Second Edition
FINITE ELEMENT ANALYSIS Thermomechanics of Solids
ß 2008 by Taylor & Francis Group, LLC.
ß 2008 by Taylor & Francis Group, LLC.
Second Edition
FINITE ELEMENT ANALYSIS Thermomechanics of Solids David W. Nicholson
Boca Raton London New York
CRC Press is an imprint of the Taylor & Francis Group, an informa business
ß 2008 by Taylor & Francis Group, LLC.
CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2008 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business 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-13: 978-1-4200-5095-0 (Hardcover) 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. Except as permitted under U.S. Copyright Law, 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 Nicholson, D. W. Finite element analysis : thermomechanics of solids / by David W. Nicholson. -- 2nd ed. p. cm. Includes bibliographical references and index. ISBN-13: 978-1-4200-5095-0 ISBN-10: 1-4200-5095-8 1. Thermal stresses--Mathematical models. 2. Finite element method. I. Title. TA418.58.N53 2008 620.1’121--dc22 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
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2007042632
Dedication To Linda and Mike, with the deepest love
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Contents Preface to the Second Edition Preface to the First Edition Author Acknowledgments Chapter 1 Introduction to the Finite Element Method 1.1 Introduction 1.2 Overview of the Finite Element Method 1.3 Mesh Development Chapter 2 Mathematical Foundations: Vectors and Matrices 2.1 Introduction 2.1.1 Range and Summation Convention 2.1.2 Substitution Operator 2.2 Vectors 2.2.1 Notation: Scalar and Vector Products 2.2.2 Gradient, Divergence, and Curl in Rectilinear Coordinates 2.3 Matrices 2.4 Eigenvalues and Eigenvectors 2.5 Coordinate Transformations 2.5.1 Transformations of Vectors 2.6 Orthogonal Curvilinear Coordinates 2.7 Gradient Operator in Orthogonal Coordinates 2.8 Divergence and Curl of Vectors in Orthogonal Coordinates 2.9 Appendix I: Divergence and Curl of Vectors in Orthogonal Curvilinear Coordinates Chapter 3 Mathematical Foundations: Tensors 3.1 Tensors 3.2 Divergence of a Tensor 3.3 Invariants 3.4 Positive Definiteness 3.5 Polar Decomposition Theorem 3.6 Kronecker Products of Tensors 3.6.1 VEC Operator and the Kronecker Product 3.6.2 Fundamental Relations for Kronecker Products 3.6.3 Eigenstructures of Kronecker Products 3.6.4 Kronecker Form of Quadratic Products 3.6.5 Kronecker Product Operators for Fourth-Order Tensors
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3.6.6 3.6.7
3.7
Transformation Properties of VEC, TEN22, TEN21, and TEN12 Kronecker Expressions for Symmetry Classes in Fourth-Order Tensors 3.6.8 Differentials of Tensor Invariants Examples
Chapter 4 Introduction to Variational Methods 4.1 Introductory Notions 4.2 Properties of the Variational Operator d 4.3 Example: Variational Equation for a Cantilevered Elastic Rod 4.4 Higher Order Variations 4.5 Examples Chapter 5 Fundamental Notions of Linear Solid Mechanics 5.1 Displacement Vector 5.2 Linear Strain and Rotation Tensors 5.3 Examples of Linear Strain and Rotation Tensors 5.4 Traction and Stress 5.5 Equilibrium 5.6 Stress and Strain Transformations 5.7 Principal Stresses and Strains 5.8 Stress–Strain Relations 5.9 Principle of Virtual Work in Linear Elasticity Chapter 6 Thermal and Thermomechanical Response 6.1 Balance of Energy and Production of Entropy 6.1.1 Balance of Energy 6.1.2 Entropy Production Inequality 6.1.3 Thermodynamic Potentials in Reversible Processes 6.2 Classical Coupled Linear Thermoelasticity 6.3 Thermal and Thermomechanical Analogs of the Principle of Virtual Work and Associated Finite Element Equations 6.3.1 Conductive Heat Transfer 6.3.2 Coupled Linear Isotropic Thermoelasticity Chapter 7 One-Dimensional Elastic Elements 7.1 Interpolation Models for One-Dimensional Elements 7.1.1 Rods 7.1.2 Beams 7.1.3 Beam-Columns 7.2 Strain–Displacement Relations in One-Dimensional Elements 7.3 Stress–Strain Relations in One-Dimensional Elements 7.3.1 General 7.3.2 One-Dimensional Members
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7.4
7.5 7.6 7.7 7.8
7.9 7.10 7.11 7.12
Element Stiffness and Mass Matrices from the Principle of Virtual Work 7.4.1 Single-Element Model for Dynamic Response of a Built-in Beam Integral Evaluation by Gaussian Quadrature: Natural Coordinates Unconstrained Rod Elements Unconstrained Elements for Beams and Beam-Columns Assemblage and Imposition of Constraints 7.8.1 Rods 7.8.2 Beams Damping in Rods and Beams General Discussion of Assemblage General Discussion on the Imposition of Constraints Inverse Variational Method Two- and Three-Dimensional Elements in Linear Elasticity and Linear Conductive Heat Transfer Interpolation Models in Two Dimensions 8.1.1 Membrane Plate 8.1.2 Plate with Bending Stresses Only 8.1.3 Plate with Stretching and Bending 8.1.4 Temperature Field in Two Dimensions 8.1.5 Axisymmetric Elements Interpolation Models in Three Dimensions Strain–Displacement Relations and Thermal Analogs 8.3.1 Strain–Displacement Relations: Two Dimensions 8.3.2 Axisymmetric Element 8.3.3 Thermal Analog for Two-Dimensional and Axisymmetric Elements 8.3.4 Three-Dimensional Elements 8.3.5 Thermal Analog in Three Dimensions Stress–Strain Relations 8.4.1 Two-Dimensional Elements 8.4.1.1 Membrane Response 8.4.1.2 Two-Dimensional Members: Bending Response of Thin Plates 8.4.1.3 Element for Plate with Membrane and Bending Response 8.4.2 Axisymmetric Element 8.4.3 Three-Dimensional Element 8.4.4 Elements for Conductive Heat Transfer Stiffness and Mass Matrices and Their Thermal Analogs Thermal Counterpart of the Principle of Virtual Work Conversion to Natural Coordinates in Two and Three Dimensions Assembly of Two- and Three-Dimensional Elements
Chapter 8 8.1
8.2 8.3
8.4
8.5 8.6 8.7 8.8
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Chapter 9 Solution Methods for Linear Problems: I 9.1 Numerical Methods in FEA 9.1.1 Solving the Finite Element Equations: Static Problems 9.1.2 Matrix Triangularization and Solution of Linear Systems 9.1.3 Triangularization of Asymmetric Matrices 9.2 Time Integration: Stability and Accuracy 9.3 Properties of the Trapezoidal Rule 9.4 Integral Evaluation by Gaussian Quadrature 9.5 Modal Analysis by FEA 9.5.1 Modal Decomposition 9.5.2 Comments on Eigenstructure Computation in Large Finite Element Systems Chapter 10 Solution Methods for Linear Problems: II 10.1 Introduction 10.2 Solution Method for an Inverse Problem 10.2.1 Inverse Problem in Elasticity 10.2.2 Existence of a Unique Solution 10.2.3 Nonsingularity Test 10.3 Accelerated Eigenstructure Computation in FEA 10.3.1 Introduction 10.3.2 Problem Statement 10.3.3 Hypersphere Path of Steepest Descent 10.3.4 Hypercircle Search 10.3.5 Eigenvalue Replacement Procedure 10.3.6 Example: Minimum Eigenvalue of the 3 3 3 Hilbert Matrix 10.4 Fourth-Order Time Integration 10.4.1 Introduction 10.4.2 Error Growth in the Newmark Method 10.4.2.1 Undamped Free Vibration 10.4.3 Adams–Moulton Formula 10.4.4 Stepwise and Cumulative Error in the Adams–Moulton Method 10.4.5 Stability Limit on Time Step in the Adams–Moulton Method 10.4.6 Introducing Numerical Damping into the Adams–Moulton Method 10.4.7 AMX: Adams–Moulton Method Applied to Systems with Acceleration 10.4.8 Comments on Filtering to Remove High-Order Modes Chapter 11 Additional Topics in Linear Thermoelastic Systems 11.1 Transient Conductive Heat Transfer in Linear Media 11.1.1 Finite Element Equation 11.1.2 Direct Integration by the Trapezoidal Rule 11.1.3 Modal Analysis in Linear Thermoelasticity
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11.2
11.3 11.4 11.5
11.6
Coupled Linear Thermoelasticity 11.2.1 Finite Element Equation 11.2.2 Thermoelasticity in an Elastic Conductor Incompressible Elastic Media Torsion of Prismatic Bars 11.4.1 Basic Relations Buckling of Elastic Beams and Plates 11.5.1 Euler Buckling of Beam Columns 11.5.1.1 Static Buckling 11.5.1.2 Dynamic Buckling 11.5.2 Euler Buckling of Plates Introduction to Contact Problems 11.6.1 Gap 11.6.2 Point-to-Point Contact 11.6.3 Point-to-Surface Contact
Chapter 12 Rotating and Unrestrained Elastic Bodies 12.1 Finite Elements in Rotation 12.1.1 Angular Velocity and Angular Acceleration Vectors 12.1.2 Velocity and Acceleration in Rotating Coordinates 12.2 Critical Speeds in Shaft Rotor Systems 12.3 Finite Element Analysis for Unconstrained Elastic Bodies 12.3.1 Body Axes 12.3.2 Euler Equations of a Rigid Body 12.3.3 Variational Equations of an Unconstrained Elastic Body 12.3.4 Principle of Virtual Work in Body Coordinates 12.3.5 Numerical Determination of the Current Position of the Body Axes 12.4 Appendix: Angular Velocity Vector in Spherical Coordinates Chapter 13 Aspects of Nonlinear Continuum Thermomechanics 13.1 Introduction 13.2 Nonlinear Kinematics of Deformation 13.2.1 Deformation Gradient Tensor 13.2.2 Lagrangian Strain Tensor 13.2.3 Velocity Gradient Tensor, Deformation Rate Tensor, and Spin Tensor 13.2.4 Differential Volume Element 13.2.5 Differential Surface Element 13.3 Mechanical Equilibrium and the Principle of Virtual Work 13.3.1 Traction Vector and Stress Tensors 13.3.2 Stress Flux 13.3.3 Balance of Mass, Linear Momentum, and Angular Momentum
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13.4 13.5
Principle of Virtual Work under Large Deformation Nonlinear Stress–Strain–Temperature Relations: The Isothermal Tangent Modulus Tensor 13.5.1 Classical Elasticity 13.5.2 Compressible Hyperelastic Materials 13.5.3 Incompressible and Near-Incompressible Hyperelastic Materials 13.5.3.1 Incompressibility 13.5.3.2 Near-Incompressibility 13.5.4 Nonlinear Materials at Large Deformation
Chapter 14 Introduction to Nonlinear FEA 14.1 Introduction 14.2 Types of Nonlinearity 14.3 Newton Iteration 14.4 Combined Incremental and Iterative Methods: A Simple Example 14.5 Finite Stretching of a Rubber Rod under Gravity 14.5.1 Model Problem 14.5.2 Nonlinear Strain–Displacement Relations 14.5.3 Stress and Tangent Modulus Relations 14.5.4 Incremental Equilibrium Relation 14.5.5 Single Element Built-in at One End 14.5.6 On Numerical Solution by Newton Iteration 14.5.7 Assembled Stiffness Matrix for a Two-Element Model of the Rubber Rod under Gravity 14.6 Newton Iteration near a Critical Point 14.7 Introduction to the Arc Length Method Chapter 15 Incremental Principle of Virtual Work 15.1 Incremental Kinematics 15.2 Stress Increments 15.3 Incremental Equation of Balance of Linear Momentum 15.4 Incremental Principle of Virtual Work 15.5 Incremental Finite Element Equation 15.6 Contributions from Nonlinear Boundary Conditions 15.7 Effect of Variable Contact 15.8 Interpretation as Newton Iteration 15.9 Buckling Tangent Modulus Tensors for Thermomechanical Response of Elastomers Introduction Compressible Elastomers
Chapter 16 16.1 16.2
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16.3
16.4 16.5 16.6
16.7
16.8
Incompressible and Near-Incompressible Elastomers 16.3.1 Examples of Expressions for the Helmholtz Potential 16.3.1.1 Invariant-Based Incompressible Models: Isothermal Problems 16.3.1.2 Invariant-Based Models for Compressible Elastomers under Isothermal Conditions 16.3.1.3 Thermomechanical Behavior under Non-Isothermal Conditions Stretch-Ratio-Based Models: Isothermal Conditions Extension to Thermohyperelastic Materials Thermomechanics of Damped Elastomers 16.6.1 Balance of Energy 16.6.2 Entropy Production Inequality 16.6.3 Dissipation Potential 16.6.4 Thermal Field Equation for Damped Elastomers Constitutive Model in Thermoviscohyperelasticity 16.7.1 Helmholtz Free Energy Density 16.7.2 Dissipation Potential Variational Principles and Finite Element Equations for Thermoviscohyperelastic Materials 16.8.1 Mechanical Equilibrium 16.8.2 Thermal Equilibrium Equation Tangent Modulus Tensors for Inelastic and Thermoinelastic Materials Plasticity Tangent Modulus Tensor in Small Strain Isothermal Plasticity Plasticity under Finite Strain 17.3.1 Kinematics 17.3.2 Plasticity Thermoplasticity 17.4.1 Balance of Energy 17.4.2 Entropy Production Inequality 17.4.3 Dissipation Potential 17.4.4 Thermoinelastic Tangent Modulus Tensor Tangent Modulus Tensor in Viscoplasticity 17.5.1 Mechanical Field 17.5.2 Thermoinelasticity: Thermal Field Continuum Damage Mechanics
Chapter 17 17.1 17.2 17.3
17.4
17.5
17.6
Chapter 18 Selected Advanced Numerical Methods in FEA 18.1 Iterative Triangularization of Perturbed Matrices 18.1.1 Introduction 18.1.2 Incremental Finite Element Equation 18.1.3 Iterative Triangularization Procedure
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18.2
18.3
Stiff Arc Length Constraint in Nonlinear FEA 18.2.1 Introduction 18.2.2 Newton Iteration for Nonlinear Finite Element Equations 18.2.3 Newton Iteration without Arc Length Constraint 18.2.4 Arc Length Method 18.2.5 Stiff Arc Length Constraint 18.2.5.1 Stiffness of K* 18.2.5.2 Arc Length Vector Which Maximizes Stiffness: Examples 18.2.5.3 Arc Length Vector Which Maximizes Stiffness: General Argument 18.2.5.4 Numerical Determination of the Optimal Arc Length Vector 18.2.6 Solution Procedure 18.2.6.1 Block Triangularization 18.2.6.2 Solution of the Outer Problem 18.2.6.3 Solution of the Inner Problem Non-Iterative Solution of Finite Element Equations in Incompressible Solids 18.3.1 Introduction 18.3.2 Finite Element Equation for an Incompressible Medium 18.3.3 Uzawa’s Method 18.3.4 Modification to Avoid Iteration
References
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Preface to the Second Edition The first edition of Finite Element Analysis: Thermomechanics of Solids was intended to give a unified but very concise presentation of the finite element method applied to thermomechanics of solids, together with supporting mathematics and continuum mechanics. Of necessity the presentation was selective, attempting to include selected important results while remaining concise and focused on the unity of the method. Coverage was provided on variational and incremental variational principles ensuing from mechanical and thermal field equations and from internal constraints, together with their realization in element formulations. Emphasis was placed on the role of tensors, to which end Kronecker product notation was used throughout. The scope embraced mechanical, thermal, and coupled thermomechanical problems; compressible, incompressible, and near-incompressible materials; static and dynamic problems as well as rotational effects; linear problems as well as problems with material, geometric, and boundary condition nonlinearity; and important numerical methods. The second edition is nearly 170% as long as the first. It is intended to achieve greater integration and balance between introductory and advanced levels, with many more fully solved examples and with advanced materials more concentrated in the latter portion of the monograph. Of course there are a number of worthy topics that have not been possible to include, for example, meshless methods and reduced integration. New coverage includes selected developments in numerical methods, detailing accelerated computations in eigenstructure extraction, time integration, and stiffness matrix triangularization. There is a much more extensive coverage of the arc length method for nonlinear problems. The treatment of rotating bodies and of buckling has been significantly expanded and enhanced. As in the first edition, the intent still has been on presenting and explicating topics in a way that shows the highly unified structure of the finite element method. The preface to the first edition listed monographs that have excellent coverage of many topics not addressed or done justice to. Since then the author has become familiar with two additional monographs that the reader may find beneficial: M.A. Crisfield (1991) and J.N. Reddy (2004). The finite element method has matured to a point that it can accurately and reliably be used, by a careful analyst, for an amazingly wide range of applications. Nonlinear problems with nonlinearities due to geometry, material properties, or boundary conditions such as contact are in many cases accessible to analysis. Problems with instabilities, such as buckling, can be treated. Dynamic problems can be solved, with the caution that eigenvalues and eigenvectors beyond the lowest few natural frequencies should be viewed with skepticism. Enormous progress has been made of the vexatious problem of sliding contact. Several areas in which additional progress would be welcome include more accurate treatment of singularities such as occurring at corners, and integrated rather than staggered treatments of
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mixed field problems. The impact of FEA will accelerate as integration with computeraided design and solid modeling systems progresses and as algorithmic and computational resources permit ‘‘attacking’’ ever more complex and larger-scale applications while insisting on high resolution.
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Preface to the First Edition Thousands of engineers use finite element codes such as ANSYS for thermomechanical and nonlinear applications. Most academic departments offering advanced degrees in mechanical engineering, civil engineering and aerospace engineering offer a first course in the finite element method, and by now almost undergraduates of such programs have some exposure to the finite element method. A number of departments offer a second course. It is hoped that the current monograph will appeal to instructors of such courses. Of course it hopefully will also be helpful to engineers engaged in self-study on nonlinear and thermomechanical finite element analysis. The principles of the finite element method are presented for application to the mechanical, thermal and thermomechanical response, both static and dynamic, of linear and nonlinear solids. It is intended to provide an integrated treatment of . . .
basic principles, material models and contact models (for example linear elasticity, hyperelasticity and thermohyperelasticity) computational, numerical and software design aspects (such as finite element data structures) modeling principles and strategies (including mesh design)
The text is designed for a second course, as a reference work and for self study. Familiarity is assumed with the finite element method at the level of a first graduate or advanced undergraduate course. A first course in the finite element method, for which many excellent books are available, barely succeeds in covering static linear elasticity and linear heat transfer. There is virtually no exposure to nonlinear methods, which are considered topics for a second course. Nor is there much emphasis on coupled thermomechanical problems. However, it is believed that many engineers would benefit from a monograph culminating in nonlinear problems and the associated continuum thermomechanics. Such as text may be used in a formal class or for self study. Many important applications have significant nonlinearity, making nonlinear finite element modeling necessary. As a few examples we mention polymer processing, metalforming, rubber components such as tires and seals, biomechanics and crashworthiness. Many applications combine thermal and mechanical response, such as rubber seals in hot engines. Engineers coping with such applications have access to powerful finite element codes and computers. But they often lack and urgently need an in-depth but compact exposition of the finite element method, providing a foundation for addressing problems. It is hoped the current monograph fills this need. Of necessity a selection to topics has been made and topics are given coverage proportional to the author’s sense of their importance to the understanding of the reader. Topics have been selected with the intent of giving a unified and complete but still compact and tractable presentations. Several other excellent texts and
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monographs have appeared over the years, from which the author has benefited. Four previous texts to which the author is indebted are: 1. Zienkiewicz and Taylor, The Finite Element Method Vols 1,2, McGrawHill, 1989. 2. Kleiber, M., Incremental Finite Element Modeling in Nonlinear Solid Mechanics, Ellis Horwood, Ltd, 1989. 3. Bonet, J. and Wood, R.D., Nonlinear Continuum Mechanics for Finite Element Analysis, Cambridge University Press, 1997. 4. Belytschko, T., Lui, W.K, and Moran, B., Nonlinear Finite Elements for Continua and Structures, J. Wiley and Sons, 2000. The current monograph has the following characteristics: 1. emphasis on use of Kronecker product notation instead of tensor, tensorindicial, Voigt, or traditional finite element matrix–vector notation; 2. emphasis on integrated and coupled thermal and mechanical effects; 3. inclusion of elasticity, hyperelasticity, plasticity and viscoelasticity with thermal effects; 4. inclusion of nonlinear boundary conditions, including contact, in an integrated incremental variational formulation. Regarding (1) Kronecker product algebra (KPA) has been widely used in control theory for many years (Graham, (1982)). It is very compact and satisfies very simple rules: for example the inverse of a Kronecker product of two nonsingular matrices is the Kronecker product of the inverses. Recently a number of extensions of KPA have been introduced and shown to permit compact expressions for otherwise elaborate quantities in continuum and computational mechanics. Examples include: 1. compact expressions for the tangent modulus tensors in hyperelasticity (invariant based and stretch-based; compressible, incompressible and near-incompressible), thermohyperelasticity and finite strain plasticity; 2. a general, compact expression for the tangent stiffness matrix in nonlinear FEA, including nonlinear boundary conditions such as contact. KPA with recent extensions can completely replace other notations in most cases of interest here. In the experience of the author, students experience little difficulty in gaining a command of it. The first three chapters concern mathematical foundations, and Kronecker product notation for tensors is introduced. The next four chapters cover relevant linear and nonlinear continuum thermomechanics, to enable a unified account of the finite element method. Chapters 8 through 15 represent a compact presentation of the finite element method in linear elastic, thermal and thermomechanical media, including solution methods. The final five chapters address nonlinear problems, based on a unified set of incremental variational principles. Material nonlinearity is treated, as is geometric nonlinearity and nonlinearity due to boundary conditions. Several numerical issues in nonlinear analysis are discussed, such as iterative triangularization of stiffness matrices.
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Author David W. Nicholson, PhD, is a professor in the Department of Mechanical, Materials, and Aerospace Engineering at the University of Central Florida (UCF), Orlando, Florida, where he has served as chair (1990–1995) and interim chair (2000–2002). He also holds an appointment in the UCF Department of Mathematics. His teaching has included the finite element method at introductory and advanced levels, continuum mechanics, and advanced dynamics. He earned his SB degree at Massachusetts Institute of Technology in 1966 and his PhD at Yale University in 1971. His 36 years of experience since his doctorate have included research in industry (Goodyear, seven years) and government (the Naval Surface Weapons Center, six years), and faculty positions at Stevens Institute of Technology (1984–1990) and UCF. He has authored over 155 articles on the finite element method, continuum mechanics, fracture mechanics, and dynamics, has authored two monographs, and has served as coeditor of four proceedings volumes. He has served as a technical editor for Applied Mechanics Reviews, as an associate editor of Tire Science and Technology, and as a member of scientific advisory committees for a number of international conferences. He has been an instructor in over 25 short courses and workshops. In 2002, he served as executive chair of the XXIst Southeastern Conference on Theoretical and Applied Mechanics, and as coeditor of the proceedings volume Developments in Theoretical and Applied Mechanics, volume XXI.
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Acknowledgments Finite Element Analysis: Thermomechanics of Solids, second edition, has been made possible through the love and patience of my wife Linda and son Michael, to whom it is dedicated. To my profound love I cheerfully add my immense gratitude. Most deserving of acknowledgment is Mr. Ashok Balasubramaniam, who developed many of the examples appearing in the second edition. His careful attention to the first edition greatly helped to make it achieve my objectives. Much of my understanding of the advanced topics addressed arose through the highly successful research of my then doctoral student, Dr. Baojiu Lin. He continued his invaluable support by providing comments and corrections on the manuscript. A number of new topics in the second edition are the result of our continuing collaboration. Thanks are due to hundreds of graduate students in my courses in continuum mechanics and finite elements over the years. I tested and refined the materials through the courses and benefited immensely from their strenuous efforts to gain command of the course materials. Finally, I would like to express my deep appreciation to Taylor & Francis=CRC Press for giving me the opportunity to prepare this second edition, which I regard as the culminating achievement of my academic career.
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1
Introduction to the Finite Element Method
1.1 INTRODUCTION This monograph is intended to present a concise and unified two-part treatment of finite element analysis (FEA) in thermomechanics. The first part encompasses topics typically found in a first course in FEA. Included are elementary mathematical foundations, an introduction to linear variational principles, the stiffness and mass matrices in linear mechanical and thermal elements, assemblage, eigenstructure determination, and numerical procedures. The second part continues into topics which are appropriate for a second course in FEA. It addresses nonlinearity due to material behavior, geometry and boundary conditions, as well as associated advanced mathematical and numerical topics, incremental variational principles including thermal effects and incompressibility constraints, and accommodation of hyperelasticity, plasticity, viscoplasticity, and damage mechanics. In thermomechanical analysis of members and structures, FEA is an essential resource for computing displacement and temperature fields from known applied loads and heat fluxes. FEA has emerged in recent decades as critical to mechanical and structural designers. Its use is often mandated by standards such as the ASME Pressure Vessel Code, by insurance requirements, and even by law. Its pervasiveness has been promoted by rapid progress in related computer hardware and software, especially computer-aided design (CAD) systems. A large number of comprehensive ‘‘user friendly’’ finite element codes are available commercially. In FEA practice, a design file developed using a CAD system is often ‘‘imported’’ into finite element codes, from which point little or no additional effort often suffices to develop the finite element model consisting of a mesh together with material, constraint, boundary condition, and initial value data. The model is communicated to an analysis module to perform sophisticated thermomechanical analysis and simulation. CAD integrated with an analysis tool such as FEA is an example of computer-aided engineering (CAE). CAE possesses the potential of identifying design problems and improvements much more efficiently, rapidly, and ‘‘costeffectively’’ than purely by ‘‘trial and error.’’ A major FEA application is the determination of stresses and temperatures in a component or member in locations where failure is thought most likely. If the stresses or temperatures exceed allowable or safe values, the product can be redesigned and then reanalyzed. Analysis can also be diagnostic, supporting
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interpretation of product failure data. Analysis can be used to assess performance, for example, by determining whether the design stiffness coefficient for a rubber spring is attained. FEA can serve to minimize weight and cost without loss of structural integrity or reliability.
1.2 OVERVIEW OF THE FINITE ELEMENT METHOD Consider a thermoelastic body with force and heat applied to its exterior boundary. The finite element method serves to determine the displacement vector u(X, t) and the temperature T(X, t) as functions of the undeformed position X and time t. The process of creating a finite element model to support design of a mechanical system may be viewed as having (at least) eight steps: 1. The body is first discretized, i.e., it is modeled as a mesh of finite elements connected at nodes. 2. Within each element interpolation models are introduced to provide approximate expressions for the unknowns, typically u(X, t) and T(X, t), in terms of their nodal values, which now become the unknowns in the finite element model. 3. The strain–displacement relation and its thermal analog are applied to the approximations for u and T to furnish approximations for the (Lagrangian) strain and the thermal gradient. 4. The stress–strain relation and its thermal analog (Fourier’s Law) are applied to obtain approximations to stress S and heat flux q in terms of the nodal values of u and T. 5. Equilibrium principles in variational form are applied using the various approximations within each element, leading to element equilibrium equations. 6. The element equilibrium equations are assembled to provide a global equilibrium equation. 7. Prescribed kinematic and temperature conditions on the boundaries (constraints) are applied to the global equilibrium equations, thereby reducing the number of degrees of freedom and eliminating ‘‘rigid body’’ modes. 8. The resulting global equilibrium equations are then solved using computer algorithms. The output is postprocessed. Initially the output should be compared to data or benchmarks or otherwise validated, to establish that the model correctly represents the underlying mechanical system. If not satisfied, the analyst may revise the finite element model and repeat the computations. When the model is validated, postprocessing, with heavy reliance on graphics, serves to interpret the results, for example, determining whether the underlying design is satisfactory. If problems with the design are identified, the analyst may then choose to revise the design. The revised design is then modeled, and the process of validation and interpretation is repeated.
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1.3 MESH DEVELOPMENT Finite element simulation has classically been viewed as having three stages: preprocessing, analysis, and postprocessing. The input file developed at the preprocessing stage consists of several elements: 1. 2. 3. 4. 5. 6.
Control information (type of analysis, etc.) Material properties (e.g., elastic modulus) Mesh (element types, nodal coordinates, connectivities) Applied force and heat flux data Supports and constraints (e.g., prescribed displacements) Initial conditions (dynamic problems)
In problems without severe stress concentrations, much of the mesh data can be developed conveniently using automatic mesh generation. With the input file developed, the analysis processor is activated. ‘‘Raw’’ output files are generated. The postprocessor module typically contains (interfaces to) graphical utilities, facilitating display of output in the form chosen by the analyst, for example, contours of the Von Mises stress. Two problems arise at this stage: validation and interpretation. The analyst may use benchmark solutions, special cases, or experimental data to validate the analysis. With validation, the analyst gains confidence in, say, the mesh. He=she still may face problems of interpretation, particularly if the output is voluminous. Fortunately, current graphical display systems make interpretation easier and more reliable, such as by displaying high stress regions in vivid colors. Postprocessors often allow the analyst to ‘‘zoom in’’ on regions of high interest, for example, where rubber is highly confined. More recent methods based on virtual reality technology enable the analyst to ‘‘fly through’’ and otherwise become immersed in the model. The goal of mesh design is to select the number and location of finite element nodes and element types so that the associated analyses are sufficiently accurate. Several methods include automatic mesh generation with adaptive capabilities which serve to produce and iteratively refine the mesh, based on a user-selected error tolerance. Even so, satisfactory meshes are not necessarily obtained, so that model editing by the analyst may be necessary. Several practical rules are given below: 1. Nodes should be located where concentrated loads and heat fluxes are applied. 2. Nodes should be located where displacements and temperatures are constrained or prescribed in a concentrated manner, for example, where ‘‘pins’’ prevent movement. 3. Nodes should be located where concentrated springs and masses and their thermal analogues are present. 4. Nodes should be located along lines and surface patches over which pressures, shear stresses, compliant foundations, distributed heat fluxes, and surface convection are applied.
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5. Nodes should be located at boundary points where the applied tractions and heat fluxes experience discontinuities. 6. Nodes should be located along lines of symmetry. 7. Nodes should be located along interfaces between different materials or components. 8. Element aspect ratios (ratio of largest to smallest element dimensions) should be no greater than, say, five. 9. Symmetric configurations should have symmetric meshes. 10. The density of elements should be greater in domains with high gradients. 11. Interior angles in elements should not be excessively acute or obtuse, for example, less than 458 or greater than 1358. 12. Element density variations should be gradual rather than abrupt. 13. Meshes should be uniform in subdomains with low gradients. 14. Element orientations should be staggered to prevent ‘‘bias.’’ In modeling a configuration, a good practice is initially to develop the mesh locally in domains expected to have high gradients, and thereafter to develop the mesh in the intervening low gradient domains, thereby ‘‘reconciling’’ the high gradient domains. There are two classes of errors in FEA: Modeling error ensues from inaccuracies in such input data as the material properties, boundary conditions, and initial values. In addition, there often are compromises in the mesh, for example, modeling sharp corners as rounded. Numerical error is primarily due to truncation and roundoff. As a practical matter, error in a finite element simulation is often assessed by comparing solutions from two meshes, the second of which is a refinement of the first. The sensitivity of finite element computations to error is to some extent controllable. If the condition number of the stiffness matrix (the ratio of the maximum to the minimum eigenvalue) is modest, sensitivity is reduced. Typically, the condition number increases rapidly as the number of nodes in a system grows. In addition, highly irregular meshes tend to produce high condition numbers. Models mixing soft components, for example, made of rubber, with stiff components, such as steel plates, are also likely to have high condition numbers. Where possible, the model should be designed to reduce the condition number.
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2
Mathematical Foundations: Vectors and Matrices
2.1 INTRODUCTION This chapter gives a review of mathematical relations which will prove to be useful in the subsequent chapters. A more complete development is given in Chandrasekharaiah and Debnath (1994).
2.1.1 RANGE
AND
SUMMATION CONVENTION
Unless otherwise noted, repeated Latin indices will imply summation over the range 1 to 3. For example, ai b i ¼
3 X
ai bi ¼ a1 b1 þ a2 b2 þ a3 b3
(2:1)
i¼1
aij bjk ¼ ai1 b1k þ ai2 b2k þ ai3 b3k
(2:2)
The repeated index is ‘‘summed out’’ and therefore ‘‘dummy’’. The quantity aijbjk in Equation 2.2 has two free indices i and k (and later will be shown to be the ikth entry of a second-order tensor). Note that Greek indices do not imply summation. Thus aaba ¼ a1b1 if a ¼ 1.
2.1.2 SUBSTITUTION OPERATOR The quantity dij, later to be called the Kronecker tensor, has the property that ( dij ¼
1,
i¼j
0,
i 6¼ j
For example, dijvj ¼ 1 3 vi, thereby illustrating the substitution property.
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(2:3)
2.2 VECTORS 2.2.1 NOTATION: SCALAR
AND
VECTOR PRODUCTS
Throughout this and the following chapters, orthogonal coordinate systems will be used. Figure 2.1 shows such a system, with base vectors e1, e2, and e3. The scalar product of vector analysis satisfies ei ej ¼ dij
(2:4)
The vector product satisfies 8 ek , i 6¼ j and ijk in right-handed order > > < 6 j and ijk not in right-handed order ei ej ¼ ek , i ¼ > > : 0, i ¼ j
(2:5)
The vector cross product enables introducing the alternating operator «ijk, also known as the ijkth entry of the permutation tensor: «ijk ¼ [ei ej ] ek 8 1, ijk distinct and in right-handed order > > < ¼ 1, ijk distinct but not in right-handed order > > : 0, ijk not distinct
3 v3
v e3 e2 e1 v1 1
FIGURE 2.1 Rectangular coordinate system.
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v2
2
(2:6)
EXAMPLE 2.1 For three 3 3 1 vectors a, b, c, we prove that a (b 3 c) ¼ b (c 3 a).
SOLUTION Writing the triple product in indical notation gives a b c ! ai 2ijk bj ck But 2ijk ¼ 2jki (cyclic order), and hence ai 2ijk bj ck ¼ ai 2jki bj ck ¼ bj 2jki ck ai ¼ b ðc aÞ Similarly a 3 b c ¼ a b 3 c
EXAMPLE 2.2 Verify that 2kij 2klm ¼ dil djm dim djl
SOLUTION A moment’s reflection shows that ij and ‘m must both differ from k and each other in any nonvanishing instances of 2kij2klm. But dildjm dimdjl likewise vanishes under these conditions. Furthermore, if i ¼ ‘ and j ¼ m but i 6¼ j, 2kij2klm ¼ 1, and dildjm dimdjl ¼ 1 0 ¼ 1. If i ¼ m and j ¼ ‘ but i 6¼ j, then 2kij ¼ 2klm and 2kij2klm ¼ 1. But in this case dildjm dimdjl ¼ 0 1 ¼ 1. The relation to be verified is thereby satisfied in conditions exhausing all cases.
Now consider two vectors v and w. It will prove convenient to use two different types of notation. In tensor-indicial notation, denoted by (*T), v and w are represented as
*T)
v ¼ v i ei ,
w ¼ wi ei
(2.7)
Occasionally base vectors are not displayed, so that v is denoted by vi. By displaying base vectors tensor-indicial notation is explicit and minimizes confusion and ambiguity. However, it is also cumbersome. In the current text, the ‘‘default’’ is matrix–vector (*M) notation, illustrated by 0 1 0 1 w1 v1 (2.8) *M) v ¼ @ v2 A, w ¼ @ w2 A v3 w3
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It is very compact, but also risks confusion by not displaying the underlying base vectors. In *M notation the transposes vT and wT are also introduced; they are displayed as ‘‘row vectors’’ *M)
vT ¼ f v1
v2
v3 g,
wT ¼ f w1
w2
w3 g
(2.9)
The scalar product of v and w is written as
*T)
v w ¼ ðvi ei Þ wj ej ¼ vi wj ei ej ¼ vi wj dij ¼ vi w i
(2.10)
The magnitude of v is defined by j vj ¼
*T)
pffiffiffiffiffiffiffiffiffi vv
(2.11)
The scalar product of v and w satisfies *T)
v w ¼ jvjjwj cos uvw
(2.12)
in which uvw is the angle between the vectors v and w. The scalar or dot product is written in matrix–vector notation as v w ! vT w
*M)
(2.13)
The vector or cross product is written as *T)
v w ¼ vi wj ei ej ¼ «ijk vi wj ek
(2.14)
Additional results on vector notation will be presented in the next section, which introduces matrix notation. Finally, the vector product satisfies *T)
jv wj ¼ jvjjwj sin uvw
(2.15)
and n is the unit normal vector perpendicular to the plane containing v and w. The area of the triangle defined by the vectors v and w is given by 12jv 3 wj.
2.2.2 GRADIENT, DIVERGENCE,
AND
CURL
IN
RECTILINEAR COORDINATES
The derivative dw=dx of a scalar f(x) with respect to a vector x is defined implicitly by *M)
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dw ¼
dw dx dx
(2.16)
and it is a row vector whose ith entry is df=dxi. In three-dimensional rectangular coordinates the gradient and divergence operators are defined by 0 1 @ð Þ B @x C B C B @ð Þ C B C (2.17) *M) rð Þ ¼ B C B @y C @ @ð Þ A @z and clearly
*M)
d dx
T ð Þ ¼ rð Þ
(2.18)
The gradient of a scalar function f satisfies the following integral relation: ð
ð rf dV ¼ nf dS
(2:19)
The expression rvT will be seen to be a tensor (see Chapter 3). Clearly rvT ¼ ½rv1 rv2 rv3
(2:20)
from which Equation 2.19 may be invoked to obtain the integral relation ð
ð rvT dV ¼ nvT dS
(2:21)
Next, a most important relation is the divergence theorem. Let V denote the volume of a closed domain, with surface S. Let n denote the exterior surface normal to S and let v denote a vector-valued function of x, the position of a given point within the body. The divergence of v satisfies the integral relation þ
þ *M)
r v dV ¼ T
nT v dS
(2:22)
The curl of the vector v, r 3 v, is expressed by ðr vÞi ¼ «ijk
@ vk @xj
(2:23)
which is the conventional cross product except that the first vector is replaced by the divergence operator. The curl satisfies an integral theorem, analogous to the divergence theorem (Schey, 1973), namely:
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ð
ð r v dV ¼ n v dS
(2:24)
Finally, the reader may verify with some effort that, for a vector v(X), and a path X(S) in which S is the arc length along the path ð
ð v dXðSÞ ¼ n r v dS
(2:25)
The integral between fixed endpoints is single valued if it is path-independent, in which case n r 3 v must vanish. But n is arbitrary since the path is arbitrary, giving the condition for v to have a path-independent integral as rv¼0
(2:26)
EXAMPLE 2.3 Verify that the relation r2v ¼ r(r v) r 3 r 3 v is satisfied in rectangular coordinates. Here r2v ¼ (r r)v is called the Laplacian of v, and r(r v) is recognized as the gradient of the divergence of v.
SOLUTION First note that r(r v) ¼
@ @vj : Next @xi @xj r r v ¼ 2ijk
@ @vm 2klm @xl @xj
¼ 2ijk 2klm
@ @vm @xj @xl
¼ 2kij 2klm
@ @vm @xj @xl
But Example 2.2 presents the relation 2kij2klm ¼ dildjm dimdjl. Accordingly, 2kij 2klm
@ @vm @ @vm ¼ dil djm dim djl @xj @xl @xj @xl @ @vj @ @vi ¼ @xi @xj @xj @xj
Verification is seen by recognizing that r(r v) ¼
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@ @vj @ @vi and r2 v ¼ : @xi @xj @xj @xj
EXAMPLE 2.4 Verify the divergence theorem using the square plate in Figure 2.2, and using v¼
xy xþy
SOLUTION divðvÞ ¼ r v ¼ rT v ¼
@ @ xy ¼2 xþy @x @y
and ð
ð r v dV ¼ 2 dx dy ¼ 2
V
V
For faces (1) and (3), n1 ¼ e1, n3 ¼ e1. ð
ð ðv n1 Þ dS þ S1
ð ðv n3 Þ dS ¼
½ðx yÞe1 þ ð x þ yÞe2 e1 dy x ¼ 1, 0
S3
ð
þ
½ð x yÞe1 þ ð x þ yÞe2 ðe1 Þ dy x ¼ 0, 0
ð1
ð1
¼ ð1 yÞ dy þ y dy ¼ 1 0
0
For faces (2) and (4), n2 ¼ e2, n4 ¼ e2.
Y
1
1
FIGURE 2.2 Test figure for the divergence theorem.
ß 2008 by Taylor & Francis Group, LLC.
X
ð
ð ðv n2 Þ dS þ
S2
ð ðv n4 Þ dS ¼
½ð x yÞe1 þ ð x þ yÞe2 e2 dx 0<x<1, y ¼ 1
S4
ð
½ð x yÞe1 þ ð x þ yÞe2 ðe2 Þ dx
þ 0<x<1, y ¼ 0
ð1
ð1
¼ ð x þ 1Þ dx x dx ¼ 1 0
0
Consequently, ð
ð ðv nÞ dS ¼
S
ð ðv n1 Þ dS þ
S1
ð ðv n2 Þ dS þ
S2
ð ðv n3 Þ dS þ
S3
ðv n4 Þ dS ¼ 2 S4
and the divergence theorem has been verified in the case in question.
EXAMPLE 2.5 In the tetrahedron shown in Figure 2.3, A1, A2, and A3 denote the areas of the faces whose normal vectors point in the e1, e2, and e3 directions, and let A and n denote the area and normal vector of the inclined face. Prove that n¼
A1 A2 A3 e1 þ e2 þ e3 A A A
3
A n e3 A1
A2
e1
A3
e2
1
FIGURE 2.3 Geometry of a tetrahedron.
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2
3
v 2 u 1
FIGURE 2.4 Illustration of the base vectors.
SOLUTION From the definition of the normal vector and referring to the Figure 2.4, we have n¼
uv ju vj
Let a, b, and c represent the length of the edges of the tetrahedron along 1-, 2-, and 3-axes, respectively. Now u ¼ ae1 þ be2
and
v ¼ ae1 þ ce3
; u v ¼ bce1 þ cae2 þ abe3 But bc ¼ 2A1; ca ¼ 2A2; ab ¼ 2A3. Also ju 3 vj ¼ 2A. Hence n¼
A1 A2 A3 e1 þ e2 þ e3 A A A
EXAMPLE 2.6 Prove that if s is a symmetric matrix with entries sij, that «ijk sjk ¼ 0, i ¼ 1, 2, 3
SOLUTION It is given that s is a symmetric tensor. Therefore sjk ¼ skj. We know that,
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«ijk
8 > < 1 if ijk take values in the cyclic order and are distinct ¼ 1 if ijk take values in the acyclic order but are distinct > : 0 two or all of i, j, k take same values
Hence, «ijk sjk ¼ «111 s11 þ «112 s12 þ «113 s13 þ «121 s21 þ «122 s 22 þ «123 s23 þ «131 s31 þ «132 s32 þ «133 s33 þ «211 s11 þ «212 s12 þ «213 s13 þ «221 s21 þ «222 s22 þ «223 s23 þ «231 s31 þ «232 s32 þ «233 s33 þ «311 s11 þ «312 s12 þ «313 s13 þ «321 s21 þ «322 s22 þ «323 s23 þ «331 s31 þ «332 s32 þ «333 s33 ¼ ðs23 s32 Þ ðs13 s31 Þ þ ðs12 s21 Þ ¼0
EXAMPLE 2.7 If v and w are 3 3 1 vectors, prove that v 3 w may be written as v w ¼ Vw in which V is an antisymmetric tensor and v is called the axial vector of V. Derive the expression for V.
SOLUTION Now v w ¼ «ijk vj wk ¼ Vik wk where Vik ¼ «ijk vj ,
½Vik ¼ Vik
Now V11 ¼ «1j1vj ¼ 0. Similarly V22 ¼ V33 ¼ 0. Also V12 ¼ «132 v3 ¼ v3 V21 ¼ «231 v3 ¼ v3 V23 ¼ «213 v1 ¼ v1 V32 ¼ «312 v1 ¼ v1 V31 ¼ «321 v2 ¼ v2 V13 ¼ «123 v2 ¼ v2
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and so
2
0 V ¼ 4 v3 v2
v3 0 v1
3 v2 v1 5 0
Also, V is antisymmetric, i.e., V ¼ VT.
2.3 MATRICES An n 3 n matrix is simply an array of numbers arranged in rows and columns, in which case we call it a second-order array. For the matrix A the entry aij occupies the intersection of the ith row and the jth column. We may also introduce the n 3 1 firstorder array a in which ai denotes the ith entry. We likewise refer to the 1 3 n array aT as first order. In the current context, a first-order array is not a vector unless it is associated with a coordinate system and certain transformation properties to be introduced shortly. In the following all matrices will be real unless otherwise noted. Several properties of first- and second-order arrays are now introduced: (i) Sum of two n 3 n matrices A and B is a matrix C in which cij ¼ aij þ bij. (ii) Product of a matrix A and a scalar q is a matrix C in which cij ¼ qaij. (iii) Transpose of a matrix A, denoted AT, is a matrix in which aijT ¼ aji. A is called symmetric if A ¼ AT, and it is called antisymmetric (or skew symmetric) if A ¼ AT. (iv) Product of two matrices A and B is the matrix C for which *T)
cij ¼ aik bkj
(2:27)
Matrix multiplication may be easily visualized as follows. Let the first-order array aiT denote the ith row of A, while the first-order array bj denotes the jth column of B. Then cij may be written as *T)
cij ¼ aTi bj
(2:28)
(v) Product of a matrix A and a first-order array c is the first-order array d in which the ith entry is di ¼ aij cj. (vi) ijth entry of the identity matrix I is dij. Thus it exhibits ones on the diagonal positions (i ¼ j) and zeroes off-diagonal (i 6¼ j). Clearly I is the matrix counterpart of the substitution operator. (vii) Determinant of A is given by *T)
detðAÞ ¼ 16 «ijk «pqr aip ajq akr
(2:29)
Suppose a and b are two nonzero first-order n 3 1 arrays. If det(A) ¼ 0 the matrix A is singular, in which case there is no solution to equations of the form Aa ¼ b. However, if b ¼ 0, there may be multiple solutions. If det(A) 6¼ 0, then there is a unique nontrivial solution a.
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(viii) Let A and B be n 3 n nonsingular matrices. The determinant has the following useful properties: detðABÞ ¼ detðAÞ detðBÞ det AT ¼ detðAÞ detðIÞ ¼ 1
*M)
(2:30)
(ix) If det(A) 6¼ 0, then A is nonsingular, and there exists an inverse matrix, A1, for which AA1 ¼ A1 A ¼ I
*M)
(2:31)
(x) The transpose of a matrix product satisfies ðABÞT ¼ BT AT
*M)
(2:32)
(xi) The inverse of a matrix product satisfies ðABÞ1 ¼ B1 A1
*M)
(2:33)
(xii) If c and d are two 3 3 1 vectors the vector product c 3 d generates the vector c 3 d ¼ Cd in which C is an antisymmetric matrix given by 2 3 0 c3 c2 *M) C ¼ 4 c3 0 c1 5 (2:34) c2 c1 0 Recalling that c 3 d ¼ «ikjckdj and noting that «ikjck denotes the (ij)th component of an antisymmetric tensor, it is immediate that [C]ij ¼ «ikjck. (xiii) If c and d are two vectors the outer product cdT generates the matrix C given by 2
c1 d1 4 C ¼ c2 d1 c3 d1
*M)
c1 d2 c2 d2 c3 d2
3 c1 d3 c2 d3 5 c3 d3
(2:35)
We will see later that C is a second-order tensor if c and d have the transformation properties of vectors. (xiv) An n 3 n matrix A may be decomposed into symmetric and antisymmetric matrices using A ¼ As þ Aa ,
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As ¼ 12 A þ AT , Aa ¼ 12 A AT
(2:36)
EXAMPLE 2.8 Find the transposes of the matrices A¼
1 1=3
1=2 , 1=4
B¼
1=3 , 1=4
BT ¼
1 1=3 1=2 1=4
(a) Verify that AB 6¼ BA. (b) Verify that (AB)T ¼ BTAT.
SOLUTION Here AT ¼
1 1=2
1 1=2 1=3 1=4
Also AB ¼
1
1=2
1=3
BA ¼
1=3
1 1=3
1=2 1=4
1=4
1 1=3 1=2 1=4
1
¼
3=4
5=24
5=24
7=144
8=9 5=12 1=2 ¼ 5=12 3=16 1=4
and clearly AB 6¼ BA. Now " ðABÞ ¼ T
" B A ¼ T
T
3=4
5=24
5=24
7=144 #" 1 1=2
1 1=3
1=4
#
1=2
1=3
#
1=4
" ¼
3=4
5=24
5=24
7=144
and it is seen that (AB)T ¼ BTAT.
EXAMPLE 2.9 For the matrices in Example 2.8 find the inverses and verify that ðABÞ1 ¼ B1 A1
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#
SOLUTION # " # " 3 6 1=4 1=2 1 A ¼ ¼ 1=12 1=3 1 4 12 # " # " 3 4 1=4 1=3 1 1 B ¼ ¼ 1=12 1=2 6 12 1 " # " # 7=144 5=24 3=4 5=24 1 1 AB ¼ ! (AB) ¼ 1=144 5=24 3=4 5=24 7=144 " # 7 30 ¼ 30 108 1
Now " 1
B A
1
¼
4
3 6
#"
12
) (AB)
1
3
6
4
12
1
#
" ¼
7
30
30
108
1
¼B A
EXAMPLE 2.10 Consider a matrix C given by C¼
a b c d
Verify that its inverse is given by C1 ¼
1 ad bc
d c
b a
SOLUTION The cofactor matrix is given by cof C ¼
d c b a
The adjoint matrix is given by adj C ¼
d b c a
The determinant of the matrix is jCj ¼ ad bc. Hence C1 ¼
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adj C 1 d ¼ jCj ad bc c
b a
#
as expected. Alternatively, note that CC
1
1 1 1 0 d b a b ad bc 0 ¼ ¼ ¼ 0 1 a c d 0 ad bc ad bc c ad bc
EXAMPLE 2.11 Using the matrix C from Example 2.10, and introducing the vectors (one-dimensional arrays) a¼
q , r
b¼
s t
verify that aT Cb ¼ bT CT a
SOLUTION aT Cb ¼ ðq r Þ
a
b
s
¼ ðq r Þ
c d t ¼ aqs þ bqt þ crs þ drt q ¼ ð as þ bt cs þ dt Þ r a c q ¼ ðs t Þ b d r ¼ bT C T a
and accordingly aTCb ¼ bTCTa.
EXAMPLE 2.12 For the geometry of Example 2.4, verify that ð
ð n A dS ¼ r AT dV
using a11 ¼ x þ y þ x2 þ y2 a12 ¼ x þ y þ x2 y2 a21 ¼ x þ y x2 y2 a22 ¼ x y x2 y2
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as þ bt cs þ dt
SOLUTION Given A¼
a11 a21
let
A ¼ ½ b1
b2 , where b1 ¼
a12 a22
a11 a21
and
b2 ¼
a12 a22
Now
@ @ e1 þ e2 ½a11 e1 þ a21 e2 ¼ 2ðx þ yÞe3 @x @y @ @ e1 þ e2 ½a12 e1 þ a22 e2 ¼ 2ðx yÞe3 r b2 ¼ @x @y 2 3 2 3 0 0 0 0 ð ð ð 6 7 6 7 ; r AT dV ¼ ½r b1 r b2 dV ¼ 4 0 0 5 dx dy ¼ 4 0 0 5 r b1 ¼
V
2(x þ y) 2(x y)
V
2 0
For faces (1) and (3), n1 ¼ e1, n3 ¼ e1. Therefore
ð S1
ð ðn1 AÞdS þ ðn3 AÞdS ¼
ð
S3
x¼1, 0
ð e1 ½ b1 b2 dS
ð ¼
2 6 6 4
e1 ½ b1 b2 dS x¼0, 0
0
0
0
0
3 7 7 dy 5
x þ y x2 y2 x y x2 y2 3 2 0 0 ð 7 6 7 dy 6 0 0 5 4 2 2 x¼0, 0
0
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y y2 y y2
0
y y2 y y2
For faces (2) and (4), n2 ¼ e2, n4 ¼ e2. Hence ð
ð ðn2 AÞdS þ ðn4 AÞdS ¼
S2
S4
ð
ð e2 ½ b1 b2 dS
0<x<1, y¼1
0<x<1, y¼0
2
ð 0<x<1, y¼1
2
ð
0
0
0
x þ y þ x2 þ y2
x y þ x2 y2
6 4
þ
3
0
6 4
¼
e2 ½ b1 b2 dS
0
0
0
0
7 5 dx 3 7 5 dx
x þ y þ x2 þ y2 x y þ x2 y2 3 2 0 0 0 0 ð1 7 6 0 0 5 dx þ 4 0 0
0<x<1, y¼0
ð1
2
6 ¼ 4
x þ x2 þ 2 x þ x2
0
0
x þ x2
3
2
0 0
3
7 6 7 5 dx ¼ 4 0 0 5
x þ x2
2 0
Therefore ð
ð
ð
ðn AÞ dS ¼ S
ðn1 AÞ dS þ S1
ð
S2
2
ðn3 AÞ dS þ S3
3 0 0 4 ¼ 0 05 2 0 ð ð Hence ðn AÞ dS ¼ r AT dV is verified. S
ð
ðn2 AÞ dS þ
ðn4 AÞ dS S4
V
2.4 EIGENVALUES AND EIGENVECTORS Again, A is an n 3 n matrix. We now introduce the eigenvalue equation
A lj I xj ¼ 0
(2:37)
The solution for xj is trivial unless A ljI is singular, in which event det (A ljI) ¼ 0. There are n possibly complex roots. If the magnitude of the eigenvectors is set to unity, they may likewise be determined. As an example consider
2 1 A¼ 1 2
(2:38)
The equation det (A ljI) ¼ 0 is expanded as (2 lj)2 1, with roots l1,2 ¼ 1, 3, and A l1 I ¼
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1 1
1 , 1
A l2 I ¼
1 1 1 1
(2:39)
Note that in each case the rows are multiples of each other, so that only one row can be considered independent. We next determine the eigenvectors. It is easily seen that magnitudes of the eigenvectors are arbitrary. For example, if x1 is an eigenvector, so is 10x1. Accordingly, the magnitudes are arbitrarily set to unity. For x1 ¼ fx11 x12 gT , x11 þ x12 ¼ 0,
x211 þ x212 ¼ 1
(2:40)
pffiffiffi from which wepconclude that x1 ¼ f1 1gT= 2. A parallel argument furnishes ffiffi ffi x2 ¼ f1 1gT = 2. If A is symmetric, the eigenvalues and eigenvectors are real and the eigenvectors are orthogonal to each other: xiTxj ¼ dij. The eigenvalue equations can be ‘‘stacked up’’ as follows: 2
l1 60 6 6 A½x1 : x2 : . . . xn ¼ ½x1 : x2 : . . . xn 6 : 6 4 : :
0 l2 : : :
: : : : : : : ln1 : 0
3 : : 7 7 7 : 7 7 05 ln
(2:41)
With obvious identifications, AX ¼ XL
(2:42)
and X may be called the modal matrix. Let yij is the ijth entry of Y ¼ XTX. Now yij ¼ xTi xj ¼ dij
(2:43)
so that Y ¼ I. We conclude that X is an orthogonal tensor: XT ¼ X1. Further XT AX ¼ L,
A ¼ XLXT
(2:44)
and X may be interpreted as representing a rotation from the reference axes to the principal axes.
2.5 COORDINATE TRANSFORMATIONS Suppose that the vectors v and w are depicted in a second coordinate system whose base vectors are denoted by e0j . Now e0j may be represented as a linear sum of the base vectors ei: *T)
e0j ¼ qji ei
(2:45)
But then ei ej0 ¼ qij ¼ cos(uij0 ). It follows that dij ¼ ei0 ej0 ¼ (qik ek) (qjl el) ¼ qikqjldkl, so that
ß 2008 by Taylor & Francis Group, LLC.
qik qjk ¼ qik qTkj
*T)
¼ dij
In *M) notation this is written as QQT ¼ I
*M)
(2:46)
in which case the matrix Q is called orthogonal. An analogous argument serves to prove that QTQ ¼ I. From Equation 2.30, 1 ¼ det(QQT) ¼ det(Q) det(QT) ¼ det2(Q). Righthanded rotations satisfy det(Q) ¼ 1, in which case Q is called proper orthogonal.
EXAMPLE 2.13 Consider the matrix
cos u sin u Q¼ sin u cos u
Verify that
(a) QQT ¼ QTQ (b) QT ¼ Q1 (c) For any 2 3 1 vector a jQaj ¼ jaj (The relation in (c) is general, and the vector Qa represents a rotation of a.)
SOLUTION First consider
QQT ¼ ¼
cos u sin u cos u sin u cos u sin u " cos2 u þ sin2 u
cos u(sin u) þ sin u cos u
sin u cos u þ cos u sin u
QT Q ¼ " ¼
cos u
sin u
sin u
cos u
sin u
sin u
cos u
cos2 u þ sin2 u sin u cos u þ cos u(sin u)
T
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#
cos2 u þ sin2 u
cos u
and QQ ¼ Q Q ¼ I, as expected. T
sin u cos u
¼
1 0 0 1
cos u sin u sin u cos u cos2 u þ sin2 u
#
¼
1 0 0 1
Next, from the foregoing relations Q
1
1 cos u ¼ cos u cos u þ sin u sin u sin u
sin u ¼ QT cos u
Let a¼
a1 a2
Now Qa ¼
cos u sin u sin u cos u
a1 a2
¼
a1 cos u þ a2 sin u a1 sin u þ a2 cos u
from which jQaj ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða1 cos u þ a2 sin uÞ2 þða1 sin u þ a2 cos uÞ2 ¼ a21 þ a22 ¼ jaj
showing that jQaj ¼ jaj.
2.5.1 TRANSFORMATIONS
OF
VECTORS
The vector v0 is the same as the vector v except that v0 is referred to ej0 while v is referred to ei. Now v0 ¼ v0j e0j ¼ v0j qji ei
*T)
¼ vi ei
(2:47)
It follows that vi ¼ vj0 qji, and hence *M)
v ¼ QT v0 ,
v0 ¼ Qv
(2:48)
in which qij is the jith entry of QT. We now state an alternate definition of a vector as a first-order tensor. Let v be an n 3 1 array of numbers referred to a coordinate system with base vectors ei. It is a vector if and only if, upon a rotation of the coordinate system to base vectors ej0 , v0 transformsaccording 0 0 to Equation 2.48. Since df dx dx is likewise equal to df, *M)
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0 df df ¼ QT dx dx
(2:49)
for which reason df=dx is called a covariant vector in a more general presentation, while v is properly called a contravariant vector. Finally, to display the base vectors to which the tensor A is referred (i.e., in tensor-indicial notation), we introduce the outer product ei ^ e j
(2:50)
and it is recognized as the matrix–vector counterpart eiejT. Now A ¼ aij ei ^ ej
(2:51a)
ei ^ ej ek ¼ ei djk
(2:51b)
Note the useful result that
In this notation, given a vector b ¼ bkek, Ab ¼ aij ei ^ ej bk ek ¼ aij bk ei ^ ej ek ¼ aij bk ei djk ¼ aij bj ei
(2:52)
as expected.
2.6 ORTHOGONAL CURVILINEAR COORDINATES The position vector of a point P referred to a three-dimensional rectilinear coordinate system is expressed in tensor-indicial notation as RP ¼ xi ei. The position vector connecting two ‘‘sufficiently close’’ points P and Q is given by DR ¼ RP RQ dRx
(2:53)
dRx ¼ dxi ei
(2:54)
where
with differential arc length dSx ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi dxi dxi
(2:55)
Suppose now that the coordinates are transformed to coordinates yj : xi ¼ xi(yj). The same position vector, now referred to the transformed system, is
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dRy ¼
3 X
dya ga
1
ga ¼ ha ga sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dxj dxj ha ¼ dya dya
(2:56)
dxi dya ffi ei ga ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dxj dxj dya dya ¼
1 dxi ei ha dya
in which ha is called the scale factor. Recall that the use of Greek letters for indices implies no summation. Clearly ga is a unit vector. Conversely, if the transformation is reversed dRy ¼ gi dyi dyi ¼ g dxj dxj i
(2:57)
with the consequence that ej ¼
X dya dyi gi ¼ ha g a dxj dxj a
(2:58)
We restrict attention to orthogonal coordinate systems yj with the property that gTa gb ¼ dab
(2:59)
The length of the vector dRy is now pffiffiffiffiffiffiffiffiffiffiffiffiffi dSy ¼ ha dyi dyi
(2:60)
Under restriction to orthogonal coordinate systems, the initial base vectors ei may be expressed in terms of ga using ei ¼ gTj ei gj 1 @xi g hi @yj j 1 @xi @xk ¼ ek hi hj @yj @yj ¼
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(2:61)
furnishing 1 @xi @xk ¼ dik hi hj @yj @yj
(2:62)
Also of interest is the volume element: the volume determined by the vector dRy is given by the vector triple product dVy ¼ ðh1 dy1 g1 Þ ½h2 dy2 g2 h3 dy3 g3 ¼ h1 h2 h3 dy1 dy2 dy3
(2:63)
and h1h2h3 is called the Jacobian of the transformation. For cylindrical coordinates using r, u, and z as shown in Figure 2.5, x1 ¼ r cos u, x2 ¼ r sin u, and x3 ¼ z. Simple manipulation furnishes that hr ¼ 1, hu ¼ r, hz ¼ 1, and er ¼ cos u e1 þ sin u e2 ,
eu ¼ sin u e1 þ cos u e2 , ez ¼ e3
(2:64)
which of course are orthonormal vectors. Also of later interest are the relations der ¼ eu du and deu ¼ er du. Transformation of the coordinate system from rectilinear to cylindrical coordinates may be viewed as a rotation of the coordinate system through u. Thus if the vector v is referred to the reference rectilinear system and v0 is the same vector referred to a cylindrical coordinate system, in two dimensions 2 v0 ¼ QðuÞv,
cos u
6 QðuÞ ¼ 4sin u 0
sin u
0
7 cos u 0 5 0 1
x3
ez x2 er
eq
q x1
FIGURE 2.5 Cylindrical coordinate system.
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3
r
(2:65)
If v0 is differentiated, for example, with respect to time t, there is a contribution from the rotation of the coordinate system: as an illustration let v and u be functions of time t, d 0 d dQðuÞ v ¼ QðuÞ v þ v dt dt dt @ dQðuÞ T ¼ v0 þ Q ðuÞv0 @t dt
(2:66)
where the partial derivative implies differentiation with u instantaneously held fixed, and 2
3 sin u cos u 0 dQðuÞ 6 7 du ¼ 4 cos u sin u 0 5 dt dt 0 0 1 Now since
dQðuÞ T dt Q ðuÞ
(2:67)
is an antisymmetric matrix V (to be identified later as a tensor)
T dQðuÞ T d dQðuÞ T T QðuÞQ ðuÞ ¼ Q ð uÞ þ Q ð uÞ 0¼ dt dt dt
(2:68)
In fact 0
0
1
dQðuÞ T B Q ðuÞ ¼ @1 0 dt 0 0
0
1
C du 0A dt 0
(2:69)
It follows that d 0 @ v ¼ v0 þ v v0 dt @t
(2:70)
in which v is the axial vector of V. Referring to Figure 2.6, spherical coordinates r, u, f are introduced by the transformation x1 ¼ r cos u cos f,
x2 ¼ r sin u cos f,
x3 ¼ r sin f
(2:71)
The position vector is stated in spherical coordinates as r ¼ x1 e1 þ x2 e2 þ x3 e3 ¼ r cos u cos f e1 þ r sin u cos f e2 þ r sin f e3
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(2:72)
x3
ef
er
f
p f — r
f
x2 eq
q
q
x1
FIGURE 2.6 Spherical coordinate system.
Of course er has the same direction as the position vector: r ¼ rer. It follows that er ¼ cos u cos f e1 þ sin u cos f e2 þ sin f e3
(2:73)
Following the general procedure in the foregoing paragraphs: @x1 @x1 @x1 ¼ cos u cos f, ¼ r sin u cos f, ¼ r cos u sin f @r @u @f @x2 @x2 @x2 ¼ sin u cos f, ¼ r cos u cos f, ¼ r sin u sin f @r @u @f @x3 @x3 @x3 ¼ sin f, ¼ 0, ¼ r cos f @r @u @f
(2:74)
The differential of the position vector furnishes dr ¼ dr er þ r cos f du eu þ r df ef er ¼ cos u cos f e1 þ sin u cos f e2 þ sin f e3 eu ¼ sin u e1 þ cos u e2 ef ¼ sin f ½cos u e1 þ sin u e2 þ cos f e3 e1 ¼ cos u cos f er sin u eu sin f cos u ef
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(2:75)
e2 ¼ sin u cos f er þ cos u eu sin f sin u ef e3 ¼ sin f er þ cos f ef The scale factors are seen to be hr ¼ 1, hu ¼ r cos f, hf ¼ r. Consider a vector v in the rectilinear system, denoted as v0 when referred to a spherical coordinate system: v ¼ v1 e1 þ v2 e2 þ v3 e3 ,
v0 ¼ vr er þ vu eu þ vf ef
(2:76)
Eliminating e1, e2, e3 in favor of er, eu, ef and use of *M notation permits writing 2
cos u cos f Qðu, fÞ ¼ 4 sin u sin f cos u
v0 ¼ Qðu, fÞv,
sin u cos f cos u sin f sin u
3 sin f 0 5 cos f
(2:77)
Suppose now that v(t), u, and f are functions of time. As in cylindrical coordinates, d 0 @ v ¼ v0 þ v v0 dt @t where v is the axial vector of 2 dQðuÞ 6 ¼4 dt and
dQðuÞ T dt Q ðuÞ.
sin u cos f cos u cos f 0 cos u sin u sin f
3
(2:78)
After some manipulation, 2
cos u sin f sin u sin f cos f
3
7 du 6 7 df sin u 0 5 þ4 0 0 0 5 dt dt cos u sin f 0 cos u cos f sin u cos f sin f
3 2 3 2 0 0 1 0 cosf 0 dQðuÞ T du 7 df 6 7 6 Q ðuÞ ¼ 4 cosf 0 sinf 5 þ 4 0 0 0 5 dt dt dt 1 0 0 0 sinf 0
(2:79)
2.7 GRADIENT OPERATOR IN ORTHOGONAL COORDINATES The gradient operator in orthogonal coordinates is of great interest owing to its role in formulating the strain tensor, a topic to be encountered in Chapter 5. In rectilinear coordinates, let c be a scalar-valued function of x : c(x). Starting with the chain rule *T)
dc ¼
@c dxi @xi
¼ ½rc dr,
dr ¼ ei dxi , rc ¼ ei
@c @xi
(2:80)
Clearly dc is a scalar and is unaffected by a coordinate transformation. Now suppose that x ¼ x(y): dr0 ¼ gi dyi. Observe that
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@c dxi @ xi X 1 @c ¼ ha dya ha @ ya a # " # " X g @c X a ¼ hb dyb gb ha @ y a a b
dc ¼
(2:81)
implying the identification ð r cÞ0 ¼
X g @c a ha @ ya a
(2:82)
For cylindrical coordinates in tensor-indicial notation with er ¼ gr, eu ¼ gu, ez ¼ gz, @ c eu @ c @c þ þ ez r @u @r @z
(2:83)
@c eu @ c ez @ c þ þ @ r r cos f @ u r @ f
(2:84)
rc ¼ er and in spherical coordinates rc ¼ e r
2.8 DIVERGENCE AND CURL OF VECTORS IN ORTHOGONAL COORDINATES Under orthogonal transformations, the divergence and curl operators are invariant and satisfy the divergence and curl theorems, respectively. Unfortunately, the transformation properties of the divergence and curl operators are elaborate. The reader is referred to texts in continuum mechanics such as Chung (1988). The development is given in Appendix I. Here we simply list the results: let v be a vector referred to rectilinear coordinates, and let v0 denote the same vector referred to orthogonal coordinates. The divergence and curl satisfy ðr vÞ0 ¼
1 @ @ @ h2 h3 v01 þ h3 h1 v02 þ h1 h2 v03 h1 h2 h3 @y1 @y2 @y3
(2:85)
1 @ 0 @ 0 h1 h3 v 3 h2 v2 g1 ð r vÞ ¼ h1 h2 h3 @y2 @y3
# @ 0 @ 0 @ 0 @ 0 h3 v 3 h1 v1 g2 þ h3 h2 v 2 h1 v1 g3 h2 @y1 @y3 @y1 @y2 0
(2:86)
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In cylindrical coordinates: 1 @ ðrvr Þ 1 @vu @vz þ þ (2:87) @z r @r r @u
1 @vz @ ðrvu Þ @vz @vr @ ðrvu Þ @vr 0 r eu þ ez (2:88) er ðr vÞ ¼ @u @r @z @u r @z @r ð r vÞ 0 ¼
2.9 APPENDIX I: DIVERGENCE AND CURL OF VECTORS IN ORTHOGONAL CURVILINEAR COORDINATES Derivatives of Base Vectors: In tensor-indicial notation, a vector v may be represented in rectilinear coordinates as v ¼ vkek. In orthogonal curvilinear coordinates it is written as X X g v0a ga ¼ v0a a v0 ¼ ha a a A line segment dr ¼ dxi ei transforms to dr0 ¼ dyk gk. Recall that ek ¼
X @yb @yl gl ¼ hb g @xk @xk b b
@xa ga ¼ ek , @yk
(2A:1)
pffiffiffiffiffiffiffiffiffiffiffiffiffi ha ¼ ga ga
From Equation 2A.1, @ga @ 2 xa ¼ ek @yj @yk @yj X a ¼ b
b j
hb gb ,
a
b j
¼
@ 2 xa @yb @yk @yj @xk
(2A:2)
The bracketed quantities are known as Christoffel symbols. From Equations 2A.1 and 2A.2 a a dha dg ¼ ga a ¼ ha (2A:3) dyj dyj j Continuing, @ga 1 @ga ga @ha ¼ @yj ha @yj ha @yj X a 1 ¼ cajb gb , cajb ¼ 1 dab ha b
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b j
hb
(2A:4)
Divergence: The development below is based on the fact that r v ¼ r0 v0 ¼ tr
dv0 dr0
¼ tr
dv dr
(2A:5)
The differential of v0 is readily seen to be dv0 ¼ dvj gj þ vj dgj
(2A:6)
First note that dvj gj ¼ ¼
@vj g dyk @yk j X 1 @vj ha @ya
a
gj ðha dya Þ
X X 1 @vj ¼ g j ^ ga hb gb dyb ha @ya a b X 1 @vj ¼ gj ^ ga dr0 @y h a a a
(2A:7)
Similarly @g j dyk @yk X vj @gj
vj dgj ¼ vj ¼
a
ha @ya
ðha dya Þ
X X vj @gj ^ ga hb gb dyb ha @ya a b X vj @gj ^ ga dr ¼ ha @ya a " !# X vj X dr0 ¼ cjab gb ^ ga ha b a ¼
(2A:8)
Consequently, dv 1 @vj ¼ djb þ vj cjba , dr ba ha @ya
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X 1 @va rv¼ þ vj cjaa ha @ya a
(2A:9)
Curl: In rectilinear coordinates, the individual entries of the curl can be expressed as a divergence as follows. For the ith entry ½r vi ¼ «ijk ¼
@ vk @ xj
@ ðiÞ w , @ xj j
ðiÞ
wj ¼ «jki vk
¼ r wðiÞ
(2A:10)
Consequently, the curl of v may be written as 0 1 r wð1Þ B C r v ¼ @ r wð2Þ A r wð3Þ
(2A:11)
The transformation properties of the curl can be readily induced from Equation 2A.9.
EXAMPLE 2.14 Obtain the expressions for the gradient, divergence, and curl in spherical coordinates.
SOLUTION In spherical coordinates g1 ¼ er , y1 ¼ r,
g2 ¼ eu , y2 ¼ u,
g3 ¼ ef y3 ¼ f
hr ¼ 1, hu ¼ r cos f,
hf ¼ r
On substituting these relations into the expressions for the gradient and divergence operators in the text, we have @c 1 @c 1 @c er þ eu þ ef @r r cos f @u r @f @ 1 @ 2 @ r cos f vr þ ðrvu Þ þ r cos f vf rv¼ 2 r cos f @r @u @f 2 1 @ ðr vr Þ 1 @vu 1 @ cos f vf þ ¼ 2 þ r @r r cos f @u r cos f @f rc ¼
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0
1 @ @ B @u rvf @f ðr cos fvu Þ er C C B B
C C B 1 B @ @ C rv¼ 2 ðvr Þ eu C rvf B r cos f C r cos f B @r @f C B C B
A @ @ @ þr ðr cos fvu Þ ðvr Þ ef @r @u 1 0 @vf @ ðcos fvu Þ er C B @u @f C B C B
C B 1 B @ rvf @vr C ¼ B cos f eu C C r cos f B @f @r C B C B A @ @ ðr cos fvu Þ @vr þ ef @r @u
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3
Mathematical Foundations: Tensors
3.1 TENSORS We now consider two n 3 1 vectors v and w and an n 3 n matrix A such that v ¼ Aw. The important assumption is made that the underlying information in this relation is preserved under rotation of the coordinate system. In particular, simple manipulation furnishes that
*M)
v0 ¼ Qv ¼ QAw ¼ QAQT Qw 0
¼ QAQT w
(3.1)
The square matrix A is now called a second-order tensor if and only if A0 ¼ QAQT. Let A and B be second-order n 3 n tensors. The manipulations below demonstrate that AT, (A þ B), AB, and A1 are likewise tensors. T 0 T A ¼ QAQT ¼ QTT AT QT A0 B0 ¼ QAQT QBQT ¼ QA QQT BQT ¼ QABQT
(3:2)
(3:3)
ðA þ BÞ0 ¼ A0 þ B0 ¼ QAQT þ QBQT ¼ QðA þ BÞQT
(3:4)
1 A01 ¼ QAQT ¼ QT1 A1 Q1 ¼ QA1 QT
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(3:5)
Let x denote an n 3 1 vector. The outer product xxT is a second-order tensor since
xxT
0
¼ x0 x0T ¼ ðQxÞðQxÞT ¼ Q xxT QT
(3:6)
Next d w ¼ dx H dx, 2
H¼
T
d dx
T
dw dx
(3:7)
But dx0T H0 dx0 ¼ ðQdxÞT H0 Q dx ¼ dxT QT H0 Q dx
(3:8)
from which we conclude that the Hessian H is a second-order tensor. @ui Finally, let u be a vector-valued function of x. Then, du ¼ @u @x dx (dui ¼ @xj dxj ) from the chair rule of calculus, from which we conclude that du ¼ dx T
T
@u @x
T (3:9)
But, also from the chain rule, duT ¼ dxT
@uT @xT
(3:10)
We conclude that T @u @uT ¼ T @x @x
(3:11)
Furthermore, if du0 is a vector generated from du by rotation in the opposite sense from the coordinate axes (i.e., clockwise if the axes rotate counterclockwise), then du0 ¼ Q du and dx ¼ QT dx0 . Hence Q is a tensor which may be viewed the counterclockwise rotation of the axes. (Note that x and Qx are vectors, implying that Q is a 0 0 tensor.) Also, since du0 ¼ @u @x0 dx , it is apparent that @u0 @u ¼ Q QT @x @x0
(3:12)
from which we conclude that @u @x is a tensor. We may similarly show that I and 0 are tensors, albeit of a special type (isotropic) owing to the property I0 ¼ I, 00 ¼ 0.
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3.2 DIVERGENCE OF A TENSOR Suppose A is a tensor and b is an arbitrary spatially constant vector of compatible dimension. The divergence of a vector has already been defined. For later purposes we will need to extend the definition ofÐ the divergence to the tensor A. Recall the divergence Ð theorem for the vector c(x): cT n dS ¼ rT c dV. Let c ¼ ATb in which b is an arbitrary constant vector. Now ð ð bT An dS ¼ rT AT b dV ð ¼ rT AT dV b ð T ¼ bT rT AT dV (3:13) Consequently, we seek to define the divergence of A such that ð ð T *M) An dS rT AT dV ¼ 0
(3:14)
In tensor-indicial notation ð ð h T i bi aij nj dS bi rT AT dV ¼ 0
(3:15)
i
Application of the divergence theorem to the vector cj ¼ biaij furnishes ð h T i @ bi dV ¼ 0 aij rT AT i @xj
(3:16)
Since b is arbitrary, we conclude that
@ @ T r T AT i ¼ aij ¼ a @xj @xj ji
(3:17)
and hence, if we are to write r A as a (column) vector (mixing tensor and matrixvector notation) we have r A ¼ [rT AT ]T
(3:18)
It should be evident that (r ) has a more elaborate meaning when applied to a tensor as opposed to a vector. Now, suppose A is written in the form 2 T3 a1 6 T7 (3:19) A ¼ 4 a2 5 aT3
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in which aTi corresponds to the ith row of A: aTi j ¼ aij . It is easily seen that rT AT ¼ rT a1
rT a 2
rT a3 Þ
(3:20)
3.3 INVARIANTS Letting A denote a nonsingular symmetric 3 3 3 tensor, the equation det(A lI) ¼ 0 can be expanded as l3 I1 l2 þ I2 l I3 ¼ 0
(3:21)
I2 ¼ 12 tr2 A tr A2 , I3 ¼ detðAÞ
(3:22)
in which I1 ¼ tr ðAÞ,
Here tr(A) ¼ dijaij denotes the trace of A. Equation 3.21 implies the Cayley– Hamilton theorem A3 I1 A2 þ I2 A I3 I ¼ 0
(3:23)
I3 ¼ 13 tr A3 I1 tr A2 þ I2 tr A A1 ¼ I31 A2 I1 A þ I2 I
(3:24)
from which
Now the trace of any n 3 n symmetric tensor B is invariant under orthogonal transformations (rotations): tr(B0 ) ¼ tr(B) since a0pq dpq ¼ qpr qqs ars dpq ¼ ars qpr qqs ¼ ars drs
(3:25)
Likewise tr(A2) and tr(A3) are invariant since A, A2, and A3 are tensors, and hence I1, I2, and I3 are invariants. Derivatives of invariants are of interest and will be presented in Section 3.6.8.
3.4 POSITIVE DEFINITENESS In the finite element method, an attractive property of some symmetric tensors is positive definiteness, defined as follows. The symmetric n 3 n tensor A is positive definite, written A > 0, if the quadratic product V(A,x) ¼ xTAx > 0 for all nonvanishing n 3 1 vectors x. The importance of this property is shown in the following example. Let P ¼ xTAx xTf, in which f is known and A > 0. After some simple manipulation,
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"
d dx
d2 P ¼ dxT
T
# d P dx dx
¼ dxT A dx
(3:26)
It follows that P is a globally convex function which attains a (global) minimum when Ax ¼ f (dP ¼ 0). This fact can be invoked in the finite element method in classical elasticity to show that the solution of the finite element equations under static conditions represents a minimum. The foregoing definition is equivalent to the statement that the symmetric n 3 n tensor A is positive definite if and only if its eigenvalues are all positive. For the sake of demonstration, if X is the matrix (tensor) which diagonalizes A (cf. Chapter 2), xT Ax ¼ xT XLXT x ¼ yT Ly, y ¼ XT x X li y2i ¼
(3:27)
i
The last expression can be positive for arbitrary y (arbitrary x) only if li > 0, i ¼ 1, 2, . . . , n. The matrix A is semidefinite if xTAx 0, and negative definite (written A < 0) if xTAx < 0. If B is a nonsingular tensor, then BTB > 0, since V(BTB,x) ¼ xTBTBx ¼ yTy > 0 in which y ¼ Bx. If B is singular, for example, if B ¼ yyT where y is an n 3 1 vector, BTB is positive semidefinite since a nonzero eigenvector x of B can be found for which the quadratic product V(BTB,x) vanishes. Now suppose that B is a nonsingular antisymmetric tensor. Then multiplying through Bxj ¼ ljxj with BT furnishes BT Bxj ¼ lj BT xj ¼ lj Bxj ¼ l2j xj
(3:28)
Since BTB is positive definite it follows that l2j > 0 and hence lj is imaginary: pffiffiffiffiffiffi ffi lj ¼ imj using i ¼ 1. Consequently, B2xj ¼ l2j xj ¼ m2j xj, demonstrating that B2 is negative definite.
3.5 POLAR DECOMPOSITION THEOREM For an n 3 n nonsingular matrix B, BTB > 0. If the modal matrix of B is denoted by Xb, we may write BT B ¼ XTb Db Xb 1
1
¼ XTb ðDb Þ2 YYT ðDb Þ2 Xb T 1 1 ¼ XTb ðDb Þ2 Y XTb ðDb Þ2 Y
ß 2008 by Taylor & Francis Group, LLC.
(3:29a)
in which Y is an (unknown) orthogonal tensor. Accordingly, we may in general write 1
B ¼ YT ðDb Þ2 Xb
(3:29b) pffiffiffiffiffiffiffiffiffi To ‘‘justify’’ Equation 3.29b we introduce the tensor-valued square root BT B using 2 pffiffiffiffiffi 3 l1 p0ffiffiffiffiffi : : : 6 0 : 7 l2 : : pffiffiffiffiffiffiffiffiffi 1 1 6 7 7 BT B ¼ XTb D2b Xb , D2b ¼ 6 (3:29c) : : : : : 6 7 4 : : : : p0ffiffiffiffiffi 5 : : : 0 ln in roots are used. It is elementary to verify that pwhich pffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi 2 the positive square BT B ¼ B, and also that BT B > 0. Now note that pffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffiffiffi 1 T pffiffiffiffiffiffiffiffiffi 1 12 2 2 2 pffiffiffiffiffiffiffiffiffi T T T T T B B B B B B B ¼ B B B B B ¼I (3:29d) pffiffiffiffiffiffiffiffiffi 12 Thus, B BT B is an orthogonal tensor, say Z, and hence we may write pffiffiffiffiffiffiffiffiffi B ¼ Z BT B 1
(3:29e) ¼ ZXTb D2b Xb T T T T T Finally, noting that ZXb ZXb ¼ Z Xb Xb Z ¼ ZZT ¼ I, we make the identification YT ¼ ZXTb in Equation 3.29b. Equations 3.29a through 3.29e play a major role in the interpretation of strain tensors to be introduced in subsequent chapters.
3.6 KRONECKER PRODUCTS OF TENSORS 3.6.1 VEC OPERATOR
AND THE
KRONECKER PRODUCT
Let A be an n 3 n (second-order) tensor. Kronecker product notation (Graham 1981) reduces A to a first-order n 3 1 tensor (vector) as follows: VEC ðAÞ ¼ fa11
a21
a31
: : :
an,n1
ann gT
(3:30)
The inverse VEC operator (IVEC) is introduced by the obvious relation IVEC(VEC(A)) ¼ A. The Kronecker product of an n 3 m matrix A and an r 3 s matrix B generates an nr 3 ms matrix as follows: 2 3 a11 B a12 B : : a1m B 6 a21 B : : : : 7 6 7 7 AB¼ 6 : : : : : (3:31) 6 7 4 : : : : : 5 : : : anm B an1 B
ß 2008 by Taylor & Francis Group, LLC.
Now if m, n, r, and s are equal to n and if A and B are tensors, then A B transforms as a second-order n2 3 n2 tensor in a sense to be explained subsequently. Equation 3.31 implies that the n2 3 1 Kronecker product of two n 3 1 vectors a and b is written as 0
1 a1 b B a2 b C B C a b ¼ B .. C @ . A
(3:32)
an b
3.6.2 FUNDAMENTAL RELATIONS
FOR
KRONECKER PRODUCTS
Six basic relations are introduced, followed by a number of subsidiary relations. The proofs of the first five relations are based on Graham (1981). Relation 1: Let A denote an n 3 m real matrix, with entry aij in the ith row and jth column. Let I ¼ ( j 1)n þ i and J ¼ (i 1)m þ j. Let Unm denote the nm 3 nm matrix, independent of A, satisfying
uJK ¼
1, 0,
K ¼ I, K 6¼ I,
uIK ¼
1, 0,
K¼J K 6¼ J
(3:33)
Then VEC AT ¼ Unm VEC ðAÞ
(3:34)
Note that uJK ¼ uJI ¼ 1 and uIK ¼ uIJ ¼ 1, with all other entries vanishing. Hence if m ¼ n, uJI ¼ uIJ, so that Unm is symmetric if m ¼ n. Relation 2: If A and B are second-order n 3 n tensors, then tr ðABÞ ¼ VEC T AT VEC ðBÞ Relation 3: then
(3:35)
If In denotes the n 3 n identity matrix and if B denotes an n 3 n tensor, I n B T ¼ ð I n BÞ T
(3:36)
Relation 4: Let A, B, C, and D, respectively, denote m 3 n, r 3 s, n 3 p, and s 3 q matrices, then ðA BÞðC DÞ ¼ AC BD Relation 5:
(3:37)
If A, B, and C are n 3 m, m 3 r, and r 3 s matrices, then VEC ðACBÞ ¼ BT AVEC ðCÞ
ß 2008 by Taylor & Francis Group, LLC.
(3:38)
Relation 6:
If a and b are n 3 1 vectors, then a b ¼ VEC
abT
T
(3:39)
As proof of Relation (6), if I ¼ ( j 1)n þ i, the Ith entry of VEC(baT) is biaj. It is likewise the Ith entry of a b. Hence a b ¼ VEC(baT) ¼ VEC([abT]T). Symmetry of Unn was established in Relation (1): Unn ¼ Un2. Note that VEC (A) ¼ UvVEC(AT) ¼ U2n2 VEC(A) if A is n 3 n, and hence the matrix Unn satisfies U2n2 ¼ In2 ,
Un2 ¼ UTn2 ¼ U1 n2
(3:40)
Unn is hereafter called the permutation tensor for n 3 n tensors. If A is symmetric (Un2 In2)VEC(A) ¼ 0. If A is antisymmetric (Un2 þ In2)VEC(A) ¼ 0. If A and B are second-order n 3 n tensors, then tr ðABÞ ¼ VEC T ðBÞVEC AT ¼ VEC T ðBÞUn2 VEC ðAÞ ¼ ½Un2 VEC ðBÞT VEC ðAÞ ¼ VEC T BT VEC ðAÞ ¼ trðBAÞ
(3:41)
thereby recovering a well-known relation. If In is the n 3 n identity tensor and in ¼ VEC(In), VEC(A) ¼ In Ain since VEC(A) ¼ VEC(AIn). If In2 is the identity tensor in an n2-dimensional (Euclidean vector) space, In In ¼ In2 since VEC(In) ¼ VEC(InIn) ¼ In InVEC(In). But in ¼ Ini and hence In In ¼ In2. If A, B, and C denote n 3 n tensors, VEC ACBT ¼ In AVEC CBT ¼ ðIn AÞðB In ÞVEC ðCÞ ¼ B AVEC ðCÞ
(3:42)
But by a parallel argument T VEC ACBT ¼ VEC BCT AT ¼ A BVEC CT ¼ A BUn2 VEC ðCÞ
(3:43)
However, the permutation tensor Un2 arises in the n2-dimensional space in the relation T VEC ACBT ¼ Un2 VEC ACBT
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(3:44)
Consequently, if C is arbitrary, Un2 B AVEC ðCÞ ¼ A BUn2 VEC ðCÞ
(3:45)
and, upon using the relation Un2 ¼ U1 n2 , we obtain an important result B A ¼ Un2 A BUn2
(3:46)
If A and B are nonsingular n 3 n tensors, ðA BÞ A1 B1 ¼ AA1 BB1 ¼ In I n ¼ I n2
(3:47)
The Kronecker sum and difference appear frequently (e.g., in control theory) and are defined as follows: A B ¼ A In þ In B,
A B ¼ A In In B
(3:48)
The Kronecker sum and difference of two n 3 n tensors are n2 3 n2 tensors in a sense explained below.
3.6.3 EIGENSTRUCTURES
OF
KRONECKER PRODUCTS
Let aj and bk denote the eigenvalues of A and B, and let yj and zk denote the corresponding eigenvectors. The Kronecker product, sum, and difference have the following eigenstructures: Expression AB AB AB
jkth eigenvalue aj bk aj þ bk aj bk
jkth eigenvector yj zk yj zk yj zk
(3:49)
As proof, aj yj bk zk ¼ aj bk yj zk ¼ Ayj Bzk ¼ (A B)(yj zk )
(3:50)
Now the eigenvalues of A In are 1 3 aj, while the eigenvectors are yj wk in which wk is an arbitrary unit vector (eigenvector of In). The corresponding quantities for In B are bk 3 1 and vj 3 zk, in which vj is an arbitrary eigenvector of In. Upon selecting wk ¼ zk and vj ¼ yj, the Kronecker sum is seen to have eigenvalues aj þ bk with corresponding eigenvectors yj zk.
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3.6.4 KRONECKER FORM
OF
QUADRATIC PRODUCTS
Let R be a second-order n 3 n tensor. The quadratic product aTRb is easily derived: if r ¼ VEC(R) aT Rb ¼ tr baT R T ¼ VEC T baT VEC ðRÞ ¼ VEC T abT VEC ðRÞ ¼ bT aT r
3.6.5 KRONECKER PRODUCT OPERATORS
(3:51) FOR
FOURTH-ORDER TENSORS
Of course fourth-order tensors, and to a lesser extent third-order tensors, play a critical role in continuum thermomechanics and the finite element method. For example, they encompass the stiffnesses relating the stress tensor to the strain tensor. Let A and B be second-order n 3 n tensors and let C be an n 3 n 3 n 3 n matrix. Suppose that A ¼ CB, which is equivalent to aij ¼ cijklbkl in which the range of i, j, k, and l is (1,n). In this case, C is called a fourth-order tensor if there exists such that A0 ¼ C0 B0 . In indicial notation the entries of C0 are related to those of C by 0 cpqmn ¼ qpiqqjqkmqlncijkl. The TEN22 operator is introduced implicitly using VEC ðAÞ ¼ TEN22ðCÞVEC ðBÞ
(3:52)
It ‘‘collapses’’ a fourth-order tensor relating two second-order tensors into a secondorder tensor in n2 3 n2-dimensional space. Note that TEN22ðACBÞVEC ðDÞ ¼ VEC ðACBDÞ ¼ In AVEC ðCBDÞ ¼ In ATEN22ðCÞVEC ðBDÞ ¼ In ATEN22ðCÞIn BVEC ðDÞ
(3:53)
and hence TEN22(ACB) ¼ In A TEN22(C)In B. Upon writing B ¼ C1A, it is immediate that VEC(B) ¼ TEN22(C1)VEC(A). But TEN22(C)VEC(B) ¼ VEC(A) and hence VEC(B) ¼ [TEN22(C)]1VEC(A). We conclude that TEN22(C1) ¼ ˆ BT, it is immediate that Un2a ¼ TEN22(C)Un2b, TEN221(C). Also, writing AT ¼ C and hence TEN22(Cˆ ) ¼ Un2TEN22(C)Un2. The inverse of the TEN22 operator is introduced using the obvious relation ITEN22(TEN22(C)) ¼ C.
3.6.6 TRANSFORMATION PROPERTIES
OF
VEC, TEN22, TEN21, AND TEN12
Suppose that A and B are real second-order n 3 n tensors and C is a fourth-order n 3 n 3 n 3 n tensor such that A ¼ CB. All are referred to a coordinate system denoted as Y. Let the unitary matrix (orthogonal tensor) Qn represent a rotation
ß 2008 by Taylor & Francis Group, LLC.
which gives rise to a coordinate system Y0 and let A0 , B0 , and C0 denote the counterparts of A, B, and C. Now, since A0 ¼ Qn AQTn , VEC ðA0 Þ ¼ Q QVEC ðAÞ
(3:54)
But note that (Q Q)T ¼ QT QT ¼ Q1 Q1 ¼ (Q Q)1. Hence Q Q is a unitary matrix in an n2-dimensional vector space. However, not all rotations in n2-dimensional space can be expressed in the form Q Q. It follows that VEC(A) transforms as an n2 3 1 vector under rotations of the form Qn2 ¼ Q Q. Now write A0 ¼ C0 B0 and observe that Q QVEC ðAÞ ¼ TEN22ðC0 ÞQ QVEC ðBÞ
(3:55)
TEN22ðC0 Þ ¼ Q QTEN22ðCÞðQ QÞT
(3:56)
It follows that
and hence TEN22(C) transforms a second-order n2 3 n2 tensor under rotations of the form Q Q. Finally, let Ca and Cb denote third-order n 3 n 3 n tensors, respectively, which satisfy the relations of the form A ¼ Cab and b ¼ CbA. We introduce the operators TEN21(Ca) and TEN12(Cb) using VEC(A) ¼ TEN21(C)b and b ¼ TEN12(C)VEC(A). The operators satisfy the transformation properties TEN21 C0a ¼ Q QTEN21ðCa ÞQT n2 n TEN12 C0b ¼ QTEN12ðCb ÞQT QT n n2
(3:57)
for which reason we say that TEN21 and TEN12 are tensors of order (2,1) and (1,2), respectively.
3.6.7 KRONECKER EXPRESSIONS TENSORS
FOR
SYMMETRY CLASSES
IN
FOURTH-ORDER
Let C denote a fourth-order tensor with entries cijkl. If the entries observe cijkl ¼ cjikl
(3:58a)
cijkl ¼ cijlk
(3:58b)
cijkl ¼ cklij
(3:58c)
we say that C is totally symmetric. A fourth-order tensor C satisfying Equation 3.58a but not Equations 3.58b and 3.58c will be called symmetric. Kronecker product conditions for symmetry are now stated. The fourth-order tensor C is totally symmetric if and only if
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TEN 22ðCÞ ¼ TEN 22T ðCÞ
(3:59a)
Un2 TEN 22ðCÞ ¼ TEN 22(C)
(3:59b)
TEN 22ðCÞUn2 ¼ TEN 22ðCÞ
(3:59c)
Equation 3.59a is equivalent to symmetry with respect to exchange of ij and kl in C. Total symmetry also implies that, for any second-order n 3 n tensor B, the corresponding tensor A ¼ CB is symmetric. Thus, if a ¼ VEC(A) and b ¼ VEC(B), then a ¼ TEN22(C). Also Un2a ¼ TEN22(C)b. Multiplying through the later expression with Un2 implies Equation 3.58b. Next, for any n 3 n tensor A, the tensor B ¼ C1A is symmetric. It follows that b ¼ TEN22(C1)a ¼ TEN221(C)a, and Un2b ¼ TEN221(C)a. Thus, TEN22(C1) ¼ Un2TEN221(C). Also TEN22(C) ¼ [Un2TEN221(C)]1 ¼ TEN22(C)Un2. The conclusion is immediate that Un2TEN22 (C)Un2 ¼ TEN22(C) if C is totally symmetric. We next prove the following: C1 is totally symmetric, if C is totally symmetric
(3:60)
Note that TEN22(C)Un2 ¼ TEN22(C) implies that Un2TEN22(C1) ¼ TEN22(C1), while Un2TEN22(C) ¼ TEN22(C) implies that TEN22(C1) Un2 ¼ TEN22(C1). Finally, we prove the following: for a nonsingular n 3 n tensor G, GCGT is totally symmetric if and only if C is totally symmetric
(3:61)
First, Equation 3.56 implies that TEN22(GCGT) ¼ I GTEN22(C) I GT, so that TEN22(GCGT) is certainly symmetric. Next consider whether A0 given by A0 ¼ GCGT B0
(3:62)
is symmetric in which B0 is a second-order nonsingular n 3 n tensor. But we may write G1 A0 GT ¼ CG1 B0 GT
(3:63)
Now G1A0 GT is symmetric since C is totally symmetric, and therefore A0 is symmetric. Next consider whether B0 given by the following is symmetric: B0 ¼ GT C1 G1 A0
(3:64)
GT B0 G ¼ C1 G1 A0 G
(3:65)
But we may write
Since C1 is totally symmetric, it follows that GT B0 G is symmetric, and hence B0 is symmetric. We conclude that GCGT is totally symmetric. The ‘‘only if’’ argument follows as a consequence of C ¼ G1 (GCGT )G
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3.6.8 DIFFERENTIALS
OF
TENSOR INVARIANTS
Let A be a symmetric 3 3 3 tensor, with invariants I1(A), I2(A), and I3(A). For a scalar-valued function f (A), @f @f daij ¼ tr dA , df ðAÞ ¼ @aij @A
@f @f ¼ @A ij @aij
(3:66)
But with a ¼ VEC(A), we may also write df ðAÞ ¼ VEC T ¼
@f @A
T ! VEC ðdAÞ
@f da @a
(3:67)
Continuing, @I1 @ T ¼ i a ¼ iT @a @a @I2 @ 1 T 2 T ¼ i a a a @a @a 2 ¼ I1 iT aT
(3:68)
and dI3 ¼ tr A2 dA I1 tr ðA dAÞ þ I2 dA ¼ tr I3 A1 dA
(3:69)
so that @I3 ¼ I3 VEC A1 @a
3.7 EXAMPLES EXAMPLE 3.1 Given a symmetric n 3 n tensor s, prove that tr ðs tr ðsÞIn =nÞ ¼ 0
ß 2008 by Taylor & Francis Group, LLC.
(3:70)
SOLUTION tr ðs trðsÞI n =nÞ ¼ tr s tr ðsÞ trðIn Þ=n:
But tr ðI n Þ ¼ n:
tr ðs trðsÞI n =nÞ ¼ tr s tr ðsÞ ¼ 0
Hence
EXAMPLE 3.2 Verify using 2 3 2 tensors that tr ðABÞ ¼ trðBAÞ
SOLUTION Let A¼
a b , c d
B¼
e f g h
Now
ae þ bg AB ¼ ce þ dg ae þ cf BA ¼ ag þ ch
af þ bh ! tr ðABÞ ¼ ae þ bg þ cf þ dh cf þ dh be þ df ! tr ðBAÞ ¼ ae þ bg þ cf þ dh bg þ dh
Hence tr (AB) ¼ tr (BA).
EXAMPLE 3.3 Express I3 as a function of I1 and I2.
SOLUTION We know that I1 ¼ tr(A). Also I2 ¼ 12 tr2 ðAÞ tr A2 ¼ 12 I12 tr A2 from which tr(A2) ¼ I12 2I2. From the Cayley–Hamilton relation A3 I1A2 þ I2 A I3I ¼ 0, it follows that I3 ¼ 13 tr A3 I1 tr A2 þ I2 tr ðAÞ ¼ 13 tr A3 I13 þ 3I1 I2
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Now using the relations I1 ¼ a1 þ a2 þ a3, I2 ¼ a1a2 þ a2a3 þ a3a1, I3 ¼ a1a2a3, we find tr(A3 ) ¼ I13 3I3 I1 I2
EXAMPLE 3.4 Using 2 3 2 tensors and 2 3 1 vectors, verify the six relations given for Kronecker products.
SOLUTION Let A¼
a11 a21
Relation 1:
a12 , a22
B¼
b12 c c , C ¼ 11 12 , b22 c21 c22 b1 a1 , b¼ a¼ a2 b2
b11 b21
D¼
d11 d21
d12 d22
VEC(AT) ¼ Un2VEC(A) 0
a11
1
2
u11
Ba C 6u B 12 C 6 21 C¼6 B @ a21 A 4 u31 a22
u41
u12
u13
u22
u23
u32 u42
u33 u43
u14
30
a11
1
C B u24 7 7B a21 C 7B C u34 5@ a12 A u44
a22
implying that 2
U22
Relation 2: First
1 0 60 0 6 ¼6 40 1 0 0
3 0 0 1 07 7 7 0 05 0 1
tr(AB) ¼ VECT(AT)VEC(B) " AB ¼
a11 b11 þ a12 b21
a11 b12 þ a12 b22
a21 b11 þ a22 b21
a21 b12 þ a22 b22
#
tr ðABÞ ¼ a11 b11 þ a12 b21 þ a21 b12 þ a22 b22
(S2:1)
Next AT ¼
a11 a12
ß 2008 by Taylor & Francis Group, LLC.
a21 a22
! VEC T AT ¼ fa11
a12
a21
a22 g
and finally 1 b11 Bb C B 21 C a22 ÞB C @ b12 A 0
VEC T AT VECðBÞ ¼ ð a11
a12
a21
b22 ¼ a11 b11 þ a12 b21 þ a21 b12 þ a22 b22
(S2:2)
thereby verifying Relation 2. Relation 3: Here
In BT ¼ (In B)T I2 ¼
1 0 0 1
and BT ¼
b11 b12
b21 b22
from which 2
b11 6b 6 12 T In B ¼ 6 4 0 0
b21 b22 0 0
0 0 b11 b12
b21 b22 0 0
0 0 b11 b12
3 0 0 7 7 7 b21 5 b22
(S2:3)
Now 2
b11 6b 6 12 T ðIn BÞ ¼ 6 4 0 0 2 b11 6b 6 21 In B ¼ 6 4 0 0
b12 b22 0 0
0 0 b11 b21
3 0 0 7 7 7 b21 5 b22 3 0 0 7 7 7 b12 5 b22
and it is immediate that 2
b11
6 6 b12 ðIn BÞT ¼ 6 6 0 4 0 ¼ In B as expected.
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T
0
3
b21
0
b22
0
0
b11
7 0 7 7 b21 7 5
0
b12
b22
(S2:4)
Relation 4:
(A B)(C D) ¼ AC BD 2
a11 b11 6 a11 b21 a11 B a12 B AB¼ ¼6 4 a21 b11 a21 B a22 B a21 b21
a11 b12 a11 b22 a21 b12 a21 b22
a12 b11 a12 b21 a22 b11 a22 b21
c12 d11 c12 d21 c22 d11 c22 d21
3 c12 d12 c12 d22 7 7 c22 d12 5 c22 d22
3 a12 b12 a12 b22 7 7 a22 b12 5 a22 b22
Similarly, 2
c11 d11 6 c11 d21 CD¼6 4 c21 d11 c21 d21
c11 d12 c11 d22 c21 d12 c21 d22
Now 2
x1 y1 6 x1 y3 ðA BÞðC DÞ ¼ 6 4 x3 y1 x3 y3
x1 y2 x1 y4 x3 y2 x3 y4
x2 y1 x2 y3 x4 y1 x4 y3
3 x2 y2 x2 y4 7 7 x4 y2 5 x4 y4
in which x1 ¼ a11 c11 þ a12 c21 , x2 ¼ a11 c12 þ a12 c22 , x3 ¼ a21 c11 þ a22 c21 , x4 ¼ a21 c12 þ a22 c22 y1 ¼ b11 d11 þ b12 d21 , y2 ¼ b11 d12 þ b12 d22 , y3 ¼ b21 d11 þ b22 d21 , y4 ¼ b21 d12 þ b22 d22 But BD ¼
b11 b21
b12 b22
d11 d21
d12 d22
¼
y1 y3
y2 y4
and now
x BD ðA BÞðC DÞ ¼ 1 x3 BD
x2 BD x4 BD
Also AC ¼
a11 a21
a12 a22
c11 c21
c12 c22
¼
x1 x3
and hence ðA BÞðC DÞ ¼ AC BD as expected.
ß 2008 by Taylor & Francis Group, LLC.
x2 x4
Relation 5: VEC(ACB) ¼ BT AVEC(C) Here ACB ¼ ¼ ¼
a11 a12 a21 a22 a11 a12 a21 a22
c11 c12
c21 c22
b11 b12
b21 b22
b11 c11 þ b21 c12 b12 c11 þ b22 c12
b11 c21 þ b21 c22 b12 c21 þ b22 c22
a11 b11 c11 þ a11 b21 c12 þ a12 b11 c21 þ a12 b21 c22 a11 b12 c11 þ a11 b22 c12 þ a12 b12 c21 þ a12 b22 c22
a21 b11 c11 þ a21 b21 c12 þ a22 b11 c21 þ a22 b21 c22 a21 b12 c11 þ a21 b22 c12 þ a22 b12 c21 þ a22 b22 c22
Hence 2
ða11 b11 Þc11 þ ða12 b11 Þc21 þ ða11 b21 Þc12 þ ða12 b21 Þc22
3
7 6 6 ða21 b11 Þc11 þ ða22 b11 Þc21 þ ða21 b21 Þc12 þ ða22 b21 Þc22 7 7 VEC ðACBÞ ¼ 6 6 ða b Þc þ ða b Þc þ ða b Þc þ ða b Þc 7 4 11 12 11 12 12 21 11 22 12 12 22 22 5 ða21 b12 Þc11 þ ða22 b12 Þc21 þ ða21 b22 Þc12 þ ða22 b22 Þc22 30 1 a11 b11 a12 b11 a11 b21 a12 b21 c11 7B C 6 6 a21 b11 a22 b11 a21 b21 a22 b21 7B c21 C 7B C ¼6 7B C 6a b 4 11 12 a12 b12 a11 b22 a12 b22 5@ c12 A 2
" ¼
a21 b12 b11 A b21 A
a22 b12 a21 b22 # b12 A VEC ðCÞ b22 A
a22 b22
c22
¼ BT AVECðCÞ as expected. Relation 6: a b ¼ VEC([abT]T) First ab¼
a1 b a2 b
1 a1 b1 B a1 b2 C C ¼B @ a2 b1 A a2 b2 0
(S2:5)
Now abT ¼
a1 ð b1 a2
b2 Þ ¼
a 1 b1 a2 b1
a1 b2 a2 b2
and also
T T a b ab ¼ 1 1 a1 b2
serving to verify Relation 6.
ß 2008 by Taylor & Francis Group, LLC.
a2 b1 , VEC a2 b2
1 0 a 1 b1 B a1 b2 C T C ¼B abT @ a2 b1 A a2 b2
(S2:6)
EXAMPLE 3.5 Write out the 9 3 9 quantity TEN22(C) in cijkl ¼ 2m(dikdjl) þ ldijdkl, which appears in the Lamé form of the stress–strain relation in linear isotropic elasticity under small strain, namely sij ¼ cijkl«kl.
SOLUTION It is readily verified that C ¼ 2mIn In þ lin iTn ,
in ¼ VEC(In ) (9 1)
Here n ¼ 3. Expansion gives
2
1 I3 6 C ¼ ð2mÞ4 0 I3 0 I3
2
1 60 6 6 60 6 6 60 6 ¼ ð2mÞ6 60 6 60 6 60 6 6 40 0
0 I3 1 I3 0 I3
0 0 0 1 0 0 0 0 0 0
1 0 0 0
0 1 0 0
0 0 0 0 0 0 0 0 0
8 9 1> > > > > > > > > 0> > > > > > >0> > > > > > > > > > 3 > > > > 0 0 I3 > < > = 7 1 0 I3 5 þ l f1 0 > > > > > > 1 I3 0 > > > > > > > > > > > >0> > > > > > > > > 0> > > > : > ; 1 2 3 1 0 0 0 0 0 60 0 0 0 0 07 6 7 6 7 60 0 0 0 0 07 6 7 6 7 0 0 0 0 07 60 6 7 7 1 0 0 0 0 7 þ l6 61 6 7 0 1 0 0 07 60 6 7 60 0 0 1 0 07 6 7 6 7 40 0 0 0 1 05 1 0 0 0 0 1
0 0 1 0 0 0 1g
3 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 07 7 7 0 0 0 0 0 0 0 07 7 7 0 0 0 0 0 0 0 07 7 0 0 0 1 0 0 0 17 7 7 0 0 0 0 0 0 0 07 7 0 0 0 0 0 0 0 07 7 7 0 0 0 0 0 0 0 05 0 0 0 1 0 0 0 1
EXAMPLE 3.6 Prove that the rows of a 3 3 3 tensor A are row vectors.
SOLUTION First note that if f and g are 3 3 1 vectors, the scalar product fTg is invariant: f 0 Tg0 ¼ (Qf)T(Qg) ¼ fTg. Alternately, let g be a 3 3 1 vector but suppose initally that f is simply a 3 3 1 array. If the matrix-algebraic product (Qf)T(Qg) is equal to fTg for all orthogonal 3 3 3 transformation matrices Q, we conclude that the array f is in fact a 3 3 1 vector.
ß 2008 by Taylor & Francis Group, LLC.
To support this statement, suppose instead that f 0 ¼ Qf þ a in which a represents the deviation from the vectorial relationship. But then f 0 Tg0 ¼ (Qf þ a)T(Qg), and consequently it must be the case that aTQg ¼ 0. But this must hold for arbitrary Q, and there is a particular instance of Q, say Qa, which rotates g to be colinear with a. If follows that aTQag can only vanish if a ¼ 0. We now consider whether the rows are transposed vectors. By hypothesis A is a 3 3 3 tensor. We write 2
aT1
3
6 T7 7 A¼6 4 a2 5 aT3 and consider the equation Ab ¼ c in which b and c are 3 3 1 vectors. Now simple manipulation serves to verify that 2
aT1 QT
3
7 6 T T7 AQT ¼ 6 4 a2 Q 5 aT3 QT and so we may write AQT(Qb) ¼ c. Clearly, 3 2 T3 a1 aT1 QT 7 6 6 T7 0 T 6 aT Q 7b ¼ 6 a 7b 5 4 2 4 25 2
aT3 QT
aT3
Since the scalar product of two 3 3 1 vectors is invariant under orthogonal transformations we conclude that 2
aT1 QT
3
2
a0T 1
3
7 6 7 6 6 aT QT 7 ¼ 6 a0T 7 5 4 2 5 4 2 aT3 QT
a0T 3
and in consequence that the rows of a 3 3 3 tensor are row vectors. Of course by using AT we may similarly conclude that the columns of A are column vectors.
EXAMPLE 3.7 Let x denote the 3 3 1 position vector in a body. Also let g be an n 3 1 vector, while A (x) isÐ an n 3 n matrix dependent on position and P0 is a constant n 3 n matrix. Prove that gT AT (x)P0 A(x)g dV may be written in a form such that the integration does not depend on P0.
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SOLUTION From the properties of Kronecker products, gT AT (x)P0 A(x)g ¼ gT gT VEC(AT (x)P0 A(x)) ¼ gT dgT AT (x) AT (x)VEC(P0 ) The integral now reduces to ð gT gT
AT (x) AT (x) dV VEC(P0 ) ¼ dgT Kg ð
K ¼ IVEC
AT (x) AT (x)dV VEC(P0 )
and the integration in the matrix K is independent of the matrix P0. This relation will be of interest later when finite element analysis of buckling is considered.
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4
Introduction to Variational Methods
4.1 INTRODUCTORY NOTIONS In this section we introduce the central notion of the variation. Recall that Chapter 1 described one step in FEA as expressing equilibrium equations as integral equations using variational calculus. Let u(x) be a vector-valued function of position vector x, and consider a vectorvalued functional F(u(x),u0 (x),x), in which u0 (x) ¼ @u=@x. (Just like the definite integral, a functional maps functions, say of x, into numbers.) Next, let v(x) be a function such that v(x) ¼ 0 whenever u(x) ¼ 0, and also v0 (x) ¼ 0 when u0 (x) ¼ 0. Otherwise, v(x) is arbitrary. The differential dF measures how much F changes if x changes. The variation dF, defined below, measures how much F changes if u and u0 change at fixed x. Following Ewing (1985) we introduce the vector-valued function F(e:F) as follows: Fðe:FÞ ¼ FðuðxÞ þ evðxÞ,u0 ðxÞ þ ev0 ðxÞ,xÞ FðuðxÞ,u0 ðxÞ,xÞ
(4:1)
in which e is a scalar ‘‘modulating’’ the difference between u and u þ ev as well as between u0 and u0 þ ev0 . The variation dF is defined by dF dF ¼ e (4:2) de je¼0 with x fixed. Elementary manipulation using differential calculus serves to demonstrate that @F @F 0 (4:3) ev ev þ tr dF ¼ @u @u0 @F 0 @F 0 in which tr @u ¼ @u0 evij . If F ¼ u, then dF ¼ du ¼ ev. If F ¼ u0 , then 0 ev ij 0 0 dF ¼ du ¼ ev . This suggests the notational convention ev ! du and ev0 ! du0 , leading to the expression @F @F 0 du du þ tr (4:4) dF ¼ @u @u0
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4.2 PROPERTIES OF THE VARIATIONAL OPERATOR d The variational operator exhibits five important properties: 1. d() commutes with linear differential operators and integrals. For example, if S denotes a prescribed contour of integration ð
ð dð Þ dS ¼ d
ð Þ dS
(4:5)
2. d( f ) vanishes when its argument f is prescribed. 3. d() satisfies the same operational rules as d(). For example, if the scalars q and r are both subject to variation, dðqr Þ ¼ qdðr Þ þ dðqÞr
(4:6)
4. If f is a prescribed function of (scalar) x and if u(x) is subject to variation, then dð fuÞ ¼ f du
(4:7)
5. Other than for (2), the variation is arbitrary. For example, for two vectors v and w, vT dw ¼ 0 simply implies that v and w are orthogonal to each other. However, vTdw ¼ 0 implies that v ¼ 0, since only the zero vector can be orthogonal to an arbitrary vector. In the current monograph, attention will be restricted to variations with respect to position. There will be no consideration of variation with respect to time. Accordingly d[ f (t)u(x)] ¼ f (t)du(x).
4.3 EXAMPLE: VARIATIONAL EQUATION FOR A CANTILEVERED ELASTIC ROD Determine the variational principle for the system in Figure 4.1 which depicts a rod of length L, cross-sectional area A, and elastic modulus E. At x ¼ 0, it is built in while at x ¼ L the tensile force P is applied. Inertia is neglected. In terms of displacement u, stress S, and (linear) strain E, the governing equations are given by
E,A
L
FIGURE 4.1 Rod under uniaxial tension.
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P
Strain displacement E ¼
du dx
Stress strain S ¼ EE dS ¼0 Equilibrium dx
(4:8)
Combining the equations furnishes EA
d2 u ¼0 dx2
(4:9)
The steps below serve to derive a variational equation which is equivalent to the foregoing differential equation and endpoint conditions (boundary conditions and constraints). Step 1: Multiply by the variation of the variable to be determined (u) and integrate over the domain. ðL duEA
d2 u A dx ¼ 0 dx2
(4:10)
0
Differential equations to be satisfied at every point in the domain have now been replaced with an integral equation whose integrand includes an arbitrary function. Step 2: Integrate by parts as needed to render the argument in the domain integral quadratic. ðL
d du ddu du duEA EA dx ¼ 0 dx dx dx dx
(4:11)
0
To determine whether an integrand is quadratic for variational purposes, disregard the variational operator and derivatives with respect to time. If what is left is quadratic, the integrand is positive definite. We will see that in the finite element method terms with this property give rise to positive definite matrices. Now the first term in Equation 4.11 is the integral of a derivative, so that ðL 0
L ddu du du EA dx ¼ duEA dx dx dx
(4:12)
0
Step 3: Identify the primary and secondary variables. The primary variable is present in the endpoint terms (rhs) under the variational symbol and is u in the current example. The secondary variable conjugate to u is EA du dx .
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Step 4: Satisfy the constraints and boundary conditions. du At x ¼ 0, u is prescribed andhence ¼ 0.At x ¼ du duL,2the load P ¼ EA dx ddu du 1 is prescribed. Also, note that dx EA dx ¼ d 2 EA dx . The right-hand term represents the reason why the left-hand term is deemed quadratic. Step 5: Form the variational equation. The foregoing equations and boundary conditions are consolidated into one integral equation as dF ¼ 0 where 2 1 du dx PuðLÞ EA 2 dx
ðL F¼
(4:13)
0
4.4 HIGHER ORDER VARIATIONS We now consider variations of order higher than unity. The jth variation of a vectorvalued quantity F is defined by dF¼e j
j
d jF dej
(4:14) je¼0
It follows that d2u ¼ 0 and d2u0 ¼ 0. Now restricting F to being a scalar-valued function F and letting x reduce to x, we obtain
d F ¼ fdu 2
T
0T
du gH
du du0
! ,
2
@ T @ F 6 6 @u @u H¼6 6 T 4 @ @ F @u0 @u
3 @ F7 @u0 7 7 T 7 @ @ 5 F @u0 @u0
T
@ @u
(4:15)
and H is known as the Hessian matrix. Now consider G given by ð ð G ¼ F ðx,uðxÞ,u0 ðxÞÞ dV þ hT ðxÞuðxÞ dS
(4:16)
in which V again denotes the volume of a domain and S denotes its surface area. In addition, h is a prescribed (known) function on S. G is called a functional since it generates a number for every function u(x). We first limit attention to a threedimensional rectangular coordinate system and suppose that dG ¼ 0, as in the Principle of Stationary Potential Energy in elasticity. Note that @ @F @F 0 @ @ þ du ¼ tr du F du @x @u0 @u0 @x @u0
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(4:17)
The first and third terms in Equation 4.17 may be recognized as divergences of vectors. We now invoke the divergence theorem to obtain 0 ¼ dG ð ð @F @F 0 du þ tr du ¼ dV þ hT ðxÞduðxÞ dS @u @ u0 ð ð ð @F @ @F T @F du dV þ n ¼ du dS þ hT ðxÞduðxÞ dS @ u @ x @ u0 @ u0
(4:18)
For suitable continuity properties of u, arbitrariness of du implies that dG ¼ 0 is equivalent to the following Euler equation, boundary conditions and constraints (the latter two are not uniquely determined by the variational principle):
8 <
@F @ @F ¼ 0T @u @x @u0 uðxÞ prescribed,
(4:19) x on S1
:nT @F þ hT ðxÞ ¼ 0T , 1 @u0
x on S S1
Let D > 0 denote a constant positive definite and symmetric second order tensor and let p denote a vector which is a nonlinear function of a second vector u that is subject to variation. The function F ¼ 12 pT (u)Dp(u) satisfies dF ¼ duT
@p @u "
T Dp
T T ! # @p @p @ @p D du þ duT d2 F ¼ duT du Dp @u @u @u @u
(4:20)
(4:21)
Since D > 0, if only the first right-hand term were present, the expression would be quadratic with the implication that d2F > 0. However, for a general functional relation between p and u the second right-hand term is not quadratic. Accordingly, the specific vector u* satisfying dF ¼ 0 may correspond to a stationary point which does not constitute a minimum.
4.5 EXAMPLES EXAMPLE 4.1 Directly apply variational calculus to F given by ðL F¼ 0
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2 1 du dx PuðLÞ EA 2 dx
to verify that dF ¼ 0 gives rise to the Euler equation EA
d2 u ¼0 dx2
What endpoint conditions (not unique) are compatible with dF ¼ 0?
SOLUTION Given that 2 1 du EA dx PuðLÞ 2 dx
ðL F¼ 0
variational operations furnish 2 # ðL " 1 du dx PduðLÞ dF ¼ d EA 2 dx 0
ðL
¼ EAdu0 u0 dx PduðLÞ
(4:22)
0
Now consider
d 0 dx ðduu Þ
ðL
¼ du0 u0 þ duu00 . After some manipulation
ðL ðL d 0 0 0 duu EA ð Þ dx ¼ EAdu u dx þ EAduu00 dx dx
0
0
0
in which ðL
0 0
EAdu u dx ¼
L ½ EAduu0 0
0
ðL
EAduu00 dx 0
On substituting Equation 4.23 into Equation 4.22, we have ðL L dF ¼ ½ EAduu0 0 EAduu00 dx PduðLÞ 0
Now dF ¼ 0 is seen to imply that 0
0
ðL
EAu ðLÞduðLÞ EAu ð0Þduð0Þ duEAu00 dx PduðLÞ ¼ 0 0
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from which ðL ½ EAu0 ðLÞ PduðLÞ EAu0 ð0Þduð0Þ duEAu00 dx ¼ 0 0
The domain integral must vanish, and the endpoint expressions must vanish. Hence ðiÞ ½ EAu0 ðLÞ PduðLÞ ¼ 0 ðiiÞ EAu0 ð0Þduð0Þ ¼ 0 ðiiiÞ EAu00 ¼ 0 Note that (iii) states the Euler equation implied by dF ¼ 0. Finally, since du is arbitrary (i) and (ii) have the following implications: ðiÞ
EAu0 jx ¼ L ¼ P
ðiiÞ EAu0 jx ¼ 0 ¼ 0
du ¼ P=EA dx x ¼ L du from which ¼0 dx x ¼ 0
so that
EXAMPLE 4.2 The governing equation for an Euler–Bernoulli beam in Figure 4.2 below is EI
d4 w ¼0 dx4
in which w is the vertical displacement of the neutral (centroidal) axis. The shear force V and the bending moment M satisfy M ¼ EI
d2 w , dx2
V ¼ EI
d3 w dx3
z Neutral axis V0
Y E, I
x L
FIGURE 4.2 Cantilevered beam.
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Using integration by parts twice, obtain the function F such that dF ¼ 0 is equivalent to the foregoing differential equation together with the boundary conditions for a cantilevered beam of length L: wð0Þ ¼ w0 ð0Þ ¼ 0,
M ðLÞ ¼ 0,
V ðL Þ ¼ V 0
SOLUTION On multiplying EIwiv ¼ 0 by dw and integrating over the domain, 0 x L we obtain ðL
dw EIwiv dx ¼ 0
0
Also d ½dw0 EIw00 ¼ dw0 EIw000 þ dw00 EIw00 dx Combining the last two equations furnishes d d ½dwEIw000 ½dw0 EIw00 ¼ dw00 EIw00 dx dx Continuing, ðL
00
ðL
00
dw EIw dx ¼ 0
d ½dwEIw000 dx þ dx
0
ðL
d ½dw0 EIw00 dx dx
0
¼ ½dwEIw000 0 þ ½dw0 EIw00 0 L
L
Note that w(0) ¼ w0 (0) ¼ 0. This implies that dw(0) ¼ dw0 (0) ¼ 0. Also evident are the conditions EIw00 ðLÞ ¼ M ðLÞ ¼ 0 and
EIw000 ðLÞ ¼ V ðLÞ ¼ V0
Some manipulation serves to establish that ðL
dw00 EIw00 dx ¼ dwðLÞV0
0
2L 3 ð 1 2 00 d4 EI ðw Þ dx5 ¼ V0 dwðLÞ 2 0 2L 3 ð 1 2 00 d4 EI ðw Þ dx V0 wðLÞ5 ¼ 0 2 0
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Finally, since dF ¼ 0 we conclude that ðL F¼
1 2 EI ðw00 Þ dx V0 wðLÞ 2
0
EXAMPLE 4.3 Equation combining rod and beam behavior What are the primary variables and the corresponding secondary variables in the following equation? B
@4w @2w @2w A þ Cw þ D ¼0 @x4 @x2 @t 2
in which 0 x L.
SOLUTION The first step is to write ðL 0
4 @ w @2w @2w dw B 4 A 2 þ Cw þ D 2 dV ¼ 0 @x @x @t
ÐL ÐL 2 The third and fourth terms in the integrand, namely 0 dwCw dV and 0 dwD @@tw2 dV are already quadratic. The second and first terms give, respectively, ðL 0
ðL 0
ðL @2w 0 L dw A 2 dV ¼ dw(Aw )j0 þ dw0 Aw0 dV @x 0
ðL @4w 000 L 0 00 L dw B 4 dV ¼ dw(Bw )j0 (dw )(Bw )j0 þ dw00 Aw00 dV @x 0
Combining the last two equations reveals the primary variables to be w, w0 . The corresponding secondary variables are Aw0 þ Bw000 and Bw00 .
EXAMPLE 4.4 Two-dimensional heat conduction Obtain the variational equations corresponding to the following equation in two dimensions. What are the primary variables and the conjugate secondary variables?
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The first equation is for unsteady heat conduction. The second equation has the same form as the wave equation and the biharmonic equation in classical elasticity. kr rT ¼ rce
@T @t
k ¼ thermal conductivity r ¼ mass density ce ¼ specific heat at constant strain SOLUTION The first step is to write ð V
@T dT kr rT rce dV ¼ 0 @t
Applying integration by parts to the first term furnishes ð
ð ð dT[kr rT] dV ¼ r (dT[krT]) dV drT [krT] dV
V
V
V
Using the divergence theorem on the first right-hand term furnishes ð
ð r (dT[krT]) dV ¼ dTn krT dS
V
S
The primary variable is seen to be T owing to its presence under the variational operator in the surface integral. The corresponding secondary variable is n krT, which equals the negative of the normal projection of the heat flux vector q ¼ krT.
EXAMPLE 4.5 Navier equation and plate equation in elasticity Find variational forms of the following equations. What are the primary and secondary variables? (m þ l)
@ @uj @ 2 ui þ m 2 ¼ rui @xi @xj @xj
Navier’s equation
@2 @2w ¼ 0 Elastic Plate equation (static) @x2i @x2j
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SOLUTION Navier’s equation The first step is to write ð
! @ @uj @ 2 ui dui (m þ l) þ m 2 r€ ui dV ¼ 0 @xi @xj @xj
First note that the third term in the integrand, duirüi, is already quadratic. Next consider the middle term in the integrand: @ @ui @dui @ui @ 2 ui ¼ dui m m þ dui m 2 @xj @xj @xj @xj @xj and @ 2 ui @ @ui @dui @ui dui m 2 ¼ dui m m @xj @xj @xj @xj @xj Consequently, using the divergence theorem, the integral of the middle term gives ð dui m
ð ð @ 2 ui @ui @dui @ui dV ¼ du n m dS m dV i j @xj @xj @xj @x2j
The first term in the integrand may be rewritten as @ @uj @dui @uj @ @uj ¼ dui (m þ l) (m þ l) þ dui (m þ l) @xj @xi @xj @xi @xi @xj with the result that @ @uj @ @uj @dui @uj dui (m þ l) (m þ l) dui (m þ l) ¼ @xj @xi @xj @xi @xj @xi Upon integration and application of the divergence theorem, ð V
ð ð @ @uj @uj @dui @uj dui (m þ l) dS (m þ l) dV dV ¼ ni dui (m þ l) @xi @xj @xj @xi @xj V
V
Collecting the surface terms gives ð S
ð @uj @ui du dS ! duT (m þ l)(r u)n þ m n dS dui ni (m þ l) þ nj m dx @xj @xj S
We now identify u as the primary variable and (m þ l)(r u)n þ m du dx n as the conjugate secondary variable.
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We now examine the domain terms. Note that ð
@dui @ui @dui @uj þ (m þ l) þ dui r€ ui dV @xj @xj @xi @xj ! ! ð ddu T du ¼ tr m þ (m þ l)(r du)(r u) þ dur€ u dV dx dx
All three terms in the integrand are quadratic. Finally the variational statement of Navier’s equation is recapitulated as ð
! ! ddu T du T m u dV tr þ (m þ l)(r du)(r u) þ du r€ dx dx ð du ¼ duT (m þ l)(r u)n þ m n dS dx
S
SOLUTION Plate equation The first step, again, is ð dw V
@2 @2w dV ¼ 0 @x2i @x2j
Integration by parts must be performed twice. The first step is @ @xi
! @ @2w @ 2 @ 2 w @dw @ @ 2 w dw þ ¼ dw 2 @xi @x2j @xi @xi @x2j @xi @x2j
(4:23)
and the second step is @ @dw @ @w @ @dw @ @w @dw @ @ 2 w ¼ þ @xj @xi @xi @xj @xj @xi @xj @xi @xi @xi @x2j
(4:24)
Subtracting Equation 4.24 from Equation 4.23 gives ! @ @ @2w @ @dw @ @w @2 @2w @ @dw @ @w ¼ dw dw @xi @xi @x2j @xj @xi @xi @xj @xj @xi @xj @xi @x2i @x2j which becomes upon integration ! ! ð @ @dw @ @w @ @2w @dw @ @w dV ¼ nj dS þ dw ni @xj @xi @xj @xi @xi @x2j @xi @xj @xi
ð V
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S
Examination of the domain terms reveals that ð V
@ @dw @xj @xi
!T !! T ð @ @w d ddw d T dw dV ¼ tr dV @xj @xi dx dx dx dx
V
and the surface terms are rewritten as ð S
@ @2w dw ni @xi @x2j ð ¼
!
@dw þ @xi
@ @w nj @xj @xi
! dS
dw (n r)r2 w þ ðrdwÞð(n r)rwÞ dS
S
The primary variables are now identified as w, rw, and the conjugate secondary variables are respectively ((n r)r2w) and ((n r)rw).
EXAMPLE 4.6 Consider a one-dimensional system described by a sixth-order differential equation: Q
d6 q ¼ 0, dx6
Q a constant:
Consider an element from xe to xeþ1. Using the natural coordinate j ¼ 1 when x ¼ xe ¼ þ1 when x ¼ xeþ1, for an interpolation model with the minimum order that is meaningful, obtain expressions for w(j), F, and g, serving to express q as qðjÞ ¼ wT ðjÞFg
SOLUTION Since the given equation is sixth order, the lowest order interpolation model consistent with six integration constants is a fifth-order polynomial, in the form (physical coordinates), qðx,t Þ ¼ wT ð xÞFgðt Þ in which 1 qe ðt Þ C B 0 B qe ðt Þ C C B 00 B q ðt Þ C C B e T 2 C gðt Þ ¼ B B q ðt Þ C, w ð xÞ ¼ 1 x x B eþ1 C C B B q0 ðtÞ C @ eþ1 A 0
00
qeþ1 ðt Þ We seek to identify F in terms of the nodal values of q.
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x3
x4
x5
Letting qe ¼ qðxe Þ, qeþ1 ¼ qðxeþ1 Þ, q0e ¼ q0 ðxe Þ . . . furnishes the following in the matrix–vector notation: 1 2 qe ðt Þ 1 xe B q0 ðt Þ C 6 C 6 0 1 B e C 6 B B q00 ðt Þ C 6 0 0 C 6 B e C¼6 B B qeþ1 ðt Þ C 6 1 xeþ1 C 6 B C 6 B 0 @ qeþ1 ðtÞ A 4 0 1 0
q00eþ1 ðt Þ
0
x3e
x4e
2xe
3x2e
4x3e
2
6xe
12x2e
x2eþ1
x3eþ1
x4eþ1
2xeþ1
3x2eþ1
4x3eþ1
7 5x4e 7 7 20x3e 7 7 7Fgðt Þ x5eþ1 7 7 7 5x4eþ1 5
2
6xeþ1
12x2eþ1
20x3eþ1
0
x5e
3
x2e
Hence 2
1 xe 6 6 0 1 6 60 0 6 F¼6 6 1 xeþ1 6 6 4 0 1 0
0
x5e
31
x2e
x3e
x4e
2xe
3x2e
4x3e
2
6xe
12x2e
x2eþ1
x3eþ1
x4eþ1
2xeþ1
3x2eþ1
4x3eþ1
7 5x4e 7 7 20x3e 7 7 7 x5eþ1 7 7 7 5x4eþ1 5
2
6xeþ1
12x2eþ1
20x3eþ1
Now, on converting this to the natural coordinate j ¼ 1 when x ¼ xe, ¼ þ1 when x ¼ xeþ1, we have a¼
2 le
@ @ ¼a @x @j from which @6 @6 64 @ 6 ¼ a6 6 ¼ 2 6 @x le @j6 @j Hence, the governing equation becomes 64 d6 q ¼0 l2e dj6 The interpolation model now becomes qðjð xÞ,t Þ ¼ wT ðjð xÞÞFgðt Þ and wT ðjð xÞÞ ¼ 1 jðxÞ jð xÞ2 jð xÞ3 jð xÞ4 jð xÞ5 gT ðtÞ ¼ qe ðt Þ q0e ðt Þ q00e ðt Þ qeþ1 ðtÞ q0eþ1 ðt Þ q00eþ1 ðt Þ
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Substituting j(xe) ¼ 1 and j(xe) ¼ 1 yields the desired result at 3 1 1 1 1 1 1 1 6 0 1 2 3 4 5 7 7 6 7 6 60 0 2 6 12 20 7 7 F ¼6 61 1 1 1 1 17 7 6 7 6 4 0 1 2 3 4 5 5 0 0 2 6 12 20 2
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5
Fundamental Notions of Linear Solid Mechanics
In this chapter we provide a condensed review of basic solid mechanics typically presented in upper-level undergraduate courses. Our emphasis is on formal relations as well as examples.
5.1 DISPLACEMENT VECTOR Figure 5.1 depicts a body in both its reference and current configurations. The former is considered to be the undeformed configuration, and the latter is called the deformed configuration—it reflects the deformation induced by the forces applied to the undeformed configuration. Consider a material particle occupying ‘‘point’’ P in the undeformed position and point Q in the deformed position. In the undeformed configuration its position determined the undeformed position vector, given in rectilinear coordinates as X ¼ X1 i þ X2 j þ X3 k
(5:1)
while the same particle in the deformed configuration gives rise to the deformed position vector x ¼ x1 i þ x2 j þ x 3 k
(5:2)
referred to the same base vectors i, j, k. It is assumed that xi are functions of Xj and time t. The vector difference between x and X is called the displacement vector. In rectilinear coordinates u¼xX
(5:3)
and in alternate notation 8 9 8 9 < u = <x X= u¼ v ¼ yY : ; : ; w zZ
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(5:4)
Undeformed configuration
y
Deformed configuration
P u X
Q x
j x
i
FIGURE 5.1 Illustration of the displacement vector.
5.2 LINEAR STRAIN AND ROTATION TENSORS Suppose there is a slight change in the position of Q. From the chain rule of calculus, du du in the X Y plane the displacements u(X,Y) and v(X,Y) satisfy du ¼ dX dX þ dY dY dv dv dX þ dY dY. Elementary calculus furnishes and dv ¼ dX
du dv
2 du 6 ¼ 4 dX dv dX Exx ¼ Exy
du 3 dY 7 dX 5 dv dY dY Eyx vxx þ Eyy vxy
vyx vyy
dX dY
(5:5)
in which 1 du du Exx ¼ þ , 2 dX dX vxx ¼ 0,
1 dv du 1 dv dv Eyx ¼ Exy ¼ þ , Eyy ¼ þ 2 dX dY 2 dY dY 1 dv du , vyy ¼ 0 vxy ¼ vyx ¼ 2 dX dY
Of course Exx, Eyy, and Exy denote the linear strains and will be seen to comprise a measure of local stretching. Further, vxy represents rotation. Suppose, for example, that x ¼ Q(t)X þ b(t) in which Q(t) is an orthogonal tensor independent of X and the vector b(t) is likewise independent of X. Q(t) is said to represent rigid body rotation and b(t) represents rigid body translation. To first order in displacements and their derivatives, the linear strains are unaffected by the rigid body component of the deformation. However, the rotation is strongly affected by rigid body rotation. In rectilinear coordinates, the linear strains and rotations are given in three dimensions by
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8 @u > > EXX ¼ normal strain > > @X > > > @v > > normal strain EYY ¼ > > > @ Y > > > @w > > normal strain > < EZZ ¼ @ Z 1 @ ui @ uj ! Eij ¼ þ 1 @u @v > 2 @ Xj @ Xi EXY ¼ þ shear strain > > > 2 @ Y @ X > > > > 1 @y @w > > þ shear strain EYZ ¼ > > 2@ Z @ Y > > > > > 1 @w @u > : EZX ¼ þ shear strain 2 @X @Z 8 vXX ¼ vYY ¼ vZZ ¼ 0 > > > > 1 @u @v > > vYX ¼ vXY ¼ > > 2 @Y @X > < 1 @ ui @ uj 1 @v @w vij ¼ ! vZY ¼ vYZ ¼ > 2 @ Xj @ Xi > > 2 @Z @Y > > > > 1 @w @u > > ¼ v ¼ v : XZ ZX 2 @X @Z
(5:6)
(5:7)
In alternate notation we may write in general that du dX dX ¼ ½EL þ vdX
du ¼
(5:8)
in which T ! T ! 1 du du 1 du du þ , v¼ EL ¼ 2 dX dX 2 dX dX The counterpart of EL and v in tensor-indicial notation is 1 @ ui @ uj 1 @ ui @ uj , vij ¼ Eij ¼ þ þ 2 @ Xj @ Xi 2 @ Xj @ Xi Since du is a 3 3 1 vector, it follows from Chapter 3 that du T , EL and v are 3 3 3 tensors. Likewise dX
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du dX
(5:9) is a 3 3 3 tensor.
5.3 EXAMPLES OF LINEAR STRAIN AND ROTATION TENSORS EXAMPLE 5.1 The plate shown is initially square and 6 cm on a side (Figure 5.2a). It is deformed as shown in Figure 5.2b. Find the displacement and the strain fields.
SOLUTION Assume that the deformed coordinates can be expressed in terms of the undeformed coordinates using the expressions x ¼ a þ bX þ cY þ dXY
(5:10)
y ¼ e þ fX þ gY þ hXY
(5:11)
Our goal is to detect the coefficients a through h by fitting the coordinates at the four corners of the element. Along the bottom face of the block Y ¼ 0 from which x ¼ a þ bX and y ¼ e þ fX. It follows that y ¼ bf x e þ ab, implying that the lower face remains a line after deformation. On the right face X ¼ 1 so that x ¼ a þ b þ (c þ d)Y and y ¼ e þ f þ (g þ h)Y. Now Y may be eliminated from these expressions to yield a linear relation between the deformed coordinates x and y. Again the side of the element remains a line in the deformed cofiguration. Similar results are are immediate for the upper face and the left-hand face. Clearly the assumed relations (5.10, 5.11) are suitable if the sides of the square remain straight after deformation. Eight nodal relations are used to determine the eight coefficients in (5.10, 5.11). Lower left node: For (X,Y) ¼ (0,0), (x,y) ¼ (0,0) and hence a ¼ 1, e ¼ 1: Lower right node: For (X,Y) ¼ (6,0), x ¼ 7:1 and y ¼ 1:1: Hence b ¼ 6:1=6, f ¼ 0:1=6 7.2, 7.3
6 cm
1.2, 7.1
7.1, 1.1 1, 1 6 cm
(a)
(b)
FIGURE 5.2 Example of strain and rotation.
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Upper left: For (X,Y) ¼ (0,6), x ¼ 1:2 and y ¼ 7:1, giving c ¼ 1:2=6 and g ¼ 6:1=6: Apex: For (X,Y) ¼ (6,6), x ¼ 7:2 and y ¼ 7:3: Accordingly, 7:2 ¼ 1 þ b * 6 þ c * 6 þ d * 36 from which d ¼ 1:3=36 7:3 ¼ 1 þ f * 6 þ g * 6 þ h * 36 from which h ¼ þ0:1=36 The displacements are found as u ¼ x X ¼ 1 þ (0:1=6)X þ (1:2=6)Y þ (1:3=36)XY v ¼ y Y ¼ 1 þ (0:1=6)X þ (0:1=6)Y þ (0:1=36)XY Upon applying the formulae for the strains and rotations we obtain @u ¼ 0:1=6 1:3Y=36 @X @v Eyy ¼ ¼ 0:1=6 þ 0:1X=36 @Y 1 @u @v 1 Exy ¼ þ ¼ ð1:3=6 1:3X=6 þ 0:1Y=36Þ 2 @Y @X 2 1 @u @v 1 ¼ ð1:1=6 1:3X=36 0:1Y=36Þ vxy ¼ 2 @Y @X 2 Exx ¼
EXAMPLE 5.2 A rectangular block in Figure 5.3 is rotated through u degrees. Find the displacements and strains and rotations. What happens if the rotation angle is very small?
SOLUTION The vector R rotates into the vector r. Both have length R. Thus, coordinates of the endpoint of R are X ¼ R cos f, Y ¼ R sin f. The deformed coordinates, displacement, strains, and rotation are now found to be Y y x q r
f
R X
FIGURE 5.3 Strain and rotation in body experiencing rigid body motion.
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x ¼ R cos(f þ u) ¼ R[ cos f cos u sin f sin u] ¼ X cos u Y sin u u ¼ ( cos u 1)X Y sin u EXX ¼ cos u 1 EXY ¼ 0
y ¼ R sin(f þ u) ¼ R[ sin f cos u þ cos f sin u] ¼ X sin u þ Y cos u v ¼ X sin u þ ( cos u 1)Y EYY ¼ cos u 1 vXY ¼ vYX ¼ sin u
At small values of u, to first order in u EXX 0, EYY 0, vXY u. Clearly, to first order in u, the strains vanish and the nonzero rotation is given by u. Of course u represents rigid body rotation.
EXAMPLE 5.3 Find the displacements, strains and rotations in deformed body shown in Figure 5.4.
SOLUTION The deformed and undeformed positions of the nodes are given in Table 5.1. We again assume that the deformed coordinates may be expressed in terms of the undeformed coordinates using x ¼ a þ bX þ cY þ dXY,
y ¼ e þ fX þ gY þ hXY
It was shown in the previous example that this form is capable of mapping the straight sides in the undeformed configuration onto straight sides in the deformed configuration. Following the same procedures as in the previous example given Node 1:
0 ¼ a þ b0 þ c0 þ d0 0 ¼ e þ f 0 þ g0 þ h0
Node 2:
cos u ¼ an þ b1 þ c0 þ d0 sin u ¼ e þ f 1 þ g0 þ h0
: b ¼ cos u : f ¼ sin u
Node 4:
sin u ¼ an þ b0 þ c1 þ d0 cos u ¼ ne þ f 0 þ g1 þ h0
: c ¼ sin u : g ¼ cos u
cos u þ sin u ¼ an þ b1 þ c1 þ d1 ¼ cos u þ sin u þ d sin u þ cos u ¼ ne þ f 1 þ g1 þ h1 ¼ sin u þ cos u þ h
Node 3:
Y
Undeformed q
4
2
FIGURE 5.4 Illustration of shear strain.
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:d ¼ 0 :h ¼ 0
The lengths of the sides are unity in both the undeformed and deformed configurations Deformed
3
q 1
:a ¼ 0 :e ¼ 0
X
TABLE 5.1 Deformed and Undeformed Nodal Coordinates Node
Undeformed Position
Deformed Position
0, 0 1, 0 1, 1 0, 1
0, 0 cos u, sin u cos u þ sin u, sin u þ cos u sin u, cos u
1 2 3 4
We have now determined that x ¼ cos u X þ sin u Y ,
y ¼ sin u X þ cos u Y
Now assuming that u is small enough to permit neglecting quadratic and higher-order terms, u ¼ ( cos u 1)X þ sin u Y uY EXX ¼ cos u 1
v ¼ sin u X þ ðcos u 1ÞY uX EYY ¼ cos u 1
u =2
u2=2
2
EXY ¼ sin u u
vXY ¼ 0
In contrast to the case of rigid body rotation, for small angle u and to first order in u, the only nonzero strain is the shear strain, which is (approximately) equal to u. The shear strain is a measure of how much the sides rotate relative to each other.
5.4 TRACTION AND STRESS Consider a differential ‘‘brick element’’ emanating from an origin of the (X,Y,Z) coordinate system shown. The differential forces on the faces of the element can be displayed as in Figures 5.5 and 5.6, dFj(i) ¼ differential force component in the jth direction acting on face dAi dAi ¼ differential area whose normal points in the ith direction The stresses are introduced by Sij ¼
@Fj(i) @Ai
and if X, Y, and Z denote the coordinate axes, dFx(x) , dAx dFy(y) Syy ¼ , dAy dF (z) Szz ¼ z , dAz
Sxx ¼
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dFy(x) dFx(y) ¼ dAx dAy (z) (y) dF dF y Syz ¼ Szy ¼ z ¼ dAy dAz (z) dF (x) dF Szx ¼ Sxz ¼ x ¼ z dAz dAx
Sxy ¼ Syx ¼
(5:12)
z
dAz dAy
dAx
y
x
FIGURE 5.5 Differential brick element. z
z
z
dFz(z) dFy(z)
dFz(y)
dFz(x) dF (x) x
dFx(y)
y
dF (x) y
y
y
x
x
x
dFx(z)
dFy(y)
FIGURE 5.6 Forces on differential brick element.
The first index i represents the differential area on which the force acts, while the second index j represents the direction in which the force is acting. The total differential force vector acting on the ith face is given by dF(i) ¼ t(i) dAi in which the traction vector t(i) on the ith face is given by t(i) ¼
dF1(i) dF (i) dF(i) e1 þ 2 e2 þ 3 e3 dAi dAi d Ai
(5:13)
More generally, we consider a differential tetrahedron enclosing the point x in the deformed configuration (Figure 5.7). The area of the inclined (shaded) face is dS, and dSi is the area of the face on the back of the tetrahedron whose exterior normal vector is ei. Simple vector analysis serves to derive that ni ¼ dSi=dS, see Example 2.5 in Chapter 2. Next let dF denote the force on a surface element dS, and let dF(i) denote the force on area dSi on the back of the tetrahedron. The traction vector acting on the inclined face is introduced by t ¼ dF=dS. As the tetrahedron shrinks to a point, the contribution of volume forces such as inertia decays faster than that of surface forces.
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2
dF
dF (3) dF (1)
dS3
n dS1
dS 1
dF (2)
dS2 3
FIGURE 5.7 Forces on a differential tetrahedron.
Balance of forces on the tetrahedron now requires that X dF ¼ dF(i)
(5:14)
i
The traction vector acting on the inclined face is defined by t¼
dF dS
(5:15)
which together with the equilibrium Equation 5.14 furnishes X dFjðiÞ tj ¼ dS i ¼
X dFjðiÞ dSi dSi dS i
¼ Sij ni
(5:16)
in which ðiÞ
Sij ¼
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dFj dSi
(5:17)
It is readily seen that Sij can be interpreted as the intensity of the force acting in the j direction on the facet pointing in the i direction. It is the ijth entry in the tensor S. In matrix–vector notation the stress–traction relation Equation 5.16 is written as t ¼ ST n
(5:18)
Since ST appears in a physically based linear relation between vectors of the same dimension, it is a tensor, as is the stress tensor S.
EXAMPLE 5.4 The plate in Figure 5.8 is subjected to the stress field Sxx ¼ a þ bx þ cy Syy ¼ d þ ex þ fy Sxy ¼ g þ hx þ jy Find the traction vector and the total force acting on the top and right faces. Consider the moment (Mz) exerted by the tractions on these two faces about the origin. On the top face consider the interval dx at x. The total force on this interval is dFtop ¼ Syx dx ex þ Syy dx ey. The total force on the face is ðL Ftop ¼
Syx dx ex þ Syy dx ey
0
ðL ¼
(g þ hx þ jH) dx ex þ (d þ ex þ fH) dx ey
0
¼ g þ jH þ 12hL L ex þ d þ 12eL þ fH L ey
y 1
H x L z
FIGURE 5.8 Forces and moments determined by tractions.
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The moment due to the force on the interval is dMtop ¼ (xex þ Hey ) dFtop ¼ (xex þ Hey ) (g þ hx þ jH) dx ex þ (d þ ex þ fH) dx ey ¼ ½½ x(d þ ex þ fH) dx H(g þ hx þ jH) dx ek and integration gives Mtop ¼ 12(fH þ d)L2 þ 13eL3 (H(g þ h)L 12HhL2 On the right face consider the interval dy at y. The total force on the interval is dFright ¼ (Sxx dy ex þ Sxy dx ey ) The total force on the face is now ðH Fright ¼
Sxx dy ex þ Sxy dy ey
0
ðH ¼
(a þ bL þ cy) dy ex þ (g þ hL þ jy) dy ey
0
¼ a þ bL þ 12cH Hex þ g þ hL þ 12 jH Hey
EXAMPLE 5.5 At point (0,0,0) the tractions t1, t2, t3 act on planes with normal vectors n1, n2, and n3. Find the stress tensor S, given that n1 ¼ p1ffiffi3 [e1 þ e2 þ e3 ],
t1 ¼ p1ffiffi3 [2e1 5e2 þ 6e3 ]
n2 ¼ p1ffiffi2 [e1 þ e2 ],
t2 ¼ p1ffiffi2 [e1 e2 þ e3 ]
n3 ¼ p1ffiffi3 [e1 e2 e3 ],
t3 ¼ p1ffiffi3 [6e1 1e2 þ 2e3 ]
SOLUTION On applying stress–traction relation t1 ¼ STn1 we find 0 1 2 2 S11 1 @ 1 pffiffiffi 5 A ¼ pffiffiffi 4 S12 3 3 S 6 13
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S21 S22 S23
30 1 S31 1 S32 5@ 1 A S33 1
and S11 þ S21 þ S31 ¼ 2
(5:19)
S12 þ S22 þ S32 ¼ 5
(5:20)
S13 þ S23 þ S33 ¼ 6
(5:21)
Similarly, t2 ¼ STn2 implies that S11 þ S21 ¼ 1
(5:22)
S12 þ S22 ¼ 1
(5:23)
S13 þ S23 ¼ 1
(5:24)
S11 S21 S31 ¼ 6
(5:25)
S12 S22 S32 ¼ 1
(5:26)
S13 S23 S33 ¼ 2
(5:27)
Finally, t3 ¼ STn3 yields
Solving the nine equations (Equations 5.19 through 5.27) by sequential elimination gives the stress tensor as 2
3 2 2 2 S¼4 3 1 1 5 1 4 5
5.5 EQUILIBRIUM We now consider equilibrium in a more general way and include inertial effects, under the restriction that the deformation is ‘‘small enough’’ to neglect effects of deformation on the stress or the equilibrium relation. Neglecting gravity and other ‘‘body forces,’’ the total external force F Ðon a body is assumed to be exerted on its boundaries. It is given by the integral F ¼ t dS. However , recalling that t ¼ STn, the Ð must divergence theorem may be applied to obtain F ¼ r ST dV . The total force Ð equal the rate of change of the total linear momentum L of the body: L ¼ rx_ dV in which the mass density is denoted by r. Now owingÐto Rayleigh Ð ’s transport theorem d _ r x dV ¼ r€x dV . Accordingly, (see Chandrasekharaiah and Debnath, 1994), dL ¼ Ð dt dt F ¼ ddtL implies the integral (global) equation r ST r€xT dV ¼ 0T . In most cases of interest in the current monograph there is a fixed point in the body so that the undeformed position X can be taken to be independent on time. In this event ü ¼ x¨. (In Chapter 12, consideration will be given to bodies in which there is no fixed point, with the consequence that the analysis is referred to a translating and=or rotating coordinate system.) The integral equation must hold for the whole
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body or any closed sub-body within the body. We conclude that the local balance of linear momentum equation €T r ST ¼ ru
(5:28)
holds pointwise (locally) throughout the body. It is also necessary to consider balance of angular momentum. The differential moment exerted by the traction vector on a surface patch is given by dM ¼ X t dS
dMi ¼ 2ijk Xj tk dS ¼ 2ijk Xj STkl nl dS
¼ X S n dS T
(5:29)
Invoking the divergence theorem (Chapter 2), the total moment on the body is given by ð Mi ¼ ð
2ijk Xj STkl nl dS
@ 2ijk Xj STkl dV @ Xl ð @ Xj T @ STkl ¼ 2ijk dV S þ 2ijk Xj @ Xl kl @ Xl
¼
(5:30)
@ ST
@X
Now @ Xjl is recognized as djl, and 2ijk Xj @ Xkll ¼ 2ijk Xj €uk by virtue of the balance of linear momentum. We may rewrite the foregoing equation as Mi ¼
ð
2ijk STkj þ r 2ijk Xj €uk dV
(5:31)
Invoking Rayleigh’s Transport Theorem to be introduced in Chapter 13, the rate of change of the angular momentum is given by ð dH d ¼ rX u_ dV dt dt ð d € dV X u_ þ rX u ¼ r dt
(5:32)
Note that dtd X ¼ u_ and hence dtd X u_ ¼ 0. The global equation for the balance of angular momentum now becomes, in indicial notation, Mi ¼
ð dHi ¼ r 3ijk Xj €uk dV dt
(5:33)
Combining Equations 5.31 and 5.32, and invoking the fact that the result applies for arbitrary closed subvolumes we have the pointwise relation
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2ijk STkj ¼ 0
(5:34)
The tensor ST may be expressed in terms of its symmetric and antisymmetric part. From Example 2.6 in Chapter 2, it was shown that 2ijkajk ¼ 0 if the corresponding tensor A is symmetric. By a similar argument it may be shown that 2ijkajk 6¼ 0 if A is antisymmetric. Consequently, Equation 5.34 cannot be satisfied if ST possesses a nonvanishing antisymmetric part. Accordingly ST is symmetric: S ¼ ST.
EXAMPLE 5.6 Assume that the differential rectangle shown is a unit thickness (Figure 5.9). Prove from (static) moment equilibrium that Sxy ¼ Syx.
SOLUTION Take the moments about the lower left-hand corner. 1. The shear stresses on the bottom and left-hand faces do not create a moment relative to this point since the line of action goes through the origin. 2. Assuming that the forces due to the normal stresses are considered to act at the midpoint of the faces on which they act, they do not contribute to the moment since the positive forces have the same line of action as the negative forces. 3. This leaves only the shear stresses on the right-hand and top faces. Now moment balance M ¼ 0 implies 0 ¼ (Sxy dy) dx (Syx dx) dy ¼ Sxy Syx dx dy with the consequence that Syx ¼ Sxy. This solution illustrates the fact that the stress tensor is symmetric. (The linear strain tensor is symmetric since it is the symmetric part of the tensor du=dX.)
Syy Syx
Sxx
Sxy Sxx
dy dx
Sxy Syx
Syy
FIGURE 5.9 Net moment on a differential element.
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EXAMPLE 5.7 Suppose that, in a differential plate element as shown, the stresses vary slightly across the element. Find the equations for static equilibrium of forces in the x- and y-directions, treating the figure shown as a free body diagram at the differential level (Figure 5.10).
SOLUTION The total force in the x-direction is / . . / dSxx dSyx dy þ S dx S dy S dFx ¼ Sxx þ xx yx þ yx dx dy But the equilibrium condition dFx ¼ 0 requires that becomes
dSxx dx
þ
dSyx dy
dx dy ¼ 0, which
dSxx dSyx þ ¼0 dx dy The total force in the y-direction is / / / / dSxy dSyy dFy ¼ Sxy þ dx Sxy dy þ Syy þ dy Syy dx dx dx
Syy + dSyy Syx + dSyx Sxy + dSxy Sxx
Sxx + dSxx
dy Sxy
dx Syx Syy
Note: dSxy =
dSxy dx
dx, etc.
FIGURE 5.10 Balance of forces on a differential element.
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and dFy ¼ 0 immediately furnishes that dSxy dSyy þ ¼0 dx dx
5.6 STRESS AND STRAIN TRANSFORMATIONS The plate shown below is under a state of stress referred to the x- and y-axis. A small cutout is shown whose lower side is inclined at an angle u from the x-axis. We embed a rotated coordinate system X 0 Y 0 in the cutout, and seek to express the stresses in the rotated system (Figure 5.11). We already know from the relations of Chapter 3 that the stresses represent a tensor and that the stress tensor satisfies the relation S0 ¼ QSQT in which Q is the orthogonal matrix rotating the base vectors eX, eY to the eX 0 , eY 0 . Furthermore, we already know that, in the plane,
cos u Q¼ sin u
sin u cos u
(5:35)
Elementary matrix multiplications and the double-angle formulae of trigonometry suffice to verify that
Y Syy Sxy = Syx
Y⬘
Sy⬘y⬘
Sx⬘y⬘
Sx⬘x⬘ X⬘ Sxx q
X
FIGURE 5.11 Stresses referred to a rotated coordinate system.
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SXX þ Syy SXX Syy þ cos 2u þ SXY sin 2u 2 2 SXX þ SYY sXX SYY ¼ cos 2u SXY sin 2u 2 2 SXX SYY sin 2u þ SXY cos 2u ¼ 2
SX 0 X 0 ¼ SY 0 Y 0 SX 0 Y 0
(5:36)
The strains induced by the stresses likewise represent a tensor and transform exactly the same way as the stresses: EXX þ EYY EXX EYY þ cos 2u þ EXY sin 2u 2 2 EXX þ EYY EXX EYY cos 2u EXY sin 2u ¼ 2 2 EXX EYY sin 2u þ EXY cos 2u ¼ 2
EX 0 X 0 ¼ EY 0 Y 0 EX 0 Y 0
(5:37)
Several simple examples are now given illustrating the use of coordinate transformations.
EXAMPLE 5.8 Relative to the x–y axes the stresses are Sxx ¼ 0,
Syy ¼ 0, Sxy ¼ 25 ksi
What are the stresses referred to axes x0 and y0 which are rotated by 458 from x and y? Do the same for stresses referred to axes x00 and y00 rotated by 458.
SOLUTION Using the stress transformations through þ458 Sxx þ Syy Sxx Syy þ cos(2 * 45) þ Sxy sin(2 * 45) ¼ 25 ksi 2 2 Sxx þ Syy Sxx Syy ¼ cos(2 * 45) Sxy sin(2 * 45) ¼ 25 ksi 2 2 Sxx Syy ¼ sin(2 * 45) þ Sxy cos(2 * 45) ¼ 0 2
Sx0 x0 ¼ Sy0 y0 Sx0 y0
Doing the stress transformations through 458 Sxx þ Syy Sxx Syy þ cos(2 * 45) þ Sxy sin(2 * 45) ¼ 25 ksi 2 2 Sxx þ Syy Sxx Syy cos(2 * 45) Sxy sin(2 * 45) ¼ þ25 ksi ¼ 2 2 Sxx Syy ¼ sin(2 * 45) þ sxy cos(2 * 45) ¼ 0 2
Sx0 x0 ¼ Sy0 y0 Sx0 y0
Since the elements at 458 have no shear stress, their normal stresses are the principal stresses.
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y Plate is 1 in. thick B
5 in.
A
x
5 3 in.
FIGURE 5.12 Length of the diagonal in a plate under strain.
EXAMPLE 5.9 The stresses in the square plate shown (Figure 5.12) are uniform and are given by Sxx ¼ 10 ksi,
Syy ¼ 10 ksi,
Sxy ¼ 0
What is the total force acting transverse to the line AB, and what is the total moment of this force relative to point A?
SOLUTION Sxx þ Syy Sxx Syy þ cos 60 þ Sxy sin 60 ¼ 5 ksi 2 2 Sxx þ Syy Sxx Syy ¼ cos 60 Sxy sin 60 ¼ 5 ksi 2 2 pffiffiffi Sxx Syy sin 60 þ Sxy cos 60 ¼ 5 3 ksi ¼ 2
Sx0 x0 ¼ Sy0 y0 Sx0 y0
MAB ¼ 50000 5 FAB ¼ Sy0 y0 * area ¼ 5 ksi * 1 * 10 ¼ 250000 in: lb ¼ 50000 lb
EXAMPLE 5.10 A plate element is under the strains Exx 6¼ 0, Eyy 6¼ 0, Exy ¼ 0. What are the strains in an element rotated þ908, 908, and 1808?
SOLUTION
/ Exx þ Eyy Exx Eyy þ cos(180) þ Exy sin(180) ¼ «yy 2 2 / Exx þ Eyy Exx Eyy ¼ cos(180) «xy sin(180) ¼ «xx 2 2 / Exx Eyy ¼ sin(180) þ Exy cos(180) ¼ 0 2
Ex0 x0 ¼ Ey0 y0 Ex0 y0
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u ¼ 90 : / Exx þ Eyy Exx Eyy þ cos(180) þ Exy sin(180) ¼ Eyy 2 2 / Exx þ Eyy «xx «yy ¼ cos(180) Exy sin(180) ¼ Exx 2 2
Ey0 y0
Ex0 y0 ¼ u ¼ 180 :
-
Ex0 x0 ¼
/ Exx Eyy sin(180) þ Exy cos(180) ¼ 0 2
/ Exx þ Eyy Exx Eyy þ cos(360) þ Exy sin(360) ¼ «xx 2 2 / Exx þ Eyy Exx Eyy cos(360) Exy sin(360) ¼ «yy ¼ 2 2
«x0 x0 ¼ «y0 y0
«x0 y0 ¼
/ Exx Eyysin(360) þ Exy cos(360) ¼ 0 2
EXAMPLE 5.11 Transformation of stresses in two dimensions (2D) Verify the stress transformation relations using force balance in two dimensions (Figure 5.13).
ds Syy
Assume unit thickness Sxy
Sxy ds
Sxy
Sxx Sn A
Sxx
n
r
Sr B
q
Sxy ds Syy
q cos q ds
FIGURE 5.13 Transformation of stress.
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sin q ds
SOLUTION In the current and succeeding example we will make use of the relations cos2 u ¼ 12 (1 þ cos 2u), sin2 u ¼ 12 (1 cos 2u), and 2 sin u cos u ¼ sin 2u. Given the stresses on a square element, we seek the normal and tangential stresses on the line AB at angle u from the x-axis. The unit vectors defining the normal and tangential directions of the line AB are given by n ¼ sin u i þ cos u j,
t ¼ cos u i þ sin u j
The vectors representing the normal, tangential, and total force on the inclined line are in terms of the normal and tangential stresses as dFn ¼ (sin u Sn ds)i þ cos u Sn ds j dFt ¼ cos u St ds i þ sin u St ds j dF ¼ ðsin u Sn ds þ cos u Sn dsÞi þ ( cos u Sn ds þ St sin u ds)j Balance of forces in the x and y directions requires that sin u Sn ds þ cos u St ds þ sin u Sxx ds cos u Sxy ds ¼ 0 cos u Sn ds þ sin u St ds þ sin u Sxy ds cos u Syy ds ¼ 0 After some manipulations involving double-angle formulae from trigonometry, we have sin u Sn þ cos u St ¼ Sxx sin u þ Sxy cos u cos u Sn þ sin u St ¼ Sxy sin u þ Syy cos u sin u Sn sin u cos u St ¼ Sxx sin2 u Sxy sin u cos u 2
cos2 u Sn þ sin u cos u St ¼ Sxy sin u cos u þ Syy cos2 u Sn ¼ Syy cos2 u þ Sxx sin2 u 2 sin u cos u Sxy and hence the stress normal to the line AB is given by Sn ¼ 12(1 þ cos 2u)Syy þ 12(1 sin2 u)Sxx sin 2u Sxy ¼ 12(Sxx þ Syy ) 12(Sxx Syy ) cos 2u sin 2u Sxy in agreement with the previously reported transformation formulae. Similar operations are performed for the tangential stress. sin u cos u Sn þ cos2 u St ¼ Sxx sin u cos u þ Sxy cos2 u sin u cos u Sn þ sin2 u St ¼ Sxy sin2 u þ Syy sin u cos u
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The stress tangential to the line AB is now found to be st ¼ (sxx syy ) sin u cos u þ sxy ( cos2 u sin2 u) st ¼
(sxx syy ) sin 2u þ sxy cos 2u 2
likewise in agreement with the previously reported transformation formulae.
EXAMPLE 5.12 Transformations of strains in 2D. @u0 @v0 1 @u0 @v0 In two dimensions show that Ex0 x0 ¼ @X 0 , Ey0 y0 ¼ @Y 0 , and Ex0 y0 ¼ 2 @Y 0 þ @X 0 .
SOLUTION Figure 5.14 depicts the undeformed position vector X and the deformed position vector x. It also shows X0 and x0 which are X and x0 rotated through u. Recall that a positive rotation of the coordinate system has the same effect as a negative rotation of a vector relative to the same coordinate system. The undeformed and deformed coordinates in the rotated vectors are given by X 0 ¼ cos(w u)R Y 0 ¼ sin(w u)R ¼ X cos u þ Y sin u ¼ X sin u þ Y cos u and x0 ¼ x cos u þ y sin u,
y0 ¼ x sin u þ y cos u
x X
q
R X⬘
R
x⬘
j i
q
f
FIGURE 5.14 Undeformed position vector before and after rotation.
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The relations just presented may be easily inverted to furnish X ¼ X 0 cos u Y 0 sin u, Y ¼ X 0 sin u þ Y 0 cos u,
x ¼ x0 cos u y0 sin u y ¼ x0 sin u þ y0 cos u
The chain rule of calculus yields the useful relations @ @x @ @y @ þ ¼ @ x0 @ x0 @ x @ x0 @ y @ @ ¼ cos u þ sin u @x @y
@ @x @ @y @ þ ¼ @ y0 @ y0 @ x @ y0 @ y @ @ ¼ sin u þ cos u @x @y
It further follows that u0 ¼ x0 X 0 ¼ (x X ) cos u þ ( y Y ) sin u ¼ u cos u þ v sin u
v0 ¼ y0 Y 0 ¼ (x X ) sin u þ (y Y ) cos u ¼ u sin u þ v cos u
Next, @u @v E0XX ¼ cos u 0 þ sin u 0 @ X @ X @u @u @v @v ¼ cos u þ sin u cos u þ cos u þ sin u sin u @X @Y @X @Y @u @v @u @v þ sin2 u þ cos u sin u þ ¼ cos2 u @X @Y @Y @X ¼ 12(1 þ cos 2u)EXX þ 12(1 cos 2u)EYY þ sin 2uEXY EXX þ EYY EXX EYY ¼ þ cos 2u þ EXY sin 2u 2 2 in agreement with Equation 5.37. Similarly, EY 0 Y 0 ¼
@v0 @Y 0
@u @v ¼ sin u 0 þ cos u 0 @Y @Y @u @u @v @v ¼ sin u þ cos u sin u þ sin u þ cos u cos u @X @Y @X @Y @u @v @u @v ¼ sin2 u þ cos2 u sin u cos u þ @X @Y @Y @X EXX þ EYY EXX EYY cos u sin 2uEXY ¼ 2 2 again in agreement with Equation 5.37.
ß 2008 by Taylor & Francis Group, LLC.
Finally, E
X0 Y 0
1 @ u0 @ v0 ¼ þ 2 @Y 0 @X0 1 @ 1 @ ¼ (u sin u þ v cos u) ðu cos u þ v sin uÞ þ 0 2 @Y 0 2 @ X 1 @ @ 1 @ @ ¼ sin u þ cos u ðu cos u þ v sin uÞ þ cos u þ sin u (u sin u þ v cos u) 2 @X @Y 2 @X @Y @u @v 1 @v @u þ cos2 u sin2 u þ ¼ 2 sin u cos u 2 @Y @X @Y @X EXX EYY ¼ sin 2u þ EXY cos 2u 2
as expected.
5.7 PRINCIPAL STRESSES AND STRAINS In the plane, upon rotation of the coordinate axes through the angle u0 satisfying 2« tan 2u0 ¼ «xx xy«yy , the shear strain and the shear stress both vanish: «X0 Y 0 ¼ 0, sX0 Y0 ¼ 0. The corresponding normal stresses (strains) are called the principal stresses (strains), denoted by sI, sII («I, «II). Of course from Section 2.4 of Chapter 2 we know that the eigenvectors of the stress tensor form the columns of an orthogonal tensor Qo(s) which serves to diagonalize S. The diagonal entries are simply the eigenvalues, which in the current context are called the principal values. In the plane the determinant equations for the principal stresses and strains are simply
EI ,II
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s Exx Eyy 2 þ E2xy 2 ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi SXX SYY 2 þ S2XY 2
Exx þ Eyy ¼ 2
SI ,II ¼
SXX þ SYY 2
The rotated axes in the current context are called the principal axes.
EXAMPLE 5.13 The strain ellipsoid A three-dimensional body has a uniform state of strain in which Exx ¼ 0,
Eyy ¼ 0:01,
Ezz ¼ 0,
Exy ¼ 0,
Eyz ¼ 0,
Ezx ¼ 0:01
Imagine a small sub-body, which is spherical with radius r in the undeformed configuration. When the strain is imposed, it becomes an ellipsoid (Figure 5.15). What are the lengths of the three semiaxes a, b, c of the ellipsoid, and what are their orientations relative to the x–y–z axes? What are the principal strains?
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z
z
r y
c
b y
x
a x
Undeformed body
Ellipsoid
FIGURE 5.15 Deformation of a small sphere.
SOLUTION 2
3 0 0 0:01 The strain tensor E ¼ 4 0 0:1 0 5 is diagonalized by a rotation of the z–x 0:01 0 0 plane about the y-axis, namely p 32 p 32 cos 0 0 0:01 4 76 4 6 76 E0 ¼ 6 0:1 0 56 1 0 7 54 0 4 0 4 0 p p p 0:01 0 0 sin sin 0 cos 4 4 4 2 p p 32 p p3 cos 0 cos 0 sin sin 4 4 76 4 47 6 7 6 6 ¼ 0:014 0 1 0 54 0 1 0 7 5 p p p p 0 sin 0 cos cos sin 4 4 4 4 3 2 p p p p 2 sin cos 0 cos2 sin2 4 4 4 47 6 7 ¼ 0:016 0 1 0 5 4 p p 2p 2p 0 2 sin cos cos sin 4 4 4 4 2 3 1 0 0 6 7 ¼ 0:014 0 1 0 5 0 0 1 2
cos
p 4
0
sin
0 1 0
p3 47 0 7 5 p cos 4
sin
The lengths of the three semiaxes are: 1.01r, 1.0r, 0.99r. The principal strains are then relative length changes of the semiaxes. In other words, 1:01r r r ¼ 0:01,
EI ¼
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1:01r r r ¼ 0:01,
EII ¼
0:99r r r ¼ 0:01
EIII ¼
EXAMPLE 5.14 The Von Mises stress and the Tresca stress Often stresses calculated by finite element analysis are compared to failure stresses measured in the laboratory. Typically, the laboratory tests are one-dimensional, either simple tension=compression or simple shear. However, the calculated stresses are often multiaxial. In order to enable comparison with failure stresses it is common to introduce ‘‘equivalent stresses’’ of which primary examples are the Von Mises stress and the Tresca stress introduced as follows using principal stresses: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 SVM ¼ pffiffiffi ðSI SII Þ2 þ ðSII SIII Þ2 þ ðSIII SI Þ2 : Von Mises 2 : Tresca t T ¼ 12 maxðjSI SII j, jSII SIII j, jSIII SI jÞ In uniaxial tension or compression the SVM is equal to the actual stress, thereby providing ‘‘motivation’’ for regarding it as the counterpart of the uniaxial tensile= compressive failure stress. On the other hand, in simple shear tT is equal to the stress, and accordingly we view it as the counterpart of the failure stress in simple shear. In the case in which there is only one normal stress SXX and one shear stress SXY, SVM and tT, we derive the following simple formulae: SVM ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S2XX þ 3S2XY ,
tT ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 S2XX þ 4S2XY 2
The principal stresses are given by vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u0 / / 12 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 2 Syy u S þ Syy C xx Sxx Sxx B u 2 SI,II ¼ Sxx þ þ S2xy , SIII ¼ 0 t@ A þ Sxy ¼ 2 2 2 2 For the Von Mises stress, we find
2 / 2 / 1 ðSI SII Þ2 þ SII SIII þ SIII SI S2VM ¼ 2 32 2 02 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 32 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 32 1 2 2 2 1@4 Sxx S S S S xx xx xx xx 2 5 2 2 ¼ 2 þ Sxy þ4 þ þ Sxy 5 þ4 þ Sxy 5 A 2 2 2 2 2 2 " # " !#! 2 1 Sxx 2 Sxx 2 Sxx þ S2xy þ 2 þ þ S2xy 4 ¼ 2 2 2 2 ¼ S2xx þ 3S2xy For the Tresca stress, / / 1 t T ¼ max jSI SII j, SII SIII , SIII SI 2 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 S S S xx xx xx ¼ max42 þ S2xy , þ þ S2xy , 2 2 2 2
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ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi3 s 2 Sxx S xx þ S2xy 5 2 2
ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Sxx 2 Now, taking the positive square root and noting that þ S2xy S2xx , the magni2 tudes satisfy sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 S S xx xx 2 þ S2xy ¼ 2 þ S2xy 2 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi 2 2 Sxx S S S xx xx xx þ þ þ S2xy ¼ þ S2xy 2 2 2 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi s 2 2 Sxx S S Sxx xx xx þ S2xy ¼ þ S2xy 2 2 2 2 regardless of the sign of Sxx. But, also, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi1 s ffi 0 2 2 2 Sxx Sxx Sxx Sxx Sxx 2 2 A 2 @ þ Sxy þ Sxy ¼ þ Sxy þ 0 2 2 2 2 2 2 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 0 2 2 2 Sxx S S Sxx Sxx xx xx þ S2xy @ þ S2xy A ¼ þ S2xy þ þ 0 2 2 2 2 2 2 Consequently, 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 Sxx þ S2xy , max42 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2 Sxx ¼2 þ S2xy 2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2 Sxx S xx þ S2xy , 2 þ 2
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi3 2 Sxx S xx þ s2xy 5 2 2
from which we obtain the expected relation t T ¼ 12
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðSxx Þ2 þ 4S2xy
5.8 STRESS–STRAIN RELATIONS In linear elasticity, the stresses are linear functions of strain. Furthermore, if the material is isotropic, its properties are characterized by two coefficients: the elastic modulus E and the Poisson’s ratio v. To illustrate isotropy using Figure 5.16, in onedimensional tension specimens cut from a single plate of a given material, the material is isotropic if the measured stress–strain curves are the same and independent of the orientation at which they are cut. Otherwise it exhibits anisotropy, but may still exhibit limited types of symmetry such as transverse isotropy or orthotropy. The isotropic stress–strain relations of linear elasticity are given in alternate forms, one of which is the Lamé form S ¼ 2mE þ l tr(E)I
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(5:38)
T 34 1
2 1
2 3 4
E
FIGURE 5.16 Illustration of isotropy.
As indicated in Example 3.6 of Chapter 3, in the Lamé relation there exists an orthogonal tensor Qo which simultaneously diagonalizes S and E. This is a formal statement of the property of isotropy. The Lamé form may be inverted to furnish E¼
1 1 l S tr (S)I 2m 2m 2m þ 3l
(5:39)
EXAMPLE 5.15 Invert the relation in Equation 5.36 to express the strain as a function of stress.
SOLUTION Applying the trace operator to Equation 5.38 results in tr (S) ¼ 2m tr (E) þ 3l tr (E) from which we conclude that tr(E) ¼
tr (S) 2m þ 3l
Now S ¼ 2mE þ l
tr (S) I 2m þ 3l
from which it is immediate that 1 l S tr (S)I E¼ 2m 2m þ 3l Now returning to the main development, in the case of uniaxial tension the Lamé relation yields
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EXX
1 l 1 SXX , ¼ 2m 2m þ 2l
EYY ¼ EZZ ¼
1 l EXX 2m 2m þ 2l
(5:40)
þ 3l) YY , it is immediate that E ¼ m(2m , n ¼ 2(mlþ l). Letting E ¼ SXX=EXX and n ¼ EEXX mþl 1 tr (S) Another important property is the bulk modulus k ¼ 3 tr(E), which is found to be 2m þ 3l E given by k ¼ 3(12 n) ¼ 3 . It becomes infinite in the incompressible case n ¼ 1.2. 1 En , from which m ¼ 2(1Eþ n). Finally, the shear modulus is readily seen to satisfy E1 ¼ 2m The inverse Lamé form of the stress–strain relations is now written in the form E ¼ 1 þE n S En tr(S)I, and is written out in the six relations as
Exx ¼ E1 [Sxx n(Syy þ Szz )] Eyy ¼ E1 [Syy n(Szz þ Sxx )] Ezz ¼ E1 [Szz n(Sxx þ Syy )] Exy ¼
1þn Sxy E
Eyz ¼
1þn Syz E
Ezx ¼
1þn Szx E
EXAMPLE 5.16 The 2’’ 3 2’’ 3 2’’ cube shown below is confined on its sides facing the x faces by rigid frictionless walls. The sides facing the z faces are free. The top and bottom faces are subjected to a compressive force of 100 lb (Figure 5.17). Take E ¼ 107 psi and n ¼ 1=3. Find all nonzero stresses and strains. Find all principal stresses and strains.
SOLUTION Since parallel sides of the block remain parallel, there is no shear strain: Exy ¼ 0,
Eyz ¼ 0,
Ezx ¼ 0
It follows from the stress–strain relations that the shear stresses vanish. Sxy ¼ 0,
Syz ¼ 0, Szx ¼ 0
It is immediate that the normal stresses and strains are also the principal stresses and strains. SI ¼ Sxx ,
SII ¼ Syy ,
SIII ¼ Szz ,
EI ¼ Exx ,
EII ¼ Eyy , EIII ¼ Ezz
We now consider the normal stresses and strains. From the description of the problem Eyy ¼
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Fy , Ay
Ezz ¼ 0,
Exx ¼ 0
y
F
x
z
FIGURE 5.17 Stresses and strains in compression of a confined block. 1
Sxx n Syy þ Szz furnishes E /
/
Using Exx ¼
Fy 25 ¼ lb Ay 3 1 5 Next, from Eyy ¼ Syy n Szz þ Sxx ¼ 106 |{z} E 3 /
Sxx ¼ nSyy ¼ n
nSyy
Eyy ¼
1n 1 n2 Fy 2 ¼ 105 Syy ¼ 9 E E Ay 2
Similarly, 1
n (1 þ n ) Syy Szz n(Sxx þ Syy ) ¼ E E /
Ezz ¼ from which
Ezz ¼ þ
n(1 þ n) Fy 105 ¼ Ay 9 E
EXAMPLE 5.17 The elements of the 9 3 9 tangent modulus tensor C implied by the Lamé form have already been given in Example 3.6. Find the conditions on l and m, and likewise on E and n, which render it positive definite, which is a requirement for stability of the material.
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The eigenvalues of a matrix of the form A þ bI are given by lj(A) þ b. Consequently, the eigenvalues of the stiffness tensor C ¼ 2mI9 þ liiT are given by 2m þ lj(iiT). The tensor iiT is of rank 1, so that eight of its nine eigenvalues vanish. The remaining eigenvalue is given by the Rayleigh extremum principle (to be presented in Chapter 10) as max ( z T ii T z ) in which z is a 9 3 1 vector. The maximizing value of z is in fact pffiffiffiffiffiffi pffiffiffi j z j ¼ 1 . It follows that the eigenvalues of C are 2m þ 3l once and 2m i= iT i ¼ i= 3 eight times. Positive definiteness therefore now requires that m > 0, l > 23 m. We determine the stability restrictions on E and n using the inverse Lamé relation in Example 5.15. In particular VEC(E) ¼ C1 VEC(S),
C 1 ¼
1þn n I9 iiT E E
and C is called the elastic compliance tensor. By the same argument as for m and l, we conclude that the eigenvalues of C1 (i.e., the tangent compliance tensor) are 1 þE n eight times and 1 þE n 3 En ¼ 1 E2n once. The material is stable if the eigenvalues of C and hence of C1 are positive. It follows after some manipulation that E > 0 and 1 n 1=2.
5.9 PRINCIPLE OF VIRTUAL WORK IN LINEAR ELASTICITY In a continuous medium ‘‘dynamic’’ equilibrium is expressed by the local form of the balance of linear momentum. The finite element method makes use of this equation after it is expressed in variational form as the Principle of Virtual Work þ
þ þ ui dV þ dEij Sij dV ¼ dui ti dS dui r€
(5:41)
under the restriction that the body is referred to a nonrotating coordinate system and has a fixed point within this system. Unrestrained and rotating coordinates are addressed in Chapter 12. The Principle of Virtual Work is now derived and applied to the following boundary conditions. At a point on the boundary the position vector is denoted by s. The boundary S decomposes into portions S1, S2, S3: On S1, the displacement u(s) is prescribed as uo(s). On S2, the traction t(s) is prescribed as to(s). On S3, the traction satisfies t(s) ¼ t0(s) A(s)u(s) B(s)ü(s). Under the stated conditions the equation for dynamic equilibrium is given by r€ ui
@ Sij ¼0 @xj
(5:42)
Multiplying through the equation with dui and integrating over the domain results in þ þ @ Sij dui r€ ui dV dui dV ¼ 0 (5:43) @xj
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As usual in variational methods, integration by parts is invoked and furnishes þ dui
þ þ @ Sij @ @ dui dV ¼ (dui Sij ) dV Sij dV @ xj @ xj @ xj þ þ 1 @ dui @ duj 1 @ dui @ duj þ Sij dV ¼ nj (dui Sij ) dS þ @ xi @ xi 2 @ xj 2 @ xj þ þ (5:44) ¼ dui ti dS (dEij þ dvij )Sij dV
Now, dvij is antisymmetric and Sij is symmetric, implying that dvijSij ¼ 0 (Example 3.6 of Chapter 3) follows immediately. The Principle of Virtual Work is rewritten in the form þ
þ þ € dV þ tr (dEL S) dV ¼ duT t dS duT ru
(5:45)
and is rewritten yet again using Kronecker product notation as þ
þ € dV þ du ru T
þ deTL s dV
¼ duT t dS
(5:46)
in which s ¼ VEC(S) and eL ¼ VEC(EL). The constitutive relations of linear elasticity are stated using the general form s ¼ xeL
(5:47)
in which x is the positive definite tangent modulus tensor. And now þ
þ þ € dV þ deTL xeL dV ¼ duT t dS duT ru
(5:48)
Suppose a sufficiently accurate approximation exists in the form u(X,t) ¼ wT(X) Fg(t), in which wT(X) is a 3 3 n matrix, F is an n 3 n constant nonsingular matrix, and g(t) is an n 3 1 vector of parameters such as nodal displacements in a finite element model. As will be seen in Chapter 7, application of the strain–displacement relations serves to derive the subordinate approximation for the linear strains as eL bT(X)Fg(t). The Principle of Virtual Work now implies that € þ dgT Kg ¼ dgT f dgT Mg
(5:49)
in which M and K are n 3 n positive definite symmetric matrices and f is an n 3 1 vector. If g(t) represents nodal displacements in a finite element model, M and K are called the mass and stiffness matrices, respectively, and f is called the consistent force vector.
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We now consider the boundary conditions more carefully. On S1 the displacement is prescribed and hence du ¼ 0 and dg ¼ 0, and there is no contribution to the consistent Ð force. On S2 the traction is prescribed as t0 and the contribution is f 2 ¼ FT wt0 dS2 . Finally, there are both compliant and inertial supports on S3, furnishing ð
ð
ð du t dS3 ¼ F T
T
wt0 dS3 F
ð
wA(s)w dS3 Fg F
T
T
T
wB(s)w dS3 F€ g T
(5:50) Assuming the foregoing approximations and carrying terms with unknowns to the left-hand side furnishes an alternative form of Equation 5.50: dgT (M þ MS )€ g þ dgT ðK þ KS Þg ¼ dgT (f 2 þ f 3 ) in which ð
wB(s)wT dS3 F ð KS ¼ FT wA(s)wT dS3 F ð f 3 ¼ FT wt0 dS3
MS ¼ FT
(5:51)
Clearly, MS is the n 3 n matrix representing the contribution of inertial boundary conditions to the mass matrix and the n 3 n matrix KS represents the contribution to the stiffness matrix from elastic support on the surface.
EXAMPLE 5.18 The equation of static equilibrium in the presence of body forces, such as gravity, is expressed by @Sij ¼ bi (X,t) @Xj Without the body forces, the dynamic Principle of Virtual Work is derived as ð
ð ð @ 2 ui dEij Sij dV þ dui r 2 dV ¼ dui ti dS @t
How should the second equation be modified to include body forces?
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SOLUTION Let b denote non-inertial body forces, for example, due to gravity or electromagnetic effects. The Equation of the balance of linear momentum is now stated as ð
ð ð r€ u dV ¼ t dS b dV
Ð Ð Ð T As before t dS ¼ ST n dS ¼ ðrT SÞ dV. For the equation to hold over arbitrary subvolumes it is necessary that, now using indicial notation, @Sij ¼ r€ ui þ bi @Xj It is easily seen that the same steps as were used to derive the Principle of Virtual Work without the body force b now lead to a more general principle in which rüi þ bi replaces rüi. In particular, ð
ð ð deT s dV þ duT (r€ u þ b) dV ¼ duT t dS
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6
Thermal and Thermomechanical Response
6.1 BALANCE OF ENERGY AND PRODUCTION OF ENTROPY 6.1.1 BALANCE
OF
ENERGY
Sometimes called The First Law of Thermodynamics, the balance of energy principle is stated as follows: the rate of change of total energy in a body, including internal energy and kinetic energy, is equal to the corresponding rate of work done by external forces on the body together with the rate of heat added to the body. In rate form _ þ _ ¼ W _ þ Q_ K
(6:1)
in which is the internal energy with density j ð _ ¼ rj_ dV
(6:2a)
_ is the rate of mechanical work, satisfying W ð _ W ¼ u_ T t dS
(6:2b)
Q_ is the rate of heat input, with heat production h and heat flux q, satisfying ð ð Q_ ¼ rh dV nT q dS
(6:2c)
_ is the rate of increase in the kinetic energy, and K ð _ ¼ ru_ T du_ dV K dt
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(6:2d)
It has been tacitly assumed that all work of external forces is done on the boundary S, and that no work is done by body forces. Invoking the divergence theorem and balance of linear momentum furnishes ð du_ rj_ þ u_ T r rT T tr ðTDÞ rh þ rT q dV ¼ 0 dt
(6:3)
The bracketed term inside the integrand vanishes by virtue of the balance of linear momentum. The relation holds for arbitrary volumes, from which the local form of balance of energy, referred to the deformed (current) configuration, is obtained as j_ ¼ tr ðTDÞ rT q þ rh
(6:4)
T is the Cauchy stress tensor, which is the stress referred to deformed (current) coordinates, and D is the stretching-rate tensor (c.f. Chapter 13). To convert Equation 6.4 to undeformed coordinates relations to be explained in Chapter 13 result in ð
ð
n q dS ¼ qT J FT n0 dS0 ð ¼ qT0 n0 dS0 , q0 ¼ J F1 q T
(6:5)
in which the subscript ‘‘0’’ indicates undeformed coordinates and F is the deformation gradient tensor. In undeformed coordinates, Equation 6.3 is rewritten as ð
r0 _ tr SE_ r0 h þ rT0 q0 dV0 ¼ 0
(6:6a)
furnishing the local form of the balance of energy as r0 j_ tr SE_ r0 h þ rT0 q0 ¼ 0
(6:6b)
Here, S is the second Piola–Kirchhoff or nominal stress, which is also explained later in Chapter 13 and which is the stress-referred to deformed coordinates.
6.1.2 ENTROPY PRODUCTION INEQUALITY Generalizing the thermodynamics of ideal and nonideal gases, the entropy production inequality for continua is introduced as follows (Callen, 1985). ð ð ð T _H ¼ rh_ dV h dV n q dS T T
(6:7a)
in which H_ is the total entropy, h is the specific entropy per unit mass, and T is the absolute temperature. This relation provides a ‘‘framework’’ for describing the
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irreversible nature of dissipative processes such as heat flow and plastic deformation. We apply the divergence theorem to the surface integral leading to the local form of the entropy production inequality: rTh_ h rT q þ qT rT=T
(6:7b)
The counterpart of Equation 6.7b in undeformed coordinates is r 0 Th_ h rT0 q0 þ qT0 r0 T=T
6.1.3 THERMODYNAMIC POTENTIALS
IN
(6:7c)
REVERSIBLE PROCESSES
The balance of energy introduces the internal energy , which is an extensive variable—its value accumulates over the domain. The mass and volume averages of extensive variables will also be referred to as extensive variables. This contrasts with intensive or pointwise variables such as the stresses and the temperature. Another extensive variable is the entropy H. In reversible elastic systems, the entropy is completely determined by heat input according to Q_ ¼ TH
(6:8)
(In Chapters 16 and 17 we shall address several irreversible effects including plasticity, viscosity, and heat conduction.) By virtue of Equation 6.8, in undeformed coordinates the balance of energy for reversible processes may be written as r 0 j_ ¼ tr SE_ þ r0 Th_
(6:9)
As before, S is the second Piola–Kirchhoff stress and now E is the Lagrangian strain to be introduced in Chapter 13. We now invoke conditions for the right-hand side of Equation 6.9 to be uniquely integrable, which renders the internal energy dependent only on the current values of the state variables consisting of E and h. For the sake of understanding we may think of T as a thermal stress and h as a thermal strain. Clearly h_ ¼ 0 if there is no heat input across the surface or generated in the volume. Consequently the entropy is an attractive state variable for representing adiabatic processes. In Callen (1985), a development is given for the stability of thermodynamic equilibrium according to which, under suitable conditions, the strain and the entropy density assume values which maximize the internal energy. Other thermodynamic potentials, depending on alternate state variables, may be introduced by a Lorentz transformation illustrated below. Doing so is attractive if the new state variables are accessible to measurement. For example, the Gibbs free energy (density) is a function of the intensive variables S and T: r 0 g ¼ r0 j tr ðSEÞ r 0 Th
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(6:10a)
so that r0 g_ ¼ tr ES_ r0 hT_
(6:10b)
Stability of thermodynamic equilibrium requires that S and T assume values which minimize g under suitable conditions. This potential is of interest in fluids experiencing adiabatic conditions in which the pressure (stress) is accessible to measurement using, for example, pitot tubes. It is also commonly used in phase changes. In solid continua the stress is often more difficult to measure than the strain. Accordingly, for solids the Helmholtz free energy (density) f is introduced using r0 f ¼ r0 j r0 Th
(6:11a)
r0 f_ ¼ tr SE_ r0 hT_
(6:11b)
furnishing
It is evident that f is a function of an intensive variable and an extensive variable. At thermodynamic equilibrium under suitable conditions it exhibits a (stationary) saddle point rather than a maximum or a minimum. Finally, for the sake of completeness, we mention a fourth potential, known as the enthalpy r0 h ¼ r0 j tr(SE), in terms of which local balance of energy now is expressed as r0 h_ ¼ tr ES_ þ r0 Th_
(6:12)
The enthalpy likewise is a function of an extensive variable and an intensive variable and exhibits a saddle point at equilibrium. It is attractive in fluids under adiabatic conditions.
6.2 CLASSICAL COUPLED LINEAR THERMOELASTICITY The classical theory of coupled thermoelasticity in isotropic media corresponds to the restriction to the linear strain tensor, E EL, and to the stress–strain temperature relation T ¼ 2mE þ l½tr ðEÞ aðT T0 ÞI
(6:13)
Also r r0. Here, a is the volumetric coefficient of thermal expansion, typically a very small number in metals and elastomers. If the temperature increases without stress being applied, the volume strain increases according to evol ¼ tr(E) ¼ a(T T0). Thermoelastic processes are assumed in the present context to be reversible, in which event r q þ h ¼ rTh. _ It is also assumed that the specific heat at constant strain, ce, given by @h (6:14) ce ¼ T @T E
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is constant. The balance of energy may now be restated as _ Þ þ r 0 Th_ r0 _ ¼ tr ðTE
(6:15)
Recall that j is a function of the extensive variables E and h. To convert to E and T as state variables which are accessible to measurement, we again invoke the Helmholtz free energy f ¼ j Th. Since f is an exact differential to ensure path independence, we infer the Maxwell relation
@h @T ¼ r @E T @TE
(6:16)
Returning to the energy balance equation, to express it using T and E as state variables, we have ! @j @j r0 j_ ¼ tr r0 E_ þ r 0 h_ @E h @h E " # ! @j @j @h _ @j @h _ ¼ tr r0 þ r0 E þ r0 T @E h @h E @ET @hE @TE
(6:17)
Also note that @j , T¼ @Eh
@j T¼ @h E
and hence # ! @j @j @h _ @j @h _ r0 þ r 0 E þ r0 T @E h @h E @ET @hE @TE
@T _ @h _ ¼ tr T T E þ r0 T T @T E @T E "
r0 j_ ¼ tr
(6:18)
We previously identified the coefficient of specific heat, assumed constant, as ce ¼ T @h @T jE so that r0 j_ ¼ tr
@T _ TT E þ r0 ce T_ @T
The local form of the balance of energy equation now becomes
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(6:19)
_ rT q ¼ r0 j_ tr (TE)
@T _ _ E þ r0 ce T_ tr(TE) ¼ tr T T @T
@T _ ¼ tr T E þ r0 ce T_ @T
(6:20)
From Equation 6.13, approximating T as T0 yields T0 @@TT alT0 I. Now assuming the isotropic version of Fourier’s Law q ¼ krT in which the conductivity k is assumed to be constant, the thermal field equation emerges k r2 T ¼ alT0 tr E_ þ r0 ce T_
(6:21)
Equation 6.21 is subjected to variational operations in Section 6.3 to obtain the finite element equation of the thermal field. The balance of linear momentum together with the stress–strain and strain– displacement relations of linear isotropic thermoelasticity imply that "
# @ 1 @ ui @ uj @ uk @ 2 ui þl 2m þ aðT T0 Þdij ¼ r 2 @ xj 2 @ xj @ xi @ xk @t
(6:22)
from which we obtain the mechanical field equation (Navier’s equation for thermoelasticity) mr2 u þ ðl þ mÞr tr ðEÞ alrT ¼ r
@u @ t2
(6:23)
The thermal field equation depends on the mechanical field through the term _ Consequently, if E is static there is no coupling. Similarly, the mechanalT0 tr(E). ical field depends on the thermal field through alrT, which likely is quite small in, say, metals, if the assumption of reversibility is a reasonable approximation. We next derive the entropy. Since ce ¼ T @@ Th jE is constant, we conclude that it has the form r0 h ¼ r0 h0 þ r0 ce lnðT=T0 Þ þ r0 h*ðEÞ
(6:24)
where h*(E) remains to be determined. Without loss of generality, we take h0 to vanish. But r0 @@Eh jT ¼ r0 @@hE* ¼ @@TT jE ¼ alI, implying that r0 h ¼ r0 h0 þ r0 ce lnðT=T0 Þ þ al tr ðEÞ
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(6:25)
Now consider f, for which the fundamental relation Equation 6.11b implies @f ¼ h @TE
r0
@f ¼T @ET
(6:26)
Integrating the entropy, r0 f ¼ r0 f 0 r0 ce ½TðlnðT=T0 Þ 1Þ al tr ðEÞT þ f *ðEÞ
(6:27)
in which f *(E) remains to be determined. Integrating the stress, r0 f ¼ r0 f0 þ m tr E2 þ l2 tr 2 ðEÞ al tr ðEÞ(T T0 ) þ f **ðTÞ
(6:28)
in which f **(T) likewise remains to be determined. Reconciling Equations 6.27 and 6.28 now furnishes r0 f ¼ r0 f 0 þmtr E2 þ l2 tr 2 ðEÞaltr ðEÞðTT0 Þr0 ce ðT lnðT=T0 Þ1Þ (6:29)
EXAMPLE 6.1 (i) Invert the constitutive relations of classical thermoelasticity (ii) Express the linear thermal expansion coefficient aL , appearing in the uniaxial tension case, as Exx ¼
Sxx þ aL (T T0 ) E
in terms of the volumetric thermal expansion coefficient.
SOLUTION Since classical thermoelasticity assumes small strain, there is no need to distinguish between the Cauchy and the second Piola–Kirchhoff stresses. The thermoelastic version of Hooke’s law is Sij ¼ 2mEij þ lEkk dij laðT T0 Þdij Taking the trace throughout furnishes Skk ¼ (2m þ 3l)Ekk 3laðT T0 Þ and Ekk ¼
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1 fSkk þ 3laðT T0 Þg 2m þ 3l
Upon substitution, Sij ¼ 2mEij þ l
1 fSkk þ 3laðT T0 Þgdij laðT T0 Þdij 2m þ 3l
and
1 3l2 aðT T0 Þdij Skk dij l 2mEij ¼ Sij l 2m þ 3l 2m þ 3l and the inverse relation is now Sij ¼ 2mEij þ lEkk dij laðT T0 Þdij Skk ¼ (2m þ 3l)Ekk 3laðT T0 Þ Ekk ¼
1 fSkk þ 3laðT T0 Þg 2m þ 3l
1 fSkk þ 3laðT T0 Þgdij laðT T0 Þdij 2m þ 3l
1 3l2 2mEij ¼ Sij l Skk dij l aðT T0 Þdij 2m þ 3l 2m þ 3l
Sij ¼ 2mEij þ l
and so, following obvious steps, Eij ¼
1þn n 1 2n Sij Skk dij þ l aðT T0 Þdij E E E
We next express l in terms of E and n. 2m þ 3l 2 E ¼ þl 3 3 2(1 þ n) E E ¼ 3(1 2n) 3(1 þ n) so that E 1 þ n (1 2n) l¼ 3(1 þ n)(1 2n) En ¼ (1 þ n)(1 2n) The inverse is now obtained as Eij ¼
1þn n n Sij Skk dij þ aðT T0 Þdij E E 1þn
In the absence of stress Ekk ¼ 3
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n aðT T0 Þ (1 þ n)
The Kronecker Product form of the inverse relation is extended from Chapter 5 as
1þn n T na e¼ I9 ii s þ (T T0 )i E E 1þn in which e ¼ VEC(E), s ¼ VEC(S), and i ¼ VEC(I). It is elementary that
@ e 1þn n T @ e na I ii i and ¼ ¼ 9 @s T E E @T S 1 þ n Finally, the linear thermal expansion coefficient is identified as aL ¼
n a (1 þ n)
To verify this outcome we return to the Lamé form and set the stress to zero. 0 ¼ 2mEij þ lEkk dij la(T T0 )dij
-
-
3la (T T0 ) (2m þ 3l)Ekk ¼ 3la(T T0 ) ! Ekk ¼ (2m þ 3l) 3la 3(1 2n) En n ¼ !3 (2m þ 3l) E (1 þ n)(1 2n) (1 þ n)
EXAMPLE 6.2 In classical thermoelasticity, derive the specific heat at constant stress cs, rather than at constant strain.
SOLUTION In classical thermoelasticity the internal energy satisfies
@S _ E þ rce T_ rj_ ¼ tr S T @ TjE
@ Eij / @ «ij _ ¼ tr Sij þ alTdij S_ pq þ T þ rce T_ @ Spq @T @e þ rce T_ ¼ ðs þ alTiÞT @ T s The inverse thermoelastic relations of Example 6.1 are invoked to furnish 2 na 3 ðs þ alTiÞT i 6 1þn 7 cS ¼ 4ce þ 5 r
na ce þ 1þn
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trðSÞ þ T0
naE (1 þ n)(1 2n) r
EXAMPLE 6.3 Determine the adiabatic elastic modulus in uniaxial tension.
SOLUTION The issue of an adiabatic modulus arises in situations in which a test specimen is thermally insulated to prevent heat transfer. Under uniaxial tension the stress–strain– temperature of classical thermoelasticity is written as sxx ¼ «xx aL ðT T0 Þ E in which aL is the linear coefficient of thermal expansion. Under adiabatic conditions the entropy is constant so that h_ ¼ 0. Accordingly, 0 ¼ r dh ¼r ¼
@h @h dT þ r dExx @T @Exx
rce dExx dT dExx T dT
dSxx e Note that rcTe rc Accordingly, under adiabatic T0 , dT ¼ aL E. aL ET0 dT ¼ rce dExx . Of course E is the isothermal elastic modulus. Next,
dSxx ¼
@Sxx @Sxx dE þ dT xx @Exx T @T «xx
¼ E dExx aL E dT
aL 2 ET0 ¼E 1þ dExx rce and finally the adiabatic elastic modulus is
aL 2 ET0 Eadiabatic ¼ E 1 þ rce
EXAMPLE 6.4 Neglecting mechanical effects, express the thermal equilibrium equation in: (i) Cylindrical coordinates (ii) Spherical coordinates
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conditions
SOLUTION In the absence of mechanical effects the thermal equilibrium equation is given by rT kT rT þ rce T_ ¼ 0 (i) Cylindrical coordinates Referring to Section 2.7 in Chapter 2, we have the expression for the temperature gradient in cylindrical coordinates as
rT ¼
@T 1 @T @T er þ eu þ ez @r r @u @z
The divergence of the gradient (cf. Equation 2.87), still using cylindrical coordinates, now gives 1 r kT rT ¼ kT r 1 ¼ kT r T
@ @T 1 @ 1 @T @ @T r þ þ @r @r r @u r @u @z @z @T @2T 1 @2T @2T þ þ þ @ r @ r 2 r2 @ u2 @ z2
Hence in the absence of mechanical effects the thermal equilibrium equation is expressed in cylindrical coordinates as
kT
1 @T @2T 1 @2T @2T @T þ 2þ 2 þ ¼ rce r @r @r r @ u2 @ z2 @t
(ii) Spherical coordinates Referring to the Appendix in Chapter 2, the expression for the temperature gradient in spherical coordinates is recognized to be
rT ¼
@T 1 @T 1 @T er þ eu þ ef @r r cos f @ u r @f
Applying the divergence furnishes
1 @ 2 @T 1 @ 1 @T 1 @ 1 @T r þ þ cos f r 2 @r @r r cos f @u r cos f @u r cos f @f r @f 2 2 2 2 @T @ T 1 @ T tan f @T 1 @ T ¼ kT þ 2 þ þ r @r @r 2 r 2 cos2 f @u2 r @f r 2 @f2
rT kT rT ¼ kT
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Hence, in the absence of mechanical effects the thermal equilibrium equation is expressed in spherical coordinates as, kT
2 @T @2T 1 @ 2 T tan f @ T 1 @ 2 T @T þ 2þ 2 þ ¼ rce r @r @r r cos2 f @ u2 r 2 @ f r 2 @ f2 @t
6.3 THERMAL AND THERMOMECHANICAL ANALOGS OF THE PRINCIPLE OF VIRTUAL WORK AND ASSOCIATED FINITE ELEMENT EQUATIONS 6.3.1 CONDUCTIVE HEAT TRANSFER Neglecting coupling to the mechanical field and assuming the isotropic form of Fourier’s Law, the thermal field equation in an isotropic medium experiencing small deformation may be written as rT kT rT þ rce T_ ¼ 0
(6:30)
We now construct a thermal counterpart of the Principle of Virtual Work. Multiplying by dT, the variation of the unknown to be determined, and applying integration by parts, we obtain ð
ð
ð _ dr TkT rT dV þ dTrce T dV ¼ dT(nTq) dS T
(6:31)
Clearly T is regarded as the primary variable and the associated secondary variable is (nTq). Suppose that the boundary is decomposed into three segments: S ¼ SI þ SII þ SIII. On SI the temperature T is prescribed as, say, T1. It follows that dT ¼ 0 on SI. On SII the heat flux q is prescribed as q1. Consequently dT(nTq) ! dT(nTq1). On SIII the heat flux is dependent on the surface temperature through a heat transfer matrix h: q ¼ q0 h(T T0). The right-hand side of Equation 6.30 now becomes ð
ð dTnT q dS ¼
S
ð
ð
dTnT q0 dS SII
dTnT q0 dS þ SIII
dTnT hðT T0 ÞdS (6:32) SIII
Next T is approximated using an interpolation function of the form T T0 NTT ðxÞuðt Þ,
dT NTT ðxÞduðt Þ
(6:33)
from which we obtain rT BTT ðxÞuðt Þ,
drT BTT ðxÞduðt Þ
(6:34)
and BT is the thermal analog of the strain–displacement matrix bT F, to be illustrated in Example 6.5 and presented in detail in Chapter 7. Upon substitution of the
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interpolation models, the thermal field equation now reduces to the system of ordinary differential equations KT uðt Þ þ MT u_ ðt Þ ¼ fT
(6:35)
in which KT ¼ KT1 þ KT2 Ð KT1 ¼ BT ðxÞkT BTT ðxÞdV Ð KT2 ¼ SIII NT ðxÞnT hNTT ðxÞdS Ð MT ¼ NT ðxÞrce NTT ðxÞdV Ð fT ¼ SII NT ðxÞnT q0 dS Ð SIII NT ðxÞnT q1 dS
Thermal Stiffness Matrix Conductance Matrix Surface Conductance Matrix Thermal Mass Matrix; Capacitance Matrix Consistent Thermal Force; Consistent Heat Flux
6.3.2 COUPLED LINEAR ISOTROPIC THERMOELASTICITY The thermal field equation is repeated as kT r2 T ¼ alT0 tr E_ þ rce T_
(6:36)
Following the same steps used for conductive heat transfer furnishes the variational principle ð
ð ð _ _ dr TkT rT dV þ dTrce T dV þ dTalT0 tr E dV ¼ dTnT q dS ð
T
(6:37)
which is the thermal analog of the Principle of Virtual Work. For the mechanical field the Principle of Virtual Work under small strain is recalled as ð
ð ð u dV ¼ duT tðsÞ dS tr ðdESÞ dV þ duT r€
(6:38)
S
Recall that S ¼ 2mE þ l½tr(E) (T T0 )I. The Principle of Virtual Work is expanded to give ð
ð ð tr dE½2mE þ l trðEÞI dV tr dElaðT T0 ÞI dV þ duT r€ u dV ð ¼ duT tðsÞ dS
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(6:39)
Now introduce the interpolation models uðxÞ ¼ NTðxÞgðt Þ
(6:40)
from which we may derive the strain–displacement matrices B(x) and b(x) satisfying VEC ðEÞ ¼ BT ðxÞgðt Þ,
tr ðEÞ ¼ bT ðxÞgðt Þ
(6:41)
Also, N is called the shape function matrix. (The operations to obtain Equation 6.41 are illustrated in Example 6.5 and presented in detail in Chapter 7.) It follows after some manipulation that € ðt Þ VT uðt Þ ¼ f Kgðt Þ þ Mg
(6:42)
in which Ð K ¼ BT ðxÞDBðxÞ dV Ð M ¼ NðxÞrNT ðxÞ dV Ð V ¼ albT ðxÞNT ðxÞ dV Ð f ¼ NðxÞt1 ðsÞ dS
Stiffness Matrix Mass Matrix Thermoelastic Matrix
(6:43)
Consistent Force Vector
We assume that the traction t(s) is specified everywhere as t1(s) on S. Also D is the isothermal tangent modulus tensor. For the thermal field, assuming that the heat flux q is specified as q1 on the surface, the thermal counterpart Equation 6.37 of the Principle of Virtual Work, together with the interpolation models Equations 6.33 and 6.34 furnish the finite element equation Ð 1 1 1 T _ _ (6:44) T0 KT uðt Þ þ T0 MT uðt Þ þ Vgðt Þ ¼ T0 f T , f T ¼ NT ðxÞn q1 dS The combined equations for a thermoelastic medium are now written in state (first order) form as 0 1 1 0 1 0 f ðt Þ g_ ðt Þ g_ ðt Þ d (6:45) W1 @ gðt ÞA þ W2 @ gðt ÞA ¼ @ 0 A dt 1 f ð t Þ uðt Þ uðt Þ T0 T 3 2 2 3 M 0 0 0 K VT 7 0 5, W2 ¼ 6 0 5 W1 ¼ 4 0 K 4 K 0 0 0 T10 MT V 0 T10 KT 0
_ g(t) u(t)}, letting yT ¼ @ Multiplying by zT ¼ {g(t)
1 f ðt Þ 0 A, 1 T0 f T ð t Þ
and performing simple
manipulations furnish the equation
d 1 T 1 1 r W1 r ¼ rT y rT W2 þ WT2 r: dt 2 2 2
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(6:46)
Note that W1 is positive definite and symmetric, while 12 W2 þ WT2 is positive semi-definite. This implies that the magnitude of r is nonincreasing if y ¼ 0. Otherwise stated, coupled linear thermoelasticity is at least marginally stable, whereas a purely elastic system is strictly marginally stable. Thus thermal conduction has a stabilizing effect which will next be shown to be analogous to viscous dissipation.
EXAMPLE 6.5 Write down the coupled thermal and elastic equations for a one-dimensional thermoelastic member modeled with one element. Suppose that, on the left end of the member, it is built into a rigid thermal reservoir with fixed temperature T0. Illustrate the analogy between conductive heat transfer and viscous damping in the case of a constant temperature field.
SOLUTION This problem will be approached by modeling the member as a single one-dimensional thermoelastic finite element, whose left end is built into a rigid thermal reservoir. The element is supposed to have length L. The displacement is modeled using u(x,t) ¼ NT(x)gm1(t) in which NT ð xÞ ¼ wT ðxÞF, wT ð xÞ ¼ ð1 xÞ L 0 1 0 1 1 F¼ ¼ L 1 1 1 L The corresponding strain–displacement matrix B is given by BT ð xÞ ¼
@ T N ð xÞ ¼ bT ð xÞF, @x
bT ð xÞ ¼ bT ð xÞF,
b T ð x Þ ¼ ð0 1 Þ
The corresponding relations for the thermal degree-of-freedom T T0 are given by NTT ð xÞ ¼ nTT ð xÞFT , nTT ð xÞ ¼ ð1 xÞ " # " #1 1 0 1 L 0 ¼ FT ¼ L 1 1 1 L BTT ð xÞ ¼
@ T N ð xÞ ¼ bTT ð xÞFT , @x T
bTT ð xÞ ¼ ð0 1Þ
For one-dimensional problems, D ! E. Following the Principle of Virtual Work the stiffness matrix is given by ð K ¼ BðxÞEBT ð xÞ dV ðL
0 1 L 1 EAð 0 L2 0 1 1 0 1 1 EA ¼ L 1 1
¼
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1 Þ dx
L 0 1 1
and for the mass matrix ð M ¼ Nð xÞrNT ð xÞ dV ðL
1 L 1 L 1 rA ð 1 x Þ dx L2 0 1 x 1 0 rAL 1 1=2 ¼ 3 1=2 1 ¼
0
1
No heat source is indicated: i.e., h ¼ 0. The thermal stiffness matrix is now ð KT ¼ BT ð xÞkT BTT ðxÞ dV ðL
0 1 L 1 kT Að 0 2 L 0 1 1 0 1 1 kT A ¼ L 1 1
¼
1 Þ dx
L 0 1 1
The thermal mass matrix is similarly obtained: ð MT ¼ NT ð xÞrce NTT ð xÞ dV ðL
1 L 0 1 L 1 rce Að1 xÞ dx ¼ 2 L 0 1 x 1 1 0 rce AL 1 1=2 ¼ 3 1=2 1 Finally, the thermoelastic matrix is presented. ð V ¼ alNT ð xÞbT ð xÞ dV ðL
1 1 L 1 alAð0 ¼ 2 L 0 1 x 0 alA 1 1 ¼ 2 1 1
1Þ dx
L 0
1 1
Collecting the foregoing results furnishes the coupled thermal and elastic finite element equations as ! !.. " # " # " # ! uð0,tÞ uð0,tÞ Tð0,tÞ T0 1 1 EA rAL 1 1=2 alA 1 1 þ ¼f L 1 1 3 1=2 1 2 uðL,tÞ uðL,t Þ 1 1 TðL,tÞ T0 !. " # " # " # ! !. Tð0,tÞ T0 Tð0,t Þ T0 uð0,tÞ 1 1 1 kT A 1 rce AL 1 1=2 alA 1 1 1 þ þ ¼ fT T0 L 1 1 T0 3 2 1 1 T0 uðL,tÞ 1=2 1 TðL,t Þ T0 TðL,tÞ T0
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A rigid ‘‘thermal bath’’ on the left is imagined to constrain both T(0,t) T0 and u(0,t) to vanish. The constraints serve to eliminate the corresponding rows and columns in the two foregoing equations, resulting in EA rAL alA € ðTðL,t Þ T0 Þ ¼ f u(L,t) þ uðL,tÞ L 3 2 1 kT A 1 rce AL alA 1 . ðT(L,t) T0 Þ þ ðTðL,t Þ T0 Þ þ u_ ðL,tÞ ¼ fT T0 L T0 3 2 T0 If the displacement field is static there is no thermomechanical coupling through the thermal field. . If (T(L,t) T0) ¼ 0, then the thermal field equation reduces to 1 alA fT u_ ðL,tÞ T0 2 TðL,t Þ T0 ¼ 1 kT A T0 L and upon substitution the mechanical field becomes 1 1 alA _ u ð L,t Þ f T C EA rAL alA B 2 C¼ f BT0 € uðL,tÞ þ uðL,tÞ A @ 1 k A L 3 2 T T0 L 0 0
Upon reorganization,
alA 2 EA rAL alL 2 u_ ðL,t Þ þ fT uðL,t Þ þ uðL,t Þ ¼ f þ € 1 kT A L 3 2kT T0 L Mathematically the middle term has exactly the same effect as conventional viscous damping. This illustrates that convective heat transfer, by carrying energy away from a site at which it is concentrated, has a stabilizing effect. However, in many materials the thermal expansion coefficient is very small, rendering the effective damping coefficient due to thermal conduction very small.
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7
One-Dimensional Elastic Elements
This chapter introduces finite elements for one-dimensional members, including rods, shafts, beams, and beam-columns. It initially presents interpolation models in physical coordinates for the sake of simplicity and brevity. But interpolation in ‘‘natural coordinates’’ is then presented to enable the use of Gaussian quadrature for integration. Use of natural coordinates to an extent reduces the sensitivity of the elements to geometric details in the physical mesh. A number of examples are given including several illustrating the use of natural coordinates.
7.1 INTERPOLATION MODELS FOR ONE-DIMENSIONAL ELEMENTS 7.1.1 RODS The governing equation for the displacements in rods (also bars, tendons, and shafts) is EA
@2u @ 2u ¼ rA 2 2 @x @t
(7:1)
in which u(x,t) denotes the radial displacement, E, A, and r are constants, x is the spatial coordinate, and t denotes time. Since the displacement is governed by a second-order partial differential equation, in the spatial domain it requires two (timedependent) constants of integration. Applied to an element, the two constants can be supplied implicitly using two nodal displacements as functions of time. We now approximate u(x,t) using its values at xe and xeþ1, as shown in Figure 7.1. The lowest-order interpolation model consistent with two integration constants is linear, in the form uð x,t Þ ¼
wTm1 ð xÞFm1 gm1 ðt Þ,
gm1 ðt Þ ¼
ue ðt Þ , ueþ1 ðt Þ
wTm1 ð xÞ ¼ (1
x)
(7:2)
We seek to identify Fm1 in terms of the nodal values of u. Letting ue ¼ u(xe) and ueþ1 ¼ u(xeþ1) furnishes ue ðt Þ ¼ (1 xe )Fm1 gm1 ðt Þ, ueþ1 ðt Þ ¼ (1 xeþ1 )Fm1 gm1 ðt Þ
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(7:3)
ue
ue+1
xe
xe+1
x
FIGURE 7.1 Rod element.
ue (t) we conclude that But since gm1 ðt Þ ¼ ueþ1 (t) 1 xe 1 Fm1 ¼ 1 xeþ1 1 xeþ1 xe , le ¼ xeþ1 xe ¼ le 1 1
(7:4)
7.1.2 BEAMS The equation for a beam, following the classical Euler–Bernoulli theory, is EI
@4w @ 2w þ rA 2 ¼ 0 4 @x @t
(7:5)
in which w(x,t) denotes the transverse (z) displacement of the beam neutral axis, and the constant I is bending moment of area. In the spatial domain there are four constants of integration. In an element the constants can be supplied implicitly by specifying the values of w and @w=@x at each of the two-element nodes. Referring to Figure 7.2, we introduce the interpolation model for w(x,t): 0 1 we B w0e C C wð x,t Þ ¼ wTb1 ð xÞFb1 gb1 ðt Þ, wTb1 ð xÞ ¼ 1 x x2 x3 , gml ðt Þ ¼ B @ weþ1 A (7:6) w0eþ1 Enforcing this model at xe and at xeþ1 furnishes 2
Fb1
1 xe 6 0 1 ¼6 4 1 xeþ1 0 1
x2e 2xe x2eþ1 2xeþ1
31 x3e 3x2e 7 7 x3eþ1 5 3x2eþ1
w te we
w te +1
xe
xe +1
(7:7)
we +1 x
FIGURE 7.2 Beam element.
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7.1.3 BEAM-COLUMNS Beam-columns are of interest, among other reasons, to predict buckling according to the Euler criterion. The displacement w is assumed to depend only on x. Also u is viewed as given by uð x, zÞ ¼ u0 ð xÞ z
@ wð x Þ @x
(7:8)
in which u0(x) represents the stretching of the neutral axis. It is necessary to know ,t ) u0(x,t), w(x,t), and @ w(x @ x at xe and xeþ1. Combining relations for the rod and the beam element, the interpolation model is now uð x, z, t Þ ¼ ð1 x ÞFm1 gm1 z 0 1 2x 3x2 Fb1 gb1 (7:9) 0 1 we B w0e C u0,e ðt Þ C , gb 1 ¼ B gm1 ¼ @ w eþ1 A u0,eþ1 ðt Þ w0eþ1
Fm1 ¼
1 x eþ1 l e 1
2
xe , 1
Fb1
1 60 ¼6 41 0
xe 1 xeþ1 1
x2e 2xe x2eþ1 2xeþ1
31 x3e 3x2e 7 7 x3eþ1 5 3x2eþ1
7.2 STRAIN–DISPLACEMENT RELATIONS IN ONE-DIMENSIONAL ELEMENTS For the rod, the strain is given by « ¼ E11 ¼ @u @x. An estimate for « implied by the interpolation model Equation 7.3 has the form «ð x, t Þ ¼ bTm1 ð xÞFm1 gm1 ðt Þ bTm1 ¼
dwTm1 ¼ ð0 1Þ dx
(7:10) (7:11)
For the beam the corresponding relation is «ð x, z, t Þ ¼ z
@2w @x2
(7:12)
from which the consistent approximation is obtained «ð x, z, t Þ ¼ zbTb1 ð xÞFb1 gb1 ðt Þ bTb1 ¼
ß 2008 by Taylor & Francis Group, LLC.
dwTb1 ¼ 0 1 dx
2x
3x2
(7:13)
For the beam-column the strain is given by
bTmb1
«(x, z, t ) ¼ bTmb1 ð xÞFmb1 gmb1 ðt Þ F m1 gm 1 ð t Þ 0 ¼ bTm1 zbTb1 0 Fmb1 gb1 ðt Þ
(7:14)
7.3 STRESS–STRAIN RELATIONS IN ONE-DIMENSIONAL ELEMENTS 7.3.1 GENERAL We first recall the stress–strain relations of an isotropic linear elasticity. If S is the stress tensor under small deformation, the stress–strain relation for a linearly elastic isotropic solid under small strain is given in Lamé’s form by S ¼ 2mEL þ l trðEL ÞI
(7:15)
in which I is the identity tensor. The Lamé coefficients are denoted by l and m, and are given in terms of the familiar elastic modulus E and Poisson ratio n as m¼
E nE , l¼ 2ð1 þ nÞ ð1 2nÞð1 þ nÞ
(7:16)
Letting s ¼ VEC(S) and e ¼ VEC(EL), the stress–strain relations are written using Kronecker product operators as s ¼ De,
D ¼ 2mI9 þ liiT
(7:17)
and D is the 9 3 9 tangent modulus tensor introduced in the previous chapters.
7.3.2 ONE-DIMENSIONAL MEMBERS For a beam-column, recalling the foregoing strain–displacement model S11 ð x, z, t Þ ¼ EE11 ¼ EzbTmb1 ð xÞFmb1 gmb1 ðt Þ
(7:18)
The cases of a rod and a beam are recovered by setting gb1 or gm1 equal to the zero vectors, respectively.
7.4 ELEMENT STIFFNESS AND MASS MATRICES FROM THE PRINCIPLE OF VIRTUAL WORK Variational calculus was introduced in Chapter 4 and the Principle of Virtual Work was introduced in Chapter 5. It is repeated here as ð ð ð dEij Sij dV þ dui r€ ui dV ¼ dui ti dS (7:19)
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As before d represents the variational operator. We assume for present purposes that the displacement, the strain, and the stress satisfy representations of the form uðx,t Þ ¼ wT ðxÞFgðt Þ, eðx,t Þ ¼ bT ðxÞFgðt Þ,
sðx,t Þ ¼ Deðx,t Þ
(7:20)
in which e ¼ VEC(E) and s ¼ VEC(S) are written as one-dimensional arrays. Since small strain is assumed, no distinction is made between the undeformed coordinates X and the deformed coordinates x.
EXAMPLE 7.1 One-element model for a built-in rod Suppose that the rod depicted in Figure 7.3 has elastic modulus E, mass density r, area A, and length L, and is modeled using a single element. It is built in at x ¼ 0. At x ¼ L there is a concentrated mass m to which is attached a spring of stiffness k. The Principle of Virtual Work reduces to the variational equation ðL d
ðL du du EA dx þ durA€ u dx ¼ du(L,t)[P ku(L,t ) m€ u(L,t )] dx dx
0
(7:21)
0
Upon application of the foregoing linear interpolation model for rod elements and enforcement of the constraint u(0,t) ¼ 0 at x ¼ 0, the stiffness and mass matrices arising from the domain reduce to scalar values as follows: M ! rAL=3, MS ! m, K ! EA=L,rAL € ¼ f: KS ! k. The governing one-element equation is EA L þk gþ 3 þm g
EXAMPLE 7.2 One-element model for a built-in rod with a constant distributed surface stress The configuration of interest is illustrated in Figure 7.4. (a) To derive the applicable version of the Principle of Virtual Work, integrate over the domain and note that ðL
d2 u du EA 2 þ px (x) dx ¼ 0, dx
px (x) ¼ 2pr0 Sxx
0
which holds for an arbitrary increment du(x) of u, subject to du(0) ¼ 0.
P
E, A, L, r
m
k
FIGURE 7.3 Rod with inertial and compliant boundary conditions.
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E, L, r0
Srx
FIGURE 7.4 Rod with constant distributed surface stress. (b) Integrate by parts ðL
ðL ðL d du ddu du duEA dx EA dx þ dupx (x) dx ¼ 0 dx dx dx dx
0
0
0
and ðL 0
d du du L duEA dx ¼ duEA
¼ 0 dx dx dx 0
since du(0) ¼ 0 and P(L) ¼ EA ðL
du(L) ¼ 0. Accordingly dx
ðL ddu du EA dx ¼ dupx (x) dx dx dx
0
0
which expresses the Principle of Virtual Work in the current case. (c) We model this member as one finite element and use the same approximation for _ as before, namely: u(x) u(x) ¼ Lx u(L) ! du(x) ¼ Lx du(L) From before, ðL dV ¼
ddu du EA dx dx dx
0
¼ Dd(L)
EA u(L) L
Now, assuming px(x) ¼ p0, ðL dW ¼ dup0 dx 0
¼
du(L)p0 L
ðL x dx 0
¼ 12 du(L)Lp0
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Hence, dV ¼ dW from which EA 1 u(L) ¼ Lp0 L 2
and also
1 L2 p0 2 EA
u(L) ¼
Now compare the last result with the exact solution given by u(x) ¼ u0A x
p0 x2 EA 2
and
u(L) ¼
p0 L 2 2EA
The exact displacement function is quadratic, whereas the function assumed in the interpolation model is linear. Despite this difference, the finite element solution gives the exact displacement at x ¼ L.
7.4.1 SINGLE-ELEMENT MODEL
FOR
DYNAMIC RESPONSE
OF A
BUILT-IN BEAM
Next consider a one-element model of a cantilevered beam to which a solid disk is welded at x ¼ L. Also attached at L is a linear spring and a torsional spring, the latter having the property that the moment developed is proportional to the (negative of the) slope of the beam. The shear force V0 and the moment M0 act at L. The member is illustrated in Figure 7.5. The Principle of Virtual Work now reduces to the equation ðL 0
ð dw00 EIw00 dx þ dwrA€ w dx 8 9 w(L,t) V0 kw(L,t) m€ < = 2 ¼ {dw(L,t) dw0 (L,t)} mr :M0 kt (w0 (L,t)) € 0 (L,t)); ( w 2
(7:22)
The interpolation model, incorporating the constraints w(0,t) ¼ w0 (0,t) ¼ 0 a priori, is L2 wð x,t Þ ¼ x2 x3 2L
L3 3L2
1
wðL,t Þ w0 ðL,t Þ
(7:23)
z y
M0
V0 m
r k
E, I, A, L, r
x T
k
FIGURE 7.5 Beam with translational and rotational inertial and compliant boundary conditions.
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Ignoring rotary inertia, the Principle of Virtual Work and the interpolation model Equation 7.23 may now be applied to furnish the domain contributions to the stiffness and mass matrices: " # " 13 # 11 EI 12 6L 35 210 L , M ¼ rAL 11 (7:24) K¼ 3 1 2 L 6L 4L2 210 L 105 L The stiffness and mass contributions from the boundary conditions are " # " # m 0 k 0 KS ¼ , MS ¼ 2 0 mr2 0 kT The governing equation is now € ðL,t Þ V0 w wðL,t Þ þ K þ K ¼ ðM þ MS Þ ð Þ S M0 € w0 ðL,t Þ w0 ðL,t Þ
(7:25)
(7:26)
EXAMPLE 7.3 Single-element model for a built-in beam under a uniformly distributed load Assuming a static response, the Principle of Virtual Work in the present case may be derived from ðL
d4 w dw EI 4 pb (x) dx ¼ 0, dx
dw(0) ¼ 0,
dw0 (0) ¼ 0
0
in which pb ¼ Sxxa. Integrating by parts twice gives d4 w dV ¼ dw EI 4 dx dx ðL 0
ðL
¼ dw00 EIw00 dx VdwjL0 M(dw0 )jL0 0
Next apply dw(0) ¼ (dw0 ) ¼ 0; V(L) ¼ M(L) ¼ 0. From the interpolation model for w(x), ðL
dw00 EIw00 dx ¼ dV
0
¼ dgT in which g ¼
w(L) . w0 (L)
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EI 12 L3 6L
6L g 4L2
ÐL We next examine the variation of the work, i.e., dW ¼ dw½pb (x) dx, assuming pb 0
equals the constant value pb0: pb ¼ pb0. From before, enforcing the conditions w(0) ¼ 0, w0 (0) ¼ 0, w(L) ¼ q1 , w0 (L) ¼ q2 a priori, w(x) is approximated as w(x) ¼
2 3x2 2x3 x x3 w(L) þ (w0 (L)) L2 L3 L L2
and now h i 2 dW ¼ pb0 L2dw(L) þ L12 (dw0 (L)) 8 9 pb0 L > > > > < = 2 0 ¼ fdw(L) (dw (L))g 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} > > > :pb0 L > ; dgT 12 The ensuing finite element equation is
EI 12 6L L 6L 4L2
w(L) w0 (L)
8p L9 b0 > > < = 2 ¼ 2 > :pb0 L > ; 12
The solution is "
# 6L 8 L 9 > > > > w(L) L3 6L 12 < 2 = ¼ > L2 > > EI 12L2 w0 (L) > : ; 12 8 9 1 pb0 L4 > > > > < = EI 8 ¼ 3> > > : 1 pb0 L > ; EI 6
(
)
4L2
The one-element finite element solution is now compared with the exact solution. The governing equation and the general form of the solution are now EIw0000 ¼ pb0 3 w(x) ¼ wnA þ wn0A x þ w00A x2 þ w000 Ax þ
pb0 x4 EI 24
Note that the exact solution is one order higher in x (fourth order) than the one-element approximation (third order). Application of the boundary conditions serves to establish that
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V (L) ¼ 0, from which w000 A ¼
w000 (L) ¼ 0,
6w000 A þ
pb0 L ¼0 EI
pb 0 L . 6EI M (L) ¼ 0, w00 (L) ¼ 0 2 pb 0 L pb 0 L 2 ¼ 0, xw00A ¼ 2w00A þ 6w000 A þ EI 2 EI 4
The tip displacement is obtained as w(L) ¼
1 pb0 L4 8 EI
and it agrees exactly with the one-element finite element solution. The tip slope is examined next and is obtained as w0 (L) ¼
pb0 L3 EI 6
which likewise agrees exactly with the one-element finite element solution. Clearly, even though the displacement model is one order lower in x than the exact solution, the one-element model gives the correct results for the displacement and slope at the tip.
EXAMPLE 7.4 Mass and stiffness matrices for a single-element beam-column Figure 7.7 is now modified to include the axial compressive force P as shown in Figure 7.6. The member is again modeled using one element. Beam-columns are assumed to be described by the classical Euler buckling equation, given in this case by w¼0 EIwiv þ Pw00 þ rA€
(7:27)
The Principle of Virtual Work gives rise to the variational equation ðL 0
ðL ðL dw00 EIw00 dx P dw0 w0 dx þ dwrA€ w dx 0
0
9 V0 kw(L,t) m€ w(L,t) = ¼ fdw(L,t) dw (L,t)g mr2 0 0 : M0 kt (w (L,t)) € (L,t)) ; ( w 2 0
8 <
Szz a E, I, L
FIGURE 7.6 Cantilevered beam under distributed load.
ß 2008 by Taylor & Francis Group, LLC.
(7:28)
z
y
V0
M0
m
r
x
P E, I, A, L, r
k
T
k
FIGURE 7.7 Built-in beam-column with translational and rotational inertia and compliant boundary conditions.
Upon application of the interpolation model Equation 7.23, the stiffness matrix due to the domain is found after some effort to be L=10 EI 12 6L P 6=5 ^ K¼ 3 L 6L 4L2 L L=10 2L2 =15
(7:29)
and the governing equation for the member is now wðL,t Þ € ðL,tÞ w V0 ^ þ K þ K ¼ S € w0 ðL,t Þ w0 ðL,t Þ M0
ðM þ MS Þ
(7:30)
Examination of Equation 7.29 reveals the presence of the negative definite matrix h i 6=5 L=10 PL L=10 . In static problems, when P becomes large enough to attain the 2L2 =15 ^ singular, buckling is predicted to occur. In a later chapter, we value Pcrit rendering K will consider natural frequencies in systems with inertia, at which time we will find that a natural frequency vanishes at Pcrit.
7.5 INTEGRAL EVALUATION BY GAUSSIAN QUADRATURE: NATURAL COORDINATES The next step is to formulate stiffness and mass matrices in which there are no constraints, say for the eth element lying between xe and xeþ1. It is helpful to digress briefly to sketch Gaussian quadrature, which underlies the notion of natural coordinates. We then return to formulate the matrices of the unconstrained element in terms of both physical and natural coordinates. Gaussian quadrature is optimal in terms of the accuracy gained for a given number of function evaluations. It, also to a certain extent, renders the element matrices insensitive to the initial geometric details of the element. In the finite element method, computation of element stiffness and mass matrices gives rise to numerous integrations, the accuracy and efficiency of which is critical. Fortunately, a method which is optimal in an important sense, called Gaussian quadrature, has long been known. It is based on converting physical coordinates to
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Ðb natural coordinates as illustrated next. Consider a f ð xÞ dx. Let j ¼ b 1 a ½2x ða þ bÞ. Clearly,Ð j maps the interval [a,b] into the interval [1,1]. The integral now transforms 1 to b 1 a 1 f ðjÞ dj. Now represent f(z) using the power series f ðzÞ ¼ a0 þ a1 z þ a2 z2 þ a3 z3 þ a4 z4 þ a5 z5 þ
(7:31)
from which ð1
2 2 f ðzÞ dz ¼ 2a1 þ 0 þ a3 þ 0 þ a5 þ 0 þ 3 5 1
(7:32)
The advantage of integration procedure on a symmetric interval can be seen in the fact that, with n function evaluations, the integral is evaluated exactly through (2n 1)st order. Consider the first 2n 1 terms in a power series representation for a function: gðzÞ ¼ a1 þ a2 z þ þ a2n z 2n1
(7:33)
Now introduce the Gaussian integration formula based on n integration (Gauss) points ji and n weights wi: ð1 1
gðzÞ dz ¼
n X
gðzi Þwi ¼ a1
i¼1
n X
w i þ a2
i¼1
n X
wi zi þ þ a2n
i¼1
n X
wi z2n1 (7:34) i
i¼1
Comparison with Equation 7.32 reveals that n X
wi ¼ 1,
i¼1 n X
n X
wi j ¼ 0,
i¼1
wi j2n2 i
i¼1
n X
wi j2i ¼ 2=3, . . .
i¼1
2 , ¼ 2n 1
n X
wi j2n1 ¼0 i
(7:35)
i¼1
It is necessary to solve for n values ji and n values wi. These are universal quantities independent of the particular function f. With ji and wi known in general, to integrate a given function g(j) exactly through j2n1 it is necessary to perform n function evaluations, namely to compute g(ji). As an example we seek two Gauss points and two weights. Now for n ¼ 2
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w1 þ w2 ¼ 2
(7:36a)
w1 j1 þ w2 j2 ¼ 0
(7:36b)
w1 j21 þ w2 j22 ¼ 23
(7:36c)
w1 j31 þ w2 j32 ¼ 0
(7:36d)
From Equations 7.36b and 7.36d, w1 j1 j 21 j22 ¼ 0, leading topjffiffi2ffi ¼ j1. From Equations 7.36a and 7.36c it now follows that j2 ¼ j1 ¼ 1= 3. Finally, the normalization w1 ¼ 1 implies that w2 ¼ 1.
EXAMPLE 7.5 Modify the rod element to replace the physical coordinate x with the ‘‘natural coordinate’’ j in which j ¼ ax þ b, jðxe Þ ¼ 1,
jðxeþ1 Þ ¼ þ1
Rewrite the interpolation model using natural coordinates, and perform the inverse to obtain Fm1.
SOLUTION We seek a, b for which the transformation j ¼ ax þ b satisfies j(xe) ¼ 1 and j(xeþ1) ¼ þ1. Elementary manipulation furnishes that a ¼ ðxeþ12 xe Þ ¼ 2=le and
@ @ @j b ¼ xeþ1leþ xe , in which le is the length of the element. Next @x ¼ @j @x, so that @ @ @2 2 @2 4 @2 ¼ a and ¼ a ¼ . Accordingly, the governing equation for the dis2 2 2 2 @x @j @x le @j @j placements in rods becomes
4EA @ 2 u @2u ¼ rA 2 2 2 le @j @t For the interpolation model the natural coordinate now is uðjðxÞ,t Þ ¼
wTm1 Fm1 gm1 ðt Þ,
wTm1
¼ ð1 jð xÞÞ, gm1 ðt Þ ¼
u e ðt Þ ueþ1 ðt Þ
Now letting ue ¼ u(j(xe)) ¼ u(1) and ueþ1 ¼ u(j(xeþ1)) ¼ u(þ1) yields
u e ðt Þ ueþ1 ðt Þ
¼ gm1 ðt Þ ¼
1 j ðxe Þ Fm1 gm1 ðt Þ 1 jðxeþ1 Þ
Consequently, Fm1 ¼
1 jðxe Þ 1 jðxeþ1 Þ
1 ¼
1 1 1 1
1 ¼
1 1 1 2 1 1
EXAMPLE 7.6 Rewrite the Euler–Bernoulli equation for the beam using the previous transformation. Rewrite the interpolation model using natural coordinates, and perform the inverse to obtain Fb1.
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SOLUTION In the natural coordinate z satisfying j ¼ a þ bx, jðxe Þ ¼ 1, jðxeþ1 Þ ¼ þ1, a ¼
2 xeþ1 þ xe , b¼ (xeþ1 xe ) xeþ1 xe
the beam equation becomes 16EI @ 4 w ¼ 0, le ¼ xeþ1 xe l4e @z4 The interpolation model is wðjð xÞ,t Þ ¼ wTb1 Fb1 gb1 ðt Þ in which 0 wTb1
¼ 1 jð xÞ j2 ð xÞ j 3 ð xÞ ,
B B gb1 ðtÞ ¼ B B @
1
we w0e =le weþ1
C C C C A
2w0eþ1 =l2 Enforcing this model at j(xe) ¼ 1, and j(xeþ1) ¼ þ1 furnishes 2 Fb1
jðxe Þ
1
6 60 1 ¼6 6 4 1 jðxeþ1 Þ 1
0 2
1 1 6 6 0 1 ¼6 61 1 4
j2 ðxe Þ 2jðxe Þ j2 ðxeþ1 Þ
2jðxeþ1 Þ 3 1 1 1 7 2 3 7 7 1 1 7 5
j3 ðxe Þ
31
7 3j2 ðxe Þ 7 7 7 j3 ðxeþ1 Þ 5 3j2 ðxeþ1 Þ
0 1 2 3 3 2 2 1 2 1 7 6 3 1 7 1 6 3 1 7 ¼ 6 46 1 0 1 7 5 4 0 1
1 1 1
EXAMPLE 7.7 We consider the clamped-free case of a beam-column modeled using a single element. We examine the effect of the compressive load P. A model using a single element gives
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V0 P
M0 L
L
FIGURE 7.8 Buckling of a cantilevered beam-column.
SOLUTION Enforcing the constraints w(0) ¼ w0 (0) ¼ 0 a priori gives the equation 2
12 65 j 4 1 j L 6 10
3 1 6 10 j L w(L) V0 ¼ 5 2 w0 (L) M0 4 15 j L2
The generalized displacements become unbounded, and hence buckling occurs, at the value of P rendering the stiffness matrix singular (Figure 7.8). 2 3 1 12 65 j 6 10 j L 2 5¼ 0, j ¼ PL det4 1 2 EI j L 4 15 j L2 6 10 from which
2 2 1 12 65 j 4 15 j L2 6 10 j L2 ¼ 0:
On solving the above equation, we have j ¼ 32.18, 2.49. Accordingly, the P values inducing buckling are P1 ¼ 2:49
EI , L2
P2 ¼ 32:18
EI L2
The exact solution is taken from Brush and Almroth (1975) as P1 ¼
p2 EI EI ¼ 2:27 2 , 4 L2 L
9 EI EI P2 ¼ p2 2 ¼ 20:42 2 4 L L
The fundamental critical load for a single element is fairly close to the exact solution, while the second buckling load is about 50% too high. Clearly, a one-element model is not very accurate for buckling. The model becomes more accurate as additional elements are added.
7.6 UNCONSTRAINED ROD ELEMENTS An element is called unconstrained if none of the points in the element are fixed. From the Principle of Virtual Work the stiffness matrix satisfies Ð K ¼ FT bðxÞD0 bT ðxÞF dV which for the foregoing rod element in the physical coordinate becomes
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@wT ðxÞ ¼ ð 0 1Þ @x 1 1 xeþ1 xe xe ¼ xeþ1 1 ðxeþ1 xe Þ 1
wT ðxÞ ¼ ð1 F¼
x Þ,
1 1
b T ð xÞ ¼
(7:37)
Also D0 ¼ E, and dV ¼ A dx, in consequence of which ð xeþ1
"
1
xeþ1 1
#
0
!
"
xeþ1 xe
EAð0 1Þ ðxeþ1 xe Þ2 xe 1 1 1 #" #" # ð xeþ1 " x 0 0 xeþ1 xe eþ1 1 EA dx ¼ 2 le xe 0 1 1 1 xe 1 " # EA 1 1 ¼ le 1 1
Ke ¼
# dx
1
xe
(7:38)
and le ¼ xeþ1 xe is the element length. We now redo the derivation for Ke using the natural coordinate j, in which j ¼ ax þ b, with 1 ¼ axe þ b, þ1 ¼ axeþ1 þ b. Interpolation in the natural coordinate now is expressed by ( T
w ¼ f1
jð xÞg,
"
F¼
1 1 2 1
1
gð t Þ ¼
# bT ¼
, 1
ue ðt Þ
)
ueþ1 ðt Þ @wT 2 @wT ¼ ¼ ð0 @x le @j
" # 1 1 1 2 2 Ke ¼ 4 1 1 le 1 þ1 ð
0 1
!
" EAð0
1Þ
1Þ
1
1
1
1
(7:39) # dx
On substituting dx ¼ l2e dj, we obtain that Ke ¼
EA 2le
þ1 ð
1
1
1
1
1
0 1
EA 1 le 1
1 1
¼
0 0
1
1
1
1
dj
(7:40)
(7:41)
Using both the physical and the natural coordinates we next determine the mass matrix Me for the segment, in which r denotes the mass density. In the physical coordinates, the Principle of Virtual Work applied to the rob element gives the mass matrix as
ß 2008 by Taylor & Francis Group, LLC.
ð M ¼ rAFT wðxÞwT ðxÞF dx ¼
"
ð xeþ1
1
1
xeþ1
#
1
!
"
xeþ1
rAð1 x Þ ðxeþ1 xe Þ2 xe 1 x 1 # " " #ð " # xeþ1 xe rA xeþ1 1 xeþ1 1 x dx ¼ 2 le xe xe x x2 1 1 1 " #" #" 1 2 2 xeþ1 xe xeþ1 rA xeþ1 1 2 xeþ1 xe ¼ 2 1 2 2 1 3 3 le xe 1 1 2 xeþ1 xe 3 xeþ1 xe 2 3 3 1 3 1 rA 3 ðxeþ1 xe Þ 6 ðxeþ1 xe Þ 5 ¼ 2 4 le 1 ðx x Þ3 1 ðx x Þ3
xe 1
xe
6
"
¼
eþ1
e
rAle 1 1=2 3 1=2 1
#
3
eþ1
#
xe
dx
#
1
e
(7:42)
In the natural coordinate, we obtain þ1 ð
Me ¼ 1
" # ! 1 1 1 1 rAð1 4 1 1 j
" jÞ
1 1
# 1 le dj 1 2
" # þ1 ð "1 j # " 1 1# rAle 1 1 dj ¼ 8 1 1 1 1 j j2 1 " #" #" # 0 1 1 rAle 1 1 2 ¼ 8 1 1 0 2=3 1 1 " # 1=2 rAle 1 ¼ 3 1=2 1
(7:43)
The terms ‘‘bar’’ and ‘‘tendon’’ are alternate names for rods. Shafts, which support torques by twisting, are described by relations which have the same mathematical form as the relations for rods. The same holds true for one-dimensional thermal or electrical conductors (the latter being described by a potential). The mass and stiffness elements for a shaft are simply Me ¼
rJle 1 1=2 , 3 1=2 1
Ke ¼
mJ 1 le 1
1 1
(7:44)
in which m ¼ 2(1Eþ n) is the elastic shear modulus and J ¼ pr04 =2 for a solid circular shaft.
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7.7 UNCONSTRAINED ELEMENTS FOR BEAMS AND BEAM-COLUMNS We next consider the mass matrix for the beam element. It is convenient to introduce the vectors g1, g2 and g as follows: w1 (xe ,t) w2 (xeþ1 ,t) , g2 ¼ g1 ¼ w01 (xe ,t) w02 (xeþ1 ,t) 9 8 w1 (xe ,t) > > > > > (7:45) > > = < w01 (xe ,t) > g1 g ¼ ---- ¼ ------------g2 > > > > > w2 (xeþ1 ,t) > > > ; : 0 w2 (xeþ1 ,t) In physical coordinates the mass matrix of the current element is given by 8 9 1 2 3T 0 xeþ1 1> 1 xe x2e x3e ð > > > > > <x= B C 60 1 2xe 3x2e 7 C 6 7 B 2 3 Me ¼ 6 f1 x x rA x g dx B C 7 2> > A 4 1 xeþ1 x2eþ1 x3eþ1 5 @ x > > > > : 3; 2 xe x 0 1 2xeþ1 3xe 2 3 1 1 xe x2e x3e " # 60 M(e) 1 2xe 3x2e 7 M(e) 6 7 11 12 6 7 ¼ (e)T 4 1 xeþ1 x2eþ1 x3eþ1 5 M12 M(2) 22 0 1 2xeþ1 3x2e M(e) 11 ¼
rALe 78 210 11Le M(2) 22
27 rALe 11Le (e) , M ¼ 12 2L2e 210 13 2 Le rALe 78 11Le ¼ 210 11Le 2L2e
13 2 Le 3 2 2 Le
(7:46)
The stiffness matrix of a beam segment is given in physical coordinates as 8 9 1 2 3T 0 xeþ1 0> x2e x3e 1 xe ð > > > > > = < B C 6 0 1 0 2xe 3x2e 7 C 6 7 B Ke ¼ 6 {0 0 2 6x} dx EI B C 7 > > A 4 1 xeþ1 2 x2eþ1 x3eþ1 5 @ > > > > : ; 2 x 6x e 0 1 2xeþ1 3xe 2 2 3 31 1 xe xe xe 6 0 1 2xe 3x2e 7 6 7 6 7 2 3 4 1 xeþ1 xeþ1 xeþ1 5 " ¼
0
1
K(e) 11
K(e) 12
(e)T K12
K(2) 22
2xeþ1 #
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3x2e (7:47)
) K(e 11
EI 12 ¼ 3 6Le Le
6Le , 4L2e
) K(e 12
EI 12 ¼ 3 Le 6Le
6Le , 2L2e
) K(e 22
EI 12 ¼ 3 Le 6Le
6Le 4L2e
Finally, if the member serves as a beam-column the stiffness matrix is augmented to furnish the new stiffness matrix Ke(bc) given by ) ¼ Ke K(P) K(bc e e 2 1 xe 6 6 0 1 K(eP) ¼ 6 6 4 1 x eþ1
x2e 2xe x2eþ1
x3e
3T 0
7 3x2e 7 7 7 x3eþ1 5
B B BP B @
0 1 2xeþ1 3x2e 2 31 1 xe x2e x3e 6 7 2xe 3x2e 7 6 0 1 6 7 6 7 x2eþ1 x3eþ1 5 4 1 x eþ1 " ¼
0
1
2xeþ1 # (eP)
K(eP) 11
K12
K(eP)T 12
K(eP) 22
) K(eP 11 ¼
ð
x e þ1
xe
9 8 0 > > > > > > > = < 1 > 0 > > 2x > > > > > ; : 2> 3x
1 1
C C 2x 3x2 dxC C A
3x2e (7:48)
P P 6=5 6=5 Le =10 (eP) ¼ , K 12 Le Le =10 2L2e =15 Le Le =10 P 6=5 Le =10 ) ¼ K(eP 22 Le Le =10 2L2e =15
Le =10 L2e =30
7.8 ASSEMBLAGE AND IMPOSITION OF CONSTRAINTS 7.8.1 RODS Consider the assemblage consisting of two rod elements: denoted as e and e þ 1, see Figure 7.9a. There are three nodes numbered n, n þ 1, and n þ 2. We first consider assemblage of the stiffness matrices, based on two principles: (a) the forces at the nodes are in equilibrium and (b) the displacements at the nodes are continuous. Principle (a) implies that, in the absence of forces applied externally to the node, at node n þ 1 the force of element e þ 1 on element e is equal and opposite the force of element e on element e þ 1. It is helpful to carefully define global (assemblage level) and local (element level) systems of notation. The global system of forces is shown in (a) while the local system is shown in (b). At the center node (eþ1) P(e) ¼0 1 P2
since no external load is applied. Also, clearly P2(e) ¼ Pn and P1(eþ1) ¼ Pnþ2.
ß 2008 by Taylor & Francis Group, LLC.
(7:49)
Pn+2
Pn
e +1
e n +1
n (a) Forces in global system
P1(e)
P1(e +1)
e
e +1 P2(e +1)
P2(e) (b) Forces in local system
FIGURE 7.9 Assembly of rod elements.
The elements individually satisfy k ðeÞ
k
ðeþ1Þ
1 1
1 1
1
1
1
1
un
¼
un þ1
u n þ1 un þ2
ð eÞ
P2
!
ðeÞ
P1
ðeþ1Þ
¼
P2
!
(7:50)
ðeþ1Þ
P1
and in this case k(e) ¼ k(eþ1) ¼ EA=L. These relations may be written as four separate equations: ð eÞ
kðeÞ un k ðeÞ unþ1 ¼ P2
(7:51a)
ð eÞ
kðeÞ un þ k ðeÞ unþ1 ¼ P1
(7:51b) ðeþ1Þ
k ðeþ1Þ unþ1 k ðeþ1Þ unþ2 ¼ P2
ðeþ1Þ
kðeþ1Þ unþ1 þ k ðeþ1Þ unþ2 ¼ P1
(7:51c) (7:51d)
Now Equations 7.51b and 7.51c are added and Equation 7.49 is applied to obtain ð eÞ
k ðeÞ un k ðeÞ unþ1 ¼ P2
k ðeÞ un þ kðeÞ þ k ðeþ1Þ unþ1 kðeþ1Þ unþ2 ¼ 0 ðeþ1Þ
kðeþ1Þ unþ1 k ðeþ1Þ unþ2 ¼ P1
ß 2008 by Taylor & Francis Group, LLC.
(7:52a) (7:52b þ 7:52c) (7:52d)
and in matrix form 2
k ð eÞ 4 k ðeÞ 0
k ðeÞ ðeÞ k þ k ðeþ1Þ kðeþ1Þ
1 0 1 un Pn k ðeþ1Þ 5@ unþ1 A¼@ 0 A Pnþ1 unþ2 kðeþ1Þ 0
30
(7:53)
The assembled stiffness matrix shown in Equation 7.53 can be visualized as an overlay of two-element stiffness matrices, referred to global indices, in which there is an intersection of the overlay. The intersection contains the sum of the lowest right hand entry of the upper matrix and the upper left hand entry of the lowest matrix. The overlay structure for a multielement rod is illustrated in Figure 7.10. The individual element-level stiffness matrices are now rewritten to refer to the global degree-of-freedom numbering system as ~ ð eÞ , KðeÞ ! K 2 ~ ðeÞ K
~ ðeþ1Þ Kðeþ1Þ ! K 3 2 1 1 0 0 7 ~ ðeþ1Þ ðeÞ 6 ðeþ1Þ 6 ¼ k 4 1 1 0 5, K ¼k 40 0 0 0 0
0 1 1
0
3 (7:54)
7 1 5 1
It is easily recognized that the global stiffness matrix (the assembled stiffness matrix K(g) of the two-element member) is simply the direct sum of the element stiffness matrices when they are referred to the global degree-of-freedom numbering system: ~ (e) þ K ~ (eþ1). K(g) ¼ K K(1) K(2) K(3) K(4) K(5) (4) (5) k22 + k11
K(N – 2) K(N –1) K(N)
FIGURE 7.10 Assembled beam stiffness matrix.
ß 2008 by Taylor & Francis Group, LLC.
By extending the two-element example toPthe multielement rod, the global ~ ðeÞ . In the subsequent sections stiffness matrix is seen to be given by KðgÞ ¼ e K we will see that in the current notation the strain energy in the two elements may be written in the form ~ ðeÞ gðeÞ VðeÞ ¼ 12 gT K ~ ðeþ1Þ gðeÞ Vðeþ1Þ ¼ 1 gT K 2
g
ðeÞT
¼ (un
unþ1
(7:55)
un þ2 )
The total strain energy of the two elements simply is the sum: 12 gT KðgÞ g. Finally notice that in the two-rod element assemblage the matrix K(g) is singular: the sum of the rows is the zero vector as is the sum of the columns. (In fact prior to the imposition of constraints the stiffness matrix in any multielement rod is singular.) In this form an attempt to solve the system will give rise to ‘‘rigid body motion.’’ To illustrate this reasoning suppose for simplicity that k(e) ¼ k(eþ1) in which case equilibrium requires that Pn ¼ Pnþ2. If computations were performed with perfect accuracy, Equation 7.53 would pose no difficulty. However, in performing computations errors ^ n ¼ Pn þ «n and P ^ nþ1 ¼ Pnþ1 þ «nþ1, arise. For example, suppose Pn is computed as P Pnþ1 ¼ Pn ¼ P. Computationally, there is now an unbalanced force «nþ1 «n. In the absence of mass, this in principle implies infinite accelerations, viewed conventionally as rigid body motion. In the finite element method, the problem of rigid body motion can be detected if the output exhibits unrealistically large deformation. This problem of nonsingularity disappears when the constraints of the problem are enforced. In the current example, symmetry implies that unþ1 ¼ 0. Recalling Equation 7.53 we now have 2
1 EA 4 1 L 0
1 2 1
1 0 1 30 Pn 0 un 1 5@ 0 A¼@ R A Pnþ1 1 unþ2
(7:56)
in which R is a reaction force which arises to enforce physical symmetry in the presence of numerically generated asymmetry. The equation corresponding to the second equation is useless in predicting the unknowns un and unþ2 since it introduces the new unknown R. Of course R is a reaction force which arises to enforce the constraint of symmetry. It is possible to ‘‘strike out’’ the second row of the equation and the second column in the matrix, conventionally referred to as ‘‘condensation.’’ If the small errors «1 and «2 are used again, the first and third equations are now rewritten as EA 1 L 0
0 1
un P þ «n ¼ unþ1 P þ «nþ1
(7:57)
with the solution EA un ¼ ½P þ «n = EA L , unþ2 ¼ ½P þ «nþ1 = L
ß 2008 by Taylor & Francis Group, LLC.
(7:58)
To preserve symmetry it is necessary for un þ unþ1 ¼ 0. However, the sum is computed as u n þ u n þ1 ¼
«n þ «nþ1 EA
(7:59)
L
The reaction force is given by R ¼ [«n þ «nþ1], and in this case it can be considered as a measure of computational error. The same assembly arguments apply equally well to the inertial forces as to the elastic forces. This is easily seen if accelerations and mass matrix components are used in Equation 7.53 instead of displacements and stiffness components (i.e., inertial response instead of elastic response). Furthermore, if both inertial and elastic responses are present, the element-level mass matrices are assembled into the global mass matrix in the same manner as in absence of elastic forces. As will be shown formally in a subsequent section, the total kinetic energy T of the two elements is ðeÞ
~ ðeþ1Þ g_ ~ þM T ¼ 12 g_ T M " # 1 1=2 ~ ð eÞ ¼ m ð eÞ M 1=2 1
ðeÞ mðeÞ ¼ 13 rAl
(7:60)
g_ T ¼ fu_ e u_ eþ1 u_ eþ2g
EXAMPLE 7.8 Write down the assembled mass and stiffness matrices of the following three-element configuration (using rod elements) (Figure 7.11). The elastic modulus is E, the mass density is r, and the cross-sectional area is A.
SOLUTION The stiffness matrix and the mass matrix for a rod element is given by KðeÞ ¼
EA 1 L 1
1 , 1
MðeÞ ¼
rAL 1 3 1=2
1=2 1
y
L
L/ 2
FIGURE 7.11 Three-element model of a rod.
ß 2008 by Taylor & Francis Group, LLC.
L
x
Now, for the first and third elements we have EA 1 1 L 1 1 1=2 rAL 1 ¼ 3 1=2 1
Kð1Þ ¼ Kð3Þ ¼ Mð1Þ ¼ Mð3Þ But the second element satisfies Kð2Þ ¼
EA 1 1 rAL=2 1 1=2 , Mð2Þ ¼ 1=2 1 L=2 1 1 3
The assembled (global) stiffness matrix is given by 2 K
ðgÞ
ð1Þ
ð1Þ
k12
k11
6 ð1Þ 6k 6 ¼ 6 21 6 0 4 0
ð1Þ
0 ð2Þ
ð2Þ
k22 þ k11 ð2Þ
0
k12 ð2Þ
ð3Þ
k21
k22 þ k11
0
k21
ð3Þ
3
7 0 7 7 7 ð3Þ 7 k12 5 ð3Þ k22
Hence 3 1 1 0 0 6 EA 6 1 3 2 0 7 7 ¼ 7 6 L 4 0 2 3 1 5 0 0 1 1 2
KðgÞ
Continuing, the assembled (global) mass matrix is 2
MðgÞ
1 1=2 rAL 6 6 ¼ 6 3 4 0 0
3 1=2 0 0 3=2 1=4 0 7 7 7 1=4 3=2 1=2 5 0 1=2 1
EXAMPLE 7.9 Show that, for the rod under gravity, a two-element model gives the exact answer for the displacement at x ¼ L, as well as a much better approximation to the exact displacement distribution (Figure 7.12).
SOLUTION The equation for a rod under gravity is EA
ß 2008 by Taylor & Francis Group, LLC.
@2u þ rAg ¼ 0, x @x2
E A r
L
g
FIGURE 7.12 Rod stretching under gravity.
u positive downward, u(0) ¼ 0,
@u(L) ¼0 @x 2
rg x As previously stated, the exact solution is u(x) ¼ rg E ½Lx 2 , u(L) ¼ E
ONE-ELEMENT
L2 2
:
MODEL
Applying variational methods to the foregoing equation furnishes ðL
ðL du0 EAu0 dx durAg dx ¼ 0
0
0
Upon invoking the interpolation model u(x) ¼ x u(L) L , we find EA rAgL rg L2 u(L) ¼ ! u(L) ¼ L 2 E 2 This result agrees with the exact solution at x ¼ L. However, the interpolation model is linear and hence does not agree with the quadratic exact solution. The finite element L 3 rg 2 model predicts that u(L=2) ¼ rg E 4, while the exact solution at x ¼ L=2 is 8 E L .
TWO-ELEMENT MODEL Here, the weight of the rod acts (is ‘‘lumped’’) at node ‘‘3’’ and the reaction, which is rALg, acts at node ‘‘1.’’ Hence the finite element equation Kg ¼ f can be written as, EA L 2 rgL2 3=8 u2 u2 2 1 ¼ rgA ¼ ! 1=2 u3 u3 E ðL=2Þ 1 1 4 1 which is exact at both nodes.
ß 2008 by Taylor & Francis Group, LLC.
EXAMPLE 7.10 Apply the method of the previous exercise to consider a stepped rod, as shown in Figure 7.13, with each segment modeled as one element.
TWO-ELEMENT
SOLUTION
The assembled stiffness matrix is 3 E2 A2 " L2 7 7 E2 A 2 1 þ a 7¼ L2 E2 A2 5 1
2
E1 A1 E2 A2 6 L1 þ L2 6 K¼6 4 E2 A2 L2 a¼
E1 A1 L1
1
#
1
L2
= ELA
2 2 2
The assembled gravitational terms furnish the consistent gravitational force as fg ¼ g
1 2 (r1 A1 L1 þ r2 A2 L2 ) L2 1 2 r 2 A2 2
! ¼ r2 A2 L2 g
1 (b 2
þ 1) 1 2
! ,
b¼
r1 A1 L1 r2 A2 L2
The solution for the displacements is u2 u3
!
1 gr2 L22 ¼ 2 E2 a
!
2þb 2þaþb
The case of the previous problem is recovered if we set a ¼ b ¼ 1 and L2 ¼ L=2.
E1 A1 r1
L1 g
E2 A2 r2
L2
FIGURE 7.13 Two-element rod stretching under gravity.
ß 2008 by Taylor & Francis Group, LLC.
EXAMPLE 7.11 Using a single finite element in each of the rod segments shown below, obtain expressions for the natural frequencies (Figure 7.14).
SOLUTION Adding the kinetic and strain energies furnishes " " #( )# u_ 1 1 r A2 L2 2 1 1 r A1 L1 2 þ 1 u_ 2 T ¼ fu_ 1 u_ 2g 2 6 2 2 3 1 2 u_ 2 3 3 22 r2 A2 L2 r1 A1 L1 r2 A2 L2 ( ) þ u_ 7 66 3 1 3 6 7 7 1 7 6 ¼ fu_ 1 u_ 2g 6 44 2 r 2 A 2 L2 r2 A2 L2 5 u_ 2 5 6 3 " " #( )# u1 1 E2 A2 1 1 1 E1 A1 2 u þ V ¼ fu1 u2g L2 2 2 L1 1 1 1 u2 22 E A E1 A1 E2 A2 3( )3 2 2 þ u 7 66 L2 L1 L2 7 1 7 1 7 6 ¼ fu1 u2g 6 5 4 5 4 2 E2 A 2 E2 A2 u2 L2 L2 Two natural frequencies are obtained from the equation 22
E2 A2 E1 A1 6 6 L2 þ L1 6 det6 44 E2 A2 L2
3 2r A L E2 A2 r A1 L1 2 2 2 þ 1 L2 7 6 3 3 2 7v 4 n1,2 r2 A2 L2 E2 A2 5 6 L2
3 r2 A2 L2 3 7 6 7 57 ¼ 0 r2 A2 L2 5 3
Some algebra serves to obtain
v2n1,2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 0 3 þ a þ b ð3 þ a þ bÞ2 að3 þ bÞ 3E2 @ A ¼ 6 þ 2b r2 L22
in which a ¼ EL1 A1 1 =EL2 A2 2 and b ¼ r1 A31 L1 =r2 A32 L2 .
E1,A1,L1,r1
FIGURE 7.14 Natural frequencies of a two-element rod.
ß 2008 by Taylor & Francis Group, LLC.
E2,A2,L2,r2
7.8.2 BEAMS A similar argument applies for beams. The potential energy and stiffness matrix of the eth element may be written as V ðeÞ ¼ 12 gTb KðeÞ gb 2 3 ð eÞ ðeÞ K K 11 12 5 KðeÞ ¼ 4 ðeÞT ðeÞ K12 K22 gTb ¼ we w0e weþ1
(7:61) w0eþ1 g
In a two-element beam model analogous to the foregoing rod model, V (g) ¼ V (e) þ V (eþ1), implying that 2
~ ðeÞ K 12
~ ð eÞ K 11
6 6 ~ ðeÞT KðgÞ ¼ 6 K 4 12 0
~ ð eÞ þ K ~ ðeþ1Þ K 22 11 ~ ðeþ1ÞT K 12
3 0
7 ~ ðeþ1Þ 7 7 K 12 5 ð eþ 1 Þ ~ K22
(7:62)
Similarly, the global mass matrix for a two-element beam segment is given by 2
~ ð eÞ M 11
6 6 ~ ðeÞT MðgÞ ¼ 6 M 4 12 0
~ ð eÞ M 12 ~ ðeþ1Þ ~ ð eÞ þ M M 22 11 ~ ðeþ1ÞT M 12
3 0 ðeþ1Þ
M12
~ ðeþ1Þ M 22
7 7 7 5
(7:63)
EXAMPLE 7.12 For the system indicated below, find the equation governing nodal displacements and slopes (Figure 7.15). This member is modeled with two beam elements, which is the minimum possible. The global displacement vector, absent the constrained degrees of freedom, is g Tg ¼ fw1 w01 w2 w02 g. The strain energy is the sum of the strain energies of the two elements and the springs: V ¼ V1 þ V2 þ V3 þ V4 in which 2
0 1 2 1 T6 0 V3 ¼ kw2 ¼ g 6 2 2 40 0
ß 2008 by Taylor & Francis Group, LLC.
3 0 0 0 0 0 07 7g , 0 k 05 g 0 0 0
2
0 1 0 1 T6 0 V4 ¼ kT w2 ¼ g 6 2 2 40 0
0 0 0 0
3 0 0 0 07 7g 0 05 0 kT
Node 1 3 1
E1,L1,L1
2
E2,L2,L2
w1
2
V0
M0
kt
3
k
1
−w 1⬘
4
– w 2⬘ w2
FIGURE 7.15 Two-element beam model with compliant supports. Also ( ) " w1 1 1 T Kð1Þ ð1Þ 0 ¼ gg V1 ¼ f w1 w1 gKI1 2 2 w01 0T " ð2Þ # ð2Þ KI1 KI2 1 V2 ¼ gTg gg ð2ÞT ð2Þ 2 K K I2
0 0
# gg
22
The total strain energy is obtained by direct addition of the element potential energies 2 ð1Þ ð2 Þ 1 T 4 K þ KI V ¼ gg ð2ÞT 2 KII
3 ð2Þ KII 5 gg k 0 ð2Þ KIII þ 0 kT
The Principle of Virtual Work now implies that 3 E 1 I1 E 2 I 2 E1 I1 E2 I2 E2 I 2 E 2 I2 þ 6 12 6 12 3 6 L2 L22 L22 7 L32 L32 78 6 L1 1 7 w1 9 8 0 9 6 E1 I1 E2 I2 E1 I1 E2 I2 E2 I 2 E2 I 2 7 > 6 > > > > 7> > > > > > > 66 þ 2 4 6 2 2 2 7< w01 = < 0 = 6 L L L L L L 1 2 2 1 2 2 7 6 ¼ 6 > E2 I 2 E2 I2 E2 I 2 E2 I 2 7 7> > w2 > > V0 > > > 6 > 12 3 6 2 12 3 þ k 6 2 : : > ; > ; 7> 6 L2 L2 L2 L2 7 w02 6 M 0 7 6 5 4 E2 I 2 E2 I2 E2 I 2 E2 I2 6 2 2 6 2 4 þ kT L L L2 L2 2 2 2
The matrix is inverted to solve for the displacements and slopes.
7.9 DAMPING IN RODS AND BEAMS In addition to mass and stiffness matrices, rods and beams experiencing timedependent loads are thought to exhibit viscous damping whose effect is represented by a damping matrix. Damping generates a stress proportional to the strain rate.
ß 2008 by Taylor & Francis Group, LLC.
In linear problems modeled by the finite element method it leads to a vector–matrix equation of the form € þ Dg_ þ Kg ¼ f ðt Þ Mg
(7:64)
in which D is the positive definite symmetric damping matrix. At the element level the counterpart of the kinetic energy and the strain energy is T the Rayleigh Damping function D(e) given by DðeÞ ¼ 12 g_ ðeÞ DðeÞ g_ ðeÞ , and the ‘‘consistent damping force’’ on the eth element is ðeÞ
f d ðt Þ ¼
@ ðeÞ D ¼ DðeÞ g_ ðeÞ @ g_
(7:65)
Just like kinetic and strain energies, the Rayleigh damping function is additive over ^ (e) is the damping matrix of the eth element referred to the elements. Accordingly, if D the global node system, the assembled damping matrix is given by X ^ ð eÞ (7:66) D¼ D e
It should be evident that the global stiffness, mass, and damping matrices have the same bandwidth: a force on one given node depends on the displacements (velocities, accelerations) of the nodes of the elements connected at the given node, thereby determining the bandwidth. Of course in lightly damped metallic structures the damping properties can be difficult to measure. In finite element practice it is common to assume Rayleigh Damping, in which the damping matrix is assumed to be a linear combination of the mass and stiffness matrices. This has the advantage of enforcing classical normal modes in which the damping matrix may be diagonalized by transformations that also diagonalize the mass and stiffness matrices. This topic will appear again in Chapter 9.
7.10 GENERAL DISCUSSION OF ASSEMBLAGE To explain the basis of the assemblage process, it is convenient to introduce Lagrange’s equation. While the Principle of Virtual Work is based on variation of in the spatial domain, Lagrange’s equation as stated below also applies variational arguments to the time domain. The equation of interest is ! d @ @ (7:67) (T V) ¼ fj dt @ g_ j @gj Here gj is the jth entry of the nodal displacement vector in which constraints have been enforced a priori. fj is the corresponding nonconservative nodal force. T and V are the total kinetic energy and total elastic strain energy of the body, expressed in terms of gj, g_ j. In particular, ð ð 1 1 ru_ 2i dV, V ¼ Sij Eij dV (7:68) T¼ 2 2 v
ß 2008 by Taylor & Francis Group, LLC.
v
But, assuming the body is represented as N elements whose volumes are denoted by Ve, N ð 1X Sij Eij dV V¼ 2 e¼1 Vc
¼
N X
1 gT Ke ge 2 e¼1 e
(7:69a)
in which ge is the local (element-level) displacement vector, only incorporating the displacements at the nodes of element e. Also Ke is the corresponding local (element-level) stiffness matrix. For example, for a rod with 10 elementsand with the 1 1 E5 A5 T . elements numbered from left to right, g5 ¼ fu5 u6 g and K5 ¼ x6 x5 1 1 But, referred to the global displacement vector g, V¼
N 1X ~ eg gT K 2 e¼1
! N 1 T X ~e g ¼ g K 2 e¼1 1 ¼ gT Kg, K ¼ 2
N X
! ~e K
(7:69b)
e¼1
Otherwise stated, the (assembled) stiffness matrix K is the algebraic sum of the local ~ e) referred to the global nodal displacement (element-level) stiffness matrices (K vector g. A parallel argument may be applied to the kinetic energy. ð T ¼ ru_ i u_ i dV V
1 0 ð N X C B ¼ @ ru_ i u_ i dV A e¼1
¼
Ve
N 1X g_ T Me g_ e 2 e¼1 e
N 1X ~ e g_ g_ T M 2 e¼1 ! N 1 T X ~ Me g_ ¼ g_ 2 e¼1
¼
1 _ ¼ g_ T Mg, 2
ß 2008 by Taylor & Francis Group, LLC.
M¼
N X e¼1
! ~e M
(7:69c)
Applied to the expressions in Equations 7.69a through 7.69c, Lagrange’s equation furnishes the expected equation, namely Mg_ þ Kg ¼ f(t).
7.11 GENERAL DISCUSSION ON THE IMPOSITION OF CONSTRAINTS Constraints serve to remove degrees of freedom. In principle, they should be enforced a priori to reduce the displacement vector to a minimum dimension in which all entries vary independently. Alternatively, the functional, whose variation is to be set equal to zero, may be augmented with additional variables (Lagrange multipliers) subject to variation. The variational principles arising from the additional variables enable enforcing the constraints a posteriori. The degrees of freedom, without regard for the constraints, and the Lagrange multipliers are then varied independently. Of course, enforcing constraints a priori in complex problems can be very difficult, especially if, for example, the constraints are ‘‘multipoint.’’ In the finite element method based on the Principle of Virtual Work, despite the mathematical objections to doing so, in most cases the variational principle is formulated without enforcing constraints a priori or without augmenting the functional using Lagrange multipliers or penalty functions. Instead, after the stiffness matrix is formulated the constraints are imposed a posteriori. The author is not aware of any errors resulting from this practice. However, in some classes of constraint problems an augmented functional is used, for example, in incompressible media to be discussed in a subsequent chapter. There are several classes of constraints. In simple constraints one or more displacements vanish at nodes on the boundary, or else assume prescribed fixed values. In linear multipoint constraints such as symmetry, several nodes are required to remain on a given line or plane. There are also internal constraints such as incompressibility, in which a kinematic requirement is imposed at all nodes throughout the body. As stated before, incompressibility will be addressed in a subsequent chapter. We first illustrate constraint enforcement when a displacement vanishes at a boundary node, in the case of a static problem. Suppose for simplicity that the constraint occurs at the last entry in the global displacement vector (prior to imposition of constraints). The finite element equation for an n degree-of-freedom system may be written as gn1 ¼ f, K 0
f n1 , fn
Kn1 K¼ kTn
kn , knn
f¼
f n1 fn
(7:70)
The stiffness matrix K is positive semidefinite and hence singular, since the vector sum of the nodal forces must vanish. If only one nodal displacement is removed the
ß 2008 by Taylor & Francis Group, LLC.
rank of the stiffness matrix is n 1, in which case the submatrix Kn1 must be positive definite. The upper block row of Equation 7.70, obtained by ‘‘striking out’’ the nth row and column of K and nth entry of f, then implies that Kn1gn1 ¼ fn1, for which there exists a unique solution. This leaves the reaction force fn unknown, but the bottom row in Equation 7.70 implies that fn ¼ knTgn1. Finally, we illustrate constraint enforcement when a boundary displacement, say the last entry of the global displacement vector, has a prescribed value gn. Now the upper block row implies that Kn1gn1 ¼ fn1 gnkn. Again the matrix of interest is obtained from K by striking out the nth row and column. However, the force vector is now different from the case of a vanishing displacement. Now there exists a unique solution of the equation Kn1gn1 ¼ fn1 gnkn.
7.12 INVERSE VARIATIONAL METHOD Given the functional subject to variation, we derive the underlying differential equation and possible combinations of boundary conditions and constraints. For this purpose we will focus on a simple example. Suppose that a rod satisfies dC ¼ 0, in which C is given by ð L 2 1 du PuðLÞ dx C¼ EA 0 2 dx
(7:71)
We again invoke the interpolation model uð x Þ ¼ f1
xgF
uðxe ,t Þ uðxeþ1 ,t Þ
(7:72)
For an element xe < x < xeþ1, we seek the matrix Ke such that Ke
uðxe ,t Þ uðxeþ1 ,t Þ
¼
fe
feþ1
(7:73)
Applying variational operations to the given equation gives dC ¼
ðL 0
du du PduðLÞ d dx dx dx EA
(7:74)
Using integration by parts on the first term furnishes, ðL du L d2 u PduðLÞ du 2 dx dC ¼ du dx 0 0 dx EA
ß 2008 by Taylor & Francis Group, LLC.
(7:75)
Now dC ¼ 0 implies that du EA
L ð L du d2 u du EA 2 dx PduðLÞ ¼ 0 dx 0 dx 0
(7:76)
The domain integral and the endpoint expressions must vanish independently. The endpoint conditions capture possible combinations of boundary conditions and constraints as follows. At x ¼ 0, either u ¼ 0 or EA du dx ¼ 0. At x ¼ L, du(L) du(L)½EA du(L) P ¼ 0, from which either u(L) ¼ 0 or EA ¼ P. dx dx From the domain integral we conclude that ðL
du EA
0
d2 u dx ¼ 0 dx2
(7:77)
and the arbitrariness of du is now appealed to again to conclude that EA
d2 u ¼0 dx2
(7:78)
which of course is the well-known differential equation of a rod under static loading. The expression of the stiffness matrix is, of course, ð (7:79) K ¼ FT bðxÞD0 bT ðxÞF dV For a one-dimensional rod element, as before, @wT ðxÞ x Þ, b ðxÞ ¼ ¼ ð0 @x 1 xeþ1 xe ¼ 1 ðxeþ1 xe Þ 1
wT ðxÞ ¼ ð1
T
1 xe 1Þ, F ¼ 1 xeþ1
1
(7:80)
Again D0 ¼ E, and dV ¼ A dx. Now, following now familiar procedures the stiffness matrix is 0 xeþ1 EA ð 0 1 Þ 2 1 1 1 ðxeþ1 xe Þ xe xe ð xeþ1 xeþ1 1 0 0 xeþ1 xe EA ¼ 2 dx le xe 1 1 1 0 1 xe EA 1 1 ¼ le 1 1
Ke ¼
ð xeþ1
1
xeþ1
1
in which le ¼ xeþ1 xe is the element length.
ß 2008 by Taylor & Francis Group, LLC.
xe 1
dx
(7:81)
EXAMPLE 7.13 Redo this derivation for Ke in the previous exercise, using the natural coordinate j, whereby j ¼ ax þ b, in which a and b are such that 1 ¼ axe þ b, þ1 ¼ axeþ1 þ b.
SOLUTION We show that it is possible to introduce interpolation directly in the natural coordinate j. @ @ 2 @ @2u ¼ @j Since @x @x @j ¼ ‘e @j, the rod equation becomes EA @j2 ¼ 0 and the corresponding interpolation relations are ( w ¼ f 1 jð xÞ g, gðtÞ ¼ T
" # 1 1 1 , F¼ 2 1 1
bT ¼
)
ue ðtÞ ueþ1 ðt Þ
@wT 2 @wT ¼ ¼ ð0 1Þ @x le @j
Using dx ¼ l2e dj, the stiffness matrix now becomes þ1 ð
Ke ¼ 1
2 0 1 1 1 2 EAð 0 4 1 1 le 1
1 1 l2 dj 1Þ 2 1 1
and finally Ke ¼
EA 1 le 1
1 1
EXAMPLE 7.14 Next regard the nodal displacement vector as a function of t. Find the matrix Me such that Me
uðxe ,t Þ uðxeþ1 ,t Þ
. .
þ Ke
uðxe ,tÞ uðxeþ1 ,t Þ
¼
fe
feþ1
in which r is the mass density. Derive Me using both physical and natural coordinates.
SOLUTION First let us derive the mass matrix using the physical coordinates. On substituting dV ¼ A dx, and letting le ¼ xeþ1 xe we have
ß 2008 by Taylor & Francis Group, LLC.
ð Me ¼ rAFT wðxÞwT ðxÞF dx ð xeþ1
"
1
xeþ1
#
1
1
"
!
xeþ1
rAð 1 x Þ ðxeþ1 xe Þ2 xe 1 x 1 # " " #ð " # xeþ1 xe rA xeþ1 1 xeþ1 1 x dx ¼ 2 2 le xe xe x x 1 1 1 " #" 1 2 2 #" xeþ1 xe xeþ1 rA xeþ1 1 2 xeþ1 xe ¼ 2 1 2 2 1 3 3 le xe 1 1 2 xeþ1 xe 3 xeþ1 xe 2 3 3 3 1 1 rA 4 3 ðxeþ1 xe Þ 6 ðxeþ1 xe Þ 5 ¼ 2 le 1 ðx x Þ3 1 ðx x Þ3 ¼
xe
6
eþ1
e
eþ1
3
xe
# dx
1
xe
#
1
e
and this becomes Me ¼
rAle 1 1=2 3 1=2 1
For the natural coordinates, following the same procedure evident in the previous exercise gives þ1 ð
Me ¼ 1
" 1 1 4 1
1 1
#
1 j
! rAð 1
"
# 1 le dj jÞ 1 1 2 1
" # þ1 ð "1 j # " 1 1# rAle 1 1 ¼ dj 8 j j2 1 1 1 1 1 " #" #" # 1 1 rAle 1 1 2 0 ¼ 8 1 1 0 2=3 1 1 and finally Me ¼
rAle 1 1=2 3 1=2 1
As has been expected, the stiffness and mass matrices are the same in the rod whether approached using physical or natural coordinates.
ß 2008 by Taylor & Francis Group, LLC.
8
Two- and ThreeDimensional Elements in Linear Elasticity and Linear Conductive Heat Transfer
8.1 INTERPOLATION MODELS IN TWO DIMENSIONS 8.1.1 MEMBRANE PLATE Consider the unconstrained triangular plate element depicted in Figure 8.1. Suppose that there is no out-of-plane stress (plane stress) or no out-of-plane displacement (plane strain). The displacements u(x,y,t) and v(x,y,t) are to be modeled using the values ue(t), ve(t), ueþ1(t), veþ1(t), ueþ2(t), and veþ2(t). A linear model in x and y suffices for each displacement owing to providing three coefficients to match three nodal values. The interpolation model now is ! " ! #" T # uð x, y, t Þ g u ðt Þ wTm2 0T Fm2 0T ¼ (8:1) gv ðt Þ vð x, y, t Þ 0T wTm2 0T FTm2 in which 0 1 2 31 1 1 xe y e B C 6 7 B B C C gu ðt Þ ¼ @ ueþ1 ðtÞ A, gv ðt Þ ¼ @ veþ1 ðt Þ A, wm2 ¼ @ x A, FTm2 ¼ 4 1 xeþ1 yeþ1 5 ueþ2 ðtÞ veþ2 ðt Þ y 1 xeþ2 yeþ2 0
u e ðt Þ
8.1.2 PLATE
1
WITH
0
ve ðt Þ
1
BENDING STRESSES ONLY
In a plate element experiencing bending only, in classical plate theory (e.g., Wang, 1953) the in-plane displacements u and v are expressed by uð x, y, z, t Þ ¼ z
ß 2008 by Taylor & Francis Group, LLC.
@w , @x
vð x, y, z, t Þ ¼ z
@w @y
(8:2)
z
Xe + 2, Ye + 2
Y
Xe +1,Ye +1 Xe,Ye
x Middle surface
FIGURE 8.1 Triangular plate element.
in which z ¼ 0 at the middle (centroidal) plane. The out-of-plane displacement w is assumed to be a function of x and y only. Clearly this model permits no in-plane (membrane) displacements in the middle plane. An example of an interpolation model is introduced as follows to express w(x,y) @w throughout the element in terms of the nodal values of w, @w @x , and @y . Clearly w has been assumed to depend only on the in-plane coordinates x and y, and the time t. wð x, y, t Þ ¼ wTb2 ð x, yÞFb2 gb2 ðt Þ
(8:3)
wTb2 ð x, yÞ ¼ 1 x y x2 xy y2 x3 12 ð x2 y þ y2 xÞ y3 @w @w @w gTb2 ðt Þ ¼ we weþ1 @x e @y e @x eþ1 @w @w @w weþ2 @y eþ1 @x eþ2 @y eþ2 2
1
6 6 60 6 6 6 60 6 6 61 6 6 6 6 F1 b2 ¼ 6 0 6 6 60 6 6 61 6 6 6 60 6 4 0
1 x y ðx þ ye Þ 2 e e e
xe
ye
x2e
xe ye
y2e
x3e
1
0
2xe
ye
0
3x2e
0
1
0
xe
2xe
0
xe ye þ 12 y2e 12 x2e þ xe ye
x2eþ1
xeþ1 yeþ1
y2eþ1
x3eþ1
1 2 xeþ1 yeþ1 ðxeþ1 þ yeþ1 Þ
xeþ1 yeþ1 1
0
2xeþ1
yeþ1
0
3x2eþ1
0
1
0
xeþ1
2yeþ1
0
xeþ1 yeþ1 þ 12 y2eþ1 12 x2eþ1 þ xeþ1 yeþ1
x2eþ2
xeþ2 yeþ2
y2eþ2
x3eþ2
1 2 xeþ2 yeþ2 ðxeþ2 þ yeþ2 Þ
xeþ2 yeþ2 1
0
2xeþ2
yeþ2
0
3x2eþ2
0
1
0
xeþ2
2yeþ2
0
ß 2008 by Taylor & Francis Group, LLC.
xeþ2 yeþ2 þ 12 y2eþ2 12 x2eþ2 þ xeþ2 yeþ2
y3e
3
7 7 7 7 7 2 7 3ye 7 7 7 3 yeþ1 7 7 7 7 0 7 7: 7 7 3y2eþ1 7 7 7 3 yeþ2 7 7 7 7 0 7 7 5 3y2eþ2 0
It follows that 0 1 @ wTb2 Bz @ x C uð x, y, z, tÞ B C B C B @ wT C @ vð x, y, z, t Þ A ¼ B b2 CFb2 gb2 ðt Þ Bz C @ @y A wð x, y, z, tÞ 0
1
(8:4)
wTb2
8.1.3 PLATE
WITH
STRETCHING
AND
BENDING
Finally, for a plate experiencing both stretching and bending, the displacements are assumed to satisfy @ wð x, y, t Þ @x @ wð x, y, t Þ vð x, y, z, t Þ ¼ v0 ð x, y, t Þ z @y
uð x, y, z, tÞ ¼ u0 ð x, y, t Þ z
(8:5)
and note that w is a function only of x, y, and t (not z). Here z ¼ 0 at the middle surface, while u0 and v0 represent the in-plane displacements. Using the nodal values of u0, v0, and w, a combined interpolation model is obtained as 0
1 @ wTb2 z B uð x, y, z, tÞ u 0 ð x, y, t Þ @x C B C C C B B C B T C B vð x, y, z, t Þ C ¼ B v0 ð x, y, t Þ C þ B @ w A B z b2 CFb2 gb2 ðt Þ @ A @ B @y C @ A 0 wð x, y, z, tÞ 0
1
0
1
2 6 wm2 6 6 ¼6 6 0T 6 4
0T
T
wTm2
0T
0T
wTb2 3 @ wT z b2 72 @ x 7 Fm2 76 6 @wTb2 7 74 0 z 7 @y 5 0 T wb2
0 Fm2 0
0
30
gu ðt Þ
1
C 7B B C 0 7 5@ gv ðt Þ A Fb2
g w ðt Þ (8:6)
8.1.4 TEMPERATURE FIELD
IN
TWO DIMENSIONS
In the two-dimensional triangular element illustrated in Figure 8.1, the linear interpolation model for the temperature is T T0 ¼ wTm2 Fm2 u2 ,
ß 2008 by Taylor & Francis Group, LLC.
uT2 ¼ ðTe T0
Teþ1 T0
Teþ2 T0 Þ
(8:7)
8.1.5 AXISYMMETRIC ELEMENTS An axisymmetric element is displayed in Figure 8.2. It is applicable to bodies which are axisymmetric and are submitted to axisymmetric loads such as all-around pressure. The radial displacement is now denoted by u and the axial displacement is denoted by w. The tangential displacement v vanishes, while radial and axial displacements are independent of u. Also u and w depend on r, z, and t. There are two cases which require distinct interpolation models. In the first case none of the nodes are on the axis of revolution (r ¼ 0), while in the second case one or two nodes are in fact on the axis. In the first case the linear interpolation model is given by
uðr, z, t Þ wðr, z, t Þ
"
¼ 2
1
wTa1 0T
re
#
0T wTa1 ze
Fa1 0
0 Fa1
31
0
gua1 ðt Þ gwa1 ðt Þ
ue
1
(8:8) 0
we
1
7 C C 6 B B wTa1 ¼ ð1 r z Þ, Fa1 ¼ 4 1 reþ1 zeþ1 5 , gua1 ¼ @ ueþ1 A, gwa1 ¼ @ weþ1 A 1 reþ2 zeþ2 ueþ2 weþ2 Now suppose that there are nodes on the axis, and note that the radial displacements are constrained to vanish on the axis. We will see later that, to attain an integrable kernal in the stiffness matrix, it is necessary to enforce the symmetry constraints a priori in the displacement interpolation model. In particular, suppose that node e is on the axis with nodes e þ 1 and e þ 2 defined counterclockwise at the other vertices. A linear interpolation model enforcing the axisymmetry constraint a priori is now ! uðr,z,t Þ wðr,z,t Þ
" ¼
wTa2
0T
0T
wTa2
#"
0
0
Fa2
wTa2
¼ ðr
z ze Þ,
Fa2
#
Fa2
reþ1 ¼ reþ2
! gua2 ðt Þ gwa2 ðt Þ
zeþ1 ze zeþ2 ze
(8:9)
1
For later purposes note that that ratio (z ze)=r is indeterminate as a point in the element is moved toward the node on the axis of revolution.
e+2 e +1 r q
FIGURE 8.2 Axisymmetric element.
ß 2008 by Taylor & Francis Group, LLC.
e
A similar formulation can be used if two nodes are on the axis of symmetry, so that the u displacement in the element is modeled using only one nodal displacement, with a coefficient vanishing at each of the nodes on the axis of revolution.
8.2 INTERPOLATION MODELS IN THREE DIMENSIONS We next consider the tetrahedron illustrated in Figure 8.3. A linear interpolation model for the temperature may be expressed as T T0 ¼ wT3T F3T u3 wT3T ¼ ð1 x y zÞ 2 1 xe ye 6 6 1 xeþ1 yeþ1 6 F3T ¼ 6 6 1 xeþ2 yeþ2 4 1 xeþ3 yeþ3 uT3T ¼ fTe T0
Teþ1 T0
(8:10)
ze
31
7 zeþ1 7 7 7 zeþ2 7 5 zeþ3
Teþ2 T0
Teþ3 T0 g
For elasticity with displacements u, v, and w, the corresponding interpolation model is 0
uð x, y, z, t Þ
1
2
wT3
B C 6 @ vð x, y, z, t Þ A ¼ 4 0T wð x, y, z, t Þ
0
T
0T wT3 0
T
32
0T
F3
76 0T 5 4 0 wT3
0
0 F3 0
0
30
g u ðt Þ
C 7B 0 5@ gv ðt Þ A gw ðt Þ
F3
z e+3
0 e
x
e+1
FIGURE 8.3 Tetrahedral element.
ß 2008 by Taylor & Francis Group, LLC.
1
e +2
y
(8:11)
8.3 STRAIN–DISPLACEMENT RELATIONS AND THERMAL ANALOGS 8.3.1 STRAIN–DISPLACEMENT RELATIONS: TWO DIMENSIONS The (linear) strain tensor for two-dimensional deformation is given by " Eð x,yÞ ¼
Exx
E xy
Exy
Eyy
#
2
@u @x
6 6 ¼6 4 1 @u @v þ 2 @y @x
3 1 @u @v þ 2 @y @x 7 7 7 5 @v @y
(8:12)
In Chapter 5 we encountered the two important cases of plane stress and plane strain. In the latter case, Ezz vanishes and Szz is not needed to achieve a solution. In the former case Szz vanishes and Ezz is not needed for solution. As opposed to VEC notation, traditional finite element ewicz notation (cf. Zienki and Taylor, 1989) introduces the strain vector «0T ¼ Exx Eyy Exy . (Strictly speaking «0 is more properly called an array since it does not have the transformation properties of vectors (cf. Chapter 2).) Upon applying the interpolation model we obtain 0 1 Exx B C ^ m2 gu2 (8:13) «0 ¼ @ Eyy A ¼ bTm2 F gv2 Exy ^ m2 is called the strain–displacement matrix and in which bTm2 F 2
bTm2
@ wTm2 6 @x 6 6 6 ¼ 6 0T 6 6 4 1 @ wT m2 2 @y
3 T
0
7 7
7 Fm 2 7 ^ 7, F m2 ¼ @y 7 0 7 T 5 1 @ wm2 @ wTm2
0
Fm2
2 @x
Hereafter, the prime will not be displayed. For a plate with bending stresses only, a vector (array) of the strains is displayed as 0 2 1 @ w B @x2 C B C B 2 C B@ wC C (8:14) «ð x, y, z, t Þ ¼ zB B @y2 C B C B 2 C @@ wA @x@y
ß 2008 by Taylor & Francis Group, LLC.
from which 0
«0 ð x, y, z, t Þ ¼ zbTb2 ð x, yÞFb2 gb2 ðt Þ,
bb2
1 @ 2 wTb2 B @x2 C B C B 2 T C B @ wb2 C C ¼B B @y2 C B C B 2 T C @ @ wb2 A
(8:15)
@x@y
Note that « 6¼ VEC(E) and instead it represents an ad hoc traditional notation in FEA. For a plate experiencing both membrane and bending stresses, the foregoing relations can be combined to furnish
bTmb2 ¼ bTm2
(8:16) «0 ð x, y, z, t Þ ¼ bTmb2 ð x, y, zÞFmb2 gmb2 ðt Þ
Fm2 gm2 ðt Þ 0 , gmb2 ðt Þ ¼ zbTb2 , Fmb2 ¼ 0 Fb2 gb2 ðt Þ
8.3.2 AXISYMMETRIC ELEMENT For the previously considered toroidal element with a triangular cross section, it is necessary to consider two cases. If there are no nodes on the axis of revolution, then application of the strain–displacement relations to the axisymmetric interpolation model furnishes 0 1 @u B C @r B C B C u B ! C " # B C r F 0 ð t Þ g a1 ua1 B C (8:17) «ðr, z, t Þ ¼ B C ¼ bTa1 @w B C 0 Fa1 gwa1 ðt Þ B C @z B C B C @ 1 @u @w A þ 2 @z @r 2 3 0 1 0 0 0 0 61 7 6 2 1 rz 0 0 0 7 7 bTa1 ¼ 6 60 0 0 0 0 17 4 5 0
0
1 2
0
1 2
0
If element e is now located on the axis of revolution, we obtain
Fa2 0 gua2 ðt Þ «ðr, z, t Þ ¼ bTa2 0 Fa1 gwa1 ðt Þ
ß 2008 by Taylor & Francis Group, LLC.
(8:18)
2 bTa2
1
6 61 6 ¼6 60 4 0
0
0 0
zze r
0 0
0
0 0
1 2
0
1 2
0
3
7 07 7 7 17 5 0
We now can see the reason for the special interpolation model. If ze were not present in the foregoing matrix, the quantity z=r would result. Of course, this quantity approaches infinity on a path approaching the node on the axis of revolution. The quantity and its square would appear in the kernal of the integral in the stiffness matrix, rendering the kernal nonintegrable. However, with the use of ze, a path to node e produces a path-dependent finite value of (z ze )=r in the limit, for which reason the kernal is integrable. The interpolation model enforces the axisymmetry constraint a priori, in order to achieve integrability. However, wherever possible the Finite Element Method enforces constraints a posteriori, which is to say that assembled finite element equation is initially obtained without accommodating constraints, and then constraints are used to remove rows and columns from the stiffness matrix.
8.3.3 THERMAL ANALOG
FOR
TWO-DIMENSIONAL
AND
AXISYMMETRIC ELEMENTS
The thermal analog of the strain is the temperature gradient. Application of the interpolation model for the temperature furnishes the relation
0 1 0 rT ¼ bTT2 FT2 u2 , bTT2 ¼ (8:19) 0 0 1 There is no need in the axisymmetric case to enforce constraints.
8.3.4 THREE-DIMENSIONAL ELEMENTS Recalling the tetrahedral element in the previous section, the strain–displacement relation for isotropic linearly elastic materials may be written as 0 1 @u B C @x B C B C @v C 0 1 B B C Exx @y B C BE C B C @w B yy C B C B C B C B Ezz C B @z C B C B C «¼B (8:20) C ¼ B 1 @u @v C B Exy C B C þ B C B C @ Eyz A B 2 @y @x C B 1 @v @w C B C Ezx þ B C B 2 @z @y C B C @ 1 @w @u A þ 2 @x @z
ß 2008 by Taylor & Francis Group, LLC.
2
F3
6 ¼ bT3 4 0 0
30 1 gu3 0 7B C 0 5@ gv3 A F3 gw3
0 F3 0
in which 2
0 60 6 6 60 T b3 ¼ 6 60 6 6 40 0
0 1
0 0 0 0
0 0
0 0
0
3 0 07 7 7 0 0 0 0 0 17 7 0 0 0 0 0 07 7 7 0 12 0 0 12 0 5
0
0
0 0
1 2
0
1 0
0 0 0 0
0 0
0 0
0
0 0
0
0
0
1 2
0 0
1 2
0
0 0 0
0
1 2
0
0
0
8.3.5 THERMAL ANALOG IN THREE DIMENSIONS Again referring to the tetrahedral element, the relation for the temperature gradient is immediately seen to be 2 rT ¼ bT3T F3T u3 ,
0
6 bT3T ¼ 4 0 0
1 0
0
0
0 0
0
0
0
0 0 0 0
0 0
0 0
0 1 0 0
0 0
0 0
0 0
0 0
3
7 0 0 5 (8:21) 0 1
8.4 STRESS–STRAIN RELATIONS 8.4.1 TWO-DIMENSIONAL ELEMENTS 8.4.1.1
Membrane Response
In two-dimensional elements, we have previously distinguished the cases of plane stress and plane strain. In plane stress, the stress–strain relations reduce to
Exx ¼ E1 Sxx nSyy
Eyy ¼ E1 Syy nSxx Exy ¼ 1 þE n Sxy
(8:22)
The case of plane strain is retrieved by using E* ¼ 1 En2 and n* ¼ 1 n n in place of E and n. In traditional finite element notation Equation 8.22 may be written as
ß 2008 by Taylor & Francis Group, LLC.
0
1 0 1 Sxx Exx @ Syy A ¼ Dm21 @ Eyy A Sxy Exy
(8:23)
in which 2
Dm21
1 ¼ E4 v 0
v 1 0
31 2 0 1 E 4v 0 5 ¼ 1 v2 1þv 0
3 v 0 1 0 5 0 1þv
(8:24)
may be called the tangent modulus matrix under plane stress (note that, unlike the previous definition, it is not based on the VEC operator and is not a tensor). The corresponding matrix for plane strain is denoted by Dm22. However, we shall see that a slightly different quantity from Dm22 is needed in plane stress. The Principle of Virtual Work uses the strain energy density given by 1 2 Sij Eij . Elementary manipulation serves to prove that 2
1 2 Sij Eij
¼ 12 ð Sxx
Syy
1 Szz Þ4 0 0
0 1 0
1 30 Exx 0 0 5@ Eyy A Exy 2
(8:25)
Accordingly, in the Principle of Virtual Work the tangent modulus matrix in plane stress is replaced by D0mb2
2 1 E 4 ¼ n 1 n2 0
n 1 0
3 0 5 0 2ð1 þ nÞ
(8:26)
and similarly for plane strain. (The peculiarity represented by the ‘‘2’’ in the lower right-hand diagonal entry is an artifact of traditional finite element notation and does not appear if VEC notation is used.) The stresses are now given in terms of nodal displacements by 0 1 1 Exx Sxx ð x,y,t Þ ^ m2 gu2 , i ¼ 1 plane stress (8:27) @ Syy ð x,y,t Þ A ¼ D0 @ Eyy A ¼ D0 bT F m2i m2i m2 2 plane strain gv2 Sxy ð x,y,t Þ Exy 0
8.4.1.2
Two-Dimensional Members: Bending Response of Thin Plates
Thin plates experiencing only bending are assumed to be in a state of plane stress. The tangent modulus matrix is again given by Equation 8.26, and now an approximation for the stress is obtained as 0
1 0 1 Sxx Exx ^ @ Syy A ¼ Dm21 @ Eyy A zDm21 bT F b2 b2 gb2 ðt Þ Sxy Exy
ß 2008 by Taylor & Francis Group, LLC.
(8:28)
8.4.1.3
Element for Plate with Membrane and Bending Response
Plane stress is likewise applicable to the combined case, and consequently the stresses are modeled as 0
1 0 1 Exx Sxx ^ mb2 gmb2 ðt Þ @ Syy A ¼ Dm21 @ Eyy A ¼ Dm21 bT ðx, y, zÞF mb2 Sxy Exy
(8:29)
8.4.2 AXISYMMETRIC ELEMENT For the purpose of determining a stress model consistent with the underlying interpolation model, it is sufficient to consider the case in which none of the nodes of the element are located on the axis of revolution. 0
1 0 1 Srr Err
B Suu C B C ^ B C ¼ Da B Euu C ¼ Da bT Fa1 a1 @ Szz A @ Ezz A 0 Srz Erz
0 ^ a1 F
gua1 ðt Þ gwa1 ðt Þ
(8:30)
in which the tangent modulus matrix is given by 2
1 6 n 6 D a ¼ E6 4 n
n 1
n n
0 0
n
1
0
31 7 7 7 5
0 1þn 2 1n n 6 1n E 6 n ¼ 6 ð1 2nÞð1 þ nÞ 4 n n 0
0
0
0
n n
0 0
1n 0
0 1 2n
3 7 7 7 5
For use in the Principle of Virtual Work, Da is modified to furnish D0a given by 2
1n 6 n E 0 6 Da ¼ ð1 2nÞð1 þ nÞ 4 n 0
n 1n n 0
n n 1n 0
3 0 7 0 7 5 0 2(1 2n)
(8:31)
8.4.3 THREE-DIMENSIONAL ELEMENT All six stresses and strains are now present. Using traditional finite element notation we write
ß 2008 by Taylor & Francis Group, LLC.
0
0 1 1 Sxx Exx B Syy C B Eyy C 2 B B C C F3 BS C BE C B zz C B zz C T4 B C ¼ D3 B C ¼ D3 b 3 0 B Sxy C B Exy C B B C C 0 @ Syz A @ Eyz A Szx
30 1 gu3 0 0 5@ gv3 A F3 gw3
0 F3 0
(8:32)
Ezx 2
n
n
0
0
0
1
n
0
0
0
n
1
0
0
0
0
0
1þn
0
0
0
0
0
1þn
0
0
0 2
0
0
1þn
1n 6 n 6 6 n E 6 ¼ ð1 2nÞð1 þ nÞ 6 6 0 4 0 0
n 1n n 0 0 0
1
6 6 n 6 6 n 6 D3 ¼ E6 6 0 6 6 4 0 0
n n 1n 0 0 0
31 7 7 7 7 7 7 7 7 7 5
0 0 0 1 2n 0 0
0 0 0 0 1 2n 0
3 0 0 7 7 0 7 7 0 7 7 0 5 1 2n
and for the Principle of Virtual Work, the associated matrix is D03 ¼
E ð1 2nÞð1 þ nÞ 2 1n n 6 n 1n 6 6 n n 6 6 6 0 0 6 4 0 0 0 0
n n 1n 0 0 0
0 0 0 0 0 0 2(1 2n) 0 0 2(1 2n) 0 0
3 0 7 0 7 7 0 7 7 7 0 7 5 0 2(1 2n)
(8:33)
8.4.4 ELEMENTS FOR CONDUCTIVE HEAT TRANSFER Assuming the isotropic version of the Fourier Law, the heat flux vector, which may be considered the thermal analog of the stress, is obtained using 8 T b F u , > > < T1 T1 1 q ¼ k bTT2 FT2 u2 , > > : T bT3 FT3 u3 ,
ß 2008 by Taylor & Francis Group, LLC.
1D 2D 3D
(8:34)
8.5 STIFFNESS AND MASS MATRICES AND THEIR THERMAL ANALOGS Elements of variational calculus were discussed in Chapter 3 and the Principle of Virtual Work was introduced in Chapter 5. It is repeated here as ð
ð ð dEij Sij dV þ dui r€ ui dV ¼ dui ti dS
(8:35)
As before we assume that the displacement, the strain, and the stress may to satisfactory accuracy be approximated using expressions of the form u(x,t) ¼ wT (x)Fg(t),
E ¼ bT (x)Fg(t), S ¼ DE
(8:36)
in which E and S are written as one-dimensional arrays in accordance with traditional finite element notation, and of course t is the traction vector. Also, for use of traditional finite element notation in the Principle of Virtual Work, it is necessary to make use of D0 , which introduces the factor 2 into the entries corresponding to shear. We suppose that the boundary is decomposed into four segments: S ¼ SI þ SII þ SIII þ SIV. On SI, u is prescribed, in which event du vanishes. On SII the traction t is prescribed as t0. On SIII there is an elastic foundation described by t ¼ t0 A(x)u, in which A(x) is a known matrix function of x. On SIV there are inertial boundary conditions by virtue of which t ¼ t0 Bü. The right-hand term now becomes ð
ð dui ti dS ¼ dg
T SII þSIII þSIV
FT w(x)t0 dS
ð
FT w(x)AwT (x)F dS g(t)
dgT ð
SIII
€ (t) FT w(x)BwT (x)F dS g
dgT
(8:37)
SIV
The leftmost term in Equation 8.35 becomes ð
ð
dEij Sij dV ¼ dg Kg(t), T
ð
K ¼ FT b(x)D0 bT (x)F dV ð
(8:38)
dui r€ ui dV ¼ dg M€ g(t), M ¼ rF w(x)w (x)F dV T
T
T
in which K is of course the stiffness matrix and M the mass matrix. Canceling the arbitrary variation and bringing terms with unknowns to the left-hand side furnish the equation g(t) ¼ f(t) (K þ KS )g(t) þ (M þ MS )€
ß 2008 by Taylor & Francis Group, LLC.
(8:39)
ð f¼
SII þSIII
FT w(x)t0 dS
ð KS ¼
FT w(x)AwT (x)F dS SIII
ð MS ¼
FT w(x)BwT (x)F dS SIV
Clearly, elastic supports on SIII furnish a boundary contribution to the stiffness matrix, while mass on the boundary segment SIV furnishes a contribution to the mass matrix.
8.6 THERMAL COUNTERPART OF THE PRINCIPLE OF VIRTUAL WORK For current purposes we focus on the equation of conductive heat transfer as kr2 T ¼ rce
@T @t
(8:40)
Multiplying by the variation of T T0, integrating by parts and applying the divergence theorem furnishes ð
ð ð @T dV ¼ dTnT q dS drT TkrT dV þ dTrce @t
(8:41)
Now suppose that the interpolation models for temperature in the current element furnish a relation of the form T T0 ¼ wTT (x)FT u(t),
rT ¼ bTT (x)FT u(t), q ¼ kTT (x)FT u(tÞ
(8:42)
The left-hand terms in Equation 8.41 may now be written as ð
ð dr TkrT dV ! du (t)KT u(t), T
ð
T
@T _ dTrce dV ¼ duT (t)MT u(t), @t
KT ¼ kFTT bT bTT FT dV ð MT ¼
(8:43) kFTT wT wTT FT
dV
KT and MT may be called the thermal stiffness (or conductance) matrix and thermal mass (or capacitance) matrix, respectively. Next, suppose that the boundary S has four zones: S ¼ SI þ SII þ SIII þ SIV. On SI the temperature is prescribed as T1, from which we conclude that dT ¼ 0. On SII the heat flux is prescribed as nTq1. On SIII, the heat flux satisfies nTq ¼ nTq1 h1(T T0), while on SIV, nTq ¼ nTq1 h2 dT=dt. The governing finite element equation is now
ß 2008 by Taylor & Francis Group, LLC.
MTS
_ t ) þ ½KT þ KTS u(t ) ¼ f T (t ) ½MT þ MTS u( ð ð ¼ FTTT wT h2 wTT dS FT , KTS ¼ FTT wT h1 wTT dS FT ð
f T ðt Þ ¼
FTT
SIV
(8:44)
SIII
wT nT q1 dS,
V ¼ SII þ SIII þ SIV
V
8.7 CONVERSION TO NATURAL COORDINATES IN TWO AND THREE DIMENSIONS The notion of natural coordinates is applicable in two and three dimensions. It requires transforming the undeformed coordinates of the physical element to a reference element with suitable symmetry properties. As an illustration consider the quadrilateral element shown below. We seek transformations zk(Xj) and their inverses Xj(zk) such that the nodes of ) (e) ,z ¼ ( 1, 1); the physical element are mapped to the values z(e (eþ1) (eþ1) (eþ2) (eþ2) (eþ3) (eþ3) 1 2 z 1 ,z 2 ¼ (1, 1); z1 ,z2 ¼ (1,1); z1 , z2 ¼ (1, 1) in the transformed element. The coordinates zk(Xj) are of course the natural coordinates. In this instance we also require that straight lines remain straight lines. The element in the transformed coordinates is depicted as follows. The mapping is achieved by the functions 8 (e) 9 8 (e) 9 X1 > X2 > > > > > > > > > > > > > > > > > > = < X (eþ1) = < X (eþ1) > 1 2 X2 ¼ f1 z1 z2 z1 z2 gFz z1 z2 gFz > > > > > > X1(eþ2) > X2(eþ2) > > > > > > > > > > > > ; : (eþ3) ; : (eþ3) > X1 X2
X 1 ¼ f1 z 1 z 2
2
1
61 6 Fz ¼ 6 41 1
1
1
1 1
1 1
1
1
1
31
2
1
1 7 16 6 1 7 7 ¼ 6 5 4 4 1 1 1 1
1
3
1
1
1 1
1 1
1 7 7 7 1 5
1
1
1
(8:45)
Along the sides z1 ¼ 1, 1, X1 and X2 are linear functions of z2. Accordingly they are linear functions of each other and hence the top and bottom faces of the square in Figure 8.5 map into lines between the nodal values at the endpoints in Figure 8.4. A similar observation holds regarding the right and left faces. Clearly the transformation relations between the physical and the natural coordinates are reminiscent of interpolation models introduced heretofore for deformed coordinates in terms of undeformed coordinates. If the transformation model involves the same order of polynomial as the interpolation model, the element in the natural coordinates is said to be isoparametric.
ß 2008 by Taylor & Francis Group, LLC.
X2
X1(e +2), X2(e +2) X1(e+3), X2(e+3)
X1(e +1), X2(e +1) X1(e), X2(e)
X1
FIGURE 8.4 Two-dimensional element in physical coordinates.
The Jacobian matrix for the transformation is defined here as by 2
@X1 6 @X 6 @z1 ¼6 J¼ @z 4 @X2 @z1
3 @X1 @z2 7 7 7 @X2 5 @z2
(8:46)
X1 (z1 ,z2 ) z1 . (Sometimes J is defined as (dz=dX)T.) in which X ¼ and z ¼ X2 (z1 ,z2 ) z2 The reader may find our definition surprising since, as first glance, z denotes the coordinates being introduced by the transformation. However, we take the view that
z2 −1,1
1,1
z1
−1,−1
1,−1
FIGURE 8.5 Two-dimensional element in natural coordinates.
ß 2008 by Taylor & Francis Group, LLC.
the natural coordinates now represent the reference configuration, since the final matrices and vectors will be expressed in terms of the natural coordinates.) The Jacobian matrix is reminiscent of the deformation gradient tensor F. The transformation zk(Xj) is invertible and the inverse transformation Xj(zk) exists if J and hence its inverse are nonsingular. The volume (area) in the physical element dA and in the transformed element dAz are related by dAz ¼ det(J1) dA, and clearly a singular inverse Jacobian matrix would imply mapping the physical element onto a zero-volume element in natural coordinates. The determinant of the Jacobian matrix is denoted as J: J ¼ det(J). Assuming nonsingularity, the inverse of the Jacobian matrix is readily verified using the Chain Rule of calculus to be given by 2
J1
@ z1 6 @ X1 ¼6 4 @z 2 @ X1
3 @ z1 @ X2 7 7 @ z2 5 @ X2
(8:47)
We now consider referring the Principle of Virtual Work to natural coordinates. The inertial term becomes ð
ð € r dV ¼ duT u € rJ dVz du u T
(8:48)
Now suppose that the displacement vector in element e is approximated using an interpolation model in the natural coordinates: u(X,t) ! u(z,t) wTz (z)Fze gze
(8:49)
The inertial term in the eth element is now ð € rJ dVz ¼ dgTze Mze g € ze duT u
ð Mze ¼ FTz wze wTze rJ dVz Fz
(8:50)
We next consider the consistent force, assuming that the traction vector is specified at all points on the exterior boundary. In Chapter 13, we will encounter a relation between the deformed surface area element dSd and the corresponding undeformed qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi element: namely dSd ¼ m dS in which m ¼ det(F) nT0 C1 n0 and n0 is the surface normal vector of the undeformed element. We may now write qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dS ¼ mz dSz , mz ¼ det(J) nTz J1 JT nz
ß 2008 by Taylor & Francis Group, LLC.
(8:51)
in which dSz is the area of the element in the natural coordinate system and nz is the corresponding unit normal vector. We now determine the corresponding consistent force. The term in the Principle of Virtual Work which represents the virtual external work of the traction is transformed according to ð
ð du (X,t )t(X,t ) dS ¼ duT (z,t )t(z,t )mz (z,t ) dSz T
¼ dgTze f ze
(8:52)
ð f ze ¼ Fze wz (z)t(z,t )mz (z,t ) dSz Finally we address the term representing virtual internal work of the stress, and is hereafter called the stiffness term. First note that, in linear elasticity, owing to the total symmetry of the tangent modulus tensor C(cijkl), dEji Sij ¼ dEij cijkl Ekl @ duj @ ul cijkl @ Xi @ Xk
@ duj @ zm @ zn @ ul ¼ cijkl @ z m @ Xi @ Xk @ z n ! @ du T @ du ¼ tr Cz @z @z ¼
(8:53)
in which the fourth-order tangent modulus referred to natural coordinates is expressed as
@z @z ½Cz mjln ¼ JT CJ1 mjln ¼ m cijkl n @Xi @Xk The stiffness term referred to natural coordinates may now be identified. ð
ð tr(dES) dV ¼ tr
@du @z
T
! @du Cz J dVz @z
(8:54)
The interpolation model (Equation 8.49) implies a subsidiary model of the form du(z,t) ¼ bTz (z)Fze gze VEC dz
ß 2008 by Taylor & Francis Group, LLC.
(8:55)
Accordingly, the stiffness term assumes the form ð tr (dES) dV ! dgTze Kze gze ð T Kze ¼ Fze bz (z) JT (z)CJ1 (z) (z)bTz (z)J ðzÞ dVz
(8:56)
The ensuing element level finite element equation is now stated as € ze þ Kge gze ¼ f ze Mz e g
(8:57)
The transformation to natural coordinates preserves the kinetic energy and the potential energy of the element, in consequence of which assemblage of element matrices to obtain global matrices proceeds by direct addition in the manner introduced in Chapter 7. Similar arguments to the above furnish the transformations for the thermal stiffness matrix. ð ð drT TkrT T dA ¼ ðrB TdTÞkJT J1 ðrB TÞJ dAz
KTze
¼ duTze KTze uze ð ¼ FTTz kbTz JT J1 bTTz J dVz FTz
(8:58)
The transformations for the thermal mass matrix and the consistent thermal force are parallel to the mechanical field counterparts.
EXAMPLE 8.1 Find the Jacobian matrix and its determinant for the transformation shown below (Figure 8.6).
SOLUTION The transformation is achieved using 8 9 x1 > > > < > = x2 x ¼ f1 B h hBgF , x > > > : 3> ; x4 X2
0.1,1
y ¼ f1 B h
8 9 y1 > > > < > = y2 hBgF > y3 > > > : ; y4
1.1,1.2 z2 –1,1
1,1
z1 1,0.1 X1
FIGURE 8.6 Figure for determinant of a Jacobian.
ß 2008 by Taylor & Francis Group, LLC.
–1,–1
1,–1
The matrix F is given above in Equation 8.45. The Jacobian matrix is obtained as 2 8 8 9 93 (1) > > x(11) > > x1 > > > > > > > > 6 > > > >7 > > 6 < x(2) > < x(2) > = =7 7 6 1 1 6 f0 1 0 z2 gF (3) f0 0 1 z1 gF (3) 7 7 6 > > > > x x > > > > 7 6 > 1 > > 1 > > > > > 6 > > > > : (4) ; : (4 ) ; 7 7 6 x1 x1 7 6 J¼6 8 9 8 97 ( 1 ) ( 1) 7 6 > > x > x > > > > >7 6 > > > 2(2) > > > 2(2) > >7 6 > > > > < = <x =7 6 2 7 6 f0 1 0 B gF x2 0 0 1 B f gF 6 1 1 (3 ) > (3) > 7 > > 6 > x2 > > > x2 > >7 > > 4 > > > >5 > > : (4 ) > ; : (4) > ; x2 x2 The determinant J is recognized as 8 (1) 910 8 (1) 91 x2 > x1 > > > > > > > > > > > > > > >C CB B > > > > > > C B B (2) > (2 ) > < < = x2 x1 =C CB C B CBf0 1 0 z 2 gF C J ¼B 0 0 1 B f gF 1 CB C B ( 3) ( 3 ) > > > > > > > > C C B B x x > > 2 > 1 > > > > > A@ A @ > > > > > > > > : : ; ; (4) (4 ) x2 x1 8 (1 ) 91 0 8 (1) 91 0 x1 > x2 > > > > > > > > > >C B > > > > >C B > > > > > C B B (2 ) > (2) > < < = =C x1 x2 > CB C B CBf0 1 0 z2 gF C B 0 0 1 B f gF 1 CB C B ( 3 ) ( 3) > > > > > > > > C C B B x x > > 1 > 2 > > > > > A@ A @ > > > > > > > > : : ; ; (4 ) (4) x1 x2 8 9 3 8 (1) 9 28 9 x2 > > 0> 0> > > > > > > > > > > > > > > > > > > > > 7 6> > > > > > > (2) > < < = = < = 7 6 0 x (1) (2) (3) (4) T 6 1 2 7 ¼ x1 x1 x1 x1 F 6 f0 0 1 B1 g f0 1 0 B2 g7F > > 7 > 6> > > > 1> x(23) > >0> > > > > > 5 > 4> > > > > > > > > > > > > : ; : ; > > : (4) ; B1 z2 x2 3 8 (1) 9 2 x > > 0 0 0 0 > > 2 > > > > 7 > 6 > )> < x(2 = 7 > 0 0 1 z (1) (2) (3) (4) T 6 1 2 7 6 ¼ x1 x1 x1 x1 F 6 7F (3) > 6 0 1 0 z 7 > 25 > > x2 > > 4 > > > > > : (4) > ; 0 z 1 z 2 0 x2 0
and finally
x(11) x(12) x(13)
8 (1) 9 8 9 2 x2 > > 1 1 0 > > > > > > > > > > > (2) > < = < = 6 1 1 0: 1 x 1 6 2 ¼ , F¼ 6 x(14) ¼ f0 1 1:1 0:1g; (3) > > > > 4 1 1 4 1: 2 x > > > > 2 > > > > > : (4) > ; : 1 ; 1 1 x2
3 1 1 1 1 7 7 7 1 15 1 1
It is left to the reader to compute the value of J using the relations in the preceding lines.
ß 2008 by Taylor & Francis Group, LLC.
8.8 ASSEMBLY OF TWO- AND THREE-DIMENSIONAL ELEMENTS We next consider assembly of stiffness matrices for physical elements in two dimensions. Assembly in three dimensions follows the same procedures. Consider the model depicted below (Figure 8.7), consisting of four rectangular elements, denoted as element e, e þ 1, e þ 2, e þ 3. The nodes are also numbered in the global system. Locally, the nodes in an element are numbered in a counterclockwise scheme starting from the lower left-hand corner. Suppose there is one degree of freedom per node (e.g., x-displacement), and one corresponding force. In the local numbering system, the force on the center node induces displacements according to (e) (e) (e) (e) ue,1 þ k3,2 ue,2 þ k3,3 ue,3 þ k3,4 ue,4 fe,3 ¼ k3,1 (eþ1) (eþ1) (eþ1) (eþ1) feþ1,4 ¼ k4,1 ueþ1,1 þ k4,2 ueþ1,2 þ k4,3 ueþ1,3 þ k4,4 ueþ1,4 (eþ2) (eþ2) (eþ2) (eþ2) feþ2,1 ¼ k1,1 ueþ2,1 þ k1,2 ueþ2,2 þ k1,3 ueþ2,3 þ k1,4 ueþ2,4
(8:59)
(eþ3) (eþ3) (eþ3) (eþ3) feþ3,2 ¼ k2,1 ueþ3,1 þ k2,2 ueþ3,2 þ k2,3 ueþ3,3 þ k2,4 ueþ3,4
The conversion from local to global coordinates is expressed by ue,1 ! u1
ue,2 ! u2
ue,3 ! u5
ue,4 ! u6
ueþ1,1 ! u2 ueþ2,1 ! u5 ueþ3,1 ! u6
ueþ1,2 ! u3 ueþ2,2 ! u4 ueþ3,2 ! u6
ueþ1,3 ! u4 ueþ2,3 ! u9 ueþ3,3 ! u8
ueþ1,4 ! u5 ueþ2,4 ! u8 ueþ3,4 ! u7
e + 3,3
e + 3,4
e + 2,4
e+3 e + 3,1
e+2
e + 3,2
6
4
5
1
e +1 2
3
e + 1,3
e + 1,4 e+1
e e,1
e + 2,2
e+2
e e,3
e + 2,1
9
8
7 e+3
e,4
e + 2,3
e,2
FIGURE 8.7 Two-dimensional assembly process.
ß 2008 by Taylor & Francis Group, LLC.
e + 1,1
e + 1,2
(8:60)
Adding the forces of the elements on the center node gives (e)
(eþ1)
(e) (eþ1) (eþ1) (eþ2) u1 þ k3,2 þ k4,1 u3 þ k4,3 þ k1,2 u2 þ k4,2 u4 f5 ¼ k3,1 (e)
(e)
(eþ1) (eþ2) (eþ3) (eþ3) þ k3,3 þ k4,4 þ k1,1 þ k2,2 þ k2,2 u5 þ k3,4 u6
(eþ3) (eþ2) (eþ3) (eþ2) u8 þ k1,3 þ k2,4 u7 þ k1,4 þ k2,3 u9
(8:61)
Taking advantage of the symmetry of the stiffness matrix, this implies that the fifth row of the stiffness matrix is (e) (e)
(eþ1) (eþ1)
(eþ1) (eþ2) kT5 ¼ k3,1 k3,2 þ k4,1 k4,3 þ k1,2 k4,2 (e) (eþ1) (eþ2) (eþ3) þ k1,1 þ k2,2 symmetry k3,3 þ k4,4
(8:62)
The process can be repeated for all of the nodes, leading to the assembled stiffness matrix. The process is essentially the same for solid and axisymmetric elements. Of course, a conceptually easier way is to add the kinetic and strain energies of the individual elements, referred to the global numbering system for degrees of freedom, as explained in Chapter 7.
EXAMPLE 8.2 Assemble the stiffness coefficients associated with node n below, assuming plane stress elements. The modulus is E and the Poisson’s ratio is n. K(1), K(2), and K(3) denote the stiffness matrices of the elements (Figure 8.8).
SOLUTION Now suppose there are two degrees of freedom per node (x- and y-displacements), and two corresponding forces. Since the elements are three-noded, and each node has two degrees of freedom, the element stiffness matrices will be 6 3 6 and the force vectors will be 6 3 1.
K (1)
n
K (3)
K (2)
FIGURE 8.8 Assemblage of triangular elements.
ß 2008 by Taylor & Francis Group, LLC.
The element level equation 2 (1) (1) k11 k12 6 (1) (1) 6 k21 k22 6 6 (1) (1) 6 k31 k32 6 6 (1) (1) 6 k41 k42 6 6 (1) (1) 4 k51 k52 (1) k61
(1) k62
for element (1) is stated as (1) (1) (1) (1) 30 1 0 1 k13 k14 k15 k16 fx1,1 u1,1 (1) (1) (1) (1) 7B C B 7B u1,2 C k23 k24 k25 k26 C B fx1,2 C 7B C C B (1) (1) (1) (1) 7B C B 7B u1,3 C k33 k34 k35 k36 C B fx1,3 C 7B ¼ C C B (1) (1) (1) (1) 7B v C B k43 k44 k45 k46 7B 1,1 C C B fy1,1 C 7B C C B (1) (1) (1) (1) 7@ v @ fy1,2 A 1,2 A k53 k54 k55 k56 5 fy1,3 v1,3 k (1) k (1) k (1) k (1) 63
64
65
66
and similarly for two remaining elements. Following the assemblage procedure, in the local system the force on the center node ‘‘n’’ induces displacements according to (1) (1) (1) (1) (1) (1) fx1,2 ¼ k21 u1,1 þ k22 u1,2 þ k23 u1,3 þ k24 v1,1 þ k25 v1,2 þ k26 v1,3 (1) (1) (1) (1) (1) (1) fy1,2 ¼ k51 u1,1 þ k52 u1,2 þ k53 u1,3 þ k54 v1,1 þ k55 v1,2 þ k56 v1,3 (2) (2) (2) (2) (2) (2) fx2,2 ¼ k21 u2,1 þ k22 u2,2 þ k23 u2,3 þ k24 v2,1 þ k25 v2,2 þ k26 v2,3 (2) (2) (2) (2) (2) (2) fy2,2 ¼ k51 u2,1 þ k52 u2,2 þ k53 u2,3 þ k54 v2,1 þ k55 v2,2 þ k56 v2,3 (3) (3) (3) (3) (3) (3) fx3,2 ¼ k21 u3,1 þ k22 u3,2 þ k23 u3,3 þ k24 v3,1 þ k25 v3,2 þ k26 v3,3 (3) (3) (3) (3) (3) (3) fy3,2 ¼ k51 u3,1 þ k52 u3,2 þ k53 u3,3 þ k54 v3,1 þ k55 v3,2 þ k56 v3,3
Conversion to the global numbering scheme for degrees of freedom is expressed by u1,1 ! unþ1 u2,1 ! unþ1 u3,1 ! unþ2
u1,2 ! un u2,2 ! un u3,2 ! un
u1,3 ! unþ3 u2,3 ! unþ2 u3,3 ! unþ3
v1,1 ! vnþ1 v2,1 ! vnþ1 v3,1 ! vnþ2
v1,2 ! vn v2,2 ! vn v3,2 ! vn
v1,3 ! vnþ3 v2,3 ! vnþ2 v3,3 ! vnþ3
Now adding the forces of the elements on the center node gives (1)
(1)
(2)
(1)
(2) (3) (2) (3) (3) þ k22 þ k22 þ k21 þ k21 þ k23 un þ k21 unþ1 þ k23 unþ2 þ k23 unþ3 fnx ¼ k22 (1)
(2) (3) (1) (2) (2) (3) (1) (3) þ k25 þ k25 þ k25 þ k24 þ k24 þ k26 vn þ k24 vnþ1 þ k26 vnþ2 þ k26 vnþ3 (1) (1) (2) (1) (2) (3) (2) (3) (3) fny ¼ k52 þ k52 þ k52 þ k51 þ k51 þ k53 un þ k51 unþ1 þ k53 unþ2 þ k53 unþ3 (1)
(2) (3) (1) (2) (2) (3) (1) (3) þ k55 þ k55 þ k55 þ k54 þ k54 þ k56 vn þ k54 vnþ1 þ k56 vnþ2 þ k56 vnþ3 The finite element equation for the three element configuration is 2 T 3 1 0 1 kn,x 0 fnx un 7 6 kT 6 nþ1,x 7B unþ1 C B fnþ1x C 7B C B C 6 T 7B C B C 6k 6 nþ2,x 7B unþ2 C B fnþ2x C 7B C B C 6 T 6 knþ3,x 7B unþ3 C B fnþ3x C C¼B C 6 T 7B 6 k 7B vn C B fny C C B C 6 n,y 7B 7B C B C 6 T 6 knþ1,y 7B vnþ1 C B fnþ1y C 7B C B C 6 7@ A @ fnþ2y A 6 kT 4 nþ2,y 5 vnþ2 fnþ3y vnþ3 kTnþ3,y
ß 2008 by Taylor & Francis Group, LLC.
The stiffness coefficients associated with the center node are as follows: 0
kn,x
(1) (2) (3) þ k22 þ k22 k22 (1)
(2) k21 þ k21 (2) (3) k23 þ k21 (1) (3) k23 þ k23
1
C B C B C B C B C B C B C B C B C, B ¼ B (1) (2) (3) C þ k þ k k B 25 25 25 C C B (1) (2) C B þ k24 k24 C B C B (2)
(3) C B k þ k A 24 26 @ (1)
(3) k26 þ k26
kn,y
0 (1)
1 (2) (3) þ k52 k52 þ k52 C B (1) (2) C B k51 þ k51 C B
C B (2) (3) C B þ k k 51 C B 53 C B (1) (3) C B k53 þ k53 C B ¼ B (1) (2) (3) C þ k þ k k B 55 55 55 C C B (1) (2) C B þ k54 k54 C B C B (2)
(3) C B k þ k A @ 56 54 (1)
(3) k56 þ k56
The remaining rows of the stiffness matrix may similarly be obtained, and the entries of the mass matrix may be obtained by a similar process.
ß 2008 by Taylor & Francis Group, LLC.
9
Solution Methods for Linear Problems: I
9.1 NUMERICAL METHODS IN FEA 9.1.1 SOLVING
THE
FINITE ELEMENT EQUATIONS: STATIC PROBLEMS
We consider numerical solution of the linear system Kg ¼ f, in which K is the positive definite and symmetric stiffness matrix. In many problems it has a large dimension, but is also banded. The matrix may be ‘‘triangularized’’ to yield the form K ¼ LLT, in which L is a lower triangular nonsingular matrix (zeroes in all entries above the diagonal). We may introduce z ¼ LTg and obtain z by solving Lz ¼ f. Next g can be computed by solving LTg ¼ z. We now see that Lz ¼ f can be conveniently solved by forward substitution. Lz ¼ f may be expanded as 2
l11
6 6 l21 6 6 l31 6 6 6 : 6 6 4 : ln1
0
:
:
:
l22
:
:
:
l32
l33
:
:
:
:
:
:
:
:
:
:
ln2
:
:
: lnn
0
30
z1
1
0
f1
1
B C B C : 7 7B z2 C B f2 C 7B C B C B C B C : 7 7B z3 C B f3 C 7B C ¼ B C B C B C : 7 7B : C B : C 7B C B C 0 5@ : A @ : A zn
(9:1)
fn
Assuming that the diagonal entries are not too small, this equation can be solved, starting from the upper left entry, using simple arithmetic: z1 ¼ f1=l11, z2 ¼ [ f2 l21z1]=l22, z3 ¼ [ f3 l31z1 l32z2]=l33, . . . . Next the equation LTg ¼ z can be solved by back substitution. The equation is expanded as 2 6 6 6 6 6 6 6 6 6 4
l11
l12
:
:
:
0
l22
:
:
:
0
0
:
:
:
:
:
: ln2,n2
ln2,n1
:
:
:
0
ln1,n1
0
ln2
:
0
0
ß 2008 by Taylor & Francis Group, LLC.
l1n
30
g1
1
0
f1
1
7B C B C 7B g 2 C B : C 7B C B C B C B C : 7 7B g 3 C B : C 7B C ¼ B C B C B C ln2,n 7 7B : C B fn2 C 7B C B C ln1,n 5@ : A @ fn1 A :
lnn
gn
fn
(9:2)
Starting from the lower right-hand entry, solution can be achieved by simple arithmetic as g n ¼ fn =lnn , gn1 ¼ ½ fn1 ln1,1 g n =ln1,n1 g n2 ¼ ½ fn2 ln2,n g n ln2,n1 g n1 =ln2,n2 , . . . In both procedures, only one unknown is encountered in each step (row).
9.1.2 MATRIX TRIANGULARIZATION
AND
SOLUTION
OF
LINEAR SYSTEMS
We next consider how to triangularize Kj. Suppose that the upper left hand ( j 1) 3 ( j 1) block Kj1 has been triangularized to furnish Kj1 ¼ Lj1LTj1 . To determine whether the j 3 j block Kj can be triangularized, we seek lj and ljj satisfying Kj ¼
Kj1 kTj
kj kjj
¼
0 ljj
Lj1 lTj
LTj1 0T
lj ljj
(9:3)
in which kj is a ( j 1) 3 1 array of the first j 1 entries of the jth column of Kj. Simple manipulation suffices to furnish kj and ljj. kj ¼ Lj1 lj qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ljj ¼ kjj lTj lj
(9:4)
Note that lj can be conveniently computed using forward substitution. Also, note qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that ljj ¼ kjj kTj K1 j1 kj . The fact that Kj > 0 implies that ljj is real. The triangularization process proceeds to the ( j þ 1)st block and from there to the complete stiffness matrix. As an illustration, consider 2
1
6 A3 ¼ 4 12 1 3
1 2 1 3 1 4
3
1 3 17 45 1 5
(9:5)
Clearly L1 ¼ LT1 ! 1. For the second block
1 l2
from which l2 ¼ 1=2 and l22 ¼
0 l22
1 l2 0 l22
¼
1 1 2
1 2 1 3
(9:6)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 1=3 ð1=2Þ2 ¼ 1= 12. And so L2 ¼
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1 1 2
0 p1ffiffiffiffi 12
(9:7)
We now proceed to the full matrix: 2
61 42 2 6 ¼4 2 6 ¼4
1 3
1
3
1 2 1 3 1 4
1
1 3 17 45 1 5
32
1 2
0 p1ffiffiffiffi
0
6 0 7 54 0
l31
l32
l33
12
l31
1 2 p1ffiffiffiffi 12
0
1 2 1 3
1 1 2
1
¼ L3 LT3 l31
3
7 l32 5
0
(9:8)
l33
3
l31
pffiffiffiffiffi l31 =2 þ l32 = 12
pffiffiffiffiffi 7 l31 =2 þ l32 = 12 5 l231 þ l232 þ l233
pffiffiffiffiffi We conclude that l31 ¼ 1=3, l32 ¼ 1= 12, l233 ¼ 1=5 1=9 1=12 ¼ 7=180. The finite element problems so far considered are direct: They involve known tractions and unknown displacements and can be solved uniquely owing to the positive definiteness of the stiffness matrix. In Chapter 10, the solution method is extended to a type of inverse problem in which tractions and displacements are both specified for some degrees of freedom on the boundary nodes, while neither is specified for other degrees of freedom.
EXAMPLE 9.1 Verify that the triangular factors L3 and LT3 for A3 in Equation 9.5 are correct.
SOLUTION From Equation 9.5 2
1
6 A3 ¼ 4 12 1 3
3
1 2 1 3 1 4
1 3 17 4 5, 1 5
1
0 p1ffiffiffiffi
2
1
61 L3 ¼ 6 42 1 3
0 p1ffiffiffiffi
12 p1ffiffiffiffi 12
3
0
7 0 7 5 1 ffi pffiffiffiffiffi 180
Now 2
61 L3 LT3 ¼ 6 42 2
1 3
1 61 ¼42 1 3
as expected.
ß 2008 by Taylor & Francis Group, LLC.
32 0 1 76 0 0 7 6 12 54 1 ffi p1ffiffiffiffi pffiffiffiffiffi 0 180 12 3 1 1 2 1 3 1 4
3 17 45 1 5
1 2 p1ffiffiffiffi 12
0
3
1 3 7 p1ffiffiffiffi 7 12 5 1 ffi pffiffiffiffiffi 180
EXAMPLE 9.2 Invoking A3 in Equation 9.5, use forward substitution followed by back substitution to solve 0 1 1 A3 g ¼ @ 1 A 1
SOLUTION Introducing LT3 g ¼ z and recalling L3 from Equation 9.5, the foregoing equation becomes 2
1
61 42 1 3
0 p1ffiffiffiffi 12 p1ffiffiffiffi 12
3 0 0 z1 1 0 1 1 0 7 5@ z2 A ¼ @ 1 A 1 1 z3 180
Using forward substitution z1 ¼ 1,
z2 ¼
pffiffiffiffiffi pffiffiffi 12 1 12 z1 ¼ 3,
z3 ¼
pffiffiffi pffiffiffiffiffiffiffiffi ffi z2 ¼ 5 180 1 13 z1 p1ffiffiffi 12
Hence 0
1 1 pffiffiffi z ¼ @ 3A pffiffiffi 5 Next 2
1
1 2 p1ffiffiffiffi 12
0
0
60 4
1 30 g1 3 B p1ffiffiffiffi 7 g2 5 @ 12 1 g3 180
1
0
1 1 p ffiffi ffi C B C A ¼ @ 3A pffiffiffi 5
and using back substitution g3 ¼
pffiffiffiffiffiffiffiffipffiffiffi 180 5 ¼ 30,
g2 ¼
pffiffiffiffiffipffiffiffi ffi g ¼ 24, 12 3 p1ffiffiffi 12 3
We conclude that 0
1 30 g ¼ @ 24 A 3
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g 1 ¼ 1 12 g2 13 g 3 ¼ 3
EXAMPLE 9.3 Triangularize the matrix 2
3 30 18 41 23 5 23 14
36 K ¼ 4 30 18
SOLUTION Triangularizing the upper left-hand 2 3 2 block K2 in the form K2 ¼ L2LT2 gives
6 36 30 ¼ l2 30 41
0 l22
6 l2 0 l22
from which l2 ¼ 5,
l22 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 41 l22 ¼ 4
Consequently, the 2 3 2 matrix K2 triangularizes to
36 30
30 6 0 6 5 ¼ 41 5 4 0 4
Now extend the procedure to the 3 3 3 matrix K: 3 2 36 30 18 6 4 30 41 23 5 ¼ 4 5 18 23 14 l31 2
0 4 l32
32 3 0 6 5 l31 0 54 0 4 l32 5 l33 0 0 l33
After simple manipulation we obtain l31 ¼ 18=6 ¼ 3,
l32 ¼ 14 ð23 5l31 Þ ¼ 2,
l33 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 14 l231 l232 ¼ 1
Accordingly, the triangular factor of K is 2
3 6 0 0 L ¼ 45 4 05 3 2 1
EXAMPLE 9.4 For the linear system 30 1 0 1 1 36 30 24 g1 4 30 41 32 5@ g2 A ¼ @ 2 A 3 24 32 27 g3 2
triangularize the matrix and solve for g1, g2, g3.
ß 2008 by Taylor & Francis Group, LLC.
SOLUTION Triangularizing the upper left-hand 2 3 2 block A2 in the form A2 ¼ L2LT2 gives
36 30
6 30 ¼ l2 41
0 l22
6 0
l2 l22
from which l2 ¼ 5, l22 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 41 l22 ¼ 4
Accordingly A2 triangularizes as follows:
6 36 30 ¼ 5 30 41
0 4
6 5 0 4
For the 3 3 3 matrix A, 3 2 36 30 24 6 4 30 41 32 5 ¼ 4 5 24 32 27 l31 2
0 4 l32
32 3 6 5 l31 0 0 54 0 4 l32 5 l33 0 0 l33
from which l31 ¼ 24=6 ¼ 4,
l32 ¼ 14 ð32 5l31 Þ ¼ 3, l33 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 27 l231 l232 ¼ 2
The triangular factor of L of A is now 2
6 L ¼ 45 4
3 0 0 4 p0ffiffiffi 5 3 2
On introducing LTg ¼ z, we encounter 2
6 45 4
30 1 0 1 z1 0 0 1 4 p0ffiffiffi 5@ z2 A ¼ @ 2 A 3 2 z3 3
Forward substitution results in z1 ¼ 1=6,
z2 ¼ 1=4ð2 5z1 Þ ¼ 7=24,
pffiffiffi pffiffiffi z3 ¼ 1= 2ð3 4z1 3z2 Þ ¼ 35 2=48
and also 0 B z ¼@
ß 2008 by Taylor & Francis Group, LLC.
1=6
1
C 7=24 A pffiffiffi 35 2=48
Next 2
30 1 0 1 g1 6 5 4 1=6 4 0 4 3 5@ g2 A ¼ @ 7=24 A pffiffiffi pffiffiffi 0 0 2 g3 35 2=48 and using backward substitution gives pffiffiffi pffiffiffi g 3 ¼ 35 2=48 1= 2 ¼ 35=48,
g 2 ¼ 1=4ð7=24 3g3 Þ ¼ 91=192,
g1 ¼ 1=6ð1=6 5g 2 4g 3 Þ ¼ 73=1152 We conclude that 0
1 73=1152 g ¼ @ 91=192 A 35=48
9.1.3 TRIANGULARIZATION
OF
ASYMMETRIC MATRICES
The foregoing triangularization is applicable to positive definite symmetric matrices. Asymmetric matrices arise in a number of finite element problems, including problems with incompressibility, unsteady rotation, or thermomechanical coupling. If the matrix is nonsingular, it may be decomposed into the product of a lower triangular and an upper triangular matrix, followed by forward and back substitution. K ¼ LU
(9:9)
(It is also possible to triangularize a singular matrix, but L will then be singular, preventing the use of forward substitution.) Assuming the ( j 1)st diagonal block has been triangularized, we consider whether the jth block admits the decomposition " Kj ¼ " ¼
Kj1
k1j
kT2j
kjj
Lj1 Uj1 lTj Uj1
#
" ¼
Lj1
0
lTj
ljj #
Lj1 uj T lj uj þ ujj ljj
#
Uj1
uj
0T
ujj
(9:10)
Now uj is obtained by forward substitution using Lj1uj ¼ k1j, and lj is obtained by back substitution using UTj1 lj ¼ k2j. Finally, ujjljj ¼ kjj lTj uj, for which purpose ujj may be arbitrarily set to unity. After the triangularization process is completed, an equation of the form Ku ¼ f can now be solved by forward substitution applied to Lz ¼ f, followed by back substitution applied to Uu ¼ z.
ß 2008 by Taylor & Francis Group, LLC.
EXAMPLE 9.5 Triangularize the asymmetric matrix 2
2mA=L 0 4 0 4mAL=Y 2 A 2mA=L
3 A 2mA=L 5 0
(This matrix will be seen later in Chapter 11 concerning incompressible materials.)
SOLUTION Triangularizing the upper left-hand 2 3 2 block B2 in the form B2 ¼ L2U2 gives
2mA=L 0 0 4mAL=Y 2
¼
l11 l21
0 l22
u11 0
u12 u22
On setting u11 and u22 as unity, l11 ¼ 2mA=L, l21 ¼ 0,
u12 ¼ 0,
l22 ¼ 4mAL=Y 2
Now consider the 3 3 3 matrix B: 2
2mA=L 4 0 A
0 4mAL=Y 2 2mA=L
3 2 2mA=L A 2mA=L 5 ¼ 4 0 0 l31
0 4mAL=Y 2 l32
32 0 1 0 0 54 0 1 l33 0 0
3 u13 u23 5 u33
On setting u33 ¼ 1, simple manipulation furnishes l31 ¼ A,
l32 ¼ 2mA=L,
u13 ¼ L=2m, u23 ¼ Y 2 =2L2 , l33 ¼ AL=2m þ mAY 2 =L3
Finally 2
2mA=L B ¼ LU ¼ 4 0 A
0 4mAL=Y 2 2mA=L
32 3 1 0 L=2m 0 54 0 1 Y 2 =2L2 5 0 ð AL=2m þ mAY 2 =L3 Þ 0 0 1
The decomposition is not unique since it is based on setting the diagonal entries of U to unity.
9.2 TIME INTEGRATION: STABILITY AND ACCURACY Much insight may be gained from considering the model equation dy ¼ ly dt
ß 2008 by Taylor & Francis Group, LLC.
(9:11)
in which l is complex. If Re(l) > 0, for the initial value y(0) ¼ y0 the solution is y(t) ¼ y0 exp(lt), and clearly y(t) ! 0. In this event the system is called asymptotically stable. We now ask whether numerical integration schemes to integrate Equation 9.11 have stability properties corresponding to asymptotic stability. In other words, does the numerical solution decay when the exact solution decays, and diverge when the exact solution diverges? For this purpose we consider the trapezoidal rule, the properties of which will be discussed in Section 9.3. Consider time steps of duration h, and suppose that the solution has been calculated through the nth time step. We seek to compute the solution at the (n þ 1)st time step. The trapezoidal rule (see Section 9.3) is given by d y y n þ1 y n , h dt
ly l2 ½ynþ1 þ yn
(9:12)
Consequently, 1 lh=2 yn 1 þ lh=2 1 lh=2 n ¼ y0 (9:13) 1 þ lh=2 lh=2 1 lh=2 Clearly, ynþ1 ! 0 if 11 þ lh=2 < 1, and ynþ1 ! 1 if 1 þ lh=2 > 1, in which jj implies the magnitude. If the first inequality is satisfied the numerical method is called A-stable (Dahlquist and Bjork, 1974). We next write l ¼ lr þ ili, and now A-stability requires that yn þ 1 ¼
2 lr h 2 li h 1 þ 2 2
2 2 < 1 lr h li h 1þ þ 2 2
(9:14)
A-stability obtains if lr > 0, which is precisely the condition for asymptotic stability. Next consider the matrix–vector system arising in the finite element method. € þ Dg_ þ Kg ¼ 0, gð0Þ ¼ g0 , Mg
g_ ð0Þ ¼ g_ 0
(9:15)
in which M, D, and K are positive definite. Elementary manipulation serves to establish that d 1 T 1 T g_ Mg_ þ g Kg ¼ g_ T Dg_ < 0 dt 2 2
(9:16)
It follows that g_ ! 0 and g ! 0. We conclude that the system is asymptotically stable.
ß 2008 by Taylor & Francis Group, LLC.
_ the n-dimensional second-order system is written Introducing the vector p ¼ g, in state form as the (2n)-dimensional first-order system of ordinary differential equations:
M 0 0 I
. D p þ I g
K 0
p f ¼ g 0
(9:17)
We next apply the trapezoidal rule to the system:
M 0
0 I
1 h pnþ1 1 h gnþ1
! pn D K þ I 0 gn
1 2 pnþ1 1 2 gnþ1
!
1 þ pn ¼ 2 ðf nþ1 þ f n Þ 0 þ gn (9:18)
From the equation in the lower row, pnþ1 ¼ 2h gnþ1 gn pn . Eliminating pnþ1 in the upper row furnishes a formula underlying the classical Newmark method: h i 2 KD gnþ1 ¼ rnþ1 , KD ¼ M þ h2 D þ h4 K h i
2 2 rnþ1 ¼ M þ h2 D h4 K yn þ M þ h2 D h2 pn þ h4 ðf nþ1 þ f n Þ
(9:19)
and KD may be called the dynamic stiffness matrix. Equation 9.19 may be solved by tiangularization of KD, followed by forward and backward substitution.
9.3 PROPERTIES OF THE TRAPEZOIDAL RULE We consider the accuracy of the trapezoidal rule, and by extension of Newmark’s method. To fix important notions consider the model equation dy ¼ f ð yÞ dx
(9:20)
Suppose this equation is approximated as aynþ1 þ byn þ h½gfnþ1 þ dfn ¼ 0
(9:21)
We now use the Taylor series to express ynþ1 and fnþ1 in terms of yn and fn. Noting that y0n ¼ fn and y00n ¼ fn0 , we obtain
0 ¼ a yn þ y0n h þ y00n h2 =2
þ byn þ hg y0n þ y00n h þ hdy0n
ß 2008 by Taylor & Francis Group, LLC.
(9:22)
For exact agreement through h2, the coefficients must satisfy a þ b ¼ 0,
a þ g þ d ¼ 0,
a=2 þ g ¼ 0
(9:23)
We also introduce the convenient normalization g þ d ¼ 1. Simple manipulation serves to derive that a ¼ 1, b ¼ 1, g ¼ 1=2, d ¼ 1=2, furnishing ynþ1 yn 1 ¼ ½ f ðynþ1 Þ þ f ðyn Þ h 2
(9:24)
which may be recognized as the trapezoidal rule. In fact the trapezoidal rule is unique and optimal in having the following three characteristics: (a) It is a one-step method, using only the values at the beginning of the current time step. (b) It is second-order accurate—it agrees exactly with the Taylor series through h2. (c) Applied to dy=dt þ ly ¼ 0 with initial condition y(0) ¼ y0, it is A-stable, i.e., numerically stable whenever system described by the equation is asymptotically stable.
EXAMPLE 9.6 For the model equation dy=dx ¼ f (y), develop a two-step numerical quadrature formula: ðaynþ1 þ byn þ gyn1 Þ þ h½ df ðynþ1 Þ þ «f ðyn Þ þ zf ðyn1 Þ ¼ 0 What is the order of the integration method (highest power in h with exact agreement with the Taylor series)?
SOLUTION Expressing ynþ1, yn1, f(ynþ1), and f(yn1) using the Taylor expansion gives ynþ1 ¼ yn þ hy0n þ
h2 00 h3 000 h4 iv y þ y þ yn þ 2! n 3! n 4!
yn1 ¼ yn hy0n þ
h2 00 h3 000 h4 iv y y þ yn 2! n 3! n 4!
f ðynþ1 Þ ¼ fn þ hfn0 þ
h2 00 h3 000 f þ fn þ 2! n 3!
f ðyn1 Þ ¼ fn hfn0 þ
h2 00 h3 000 f fn þ 2! n 3!
ß 2008 by Taylor & Francis Group, LLC.
But fn ¼ y0n also implies h2 000 h3 iv y þ yn þ 2! n 3! 2 3 h h f ðyn1 Þ ¼ y0n hy00n þ y000 yiv þ n 2! 3! n f ðynþ1 Þ ¼ y0n þ hy00n þ
On substitution into the quadrature formula, we have h2 00 h3 000 h4 iv 0 a yn þ hyn þ yn þ yn þ yn þ byn 2! 3! 4! 2 3 h h h4 iv y þ g yn hy0n þ y00n y000 þ 2! 3! n 4! n 3 2 2 3 h 000 h iv 0 00 0 þ hy þ þ d y y y þ «y 6 n n n7 2! n 3! n 7 6 þ h6 7 ¼ 0 þ 0 h5 2 3 5 4 h h þz y0n hy00n þ y000 yivn 2! n 3! For exact agreement through h4, the coefficients must satisfy a þ b þ g ¼ 0, a g þ d þ « þ z ¼ 0, a=2 þ g=2 þ d z ¼ 0 a=6 g=6 þ d=2 þ z=2 ¼ 0, a=24 þ g=24 þ d=6 z=6 ¼ 0 We now introduce the convenient normalization d þ « þ z ¼ 1. Simple manipulation serves to derive a ¼ 1=2,
b ¼ 0, g ¼ 1=2,
d ¼ z ¼ 1=6, « ¼ 2=3
The quadrature formula is now stated as ½1=2ynþ1 þ 0 þ 1=2yn1 þ h½1=6f ðynþ1 Þ þ 2=3f ðyn Þ þ 1=6f ðyn1 Þ ¼ 0 On rearranging ynþ1 yn1 1 ¼ 3 ½ f ðyn1 Þ þ 4f ðyn Þ þ f ðyn1 Þ h which is a the two-step numerical integration model. Since the relations are exact through h4, this is a fourth-order accurate and its order of integration is four. However, it is not A-stable, as discussed in Chapter 10.
EXAMPLE 9.7 In the damped linear mechanical system M€ g þ Dg_ þ Kg ¼ f ðt Þ
ß 2008 by Taylor & Francis Group, LLC.
suppose that g(t) ¼ gn at the nth time step. Derive KD and rnþ1 such that g at the (n þ 1) st time step satisfies KD gnþ1 ¼ rnþ1
SOLUTION Introducing the state form relation p ¼ g_ in the equation, we have
M 0
0 I
. f p p D K ¼ þ 0 g g I 0
This expression may be rewritten using the trapezoidal rule as
M
0
0
I
1 h pnþ1 1 gnþ1 h
pn gn
!
þ
D
K
I
0
1 2 pnþ1 1 gnþ1 2
þ pn
!
þ gn
¼
1 2 ðf nþ1
þ f nÞ
!
0
Now the lower row implies pnþ1 ¼ 2h gnþ1 gn pn Now consider the upper row:
1 hM
pnþ1 pn þ 12 D pnþ1 þ pn þ 12 K gnþ1 þ gn ¼ 12 ðf nþ1 þ f n Þ
Eliminating pnþ1 results in
1 2 hM h
gnþ1 gn 2pn þ 12 D 2h gnþ1 gn þ 12 K gnþ1 þ gn ¼ 12 ðf nþ1 þ f n Þ
On multiplying throughout by h2=2 and rearranging, h h2 h h2 h2 M þ D þ K gnþ1 ¼ M þ D K gn þ hMpn þ ðf nþ1 þ f n Þ 2 2 4 4 4 Comparing the above equation with the given equation KDgnþ1 ¼ rnþ1, it follows that h h2 KD ¼ M þ D þ K 4 2 h h2 h2 rnþ1 ¼ M þ D K gn þ hMpn þ ðf nþ1 þ f n Þ 2 4 4 A higher-order time integration scheme is presented in Chapter 10.
ß 2008 by Taylor & Francis Group, LLC.
9.4 INTEGRAL EVALUATION BY GAUSSIAN QUADRATURE There are many integrations in the finite element method, the accuracy and efficiency of which is critical. Fortunately, a method which is optimal in an important sense, called Gaussian quadrature, has long been known (it was previously introduced in Chapter 7). It is basedÐ on converting physical coordinates to intrinsic b ‘‘natural’’ coordinates. Consider a f ð xÞ dx. Let j ¼ b 1 a ½2x ða þ bÞ. Clearly, j mapsÐ the interval [a,b] into the interval [1,1]. The integral now becomes 1 1 b a 1 f ðjÞ dj . Now consider the power series f ðjÞ ¼ a0 þ a1 j þ a2 j2 þ a3 j3 þ a4 j4 þ a5 j5 þ
(9:25)
from which ð1 1
f ðjÞ dj ¼ 2a1 þ 0 þ 23 a3 þ 0 þ 25 a5 þ 0 þ
(9:26)
The advantages for integration on a symmetric interval, evident in the zeroes in Equation 9.26, can be seen in the fact that, with n function evaluations, an integral can be evaluated exactly through (2n 1)st order in the Taylor series. Consider the first 2n 1 terms in a power series representation for a function: gðjÞ ¼ a1 þ a2 j þ þ a2n j 2n1
(9:27)
Assume that n integration (Gauss) points ji and n weights are used as follows: ð1 1
gðjÞ dj ¼
n X
gðji Þwi ¼ a1
n X
wi þ a 2
n X
i ¼1
i¼1
wi ji þ þ a2n
n X i ¼1
i¼1
1 wi j2n i
(9:28) Comparison with Equation 9.26 implies that n X
n X
wi ¼ 2,
i ¼1
i¼1 n X
wi j2i n2 ¼
i¼1
2 , 2n 1
wi ji ¼ 0,
n X
wi j2i ¼ 2=3, . . .
i¼1 n X i ¼1
wi j2i n1 ¼ 0
(9:29)
It is necessary to solve for n integration points ji and n weights wi. These are universal quantities. Thereafter, to integrate a given function g(j) exactly through j2n1, it is necessary to perform n function evaluations to compute g(ji). As an example, we seek two Gauss points and two weights (n ¼ 2). After simple manipulation w1 þ w2 ¼ 2 w1 j21
þ
ß 2008 by Taylor & Francis Group, LLC.
w2 j22
¼
2 3
(a),
w1 j 1 þ w 2 j 2 ¼ 0
(c),
w1 j31
þ
w2 j32
(b)
¼ 0 (d)
(9:30)
From Equations 9.30b and 9.30d, w1 j1 j 21 j22 ¼ 0, leading topjffiffi2ffi ¼ j1. From Equations 9.30a and 9.30c it now follows that j2 ¼ j1 ¼ 1= 3. Finally, the normalization w1 ¼ 1 implies that w2 ¼ 1.
EXAMPLE 9.8 Find the integration (Gauss) points and weights for n ¼ 3.
SOLUTION Owing to complexity the steps in the solution are given in detail. On substituting n ¼ 3 in Equation 9.29 w1 þ w2 þ w3 ¼ 2
(9:31)
w1 j1 þ w2 j2 þ w3 j3 ¼ 0
(9:32)
w1 j21 þ w2 j22 þ w3 j23 ¼ 2=3
(9:33)
w1 j31 þ w2 j32 þ w3 j33 ¼ 0
(9:34)
w1 j41 þ w2 j42 þ w3 j43 ¼ 2=5
(9:35)
w1 j51 þ w2 j52 þ w3 j53 ¼ 0
(9:36)
Multiplying Equations 9.32 and 9.34 by j22 furnishes w1 j1 j22 þ w2 j32 þ w3 j22 j3 ¼ 0
(9:37)
w1 j31 j22 þ w2 j52 þ w3 j22 j33 ¼ 0
(9:38)
Now, on subtracting Equation 9.34 from Equation 9.37, and Equation 9.36 from Equation 9.38, w1 j1 j22 j21 þ w3 j3 j22 j23 ¼ 0 w1 j31 j22 j21 þ w3 j33 j22 j23 ¼ 0
(9:39) (9:40)
On multiplying Equation 9.39 by j22 , w1 j31 j22 j21 þ w3 j21 j3 j22 j23 ¼ 0
(9:41)
Equation 9.40 is subtracted from Equation 9.41 to furnish w3 j3 j21 j23 j22 j23 ¼ 0
(9:42)
Assuming that the integration points are equally spaced, Equation 9.42 implies j1 ¼ j3
ß 2008 by Taylor & Francis Group, LLC.
(9:43)
On substituting Equation 9.43 in Equation 9.39, we conclude that w1 ¼ w3
(9:44)
On substituting Equations 9.43 and 9.44 in Equation 9.32, we learn that w2 j2 ¼ 0 and also j2 ¼ 0
(9:45)
Next substituting Equations 9.43 through 9.45 in Equation 9.33 and 9.35 leads to w1 j21 ¼ 1=3
(9:46)
w1 j41 ¼ 1=5
(9:47)
Equations 9.43, 9.46, and 9.47 serve to derive that pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi j1 ¼ 3=5, j3 ¼ 3=5 Simple manipulations furnish the remaining unknowns as w1 ¼ 5=9 ¼ w3 ,
w2 ¼ 8=9
The results are summarized as w1 ¼ 5=9, w2 ¼ 8=9, pffiffiffiffiffiffiffiffi j1 ¼ 3=5, j2 ¼ 0,
w3 ¼ 5=9 pffiffiffiffiffiffiffiffi j3 ¼ 3=5
9.5 MODAL ANALYSIS BY FEA 9.5.1 MODAL DECOMPOSITION In the absence of damping the finite element equation for a linear mechanical system, which is unforced but has nonzero initial values, is described by M€ g þ Kg ¼ 0,
gð0Þ ¼ g0 , g_ ð0Þ ¼ g_ 0
(9:48)
^ exp(lt): it furnishes upon substitution Assume a solution of the form g ¼ g
^¼0 K þ l2 M g
(9:49)
The jth eigenvalue lj is obtained by solving det(K þ l2j M) ¼ 0, and a corresponding eigenvector vector gj may likewise be computed (see example below). For the sake
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of generality suppose that lj and gj are complex. Let gjH denote the complex gH Kg
conjugate (Hermitian) transpose of gj. Now l2j satisfies l2j ¼ gHj Mgj . Since M and j
j
K are real and positive definite, it follows that lj is pure imaginary: lj ¼ ivj. Also, without loss of generality, we may take gj to be real and orthogonal with respect to both M and K.
EXAMPLE 9.9 We seek the modes of the system governed by the equation
2 0 0 1
g1 g2
.. þ
k 2 1 0 g1 ¼ g2 0 m 1 1
pffiffiffi 2 2 2 Let z2 ¼ vj2=v p0ffiffiffi , v0 ¼ k=m. For the determinant to vanish, 1 z ¼ 1= 2. Using 2 1 zþ ¼ 1= 2, the first eigenvector satisfies ! pffiffiffi ð1Þ h i2 h i2 0 g1 2 1 ð1Þ ð1Þ pffiffiffi , g1 þ g2 ¼ 1 ¼ ð1Þ 0 1 1= 2 g2 pffiffiffi ð1Þ pffiffiffi pffiffiffi ð1Þ implying that g1 ¼ 1= 3, g2 ¼ 2=pffiffiffi3. The corresponding procedures for the pffiffiffi pffiffiffi ð2Þ ð2Þ second eigenvalue furnish that g1 ¼ 1= 3, g 2 ¼ 2= 3. It is readily verified that gð2ÞT Mgð1Þ ¼ gð1ÞT Mgð2Þ ¼ 0, gð2ÞT Kgð1Þ ¼ gð1ÞT Kgð2Þ ¼ 0,
m1 ¼ 4=3, m2 ¼ 4=3 pffiffiffi pffiffiffi k1 ¼ 43 1 1= 2 , k2 ¼ 43 1 þ 1= 2
Returning to the general development, the modal matrix X is now defined as X ¼ ½ g1
g2
g3
gn
(9:50)
Since the jkth entry of XTMX and XTKX are gjTMgk and gjTKgk, respectively, it follows that 2
m1 6 0 6 XT MX ¼ 6 : 4 : :
0 m2 : : :
: : : : : : : : : :
3 : : 7 7 : 7, : 5 mn
2
k1 6 0 6 XT KX ¼ 6 : 4 : :
0 k2 : : :
: : : : :
3 : : : : 7 7 : : 7 : : 5 : kn
(9:51)
The modal matrix is said to be orthogonal with respect to both M and K, but it is not simply orthogonal since X1 6¼ XT. The governing equation is now rewritten as XT MX€ j þ XT KXj ¼ g,
j ¼ X1 g,
g ¼ XT f
(9:52)
implying the (uncoupled) modes mj j€j þ kj jj ¼ gj ðt Þ
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(9:53)
Suppose that gj(t) ¼ gj0(t) sin vt. Neglecting transients, the steady state solution for the jth mode is jj ¼
gj0 sinðvt Þ k j v2 mj
(9:54)
It is evident that, if v2 e vj2 ¼ kj=mj (resonance), the response amplitude for the jth mode is much greater than for the other modes, so that the structural motion under this excitation frequency illustrates the mode. For this reason the modes can easily be animated.
EXAMPLE 9.10 For the system shown below (Figure 9.1), (a) (b) (c) (d) (e) (f) (g) (h)
Find the eigenvalues 2 2 M and of K vn2 M are linearly dependent Verify that the rows of K vn1 Find the eigenvectors Verify their orthogonality with respect to the mass and stiffness matrices Find the modal matrix Find the modal stiffnesses and modal dampers Find the modal masses Verify that the natural frequencies in the modal equations are the same as the system eigenvalues (i) Find the modal coordinates (j) Find the modal forces (k) Find the solutions for the two modes under the conditions _ f(t) ¼ 0, x(0) ¼ 0, x(0) ¼ x_ 0 (l) Transform the modal solutions back to the physical coordinates
SOLUTION The governing equation is m
3 0 0 2
€x1 €x2
x_ 1 x1 3 2 3 2 F1 þc þk ¼ x2 F2 2 2 2 2 x_ 2
c k
2c 3m
x1
2k F1
FIGURE 9.1 Two-degree-of-freedom vibrating system.
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2m
x2
F2
The natural frequencies are obtained from: v2 3 0 3 2 2 det ¼ 0, v20 ¼ k=m 2 2 v0 0 2 pffiffiffiffiffiffiffiffi resulting in the eigenvalues v2n1,2 ¼ 1 2=3 v20 . pffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffi
If the first eigenvalue is used, the two rows become f3 2=3 2g and f2 2 2=3g. pffiffiffiffiffiffiffiffi The second row is 3=2 first row, so that the two rows are linearly dependent. pffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffi
The rows corresponding to the second eigenvalue are f3 2=3 2g and f2 2 2=3g. pffiffiffiffiffiffiffiffi The second row is 3=2 first row. We now seek the eigenvectors. Using the first eigenvalue gives qffiffi pffiffiffiffiffiffiffiffi (1) (1) 3 (1) 3 1 1 þ 2=3 x(1) 2x ¼ 0 ! x ¼ 1 2 2 2x1
Similarly, x(2)
2 (1) 2 1 1 ¼ x(1) þ x2 ! x(1) ¼ pffiffiffi 1 5 pffiffiffi 1 2 ¼ pffiffiffi pffiffiffi . 3 5
pffiffiffi p2ffiffiffi 3
The eigenvectors are orthogonal with respect to M and K since pffiffiffi 1 npffiffiffipffiffiffio 3 0 1 p2ffiffiffi ¼ (6 6) ¼ 0 2 3 0 2 3 5 5 p ffiffi ffi pffiffiffi pffiffiffi 1 npffiffiffipffiffiffio 3 2 1 p2ffiffiffi ¼ (6 2 6 þ 2 6 6) ¼ 0: 2 3 2 2 3 5 5 pffiffiffi pffiffiffi 1 2 p2ffiffiffi . The modal masses satisfy The modal matrix is X ¼ pffiffiffi pffiffiffi 3 3 5 ( pffiffiffi ) 1 pffiffiffi pffiffiffi 3 0 12 2 m1 ¼ m pffiffiffi ¼ m 2 3 5 5 0 2 3 ( pffiffiffi ) pffiffiffi 3 0 1 pffiffiffi 12 2 m2 ¼ m pffiffiffi ¼ m 2 3 5 5 0 2 3 pffiffi pffiffi Similarly, the modal stiffnesses are k1 ¼ p12ffiffi54 6 k and k2 ¼ 12 þ54 6 k, and the modal pffiffi dampers are d1 ¼ 12 54 6 c and d2 ¼ 12 þ54 6 c. Finally, the modal damping factors are 1 ffi 2 ffi pdffiffiffiffiffiffiffi z1,2 ¼ 2pdffiffiffiffiffiffiffi k1 m , 2 k2 m . 1
2
The Modal Equations imply the natural frequencies pffiffiffi 12 4 6 pffiffiffiffiffiffiffiffi k1 k k 5 1 2=3 ¼ v2n2 , ¼ ¼ 12 m m1 m 5
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pffiffiffi 12 þ 4 6 pffiffiffiffiffiffiffiffi k2 k k 5 ¼ ¼ 1 þ 2=3 ¼ v2n1 12 m2 m m 5 in exact agreement with the relations obtained from the original (coupled) system equation. pffiffi h pffiffiffi pffiffiffi i Since X1 ¼ 2p5ffiffi6 pffiffi33ffi p2ffiffi2ffi , the modal coordinates satisfy
y1 y2
¼ X1
x1 x2
pffiffiffi pffiffiffi pffiffiffi 5 3x þ 2x ¼ pffiffiffi pffiffiffi 1 pffiffiffi 2 3x1 2x2 2 6
The modal forces are obtained as
g1 g2
¼ XT
f1 f2
1 ¼ pffiffiffi 5
pffiffiffi pffiffiffi p2ffiffiffif1 þ p3 ffiffiffif2 2f1 3f2
We now seek the solutions of the modal equations. Initial conditions in terms of modal coordinates are given by pffiffiffi pffiffiffi pffiffiffi 5 3x_ (0) þ pffiffi2ffix_ 2 (0) ¼ pffiffiffi pffiffiffi 1 3x_ 1 (0) 2x_ 2 (0) 2 6 n o n o qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and the two modal forces vanish: gg1 ¼ 00 . Introducing vd1,2 ¼ vn1,2 1 j21,2 , the
y1 (0) y2 (0)
0 ¼ 0
y_ 1 (0) y_ 2 (0)
2
foregoing reduces to two initial single-degree-of-freedom initial value problems with solutions y_ 1 (0) sinðvd1 t Þ vd1 y_ 2 (0) sinðvd2 t Þ y2 (t) ¼ exp(z2 vn2 t) vd2 y1 (t) ¼ exp(z1 vn1 t)
The solutions in physical coordinates are recovered using
x1 x2
pffiffiffi 1 y 2[y (t) þ y2 (t)] ¼ X 1 ¼ pffiffiffi pffiffiffi 1 y2 3[y1 (t) y2 (t)] 5
EXAMPLE 9.11 Figure 9.2 shows a clamped–clamped beam, modeled as two finite elements. Show that the finite element equations decompose into two uncoupled modes: One representing symmetric response and the other representing antisymmetric response. L/2
L/2
E, I, A, r
FIGURE 9.2 Clamped–clamped beam modeled as two elements.
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SOLUTION The mass and stiffness matrices of a beam element with no constraints are given by
(e) M(e) K11 K(e) M(e) 11 12 , K(e) ¼ 12 MT(e) M(e) KT(e) K(e) 12 22 12 22 rALe 156 22Le rALe rALe 156 22Le 54 13Le (e) (e) ¼ ¼ ¼ , M , M M(e) 11 12 22 420 22Le 4L2e 420 13Le 3L2e 420 22Le 4L2e EI EI 12 6Le EI 12 6Le 12 6Le (e) (e) , K , K K(e) ¼ ¼ ¼ 11 12 22 L3e 6Le 4L2e L3e 6Le 2L2e L3e 6Le 4L2e M(e) ¼
Using Le ¼ L=2, assembling the stiffness and mass matrices, and imposing the clamped constraints results in
192EI=L3 0
0 16EI=L
w2 w02
þ
13rAL=35 0
0 rAL3 =420
€2 w € w02
¼
0 0
in which w2 is the transverse displacement of the mid-node. This equation represents two separate single-degree-of-freedom systems, with the natural frequencies
vn1
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisffiffiffiffirffiffiffi 35 192 E I 1 ¼ , 13 r A L2
vn2
sffiffiffiffirffiffiffi E I 1 ¼ 16 420 r A L2
To consider the question of symmetry and antisymmetry, consider the case in which the right-hand side satisfies M ¼ 0, but V ¼
0, V0 ,
t<0 t0
. This is to say that there is a shear
force imposed at t ¼ 0 at the midpoint but no bending moment. If initially (w2(L=2,0) ¼ w_ 2(L=2,0) ¼ 0, w20 (L=2,0) ¼ w_ 20 (L=2,0) ¼ 0), then w0 (L=2,t) ¼ 0 and the deformation is symmetric with natural frequency vn1. On the other hand, if V ¼ 0, n 0 but V ¼ M0, , tt < , then w2(L=2, t) ¼ 0 and the deformation is antisymmetric with 0 0
natural frequency vn2.
EXAMPLE 9.12 As an example of eigenvalue determination, consider M¼
k 1 0 , K ¼ 11 0 1 k12
k12 k22
Now det [K þ l2I] ¼ 0 reduces to 2 2
2 l þ½k11 þ k22 l2 þ k11 k22 þ k12 ¼0
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with the roots l2þ,
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2 2 ¼ ½k11 þ k22 ½k11 þ k22 4 k11 k22 þ k12 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ¼ ½k11 þ k22 ½k11 k22 2 2
so that both l2þ and l2 are negative (since k11 and k22 are positive). Returning to the general development, we now consider eigenvectors. The eigenvalue equations for the ith and jth eigenvectors are written as h i
K þ v2j M gj ¼ 0, K þ v2k M gk ¼ 0
(9:55)
It is easily seen that the eigenvectors have arbitrary magnitudes, and for convenience we assume that they have unit magnitude: gTj gj ¼ 1. Simple manipulation furnishes that h i gTk Kgj gTj Kgk v2j gTk Mgj v2k gTj Mgk ¼ 0
(9:56)
Symmetry of K and M imply that gTk Kgj gTj Kgk ¼ 0,
h
i v2j gTk Mgj v2k gTj Mgk ¼ v2j v2k gTk Mgj ¼ 0 (9:57)
Assuming for convenience that the eigenvalues are all distinct, it follows that gTj Mgk ¼ 0,
gTj Kgk ¼ 0,
j 6¼ k
(9:58)
The eigenvectors are thus said to be orthogonal with respect to M and K. The quantities mj ¼ gTj Mgj and kj ¼ gTj Kgj are called the ( jth) modal mass and ( jth) modal stiffness.
EXAMPLE 9.13 (a) Find the modal masses m1 and m2 and the modal stiffnesses k1 and k2 of the system 3
1 0 0 2
g€1 g2
þ 27
1 1 1 2
g1 g2
¼
10 20
sinð10tÞ
(b) Determine the steady state response of the system (i.e., particular solution to the equation).
SOLUTION Consider the homogeneous equation
M
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g€1 g€2
þK
g1 g2
¼0
1 0 M¼3 0 2
g€1 g1 1 1 , K ¼ 27 g€2 g2 1 2
pffiffiffi pffiffiffi For det[K z2M] to vanish 9 z2 ¼ 9= 2. For 9 z2þ ¼ þ9= 2, the eigenvector is obtained from
! pffiffiffi ð1Þ h i2 h i2 0 g1 ð1Þ ð1Þ 27= 2 27 pffiffiffi , g1 þ g2 ¼ 1 ¼ ð1Þ 0 27 54= 2 g2
Simple manipulation furnishes ð1Þ
g1 ¼
pffiffiffiffiffiffiffiffi 2=3
ð1Þ
and g2 ¼
pffiffiffiffiffiffiffiffi 1=3
pffiffiffi The corresponding procedure for 9 z2 ¼ 9= 2 furnishes ð2Þ
g1 ¼
pffiffiffiffiffiffiffiffi 2=3
ð2Þ
and g2 ¼
pffiffiffiffiffiffiffiffi 1=3
The modal matrix X is given by
X¼ g
ð1Þ
ð2Þ
g
pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 2=3 p2=3 ffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffi 1=3 1=3
Hence, the modal masses are given by pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 2=3 p1=3 ffiffiffiffiffiffiffiffi 3 1 0 p2=3 ffiffiffiffiffiffiffiffi p2=3 ffiffiffiffiffiffiffiffi ¼ 4 0 XT MX ¼ pffiffiffiffiffiffiffiffi 0 2 0 4 2=3 1=3 1=3 1=3 so that m1 ¼ m2 ¼ 4. The modal stiffnesses are obtained using pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 1 1 p2=3 2=3 1=3 ffiffiffiffiffiffiffiffi p2=3 ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi ffi p ffiffiffiffiffiffiffi ffi X KX ¼ 27 1 2 1=3 1=3 2=3 1=3 pffiffiffi 36 18 2 0 pffiffiffi ¼ 0 36 þ 18 2 T
pffiffiffi pffiffiffi and so k1 ¼ 36 18 2 and k2 ¼ 36 þ 18 2. The steady state response of the system satisfies jj ¼
Since g0 sinðvt Þ ¼ gðt Þ ¼ j1 ¼
10 20
gj0 sinðvtÞ k j v2 mj
sinð10t Þ, simple substitution furnishes
10 sinð10t Þ pffiffiffi 36 18 2 100ð4Þ
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and j2 ¼
20 sinð10t Þ pffiffiffi 36 þ 18 2 100ð4Þ
from which the steady state solution results as 1 1 p ffiffi ffi B 182 9 2 C C5 sinð10t Þ j(t) ¼ B A @ 1 pffiffiffi 182 þ 9 2 0
EXAMPLE 9.14 Express the following equations in state form, apply the trapezoidal rule, and triangularize the ensuing dynamic stiffness matrix: M€ g þ Kg Sp ¼ f, ST g ¼ 0 (Equations in this form will be seen to arise in finite element models of incompressible elastic bodies.)
SOLUTION The foregoing equation is expressed in state form as follows: 2
M 4 0T 0T
30 1. 2 g_ ðt Þ 0 M 0 5@ gðt Þ A þ 4 K 0 p ðt Þ 0T
0 K 0T
K 0 ST
30 1 0 1 g_ ðtÞ f ðt Þ S 0 5@ gðtÞ A ¼ @ 0 A p ðt Þ 0 0
This expression is rewritten using the trapezoidal rule and p ¼ g_ as 2
M
6 T 40 0T
0 K 0T
1 2 30 1 M K h pnþ1 pn 7B C 6 0 5@ 1h gnþ1 gn A þ 4 K 0 1 0T ST 0 h ðpnþ1 p n Þ
0
01 B ¼@
2 ðf nþ1
þ fnÞ
0
1 30 1 S 2 pnþ1 þ pn 7B C 0 5@ 12 gnþ1 þ gn A 1 0 2 ðpnþ1 þ pn Þ
1 C A
0 The second row implies that pnþ1 ¼ rewritten as
2 h
gnþ1 gn pn , enabling the first row can be
1 1 2 1 1 M gnþ1 gn 2pn þ K gnþ1 þ gn Sðpnþ1 þ pn Þ ¼ ðf nþ1 þ f n Þ h h 2 2 2 Multiplying throughout by h2=2 and rearranging gives Mþ
h2 h2 h2 K gnþ1 Spnþ1 ¼ ½f nþ1 þ f n Kgn þ Spn þ M½gn þ hpn (9:59) 4 4 4
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The third row, after multiplying by h2=2, is now h2 T h2 S gnþ1 ¼ ST gn 4 4
(9:60)
Equations 9.59 and 9.60 are written in matrix–vector notation as 2
3 2 2 M þ h4 K h4 S g nþ1 4 5 ¼ pnþ1 h2 T 0 4 S
!
gnþ1 h4 ST gn 2
in which 2
gnþ1 ¼ h4 ½f nþ1 þ f n Kgn þ Spn þ M½gn þ hpn The dynamic stiffness matrix emerges as 2 KD ¼ 4
2
2
h4 S
M þ h4 K h2 4
ST
3 5
0
Now KD is decomposed into a product of a lower triangular and an upper triangular matrix KD ¼ LU as 2 4
2
2
h4 S
M þ h4 K h2 4
ST
0
3
5 ¼ L11 L21
0 L22
U11 0
U12 U22
On setting U11 ¼ LT11 and L22 ¼ I, we find that 2
L11 LT11 ¼ M þ h4 K which can be triangularized to find L11 since, M and K are positive definite. In addition U12 ¼ h4 L1 11 S, 2
L21 ¼ h4 ST LT 11 , 2
1 h U22 ¼ L21 U12 ¼ 16 ST LT 11 L11 S 4
and finally " KD ¼ LU ¼
L11 h2 4
#2 T L11 4 I 0T
h4 L1 11 S 2
0
S LT 11 T
3 5
h4 T T 1 16 S L11 L11 S
EXAMPLE 9.15 Derive general expressions for the modal decomposition of the two-degree-of-freedom system
k11 k12
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k12 k22
x1 x2
þ
m11 m12
m12 m22
€x1 €x2
¼
0 0
SOLUTION (a) Eigenvalues k11 v2 m11 k12 v2 m12 0 ¼ det k12 v2 m12 k22 v2 m22 ¼ k11 v2 m11 k22 v2 m22 k12 v2 m12 k12 v2 m12 2 2 v2 (m11 k22 þ k11 m22 2k12 m12 ) þ v2 m11 m22 m212 ¼ k11 k22 k12 Letting 2 Dkk ¼ k11 k22 k12 , Dkm ¼ (m11 k22 þ k11 m22 2k12 m12 ), Dmm ¼ m11 m22 m212 the eigenvalues are obtained as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 v 1:2 ¼ Dmk ðDmk Þ2 4Dk 2Dm We know from other considerations that (Dmk)2 4Dk > 0. (b) Eigenvectors The eigenvector paired with the eigenvalue vj satisfies
2 2 k11 v2j m11 x(1j) þ k12 v2j m12 x(2j) ¼ 0, x(1j) þ x(2j) ¼ 1
from which
1 x(1j) ¼ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !2ffi , u 2 u k v m 11 11 j t1 þ k12 v2j m12
x(2j)
k11 v2j m11 k12 v2j m12 ¼ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !2ffi u 2 u k v m 11 11 j t1 þ k12 v2j m12
(c) Modal matrix 2
X¼
x11 x21
1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 k11 v21 m11 2 6 6 1 þ k12 v2 m12 1 6 x12 k11 v21 m11 ¼6 6 x22 2 6 k12 v1 m12 6 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k v2 m 2 4 11 11 1 þ k v12 m 12
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1
12
3 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7 k11 v2 m11 2 7 1 þ k v22 m 7 12 12 2 7 7 k11 v22 m11 7 k v2 m 7 12 2 12 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k v2 m 2 7 5 11 11 1 þ k v22 m 12
2
12
(d) Modal masses and stiffnesses
0 k1 T , X KX ¼ m2 0
m1 X MX ¼ 0 T
0 k2
m1 ¼ x211 m11 þ 2x11 x21 m12 þ x221 m22 ,
k1 ¼ x211 k11 þ 2x11 x21 k12 þ x221 k22
m2 ¼ x212 m11 þ 2x12 x22 m12 þ x222 m22 ,
k2 ¼ x212 k11 þ 2x12 x22 m12 þ x222 m22
(e) Modal coordinates The modal coordinate vector y is obtained from the physical coordinate vector x using
y1 y2
¼
x11 x21
x12 x22
1
x1 x2
Carrying out the manipulations gives
x1 x2
¼
x12
x22
x21 x11 x11 x22 x21 x12
1 ¼ x11 x22 x21 x12
y1
y2 x22 y1 x12 y2 x21 y1 þ x11 y2
(f) Modal force and physical force The modal force vector g is now obtained from the physical force vector f:
g1 g2
f1 ¼ XT f2 x11 f1 þ x21 f2 ¼ x12 f1 þ x22 f2
(g) Modes Using the foregoing relations, the system is expressed as modes, that is to say, as two uncoupled second-order single-degree-of-freedom systems. m1€y1 þ k1 y1 ¼ g1 (t) m2€y2 þ k2 y2 ¼ g2 (t)
EXAMPLE 9.16 Consider the effect of the mesh on the eigenvalues of the simple system illustrated (Figure 9.3).
SOLUTION Formulating and assembling the stiffness and mass matrices and imposing the constraints at the two ends results in a single-degree-of-freedom system governed by
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E, A, r L1 L
FIGURE 9.3 Two-element model of a clamped–clamped rod.
EA
1 1 rAL € u(L1 ,t ) ¼ 0 u(L1 ,t) þ þ L1 L L1 3
The eigenvalue is given by sffiffiffiffi pffiffiffi E1 3 , vn ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a(1 a) r L
a¼
L1 L
qffiffiffi qffiffiffi pffiffi qffiffiffi E 4 E If a ¼ 1=2, vn ¼ 2 L 3 Er ¼ 3:42 L r . However, If a ¼ 1=4, vn ¼ L r . These two results are significantly different from each other and from the exact solution, which qffiffiffi is vn ¼ pL Er . However, accurate results may be obtained for this minimum eigenvalue by using a sufficient number of elements.
9.5.2 COMMENTS ON EIGENSTRUCTURE COMPUTATION ELEMENT SYSTEMS
IN
LARGE FINITE
Jacobi and Subspace Iterations (cf. Bathe, 1996) are prominent among the many methods which have been formulated to compute the eigenvalues and eigenvectors of a large finite element system. The first finds all eigenvalues of the system, and then finds the eigenvectors using the eigenvalues. The second selectively finds the lowest eigenvalues and corresponding eigenvectors. One difficulty with using the finite element method for eigenstructures is that the mesh makes a contribution to the values obtained. This is a relatively minor problem for the lowest modes, but becomes progressively more bothersome as the mode number increases. In Chapter 10, we describe an easily visualized method, which we call the hypercircle method (Nicholson and Lin, 2006). It represents a kind of steepest descent optimization method. For present purposes, we describe the minimum properties of eigenvalues and eigenvectors. For finite element equation of a linear € þ Kg ¼ 0 with positive definite symmetric matrices K and M, the elastic system Mg minimum eigenvalue and corresponding eigenvector (called an eigenpair) satisfy the Rayleigh quotient relation ln ¼ min lj ¼ j
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xTn Kxn xT Kx ¼ min xTn Mxn xT x¼1 xT Mx
(9:61)
Since the vector iterates in the minimization process are constrained to have unit magnitude, all such vectors are related to each other through a proper orthogonal matrix. If x(0) is the initial iterate, then the vth iterate may be expressed as x(v) ¼ Qvx(0). It is readily seen that this reduces the iteration process to determining a sequence of rotation matrices which progressively reduce the magnitude of the Rayleigh quotient. However, once an eigenvalue and an eigenvector have been determined, the matrices must be altered in such a way that the subsequent minimization steps do not return to the same pair. If the eigenpair is actually removed the process is called deflation. In contrast in Chapter 10, a replacement process is described in which the lowest eigenvalue is replaced with a higher value without altering the eigenvectors.
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ß 2008 by Taylor & Francis Group, LLC.
10
Solution Methods for Linear Problems: II
10.1 INTRODUCTION Chapter 9 presented a number of conventional numerical methods in the linear FEA. This chapter introduces three comparatively advanced topics. The first is solution on an inverse problem, while the second addresses use of fourth-order time integration method, and the third presents an optimization-based method for computing eigenvalues and eigenvectors.
10.2 SOLUTION METHOD FOR AN INVERSE PROBLEM 10.2.1 INVERSE PROBLEM
IN
ELASTICITY
Many practical applications involving response of elastic bodies give rise to inverse problems. In FEA, for a given mesh and set of physical properties, even though a well-posed direct problem generally possesses a unique solution in classical linear elasticity, a corresponding inverse problem (based on the same stiffness matrix) may not. Furthermore, even when the inverse problem possesses a unique solution when modeled ‘‘exactly’’ using the classical linear theory of elasticity, an unfortunate choice of a mesh may cause the finite element version of the inverse problem to fail to do so. The current formulation addresses a particular example of an inverse problem and exploits a matrix nonsingularity criterion for assuring that the finite element model possesses a unique solution. A numerical test is applied to verify satisfaction of the criterion. The test is based on the linear independence of the rows of a nonsingular matrix. If the nonsingularity condition is violated, the mesh can be modified and the nonsingularity condition applied again. In a linearly elastic body which has known physical properties and which is experiencing small strains under statically applied loads, the finite element equation may be written as 8 9 8 9 < u1 = < f 1 = K u2 ¼ f 2 , : ; : ; ui 0
2
K11 K ¼ 4 K21 K31
K12 K22 K32
3 K13 K23 5 K33
(10:1)
Here the ni 3 1 vector ui denotes the displacement degrees of freedom at interior nodes, while the n1 3 1 vector and the n2 3 1 vector u2 denote displacement degrees
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of freedom at two different sets of boundary nodes. It is assumed that external forces are applied only to boundary nodes, with the n1 3 1 force vector f1 and the n2 3 1 force vector f2 referred to the two sets. All matrices and vectors are real. Furthermore, the finite element stiffness matrix K is positive definite, written K > 0. The system has n degrees of freedom in which n ¼ n1 þ n2 þ ni. It is assumed that Equation 10.1 reflects degrees of freedom remaining after any simple constraints on the body have been applied. Here, in a simple constraint a displacement degree of freedom is specified and the corresponding reaction force is an unknown. (Direct problems only exhibit simple constraints.) This contrasts with complex (overspecified) constraints appearing in inverse problems in which both the displacement and force are prescribed for a degree of freedom. In finite element models of static problems in linear elasticity, the n 3 n stiffness matrix K is symmetric and positive definite (after simple constraints have been enforced). In the direct problem, which is considered ‘‘well posed,’’ the force vectors f1 and f2 are prescribed, and the corresponding displacement vectors u1 and u2 are unknowns to be determined. Accordingly, at each boundary degree of freedom only one quantity (the force) is specified. Positive definiteness of K implies that the solution of the direct problem exists and is unique. In contrast, in the particular type of inverse problem (e.g., Dennis et al., 2004) of interest here, the first node set is ‘‘overspecified’’ in that displacements and tractions are specified at the same nodal degrees of freedom (complex constraints). Correspondingly, the second node set is ‘‘underspecified’’ in that neither displacements nor tractions are specified at the degrees of freedom. As shown subsequently there is no assurance that a unique solution exists for the finite element inverse problem even when it does in the direct problem.
10.2.2 EXISTENCE OF
A
UNIQUE SOLUTION
We first employ an example to demonstrate that in inverse problems a unique solution may not exist even though the solution of the corresponding direct problem does.
EXAMPLE 10.1 Let K11 and K22 be two n 3 n positive definite symmetric matrices, and let K12 denote a singular n 3 n matrix. Next, introduce the matrix H given by H¼
Kp11ffiffiffiffi KT12 K11
pffiffiffiffi pffiffiffiffi K11 K12 K11 ¼ KT12 KT12 K12 þ K22
0 pffiffiffiffi K22
pffiffiffiffi K11 0T
pKffiffiffiffi12 K22
(10:2)
Of course, under the stated conditions H is positive definite and symmetric, written as H > 0. of H Direct Problem: Letting ff k1 be a known 2n 3 2n vector, positive definiteness k2
implies thatthere exists a unique solution for the unknown vector H
uu1 uu2
¼
f k1 f k2
.
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uu1 uu2
satisfying
Inverse Problem: Now suppose that the right-hand side contains the vector is
f k1 f u2
in
which fk1 is known but fu2 is unknown. Also, suppose that the left-hand side now
contains the vector
uk 1 uu2
in which uk1 is known but uu2 is unknown. We are now
confronted with the inverse problem " K11 pffiffiffiffi KT12 K11
# pffiffiffiffi uk 1 f k1 K11 K12 ¼ uu2 f u2 KT12 K12 þ K22
(10:3)
The upper block row implies that pffiffiffiffi K11 K12 uu2 ¼ f k1 K11 uk1
(10:4)
pffiffiffiffi But K12 and hence K11 K12 are singular. Hence, either there is no unique solution for uu2 or there are many solutions. The second block row does not mitigate this difficulty since it introduces the additional unknown vector fu2. Again the inverse problem involving H does not possess a unique solution although the direct problem does. Returning to the general development, a sufficient condition is now presented for the existence and uniqueness of the solution to a finite element model of the inverse problem of interest. However, as will be demonstrated, it is quite possible that the stiffness matrix in one mesh will not satisfy the sufficient condition, while the stiffness matrix in another mesh for the same physical problem will do so. From the third row in Equation 10.1, we have K31 u1 þ K32 u2 þ K33 ui ¼ 0
(10:5)
u3 ¼ K1 33 ðK31 u1 þ K32 u2 Þ
(10:6)
and, recalling that K33 > 0,
Upon substitution, the upper two rows provide the equations
1 K11 K13 K1 33 K31 u1 þ K12 K13 K33 K32 u2 ¼ f 1 1 K21 K23 K1 33 K31 u1 þ K22 K23 K33 K32 u2 ¼ f 2
(10:7) (10:8)
Positive definiteness of K implies that its principal minors are positive definite. It follows that K11 K13 K1 33 K31 > 0 K22 K23 K1 33 K32 > 0 and "
K11 K13 K1 33 K31 K11 K13 K1 33 K31
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K12 K13 K1 33 K32 K22 K23 K1 33 K32
# >0
(10:9)
Now suppose that u1 ¼ uk and f1 ¼ fk are known, while u2 ¼ uu and f2 ¼ fu are unknown, thereby introducing an inverse problem of the type of interest here. Equations 10.7 and 10.8 are now rewritten as
1 K11 K13 K1 33 K31 uk þ K12 K13 K33 K32 uu ¼ f k 1 1 K21 K23 K 33 K31 uk þ K22 K23 K33 K32 uu ¼ f u
(10:10) (10:11)
Equation 10.10 immediately furnishes
1 K12 K13 K1 33 K32 uu ¼ f k K11 K13 K33 K31 uk
(10:12)
For the solution of the inverse problem expressed by Equations 10.10 and 10.11 to exist 1 and be unique it is necessary and sufficient that K12 K13K 33 K32 be nonsingular, regardless of uk and fk. Once uu is obtained by solving Equation 10.12, fu is immediately found from Equation 10.11.
EXAMPLE 10.2 Inverse problems in two-element cantilevered beams Figure 10.1 depicts a cantilevered beam modeled by two elements. The elastic modulus E and the bending moment of area I are the same in the two elements, but their lengths L1 and L2 differ. The (unclamped) nodes are denoted as 1 and 2. The vertical (z) displacement and slope are denoted by w, w0 , the shear force by V, and the bending moment by M. The finite element equation for the two-element beam configuration is given by
3 12 12 6 6 2 7 3 6 6 L3 þ L3 L L2 7 66 1 2
1
7 66 4 4 5 64 6 6 6 2 þ 2 6 L1 L 2 L1 L 2 EI 6 2 12 6 3 6 6 6 6 L32 L22 7 6 6 7 6 4 6 45 4 2 L2 L2 22
2 12 6 L32 6 4 6 L22 2 12 6 L32 6 46 L22
3 63 2 L2 7 7 7 78 9 8 9 wðL1 Þ V ðL1 Þ > 4 57 > > 7> > > > > > > > 7> L2 7< w0 ðL1 Þ = < M ðL1 Þ = 7 ¼ (10:13) > wðL1 þ L2 Þ > > 63 7 V ðL1 þ L2 Þ > > > > 7> > > > > : : ; ; 7 0 L22 7 M ðL1 þ L2 Þ 7 7 7 w ðL1 þ L2 Þ 45 5
L2
z
y E, I
1
E, I
2
FIGURE 10.1 Inverse problem for a two-element beam problem.
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x
Four distinct inverse problems are now considered.
Case I:
w(L1), V(L1), w(L1 þ L2), V(L1 þ L2) are prescribed.
The solution exists and is unique if fact, it is with determinant equalling
6=L21 6=L22 6=L22 36=L21 L22 .
6=L22 6=L22
is nonsingular. In
Case II: w(L1), V(L1), w0 (L1 þ L2), M(L1 þ L2) are prescribed.
6=L2 1 6=L2 , which is the transpose of the The matrix of interest is 6=L6=L 2 6=L22 2 matrix in the first case and has the same nonvanishing determinant. 2
2
2
Case III: w0 (L1), M(L1), w0 (L1 þ L2), M(L1 þ L2) are prescribed.
12=L31 þ 12=L32 6=L22 arises 6=L22 4=L2 equalling 48=L31 L2 þ 12=L42 .
The matrix is minant
which is nonsingular with deter-
Case IV: w0 (L1), M(L1), w(L1 þ L2), V(L1 þ L2) are prescribed.
The matrix 72=L31 L22 .
12=L31 þ 12=L32 6=L22
12=L32 6=L22
is nonsingular, with determinant
10.2.3 NONSINGULARITY TEST We next introduce a test for nonsingularity of K12 K13 K1 33 K32 . Of course it is necessary that n1 ¼ n2, since otherwise this matrix is not square and hence is singular. The first result, which is easily proved, is that K12 K13 K1 22 K32 is nonsingular if, K12 K23 and only if, K K is nonsingular. 32
33
Next a test of nonsingularity of written in the form
K12 K32
K23 K33
is given. An n 3 n matrix A may be
2
3 aT1 6 aT 7 6 2 7 6 7 6 : 7 6 T 7 7 A¼6 6 aj 7 6 7 6 : 7 6 T 7 4 an1 5 aTn
(10:14)
in which the ith row of the matrix A is written as the row vector ajT. Using GramSchmidt orthogonalization, we may construct a set of orthonormal base vectors ei as follows. The n base vectors ej are given by
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^e2 ¼ a2 (aT2 e1 )e1 , : : n1 X ^en1 ¼ an1 (aTn1 ej )ej ,
e1 ¼ a1 =ja1 j e2 ¼ ^e2 =j^e2 j : : en1 ¼ ^en1 =j^en1 j
(10:15)
j¼1
^en ¼ an
n1 X
en ¼ ^en =j^en j
(aTn ej )ej ,
j¼1
If A is nonsingular, the jth row vector cannot be a linear combination of the foregoing j 1 row vectors. Accordingly, the jth row vector aj exists in the j-dimensional subspace spanned by orthonormal base vectors e1, e2, . . . , ej. For the moment, suppose instead that the matrix A is singular with unit rank deficiency, and that aTn is a linear combination of the foregoing row vectors and hence exists in an n 1-dimensional subspace spanned by the base vectors e1, e2, . . . , ej, . . . , en1. But, if ên simultaneously (i) lies in the n 1-dimensional subspace and (ii) is orthogonal to the base vectors of the subspace, it must equal the null vector: ên ¼ 0. Accordingly, the condition for the matrix A to be nonsingular is ^ek 6¼ 0,
k ¼ 1, 2, 3, . . . , n
(10:16)
Conversely, if any of the vectors êk vanish, A is singular. Simple examples are now introduced to illustrate the application and performance of the nonsingularity test expressed in Equation 10.16.
EXAMPLE 10.3 We first consider the matrix A¼
1 2 2 4þ«
(10:17)
in which « << 1. Following the operations in Equation 10.15 the base vectors are found to be ( ) 1 1 e1 ¼ pffiffiffi 5 2 ( )# ( ) ( ) " 1 1 2 1 1 ^e2 ¼ pffiffiffi pffiffiffi f 2 4 þ « g 2 4 5 5 2 ( ) 1 « ¼ (10:18) 5 4« In fact, this matrix A is nonsingular if « > 0 and becomes singular when « # 0. But also note that, in the current test, ê2 6¼ 0 if « > 0 and ê2 ! 0 as « # 0, consistent with Equation 10.16.
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EXAMPLE 10.4 Next, consider 2
1
6 A ¼ 42 3
2
3
3
7 4 6 þ «1 5
(10:19)
6 9 þ «2
We first seek to determine whether the current test correctly identifies values of «1 and «2 for which A is singular. We first check ê2. 8 9 > >1> > 1 < = e1 ¼ pffiffiffiffiffi 2 > 14 > > : > ; 3 8 9 8 9 8 9 1> 2 > > > > > > >1> > > > < = < = 1 1 < = ^e2 ¼ f 2 4 6 þ «1 g pffiffiffiffiffi 2 pffiffiffiffiffi 2 4 > 14 > > > > 14 > > > > > : > ; : > ; : ; 3 3 6 þ «1 8 9 > > 3 > > = 1 < ¼ 6 «1 > 14 > > > : ; 5
(10:20)
It is evident that A is singular if «1 ¼ 0. The current procedure furnishes that ê2 6¼ 0 if «1 > 0, but ê2 ! 0 as «1 # 0. Assume for the moment that «1 > 0, so that 9 8 3 > > = < 1 e2 ¼ pffiffiffiffiffi 6 > 70 > ; : 5
(10:21)
Note that «1 does not appear in e2. We next determine the third base vector, assuming «1 6¼ 0. 8 93 8 9 1> > > > > >1> > < = 7 1 < = 6 1 7 pffiffiffiffiffi 2 ^e3 ¼ p ffiffiffiffiffi 6 f g 6 3 6 9 þ « 2 2 4 > >5 14 > > > 14 > > > > > ; : > ; : > ; : 9 þ «2 3 3 8 8 93 9 8 9 2 3 > 3 > > > >0> > > > > < < = = > < > = 7 1 6 1 7 6 4f 3 6 9 þ «2 g pffiffiffiffiffi 6 5 pffiffiffiffiffi 6 ¼ 0 > > > > 70 > 70 > > > > > : : ; ; > : > ; 5 5 0 8 > > <
3
9 > > =
2
(10:22)
The test in Equation 10.16 indicates that the matrix A is singular regardless of «2. That this result is correct is easily seen by recognizing that the second column in A is proportional to the first column regardless of «2 (or «1).
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10.3 ACCELERATED EIGENSTRUCTURE COMPUTATION IN FEA 10.3.1 INTRODUCTION Calculation of lowest modes is a common task in finite element modeling of large systems. The primary established method in FEA is Subspace Iteration (Bathe, 1996). Here, an optimization algorithm (Nicholson and Lin, 2006) is described which exploits the fact that (i) the eigenvectors terminate on a unit hypersphere and (ii) that the minimum eigenvalue and associated eigenvector satisfy the Rayleigh minimum principle. At the current estimate for the minimizing eigenvector, a direction on the hypersphere is found analytically in which the Rayleigh quotient experiences ‘‘steepest descent.’’ Kronecker product algebra is instrumental in the derivation. The current eigenvector and a unit vector representing the direction of steepest descent define a plane intersecting the hypersphere along a unit hypercircle. An analytical solution is found for the vector minimizing the Rayleigh quotient on the hypercircle, constituting a hypercircle counterpart of a ‘‘line search.’’ At this last vector a new steepest descent vector is determined and the process is repeated. In a numerical example, the algorithm converges very rapidly. Also introduced is a counterpart of ‘‘deflation’’ to replace the smallest eigenvalue while leaving the eigenvectors and the dimension of the matrix unchanged. The deflation procedure currently is based on the assumption that none of the eigenvalues in the lowest modes are repeated.
10.3.2 PROBLEM STATEMENT For dynamic response of a linear elastic system, the finite element equation is conventionally written as ^€ ^ x ¼ ^f(t) ^ M x þ K^
(10:23)
^ and K ^ are n 3 n, real, positive definite, and symmetric, in which the matrices M ^ while ^x and f(t) are real n 3 1 vectors. For convenience the foregoing system is rewritten as € x þ Kx ¼ f(t)
(10:24)
^ 1=2 and x ¼ M1=2^ x. The modes of the system are determined in which K ¼ M1=2 KM by the eigenvalues lj and the real orthonormal eigenvectors xj, with the assumed magnitude ordering l1 > l2 > > ln. The primary interest is in the low-frequency modes, say with lj having magnitudes less than a user-specified threshold value. The foundation of the method being introduced is minimization of the Rayleigh quotient (e.g., Dahlquist and Bjork, 1976). Namely, the minimum eigenvalue ln and the corresponding minimizing eigenvector xn satisfy ln (K) ¼ min lj (K) 1 jn
¼ xTn Kxn ¼ min (xT Kx) T x x¼1
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(10:25)
pffiffiffiffiffiffiffiffi To enforce unit magnitude of n ¼ x= xT x a priori, we replace n with n ¼ Qy
(10:26)
in which y is a unit vector chosen to give the initial estimate of the minimizing eigenvector. Also Q is an orthogonal transformation rotating y to n, which of course is also of unit magnitude. Clearly, the tip of n lies on a unit hypersphere (a sphere in n-space). As Q changes during minimization, the tip of n prescribes a path on the hypersphere. Otherwise stated, the task is to determine the hyperspherical path which is optimal in a meaningful sense and which terminates at the tip of the desired eigenvector.
10.3.3 HYPERSPHERE PATH
OF
STEEPEST DESCENT
We now determine the differential dQ corresponding to the most rapid rate of decrease of F1 ¼ yTQT KQy. Letting q denote VEC(Q), Kronecker Product Algebra (presented in Chapter 3) furnishes the following relations: dF1 (Q) ¼ yT dQT KQy þ yT QT K dQy ¼ yT yT VEC(dQT KQ þ QT K dQ) ¼ yT yT [(QT K) IU þ I (QT K)] dq ¼ [(yT QT K) yT U þ yT (yT QT K)] dq ¼ mT dq
(10:27)
in which dq ¼ VEC(dQ) mT ¼ (I þ U)yT (nT K),
n ¼ Qy
However, we may also write dF1 ¼ t(QMT dQ)
(10:28)
in which M ¼ IVEC(m) and t denotes the trace. Invariance of the trace under similarity transformations implies dF1 ¼ t(MT dQ QT )
(10:29)
Note that the matrix dQ QT is antisymmetric. Let B ¼ QMT and let C ¼ 12 (B BT ). The trace vanishes for the product of a symmetric and an antisymmetric matrix: dF1 ¼ t(B dQ QT ) ¼ t(C dQ QT )
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(10:30)
Note that C2 is negative definite if the antisymmetric matrix C is nonsingular, so that steepest descent is attained if dQQT is proportional to C. Accordingly, we seek the path defined by the relation dQ ¼ dL
C(Q)Q t1=2 ( C2 (Q))
(10:31)
and dL may be viewed as a differential arc length since dL ¼ t 1=2 (dQ dQT )
(10:32)
Away from the minimum, the procedure coerces F to decrease since dF ¼ dLt 1=2 ( C2 (Q)) 0
(10:33)
Of course the current position vector after j iterations is given by nj ¼ Qjy, in which Qj denotes the current value of Q. The method for identifying an orthogonal matrix Q (i.e., Qj) rotating y to n(nj) is described in Nicholson and Lin (2006).
10.3.4 HYPERCIRCLE SEARCH Suppose that j iterations have occurred leading to the current vector nj ¼ Qjy. We introduce the incremental position vector dnj ¼ dQj nj. It follows that to first order in the increments nTj dnj ¼ yT QTj dQj y ¼0
(10:34)
since QT dQ is antisymmetric. It also follows that the unit vector r(nj) given by dnj r nj ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dnTj dnj
(10:35)
is orthogonal to nj and hence is tangent to the hypersphere. We say that the steepest descent from position nj occurs in the direction represented by the unit vector r(nj). The vectors nj and r(nj) define a hyperplane whose intersection with the hypersphere is a hypercircle, as illustrated in Figure 10.2. Any vector p terminating on the hypercircle may be expressed in terms of nj, r(nj), and an angle h, illustrated in Figure 10.3. Likewise, on the hypercircle F may be expressed as a simple function of nj, r(nj), and h. The hypercircle search simply consists of determining the angle h0 which minimizes F1 on the hypercircle. It will now be seen that determination of h0 reduces to a simple algebraic problem with an analytical solution. Finally, if we designate the value of Q minimizing the Rayleigh quotient as Qmin, the minimizing eigenvector is obtained immediately as nmin ¼ Qminy, and of course, the minimum eigenvalue satisfies lmin ¼ F(Qmin).
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3
n 2
r
1
FIGURE 10.2 Hypersphere and hyperplane determined by n and r.
Minimizing F on the hypercircle reduces to determining an angle which can be expressed analytically in terms of an inverse tangent. The vector p(h) illustrated below in Figure 10.3 is given by pðhÞ ¼ cos hn þ sin hr On the hypercircle the function F is expressed in terms of the angle h by
p(h)
r
h
n
FIGURE 10.3 Vector terminating on the hypercircle.
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(10:36)
F ¼ a cos2 h þ b sin2 h þ 2g sin h cos h ¼
aþb ab þ cos 2h þ g sin 2h 2 2
a ¼ nT Kn, b ¼ rT Kr,
(10:37)
g ¼ nT Kr
The extrema of F occur at the values of h, say h*, satisfying
ab sin 2h* þ g cos 2h* ¼ 0 2
(10:38)
from which
1 2g 1 h* ¼ tan 2 ab
(10:39)
2g There are two such values of h*, which are in the first and third quadrants if ab > 0, 2g but the second or fourth quadrants if ab < 0. One value minimizes F1 on the hypercircle, and the second maximizes F1. For the present extremum to be a minimum, h* must satisfy
ab cos 2h* g sin 2h* > 0 2
(10:40)
Simple manipulation furnishes "
#
1 ab 2 þ1 sin 2h* > 0 g 2 g 2
(10:41)
Consequently, h* is found in the first or fourth quadrants as follows: if g 0, if g < 0,
p=2 h* 0 0 < h* p=2
(10:42)
10.3.5 EIGENVALUE REPLACEMENT PROCEDURE Once the minimum eigenvalue and the corresponding (minimizing) eigenvector are determined, it is necessary to remove the eigenvalue from the matrix while leaving the remaining eigenvalues and all of the eigenvectors unchanged, a process we refer to as eigenvalue replacement. The scheme presented below replaces the most recently computed minimum eigenvalue with magnitude less than a threshold value with a value above the threshold, without altering the eigenvectors or reducing the dimensions of (deflating) the matrix. Once the minimum eigenvalue is replaced, the minimization process is repeated to compute the next largest eigenvalue and corresponding eigenvector, and continues until an eigenvalue is obtained whose
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magnitude equals or exceeds a user-specified threshold value. Currently, the foregoing deflation procedure is based on the restriction that none of the eigenvalues of the modes of interest has multiplicity greater than unity. Suppose ln and nn have been computed. We may construct a set of vectors (n1) (n1) pn1 , pn2 , . . . , p1(n1) which are orthonormal to each other and to nn. They are likewise orthogonal to Knn since it coincides in direction with nn. In particular, (n1) (n1) pn1 , pn2 , . . . , p1(n1) are obtained sequentially using the Gram-Schmidt scheme nj1 X
^ j(n1) ¼ qj(n1) p
(n1) (n1) (qj(n1)T pjþk )pjþk (qj(n1)T nn )nn
k¼1
j ¼ n 1, n 2, . . . , 1,
pj(n1)
¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ^j(n1)T p ^ j(n1) p
(10:43)
^j(n1) p
in which qj(n1) are ‘‘judiciously chosen’’ ‘‘trial’’ vectors. The matrix Rðnn1Þ given by (n1) (n1) (n1) (n1) Rn(n1) ¼ Rn1 nn , Rn1 ¼ p1(n1) p2(n1) pn2 pn1
(10:44)
is orthogonal and gives rise to the singularity transformation " Rn(n1)T KRn(n1) ¼
* Kn1 0T
# 0 ln
(10:45)
In Equation 10.45 note that (i) Rn(n1)T KRn(n1) jn ¼ pj(n1)T (Knn ) ¼ 0, (ii) h i (n1)T ðn1ÞT (n1) Rn KRn(n1) nj ¼ nTn Kpj ¼ pj(n1)T (Knn ) ¼ 0, and (iii) K*n1 ¼ pi Kpj(n1) . ij * are l , l , . . . , l , which are the largest n1 The eigenvalues of Kn1 n1 n2 1 eigenvalues of K. Of course, the eigenvectors of K*n1 are also eigenvectors of K. Suppose the largest eigenvalue of interest is no greater in magnitude than a threshold value lth, corresponding to the mode with the highest natural frequency of interest. We may introduce the matrix " (n1)
K
¼
Rn(n1)
K*n1 0T
# 0 Rn(n1)T lth
(10:46)
Now the eigenvalues of K(n1) are ln1, ln2, . . . , lth, . . . , l1. It is next demonstrated that K(n1) has the same eigenvectors as K. Since K*n1 is symmetric and positive definite there exists an orthogonal matrix Vn1 (which need not be computed) such that (n1)
K
¼
Rn(n1)
Vn1 0T
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0 1
" * L Kn1 0T
0 ln þ lth
#
VTn1 0T
0 (n1)T R (10:47) 1 n
in which, assuming obvious eigenvalue ordering, 2 l1 0 : 60 l 0 2 6 6 * 6 : : L Kn1 ¼ 6 : 6 4 : : : : It follows that 2 L K*n1 4 0
T
0
3
""
5¼
0T
lth
VTn1 0 1
:
: : :
:
0
Rn(n1)T
K
(n1)
3
: 7 7 7 : 7 7 7 0 5
:
#
#
:
(10:48)
ln1 "
" Rn(n1)
Vn1 0 0T
## (10:49)
1
0 is also recognized as an orthogonal matrix which diagHowever, Rn(n1) V0n1 T 1 onalizes K: 2 3 "" # " ## # " L K*n1 0 Vn1 0 VTn1 0 (n1)T (n1) 4 5¼ Rn (10:50) K Rn 0T 1 1 0T ln 0T
But matrices which are diagonalized by the same similarity transformation, in particular K and K(n1), have the same eigenvectors, Nicholson and Lin (1996). The replacement process continues as follows. The next largest eigenvalue is ln1 with corresponding eigenvector nn1, both of which are computed by minimizing the Rayleigh quotient applied to K(n1). A new set of unit vectors pj(n2) , j ¼ n2, n1, . . . , 1 is generated, which are orthogonal to each other as well as nn1 and nn. Extending the steps shown above furnishes a matrix of the form 2 * 3 0 0 Kn2 6 7 Rn(n2)T K(n2) Rn(n2) ¼ 4 0T (10:51) 5 ln1 þ lth 0 0T
ln þ lth
0
The deflation process is repeated to furnish the desired eigenvalues (with magnitude not exceeding lth) in order of increasing magnitude, and to furnish the corresponding eigenvectors.
10.3.6 EXAMPLE: MINIMUM EIGENVALUE
OF THE
3 3 3 HILBERT MATRIX
We illustrate the hyperspherical method by applying it to determine the minimum eigenvalue lmin and corresponding eigenvector nmin of the 3 3 3 Hilbert matrix 3 2 1 1=2 1=3 7 6 (10:52) H3 ¼ 4 1=2 1=3 1=4 5 1=3 1=4
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1=5
Hilbert matrices are notorious for being ill-conditioned even though positive definite and symmetric. A simple program has been written in high precision to compute l3 and n3. The initial vector is assumed using the diagonal terms of H3: 8 9 < 1=2 = 1 1=2 v ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (1=2)2 þ (1=3)2 þ (1=5)2 : 1=5 ;
(10:53)
The procedure generates and makes use of two vectors in each step after the first step. To use the hypercircle method in the first step, a second vector q must be introduced at the outset. We used the vector 8 9 <1= q¼ 0 : ; 0
(10:54)
The computed iterates for the lowest eigenvalues and the relative errors are shown in Table 10.1 below. Even though the errors are very high in the initial estimate, convergence still occurs rapidly and furnishes extremely accurate values. Convergence appears to be much more rapid than in a linear convergence scheme. (Subspace Iteration exhibits linear convergence.) The eigenvectors likewise converge very rapidly.
10.4 FOURTH-ORDER TIME INTEGRATION 10.4.1 INTRODUCTION In FEA of elastic systems with light viscous damping, the widely used Newmark method is the second order, one step, and A-stable. A systematic presentation of established time integration methods for FEA is given in Zienkiewicz and Taylor (1989), covering the well-known methods of Newmark, Houbolt, Wilson, Hibler, and others. The Newmark method is a reformulation of the classical Trapezoidal TABLE 10.1 Convergence of Iterations for Lowest Eigenvalue Computed Minimum Eigenvalue 1.3422136422136 0.1126113206822 0.0047836190059 0.0026907039866 0.0026873457448 0.0026873403644 0.0026873403558 0.0026873403558
ß 2008 by Taylor & Francis Group, LLC.
Percent Error (%) 49845.800104181800000 4090.437599027870000 78.005699785540800 0.125165789790772 0.000200532865081 0.000000321273041 0.000000000515009 0.000000000001485
Rule for systems with accelerations. Even though it is A-stable, time step sizes are severely limited by considerations of accuracy. Also, ‘‘numerical damping,’’ for example in the Wilson-Theta method, has been introduced to attenuate higher order modes. It does so at a modest cost in accuracy (near-second order). In this section a fourth-order counterpart of the Newmark method is described which extends the fourth-order three-step Adams–Moulton (AM) method to systems with acceleration, referred to below as AMX. (AM refers to first-order systems while AMX refers to second-order systems.) The AMX method is three-step. No three-step method can be A-stable, by virtue of a classical theorem of Dahlquist (cf. Dahlquist and Bjork, 1974; Gear, 1971). However if, after every time step (or set of several steps) numerically unstable higher order modes are filtered from the response using, for example, the Wavelet Packet transform (Kaplan, 2002), the stability-based restrictions on the time step size can be comparable to the restrictions ensuing from accuracy (Nicholson and Lin, 2006). Also, a modification of the AMX method is given to incorporate numerical damping, rendering the modification near-fourth order. AMX gives rise to a linear system involving a dynamic stiffness matrix. Solution using triangularization followed by forward and back substitution is seen to require exactly the same computational effort as the Newmark method.
10.4.2 ERROR GROWTH IN
THE
NEWMARK METHOD
The errors and error growth in the Newmark method have been extensively described by Nicholson and Lin (2005) for free and forced response, and results are quoted here on undamped free response. 10.4.2.1
Undamped Free Vibration
To illustrate the error properties of the Newmark method, suppose that the highest mode to be computed is the Jth mode with natural frequency vJ, and suppose there is no damping. The Newmark method does not produce any magnitude error, but there is a phase angle error since the Newmark method evidently approximates vJh=2 as tan1(vJh=2). The phase error in a step is given by
p2 h 2 «h 3 TJ
(10:55)
in which TJ is the time period of the model. For illustration we choose h to attain a 1 stepwise relative error of 300 , in which instance the number of time steps per period is TJ found to be h ¼ 10p 32. Furthermore, we determine the effect of stepwise error on cumulative error. The exact solution after M time steps is ynþ1 ¼ exp(Mlh)y0 ¼ [exp(lh)]Mþ1y0. Recalling Equation 10.55, the numerical solution is ~ynþ1 ¼ [(1 þ «h) exp(lh)]Mþ1 ynþ1 y0 ¼ (1 þ «h)Mþ1ynþ1. The cumulative error after M þ 1 steps is «T ¼ ~ynþ1y ¼ nþ1 Mþ1 1. The number of time steps after N time periods is given by (1 þ «h ) ppN ffiffiffiffi ffiffiffiffiffi. Now (1 þ «h ) 3«h ¼ 1 þ «T , and ppN ffiffiffiffiffi ln (1 þ «h ) ¼ ln (1 þ «T ). M þ 1 ¼ ThJ N ¼ ppN 3« 3« h
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h
We assume that both relative errors are much less than unity in magnitude, and take the first nonvanishing terms in the Taylor expansion of the natural logarithms. Now pffiffiffiffiffi pN pffiffi «h «T , so that «h 23 2 «2 . Otherwise stated, the time step is to be selected to p N T 3 2 satisfy ThJ ¼ p3 «NT . As an example, suppose that N ¼ 10 and «T ¼ 1=10. This implies the number of time steps per period is 333.
10.4.3 ADAMS–MOULTON FORMULA Consider the differential equation dy ¼ f (y), dt
y(0) ¼ y0
(10:56)
in which f(y) is a known function. A three-step integration formula has the form (Gear, 1971) a0 ynþ1 þ a1 yn þ a2 yn1 þ a3 yn2 þ hb0 fnþ1 þ hb1 fn þ hb2 fn1 þ hb3 fn2 ¼ 0 (10:57) in which the coefficients aj and bj are to be determined from accuracy and stability considerations. Elements of the derivations are reproduced here from, for example, Gear (1971), to set the stage for a stability and accuracy analysis and for the introduction of damping. Taylor expansion if ynþj and fnþj, j ¼ 1, 0, 1, 2 through fifth order furnishes the following relations:
7 a 12 0 5 12
þ
b0 þ b1 þ b2 þ b3 ¼ 1
(10:58a)
a0 þ a1 þ a2 þ a3 ¼ 0 : d(h(0) )
(10:58b)
a1 þ 2a2 þ 3a3 b0 b1 b2 b3 ¼ 0 : d(h)
(10:58c)
1 a 12 1
1 6
a1 þ a2 þ 49 a0 144
þ
13 a1 144
17 a 12 2
11 a3 4
þ
23 a2 144
47 a 12 3
7 b 12 0
þ
þ b1 þ 2b2 þ 3b3 ¼ 0 : d(h2 )
þ
157 a3 144
1 b 12 1
þ
5 b 12 1
17 b 12 2
1 6
47 b 12 3
b2
¼ 0 : d(h3 )
11 b 4 3
¼ 0 : d(h4 )
(10:58d) (10:58e) (10:58f)
The error coefficient (coefficient of h5) is given by C5 ¼
7 5 1 a0 þ a1 a2 48 48 48 23 157 þ b b 144 2 144 3
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þ
11 a3 48
49 b 144 0
13 b 144 1
(10:59)
Equations 10.58a through 10.58f represent six equations in eight unknowns. Accordingly, a0, a1, a2, a3, b1, and b2 may be expressed in terms of b0 and b3. Simple calculation serves to obtain 9 8 9 8 9 9 8 8 3=4 > a1 > 9=2 > 3=2 > > > > > > > > > > > > > > > > > > > > > > > a2 > > > = < 3=4 > = < = = < 3=2 > < 9=2 > a3 ¼ 1=12 þ b0 1=2 þ b3 5=2 > > > > > > > > > > > 1 > 2 > b > > > 1=2 > > > > > > > > > > > > > > ; : ; > : 1> ; ; : : 1=2 b2 2 1
(10:60)
and a0 ¼ 1=12 5=2b0 þ 1=2 b3 There are several types of errors in a multistep method. One is primary error associated with the primary eigenvalue lJ, which in the current context determines magnitude error and phase error. The primary eigenvalue should match the exact eigenvalues in the physical mode. In addition there are ‘‘extraneous errors,’’ also called ‘‘parasitic errors,’’ which are associated with secondary eigenvalues arising in Equation 10.57 and are purely numerical in origin. It is reasonable to choose b0 and b3 to minimize the magnitudes of the two extraneous eigenvalues at vanishing values of h. With this choice, a finite time step is required before the extraneous eigenvalues (as a function of the time step) approach unity in magnitude and thereby imply instability. The secondary eigenvalues vanish at h ¼ 0 if b0 and b3 are chosen 1 . In fact, doing so yields the classical fourth-order threesuch that b0 ¼ 38 and b3 ¼ 24 step Adams–Moulton (AM) formula 5 1 ynþ1 yn þ lh 38 ynþ1 þ 19 24 yn 24 yn1 þ 24 yn2 ¼ 0
(10:61)
and more generally 5 1 ynþ1 yn h 38 fnþ1 þ 19 24 fn 24 fn1 þ 24 fn2 ¼ 0
(10:62)
AM is next seen to preserve magnitude and compute phase angle with significantly larger time steps than Newmark for comparable levels of accuracy. However, recall that Newmark is A-stable, while AM exhibits numerical stability only up to a critical ratio (to be discussed) of the time step to the period. AM is not suitable if highfrequency modes are present—a likely occurrence in large finite element systems. As stated previously, one possible way to remove high-frequency modes is filtering using, for example, the Wavelet Packet transform.
10.4.4 STEPWISE AND CUMULATIVE ERROR IN
THE
ADAMS–MOULTON METHOD
We consider errors in free vibration of the undamped Jth mode. Again there is no magnitude error, but as in Newmark there is a phase error. Nicholson and Lin (2005) reported that the stepwise error estimate
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«h
4 p4 h 4 h 2:165 TJ 45 TJ
(10:63)
For comparison with Newmark we again choose the stepwise relative phase angle error to be 1=300. The result is TJ=h 4, or four time steps per period. This contrasts with 32 time steps per period in the Newmark method. The AM method shows an even greater advantage when cumulative error is considered. From Equation 10.63 the error measure «h ¼ (h=TJ)4 is representative of the relative phase error per step in the AM method applied to the transient solution. The number of time steps in N periods satisfies M þ 1 ¼ NTJ=h ¼ N=(«h)1=4. Now 1=4 (1 þ «h)N=(«h ) ¼ 1 þ «T. Use of the natural logarithm and the first term in the Taylor series leads to N(« h)3=4 ¼ «T, implying that the number of time steps per period is 1=3 . As before, consider N ¼ 10 and «T ¼ 1=10. The number of given by ThJ ¼ «NT time steps per period is now 4.7, much lower than 333 for Newmark.
10.4.5 STABILITY LIMIT
ON
TIME STEP IN
THE
ADAMS–MOULTON METHOD
For a finite time step h the eigenvalues of the AM formula are obtained from (Gear, 1971)
1 þ 38 lJ h h3 (1
19 2 24 lJ h)h
5 1 24 lJ hh þ 24 lJ h ¼ 0
(10:64)
We already know that the primary eigenvalue is h ¼ exp(lJ h) to fourth order in h. This permits approximating Equation 10.64 as (h exp(lJ h)) h2 þ g 1 h þ g 2 ¼ 0
lJ h 5lJ h lJ h exp(2lJ h) þ exp(lJ h) exp(lJ h) 24 24 24 (10:65) , g2 ¼ g1 ¼ 3 3 1 þ lJ h 1 þ lJ h 8
8
Numerical stability requires that the extraneous eigenvalues be interior to the unit circle in the complex plane. They are given by qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h1 ,h2 ¼ g1 g 21 4g2 (10:66) Clearly, the complex values and magnitudes can be readily computed; they are presented and displayed in the following paragraphs. We concern ourselves with underdamped media, to which end we write qffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi lJ h ¼ vJ h zJ þ i 1 z2J , i ¼ 1, Re(lJ h) ¼ vJ hzJ , Im(lJ h) ¼ vJ h 1 z2J (10:67)
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Stability limit
(h/Tj)max
0.6 0.4 0.2 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
zj
FIGURE 10.4 Maximum time step ratio vs. damping factor.
Following the language of vibration theory, zJ is called the damping factor and vJ the 2p undamped natural frequency (ofqthe Jth ffiffiffiffiffiffiffiffiffiffiffiffi ffi mode). The undamped period is TJ ¼ vJ . 2ph 2 Upon writing lJ h ¼ T zJ þ i 1 zJ , the goal becomes numerical evaluation of the critical time step relative to the undamped period, denoted (h=TJ)max, as a function of zJ. The results are depicted in Figure 10.4. Computations are performed over the ‘‘underdamped’’ range 0 < zJ < 1. Two important observations may be made from Figure 10.4. .
.
The numerical stability limits on the time step are comparable to the previously obtained accuracy limits. For example, a critical ratio of 0.2 corresponds to five time steps in a period, while a phase error of 1=300 in a step involves four time steps per period. Subcritical damping (i.e., zJ < 1), whether numerical or viscous, may triple the critical time step for stability compared to the undamped case. But, for a system whose frequencies span a large spectrum, in the high-frequency modes numerical instability associated with extraneous roots is not necessarily suppressed by subcritical damping, whether viscous or ‘‘numerical.’’
Of course, in large finite element systems there are many modes. Suppose the time step is chosen for accuracy in the lower modes but is otherwise limited from below by considerations of computational effort. .
.
.
Above the modes of interest, say the lowest 10, in a large finite element system there likely are modes for which the time step ratio violates the AM stability criteria if the time step is chosen for moderate computational effort. Newmark is A-stable. Even so high-frequency modes are still calculated inaccurately in consequence of the high value of h=TK, K J. Fortunately, in lightly damped systems, at a slight cost in accuracy it is possible to employ numerical damping to attenuate higher-order modes to remove the inaccuracy in the Newmark method. But, as shown above, numerical damping does not appear to be capable of removing the numerical instability in the AM method. Instead, it appears necessary to remove the high-frequency content from the output by another means, such as filtering.
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10.4.6 INTRODUCING NUMERICAL DAMPING METHOD
INTO THE
ADAMS–MOULTON
Of course it may not be wise to insist that, in the absence of (physical) damping (i.e., ‘‘marginal stability’’), the magnitude be preserved in each time step, even in the lower modes. There is a risk that, owing to round-off or truncation errors, the magnitude will grow slightly in a time step. Accordingly, in lightly damped elastic systems there is good reason to introduce ‘‘numerical damping’’ (cf. Zienkiewicz and Taylor, 1989), to ensure that no such growth can occur. A formulation for numerical damping in AM is given below, to ensure that magnitude decay in one time step occurs in fourth order. In terms of the primary eigenvalue the AM formula is equivalent to the Pade approximation (Nicholson and Lin, 2005) 1 4 1 þ ax þ bx2 1 x þ 12 x2 16 x3 þ 24 x 1 þ cx þ dx2
(10:68)
in which the four coefficients a, b, c, d serve to enforce accuracy through fourth order. Now exact agreement through h3 rather than h4 requires (i) h(1): a ¼ c 1, (ii) hð2Þ : b ¼ d c þ 12, (iii) hð3Þ : 0 ¼ d þ 12 c 16. We next require that the remain1 . As will be seen, der at fourth order equal xh4, in which event (iv) x ¼ 12 d 16 c þ 24 the parameter x controls magnitude decay and is selected to represent ‘‘numerical damping.’’ The coefficients a, b, c, d are now expressed in terms of x as 1 1 6x, ðiiiÞ c ¼ 12 þ 12x, ðivÞ d ¼ 12 þ 6x ðiÞ a ¼ 12 þ 12x, ðiiÞ b ¼ 12
(10:69)
We determine the effect of x on magnitude under undamped oscillation. The 1 1 12x ivh þ 1 6x ðivhÞ2 amplitude decreases in a given step if 2 12 < 1, 1 þ 1 þ 12x ivh þ 1 þ 6x ðivhÞ2 2 12 which is equivalent to 0
(10:70)
Consequently, the magnitude decreases if 0 < A þ B ¼ ðvhÞ2 ½24x 24x þ 2xðvhÞ4 ¼ 2x ðvhÞ4
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(10:71)
The numerical value of the magnitude decreases in each time step by Dmag given by AþB 2x ðvhÞ4 þ d h5 (10:72) Dmag ¼ 1 2 1 þ 6x ðivhÞ 1 þ 2 þ 12x ivh þ 12 in which the Taylor approximation has again been used. The coefficients in the numerical integration formula, Equation 10.58a through 10.58f, are now modified to satisfy
7 12 a0
b0 þ b1 þ b2 þ b3 ¼ 1
(10:73a)
a0 þ a1 þ a2 þ a3 ¼ 0 : d (h(0) )
(10:73b)
a1 þ 2a2 þ 3a3 b0 b1 b2 b3 ¼ 0 : d (h)
(10:73c)
1 47 2 þ 12 a1 17 12 a2 12 a3 þ b1 þ 2b2 þ 3b3 ¼ 0 : d (h )
(10:73d)
5 7 1 17 47 3 a1 þ 16 a2 þ 11 12 4 a3 þ 12 b0 þ 12 b1 12 b2 12 b3 ¼ 0 : d (h )
(10:73e)
13 23 5 1 11 4 þ 144 a1 144 a2 þ 157 144 a3 þ 12 b1 6 b2 4 b3 ¼ x : d (h )
(10:73f)
49 144 a0
from which 9 8 9 8 9 9 9 8 8 8 3=4 > a1 > 3 > 9=2 > 3=2 > > > > > > > > > > > > > > > > > > > > > > > > 3=4 > > > > > > 3 > > > > > > > > > > > = < = > < a2 > = < = = > < 3=2 > < 9=2 > þ 3 x a3 ¼ 1=12 þ b0 1=2 þ b3 5=2 > > > > > > > > > > > > > > > > > > > b1 > > > > > > > 1=2 > 6 > 1 > 2 > > > > > > > > > > > > > > > > > > > > > ; : ; : ; ; ; : : : 1=2 6 2 1 b2 and
(10:74)
a0 ¼ 1=12 5=2b0 þ 12 b3 þ 3x
Again choosing the AM values b0 ¼ 3=8, b3 ¼ 1=24, the coefficients are now 5 a0 ¼ 1 þ 3x, a1 ¼ 1 3x , a2 ¼ 3x , a3 ¼ 3x, b1 ¼ 19 24 þ 6x , b2 ¼ 24 6x
(10:75) The desired numerical integration formula, representing the AM formula, modified to incorporate numerical damping, now emerges as 3x ynþ1 yn þ 13x (yn1 yn2 )
h
¼
1 3 19 fnþ1 þ þ 6x fn 1 3x 8 24
5 1 þ 6x fn1 þ fn2 24 24
(10:76)
Clearly, if x is set equal to zero, the AM formula (Equation 10.61) is recovered.
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To illustrate attractive values of x, suppose that the Jth mode is the highest mode for which accurate values are sought. Recall that amplitude reduction is given by 4 4 2x(vj h)4 ¼ 2x 2Tpjh 3200x Thj . To attain low attenuation of this mode, for example, for a magnitude reduction of 1=10000 per time step at 10 steps per period, we obtain x 3 3 104.
10.4.7 AMX: ADAMS–MOULTON METHOD APPLIED WITH ACCELERATION
TO
SYSTEMS
For an elastic system with viscous damping, the conventional finite element equation is written as M€ x þ Dx_ þ Kx ¼ f (t )
(10:77)
in which, as usual, M, D, and K are the n 3 n mass, damping, and stiffness matrices, assumed positive definite and symmetric, x is the n 3 1 global displacement vector, and f(t) is the n 3 1 force vector, which is prescribed as a function of time. To attain a counterpart of the Newmark formula, Equation 10.77 is expressed in state form as "
#( ) ( ) #( ): " D K x_ x_ f þ ¼ I 0 I x x 0
M
0
0
(10:78)
Equation 10.76 implies the substitutions 9 3x > (x_ n1 x_ n2 ) > > = 1 3x 1 > h> 3x x > > > (xn1 xn2 ) > ; : xnþ1 xn þ 1 3x 19 5 ) ( ) (3 1 _ _ _ _ x_ 8 xnþ1 þ 24 þ 6x xn 24 þ 6x xn1 þ 24 xn2 19 5 3 1 x 8 xnþ1 þ 24 þ 6x xn 24 þ 6x xn1 þ 24 xn2 ( ) (3 ) 19 5 1 f 1 8 f nþ1 þ 24 þ 6x f n 24 þ 6x f n1 þ 24 f n2 1 3x 0 0
( ): x_
8 > > > < x_ nþ1 x_ n þ
(10:79)
Obvious reorganization furnishes " 1 M h 0
0 I
#(
x_ nþ1
)
xnþ1
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þ
3 8
"
D
K
1 3x I
0
#(
x_ nþ1 xnþ1
)
( ¼
rnþ1 snþ1
) (10:80)
in which (
rn þ1
)
snþ1
9 8 3x > > > #> _ _ _ x x ) ( x > > n n 1 n 2 = 1 3x 1 M 0 < ¼ > h 0 I > 3x > > > (xn1 xn2 ) > ; : xn 1 3x 5 " #( 19 ) 1 _ D K 24 þ 6x x_ n þ 24 x n 2 þ 6x x_ n1 24 1 þ 5 1 1 3x I 0 19 24 þ 6x xn þ 24 þ 6x xn1 24 xn2 19 5 (3 ) 1 1 8 f nþ1 þ 24 þ 6x f n 24 þ 6x f n1 þ 24 f n2 þ 1 3x 0 (10:81) "
Of course rsnþ1 is known from the solutions at the previous time steps. Use of nþ1 the second block row in Equation 10.80 provides the identification x_ nþ1 ¼ 83 (1 3x ) 1h xnþ1 snþ1 . Upon substitution in the first block row the desired fourth-order counterpart AMX of the Newmark formula, modified to incorporate numerical damping, is now 1 3h2 1 8 8 KD xnþ1 ¼ rnþ1 þ M (1 3x ) þ D snþ1 (1 3x) 8 h 3 3
2 3 h 3 h K KD ¼ M þ Dþ 8 1 3x 8 1 3x
(10:82)
Solution of the linear system Equation 10.82 may be accomplished by conventional finite element procedures consisting of triangularization of the positive definite, symmetric, banded dynamic stiffness matrix KD, together with forward and backward substitution. In fact, the effort to do so is exactly the same as in the Newmark method.
10.4.8 COMMENTS
ON
FILTERING
TO
REMOVE HIGH-ORDER MODES
By being confined to one step, the Newmark and other early methods (Zienkiewicz and Taylor, 1989) give rise only to the primary eigenvalue and avoid numerical instability ensuing from extraneous eigenvalues. However, high-frequency modes may still be computed inaccurately since h=TK becomes large when K > J. To ‘‘attenuate’’ inaccuracy in high-frequency modes in one-step methods, numerical damping has often been introduced at a slight loss of accuracy, for example the Wilson-theta method. We refer to such modified Newmark methods as near-second order. The AMX method attains high accuracy in the lower modes with appreciably larger and=or fewer time steps than in the Newmark method. However, in systems with a broad eigenvalue spectrum, it appears wise to introduce a technique to remove the potentially unstable high-frequency modes. Recall that subcritical damping,
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viscous or numerical, cannot be relied upon to obviate the numerical instability associated with extraneous eigenvalues in AM. Fortunately, even in the presence of transients, in the last two decades high-frequency mode removal has become possible using filters based on the fast and discrete versions of the Wavelet transform (cf. Kaplan, 2002; Bettayeb et al., 2004). The classical Fourier transform introduces one parameter (the frequency) and maps the time domain onto the frequency domain. The Fourier components are globally regular functions such as sinusoids. The Wavelet transform has two parameters, which are scale and time, scale being similar to the reciprocal of the frequency. Wavelets are locally regular functions. In simple wavelet filters, the scale function represents a ‘‘low pass filter’’ covering the lower half of the frequency spectrum, while the wavelet function represents a ‘‘high pass filter’’ covering the upper half of the spectrum (Bettayeb et al., 2004). In the current context the discrete signal of interest corresponds to displacement values computed during the last few time periods, assuming filtering was applied prior to these periods. Doing so sets the present values of the modal amplitudes to zero in the upper half of the spectrum. The reduced signal is then reconstructed in the time domain. The filter is now again applied to the reconstructed signal to partition it into low pass and high pass segments. The process is continued until the reconstructed signal has a spectrum which (a) is in the stable range, and (b) contains the highest natural frequency (mode) for which accurate computations are sought (assuming (b) is compatible with (a)).
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11
Additional Topics in Linear Thermoelastic Systems
Topics addressed in this chapter include linear conductive heat transfer, linear thermoelasticity, incompressible elastic materials, elastic torsion, and buckling.
11.1 TRANSIENT CONDUCTIVE HEAT TRANSFER IN LINEAR MEDIA 11.1.1 FINITE ELEMENT EQUATION The governing equation for conductive heat transfer without heat sources in an isotropic medium is kr2 T ¼ rce T_
(11:1)
in which T is the (absolute) temperature, k is the thermal conductivity, and ce is the coefficient of specific heat at constant strain. We invoke the interpolation model T(t) T0 ¼ wTT (x)FTu(t) in which u(t) is the vector of nodal temperatures (minus T0), while wTT (x) and FT are the thermal counterparts of wT(x) and F in mechanical fields. Also application of the gradient leads to a relation of the form rT ¼ bTT FT u(t), and the finite element equation assumes the form KT u þ MT u_ ¼ q(T) ð ð KT ¼ FTT bT kbTT FT dV, MT ¼ FTT wT rce wTT FTT dV
(11:2)
This equation is parabolic (first order in the time rates), and implies that the temperature changes occur immediately at all points in the domain, but at smaller initial rates away from where the heat is added. This contrasts with the hyperbolic (second order in time rates) solid mechanics equations, in which information propagates into the unperturbed medium as finite velocity waves, and in which oscillatory response occurs in response to a perturbation.
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11.1.2 DIRECT INTEGRATION
BY THE
TRAPEZOIDAL RULE
Equation 11.1 is already in state form since it is first order, and the trapezoidal rule can be applied directly. MT
q þ qn unþ1 un unþ1 þ un þ KT ¼ nþ1 h 2 2
(11:3)
from which KDT unþ1 ¼ rnþ1
(11:4)
KDT ¼ MT þ h2KT rnþ1 ¼ MT un h2KT un h2(qnþ1 þ qn ) For the assumed conditions the dynamic thermal stiffness matrix is positive definite and for the current time step the foregoing equation can be solved in the same manner as in the static counterpart, namely triangularization followed by forward substitution.
11.1.3 MODAL ANALYSIS
IN
LINEAR THERMOELASTICITY
Modes are not of much interest in thermal problems since the modes are not oscillatory or useful to visualize. However, the foregoing equation can still be decomposed into independent single degree of freedom systems. First we note that the thermal system is asymptotically stable. In particular, suppose the inhomogeneous term vanishes and that u at t ¼ 0 does not vanish. Multiplying the foregoing equation by uT and elementary manipulation furnishes that d uT M T u ¼ uT KT u < 0 dt 2
(11:5)
Clearly the product uTMT u decreases continuously. But it vanishes only if u vanishes. Next, to examine the modes assume a solution of the form u(t) ¼ u0j exp(lj t). The eigenvectors u0j satisfy mTj , j ¼ k kTj , j ¼ k T T , u0j KT u0k ¼ (11:6) u0j MT u0k ¼ 0, j 6¼ k 0, j 6¼ k and we call mTj and kTj the jth modal thermal mass and jth modal thermal stiffness. We may also form the modal matrix Q ¼ [u01 u0n], and again 2
mTj
6 6 0 Q MT Q ¼ 6 6 : 4 : 0 T
0 mTj : : :
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: : : : :
3 : 0 : :7 7 : :7 7, : :5 : :
2
kTj
6 6 0 Q KT Q ¼ 6 6 : 4 : 0 T
0 kTj : : :
: : : : :
3 : 0 : :7 7 : :7 7 (11:7) : :5 : :
Let j ¼ Q1u and g(t) ¼ QTq(t). Pre- and post-multiplying Equation 11.2 with QT and Q, respectively, furnishes the decoupled equations mTj j_j þ kTj jj ¼ gj
(11:8)
Supposing for convenience that gj is a constant, the general solution is of the form kTj jj (t ) ¼ jj 0 exp t mTj
! þ
ðt 0
! kTj exp ðt t Þ gj (t ) dt mTj
(11:9)
illustrating the monotonically decreasing nature of the free (g ¼ 0) response. Now there are n uncoupled single degrees of freedom.
11.2 COUPLED LINEAR THERMOELASTICITY 11.2.1 FINITE ELEMENT EQUATION The classical theory of coupled thermoelasticity accommodates the fact that the thermal and mechanical fields are coupled. For isotropic materials, assuming that temperature only affects the volume of an element, the stress–strain relation is Sij ¼ 2mEij þ l(Ekk a(T T0 ))dij
(11:10)
in which a denotes the volumetric thermal expansion coefficient. The equilibrium @S uj . The Principle of Virtual Work (Chapter 5) implies equation is repeated as @xiji ¼ r€ that ð
ð ð ð dEij 2mEij þ lEkk dij dV0 þ dui r€ ui dV0 al dEij (T T0 )dij dV0 ¼ dduj tj dS0 (11:11)
in which, as before, tj refers to the traction vector. Now consider the interpolation models u(x,t) ¼ NT(x)g(t), Eij ! e ¼ BT(x)g(t), T T0 ¼ vT(x)u(t), rT ¼ BTT (x)u(x) (11:12) in which e is the strain written in conventional finite element notation as a column vector. As before, N is the shape function matrix, its thermal counterpart is v(x), B is the strain–displacement matrix, and BT is its thermal counterpart. Now familiar procedures furnish the finite element equation M€ g(t) þ Kg(t) Su(t) ¼ f(t),
ß 2008 by Taylor & Francis Group, LLC.
ð S ¼ al BvT dV0
(11:13)
The quantity S may be called the thermomechanical stiffness matrix. If there are nm displacement degrees of freedom and nt temperature degrees of freedom, the quantities appearing in the equation have dimensions according to M, K : nm nm , g(t), f(t) : nm 1,
S : nm nt ,
u(t) : nt 1
We next address the thermal field. The energy balance equation, including coupling to mechanical effects, is given by _ kr2 T ¼ rce T_ þ alT0 tr(E)
(11:14)
Application of the usual variational methods and interpolation models implies that _ þ T 0 ST g(t) _ ¼ q, KT u(t) þ MT u(t)
ð q ¼ vn q dS
(11:15)
Now consider the special case in which T is constant. Then, at the global level, T _ u(t) ¼ T0 K1 The thermal field is thus eliminated at the global level, giving T S g(t). the new governing equation as T _ þ Kg(t) ¼ f(t) M€ g(t) þ T0 SK1 T S g(t)
(11:16)
Conductive heat transfer is thereby seen to be analogous to damping. The thermomechanical system is now asymptotically stable (positive effective damping) rather than asymptotically marginally stable (no effective damping). We next express the global equations in state form as Q1 z_ þ Q2 z ¼ f
(11:17)
in which 2
3 0 1 g_ M 0 0 0 5, z ¼ @ g A Q1 ¼ 4 0 K u 0 0 MT =T0 2 3 0 1 0 K S f Q2 ¼ 4 K 0 0 5, f ¼ @ 0 A q=T0 ST 0 KT =T0 Clearly Equation 11.17 can be integrated numerically using the trapezoidal rule: Q1 þ h2 Q2 znþ1 ¼ Q1 h2 Q2 zn þ 12 ½f nþ1 þ f n
(11:18)
We consider asymptotic stability, for which purpose it is sufficient to take f ¼ 0, z(0) ¼ z0. Upon pre-multiplying Equation 11.17 by zT, we obtain
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d 1 T z Q1 z ¼ zT Q2 z dt 2 ¼ zT 12 Q2 þ QT2 z ¼ uT KT u
(11:19)
and z must be real. Assuming that u 6¼ 0, it follows that z # 0, and hence the system is asymptotically stable.
EXAMPLE 11.1 Find the exact solution for a circular rod of length L, radius r, mass density r, specific heat ce, conductivity k, and cross-sectional area A ¼ pr2. The initial temperature is T0, and the rod is built into a large wall at fixed temperature T0 (see Figure 11.1). However, at time t ¼ 0, the temperature T1 is imposed at x ¼ L. Compare the exact solution to the one- and two-element solutions. Note that for a one-element model kA rce AL _ u(L,t) þ u(L,t) ¼ q(L) L 3
SOLUTION (i) Exact solution The governing equation for conductive heat transfer in one dimension is given by k
@2T @T ¼ rce @x2 @t
We seek to solve this equation using ‘‘Separation of Variables.’’ If a ¼ k=rce . @ 2 T 1 @T ¼ @x2 a @t 2 #
#
@ T 1 @T Now let T# ¼ T [T0 þ x(T1 T0)=L]. Now with @x2 ¼ a @t , T# ð0Þ ¼ T# ðLÞ ¼ 0. ^ with the Next assume the spatial-temporal decomposition T# (x,t) ¼ X (x)T(t), consequence that ^_ X 00 1 T ¼ l2j , lj the jth eigenvalue ¼ ^ aT X
The function X(x) corresponding to lj is now denoted as Xj(x) and similarly for Tj(t). The two functions satisfy
L r T0
FIGURE 11.1 One-element thermal conductor.
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T1, t > 0
Xj00 þ l2j Xj ¼ 0 ! Xj ¼ Aj cos(lj x) þ Bj sin(lj x)
T0j þ al2j Tj ¼ 0 ! Tj ¼ Cj exp al2j t in which Aj, Bj, and Cj remain to be determined. The eigenvalues satisfy sin(ljL) ¼ 0 which implies that lj ¼ jp=L. With Dj ¼ AjCj and Ej ¼ BjCj, the solution assumes the form Tðx,t Þ ¼
X
Dj cos( jpx=L) þ Ej sin( jpx=L) exp aj2 p2 t=L
j
þ T0 þ
x(T1 T0 ) L
The initial condition T(x,0) ¼ T0 and standard application of the orthogonality properties of the eigenfunctions (e.g., Hildebrand, 1976) cos( jpx=L) and sin(jpx=L) permit determination of the coefficients Dj and Ej. (ii) Finite element solution For a thermal element with natural coordinates j ¼ 1 at x ¼ xe, and j ¼ þ1 at x ¼ xeþ1, 1 1 1 1 1 1 ¼ fTT ¼ ( 1 j ), FT ¼ 1 1 2 1 1 Also dx ¼ L2 dj, and
d dx
¼ L2
d dj .
Consequently, the thermal stiffness (conductance)
matrix is given by ð KT ¼ FTT bT kbTT FT dV þ1 ð 1 1 1 1 1 2 0 2 L k ( 0 1 )A dj 4 1 1 L 1 L 2 1 1 1 kA 1 1 ¼ L 1 1 ¼
Continuing, the thermal mass (capacitance) matrix is now ð MT ¼ FTT wT rce wTT FT dV þ1 ð 1 1 1 rAce L 1 1 ( 1 j ) dj 8 1 1 j 1 1 1 rAce L 1 1=2 ¼ 3 1=2 1
¼
(iia) One-element solution: conductor built on the right-hand side (rhs) Here, the interpolation model is T T0 ¼ xu(L,t)=L. Consequently, the thermal stiffness and mass matrices degenerate to scalars: KT ¼
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kA , L
MT ¼
rAce L 3
Substitution into Equation 11.2 furnishes the one-element equation kA rce AL _ u(L,t) þ u(L,t) ¼ q(L) L 3 This equation is to be solved for u(L,t) with an assumed value of q(L) adjusted to enforce the constraint T(L) ¼ T1. (iib) Two-element solution Now the length of each element is L=2, and so kA rAce (L=2) 1 1 1 1=2 , MT ¼ KT ¼ 1=2 1 (L=2) 1 1 3 After simple manipulation the assembled global finite element equation emerges as 1 1: 2 30 2 30 u(0,t) u(0,t) 1 1 0 1 1=2 0 kA 6 rAc (L=2) C C 7B 6 7B e 4 1 2 1 5@ u(L=2,t) A þ 4 1=2 2 1=2 5@ u(L=2,t) A (L=2) 3 u(L,t) u(L,t) 0 1 1 0 1=2 1 0 1 q(L) B C ¼@ 0 A q(L) But u(0,t) ¼ 0, and in consequence 2 1 u(L=2,t) u(L=2,t) : kA rAce (L=2) 2 1=2 þ (L=2) 1 1 3 u(L,t) 1=2 1 u(L,t) 0 ¼ q(L) as expected.
11.2.2 THERMOELASTICITY
IN AN
ELASTIC CONDUCTOR
Consider a thermoelastic rod which is built into a large rigid, nonconducting temperature reservoir at x ¼ 0. The force f0 and heat flux q0 are prescribed at x ¼ L. A single element is used to model the rod. Now u(x,t) ¼ xg(t)=L, E(x,t) ¼ g(t)=L, T T0 ¼ xu(t)=L, dT=dx ¼ u(t)=L (11:20) Ð The thermoelastic stiffness matrix becomes S ¼ al BvT dV ! S ¼ alA=2. The governing equations are now rAL EA 1 g alAu ¼ f0 g€ þ 3 L 2 1 rce AL _ 1 kA 1 uþ u þ alAg_ ¼ q0 T0 3 T0 L 2
EXAMPLE 11.2 State the equations of a thermoelastic medium in state form.
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(11:21)
SOLUTION The equations of a thermoelastic medium are: M€ g(t) þ Kg(t) Su(t) ¼ f(t) 1 1 1 _ þ ST g(t) _ ¼ q KT u(t) þ MT u(t) T0 T0 T0 M!
rAL EA , K! , 3 L
KT !
kA , L
MT !
rce AL , 3
S¼
laA 2
_ we have On converting state form using p ¼ g, 2 30 1 : 2 30 1 0 1 f p p M 0 0 0 K S 6 7B C 6 7B C B C 0 5@ g A þ4 K 0 0 5@ g A ¼ @ 0 A 40 K T q=T0 u u 0 0 MT =T0 S 0 KT =T0 If the trapezoidal rule with timestep h is applied, an algebraic equation arises in which appears the dynamic thermoelastic stiffness matrix 2 h h 3 M 2K 2S 6 0 7 KDT ¼ 4 h2 K K 5 h T 2S
0T
KT
Some manipulation serves to verify that KDT satisfies the triangularization 3 2 M1=2 0T " #" # " # 7 6 1=2 K 0 KDT ¼6 7 KM1=2 h2 KM 5 4h 1=2 1=2 þ M K M S 4 ST M1=2 2 ST M1=2 0T KT 2 3 h 1=2 M1=2 K M1=2 S M 5 2 4 T 0 I
11.3 INCOMPRESSIBLE ELASTIC MEDIA Rubber and polymer-based plastics, as well as biological tissue, frequently exemplify the internal constraint of incompressibility. For a compressible elastic material, the isotropic stress Skk and the volume (dilatational) strain Ekk are related by Skk ¼ 3kEkk, in which k ¼ E=[3(12v)] is recognized as the bulk modulus. Clearly, as v !1=2, the (hydrostatic) pressure p ¼ Skk=3 needed to attain a finite compressive volume strain (Ekk < 0) becomes infinite using elastic relations in their current form. However, we will see that this difficulty is avoided by correctly incorporating incompressibility through a limiting process as v ! 1=2. Consider the case of plane strain implying that Ezz ¼ 0. The tangent modulus matrix D is now readily found as
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1 1 0 Exx Sxx C C B B @ Syy A¼ D@ Eyy A,
2
0
Szz
Ezz
D¼
1 v2
E 6 4 v(1 þ v) (1 þ v)(1 2v) 0
v(1 þ v) 1 v2 0
3 0 7 0 5 1 2v (11:22)
Clearly, in the current form D becomes unbounded as v ! 1=2. Further, suppose that for a material through to be nearly incompressible v is estimated as 0.495 while the correct value is 0.49. It might be supposed that the estimated value is a good approximation for the correct value. However, for the correct value (12v)1 ¼ 50, but for the estimated value (12v)1 ¼ 100, implying 100% error. The problem of unbounded magnitude is addressed as follows. In an incompressible material a pressure field p(x) arises which serves to enforce the incompressibility constraint. Since the trace of the strains vanishes everywhere, the strains are not sufficient to determine the stresses. However, the strains together with the pressure are sufficient. In FEA, a general interpolation model is used at the outset for the displacement field. Another interpolation model must be introduced for the pressure field. Owing to the fact that pressures are stress variables, the order of the interpolation should be one degree lower than for the displacement interpolation model (Hughes, 2000). The Principle of Virtual Work is now expressed in terms of the displacements and pressure, and an adjoining equation is introduced to enforce the incompressibility constraint a posteriori. The pressure may be shown to serve as a Lagrange multiplier, in which event the displacement vector and the pressure are varied independently. In incompressible materials, to preserve finite stresses we suppose that the second Lamé coefficient satisfies l ! 1 as tr(E) ! 0, in such a way that the product is an indeterminate quantity denoted by p: l tr(E) ! p(x,t)
(11:23)
The Lamé form of the elastic constitutive relations is replaced by Sij ¼ 2mEij pdij
(11:24)
together with the incompressibility constraint Eij dij ¼ 0. There now are two independent principal strains and the pressure with which to determine the three principal stresses. @w , In a compressible elastic material the strain energy function w satisfies Sij ¼ @E ij and the domain term in the Principle of Virtual Work may be rewritten as Ð Ð dEij Sij dV ¼ dw dV. The elastic strain energy is given by w ¼ mEij Eij þ l2 E2kk . For reasons to be seen shortly, we introduce the augmented strain energy function w0 ¼ mEij Eij pEkk and assume the variational principle ð ð ð 0 T u dV0 ¼ duT t dS0 dw dV0 þ du r€
(11:25)
(11:26)
Now considering u and p to vary independently, the integrand of the first term becomes dw0 ¼ dEij[2mEij pdij] dpEkk, furnishing two variational relations
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ð
ð
ð
u dV0 ¼ duT t dS0 tr ðdESÞ dV0 þ du r€ ð dpEkk dV0 ¼ 0 T
(11:27a) (11:27b)
The first relation may be recognized as a counterpart the Principle of Virtual Work (variational principle for the displacement field), and the second equation serves to enforce the internal constraint of incompressibility (variational principle for the pressure field). Next introduce the interpolation models u ¼ NT (x)g(t), e ¼ BT (x)g(t) Ekk ¼ bT (x)g(t),
p(x,t) ¼ jT (x)p(t)
(11:28)
Substitution serves to derive that M€ g(t) þ Kg(t) Sp(t) ¼ f(t) ð S ¼ bjT dV0 , ST g(t) ¼ 0 Assuming these equations apply at the global 2 3 0 1 2 _ g(t) M K M 0 0 d 4 0 K 0 5 @ g(t) A þ 4 K 0 dt p(t) 0 T 0T 0 ST 0T
(11:29)
level, state form is expressed as 30 1 0 1 _ S g(t) f(t) 0 5@ g(t) A ¼ @ 0 A (11:30) p(t) 0 0
The second matrix is antisymmetric except for the upper left-hand diagonal term. Further, the system exhibits marginal asymptotic stability; namely, if f(t) ¼ 0 while _ g(0), g(0), and p(0) do not all vanish, then 2 30 2 13 _ g(t) M 0 0 d 41 T (11:31) g_ (t) gT (t) pT (t) 4 0 K 0 5@ g(t) A5¼ 0 dt 2 p(t) 0T 0T 0
EXAMPLE 11.3 Put the following equations in state form, apply the trapezoidal rule, and triangularize the ensuing dynamic stiffness matrix assuming that the triangular factors of M and K are known. M€ g þ Kg Sp ¼ f,
ST g ¼ 0
SOLUTION The given equations are expressed in state form as follows: 1: 2 1 0 1 2 30 30 _ _ g(t) g(t) f(t) M 0 0 M K S 4 0 K 0 5@ g(t) A þ 4 K 0 0 5@ g(t) A ¼ @ 0 A p(t) p(t) 0 0 0 0 0 ST 0
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Using the trapezoidal rule and p ¼ g_ gives 1 2 1 0 30 1 30 1 1 1 M K S h (pnþ1 pn ) 2 (pnþ1 þ pn ) 2 (f nþ1 þ f n ) C C B B 7 1 6 7B 1 C C 6 C B 6 0 K 0 7B C (g gn ) C þ 6 (g þ gn ) C ¼ B K 0 0 7 0 5B 4 5B A A 4 A @ @ h nþ1 @ 2 nþ1 T 1 1 0 0 0 0 S 0 0 h (pnþ1 pn ) 2 (pnþ1 þ pn ) 2
M 0 0
The second row implies that pnþ1 ¼ 2h (gnþ1 gn ) pn . Thence the first row becomes
1 2 h M h (gnþ1
gn ) 2pn þ 12 K(gnþ1 þ gn ) 12 S(pnþ1 þ pn ) ¼ 12 (f nþ1 þ f n )
Multiplying throughout by h2=2 and rearranging results in h
i 2 2 2 M þ h4 K gnþ1 h4 Spnþ1 ¼ h4 ½f nþ1 þ f n Kgn þ Spn þ M½gn þ hpn (11:32a)
The third row, after multiplying by h2=2, becomes h2 4
ST gnþ1 ¼ h4 ST gn 2
(11:32b)
The equations are restated in vector–matrix notation as "
2
M þ h4 K h2 4
ST
2
h4 S
#
gnþ1
!
pnþ1
0
!
gnþ1
¼
h4 ST gn 2
in which gnþ1 ¼
h2 ½f nþ1 þ f n Kgn þ Spn þ M½gn þ hpn 4
The dynamic stiffness matrix now is " KD ¼
2
M þ h4 K h2 4
ST
2
h4 S
#
0
Next, KD is decomposed into a product of a lower triangular and an upper triangular matrix. We first write "
2
M þ h4 K h2 4
ST
2
h4 S
"
#
L11
0
L21
L22
L11 LT11 ¼ M þ
h2 K 4
0
¼
#"
U11
U12
0
U22
#
Setting U11 ¼ LT11 , and L22 ¼ I gives
which may be triangularized to find L11 since, the triangular factors of M and K are known. Further ß 2008 by Taylor & Francis Group, LLC.
U12 ¼ L21 ¼
h2 1 L S 4 11
h2 T T S L11 4
U22 ¼ L21 U12 ¼
h4 T T 1 S L11 L11 S 16
Accordingly, the triangularization is expressed as " #2 T 0 L L11 4 11 KD ¼ LU ¼ 2 T T h I 0 4 S L11
h4 L1 11 S 2
h4 T T 1 16 S L11 L11 S
3 5
EXAMPLE 11.4 In an element of an incompressible square rod of cross-sectional area A, it is necessary to consider the displacements v and w. Suppose the length is L, the lateral dimension is Y, and the interpolation models are linear for the displacements (u linear in x, with v, w linear in y) and constant for the pressure. Show that the finite element equation assumes the form 2
2mA=L 4 0 A
1 0 1 30 A f u(L) 2AL=Y 5@ v(Y) A ¼ @ 0 A 0 p 0
0 4mAL=Y 2 2AL=Y
and that this implies the relation 3m u(L) L ¼ f.
SOLUTION The interpolation models are given by x u ¼ u(L), L
v¼w¼
y v(Y) Y
Now consider the virtual work term for virtual strain energy. ð d«ij 2m«ij dV ¼ 2mALd«ij «ij ¼ 2mAL d«xx «xx þ d«yy «yy þ d«zz «zz du(L) u(L) dv(Y) v(Y) þ2 ¼ 2mAL L L Y Y " # ! u(L) 1=L2 0 ¼ 2mAL(du(L) dv(Y)) v(Y) 0 2=Y 2 Ð But d«ij 2m«ij dV ¼ dgT Kg. Accordingly, the foregoing equation implies g¼
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u(L) , v(Y)
K¼
2mA=L 0
0 4mAL=Y 2
Next ð d«kk p dV ¼ ALp
du(L) dv(Y ) þ2 ¼ ( du(L), L Y
u(L) dv(Y ))AL p v(Y )
Ð But, d«kk p dV ¼ dgT Sp. Also, if we assume that p is a constant, we have A p ¼ p, S ¼ 2AL=Y After neglecting the inertia term in Equation 11.29, the equations of interest become ST g ¼ 0
Kg Sp ¼ f, Use of state form furnishes
K ST
S 0
g p
f ¼ 0
Accordingly 2
2mA=L 4 0 A
0 4mAL=Y 2 2AL=Y
30 1 0 1 A u(L) f 2AL=Y 5@ v(Y) A ¼ @ 0 A 0 p 0
The foregoing matrix product is equivalent to the three relations 2mA u(L) Ap ¼ f L 4mAL 2AL v(Y) p¼0 Y2 Y Au(L) þ
2AL v(Y) ¼ 0 Y
(11:33a) (11:33b) (11:34)
Multiplying Equation 11.34 by 2m Y gives rise to 2mA 4mAL u(L) þ v(Y) ¼ 0 Y Y2
(11:35)
Next subtract Equation 11.35 from Equation 11.33b. u(L) þ
2AL p ¼ 0, Y
m p ¼ u(L) L
Substituting the last expression in Equation 11.34 now produces the relation 3mA
u(L) ¼f L
This agrees exactly with the uniaxial result EA u(L) L ¼ f in the incompressible case since n ¼ 1=2 and m ¼ E=2(1 þ n).
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11.4 TORSION OF PRISMATIC BARS 11.4.1 BASIC RELATIONS Figure 11.2 illustrates a member experiencing torsion. The member in this case is cylindrical with length L and radius r0. The base is fixed and a torque is applied at the top surface which causes the member to twist. The twist at height z is u(z), and at height L it is u0. Ordinarily, in finite element problems so far considered, the displacement is the basic unknown. It is approximated by an interpolation model, from which an approximation for the strain tensor is obtained. Then an approximation for the stress tensor is obtained using the stress–strain relations. The nodal displacements are then solved by an equilibrium principle, in the form of the Principle of Virtual Work. In the current problem an alternative path is followed, in which the stress tensor, or more precisely a stress potential, is the unknown. The strains are determined from the stresses. However, for arbitrary stresses satisfying equilibrium the strain field may not be compatible, enabling it to determine a displacement field which is unique to within a rigid body translation and rotation. The compatibility condition (cf. Chapter 5) is now enforced, furnishing a partial differential equation known as the Poisson equation. A variational argument is applied to furnish a finite element expression for the torsional constant of the section. For the member before twist, consider points X and Y at angle f and at radial position r. Clearly, X ¼ r cos f and Y ¼ r sin f. Twist induces a rotation through angle u(z) but it does not affect the radial position of a given point. Now x ¼ r cos (f þ u), y ¼ r sin (f þ u). Use of double angle formulae furnishes the displacements, and restriction to small angles u furnishes, to first order, z T q0 r0 Section after twist Section before twist
L
q
x
z y f x
FIGURE 11.2 Twist of a prismatic rod.
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u ¼ Yu, v ¼ Xu
(11:36)
It is also assumed that torsion does not increase the length of the member, which is attained by requiring that w only depends on X and Y. The quantity w(X,Y) is called the warping function. Elementary manipulation serves to verify that all strains vanish except Exz and Eyz, for which Exz ¼
1 @w @u y , 2 @x @z
Eyz ¼
1 @w @u þx 2 @y @z
(11:37)
Equilibrium requires that @Sxz @Syz þ ¼0 @x @y
(11:38)
with all other stresses vanishing owing to the assumed displacement field. The equilibrium relation may be identically satisfied by a potential function c for which Sxz ¼
@c , @y
Syz ¼
@c @x
(11:39)
It remains to satisfy the compatibility condition, to ensure that the strain field arises from a displacement field which is unique to within a rigid body translation and rotation. (Compatibility is automatically satisfied if the displacements are considered the unknowns and are approximated by a continuous interpolation model. Here the stresses are the unknowns.) From the stress–strain relation Exz ¼
1 1 @c Sxz ¼ , 2m 2m @y
Eyz ¼
Compatibility (integrability) now requires that
1 1 @c Syz ¼ 2m 2m @x
@2 w @x @y
(11:40)
@ w ¼ @y @x, from which 2
@ 1 @c 1 du @ 1 @c 1 du þ y þ x ¼0 @y 2m @y 2 dz @x 2m @x 2 dz
(11:41)
furnishing Poisson’s equation for the potential function c: @2c @ 2c du þ ¼ 2m @x2 @y2 dz
(11:42)
For boundary conditions, it is assumed that the lateral boundaries of the member are traction free. Now the assumed displacement fields already imply that tx ¼ 0 and ty ¼ 0 on the lateral boundary S. For traction tz to vanish requires that
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tz ¼ nx Sxz þ ny Syz ¼ 0
on S
(11:43)
Upon examining Figure 11.3 it may be seen that nx ¼ dy=ds and ny ¼ dx=ds, with s the arc length along the boundary at z. Consequently, dy dx Sxz Syz ds ds dy @c dx @c ¼ þ ds @y ds @x dc ¼ ds
tz ¼
(11:44)
Now dc ds ¼ 0 on S, and therefore c is a constant on S, and it may, in general, be taken as zero. Since c is elsewhere the unknown, the vanishing values on the exterior boundary are analogous to constraints in conventional displacement-based problems. We next consider the total torque T on the member. Figure 11.4 depicts the cross section at z in which the torque dT on the element at x and y is given by dT ¼ xSyz dx dy ySxz dx dy dc dc ¼ x y dx dy dx dy
(11:45)
Integration furnishes ð
dc dc T ¼ x þy dx dy dx dy ð d(xc) d(yc) dx dy þ c þ dx dy ¼ dx dy dx dy ð ð xc ¼ r dx dy þ 2 c dx dy yc
y dy
dy
z
n ds
–dx
y
FIGURE 11.3 Illustration of geometric relation.
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nx = cos χ = dy/ds ny = s i n χ = dx/ds
(11:46)
z
dy Syz
dx
Sxz
x
y
y x
FIGURE 11.4 Evaluation of twisting moment.
Ð Application of the divergence theorem to the first term leads to c xnx þ yny ds, which vanishes since c vanishes on S. Finally ð T ¼ 2 c dx dy
(11:47)
We now apply variational methods to the Poisson equation, considering the stress potential function c to be the unknown. ð
dc½r rc þ 2mu0 dx dy ¼ 0
(11:48)
in which u0 ¼ du=dx. Integration by parts, use of the divergence theorem and imposition of the ‘‘constraint’’ c ¼ 0 on S furnishes ð ð ðrdcÞ rc dx dy ¼ dc2mu0 dx dy (11:49) The integrals are to be evaluated over a set of small elements. In the eth element approximate c as vTc (x,y)Ce he in which vw is a vector with dimension (number of rows) equal to the number of nonvanishing nodal values of c. The gradient rc has a corresponding (derived) interpolation model rc ¼ bTc (x,y)Ce he , in which bw is a matrix. The finite element counterpart of the Poisson equation at the element level now is written as 0 (e) K(e) c he ¼ 2mu f c
ð T T ¼ C K(e) e bc bc dx dy Ce , c
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ð T f (e) ¼ C e vc (x,y) dx dy c
(11:50)
and the stiffness matrix should be nonsingular since the constraint c ¼ 0 on S has already been used. It follows that, globally, hg ¼ 2mu0 Kc(g)1 f (g) c . Next, the torque satisfies ð T ¼ 2c dx dy ¼ 2hTg f (g) c ¼ 4mu0 f c(g)T Kc(g)1 f (g) c
(11:51)
In the theory of torsion it is common to introduce the torsional constant J for which T ¼ mJu0 . It follows immediately that J ¼ 4fc(g)T Kc(g)1 fc(g).
EXAMPLE 11.5 Figure 11.5 depicts a single triangular element in the cross section of a shaft experiencing torsion. Assuming the interpolation model 2
1 x1 c(x,y) ¼ ( 1 x y )4 1 x2 1 x3
31 0 1 y1 c1 y2 5 @ c2 A y3 c3
find Kc, fc, and the torsional constant J for this element.
y
(2,3)
(3,2)
(1,1) x
FIGURE 11.5 Triangular element in shaft cross section.
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SOLUTION Here vTT (x,y) ¼ ( 1 x y ) 2 3 31 2 31 2 5 1 1 1 1 1 1 x1 y1 16 7 6 7 6 7 C ¼ 4 1 x2 y2 5 ¼ 4 1 3 2 5 ¼ 4 1 2 1 5 3 1 1 2 1 2 3 1 x3 y3 0 1 c1 0 1 0 h ¼ @ c2 A, bTT ¼ rvTT (x,y) ¼ 0 0 1 c3 Also ð KT ¼ CT bT bTT dx dy C 2 3 2 32 5 0 0 0 ð 5 1 1 16 6 7 76 ¼ 4 1 2 1 54 0 1 0 5 dx dy4 1 9 1 0 0 1 1 1 2
1 2 1
3 1 7 1 5 2
But ð
1 x1 1
dx dy ¼ det 1 x2 2 1 x3
y1
3 y2
¼ Area of the triangle ¼ 2
y3
Consequently, 2 2 1 14 KT ¼ 1 5 6 1 4
3 1 4 5 5
Further, ð fc ¼ C
T
2 5 14 vc (x,y) dx dy ¼ 1 3 1
1 2 1
3 0 1 1 ð 1 1 5 @ x Adx dy 2 y
and ð
ð x dx dy ¼ (x x0 þ x0 ) dx dy
in which (x0,y0) is the centroid of the triangle and is given by x0 ¼
x1 þ x2 þ x3 ¼ 2, 3
ß 2008 by Taylor & Francis Group, LLC.
y0 ¼
y1 þ y2 þ y3 ¼2 3
Continuing, ð
ð 3 (x x0 ) dx0 dy0 ¼ 0, and dx0 dy0 ¼ Area of the triangle ¼ 2
Ð Ð Finally x dx dy ¼ 3, y dx dy ¼ 3, and
0 1 1 1@ A fT ¼ 1 2 1
The torsional constant J for the element is now found to be given by J ¼ 4f Tc K1 c fc ¼ 0:6
EXAMPLE 11.6 Find the torsional constant in the circular shaft shown below using four axisymmetric elements, as depicted in Figure 11.6.
SOLUTION Since the configuration is axisymmetric and the constraints c2 ¼ c3 ¼ c4 ¼ c5 ¼ 0 are enforced a priori, the linear interpolation model in each element is the same and may be taken as c(r) ¼
r0 r c1 r0
We conclude that vw ! r0 r, Cw ¼
1 , r0
hw ¼ c1 ,
r¼
d , dr
bw ¼ 1:
r0 y 5
4 2
3
r
c
1 2
1
3
FIGURE 11.6 Torsional constant of a circular shaft.
ß 2008 by Taylor & Francis Group, LLC.
4
x
Ð pr02 (g) 1 Simple manipulation yields K(g) w ¼ p and f w ¼ r0 (r0 r)r dr dx ¼ 3 , from which (g)1 (g) J ¼ 4f (g)T fw w Kw
¼ 49 pr04 The exact answer is J ¼ 12 pr04 , so that the one degree of freedom finite element model is accurate to within 10%.
11.5 BUCKLING OF ELASTIC BEAMS AND PLATES 11.5.1 EULER BUCKLING 11.5.1.1
OF
BEAM COLUMNS
Static Buckling
Under in-plane compressive loads, the resistance of a thin member (beam or plate) can be reduced progressively, culminating in buckling. There are two equilibrium states that the member potentially can sustain—compression only, or compression with bending. The member will ‘‘snap’’ to the second state if it involves less ‘‘potential energy’’ than the first state. The notions explaining buckling are addressed in detail in subsequent sections. For present purposes we focus on beams and plates, using classical equations in which, by retaining lowest order ‘‘linearized’’ corrections for geometric nonlinearity, in-plane compressive forces appear. For the beam shown in Figure 11.7, the classical Euler buckling equation is EIwiv þ Pw00 þ rA€ w¼0
(11:52)
and P is the axial compressive force. The interpolation model for w(x) has the form w(x) ¼ wT(x)Fg. Following the usual variational procedures (integration by parts) furnishes ð dwrA€ w dx ! dgT M€ g, ð
ð M ¼ FT f(x)rAfT (x) dV F
(11:53)
ð ð dw EIwiv þ Pw00 dx ¼ dw00 EIw00 dx dw0 Pw0 dx ½(dw)(Pw0 EIw000 ) 0 ½(dw0 )(EIw00 ) 0 L
z
L
y E, I, A, L, ρ
Q0 M0
FIGURE 11.7 Euler buckling of a beam column.
ß 2008 by Taylor & Francis Group, LLC.
P
x
At x ¼ 0, both dw and dw0 vanish. Note that the shear force V and the bending 0 00 moment M are identified as V ¼ EIw and M ¼ EIw’’. The ‘‘effective shear 00 0 force’’ Q is defined as Q ¼ Pw0 EIw . For the specific case illustrated in Figure 11.7, for a one-element model we may use the interpolation model w(x) ¼ x2
x
3
L2 2L
L3 3L2
1
w(L) g(t), g(t ) ¼ w0 (L)
(11:54)
As seen in Example 11.7, the mass matrix is shown after some algebra to be " M ¼ rALK0 ,
^0 ¼ K
13 35 11 210 L
11 210 L 1 2 105 L
# (11:55)
Similarly, ð
ð
2 P^ dw0 Pw0 dx ¼ dgT K 1 g, L
dw00 EIw00 dx ¼ dgT
^1 ¼ 4 K
EI ^ ^2 ¼ K2 g, K L3
6 5
1 10 L
1 10 L
2 2 15 L
"
3 5
#
12
6L
6L
4L2
(11:56)
The governing equation is written in finite element form as
EI ^ P^ € ¼ f, K K 2 1 g þ r ALK0 g L3 L
f¼
Q0 M0
(11:57)
€ ¼ 0 in which case the solution has the form In a static problem, g
^ 2 PL2 K ^1 cof K EI g¼ f ^1 ^ 2 PL2 K det K EI
(11:58)
in which cof() denotes the cofactor matrix, and clearly g ! 1 for values of ^ 2 PL2 K ^ 1 ¼ 0. which render det K EI 11.5.1.2
PL2 EI
Dynamic Buckling
In a dynamic problem, it may be of interest to determine the effect of P on the resonance frequency. Suppose that f(t) ¼ f0 exp(ivt), in which f0 is a known vector. The displacement function satisfies g(t) ¼ g0 exp(ivt), in which the amplitude vector g0 satisfies
ß 2008 by Taylor & Francis Group, LLC.
EI ^ P^ 2 ^ K2 K1 v rALK0 g0 ¼ f 0 L3 L
(11:59)
Resonance occurs at a frequency v0 for which EI ^ P^ 2 ^ det 3 K2 K1 v0 rALK0 ¼ 0 L L
(11:60)
1 ^ K0 Clearly, v02 is an eigenvalue (often the minimum) of the matrix rAL EI 1=2 P 2 ^ ^ ^ . The resonance frequency v0 is reduced by the presence of P L3 K2 L K1 K0 and vanishes precisely at the critical value of P. 1=2
EXAMPLE 11.7 ^ 0, K ^ 1, K ^ 2 for a cantilevered beam modeled as one element. Derive the matrices K
SOLUTION Enforcing the clamped constraints a priori, the interpolation model is
w (x) ¼ T
(x2
x
L2 F¼ 2L
3 ),
L3 3L2
1
1 3L L2 ¼ 3 L 2 L
The mass matrix satisfies ð M ¼ FT w(x)rAwT (x) dx F # 0 1 2 ðL x2 @ ArA( x2 L 0 x3
" 1 3L ¼ 6 L L2 2 ¼ rAL4
13 35
11 210 L
11 210 L
1 2 105 L
2 x3 ) dx4
3L
2 L
3 5
^ 0, and so But M ¼ rALK " ^0 ¼ K
13 35
11 210 L
11 210 L
1 2 105 L
#
Continuing, w0 ¼ bT Fg,
ß 2008 by Taylor & Francis Group, LLC.
L2
bT ¼ @=@x wT (x) ¼ (2x 3x2 )
3 5
ð
ð dw0 Pw0 dx ¼ dgT FT bPbT dx Fg " P 3L ¼ dg 6 L L2 T
¼ dgT
P L
"
#L 2 ð
2x
L
3x2
0
6 5
1 10 L
1 10 L
2 2 15 L
!
2
2x 3x
" dx
3L
L2
2 L
# g
# g
Ð ^ 1 g, we conclude that But since dw0 Pw0 dx ¼ dgT PL K " ^1 ¼ K
6 5 1 10 L
1 10 L 2 2 15 L
#
Finally, ð
00
ðL
00
dw EIw dx ¼ dg F T
T 0
!
2 6x
" EI 3L ¼ dg 6 L L2 " 12 T EI ¼ dg 3 L 6L
EI( 2 2
T
L 6L 4L2
6x ) dx Fg
#"
#
4L
6L2
6L2
12L3
#"
3L
L2
2 L
# g
g
Ð ^ But dw00 EIw00 dx ¼ dgT EI L3 K2 g, from which ^2 ¼ K
12
6L
6L 4L2
EXAMPLE 11.8 Determine the two buckling load of a cantilevered beam modeled as one element.
SOLUTION
^ 2 PL2 K ^ 1 ¼ 0, leading to The two critical values can be obtained using det K EI "
# 1 12 65 j 6 10 j L PL2 det 2 ¼ 0, j ¼ 1 2 EI 6 10 j L 4 15 j L and so 2 2 1 j L2 6 10 j L2 ¼ 0 12 65 j 4 15
ß 2008 by Taylor & Francis Group, LLC.
On solving the above equation, we have j ¼ 2.49, 32.18, with the corresponding critical values P ¼ 2:49
EI EI , 32:18 2 L2 L
The lower value of the coefficient of EI=L is very close to the exact value of p2=4, but the higher value is not very close to the exact value, given by 9p2=4 ¼ 22.2.
EXAMPLE 11.9 Interpretation of Buckling Modes Next consider static buckling of a clamped–clamped beam as shown in Figure 11.8. If we impose clamped constraints from both the left and right equations, the first and second elements contribute the following stiffness matrices, using notation introduced in Chapter 7. EI 12 3L 3L (2) ¼ , K 11 L2 l3e 3L L2 2P 6=5 L=20 2P 6=5 L=20 (1) ¼ ¼ , K 11 L L=20 L2 =30 L L=20 L2 =30 K(1) 22 ¼
K(1) 22
8EI 12 L3 3L
The assembled stiffness matrix is now obtained by direct addition: (1) (2) (2) K ¼ K(1) 22 þ K11 K22 þ K11 8EI 24 0 2P 12=5 0 ¼ 3 0 2L2 0 L2 =30 L L and the load–deflection relation is "
8EI 24 0 2P 12=5 2 0 L3 0 L L
0 L2 =30
#
w2 V2 ¼ 0 w2 M2
The two critical buckling loads can be obained analytically since the two buckling modes are uncoupled. P1 L2 ¼ 40 4p2 , EI
P2 L2 ¼ 120 12p2 EI
V0 P
M0 L
L
FIGURE 11.8 Buckling of a clamped–clamped beam.
ß 2008 by Taylor & Francis Group, LLC.
Note that the two element values of the critical load correspond to the symmetric and antisymmetric behaviors. (i) In particular, if M0 ¼ 0 there is only one critical load and it is P1. Furthermore, w02 ¼ 0 and w(L=2 z) ¼ w(L=2 þ z). Alternatively stated, the deformation is symmetric about L=2. (ii) However, if V0 ¼ 0, there is only one critical load and it is P2. In this case, w2 ¼ 0 and w(L=2 z) ¼ w(L=2 þ z). This of course represents antisymmetry. Returning to the main development, but referring to Figure 11.8, we now compare the finite element approach with the exact method, assuming static conditions. Consider the symmetric case. Let w(x) ¼ wc(x) þ wp(x), in which wc(x) is the characteristic solution and wp(x) is the particular solution reflecting the perturbation. From the Euler buckling Equation 11.52 wc(x) has a general solution of the form wc(x) ¼ a þ qffiffiffiffi P bx þ g cos kx þ d sin kx, in which k ¼ EI . Now w ¼ w0 ¼ 0 at x ¼ 0, w0 (L) ¼ 0 and EIw000 L2 ¼ V1 , expressed as the conditions 1a þ 0b þ 1g þ 0d ¼ wp (0) 0a þ 1b þ 0g þ kd ¼ w0p (0)
(11:61)
0a þ 1b gk sin(kL=2) þ dk cos(kL=2) ¼ w0p (L=2) 000
0a þ 0b þ gk3 sin(kL=2) dk3 cos(kL=2) ¼ EIwp (L=2) þ V1 or, in matrix–vector notation, 2 1 wp (0) 1 60 C B wp0 (0) 6 C B Bz ¼ B C, B ¼ 6 40 A @ wp0 (L=2) 000 EIwp (L=2) þ V1 0 0
0 1 1 0 1 k sin(kL=2)
0 k k cos(kL=2)
0 k3 sin(kL=2) k3 cos(kL=2)
3 7 7 7, 5
0 1 a BbC B C z¼B C @gA d (11:62)
For the solution to ‘‘blow up’’ it is necessary for the matrix B to be singular, which occurs if the corresponding homogeneous problem in fact possesses a unique solution. Accordingly, we seek conditions under which there exists a nonvanishing vector z for which Bz ¼ 0. Direct elimination of a and b furnishes a ¼ g and b ¼ kd. The remaining coefficients must satisfy
sin(kL=2) sin(kL=2)
cos(kL=2) 1 cos(kL=2)
0 g ¼ 0 d
(11:63)
A nonvanishing solution is possible only if the determinant vanishes, which reduces to the requirement sin kL ¼ 0. This equation has many solutions for kL=2, including kL=2 ¼ 0. The lowest nontrivial solution is kL=2 ¼ p, from which Pcrit ¼ 4p2 EI=L2 ¼ 39.88 EI=L2. Clearly, the symmetric solution in the foregoing two-element model (Pcrit ¼ 40 EI=L2) gives a very accurate result.
ß 2008 by Taylor & Francis Group, LLC.
For the antisymmetric case the corresponding result is that tan kL=2 ¼ kL=2. The lowest meaningful root of this equation is kL ¼ 4.49 (Brush and Almroth, 1975), giving Pcrit ¼ 80.76 EI=L2. Clearly, unlike the symmetric solution, the axisymmetric solution from the two-element model (Pcrit ¼ 120 EI=L2) is not very accurate. To this point it has been implicitly assumed that the beam column is initially perfectly straight. This assumption can lead to overestimates of the critical buckling load. Now consider that there is a known initial distribution w0(x). The governing equation is now d2 d2 d2 EI 2 (w w0 ) þ P 2 (w w0 ) ¼ 0 2 dx dx dx
(11:64)
d2 d2 d2 d2 d2 d2 EI w þ P w ¼ EI w þ P w0 0 dx2 dx2 dx2 dx2 dx2 dx2
(11:65)
or equivalently
Now crookedness is modeled as a perturbation. Similarly, if the cross-sectional properties of the beam column exhibit a small amount of variation, say EI(x) ¼ EI0[1 þ q sin(px=L)], the effect of the variation may likewise be modeled as a perturbation.
EXAMPLE 11.10 In the clamped–clamped beam column use four equal length elements to determine how much improvement, if any, occurs in the symmetric and antisymmetric cases.
SOLUTION Consider the right half of the configuration, which has two beams of length L=2. ^ we assume an interpolation model w(x) ¼ wT(x)Fg in which Denoting L=4 as L, wT (x) ¼ ( 1 x x2 x3 ) 2 ^3 31 2 1 0 0 0 L 6 7 6 0 1 0 0 7 16 0 6 F¼6 ¼ 6 ^ ^2 ^3 7 ^3 4 3L ^ 5 41 L L L L ^ 3L ^2 0 1 2L 2
3
1 we 7 B w0 C 7 B e C 7, g ¼ B C 5 @ weþ1 A
0 ^3 L ^2 2L
0 0 ^ 3L
0 0 ^2 L
^ L
^ 2 L
0
w0eþ1
The interpolation model also involves the relations w0 ¼ bT Fg,
bT ¼
ß 2008 by Taylor & Francis Group, LLC.
@wT (x) ¼ (0 @x
1 2x 3x2 ), w00 ¼ ( 0
0 2 6x )Fg
Now ð ^ 1 ¼ FT bLbT dx F K 2
^3 L
^ 3L
0
6 16 60 ¼ 6 5 ^ L 6 40
2
3
2 L 0 ð 7 6
0
^7 60 1 L 7 6 7 6 6 ^ 2 7 0 3L 5 4 0 2x ^ 0 0 3x2 ^ 2 L 0 0 L 3 2 6 1 ^ 1 ^ 10 L 65 10 L 5 7 6 6 1 L 2 ^2 1 ^ 1 ^2 7 L 6 10 ^ 15 10 L 30 L 7 7 ¼6 6 1 ^ 6 1 ^ 7 7 6 65 L L 10 5 10 5 4 1 ^ 1 ^2 1 ^ 2 ^2 L 30 L 10 L 15 L 10 ^2 2L
^3 L
0 2x 4x2 6x3
0
3 2
^3 L
0
7 6 6 ^3 3x2 7 L 7 6 0 dx6 7 6 ^ 2L ^2 6x3 7 5 4 3L ^ 9x4 2 L
0
3
0
7 0 7 7 7 ^ L ^2 7 3L 5 ^ 2 L 0
Continuing,
2
ð
2 ^3 ^ 2 3 L0 0 1 L 3L 0 7 B C 6 2 ^ ^ ^3 6 0 2L L 7 B 0 C L 7 B C( 0 0 2 6x ) dx6 7 6 C B ^ 2 5 @ 2 A ^ 2L ^2 0 3L 43L ^ 0 ^ ^ 2 L 6x 2 L 0 L 3 ^ 12 6L ^ 6L ^2 ^ ^2 7 7 4L 6L 2L 7 ^ ^ 7 6L 12 6L 5 2 2 ^ ^ ^ 6L 4L 2L
2 ^3 L 6 60 ^2 ¼ 1 6 K ^5 6 L 40
0 ^3 L
0
12 6 ^ 6 6L ¼6 6 4 12 ^ 6L
3
0
0
0 ^ 3L
0 ^ L2
7 7 7 7 5
^ 2 L
Since the element length is L=4, the element matrices become 2
3 5
6 1 6 L 6 40 ½K1 1 ¼ ½K1 2 ¼ 6 6 3 4 5
1 40 L 35 1 2 120 L
1 40 L
1 40 L
3 5
1 1 2 L 480 L 40
1 40 L
1 40 L
3
2
3L=2 12 3L=2
12
3
7 6 7 1 27 6 3L=2 L2 =4 3L=2 L2=8 7 480 L 7 6 7 7, ½K2 1 ¼ ½K2 2 ¼ 6 7 1 7 6 12 3L=2 12 3L=2 7 L 5 4 5 40 3L=2
1 2 120 L
L2=8
3L=2 L2 =4
The assembled matrices, after enforcing the constraints w(0) ¼ w0 (0) ¼ 0 are 2
1 5
6 6 0 6 ^ K1 ¼ 6 6 6 4 5 1 40 L
0
65
1 2 60 L
1 40 L
1 40 L
6 5
1 480 L2
1 40 L
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1 40 L
3
7 1 480 L2 7 7 7, 1 7 L 5 40 1 2 120 L
2
24
6 6 0 6 ^ K2 ¼ 6 6 12 4 3L=2
0
12
L2 =2
3L=2
3L=2
12
L =8
3L=2
2
3L=2
3
7 L2=8 7 7 7 3L=2 7 5 L2 =4
The governing equation is written in finite element form as 2
2 24 6 6 6 EI 6 0 6 6 6 6 6(L=4)3 6 12 4 4
0
12 3L=2
3
2
12 5
1 65 40 L
0
6 7 L2 =2 3L=2 L2=8 7 P 6 6 0 7 6 7 6 6 7 (L=4) 3L=2 12 3L=2 5 4 5
1 2 240 L
1 40 L
1 40 L
6 5
1 1 2 40 L 480 L
3L=2 L2=8 3L=2 L2 =4
1 w(L=4) C B B w0 (L=4) C C, B g3 ¼ B C @ w(L=2) A 0
0
0 B B 0 f3 ¼ B BV @ 1
w0 (L=2)
1 40 L
33
77 1 2 77 480 L 77 7 7g3 ¼ f 3 1 77 40 L 5 5 1 2 120 L
1 C C C C A
M1
Symmetric Case: M0 ¼ 0; w0 (L) ¼ 0 The governing equation reduces to 3 2 2 12 24 0 12 5 7 6 EI 6 P 6 7 6 6 6 2 L =2 3L=2 7 6 0 6 6 0 5 (L=4) 4 4(L=4)3 4 65 12 3L=2 12 2
0 1 2 60 L 1 40 L
1 0 1 0 w(L=4) C B C 7 7B C B C 7 7B 0 1 7Bw (L=4)C ¼ B 0 C 40 L 7 A @ A 5 5@ 6 V1 w(L=2) 5 65
3 30
Buckling occurs if 2 6 6 det6 4
24 12 5 j 0 12 þ 65 j
0
1 j L2 =4 2 15 1 3 20 j L=2
3 12 þ 65 j 7 7 1 3 20 j L=2 7 ¼ 0, 5 12 65 j
j¼
P=(L=4) PL2 ¼ 16 EI EI=(L=4)3
The roots of the above equation are j ¼ 10, 2.486, 32.1807, independently of L. The lowest nontrivial root is j ¼ 2.486 and so P1 ¼ 39:736
EI L2
which is very close to the exact solution (4p2 EI=L2). Antisymmetric Case: V0 ¼ 0; w(L) ¼ 0 Following the same steps as before, the governing equation is found to be 3 30 2 12 3 1 0 1 2 1 24 0 3=2 0 40 w(L=4) 0 5 7 6 7 6 7 C B C B 7 6 EI 6 7 6 P B 7 C 6 B 1 1 7 Lw0 (L=4) C ¼ B 0 C 6 6 0 0 L=2 18 L 7 480 L7 C 60 L 6(L=4)3 6 7 7B 6 5 A @ A 4 (L=4) 4 5 5@ 4 1 1 1 L 3=2 18 L 40 480 L 120 L M1 Lw0 (L=2) 2
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The critical loads are obtained from 3 1 3 þ 20 j 7 6 1 1 1 7 det6 0 2 15 j L=4 2 þ 120 j L=4 5 ¼ 0 4 1 1 1 1 1 1 30 j L=4 2 3 þ 20 j L 2 þ 120 j L=4 2
24 12 5 j
1 2
0
The roots are found with a little effort to be j ¼ 5.18, 18.78, 49.38, giving the lowest nontrivial buckling load as P1 ¼ 82:88
EI L2
which is very close to the exact solution (80.76 EI=L2). Clearly, a four-element model gives a much better approximation than the twoelement model in the antisymmetric case. The percentage of error decreases from 48.6% to 2.6%.
11.5.2 EULER BUCKLING
OF
PLATES
The governing equation for an isotropic plate element subject to in-plane loads is (Wang, 1953) Eh2 @2w @ 2w @2w 4 w þ P þ P þ P r ¼0 x y xy 12ð1 n2 Þ @ x2 @ y2 @ x@ y
(11:66)
in which the loads are illustrated in Figure 11.9. The variational methods in Chapter 4 furnish ð
ð ð ð dwr4 w dA ¼ tr(dWW) dA þ dw(n r)r2 w dS drw r(n rw) dS (11:67)
z
Y Pxy Pyx
x
h
Px
FIGURE 11.9 Plate element with in-plane compressive loads.
ß 2008 by Taylor & Francis Group, LLC.
Py
in which W ¼ rrT w (a matrix!). In addition ð
@2w @2w @2w dA dw Px 2 þ Py 2 þ Pxy @x @y @ x@ y ð ð T ¼ dwn p dS (rdw)T P(rw) dA
0
1 @w 1 @w þ P P B x @ x 2 xy @ y C C, p¼B @ @w @w A 1 2 Pxy @ x þ Py @ y
P¼
Px 1 2 Pxy
(11:68)
1 2 Pxy
Py
We recall the interpolation model w(x,y) ¼ wTb2 Fb2 gb2 , from which we may obtain the relations of the form rw ¼
wx wy
¼ bT1b2 Fb2 gb2 ,
VEC (W) ¼ bT2b2 Fb2 gb2
(11:69a)
2
2
bT1b2
3 @ T wb2 6 @x 7 7, b T ¼6 2b2 4@ 5 T wb2 @y
3 @ 2 wTb2 6 @ x2 7 6 7 6 2 T 7 6 @ wb2 7 6 7 6 @ x@ y 7 6 7 ¼6 7 6 @ 2 wT 7 b2 7 6 6 @ x@ y 7 6 7 6 7 4 @ 2 wT 5
(11:69b)
b2
@ x2
We also assume that the secondary variables 12(1Eh v2 ) (n r)r2 w, 12(1Eh v2 ) (n r)rw, and p are prescribed on S. Doing so serves to derive 2
2
½Kb21 Kb22 gb2 ¼ f Kb21 ¼
ð Eh2 T b bT dA Fb2 , F 12(1 v2 ) b2 2b2 2b2
(11:70)
ð Kb22 ¼ FTb2 b1b2 PbT1b2 dV Fb2
and f reflects the secondary variables prescribed on S. As illustrated in Figure 11.10 we now consider a three-dimensional loading space in which Px, Py, and Pxy correspond to the axes in terms of which we seek to determine a surface of critical values at which buckling occurs. In this space a straight line emanating from the origin represents a proportional loading path. Let the load intensity l denote the distance to a given point on this line. By analogy with spherical coordinates, there exist two angles u and f such that Px ¼ l cos u cos f,
ß 2008 by Taylor & Francis Group, LLC.
Py ¼ l sin u cos f,
Pxy ¼ l sin f
(11:71)
Pxy
l f Py q Px
FIGURE 11.10 Loading space for plate buckling.
Now ^ b22 (u,w) Kb22 ¼ lK ð T ^ ^ w)bT dV Fb2 Kb22 (u,w) ¼ Fb2 b1b2 P(u, 1b2 cos u cos f sin f ^ P(u,f) ¼ sin f sin u cos f
(11:72)
For each pair (u,f), buckling occurs at a critical load intensity lcrit(u,f), satisfying ^ b22 ¼ 0 (11:73) det Kb21 lcrit ðu,fÞK A surface of critical load intensities lcrit(u,f) can be drawn in the loading space of Figure 11.10 by evaluating lcrit(u,f) over all values of (u,f) and discarding values which are negative. Recalling Example 3.8 in Chapter 3, we may write ð ^ Kb22 ¼ IVEC FTb2 FTb2 bT1b2 (x) bT1b2 (x) dV VEC (P) (11:74) Accordingly, in computing the part of the stiffness matrix incorporating the in-plane load, volume integration need only be performed once, independently of l, u, and w.
11.6 INTRODUCTION TO CONTACT PROBLEMS 11.6.1 GAP In many practical problems the information required to develop a finite element model, for example, the geometry of a member and the properties of its constituent materials, can be determined with little uncertainty or ambiguity. However, often the loads experienced by the member are not easily identified. This is especially true if loads are transmitted to
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P
Contactor
d A k
B
C
k
k g
Target
FIGURE 11.11 Simple contact problem.
the member along an interface with a second member. This class of problems has been called contact problems, and they are arguably the most common boundary conditions encountered in ‘‘the real world.’’ The finite element community has devoted and continues to devote a great deal of effort to contact problems, culminating in gap and interface elements for contact. Here, we provide a simple introduction to gap elements. First consider the three spring configuration in Figure 11.11. All springs are of stiffness k. Springs A and C extend from the top plate, called the contactor, to the bottom plate, called the target. The bottom of spring B is initially remote from the target by a gap g. The exact stiffness of this configuration is bilinear: 2k, d < g (11:75) kc ¼ 3k, d g From the viewpoint of the finite element method Figure 11.11 poses the following difficulty. If a node is set at the lowest point on Spring B and at the point directly below it on the target, these nodes are not initially connected, but may later be connected in the physical problem after contact is established. Further, it is necessary to satisfy the nonpenetration constraint, which may be stated as an inequality, whereby the middle spring does not move through the target after contact is established. If the nodes are considered unconnected in the finite element model, there is nothing to enforce the nonpenetration constraint. If, however, the nodes are considered connected, the nonpenetration constraint can be satisfied but the stiffness is artificially high. This difficulty is overcome in an approximate sense by a bilinear contact element. In particular, we introduce a new spring kg as shown in Figure 11.12. The stiffness of the middle spring pair (B in series with the contact spring) is now denoted as km and is given by 1 1 1 (11:76) km ¼ þ k kg It is desirable for the middle spring to be soft when the gap is open (g > d), and to be stiff when the gap is closed (g < d). For the purpose of illustration we make the selection
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P
Contactor
d A
B
C
k
k
k kg
Target
FIGURE 11.12 Spring representing contact element.
kg ¼
k =100, 100k ,
g>d g d
(11:77)
Elementary algebra suffices to demonstrate that kc
2k þ 0:01k , 2k þ 0:99k ,
g>d g d
(11:78)
Consequently, the model with the contact element is too stiff by 0.5% when the gap is open and is too soft by 0.33% when the gap is closed (contact). One observation from this example is that the stiffness of the gap element should be related to the stiffnesses of the contactor and the target in the vicinity of the contact point. In finite element modeling, ‘‘slave’’ nodes placed at points ‘‘i’’ on the target and ‘‘master’’ nodes placed at points ‘‘j’’ on the contactor are not initially connected, but may later be connected in the physical problem when contact is established. If, however, in the finite element model the nodes are always connected by the gap element, penetration is prevented, and the bilinear spring serves to overcome the difficulty of overstiffness before closure and understiffness after closure by rendering the effective stiffnesses close to the correct stiffnesses.
11.6.2 POINT-TO-POINT CONTACT Of course, in the more general case, it is not known what points, say on the target, will come into contact with the contactor, and there is no guarantee that target nodes will come into contact with contactor nodes. The gap elements can be used to account for the unknown contact area, as follows. Figure 11.13 shows a contactor and a target on which are indicated candidate contact areas dSc and dSt, containing nodes c1, c2, . . . , cn, t1, t2, . . . , tn. The candidate contact areas must contain all points for which there is a possibility of establishing contact. The gap (i.e., the distance in the undeformed configuration) from the ith node of the contactor to the jth node of the target is denoted by gij. In point-to-point contact,
ß 2008 by Taylor & Francis Group, LLC.
Candidate contactor contact surface dSc c1
c3
c2
a 31 k(g31)
k
t2 dSt
t1
x
t3
Candidate target contact surface
FIGURE 11.13 Point-to-point contact.
in preselected candidate contact zones, each node on the contactor is connected to each node of the target by a spring with a bilinear stiffness. (Clearly, this element may miss the edge of the contact zone when the edge does not occur at a node.) The angle between the spring and the normal at the contactor node is aij, while the angle between the spring and the normal to the target is aji. Under load, the ith contactor node experiences displacement uij in the direction of the jth target node, and the jth target node experiences displacement uji. For illustration, the spring connecting the ith contactor node with the jth target node has stiffness kij given by kijlower , dij < gij (11:79) kij ¼ kijupper , dij gij in which dij ¼ uij þ uji is the relative displacement. The force in the spring connecting the ith contactor node and jth target node is fij ¼ kij(gij)dij (no Psummation). The total normal force experienced by the ith contactor node is fi ¼ fij cos(aij ). As an example of how the spring stiffness might be takenj to depend on the gap to achieve a continuous approximation to a bilinear function, consider the expression "
!# qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 1 a 2 (gij dij g) (gij dij g )
kij (gij dij ) ¼ k0 « þ (1 2«) tan
p 2 (11:80) in which g, a, and « are positive parameters selected as follows. When gij dij g > 0, kij attains the lower shelf value k0«, and we assume that « 1. If gij dij g < 0, kij approaches the upper shelf value k0(1 «). We choose g to be a small value to attain a narrow transition range from the lower to the upper shelf values. In the range 0 < gij dij < g there is a rapid but continuous transition from the lower shelf (soft) value to the upper shelf (stiff) value. If we now choose a such that ag ¼ 1, kij equals k0=2(þO(«)) when the gap initially closes (gij ¼ dij). The spring characteristic is illustrated in Figure 11.14.
ß 2008 by Taylor & Francis Group, LLC.
kij ek0
k0
k0 /2
dij – gij
ek0
FIGURE 11.14 Illustration of a gap stiffness function.
The total normal force on a contactor node is the sum of the individual contact element forces. For example, for the jth contactor node, fnj ¼
Nc X
kij (gij dij )dij cos(aij )
(11:81)
i
Clearly, significant forces are exerted only by the contact elements which are ‘‘closed.’’
11.6.3 POINT-TO-SURFACE CONTACT We now briefly consider point-to-surface contact, illustrated in Figure 11.15 using a triangular element. Here target node t3 is connected via a triangular element to contactor nodes c1 and c2. The stiffness matrix of the element can be written as ^ in which g1 d1 is the gap between nodes t1 and c1, and K ^ kð[g1 d1], [g2 d2]ÞK, Candidate target contact surface
t1
dS t t2
t3
Element connecting node t3 with nodes c1 and c2
∗
∗
c2 dSc
c1
∗c
Candidate contactor contact surface
FIGURE 11.15 Element for point-to-surface contact.
ß 2008 by Taylor & Francis Group, LLC.
3
is the geometric part of the stiffness matrix of a triangular elastic element. The stiffness matrix of the element may be made a function of both gaps. Total force normal to the target node is the sum of the forces exerted by the contact elements to the candidate contactor nodes. In some finite element codes schemes such as illustrated in Figure 11.15 are also used to approximate the tangential force in the case of friction. Namely, an ‘elastic friction’ force is assumed in which the tangential tractions are assumed to be proportional to the normal traction through a friction coefficient. Elastic friction models do not appear to consider sliding and may be considered bonded contact. Advanced models address sliding contact and incorporate friction laws not based on the Coulomb model.
EXAMPLE 11.11 1. Consider a finite element model for a set of springs, illustrated below (Figure 11.16). A load moves the left-hand plate toward the fixed right-hand plate. (a) What is the load–deflection curve of the configuration? (b) For a finite element model, suppose a bilinear spring is supplied to bridge the gap H. What is the load–deflection curve of the finite element model? (c) Identify a kg value for which the load–deflection behavior of the finite element model is close to the actual configuration. (d) Why is the new spring needed in the finite element model? 2. Suppose a contact element is added in the foregoing problem, in which the stiffness (spring rate) satisfies "
!# qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a
2 (gij dij g) (gij dij g)
2
2 kij (gij dij ) ¼ k0 « þ (1 2«) tan1 p
with ag ¼ 1, kL ¼ k=100, and ku ¼ 100k. Compute the stiffness=k for the configuration as a function of the deflection.
F
k
k k
k k H L
FIGURE 11.16 Set of springs with gap.
ß 2008 by Taylor & Francis Group, LLC.
SOLUTION 1. The exact stiffness of this configuration is kc ¼
3k, d < H, gap is open 3:5k, d H, gap is closed
in which when d Hkc is calculated as follows: kc ¼ 3k þ
1 1 1 ¼ 3:5k þ k k
The exact load–deflection curve is plotted as in Figure 11.17 Now let us introduce a gap element of stiffness kg as illustrated in Figure 11.18.
3.5k
F
3k
H
d
Gap
FIGURE 11.17 Load–deflection curve of the actual configuration. F
k A k
B k
i
kg
C k D H L
FIGURE 11.18 Illustration of gap element.
ß 2008 by Taylor & Francis Group, LLC.
j
k
It is desirable for the spring C to be soft when the gap is open (d < H), and to be stiff when the gap is closed (d > H). Suppose kg ¼
k=100, 100k,
d
When the gap is open, i.e., d < H, the stiffness of the configuration is given by
1 1 1 1 kc ¼ 3k þ þ þ ¼ 3k þ 0:0098k k kg k And when the gap is closed, i.e., d > H, the stiffness of the configuration is given by
1 1 1 1 þ þ ¼ 3k þ 0:4975k kc ¼ 3k þ k kg k Accordingly kc ¼
3k þ 0:0098k, d < H 3k þ 0:4975k, d H
The model with the gap element is overly stiff by 0.33% when the gap is open, and is overly soft by 0.07% when the gap is closed. The exact and approximate values are illustrated as in Figure 11.19. Clearly, the assumed value of kg gives a close approximation to the actual configuration. 2. Case (1): gap is open: When the gap is open, g > d þ g, i.e., g d g > 0. Hence we have
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(gij dij g)2 (gij dij g) ¼ 0
) kij (gij dij ) ¼ kL ¼ k0 « With kg Actual 3.5k
F
3.4975k 3.0098k 3k H
d
Gap
FIGURE 11.19 Load–deflection curve of the actual and the finite element configuration.
ß 2008 by Taylor & Francis Group, LLC.
Case (2) gap is closed: When the gap is closed, g < d þ g, i.e., (g d g) > 0. Now a
2 kij (gij dij ) ¼ k0 « þ (1 2«) tan1 2(d þ g g) p 2 Now tan1(a(d þ g g)) asymptotically approaches p=2, for which kij (gij dij ) ¼ kU ¼ k0 (1 «) Case (3) gap just closes: When the gap just closes, g < d þ g, i.e., g d g < 0. Now, a
tan1 2(d þ g g) ¼ tan1ða(d þ g g)Þ 2 On substituting ag ¼ 1, we obtain 2 kij (gij dij ) ¼ k ¼ k0 « þ (1 2«) tan1ða(d g) þ 1Þ p Since d ¼ g under the stated condition 1« k0 kij (gij dij ) ¼ k ¼ k0 « þ 2 2 The results can be summarized as 8 < k0 «, gap is open kij (gij dij ) ¼ k0 =2, gap just closes : k0 , gap is closed
ß 2008 by Taylor & Francis Group, LLC.
12
Rotating and Unrestrained Elastic Bodies
12.1 FINITE ELEMENTS IN ROTATION 12.1.1 ANGULAR VELOCITY
AND
ANGULAR ACCELERATION VECTORS
We consider a vector b which is referred to an instantaneous coordinate system with base vectors ex, ey, and ez. Suppose that in a time increment dt there is rotation of the x–y plane clockwise about the z-axis through a small angle dc, generating a new vector b0 , and giving rise to the new coordinates x 0 , y 0 , z 0 in which z 0 coincides with z. This rotation is depicted in Figure 12.1. Note: A negative rotation of the coordinate system is equivalent to a positive rotation of the vector about the instantaneously fixed coordinate system. Prior to the rotation, the vector may be expressed as bT ¼ {bx by bz}. For the moment we assume that bx, by, and bz are constants. The rotation gives rise to b0 which satisfies 2
cos( dc)
6 b0 ¼ 6 4 sin( dc)
sin( dc) cos( dc)
0
0
0
22
3
1 0
66 7 66 07 5b 4 4 0 1 0 0
1
0
3
2
0
6 7 6 07 5 þ dc4 1 1
0
1
0
33
0
77 7 07 55b
0
0 (12:1)
from which 2
0
6 b0 ¼ b þ dc4 1 0
1 0 0
0
3
7 0 5b 0
Next, suppose that the y0 –z0 plane is rotated about the x-axis through the incremental angle du, corresponding to a positive rotation of the vector with respect to the instantaneously fixed x 0 y 0 z 0 coordinate system (cf. Figure 12.2). We now have the vector b00 and the directions x00 , y00 , z00 , in which b00 is given by
ß 2008 by Taylor & Francis Group, LLC.
z, z ⬘
y dy
dy
y⬘
x
x⬘
FIGURE 12.1 Differential clockwise rotation about the z-axis.
z⬙
z⬘ dq
y⬘ dq
x ⬘,x ⬙
FIGURE 12.2 Differential rotation about the x0 -axis.
ß 2008 by Taylor & Francis Group, LLC.
y⬙
2
1 b00 ¼ 4 0 0
3 2 0 1 sin(du) 5b0 4 0 cos(du) 0
0 cos(du) sin(du)
0 1 du
3 0 du 5b0 1
(12:2)
But eliminating b0 furnishes 22
2
3
1
0
0
0
0
66 6 b00 ¼ 6 44 0
1
6 7 6 07 5 þ du4 0
0
0
0 1 2 2
0 0
6 6 6 bþ6 4du4 0 0
0
0
1 3
3 32 2
0
2
0 1
0
33
0
0
7 76 6 76 6 1 7 5 54 4 0
1
6 7 6 07 5 þ dc4 1
0
77 7 07 55b
0
0
0
0
0
1
0
0 2
0 7 6 6 0 1 7 5 þ dc4 1 1
3
1
0
0
1 33
0
77 7 07 55b
0
0
(12:3)
Finally, referring to Figure 12.3, the z00 x00 plane is rotated around the y00 -axis clockwise through an angle df, furnishing the new axes x000 , y000 , z000 , and the new vector b000 . The vector b000 is given by
z⬙
z df
y ⬙, y
df
x
x⬙
FIGURE 12.3 Differential clockwise rotation about the y00 -axis.
ß 2008 by Taylor & Francis Group, LLC.
2 6 b000 ¼6 4
cos(df) 0 sin(df) 0
1
0
22
3
1 0 0
3
2
0 0 1
33
7 00 66 7 6 77 7b 66 0 1 0 7 þ df6 0 0 0 77b00 5 5 4 55 44
sin(df) 0 cos(df) 0 0 1 1 0 0 3 2 2 3 33 2 2 0 0 1 0 0 0 0 1 0 7 6 7 6 77 6 6 6 7 6 7 6 77 b þ6 4df4 0 0 0 5 þ du4 0 0 1 5 þ dc4 1 0 0 55b 2
1 0 0 dc df
0 6 bþ6 4 dc
0 0 0
7 du 7 5b
0
df
0 1 0
3
(12:4)
0
du
The derivative of b is equated with 2
0 db 6 ¼4 c_ dt f_
c_ 0 u_
b000 b dt ,
and now
3 f_ 7 u_ 5b 0
_ z cb _ x )ez _ y )ex þ (cb _ x ub _ z )ey þ (ub _ y fb ¼ (fb ¼vb
(12:5)
in which the instantaneous angular position vector is identified as u ¼
8 9
uex þ fey þ cez , and the angular velocity vector is identified as v ¼ 8 9 < u_ = f_ : _; c
¼
du ¼ dt
_ y þ ce _ x þ fe _ z . The angular velocity vector describes the rate at which ¼ ue
the vector b rotates in a counterclockwise sense about the instantaneous axes. In the appendix, the vector u is expressed in terms of angles used in spherical coordinates. The angular position vector presented above is one version of what are called the Euler angles.
12.1.2 VELOCITY
AND
ACCELERATION
IN
ROTATING COORDINATES
More generally, again consider rotation of a body about a fixed axis in which at least one point in the body is located on the axis. The coordinate system is embedded in the fixed point and rotates with the body. The undeformed position vector X0 in the rotated system is related to its counterpart X in the unrotated system by X0 ¼ Q(t)X, in which Q is a proper orthogonal tensor. If the deformed body is viewed in the same coordinate system, the counterpart for the deformed position is x0 ¼ Q(t)x. The displacement likewise satisfies u0 ¼ Q(t)u. Time differentiation gives
ß 2008 by Taylor & Francis Group, LLC.
du0 (X,t ) du(X,t ) _ ¼ Q(t ) þ Q(t )u(X,t ) dt dt du(X,t ) _ þ Q(t )QT (t )u0 (X,t ): ¼ Q(t ) dt
(12:6)
We make the identification @ u0 (X,t ) du(X,t ) ¼ Q (t ) @t dt
(12:7)
0
with the interpretation that @ u @(Xt ,t) denotes the derivative of u0 with the rotation tensor Q(t) held fixed at its current position, and vanishes if u is unchanging relative to the _ T þ QQ _ T, V ¼ QQ _ T is antirotating system. In addition, since 0 ¼ ddt QQT ¼ QQ symmetric. Finally, from Chapter 3 for the 3 3 3 antisymmetric tensor V and for any 3 3 1 vector b there exists a 3 3 1 vector v satisfying Vb ¼ v3b. Of course v is the angular velocity vector of dynamics, and it is referred to the instantaneous (rotating) coordinate system; its time derivative @@vt is the angular acceleration vector, hereafter denoted by a and referred to the instantaneous coordinate system. Corresponding to a there is an antisymmetric tensor A ¼ @ V=@ t such that Ab ¼ a 3 b. Now with the prime on u no longer displayed, the total time derivative of the displacement vector, i.e., the velocity vector v, may be expressed in the rotating system as v¼
du ¼ dt
@ @u þ v u ¼ þvu @t @t
(12:8)
The acceleration vector is similarly expressed as a ¼ dv=dt ¼ (@=@t þ v )(@u=@t þ v u) ¼ @ 2 u=@t 2 þ 2v @u=@t þ v v u þ a u
(12:9)
The four right-hand terms in Equation 12.9 are, respectively, called the translational, Coriolis, centrifugal, and angular accelerations, respectively. In the Principle of Virtual Work, for reasons to be explained in a subsequent sections let dû denote the variation of u with the coordinate system held fixed at its instantaneous position. The corresponding inertial term becomes Ð T d2 0 d^ u r 2 (u þ X0 )dV. Now assuming the usual interpolation model u0 (X0 ,t) ¼ dt wT (X0 )Fg(t), the inertial term becomes ð d^ uT r
d2 u 0 d2 g dg T þ (G dV ¼ dg M þ G þ A)g 1 2 dt 2 dt dt 2
ß 2008 by Taylor & Francis Group, LLC.
(12:10)
ð M¼F
ð rww dV F,
T
T
G1 ¼ F
T
rwVwT dV F
ð ð G2 ¼ FT rwV2 wT dV F, A ¼ FT rwAwT dV F The matrix M is the conventional positive definite and symmetric mass matrix in a nonrotating system. The Coriolis matrix G1 is antisymmetric, the centrifugal matrix G2 is negative definite, and the angular acceleration matrix A is antisymmetric. (G2 involves the square of an antisymmetric matrix. Since an antisymmetric matrix has pure imaginary eigenvalues [Chapter 2], its square is symmetric with negative real eigenvalues and hence is negative definite.) There is also a rigid body force term due to centrifugal and angular accelerations. ð ^T r du
ð f rot ¼ FT rw V2 þ A X0 dV
d2 0 X dV ¼ dgT f rot , dt2
(12:11)
Consolidating the terms, the ensuing governing finite element equation is now M
d2 g dg þ G1 þ [K þ G2 þ A]g ¼ f f rot dt2 dt
(12:12)
EXAMPLE 12.1 Compare a one-element model of a rotating rod to the exact model. The rod is depicted as follows (Figure 12.4). We consider the member a rod and accordingly neglect effects in direction transverse to the motion. The radial displacement is denoted as u(r). The acceleration is obtained as u ¼ uer ,
v ¼ vez ,
a ¼ aez
@2u @u € ¼ 2 er þ 2vez er þ vez vez uer þ aez uer u @t @t The Coriolis and angular accelerations terms are aligned with eu and are therefore not of interest in a rod. The centrifugal acceleration points back radially toward the origin. The acceleration vector therefore reduces to €¼ u
@2u 2 v u er @t2
We also obtain the rigid body contribution €r0 ¼ v2 r0 er . Enforcing the constraint at the shaft a priori, we introduce the interpolation model u(r0 ,t) ¼ rL0 g(t ). The Principle of Virtual Work (using dû ! du) now implies
ß 2008 by Taylor & Francis Group, LLC.
z w
y
r E, A, L, r
q x
f0 sin wt
FIGURE 12.4 Steadily rotating elastic rod.
ðL dur 0
@2u rAL rAL 2 rAL2 2 2 € v r v u A dr ¼ dg g g v v 0 0 @t2 3 3 3
The stiffness term is found as ðL dErr Srr A dr0 ¼ dg
EA g L
0
The one-element finite element equation is now
EA rAL 2 rAL rAL2 v v gþ g€ ¼ 3 L 3 3
In the steady state g€ ¼ 0. Under steady state rotation, g ! 1 if v ! vcr ¼ qffiffiffiffiffiffiffiffiffi qffiffiqffiffiffi EA=L 3 E , and vcr is called the critical speed. Note that the critical speed is rAL=3 ¼ L r nothing but the natural frequency of the nonrotating rod. To assess accuracy the one-element finite element model is now compared to the exact solution. The exact governing differential equation is expressed as EA
d2 u þ rv2 u ¼ rv2 r0 dr02
The particular solution is up ¼ r0, in which event the characteristic solution satisfies EA
ß 2008 by Taylor & Francis Group, LLC.
d2 uc þ rv2 uc ¼ 0 dr02
qffiffiffiffiffiffi qffiffiffiffiffiffi rv2 rv2 þ b cos The characteristic solution has the form uc ¼ a sin r 0 E E r0 , and the boundary conditions qffiffiffiffiffiffi qffiffiffiffiffiffi rv2 rv2 a¼L E cos E L .
are
u(0) ¼ 0,
EA du(L) dr0 ¼ 0,
giving
b ¼ 0,
Finally, the displacement function emerges as rffiffiffiffiffiffiffiffi ! rffiffiffiffiffiffiffiffi ! 3 2 rv2 rv2 r0 L L sin tan 7 6 E E 7 6 7 ffiffiffiffiffiffiffiffi r u(r0 ;v2 ) ¼ rffiffiffiffiffiffiffiffi g ¼ L6 1 rffiffiffiffiffiffiffiffi ! r0 , 7 6 2 rv 5 4 rv2 rv2 cos L E E E qffiffiffiffiffiffi p 2 The nodal displacement g becomes unbounded when rvE L ¼ , and hence when 2 qffiffiffi p E v ¼ 2L , which is the exact natural frequency of the nonrotating rod. The critical r speed in the single element finite element model is within 10% of the exact value.
EXAMPLE 12.2 Unsteady rotation of a beam column about an axis Consider a thin beam column that is rotating unsteadily around a shaft. Its thin (local z) direction points in the direction of the motion, giving rise to Coriolis effects in bending. Derive the ensuing one-element model. Note that a beam column couples extension and bending.
SOLUTION Enforcing the clamped constraints at the axis of revolution, the interpolation model for a beam column (Figure 12.5) is given by u(x,y,t) ¼ u0 (x,t) y @n(x,t) @x , with
Z w, a
Y
z
Beam column X
FIGURE 12.5 Beam column in unsteady rotation about an axis.
ß 2008 by Taylor & Francis Group, LLC.
q, y
r
x u0 (x,t ) ¼ u0 (L,t), v ¼ wTb Fb gb L 2 1 L3 L3 L 1 3L2 wTb ¼ (x2 x3 ), Fb ¼ ¼ L4 2L L2 2L 3L2 v(L) @ v(x,t) gb ¼ , v0 (x,t ) ¼ ¼ bTb Fb gb , bTb ¼ (2x 3x2 ) @x v0 (L) The stiffness matrix not affected by the motion and is given by (cf. Chapter 7)
ð d«ij s ij dV ¼ dgT Kg, K ¼ Km ¼
EA EI 12 , Kb ¼ 3 6 L L
6 , 4
Km 0
0 Kb
g¼
u0 (L,t) gb
The inertial terms now are stated as d2 u @ 2 u @u ¼ 2 þ 2v þvvuþau dt 2 @t @t 1 0 2 @ (u0 yv0 ) @v 2 0 ð av v (u yv ) 2v 0 C B @t 2 @t Cr dV ¼ (du0 ydv0 dv)B A @ 0 2 @(u0 yv ) @ v 0 2 þ a(u0 yv ) þ 2 v v 2v @t @t 2 0 2 ð ð @ u0 @v @ v 2 0 2 0 v u0 2v av r dV þ ydv y 2 v yv r dV ¼ du0 @t @t 2 @t ð 2 @u0 @ v þ dv 2v þ au0 þ 2 v2 v r dV @t @t Performing the indicated manipulations results in the following relations: ð du0
@ 2 u0 r dV ¼ du0 (L,t) @t 2
ðL 0
x x @ 2 u0 (L,t) rA dx LL @t 2
rAL @ 2 u0 (L,t) ¼ du0 (L,t) 3 @t 2 ð rAL Similarly, du0 (v2 )u0 r dV ¼ du0 (L,t) v2 u0 (L,t) 3 ðL ð @v x (2v)fTb Fb rA dx g_ b du0 (2v) r dV ¼ du0 (L,t) @t L 0
# " 2 L3 3L 1 L ¼ du0 (L,tÞ(2vrA) g_ 4 5 2L L2 b rAL ¼ du0 (L,t) 2v (7 L) g_ b 20
ß 2008 by Taylor & Francis Group, LLC.
ð
ðL du0 (av)r dV ¼ du0 (L,t )
x (a)fTb Fb rA dx gb L
0
" 3L2 rA 1 L ¼ du0 (L,t ) a L 4 5 2L
#
L3 L2
gb
rAL ¼ du0 (L,t ) a (7 L ) g b 20 ð ð @ 2 v0 €b y2 dv0 2 r dV ¼ dgTb FTb bb rI bTb dx Fb g @t ðL " 4x2 6x3 # T T €b dx Fb g ¼ dgb Fb rI 6x3 9x4 0 ¼
"
rI L
dgTb
6 5
1 10 L
1 10 L
2 2 15 L
# €b g
^ 1 encountered The matrix in the preceding expression is recognized as proportional to K in Chapter 10 in conjunction with buckling. Continuing, ð
0 dvau0 r dV ¼ dgTb arAL@
ð dv
1
7 20
Au0 (L,t)
1 20 L
ð @2v €b r dV ¼ dgTb FTb wb rAwTb dx Fb g 2 @t ðL ¼
dgTb FTb
2 r A4
0
2 rA ¼ dgTb 8 4 L
ð
dgTb rAL4
0
0
x5
x5
x6
3 L2
2 ¼
x4
3
L
ß 2008 by Taylor & Francis Group, LLC.
2
L
5 dx Fb g €b
32 1 54
2
13 35
11 210 L
11 210 L
1 2 105 L
y dv (v )v r dV ¼ 2
2L
3
5
1 6 6L
1 6 6L
1 7 7L
5L
32 54
L3
2L
L
2
3 5g €b
rI dgTb (v2 ) L
"
6 5 1 10 L
3
3L2
1 10 L 2 2 15 L
# €b g
5g €b
ð dv(2v)
ð @u0 x r dV ¼ dgTb FTb (2v)wb rA dx u_ 0 (L,t) L @t " 2 # 1 4! 3L 2L 4L T 2vrA ¼ dgb u_ 0 (L,t) 1 5 L L3 L2 5L 7 ! ¼ dgTb 2vrAL
20
u_ 0 (L,t)
1 20 L
Finally, "
ð
dv(v2 )vr dV ¼ dgTb (v2 )rAL
13 35 11 210 L
11 210 L 1 2 105 L
# €b g
Upon combining gm ¼ u0 and gb into the vector g, the inertial terms in the Principle of Virtual Work are now stated as 3 2 1 0 0 7 @ u0 rAL 6 6 0 0 0 7g du0 2 r dV ¼ dgT 5€ @t 3 4 0 0 0 3 2 1 0 0 ð 7 rAL 6 6 0 0 0 7g du0 (v2 )u0 r dV ¼ dgT v2 5 3 4 0 0 0 3 2 0 7 L ð 7 @v rAL 6 6 0 0 0 7g_ du0 (2v) r dV ¼ dgT 2v 5 @t 20 4 0 0 0 3 2 0 7 L ð 7 rAL 6 6 0 0 0 7g du0 (av)r dV ¼ dgT a 5 20 4 0 0 0 ð
ð
ð
2
2 y2 dv0
0
@ 2 v0 rI 6 0 r dV ¼ dgT 6 2 @t L4 0
0
0
6 5
1 10 L
1 L 10
2 2 L 15
2 0 6 rI 60 y2 dv0 (v2 )v0 r dV ¼ dgT v2 L 4 0
ß 2008 by Taylor & Francis Group, LLC.
3 7 7g 5€
0
0
6 5
1 10 L
1 10 L
2 2 15 L
3 7 7g 5
3 2 0 0 0 ð 7 @u0 rAL 6 7 6 r dV ¼ dgT 2v dv(2v) 6 7 0 0 7g_ 5 20 4 @t L 0 0 3 2 0 0 0 6 ð 7 rAL 6 7 dvau0 r dV ¼ dgT a 6 7 0 0 7g 5 20 4 L 0 0 3 2 0 0 0 ð 2 7 6 @ v 13 11 7€ dv 2 r dV ¼ dgT rAL6 35 210 L 5g 40 @t 0 ð
11 210 L
2
1 2 105 L
0
0
0
6 dv(v2 )vr dV ¼ dgT (v2 rAL)6 40
13 35
11 210 L
11 210 L
1 2 105 L
0
3 7 7g 5
and the stiffness term becomes 2
EA 0 6 L 6 ð 6 EI d«ij sij dV ¼ dgT 6 6 0 12 L3 6 4 EI 0 6 3 L
3 0 7 7 EI 7 g 6 37 L 7 7 EI 5 4 3 L
The one-element model for the rotating beam column emerges as 2 rAL 6 3 6 6 6 0 6 4 0
3
3 2 0 7 L 7 7 7 rAL 6 7 7 6 € þ 2v 7g 6 7 0 0 7g_ 13 6 rI 11 2 1 rAL þ rAL þ rI 7 5 20 4 35 5 L 210 10 5 L 0 0 11 2 1 1 3 2 210 rAL þ 10 rI 105 rAL þ 15 rIL
2 EA
0
v2
rAL 3
6 L 6 6 6 7 þ6 6 20 arAL 6 6 4 1 2 20 arAL
0
7 arAL 20
1 20 arAL2
3
0 1 7 fu 7 7 C B
11 7 EI EI B C 6 rI 1 g ¼ B fv C 6 3 v2 210 12 3 v2 13 rAL2 þ 10 rI 7 7 35 rAL þ 5 L @ A L L 7 7 5 Mv
11 EI
1 EI 1 2 6 3 v2 210 rAL2 þ 10 rI 4 3 v2 105 rAL3 þ 15 rIL L L
We observe that unsteady rotation of a beam column gives rise to Coriolis and angular accelerations effects serving to couple the membrane with the bending response.
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Returning to the main development, in two-dimensional steady rotation of an undamped elastic medium without Coriolis effects, the governing equation reduces to € þ (K v2 M)g ¼ f f v Mg
(12:13)
We consider the eigenvalues and eigenvectors implied by this equation. The eigenvalues v2j and eigenvectors xj satisfy the generalized eigenvalue equation 0 ¼ (K v2 M) v2j M xj
¼ K v2 þ v2j M xj (12:14) Now v2 þ v2j ¼ v2nj in which vnj are the natural frequencies found in the nonrotating system with the same stiffness and mass matrix. It follows that the eigenvalues of the rotating system satisfy v2j ¼ v2nj v2
(12:15)
However, the eigenvectors xj simultaneously diagonalize K and M regardless of v. We conclude that the eigenvectors are not affected by rotation. On the other hand, if the Coriolis and angular acceleration matrices need to be taken into account, the eigenvalues may in general have nonzero real and imaginary parts.
12.2 CRITICAL SPEEDS IN SHAFT ROTOR SYSTEMS There is instability in elastic bodies under steady rotation, associated with the notion of the critical speed, already touched upon Example 12.1. Critical speeds are illustrated through several additional examples.
EXAMPLE 12.3 Find the critical speeds in a beam rotating about its axis, in the symmetric and antisymmetric cases. Case 1: Transverse load as shown in Figure 12.6
SOLUTION The member under study is called a shaft in power generation applications. But here we refer to it as a beam since we are concerned with the transverse displacements induced by rotation. The force Fr rotates with the beam. An imperfection (mass unbalance) in the beam will cause it to bend away from the axis, and the bent configuration will
E, A, r
L/ 2
Fr
w
L/2
FIGURE 12.6 Rotating shaft with a transverse load.
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likewise rotate with the shaft. The more the beam bends the greater the centrifugal forces, thereby tending to cause the beam to bend even further. Using two elements and imposing clamped constraints on the right and left bearings gives the stiffness in terms in the Principle of Virtual Work as
(2) K(1) 22 þ K11
w2 w02
¼
192EI=L3 0
0 16EI=L
w2 w02
in which the subscript 2 denotes the central node. (The bearings are considered rigid.) The inertial terms of the beam are similarly found. Under the assumed steady rotation and assuming that the deformed profile is constant in the radial (outward) direction, the Coriolis and angular acceleration matrices vanish, leaving
w
(1) w2 €2 (2) 2 M þ M v 22 11 w02 w02 " #
13 w2 w_ 2 rAL 0 2 v ¼ 35 1 3 w_ 02 w02 0 420 rAL
(2) M(1) 22 þ M11
The ensuing finite element equation is therefore " 13
35 rAL
0
0
1 2 420 rAL
#(
€ w02
" 13 v
2
35 rAL
0
0
1 2 420 rAL
"
)
€2 w
#
þ
!
(
192EI=L3
0
0
16EI=L
w2 w02
)
( ¼
Fr
#
) (12:16)
0
Of course the equations represented by the rows of Equation 12.16 are uncoupled, leading to the two critical speeds sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 192EI=L3 16EI=L , vc2 ¼ vc1 ¼ 13rAL=35 rAL3 =420 If v ¼ vc1, the symmetric bending profile (about the midpoint) grows in a unstable fashion. This happens in the current case since the force Fr induces a symmetric deformation. In either of these cases the stiffness matrix K is singular. However, only low amplitude motion will happen in the antisymmetric mode unless there is a moment at the midpoint, as discussed in Case 2. Case 2: Suppose instead that at the midpoint there is a moment which rotates with the shaft (Figure 12.7).
E, A, r
L/ 2
Mr
w
L/ 2
FIGURE 12.7 Rotating beam with bending moment.
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Symmetric profile
Antisymmetric profile
FIGURE 12.8 Symmetric and antisymmetric deformations. Now the response is antisymmetric. If v ¼ vc2, an unstable antisymmetric deformation occurs. However, in this case if v ¼ vc1, there is low amplitude response since the applied moment only induces an antisymmetric motion. The symmetric and antisymmetric deformations are illustrated in Figure 12.8.
EXAMPLE 12.4 Shaft rotor systems neglecting the mass of the shaft Power transmission shafts use rotors to store kinetic energy. We seek the critical speeds in the shaft rotor system depicted in Figure 12.9 below. Suppose the shaft is balanced such that its center of mass is located at its geometric center. The rotor does not induce bending if it is balanced side-to-side as well as top-to-bottom. Top-to-bottom symmetry and side-to-side symmetry are shown below (Figure 12.10). In top-to-bottom balance, each half has the same mass, and the centers of mass are located at the geometric centers of the semicircular disks. In side-to-side balance, the right half disk has its center of mass on the midpoint along the axis, and similarly for the left-hand side (Figure 12.11). Now suppose that the rotor is not balanced. For simplicity suppose that the centers of mass of the top and bottom of the right and left halves lie in the same plane, which of course rotate with the shaft. The unbalance can be modeled as follows (Figure 12.12). Here e1 and h1 represent the horizontal and vertical offsets of the center of mass of the upper right portion of the disk from the geometric center of the rotor, and similarly for e2, e3, e4, h2, h3, h4. The disk imposes a radial force and a radial moment on the shaft at its midpoint. Taking account of the deformation of the shaft, the kinetic energy of the disk is 1 mr 2 ma2 2 Td ¼ þ v 2 4 12 1 m 2 v (w2 þ h1 (w02 )e1 )2 þ (w2 h2 (w02 )e2 )2 þ 2 4 þ (w2 þ h3 þ (w02 )e3 )2 þ (w2 h4 þ (w02 )e4 )2 m, r
E, A, I, r
w a
L/ 2
FIGURE 12.9 Simple shaft rotor system.
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L/ 2
m /2 •
4r0 / 3p 4r0 / 3p
m /2 •
m
FIGURE 12.10 Top-to-bottom balance; rotor has mass m and radius r.
Centers of mass of the left and right half disks are located at their geometric centers
+
FIGURE 12.11 Side-to-side balance.
u m/4
m/4 r
e3 e 1 h3 h4 e4 e2 m/4
h1 h2
m/4 m/4 h3 e3 e1 h1 h4 e4 e2 h2 m/4
m/4
m/4
FIGURE 12.12 Rotor with top-to-bottom and side-to-side unbalance. The first term represents the rigid body rotational kinetic energy of the balanced rotor while the second term represents the unbalance contributions to the (translational) kinetic energy. Neglecting the kinetic energy of the beam, the elastic strain energy is the same as in Cases 1 and 2 of Example 12.3. The two-element finite element model is expressed as
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0 B 192EI =L3 B @ 0 ¼
2
6 0 mv2 6 4 16EI =L
31 e1 þ e2 e3 e4
7C w2 4 7C e21 þ e22 þ e23 þ e24 5A w02 4
1 e1 þ e2 e3 e4 4
h1 h2 þ h3 h4 mv2 4 h1 e1 þ h2 e2 þ h3 e3 h4 e4
The right-hand side represents a force and a moment induced by rigid body motion. If the rotors are perfectly balanced side-to-side (e1 ¼ e2 ¼ e3 ¼ e4 ¼ a=2) the off-diagonal entries of the second matrix vanish. If the rotor is perfectly balanced top-to-bottom, then h1 ¼ h2 ¼ h3 ¼ h4 ¼ 0 and the rigid body contribution vanishes. The critical speeds are obtained by solving 0 B 192EI =L3 detB @ 0
2
6 0 mv2 6 4 16EI =L
1
e1 þ e2 e3 e4 4
31 e1 þ e2 e3 e4 7C 4 7C 2 2 2 2 5A ¼ 0 e1 þ e2 þ e3 þ e4 4
once numerical values have been established for the offsets.
Example 12.4 is now extended to include the effects of the shaft (beam) kinetic energy. Using the previously developed mass matrix for the beam elements (cf. Chapter 11), we have the two-element finite element equation, and again the critical speeds are obtained by finding the determinant of the matrix on the left-hand side of the equation. 0 31 2 e1 þ e2 e3 e4 13 rAL
1 þ 35 m B 192EI =L3 7 C w2 6 0 4 26 B 7C m v 2 2 2 2 @ 5A w0 4 e1 þ e2 e3 e4 e þ e e e 0 16EI =L 1 2 3 4 1 rAL2 2 þ 420 m 4 4
h1 h2 þ h3 h4 mv2 ¼ 4 h1 e1 þ h2 e2 h3 e3 h4 e4
Computation of the two critical speeds is straightforward and is left to the reader as an exercise.
12.3 FINITE ELEMENT ANALYSIS FOR UNCONSTRAINED ELASTIC BODIES 12.3.1 BODY AXES We next consider the three-dimensional response of an elastic body which has no fixed points, as in spacecraft, for example as depicted in Figure 12.13. It is assumed that the tractions are prescribed on the undeformed surface of the elastic body, and that they rotate and translate with the body. The response is thereby referred to the ‘‘body axes,’’ i.e., axes embedded in the corresponding rigid body. An example of body axes is the principal axes of the moment of inertia tensor in the body in Figure 12.13, assuming it is rigid. Otherwise stated, it is assumed that the points of the rigid body coincide with the points of the elastic body in its undeformed configuration.
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u (x,t)
z
x(t)
rc (t)
y x
FIGURE 12.13 Three-dimensional unconstrained 3-D element.
The position vector r of a point in the rotating and translating elastic body may be decomposed as r ¼ rc þ j þ u
(12:17)
in which rc is the position vector to the center of mass, j is the relative position vector from the center of mass to the undeformed position of the current point, in the body system, and u is the displacement from the undeformed to the deformed position, likewise in the body system. It is necessary to develop expressions for the variations of rc, j, and u. To this end we regard rc as being referred to the absolute coordinate system (xyz), in which case we need only write drc. Next j is fixed in the rotating and translating system, so that its variation comes purely from the coordinate system. In analogy with j_ ¼ v j with v ¼ @u @t , we now have dj ¼ du 3 j. Finally, the variation of the displacement vector has contributions from the coordinate system as well as from the fact that it changes relative to the @u rotating system. In analogy with du dt ¼ @t þ v u, the variation of u is expressed as du ¼ dû þ du 3 u, in which dû is the variation of u with the coordinate system instantaneously constrained not to rotate. In order to conduct FEA of an elastic body undergoing unconstrained motion (rigid body translation and rotation), it is necessary to compute the motion of the body axes, which is achieved by integrating the Euler equations presented next.
12.3.2 EULER EQUATIONS
OF A
RIGID BODY
We first restrict attention to rigid bodies for which u ¼ 0. The position vector now has the decomposition r ¼ rc þ j,
dj ¼ du j
(12:18)
Neglecting body forces, equilibrium (referred to the translating=rotating frame) is @S € expressed as @jijj r[€ h i rci þ ji ] ¼ 0. Variation with respect to rc, namely Ð @Sij € drci @jj r[€rci þ ji ] dV ¼ 0, results in ß 2008 by Taylor & Francis Group, LLC.
ð drci
ð @ Sij dV ¼ drci [nj Sij ] dS @ Bj ð ¼ drci ti dS
(12:19)
Ð and we note that F ¼ t dS is the total force exerted by the surface tractions. For the inertial terms ð ð ð (12:20) drci [r[€rci þ €Bi ]] dV ¼ drci r dV €rci þ drci rj€i dV |fflfflffl{zfflfflffl} m T Ðand so drc [F m€rc ] ¼ 0 since the definition of the center of mass implies that € rji dV ¼ 0, and m is the total mass. The first Euler equation may now be stated as
(12:21)
F ¼ m€rc We next consider the effect of the variation dj, ð @ Sij € r[€rci þ j] dV 0 ¼ dji @ jj ð @Sij ¼ 2ipq dup jq r[€rci þ €Bi ] dV @jj
(12:22)
and 2ipq is recognized as the (third order) permutation tensor (cf. Chapter 3). Observe that ð ð 2ipq dup Bq€rci r dV ¼ 2ipq dup jq r dV €rci ¼ 0 (12:23) Next, ð
ð @Sij @ @ dV ¼ 2ipq dup Bq [2ipq dup Bq Sij ] [2ipq dup jq ]Sij dV @jj @jj @jj ð ð @jq Sij dV ¼ nj [2ipq dup jq Sij ] dS 2ipq dup @jj ð ð ¼ dup 2pqi jq ti dV [2ipq dup dqj ]Sij dV
(12:24)
Ð We observe that Mc ¼ j 3 t dS is the total moment of the tractions about the center of Ð mass. Also, [2ipq dup dqj ]Sij dV ¼ 0 since aij ¼ 2ipq dup dqj is antisymmetric. Ð It remains to consider the inertial terms 2ipq dup Bq [r[€rci þ j€i ]] dV. First ð
ð
2ipq dup jq€rci r dV ¼2ipq dup j€q r dV ¼ 0
ß 2008 by Taylor & Francis Group, LLC.
(12:25)
Continuing, ð
ð j dS 2ipq dup jq j€i dV ¼ duT rj € ð ¼ duT rj [v v j þ a j] dV ð ð ¼ duT [(v j) (j v)]r dV þ (j j a)r dV ¼ duT [Ja þ v Jv]
(12:26)
Ð in which J ¼ ZTZr dV is the positive definite (moment of) inertia tensor, and Z is the the antisymmetric tensor satisfying Zb ¼ j 3 b for any 3 3 1 vector b. The second Euler equation now arises as M ¼ Ja þ v Jv If we write jT ¼ {j1 given by
j2
j3} the inertia tensor is found with routine effort to be
ð 2 j 2 þ j2 2
J¼
(12:27)
3
j1 j2
j1 j3
3
6 4 j1 j2
j23 þ j21
7 j2 j3 5r dV
j3 j1
j2 j3
j21
12.3.3 VARIATIONAL EQUATIONS
OF AN
þ
(12:28)
j22
UNCONSTRAINED ELASTIC BODY
The relations below for an elastic body are developed on the assumption that the reference undeformed configuration in the current coordinate system coincides with the rigid body. The balance of linear momentum expressed in the body system is @Sij ¼ r[€rci þ j€i þ €ui ] @jj
(12:29)
Recall that dr ¼ drc þ du 3 (j þ u) þ du0 , in which du0 is the variation of u with the axes instantaneously held fixed. The quantities rc, y, and u0 may be varied independently since there is no constraint relating them. For drc, the variational statement is
ð drci
From the previous section
Ð
@Sij r[€rci þ j€i þ € ui ] dV ¼ 0 @jj h
drci
@Sij @j j
(12:30)
i r[€rci þ j€i ] dV ¼ 0. The term involving the
divergence of the stress becomes the traction term on application of the Divergence
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Theorem. But recall that the tractions are considered to be specified on the undeformed body, the same as in the rigid body. We Ð Ð which renders the tractions ui dV ¼ 0 andÐ also r€ ui dV ¼ 0. Assuming vanishing initial conclude that drci r€ _ it follows that ru dV ¼ 0 with the consequence that the center values of u and u, of mass in the elastic body coincides with that in the rigid body. It likewise follows that in finite element modeling we may impose the constraint u(0,t) ¼ 0 at j ¼ 0. Of course the constraint is applied in the body axis system. Ð the effect of du, namely 0 ¼ 2ipq dup (jq þ uq ) h Secondly, consider i @Sij r[€rci þ j€i þ € ui ] dV. @jj
The assumption that the tractions are prescribed in the undeformed configuration but referred to the current coordinate system implies that the moments are the same in the rigid body and undeformed configuration of the elastic body, and hence that Ð @S 2ipq dup uq @jijj dV ¼ 0. The implication is that ð ð 2ipq dup (jq þ uq )r[j€i þ € ui ] dV ¼ 2ipq dup jq rj€i dV
(12:31)
Assuming small displacements and thereby neglecting products of u and its time derivatives results in the relation ð
ð j r€ u dV þ u r€ j dV ¼ 0
(12:32)
Upon expansion of the accelerations, ð ð 0 ¼ j r€ u dV þ u r€ j dV ð ð ¼ u (v v þ a) jr dV þ j (v v þ a) ur dV ð ð @2u @u þ j 2 r dV þ j 2v r dV @t @t
(12:33)
Ð Ð But j (v v þ a) ur dV ¼ u (v v þ a) jr dV has the consequence that ð j
ð @2u @u r dV ¼ 0 r dV 2v j @t 2 @t
(12:34)
Now the partial time derivative at fixed body axes in general has no relation to the angular velocity vector v describing the rotation of the body axes. We conclude that Ð @ j ur dV ¼ 0, and with suitable initial conditions that @t ð j ur dV ¼ 0
ß 2008 by Taylor & Francis Group, LLC.
(12:35)
Ð Ð The relations ru dV ¼ 0 and j ru dV ¼ 0 have classically been derived by Truesdell and Noll (1965) by assuming that the deformed configuration minimizes kinetic energy.
EXAMPLE 12.5 Ð Write down the proof that, for an unconstrained elastic body, if j ur dV ¼ 0 ð r
ð
d2 j u dV ¼ 0, dt2
rj
d2 u dV ¼ 0 dt 2
Comment: InÐ Chapter 13 Ð we will encounter the Reynolds Transport Theorem which states that dtd rb dV ¼ r ddtb dV , for any vector b. It is used in the solution below.
SOLUTION
Ð Ð Ð d2 j First u dV ¼ [(v v) þ a] j ru dV ¼ 0. Secondly, 0 ¼ dtd j dt 2 r Ð Ð term vanishes by virtue ru dV ¼ v j ru dV þ j r du dt dV . The first right-hand Ð of the problem statement, and we conclude that j r du dt dV ¼ 0. Finally, Ð Ð Ð 2 Ð 2 0 ¼ dtd 2 j ru dV ¼ (v v þ a) j ru dV þ v j r du dV þ j r dtd 2 u dV . dt Ð 2 The first two right-hand terms vanish, in consequence of which j r dtd 2 u dV ¼ 0.
Ð Returning to the general presentation, the rotational relation j ru dV ¼ 0 together with u(0, t) ¼ 0 represents a set of constraints that prevent the center of mass of the elastic body from displacing from that the rigid body, and also prevents the body axes in the elastic body from rotating relative to the body axes of the rigid body. The rotational constraint is global while the center-of-mass constraint can be enforced at a point. To examine the treatment of the rotational constraint we assume the global interpolation model (referred to the body axes) in the form uk ¼ wkl (j)Fmn g n (t ). The constraint is now expressed in the form Yg ¼ 0,
ð Y ¼ 2ijk jj wkl Fmn r dV
(12:36)
But Equation 12.36 may be restated as [Y3,n3
gn3 Y3,3 ] ¼ 0, with g3 ¼ Y1 3,3 Y3,n3 gn3 g3
(12:37)
provided that Y3,3 is nonsingular. We therefore use a global constraint to remove three degrees of freedom. For illustration consider a finite element model in which
Kn3,n3 Kn3,3 Kn3,3 K3,3
gn3 g3
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¼
fn3 f3
and g3 ¼ Y1 3,3 Y3,n3 gn3 (12:38)
It appears that f3 must be an as-yet-unknown reaction force arising to enforce the rotational constraint. If the constraint is used to eliminate g3 in terms of gn–3, we encounter gn–3. The finite element equation to be solved is now [Kn3,n3 Kn3,3 Y1 3,3 Y3,n3 ]gn3 ¼ f n3
(12:39)
and note that the resulting stiffness matrix is generally nonsymmetric. If the reaction forces enforcing the constraints are of interest, they can be computed using f 3 ¼ [Kn3,3 K3,3 Y1 3,3 Y3,n3 ]gn3
12.3.4 PRINCIPLE
OF
VIRTUAL WORK
IN
(12:40)
BODY COORDINATES
h i Ð @S To consider the effect of du0 , recall that du0i @jijj r[€rci þ j€i þ u€i ] dV ¼ 0. First Ð Ð note that du0i r dV €rci ¼ 0 since du0i r dV ¼ 0. The remaining terms are formally the same as for a body with a fixed point, and consequently the variational equations reduce to d2 g dg þ G1 þ [K þ G2 þ A]g ¼ f f rot dt2 dt ð ð T T T G1 ¼ F rwVw dV F, G2 ¼ F rwV2 wT dV F, ð ð A ¼ FT rwAwT dV F, f rot ¼ FT rw[V2 þ A]X 0 ] dV M
(12:41)
The terms in Equation 12.41 appeared in the first section of this chapter and their names and meanings were given.
12.3.5 NUMERICAL DETERMINATION OF THE BODY AXES
OF THE
CURRENT POSITION
For a complete finite element solution, it is necessary to be able to compute the current position of the body axes using the Euler equations, although this is not actually accomplished using finite element equations. Recall that it has been assumed that the tractions, and hence the total force F(t) and moment M(t), are prescribed as functions of time on the undeformed configuration of the elastic body, referred to rotating axes. The current position of the center of mass in the absolute coordinate system is obtained with little effort by integrating m€rc ¼ F(t). The current value of the angular velocity is obtained by integrating the equations dv ¼ J1 [M(t) v Jv] dt dQ ¼ V(t)Q(t) dt
ß 2008 by Taylor & Francis Group, LLC.
(12:42)
the first of which is nonlinear. Recall that v is the axial vector of Q. Suppose the time step in numerical time integration is h, that the time after n steps is denoted by tn ¼ nh, and that the solution has been computed at and before tn. Application of the Trapezoidal Rule and some manipulation serve to derive the integration formula h h h I þ J1 V(tnþ1 J) v(tnþ1 ) ¼ I J1 V(tn )J v(tn ) þ J1 [M(tnþ1 ) þ M(tn )] 2 2 2 1 h h I V(tn ) Q(tn ) (12:43) Q(tnþ1 ) ¼ I þ V(tnþ1 ) 2 2
12.4 APPENDIX: ANGULAR VELOCITY VECTOR IN SPHERICAL COORDINATES This appendix gives an interpretation of the angular velocity vector in terms of the angles appearing in spherical coordinates, as depicted in Figure 12.14. We assume that the spherical coordinate system coincides with the body axes. The xyz axes are a rectilinear inertial (nonchanging) system. The base vector er is collinear with the position vector r. The second vector ej is perpendicular to r and lies in a plane parallel to the x–y plane. The vector ez completes the right-handed rjz system and points toward the z-axis. The base vectors of the body system may be derived as er ¼ cos z(cos jex þ sin jet ) þ sin zeZ ej ¼ sin jex þ cos jet ez ¼ sin z(cos jex þ sin jet ) þ cos zeZ z
ez
er ex
r z
y x x
FIGURE 12.14 Angles used in spherical coordinates.
ß 2008 by Taylor & Francis Group, LLC.
(12A:1)
Let b denote the fixed length vector given by b ¼ aex þ bey þ ced. It may be rotated to the spherical system as b0 ¼ aer þ bej þ cez, which upon making the substitutions according to Equation 12A.1 gives b0 ¼ (cos j cos za sin ja sin z cos jc)ex þ (sin j cos za þ cos jb sin j sin zc)ey þ (sin za þ cos zc)ez
(12A:2)
8 9 8 09 2 cos j cos z a > a> > > > < > = < > = > 6 ¼ Q b ¼ b0 , Q ¼ 6 4 sin j cos z > > > > > : 0> ; : > ; > sin z c c
sin j cos j
sin z cos j
3
7 sin j sin z 7 5 cos z
0
Also 2
sin j cos z cos j
6 _ ¼ 6 cos j cos z Q 4 0
sin z sin j
2
3
cos j sin z 0 cos z cos j
3
7 6 7 _ 6 7_ sin j cos j sin z 7 5j þ 4 sin z sin j 0 sin j cos z 5z 0
cos z
0
sin z (12A:3)
0
But note also that 2
sin j cos z cos j
sin z sin j
32
6 76 6 cos j cos z sin j cos j sin z 76 4 54 0 2
0
32
0
sin z
sin j
cos j cos z
6 76 6 sin z sin j 0 sin j cos z 76 4 54 cos z
sin j cos z sin z
3
2
sin j cos z sin z
sin j
3
3
2
0 0 0 0
7 6 6 0 7 5¼4 0
cos j
0 1 0
7 6 7 6 7 0 7 5 ¼ 41 0 05
cos j
sin z cos j sin j sin z cos z
0
cos j sin z 0 cos z cos j
cos j cos z
sin z cos j sin j sin z cos z
0 0
cos j sin j
cos j
3
7 sin j 7 5 0
_ ¼ Vb0 , Now we regard the angles j, z to be time dependent, in which case b_ 0 ¼ Qb T _ V ¼ QQ . After some manipulation 2 6 V¼4
0 j_ cos jz_
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j_ 0 sin jz_
3 cos jz_ 7 sin jz_ 5 0
(12A:4)
The corresponding angular velocity vector is vT ¼ {sin jz_ cos jz_ j_}. Of course it represents the counterclockwise rotation of the body about its instantaneous position. Referring to the angles introduced in Section 12.1, we now make the identifications 8 9 8 9 < u_ = < sin jz_ = f_ ¼ cos jz_ : _; : _ ; c j
ß 2008 by Taylor & Francis Group, LLC.
(12A:5)
13
Aspects of Nonlinear Continuum Thermomechanics
13.1 INTRODUCTION The first 12 topics of the current monograph have been concerned with linear FEA. The underlying notions of linear solid mechanics and conductive heat transfer were addressed in Chapter 5 and 6. Here, we briefly present several nonlinear topics from continuum thermomechanics, which will enable treating nonlinear finite element techniques thereafter. A more complete account is given, for example, in Chandrasekharaiah and Debnath (1994).
13.2 NONLINEAR KINEMATICS OF DEFORMATION 13.2.1 DEFORMATION GRADIENT TENSOR Displacement: In FEA for finite deformation, it is necessary to carefully distinguish between the current (or ‘‘deformed’’) configuration (i.e., at the current time or load step) and a reference configuration which is usually considered strain-free. Here, both configurations are referred to the same orthogonal coordinate system characterized by the base vectors e1, e2, e3 (see Figure 2.1). Consider a body with volume V and surface S in the current configuration. The particle P occupies a position represented by the position vector x, and experiences (empirical) temperature T. In the corresponding undeformed configuration, the position of P is described by X, and the temperature has the value T0 independent of X. It is now assumed that x is a function of X and t, and T is also a function of X and t. The relations are written as x(X,t) and T(X,t) and it is assumed that x and T are continuously differentiable in X and t through whatever order needed in the subsequent development (Figure 13.1). Displacement Vector: The vector u(X,t) in Equation 13.1 represents the displacement from positions X to x: u(X,t ) ¼ x X
(13:1)
Now consider two close points P and Q in the undeformed configuration. The vector difference XP XQ is represented as a differential dX with squared length
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Y
Y
d c
g h
1 e
X
f a
1 Element at time t
b
X
Element at time t + dt
FIGURE 13.1 Position vectors in deformed and undeformed configurations.
dS2 ¼ dXT dX. The corresponding quantity in the deformed configuration is dx, with ds2 ¼ dxT dx (Figure 13.2). Deformation Gradient Tensor: The deformation gradient tensor F is introduced as dx ¼ F dX,
F¼
@x @X
(13:2)
F satisfies the polar decomposition theorem F ¼ USVT
(13:3)
Deformed
e2
x
x
Undeformed
e1
FIGURE 13.2 Deformed and undeformed distances between adjacent points.
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in which U and V are orthogonal and pffiffiffiffiffiffiffiffi ffi S is a positive definite diagonal tensor whose entries lj, the eigenvalues of FT F, are called the principal stretches. 2
l1 S¼4 0 0
0 l2 0
3 0 0 5 l3
(13:4)
From Equation 13.3 F may be visualized as representing a rotation, followed by a stretch, followed by a second rotation.
13.2.2 LAGRANGIAN STRAIN TENSOR The deformation-induced change in squared length is given by ds2 dS2 ¼ dXT E dX, E ¼ 12 [FT F I]
(13:5)
in which E denotes the Lagrangian strain tensor. Also of interest is the right Cauchy–Green strain C ¼ FTF ¼ 2E þ I. Note that F ¼ I þ @ u=@ X. If quadratic terms in @ u=@ X are neglected the linear strain tensor EL introduced in Chapter 5 is recovered as " T # 1 @u @u þ EL ¼ 2 @X @X
(13:6)
Upon application of Equation 13.3, E is rewritten as h i E ¼ V 12 (S2 I)VT
(13:7)
Under pure rigid body translation and rotation x ¼ b(t) þ QX, we obtain F ¼ Q and E ¼ 12 [QT Q I] ¼ 0, implying that the Lagrangian strain identically vanishes under rigid body motion, unlike the linear strain (see Chapter 5).
EXAMPLE 13.1 F, E, and u in cylindrical coordinates Cylindrical Coordinates: In cylindrical coordinates, the position vector in the undeformed (reference) configuration is given by R ¼ ReR þ ZeZ, with eR ¼ cos Qex þ sin Qey and eQ ¼ sin Qex þ cos Qey. In the deformed (current) configuration, the position vector is given by r ¼ rer þ zez, er ¼ cos uex þ sin uey and eu ¼ sin uex þ cos uey. We first seek F.
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dr ¼ drer þ r du eu þ dz ez @r 1 @r @r ¼ dR þ R dQ þ dZ er @R R @Q @Z @u r @u @u dR þ R dQ þ r dZ eu þ r @R R @Q @Z dz 1 dz @z þ dR þ R dQ þ dZ ez dR R dQ @Z 3 2 @r 1 @r @r 8 90 6 @R R @Q @Z 7 > 7 6 < dR > = 6 @u r @u @u 7 7 ¼ F0 R dQ , F0 ¼ 6 r r 6 @R R @Q > > @Z 7 : ; 7 6 dZ 4 dz 1 dz @z 5 dR
R dQ
@Z
in which the prime calls attention to the fact that the base vectors are still er, eu, ez. Rotation of the er, eu, ez system to coincide with the base eR, eQ, eZ system gives rise to the orthogonal tensor 2
cos(u Q) Q ¼ 4 sin(u Q) 0 T
3 sin(u Q) 0 cos(u Q) 0 5 0 1
We conclude that the deformation gradient tensor is given by F ¼ QTF0 . The displacement vector is now u ¼ [r cos(u Q) R]eR þ r sin ueQ þ (z Z)eZ We now apply the chain rule to ds2 in cylindrical coordinates to obtain the right Cauchy–Green strain tensor C. ds2 ¼ dr dr ¼ dr 2 þ r 2 du2 þ dz2 1 0 dR C B ¼ {dR R dQ dZ}C@ R dQ A dZ in which
2 dr 2 du 2 dz 1 þ r þ , eRR ¼ (cRR 1) dR dR dR 2 2 2 2 1 dr r du 1 dz 1 ¼ þ þ , eQQ ¼ (cQQ 1) R dQ R dQ R dQ 2 2 2 dr du 2 dz 1 ¼ þ r þ , eZZ ¼ (cZZ 1) dZ dZ dZ 2 dr du du r du dz 1 dz 1 ¼ r þ r þ , eRQ ¼ cRQ dR dR dR R dQ dR R dQ 2
cRR ¼ cQQ cZZ cRQ
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cQZ cZR
1 dr dr r du du 1 dz dz þ r þ , ¼ R dQ dZ R dQ dZ R dQ dZ dr dr du du dz dz þ r r þ , ¼ dZ dR dZ dR dZ dR
1 eQZ ¼ cQZ 2 1 eZR ¼ cZR 2
Of course the Lagrangian strain tensor is obtained from E ¼ 12 (C I), with Q not appearing.
13.2.3 VELOCITY GRADIENT TENSOR, DEFORMATION RATE TENSOR, AND SPIN TENSOR We now introduce the particle velocity v ¼ @x=@t and assume that it is an explicit function of x(t) and t. The velocity gradient tensor L is introduced using dv ¼ L dx, from which dv dx dv dX ¼ dX dx _ 1 ¼ FF
L¼
(13:8)
Its symmetric part, called the deformation rate tensor, is D ¼ 12 L þ LT
(13:9)
It may be regarded as a strain rate referred to the current configuration. The corresponding strain rate referred to the undeformed configuration is the Lagrangian strain rate: T E_ ¼ 12 FT F_ þ F_ F n o _ 1 þ FT F_ T F ¼ FT 1 FF 2
¼ FT DF
(13:10)
The antisymmetric portion of L is called the spin tensor W: W ¼ 12 [L LT ]
(13:11)
Suppose the deformation consists only of a time-dependent rigid body motion expressed by x(t) ¼ Q(t)X þ b(t),
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QT (t)Q(t) ¼ I
(13:12)
_ T, and recall from Chapter 12 Clearly, F ¼ Q and E ¼ 0, D ¼ 0 and L ¼ W ¼ QQ T _ that QQ is antisymmetric.
EXAMPLE 13.2 v, L, D, and W in cylindrical coordinates The velocity vector in cylindrical coordinates is @r @u @z er þ r eu þ ef @t @t @t ¼ vr e_ r þ vu eu þ vf ef
v¼
Observe that dv ¼ dvr er þ dvu eu þ dvz ez þ vr der þ vu deu h i h i vu vr ¼ dvr r du er þ dvu þ r du eu þ dvz ez r r Now converting to matrix–vector notation, 1 0 dvr dr þ 1 dvr r du þ dvr dz vu r du r r du dz C B dr C B dv dv d v v 1 u u u r B dv ¼ B dr þ r du r du þ dz dz r r du C C d r A @ dvz dv dv 1 z z dr dr þ r du r du þ dz dz 0
2
1
dvr dr 6 dr 6 dv B C u ¼ L@ r du A, L ¼ 6 6 dr 4 dz dvz dr
1 r 1 r
dvr vu r du dvu þ vr r du d v 1 z r du
3 dvr dz 7 dvu 7 7 dz 7 5 dvz dz
Of course D and W are obtained as the symmetric and antisymmetric portions of L.
13.2.4 DIFFERENTIAL VOLUME ELEMENT The volume implied by the differential position vector dR is given by the vector triple product dV0 ¼ dX1 dX2 dX3 ¼ dX1 dX2 dX3 dX1 ¼ dX1 e1 , dX2 ¼ dX2 e2 , dX3 ¼ dX3 e3
(13:13)
dx
The vectors dXi deform into dxj ¼ dXji ej dXi . The deformed volume is now readily verified as dV ¼ dx1 dx2 dx3 ¼ JdV0 ,
and J as before is called the Jacobian.
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1
J ¼ det(F) ¼ det2 (C)
(13:14)
The time derivative of J is p prom ffiffiffiffi inent in incremental formulations in continuum mechanics. Recalling that J ¼ I3 , we have d 1 dI3 J¼ dt 2J dt ¼ J2 tr (C1 C_ ) _ þ tr (F1 FT F_ F)] ¼ J2 [tr(F1 F) ! T F1 F_ þ FT F_ ¼ J tr 2 T
¼ J tr(D)
(13:15)
EXAMPLE 13.3 Relation of D to E_ _ E, and F. First differentiate E to find We now express D and its trace in terms of E, T E_ ¼ 12 (FT F_ þ F_ F)
from which _ 1 ¼ 1 (FF _ 1 þ FT F_ T ) FT EF 2 ¼D Next, _ 1 ) tr (D) ¼ tr(FT EF 1 _ FT ) ¼ tr(EF ¼ tr(E_ (I þ 2E)1 )
13.2.5 DIFFERENTIAL SURFACE ELEMENT Let dS denote a surface element in the deformed configuration, with exterior unit normal n illustrated in Figure 13.3. The counterparts from the reference configuration are dS0 and n0. A surface element dS obeys Nanson’s theorem (cf. Chandrasekharaiah and Debnath, 1994) n dS ¼ JFT n0 dS0 Taking the magnitude of n dS, we conclude that
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(13:16)
Q⬘ P⬘
ds
Q∗ P∗
dS
FIGURE 13.3 Differential length changes.
dS ¼ J
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nT0 C1 n0 dS0 ,
FT n0 n ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nT0 C1 n0
(13:17)
Of course during deformation the surface normal changes direction (cf. Figure 13.4), a fact which is important, for example, in contact problems. For later use, in incremental variational methods we consider the differential dn dt and d(n dS): d dJ dFT n0 dS0 [n dS] ¼ FT n0 dS þ J dt dt dt
(13:18)
But, recalling Equation 13.15, dJ ¼ J tr(D) dt dJ T F n0 dS0 ¼ tr(D)JFT n0 dS0 dt ¼ tr(D)n dS
(13:19)
n n0
dX2
dS0 dX1
dx2 dS dx1
FIGURE 13.4 Undeformed and deformed surface patches.
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Also, since d(FTFT) ¼ 0, dFT dFT T ¼ FT F dt dt ¼ LT FT
(13:20)
Finally, we have d[n dS] ¼ [tr(D)I LT ]n dS dt
(13:21)
EXAMPLE 13.4 Referring to Figure 13.5, determine u, F, J, and E as functions of X and Y: use H ¼ 1, W ¼ 1, a ¼ 0.1, b ¼ 0.1, c ¼ 0.3, d ¼ 0.2, e ¼ 0.2, f ¼ 0.1. Assume a unit thickness in the Z-direction in both the deformed and undeformed configurations.
SOLUTION A deformation model which captures the fact that straight sides remain straight is given in the form assumed in the form x ¼ aX þ bY þ gXY,
y ¼ dX þ «Y þ zXY
It is necessary to determine a, b, g, d, «, and z from the coordinates of the vertices in the deformed configuration. At (X, Y) ¼ (W, 0): At (X, Y) ¼ (0, H):
W þ a ¼ aW, e ¼ bH,
b ¼ dW H þ f ¼ «H
At (X, Y) ¼ (W, H):
W þ c ¼ aW þ bH þ gWH,
H þ d ¼ dW þ «H þ zWH
Y
Y
c e
d
H
H f
b W Undeformed plate
X
W
X a
Deformed plate
FIGURE 13.5 Plate elements in undeformed and deformed states.
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After elementary manipulations, a e b ¼ 1:1, b ¼ ¼ 0:2, d ¼ ¼ 0:1 W H W f W þ c (W þ a) e H þ d b (H þ f ) « ¼ 1 þ ¼ 1:2, g ¼ ¼ 0:4, z ¼ ¼0 H WH WH
a¼1þ
The 2 3 2 deformation gradient tensor F and its determinant now are: " F¼
a þ gY
b þ gX
d þ zY
« þ zX
"
# ¼
1:1 þ 0:4Y
0:2 þ 0:4X
0:1
1:2
#
J ¼ 1:3 þ 0:48Y 0:04X The displacement vector is u¼
xX yY
¼
(a 1)X þ bY þ gXY dX þ (« 1)Y þ zXY
¼
0:1X þ 0:2Y þ 0:4XY 0:1X þ 0:2Y
The Lagrangian strain E ¼ 12 [FT F I] is found to be E¼ ¼
" #" 1 1:1 þ 0:4Y 0:1 1:1 þ 0:4Y 2 0:2 þ 0:4X 1:2 0:1 " 0:11 þ 0:44Y þ 0:08Y 2
0:2 þ 0:4X 1:2
#
" # 1 1 0 2 0 1
0:17 þ 0:22X þ 0:04Y þ 0:08XY
0:17 þ 0:22X þ 0:04Y þ 0:08XY
#
0:24 þ 0:08X þ 0:08X 2
EXAMPLE 13.5 Figure 13.6 shows a square element at time t and at t þ dt. Estimate L, D, and W at time t. Use a ¼ 0.1 dt, b ¼ 1 þ 0.2 dt, c ¼ 0.2 dt, d ¼ 1 þ 0.4 dt, e ¼ 0.05 dt, f ¼ 0.1 dt, Y
Y
d c
g h
1
X 1 Element at time t
e
f a
b Element at time t + dt
FIGURE 13.6 Element experiencing rigid body motion and deformation.
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X
g ¼ 1 0.1 dt, h ¼ 1 þ 0.5 dt. Assume a unit thickness in the Z-direction in both the deformed and undeformed configurations.
SOLUTION First represent the deformed position vectors in terms of the undeformed position vectors using eight coefficients to be determined using the given geometry. In particular, x ¼ a þ bX þ gY þ dXY ,
y ¼ « þ zX þ hY þ uXY
This expression likewise captures the fact that straight sides remain straight. Following procedures analogous to Example 13.3, we find a ¼ 0:1 dt , « ¼ 0:05 dt b ¼ 1 þ 0:2 dt, z ¼ 0:1 dt g ¼ 0:2 dt , h ¼ 1 0:1 dt d ¼ 0:5 dt , u ¼ 0:05 dt The velocities may be estimated using vx
xX dt
and vy
yY dt ,
from which
vx ¼ 0:1 þ 0:2X þ 0:2Y 0:5XY vy ¼ 0:05 þ 0:1X 0:1Y 0:05XY The tensors L, D, and W are now readily found as " L¼ " D¼ " W¼
0:2 0:5Y
0:2 0:5X
0:1 0:05Y
0:1 0:05X
#
0:2 0:5Y
0:15 0:25X 0:025Y
0:15 0:25X 0:025Y
0:1 0:05X
0
0:4 0:25X þ 0:025Y
0:4 þ 0:25X 0:025Y
0
# #
13.3 MECHANICAL EQUILIBRIUM AND THE PRINCIPLE OF VIRTUAL WORK 13.3.1 TRACTION VECTOR
AND
STRESS TENSORS
A stress tensor was previously introduced in Chapter 5. However, it was in the context of small deformation, in which case it was not indicated how the stress changes when referred to the undeformed as opposed to the current (deformed) configuration. The distinction is central to the development below, leading to the Cauchy stress tensor (current configuration) and the first and second Piola–Kirchoff stress tensor. Cauchy Stress: We consider a differential tetrahedron enclosing the point x in the deformed configuration, as illustrated in Figure 13.7. The area of the inclined face
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x3
dx3
x dx2
x2
dx1
x1
FIGURE 13.7 Differential tetrahedron.
is dS, and dSi is the area of the face whose exterior normal vector is ei. Simple vector analysis serves to derive that ni ¼ dSi=dS, see Example 2.5. Now referring to Figure 13.8, let dF denote the force on the surface element dS, and let dF(i) denote the force on area dSi. As the tetrahedron shrinks to a point, the contribution of volume forces suchPas inertia decays faster than surface forces. Balance of forces requires that dFj ¼ dFj(i) . i
3
dF n dS (2) dF (2)
dS
dF (1) dS (1) 2
dS (3)
dF (3)
1
FIGURE 13.8 Forces applied to a differential tetrahedron.
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The traction vector acting on the inclined face is defined by t¼
dF dS
(13:22)
from which tj ¼
X dFj(i) dSi dSi dS i
¼ Tij ni
(13:23)
in which Tij ¼
dPj(i) dSi
(13:24)
It is readily seen that Tij can be interpreted as the intensity of the force acting in the j direction on the facet pointing in the i direction, and is recognized as ijth entry of the Cauchy stress T. In matrix–vector notation the stress–traction relation is written as (13:25)
t ¼ TT n
In Section 13.3.2 it will be seen that T is symmetric by virtue of the balance of angular momentum. Equation 13.24 implies that TT is a tensor, from which it follows that T is a tensor. In traditional depictions the stresses on the back faces are represented by arrows pointing in negative directions. However, this depiction can be confusing— the arrows actually represent the directions of the traction components. Consider the one-dimensional member in Figure 13.9. The traction vector te1 acts at x ¼ L, while the traction vector te1 at x ¼ 0. At x ¼ L, the corresponding stress is t11 ¼ te1 e1 ¼ t. At x ¼ 0 the stress is given by (te1) (e1) ¼ t. Clearly, the stress at both ends, and in fact throughout the member, is positive (tensile).
Y
t
t
z
FIGURE 13.9 Tractions on a bar experiencing uniaxial tension.
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x
We will see later that the stress tensor is symmetric by virtue of the balance of angular momentum. First Piola–Kirchhoff Stress Tensor: Transformation to undeformed coordinates is now considered. From the transformation properties of a surface element, we have t dS ¼ TT n dS ¼ TT JFT n0 dS0 n0 dS0 , ¼S T
¼ JF1 T S
(13:26)
is known as the first Piola–Kirchhoff stress tensor and it is not symmetric. S Second Piola–Kirchhoff Stress: We next derive the stress tensor which is conjugate to the Lagrangian strain rate, i.e., gives the correct amount of work per unit undeformed volume. At a segment dS at x on the deformed boundary, assuming _ of the tractions is static conditions the rate of work dW _ ¼ dFT u_ dW ¼ tT u_ dS
(13:27)
Over the surface S, shifting to tensor-indicial notation and invoking the divergence theorem, we find ð _ ¼ tT u_ dS W ð ¼ Tij ni u_ j dS ð ð @ u_ j @Tij Tij dV þ u_ j dV ¼ @xi @xi
(13:28)
@T
We will shortly see that static equilibrium implies that @xiji ¼ 0, which will enable us to conclude that ð ð ð _ ¼ @ u_ j Tij dV ¼ tr(TL) dV ¼ tr(TD) dV (13:29) W @xi To convert to undeformed coordinates, note that ð _ ¼ tr(TD) dV W ð _ T ) dV0 ¼ tr(JTF1 EF ð _ dV0 ¼ tr(JFT TF1 E) ð _ dV, ¼ tr(SE) S ¼ JFT TF1
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(13:30)
and the tensor S is called the second Piola–Kirchhoff stress tensor. It is symmetric if T is symmetric, which we will shortly see to be the case.
EXAMPLE 13.6 At point (0,0,0) the tractions t1, t2, t3 act on planes with normal vectors n1, n2, and n3. Find the Cauchy stress tensor T. Given: n1 ¼ p1ffiffi3 [e1 þ e2 þ e3 ],
t1 ¼ p1ffiffi3 [6e1 þ 9e2 þ 12e3 ]
n2 ¼ p1ffiffi3 [e1 þ e2 e3 ],
t2 ¼ p1ffiffi3 [0e1 þ 1e2 þ 3e3 ]
n3 ¼ p1ffiffi3 [e1 e2 e3 ],
t3 ¼ p1ffiffi3 [4e1 þ 5e2 þ 6e3 ]
SOLUTION This problem requires application of the stress–traction relation. Now 0 1 2 6 T 1 @ A 4 11 pffiffiffi 9 ¼ T21 3 12 T 31
T12 T22 T32
3 0 1 T13 1 1 T23 5 pffiffiffi @ 1 A 3 1 T33
which gives T11 þ T12 þ T13 ¼ 6
(13:31)
T21 þ T22 þ T23 ¼ 9
(13:32)
T31 þ T32 þ T33 ¼ 12
(13:33)
Similarly, 0 1 2 0 T 1 @ A 4 11 pffiffiffi 1 ¼ T21 3 2 T 31
T12 T22 T32
3 0 1 1 T13 1 T23 5 pffiffiffi @ 1 A 3 1 T33
from which T11 þ T12 T12 ¼ 0
(13:34)
T21 þ T22 T23 ¼ 1
(13:35)
T31 þ T32 T33 ¼ 2 0 1 2 3 4 T11 T12 T13 1 1 @ 1 pffiffiffi 5 A ¼ 4 T21 T22 T23 5 pffiffiffi @ 1 A 3 6 3 1 T31 T32 T33
(13:36)
0
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1
and now T11 T12 T13 ¼ 4
(13:37)
T21 T22 T23 ¼ 5
(13:38)
T31 T32 T33 ¼ 6
(13:39)
It is elementary to attain the solution, which 2 1 T ¼ 42 3
is 3 2 3 3 45 4 5
13.3.2 STRESS FLUX Consider two deformations x1 and x2 differing only be a rigid body motion: x2 ¼ V(t)x1 þ b(t)
(13:40)
in which V(t) is an arbitrary orthonormal tensor. A tensor A(x) is objective (cf. Eringen, 1962) if A(x2 ) ¼ VA(x1 )VT
(13:41)
If a tensor is objective, the differences seen by observers at x1 and x2 are accounted for by the transformations relating the two associated motions. _ is not It turns out that the matrix of time derivatives of the Cauchy stress, T, objective, while the deformation rate tensor D is. Accordingly, if j is a fourth-order tensor, a constitutive equation of the form T_ ¼ jD would be senseless. Instead, the time derivative of T is replaced with an objective stress flux, as explained below. First note that F2 ¼ VF1 L2 ¼ F_ 2 F1 2 T T _ ¼ VV VF1 þ VF_ 1 F1 1 V _ T ¼ V þ VL1 VT , V ¼ VV
(13:42)
The tensor V is antisymmetric since dI=dt ¼ 0 ¼ V þ VT. Clearly, the tensors F and L are not objective. We seek a stress flux affording the simplest conversion from deformed to undeformed coordinates. To this end we examine the time derivative of the second Piola–Kirchhoff stress. dS d ¼ [JFT TF1 ] dt dt
T _ 1 þ JFT TF_ 1 ¼ Jtr(D)FT TF1 þ J F_ TF1 þ JFT TF
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(13:43)
_ ¼ 0 so that F_ 1 ¼ F1 FF _ 1 and F_ T ¼ FT F_ T FT . Continuing, But (F1 F) dS ¼ JFT T˚ F1 dt
(13:44)
˚ ¼ T_ þ tr(D)T LT TLT T
(13:45)
in which
is known as the Truesdell stress flux. Under pure rotation F ¼ Q and S_ ¼ JQT˚ QT . To prove the objectivity of T˚ , note that ˚ 2 ¼ T_ 2 þ T2 tr(D2 ) L2 T2 T2 LT T 2 T . T ¼ VT1 V þ VT1 V tr(D1 ) VL1 VT þ V VT1 VT VTT1 VT VL1 T VT V ¼ VT_ 1 VT þ VVT1 VT VT1 VT V þ VT1 VT tr(D1 ) VL1 T1 VT VT1 L1 T VT VVT1 VT þ VT1 VT V ¼ VT˚ 1 VT
(13:46)
as desired. The choice of stress flux is not unique. For example, the stress flux given by ˚ T tr(D) T˚ ¼ T ¼ T_ LT TLT
(13:47a)
is also objective, as is the widely used Jaumann stress flux 4
T ¼ T_ þ TW WT,
W ¼ 12 (L LT )
(13:47b)
EXAMPLE 13.7 Relate the Jaumann stress flux to the Truesdell stress flux and to the rate of the second Piola–Kirchhoff stress
SOLUTION Expanding the velocity gradient tensor in the Truesdell stress flux gives ˚ ¼ T_ þ T trD LT TLT T ¼ T_ þ T trD (D þ W)T T(D W)
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¼ T trD DT TD þ (T_ þ TW WT) 4
¼ T trD DT TD þ T Secondly, _ _ þ SE(2E _ JF1 T˚ FT ¼ S_ JS tr E(2E þ I)1 {(2E þ I)1 ES þ I)1 } We will discuss this result further in conjunction with incremental stress–strain relations. It is clear that, if the Jaumann stress flux is used, conversion to undeformed coordinates introduces increments of strain not being proportional to increments of stress.
13.3.3 BALANCE
OF
MASS, LINEAR MOMENTUM,
AND
ANGULAR MOMENTUM
Balance of Mass: Balance of mass requires that the total mass of an isolated body not change: d dt
ð r dV ¼ 0
(13:48)
in which r(x,t) is the mass density. Since dV ¼ J dV0, it follows that rJ ¼ r0. Reynolds Transport Theorem: This useful principle is a consequence of balance of mass. Let w(x,t) denote a vector-valued function. Conversion of the volume integral from deformed to undeformed coordinates is simply achieved as Ð Ð rw(x,t) dV ¼ r0 w(x,t) dV0. The Reynolds Transport Theorem follows as V
d dt
ð
V0
ð d 1 d r w(x, t) þ (rJ) w(x, t) dV dt J dt ð d ¼ r w(x, t) dV dt
rw(x,t) dV ¼
(13:49)
Balance of Linear Momentum: In a fixed coordinate system, balance of linear momentum requires that the total force on a body with volume V and surface S be equal to the rate of change of linear momentum. Assuming that all force is applied on the exterior surface, the equation of interest is ð ð d du F ¼ t dS ¼ r dV dt dt
(13:50)
Invoking the Reynolds Transport Theorem yields ð
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ð 2 du t dS ¼ r 2 dV dt
(13:51)
In current coordinates, application of the divergence theorem (Equation 3.18) furnishes the equilibrium equation ð
ð t dS ¼ TT n dS ð ¼ [rT T]T dV
(13:52)
Equation 13.51 now becomes ð
d2 uT r T r 2 dV ¼ 0T dt T
(13:53)
Since this equation applies not only to the whole body but to arbitrary subdomains of the body, the argument of the integral in Equation 13.53 must vanish pointwise: rT T ¼ r
d 2 uT dt 2
(13:54)
To convert to undeformed coordinates the first Piola–Kirchhoff stress is invoked to furnish ð
ð 2 T n0 dS0 ¼ r0 d u dV0 S dt 2
(13:55)
and the divergence theorem furnishes ¼r rT0 S 0
d2 uT dt 2
(13:56)
in which r0 denotes the divergence operator referred to undeformed coordinates. This equation will later be the starting point in the formulation of incremental variational principles. Balance of Angular Momentum: Assuming that only surface forces are present, relative to the origin and a fixed coordinate system the total moment of the traction is equal to the rate of change of angular momentum: ð x t dS ¼
d dt
ð x rx_ dV
(13:57)
To examine this principle further it is convenient to use tensor-indicial notation. First note using the divergence theorem that
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ð
ð x t dS ¼ «ijk xj tk dS ð ¼ «ijk xj Tlk nl dS ð @ («ijk xj Tlk ) dV ¼ @x ð l ð @ ¼ «ijk djl Tlk dV þ «ijk xj Tlk dV @ xl ð ð @ Tlk dV ¼ «ijk Tjk dV þ «ijk xj @ xl
(13:58)
and of course «ijk is the third-order permutation tensor introduced in Chapter 3. Continuing, d dt
ð
ð
ð
x rx_ dV ¼ x r€ x dV þ x_ rx_ dV ð ¼ x r€ x dV ð ¼ «ijk xj r€xk dV
(13:59)
Balance of angular momentum may thus be restated as
@ 0 ¼ «ijk Tjk dV þ «ijk xj Tlk r€xk dV @ xl ð
ð
(13:60)
The second term vanishes by virtue of balance of linear momentum (Equation 13.54), leaving «ijk Tjk ¼ 0
(13:61)
which implies that T is symmetric: T ¼ TT (Example 2.6). Note also that S is also is not symmetric. symmetric but that S
13.4 PRINCIPLE OF VIRTUAL WORK UNDER LARGE DEFORMATION The balance of linear and the balance of angular momentum lead to auxiliary variational principles that are fundamental to the finite element method. Variational methods were introduced in Chapter 4. We recall the balance of linear momentum in rectilinear coordinates as @ Tkl r€ uk ¼ 0 @xl
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(13:62)
and Tlk ¼ Tkl by virtue of the balance of angular momentum. We have tacitly assumed that €x ¼ €uk which is to say that deformed positions are referred to a coordinate system that does not translate or rotate. A variational principle is sought from
ð duk
@ Tkl r€ uk dV ¼ 0 @xl
(13:63)
in which duk is an admissible (i.e., consistent with constraints) variation of uk. We consider the spatial dependence of uk to be subjected to variation, but not the temporal dependence. For example, if uk can be represented, at least locally, as u ¼ NT (x)g(t) then duk ¼ [NT (x)]kl dgl (t)
(13:64)
Ð The second term in the variational equation simply remains as duk r€uk dV. The first term Ð is integrated by parts Ð konce for reasons which will be identified shortly: it becomes @x@ l [duk Tlk ] dV @du @xl Tlk dV. From the divergence theorem, ð
ð @ [duk Tlk ] dV ¼ nl [duk Tlk ] dS @xl ð ¼ duk tk dS
(13:65)
which may be interpreted as the virtual work of the tractions on the exterior boundary. Next, since T is symmetric, ð
ð @duk Tlk dV ¼ d 3kl Tlk dV @xl
1 @uk @ul þ 3kl ¼ 2 @xl @xk
(13:66)
Ð and we call 3kl the Eulerian strain. The term d3kl Tlk dV may be called the virtual Ð work of the stresses. Next, to evaluate duk Tk dS we suppose that the exterior boundary consists of three zones: S ¼ S1 þ S2 þ S3. On S1 the displacement uk is prescribed, causing the integral over S1 toÐ vanish. On S2 suppose that the traction is prescribed as tk(s). The contribution is S2duk tk (s) dS. Finally on S3 suppose that Ð Ð tk ¼ tk (s) [A(s)]kl ul (s), furnishing S3duk tk (s) dS S3duk [A(s)]kl ul (s) dS. The various contributions are consolidated into the large deformation form of the principle of virtual work (Zienkiewicz and Taylor, 1989) in current (deformed) coordinates as
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ð
ð
uk dV d 3kl Tlk dV þ duk r€ ð ð ð ¼ duktk (s) dS þ duktk (s) dS duk [A(s)]kl ul (s) dS S2
S3
(13:67)
S3
Now consider the case in which, as in classical elasticity, Tlk ¼ dlkmn 3mn, T ¼ D 3
(13:68)
in which dlkmn are the entries of a fourth-order constant positive definite tangent modul us tensor D. The first term in Equation 13.67 now becomes Ð d 12 3kl dlkmn 3mn dV , with a positive definite integrand. Achieving this outcome is the motivation behind integrating by parts once. In the language of Chapter 4, uk is the primary variable and tk is the secondary variable: on any boundary point either uk or tk is typically specified, and tk may be specified as a function of uk. Finally, application of the interpolation model (Equation 13.64) and cancellation of dgT furnishes the ordinary differential equation € þ [K þ H]g ¼ f (t ) Mg
(13:69)
in which ð ð K ¼ BDBT dV M ¼ N(x)NT (x)r dV , ð ð H ¼ N(x)ANT (x) dV , f (t ) ¼ N(s)tk (s) dS
(13:70)
S2 þS 3
S3
and in which VEC(T) ¼ xVEC(3), x ¼ TEN22(D). Also VEC(3) ¼ BT(x)g, and B is derived from the strain–displacement relations, M is the positive definite mass matrix, K is the positive definite matrix representing the domain contribution to the stiffness matrix, H is the boundary contribution to the stiffness matrix, and f is the consistent force vector. These notions will be addressed in greater detail in subsequent chapters. Owing to its importance in nonlinear FEA we go into considerable detail to convert the foregoing large deformation form of the Principle of Virtual Work to undeformed coordinates. First ð ð uk dV0 duk r€uk dV ! duk r0 € ð ð ð ð 0 duk tk (s) dS þ duk tk (s) dS ! duk tk (s0 ) dS0 þ duktk0 (s0 ) dS0 (13:71) S2
S3
S20
S30
using tk0 ¼ mtk
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m¼J
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nT0 C1 n0
The traction relation t0 ¼ mt is seen from the fact that t0 dS0 ¼ t dS ¼ tm dS0. Next, ð
ð duk [A(s)]kl ul (s) dS ! S3
duk [mA(s)]kl ul (s0 ) dS0
(13:72)
S3
Some manipulation is required to convert the virtual work of the stresses. Observe that " # 1 @du @du T d3 ¼ þ 2 @x @x " T # 1 @du @X @du @X ¼ þ 2 @X @x @X @x ¼ 12 [dFF1 þ FT dFT ] ¼ FT 12 [FT dF þ dFT F] F1 ¼ FT dEF1
(13:73)
Third ð
ð d 3kl Tlk dV ¼ tr(d 3 T) dV ð ¼ tr(FT dEF1 T)J dV0 ð ¼ tr(dEJF1 TFT ) dV0 ð ¼ tr(dES) dV0 ð ¼ dEji Sij dV0
(13:74)
Consolidating the foregoing terms the Principle of Virtual Work in undeformed coordinates is ð
ð
ð
dEji Sij dV0 þ duk r 0 € uk dV0 ¼ S20
duktk0 (s0 ) dS0
ð
þ S30
duktk0 (s0 ) dS0
duk [mA(s0 )]kl ul (s0 ) dS0 S30
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ð
(13:75)
EXAMPLE 13.8 Given the Cauchy stress tensor T, find the first and second Piola–Kirchhoff stress tensors if x(t) ¼ Q(t)D(t)X, in which 2
1 þ at D(t) ¼ 4 0 0
3 0 0 5 1 þ ct
0 1 þ bt 0
Assume Q represents a plane rotation about the z-axis.
SOLUTION We know that the deformation gradient tensor is given by F ¼ dx(t) dX . Hence, 2
cos u
6 F ¼ Q(t)D(t) ¼ 4 sin u 0
sin u
0
32
76 cos u 0 54 0 1
1 þ at
0
0 0
1 þ bt 0
2
(1 þ at) cos u (1 þ bt) sin u 6 ¼ 4 (1 þ at) sin u (1 þ bt) cos u 0 0
3
0
7 0 5 1 þ ct
3 0 7 0 5 (1 þ ct)
and also 2
1
F
cos u 6 (1 þ at) 6 6 sin u ¼6 6 (1 þ at) 6 4 0
sin u (1 þ bt) cos u (1 þ bt)
0
3 0 0 1 (1 þ ct)
7 7 7 7 7 7 5
The Jacobian J ¼ det F is given by J ¼ det(F) ¼ (1 þ at)(1 þ bt)(1 þ ct) The first Piola–Kirchhoff stress is now given by 2 cos u sin u 6 (1 þ at) (1 þ bt) 6 sin u cos u 6 1 S ¼ JF T ¼ (1 þ at)(1 þ bt)(1 þ ct)6 6 (1 þ at) (1 þ bt) 4 0 0 and the second Piola–Kirchhoff stress is obtained immediately as
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3 0 0 1 (1 þ ct)
7 7 7 7T 7 5
S ¼ JF1 TFT
2 cos u sin u 6 (1 þ at ) ( 1 þ bt ) 6 6 sin u cos u 6 ¼ (1 þ at)(1 þ bt )(1 þ ct)6 6 (1 þ at ) (1 þ bt ) 4 0 0 3 2 cos u sin u 0 7 6 (1 þ at ) (1 þ bt) 7 6 7 6 sin u cos u 7 T6 0 7 6 (1 þ at ) (1 þ bt) 7 6 4 1 5 0 0 (1 þ ct )
3 0 0 1 (1 þ ct )
7 7 7 7 7 7 5
13.5 NONLINEAR STRESS–STRAIN–TEMPERATURE RELATIONS: THE ISOTHERMAL TANGENT MODULUS TENSOR 13.5.1 CLASSICAL ELASTICITY Under small deformation, the fourth-order tangent modulus tensor D in linear elasticity is defined implicitly by dT ¼ D dEL
(13:76)
in which EL is the small strain tensor. In linear isotropic elasticity, the stress–strain relations are written in the Lamé form as T ¼ 2mEL þ l tr (EL )I
(13:77)
Using Kronecker product notation from Chapter 3, Equation 13.77 may be rewritten as VEC (T) ¼ 2mI IEL þ liiT VEC (EL ) (13:78) from which we conclude that D ¼ ITEN 22 2mI I þ liiT
(13:79)
13.5.2 COMPRESSIBLE HYPERELASTIC MATERIALS In isotropic hyperelasticity, which is descriptive of compressible rubber, the second Piola–Kirchhoff stress is taken to be derivable from a strain energy function w which depends on the principal invariants I1, I2, I3 of the right Cauchy–Green strain tensor, introduced in Chapter 3. S¼
dw dw dw dw ¼2 , sT ¼ ¼2 dE dC de dc
s ¼ VEC(S),
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e ¼ VEC(E),
c ¼ VEC(C)
(13:80a) (13:80b)
Simple manipulation serves to obtain s ¼ 2fi ni ,
fi ¼
@w , @ Ii
nTi ¼
@ Ii @c
(13:81)
From Chapter 3, Section 3.6.8, n1 ¼ i, n2 ¼ I1 i c, n3 ¼ VEC (C1 )I3
(13:82)
The tangent modulus tensor D0 referred to the undeformed configuration is given by dS ¼ D0 d E,
ds ¼ TEN 22(D0 ) de
(13:83)
and furthermore TEN 22(D0 ) ¼ 4fij ni nTj þ 4fi Ai ,
Ai ¼
dni dc
(13:84)
Finally, recalling Chapter 3, dn1 ¼0 dc dn2 A2 ¼ dc d ¼ [I1 i c] dc ¼ iiT I9 A1 ¼
d VEC (C1 )I3 dc d ¼ I2 i I1 c þ VEC (C2 ) dc ¼ inT2 ciT þ C C ¼ I1 iiT I9 icT þ ciT þ C C
(13:85a)
(13:85b)
A3 ¼
13.5.3 INCOMPRESSIBLE
AND
(13:85c)
NEAR-INCOMPRESSIBLE HYPERELASTIC MATERIALS
Polymeric materials such as natural rubber are often nearly incompressible. For some applications they may be idealized as incompressible. But for applications involving confinement, such as in the corners of seal wells, it may be necessary to accommodate the small degree of compressibility to achieve high accuracy in the stresses. Incompressibility and near-incompressibility represent internal constraints. The principal (Eulerian) strains are not independent, and the (Cauchy) stresses are not determined completely by the strains, instead, differences in the principal stresses are determined by differences in principal strains (Oden, 1972).
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An additional field is introduced to enforce the internal constraint, and we will see that this internal field may be identified as the hydrostatic pressure (referred to the current configuration). 13.5.3.1
Incompressibility
The constraint of incompressibility is expressed by the relation J ¼ 1 and note that J ¼ det F qffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ det2 (F) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ det(F) det(FT ) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ det2 (FT F) pffiffiffiffi ¼ I3
(13:86)
and consequently the constraint of incompressibility may be restated as I3 ¼ 1. The constraint I3 ¼ 1 may be enforced using a Lagrange multiplier (Oden, 1972) denoted here as p. The multiplier depends on X and is in fact the additional field just mentioned. Oden (1972) proposed introducing an augmented strain energy function w0 similar to 1 w0 ¼ w I10 ,I20 p(J 1), 2
I10 ¼
I1 1=3 I3
, I20 ¼
I2
(13:87)
2=3
I3
in which w is interpreted as the conventional strain energy function but with dependence on I3(¼1) removed. I10 and I20 are called the deviatoric invariants. For reasons to be explained in Chapter 15 which presents additional variational principles to address global constraints, this form serves to enforce incompressibility, with S now given by @ w0 @e ¼ 2f01 n1 0 þ 2f02 n02 1J pn3
sT ¼
f01
@w ¼ 0, @ I1
f02
@w ¼ 0, @ I2
0 T @ I1 n1 ¼ , @c 0
n02
(13:88)
@ I20 ¼ @c
T
To convert to deformed coordinates, recall that S ¼ JF1TFT. An example presented below serves to derive c01 and c02 and also t ¼ VEC (T) ¼ 2c01 m01 þ 2c02 m02 pi
(13:89)
The example will also establish that iT m01 ¼ 0 and iT m02 ¼ 0. Note that t does not denote the traction vector in the present context. We find that p ¼ tr(T)=3 since
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tr(T) ¼ iT t ¼ 2c01 iT m01 þ 2c02 iT m02 piT i ¼ iT ip ¼ 3p
(13:90)
Evidently the Lagrange multiplier enforcing incompressibility is the ‘‘true’’ hydrostatic pressure. Finally, the tangent modulus tensor is somewhat more complicated, because dS depends on dE and dp. We will see in a subsequent chapter that the tangent modulus tensor may be defined as D* using 2
ds 6 de * TEN22(D ) ¼ 4 T ds dp
3 ds dp 7 5 0
h T T ds ¼ 4 f01 A01 þ f02 A02 þ f011 n01 n01 þ f012 n01 n02 de T T i þ f021 n02 n01 þ f022 n02 n02
(13:91)
ds 1 ¼ n3 dp J Note the negative sign in the lower left entry of D*.
EXAMPLE 13.9 In undeformed coordinates and Kronecker product (VEC) notation, the second Piola– Kirchhoff stress for an incompressible hyperelastic materials may be written as s¼
0 T @w 1 ¼ 2w01 n01 þ 2w02 n02 p n3 @e J
Find the corresponding expression in deformed coordinates. Derive c01 and c02 , in which direct transformation furnishes t ¼ 2c01 m01 þ 2c02 m02 pi
SOLUTION The Cauchy stress is related to the second Piola–Kirchhoff stress by the relation T ¼ 1J FSFT . Upon invoking Kronecker product relations t ¼ VEC(T) ¼ 1J F FVEC(S) ¼ 1J F Fs
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Applying the transformation to the stress–strain relation results in 2w0 1 2w0 1 t ¼ F F 2w01 n01 þ 2w02 n02 pI3 n3 ¼ 1 F Fn01 þ 2 F Fn02 p 2 F Fn3 J J J J 1 1 I3 Note that 2 F Fn3 ¼ i, since IVEC 2 F Fn3 ¼ 2 FC1 FT ¼ I: J J J Finally, let c01 ¼
w01 0 w0 , c2 ¼ 2 , m01 ¼ F Fn01 , m02 ¼ F Fn02 : J J
EXAMPLE 13.10 Uniaxial tension Consider the Neo-Hookean elastomer satisfying w ¼ a[I1 1] subject to I3 ¼ 1 We seek the relation between s1 and E1. The solution will be obtained twice, once by enforcing the incompressibility constraint a priori and the second by enforcing the constraint a posteriori. A priori: We assume for the sake of brevity that E2 ¼ E3. Now I3 ¼ 1 implies that pffiffiffiffiffi c2 ¼ 1= c1 . The strain energy function now is w ¼ a c1 þ p2ffiffiffi c1 3 . The stress S1 is now found as
" # dw 1 s1 ¼ 2 ¼ 2a 1 3=2 dc1 c 1
A posteriori: Now use the augmented function w0 ¼ a[II1 1] p2 [I3 1] and dw0 ¼ 2a p=c1 dc1 dw0 0 ¼ s2 ¼ 2 ¼ 2a p=c2 dc2 0 dw 0 ¼ s3 ¼ 2 ¼ 2a p=c3 dc3 0 dw 0¼ ¼ 0 ! I3 ¼ 1 dp pffiffiffiffiffi It follows that c2 ¼ c3 ¼ 1= c1 and p=c2 ¼ 2a. We conclude that s1 ¼ 2
s1 ¼ 2a p=c1 c2 ¼ 2a p=c2 c 1
c2 ¼ 2a 1 c1 " # 1 ¼ 2a 1 3=2 c1 which agrees with the relation obtained by the a priori argument.
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A posteriori with deviatoric invariants: Finally consider the augmented function with deviatoric invariants: w0 ¼ a[I1 =I3 3] p2 [I3 1] " # dw0 1 1 I1 I3 I3 p s1 ¼ 2 ¼ 2a 1=3 4=3 dc1 c1 3 I c1 I 1=3
3
(13:92) (13:93)
3
" dw0 1 1 ¼ 2a 1=3 0 ¼ s2 ¼ 2 dc2 3 I3 " dw0 1 1 ¼ 2a 1=3 0 ¼ s3 ¼ 2 3 dc3 I 3
# I1 I3 I3 p 4=3 c c2 2 I
(13:94)
3
# I1 I3 I3 p 4=3 c c3 3 I
(13:95)
3
Note that Equations 13.94 and 13.95 imply that c2 ¼ c3. Further, dw0 ¼ 0 ! I3 ¼ 1 dp demonstrating that incompressibility is satisfied. Now I10 ¼ I1 ¼ c1 þ 2c2 , and using Equation 13.94 gives
2a c2 c1 p ¼ 3 c2 c2 Substitution into Equation 13.92 results in " s1 ¼ 2a 1
1
#
3=2
c1
as in the a priori case and in the first a posteriori case. The Lagrange multiplier p is again revealed as the hydrostatic pressure referred to current coordinates.
13.5.3.2
Near-Incompressibility
As will be seen in Chapter 15, in the augmented strain energy function w00 ¼ w(I10 ,I20 ) p[J 1]
p2 2k
(13:96)
the pressure p serves to enforce the generalized constraint p ¼ k[J 1]
(13:97)
Here k is the bulk modulus (@p=@J)I10 ,I20 , and it is assumed to be very large compared to, for example, the small strain shear modulus. The tangent modulus tensor is now
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2
ds de 4 T TEN 22(D*) ¼ ds dp
13.5.4 NONLINEAR MATERIALS
AT
ds 3 dp 5 1 k
(13:98)
LARGE DEFORMATION
Suppose that the constitutive relations are measured at constant temperature in the current configuration and are found to obey the form T˚ ¼ DD
(13:99)
in which the fourth-order tangent modulus tensor D may in general be a function of stress, strain, temperature, and internal state variables to be discussed in subsequent ˚ is the Truesdell stress flux. The form in Equation 13.99 is chapters. Recall that T ˚ sensible since T and D are both objective, and it encompasses the rate-constitutive models typical of hypoelasticity and plasticity. Conversion to undeformed coordinates is realized by S_ ¼ J F1 DDFT _ 1 F T ¼ J F1 DFT EF
(13:100)
With s ¼ VEC(S) and e ¼ VEC(E), _ 1 FT ) s_ ¼ J I F1 VEC (DFT EF _ 1 ) ¼ J I F1 (F1 I)TEN 22(D)VEC (FT EF _ ¼ J F1 F1 TEN 22(D)(FT I)(I FT )VEC (E) ¼ J F1 F1 TEN 22(D)FT FT e_ ¼ TEN 22(D0 )e_
(13:101)
in which D0 ¼ ITEN 22(J F1 F1 TEN 22(D)FT FT )
(13:102)
Since s_ ¼ TEN 22(D0 )_e
(13:103)
we may regard D0 as the tangent modulus tensor referred to undeformed coordinates. The use of the Jaumann stress flux, or any flux other than the Truesdell flux, will not lead to a counterpart of D0 which is proportional only to a tangent modulus tensor measured in the current configuration. Recall the relation derived in Example 13.5:
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_ JF1 T˚ FT ¼ S_ JS tr E(2E þ I)1 _ þ SE(2E _ {(2E þ I)1 ES þ I)1 }
(13:104)
˚ ¼ DD, ^ some effort suffices to obtain a relation of the form If we assume that T s_ ¼ TEN22(D01 þ D02 )_e
(13:105)
in which ^ T FT D01 ¼ ITEN22 JF1 F1 TEN22(D)F D02 ¼ ITEN22 J sVEC T (2E þ I)1 þ [S (2E þ I)1 þ (2E þ I)1 S] ^ but that D02 is not. Observe that D01 is proportional to D,
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14
Introduction to Nonlinear FEA
14.1 INTRODUCTION Chapters 2 through 12 addressed linear solid mechanics and heat transfer, and corresponding finite element methods for linear problems. Applications that linear methods serve to analyze include structures under mild loads, disks and rotors spinning at modest angular velocities, and heated plates. However, a large number of problems of interest are nonlinear. For one example, plasticity is a nonlinear materials theory suited for metals in metal forming, vehicle crash, and ballistics applications. In problems with high levels of heat input, mechanical properties such as the elastic modulus, and thermal properties such as the coefficient of specific heat, may be strongly temperature dependent. Rubber seals and gaskets commonly experience strains exceeding 50%. Soft biological tissues typically are modeled as rubberlike. Many problems involve variable contact, for example, meshing gear teeth. Heat conducted across electrical contacts may be strongly dependent on normal pressures. Fortunately, much of the linear finite element method can be extended to nonlinear problems, as explained in this chapter. In the following sections, attention is confined to isothermal problems. The extension to thermomechanical problems will be presented in the subsequent chapter. In particular, we will see that the finite element equations in the nonlinear case are formally similar to the linear finite element equations if the displacements and forces (and=or temperatures and heat fluxes) are replaced by incremental counterparts. The tangent stiffness matrix is now a function of the nodal displacements (and temperatures). In addition, it will be seen to possess the extremely important property of serving as the Jacobian matrix in Newton iteration, which is an extremely attractive (arguably optimal) method for solving nonlinear algebraic equations.
14.2 TYPES OF NONLINEARITY There are three major types of nonlinearity in thermomechanical boundary value problems: (i) material nonlinearity, (ii) geometric nonlinearity, and (iii) boundary condition nonlinearity. Of course these effects can and do occur in together. Nonlinearities may also be present if the formulation is referred to deformed coordinates, possibly introducing stress fluxes and convected coordinates. Material nonlinearity may occur through nonlinear dependence of the stress on the strain and=or temperature, including strain and temperature dependence
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of the tangent modulus tensor. Nonlinear material behavior may also ensue from history dependence, for example, a nonlinear dependence on the total ‘‘plastic work.’’ Geometric nonlinearity occurs because of large deformation, especially in problems referring to undeformed coordinates. Rubber components typically exhibit large deformation and require nonlinear kinematic descriptions. In this situation, a choice needs to be made of a strain measure and of the stress conjugate to it. Boundary condition nonlinearity occurs because of nonlinear mechanical and inertial supports on the boundary. An example of a nonlinear support is a rubber pad under a machine, to absorb vibrations. It may also occur if the contact area between two bodies is an unknown to be determined as part of the solution. For finite element methods for nonlinear problems, the loads are often viewed as applied in increments. Then incremental variational principles together with interpolation models for incremental displacements and incremental temperatures (the primary variables) furnish algebraic (static) or ordinary algebraic-differential equations (dynamic) in terms of vector-valued incremental displacements and=or incremental temperatures. For mechanical systems a typical equation is € ¼ Df K(g)Dg þ M(g)Dg
(14:1)
in which Dg is the incremental displacement vector (to be defined shortly), Df is the incremental force vector, K(g) is the tangent stiffness matrix, and M(g) is the (tangent) mass matrix. It will be seen that this type of equation is a realization of Newton iteration method for nonlinear equations.
14.3 NEWTON ITERATION By virtue of its property of quadratic convergence, Newton iteration is arguably an optimal method for solving nonlinear algebraic equations. It is introduced in the current chapter, and its application to finite element analysis is scrutinized. In recent years very effective methods for solving nonlinear finite element problems, known as arc length methods, have been introduced. In Chapter 18 we present an arc length method which in fact is a particular realization of Newton iteration. In it, arc length constraints are accommodated by expanding the solution space to a dimension greater than in the conventional finite element method (e.g., greater than the number of incremental displacement degrees of freedom). Letting f and x denote scalars, consider the nonlinear algebraic equation f(x; l) ¼ 0 in which l is a parameter we will call the load intensity. Such equations are often solved numerically by a two-stage process. The first is load incrementation: the load intensity l is increased progressively using small increments. The second is iteration: at each increment the unknown x is computed using an iteration procedure. Suppose that at the nth increment of l an accurate solution has been achieved as xn. We now further suppose for simplicity that xn is ‘‘close’’ to the actual solution xnþ1 at the (n þ 1)st increment of l. Using x(0) nþ1 ¼ xn as the starting iterate, Newton iteration provides subsequent iterates according to the scheme
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x( jþ1) ¼ x( j)
1 df f (x( j) ) dx jx( j)
(14:2)
( jþ1) j) Let Dnþ1, j denote xnþ1 x(nþ1 , which is the difference between two iterates. Then, to first order in the Taylor series
Dnþ1, j Dnþ1, j1
" # 1 df 1 ( j) df ( j1) ¼ f (x ) f (x ) dx jx( j) dx jx( j1) 1 ( j) df f (x ) f (x( j1) ) dx jx( j)
1 df df Dnþ1, j1 þ 02 dx jx( j) dx jx( j)
Dnþ1, j1 þ 02
(14:3)
in which 02 refers to second-order terms in increments. It follows that Dnþ1, j 02. For this reason Newton iteration is observed to converge quadratically (presumably to the correct solution if the initial iterate is ‘‘sufficiently close’’). After the iteration scheme has converged to the solution, the load intensity is incremented again.
EXAMPLE 14.1 Consider f(x) ¼ (x 1)2. Newton iteration is realized as the iteration scheme 1 2
x( jþ1) ¼ x( j) (x( j) 1) The solution is unity, twice. If the initial iterate x(0) is taken to be 1=2, the iterates become 1=2, 3=4, 7=8, 15=16, . . . , converging to unity. If x(0) ¼ 2, the iterates are 3=2, 5=4, 9=8, 17=16, likewise converging to unity. In both cases the error is halved in each iteration. Returning to the main development, the nonlinear finite element method under static conditions will frequently be seen to pose nonlinear algebraic equations of the form duT [w(u) lv] ¼ 0
(14:4)
in which u and w are n 3 1 vectors, v is a known constant n 3 1 unit vector, and l represents ‘‘load intensity.’’ The Newton iteration scheme provides the ( j þ 1)st iterate for unþ1 as Dnþ1, j
" #1 h i @w j) ¼ w(u(nþ1 ) ljþ1 v , @u ju( j) nþ1
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jþ1) j) u(nþ1 u(nþ1 ¼ Dnþ1, j
(14:5)
in which, for example, the initial iterate is u(n0) . The use of an explicit matrix inverse is avoided by solving the linear system
@w ( j) Dnþ1, j ¼ w(unþ1 ) vjþ1 , @ u u( j)
( j) u(njþ1) þ1 ¼ unþ1 þ Dnþ1, j
(14:6)
nþ1
Note that
h i @w @ u u( j) in Newton iteration is a Jacobian matrix. n þ1
14.4 COMBINED INCREMENTAL AND ITERATIVE METHODS: A SIMPLE EXAMPLE As illustrated in Figure 14.1, consider a one-dimensional rod of nonlinear material under small deformation, in which the elastic modulus is a linear function of strain: E ¼ E0(1 þ a«), « ¼ E11. The cross-sectional area A0 and the length L0 are constants, on the understanding that the deformation is expressed in terms of the undeformed configuration. Suppose that under static loading the equilibrium equation is E0 A0 g g¼P 1þa L0 L0
(14:7)
The load is applied in increments DjP ¼ Pjþ1 Pj and the load after the (n 1)st increment is applied is denoted as Pn. Suppose next that the solution gn has been computed accurately at Pn. We now consider the actions necessary to determine the solution gnþ1 at load Pnþ1. This is done by first computing the value of Dng ¼ gnþ1 gn. Subtracting Equation 14.2 at the nth increment from the same equation but at the nth increment, the incremental equilibrium equation is now E0 A0 2g n þ Dn g 1þa Dn g ¼ Dn P L0 L
(14:8)
Equation 14.8 is quadratic in the increment Dn g-in fact geometric nonlinearity generally leads to a quadratic function of increments. The error of neglecting the quadratic term may be small if the load increment is sufficiently small. However, we will retain the nonlinear term for the sake of accuracy and of illustrating the use of iterative procedures. In particular, the foregoing equation may be written in the form
E0, A, L
P g
FIGURE 14.1 Stretching of a rod of nonlinear material.
ß 2008 by Taylor & Francis Group, LLC.
g(x) ¼ bx þ zx2 h ¼ 0
(14:9)
in which x ¼ Dn g, h ¼ LE00DAn0P , b ¼ 1 þ 2agn =L0 , and z ¼ La0 . Newton iteration furnishes the quadratically convergent iteration scheme for the nth iterate, namely: xnþ1 ¼ xn
[bxn þ zxn2 h] , x0 ¼ h=b b þ 2zxn
(14:10)
As an example we use the values E ¼ 107 psi, A ¼ 1 in2, L ¼ 10 in, a ¼ 2, gn ¼ 1, and DnP ¼ 3. These values imply significant nonlinearity and 10% strain. A simple program written in double precision produces the following values, which demonstrate a convergence in two iterations. Iterate 0 1 2 3 4
Value 0 0.21428571e-5 0.21428565e-5 0.21428565e-5 0.21428565e-5
EXAMPLE 14.2 The efficiency of this scheme is addressed as follows. Consider z ¼ 1, b ¼ 12, and h ¼ 1. The correct solution is x ¼ 0.0828. Starting with the initial value x ¼ 0, the first 1 1 1 ¼ 0:833, and 12 1 144 two iterates are, approximately, 12 ¼ 0:0828.
14.5 FINITE STRETCHING OF A RUBBER ROD UNDER GRAVITY 14.5.1 MODEL PROBLEM Figure 14.2 shows a rubber rod element under gravity—it is assumed to attain finite strain and to experience uniaxial tension. The figure exhibits the undeformed configuration, with an element occupying the interval (Xe, Xeþ1), with X denoting the
Pe xe, ue
Pe+1
xe+1, ue+1
FIGURE 14.2 Rubber rod element under gravity.
ß 2008 by Taylor & Francis Group, LLC.
g
downward direction. The element’s length is le, its cross-sectional area is Ae, and its mass density is r. It is composed of rubber and is stretched axially by the loads Pe and Peþ1. Prior to stretching, a given material particle is located at X. After deformation it is located at x(X), and the displacement u(X) is given by u(X) ¼ x(X) X.
14.5.2 NONLINEAR STRAIN–DISPLACEMENT RELATIONS The element in Figure 14.2 is assumed to be short enough that a satisfactory approximation for the displacement u(X) is provided by the linear interpolation model u(X ) ¼ ue þ (X Xe )(ueþ1 ue )=le 1 ¼ NT ae , NT ¼ {xe x x xe } le
(14:11)
in which aTe ¼ {ue ueþ1 }. Now u(Xe) ¼ ue and u(Xeþ1) ¼ ueþ1 are viewed as the unknowns to be determined using the finite element method. The Lagrangian strain E ¼ Exx is approximated in the element as @u 1 @u 2 þ @x 2 @x ueþ1 ue 1 ueþ1 ue 2 ¼ þ le le 2 " # 1 1 T 1 1 1 ¼ {1 1}ae þ ae 2 ae le 2 l e 1 1
E¼
(14:12)
In alternative notation Equation 14.12 is written as dExx ¼ BT dae , B ¼ BL þ BNL ae 1 1 1 1 1 , BNL ¼ 2 BL ¼ 1 le l e 1 1
(14:13)
The vector BL and the matrix BNL are called the linear and nonlinear strain– displacement matrices.
14.5.3 STRESS AND TANGENT MODULUS RELATIONS The Neo–Hookean strain energy density function w was previously encountered in Chapter 13. It accommodates incompressibility and is stated in terms of the eigenvalues c1, c2, and c3 of C ¼ 2E þ I as follows: w¼
D (c1 þ c2 þ c3 3), subject to c1 c2 c3 1 ¼ 0 6
(14:14)
in which D is the (small strain) elastic modulus. We saw in Chapter 13 that c2 ¼ c3.
ß 2008 by Taylor & Francis Group, LLC.
We first enforce the incompressibility constraint a priori by using the substitution 1 c2 ¼ c3 ¼ pffiffiffiffiffi c1
(14:15)
After elementary manipulations and using c1 ¼ 2Exx þ 1, the strain energy function emerges as w¼
D 2Exx þ 1 þ 2=(2Exx þ 1)1=2 6
(14:16)
The (second Piola–Kirchhoff) stress Sxx, defined in general in Chapter 13, is now obtained as @w @ Exx D ¼ 1 (2Exx þ 1)3=2 3
Sxx ¼
(14:17)
For small strains (2Exx þ 1)3=2 1 3Exx in which case Sxx DExx. The tangent modulus DT is also required: @ Sxx @ Exx ¼ D(2Exx þ 1)5=2
DT ¼
(14:18)
If the strain Exx is small compared to unity, DT reduces to D. We next satisfy the incompressibility constraint a posteriori, which we will see is how the constraint is generally satisfied in finite element analysis. An augmented strain energy function w* is introduced by w* ¼
D p [c1 þ c2 þ c3 3] (c1 c2 c3 1) 6 2
(14:19)
in which the Lagrange multiplier p can be shown to be the (true) hydrostatic pressure, as was shown in a similar situation in Chapter 13. The augmented energy is stationary with respect to p as well as c1, c2, and c3, from which it follows that c1c2c3 1 ¼ 0 (incompressibility). The second Piola–Kirchhoff stresses satisfy Sxx ¼
@w* @w* D p c1 c2 c3 ¼2 ¼ @Exx @c1 3 2 c1
Syy ¼
@w* @w* D p c1 c2 c3 ¼2 ¼ ¼0 @Eyy @c2 3 2 c2
Szz ¼
@w* @w* D p c1 c2 c3 ¼2 ¼ ¼0 @Ezz @c3 3 2 c3
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(14:20)
The second and third equations in Equation 14.20 are immediately seen to imply c2 ¼ c3, as previously stated. Enforcement of the stationarity condition (dw* ¼ 0) for pffiffiffiffiffi p(c1c2c3 1 ¼ Syy ¼ Szz ¼ 0 ¼ 0 now furnishes that p ¼ 23 D= c1 . It follows 0) with 3=2 =3, in agreement with Equation 14.7. that Sxx ¼ D 1 c1
14.5.4 INCREMENTAL EQUILIBRIUM RELATION The Principle of Virtual Work states the condition for static equilibrium of the rod as Xð eþ1
f(ae ; Pe ) ¼
2 6 BSxx Ae dX Pe 4rg
Xe
Xð eþ1
3 7 NT Ae dX 5ae ¼ 0,
Pe ¼
Xe
Pee
Peeþ1 (14:21)
in which the third left-hand term represents the weight of the element, while Pe represents the forces from the adjacent elements. In an incremental formulation, we replace the loads and displacements by their differential forms. In particular Xð eþ1
Xð eþ1
B dSxx Ae dX þ
0 ¼ df ¼ Xe
dB SAe dX dPe Xe
¼ Ke dae dP ¼ K1e þ K2e þ K3e þ K4e dae dPe e
(14:22)
in which Xð eþ1
K1e ¼ Ae
BL DT BTL dX Xe
DT A e 1 ¼ le 1
1 1
(14:23a)
Xð eþ1
K2e ¼ Ae
DT (BL aTe BNL þ BNL ae BTL ) dX Xe
DT Ae [ueþ1 ue ] 1 2 ¼ le le 1
1 1
(14:23b)
Xð eþ1
K3e ¼ Ae
BNL ae aTe BTNL dX Xe
DT Ae ueþ1 ue 2 1 ¼ le le 1
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1 1
(14:23c)
Xð e þ1
K4e ¼ Ae
BL S dX Xe
Sxx Ae 1 ¼ le 1
1
(14:23d)
1
and 1 DT ¼ le
Xð e þ1
Xð eþ1
1 DT dX , Sxx ¼ le
Sxx dX
Xe
(14:23e)
Xe
Combining Equations 14.23a through 14.23e produces a simple relation for the tangent stiffness matrix: Ke ¼ k
1
1
1
1
,
( DT Ae k¼ le
) ueþ1 ue ueþ1 ue 2 Sxx Ae þ 1þ2 þ le le le (14:24)
Suppose (ue þ 1 ue)=le is small compared to unity, with the consequence that Exx is also small compared to unity. It follows in this case that DT D, Sxx Sxx. In consequence, Ke reduces to the stiffness matrix for a rod element of linearly elastic material experiencing small strain: Ke ¼
EA e 1 le 1
1 1
(14:25)
Returning to the nonlinear problem, several special cases are now used to illuminate additional aspects of finite element modeling.
14.5.5 SINGLE ELEMENT BUILT-IN
AT
ONE END
Figure 14.3 depicts a single element model of a rod which is built in at X ¼ 0: Xe ¼ X0 ¼ 0. At the opposite end, Xeþ1 ¼ X1 ¼ 1, it is submitted to the load P. The displacement at X ¼ 0 is subject to the constraint u(0) ¼ u0 ¼ 0, so that Equation 14.24 becomes k(u1 )
1 1
1 1
0 du1
¼
dPr dP
(14:26)
in which dPr is an incremental reaction force which is considered unknown (of course from equilibrium dPr ¼ dP). Enforcing the constraint at the top of the rod causes Equation 14.26 to ‘‘condense’’ to k du1 ¼ dP
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(14:27)
g
A, ρ
X1
u1
X P
FIGURE 14.3 Rubber rod element under gravity: built in at top.
Also, the current shape function degenerates to the expression N ! N ¼ X=X1. The (Lagrangian) strain is given by E(u1 ) ¼
2 u1 1 u1 þ X1 2 X1
(14:28)
and the strain–displacement matrix reduces to B!B¼
1 u1 þ 2 X1 X1
(14:29)
The (second Piola–Kirchhoff) stress Sxx is now obtained as a function of u1: 2
" (
1 u1 1 u1 þ Sxx (u1 ) ¼ D41 2 X1 2 X1 3
14.5.6 ON NUMERICAL SOLUTION
BY
2 )
#32 3 1 5
(14:30)
NEWTON ITERATION
We see to solve Equation 14.21 numerically by Newton iteration, sometimes called ‘‘load balancing’’ in this context, as explained below. The Principle of Virtual Work implies that Xð1
f ( u1 ) ¼
2 BSxx A dX P 4rg
0
Xð1
3 NT A dX 5ae
0
0 2 " ( #32 31 2 ) 1 u1 1 u1 1 u1 X1 AX1 @ D41 2 1 5A P rgA u1 ¼ þ þ le l2e 3 X1 2 X1 2
¼0
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(14:31)
in which the residual function f(u1) has been introduced. Consider an iteration process in which the jth iterate u j1 has been determined. Newton iteration determines the next iterate u1jþ1 using k(u j1 )Dj u1 ¼ w(u j1 ) u 1jþ1 ¼ u j1 þ Dj u1
(14:32)
in which k(u j1 ) ¼ @ f(u j1 )=@ u j1 has already been given in Equation 14.24 and is now seen to be the Jacobian matrix of Newton iteration. Convergence to the correct value u1 is usually rapid provided that the initial iterate u01 is sufficiently close to u1. A satisfactory starting iterate for the current load step can be obtained by extrapolating from the previous solution values. As illustration, suppose the solution, say, is known at the load Pj ¼ jDP, in which DP is a load increment. The load is now incremented to produce Pjþ1 ¼ ( jþ1)DP, and the a starting iterate is needed to converge to the solution u1, jþ1 at this load using Newton iteration. It is frequently satisfactory to use u01, jþ1 ¼ u1, j . Another possibility is to use u01, jþ1 ¼ 2u1, j u1, j1 , or more generally ‘‘line search’’ using two or more previous solution values.
14.5.7 ASSEMBLED STIFFNESS MATRIX FOR A TWO-ELEMENT MODEL OF THE RUBBER ROD UNDER GRAVITY A two-element model of the rubber rod under gravity is shown in Figure 14.4. Assemblage procedures are now illustrated for combining the element equilibrium relations to obtain the global equilibrium relation holding for an assemblage of elements. We will see that they are simple extensions of assembly procedures in linear finite element analysis. For elements ‘‘e’’ and ‘‘e þ 1’’ the element equilibrium relations are expanded as Pee
xe,ue
P,Ae
e Pe+1
xe+1,ue+1
e+1 Pe+1
P, Ae+1 xe+2,ue+2
e+1 Pe+2
x
FIGURE 14.4 Two rubber rod elements under gravity.
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g
e e k11 due þ k12 dueþ1 ¼ dPee
due þ
(14:33a)
dueþ1 ¼
dPeeþ1
(14:33b)
dueþ2 ¼
dPeþ1 eþ1
(14:33c)
eþ1 eþ1 k21 dueþ1 þ k22 dueþ2 ¼ dPeþ1 eþ2
(14:33d)
e k21 eþ1 k11
dueþ1 þ
e k22
eþ1 k12
The superscript indicates the element index, and the subscript indicates the node index. If no external force is applied at xeþ1, the interelement incremental force balance is expressed as dPeeþ1 þ dPeþ1 eþ1 ¼ 0
(14:34)
Adding Equations 14.33b and 14.33c furnishes e
e eþ1 eþ1 k21 due þ k22 þ k11 dueþ2 ¼ 0 dueþ1 þ k12
(14:35)
Equations 14.33a, 14.33d, and 14.35 are expressed in matrix form as 2
e k11 e 4 k21 0
e k12 eþ1 e k22 þ k11 eþ1 k21
1 1 0 dPee due C B eþ1 5@ k12 dueþ1 A ¼ @ 0 A eþ1 dPeþ1 k22 dueþ2 eþ2 0
30
(14:36)
Equation 14.36 illustrates that the (incremental) global stiffness matrix is formed by ‘‘overlaying’’ Ke and Keþ1, with the entries added in the intersection (the 2–2 element). Two identical elements under gravity, under equal end loads: If Ke ¼ Keþ1 and dPee ¼ dPeþ1 eþ1 ¼ dP, overlaying the element matrices leads to the global (two element) relation 1 0 1 2 30 dP 1 1 0 du1 (14:37) k4 1 2 1 5@ du2 A ¼ @ 0 A du3 dP 0 1 1 Note that Equation 14.37 has no solution since the global stiffness matrix has no inverse: the second row is the negative of the sum of the first and last rows. This suggests that, owing to numerical errors in the load increments, the condition for static equilibrium is not satisfied numerically, and therefore the body is predicted to accelerate indefinitely (undergoes rigid body motion). However, we also know that this configuration, under equal and opposite incremental loads, is symmetric, implying a constraint du2 ¼ 0. This constraint permits ‘‘condensation,’’ that is, reducing Equation 14.37 to a system with two unknowns by eliminating rows and columns associated with the middle incremental displacement. In particular,
1 k 1
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1 1
du1 du3
¼
dP dP
(14:38)
The condensed matrix is now proportional to the identity matrix, and the system has a solution. More generally, stiffness matrices may easily be singular or nearly singular (with a large condition number) unless constraints are used or introduced to suppress ‘‘rigid body modes.’’
14.6 NEWTON ITERATION NEAR A CRITICAL POINT Now the Jacobian matrix is likely to be ill-conditioned whenever the tangent modulus tensor is nearly singular, which is said to occur at critical points. Unfortunately, since elastomers or metals experiencing plasticity are typically very compliant at large deformation, the analyst attempting to perform computations into large strain ranges may well encounter a critical point. Buckling also represents an example of a critical point. Since the tangent stiffness matrix is the Jacobian matrix for Newton iteration, convergence problems arise when the tangent stiffness becomes singular or ill-conditioned, and special methods are invoked to continue computations near and past critical points. Several ways to continue computation in the vicinity of critical points are now listed. 1. 2. 3. 4.
Increase stiffness, such as by introducing additional constraints if available Reduce load step sizes, and reform the stiffness matrix after each iteration Switch to displacement control rather than load control Using an arc length method, of which a particular method is described in Chapter 17
We now illustrate such a convergence problem using an equation of the form c(x) ¼ 0. As previously stated, Newton iteration seeks a solution through an iterative process given by
xjþ1
dc(xj ) 1 ¼ xj c(xj ) dx
(14:39)
( jþ1) j) x(nþ1 , Clearly, recalling Equation14.3 using sequential iterates, with Dnþ1, j ¼ xnþ1
Dnþ1, j Dnþ1, j1
1 dc dc Dnþ1, j1 þ 02 dx jx( j) dx jx( j)
(14:40)
( j) Strictly speaking the argument of dc dx is not x . Rather, thanks to the Mean Value Theorem, the argument value is an unknown value x* between xn and theconverged dc 1 does not cancel the ill-conditioned matrix , in which of xnþ1. Clearly, dc ( j) dx x* dx jx case the iterates will grow rapidly.
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EXAMPLE 14.3 The performance of the Newton iteration procedure near a critical point is now illustrated using a simple example showing a slope (analogous to the tangent modulus tensor) which is asymptotically approaching zero. Consider 2 p
f(x) ¼ tan1 (x) y ¼ 0
(14:41)
Equation 14.41 is depicted in Figure 14.5. The goal is to find the solution of x as y is incremented in the range (0,1). Clearly x approaches infinity as y approaches unity, so that the goal is to generate the x(y) relationship accurately as close as possible to y ¼ 1. The curve will appear as in Figure 14.5. When y is plotted against x, the curve asymptotically approaches unity, with the slope (stiffness) approaching zero. As y is incremented by small amounts just below unity, the differences in x due to the increment are large, so that the solution value at a nth load step, if used as the initial iterate, is not close to the solution at the (n þ 1)st load step. Suppose that y is incremented such that the nth value of y is yn ¼ nDy. To obtain the solution of the (n þ 1)th step, the Newton iteration procedure generates the (n þ 1)st iterate from the vth iterate as follows: "
vþ1) x(nþ1
) #1 )
df x(nvþ1 ¼ f x(nvþ1 dx ) 2 1 (v) p ) tan xnþ1 ynþ1 ¼ x(nvþ1 1 þ x(nvþ1 2 ) x(nvþ1
(14:42)
) is the vth iterate for the solution xnþ1. Of course a starting iterate x0nþ1 is in which x(nvþ1 needed. An attractive candidate is xn. However, this may not be good enough when convergence difficulties appear. Newton iteration for Equation 14.42 was implemented for this example in a simple double precision program. Numerical results are shown in Table 14.1. The increment Dy was reduced significantly as the asymptote was approached. Despite this reduction the iteration count increased noticeably near y ¼ 0.9999, illustrating the onset of convergence difficulties.
Y
Encounter difficulties with convergence
1
X
FIGURE 14.5 Illustration of the inverse tangent function.
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TABLE 14.1 Convergence of Newton Iteration for y ¼ 2tan1(x)=p Iteration Count 2 3 3 3 3 4 4 4 5 6
x
y
Dy
12.7062083 63.6568343 70.7309329 90.9422071 127.321711 212.206062 636.628644 909.475631 1591.60798 6367.13818
0.95 0.99 0.991 0.993 0.995 0.997 0.999 0.9993 0.9996 0.9999
0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001
14.7 INTRODUCTION TO THE ARC LENGTH METHOD In, say, uniaxial tension, the Principle of Virtual Work leads to an equation of the form f(u) ¼ f, df(u) du ¼ k(u) (incremental stiffness matrix). The arc length method to be discussed here is helpful if k(u) is near zero (i.e., near a critical point). In nonlinear problems we increment the loads using a load intensity parameter l, f ¼ lf0 ,
0l1
(14:43)
in which f0 is the final load to be attained. The equilibrium equation thereby becomes f(u) ¼ lf0
(14:44)
Suppose the solution has been attained at ln f0, denoted by un, and the solution is sought for the subsequent load increment, satisfying f(unþ1 ) ¼ lnþ1 f0
(14:45)
df Introducing the incremental relations f(unþ1 ) f(un ) 12 df du (unþ1 ) þ du (un ) Dn u ¼ 12 (k(unþ1 ) þ k(un ))Dn u, an approximate solution may be obtained from the equation 1 (k(unþ1 ) þ k(un ))Dn u 2
¼ (Dn l)f0
(14:46)
in which Dn l ¼ lnþ1 ln , Dn u ¼ unþ1 un An equation in this form is said to represent load control since the solution is obtained at prescribed load levels.
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If we make the additional approximation that 1 (k(unþ1 ) þ k(un )) 2
k(un )
(14:47)
the equation is ‘‘explicit’’ and its solution requires no iteration. Otherwise, the method is implicit and iteration is required. In problems in buckling, plasticity and hyperelasticity, it is quite possible for 1 (k(u nþ1 ) þ k(un )) to be singular or nearly singular in which case there are severe 2 difficulties achieving convergence and accuracy. An alternative is ‘‘displacement control,’’ in which increments of u are specified, and the conjugate force increments become reaction forces to be computed as part of the solution. We next discuss a combination of load control and displacement control, known as arc length control. The arc length method is helpful in situations in which load control fails owing to a singular tangent stiffness matrix. It augments the Principle of Virtual Work with an additional equation imposing a constraint on a function of the magnitudes of the load and displacement increments. In the uniaxial tension case being considered here, an example of the resulting equations is _
f (u, l) ¼ f(u) lf0 ¼ 0 equilibrium c(u, l) ¼ a(u un ) þ b(l ln ) DS ¼ 0
arc length constraint
(14:48)
Here DS is a small positive number representing the ‘‘length’’ of the increment in the ul plane, and the coefficients a and b will be chosen subsequently to promote convergence. Newton iteration is now applied to this pair of equations
_ (v)
31 _ (v) @f u(v) @f unþ1 , l(v) 8 _ nþ1 , lnþ1 nþ1
9 (v) (v) < = 6 7 f u , l nþ1 nþ1 @u @l 6 7 ¼ 6 7
: l(vþ1) ; : l(v) ; 4 @c u(v) , l(v) 5 : c u(v) , l(v) ; (v) @c u(v) nþ1 nþ1 nþ1 nþ1 nþ1 , lnþ1 nþ1 nþ1 @u @l 8 9 "
9
#1 8 _ (v) < u(v) = < f unþ1 , l(v) = k u(v) f nþ1 0 nþ1 nþ1 ¼ (14:49) : l(v) ; : c u(v) , l(v) ; a b nþ1 nþ1 nþ1 8 9 (vþ1) < unþ1 =
2
8 9 < u(v) = nþ1
and, consequently, Newton iteration with the arc length constraint is given by the iteration scheme " (v)
k unþ1 a
f0
#(
b
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(vþ1) unþ1 u(v) nþ1 (vþ1) lnþ1 l(v) nþ1
)
( _ (v)
) f unþ1 , l(v) nþ1 ¼ (v)
c u(v) nþ1 , lnþ1
(14:50)
Suppose k u(v) nþ1 ¼ 0, and choose a ¼ f0. The eigenvalues m1,2 of the matrix A are now given by the equation m2j bmj þ f02 ¼ 0
(14:51)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 b b2 4f02 ¼ 2
(14:52)
with the two roots m1, 2
The choice of b which maximizes the smaller eigenvalue is b ¼ 2f0. This choice renders the two eigenvalues equal and positive with
the value m1 ¼ m1 ¼ f0, and thereby circumvents the singularity arising if k u(v) nþ1 ¼ 0.
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15
Incremental Principle of Virtual Work
15.1 INCREMENTAL KINEMATICS The Principle of Virtual Work is extended to nonlinear solid mechanics by restating the kinematic and equilibrium relations in incremental form and by applying variational principles using the displacement increments as the primary variables. Issues such as thermal effects and incompressibility are addressed in Chapters 16 and 17 dealing with thermohyperelasticity and thermoinelasticity. Recall that the displacement vector u(X) has been assumed to admit a satisfactory approximation at the element level in the form u(X) ¼ wT(X) Fg(t). Also recall that the deformation gradient tensor is given by F ¼ @@ux. Suppose that the body under study is subjected to a load vector P, which is applied incrementally via load increments DjP ¼ Pjþ1 Pj. The load after the (n 1)th load step is denoted as Pn. The solution Pn is considered to have been determined computationally, and the solution of the current displacement increments is sought. Let Dnu ¼ unþ1 un, implying the incremental interpolation model Dn u ¼ wT (X)FDn g
(15:1)
By suitably arranging the derivatives of Dnu with respect to X, a matrix M(X) may easily be determined for which VEC(DnF) ¼ M(X)Dng, to be demonstrated in Example 15.1. We next consider the Lagrangian strain tensor E(X) ¼ 12 (FT F I). Using Kronecker product algebra introduced in Chapter 3, we readily find that to first order in increments Dn e ¼ VEC(Dn E) ¼ VEC 12 FT Dn F þ Dn FT F ¼ 12 I FT þ F IU VEC(Dn F) ¼ GT Dn g, GT ¼ 12 I FT þ F IU M(X)Dn g
(15:2)
The form in Equation 15.2 shows the convenience of Kronecker product notation. Namely it enables moving the incremental displacement vector to the end of the
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expression, with the consequence that it can be placed outside the domain integrals we will encounter subsequently. Alternatively, for the current configuration an alternative strain measure is the Eulerian strain 9 ¼ 12 I FT F1 , which refers to deformed coordinates. Note that, since Dn (FF1 ) ¼ 0, Dn F1 ¼ F1 Dn FF1 . Similarly, Dn FT ¼ FT Dn FT FT . Simple manipulation furnishes the incremental strain–displacement relation for 9 as VEC (Dn 9) ¼ 12 FT F1 FT U þ F1 FT F1 MDn g
(15:3)
There also are geometric changes for which an incremental representation is useful. For example, since the Jacobian J ¼ det(F) satisfies dJ ¼ J tr(F1dF), we obtain the approximate formula Dn J ¼ J tr(F1 Dn F) ¼ JVEC T (FT )VEC (Dn F) ¼ VEC T (FT )J MDn g
(15:4)
Also of interest are the incremental counterparts for the directed area of surface element, the surface normal vector, and the surface area of an element. Chapter 13 reported the relations d[n dS] ¼ tr (D)I LT n dS dt dn T ¼ n Dn I LT n dt d dS ¼ tr (D) nT Dn dS dt
(15:5a) (15:5b) (15:5c)
and we directly obtain the incremental counterparts as Dn [n dS] ¼ dS nVEC T F1 nT FT U MDn g Dn n ¼ n nT FT nT nT FT U MDn g Dn dS ¼ dS nVEC T FT n nT FT nT MDn g
(15:6a) (15:6b) (15:6c)
EXAMPLE 15.1 The derivation of Equation 15.6b from Equation 15.5b is now presented. First note that Dn n dn dt Dn t. Next
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(nT Dn)nDn t n nT Dn FF1 n ¼ n nT nT VEC Dn FF1 ¼ n nT nT FT IVEC(Dn F) ¼ n nT FT nT MDn g Finally, LT nDn t ¼ FT DFT n ¼ VEC FT Dn FT n ¼ nT FT UMDn g Several classical texts (e.g., Zienkiewicz and Taylor, 1991) on FEA have decomposed the strain–displacement relations into a linear portion and a residual nonlinear portion. In the current notation this is written as G ¼ BL þ BNL. If we write F ¼ I þ Fu in which Fu ¼ @@uX, it is immediate from Equation 15.3 that de ¼ BTL þ BTNL (g) dg BTL ¼ 12 (I I þ I IU)M(X) BTNL
¼
1 2 (I
FTu
(15:7)
þ Fu IU)M(X)
EXAMPLE 15.2 Assuming linear interpolation models for u, v in a plane triangular membrane element with vertices (0,0), (1,0), (0,1), obtain the matrices M, G, BL, and BNL.
SOLUTION The interpolation model is
8 9 u1 > > > > > > > > > u > > 2> > > > > u 1 x y 0 0 0 F 0 < u3 = ¼ v 0 0 0 1 x y 0 F > v1 > > > > > > > > > > > v > > 2 > ; : > v3 2 31 2 3 1 0 0 1 0 0 6 7 6 7 F ¼ 4 1 1 0 5 ¼ 4 1 1 0 5 1 0 1 1 0 1 We first formulate the deformation gradient tensor F.
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Let g1 ¼
8 9 < u1 = u : 2; u3
and g2 ¼
effort we may write " B¼
@u @x
"
8 9 < v1 = v . : 2; v3
Now F ¼ I þ @@ux, and
@u @x
¼
@u @x @v @x
@u @y @v @y
# . After some
¼ I þ BYG and Dn F ¼ BYDn G in which
bT11
bT12
0T
0T
0T
0T
bT21
bT22
2
# ,
3
F
0
0
0
60 6 Y¼6 40
F
0
0
F
07 7 7, 05
0
0
0
F
2
g1
6 0 6 G¼6 4 g2 0
0
3
g1 7 7 7 0 5 g2
with bT11 ¼ {0 1 0},
bT12 ¼ {0 0 1},
bT21 ¼ {0 1 0},
bT22 ¼ {0 0 1}
Now ( VEC(Dn G) ¼ (I BY)
Dn g1
)
Dn g2
VEC(Dn F) ¼ (I BY)VEC(Dn G) ¼ (I BY)IDn g in which 2
I 60 6 60 6 60 I¼6 60 6 6I 6 40 0
3 0 07 7 I7 7 07 7 07 7 07 7 05 I
We conclude that M ¼ (I BY)I. Next the strain–displacement matrix G is given by GT ¼ 12 I (I þ BYG)T þ ½(I þ BYG) IU (I BY)I Finally, the linear and nonlinear portions of the strain–displacement matrices are obtained as BTL ¼ 12 [I I þ (I I)U](I BY)I BTNL ¼ 12 I (BYG)T þ [(BYG) I]U (I BY)I
EXAMPLE 15.3 Repeat Example 15.2 with linear interpolation models for u, v, and w in a tetrahedral element with vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1).
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SOLUTION The interpolation model is u(x, y, z) ¼ {1 x y
8 9 u1 > > > = 2 , z}F > u > > : 3> ; u4
v(x, y, z) ¼ {1 x
w(x, y, z) ¼ {1 x y 2
1 61 F¼4 1 1
0 1 0 0
z}F
8 9 w1 > > > <w > = 2
> w > > : 3> ; w4
2 3 1 0 1 07 6 1 ¼4 5 1 0 1 1
0 0 1 0
8 9 v1 > > > = 2 y z}F > v > > : 3> ; v4
0 1 0 0
0 0 1 0
3 0 07 05 1
Following Example 15.2, we again formulate the deformation gradient tensor. 3 2 @u @u @u 6 @x @y @z 7 7 6 @u @Dn u @u 6 @v @v @v 7 7 ¼ BYG F¼Iþ , Dn F ¼ , ¼6 7 @x @x @x 6 6 @x @y @z 7 4 @w @w @w 5 g1 6 0 6 6 0 6 6 g2 6 G¼6 6 0 6 0 6 6g 6 3 4 0 0
0 g1 0 0 g2 0 0 g3 0
@x 3 0 0 7 7 g1 7 7 0 7 7 0 7 7 g2 7 7 0 7 7 0 5 g3
bT12
bT13
0T
0T
0T
0T
0T
0T
0T
bT21
bT22
bT23
0T
0T
0T
0T
0T
0T
0T
bT21
bT32
2
@y
@z
Now 2
bT11
6 T B¼6 40 0T
2
F 0 60 F 60 0 6 60 0 6 Y¼6 60 0 60 0 6 60 0 4 0 0 0 0
0 0 F 0 0 0 0 0 0
0 0 0 F 0 0 0 0 0
bT11 ¼ bT21 ¼ bT31 ¼ {0 1 0 0},
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0 0 0 0 F 0 0 0 0
0 0 0 0 0 F 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 F 0 0 F 0 0
0T
3
7 0T 7 5 bT33
3 0 07 07 7 07 7 07 7 07 7 07 5 0 F
bT12 ¼ bT22 ¼ bT32 ¼ {0 0 1 0}
bT13 ¼ bT23 ¼ bT33 ¼ {0
0 1 0}
Finally, the incremental displacements are 9 8 < Dn g1 = Dn G ¼ I Dn g2 in which I is given below: ; : Dn g3 As in Example 15.2, M ¼ (I BY)I GT ¼ 12 I (I þ BYG)T þ (I þ BYG) I U (I BY)I BTL ¼ 12 [I I þ (I I)U](I BY)I BTNL ¼ 12 I (BYG)T þ [(BYG) I]U (I BY)I and the matrix I is now obtained as 2
I 60 6 6 60 6 60 6 60 6 60 6 6 60 6 60 6 60 6 60 6 6 6I 6 60 6 60 6 6 I ¼ 60 6 60 6 60 6 60 6 6 60 6 60 6 60 6 6I 6 6 60 6 60 6 60 6 6 60 6 40 0
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0 0 0 I 0 0 0 0 0 0 0 0 0 I 0 0 0 0 0 0 0 0 0 I 0 0 0
3 0 07 7 7 07 7 07 7 07 7 07 7 7 I7 7 07 7 07 7 7 07 7 07 7 07 7 07 7 7 07 7 07 7 07 7 I7 7 7 07 7 07 7 07 7 07 7 7 07 7 07 7 07 7 7 07 7 05 I
15.2 STRESS INCREMENTS For the purposes of deriving an incremental variational principle we shall see that the is the starting point. However, to formuincremental 1st Piola–Kirchhoff stress DnS late mechanical properties , the objective increment of the Cauchy stress, based on the Truesdell stress flux, Dn T, is the starting point. Furthermore, in the resulting variational statement, which we called the Incremental Principle of Virtual Work, we shall find that quantity which appears is the increment of the 2nd Piola–Kirchhoff stress, DnS. ¼ SFT , from which to first order From Chapter 5, S ¼ Dn SFT þ SDn FT Dn S
(15:8)
For the Cauchy stress, the increment must take account of the rotation of the underlying coordinate system and thereby be objective. We recall the objective Truesdell stress flux @ T=@ t introduced in Chapter 13:
@ T=@t ¼ @T=@t þ T tr(D) LT TLT
(15:9)
Among the possible stress fluxes, it is unique in being proportional to the rate of the 2nd Piola–Kirchhoff stress, namely
@S=@t ¼ JF1 @ T=@t FT
(15:10)
An objective Truesdell stress increment Dn T dt(@ T=@t) is readily obtained as
1 VEC Dn T ¼ FT FVEC(Dn S) J
(15:11)
Further, once VEC Dn T has been determined, the (nonobjective) increment DnT of the Cauchy stress may be computed using Dn T ¼ Dn T þ T tr (Dn F)F1 Dn FF1 T TFT Dn FT
(15:12)
from which
VEC(Dn T) ¼ VEC Dn T TVECT FT TFT I I TFT MDn g (15:13)
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15.3 INCREMENTAL EQUATION OF BALANCE OF LINEAR MOMENTUM We now formulate the incremental equilibrium equation of nonlinear solid mechanics (assuming that the body has a fixed point). In the deformed (Eulerian) configuration, equilibrium at tn requires ð ð € dV TT n dS ¼ ru (15:14) Referred to the undeformed (Lagrangian) configuration, this equation becomes ð ð T n0 dS0 ¼ r u (15:15) S 0 € dV0 S0 denotes the surface (boundary) in the undeformed configuration, and n0 is the is surface normal vector in the undeformed configuration. Suppose the solution for S known as Sn at time tn and is sought at tnþ1. As usual, we introduce the increment to denote S nþ1 S n. Now subtracting the equilibrium equation at time tn from DnS that at tnþ1 furnishes the incremental equilibrium equation ð ð T n0 dS0 ¼ r Dn u € dV0 Dn S (15:16) 0 Application of the divergence theorem furnishes the differential equation
T T ¼ r Dn u € r T Dn S 0
(15:17)
which is the local form of the incremental equilibrium equation.
15.4 INCREMENTAL PRINCIPLE OF VIRTUAL WORK To derive a variational principle for the current formulation, the quantity to be varied is the incremental displacement vector which accordingly is the primary variable. Following Chapter 4, Equation 15.17 is multiplied by (dDn u)T Integration is performed over the domain The divergence theorem is invoked once Terms appearing on the boundary are identified as primary and secondary variables (v) Boundary conditions and constraints are applied
(i) (ii) (iii) (iv)
The reasoning process is very similar to that in the derivation of the Principle of Virtual work in finite deformation in which u is the unknown, and furnishes ð ð ð ð € dV0 ¼ dDn uT Dn t0 dS0 tr(dDn ET Dn S) dV0 þ dDn FSDn FT dV0 þ dDn uT r0 Dn u (15:18)
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in which t0 is the traction experienced by dS0. The increment Dnt0 is momentarily assumed to be specified on the undeformed boundary. However, in a subsequent section it will be derived in the case in which the traction increment is specified in the current configuration. The fourth term describes the virtual external work of the traction increments. The first term may be said to describe the virtual internal work of the stress increments, referred to the undeformed configuration. The third term describes the virtual internal work of the inertial force increments. The second term has no counterpart in the previously formulated Principle of Virtual Work in Chapter 13 and arises because of nonlinear geometric effects in the incremental formulation. We simply call it the geometric stiffness integral. Owing to the importance of the Incremental Principle of Virtual Work, it is derived in detail below. It is convenient to perform the derivation using tensorindicial notation. The incremental equilibrium equation referred to the undeformed configuration is restated as ð ð ð @ ij ) dV0 ¼ @ [dDn ui (Dn S ij )] dV0 @ ½dDn ui Dn S ij dV0 ( Dn S dDn ui @ Xj @ Xj @ Xj ð ui dV0 ¼ dDn ui r0 Dn € (15:19) The divergence theorem is invoked to convert the first right-hand term to ð ð @ ij )] dV0 ¼ dDn ui (nj Dn S ij ) dS0 [dDn ui (Dn S @ Xj ð ¼ dDn ui Dn t0j dS0
(15:20)
which is recognized as the fourth term in Equation 15.18. To first order in increments and now using tensor notation, the second right-hand term is written as ð ð @ dV0 [dDn ui ]Dn Sij dV0 ¼ tr(dDn FDn S) @Xj ð ¼ tr dDn F Dn SFT þ SDn FT dV0 ð ð ¼ tr FT dDn FDn S dV0 þ tr dDn FSDn FT dV0 (15:21) The second term is recognized as the second term in Equation 15.18. Recalling that the variational operator is applied to the primary variable which is an incremental displacement vector, the first term now becomes ð ð tr FT dDn FDn S dV0 ¼ tr 12 FT dDn F þ dDn FT F Dn S dV0 ð ¼ tr dDn ET Dn S dV0 (15:22)
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which is recognized as the first term in Equation 15.18. Finally, the third term in Equation 15.18 is recognized as the incremental virtual work of the incremental inertial forces.
15.5 INCREMENTAL FINITE ELEMENT EQUATION For present purposes let us suppose constitutive relations in the form Dn S ¼ D0 (X, gn )Dn E þ VDn t
(15:23)
in which D0 (X, g) is the fourth-order tangent modulus tensor. Equation 15.23 is capable of describing combined rate-independent and linearly rate-dependent response, for example, combined plasticity and viscoplasticity. To take advantage of Kronecker product notation, the equation is rewritten as Dn s ¼ x0 (X, gn )Dn e þ vDn t s ¼ VEC (S), e ¼ VEC (E),
(15:24)
x0 ¼ TEN 22(D0 ), v ¼ VEC (V)
Using the interpolation model (Equation 15.1), Equation 15.18 becomes € [Dn f f v Dn t ] ¼ 0 dDn gT ½ðKT þ KG ÞDn g þ MDn g Ð T Ð KT ¼ M Gx0 GT M dV0 , KG ¼ MT S M dV0 Ð Ð M ¼ r0 FT wwT F dV , Dn f ¼ FT wDt0 dS0 Ð f v ¼ MT Gv dV0
(15:25)
KT is now called the tangent modulus matrix, KG the geometric stiffness matrix, M the (incremental) mass matrix, Dnf is called the incremental (surface) force vector, and we call Dnfv the incremental viscous force.
15.6 CONTRIBUTIONS FROM NONLINEAR BOUNDARY CONDITIONS Recalling Chapter 13, let Ii denote the principal invariants of C, and let i ¼ VEC(I), c2 ¼ VEC (C2 ), nTi ¼ @ Ii =@ c, and Ai ¼ @ ni=@ c. Recall from Chapter 13 that n1 ¼ i,
n2 ¼ I1 i c,
n3 ¼ I2 i I1 c þ c2 ¼ I3 VEC(C1 ), I 9 ¼ I I
A1 ¼ 0, A2 ¼ iiT I9 , A3 ¼ I C þ C I (icT þ ciT ) þ I1 (iiT I9 )
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(15:26)
Equation 15.18 can be applied if increments of tractions are prescribed on the undeformed surface S0. We now consider the more complex situation in which Dn is referred to the deformed surface S, on which they are prescribed functions of u. This situation may occur under, for example, ‘‘dead loading.’’ Using the relations of Chapter 13 conversion to undeformed coordinates is obtained using the relations t dS ¼ t0 dS0 ,
dS ¼ m dS0 ,
m¼J
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi nT0 C1 n0 ¼ nT0 nT0 n3
(15:27)
and from Nicholson and Lin (1997b) Dn m dm ¼ mT dc mT Dn c,
mT ¼ nT0 nT0 A3 =2m
(15:28)
Suppose that Dnt is expressed on S as follows: Dn t ¼ Dn t ATM Dn u
(15:29)
Here Dnt is prescribed, while AM is a known function of u. Also S0 is the undeformed counterpart of S. Owing to the presence of AM the expression in Equation 15.29 is capable of modeling boundary conditions such as support by a nonlinear elastic foundation. From the fact that t dS ¼ t0 dS0, we conclude that t ¼ t0=m. It follows that Dn t0 ¼ mDn t þ tmT Dn c ¼ m(Dn t ATM Dn u) þ tmT Dn c
(15:30)
From the Incremental Principle of Virtual Work, the right-hand term may be written as ð
ð dDuT Dt0 dS0 ¼ dDuT [m(Dt ATM Du) tmT Dc] dS0
(15:31)
Now recalling the interpolation models for the increments we obtain an incremental force vector plus two boundary contributions to the stiffness terms. In particular ð dDn uT Dn t0 dS0 ¼ dDn gT Dn f dDn gT [KBF þ KBN ]Dn g ð ð Dn f ¼ mDn t dS0 , KBF ¼ FT wmATM wT F dS0 ,
(15:32)
ð KBN ¼ 2 FT w(tmT )GT M dS0
The first boundary contribution is from the nonlinear elastic foundation coupling the traction and displacement increments on the boundary. The second arises
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from geometric nonlinearity in which the traction increment is prescribed on the current configuration.
15.7 EFFECT OF VARIABLE CONTACT In many, if not most, ‘‘real world’’ problems, loads are transmitted to the member of interest via contact with other members, for example in gear teeth. The extent of the contact zone is now an unknown to be determined as part of the solution process. Solution of contact problems, discussed in an introductory way in Chapter 12, is a difficult problem that has absorbed the attention of many investigators. Some algorithms are suited primarily for linear kinematics. Here a development is given of an example which explicitly addresses the effect of large deformation. Figure 15.1 shows a contactor moving toward a foundation, assumed to be rigid and fixed. We seek to follow the development of the contact area and the tractions arising throughout it. Recalling Chapter 12, corresponding to a point x on the contactor surface there is a target point y(x) on the foundation to which the normal n(x) at x points. As the contactor starts to deform n(x) rotates and points toward a new value y(x). As the point x approaches contact, the point y(x) approaches the foundation point which will come into contact with the contactor point at x. We define a gap function g(u,x) using y(x) ¼ x þ gn. Let m be the surface normal vector to the target (foundation) at y(x). Also, let Sc be the candidate contact surface on the contactor, whose undeformed counterpart is S0c. There likewise is a candidate contact surface Sf on the foundation. We limit attention to bonded contact, in which particles coming into contact with each other remain in contact and do not slide away from each other. Algorithms for sliding contact with and without friction are available. For simplicity, for the moment we also assume that shear tractions, in the osculating plane of point of interest, are
Contactor
n
g
m y(x)
Foundation
FIGURE 15.1 Contact scenario.
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negligible. Suppose now that the interface can be represented by an elastic foundation satisfying the incremental relation Dn tn ¼ k(g)Dn un
(15:33)
Here tn ¼ nTt and un ¼ nTu are the normal components of the traction and displacement vectors at x. Since the only traction being considered is the normal traction (to the contactor surface), the transverse components of Du are not needed (do not result from work). Also, k(g) is a nonlinear stiffness function given in terms of the gap by, for example, k(g) ¼
i kH hp arctan(ak g «r ) þ kL , p 2
kH =kL 1
(15:34)
As in Chapter 12, when g is positive, the gap is open and k approaches kL which should be chosen as a small number, theoretically zero. As g becomes negative, the gap is closing and k thereafter rapidly approaches kH which should be chosen as a large number (theoretically infinite to prevent penetration of the rigid body). A function similar to this was also discussed in Chapter 14, and its characteristics were illustrated graphically. Under the assumption that only the normal traction on the contactor surface in important, it likewise follows that we may use the relation t ¼ tnn, with the consequence that Dn t ¼ Dn tn n þ tn Dn n
(15:35)
The contact model contributes the matrix Kc to the total stiffness matrix as follows (Nicholson and Lin, 1997b): ð ð T dDu Dtm dS0 ¼ dDuTn Dtn m dS0 ¼ dDgT Kc Dg ð ð Kc ¼ 2 FT wntn mT bT F dSc0 þ kc (g)FT wnnT wT Fm dSc0 ð þ FT wtn mhT dSc0
(15:36)
in which hT is presented below. To update the gap, use may be made of the following relations reported in Nicholson and Lin (1997b). The differential vector dy is tangent to the foundation surface, and hence mT dy ¼ 0. It follows that 0 mT Dn u þ gmT Dn n þ mT nDn g Dg ¼
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mT Du þ gmT Dn mT n
(15:37)
Using Equations 15.7 and 15.37 we may derive with some effort that mT FT w þ ghT mT n hT ¼ mT n (nT F1 ) nT nT FT MT
Dg ¼ GT Dg,
GT ¼
(15:38)
15.8 INTERPRETATION AS NEWTON ITERATION The (nonincremental) Principle of Virtual Work is restated in Lagrangian coordinates as ð
ð ð € dV0 ¼ duT t0 dS0 tr (dES) dV0 þ duT ru
(15:39)
We assume for convenience that t0 is prescribed on S0.The interpolation model for e was shown in Equation 15.8 to have the form de ¼ BTL þ BTNL (g) dg. Upon cancellation of the variation dgT, an algebraic equation in g is obtained as ð ð ð € dV0 , f ¼ FT wt0 dS0 , s ¼ VEC (S) 0 ¼ C(g, f ) ¼ ½BL þ BNL (g)s dV0 þ FT wr0 u (15:40) At the nth load step Newton iteration is expressed as (n) (nþ1) (n) 1 gnþ 1 ¼ gnþ1 J F gnþ1 , f nþ1 ,
J¼
@ F (n) g n þ1 , f n þ1 @g
(15:41)
or alternatively as a linear system (n) (n) J gn(nþþ11) gnþ 1 ¼ F gnþ1 , f nþ1 (nþ1) (n ) gn(nþþ11) ¼ gn(nþ) 1 þ gnþ 1 gnþ1
(15:42)
If the load increments are small enough, the starting iterate may be estimated as the solution from the nth load step. Also a stopping (convergence) criterion is needed to determine when the effort to generate additional iterates is not rewarded by increased accuracy. Careful examination of the relations from this and the incremental formulations uncovers that J ¼ KT þ KG (static) ¼ KG þ KT þ 2M=h2 (dynamic)
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(15:43)
presuming, for example, that the trapezoidal rule is used in the dynamic case to model the time derivatives. Clearly the incremental stiffness matrix is the same as the Jacobian matrix in Newton iteration. Equation 15.43 is, of course, a very satisfying result—it reveals that the Jacobian matrix of Newton Iteration may be calculated by conventional finite element procedures at the element level followed by conventional assembly procedures. If the incremental equation is only solved once at each load increment, the ensuing solution may be viewed as the first iterate in a Newton iteration scheme. The one-time incremental solution can potentially be improved by additional iterations following Equation 15.44, but at the cost of additional effort. However, it is also possible that the use of multiple iterations will enable larger load steps, thereby compensating for the computational effort.
15.9 BUCKLING Finite element equations based on classical linear buckling equations for beams and plates were treated in Chapter 11. In the classical equations, what strictly are geometrically nonlinear terms appear through what may be considered a linear correction term, furnishing linear equations. Here, in the absence of inertia and nonlinearity in the boundary conditions, we briefly present a more general viewpoint based on the incremental equilibrium equation (KT þ KG )Dn g ¼ Dn f
(15:44)
This solution will predict a very large incremental displacement if the stiffness matrix KT þ KG is ill-conditioned or outright singular. Of course, in elastic media, KT is positive definite. However, in the presence of in-plane compression, we will see that KG may have a negative eigenvalue whose magnitude is comparable to the smallest positive eigenvalue of KT. To see this recall that ð
ð
KT ¼ M GxG M dV0 , KG ¼ MT S IM dV0 T
T
(15:45)
We suppose that the element in question is thin in a local z (out-of-plane direction), corresponding to plane stress. Now in plate and shell theory, it is a common practice to add a transverse shear stress on the element boundaries to allow the element to support transverse loads. We assume that the transverse shear stresses only appear in the incremental force term and the tangent stiffness term, and that the geometric stiffness term strictly satisfies the plane stress assumption. It follows that, if the threedirection is out of the plane, the geometric stiffness term KG contains the expression 2
S11 I S I ¼ 4 S12 I 0I
3 S12 I 0I S22 I 0I 5 0I 0I
(15:46)
in which S is of course the 2nd Piola–Kirchhoff stress. In classical (linear) buckling theory loads which are applied proportionately induce proportionate in-plane stresses
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(i.e., the stress components have constant ratios). Accordingly, for a given load path, only one parameter, the length of the straight line the stress point traverses in the space of in-plane stresses, arises in the eigenvalue problem for the critical buckling load. In nonlinear problems, there is no assurance that the stress point follows a straight line even if the loads are proportionate. Instead, if l denotes the distance along the line followed by the load point in proportional loading in load space, the stresses become numerical functions of l. As a simple alternative to the classical case addressing a complete member, we consider buckling of a single element, and suppose that the stresses appearing in Equation 15.46 are applied in a compressive sense along the faces of the element and in a proportional manner, i.e., 2
3 0I 0I 5 0I
^12 )I ( S ^22 )I ( S 0I
^11 )I ( S 4 ^12 )I S I ! l ( S 0I
(15:47)
in which the circumflex implies a reference value along the stress path at which l ¼ 1. The negative signs on the stresses are present since buckling is associated with compressive stresses, although the sign is not needed for the shear stress. At the element level, the equation now becomes
^ G Dgnþ1 ¼ Df nþ1 KT lK
(15:48)
At a given load increment, the critical stress intensity for the current stress path, as a function of an two (spherical) angles determining the path in the stress space illustrated in Figure 11.11, is obtained by computing the l value rendering (KT ^ G) singular. To an extent, the integration in computing K ^ G can be made indelK pendent of the stress path by adapting Equation 11.70 as follows: ^ G ¼ IVEC K
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ð
M M dV0 T
T
^I VEC S
(15:49)
16
Tangent Modulus Tensors for Thermomechanical Response of Elastomers
16.1 INTRODUCTION Elastomeric materials embrace natural and synthetic rubber as well as biological tissues. Attention in this chapter is restricted to isotropic elastomers. Their characteristic is that deformation is recoverable even up to very large strains, and the stress is a nonlinear function of strain. Accordingly, they pose issues of material and geometric nonlinearity, potentially also of boundary condition nonlinearity. They pose two additional issues not addressed in Chapter 15. One is the presence of thermal fields coupled (weakly) to the mechanical field, and the second is the presence of a pressure field arising to enforce the constraint of incompressibility or near-incompressibility and serving as an additional primary variable. Within an element the finite element method makes use of interpolation models for the displacement vector u(X,t) and temperature T(X,t), and pressure p ¼ tr(t)=3 in incompressible or near-incompressible materials: u(X,t ) ¼ N T (X)g(t ), T(X,t ) T0 ¼ nT (X)u(t ),
p ¼ jT (X)c(t )
(16:1)
in which T0 is the temperature in the reference configuration, assumed constant. Here N(X) ¼ wT(X)F, n(X), and j(X) are shape functions and g, u, and c are vectors of nodal values of displacement, temperature (T T0), and pressure, respectively. Application of the strain displacement relations and their thermal analogs furnishes f 1 ¼ VEC(F I) ¼ M1 g ¼ UT M2 g, f 2 ¼ VEC(FT I) ¼ M2 g, b ¼ M2 GT ,
GT ¼ 12 (FT I þ I FT U),
r0 T ¼ bTT u
de ¼ bT dg, (16:2)
in which e ¼ VEC(E) is the Lagrangian strain vector. Also, r0 ¼ FTr is the gradient operator referred to the deformed configuration. Of course, the matrix b and the vector bT are typically expressed in terms of natural coordinates.
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16.2 COMPRESSIBLE ELASTOMERS The Helmholtz potential was introduced in Chapter 6 and shown to underlie the relations of classical coupled thermoelasticity. The thermohyperelastic properties of compressible elastomers may likewise be derived from the Helmholtz free energy density f (per unit mass), which is a function of T and E. Under isothermal conditions it is conventional to introduce the strain energy density w(E) ¼ r0f (T,E) (T constant), in which r0 is the density in the undeformed configuration. Typically, the elastomer is assumed to be isotropic, in which case f can be expressed as a function of T, I1, I2, and I3. Alternatively, it may be expressed as pffiffiffiaffi function of T and the stretch ratios l1, l2, and l3, which are the eigenvalues of C, the square root of the right Cauchy–Green strain tensor. With f specified as a function of T, I1, I2, and I3, the entropy density h per unit mass and the specific heat ce at constant (Lagrangian) strain are quoted from Chapter 6 as @ h @ f , h ¼ (16:3) ce ¼ T @ T E @ T E The second Piola–Kirchhoff stress satisfies the relation (cf. Chapter 6) sT ¼ VEC T (S) ¼ r0 @@fe T and is obtained as sT ¼ 2
X i
r 0 fi nTi ,
fi ¼
@f @ Ii
(16:4)
Also of importance is the (isothermal) tangent modulus matrix DT ¼
XX X @ s @2f ¼4 r0 fij ni nTj þ 4 r0 fi Ai , fij ¼ @e T @ Ii @ Ij i j i
(16:5)
An expression for DT has also been derived by Nicholson and Lin (1997c) for compressible, incompressible, and near-incompressible elastomers described by strain energy functions (Helmholtz free energy functions) based on the use of stretch ratios rather than invariants.
16.3 INCOMPRESSIBLE AND NEAR-INCOMPRESSIBLE ELASTOMERS When the temperature T is held constant, elastomers can often be considered to satisfy the internal constraint of incompressibility or near-incompressibility. To satisfy the constraint a posteriori, f is augmented with terms involving a new parameter playing the role of a Lagrange multiplier. Typically, this new parameter may be interpreted as the pressure p, referred to the undeformed configuration. Consequently, the thermohyperelastic properties of incompressible and near-incompressible elastomers may be
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derived from the augmented Helmholtz free energy, which is a function of E, T, and p. The constraint introduces additional terms into the governing finite element equations and requires an interpolation model for the new primary variable p. If the elastomer is incompressible at constant temperature, the augmented Helmholtz function f may be written as f ¼ fd (J1 ,J2 ,T) lj(J ,T)=r 0 ,
1=3
J1 ¼ I1 =I3 ,
2=3
J2 ¼ I2 =I3
(16:6)
where j is a material function satisfying the constraint j(J,T) ¼ 0 and 1=2 J ¼ I3 ¼ det(F). It is easily shown that fd depends on the deviatoric Lagrangian 1=3
strain Ed ( ¼ 12 (C=I3 I) owing to the introduction of the deviatoric invariants J2 and J3, previously encountered in Chapter 13, Equation 13.76. The ‘‘thermodynamic’’ pressure is given by @ j l ¼ p ¼ tr (T)=3 ¼ (16:7) @ J T For an elastomer which is near-incompressible at constant temperature, f may be written as r 0 f ¼ r0 fd (J1 ,J2 ,T) pj(J ,T) p2 =2k0
(16:8)
in which k0 is a constant. The near-incompressibility constraint is expressed by @ f=@ p ¼ 0, which implies p ¼ k0 j(J ,T)
(16:9)
The bulk modulus k defined by k ¼ @@ pJ T , and we conclude that @ j k ¼ k0 @J T
(16:10)
Chen et al. (1997) presented sufficient conditions under which near-incompressible models reduce to the incompressible case as k ! 1. Nicholson and Lin (1996) proposed the relations j(J,T) ¼ f 3 (T)J 1,
fd ¼ f1 (J1 ,J2 ) þ f2 (T),
f2 (T) ¼ ce T(1 ln(T=T0 )) (16:11)
with the consequence that p ¼ k0 ( f 3 (T)J 1), k ¼ f 3 (T)k0
(16:12)
Equation 16.12 provides a linear pressure–volume relation in which thermomechanical effects are confined to thermal expansion expressed using a constant volume
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coefficient a. It directly generalizes the pressure–volume relation of classical linear isotropic elasticity. If the near-incompressibility constraint is assumed to be satisfied a priori, the Helmholtz free energy is recovered as f(I1 ,I2 ,I3 ,T) ¼ fd (J1 ,J2 ,T) þ k0 ( f 3 (T) 1)2 =2r0
(16:13)
The last term in Equation 16.13 results from retaining the lowest nonvanishing term in a Taylor series representation of f about f 3(T)J1. Assuming Equation 16.13, the entropy density now includes a term involving p: h¼
@f @f ¼ d þ pa f 4 (T)=r0 , @T @T
p ¼ p=f 3 (T)
(16:14)
The stress and the (9 3 9) tangent modulus tensor are correspondingly modified from the compressible case to accommodate near-incompressibility @fd pf 3 (T)nT3 =J s ¼ r0 @e T,p T @s @ @fd DTP ¼ ¼ r0 pf 3 (T) 2A3 =J n3 nT3 =J 3 @e T,p @e @e T
16.3.1 EXAMPLES OF EXPRESSIONS
FOR THE
(16:15)
HELMHOLTZ POTENTIAL
There are two broad approaches to the formulation of Helmholtz potential: 1. To express f as a function of I1, I2, and I3, and T (and p) 2. To express f as a function of the principal stretches l1, l2, and l3, and T (and p) The latter approach is thought to possess the convenient feature of allowing direct use of test data, say from uniaxial tension. We distinguish several cases. 16.3.1.1
Invariant-Based Incompressible Models: Isothermal Problems
In the entitled case, the strain energy function depends only on I1, I2 and incompressibility is expressed by the constraint I3 ¼ 1, assumed to be satisfied a priori. In this category, the most widely used models include the Neo-Hookean material (a): f ¼ C1 (I1 3),
I3 ¼ 1
(16:16)
and the (two-term) Mooney–Rivlin material (b): f ¼ C1 (I1 3) þ C2 (I2 3),
I3 ¼ 1
(16:17)
in which C1 and C2 are material constants. Most finite element codes with hyperelastic elements support the Mooney–Rivlin model. In principle, Mooney–Rivlin
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coefficients C1 and C2 can be determined independently by ‘‘fitting’’ suitable load– deflection curves, for example, uniaxial tension. Values for several different rubber compounds are listed in Nicholson and Nelson (1990). 16.3.1.2
Invariant-Based Models for Compressible Elastomers under Isothermal Conditions
Two widely studied strain energy functions are due to Blatz and Ko (1962). Let G0 be the shear modulus and n0 the Poisson’s ratio, referred to the undeformed configuration. The two models are: n0 1 1 2n0 1 þ n0 12n0 r 0 f 1 ¼ G0 I 1 þ I3 I3 n0 n0 2 1 I2 r 0 f 2 ¼ G0 þ 2I3 5 I3 2
(16:18)
Let w denote the Helmholtz free energy evaluated at a constant temperature, in which case it reduces to the strain energy. We note a general expression for w which is implemented in several commercial finite element codes (e.g., ANSYS, 2000): w(J1 ,J2 ,J ) ¼
XX i
Cij (J1 3)i (J2 3) j þ
j
Jr ¼ J =(1 þ Eth )
X
(Jr 1)k =Dk ,
k
(16:19)
in which Eth is called the thermal expansion strain, while Cij and Dk are material constants. Several codes provide software routines for estimating the model coefficients from user-supplied data. Several authors have attempted to uncouple the response into isochoric (volume conserving) and volumetric parts even in the compressible range, giving rise to functions of the form w ¼ w1 (J1 ,J2 ) þ w2 (J ), and recall that J1, J2 are the deviatoric invariants. A number of proposed forms for 2 are discussed in Holzappel (1996). 16.3.1.3
Thermomechanical Behavior under Non-Isothermal Conditions
Finally we illustrate the accommodation of coupled thermomechanical effects. A more detailed presentation is given in Section 16.5. Simple extensions of, say, the Mooney–Rivlin material have been proposed by Dillon (1962), Nicholson and Nelson (1990), and Nicholson (1995) for compressible elastomers, and in Nicholson and Lin (1996) for incompressible and near-incompressible elastomers. From the latter reference, the model for near-incompressible elastomers is r0 f ¼ C1 (J1 2) þ C2 (J2 3) þ r0 ce T(1 ln(T=T0 )) ( f 3 (T)J 1)p p2 =2k0 (16:20) in which, as before, p ¼ p=f 3(T). The model assumes that the coefficient of specific heat at constant strain is a constant. As previously mentioned a model similar to
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Nicholson and Lin (1996) has been proposed by Holzappel and Simo (1996) for compressible elastomers described using stretch ratios.
16.4 STRETCH-RATIO-BASED MODELS: ISOTHERMAL CONDITIONS For compressible elastomers, Valanis and Landel (1967) proposed a strain energy function based on the decomposition f(l1 , l2 , l3 , T) ¼ f(l1 ,T) þ f(l2 ,T) þ f(l3 ,T), T fixed
(16:21)
Ogden (1986) has proposed the form r0 f(l,T) ¼
N X
mp (lap 1), T fixed
(16:22)
1
In principle, in incompressible isotropic elastomers stretch-ratio-based models have the advantage of permitting direct use of ‘‘archival’’ data from single stress tests, for example uniaxial tension. We now illustrate the application of Kronecker product algebra to thermohyperelastic materials under isothermal conditions. We then accommodate thermal effects. From Nicholson and Lin (1997c), we invoke the expression for the differential of a tensor-valued isotropic function of a tensor. Namely let A denote a nonsingular n 3 n tensor with distinct eigenvalues, and let F(A) be a tensor-valued isotropic function of A, admitting representation as a convergent polynomial: F(A) ¼
/ X
fj Aj
(16:23)
0
Here fj are constants. A compact expression for the differential dF(A) is presented using Kronecker product notation. The reader is referred to Nicholson and Lin (1997c) for the derivation of the following expression. With f ¼ VEC(F) and a ¼ VEC(A), df(a) ¼ 12 F0 T F0 da þ W dv F0 (A) ¼
/ X
jfj Aj1 ,
0
dF df ¼ ITEN22 dA da
(16:24)
W ¼ (F AF0=2)T (F AF0=2) þ 12 (AT F0 F0 T A) Also, dv ¼ VEC(dV) in which dV is an antisymmetric tensor representing the rate of rotation of the principal directions. The critical step is to determine a matrix J such that W dv ¼ J da. It is shown in Nicholson and Lin (1997c) that J ¼ [AT A]I W,
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in which [AT A]I is the Morse–Penrose inverse (Dahlquist and Bjork, 1974). Accordingly df=da ¼ F0T F0=2 [AT A]I W
(16:25)
We now apply the tensor derivative to elastomers modeled using stretch ratios, especially in the model due to Ogden (1972). In particular a strain energy function w was proposed which for compressible elastomers and for isothermal response is equivalent to the form ! X zi w ¼ tr ji C I (16:26) i
in which ji, zi are material properties. The (9 3 9) tangent modulus tensor x0 appearing in Chapter 15 for the incremental form of the Principle of Virtual Work is obtained as x0 ¼ 4
X
zi ji (zi 1)Czi2 Czi2 =2 þ 4
i
Wi ¼
X
zi ji [AT A]I Wi
(16:27)
i
3 zi zi z 1 z i 2 C Cz i þ i C C C z i 2 ] [C 2 2
(16:28)
16.5 EXTENSION TO THERMOHYPERELASTIC MATERIALS A development of the thermohyperelastic model in Equation 16.20 is now given, following Nicholson and Lin (1996). The body initially experiences temperature T0 uniformly. It is assumed that temperature effects occur primarily as thermal expansion, that volume changes are small, and that volume changes depend linearly on temperature. Thus materials of present interest may be described as mechanically nonlinear but thermally linear. Owing to the role of thermal expansion, it is desirable to uncouple dilatational and deviatoric effects as much as possible. To this end we invoke the deviatoric 1=3 principal invariant of C. Cauchy–Green strain C^ ¼ C=I3 in which I3 is the third 1=2 Upon modifying w and expanding it in J 1, J ¼ I3 and retaining lowest order terms, we obtain ! X 1 ji C^ zi I þ k(J 1)2 (16:29) w ¼ tr 2 i in which k is the bulk modulus. The expression for x0 in Equation 16.27 is affected by these modifications. To accommodate thermal effects it is necessary to recognize that w is simply the Helmholtz free energy density r0f under isothermal conditions, in which r0 is the mass density in the undeformed configuration. It is assumed that f ¼ 0 in the undeformed configuration. As for invariant-based models, we may obtain a function
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f with three terms: a purely mechanical term fM, a purely thermal term fT, and a mixed term fTM. Now with entropy denoted by h, f satisfies the relations sT ¼ r 0
@ w , @ ejT
h¼
@w @ Tje
(16:30)
As previously stated, the specific heat at constant strain, ce ¼ T@ h=@ Tje , is assumed to be constant, from which we obtain fT ¼ ce T[1 ln(T=T0 )]
(16:31)
On the assumption that thermal effects in shear (i.e., deviatoric effects) can be neglected relative to thermal effects in dilatation, the purely mechanical effect is equated with the deviatoric term in Equation 16.29. ! X zi ^ fM ¼ tr ji C I (16:32) i
Of greatest present interest in the current context is fTM. The development of Nicholson and Lin (1996) furnishes fTM ¼
k[b3 (T)J 1]2 , 2r
b(T) ¼ (1 þ a(T=T0 )=3)1
(16:33)
The tangent modulus tensor x0 ¼ @ s=@ e now has two parts: xM þ xTM , in which xM is recognized as x0, derived in Equation 16.27. Omitting the details, Kronecker product algebra serves to derive the following expression for the thermomechanical position of the tangent modulus tensor.
k 3 b3 n3 nT3 T 3 xTM ¼ b 2 n3 n3 þ (b J 1) A3 2 (16:34) Jr J J The foregoing discussion of thermohyperelastic models has been limited to compressible elastomers. However, many elastomers used in applications such as seals are incompressible or near-incompressible. For such applications, as we have seen that an additional field variable is introduced, namely the hydrostatic pressure (referred to deformed coordinates). It serves as a Lagrange multiplier enforcing the incompressibility and near-incompressibility constraints. Following the approach for invariant-based models, Equation 16.34 may be extended to incorporate the constraints of incompressibility and near-incompressibility. The tangent modulus tensor presented here purely addresses the differential of stress with respect to strain. However, if coupled heat transfer (conduction and radiation) is considered, a more general expression for the tangent modulus tensor is required, expressing increments of stress and entropy in terms of increments of strain and temperature. A development accommodating heat transfer for invariant based elastomers is given in Nicholson and Lin (1997a).
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EXAMPLE 16.1 Derive explicit forms of the stress and tangent modulus tensors using the Helmoltz potential for a near-incompressible thermohyperelastic material.
SOLUTION Enforcing the near-incompressibility constraint a priori, the Helmholtz potential function of interest is f(T,J1 ,J2 ,p) ¼
1 1 k0 ( f 3 (T)J 1)2 [C1 (J1 2) þ C2 (J2 3)] þ ce T[1 ln (T=T0 )] þ 2 r0 r0
Now the stress is to be obtained using the relations sT ¼ r 0
@ f dJ1 dJ2 dJ ¼ C1 þ C2 pf 3 (T) , p ¼ k0 ( f 3 (T)J 1) @ e T,p de de de
In terms of m1 and m2 presented subsequently in Equation 16.63 dJ1 ¼ mT1 , de
dJ2 ¼ mT2 de
Also dJ d pffiffi 1 dI3 nT3 ¼2 I 3 ¼ pffiffi ¼ de dc J I 3 dc Consequently, sT ¼ C1 mT1 þ C2 mT2 kpf 3 (T)
nT3 J
The tangent modulus tensor may now be stated as DT ¼
@ s @ m1 @ m2 A3 n3 nT3 3 þ C p f ¼ C (T) 2 1 2 @e @e J J3 @ e T,p
1=3 Since m1 ¼ 2 i 13 II13 n3 =I3 and m2 ¼ 2 n2 23 manipulation serves to derive that
I2 I3
2=3 n3 =I3 (Equation 16.63) some
dm1 dm1 4 1 I1 m1 nT3 þ 1=3 n3 iT n3 nT3 þ I1 A3 ¼2 ¼ de dc I3 3I3 I3 dm2 8 m2 nT3 4 2 I2 2 n3 nT2 2 I2 T þ 2=3 A2 þ n n A ¼ 3 3 3 de 3 I3 3 I3 2 3 I3 3 I3 I3
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EXAMPLE 16.2 Recover ce upon differentiating frt ¼ ce T[1 ln(T=T0 )] twice with respect to T.
SOLUTION Differentiation once gives
@frt 1 1 ¼ ce [1 ln(T=T0 )] þ ce T T=T0 T0 @T ¼ ce ln(T=T0 ) Differentiating a second time gives @ 2 frt 1 1 ce ¼ ce ¼ 2 T=T T T @T 0 0 Hence ce ¼ T
@ 2 frt @T2
@h rt But we know that ce ¼ T @T , in which h ¼ @f @T . It follows that E ce ¼ T
@ 2 frt @T2
verifying that ce is recovered from ce T[1 ln(T=T0 )].
16.6 THERMOMECHANICS OF DAMPED ELASTOMERS Thermoviscohyperelasticity is a topic central to important applications such as rubber mounts used in hot engines for vibration isolation. The current section describes a simple thermoviscohyperelastic constitutive model thought to be suitable for near-incompressible elastomers exhibiting modest levels of viscous damping following a Voigt-type of model. Two potential functions are used to provide a systematic treatment of reversible and irreversible effects. One is the familiar Helmholtz free energy in terms of the strain and the temperature; it describes reversible, thermohyperelastic effects. The second potential function, based on the model of Ziegler and Wehrli (1987), incorporates elements for modeling viscous dissipation and arises directly from the entropy production inequality. It provides a consistent thermodynamic framework for describing damping in terms of a viscosity tensor depending on strain and temperature. The formulation leads to a simple energy balance equation, which is used to derive a rate variational principle. Together with the Principle of Virtual Work, variational equations governing coupled thermal and mechanical effects are presented. Finite element equations are derived from the thermal equilibrium equation
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and from the Principle of Virtual Work. Several quantities such as internal energy density x have reversible and irreversible portions, indicated by the subscripts r and i: x ¼ xr þ xi. The thermodynamic formulation in the succeeding paragraphs is referred to undeformed coordinates. There are several types of viscoelastic behavior in elastomers, especially if they contain fillers such as carbon black. For example, under load elastomers experience stress softening and compression set, which are long-term viscoelastic phenomena. Of interest here is the type of damping which is usually assumed in vibration isolation, in which the stresses have an elastic and a viscous portion reminiscent of the classical Voigt model, and the viscous portion is proportional to strain rates. The time constants are small. It is viewed as arising in small motions superimposed on the large strains which already reflect long-term viscoelastic effects.
16.6.1 BALANCE
OF
ENERGY
The conventional equation for the balance of energy is expressed as r0 x_ ¼ sT e_ rT0 q0 þ r0 h ¼ sTr e_ þ sTi e_ rT0 q0 þ r0 h
(16:35)
in which s ¼ VEC(S) and e ¼ VEC(E). Here x is the internal energy per unit mass, q0 is the heat flux vector referred to undeformed coordinates, r0 is the divergence operator referred to undeformed coordinates, and h is the heat input per unit mass, for simplicity assumed independent of temperature. The state variables are recognized to be e and T. The Helmholtz free energy fr per unit mass (which is regarded as reversible) and the entropy h per unit mass are introduced using fr ¼ x Th
(16:36)
rT0 q0 r0 h ¼ sTr e_ þ sTi e_ r0 Th_ r0 hT_ r0 f_ r
(16:37)
Upon obvious rearrangement,
16.6.2 ENTROPY PRODUCTION INEQUALITY The entropy production inequality is stated as r0 Th_ rT0 q0 þ r0 h þ qT0 rT=T r0 f_ r sTr e_ sTi e_ þ r 0 Th_ þ r0 hT_ þ qT0 rT=T
(16:38)
As previously indicated, the Helmholtz potential is assumed to represent only reversible thermohyperelastic effects. However, we decompose h into reversible and irreversible portions: h ¼ hr þ hi. Also, fr, hr, and hi are assumed to be differentiable functions of E and T. We further suppose that hi ¼ hi1 þ hi2 in which
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r0 Th_ i2 ¼ rT0 q0 þ r0 h i
(16:39)
This may be interpreted as saying that part of the viscous dissipation is to ‘‘absorbed’’ as heat. We finally suppose that reversible effects are absorbed as a reversible portion of the heat input, as follows: r0 Th_ r ¼ rT0 q0 þ r0 h r
(16:40)
In addition, from conventional arguments using Maxwell relations, r@fr =@e ¼ sTr ,
@fr =@T ¼ hr
(16:41)
The consequence of the foregoing assumptions now emerges as the inequality sTi e_ qT0 r0 T=T r0 hi1 T_
(16:42)
Inequality (Equation 16.42) is satisfied if r0 hi1 T_ 0 and also if sTi e_ 0
(16:43a)
qT0 rT=T 0
(16:43b)
Inequality (Equation 16.43b) is conventionally assumed to express the fact that heat flows irreversibly from hot to cold zones. Inequality (Equation 16.43a) states that ‘‘viscous work’’ is dissipative. The statement r0hi1 T_ 0 is difficult to justify except by appealing to its consequence in the form of the physically appealing inequalities in Equation 16.43. Of course it may be true that hi2 ¼ 0, in which case the irreversible entropy is related to the irreversible heat input in precisely the same way as the reversible entropy is related to the reversible heat input.
16.6.3 DISSIPATION POTENTIAL Following Ziegler and Wehrli (1987), the specific dissipation rate potential C(q0,_e,e,T) ¼ r0hiT_ is introduced and assumed to serve as a rate potential for the irreversible stress and temperature gradient as follows: sTi ¼ r0 Li @C=@ e_
(16:44a)
rT0 T=T ¼ Lt r0 @C=@q0
(16:44b)
The function C is selected such that Li and Lt are positive scalars, in which case inequalities (Equation 16.44a and 16.44b) require that
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(@C=@ e_ )_e 0
(16:45a)
(@C=@q0 )q0 0
(16:45b)
This may be interpreted as indicating the convexity of a dissipation surface in (e_ ,q0) space. Finally, to state the constitutive relations for a thermoviscohyperelastic material it is sufficient to specify fr and C. A simple example is now presented to illustrate how inequality (Equation 16.45) provides a ‘‘framework’’ for describing dissipative effects. On the expectation that properties governing heat transfer are not affected by strain, we introduce the decomposition C ¼ Ci þ Ct , r0 Ct ¼ 12 Lt qT0 q0
(16:46)
Here, Ct represents thermal effects. Equation 16.44b implies that r0 T=T ¼ Lt q0
(16:47)
This is essentially the conventional Fourier law of heat conduction, with 1=Lt recognized as the thermal conductivity. As an elementary example of viscous dissipation, suppose that Ci ¼ m(T, J1 , J2 )e_ T e_ =2, Di ¼ 1
(16:48)
in which m(T,J1,J2) is the viscosity. Application of Equation 16.44a now gives si ¼ m(T,J1 ,J2 )e_
(16:49)
and inequality (Equation 16.45a) requires that the viscosity function m be positive.
16.6.4 THERMAL FIELD EQUATION
FOR
DAMPED ELASTOMERS
The energy balance equations of thermohyperelasticity (i.e., the reversible response) are now reappearing in terms of a balance law among reversible portions of the stress, entropy, and internal energy. Equation 16.41 implies that r0 f_ r ¼ sTr e_ r0 hr T_
(16:50)
The ensuing Maxwell reciprocity relation is @sTr =@T ¼ r0 @hr =@e
(16:51)
Now familiar operations furnish the reversible part of the equation of thermal equilibrium (balance of energy). _ ce ¼ T@hr =@T rT0 q0 þ r0 h r ¼ T @sTr =@T e_ þ r0 ce T,
(16:52)
For the irreversible part, we recall the relation rT0 q0 r0 h i ¼ sTi e_ þ r0 ci T_
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(16:53)
and r0 Th_ i2 ¼ r0 T
@ hi2 @h e_ þ r0 T i2 T_ @e @T
(16:54)
and sTi ¼ r0 T
@ hi2 , @e
ci ¼ T
@ hi2 @T
(16:55)
Upon adding the reversible and irreversible portions of the heat input we obtain the thermal field equation rT0 q0 þ r0 h ¼ T@ sTr =@ Te_ sTi e_ þ r0 (ce þ ci )T_
(16:56)
It is easily seen that Equation 16.62 directly reduces to a well-known expression in classical linear thermoelasticity when irreversible terms are suppressed. In addition, under adiabatic conditions in which rT0 q0 þ r0 h ¼ 0, most of the ‘‘viscous’’ work sTi e_ is absorbed as a temperature increase controlled by r0(ce þ ci), while a smaller portion is absorbed into the elastic strain energy field. Finally, note that the specific heat at constant strain possesses a reversible and an irreversible portion.
16.7 CONSTITUTIVE MODEL IN THERMOVISCOHYPERELASTICITY If the current formulation is followed, it is sufficient to introduce the Helmholtz free energy density and the dissipation potential to characterize thermal, mechanical, and viscous behavior
16.7.1 HELMHOLTZ FREE ENERGY DENSITY In the moderately damped thermohyperelastic material the reversible stress is assumed to satisfy a thermohyperelastic constitutive relation suitable for nearincompressible elastomers. Following the earlier development for an undamped elastomer, the Helmholtz free energy is introduced using fr ¼ frm (J1 ,J2 ,I3 ) þ frt (T) þ frtm (T,I3 ) þ fro
(16:57)
Here frm represents the purely mechanical response and can be identified as the conventional isothermal strain energy density function associated, for example, with the Mooney–Rivlin model. The formulation can easily be adapted to stretch ratiobased models such as the Ogden model (Ogden, 1986). The function frt(T) represents the purely thermal portion of the Helmholtz free energy density. Finally, frtm(T,I3) represents thermomechanical effects, again based on the assumption that the primary coupling is through volumetric expansion. The quantity fro represents the Helmholtz free energy in the reference state, and for simplicity it is assumed to vanish. The forms of frt and frtm developed in the previous sections of this chapter are recalled.
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frt (T) ¼ ce T[1 ln(T=T0 )]
(16:58)
k 3 [f (T)J 1]2 2r0 h i1 a f (T) ¼ 1 þ (T T0 ) 3
frtm (T,I3 ) ¼ 1=2
J ¼ I3
¼ det(F),
(16:59)
and a is of course the volumetric coefficient of thermal expansion. For the sake of illustration, for frm(J1,J2,I3) we display the classical two-term Mooney–Rivlin model (Gent, 1992): frm (J1 ,J2 ,I3 ) ¼ C1 (J1 1) þ C2 (J2 1),
I3 ¼ 1
(16:60)
The reversible stress is now stated as sr ¼ 2fj nj , n1 ¼ i, i ¼ VEC (I),
fj ¼ @ fr =@ Ij
n1 ¼ I1 i c,
c ¼ VEC (C),
(16:61) n3 ¼ I3 VEC (C1 )
16.7.2 DISSIPATION POTENTIAL Fourier’s law of heat conduction is recalled as from r Ct ¼
1 T [q q ] 2kt 0 0
(16:62)
The viscous stress si depends on the shear part of the strain rate as well as the temperature. However, since the elastomers of interest are nearly incompressible, to good approximation si can be taken as a function of the (total) Lagrangian strain rate. The current framework admits several possible expressions for Ci, of which an example was already given in Section 16.6.3. Here, taking a more general viewpoint, we seek expressions of the form Ci ¼ 12 e_ T Dv (e,T)e_ in which Dv(e,T) is called the viscosity tensor; it is symmetric and positive definite to satisfy Equation 16.45b. (Of course, the correct expression is determined by experiments.) The simple example in Section 16.6.3 corresponds to Dv(e,T) ¼ m(e,T)I. As a second example, to ensure isotropy suppose that Ci depends on J_ 1 and J_ 2 through a relation of the form Ci(J_1,J_2,e,T), and note that
1 I1 T . 1=3 _ 2 I2 T . 2=3 J_ 1 ¼ mT1 e_ , mT1 ¼ 2 iT n3 I3 , J2 ¼ mT2 e_ , mT2 ¼ 2 nT2 n3 I 3 3 I3 3 I3 (16:63) For an expression reminiscent of the two-term Mooney–Rivlin strain energy function let us consider the specific form: Ci ¼ m1 (T) C1v J_ 12 =2 þ C2v J_ 22
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(16:64)
in which C1v and C2v are constant positive material coefficients. We now obtain the viscosity tensor Dv ¼ m(T) C1v m1 mT1 þ C2v m2 mT2 (16:65) Unfortunately, this tensor is problematic since it is only positive semi-definite. To see this consider whether there exists a nonvanishing vector b for which (C1v m1 mT1 þ C2v m2 mT2 )b ¼ 0. But this is certainly true since b need only lie in a subspace exterior to the twodimensional subspace spanned by m1 and m2. As a third example, suppose that the dissipation potential is expressed in terms of the deformation rate tensor D as Ci ¼ m(T)tr(D2 )=2, which has the advantage that the deformation rate tensor D is in the observed (current) configuration in which measurements are performed. But now, with d ¼ VEC(D), d ¼ FT FT e_ Dv ¼ m(T)F1 F1 FT FT ¼ m(T)(F1 FT ) (F1 FT ) ¼ m(T)(2E þ I)1 (2E þ I)1
(16:66)
which is always positive definite if m(T) > 0.
16.8 VARIATIONAL PRINCIPLES AND FINITE ELEMENT EQUATIONS FOR THERMOVISCOHYPERELASTIC MATERIALS 16.8.1 MECHANICAL EQUILIBRIUM In this section we present one of the several possible formulations for the finite element equations of a thermoviscohyperelastic medium, neglecting inertia. Application of variational methods to the mechanical field equation (Balance of Linear Momentum) furnishes the Principle of Virtual Work in the form ð ð ð (16:67) tr(dESi ) dV0 ¼ duT t0 dS0 tr(dESr ) dV0 in which t0 as usual denotes the traction vector on the undeformed surface S0. As illustrated in the examples of the foregoing section, we expect that the dissipation potential has the form Ci ¼ 12 e_ T Dv (e,T)_e, from which si ¼ Dv e_
(16:68)
and Dv is of course the viscosity tensor and is symmetric. Furthermore, it is positive definite since sTi e_ 0 for all e_ . Equation 16.67 is thus rewritten as ð
ð
ð
de Dv e_ dV0 ¼ du t0 dS0 deT sr dV0 T
ß 2008 by Taylor & Francis Group, LLC.
T
(16:69)
_ t ), u(X,t ) ¼ wT (X)Fg(t ), With interpolation models of the form e_ (X,t ) ¼ bT (X)Fg( T and T(X,t ) T0 ¼ n (X)Cu(t ), at the element level the finite element equation for the mechanical field now becomes ð ð Kv (g,T)g_ ¼ f f r (g), Kv ¼ FT bDv (X,g,u)bT F dV0 , f r (g,t ) ¼ FT bsr (X,t ,g)dV0 (16:70) The tangent viscous matrix Kv now plays the role of the tangent modulus matrix in hyperelasticity. Hyperelastic effects now appear in the force term fr(g,t).
16.8.2 THERMAL EQUILIBRIUM EQUATION The equation for thermal equilibrium is rewritten as r0 (h_ r þ h_ i2 ) ¼ rT0 q0 þ r0 h =T
(16:71)
With some effort using Equation 16.56, a rate (incremental) variational principle may be obtained in the form ð
dT T@ sTr =@ Te_ sTi e_ þ r0 (ce þ ci )T_ =T dV0 ð ¼ dT rT0 q0 þ r0 h =T dV0
(16:72)
Upon approximating T in the denominator by T0, letting k denote the thermal conductivity, and invoking the isotropic Fourier law of heat conduction, we may obtain the thermal equilibrium equation in the form ð
ð dT @ sTr =@ Te_ sTi e_ þ r0 (ce þ ci )T_ dV0 þ k (r0 dT)T r0 T dV0 ð ð ¼ dTr0 h dV0 dTnT q0 dS0
(16:73)
Using the usual interpolation models for displacement and temperature, Equation 16.73 reduces directly to finite element equation for the thermal field. The equation assumes the form MT u_ þ STM (g,u)g_ þ KT u ¼ f h þ f q ð STM (g,u) ¼ CT n @sTr =@T sTi bT F dV0
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(16:74)
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17
Tangent Modulus Tensors for Inelastic and Thermoinelastic Materials
17.1 PLASTICITY The theories of plasticity and thermoplasticity model material behavior in important applications such as metal forming, ballistics, and welding. The main goals of the current section are to present an example of constitutive models in plasticity, viscoplasticity, thermoplasticity, and damage mechanics, to derive the corresponding tangent modulus tensors, and to formulate variational and finite element statements, all while accommodating the challenging problems of finite strain and kinematic hardening. But we first start with a presentation of a constitutive model in small strain isothermal plasticity.
17.2 TANGENT MODULUS TENSOR IN SMALL STRAIN ISOTHERMAL PLASTICITY The basic assumptions are given below. 1. Kinematic Decomposition The strain rate decomposes into an elastic (recoverable, reversible) portion and an inelastic (permanent, irreversible) portion as follows: E_ ¼ E_ e þ E_ i
(17:1)
in which E_ e satisfies small strain isotropic elastic relations (inverse of the Lamé form) l _ _Ee ¼ 1 S_ S_ * tr(S)I 2m 2m þ 3l
ß 2008 by Taylor & Francis Group, LLC.
(17:2a)
in which S*d is a traceless reference stress (backstress) to be introduced shortly. The dilatational and deviatoric portions of this relation satisfy 1 _ E_ ed ¼ Sd S_ * , 2m
tr(E_ e ) ¼
1 _ tr(S)I 2m þ 3l
(17:2b)
In VEC notation, Equation 17.2 is restated as e_ e ¼ Ce (_s s_ *),
Ce ¼
1 l I9 iiT 2m 2m þ 3l
(17:3a)
and Ce is the elastic compliance tensor (inverse of the elastic modulus tensor). For the deviatoric and dilatational parts the corresponding relations are e_ ed ¼ C0e ðs_ d s_ *Þ, iT e_ e ¼
C0e ¼
I9 2m
1 iT s_ 2m þ 3l
(17:3b)
Typically, plastic strain is viewed as permanent strain. As illustrated in Figure 17.1, in a uniaxial tensile specimen the stress S11 may be increased to the point A, and then unloaded along the path AB. The slope of the unloading portion is E, the same as that of the initial elastic portion. When the stress becomes zero, there still is a residual strain Ei, which may be identified as the inelastic strain. However, if the stress had instead been increased to point C, it would encounter reversed loading at point D, which reflects the fact that the elastic region need not include the zero-stress value.
C S11
A E
E D
E B Ei
FIGURE 17.1 Illustration of inelastic strain.
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F
E11
2. Inelastic Incompressibility The inelastic strain does not contribute to the volume strain. tr (E_ i ) ¼ iT e_ i ¼ 0
(17:4)
It follows that inelastic strain only contributes to the shear (deviatoric) portion of the strain, and hence is said to occur in shear. 3. Independence of the Hydrostatic Stress In plasticity inelastic strain is assumed not to be affected by the hydrostatic (isotropic) portion of the stress, tr(S), and to depend only on the deviatoric (shear) portion of the stress, Sd ¼ S 13 tr (S)I or sd ¼ I9 13 iiT s 4. Yield and Loading Conditions There exists a function C(sd s*d ,ei ,k) ¼ 0, which represents a closed convex surface in stress space, called the yield surface. Here s* is a point interior to the yield surface, sometimes called the backstress. It changes if inelastic strain changes, referred to as inelastic flow, and the elastic strain vanishes when s ¼ s*. If flow is occurring the current stress point must be located on this surface. In Section 17.5, we will consider viscoplasticity, in which the notion of the yield surface is modified such that the current stress point is now exterior to the yield surface during flow. The current stress point may then be viewed as being on a loading surface while the yield surface is now a reference surface. In plasticity, there are three possibilities referred to as the loading conditions. (a)
C(sd s*d ,ei ,k) < 0
The stress point is interior to the yield surface and only elastic strain is changing. The stress changes are proportional to strain changes through elastic relations. (b)
C(sd s*d ,ei ,k) ¼ 0
and
@C s_ d ¼ 0 @sd
The stress point is located on the yield surface, but is moving tangentially to the surface. In this event inelastic strain is unchanging, and the stress changes are related to the strain changes by elastic relations. This situation is referred to as neutral loading. (c)
C(sd s*d ,ei ,k) ¼ 0
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and
@C s_ d > 0 @sd
The stress point is located on the yield surface and is moving toward the exterior of the surface. Inelastic flow occurs in this case, which is referred to as loading. 5. Hardening The vector k represents the ‘‘hardening’’ effect of inelastic deformation history, reflecting irreversibility of plastic deformation. Visually hardening occurs as shape, size, and location changes of the yield surface. The hardening vector k is assumed to satisfy an evolution relation of the form k_ ¼ H1 (sd s*d ,ei ,k)_ei
(17:5)
in which H1 is an experimentally determined matrix. Work hardening is a commonly made assumption. Here it is expressed by the relation k_ ¼ k1 (s s*)_ei
(17:6a)
in which k1 is a material constant. An example of a work hardening model for evolution of the backstress is given by s_ * ¼ k2 (s s*)_ei
(17:6b)
and k2 is a material constant. A relation of this form will play a role in our later treatment of kinematic hardening. 6. Associated Flow Rule The inelastic strain rate vector in plasticity is assumed to be normal to the yield surface at the current stress point. T @C e_ i ¼ L @s
(17:7)
T . If L > 0, the associated flow rule then implies Note that s_ T e_ i ¼ L_sT @C @s that s_ T e_ i > 0 during plastic deformation, which is known as Drucker’s criterion for stability in the small (cf. Rowe et al., 1991). It also follows from the assumption that the yield surface is convex that (s s*)_ei 0. 7. Consistency Condition The consistency condition states that the stress point remains on the yield surface during loading, in which event the yield function satisfies the relation _ ¼ C
@C @C @C (_sd s_ *d ) þ H1 e_ i ¼ 0 þ @ei @k @(sd s*d )
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(17:8)
Thanks to hardening, the yield surface deforms or moves such that the stress point remains on it if plastic flow is occurring, even as the stress point is moving toward the exterior of the yield surface. 8. Constitutive Relation for the Inelastic Strain Rate Equations 17.7 and 17.8 imply that T @C @C @C @C þ ¼0 (sd s*d ). þ L H1 @ei @k @sd @(sd s*d )
(17:9)
The parameter L is immediately seen to be @C (_sd s_*d ) @(sd s*d ) L¼ T @C @C @C þ H1 @ei @k @(sd s*d )
(17:10)
The inelastic strain rate is now seen to be given by e_ i ¼ Ci (_sd s_ *d ) in which
T @C @C s_ d * @(sd sd ) @(sd s*d ) Ci ¼ (17:11) T @C @C @C þ H1 @ei @k @(sd s*d ) T @C @C The requirement that L > 0 implies that @C þ H i @ei @k @(sd s*) > 0, in d
which event Ci becomes positive semidefinite. In terms of the full (as opposed to deviatoric) stress tensor S ¼ IVEC(s), the relation governing the inelastic strain rate is now (17:12) e_ i ¼ Ci I9 13 iiT (_s s_ *) 9. Tangent Modulus Tensor We first suppose that the evolution of the backstress follows a hardening model of the form s_*d ¼
n
H2 e_ i 0
loading otherwise
(17:13)
But now e_ i ¼ Ci (_sd s_ *d ) ¼ Ci s_ d Ci H2 e_ i , with the consequence that e_ i ¼ (I þ Ci H2 )1 Ci s_ d
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(17:14)
assuming that I þ CiH2 is nonsingular, as seems very reasonable owing to the typically small magnitude of Ci. It follows that s_ * ¼ H2 e_ i ¼ H2 Ci (_sd s_ *), and hence s_ * ¼ (I þ H2 Ci )1 H2 Ci s_ d
(17:15)
provided that plastic flow is occurring. We now add the deviatoric elastic and inelastic strain rates to obtain the total deviatoric strain rate. e_ d ¼ e_ ed þ e_ i ¼ (C0e þ Ci )(_sd s_ *) I9 þ Ci (I þ H2 Ci )1 s_ d (17:16) ¼ 2m 1 I9 It follows that s_ d ¼ (I þ H2 Ci ) 2m þ Ci e_ d during plastic flow. The dilatational portion of the strain rate is elastic and satisfies Equation 17.3b: iT s_ ¼ (2m þ 3l)iT e_ . The relation between the total stress rate and the total strain rate may be derived to obtain the elastic–plastic tangent modulus tensor Dep: s_ ¼ s_ d þ iiT s_ ¼ Dep e_ 1 I9 iiT þ Ci þ (2m þ 3l)iiT Dep ¼ (I þ H2 Ci ) I9 2m 3
(17:17)
Recall that the Incremental Principle of Virtual Work requires the tensor relating the stress increment to the strain increment. We may now say that Dn s Dep Dn e
(17:18)
Equation 17.18 indicates the rate independent (inviscid) nature of plasticity since the strain increment is proportional to the stress increment no matter how rapidly or slowly the stress is applied.
EXAMPLE 17.1 Von Mises yield surface with kinematic and isotropic hardening The entitled yield function is given by Ci ¼
ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (sd k1 ei )T (sd k1 ei ) k0 þ k2 sTd e_ i dt ¼ 0
Ð i Observe that work hardening is present since k ¼ k2 sTd e_ i dt. Also @C @k ! k2 and T H1 ! k2 sd . In addition, s* ¼ k1ei, so that H2 ! k1. After straighforward manipulation,
ß 2008 by Taylor & Francis Group, LLC.
@ Ci ¼ nT , @ sd
(sd k1 ei )T nT ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (sd k1 ei )T (sd k1 ei )
@ Ci @ Ci H1 ¼ k1 nT þ k2 sTd þ @ ei @k from which Ci ¼
nnT k1 þ k2 sTd n
This matrix is positive semidefinite assuming that k1 þ k2 sTd n > 0. Substitution into Equation 17.17 immediately yields Dep.
EXAMPLE 17.2 Yield surface with strain hardening: small deformation and uniaxial loading In isothermal plasticity, assuming the following yield function, find the stress–strain curve under uniaxial loading. Ci ¼
ð t qffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e_ Ti e_ i dt (sd k1 ei )T (sd k1 ei ) k0 þ k2 0
We again assume that plastic strain is incompressible: tr(Ei) ¼ 0. This yield surface exhibits strain hardening in that the radius of the yield surface depends on the arc length traversed in inelastic strain space.
SOLUTION In the case of uniaxial stress and isotropy, Syy ¼ Szz ¼ 0, with the consequence that Sdxx ¼ Sxx 13 (Sxx þ 0 þ 0) ¼ 23 Sxx
Sdyy ¼ Sdzz ¼ 13 Sxx For plastic incompressibility, tr(Ei) ¼ 0. Hence, Epxx þ Epyy þ Epzz ¼ 0. And so for uniaxial loading Epyy ¼ Epzz ¼ 12 Epxx . The consistency condition Ci ¼ 0 now implies that ð t qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 ffi 2 2 2 p 2 1 p 2 e_ pxx þ 12 e_ pxx þ 12 e_ pxx dt ¼ k s k e þ s þ k e þ k 1 xx 0 2 3 xx 3 xx 2 1 xx 0
and hence qffiffiffi qffiffiffi 2 3 p ¼ k0 þ k2 32 epxx 3 s xx 2 k1 exx
so that
sxx ¼
qffiffiffi 3 2 k0
þ 32 (k1 þ k2 )epxx
The ensuing uniaxial stress–strain curve is depicted in Figure 17.2.
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Sxx
3 Slope = 2 (k1+ k2)
3 k 2 0
E
exx
Unloading curve
FIGURE 17.2 Stress–strain path with strain hardening.
17.3 PLASTICITY UNDER FINITE STRAIN 17.3.1 KINEMATICS The deformation rate tensor admits an additive decomposition into elastic and inelastic portions. D ¼ De þ Di
(17:19)
The Lagrangian strain tensor E satisfies the relation E_ ¼ FTDF, from which we formally introduce the elastic and inelastic strains (for large deformation) as ð
ð
Ee ¼ F De F dt, T
Ei ¼ FT Di F dt
(17:20)
Of course there are alternatives to this type of decomposition, for example, the logarithmic plastic strain (Xiao et al., 1997).
17.3.2 PLASTICITY We present an example of a constitutive equation for plasticity at large deformation to illustrate how the tangent modulus tensor is formulated. For simplicity we ignore the difference between the stress and strain and their deviatoric counterparts (i.e., ignore elastic strain), and we also assume that the backstress remains at the origin in stress space (no kinematic hardening). With xe the tangent modulus tensor relating s_ to e_ under elastic conditions (retaining the assumption of rate independence), the constitutive equation of interest is obtained from the previous section using
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1 _ _i Ce ¼ x e , k ¼ H(s,ei ,k)e " # @ Ci T @ C i @ Ci @ Ci @ Ci T h, h ¼ H Ci (s,ei ,k) ¼ þ @s @s @ ei @k @s
e_ i ¼ Ci s_ ,
e_ e ¼ Ce s_ ,
(17:21)
As before Ci is called the yield function, but now it is a function of the inelastic portion of the Lagrangian strain, as well as of the history of inelastic deformation represented by, say, work hardening. Combining the elastic and inelastic portions furnishes the tangent modulus tensor as
1 1 x ¼ x ¼ [I þ xe Ci ]1 xe e þ Ci
(17:22)
Suppose that in uniaxial tension the elastic portion of the tangent modulus is xe ! Ee, and that the inelastic portion relating the stress increment and the inelastic strain 1 increments is C i ! Ei . Typically Ei Ee. The total uniaxial tangent modulus is Ei then (1 þ Ei =Ee ). For the sake of visualization we illustrate several possible behaviors of the yield surface. It is distorted by the history of plastic strain through hardening. In Figure 17.3, the conventional model of isotropic hardening is illustrated in which the yield surface expands as a result of plastic deformation. The principal values of the second Piola–Kirchhoff stress are shown on the axes, and the yield surface is shown in the SI SII plane. This model is unrealistic in predicting a growing elastic region— reversed plastic loading is typically encountered at much higher stresses than isotropic hardening predicts. An alternative is kinematic hardening (Figure 17.4), in which the yield surface moves with the stress point. After a few percent of plastic strain have developed, the yield surface may cease to encircle the origin. A reference point interior to the yield surface, previously encountered as the backstress, is assumed to serve as the point at which the elastic strain vanishes. SII
Path of stress point
SI
Principal stresses SI SII SIII SIII
FIGURE 17.3 Illustration of yield surface expansion under isotropic hardening.
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SII
Path of stress point
SI
SIII
FIGURE 17.4 Illustration of yield surface motion under kinematic hardening.
Combined isotropic and kinematic hardening is shown in Figure 17.5. However, this figure shows that the yield surface contracts, which is consistent with actual observations (e.g., Ellyin, 1997). The rate of movement must in some sense exceed the rate of contraction for the material to remain stable, with a positive definite tangent modulus tensor.
17.4 THERMOPLASTICITY If plastic work occurs over a sufficiently short time period there is insufficient time for heat to escape, with the consequence that some or all of the plastic work is converted into heat and gives rise to an increased temperature. Of course, it is also
SII
Path of stress point
SI
SIII
FIGURE 17.5 Illustration of combined kinematic and isotropic hardening.
ß 2008 by Taylor & Francis Group, LLC.
possible that plastic work occurs in the presence of externally introduced heat. Both effects represent instances of thermoplasticity. As in Chapter 16, following Ziegler and Wehrli (1987) two potential functions are introduced to provide a systematic way to give separate descriptions to reversible and dissipative (irreversible) effects. The first is interpreted as the Helmholtz free energy density and the second is a dissipation potential. To accommodate kinematic hardening we also assume an extension of the Green and Naghdi (GN) (1965) formulation, in which the Helmholtz free energy decomposes into reversible and irreversible parts, with the irreversible part depending on the plastic strain. Here it also depends on the temperature and a workless internal state variable. Application of thermodynamic concepts to inelastic deformation is much more challenging than for damped elastomers treated in Chapter 16. The reason is that inelasticity, for example, via the yield surface, is usually described in stress space while the Helmholtz free energy uses the strain as a state variable.
17.4.1 BALANCE
OF
ENERGY
The conventional equation for energy balance is augmented using a vectorvalued workless internal variable a0 regarded as representing ‘‘microstructural rearrangements.’’ r0 x_ 0 ¼ sT e_ r þ sT e_ i rT0 q0 þ r0 h þ bT0 a_ 0
(17:23)
where x0 is the internal energy per unit mass in the undeformed configuration, s ¼ VEC(S), e ¼ VEC(E), and b0 is the ‘‘flux’’ per unit mass associated with a0. However, note that b0 ¼ 0 for a0 to be workless, and hence its reversible and irreversible portions are related by b0i ¼ b0r. Also q0 is the heat flux vector referred to undeformed coordinates and h is the heat input per unit mass, for simplicity assumed to be independent of temperature. For use in the Helmholtz free energy, the state variables are recognized to be Er, Ei, T, and a0. The next few paragraphs will go over some of the same ground as for damped elastomers in Chapter 16, except for two major points. In Chapter 16, the stress was assumed to decompose into reversible and irreversible portions, in the spirit of elementary Voigt models of viscoelasticity. In the current context, the strain shows a corresponding decomposition, in the spirit of the classical Maxwell models of viscoelasticity. In addition, introducing a workless internal variable a0 will be seen to give the model the flexibility to accommodate phenomena such as kinematic hardening. The Helmholtz free energy f0 per unit mass and the entropy h per unit mass are again introduced using f0 ¼ x 0 Th0
(17:24)
The balance of energy, which is Equation 17.23 governing the thermal field, is now rewritten as rT0 q0 r0 h ¼ sT e_ r þ sT e_ i r0 T_ 0 r0 0 T_ r0 f_ 0 þ bT0 a_ 0
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(17:25)
17.4.2 ENTROPY PRODUCTION INEQUALITY Entropy production now is governed by the inequality r0 T_ 0 rT0 q0 þ r0 h þ qT0 rT=T r0 f_ 0 sT e_ r sT e_ i þ r0 T_ 0 þ r0 0 T_ þ qT0 rT=T bT0 a_ 0
(17:26)
Now viewing f0r as a differentiable function of er, T, and a0, we conclude that @ f0r =@ er ¼ sT ,
@ f0r =@ T ¼ r0 hr0 ,
@ f0r =@ a0 ¼ bT0r
(17:27)
Extending the GN formulation, we introduce the stress s* using s*T ¼ r0@ fi=@ ei and assume that h0 ¼ @ f0=@ T and r0 @ f0i =@ a0 ¼ bT0i : s*T ¼ r0 @ fi =@ ei ,
h0i ¼ @ fi =@ T,
r0 @ f0i =@ a0 ¼
h0r ¼ @ fr =@ T
bT0i
(17:28)
The entropy production inequality (Equation 17.26) now reduces to (sT s*T )e_ i qT0 r0 T=T 0
(17:29)
Inequality (Equation 17.29) is satisfied if (sT s*T )e_ i 0
(17:30a)
qT0 r0 T=T 0
(17:30b)
Inequality Equation 17.30 states that the inelastic work done by the reduced stress sT s*T is positive and that heat flows from hot to cold. The first inequality exhibits the quantity s* ¼ VEC(S*) with dimensions of stress. In the subsequent sections s* will be viewed as the previously mentioned backstress: it is interior to a yield surface and can be used to characterize the motion of the yield surface in stress space. Clearly, the present formulation gives a thermodynamic interpretation to the back stress. In classical kinematic hardening in which the hyperspherical yield surface does not change size or shape but just moves, the reference stress is simply the geometric center. If kinematic hardening occurs, as stated before, the yield surface need not include the origin even with small amounts of plastic deformation. Thus, there is no reason in general to regard e_ r as vanishing at the origin. Instead e_ r ¼ 0 is now assumed to hold when s ¼ s*.
17.4.3 DISSIPATION POTENTIAL Following the approach developed in Chapter 16 we introduce a specific irreversible potential C for which e
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rT0 T=T ¼ Lt r0 @C=@q0 ,
e
e_ Ti ¼ r0 Li @C=@ ,
¼ s s*
(17:31a)
from which, with Li > 0 and Lt > 0, Equation 17.30 becomes r0 Li (@C=@ ) þ r0 Lt (@C=@q0 )q0 > 0
(17:31b)
ee
Partly on the expectation that properties governing heat transfer are not affected by strain, we introduce the following decomposition into inelastic and thermal portions: C ¼ Ci þ Ct ,
r 0 Ct ¼
Lt T q q0 2 0
(17:32)
and Ci will be seen in subsequent sections to represent mechanical effects. The thermal constitutive relation derived from Ct implies Fourier’s law: r0 T=T ¼ Lt q0
(17:33)
17.4.4 THERMOINELASTIC TANGENT MODULUS TENSOR The elastic strain rate is assumed to correspond to small elastic deformation superimposed on the finite inelastic deformation, and to satisfy a linear thermohypoelastic constitutive relation e_ r ¼ Cr (s s*). þ ar T_
(17:34)
Of course Cr is a 9 3 9 second-order elastic compliance tensor, and ar is the 9 3 1 thermoelastic expansion vector, with both presumed to be constant and known from measurements. Analogously, for rate-independent thermoplasticity we seek tensors Ci and ai, depending on , ei, and T, such that e
e_ i ¼ Ci (s s*). þ ai T_
(17:35a)
. e_ ¼ [Cr þ Ci ](s s*) þ (ar þ ai )T_
(17:35b)
During thermoplastic deformation the stress and temperature satisfy a thermoplastic yield condition of the form Pi ( , ei , k, T, h0i2 ) ¼ 0
(17:36)
e
and Pi is called the yield function. Here the vector k is introduced to represent the effect of the history of inelastic strain ei, for example, through work hardening. To embrace dependence on the temperature, it is now assumed to be given by a relation of the form, k_ ¼ K(ei , k, T)_ei
(17:37)
_ i ¼ 0 during thermoplastic flow, and The ‘‘consistency condition’’ requires that P accordingly dPi dPi dPi _ dPi _ dPi _ þ kþ e_ i þ h_ ¼ 0 Tþ d dei dk dT dh0i2 0i2 e
e
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(17:38)
We introduce a thermoplastic extension of the conventional associated flow rule, whereby the inelastic strain rate vector is normal to the yield surface at the current stress point. Here we add an analogous assumption regarding the entropy.
dPi e_ i ¼ Li d
T (17:39a)
e
h_ 0i 2 ¼ Li
dPi dT
(17:39b)
Equation 17.39 suggests that the yield function may be identified as the dissipation potential: Pi ¼ r0Ci. Upon making this identification, standard manipulation furnishes _ e_ i ¼ Ci _ þ ai T, h_ 0i2 ¼ bTi _ þ ci T_ T @Ci @Ci . @Ci T @Ci . H, ai ¼ bi ¼ H Ci ¼ @ @ @ @T " # 2 T @Ci . @Ci @Ci @Ci @Ci @Ci þ þ ci ¼ K H, H¼ @T @ei @K @ @h0i2 @T e
e
(17:40)
e
e
e
e
and H must be positive for Li to be positive. Note that the dependence of the yield function on temperature accounts for ci in the current formulation. The dissipation inequalities (Equation 17.30) are now satisfied if H > 0. Next, recall that s* depends on ei, T, and a0 since s*T ¼ r0 @f0i =@ei . For simplicity we now neglect dependence on a0 and assume that a relation of the following form can be measured for s*: _ G¼ s_ * ¼ G_ei þ qT,
T @ @ Ci , @ei @ei
qT ¼ @ 2 Ci =@ei @T
(17:41)
The thermoinelastic tangent compliance tensor and thermomechanical vector are obtained after simple additional manipulation as e_ ¼ C_s þ aT_
C ¼ (Cr þ Ci ) I (I þ GCi )1 GCii
a ¼ ar þ ai þ (Cr þ Ci )Gai (Cr þ Ci )(I þ GCi )1 q
(17:42a) (17:42b)
Of course the tangent modulus tensor is the inverse of C. The foregoing formulation may be extended to enforce plastic incompressibility.
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EXAMPLE 17.3 Thermoplastic Helmholtz free energy and dissipation function We now provide a simple example using the Helmholtz free energy density function and the dissipation potential function to derive constitutive relations. The expression assumed below involves a Von Mises yield function, linear kinematic hardening, linear work hardening, and linear thermal softening. 1. Helmholtz free energy density f0 ¼ f0r þ f0i , r0 f0i ¼ k3 eTi ei T 1 0 r0 f0r ¼ eTr C1 r er =2 a Cr (T T0 )er þ r 0 cr T(1 ln (T=T0 ))
in which c0r is a known constant. Applying the previous relations furnishes 1 ¼ r0 (@ f0r =@ er )T ¼ C r [er ar (T T0 )]
e and
cr ¼ T
@ 2 f0r ¼ c0r @ T2
Of course the last two relations between the reduced stress, the elastic strain, and the temperature are the same as in linear thermoelasticity except for the presence of the backstress. 2. Dissipation potential The dissipation potential is again assumed to have a decomposition into mechanical and thermal portions, and the following specific forms are introduced:
pffiffiffiffiffiffiffiffiffi T
e e
Ci ¼
r0 Ct ¼
Lt T q q 2 0 0
[k0 þ k1 k k2 (T T0 )] ¼ 0,
k_ ¼
T
e
C ¼ Ci þ Ct ,
e_ i
Straightforward manipulations serve to derive pffiffiffiffiffiffiffiffiffi T ¼ k1 [K0 þ k1 k k2 (T T0 )] H ¼ k1 e e
e e ee T
. H,
. ci ¼ k22 H,
k2 . ai ¼ bi ¼ pffiffiffiffiffiffiffiffiffi H T
e e
T
e
Ci ¼
Consider a two-stage thermomechanical loading illustrated schematically in Figure 17.6. Let SI, SII, and SIII denote the principal values of the second Piola–Kirchhoff stress, and suppose that SIII ¼ 0. In the first stage, with the temperature held fixed at T0 the stresses are applied proportionally well into the plastic range. The center of the yield surface moves along a line in the (SI, SII) plane, and the yield surface expands as it moves. In the second stage, suppose that the stresses S1 and S2 are fixed but that the temperature increases to T1 and then to T2, T3, and T4. The plastic strain must increase and hence the center of the yield surface moves. In addition, in the assumed yield function strain hardening tends to cause the yield surface to expand isotropically around s* while the
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S II T1 T0
T3
T2 T4
T0 T0 T0
T0
SI T0
S III
FIGURE 17.6 Effect of load and temperature on yield surface. increased temperature tends to make it contract. However, in this case thermal softening must dominate strain hardening and contraction must occur since the center of the yield surface must move further along the path shown even as the yield surface continues to ‘‘kiss’’ the fixed stresses SI and SII. Unfortunately, accurate finite element computations in plasticity and thermoplasticity often require close attention to the location of the front of the yielded zone. This front will usually occur interior to elements, essentially reducing the continuity order of the fields (discontinuity in strain gradients). In addition in computations the stress point will initially deviate from the yield surface. Special procedures such as Return Mapping (cf. Belytschko et al., 2000) have been developed in some codes to coerce the stress point onto the yield surface. The shrinkage of the yield surface with temperature may provide the explanation of the phenomenon of adiabatic shear banding, which is commonly encountered in some materials during impact or metal forming. In rapid processes such as high-speed metalworking, plastic work is mostly converted into heat thereby causing high temperatures— there is not enough time for the heat to flow away from a spot experiencing high plastic deformation. However, under some conditions the process is unstable even as the stress level is maintained. In particular, as the material gets hotter the rate of plastic work accelerates, thanks to the softening evident in Figure 17.6. The instability is manifested in small periodically spaced bands in the center of which the material has melted and resolidified, usually in a much more brittle form than before. The adiabatic shear bands thereby formed can nucleate brittle fracture.
17.5 TANGENT MODULUS TENSOR IN VISCOPLASTICITY 17.5.1 MECHANICAL FIELD The thermodynamic discussion of Section 17.4 applies to thermoinelastic deformation, for which the first example given concerned quasi-static plasticity and thermoplasticity. However, it is equally applicable when rate sensitivity is present, in which
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case viscoplasticity and thermoviscoplasticity are attractive models. An example of a constitutive model, for example, following Perzyna (1971), is given in undeformed coordinates as @ Ci T
k @ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mv e_ i ¼ 1 T Ci @C @C e
e
8
< k k 1 1 ¼ Ci : Ci 0
i
@
e
i
@
k 0 Ci otherwise
(17:43)
if 1
and the dissipation function Ci( ,ei,k,T,h0i) now is also a loading surface function (to be explained shortly); mv is called the viscosity. The inelastic strain rate vanishes if 1 Ck i < 0, which is interpreted to mean that the stress point (i.e., ) is interior to the reference surface determined by points ^s satisfying Ci(^ss*,ei,k,T,h0i) ¼ 0. Inelastic flow is occurring if 1 Ck i > 0, in which case the stress point is exterior to the reference surface. The equation of the loading surface is qffiffiffiffiffiffiffiffi . Ci ( ,ei ,k,T,h0i ) ¼ k 1 mv e_ Ti e_ i , and clearly the model must be restricted to qffiffiffiffiffiffiffiffi strain rates satisfying mv e_ Ti e_ i < 1. e
e
e
e
The elastic response is still considered linear and in the form 1 _ e_ r ¼ x r _ þ ar T
(17:44)
e
Also from thermoplasticity we retain the relations _ s_ * ¼ Ge_ i þ qT,
G¼
T @ @ Ci , qT ¼ @ 2 Ci =@ ei @ T @ ei @ ei
(17:45)
Corresponding to there is a new reference stress s0 and a corresponding reference vector 0 ¼ s0 s* defined as follows. The vector 0 lies on the previously mentioned quasi-static reference yield surface such the vectors and 0 have the same origin and direction. The latter terminates on the reference surface while the former terminates outside the reference surface (at the current value of ) if inelastic flow is occurring. Alternatively stated, 0 is located at the intersection of the reference yield surface and the line correcting s* to s. This situation is illustrated in Figure 17.7. Figure 17.7 suggests that viscoplasticity and thermoviscoplasticity can be formulated to accommodate phenomena such as kinematic hardening and thermal shrinkage of the reference yield surface. As in plasticity we assume that the backstress has an evolution law of the form s_ * ¼ Hbe_ i, from which we find e
e
e
e
e
e
e
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S II
Reference yield surface Loading surface .T.
ψ = k /(1 – mv ei ei ) i ψi = k
Stress
Backstress
SI
SIII
FIGURE 17.7 Illustration of loading surface and reference surface in viscoplasticity.
k (@C=@ )T s_ * ¼ mv 1 Hb pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiTffi C (@C=@ )(@C=@ ) e
(17:46)
e
e
The tangent modulus tensor now reduces to the constant elastic tensor xr, and viscoplastic effects appear in a force denoted fv, which depends on the current values of the state variables but not the current values of their rates. From Equation 17.44 s_ s_ * ¼ xr e_ r xr ar T_ and so s_ ¼ H_ei þ xr (_e e_ i ) xr ar T_
k (@C=@ )T _ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ xr e_ xr ar T þ f v , f v ¼ mv (Hb xr ) 1 Ci (@C=@ )(@C=@ )T e
e
e
(17:47) The Incremental Principle of Virtual Work and the corresponding finite element equation are now stated to first order in increments as ð ð ð € dV0 dDn eT xr Dn e dV0 þ dDn eT xr ar Dn T dV0 þ dDn uT r0 Dn u ð ð ¼ dDn uT Dn t dS0 dDn eT f v dV0 (17:48) and
STM
€ ¼ F Fv KDn g þ STM Dn u þ MDn g ð ð T ¼ (F b)(xr ar )nC dV0 , Fv ¼ FT bf v dV0
in which b appears in the incremental strain–displacement relation.
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EXAMPLE 17.4 Find the stress–strain curve under uniaxial tension if a constant strain rate is imposed in the linear, small strain, isothermal viscoplasticity model in which the yield surface is given by Ci ¼
ð t qffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e_ Ti e_ i dt (sd k1 ei )T (sd k1 ei ) k0 þ k2 0
Of course this surface reflects strain hardening. Referring to the previous exercise in plasticity Ci may be written as Ci ¼
qffiffiffi qffiffiffi vp 2 3 3 Sxx 2 (k1 þ k2 ) Exx k0
To formulate the finite element relations, use is made of the fact that @Ci @Sxx
,sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi @Ci @Ci T ¼1 @Sxx @Sxx
The constitutive model for viscoplasticity reduces to the uniaxial relations qffiffiffi qffiffiffi
vp 3 mv E_ xx ¼ Ci k ¼ 23 Sxx 32 (k1 þ k2 )Evp xx 2 (k0 þ k) Of course the strain rate may be decomposed into elastic and viscoelastic parts as e e_ ¼ e_ e þ e_ vp . Also E_ xx ¼ S_ xx =E. Accordingly qffiffiffi qffiffiffi S_ xx 1 2
3 þ k) E_ xx ¼ Sxx 32 (k1 þ k2 )Evp þ xx 3 2 (k0 E mv Now
d dExx d , and so ¼ dt dExx dt
qffiffiffi qffiffi E_ xx dSxx 1 2
vp 3 3 E_ xx ¼ þ (k þ k )E S xx 1 2 xx 2 2 (k0 þ k) E dExx mv 3
from which emerges a simple differential equation for the uniaxial stress–strain relation at the constant strain rate E_ xx: rffiffiffi rffiffiffi dSxx 2 E 3 (k1 þ k2 )E vp (k0 þ k) þ Sxx ¼ Exx þ E 1 þ dExx 3 mv E_ xx 2 mv E_ xx mv E_ xx rffiffiffi 3 (k1 þ k2 )E Sxx (k0 þ k) Exx ¼ þE 1þ 2 mv E_ xx E mv E_ xx so that dSxx þ dExx
! rffiffiffi rffiffiffi rffiffiffi 2 E 3 (k1 þ k2 ) 3 (k1 þ k2 )E (k0 þ k) Sxx ¼ þ Exx þ E 1 þ 3 mv E_ xx 2 mv E_ xx 2 mv E_ xx mv E_ xx
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Exx Sxx
3 Slope = 2 (k1+ k2)
k
Exx
FIGURE 17.8 Stress vs. strain at constant strain rate in a viscoplastic material.
The transient part of the solution is given by ! ! rffiffiffi rffiffiffi 2 E 3 (k1 þ k2 ) Sxx ¼ A exp þ Exx 3 mv E_ xx 2 mv E_ xx in which A is determined by imposing initial values on the combined transient and steady state solutions. The steady state part of the solution is given by Sxx ¼
þ k2 ) Exx þ 2 (k1 þ k2 ) 1þ 3 E 3 2 (k1
rffiffiffi 3 [mv E_ xx þ (k0 þ k )] 3 (k1 þ k2 ) 2 1þ 2 E
The slope of the asymptote is approximately 32(k1 þ k2) and its intercept is approximately qffiffi 3 _ 2[(k0 þ k ) þ mv Exx ]. The ensuing uniaxial stress–strain curve is illustrated in Figure 17.8.
17.5.2 THERMOINELASTICITY: THERMAL FIELD Of course a variational equation and a finite element equation are needed for the thermal field Equation 17.25, and are formulated as was done in Chapter 16. To illustrate the process of formulating the finite element equation for the thermal field, we make the simplifying assumption that the irreversible entropy only depends on temperature. The equation for the reversible portion of the energy balance is @ (s s*)T e_ r rT0 q0 þ r0 h r ¼ r0 ce T_ T @T
(17:49)
which of course reduces to the thermal field equation of classical thermoelasticity when the backstress is removed and the reversible strain is equated with the total strain.
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The equation for the irreversible portion of the energy balance is now
rT0 q0 þ r0 h i ¼ r0 ci T_ (s s*)e_ i
(17:50)
Adding the reversible and irreversible portions now gives rT0 q0 þ r0 h ¼ r0 (ce þ ci )T_ (s s*)e_ i T
@ (s s*) @T
(17:51)
In adiabatic situations such as short-term response after impact, the left-hand side of Equation 17.51 vanishes. The right side then suggests that the inelastic work _ (s s*)e_ i is converted into temperature increase in accordance with r0(ce þ ci)T, except for a small portion (proportional to the thermal expansion coefficient) represented by T @ (ss*). @T
We now wish to formulate the incremental finite element equation for the thermal field. We use the constitutive model introduced in Section 17.5.1 for thermoplasticity, and the counterparts for thermoviscoplasticity can be recovered by obvious substitutions of the constitutive relations for the inelastic strain rate. For simplicity we assume that there is no internal generation of heat, i.e., h_ ¼ 0. We also assume Fourier’s law in the form q0 ¼ k0rT, and that (ce þ ci) is constant. The governing equation now becomes _ ¼0 k0 r2 T þ r0 (ce þ ci )T_ þ la(iT e_ ) (s s*)T (Cs_ þ aT)
(17:52)
Applying Fourier’s law now gives k0 r2 T þ r0 (ce þ ci )T_ þ la(iT e_ ) (s s*)T Cxe_ (s s*)T ai T_ ¼ 0 (17:53) Equations 17.52 and 17.53 provide a way of avoiding using the inelastic strain increment explicitly in the variational principle. The finite element equation is now sought. As in Chapter 16 we use k0 r2 T k0 r2 Dn T þ k0 r2 Tn
(17:54)
In terms of increments the thermal field equation is now
hk0 r2 Dn T þ r0 (ce þ ci ) (s s*)T ai Dn T
þ laiT (s s*)T Cx Dn e ¼ hk0 r2 Tn
(17:55)
Standard finite element procedures now provide the element-level finite element equation for the thermal field. [hKu þ (Mu1 þ Mu2 )]Dn u þ ST2 Dn g ¼ Dn f T1 Dn f T2
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ð
ð
Ku ¼ C Mu1 ¼ CT v[r0 (ce þ ci )]vT C dV0 , ð ð
Mu2 ¼ CT v (s s*)T ai vT C dV0 , ST2 ¼ CT v laiT (s s*)T Cx bF dV0 , ð ð T T Dn f T1 ¼ C v n0 q0 dS0 , Dn f T2 ¼ h CT vk0 r2 Tn dS0 (17:56) T
bT k0 bTT C dV0 ,
Finally, we recapitulate the incremental finite element equations of the mechanical and thermal fields in the form € þ (KT þ KG )Dn g S1 Dn u ¼ Dn f m MDn g [hKu þ (Mu1 þ Mu2 )]Dn u þ
T S2 Dn g
: mechanical field
¼ Dn f T1 Dn f T2 : thermal field
(17:57)
17.6 CONTINUUM DAMAGE MECHANICS Ductile fracture occurs by processes which are associated with the notion of damage. A damage parameter is also introduced to explain tertiary creep, in which the strain grows rapidly at a fixed stress and temperature. An internal damage variable is introduced which accumulates with inelastic deformation. It also is manifested in reductions in properties such as the experimental values of the elastic modulus and yield stress. When the damage parameter in a given element reaches a known or assumed critical value, the element is considered to have failed. The element may then be removed from the mesh (the element is considered to be no longer supporting the load). If so, the displacement and temperature fields are recalculated to accommodate the element deletion. There are two different ‘‘schools’’ of thought on the suitable notion of a damage parameter. One, associated with Gurson (1977), Tvergaard (1981), and Thomason (1990), considers damage to involve a specific mechanism occurring in a three-stage process: nucleation of voids, their subsequent growth, and finally their coalescence to form a macroscopic defect. The coalescence event can be used as a criterion for element failure. The parameter used to measure damage is the void volume fraction f. Models and criteria for the three processes have been formulated. For both nucleation and growth, evolution of f is governed by a constitutive equation of the form f ¼ (f,ei ,T)
(17:58)
for which several specific forms have been proposed. To this point, a nominal stress is used in the sense that the reduced ability of material to support stress is not accommodated. The second school of thought is more phenomenological in nature and is not dependent on a specific micromechanical mechanism. It uses the parameter D, which is interpreted as the fraction of damaged area Ad to total area A0 that the stress (traction) acts on. Consider a uniaxial tensile specimen which has experienced
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damage and is now exhibiting elastic behavior. Suppose that damaged area Ad can no longer support a load. For a given load P, the true stress at a point in the undamaged P 1 P 1 0 0 ¼ 1D zone is S ¼ A0 A A0 ¼ 1D S . Here S is a nominal stress, but is also the D measured stress. If E is the elastic modulus measured in an undamaged specimen, the modulus measured in the current specimen will be E0 ¼ E=(1 D), demonstrating that damage is manifest in small changes in properties, in particular D ¼ 1 E=E0 . As an illustration of damage, suppose specimens are loaded into the plastic range, unloaded, and then loaded again. Without the notion of damage the stress– strain curve should return to its original path. However, owing to damage there are slight changes in the elastic slope, in the yield stress, and in the slope after yield (exaggerated in Figure 17.9). From the standpoint of thermodynamics, damage is an internal variable representing irreversible effects and as such is dissipative. In reality the amount of mechanical or thermal energy absorbed by damage is probably small, so that its role in the energy balance equation is often neglected. Even so, for the sake of a consistent framework for treating dissipation associated with damage, a dissipation potential Cd may be introduced for damage, as has been done, for example, by Bonora (1997). As an example, the contribution to the irreversible entropy production may be assumed in the rate form DD_ 0, in which D is the ‘‘force’’ associated with ‘‘flux’’ D. Positive dissipation is assured if relations are used such as D¼
@Cd 1 2 , Cd ¼ Ld (ei ,T,k)D_ , 2 @ D_
Ld (ei ,T,k) > 0
(17:59)
An example of a potentially satisfactory function tying damage to inelastic work is ð Ld (ei ,T,k) ¼ Ld0 (s s*)_ei dt,
Ld0 a positive constant
(17:60)
Specific examples of constitutive relations for damage are given, for example, in Bonora (1997). S11
Sy1
Ep1
E1
Ep3
Ep2
Sy3
Sy2 E2
E3
E11
FIGURE 17.9 Illustration of effect of damage on elastic–plastic properties.
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At the current values of the damage parameter, the finite element equations are solved for the nodal displacements, from which may be computed the inelastic strains and the inelastic work done in the current load or time increment. This information may then be used to update the damage parameter values at each element using the damage evolution equation (Equation 17.59). Upon doing so, the damage parameter values are compared to critical values. As stated previously, if the critical damage parameter value is attained, the element is deleted. In many cases the string of deleted elements may be viewed as a crack (Al-Grafi, 2003). The finite element code LS_DYNA version 9.5 (2000) incorporates a material model which includes viscoplasticity and damage mechanics. It can easily be upgraded to include thermal effects in which all viscoplastic work is turned into heat. Such a model has been shown to reproduce the location and path of a crack in a dynamically loaded welded structure (Moraes and Nicholson, 2002).
EXAMPLE 17.5 Tertiary Creep of IN 617 at high temperature An example of how a damage variable is useful in modeling material behavior is provided by tertiary creep of IN 617 (Gordon and Nicholson, 2006). Creep may be regarded as a type of viscoplasticity except that the reference yield surface is sometimes neglected. The steel alloy IN 617 is used in turbomachinery and experiences rapid growth of strain
1.0 Material: IN617
Total strain, e (%)
0.8
0.6
0.4
0.2 649 C (1200 F) 760 C (1400 F) 871 C (1600 F) 982 C (1800 F)
0.0
0
50
100
150
200
Time, t (h)
FIGURE 17.10 Tertiary creep of IN 617. (From Gordon, A.P. and Nicholson, D.W., Finite Element Analysis of IN 617 Tertiary Creep, Report, University of Central Florida, Orlando, FL, 2006.)
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under fixed stress and at high temperature. The tertiary creep behavior is shown experimentally in Figure 17.10, in which lower stresses were used at the higher temperatures. A Norton-type isotropic constitutive model has been formulated in which the Von Mises creep strain rate «_ cr is modeled as s n «_ cr ¼ B(exp 9=RT) 1D
(17:61)
Here s is the Von Mises stress and 9 is the activation energy, while the constants B and n are selected for a best fit up to 5% strain. The damage evolution equation, based on the classical treatment of Rabotnov (1969), is given by D_ ¼
Msx (1 D)f
(17:62)
in which M, x, and f are likewise constants to be chosen to match experiments up to 5% strain. Equations 17.61 and 17.62 were implemented by Gordon and Nicholson (2006) in ANSYS in a user-material model and used to find the best-fit values of B, M, n, x, and f. An example of the comparison between the model and the experimental data, up to 5% strain, is shown in Figure 17.11.
Secondary only 871C - 62 MPa 871C - 62 MPa 871C - 69 MPa 871C - 69 MPa Secondary and tertiary
30
Total strain, e (%)
25
Material: IN617
20 15 10 5 0 0
200
400
600
800
1000
1200
Time, t (h)
FIGURE 17.11 Comparison of computations and experimental values up to 5% strain. (From Gordon, A.P. and Nicholson, D.W., Finite Element Analysis of IN 617 Tertiary Creep, Report, University of Central Florida, Orlando, FL, 2006.)
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18
Selected Advanced Numerical Methods in FEA
In nonlinear finite element analysis, a solution is typically sought using Newton iteration, in classical form or augmented as an arc length method to bypass critical points in the load–deflection behavior. Another important topic is the treatment of incompressibility in nonlinear problems. Here three additional treatments of numerical methods are presented.
18.1 ITERATIVE TRIANGULARIZATION OF PERTURBED MATRICES 18.1.1 INTRODUCTION The finite element method applied to nonlinear problems typically gives rise to a large linear system of the form K0(g0)Dg0 ¼ Df0 in which the stiffness matrix K0 is positive definite and symmetric and banded. Also, Dg0 is the incremental nodal displacement vector and Df0 is the incremental nodal force vector. The stiffness matrix depends on nodal displacements and is updated during the incremental solution process, leading to a perturbed matrix K ¼ K0 þ DK, in which DK is assumed to be very small in norm (e.g., magnitude of largest eigenvalue) compared to K0. Given the fact that triangular factors L0 and LT0 have been obtained for K0, it is attractive to formulate and employ an iteration procedure for the perturbed matrix K using L0 as the initial iterate. The procedure should not involve solving intermediate linear systems except by forward and backward substitution using already known triangular factors. A procedure is presented in the section below following Nicholson (2005a) and shown, in three simple examples, to produce accurate estimates within a few iterations. In the scalar case, the iterates examined ‘‘track’’ the Taylor series exactly. The author is unaware of any previously established and widely implemented iterative procedure for matrix triangularization. It should be noted that updating triangular factors is relevant to important finite element applications other than the nonlinear solid mechanics. For example, in fracture mechanics suppose that the crack front advances. There is a local change in the mesh in the vicinity of the crack tip, corresponding to a perturbation of the stiffness matrix. As a second example, if the finite element method is used in an
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optimal design study, the search procedure sets the design parameters and an FEA is performed. Then the search procedure moves the search point locally in space of design parameters, thereby perturbing the structural model and the stiffness matrix.
18.1.2 INCREMENTAL FINITE ELEMENT EQUATION To set the problem under study in the appropriate context, we consider dynamic response of a nonlinear solid. Application of the Incremental Principle of Virtual Work and introduction of suitable interpolation models, following Chapter 15, furnish incremental finite element relations: € þ K(g)Dg ¼ Df MDg
(18:1)
M mass matrix, n 3 n and positive definite, assumed constant K(g) incremental stiffness matrix, n 3 n and positive definite Dg incremental nodal displacement vector Df incremental consistent nodal force vector We assume that Equation 18.1 is integrated using a one-step procedure based on the trapezoidal rule (Newmark’s method). Let h denote the time step and let gn ¼ g(tn). At time tnþ1 ¼ (nþ1)h, Equation 18.1 becomes, following (Zienkiewicz and Taylor, 1989), h
i 2 M þ h4 K(gn ) Dnþ1 g ¼ Dnþ1 g
(18:2)
in which Dnþ1 g ¼ h4 (Dnþ1 f þ Dn f KDn g) þ M(Dn g þ hDn q), Dn q ¼ Dn g_ 2
Dnþ1 g ¼ gnþ1 gn We say that the stiffness matrix at the (n þ 1)st load or time step is perturbed relative to the stiffness matrix at the nth load step. The solution of perturbed linear systems has been the subject of many investigations. Schemes based on explicit matrix inversion include the Sherman–Morrison–Woodbury formulae (cf. Golub and Van Loan, 1986). An alternate method is to carry bothersome terms to the right-hand side and then to iterate. For example, the perturbed linear system may be approximated to first order in increments as K0 Dg ¼ Df DKg0
(18:3)
and an iterative solution procedure, assuming convergence, may then be employed as K0 Dg( jþ1) ¼ Df DK g( j) g0 ,
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g( jþ1) ¼ g0 þ Dg( jþ1)
(18:4)
Unfortunately, in a typical nonlinear problem, especially in systems with decreasing stiffness, it is necessary to update the triangular factors after several increments. If plasticity, hyperelasticity, or buckling is of concern, it is often wise to update the stiffness matrix after each increment.
18.1.3 ITERATIVE TRIANGULARIZATION PROCEDURE A square matrix is said to be lower triangular if all super-diagonal entries vanish. Similarly a square matrix is said to be upper triangular if all subdiagonal entries vanish. Consider a nonsingular real matrix A. It may be decomposed as A ¼ Al þ diag(A) þ Au
(18:5)
in which diag(A) consists of the diagonal entries of A, with zeroes elsewhere, Al coincides with A below the diagonal with all other entries set to zero, and Au coincides with A above the diagonal with all other entries set to zero. Now limiting attention to symmetric matrices, for later use we introduce the matrix functions lower(A) ¼ Al þ 12 diag(A),
upper(A) ¼ Au þ 12 diag(A)
(18:6)
The reader may readily verify that 1. The product of two lower (upper) triangular matrices is also lower (upper) triangular 2. The inverse of a nonsingular lower (upper) triangular matrix is also lower (upper) triangular To formulate the iteration procedure, let K0 denote a symmetric positive definite matrix for which the unique triangular factors L0 and LT0 have already been computed. If K0 is banded, the maximum width of its rows (the bandwidth) equals 2b 1, in which b is the bandwidth of L0. The factors of the perturbed matrix K may be written as [K0 þ DK] ¼ [L0 þ DL] LT0 þ DLT
(18:7)
We may rewrite Equation 18.7 as T T T ¼ L1 I þ L1 0 DL I þ DL L0 0 [K0 þ DK]L0
(18:8)
T T 1 T 1 T T L1 0 DL þ DL L0 ¼ L0 DKL0 L0 DLDL L0
(18:9)
from which
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Note that L1 0 DL is lower triangular. It follows that T 1 T T DL ¼ L0 lower L1 0 DKL0 L0 DLDL L0
(18:10)
The factor of 1=2 in the definition of lower(*) and upper(*) functions is motivated by T T the fact that the diagonal entries of L1 0 DL and DL L0 are the same. An iteration procedure based on Equation 18.10 is proposed as T 1 ( j) ( j)T T L0 DL( jþ1) ¼ L0 lower L1 0 DKL0 L0 DL DL T DL(1) ¼ L0 lower L1 0 DKL0
(18:11)
Note that the computations in Equation 18.11 may be performed using only forward substitution involving L0. As demonstration, we introduce the matrix C using L0C ¼ DKDLDLT. (C is not the right Cauchy–Green strain tensor.) Clearly C may be computed using forward substitution. Now introduce D using DLT0 ¼ C, so that L0DT ¼ CT. Of course DT and hence also D are computed using forward substitution. The last step is to compute DL ¼ L0lower(D). We now introduce an approximate convergence argument. For an approximate convergence criterion, we study the similar relation DA ¼ A1 DK (DA)T1 (DA)
(18:12)
in which (DA)1 is the solution (converged iterate) for DA. Consider the iteration procedure DA( jþ1) ¼ A1 DK (DA)T1 DAj
(18:13)
which is very similar to Equation 18.11. Subtraction of two successive iterates and application of matrix norm inequalities furnish h i DA( jþ2) DA( jþ1) ¼ A1 DA1 DA( jþ1) DA( j)
(18:14)
An example of a matrix norm is the Euclidean norm norm(A) ¼ tr1=2(ATA). Application of matrix norm properties furnishes norm DA( jþ2) DA( jþ1) norm A1 DA1 norm DA( jþ1) DA( j)
(18:15)
Convergence is assured in this example if s(A1 DA1 ) < 1, in which s denotes the spectral radius (e.g., Dahlquist and Bjork, 1974; magnitude of the largest eigenvalue) and serves as a greatest lower bound on matrix norms (Varga, 1962). But then, recalling that A is positive definite, we recognize that s(DA1 ) s A1 DA1 s A1 s(DA1 ) ¼ lmin (A)
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(18:16)
A (conservative) convergence criterion is now revealed as maxj jlj (DA1 )j < mink jlk (A)j
(18:17)
in which lj(DA1) denotes the jth eigenvalue of an n 3 n matrix DA1. Clearly, convergence is expected if the perturbation matrix DA1 has a sufficiently small norm. Applied to the current problem, we likewise expect convergence to occur if maxj jlj (DL1 )j < mink jlk (L0 )j. The convergence criterion in Equation 18.17 appears to be discouraging if K0 is ill-conditioned since mink jlk (L0 )j is then very small. Ill-conditioned stiffness matrices are a common occurrence in the nonlinear finite element method, for example, when the structural material exhibits plasticity or hyperelasticity, or when buckling occurs. However, in such situations the equilibrium equation can usually be adjoined to an ‘‘arc length’’ constraint of which an example is presented in subsequent sections. Doing so results in an augmented linear system in which the matrix is no longer ill-conditioned (Kleiber, 1989).
EXAMPLE 18.1 Demonstrate convergence in the scalar equation 2LDL þ DL2 ¼ DK
(18:18)
SOLUTION A Taylor series representation for the solution DL exists in the form DL ¼ a0 DK þ a1 DK 2 þ a2 DK 3 þ a3 DK 4 þ
(18:19)
Upon substituting (18.18) into (18.19) and making suitable identifications, the first few terms of the Taylor series are obtained as 3 1 1 1 5 DK 3 3 DL ¼ DK DK þ O DK 3 DK 2 þ L9 2L 8L 16L5 128L7 The scalar version of the iteration procedure (Equation 18.11) is 2 2LDL( jþ1) ¼ DK DL( j) , DL(0) ¼ 0 Omitting the manipulations, the first three iterates are obtained as 1 DK 2L 1 1 DK 3 DK 2 DL(2) ¼ 2L 8L |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} DL(1) ¼
Taylor series 1 1 1 5 DK 3 DK 4 DK 3 DK 2 þ DL(3) ¼ 2L 8L 16L5 128L7 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Taylor series and are seen to ‘‘track’’ the Taylor series exactly.
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EXAMPLE 18.2 Generate the first two iterates for the matrices 2
a
6 K0 ¼ 4 b d
0 0
32
a b d
c
76 0 54 0
e
f
c
0 0
2
3
a2
3
7 6 e 5 ¼ 4 ab
ab
ad
b2 þ c2
bd þ ce
f
bd þ ce
d 2 þ e2 þ f 2
ad
and 2
a
0
6 K0 þ DK ¼ 6 4b 2
0
a b
76 6 07 54 0
c eþg
d
32
2
7 e þ g7 5
c
0 0
f
3
f
3
ab
ad
b2 þ c2
bd þ c(e þ g)
a
6 ¼6 4 ab
d
7 7 5
bd þ c(e þ g) d 2 þ (e þ g)2 þ f 2 3 0 0 0 7 6 7 cg DK ¼ 6 5 40 0 0 cg g(2e þ g) 2
ad
SOLUTION The initial triangular factor satisfies. 2
a 6 L0 ¼ 4 b d
3 0 0 7 c 0 5, e f
2
cf 1 6 ¼ bf 4 acf be cd
L1 0
0 af ae
3 0 7 05 ac
The exact (converged) triangular factor is 2
L0 þ DL1
a 6 ¼ 4b d
0 c eþg
3 0 7 05 f
from which 2
DL1
3 0 0 0 6 7 ¼ 40 0 05 0 g 0
reflecting the error in L0 as the initial approximation to L0 þ DL1.
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7 5
Upon performing the iteration procedure we find 32 0 cf 0 0 1 6 76 ¼ af 0 54 0 4 bf (acf )2 0 be cd ae ac 2 3 0 0 0 1 6 7 2 2 0 a c fg 5 ¼ 40 (acf )2 2 2 2 2 2 0 a c fg a c g 2
T L1 0 DKL0
0 0
32
bf af
cf 76 54 0
0 cg
cg g(2e þ g)
0
0
3 be cd 7 ae 5 ac
Accordingly, 2
0 0 60 0 T DKL Þ ¼ lowerðL1 4 0 0 0 gf
3 0 0 7 5
1 g2 2 f2
and we obtain the first iterate as 2
DL(1)
3 0 0 7 5
0 0 60 0 1 T ¼ L0 lowerðL0 DKL0 Þ ¼ 4 0 g
1 g2 2 f
In terms of the Euclidean matrix norm, a measure of relative error is introduced as normðDL1 DL(1) Þ 1 g ¼ errorðDL Þ ¼ norm(DL1 ) 2 f (1)
Suppose gf ¼ 0:1. The relative error for the initial iterate is then error(DL(1)) ¼ 5%. Otherwise stated, in this example, the error in the triangular factors was reduced 95% in one iterate. We seek the second iterate and examine the additional error reduction. Now 2 (1) (1)T T L1 L0 ¼ 0 DL DL
cf
0
0
32
0 0
76 1 6 6 6 bf af 07 54 0 (acf )2 4 0 be cd ae ac 3 2 2 0 cf bf be cd 7 6 6 6 6 af ae 7 5 ¼ 40 40 0 0 0 ac
0
32
76 60 0 0 7 54 2 g g 12 f 0 0 3 0 0 7 7 0 0 5 g2 1 g2 0 f 2 (1 þ 4 f 2 ) 0
Continuing, 2
T 1 (1) (1)T T L1 L0 0 DKL0 L0 DL DL
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0 0
0 0 60 0 ¼6 4 0 gf
3
0 g f
14
4
g f4
7 7 5
0
3
7 g 7 5
1 g2 2 f
2
lower(L1 DKLT
0 60 1 (1) (1)T T L DL DL L ) ¼ 4 0
3
0 0
0 0
g f
4 18 gf 4
7 5
The second iterate is thereby given by 1 T 1 (1 ) (1)T T L0 ) DL(2) ¼ L0 lower(L 0 DKL0 L0 DL DL 3 2 32 0 0 0 a 0 0 7 6 76 0 0 0 7 ¼ 4 b c 0 56 5 4 4 0 gf 18 gf 4 d e f 3 2 0 0 0 7 6 0 7 ¼6 5 40 0 4 0 g 18 gf 3
The relative error for the second iterate now is 1 g4 norm DL1 DL(2) 8 f3 error DL(2) ¼ ¼ 0:0125% ¼ g norm(DL1 ) Evidently, in the first iterate the error is reduced to 1=20th of the initial value, and in the second iterate to 1=8000th of the initial value, representing a 400-fold improvement from the first to the second iterate. The convergence rate is much faster than linear convergence characteristic of ‘‘fixed point iteration.’’
EXAMPLE 18.3 Illustrate the effect of the perturbation on the rate of convergence.
SOLUTION Suppose " K¼
1 1 2
1 2 10 þ n 30
# ,
" K0 ¼
1 1 2
1 2 1 3
#
" , DK ¼
0
0
0
n 30
#
in which n > 5=2. The exact solutions are " L¼
1 1 2
# 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 10 þ n 1 , 30 4
" DL1 ¼
0 0
0 ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 þ n 1 p1ffiffiffiffi 30 4 12
#
Clearly, the method fails if n 5=2, in which case K is no longer positive definite. To ensure real numbers attention is restricted to the range 5=2 n 5=2. Following the procedures in Example 18.2, the first two iterates are obtained as
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TABLE 18.1 Error Reduction with First and Second Iterations n 0.5 0.5 þ2.0 2.0
Error(DL(1))
1 Error(DL(1))
Error(DL(2))
1 Error(DL(2))
0.053 0.057 0.176 0.277
0.947 0.943 0.824 0.75
0.0042 0.0094 0.059 0.132
0.9958 0.9916 0.941 0.916
0 D L1 ¼ 0
0
npffiffi 10 3
,
0 DL2 ¼ 0
npffiffi 10 3
0 2 100n pffiffi
3
The previously defined relative error measures are now given by ! rffiffiffiffiffiffiffiffiffiffiffiffiffiffi n 2n 1þ 1 5 (1 ) 5 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi , ¼ error DL 2n 1 þ 1 5
! rffiffiffiffiffiffiffiffiffiffiffiffiffiffi n n2 2n 1þ 1 5 50 5 rffiffiffiffiffiffiffiffiffiffiffiffiffiffi error(DL(2) ) ¼ 2n 1 þ 1 5
The effect of the perturbation is thus reduced as illustrated in Table 18.1. For the relatively small perturbation represented by n ¼ 0.5, the first iteration removes approximately 95% of the initial error. The second iteration removes over 99% of the error, for at least a fivefold improvement over the first iteration. For the relatively large perturbation represented by n ¼ 2.0, the first iterate removes more than 70% of the initial error. Over 90% is removed by the second iteration, for at least a twofold improvement over the first iteration.
18.2 STIFF ARC LENGTH CONSTRAINT IN NONLINEAR FEA 18.2.1 INTRODUCTION Arc length constraints enable iterative solution procedures in nonlinear FEA to converge even at critical points, such as occur in buckling problems, and enable computation to continue beyond the critical points, for example, in postbuckling. They were briefly introduced in a simple example in Chapter 14. The arc length constraint replaces the conventional m 3 m stiffness matrix with an augmented (m þ 1) 3 (m þ 1) stiffness matrix. Its use is referred to as arc length control, in contrast to load control which furnishes the conventional stiffness matrix. It also contrasts with displacement control in which displacements are introduced in increments and the corresponding forces are computed as reactions. In the current chapter, an example of an arc length constraint is introduced following Nicholson (2005b). It identifies arc length parameters maximizing the stiffness (absolute value of the determinant) of the augmented matrix. The parameters, viewed as a vector, must be perpendicular to the rows of the stiffness matrix, likewise considered vectors. The augmented stiffness matrix is nonsymmetric and lacks the small bandwidth of the
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conventional stiffness matrix. However, using a block triangularization, it is demonstrated that solution may be attained by standard finite element operations, namely triangularization of a banded nonsingular portion of the stiffness matrix followed by forward and backward substitutions involving banded lower and upper triangular matrices. The proposed constraint is expected to permit convergence under longer arc lengths than currently implemented methods. A simple example is presented to illustrate application of the constraint. In conventional load-control-based finite element modeling of nonlinear problems in solid mechanics, a combined incremental and iterative solution procedure is typically followed. The iteration procedure is a realization of Newton iteration, in which the stiffness matrix (combined geometric and tangent) serves as the Jacobian matrix. Frequently, due to nonlinear geometry under compression or to softening material behavior such as plasticity or hyperelasticity, the stiffness matrix appearing in the incremental equations exhibits a ‘‘critical point’’ at (near) which it is singular (ill-conditioned). Arc length constraints, implementing arc length control, have been introduced to permit calculation at critical points. A recent review of arc length methods has been authored by Memon and Su (2004). Three different implementations have been compared in Ragon et al. (2001). The initial method is attributed to Riks (1979) and Wempner (1971), and modified by Crisfield (1981), Ramm (1981), and Fafard and Massicotte (1993). Arc length constraints are widely implemented in finite element codes, e.g., Moharir (1998). Arc length methods adjoin a constraint to the conventional incremental finite element equation arising under load control to furnish an augmented stiffness matrix. Doing so enables iterations to converge even at critical points. The goal of the current investigation is to identify parameters of the arc length method ensuring that the augmented stiffness matrix is not only nonsingular but optimally stiffened. It will be seen that the best choice is to select a vector arising in the arc length constraint, appearing in the bottom row of the augmented stiffness matrix, such that it is orthogonal to the rows (considered as vectors) of the conventional stiffness matrix. One issue raised by previous investigators (e.g., Crisfield, 1981) has been that the augmented stiffness matrix is nonsymmetric and unbanded. However, by introducing a block triangularization, the solution procedure will be shown to reduce to conventional finite element operations, namely triangularization of a banded symmetric nonsingular matrix, followed by forward and backward substitution involving banded lower and upper triangular matrices. The notion of a critical point is illustrated in Figure 18.1. Consider a body submitted to loads following a path in load space to a final load f0, the magnitude of which is the maximum attained on the path. Along the path the magnitude of the load may be written as ljf0j, in which the load intensity l satisfies 0 l 1. The magnitude of the global displacement vector is denoted as jgj. Figure 18.1 illustrates a critical point followed by a zone of decreasing load intensity and negative stiffness, such as occurs in postbuckling. In the current investigation, critical points may be maxima, minima, or saddle points.
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l Critical points Stiff
Stiff Postbuckled |g |
FIGURE 18.1 Load–deflection characteristic exhibiting critical points and buckling.
18.2.2 NEWTON ITERATION
FOR
NONLINEAR FINITE ELEMENT EQUATIONS
For the sake of explaining how, in the instance being presented, the arc length method is a special case of Newton iteration, we briefly recapitulate Newton iteration in nonlinear FEA, expanding on the presentation in Chapter 14. Consider a solid body referred to the undeformed configuration, with volume V and boundary S. It experiences large deformation and nonlinear material behavior under boundary tractions t prescribed on S. Balance of linear momentum (load balancing) is fulfilled by the Principle of Virtual Work (Chapter 15) as c(g(t )) ¼ 0,
ð ð ð € dV duT t dS dgT c(g) ¼ trace(d«s) dV þ duT ru
(18:20)
c m 1 vector representing unbalanced loads when nonvanishing g(t ) m 1 time dependent vector of nodal displacements u(X,t ) 3 1 displacement vector X «
3 1 position vector in undeformed configuration 3 3 Lagrangian strain tensor
s r
3 3 second Piola---Kirchhoff stress tensor mass density
and d denotes the variational operator. The goal is to compute the displacement vector u(X,t). The body is assumed to be ‘‘discretized’’ using a mesh of small elements connected at nodes. In FEA, it is assumed that the displacement vector in the eth element may be approximated to satisfactory accuracy using an interpolation model of the form u(X,t) wTe (X)Fe ge (t)
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(18:21)
in which we(X) ¼ me 3 1 vector-valued function of position Fe ¼ me 3 me matrix of constants reflecting element geometry ge(t) ¼ me 3 1 nodal displacement vector for the element Suppose that the solution process has been invoked to determine gn, the global nodal displacement vector at time tn ¼ nh, with ‘‘small’’ time step h. The task is now to formulate a scheme for iteratively computing gnþ1 at tnþ1 ¼ (n þ 1)h. For this purpose, we first describe Newton iteration under load control, which is wellknown to experience convergence difficulties near ‘‘critical points’’ at which the dynamic stiffness matrix becomes singular. We then introduce a recently introduced arc length method Nicholson (2005b) with what will be called a stiff constraint. It will be seen to enable circumventing the critical point.
18.2.3 NEWTON ITERATION
WITHOUT
ARC LENGTH CONSTRAINT
Let Dn g ¼ gnþ1 – gn denote the incremental nodal displacement vector, with similar definitions for stress, strain, displacement, and traction. From Chapter 15, we know that at dynamic equilibrium € Dn f m ¼ 0 Dn c ¼ [KG þ KT ]Dn g þ MDn g ð ð T T KT (gnþ1 ) ¼ M GxG M dV0 , KG (gnþ1 ) ¼ MT s IM dV0 ð ð T T M ¼ r0 F ww F dV0 , Dn f m ¼ r0 FT wDt0 dS0 VEC (Dn F) ¼ M(X)Dn g,
F KT KG x U VEC
(18:22)
GT ¼ 12 [I FT þ F I]M(X)
3 3 3 deformation gradient tensor m 3 m tangent stiffness matrix m 3 m geometric stiffness matrix 9 3 9 tangent modulus tensor 9 3 9 permutation tensor VEC operator Kronecker product symbol
€ in terms of Dng and thereby furnishes The Newmark method serves to express Dng Dn c ¼ KG þ KT þ h42 M Dn g Dn f m Dn rm ¼ 0 Dn rm ¼ Dn1 f m þ 2hMDn1 g_ KG þ KT h42 M Dn1 g
(18:23)
Of course the residual vector Dnrm is known from the previous (nth) time step.
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The Newton iteration scheme may now be stated as þ1 ) (y) (y ) (0) KD g(y nþ1 gnþ1 ¼ c gnþ1 , gnþ1 ¼ gn KD ¼ KG þ KT þ h42 M
(18:24)
Note that KD is real and symmetric. Under load control, Equation 18.24 is the equation to solved. The matrix KD(gn) in many problems exhibits critical points, at (near) which it becomes singular (ill-conditioned). This difficulty may, for example, be produced (a) by geometric nonlinearity under compression (e.g., buckling), associated with the geometric stiffness matrix KG (gn) or (b) by material softness, such as plasticity or hyperelasticity, associated with the tangent stiffness matrix KT (g). Beyond the critical point the stiffness matrix may be indefinite, as in postbuckling (cf. Figure 18.1). Clearly, computation is impossible (or very difficult) in such cases if direct solution of Equation 18.5 is attempted.
18.2.4 ARC LENGTH METHOD However, as stated previously arc length methods have been used, by now for several decades, to circumvent the numerical difficulties posed by critical points. They involve two major features: (a) Writing Dnfm ¼ (Dnl)fm, in which fm is the final load vector to be attained, and Dnl is the incremental load intensity (b) Adjoining an arc length constraint to Equation 18.24 in the form z(gnþ1 , lnþ1 ) ¼ gTm Dn g þ gDn l DS ¼ 0
(18:25)
gm an m 3 1 vector to be determined g a scalar to be determined DS arc length, a positive scalar parameter
gm We refer to as the arc length vector. g It appears that the most commonly implemented methods (e.g., Riks, 1979) involve the conventional choices gm ¼ 12 Dn g g ¼ 12 Dn l
(18:26)
and the arc length constraint is expressed using DS2 : 12 Dn gT Dn g þ 12 (Dn l)2 DS2 ¼ 0. With this choice, the arc length constraint can be visualized in terms of an m þ 1 dimensional hypersphere surrounding the solution points at the nth load step.
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A search procedure is then followed on the hypersphere until the equilibrium equation is satisfied (load is balanced). Returning to the general case represented by Equation 18.25, the m 3 m incremental finite element equation is now supplanted by an (m þ 1) 3 (m þ 1) augmented system of equations. To apply Newton iteration to the augmented system, note that dc(gnþ1 , lnþ1 ) ¼ KD dgnþ1 f 0 dlnþ1
(18:27)
dz(gnþ1 , lnþ1 ) ¼ gTm dgnþ1 þ g dlnþ1
Newton iteration applied to Equation 18.27 now gives rise to the m 3 1 linear system ( K*
(y) g(yþ1) nþ1 gnþ1
l(yþ1) nþ1
l(y) nþ1
(
) ¼
) (y) c g(y) nþ1 , lnþ1 , (y) , l z g(y) nþ1 nþ1
" K* ¼
KD g
T
f m g
# ,
l(0) nþ1 ¼ ln (18:28)
Choices gm and g which ensure that K* is nonsingular, and even more which maximize the magnitude of the determinant of K*, will be said to render the arc length constraint stiff. Our primary task now is to consider choices for gm and g to maximize stiffness (defined below as the magnitude of the determinant). Doing so is expected to permit convergence using larger values of the arc length parameter than in the current method. To find the optimal parameters we restrict attention to the case in which KD has rank m 1 (rank deficiency of unity). The case of rank lower than m 1 will be addressed in a subsequent investigation.
18.2.5 STIFF ARC LENGTH CONSTRAINT 18.2.5.1
Stiffness of K*
We assume that KD is of rank m 1 (rank deficiency ¼ 1), and rewrite it as " KD ¼
(m1) KD kTm1
km1 km
# (18:29)
Owing to unit rank deficiency, any row of KD can be expressed as a linear combination of the other rows. Furthermore the rows and columns can be ordered such that the upper (m 2 1) 3 (m 1) block is3 nonsingular. To illustrate this fact 2 3 consider the reordering
1 0 40 0 3 0
3 05 6
!
1 3 4 3 6 0 0
0 0 5. 0
The 3 3 3 matrix is of rank 2. The
upper left-hand 2 3 2 block of the reordered matrix is nonsingular. (It appears that reordering will not be necessary if none of the rows of KD is null; for example, no 2 3 1
reordering is needed in 4 1 0
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1 2 1
0 1 5:Þ 1
We assume that the above-mentioned ordering either is not needed or has been is nonsingular and kTm1 is a linear combination of performed. Accordingly, K(m1) D (m1) the m 1 rows of KD : there exists an (m 1) 3 1 vector a such that _ a km1 ¼ K(m1) D km ¼ kTm1 a (m1) 1 km1 ¼ kTm1 KD ¼ aT K(m1) a D
(18:30)
If KD is positive semidefinite, K(m1) must be positive definite and km must be a D positive number. Now consider the augmented stiffness matrix incorporating the arc length constraint. 2
K(m1) D
6 T K* ¼ 6 4 km1 gTm1
km1
f m1
km
f m
gm
g
3 7 7, gm ¼ 5
(
gm1
)
( , fm ¼
gm
f m1 fm
) (18:31)
The matrix is singular if the third column is a linear combination of the first two (block) columns. If so there exist a vector m and a scalar n for which _ m þ nkm1 f m1 ¼ K(m1) D
(18:32a)
fm ¼ kTm1 m þ nkmm
(18:32b)
(m1) ¼ aT KD m þ naT km1
(18:32c)
¼ aT f m1
(18:32d)
g ¼ gTm1 m þ ngm
(18:32e)
and
1
In the particular situation in which fm ¼ kTm1 K(m1) f m1 , the matrix is singular D regardless of the choice of gTm1 , gm , and g, since the second row (of blocks) is a linear combination of the first row. Since the finite element code contains the matrix and a solver, this difficulty can readily be detected. Suppose, for example, K(m1) D (m1) that KD is positive semidefinite, in which case KD is positive definite. The linear (m1) system KD h ¼ f m1 may be solved numerically using triangularization followed by forward and backward substitution. Next kTm1 h is compared to fm. If they are equal, a different path in load space should be followed to attain the final load. Alternatively, in the much more likely situation in which fm 6¼ kTm1 (m1)1 KD f m1 , the matrix is nonsingular provided that gTm1 , gm , and g are chosen
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such that the bottom row is not a linear combination of the upper two (block) rows: ^ and a scalar n^ for which i.e., such that there do not exist a vector m (m1) ^ T KD þ n^kTm1 gTm1 ¼ m
^ T km1 þ n^km gm ¼ m (18:33)
(m1) ^ T KD ¼m a þ n^kTm1 a
¼ gTm1 a ^ T f m1 n^fm g ¼ m We assume that the choice of gTm1 , gm , and g does not satisfy Equation 18.33, and proceed to consider the choice which renders K* stiff. 18.2.5.2
Arc Length Vector Which Maximizes Stiffness: Examples
The stiffness x(A) of a square matrix A is defined as the magnitude of its determinant: x(A) ¼ jdet(A)j. The matrix is said to be stiff if x(A) > 0 (i.e., is nonsingular). Clearly, if stiffness is near zero, the matrix is nearly singular. If the arc length parameters are chosen to maximize the stiffness, the matrix A is ‘‘farther away’’ from being singular than for other choices. The first m rows of K*, considered as row vectors, span an m dimensional space, whose (possibly complex) base vectors we denote by ej, j ¼ 1, 2, . . . , m. It will be seen that g does not affect stiffness, and that the magnitudes of gm ¼ ggm1 and g can m
both be set to unity since it is later necessary to manipulate the arc length DS to attain convergence. (It can easily be easily seen in Equation 18.26 that a similar magnitude choice is implicit in the conventional arc length formulation.) The direction of the lower row, viewed as a (row) vector, is of greatest interest. As will be shown in the following, it should be chosen to coincide with the null eigenvector of KD, and thereby to be orthogonal to the m rows of KD.
EXAMPLE 18.4 As a simple example, consider the matrix 2 6 6 A¼6 4
1
0
0
3
7 7 0 1 0 7 g1 g2 g3 5 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 2 2 2 2 2 g1 þ g2 þ g3 g1 þ g2 þ g3 g1 þ g2 þ g3
g3 The determinant of A is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . The stiffness attains its maximum value if 2 g1 þ g22 þ g23 g1 ¼ 0, g2 ¼ 0, g3 ¼ 1. Considering the rows as row vectors, with this choice the third row is orthogonal to the first two rows.
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EXAMPLE 18.5 As another simple example, consider the matrix 2
a11
a12
3
0
7 7 a22 0 a12 7 5 g1 g2 g3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 2 2 2 2 2 g1 þ g2 þ g3 g1 þ g2 þ g3 g1 þ g2 þ g3 8 9 g1 > > 2 ^ 3 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > " # > > A 0 < a11 a12 g21 þ g22 þ g23 = ^ 4 5 g3 ¼ , g¼ , A¼ T pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g2 g > > a12 a22 > > > pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > g21 þ g22 þ g23 : ; g21 þ g22 þ g23
6 6 A¼6 4
Since  is symmetric, there exists an orthogonal matrix Q for which ^ ^ T ¼ L ¼ l1 (A) QAQ 0
0 ^ , l2 (A)
l1,2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r a11 a22 2 þ a212 2
a11 þ a22 ¼ 2
We now subject A to a similarity transformation as follows: " G¼
Q 0T
#2 ^ A 4 T g 1
0
3" 0 QT 5 g3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0T g21 þ g22 þ g23
0
2
#
1
L
¼ 4 gT QT
3 0 g3 5 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g21 þ g22 þ g23
g3 . and G possesses the same eigenvalues as A, namely l1, l2, and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 g1 þ g22 þ g23 Clearly, the choice which maximizes the stiffness is, again, g1 ¼ 0, g2 ¼ 0, g3 ¼ 1. With this choice, considering the rows of A to be (row) vectors, the bottom row of A is again orthogonal to the first two rows.
EXAMPLE 18.6 Finally, consider the matrix A¼
cos u sin u cos c sin c
The determinant of A is given by det(A) ¼ cos u sin c sin u cos c ¼ sin(c u) and the maximum stiffness clearly is attained when c u ¼ p=2. With þp=2, cos u sin u , and note that the bottom row is orthogonal to the top row. A¼ sin u cos u cos u sin u . The second row is again orthogonal to With p=2, we obtain A ¼ sin u cos u the first row, but points in the opposite direction from the positive choice.
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18.2.5.3
Arc Length Vector Which Maximizes Stiffness: General Argument
A proof is now given generalizing the observations from examples (18.4 through 18.6). The symmetric matrix KD is again assumed to be m 3 m and to have rank m 1. Given an m3 1 vector fm, as stated previously our task is to determine the m 3 1 vector ggm , with gm and gmþ1 both of unit magnitude, to maximize x(K*). mþ1
Since KD is symmetric, there exist m orthonormal eigenvectors jw, one of which corresponds to the null eigenvalue. We set the order such that jm corresponds to the null eigenvalue and form the proper orthogonal matrix Q ¼ [j1 j2 jm]. The determinant of K* is unaffected by the similarity transformation, giving rise to K# as follows: " #
K ¼ " ¼
Q
0
0T
1
#"
Lm
^f
^gTm
^ g
KD
f
gTm #
g
,
#"
QT
0
0T
1
#
Lm ¼ QKD QT , ^f ¼ Qf, ^gm ¼ Qgm , ^g ¼ g
(18:34)
Since the eigenvectors serve to diagonalize KD, K# now becomes 2
l1 (KD ) 6 0 6 6 6 0 6 6 : 6 6 : K# ¼ 6 6 : 6 6 : 6 6 6 : 6 4 0 ^g1
0 l2 (KD ) 0 : : : : : 0 ^ g2
0 0 l3 (KD ) : : : : : 0 ^ g3
: : : : : : : : : :
: : : : : : : : : :
: : : : : : : : : :
: : : : : : : : : : : : : : : lm1 (KD ) : : : g^m1
0 0 0 : : : : : 0 ^gm
3 ^f1 ^f2 7 7 7 ^f3 7 7 : 7 7 : 7 7 : 7 7 7 : 7 7 ^fm1 7 7 ^fm 5 ^g (18:35)
The determinant of K# is given by det(K# ) ¼ det(K*) ¼ ^ gm ^fm
m1 Y
lj (KD )
(18:36)
j¼1
Note that it is independent of g ¼ ^ g, which has been previously set to unit magnitude. Recalling that {gTm1 gm } has unit magnitude, the stiffness x(K*) is maximized by the choice ^gm ¼ 1
and
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^ gj ¼ 0
if j ¼ 1, 2, 3, . . . , m 2, m 1
(18:37)
K# is now given by 2
l1 (Km ) 0 0 6 l2 (Km ) 0 6 0 6 6 0 0 l (K 3 m) 6 6 : : 6 : 6 6 : : : K# ¼ 6 6 : : : 6 6 6 : : : 6 6 : : : 6 6 4 0 0 0 0 0 0
: : : : : : : : : :
: : : : : : : : : :
: : : : : : : : : :
: : : : : : : : : : : : : : : lm1 (Km ) : : : 0
3 0 ^f1 7 0 ^f2 7 7 0 ^f3 7 7 : : 7 7 7 : : 7 7 : : 7 7 7 : : 7 7 : ^fm1 7 7 7 0 ^fm 5 1 1
(18:38)
If the rows of K* are viewed as vectors, it is clear that g^m is orthogonal to the first m 1 rows of KD. In fact, it is orthogonal to all m rows since the mth row is a linear combination of the previous rows owing to the unit rank deficiency of KD. Hence QKDQT^gm ¼ 0. But note that gm 0 ¼ QKD QT ^ ¼ QKD QT Qgm ¼ QKD gm
(18:39)
from which we conclude that KDgm ¼ 0, that gm is orthogonal to the rows of KD, and finally that that gm is collinear with the null eigenvector of KD. 18.2.5.4
Numerical Determination of the Optimal Arc Length Vector
The process of identifying a unit vector orthogonal to m 1 vectors is well known as Gram–Schmidt orthogonalization and is briefly summarized pffiffiffiffiffiffiffiffiffiffi in the current context. Let aT1 denote the first row of K*, and let a01 ¼ a1 = aT1 a1 . A sequence of m 1 orthonormal eigenvectors a0j is then generated sequentially according to 00
aj ¼ aj
j1 X
aTj a0i a0i
(18:40)
i¼1
a0i
qffiffiffiffiffiffiffiffiffiffiffi 00 00 ¼ ai = ai T ai 00
in which aTj denotes the jth row of K*. Clearly, 0
00
0
akT aj ¼ akT aj
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j1 X 1, aTj a0i dik ¼ 0, dik ¼ 0, i¼1
k¼i k 6¼ i
(18:41)
The arc length vector is now found as g0m ¼ bm
m1 X i ¼1
bTm a0i a0i
and gm ¼ g0m
.qffiffiffiffiffiffiffiffiffiffiffi g0mT g0m
(18:42)
in which bm is a trial vector, for example, fm. Note that the arc length vector gm can point in a positive or negative sense along the underlying base vector, and g may also be positive or negative. To determine the sense of the gm and g, we recall Figure 18.1. For consistency with the positive arc length DS, the sum of gTm Dn g and gDn l must each be nonnegative. If these two terms were of opposite sign, an unstable numerical process would ensue. For example, if gDnl is negative and grows in magnitude, the arc length constraint requires gTm Dn g to be positive and to likewise grow in magnitude. There is no impediment to this process continuing until both terms attain very large magnitudes, despite the small value of DS. Consequently the arc length relation (Equation 18.24) serves as a constraint only if both terms are nonnegative. In the stiff segments of the load– deflection curve, in which the load intensity and the magnitude of the displacement vector are both increasing, this requires that g 0 and that p2 cos1
gm Dn g pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi T T
(Dn g) Dn g
p2 . In the postbuckled segments, in which the load inten-
sity decreases butthe magnitud e of the displacements increases, it follows that g < 0 and p2 cos1
gm D n g pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T T
(Dn g) Dn g
p2 . If the (n þ 1)st load step occurs at a critical point,
then Dn l ¼ 0 and gTm Dn g ¼ DS. Near a critical point, an appealing way to predict the sense of gm and g before computation at the current step is select them to satisfy gDn1 l > 0,
p gTm Dn1 g ffi cos1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 (Dn1 g)T Dn1 g
!
p 2
(18:43)
thereby making use of the solution at the previous time step. Doing so should be safe away from critical points. Across the critical point, the sign of Dnþ1l may or may not change from that Dn1l of depending on whether it is an extremum or a saddle point. In this case it appears wise to compute one iterate with g > 0 and one with g < 0, and to continue with the value for which gDnþ1l > 0.
18.2.6 SOLUTION PROCEDURE At first glance, the matrix K* appears problematic in that it is asymmetric and unbanded, as is true in the conventional arc length method. However asymmetry
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and unbandedness are consequences of the (m þ 1)st row and column. The upper left m 3 m block is symmetric and can be stored accordingly. As will be seen, after a block triangularization is invoked the solution procedure can be reduced to conventional finite element operations including triangularization of the nonsingular sym(m1) , followed by forward and backward substitutions using banded metric matrix KD lower and upper triangular matrices, respectively. The procedure is established below. 18.2.6.1
Block Triangularization
The reader may verify the block triangularization 2
km f m1 K(m1) D
6 6 T 6 km 4 gTm1
km
fm
gm
g
3 32 Um1 L1 0 Lm1 m1 {km f m1 } 7 6( 7 " # 76 # 6( T ) " 7 6 kT ) 7 7 1 0 k f 6 7¼4 7 0 m m m 54 5 U1 G 5 m1 0 1 gTm1 gm g 0T 3
2
(18:44)
kTm gTm1
in which G ¼
1 U1 m1 Lm1 {km f m1 }. Also conventional LU triangulariza-
¼ Lm1 Um1 , in which Lm1 is lower triangular tion is invoked to furnish K(m1) D and Um1 is upper triangular. (m1) If KD is symmetric and positive definite (i.e., KD is positive semidefinite with rank m 1), then Um1 ¼ LTm1 . If we introduce {w11 w12} obtained from UTm1 {w11 w12 } ¼ {km f m1 }, forward substitution and transposition furnishes
T w11 wT12
kTm gTm1
U1 m1 . Also, writing Lm1 {w21
w22 } ¼ {km f m1 } and using
forward substitution furnishes the vector {w21
w22 } ¼ L1 m1 {km f m1 }. Note
¼
that G ¼ 18.2.6.2
wT11 wT12
{w21
w22 } is a 2 3 2 matrix.
Solution of the Outer Problem
We first describe the solution process for the outer problem expressed as 2
Lm1 6 ( wT ) 4 11 wT12
8 9 9 38 cm1 > {0 0} > zm1 > > > > <( )= " # 7<( )= zm 1 0 5 cm ¼ > > > > > > ; : : zmþ1 0 1 z ;
(18:45)
Forward substitution is used to solve Lm1 zm1 ¼ cm1 for zm1 , from which (
zm zmþ1
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) ¼
( ) cm z
( ) wT11 wT12
zm1
(18:46)
18.2.6.3
Solution of the Inner Problem
The inner problem is now expressed as 2
Um1 6 ( 0T ) 4 0T
9 9 8 38 {w21 w22 } > Dnþ1 gm1 > > zm1 > )= " # 7<( ) = <( zm km fm 5 Dnþ1 g m ¼ > > > ; : : ; > zmþ1 gm 1 Dnþ1 z
(18:47)
Next
Dnþ1 g m Dnþ1 z
1 fm
zm gm km ¼ km þ fm gm zmþ1
(18:48)
Finally backward substitution serves to solve for Dnþ1gm1 using
Um1 Dnþ1 gm1 ¼ zm1 {w21
1 fm
zm gm km w22 } k m þ f m gm zmþ1
(18:49)
EXAMPLE 18.7 Illustrate the orthogonalization and solution procedure using 8 9 1> > > < > = 7 6 7 KD ¼ 6 4 1 2 1 5, f ¼ > 2 > > : > ; 3 0 1 1 8 9 9 3 8 2 Dn c1 > 4> 1 1 0 1 > > > > > > > > > > 7 > > > 6 > > > > > > > 6 1 2 1 2 7 > <3> < Dn c2 = = 7 6 * 3 7, K ¼6 ¼ 10 7 > 6 > > > > 6 0 1 1 3 7 > 2> Dn c3 > > > > > > > > 5 > 4 > > > > > > > > : : ; ; 1 Dn B g1 g 2 g 3 g 4 2
in which jg4j ¼ 1 and
1
1
0
3
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g21 þ g22 þ g23 ¼ 1.
SOLUTION Elementary algebra serves to establish that det(K*) ¼ 6(g1þg2þg3), independently of g4. It remains to determine g1, g2, and g3. We first analytically determine the values of g1, g2, and g3, which maximize the determinant subject to the magnitude restriction. This is equivalent to maximizing the augmented function P ¼ 6(g1 þ g2 þ g3 ) þ 9(g21 þ g22 þ g23 1)
ß 2008 by Taylor & Francis Group, LLC.
in which 9 is a Lagrange multiplier. Elementary manipulation furnishes both a maximum and a minimum: @P ¼ g21 þ g22 þ g23 1 ¼ 0 @L @P ¼ 6 þ 29g1 ¼ 0 @g1 @P ¼ 6 þ 29g2 ¼ 0 @g2 @P ¼ 6 þ 29g3 ¼ 0 @g3 pffiffiffi Clearly, the extrema satisfy the magnitudep condition, and g1 ¼ g2 ¼ g3 ¼ 1= ffiffiffi pffiffiffi 3. The maximum value pffiffiffi of the determinant is 6 3, the minimum value is 6 3, and the stiffness is 6 3. To8determine g1, g2, and g3 using the orthogonalization procedure, we set g4 ¼81 and 9 9 < g1 = g2 to lie exterior : ; g3 8 9 <0= a2 ¼ 1 . First, : ; 1
seek and
<1=
to the subspace spanned occupied by the vectors a1 ¼ 1 :
0
;
8 9 <1= 1 a01 ¼ pffiffiffi 1 2: 0 ;
and
8 9 2 <0= 1 a2 ¼ 1 4{1 1 : ; 2 1 00
8 9 8 938 9 < 0 = < 1 = 1 <1= 1 0} 1 5 1 ¼ : ; : ; 2: ; 0 2 1
Consequently, 8 9 <1= 1 a02 ¼ pffiffiffi 1 6: 2 ; Next, using f as a trial vector, 8 9 8 9 8 938 8 93 8 9 9 2 2 g1 > > > > <^ = > <1> = 1 <1> = > < 1 > = 1 <1> = > < 1 > = 6 6 7 7 ^ g2 ¼ 2 4{1 1 0} 2 5 1 4{1 1 2} 2 5 1 > > > > > : > ; > : > ; 2 : > ; > : ; 6 : > ; > : ; ^ 3 3 0 3 2 g3 8 9 > <1> = ¼2 1 > : > ; 1
ß 2008 by Taylor & Francis Group, LLC.
and, setting the magnitude equal to unity, the desired vector is found as 8 9 > < g1 > = gm ¼ g2 > : > ; g3 8 9 >1> 1 < = ¼ pffiffiffi 1 3> : > ; 1 in agreement with the analytical maximum. Taking the positive sense, the augmented stiffness matrix is now 2
1
1
1
0
3
7 6 6 1 2 1 2 7 7 6 * K ¼6 7 6 0 1 1 3 7 5 4 p1ffiffi p1ffiffi p1ffiffi 1 3 3 3 To illustrate the solution procedure, a glance at Equation 18.43 reveals the identifications " K(m1) D
¼
1
1
#
1 2 ( ) 0 , km ¼ 1
" , Lm1 ¼
gm1
1
0
#
1 1 ( ) 1 1 , ¼ pffiffiffi 3 1
" , Um1 ¼ f m1 ¼
1 1
#
0 1 ( ) 1 2
Elementary manipulation gives U1 m1
( T ) 1 1 1 0 0 1 km 1 1 p ffiffi ffi p ffiffi ffi U ¼ , Lm¼1 ¼ , ¼ m1 0 1 1 1 1= 3 2= 3 gTm1
L1 m1 {km f m1 } ¼ and
( G¼
kTm
0 1 1 3
)
gTm1
1 U1 m1 Lm1 {km f m1 } ¼
1 pffiffiffi 2= 3
3 pffiffiffi 7= 3
The block triangularization now results in 2
1 0 6 1 1 6 " # K* ¼ 6 6 0 1 4 p1ffiffi p2ffiffi 3
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3
0 0 1 0
3 32 1 1 0 1 0 7 6 1 3 0 7 76 0 1 7 " #7 6 7 7 6 0 0 0 6 0 7 54 5 pffiffiffi 8ffiffi p 3 1 þ 0 0 1 3
The outer and inner problems are expressed as 38 9 8 9 1 0 0 0 Dn z1 > 4> > > > > > > > > > 7 6 1 1 <3> = 0 0 7< Dn z2 = 6 3 # 7 6" ¼ 10 7 6 0 1 > 1 0 5> Dn z 3 > 2> > > > > 4 > > > : ; : > ; p1ffiffi p2ffiffi 0 1 Dn z 1 1 3 3 2
2
1 1
6 0 1 6 6 6 0 0 4 0 0
38 9 8 9 D g Dn z1 > > > > n 1> > > > > > > > 7 = < = 1 3 7< " # 7 Dn g2 ¼ Dn z2 7 0 6 > Dn g 3 > > > > Dn z3 > > 5> pffiffiffi > > > : ; > : ; 3 1 þ p7ffiffi3 Dn § Dn z1
0
1
Upon applying forward substitution in the outer problem followed by backward substitution in the inner problem, the solution is determined to be 8 9 8 9 4 D n z1 > > > > > > > > > > > > > > > < Dn z2 > = < = 7 3 ¼ 10 , > > > Dn z3 > 9 > > > > > > > > > > > > pffiffiffi ; : ; : Dn z4 1 þ 18 3
pffiffiffi 9 8 1 8 9 (1 þ 5= 3) > > 2 > > D g n 1> > > > > > > > > > > > > > < Dn g > = < 5 1 þ p1 ffiffi > = 2 2 3 3 ¼ 10 pffiffiffi > > > Dn g3 > 5 ffiffi > > > p > > > > 1 þ 3 > > > > > 2 3 : ; > > > > : ; Dn § 3=2
In the current example, it is evident that the stiffness matrix KD arising under load control is singular. But the augmented matrix arising under stiff arc length control is well behaved and a solution is readily attained by a procedure combining triangularization, forward substitution, and backward substitution.
18.3 NON-ITERATIVE SOLUTION OF FINITE ELEMENT EQUATIONS IN INCOMPRESSIBLE SOLIDS 18.3.1 INTRODUCTION Finite element equations for incompressible and near-incompressible media give rise to a matrix with a diagonal block of zeroes or very small numbers. The matrices are not amenable to conventional techniques involving pivoting on diagonal entries. Uzawa methods (Arrow et al., 1959) have been applied to the associated linear systems. They are iterative and converge when the matrix is nonsingular. In the current study an alternate form of the matrix is used which is amenable to solution without iteration. It likewise is applicable whenever the matrix is nonsingular. The solution process consists of a block LU factorization, followed by Cholesky decomposition (triangularization) of a positive definite diagonal block together with several forward and backward substitution operations. Two illustrative examples are developed. In compressible solids, the finite element stiffness matrix typically is positive definite. The governing equation in finite element form can frequently be solved by triangularization, consisting of Cholesky decomposition followed by forward and
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backward substitution. However, suppose the material is incompressible or nearincompressible. The strains are now subject to an internal constraint (are not independent), and serve to determine stresses only to within an indeterminate pressure. The pressure field serves as an unknown Lagrange multiplier in an auxiliary finite element equation to satisfy the incompressibility constraint. The structure of the finite element equations (equilibrium plus constraint) at first glance poses a computational problem, since a block of the stiffness matrix is null. Traditional triangularization and solution methods based on pivoting are not directly applicable to such a matrix. Much of the literature on this problem is based on the Uzawa method, which attains the solution through an iteration scheme. (cf. In the current investigation, a block triangularization solution is formulated in which the blocks are obtained by Cholesky decomposition, as well as forward and backward substitutions. This scheme obviates the need for iteration while using real variables.)
18.3.2 FINITE ELEMENT EQUATION
FOR AN INCOMPRESSIBLE
MEDIUM
To set the problem under study in a context, we consider dynamic response of a nearincompressible nonlinear solid. The special cases of static response, incompressible media, and linear behavior can be retrieved from this case. Application of the Incremental Principle of Virtual Work and introduction of suitable interpolation models (Chapter 15) furnish the finite element relations € þ K(g)Dg VDp ¼ Df MD g VT Dg þ
Dp ¼0 k
(18:50)
M mass matrix, n 3 n and positive definite K incremental stiffness matrix, n 3 n and positive definite V pressure–displacement matrix, n 3 p of rank p, p < n Dg incremental nodal displacement vector Dp incremental nodal pressure vector Df incremental consistent nodal force vector In particular, as usual Dg is the difference between the nodal displacement vectors at two load or time steps. We assume that the mass and stiffness matrices are positive definite and n 3 n. The pressure–displacement matrix V is n 3 p and is restricted to have rank p, p < n. The first equation is a realization of the balance of linear momentum, while the second represents the a posteriori enforcement of the near-incompressibility constraint. If k ! 1, the incompressible case is recovered. If M ¼ 0, the static case is recovered. Finally, to specialize to the linear case, the incremental symbol D may be removed. We assume that Equation 18.50 is integrated using a one-step procedure based on the trapezoidal rule (Newmark’s method). As before let h denote the time step and let gn ¼ g(tn). At time tnþ1 ¼ (n þ 1)h Equation 18.50 becomes, following Zienkiewicz and Taylor (1989),
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( A
Dnþ1 g Dnþ1 p
)
( ¼
Dnþ1 g
"
)
0
,
A¼
2
2
2
h2
M þ h4 K(gn ) h2 V h2 VT
2
#
Ip k
(18:51)
in which Ip is the p 3 p identity matrix Dnþ1 g ¼
h2 (Dnþ1 f þ Dn f KDn g) þ M(Dn g þ hDn q),Dn q ¼ Dn g_ 4
Dnþ1 g ¼ gnþ1 gn : Note that A is symmetric, with the advantage of saving computer storage. A comment is in order on the restriction that the n 3 p matrix V have rank p. We K V consider whether the matrix VT 0 is singular, if K is n 3 n and positive definite while V is n 3 p but of rank p 1 (or less). The matrix is singular if, and only if, I K1=2 V is singular. But this new matrix is singular if, and only if, there VT K1=2 0 exists a nontrivial vector whose product with this matrix vanishes. This is simply the ¼ 0. Such a condition that there exist a nonzero p 3 1 vector x for which VTK1Vx T 1 K V vector exists since V K V at most has rank p 1. It follows that VT 0 is singular if V has rank less than p. K Next consider the matrix VT
V Ip =k
in which the bulk modulus is a very large
positive number (for near-incompressibility). Again V has rank p 1. The matrix cannot be singular since only the zero vector satisfies VT K1 V þ Ip =k x ¼ 0. The smallest eigenvalue of this latter matrix is 1=k, which is a very small number. But the condition number of this matrix is k max [lj VK1 VT ] þ 1, and will typically be a j¼1,p
very large number for near-incompressible materials. Consequently, if V has rank p 1 (or less), the matrix is expected to be ill-conditioned and convergence will be very difficult to achieve.
18.3.3 UZAWA’S METHOD To explain Uzawa’s method we follow the development in Zienkiewicz and Taylor (1989) for the case in which
K V f g (18:52) , C¼ ¼ C 0 VT 0 p There is an extensive and continuing literature on the Uzawa method (Hu and Zou, 2001). The term pr is subtracted from both sides to furnish f K V g ¼ (18:53) I pr VT rp p
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The Uzawa method is realized as the iteration scheme 8 9 2
jþ1 < f = K g pj , B ¼ 4 ¼ B VT : ; p r
3 V Ip 5 r
(18:54)
in which r > 0 is an acceleration parameter. Successive iterates of p satisfy (18:55) Ip þ rVT K1 V p jþ1 ¼ p j rVT K1 f in which the superscript is a counter for the iterate. This sequence converges if the maximum eigenvalue of [Ip þ rVT K1 V]1 is less than unity. But note that VTK1V is positive definite and of rank p. Hence the eigenvalues of Ip þ rVT K1 V all exceed unity. It follows that the eigenvalues of 1 are less than unity, implying convergence. This scheme repreIp þ rVT K1 V sents ‘‘fixed point iteration,’’ for which the convergence rate is linear. We note that B can be triangularized using complex variables as follows: 3 " #2 T L1 V L L 0 h i5 4 B¼ (18:56) I 0 i rp þ WT W VT LT iIp LLT ¼ K,
i¼
pffiffiffiffiffiffiffi 1,
and W is obtained by solving the linear system LW ¼ V using forward substitution. The triangularization is only performed once in linear problems. Forward and backward substitution then are repeated at each iteration to attain the solution. In particular, the decomposition is used #( " #( jþ1 ) ( ) " T ) ( jþ1 ) L1 V i L f L 0 z1 z1 g jþ1 h ¼ ¼ Ip T pj , T T jþ1 jþ1 0 i r þW W r V L iIp p z2 z2jþ1 (18:57a) followed by readily performed operations on block submatrices: Lz1jþ1 ¼ f p z2jþ1 ¼ i iVT LT z1jþ1 r h i Ip i r þ WT W p jþ1 ¼ z2jþ1
forward substitution
(18:57b)
backward substitution
(18:57c)
p p triangularization
(18:57d)
backward substitution
(18:57e)
j
LT g jþ1 ¼ h Note that
Ip r
z1jþ1
þ L1 Vp jþ1
i þ WT W in Equation 18.57d is positive definite, so that Cholesky
decomposition using real mathematics applicable.
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We note that it is also problematic for the Uzawa method if the rank of V equals p 1 (or less). Recall from Equation 18.54 the convergence criterion that the eigenvalues of ½Ip þ rVTK1V 1 must be less than unity. However, if rank V ¼ p 1, VTK1V is singular (rank less than p) and ½Ip þ rVTK1V 1 has an eigenvalue equal to unity. Consequently, convergence will not occur. (The existence of a difficulty is not surprising since the matrix in Equation 18.52 is singular if rank (V) < p.)
18.3.4 MODIFICATION
AVOID ITERATION
TO
A simple modification to the Uzawa method is introduced which furnishes solution without iteration while using real mathematics. Returning to the incremental finite element formulation, Equation 18.50 may be rewritten in the equivalent form ( C
Dnþ1 g
(
) ¼
Dnþ1 p
Dnþ1 g
"
) ,
0
C¼
2
2
M þ h4 K(gn ) h2 V h2 2
# (18:58)
h2 Ip 2 k
VT
Note that the sign of the lower row in the matrix has been changed. This forfeits the symmetry of the matrix (which has no real significance for computer storage), but will prove to permit a non-iterative solution based on triangularization of positive definite submatrices (and real number operations). Observe that block triangular factorization in Equation 18.58 furnishes "
2
2
M þ h4 K(gn ) h2 V h2 2
# ¼
h2 Ip 2 k
VT
"
Ls LTs
Ls T T h2 V Ls 2
¼Mþ
h2 4
0 Ip
#"
0T
#
h L1 s 2 V 2
LTs h4 4
WT W þ h2
2
Ip k
(18:59)
K(gn )
The solution is attained by the following decomposition and readily performed operations of block submatrices. "
Ls h2 2
LTs T 0
VT LT s
0 Ip
h2 L1 s 2 V h4 T h2 Ip 4 W Wþ 2 k
Ls Dnþ1 z1 ¼ Dnþ1 g w¼
LTs Dnþ1 z1 h2 T
#
Dnþ1 z1 Dnþ1 z2 Dnþ1 g Dnþ1 p
¼
¼
Dnþ1 g 0
Dnþ1 z1 Dnþ1 z2
forward substitution backward substitution
Dnþ1 z2 ¼
V w 2 Ls W ¼ V
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forward substitution
(18:60)
h
2 h4 T W W þ h2 Ip =k 4
i
Dnþ1 p ¼ Dnþ1 z2 2
LTs Dnþ1 g ¼ Dnþ1 z1 þ h2 WDnþ1 p
triangularization
backward substitution
4 This solution procedure is enabled by the fact that h4 WT W þ Ip =k is positive definite and hence can be triangularized. It is worth noting that a non-iterative solution can be achieved using the symmetric form in the Uzawa method (Equation 18.54), following the procedure presented in Equation 18.57 involving complex numbers.
EXAMPLE 18.8 Demonstrate that the method works in the following example. Consider a single element model for an incompressible isotropic elastic rod, shown in Figure 18.2. The rod is of length L with a square Y by Y cross section, and A ¼ Y2. Shear modulus is m. Interpolation models are assumed in the form u(x) ¼ xu(L)=L,
v(y) ¼ w(y) ¼ yv(Y )=Y ,
p ¼ p0 (constant)
Omitting the details, the Principle of Virtual Work together with the variational form of the incompressibility constraint may be readily shown to furnish 2 2mA L
6 6 0 4 A
0 4mAL Y2 2 AL Y
38 9 8 9 u(L) > > f > > = < = < 7 2A YL 7 5> v(Y ) > ¼ > 0 > ; : ; : 0 p0 0 A
This equation was encountered and solved in Chapter 11. Simple manipulation following the foregoing procedures furnishes f ¼ 3mLA. This is the exact answer for incompressible isotropic linear elasticity, since the Young’s modulus E in this case E , and the Poisson’s ratio n equals 1=2. satisfies m ¼ 2(1þn)
y
m L
FIGURE 18.2 Rod of incompressible elastic material.
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f
x
Y Y
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