. A two-node Euler-Bernoulli beam element of length L is shown in Figure 7.10. As shown in the ﬁgure, each local node has two degrees of freedom, a vertical deﬂection in the Y direction, and a rotation. Hence the displacement vector, de , for this element can be written

Structural Finite Elements

225 ⎧ ⎫ ⎪ ⎪ v1 ⎪ ⎪ ⎨ ⎬ θ1 de = ⎪ ⎪ v2 ⎪ ⎪ ⎩ ⎭ θ2

as

(7.58)

where va and θa are the vertical deﬂection and rotation at local node a. In the derivation of the ﬁnite element shape functions for the two-node EulerBernoulli beam element, we insist that both the vertical deﬂection and the rotation be continuous across element boundaries. It is necessary that the rotation be continuous to avoid violating the assumption that planes remain perpendicular to the neutral surface. Under the present assumption of small rotations, this is equivalent to insisting that both v and v,X be continuous across element boundaries. It turns out that the lowest order polynomial that assures continuity of both v and v,X is cubic, i.e., v˜(X) = α1 + α2 X + α3 X 2 + α4 X 3

(7.59)

where α1 , . . . , α4 are constants. In order to derive the shape functions, it is necessary to solve for these constants in terms of the nodal quantities (7.58). To get the four equations needed, we insist that v˜(X = 0) = v1 v˜,X (X = 0) = θ1 v˜(X = L) = v2 v˜,X (x = L) = θ2 By using equation (7.59) ⎡ 1 ⎢0 ⎢ ⎣1 0

and invoking the above boundary conditions we get ⎤ ⎧ ⎫ ⎧ ⎫

0 0 0 ⎪ v1 ⎪

⎪ α1 ⎪ ⎪ ⎪ ⎬ ⎨ ⎪ ⎬ ⎨ ⎪ 1 0 0 ⎥ α θ1 2 ⎥ = (7.60) 2 3 ⎦ L L L ⎪ ⎪ ⎪ v2 ⎪ ⎪ ⎪ α3 ⎪ ⎩ ⎭ ⎭ ⎪ 2 ⎩ 1 2L 3L α4 θ2

Finally, after solving the system of equations (7.60) and substituting the result back into (7.59) we obtain v˜(X) = N1 (X)v1 + N2 (X)θ1 + N3 (X)v2 + N4 (X)θ2

(7.61)

where N1 , . . . , N4 are the ﬁnite element shape functions. Introducing the dimensionless variable, ξ = X/L, the four shape functions can be written as N1 = 1 − 3ξ 2 + 2ξ 3

N2 = Lξ(1 − 2ξ + ξ 2 )

N3 = ξ 2 (3 − 2ξ)

N4 = Lξ 2 (ξ − 1)

(7.62)

226

Finite Element Method 1 N3

N1

N2 0

N4 0

ξ

1

FIGURE 7.11 Shape functions for two-node Euler-Bernoulli beam element. The four shape functions deﬁned in equation (7.62) are plotted in Figure 7.11. They give rise to trial functions that have continuous ﬁrst derivatives and piecewise continuous second derivatives. Such shape functions are classiﬁed as C 1 continuous. The shape functions given in (7.62) are known as Hermite polynomials. To end this subsection, we now derive the B-matrix for the two-node EulerBernoulli beam element. Recall that the element B-matrix provides the link between the state of strain in the element and the element nodal unknowns. For the Euler-Bernoulli beam element, the only relevant strain component is the extensional strain XX . The ﬁnite element approximation for the exten sional strain is obtained as follows: XX = −Y θ,X = −Y v,XX ⇒ {˜XX }e = −Y Ne,XX de = −Y Be de

(7.63)

The matrix, Be , in the last line of equation (7.63) contains the second deriva tives of the shape functions with respect to X. These derivatives can be easily evaluated using the deﬁnition of the shape functions (7.62) and by employing the chain rule, e.g., dNe dξ dξ dX Le Ne,ξ = 2

Ne,X =

(7.64)

After evaluating all the derivatives, we get Be

1×4

=

1 12ξ − 6 Le (6ξ − 4) −12ξ + 6 Le (6ξ − 2) L2e

(7.65)

Structural Finite Elements

227

As a check on the B-matrix reported in equation (7.65), in the event that the beam element undergoes a rigid body translation in the Y direction, no extensional strain should develop in the element. If the element undergoes a rigid-body translation of magnitude 1, the resulting nodal displacement vector would be ⎧ ⎫ ⎪ ⎪1⎪ ⎪ ⎨ ⎬ 0 (7.66) de = ⎪ ⎪1⎪ ⎪ ⎩ ⎭ 0 The resulting extensional strain is then ⎧ ⎫ ⎪ ⎪1⎪ ⎪ ⎨0⎬ 1 {˜XX }e = −Y 2 12ξ − 6 Le (6ξ − 4) −12ξ + 6 Le (6ξ − 2) 1⎪ ⎪ Le ⎪ ⎩ ⎪ ⎭ 0 Y = − 2 [(12ξ − 6)(1) + (−12ξ + 6)(1)] = 0 (7.67) Le which gives the desired result. One would come to the same conclusion (that no extensional strain develops) if the element undergoes a rigid-body rotation. This can be accomplished, for example, by letting ⎧ ⎫ ⎪ ⎪0⎪ ⎪ ⎨ ⎬ 1 de = (7.68) ⎪ ⎪0⎪ ⎪ ⎩ ⎭ 1

7.2.3

Element matrix

Like the two-node truss element, we derive the element matrix and vector for the Euler-Bernoulli beam element by starting with the principle of virtual work. We begin by writing the internal virtual work over a single beam element as follows: δWint e =

Ωe

σij wi,j dΓ

(7.69)

where σij is the Cauchy stress, wi are the components of a vector-valued test function, and Ωe is the volume of the beam element. In the present two-dimensional formulation, the test function has two com ponents, wX and wY . Let us now restrict our attention to test functions that are C 1 continuous and obey the same kinematic assumptions as the displace ment components. Doing this we write wX = −Y δθ(X) wY = δv(X)

(7.70)

228

Finite Element Method

where the test function δθ can be thought of as a virtual rotation, and δv(X) can be viewed as a virtual deﬂection in the Y direction. Assuming that the virtual rotations are small, the two test functions, δv and δθ, are related as follows: δθ = δv,X Because the only nonzero stress component in the problem is σXX , the internal virtual work statement simpliﬁes to the following: δWint = σXX wX,X dΩ e Ωe # " =−

Le

Y Ae

σXX δθ,X dX

0

dA

(7.71)

Here in equation (7.71), we have assumed that the beam has a constant crosssectional area along the length, X, of the beam. Hence, the diﬀerential volume becomes dΩe = Ae dX. The next step is to recall the relationship (7.57) that provides the connection between the normal stress and the slope of the deﬂection curve, i.e., σXX = −EY θ,X . By substituting this relationship into (7.71) we obtain Le int 2 Y dA θ,X δθ,X dX (7.72) δWe = Ee Ae

0

The ﬁrst integral on the right-hand side of (7.72) is, by deﬁnition, the area moment of inertia of the beam’s cross section about the neutral axis, i.e., Y 2 dA = IZ (7.73) Ae

As an example, for a beam with a rectangular cross section of width, b, and height, h, the area moment of inertia about the neutral axis is IZ =

bh3 12

(7.74)

The internal work statement can now be written as Le δWint = E I θ,X δθ,X dX e Z e

(7.75)

0

Substituting the following ﬁnite element approximations θ,X = Be de δθ,X = Be we

(7.76)

into equation (7.75) yields δWint e

=

weT

Ee IZ Le

1 0

BTe Be

dξ de

(7.77)

Structural Finite Elements

229 Z

Γq

Y

X FIGURE 7.12 Traction distribution acting on Γq .

where we have performed the change of variables, X = Le ξ, and hence, dX = Le dξ. The element matrix is the 4 × 4 matrix inside the curly brackets (sandwiched in between the nodal displacement vector and the vector contain ing the nodal values of the test functions) on the right-hand side of (7.77). Upon performing the required matrix multiplications and integrating over the length of the element, the element matrix is ﬁnally obtained as ⎡ ⎤ 12 6Le −12 6Le 2 2 ⎥ Ee IZ ⎢ ⎢ 6Le 4Le −6Le 2Le ⎥ (7.78) Ke = ⎣ 3 −12 −6Le 12 −6Le ⎦ Le 6Le 2L2e −6Le 4L2e

7.2.4

Element vector

Let us now focus on the derivation of the element vector for the two-node Euler-Bernoulli beam element. Recall that the element vector arises from the right-hand side of the principle of virtual work statement (the external virtual work). In the absence of body forces, the external virtual work can be written as ext ti wi dΓ (7.79) δW = Γ

where ti are the components of the traction vector acting on the outer surface, Γ, of the beam. In the present formulation, we have assumed that the beam is subjected to loads that act only in the X−Y plane. Hence, the surfaces on which tractions can act are the top and bottom surfaces of the beam, as well as the right and left ends. Note that tractions can be applied on the right and left ends as long as the resultant force does not have a component in the axial, X, direction. Let us for the moment focus on the top surface of the beam, which is labeled Γq , in Figure 7.12. We assume that the traction distribution acting on this surface can vary in the X direction, but remains constant in the Z direction.

230

Finite Element Method Z ΓM X

Y FIGURE 7.13 Traction distribution acting on ΓM .

If we let b be the thickness of the beam in the Z direction, then the force per unit length, q(X), acting on the beam is just the vertical traction tY times the thickness, b, i.e., q(X) = btY (X)

(7.80)

Notice that the force per unit length, q(X), is only a function of X. It is commonly referred to as a distributed load. The contribution to the external work from the distributed load acting on the top surface can be written as Γq

ti wi dΓ = b =

l 0

tY wY dX

L

q(X)δv(X) dX

(7.81)

0

where L is the length of the beam. Next, let us focus on the traction distribution acting on the right face of the beam as shown in Figure 7.13. Here we assume that the traction distribution is such that it produces no net force in the axial direction of the beam. This requirement can be stated mathematically as follows:

ΓM

tX dΓ = 0 ⇒

h/2

b −h/2

σXX dY = 0

(7.82)

where we have assumed that the shape of the beam’s cross section is rectan gular, and h is the total height of the beam in the Y direction. Let us now investigate the external virtual work of the traction distribution acting on the right surface of the beam, ΓM . The external virtual work can

Structural Finite Elements

231

be written as

tX wX dΓ =

ΓM

ΓM

(σXX )(−Y δθ) dΓ

= −b

h/2

−h/2

σXX Y δθ dY

= M ∗ δθ

(7.83)

where M ∗ is the net bending moment that is prescribed on ΓM . If shear tractions are prescribed on one of the ends of the beam, we will call this end ΓV . The contribution to the external virtual work due to shear tractions acting on ΓV can be written as ΓV

tY wY dΓ = b

h/2

−h/2 ∗

tY w, dY

=V w

(7.84)

where V ∗ is the net shear force acting on ΓV . Finally, the external virtual work due to all of the tractions acting on the surface of the beam combined can be written as L δWext = qδv dX + M ∗ δθ + V ∗ δv (7.85) 0

We end this subsection by deriving the element vector for the case where the element is subjected to a uniform distributed load of magnitude q. For now we will assume that the element does not touch either of the boundaries ΓM or ΓV . Under these circumstances, the external virtual work reduces to = δWext e

Le

qδv dX

(7.86)

0

Next, we interpolate the test function δv over the length, Le , of the element using the same shape functions that we used to interpolate the trial function. Doing this gives ˜ = Ne we (7.87) δv e

where Ne is the 1×4 shape function matrix that containes the shape functions deﬁned in equation (7.62), and we is a 4 × 1 vector containing the values of the test functions at the nodal points, i.e., ⎧ ⎫ δv1 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ δθ1 (7.88) we = δv2 ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ δθ2

232

Finite Element Method

After plugging the ﬁnite element approximation (7.87) into the external work expression (7.86) we get δWext e

=

weT

qLe

0

1

NeT

dξ

(7.89)

where, again, we have used the natural coordinate ξ = X/Le . We now recog nize that the 4 × 1 vector inside the curly brackets on the right-hand side of equation (7.89) is the element vector, i.e, fe = qLe

0

1

NTe dξ

(7.90)

Evaluating the integral on the right-hand side of equation (7.90) above yields ⎧ ⎫ ⎪ ⎪ 1/2 ⎪ ⎪ ⎨ ⎬ Le /12 fe = qLe (7.91) ⎪ ⎪ 1/2 ⎪ ⎪ ⎩ ⎭ −Le /12 We observe from the element vector (7.91) that in order to correctly model the situation of a uniform distributed load acting over the span of an element, we must apply both nodal forces in the Y direction as well as equal and opposite concentrated couples at the nodal locations. As a closing remark, if an element does in fact touch one of the boundaries (either ΓM or ΓV ) then the element vector for that element is simply augmented by the concentrated couple or concentrated force at the local node that touches the boundary. The entire procedure will be demonstrated in the following example.

Example 7.3 In order to demonstrate the use of the Euler-Bernoulli beam element, let us consider an example of a cantilever beam of length L = 10 m loaded by a uniform distributed load, q = 1000 N/m. A concentrated force, P = 1000 N, acts at the right end as shown in Figure 7.14. The concentrated force can be viewed as the resultant force of a shear traction distribution applied at the right end of the beam. The cross section of the beam is rectangular with base b = 100 mm, and height h = 400 mm. We will assume that the beam is made of steel with Young’s modulus E = 200 GPa. To get an idea of how many elements to use along the length of the beam, it is instructive to look at the bending moment diagram. The bending moment diagram for the present example is shown in Figure 7.15. As shown in the ﬁgure, the maximum bending moment, M max = −60, 000 N · m, occurs at the left end of the beam, and the bending moment goes to zero at the right end as expected. From elementary mechanics of materials theory, the bending

Structural Finite Elements

233 P = 1000 N q = 1000 N/m X

Y FIGURE 7.14 Cantilever beam subjected to uniform distributed load and concentrated end load.

moment at position X along the length of the beam is related to the vertical deﬂection at that location through the relation EIZ v,XX = −M (X)

(7.92)

where E is Young’s modulus, and IZ is the area moment of inertia of the cross section about the neutral axis. Employing the ﬁnite element interpolation function (7.61), and diﬀerentiating the trial function twice with respect to X

0 M (N·m)

-20,000

-40,000

-60,000 0 FIGURE 7.15 Bending moment diagram.

2

4

6 X (m)

8

10

234

Finite Element Method

gives {v˜,XX }e = Be de 1 = 2 [(12ξ − 6)v1 + Le (6ξ − 4)θ1 Le + (−12ξ + 6)v2 + Le (6ξ − 2)θ2 ]

(7.93)

where ξ = X/Le . We see from equation (7.93) above that the ﬁnite element approximation for v,XX is capable of capturing a linearly varying bending moment. For this reason, we can get away with using only one beam element in regions where the bending moment varies linearly with X. In the present example, the bending moment varies quadratically along the length of the beam, and hence, more than one element will be necessary to obtain good accuracy. Let us now begin the process of obtaining an approximate ﬁnite element solution by discretizing the beam along its length with two Euler-Bernoulli beam elements (each of equal length) as shown in Figure 7.16. The elements are identiﬁed by numbers inside rounded boxes. The global node numbers are labeled in bold, and the global equation numbers associated with each node are listed below the global node numbers. Following the ﬁnite element assembly process described in Chapter 3, we begin by writing down the element matrices and indicating the local-to-global equation numbering to the right of these matrices. Doing this we obtain ⎡ ⎤ ⎡ 12 6L −12 6L 1 12 6L −12 6L ⎢ 6L 4L2 −6L 2L2 2 ⎥ ⎢ 6L 4L2 −6L 2L2 ⎥ EIZ ⎢ EIZ ⎢ −12 −6L 12 −6L 3 ⎥ −12 −6L 12 −6L K2 = 3 ⎢ K1 = 3 ⎢ ⎢ ⎥ L ⎣ 6L 2L2 −6L 4L2 4 ⎦ L ⎢ ⎣ 6L 2L2 −6L 4L2 1 2 3 4 3 4 5 6 Upon assembly of the element matrices, the 6 × 6 global system matrix turns out to be ⎡ ⎤ 12 6L −12 6L 0 0 ⎢ 6L 4L2 −6L 2L2 0 0 ⎥ ⎢ ⎥ ⎥ EIZ ⎢ −12 −6L 24 0 −12 6L ⎢ ⎥ (7.94) K = 3 ⎢ L ⎢ 6L 2L2 0 8L2 −6L 2L2 ⎥ ⎥ ⎣ 0 0 −12 −6L 12 −6L ⎦ 0 0 6L 2L2 −6L 4L2

1 1,2

� � �1 �

2 2,3

FIGURE 7.16 Two-element model of cantilever beam.

� � �2 �

3 3,4

⎤ 3 4⎥ ⎥ 5⎥ ⎥ 6⎦

Structural Finite Elements

235

Next, from equation (7.90), we can write down the element vectors as follows: ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ 0 3⎪ 1/2 1 ⎪ 1/2 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎨ ⎬ ⎨ ⎨ L/12 2 L/12 4 0 4 f2 = qL + f1 = qL 1/2 3 ⎪ P 5⎪ ⎪ ⎪ ⎪ ⎪ 1/2 5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎭ ⎩ ⎩ 0 6 −L/12 4 −L/12 6 Note that we have augmented the element vector f2 by the concentrated load, P , acting at local node 2 of the second element. Assembling the element vectors yields ⎧ ⎫ ⎪ ⎪ qL/2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ qL2 /12 ⎪ ⎪ ⎪ ⎨ ⎬ qL (7.95) f= 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ qL/2 + P ⎪ ⎪ ⎪ ⎭ ⎩ −qL2 /12 The global system of equations can now be written as ⎫ ⎡ ⎤⎧ ⎫ ⎧ qL/2 ⎪ v1 ⎪ 12 6L −12 6L 0 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ 6L 4L2 −6L 2L2 0 ⎪ ⎪ θ1 ⎪ qL2 /12 ⎪ 0 ⎥ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎨ ⎬ ⎬ ⎥ v2 EIZ ⎢ −12 −6L 0 −12 6L 24 qL ⎢ ⎥ = 2 θ2 ⎪ ⎪ 0 8L2 −6L 2L2 ⎥ 0 ⎪ L3 ⎢ ⎪ ⎪ ⎪ ⎢ 6L 2L ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎣ 0 0 −12 −6L 12 −6L ⎦ ⎪ v qL/2 + P ⎪ ⎪ ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎭ ⎭ 2 2 2 θ3 6L 2L −6L 4L −qL /12 0 0

(7.96)

Before solving the global system of equations (7.96), it is necessary to en force the essential boundary conditions at global node 1. Because node 1 is located at a clamped end, both the deﬂection, v1 , and the rotation, θ1 , are zero at this location. These boundary conditions can be enforced by crossing out the ﬁrst and second rows and columns of the global system of equations. Doing this, we obtain the following reduced system to be solved: ⎫ ⎡ ⎤⎧ ⎫ ⎧ qL 24 0 −12 6L ⎪ ⎪ ⎪ ⎪ v2 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ 2 2 ⎥⎨ EIZ ⎢ 0 −6L 2L θ 0 8L 2 ⎢ ⎥ = (7.97) ⎪ v3 ⎪ ⎪ ⎪ qL/2 + P ⎪ ⎪ L3 ⎣ −12 −6L 12 −6L ⎦ ⎪ ⎩ ⎪ ⎭ ⎩ ⎭ 2 2 2 −qL /12 6L 2L −6L 4L θ3 Solving the reduced system (7.97) yields (to eight digits beyond the decimal point) ⎧ ⎫ ⎧ ⎫ .00000000 ⎪ v1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ .00000000 θ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎬ ⎨ ⎪ ⎬ ⎪ ⎨ .00512695 v2 (7.98) = θ2 ⎪ ⎪ ⎪ ⎪ .00171875 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ .01484375 ⎪ v3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎪ ⎭ ⎪ ⎩ .00203125 θ3

236

Finite Element Method

Having obtained the nodal displacements and rotations, let us now calculate the bending moment along the length of each element from the ﬁnite element solution. Utilizing equation (7.92), the bending moment at location X along the length of the beam can be written as M (X) = −EIZ v,XX = −EIZ θ,X ⇒ Me = −Ee IZ Be de

(7.99)

where in the last line we have plugged in the ﬁnite element approximation θ,X = Be de . The bending moment along the length of element 1 then becomes M1 (ξ) = −EIZ B1 d1

⎧ ⎫ ⎪ ⎪ v1 ⎪ ⎪ ⎨ θ1 ⎬ −EIZ 12ξ − 6 L(6ξ − 4) −12ξ + 6 L(6ξ − 2) = v2 ⎪ ⎪ L2 ⎪ ⎪ ⎩ ⎭ θ2

Next, noting that the length of element 1 is 5 m, and the local displacement vector is ⎧ ⎫ ⎪ ⎪ ⎪ 0.00000000 ⎪ ⎨ ⎬ 0.00000000 d1 = (7.100) ⎪ ⎪ ⎪ 0.00512695 ⎪ ⎩ ⎭ 0.00171875 we get the following equation for the bending moment along the length of element 1: (7.101) M1 (ξ) = 42499.8400ξ − 57916.5866 When evaluating the bending moment along the length of the element, we must remember that the natural coordinate, ξ = X/l, is measured from the left end of the element. Following the same procedure, we get the following for the bending moment along the length of the second element: M2 (ξ) = 17500.1600ξ − 15416.7467

(7.102)

The bending moment obtained from the ﬁnite element method is plotted versus distance, X, along the length of the beam in Figure 7.17. The ﬁnite element approximation is represented by the dashed line in the ﬁgure, and the exact solution is depicted by the solid line. Notice that the ﬁnite element approximation predicts a linearly varying bending moment along the length of each element. Also notice that the natural boundary condition, M ∗ = 0, at the right end of the beam is only approximately satisﬁed by the present ﬁnite element solution. If instead, we wanted to obtain the normal stress distribution in each beam element, we would follow essentially the same procedure that we used to compute the bending moment, i.e., σe = −Ee Y Be de . The maximum tensile

Structural Finite Elements

M (N · m)

237

0

-20,000

-40,000

-60,000 0

2

4 6 X (m)

8

10

FIGURE 7.17 Bending moment versus distance, X, along the length of the beam. The dashed line represents the ﬁnite element approximation.

and compressive stress occurs at the top and bottom of the beam at the left end, where the bending moment is maximum. As a closing remark, it is worth mentioning that very accurate ﬁnite solu tions to beam and frame problems can be obtained using just a few EulerBernoulli beam elements. In the present two- element example, the ﬁnite element solution for the end deﬂection agrees with the exact solution to eight digits!

7.3

Mindlin-Reissner plate theory

Due to the importance of plates and shells in engineering analysis and design, in this section we provide an introduction to the rich subject of plate and shell element technology by presenting Mindlin-Reissner plate theory and its ﬁnite element implementation. One of the big advantages of the Mindlin-Reissner theory is that it does not neglect transverse shear strains (as was the case in Euler-Bernoulli beam theory). Hence, the theory allows for the analysis of both thin and thick plates using low-order elements such as four-node quadri laterals. After presenting the theory and ﬁnite element implementation, we end the section with a case study in which we analyze a simply supported plate with a uniform distributed load acting over the top surface. In the case study we explore the performance of the Mindlin-Reissner plate element and point out some numerical diﬃculties that can arise when using the element and how to avoid them in practice. For additional information on this subject,

238

Finite Element Method Z Y

X

FIGURE 7.18 Mindlin-Reissner plate in undeformed and deformed conﬁgurations. the reader is encouraged to visit Hughes and Tezduyar [12].

7.3.1

Assumptions

To help understand the assumptions that are made in Mindlin-Reissner plate theory, let us focus our attention on Figure 7.18. The plate with thickness t originally lies in the X−Y plane. The midsurface of the plate is the shaded surface in the undeformed (upper) conﬁguration. The plate can be thought of as being composed of inﬁnitely many inextensible ﬁbers, or line segments, that are originally perpendicular to the midsurface. Each ﬁber goes from the original conﬁguration to the deformed conﬁguration (below) by translating in the Z direction and rotating about its center. The midsurface of the plate is also shaded in the deformed conﬁguration. Notice that the midsurface in the deformed conﬁguration is not in general planar, and the ﬁbers do not have to remain perpendicular to it. The kinematics of an individual ﬁber is illustrated in Figure 7.19. After translating in the Z direction, the ﬁber gets into the deformed conﬁguration by ﬁrst undergoing a positive (right-hand rule) rotation about the Y axis, followed by a negative rotation about the X axis. The resulting displacement components in the X and Y directions respectively are labeled uX and uY in the ﬁgure. Let us now focus our attention on a point that lies somewhere along the length of the ﬁber. If the rotations θX and θY are assumed to be small, then the displacement components uX and uY at this point can be expressed as uX = ZθX

uY = ZθY

(7.103)

where Z is the distance along the ﬁber from the ﬁber center to the point of interest. The coordinate Z is positive at points above the midsurface and

Structural Finite Elements

239 Z

uX

uY

Y X FIGURE 7.19 Kinematics of a single ﬁber. negative at points below. The displacement component uZ can be written as uZ = v(X, Y )

(7.104)

Equation (7.104) states that the vertical displacement, v, of any point along an individual ﬁber, is a function of X and Y only. Another important assumption in Mindlin-Reissner plate theory is that the normal stress in the Z direction is zero, i.e., σZZ = 0. This assumption does not follow from the kinematic assumptions stated above. We will see later, when we introduce Hooke’s law into the formulation, that assuming σZZ = 0 gives rise to a nonzero extensional strain component in the Z direction. This is in contradiction to equation (7.104), which implies ZZ = ∂v/∂Z = 0! Again, such contradictions are common in the development of structural ele ments. They are justiﬁable when they simplify the theory and good accuracy is still achieved between the theory and experiment. Note that the assump tion σZZ = 0 alone is not identical to the plane-stress assumption discussed in Chapter 4. The plane-stress assumption assumes that the transverse shear stresses, σXZ and σY Z , are also zero. In Mindlin-Reissner plate theory, equa tion (7.103) allows for the possibility of transverse shear strains, and the associated transverse shear stresses are not ignored.

7.3.2

Strain-displacement relations and Hooke’s law

In the linearized theory of elasticity discussed in Chapter 4, the small-strain components, ij , are related to the displacement components, ui , through the

240

Finite Element Method

relation ij =

1 (ui,j + uj,i ) 2

(7.105)

Assuming the small-strain kinematics (7.105), the relevant strain components in Mindlin-Reissner plate theory are obtained by diﬀerentiating the displace ment components given in equations (7.103) and (7.104) with respect to X and Y , respectively. Doing this gives XX = uX,X = ZθX,X Y Y = uY,Y = ZθY,Y γY Z = uY,Z + uZ,Y = θY + v,Y γXZ = uX,Z + uZ,X = θX + v,X γXY = uX,Y + uY,X = Z (θX,Y + θY,X )

(7.106)

Next, assuming linearly elastic and isotropic material behavior, the nonzero stress components can be expressed in terms of the strain components deﬁned in (7.106) above using the generalized Hooke’s law. To do this, we store the ﬁve nonzero stress components in two vectors as follows: ⎧ ⎫ ⎨ σXX ⎬ σ b = σY Y ; ⎩ ⎭ σXY

s

σ =

σXZ σY Z

(7.107)

The 3 × 1 stress vector σ b contains the in-plane components of the stress, and σ s contains the transverse shear stress components. The in-plane stress components can be written in terms of the in-plane strain components as ⎧ ⎫ ⎡ E ⎫ ¯ νE ¯ 0 ⎤⎧ ⎨ σXX ⎬ ⎨ XX ⎬ ⎢ ¯ ¯ ⎥ σY Y E 0 ⎦ Y Y = ⎣ νE ⎩ ⎭ ⎭ ⎩ σXY γXY 0 0 G ⎡ ¯ ⎫ ¯ 0 ⎤⎧ E νE θX,X ⎬ ⎨ ⎢ ¯ ¯ ⎥ E 0⎦ θY,Y = Z ⎣ νE ⎩ ⎭ θX,Y + θY,X 0 0 G = ZCb κ

(7.108)

¯ = E/ 1 − ν 2 , Cb is the 3 × 3 plane stress elasticity matrix, and κ is where E a 3 × 1 vector containing the partial derivatives of the rotations. The vector κ is called the curvature vector. The transverse shear stress components can

Structural Finite Elements

241 Z 4

η 3

1 2

ξ FIGURE 7.20 Mindlin-Reissner plate element in parent domain.

be written in terms of the transverse shear strain components as

σXZ σY Z

G 0 γXZ = 0 G γY Z G 0 θX + v,X = θY + v,Y 0 G = Gγ

(7.109)

where G is the shear modulus, and γ is a 2×1 vector containing the transverse shear strains. Having deﬁned both the strain-displacement and stress-strain relationships, in the following sections we derive the ﬁnite element matrices and vectors associated with the four-node Mindlin-Reissner plate element.

7.3.3

Finite element approximations

A Mindlin-Reissner plate element in the parent domain is shown in Fig ure 7.20. The neutral surface in the parent domain is a planar bi-unit square that lies in the plane Z = 0. The four nodes are located at the corners of the bi-unit square, and the local node numbering is chosen such that the nodes are numbered counterclockwise from 1 to 4 as shown in the ﬁgure. The element has thickness t, and all of the ﬁbers in the parent domain are perpendicular to the neutral surface. There are three unknowns associated with each node: vertical deﬂection, va , and two rotations, θXa and θY a . The two rotations are often referred to as nodal rotations. They really refer to the rotation of the ﬁbers passing through the nodes about the coordinate axes. Having said this, we can now deﬁne a 3 × 1 vector, dea , containing the nodal unknowns

242

Finite Element Method

(or degrees of freedom) at node a as follows: ⎧ ⎫ ⎨ va ⎬ dea = θXa 3×1 ⎩ ⎭ θY a

(7.110)

The 12 × 1 vector containing the nodal unknowns associated with all four nodes can then be written as ⎧ ⎫ ⎪ ⎪ de1 ⎪ ⎪ ⎨ ⎬ de2 de (7.111) = 12 × 1 ⎪ ⎪ de3 ⎪ ⎪ ⎩ ⎭ de4 Next, we employ the bi-linear shape functions to interpolate the vertical de ﬂection and nodal rotations over the element. Doing this yields the following ﬁnite element approximations: v˜e (X, Y ) = Na (ξ, η)va ˜ θXe (X, Y ) = Na θXa θ˜Y e (X, Y ) = Na θY a ,

(7.112)

where Na (ξ, η) are the bilinear shape functions, i.e., Na (ξ, η) =

1 (1 ± ξ)(1 ± η) 4

(7.113)

The ﬁnite element approximation for the curvature vector deﬁned in equation (7.106) can now be written as ⎧ ⎫ Na,X θXa ⎨ ⎬ ˜e = Na,Y θY a κ (7.114) ⎩ ⎭ Na,Y θXa + Na,X θY a or ˜ e = Bbe de κ

(7.115)

where Bbe is a 3 × 12 matrix widely referred to as the bending B-matrix. The bending B-matrix is obtained by writing equation (7.114) in matrix notation. The result is Bbe = B B B B (7.116) e1

3 × 12

where

e2

e3

e4

⎡

Bbea

⎤ 0 Na,X 0 = ⎣ 0 0 Na,Y ⎦ 0 Na,Y Na,X

(7.117)

Structural Finite Elements

243

We end this section by writing down the ﬁnite element approximation for the vector γ containing the transverse shear strains. From equation (7.109) we write θ˜X + v˜,X ˜ = ˜ ⇒ γ θY + v˜,Y Na θXa + Na,X va ˜e = γ (7.118) Na θY a + Na,Y va ˜ , can be written in matrix notation as The ﬁnite element approximation, γ follows: ˜ = Bse de γ (7.119) where Bse is called the shear B-matrix. The shear B-matrix is deﬁned as Bse

2 × 12

= Bse1 Bse2 Bse3 Bse4

where Bsea

Na,X Na 0 = Na,Y 0 Na

(7.120)

(7.121)

is the contribution to the shear B-matrix from local node a.

7.3.4

Element matrix

From our study of Chapter 6, we recall that the element matrix (stiﬀness matrix) comes from the internal virtual work statement written over a sin gle element deﬁned by the domain Ωe . For a plate element with constant thickness, t, the internal virtual work statement can be written as δWint e

t/2

= −t/2

Ae

σij wi,j dA

(7.122)

where Ae is the area of the element, and wi are the components of a vectorvalued test function that will be deﬁned below. By expanding the integrand in (7.122) we get σij wi,j = σXX wX,X + σY Y wY,Y + σXY (wX,Y + wY,X ) +σXZ (wX,Z + wZ,X ) + σY Z (wY,Z + wZ,Y )

(7.123)

where we have used the that fact that the stress tensor is symmetric. Notice that only ﬁve out of the six stress components appear in (7.123), because we have assumed that σZZ =0. In deﬁning the test function, as usual, we insist that the test function be continuous and identically zero at points on the boundary of the domain where the displacement components are prescribed. In our present theory for plates,

244

Finite Element Method

we also insist that the test function obey the same kinematic assumptions as the displacement ﬁeld. Doing this we can write wX = ZδθX (X, Y ) wY = ZδθY (X, Y ) wZ = δv(X, Y )

(7.124)

Here in equation (7.124), Z is the vertical distance from the midsurface of the plate to the point of interest. The quantities δθX , δθY , and δv(X, Y ) are scalar test functions. They obey the following conditions at the essential boundary points: ⎧ ⎪δθ ⎨ X =0 : δθY = 0 : ⎪ ⎩ δv = 0 :

at boundary points where θX is speciﬁed at boundary points where θY is speciﬁed at boundary points where v is speciﬁed

Diﬀerentiating the test functions (7.124) with respect to X and Y gives wX,X = ZδθX,X wY,Y = ZδθY,Y wX,Y + wY,X = Z (δθX,Y + δθY,X ) wX,Z + wZ,X = δθX + δv,X wY,Z + wZ,Y = δθY + δv,Y

(7.125)

It is convenient to store the derivatives of the test function deﬁned in (7.125) in two vectors as follows: ⎧ ⎫ ⎧ ⎫ wX,X δθX,X ⎬ ⎨ ⎬ ⎨ wY,Y δθY,Y =Z ⎩ ⎭ ⎩ ⎭ wX,Y + wY,X δθX,Y + δθY,X = Zδκ and

wX,Z + wZ,X wY,Z + wZ,Y

=

δθX + δv,X δθY + δv,Y

= δγ where the 3 × 1 vector, δκ, can be viewed as a virtual curvature vector, and the 2 × 1 vector, δγ, can be viewed as a virtual transverse shear strain vector. The virtual internal work expression can now be written in matrix notation in terms of the stress vectors σ b and σ s as follows: t/2 T δκT σ b δγ σ s 2 δWint = Z dA + t dA (7.126) e 1×2 2×1 1×3 3×1 −t/2

Ae

Ae

Structural Finite Elements

245

where the ﬁrst integral on the right-hand side is the bending contribution, and the second integral is the shear contribution. In order to complete the derivation of the element matrices for the MindlinReissner plate element, it is necessary to substitute the ﬁnite element approx imations into the internal virtual work (7.126). The ﬁnite element approx imations for the virtual curvature and shear strain vectors are obtained by deﬁning a vector, we , containing the nodal values of the components of the test function. This can be done as follows: ⎧ ⎫ ⎪ ⎪ ⎪ we1 ⎬ ⎪ ⎨ we2 we (7.127) = 12 × 1 ⎪ ⎪ ⎪ we3 ⎭ ⎪ ⎩ we4 where wea

⎧ ⎫ ⎨ δva ⎬ = δθXa ⎩ ⎭ δθY a

(7.128)

is the 3 × 1 contribution from each node a. If we now interpolate the com ponents of the test function over the element using the same bilinear shape functions that were used to interpolate the displacement ﬁeld we get ⎫ ⎧ ⎫ ⎧ ˜ ⎬ ⎨ Na δva ⎬ ⎨ δv ˜ = Na δθXa (7.129) δθ ⎩ ⎭ ⎩ ˜X⎭ Na δθXa δθY e

The ﬁnite element approximations for the virtual curvature vector and virtual shear strain vector are then written as

and

˜ = Bb we δκ e

(7.130)

˜ = Bs we δγ e

(7.131)

where Bbe and Bes are the bending B-matrix and the shear B-matrix deﬁned in equations (7.116) and (7.120). Next, plugging in the ﬁnite element approx imations into the internal virtual work statement (7.126) yields " # T δWint e = we

t/2

−t/2

Ae

b b Z 2 BbT e Ce Be dA + tG

Ae

s BsT e Be dA de

(7.132)

We recognize that the element stiﬀness matrix, Ke , is the 12 × 12 matrix inside the curly brackets in equation (7.132). Carrying out the integration with respect to Z in closed-form, the stiﬀness matrix can be written as t3 b b s s BbT C B dA + t BsT (7.133) Ke = e Ce Be dA 12 Ae e e e Ae

246

Finite Element Method

The ﬁrst integral on the right-hand side of equation (7.133) above is referred to as the bending stiﬀness matrix. The second integral is the shear stiﬀness matrix. Both integrals can be fully integrated by employing a 2×2 quadrature rule on the integrals expressed in the parent domain. It turns out that fully integrating the shear stiﬀness matrix leads to numerical diﬃculties when the thickness of the plate is much less than the characteristic in-plane width. This issue is explored at the end of this chapter, where we consider the problem of a square plate with clamped edges subjected to uniform pressure loading on the top surface.

7.3.5

Element vector

In the absence of body forces, the external virtual work statement can be written as ext δW = t∗i wi dΓ (7.134) Γt

The element vector for a Mindlin-Reissner plate element is obtained by writing the external virtual work expression over a single element and then substitut ing the ﬁnite element approximations into the resulting expression. In doing this, we recognize that, in general, the traction distribution can act over the top and bottom surfaces of the plate element. In addition, if the element has an edge that is contiguous to the boundary of the plate as shown in Fig ure 7.21, then that edge may also have surface tractions acting upon it. In the present theory we assume that the edge tractions are distributed in such a way that the resultant force per unit length in the direction perpendicular to the boundary is zero. This means that there can be no in-plane extension or compression of the neutral surface. In the present derivation we assume that the tractions acting on the top and bottom surfaces of the plate act normal to the surface, i.e., no shear tractions act on the top and bottom. Letting q be the traction component in the Z direction, the external virtual work statement can now be written over a single element as ext ∗ qδv dA + M δθn dΓ + V ∗ δv dΓ (7.135) δWe = Ae

Γe

Γe

Here in equation (7.135), the ﬁrst integral on the right-hand side gives the virtual work contribution from the normal traction distribution acting over the top and bottom surfaces of the element. Recall that the quantity δv is the component of the test function in the Z direction. Hence, the product qδv gives force per unit length. The second integral on the right-hand side is the external work contribution for the external normal tractions acting on the element edge Γe . The quantity M ∗ is the moment per unit length obtained by integrating the normal traction distribution through the thickness of the plate. The moment per unit length is obtained by multiplying the normal traction tn by the wn = Zδθn and integrating through the thickness of the

Structural Finite Elements

247 Z s n Ae

Γe M∗ V∗

FIGURE 7.21 Traction distribution acting on plate. plate, i.e., Γe

t/2

−t/2

t∗n wn

t/2

dZdΓ = Γe

=

Γe

−t/2

Zt∗n δθn dZdΓ

M ∗ δθn dΓ

(7.136)

Here in equation (7.136) above, the quantity δθn can be viewed as a virtual right-hand-rule rotation about the n axis labeled in Figure 7.21. The last integral on the right-hand side of (7.135) comes from multiplying the vertical shear traction acting on the edge by the test function, δv, and integrating over the edge of the plate element. Doing this we get t/2 t∗Z δv dZdΓ = V ∗ δv dΓ (7.137) Γe

−t/2

Γe

where V ∗ is the vertical shear force per unit length acting on the element edge. The element vector can now be obtained by substituting the ﬁnite element approximations into the external virtual work expression (7.135). To do this, we need to write the ﬁnite element approximations for the test function δv and δθn in terms of shape function matrices and the 12 × 1 vector, we , which contains the nodal values of the test function. The ﬁnite element approximation for δv can be written as ˜ = Nv we (7.138) δv e where Nev is a 1 × 12 shape function matrix which is given as Nve = N1 0 0 N2 0 0 N3 0 0 N4 0 0

(7.139)

248

Finite Element Method

The ﬁnite element approximation for the component of the test function, wn , can be obtained as follows. The normal component of the test function is given as (7.140) wn = en · w, where en is a unit base vector pointing in the n direction labeled in Figure 7.21. Because the n axis lies in the X−Y plane, the dot product operation (7.140) above is simply wn = mm wX + nn wY wX = mm nn wY

(7.141)

where mm and nn are the direction cosines between the local n axis and the X and Y axes, i.e., mm = en · eX and nn = en · eY . Next, substituting wX = ZδθX and wY = ZδθY into equation (7.141) yields

wn = Z mm nn

δθX δθY

(7.142)

Finally, because wn = Zδθn , where δθn corresponds to a positive (right-hand rule) rotation about the s axis shown in Figure 7.21, we conclude that δθn = mm nn

δθX δθY

(7.143)

Equation (7.143) can be written in matrix notation as follows: δθn = ANθe we

(7.144)

where A = mm nn is a 1 × 2 matrix containing the direction cosines, Nθe

0 N1 0 0 N2 0 0 N3 0 0 N4 0 = 0 0 N1 0 0 N2 0 0 N3 0 0 N4

(7.145)

is a matrix containing the shape functions, and ⎧ ⎫ ⎪ we1 ⎪ ⎪ ⎪ ⎨ ⎬ we2 we = ⎪ we3 ⎪ ⎪ ⎪ ⎩ ⎭ we4 where

⎧ ⎫ ⎨ δva ⎬ we a = δθXa ⎩ ⎭ δθY a

(7.146)

Structural Finite Elements

249

The last step in the derivation of the element vector for the four-node Mindlin-Reissner plate element is to substitute the following ﬁnite element approximations: δv˜e = weT NvT e T δθ˜ne = weT NθT e A

(7.147)

into the external virtual work statement. Doing this yields T vT θT T ∗ vT ∗ = w N q dA + N A M dΓ + N V dΓ δWext e e e e e Ae

Γe

Γe

The element vector is the quantity inside the curly brackets in the equation above. Hence, θT T ∗ ∗ fe = NvT q dA + N A M dΓ + NvT (7.148) e e e V dΓ Ae

Γe

Γe

To end this section, we note that the area integral and the line integrals that appear in equation (7.148) can be readily calculated as equivalent area and line integrals in the parent domain using the techniques covered in Chapter 6 (see Section 6.2.6 and Example 6.2).

7.4

Deﬂection of a clamped plate

To illustrate the practical use of the Mindlin-Reissner plate element, let us consider the problem of a 2 m × 2 m square plate with clamped edges as shown in Figure 7.22. The plate is loaded by a uniform traction distribution acting in the negative Z direction and having magnitude q = 100 kPa. We will begin the analysis by taking the thickness of the plate to be t = 0.1 m, and we will assume that the plate is linearly elastic and isotropic with E = 200 GPa and ν = 0.3. It is helpful to refer to analytical solutions, if available, to study the per formance and convergence of ﬁnite element types. An analytical solution to the problem of interest, based on Poisson-Kirchhoﬀ (thin plate) theory, can be obtained by Hencky’s method.[13] The result for the deﬂection, v c , at the center of the plate as reported by Taylor and Govindjee [14] can be written as qa4 (7.149) vc = α D where, to ten digits, α = 1.265319087 × 10−3 (7.150)

250

Finite Element Method

q = 100 kPa

Z Y

X

FIGURE 7.22 Deﬂection of a square plate with clamped edges.

and D=

E t3 2 1 − ν 12

(7.151)

is the stiﬀness of the plate. Because the plate has clamped edges, the boundary conditions along the lines Y = 0, Y = 2 m, X = 0, and X = 2 m are simply θX = 0;

v = 0;

and

θY = 0

(7.152)

The ﬁnite element mesh employed in the analysis is a 20 × 20 uniform grid of four-node quadrilaterals as shown in Figure 7.23.

Y

v = 0, θX = 0, θY = 0

X FIGURE 7.23 Mesh employed for clamped plate: 20 × 20 uniform grid.

251

1

1

0.8

0.8

v/v ˜ exact

v/v ˜ exact

Structural Finite Elements

0.6

0.6

0.4

0.4

0.2

0.2

0

0

0.5

1

X (m)

1.5

2

0

0

0.5

1

1.5

2

X (m)

(a) (b)

FIGURE 7.24 Deﬂection along the line Y = 1 m, t = 0.1 m (a) and t = 0.01 m (b).

Numerical Results The numerical results for the vertical deﬂection along the center of the clamped plate are illustrated in Figure 7.24. Numerical results for the case when the plate thickness is t = 0.1 m are shown in Figure 7.24(a). The dashed line in the ﬁgure shows the vertical deﬂection normalized with respect to the ana lytical solution (7.149) along the line Y = 1 m. Here, full integration (2 × 2) quadrature was employed for both the bending and shear stiﬀness matrices given in equation (7.133). The solid line in Figure 7.24(a) shows the vertical deﬂection along the line Y = 1 m for the case when the bending stiﬀness ma trix is fully integrated, and one-point quadrature was employed to integrate the shear stiﬀness matrix. The practice of using diﬀerent quadrature rules on various parts of the stiﬀness matrix is known as selective reduced integration (SRI). Numerical results for the vertical deﬂection of the plate along the line Y = 1 m are shown for the case when the plate thickness is reduced to t = 0.01 m in Figure 7.24 (b). Again, the vertical deﬂection is normalized with respect to the analytical solution (7.149). Here the solid line represents the vertical deﬂection using the technique of selective reduced integration, i.e., 2 × 2 quadrature for the bending stiﬀness and 1 × 1 quadrature for the shear stiﬀness matrix. The dashed line in the ﬁgure represents the case when both stiﬀness matrices are fully integrated. We observe from the ﬁgure that the response of the fully integrated element is extremely stiﬀ in bending. This behavior is referred to as shear locking. On the other hand, the element that employs selective reduced integration performs very well. It turns out that under most circumstances, the Mindlin-Reissner plate ele ment with selective reduced integration gives excellent results. Unfortunately, the underintegrated element can still, under certain circumstances, undergo zero-energy modes as discussed in Chapter 6, Section 6.7. Hourglass control can be added to the element to make it more robust (see, e.g., Flanagan and Belytscko [8], Hughes [5], and Belytschko et al. [9] for more details). For the

252

Finite Element Method

FIGURE 7.25 Contour plot of vertical deﬂection.

sake of interest, a contour plot of the vertical deﬂection superposed on top of the deformed geometry (magniﬁed 50×) is shown in Figure 7.25.

7.5

Problems

Problem 7.1 P

Y

element 1

1

X

3 element 2

2

FIGURE 7.26 Problem 7.1. Truss assemblage.

Consider the truss assemblage shown in Figure 7.26. The cross section of each element is rectangular with dimensions 2 in.×1 in.. The nodal coordinates are give in Table 7.4, and the material properties (Young’s modulus and initial yield stress) are provided in Table 7.5. Model the problem with two truss elements and perform the following:

Structural Finite Elements

253

TABLE 7.4

Problem 7.1: Nodal coordinates. Node X(in.) Y (in.)

1 2 3

0 48 24

0

0 36

TABLE 7.5

Problem 7.1. Element properties. Element E(psi) σY (psi) 1 2

30 × 106 10 × 106

30,000 10,000

(a)

If the load P = 1000 lb, determine the displacement components d3X and d3Y .

(b)

Determine the normal stress in each element.

(c)

Determine the critical load Pc that can be applied to the structure before the onset of yield in any one of the elements.

Problem 7.2 A two-dimensional truss assembly loaded by a force of magnitude P = 1000 lb is supported by a cable as shown in Figure 7.27. The force vector is given as ' &√ √ 2 2 e1 + e2 . P = 1000 2 2 The cable has a diameter of D = 0.25 in. and is made of steel with Young’s modulus E = 30 × 106 psi. Each truss is made of aluminum with Young’s modulus 10 × 106 psi and cross-sectional area A = 1.0 in.2 . The coordinates of nodes 1–4 labeled in the ﬁgure are given in Table 7.6. Model the problem with three truss elements and determine the following: (a)

The displacement components d3X and d3Y

(b)

The tension in the cable

254

Finite Element Method

Y 4

2

1

X

3 P

FIGURE 7.27 Problem 7.2. Truss assemblage supported by a cable.

TABLE 7.6

Problem 7.2. Nodal coordinates. Node X(in.) Y (in.)

1 2 3

0 20 3

0

0 10

Problem 7.3

Consider the problem of a simply supported beam as shown in Figure 7.28. The beam is loaded by a uniform distributed load of magnitude q = 1000 kN/m. The beam is 8 m long and has a square 100 mm×100 mm cross section. Model the problem with two Euler-Bernoulli beam elements, each of equal length, to estimate the center deﬂection of the beam. Also obtain an estimate for the bending moment at the center of the beam.

Structural Finite Elements

255

q = 1000 kN/m

FIGURE 7.28 Problem 7.3. Simply supported beam.

8 Linear Transient Analysis

The ﬁrst pull on the cord always sends the drapes in the wrong direction. —Boyle’s Other Law An important decision that must be made during the analysis of solids and structures is whether to perform a static analysis, a quasi-static analysis, or a dynamic analysis. In a static analysis, all variables in the problem, such as applied load and material properties, are assumed to be independent of time. Inertia is also neglected. In a quasi-static analysis, the loads and material properties can depend on time, but again inertia is ignored. Viscoelastic behavior and creep are examples of time-dependent material models. In a dynamic analysis all variables in the problem can depend on time, and inertia eﬀects are included. In general, it is appropriate to perform a static analysis when the natural frequency of the external loading is much lower than the lowest natural fre quency of the structure. A rule of thumb is that a static analysis is suﬃcient when the frequency of the loading is less than one-tenth of the lowest natural frequency of the structure. When the frequency of the loading is above this threshold, inertia eﬀects become important and should not be ignored. Transient problems in solids and structures can be separated into two cat egories: wave propagation and structural dynamics. In wave-propagation problems, the frequency of the external loading is very high. Blast and im pact loading are common examples. In wave-propagation analyses, the time duration of the event is, in general, very small — on the order of milliseconds to microseconds. Hence, it may be important to capture the high-frequency information such as stress wave propagation that occurs within these small time scales. In structural dynamics, the main concern is the overall motion of the structure without regard to the high-frequency (small time scale) in formation. Here the events are on the order of seconds to minutes, or longer. At the end of this chapter, we also include a discussion on transient heat conduction. Given an initial temperature distribution, it is often necessary to determine how long it takes for the temperature distribution in a solid body to reach steady state. In addition, heat transfer phenomena may be coupled with structural analysis, such as thermal stress problems where heat transients are important and stress information is desired during the transient heat-conduction event.

257

258

Finite Element Method ti b Ω

Γt

Γu FIGURE 8.1 Newton’s second law applied to a continuum.

8.1

Derivation of the equation of motion

The derivation of the equation of motion is very similar to the derivation of the equilibrium equation derived in Chapter 4. To begin, consider the solid body shown in Figure 8.1. The body is loaded by surface tractions and a body force distribution. The equation of motion is obtained by applying Newton’s second law to the continuum. Assuming that the mass of the body does not change with time, the balance statement says that the net external force vector acting on the body is equal to its total mass times acceleration. Mathematically, the statement can be written as ti dΓ + bi dΩ = ρu ¨i dΩ (8.1) Γ

Ω

Ω

where ρ is the mass density (mass per unit volume) of a point in the contin uum, and u ¨i are the components of acceleration at that point. Throughout this chapter, superposed dots indicate diﬀerentiation with respect to time. The equation of motion can be obtained by converting all integrals in the balance statement (8.1) above into volume integrals, and then invoking the fact that the domain of integration, Ω, is arbitrary as discussed in Section 4.3. Recall that the surface integral on the left-hand side can be converted into a volume integral by utilizing Cauchy’s law and then employing the divergence theorem. Doing this gives (σij nj + bi − ρu ¨i ) dΩ = 0 (8.2) Ω

and hence the equation of motion emerges as

¨i σij,j + bi = ρu

(8.3)

Linear Transient Analysis

259

It is worth mentioning here that the partial diﬀerential equation (8.3) is also known as the wave equation. We seek a displacement ﬁeld, ui = ui (Xi , t), that is a function of both space and time and satisﬁes both the partial diﬀerential equation and the boundary conditions. To describe the boundary conditions, the displacement components can be prescribed functions of time on Γu . This is the essential boundary condition, i.e., ui (Xi , t) = u∗i (Xi , t) on Γu . The traction distribution can also be speciﬁed functions of time on Γt . This is the natural boundary condition. The essential and natural boundary conditions can be summarized as follows: ui (Xi , t) = u∗i (Xi , t) on Γu ti (Xi , t) = t∗i (Xi , t) on Γt

(8.4)

The partial diﬀerential equation (8.3) along with the boundary conditions (8.4) is known as the strong form of the equation of motion.

8.1.1

Weak form

The weak form of the equation of motion is obtained by multiplying the partial diﬀerential equation (8.3) by a vector-valued test function, wi , and then integrating over the domain of the body. The process of deriving the weak form can be written down as follows: σij,j wi dΩ + bi wi dΩ = ρu ¨i dΩ ⇒ Ω Ω Ω (σij wi ),j dΩ − σij wi,j dΩ + bi wi dΩ = ρu ¨i dΩ (8.5) Ω

Ω

Ω

Ω

The last step is to apply the divergence theorem to the ﬁrst integral on the second line of (8.5) above and then to recognize that σij nj = t∗i on Γt from Cauchy’s law. Finally, the weak form can be written as σij wi,j dΩ + ρu ¨i wi dΩ = bi wi dΩ + t∗i wi dΓ (8.6) Ω

Ω

Ω

Γt

Note that it is only necessary to carry out the integration over Γt , because the test function is deﬁned to be zero on Γu .

8.2

Semi-discrete equations of motion

Let us now proceed and substitute the usual ﬁnite element approximations into the weak form. It is convenient to begin the process by writing down the weak form over a single element and then substituting the approximate

260

Finite Element Method

ﬁelds into that statement. This procedure provides the element matrices and element vectors that are needed to form the global discretized system of equations. The ﬁrst step in obtaining the discretized equations is to approximate the vector-valued test function, w, as follows: ˜ = Ne we = weT NTe w

(8.7)

where, again, we contains the values of the test function at the nodal points. We emphasize that the test function does not depend on time; it only depends ¨ , within the element is on the spatial variables. Next, the acceleration ﬁeld, u approximated as ¨e ¨ = Ne d ˜ (8.8) u ¨ e is a vector containing the nodal accelerations for the element. That where d is, ⎧ ⎫ ⎨ d¨aX ⎬ ¨ ea = d¨aY d (8.9) ⎩¨ ⎭ daZ where a ranges from 1 to the number of nodes within the element. Substituting equations (8.7) and (8.8) into the weak form written over a single element yields T T T ¨ Be σ dΩ + ρNe Ne dΩ de we Ωe Ωe NTe b dΩ + NTe t∗ dΓ (8.10) = weT Ωe

Γt

For general three-dimensional solid mechanics problems, the quantity σ in (8.10) above is a 6 × 1 vector containing the stress components σ11 , σ22 , σ33 , σ23 , σ13 , and σ12 . The quantities Be and Ne are 6 × 3a and 3 × 3a matrices respectively. For example, for the case of the eight-node brick element, the order of the B-matrix would be 6 × 24. Invoking the arbitrariness of we in equation (8.10) above yields ¨ e + f int = f ext Me d e e where

Me =

Ωe

ρNTe Ne dΩ

is called the consistent mass matrix. In addition, the vector feint = BTe σ dΩ Ωe

(8.11)

(8.12)

(8.13)

Linear Transient Analysis

261

is an element vector that contains the internal nodal forces, and feext = NeT b dΩ + NeT t∗ dΓ Ωe

(8.14)

Γt

is the element vector that contains the equivalent external nodal forces. We note that the element vector containing the internal nodal forces is not com pletely discretized, because we have not yet made any assumptions as to how the stress ﬁeld varies within the element. If we assume linearly elastic behav ior, i.e., σ = C (8.15) ˜ , would where C is the elasticity matrix, then the approximate stress ﬁeld, σ be ˜ = Ce Be de (8.16) σ Hence, for linearly elastic materials, the internal force vector is equal to the element stiﬀness matrix times the vector containing the nodal displacements. In other words, int T Be Ce Be dΩ de = Ke de (8.17) fe = Ωe

The element matrices and vectors that are needed in dynamic analysis of solids and structures are summarized in Table 8.1. After matrix and vector assembly, the global system of equations can be written as ¨ + f int = f ext (8.18) Md For linear dynamic problems the equation is written as ¨ + Kd = f ext Md

(8.19)

The system of equations (8.18), or equivalently (8.19) for linear problems, is referred to as the semi-discrete form. This means that the equations have been discretized in space using the ﬁnite element shape functions, but they are still continuous in time. The semi-discrete form is a system of second-order, ordinary diﬀerential equations that must be integrated in time to obtain the desired time history. TABLE 8.1

Element matrices and vectors for dynamic analysis. Me #

ρNTe Ne dΩ Ωe

feint

# Ωe

BTe σ dΩ

Ke de

feext or

if linear

# Ωe

NTe b dΩ +

# Γt

NTe t∗ dΓ

262

8.2.1

Finite Element Method

Properties of the mass matrix

In the study of linear algebra, a quadratic form is deﬁned as α = xT Qx

(8.20)

where x is an n × 1 vector, Q is an n × n matrix, and α is the scalar that results after performing the required matrix multiplication. Having deﬁned the quadratic form, the matrix Q is deﬁned as positive deﬁnite if the quantity α is greater than zero for all vectors, x, that are not identically zero. In other words, Q is deﬁned to be positive deﬁnite if xT Qx > 0 for all x = 0 The kinetic energy of a body in motion is deﬁned as 1 ρu˙ 2 dΩ KE = 2 Ω

(8.21)

(8.22)

where u˙ is the velocity ﬁeld. The ﬁnite element approximation for the velocity vector over a single element can be written as ˜˙ u = Ne d˙ e (8.23) e

where d˙ e are the nodal velocities. Substituting the ﬁnite element approxima tion for the velocity ﬁeld into the kinetic energy expression (8.22) yields the following expression for the kinetic energy of an element: %$ % $ ˜E = 1 K ρ Ne d˙ e Ne d˙ e dΩ 2 Ωe e 1 ˙T T = de ρNe Ne dΩ d˙ e 2 Ωe 1 T = d˙e Me d˙ e (8.24) 2 where we recognize that the last line in equation (8.24) above is a quadratic form involving the nodal velocity vector and the element mass matrix. Be cause the kinetic energy of an element is always greater than zero (except for the special case when the nodal velocities are all zero), we conclude that the element mass matrix is positive deﬁnite. An important consequence of the fact that the element mass matrix is positive deﬁnite is that it is nonsingular. This means that the global mass matrix that results after assembly is also nonsingular. Thus, the dynamic analysis of an unconstrained body does not cause numerical diﬃculties — unlike static analysis, which requires degrees of freedom to be constrained after assembly to prevent the system matrix from being singular.

Linear Transient Analysis

8.2.2

263

Natural frequencies and normal modes

In the absence of external forces acting on a body, the right-hand side of the semi-discrete equations of motion is zero, and the system of equations is called homogeneous. The homogeneous system of equations is also often referred to as a free-vibration problem. When the system of equations is linear, i.e., f int = Kd, then the global homogeneous system can be written as ¨ + Kd = 0 Md

(8.25)

where we emphasize that the vector, d = d(t), is a continuous function of time. Let us now assume that the solution, d(t), to the system of equations (8.25) is oscillatory in time and has the form d(t) = Aeiωt

(8.26)

where i is the imaginary number, and ω is called the frequency of oscillation (having units of rad/s). The quantity A in equation (8.26) above is a m × 1 vector containing unknown real constants, where m is the number of equa tions in the linear system. The vector A can be viewed as the amplitude of oscillation. It turns out that the time-dependent function, eiωt , can be written in terms of the sine and cosine functions as follows: eiωt = cos ωt + i sin ωt

(8.27)

Equation (8.27) is called Euler’s equation. It can be derived by expanding the functions cos ωt, sin ωt, and eiωt as Taylor series and then judiciously combin ing the resulting expressions (see Chapter 2, Problem 2.12). Diﬀerentiating equation (8.26) ﬁrst once, and then twice, with respect to time yields d˙ = Aiωeiωt ¨ = −Aω 2 eiωt d

(8.28)

After inserting the derivatives (8.28) back into equation (8.25) we obtain K − ω 2 M Aeiωt = 0 (8.29) We now observe that equation (8.29) above is beginning to look like an eigenvalue problem introduced in Chapter 2. In order to get this equation into standard form, we can pre-multiply both sides by the inverse of the mass matrix, M−1 . Doing this yields −1 M K − ω 2 I Aeiωt = 0 (8.30) where I is an m × m identity matrix. We now seek values of ω that satisfy equation (8.30). The only way this equation can be satisﬁed for nontrivial vec tors Aeiωt is when the determinant of the resulting matrix inside the brackets is zero, i.e., M−1 K − ω 2 I = 0 (8.31)

264

Finite Element Method

Hence, ﬁnding the natural frequencies of the linear system (8.25) boils down to ﬁnding the eigenvalues of the resulting m × m matrix M−1 K. In prac tice, the eigenvalues can be obtained by an eigenvalue extraction routine such as the Lanczos algorithm [15]. Once the eigenvalues are obtained, they can be substituted back into equation (8.28), one by one, to obtain the individ ual eigenvectors, A(i) , that correspond to each eigenvalue ωi . The resulting vectors, A(i) , are widely referred to as normal modes or mode shapes.

Example 8.1 To illustrate the process of calculating natural frequencies and normal modes by the ﬁnite element method, let us consider an unconstrained, two-node linear-u element. Recall from our work in Chapter 3 that the element stiﬀness matrix for the linear-u element is A e Ee 1 −1 (8.32) Ke = −1 1 Le where Ae , Le , and Ee are the cross-sectional area, length, and Young’s mod ulus, respectively. The element mass matrix is obtained by evaluating the following integral over the length of the element: Le NTe Ne dX (8.33) Me = Ae ρe 0

where ρe is the mass density of the element (assumed to be constant along its length), and Ne is the 1 × 2 matrix containing the shape functions, i.e., X X Ne = 1− 1×2 Le Le If we now perform the requisite matrix multiplications and integrate the re sulting matrix we obtain ρe Ae Le 2 1 (8.34) Me = 1 2 6 It is interesting that the sum of all of the elements inside the mass matrix (8.33) above equals the total mass of the element. Note that the total mass of the element equals the mass density of the element times the volume of the element. Hence, the total mass is equal to ρe Ae Le . Armed with the element stiﬀness and mass matrices, we can now go ahead 1 and compute the eigenvalues of M− e Ke . Multiplying the element stiﬀness matrix by the inverse of the element mass matrix and then subtracting ω 2 I yields ⎡ ⎤ 6Ee 6Ee 2 − ω − −1 ⎢ ρe L2e ρe L2e ⎥ ⎥ Me Ke − ω 2 I = ⎢ (8.35) ⎣ ⎦ 6Ee 6Ee 2 − −ω ρe L2e ρe L2e

Linear Transient Analysis

265

Evaluating the determinant of the matrix (8.35) above and setting the result equal to zero gives the following characteristic equation: (8.36) ω 2 −12Ee + ω 2 ρe L2e = 0 Solving equation (8.36) for ω gives ω1 = 0 √ 2 3c ω2 = Le

(8.37)

The ﬁrst natural frequency, ω1 = 0, corresponds to a rigid-body translation of an oscillatory the element. The second natural frequency, ω2 , corresponds to extension/compression mode of vibration. The quantity, c = Ee /ρe , that appears on the second line of equation (8.37) is the longitudinal wave speed. The longitudinal wave speed refers to the velocity at which a longitudinal stress wave propagates in a linearly elastic medium with Young’s modulus Ee and mass density ρe . The wave speed in structural materials is very high. For example, for steel with E = 200 GPa and ρ = 7850 kg/m3 , the wave speed is approximately 5048 m/s! The normal modes of vibration for the unconstrained two-node linear-u element can now be obtained by substituting the natural frequencies, ω1 and ω2 , back into equation (8.30) and solving for the normal modes, A(1) and A(2) . For the ﬁrst natural frequency, ω1 = 0, we obtain ⎡ ⎤ 6Ee 6Ee

− (1) 2 ⎥ ⎢ ρe L2e ρ L 0 A e e ⎥ 1 ⎢ = (8.38)

⎣ 6Ee 6Ee ⎦ A(1) 0 2 − ρe L2e ρe L2e Expanding the ﬁrst row in (8.38) above yields 6Ee (1) 6Ee (1) A1 − A =0⇒ 2 ρe Le ρe L2e 2 (1)

(1)

A1 = A2

(8.39)

We come to the same conclusion after expanding the second row. We see from equation (8.39) that the components of the eigenvector A(1) are equal, but the magnitude is arbitrary. It is customary to normalize each eigenvector (normal mode) so that the magnitude is unity. Doing this for A(1) gives

A(1)

⎧√ ⎫ ⎪ ⎪ 2⎪ ⎪ ⎨ ⎬ 2 = √ ⎪ ⎪ 2⎪ ⎪ ⎩ ⎭ 2

(8.40)

266

Finite Element Method

Following the same procedure, the normalized components of the second nor mal mode are found to be ⎧ √ ⎫ ⎪ ⎪ ⎪ 2 ⎬ ⎪ ⎨ 2 (2) A = (8.41) √ ⎪ 2⎪ ⎪ ⎪ ⎩ ⎭ − 2

8.3

Central diﬀerence method

The central diﬀerence method is a widely used method for integrating the semi-discrete equations of motion derived in Section 8.2. The main idea is to “break up” or discretize the equations in time leading to a system of equations that must be solved at each time step. Marching forward in time, from time step to time step, then provides an approximation of the time history of the ﬁeld variable in question. In order to describe the central diﬀerence method, let us focus our attention on Figure 8.2. A generic plot of velocity, v, at a given node in the model versus time, t, is shown in Figure 8.2 (a). The discrete instants of time n − 1, n, and n + 1 are labeled in the ﬁgure. Intermediate instants of time, n − 1/2 and n + 1/2, are also labeled in the ﬁgure. The acceleration of a nodal point is the derivative of the velocity of that node with respect to time. Keeping this in mind, the ﬁrst step in the central diﬀerence method is to approximate the acceleration of the node at time n by the slope of the straight line drawn between the velocity at times n − 1/2 and n + 1/2 as illustrated in the ﬁgure.

v

d Δt

n−1

n − 1/2

n

n+1

n + 1/2

n−1

n − 1/2

n

n+1

n + 1/2

(a) (b) FIGURE 8.2 Central diﬀerence approximations for velocity (a) and displacement (b) versus time.

Linear Transient Analysis

267

Mathematically, this approximation can be written as a = d¨ =

vn+1/2 − vn−1/2 dv ≈ dt Δt

(8.42)

where Δt is the time step (the width of the time interval between time n −1/2 and n + 1/2). Next, we need to use the fact that the nodal velocity is the derivative of the displacement with respect to time. This will allow us to obtain an approximation for the acceleration at time n in terms of the displacement at discrete instants of time. To do this, we now focus on Figure 8.2 (b). Here, a generic plot of the displacement versus time of the same node is illustrated. The velocity at time n + 1/2 is approximated as the slope of the straight line from dn to dn+1 . Hence we can write vn+1/2 ≈

dn+1 − dn Δt

(8.43)

Similarly we make the approximation that the velocity at time n − 1/2 is the slope of the straight line from dn−1 to dn , i.e., vn−1/2 ≈

dn − dn−1 Δt

(8.44)

Finally, substituting equations (8.43) and (8.44) into equation (8.42) yields dn − dn−1 dn+1 − dn − Δt Δt an = d¨n =≈ Δt or dn+1 − 2dn + dn−1 an = (8.45) Δt2 The semi-discrete form at some time n is written as ¨ n = Fn − fn Md

(8.46)

where Fn is the external force vector at time n, and fn is the internal force vector. Note that for linear dynamic problems, the internal force vector is equal to Kdn . Substituting the central diﬀerence approximation (8.45) for the nodal accelerations at time n yields 1 1 Mdn+1 = Fn − fn + M(2dn − dn−1 ) Δt2 Δt2

(8.47)

Next, multiplying both sides of equation (8.47) by Δt2 M−1 gives dn+1 = Δt2 M−1 (Fn − fn ) + 2dn − dn−1

(8.48)

Notice that in equation (8.48) above, if the mass matrix is diagonal, the calculation of the nodal displacement vector at the future instant of time,

268

Finite Element Method

TABLE 8.2

Central diﬀerence algorithm. 1.

Set up initial conditions, dn−1 = dn , d˙ n−1/2 = d˙ n

2.

Set up thetime independent mass matrix, & # M = e Ωe ρe NTe Ne dΩ

3.

Loop over time steps

& #

BTe σ dΩ

3.1

Compute the internal force vector, fn =

3.2

Compute the external force vector, # & # Fn = e Ωe NTe b dΩ + Γe NT t∗ dΓ

3.3

Solve for dn+1

3.4

Compute nodal velocities, d˙ n+1/2 = (dn+1 − dn ) /Δt $ % ¨ n = d˙ n+1/2 − d˙ n−1/2 /Δt Compute nodal accelerations, d

3.5

e

Ωe

3.6 Update displacement and velocity vectors,

dn = dn+1 , d˙ n−1/2 = d˙ n+1/2

4.

Continue time loop

n + 1, does not involve solving a matrix problem at each time step. The calculation simply requires knowledge of the external force vector, the internal force vector, and the nodal displacement vector at the historical instants of time n and n − 1. For this reason, the central diﬀerence time-integration scheme is referred to as an explicit method. A ﬂow chart of the computational procedure for implementation of the central diﬀerence method is given in Table 8.2. The central diﬀerence timeintegration scheme turns out to be a very robust and eﬃcient method for integrating the semi-discrete equations in time. The method can even handle nonlinear material behavior without any diﬃculty, because the method only requires that the internal force vector be calculated at the known equilibrium conﬁguration of the body at time n. On the other hand, one of the major drawbacks of the method is that it is conditionally stable. This means that we must pay particular attention to time-step size selection. If the time step is chosen to be above a critical value, the solution becomes unstable, i.e., the approximate ﬁnite element solution for the nodal displacements tends toward inﬁnity as time marches on. This, of course, is physically unrealistic. In the following subsections we provide some examples to illustrate the use of the central diﬀerence method and to discuss accuracy of the numerical results. We also provide a discussion on the stability of the method and provide guidelines for proper time-step selection.

Linear Transient Analysis

269

θ l

eθ mg

FIGURE 8.3 Simple pendulum.

8.3.1

Dynamic response of a simple pendulum

To demonstrate the use of the central diﬀerence method to obtain the dynamic response of mechanical systems, we consider the example of a simple pendulum as shown in Figure 8.3. A point mass, m, is attached to the end of a massless, rigid link of length l. Given the initial angle, θ0 , between the rigid link and the vertical dashed line shown in the ﬁgure, it is desired to obtain the response of the mass subjected only to the force of gravity. The equation of motion for the point mass can be obtained by summing forces in the direction of the unit vector eθ labeled in the ﬁgure (tangential direction) and setting the resultant force equal to the mass times the acceleration in the tangential direction. Doing this yields −mg sin θ = ms¨

(8.49)

where g is the gravitational constant, and s represents the distance traveled by the mass from some initial starting location along the circular path depicted by the dashed arc. Given the relationship between arc length and the angle θ, i.e., s = lθ, equation (8.49) can be recast as mlθ¨ + mg sin θ = 0

(8.50)

When the angle θ is much less than one radian, then sin θ ≈ θ, and equation (8.50) reduces to g (8.51) θ¨ + θ = 0 l Equation (8.51) above is a second-order ordinary diﬀerential equation in time. We now seek the time history, θ(t), subjected to a set of initial condi-

270

Finite Element Method

tions. In the present example we will take the initial starting angle at time t = 0 to be θ0 = π/30, and we will assume that the pendulum is initially at rest. We will take the gravitational constant and length of the pendulum arm to be g = 9.81 m/s2 , and l = 1 m, respectively. To obtain an approximate solution to equation (8.51) using the central diﬀerence method, we begin by writing down the central diﬀerence approx imation for θ¨ at some discrete instant of time n, and then substituting the approximation back into equation (8.51). The equation of motion at time n is written as g (8.52) θ¨n + θn = 0 l Substituting the central diﬀerence approximation, 1 (θn+1 − 2θn + θn−1 ) θ¨n ≈ Δt2 into equation (8.52) above and simplifying gives gΔt2 − 2 θn + θn−1 = 0 θn+1 + l

(8.53)

(8.54)

By selecting a time step size and taking n = 0 to represent the orientation of the pendulum at time zero, we can now solve equation (8.54) above for the new angle θn+1 . The central diﬀerence equation to be solved can be written as gΔt2 − 2 θn − θn−1 (8.55) θn+1 = − l Notice that the new angle, θn+1 , depends only on historical information on the right-hand side of equation (8.55). Once the new angle is calculated, the new angle then becomes a historical value, and the entire process can be repeated, yielding θn+2 , θn+3 , etc. as the solution marches forward in time. The numerical solution is plotted versus the exact solution of equation (8.51) for four diﬀerent time steps in Figure 8.4. As shown in the ﬁgure, the solution is obtained up to t = 5 s for Δt = 0.05 s, Δt = 0.1 s, Δt = 0.4 s, and Δt = 0.6386 s. In each plot, the numerical result is represented by the circular data points, and the exact solution is depicted by the solid line. Notice that for the smallest time step considered, the agreement between the numerical results and the exact solution is excellent. As the time step is subsequently increased, however, notice that the period (time between suc cessive peaks) becomes shorter. This is known as period error. Notice also that for the largest time step selected, Δt = 0.6386, the angle, θ, becomes unbounded as time marches forward. It turns out that the critical time step in the present problem (to eight digits beyond the decimal point) is 2 = 0.63855086 Δtcrit = g/l

(8.56)

271

0.2

0.2

0.1

0.1 θ (rad)

θ (rad)

Linear Transient Analysis

0 -0.1

0 -0.1 Δt = 0.1

Δt = 0.05 -0.2 0

1

2

3

4

-0.2 0

5

1

2

0.2

2

0.1

1

0

4

5

0 -1

-0.1 Δt = 0.4 -0.2 0

3 t (s)

θ (rad)

θ (rad)

t (s)

1

2

3 t (s)

Δt = 0.6386 4

5

-2

0

1

2

3 t (s)

4

5

FIGURE 8.4 Angle θ plotted versus time for four time steps: Δt = 0.05 s, Δt = 0.1 s, Δt = 0.4 s, and Δt = 0.6386 s.

If we choose a time step that is above the critical time step (even just slightly above) the solution is unstable. A discussion on the stability of the central diﬀerence method and the cal culation of the critical time step is given in the following subsection.

8.3.2

Stability of central diﬀerence method

In order to assess the stability of the central diﬀerence method, it is only necessary to analyze the following single diﬀerential equation: d¨ + ω 2 d = 0

(8.57)

The exact solution to the diﬀerential equation (8.57) above can be written as d(t) = Aeiωt = A cos ωt + B sin ωt

(8.58)

where ω is the natural frequency, and A and B are constants that, in general, can be real or complex.

272

Finite Element Method

Let us now consider the ﬁnite diﬀerence equation corresponding to equation (8.57) at time n as follows: d¨n + ω 2 dn = 0

(8.59)

Substituting the central diﬀerence approximation (8.45) into (8.59) yields dn+1 + ω 2 Δt2 − 2 dn + dn−1 = 0 (8.60) Next, letting Ω2 = ω 2 Δt2 and simplifying, equation (8.60) above can be written as dn+1 + Ω2 − 2 dn + dn−1 = 0 (8.61) where the quantity Ω = ωΔt is called the sampling frequency. The stability of the central diﬀerence method can be examined by assuming that the solution to the diﬀerence equation (8.61) has the form dn = eαnΔt = φn

(8.62)

where Δt is the time step size, and n is the discrete time counter. By com paring equations (8.26) and (8.62), we observe that in order for the diﬀerence solution (8.62) to have the same oscillatory nature as the exact solution, the parameter α needs to be a complex constant. Substituting the assumed form (8.62) into the diﬀerence equation (8.61) gives φn+1 + Ω2 − 2 φn + φn−1 = 0 (8.63) Next, dividing both sides of (8.63) by φn−1 yields the following characteristic equation: φ2 + Ω2 − 2 φ + 1 = 0 (8.64) The characteristic equation (8.64) is simply a quadratic equation. The solu tion for φ is then ' 2 (Ω2 − 2) − 4 2 φ=− Ω −2 ± (8.65) 2 In order for the solution to the diﬀerence equation to have the desired oscillatory behavior, the quantity φ must have both real and imaginary parts. The ﬁrst term on the right-hand side of equation (8.65) is real, because Ω2 is a real positive constant. Hence, in order for φ to have an imaginary part, the quantity inside the radical must be less than zero. Thus,

2 Ω2 − 2 < 4 ⇒ Ω2 − 2 < 2 Ω