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Finite Element Analysis Thermomechanics of Solids
David W...
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Finite Element Analysis Thermomechanics of Solids
David W. Nicholson
CRC PR E S S Boca Raton London New York Washington, D.C.
© 2003 by CRC CRC Press LLC
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Library of Congress Cataloging-in-Publication Data Nicholson, D.W. Finite element analysis : thermomechanics of solids / David W. Nicholson. p. cm. Includes bibliographical references and index. ISBN 0-8493-0749-X (alk. paper) 1. Thermal stresses—Mathematical models. 2. Finite element method. I. Title. TA418.58.N53 2003 620.1′121—dc21
2002041419
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. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from CRC Press LLC for such copying. Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation, without intent to infringe.
Visit the CRC Press Web site at www.crcpress.com © 2003 by CRC CRC Press LLC No claim to original U.S. Government works International Standard Book Number 0-8493-0749-X Library of Congress Card Number 2002041419 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper
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Dedication To Linda and Mike, with profound love
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Preface 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-level course in the finite-element method, and by now, almost all undergraduates of such programs have some exposure to the finite-element method. A number of departments offer a second-level course. It is hoped that this text 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 provides 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 finiteelement data structures). • Modeling principles and strategies (including mesh design). The text is designed for a second-level course, as a reference work, or for self study. Familiarity is assumed with the finite-element method at the level of a firstlevel graduate or advanced undergraduate course. A first-level 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 topics for a secondlevel course. Nor is there much emphasis on coupled thermomechanical problems. However, many engineers could benefit from a text covering nonlinear problems and the associated continuum thermomechanics. Such a text could 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; metal forming; 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. However, they often lack and urgently need an in-depth but compact exposition of the finite-element method, which provides a foundation for addressing future problems. It is hoped this text also fills this need. Of necessity, a selection of topics has been made, and topics are given coverage proportional to the author’s sense of their importance to the reader’s understanding. Topics have been selected with the intent of giving a unified and complete, but still compact and tractable, presentation. Several other excellent texts and monographs
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have appeared over the years, from which the author has benefited. Four texts to which the author is indebted are: 1. Zienkiewicz, O.C. and Taylor R.L., The Finite Element Method, Vols. 1 and 2, McGraw Hill, London, 1989. 2. Kleiber, M., Incremental Finite Element Modeling in Nonlinear Solid Mechanics, Chichester, Ellis Horwood, Ltd., 1989. 3. Bonet, J. and Wood, R.D., Nonlinear Continuum Mechanics for Finite Element Analysis, Cambridge, Cambridge University Press, 1997. 4. Belytschko, T., Lui, W.K, and Moran, B., Nonlinear Finite Elements for Continua and Structures, New York, John Wiley and Sons, 2000. This text has the following characteristics: 1. Emphasis on the use of Kronecker Product notation instead of tensor, tensor-indicial, 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. Kronecker Product algebra (KPA) has been widely used in control theory for many years (Graham, 1982). It is highly compact and satisfies 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 nearincompressible), 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 author’s experience, 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 also, 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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Acknowledgment Finite Element Analysis: Thermomechanics of Solids is the culmination of many years of teaching and research, made possible through the love and patience of my wife Linda and son Michael, to whom this is dedicated. To my profound love, I cheerfully add my immense gratitude. Much of my understanding of the advanced topics addressed in this book 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. Also deserving of acknowledgment is Ashok Balasubranamian, whose careful attention to the manuscript has greatly helped to make it achieve my objectives. 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 thank CRC Press for taking a chance on me and especially for Cindy Renee Carelli for patience and many good suggestions.
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About the Author David W. Nicholson, Ph.D. is a professor in the Department of Mechanical, Materials, and Aerospace Engineering at the University of Central Florida, Orlando, Florida, where he has served as the Chair (1990–1995) and Interim Chair (2000–2002). He earned an S.B. degree at MIT in 1966 and a doctorate at Yale in 1971. He has 31 years of experience since the doctorate, including research in industry (Goodyear, 7 years) and government (the Naval Surface Weapons Center, 6 years), and faculty positions at Stevens Institute of Technology and UCF. He has authored over 140 articles on the finite-element method, continuum mechanics, fracture mechanics, and dynamics. He has served as a technical editor for Applied Mechanics Review, 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 the instructor in over 20 short courses and workshops. Recently, he served as Executive Chair of the XXIst Southeastern Conference on Theoretical and Applied Mechanics and as co-editor of Developments in Theoretical and Applied Mechanics, Volume XXI.
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Table of Contents Chapter 1 1.1
1.2
1.3
1.4
Introduction....................................................................................................1 1.1.1 Range and Summation Convention ................................................1 1.1.2 Substitution Operator ......................................................................1 Vectors............................................................................................................2 1.2.1 Notation ...........................................................................................2 1.2.2 Gradient, Divergence, and Curl ......................................................4 Matrices..........................................................................................................5 1.3.1 Eigenvalues and Eigenvectors.........................................................8 1.3.2 Coordinate Transformations............................................................9 1.3.3 Transformations of Vectors .............................................................9 1.3.4 Orthogonal Curvilinear Coordinates.............................................11 1.3.5 Gradient Operator..........................................................................16 1.3.6 Divergence and Curl of Vectors....................................................17 Appendix I: Divergence and Curl of Vectors in Orthogonal Curvilinear Coordinates................................................................................18 Derivatives of Base Vectors ..........................................................18 Divergence.....................................................................................19 Curl ................................................................................................20 Exercises ......................................................................................................20
Chapter 2 2.1 2.2
2.3 2.4 2.5 2.6
Mathematical Foundations: Vectors and Matrices...............................1
Mathematical Foundations: Tensors ..................................................25
Tensors .........................................................................................................25 Divergence, Curl, and Laplacian of a Tensor .............................................27 2.2.1 Divergence.....................................................................................27 2.2.2 Curl and Laplacian ........................................................................28 Invariants......................................................................................................29 Positive Definiteness....................................................................................30 Polar Decomposition Theorem....................................................................31 Kronecker Products on Tensors...................................................................32 2.6.1 VEC Operator and the Kronecker Product ...................................32 2.6.2 Fundamental Relations for Kronecker Products...........................33 2.6.3 Eigenstructures of Kronecker Products ........................................35 2.6.4 Kronecker Form of Quadratic Products........................................36 2.6.5 Kronecker Product Operators for Fourth-Order Tensors..............36 2.6.6 Transformation Properties of VEC and TEN22 ............................37 2.6.7 Kronecker Product Functions for Tensor Outer Products ............38
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2.6.8
2.7
Kronecker Expressions for Symmetry Classes in Fourth-Order Tensors ................................................................40 2.6.9 Differentials of Tensor Invariants .................................................41 Exercises ......................................................................................................42
Chapter 3 3.1 3.2
3.3
Introduction to Variational Methods............................................................43 Newton Iteration and Arc-Length Methods ................................................47 3.2.1 Newton Iteration............................................................................47 3.2.2 Critical Points and the Arc-Length Method .................................48 Exercises ......................................................................................................49
Chapter 4 4.1
4.2
4.3 4.4 4.5 4.6 4.7 4.8
5.2 5.3
5.4 5.5 5.6
Kinematics of Deformation ...............................................................51
Kinematics ...................................................................................................51 4.1.1 Displacement .................................................................................51 4.1.2 Displacement Vector......................................................................52 4.1.3 Deformation Gradient Tensor .......................................................52 Strain ............................................................................................................53 4.2.1 F, E, EL and u in Orthogonal Coordinates....................................53 4.2.2 Velocity-Gradient Tensor, Deformation-Rate Tensor, and Spin Tensor.............................................................................56 Differential Volume Element .......................................................................60 Differential Surface Element .......................................................................61 Rotation Tensor............................................................................................63 Compatibility Conditions For EL and D .....................................................64 Sample Problems .........................................................................................67 Exercises ......................................................................................................69
Chapter 5 5.1
Introduction to Variational and Numerical Methods.........................43
Mechanical Equilibrium and the Principle of Virtual Work .............73
Traction and Stress ......................................................................................73 5.1.1 Cauchy Stress ................................................................................73 5.1.2 1st Piola-Kirchhoff Stress .............................................................75 5.1.3 2nd Piola-Kirchhoff Stress............................................................76 Stress Flux ...................................................................................................77 Balance of Mass, Linear Momentum, and Angular Momentum................79 5.3.1 Balance of Mass ............................................................................79 5.3.2 Rayleigh Transport Theorem ........................................................79 5.3.3 Balance of Linear Momentum ......................................................79 5.3.4 Balance of Angular Momentum....................................................80 Principle of Virtual Work ............................................................................82 Sample Problems .........................................................................................85 Exercises ......................................................................................................89
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Chapter 6 6.1 6.2
6.3
6.4 6.5
Stress-Strain Behavior: Classical Linear Elasticity ....................................95 Isothermal Tangent-Modulus Tensor...........................................................97 6.2.1 Classical Elasticity ........................................................................97 6.2.2 Compressible Hyperelastic Materials ...........................................97 Incompressible and Near-Incompressible Hyperelastic Materials..............99 6.3.1 Incompressibility ...........................................................................99 6.3.2 Near-Incompressibility ................................................................102 Nonlinear Materials at Large Deformation...............................................103 Exercises ....................................................................................................104
Chapter 7 7.1
7.2 7.3
7.4
9.2
9.3
Introduction to the Finite-Element Method.....................................117
Introduction................................................................................................117 Overview of the Finite-Element Method ..................................................117 Mesh Development ....................................................................................118
Chapter 9 9.1
Thermal and Thermomechanical Response.....................................107
Balance of Energy and Production of Entropy.........................................107 7.1.1 Balance of Energy .......................................................................107 7.1.2 Entropy Production Inequality ....................................................108 7.1.3 Thermodynamic Potentials..........................................................109 Classical Coupled Linear Thermoelasticity ..............................................110 Thermal and Thermomechanical Analogs of the Principle of Virtual Work ..........................................................................................113 7.3.1 Conductive Heat Transfer ...........................................................113 7.3.2 Coupled Linear Isotropic Thermoelasticity ................................114 Exercises ....................................................................................................116
Chapter 8 8.1 8.2 8.3
Stress-Strain Relation and the Tangent-Modulus Tensor ..................95
Element Fields in Linear Problems .................................................121
Interpolation Models..................................................................................121 9.1.1 One-Dimensional Members ........................................................121 9.1.2 Interpolation Models: Two Dimensions......................................124 9.1.3 Interpolation Models: Three Dimensions ...................................127 Strain-Displacement Relations and Thermal Analogs ..............................128 9.2.1 Strain-Displacement Relations: One Dimension ........................128 9.2.2 Strain-Displacement Relations: Two Dimensions ......................129 9.2.3 Axisymmetric Element on Axis of Revolution ..........................130 9.2.4 Thermal Analog in Two Dimensions..........................................131 9.2.5 Three-Dimensional Elements......................................................131 9.2.6 Thermal Analog in Three Dimensions........................................132 Stress-Strain-Temperature Relations in Linear Thermoelasticity.............132 9.3.1 Overview .....................................................................................132 9.3.2 One-Dimensional Members ........................................................132 9.3.3 Two-Dimensional Elements ........................................................133
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9.4
9.3.4 Element for Plate with Membrane and Bending Response .......135 9.3.5 Axisymmetric Element................................................................135 9.3.6 Three-Dimensional Element .......................................................136 9.3.7 Elements for Conductive Heat Transfer .....................................137 Exercises ....................................................................................................137
Chapter 10 Element and Global Stiffness and Mass Matrices ..........................139 10.1 10.2 10.3
10.3
Application of the Principle of Virtual Work............................................139 Thermal Counterpart of the Principle of Virtual Work.............................141 Assemblage and Imposition of Constraints ..............................................142 10.3.1 Rods.............................................................................................142 10.3.2 Beams ..........................................................................................146 10.3.3 Two-Dimensional Elements ........................................................147 Exercises ....................................................................................................149
Chapter 11 Solution Methods for Linear Problems ...........................................153 11.1
11.2 11.3 11.4 11.5
11.6
Numerical Methods in FEA ......................................................................153 11.1.1 Solving the Finite-Element Equations: Static Problems ............153 11.1.2 Matrix Triangularization and Solution of Linear Systems.........154 11.1.3 Triangularization of Asymmetric Matrices.................................155 Time Integration: Stability and Accuracy .................................................156 Newmark’s Method ...................................................................................157 Integral Evaluation by Gaussian Quadrature ............................................158 Modal Analysis by FEA ............................................................................159 11.5.1 Modal Decomposition .................................................................159 11.5.2 Computation of Eigenvectors and Eigenvalues ..........................162 Exercises ....................................................................................................164
Chapter 12 Rotating and Unrestrained Elastic Bodies.......................................167 12.1 12.2 12.3
Finite Elements in Rotation.......................................................................167 Finite-Element Analysis for Unconstrained Elastic Bodies......................169 Exercises ....................................................................................................171
Chapter 13 Thermal, Thermoelastic, and Incompressible Media ......................173 13.1
13.2
13.3 13.4 13.5
Transient Conductive-Heat Transfer..........................................................173 13.1.1 Finite-Element Equation .............................................................173 13.1.2 Direct Integration by the Trapezoidal Rule ................................173 13.1.3 Modal Analysis............................................................................174 Coupled Linear Thermoelasticity ..............................................................175 13.2.1 Finite-Element Equation .............................................................175 13.2.2 Thermoelasticity in a Rod...........................................................177 Compressible Elastic Media ......................................................................177 Incompressible Elastic Media ...................................................................178 Exercises ....................................................................................................180
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Chapter 14 Torsion and Buckling.......................................................................181 14.1 14.2
14.3
Torsion of Prismatic Bars..........................................................................181 Buckling of Beams and Plates ..................................................................185 14.2.1 Euler Buckling of Beam Columns..............................................185 14.2.2 Euler Buckling of Plates .............................................................190 Exercises ....................................................................................................193
Chapter 15 Introduction to Contact Problems....................................................195 15.1 15.2 15.3 15.4
Introduction: the Gap.................................................................................195 Point-to-Point Contact ...............................................................................197 Point-to-Surface Contact ...........................................................................199 Exercises ....................................................................................................199
Chapter 16 Introduction to Nonlinear FEA........................................................201 16.1 16.2 16.3 16.4
16.5 16.6
Overview....................................................................................................201 Types of Nonlinearity ................................................................................201 Combined Incremental and Iterative Methods: a Simple Example..........202 Finite Stretching of a Rubber Rod under Gravity: a Simple Example ....203 16.4.1 Nonlinear Strain-Displacement Relations...................................203 16.4.2 Stress and Tangent Modulus Relations.......................................204 16.4.3 Incremental Equilibrium Relation...............................................205 16.4.4 Numerical Solution by Newton Iteration....................................208 Illustration of Newton Iteration.................................................................211 16.5.1 Example.......................................................................................212 Exercises ....................................................................................................213
Chapter 17 Incremental Principle of Virtual Work ............................................215 17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9 17.10
Incremental Kinematics .............................................................................215 Incremental Stresses ..................................................................................216 Incremental Equilibrium Equation ............................................................217 Incremental Principle of Virtual Work ......................................................218 Incremental Finite-Element Equation .......................................................219 Incremental Contributions from Nonlinear Boundary Conditions ...........220 Effect of Variable Contact .........................................................................221 Interpretation as Newton Iteration.............................................................223 Buckling.....................................................................................................224 Exercises ....................................................................................................226
Chapter 18 Tangent-Modulus Tensors for Thermomechanical Response of Elastomers ...................................................................227 18.1 18.2 18.3
Introduction................................................................................................227 Compressible Elastomers ..........................................................................227 Incompressible and Near-Incompressible Elastomers ..............................228 18.3.1 Specific Expressions for the Helmholtz Potential ......................230
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18.4 18.5 18.6
18.7
18.8
18.9
Stretch Ratio-Based Models: Isothermal Conditions................................231 Extension to Thermohyperelastic Materials..............................................233 Thermomechanics of Damped Elastomers................................................234 18.6.1 Balance of Energy .......................................................................235 18.6.2 Entropy Production Inequality ....................................................235 18.6.3 Dissipation Potential ...................................................................236 18.6.4 Thermal-Field Equation for Damped Elastomers.......................237 Constitutive Model: Potential Functions...................................................238 18.7.1 Helmholtz Free-Energy Density .................................................238 18.7.2 Specific Dissipation Potential .....................................................239 Variational Principles.................................................................................240 18.8.1 Mechanical Equilibrium..............................................................240 18.8.2 Thermal Equilibrium ...................................................................240 Exercises ....................................................................................................241
Chapter 19 Inelastic and Thermoinelastic Materials..........................................243 19.1
19.2
19.3 19.4 19.5 19.6
Plasticity.....................................................................................................243 19.1.1 Kinematics...................................................................................243 19.1.2 Plasticity ......................................................................................243 Thermoplasticity ........................................................................................246 19.2.1 Balance of Energy .......................................................................246 19.2.2 Entropy-Production Inequality ....................................................247 19.2.3 Dissipation Potential ...................................................................248 Thermoinelastic Tangent-Modulus Tensor ................................................249 19.3.1 Example.......................................................................................250 Tangent-Modulus Tensor in Viscoplasticity ..............................................252 Continuum Damage Mechanics ................................................................254 Exercises ....................................................................................................256
Chapter 20 Advanced Numerical Methods ........................................................257 20.1
20.2 20.3
Iterative Triangularization of Perturbed Matrices .....................................257 20.1.1 Introduction .................................................................................257 20.1.2 Notation and Background ...........................................................258 20.1.3 Iteration Scheme..........................................................................259 20.1.4 Heuristic Convergence Argument ...............................................259 20.1.5 Sample Problem ..........................................................................260 Ozawa’s Method for Incompressible Materials ........................................262 Exercises ....................................................................................................263
Monographs and Texts ........................................................................................265 Articles and Other Sources.................................................................................267
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1
Mathematical Foundations: Vectors and Matrices
1.1 INTRODUCTION This chapter provides an overview of mathematical relations, which will prove useful in the subsequent chapters. Chandrashekharaiah and Debnath (1994) provide a more complete discussion of the concepts introduced here.
1.1.1 RANGE
AND
SUMMATION CONVENTION
Unless otherwise noted, repeated Latin indices imply summation over the range 1 to 3. For example: 3
ai bi =
∑a b = a b + a b i i
1 1
2 2
+ a3 b3
(1.1)
i =1
aij bjk = ai1b1k + ai 2 b2 k + ai 3b3k
(1.2)
The repeated index is “summed out” and, therefore, dummy. The quantity aijbjk in th Equation (1.2) has two free indices, i and k (and later will be shown to be the ik entry of a second-order tensor). Note that Greek indices do not imply summation. Thus, aα bα = a1b1 if α = 1.
1.1.2 SUBSTITUTION OPERATOR The quantity, δij, later to be called the Kronecker tensor, has the property that 1 δ ij = 0
i= j i≠ j
(1.3)
For example, δijvj = 1 × vi, thus illustrating the substitution property.
1
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2
Finite Element Analysis: Thermomechanics of Solids
3 v3
v e3 e2
v2
2
e1 v1 1 FIGURE 1.1 Rectilinear coordinate system.
1.2 VECTORS 1.2.1 NOTATION Throughout this and the following chapters, orthogonal coordinate systems will be used. Figure 1.1 shows such a system, with base vectors e1, e2, and e3. The scalar product of vector analysis satisfies e i ⋅ e j = δ ij
(1.4)
e k i ≠ j and ijk in right-handed order e i × e j = − e k i ≠ j and ijk not in right-handed order 0 i = j
(1.5)
The vector product satisfies
It is an obvious step to introduce the alternating operator, εijk, also known as the th ijk entry of the permutation tensor:
ε ijk = [e i × e j ] ⋅ e k 1 = −1 0
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ijk distinct and in right-handed order ijk distinct but not in right-handed order ijk not distinct
(1.6)
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Mathematical Foundations: Vectors and Matrices
3
Consider two vectors, v and w. It is convenient to use two different types of notation. In tensor indicial notation, denoted by (*T), v and w are represented as v = vi e i
*T)
w = wi e i
(1.7)
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 this text, the “default” is matrix-vector (*M) notation, illustrated by v1 v = v2 v3
*M)
w1 w = w2 w3
(1.8)
It is compact, but also risks confusion by not displaying the underlying base T T vectors. In *M notation, the transposes v and w are also introduced; they are displayed as “row vectors”: *M)
v T = {v1
v2
v3}
w T = {w1
w2
w3}
(1.9)
The scalar product of v and w is written as v ⋅ w = (vi e i ) ⋅ ( w j e j ) = vi w j e i ⋅ e j *T)
= vi w jδ ij = vi wi
(1.10)
v = v⋅v
(1.11)
The magnitude of v is defined by *T)
The scalar product of v and w satisfies *T)
v ⋅ w = v w cos θ vw
(1.12)
in which θvw is the angle between the vectors v and w. The scalar, or dot, product is *M)
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v ⋅ w → vTw
(1.13)
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4
Finite Element Analysis: Thermomechanics of Solids
The vector, or cross, product is written as v × w = vi w j e i × e j *T)
= ε ijk vi w j e k
(1.14)
Additional results on vector notation are presented in the next section, which introduces matrix notation. Finally, the vector product satisfies v × w = v w sin θ vw
*T)
(1.15)
and vxw is colinear with n 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 12 v × w .
1.2.2 GRADIENT, DIVERGENCE,
AND
CURL
The derivative, dφ/dx, of a scalar φ with respect to a vector x is defined implicitly by dφ =
*M)
dφ dx dx
(1.16)
and it is a row vector whose i entry is dφ/dxi. In three-dimensional rectangular coordinates, the gradient and divergence operators are defined by th
*M)
∂( ) ∂x ∂( ) ∇( ) = ∂y ∂ ( ) ∂z
(1.17)
and clearly, T
*M)
d ( ) = ∇( ) dx
(1.18)
The gradient of a scalar function φ satisfies the following integral relation:
∫ ∇φ dV = ∫ nφ dS
(1.19)
The expression ∇v T will be seen to be a tensor (see Chapter 2). Clearly, ∇v T = [∇v1
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∇v2
∇v3 ]
(1.20)
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Mathematical Foundations: Vectors and Matrices
5
from which we obtain the integral relation
∫ ∇v dV = ∫ nv dS T
T
(1.21)
Another 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 *M)
∫ dx dV = ∫ n vdS dv
T
(1.22)
The curl of vector v, ∇ × v, is expressed by (∇ × v)i = ε ijk
∂ v ∂x j k
(1.23)
which is the conventional cross-product, except that the divergence operator replaces the first vector. The curl satisfies the curl theorem, analogous to the divergence theorem (Schey, 1973):
∫ ∇ × v dV = ∫ n × v dS
(1.24)
Finally, the reader may verify, with some effort that, for a vector v(X) and a path X(S) in which S is the length along the path,
∫ v ⋅ dX(S) = ∫ n ⋅ ∇ × v dS .
(1.25)
The integral between fixed endpoints is single-valued if it is path-independent, in which case n ⋅ ∇ × v must vanish. However, n is arbitrary since the path is arbitrary, thus giving the condition for v to have a path-independent integral as ∇ × v = 0.
(1.26)
1.3 MATRICES An n × n matrix is simply an array of numbers arranged in rows and columns, also known as a second-order array. For the matrix A, the entry aij occupies the intersection th th of the i row and the j column. We may also introduce the n × 1 first-order array a, th T in which ai denotes the i entry. We likewise refer to the 1 × n array, a , as first-order.
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Finite Element Analysis: Thermomechanics of Solids
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 are real unless otherwise noted. Several properties of first- and second-order arrays are as follows: The sum of two n × n matrices, A and B, is a matrix, C, in which cij = aij + bij. The product of a matrix, A, and a scalar, q, is a matrix, C, in which cij = qaij. T The transpose of a matrix, A, denoted A , is a matrix in which aijT = a ji A is T T called symmetric if A = A , and it is called antisymmetric if A = −A . The product of two matrices, A and B, is the matrix, C, for which *T)
cij = aik bkj
(1.27)
Consider the following to visualize matrix multiplication. Let the first-order th th array a iT denote the i row of A, while the first-order array bj denotes the j column of B. Then cij can be written as *T)
cij = a iT b j
(1.28)
The product of a matrix A and a first-order array c is the first-order array d th in which the i entry is di = aijcj. th The ij entry of the identity matrix I is δij. Thus, it exhibits ones on the diagonal positions (i = j) and zeroes off-diagonal (i ≠ j). Thus, I is the matrix counterpart of the substitution operator. The determinant of A is given by
*T)
det(A) =
1 ε ε a a a 6 ijk pqr ip jq kr
(1.29)
Suppose a and b are two non-zero, first-order n × 1 arrays. If det(A) = 0, the matrix A is singular, in which case there is no solution to equations of the form Aa = b. However, if b = 0, there may be multiple solutions. If det(A) ≠ 0, then there is a unique, nontrivial solution a. Let A and B be n × n nonsingular matrices. The determinant has the following useful properties: det(AB) = det(A) det(B) *M)
det(A T ) = det(A) det(I) = 1
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(1.30)
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Mathematical Foundations: Vectors and Matrices
7 −1
If det(A) ≠ 0, then A is nonsingular and there exists an inverse matrix, A , for which *M)
AA −1 = A −1A = I
(1.31)
The transpose of a matrix product satisfies *M)
(AB) T = BT A T
(1.32)
The inverse of a matrix product satisfies *M)
(AB) −1 = B −1A −1
(1.33)
If c and d are two 3 × 1 vectors, the vector product c × d generates the vector c × d = Cd, in which C is an antisymmetric matrix given by
*M)
0 C = c3 − c2
−c3 0 c1
c2 −c1 0
(1.34)
Recalling that c × d = εikjckdj, and noting that εikjck denotes the (ij) component of an antisymmetric tensor, it is immediate that [C]ij = εikjck. th
T
If c and d are two vectors, the outer product cd generates the matrix C given by
*M)
c1d1 C = c2 d1 c3 d1
c1d2 c2 d 2 c3 d2
c1d3 c2 d 3 c3 d3
(1.35)
We will see later that C is a second-order tensor if c and d have the transformation properties of vectors. An n × n matrix A can be decomposed into symmetric and antisymmetric matrices using A = As + Aa ,
© 2003 by CRC CRC Press LLC
As =
1 [A + A T ], 2
Aa =
1 [A − A T ] 2
(1.36)
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Finite Element Analysis: Thermomechanics of Solids
1.3.1 EIGENVALUES
AND
EIGENVECTORS
In this case, A is again an n × n tensor. The eigenvalue equation is ( A − λ j I )x j = 0
(1.37)
The solution for xj is trivial unless A − λjI is singular, in which event det(A − λjI) = 0. There are n possible complex roots. If the magnitude of the eigenvectors is set to unity, they may likewise be determined. As an example, consider 2 A= 1
1 2
(1.38)
The equation det(A − λjI) = 0 is expanded as (2 − λj) − 1, with roots λ1,2 = 1, 3, and 2
1 A − λ1I = 1
−1 A − λ2 I = 1
1 1
1 −1
(1.39)
Note that in each case, the rows are multiples of each other, so that only one row is 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. T Accordingly, the magnitudes are arbitrarily set to unity. For x1 = {x11 x12} , x11 + x12 = 0
2 2 x11 + x12 =1
(1.40)
from which we conclude that x1 = {1 −1}T / 2 . A parallel argument furnishes x 2 = {1 1}T / 2 . If A is symmetric, the eigenvalues and eigenvectors are real and the eigenvectors are orthogonal to each other: x iT x j = δ ij . The eigenvalue equations can be “stacked up,” as follows. A[x1 : x 2 :K x n ] = [x1 : x 2 :K x n ]
λ1
0
.
.
0
λ2
.
.
.
.
.
.
.
.
.
λ n−1
.
.
.
0
. . . (1.41) 0 λ n
With obvious identifications, AX = XΛ
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(1.42)
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9
and X is the modal matrix. Let yij represent the ij entry of Y = X X th
T
yij = x iT x j = δ ij
(1.43) −1
so that Y = I. We can conclude that X is an orthogonal tensor: X = X . Further, T
X T AX = Λ
A = X Λ XT
(1.44)
and X can be interpreted as representing a rotation from the reference axes to the principal axes.
1.3.2 COORDINATE TRANSFORMATIONS Suppose that the vectors v and w are depicted in a second-coordinate system whose base vectors are denoted by e ′j . Now, e ′j can be represented as a linear sum of the base vectors ei: e ′j = q ji e i
*T)
(1.45)
But then e i ⋅ e ′j = qij = cos(θ ij ′ ). It follows that δij = e ′i ⋅ e ′j = (qikek) ⋅ (qjlel) = qikqjlδkl, so that qik q jk = qik qkjT *T)
= δ ij
In *M) notation, this is written as QQ T = I
*M)
(1.46)
in which case the matrix Q is called orthogonal. An analogous argument proves that T T T 2 Q Q = I. From Equation (1.30), 1 = det(QQ ) = det(Q)det(Q ) = det (Q). Righthanded rotations satisfy det(Q) = 1, in which case Q is called proper orthogonal.
1.3.3 TRANSFORMATIONS
OF
VECTORS
The vector v′ is the same as the vector v, except that v′ is referred to e ′j , while v is referred to ei. Now v ′ = v ′j e ′j *T)
= v ′j q ji e i = vi e i
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(1.47)
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Finite Element Analysis: Thermomechanics of Solids
It follows that vi = v ′j q ji , and hence *M)
v = Q T v ′ (a )
th
v ′ = Qv ( b)
(1.48)
T
in which qij is the ji entry of Q . We can also state an alternate definition of a vector as a first-order tensor. Let v be an n × 1 array of numbers referring 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 e ′j , v′ transforms according to Equation (1.48). Since ( ddφx )′ dx ′ is likewise equal to dφ,
*M)
dφ ′ = dφ Q T dx dx
(1.49)
for which reason dφ/dx is called a contravariant vector, while v is properly called a covariant 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
(1.50)
with the matrix-vector counterpart e i e Tj . Now A = aij e i ∧ e j
(1.51)
Note the useful result that e i ∧ e j ⋅ e k = e iδ jk In this notation, given a vector b = bkek, Ab = aij e i ∧ e j ⋅ bk e k = aij bk e i ∧ e j ⋅ e k = aij bk e iδ jk = aij b j e i as expected.
© 2003 by CRC CRC Press LLC
(1.52)
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11
1.3.4 ORTHOGONAL CURVILINEAR COORDINATES The position vector of a point, P, referring to a three-dimensional, rectilinear, coordinate system is expressed in tensor-indicial notation as RP = xiei. The position vector connecting two “sufficiently close” points P and Q is given by ∆ R = R P − R Q ≈ dR x
(1.53)
dR x = dxi e i
(1.54)
dSx = dxi dxi
(1.55)
where
with arc length
Suppose now that the coordinates are transformed to yj coordinates: xi = xi(yj). The same position vector, now referred to the transformed system, is 3
dR y =
∑ dy g
α α
1
gα = hα γ α dx j dx j
hα =
dyα dyα dxi dyα
γα =
=
dx j dx j dyα dyα
(1.56)
ei
1 dxi e hα dyα i
in which hα is called the scale factor. Recall that the use of Greek letters for indices implies no summation. Clearly, γα is a unit vector. Conversely, if the transformation is reversed, dR y = g i dyi =
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dyi g dx dx j i j
(1.57)
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Finite Element Analysis: Thermomechanics of Solids
then the consequence is that ej =
dyi g = dx j i
∑ dx
dyα
α
hα γ α
(1.58)
j
We restrict attention to orthogonal coordinate systems yj, with the property that γ αT γ β = δ αβ
(1.59)
The length of the vector dRy is now dSy = hα dyi dyi
(1.60)
Under restriction to orthogonal coordinate systems, the initial base vectors ei can be expressed in terms of γα using
(
)
e i = γ Tj e i γ j =
1 ∂xi γ hi ∂y j j
=
1 ∂xi ∂x k e hi h j ∂y j ∂y j k
(1.61)
and furnishing 1 ∂xi ∂x k = δ ik hi h j ∂y j ∂y j
(1.62)
Also of interest is the volume element; the volume determined by the vector dRy is given by the vector triple product dVy = (h1dy1 γ 1 ) ⋅ [h2 dy2 γ 2 × h3 dy3 γ 3 ] = h1h2 h3 dy1dy2 dy3
(1.63)
and h1h2h3 is known as the Jacobian of the transformation. For cylindrical coordinates using r, θ, and z, as shown in Figure 1.2, x1 = rcos θ, x2 = rsin θ, and x3 = z. Simple manipulation furnishes that hr = 1, hθ = r, hz = 1, and e r = cos θ e1 + sin θ e2
eθ = − sin θ e1 + cos θ e2
e z = e3
(1.64)
which, of course, are orthonormal vectors. Also of interest are the relations der = eθ dθ and deθ = −er dθ.
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13
x3
ez x2 eθ
er θ x1
r
FIGURE 1.2 Cylindrical coordinate system.
Transformation of the coordinate system from rectilinear to cylindrical coordinates can be viewed as a rotation of the coordinate system through θ. Thus, if the vector v is referred to the reference rectilinear system and v′ is the same vector referred to a cylindrical coordinate system, then in two dimensions, cos θ Q(θ ) = − sin θ 0
v ′ = Q(θ )v
sin θ cos θ 0
0 0 1
(1.65)
If v′ is differentiated, for example, with respect to time t, there is a contribution from the rotation of the coordinate system: for example, if v and θ are functions of time t, d d dQ(θ ) v ′ = Q(θ ) v + v dt dt dt =
∂ dQ(θ ) T v′ + Q (θ )v ′ ∂t dt
(1.66)
where the partial derivative implies differentiation with θ instantaneously held fixed and − sin θ dQ(θ ) = − cos θ dt 0
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cos θ − sin θ 0
0 dθ 0 dt 1
(1.67)
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Finite Element Analysis: Thermomechanics of Solids
x3 eφ
er φ
π φ
φ
x2 eθ
θ
θ
x1
FIGURE 1.3 Spherical coordinate system.
Now since
dQ (θ ) dt
Q T (θ ) is an antisymmetric matrix Ω (to be identified later as a tensor)
0=
d dQ(θ ) T dQ(θ ) T Q (θ ) + Q (θ ) (Q(θ )Q T (θ )) = dt dt dt
T
(1.68)
In fact, 0 dQ(θ ) T Q (θ ) = −1 dt 0
1 0 0
0 dθ 0 dt 0
(1.69)
It follows that d ∂ v′ = v′ + ω × v′ dt ∂t
(1.70)
in which ω is the axial vector of Ω. Referring to Figure 1.3, spherical coordinates r, θ, and φ are introduced by the transformation x1 = r cos θ cos φ
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x 2 = r sin θ cos φ
x3 = r sin φ
(1.71)
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15
The position vector is given by r = x1e1 + x 2 e 2 + x3 e 3 = r cos θ cos φ e1 + r sin θ cos φ e 2 + r sin φ e 3
(1.72)
Now er has the same direction as the position vector: r = rer. Thus, it follows that e r = cos θ cos φ e1 + sin θ cos φ e 2 + sin φ e 3
(1.73)
Following the general procedure in the preceding paragraphs, ∂x1 = cos θ cos φ ∂r ∂x 2 = sin θ cos φ ∂r ∂x3 = sin φ ∂r
∂x1 = − r sin θ cos φ ∂θ ∂x 2 = r cos θ cos φ ∂θ ∂x3 =0 ∂θ
∂x1 = − r cos θ sin φ ∂φ ∂x 2 = − r sin θ sin φ ∂φ ∂x3 = r cos φ ∂φ
(1.74)
The differential of the position vector furnishes dr = dr e r + r cos φ dθ eθ + rdφ eφ
(1.75)
e r = cos θ cos φ e1 + sin θ cos φ e 2 + sin φ e 3 e1 = cos θ cos φ e r − sin θ eθ − sin φ cos θ eφ eθ = − sin θ e1 + cos θ e 2
e 2 = sin θ cos φ e r + cos θ eθ − sin φ sin θ eφ
eφ = − sin φ[cos θ e1 + sin θ e 2 ] + cos φ e 3
e 3 = sin φ e r + cos φ eφ .
The scale factors are hr = 1, hθ = rcos φ, hφ = r. Consider a vector v in the rectilinear system, denoted as v′ when referred to a spherical coordinate system: v = v1e1 + v2 e 2 + v3 e 3
v ′ = vr e r + vθ eθ + vφ eφ .
(1.76)
Eliminating e1, e2, e3 in favor of er, eθ, eφ and using *M notation permits writing
v ′ = Q(θ , φ )v,
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cos θ cos φ Q(θ , φ ) = − sin θ − sin φ cos θ
sin θ cos φ cos θ − sin φ sin θ
sin φ 0 . cos φ
(1.77)
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Finite Element Analysis: Thermomechanics of Solids
Suppose now that v(t), θ, and φ are functions of time. As in cylindrical coordinates, d ∂ v′ = v′ + ω × v′ dt ∂t dQ (θ ) dt
where ω is the axial vector of
Q T (θ ). After some manipulation,
− sin θ cos φ dQ(θ ) = − cos θ dt sin θ sin φ
cos θ cos φ
0
− sin θ
0
− cos θ sin φ
0
− cos θ sin φ + 0 − cos θ cos φ
− sin θ sin φ 0
dθ dt
cos φ
dφ dt − sin φ
0
− sin θ cos φ
cos φ
0 dQ(θ ) T Q (θ ) = − cos φ dt 0
(1.78)
0
0 dθ + 0 sin φ dt −1 0
0 − sin φ
(1.79)
0
1
0
0
0
0
dφ dt
1.3.5 GRADIENT OPERATOR In rectilinear coordinates, let ψ be a scalar-valued function of x: ψ (x), starting with the chain rule dψ = *T)
∂ψ dx ∂xi i
= [∇ψ ] ⋅ dr,
dr = e i dxi
∇ψ = e i
∂ψ ∂xi
(1.80)
Clearly, dψ is a scalar and is unaffected by a coordinate transformation. Suppose that x = x(y): dr′ = gi dyi. Observe that dψ = =
∂ψ dx ∂xi i
=
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∂ψ
∑ h1 ∂y α
∑
γ α ∂ψ ⋅ hα ∂yα
α
α
hα dyα
α
∑ β
hβ dyβ γ β
(1.81)
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17
implying the identification (∇ψ )′ =
γ α ∂ψ α ∂yα
∑h α
(1.82)
For cylindrical coordinates in tensor-indicial notation with er = γr, eθ = γθ, ez = γz, ∂ψ ∂ψ eθ ∂ψ + + ez ∂r ∂z r ∂θ
(1.83)
eθ ∂ψ e z ∂ψ ∂ψ + + ∂r r cos φ ∂θ r ∂φ
(1.84)
∇ψ = e r and in spherical coordinates ∇ψ = e r
1.3.6 DIVERGENCE
AND
CURL
OF
VECTORS
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 at the end of the chapter. Here, we simply list the results. Let v be a vector referred to rectilinear coordinates, and let v′ denote the same vector referred to orthogonal coordinates. The divergence and curl satisfy (∇ ⋅ v ) ′ =
1 h1h2 h3
∂ ∂ ∂ (h2 h3 v1′ ) + (h3 h1v2′ ) + (h1h2 v3′ ) ∂ ∂ ∂ y y y 2 3 1
(1.85)
and
(∇ × v ) ′ =
1 h1h2 h3
∂ ∂ (h3 v3′ ) − (h2 v2′ )γ 1 h1 ∂y3 ∂y2
∂ ∂ − h2 (h3 v3′ ) − (h1v1′ )γ 2 ∂y3 ∂y1 ∂ ∂ + h3 (h2 v2′ ) − (h1v1′ )γ 3 ∂y 2 ∂y1
(1.86)
and in cylindrical coordinates: (∇ ⋅ v ′ ) =
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1 ∂(rvr ) 1 ∂vθ ∂vz + + r ∂r r ∂θ ∂z
(1.87)
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Finite Element Analysis: Thermomechanics of Solids
and (∇ × v ) ′ =
1 ∂vz ∂(rvθ ) ∂(rvθ ) ∂vr ∂vz ∂vr − − − reθ + e e r − r ∂θ ∂θ z ∂z ∂z ∂r ∂r
(1.88)
APPENDIX I: DIVERGENCE AND CURL OF VECTORS IN ORTHOGONAL CURVILINEAR COORDINATES DERIVATIVES
OF
BASE VECTORS
In tensor-indicial notation, a vector v can be represented in rectilinear coordinates as v = vkek. In orthogonal curvilinear coordinates, it is written as v′ = ∑ vα′ γ α = α ∑ vα′ ghαα . α A line segment dr = dxiei transforms to dr′ = dykgk. Recall that ∂yl g = ∂x k l
ek =
gα =
∑h
∂xα e ∂yk k
β
β
∂yβ ∂x k
γβ
hα = gα ⋅ gα
(a.1)
α
(a.2)
From Equation (a.1), ∂gα ∂ 2 xα = e ∂y j ∂yk ∂y j k =
α
∑
β hβ γ β ,
j
β
j
β ∂ 2 xα ∂yβ = hβ ∂yk ∂y j ∂x k
The bracketed quantities are known as Cristoffel symbols. From Equations (a.1 and a.2), dhα dg = γα ⋅ α dy j dy j α =
Continuing,
j
α hα
(a.3)
∂γ α 1 ∂gα γ α ∂hα = − hα ∂y j hα ∂y j ∂y j =
∑
cαjβ γ β ,
β
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cαjβ =
α 1 (1 − δ αβ ) hα
j
β hβ
(a.4)
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19
DIVERGENCE The development that follows is based on the fact that dv ′ dv ∇ ⋅ v = ∇ ′ ⋅ v ′ = tr = tr dr dr ′
(a.5)
The differential of v′ is readily seen to be dv ′ = dv j γ j + v j dγ j
(a.6)
First, note that dv j γ j = =
∂v j ∂yk
γ j dyk
α
α
=
α
γ j (hα dyα )
1 ∂v j γ j ∧ γα ⋅ α ∂yα
∑ h α
=
∂v j
∑ h1 ∂y
∑ (h γ dy ) β
β
β
β
1 ∂v j γ j ∧ γ α ⋅ dr′ ∂ y α α
∑ h α
(a.7)
Similarly, vjd γ j = vj =
∂γ j
v j ∂γ j
∑ h α
=
∂yα
∧ γα ⋅
∑ (h γ dy ) β
β
β
β
v j ∂γ j
∑ α
© 2003 by CRC CRC Press LLC
α
∑ h α
=
v j ∂γ j (hα dyα ) α ∂yα
∑ h α
=
dyk
∂yk
α
v j hα
∧ γ α ⋅ dr′ ∂yα
∑ β
c jαβ γ β ∧ γ α ⋅ dr′
(a.8)
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Finite Element Analysis: Thermomechanics of Solids
Consequently, dv = 1 ∂v j δ + v c , j jβα dr βα h ∂y jβ α α
∇⋅v =
∂vα
∑ h1 ∂y α
α
α
+ v j c jαα
(a.9)
CURL In rectilinear coordinates, the individual entries of the curl can be expressed as a th divergence, as follows. For the i entry, [∇ × v]i = ε ijk =
∂vk ∂x j
∂ (i ) w ∂x j j
w (ji ) = ε jki vk
= ∇ ⋅ w (i )
(a.10)
Consequently, the curl of v can be written as ∇ ⋅ w (1) ∇ × v = ∇ ⋅ w (2) ∇ ⋅ w ( 3)
(a.11)
The transformation properties of the curl can be readily induced from Equation (a.9).
1.4 EXERCISES 1. In the tetrahedron shown in Figure 1.4, A1, A2, and A3 denote the areas of the faces whose normal vectors point in the −e1, −e2, and −e3 directions. Let A and n denote the area and normal vector of the inclined face, respectively. Prove that n=
A A1 A e + 2e + 3e A 1 A 2 A 3
2. Prove that if σ is a symmetric tensor with entries σij, that
ε ijk σ jk = 0,
i = 1, 2, 3.
3. If v and w are n × 1 vectors, prove that v × w can be written as v × w = Vw
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21
3
A n e3 A1
A2 A3
e1
e2
2
1 FIGURE 1.4 Geometry of a tetrahedron.
in which V is an antisymmetric tensor and v is the axial vector of V. Derive the expression for V. 4. Find the transposes of the matrices 1 A= −1/ 3
−1/ 2 1/ 4
1 B= 1/ 2
1/ 3 1/ 4
(a) Verify that AB ≠ BA. T T T (b) Verify that (AB) = B A . 5. Consider a matrix C given by a C= c
b d
Verify that its inverse is given by C−1 =
1 d ad − bc −c
−b a
6. For the matrices in Exercise 4, find the inverses and verify that (AB) −1 = B −1 A −1 7. Consider the matrix cos θ Q= − sin θ
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sin θ cos θ
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Finite Element Analysis: Thermomechanics of Solids
Verify that T T (a) QQ = Q Q T −1 (b) Q = Q (c) For any 2 × 1 vector a Qa = a [The relation in (c) is general, and Qa represents a rotation of a.] 8. Using the matrix C from Exercise 5, and introducing the vectors (onedimensional arrays) q a= r
s b= t
verify that a T Cb = b T C T a 9. Verify the divergence theorem using the following block, where x − y v= x + y Y
1
1
X
FIGURE 1.5 Test figure for the divergence theorem.
10. For the vector and geometry of Exercise 9, verify that
∫ n × vdS = ∫ ∇ × vdV © 2003 by CRC CRC Press LLC
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23
11. Using the geometry of Exercise 9, verify that
∫ n × AdS = ∫ ∇ × A dV T
using a11 = x + y + x 2 + y 2 a12 = x + y + x 2 − y 2 a21 = x + y − x 2 − y 2 a22 = x − y − x 2 − y 2 12. Obtain the expressions for the gradient, divergence, and curl in spherical coordinates.
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2
Mathematical Foundations: Tensors
2.1 TENSORS We now consider two n × 1 vectors, v and w, and an n × n matrix, A, such that v = Aw. We now make the important assumption that the underlying information in this relation is preserved under rotation. In particular, simple manipulation furnishes that v′ = Qv ∗
)
= QAw = QAQTQw = QAQT w ′.
(2.1)
The square matrix A is now called a second-order tensor if and only if A′ = QAQ . Let A and B be second-order n × n tensors. The manipulations that follow −1 T demonstrate that A , (A + B), AB, and A are also tensors. T
(A T )′ = (QAQT )T T
= QT ATQT
(2.2)
A ′ B′ = (QAQ T )(QBQ T ) = QA(QQ T )BQ T = QABQ T
(2.3)
(A + B)′ = A ′ + B′ = QAQ T + QBQ T = Q(A + B)Q T
(2.4)
A′ −1 = (QAQT )−1 −1
= Q T A −1Q −1 = QA −1Q T .
(2.5)
25
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Finite Element Analysis: Thermomechanics of Solids
Let x denote an n × 1 vector. The outer product, xx , is a second-order tensor since T
( xx T )′ = x ′x ′ T = (Qx )(Qx ) T = Q( xx T )Q T
(2.6)
Next, d dφ H = . dx d x T
d 2φ = dx T H dx
(2.7)
However, dx ′ T H ′ dx ′ = (Q dx ) T H ′Q dx = dx T (Q T H ′Q) dx,
(2.8)
from which we conclude that the Hessian H is a second-order tensor. Finally, let u be a vector-valued function of x. Then, du = ∂u dx, from which ∂x
∂u du T = dx T ∂x
T
(2.9)
and also T
∂ du T = dx T u T . ∂x
(2.10)
We conclude that T
∂u T ∂u = T. ∂x ∂x
(2.11)
Furthermore, if du′ is a vector generated from du by rotation in the opposite sense from the coordinate axes, then du′ = Qdu and dx = Qdx′. Hence, Q is a tensor. ′ Also, since du ′ = ∂u ′ dx ′ , it is apparent that ∂x
∂u ′ ∂u T =Q Q , ∂x ′ ∂x from which we conclude that are tensors.
© 2003 by CRC CRC Press LLC
∂u ∂x
(2.12)
is a tensor. We can similarly show that I and 0
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27
2.2 DIVERGENCE, CURL, AND LAPLACIAN OF A TENSOR Suppose A is a tensor and b is an arbitrary, spatially constant vector of compatible dimension. The divergence and curl of a vector have already been defined. For later purposes, we need to extend the definition of the divergence and the curl to A.
2.2.1 DIVERGENCE Recall the divergence theorem ∫ c Tn dS = ∫ ∇Tc dV. Let c = A T b, in which b is an arbitrary constant vector. Now
∫
b T An dS = =
∫ ∇ (A b)dV T
T
∫ ∇ A dV b T
T
∫
= b T [∇ T A T ]T dV .
(2.13)
Consequently, we must define the divergence of A such that *
∫ AndS − ∫ [∇ A ] dV = 0.
)
T
T T
(2.14)
In tensor-indicial notation,
∫ b a n dS − ∫ b [[∇ A ] ] dV = 0. T
i ij j
T T
i
i
(2.15)
Application of the divergence theorem to the vector cj = bi aij furnishes
∂ bi aij − [∇ T A T ]T ∂x j
[
∫
] dV = 0. i
(2.16)
Since b is arbitrary, we conclude that [∇ T A T ]i =
∂ ∂ aij = a T. ∂x j ∂x j ji
(2.17)
Thus, if we are to write ∇ ⋅ A as a (column) vector, mixing tensor- and matrixvector notation, ∇ ⋅ A = [∇ T A T ]T .
© 2003 by CRC CRC Press LLC
(2.18)
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Finite Element Analysis: Thermomechanics of Solids
It should be evident that ( ∇ ⋅) has different meanings when applied to a tensor as opposed to a vector. Suppose A is written in the form α 1T A = α T2 , T α3
(2.19)
th
in which α iT corresponds to the i row of A:[α iT ] j = aij . It is easily seen that ∇ T A T = (∇ T α 1
2.2.2 CURL
AND
∇ Tα 2
∇ T α 3 ).
(2.20)
LAPLACIAN
The curl of vector c satisfies the curl theorem ∫ ∇ × c dV = ∫ n × c dS. Using tensorindicial notation,
∫ n × cdS = ∫ ε
ijk
n j akl bl dS
= ε ijk n j akl dS bl
∫
=
∫ c n dSb , ij
j
l
cij = ε ijk akl bl .
(2.21)
From the divergence theorem applied to the tensor cij = εijk akl bl, n × AbdS = i
∫
∂
∫ ∂x
(ε ijk akl bl )dV j
∂a = ε ijk kl dV bl ∂x j
∫
= ∇ × A T dV b , i
∫
if [∇ × A]il = ε ijk
∂ a . ∂x j kl
(2.22)
Let α lT denote the row vector (array) corresponding to the l row of A: [α lT ]k = alk . It follows that th
[
∇ × A = ∇ × α1
© 2003 by CRC CRC Press LLC
∇ × α2
]
∇ × α3 .
(2.23)
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Mathematical Foundations: Tensors
29
th If βI is the array for the I column of A, then
[
∇ × A T = ∇ × β1
]
∇ × β2
∇ × β3 .
(2.24)
The Laplacian applied to A is defined by [∇ 2 A]ij = ∇ 2 aij .
(2.25)
It follows, therefore, that ∇ 2 A = [∇ 2 β 1
∇ 2β 2
∇ 2β 3 ].
(2.26)
The vectors β i satisfy the Helmholtz decomposition ∇ 2β i = ∇(∇ ⋅ β i ) − ∇ × ∇ × β i .
(2.27)
Observe from the following results that ∇ 2 A = ∇(∇ ⋅ A T ) − ∇ × [∇ × A T ]T .
(2.28)
An integral theorem for the Laplacian of a tensor is now found as
∫ ∇ AdV = ∫ (n∇ )AdS − ∫ n × [∇ × A ] dS. 2
T
T T
(2.29)
2.3 INVARIANTS Letting A denote a nonsingular, symmetric, 3 × 3 tensor, the equation det(A − λl) = 0 can be expanded as
λ3 − I1λ2 + I2 λ − I3 = 0,
(2.30)
in which I1 = tr(A)
I2 =
1 2 [tr (A) − tr(A 2 )] 2
I3 = det(A).
(2.31)
Here, tr (A) = δijaij denotes the trace of A. Equation 2.30 also implies the CayleyHamilton theorem: A 3 − I1A 2 + I2 A − I3 I = 0,
© 2003 by CRC CRC Press LLC
(2.32)
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Finite Element Analysis: Thermomechanics of Solids
from which 1 I3 = [tr(A 3 ) − I1tr(A 2 ) + I2 tr(A)] 3 A
−1
−1 3
(2.33)
= I [A − I1A + I2 I] 2
The trace of any n × n symmetric tensor B is invariant under orthogonal transformations (rotations), such as tr(B′) = tr(B), since a ′pqδ pq = q pr qqs arsδ pq = ars q pr qqs = arsδ rs . 2
3
(2.34) 2
3
Likewise, tr(A ) and tr(A ) are invariant since A, A , and A are tensors, thus I1, I2, and I3 are invariants. Derivatives of invariants are presented in a subsequent section.
2.4 POSITIVE DEFINITENESS In the finite-element method, an attractive property of some symmetric tensors is positive definiteness, defined as follows. The symmetric n × n tensor A is positivedefinite, written A > 0, if, for all nonvanishing n × 1 vectors x, the quadratic product T q(A, x) = x Ax > 0. The importance of this property is shown in the following T T example. Let Π = 12 x Ax − x f, in which f is known and A > 0. After some simple manipulation, d T d d 2 Π = dx T Π dx dx dx = dx T A dx.
(2.35)
It follows that Π is a globally convex function that attains a minimum when Ax = f (dΠ = 0). The following definition is equivalent to the statement that the symmetric n × n tensor A is positive-definite if and only if its eigenvalues are positive. For the sake of demonstration, x T Ax = x T XΛΛT x = y T Λy, =
∑λ y . 2 i i
i
© 2003 by CRC CRC Press LLC
(y = X T x ) (2.36)
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31
The last expression can be positive for arbitrary y (arbitrary x) only if λi > 0, i = T 1, 2,…, n. The matrix A is semidefinite if x Ax ≥ 0, and negative-definite (written T T T A < 0), if x Ax < 0. If B is a nonsingular tensor, then B B > 0, since q(B B, x) = T T T x B Bx = y y > 0 (in which y = Bx and Ω denotes the quadratic product). If B is Τ Τ singular, for example if B = yy where y is an n × 1 vector, B B is positive-semidefinite since a nonzero eigenvector x of B can be found for which the quadratic product T q(B B, x) vanishes. Now suppose that B is a nonsingular, antisymmetric tensor. Multiplying through T Bxj = λ j xj with B furnishes BT Bx j = λ j BT x j = − λ j Bx j = − λ2j x j .
(2.37)
Τ
Since B B is positive-definite, it follows that − λ2j > 0. Thus, λ j is imaginary: 2 λ j = iµ j using i = −1. Consequently, B2 x j = λ2j x j = − µ 2j x j, demonstrating that B is negative-definite.
2.5 POLAR DECOMPOSITION THEOREM For an n × n matrix B, B B > 0. If the modal matrix of B is denoted by Xb, we can write T
BT B = X Tb ∆ b X b 1
1
= X Tb ( ∆ b ) 2 YY T ( ∆ b ) 2 X b T
1 1 = X Tb ( ∆ b ) 2 Y X Tb ( ∆ b ) 2 Y ,
(2.38a)
in which Y is an (unknown) orthogonal tensor. In general, we can write 1
B = Y T (∆ b ) 2 X b .
(2.38b)
To “justify” Equation 2.38b, we introduce the square root
1 2 b
BT B = X Tb ∆ X b ,
© 2003 by CRC CRC Press LLC
1 ∆ b2 =
λ1
0
.
.
0
λ2
.
.
.
.
.
.
.
.
.
.
.
.
.
0
BT B using . . . , 0 λn
(2.38c)
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Finite Element Analysis: Thermomechanics of Solids
in which the positive square roots are used. It is easy to verify that ( BT B ) 2 = B and that BT B > 0 . Note that B B T B
(
)
−
1 2
B B T B
(
)
−
1 2
T
=
(
T
B B
)
−
1 2
T [B B]
(
T
B B
)
−
= I.
1 2
(2.38d)
Thus, B( BT B ) −1/2 is an orthogonal tensor, called, for example, Z, and hence we can write B = Z BT B 1
= ZX Tb ∆ b2 X b .
(2.38e)
Finally, noting that (ZX Tb )(ZX Tb ) T = Z(X Tb X b )Z T = ZZ T = I , we make the identification Y T = ZX Tb in Equation 2.38b. Equation 2.38 plays a major role in the interpretation of strain tensors, a concept that is introduced in subsequent chapters.
2.6 KRONECKER PRODUCTS ON TENSORS 2.6.1 VEC OPERATOR
AND THE
KRONECKER PRODUCT
Let A be an n × n (second-order) tensor. Kronecker product notation (Graham, 1981) reduces A to a first-order n × 1 tensor (vector), as follows. VEC(A) = {a11
a21
a31
L
an,n−1
ann}T .
(2.39)
The inverse VEC operator, IVEC, is introduced by the obvious relation IVEC(VEC(A)) = A. The Kronecker product of an n × m matrix A and an r × s matrix B generates an nr × ms matrix, as follows. a11B a21B A⊗B= . . a B n1
a12 B
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a1m B . . . . anm B
(2.40)
If m, n, r, and s are equal to n, and if A and B are tensors, then A ⊗ B 2 2 transforms as a second-order n × n tensor in a sense that is explained subsequently.
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33
Equation 2.40 implies that the n × 1 Kronecker product of two n × 1 vectors a and b is written as 2
a1b a2 b a⊗b = . . . an b
2.6.2 FUNDAMENTAL RELATIONS
FOR
(2.41)
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). th Relation 1: Let A denote an n × m real matrix, with entry aij in the i row and th j column. Let I = (j − 1)n + i and J = (i − 1)m + j. Let Unm denote the nm × nm matrix, independent of A, satisfying 1, u JK = 0,
K=I K≠I
1, u IK = 0,
K=J K≠J
.
(2.42)
Then, VEC(A T ) = U nm VEC(A).
(2.43)
Note that uJK = uJI = 1 and uIK = uIJ = 1, with all other entries vanishing. Hence if m = n, then uJI = uIJ, so that Unm is symmetric if m = n. Relation 2: If A and B are second-order n × n tensors, then tr(AB) = VEC T (A T )VEC(B).
(2.44)
Relation 3: If In denotes the n × n identity matrix, and if B denotes an n × n tensor, then I n ⊗ BT = (I n ⊗ B) T .
(2.45)
Relation 4: Let A, B, C, and D, respectively, denote m × n, r × s, n × p, and s × q matrices. Then, (A ⊗ B) (C ⊗ D) = AC ⊗ BD.
© 2003 by CRC CRC Press LLC
(2.46)
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Finite Element Analysis: Thermomechanics of Solids
Relation 5: If A, B, and C are n × m, m × r, and r × s matrices, then VEC(ACB) = BT ⊗ AVEC(C).
(2.47)
Relation 6: If a and b are n × 1 vectors, then a ⊗ b = VEC([ab T ]T ).
(2.48)
As proof of Relation 6, if I = ( j − 1)n + i, then the I entry of VEC(ba ) is th T T T bi aj. It is also the I entry of a ⊗ b. Hence, a ⊗ b = VEC(ba ) = VEC([ab ] ). T Symmetry of Unn was established in Relation 1. Note that VEC(A) = Unn VEC(A ) = 2 U nn VEC(A) if A is n × n, and hence the matrix Unn satisfies th
−1 . U nn = U Tnn = U nn
U 2nn = I n2
T
(2.49)
Unn is hereafter called the permutation tensor for n × n matrices. If A is symmetric, then (U nn − I n2 ) VEC(A) = 0. If A is antisymmetric, then (Unn + Inn)VEC(A) = 0. If A and B are second-order n × n tensors, then tr(AB) = VEC T (B)VEC(A T ) = VEC T (B)U nn VEC(A) = [U nn VEC(B)]T VEC(A) = VEC T (BT )VEC(A) = tr(BA),
(2.50)
thereby recovering a well-known relation. If In is the n × n identity tensor and in = VEC(In), VEC(A) = In ⊗ Ain since 2 VEC(A) = VEC(AIn). If Inn is the identity tensor in n -dimensional space, then In ⊗ In = Inn since VEC(I n ) = VEC(I n I n ) = In ⊗ InVEC(In). Now, in = Ini, thus In ⊗ In = I n 2 . If A, B, and C denote n × n tensors, then VEC(ACBT ) = I n ⊗ AVEC(CBT ) = (I n ⊗ A)(B ⊗ I n )VEC(C) = B ⊗ AVEC(C).
(2.51)
However, by a parallel argument, VEC[(ACBT )T ] = VEC(BCT A T ) = A ⊗ BVEC(CT ) = A ⊗ BU n2 VEC(C).
© 2003 by CRC CRC Press LLC
(2.52)
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35
The permutation tensor arises in the relation VEC[(ACBT )T ] = U n2 VEC(ACBT ).
(2.53)
Consequently, if C is arbitrary, U n2 B ⊗ AVEC(C) = A ⊗ BU n2 VEC(C),
(2.54)
and, upon using the relation U 2 = U −21 , we obtain an important result: n n B ⊗ A = U n 2 A ⊗ BU n 2 .
(2.55)
If A and B are nonsingular n × n tensors, then (A ⊗ B)(A −1 ⊗ B −1 ) = AA −1 ⊗ BB −1 = In ⊗ In = I n2 .
(2.56)
The Kronecker sum and difference appear frequently (for example, in control theory) and are defined as follows: A ⊕ B = A ⊗ In + In ⊗ B
A B = A ⊗ I n − I n ⊗ B.
(2.57)
The Kronecker sum and difference of two n × n tensors are n × n tensors, as explained in the following section. 2
2.6.3 EIGENSTRUCTURES
OF
2
KRONECKER PRODUCTS
Let αj and βk 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
jk th eigenvalue
jk th eigenvector
A⊗B
α j βk
y j ⊗ zk
A⊕B
α j + βk
y j ⊗ zk
α j − βk
yj ⊗ zk
A
B
© 2003 by CRC CRC Press LLC
(2.58)
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Finite Element Analysis: Thermomechanics of Solids
As proof,
α j y j ⊗ βk zk = α j βk y j ⊗ zk = Ay j ⊗ Bz k = (A ⊗ B)(y j ⊗ z k ).
(2.59)
Now, the eigenvalues of A ⊗ In are 1 × α j, while the eigenvectors are yj ⊗ wk, in which wk is an arbitrary unit vector (eigenvector of In). The corresponding quantities for In ⊗ B are βk × 1 and vj × zk , in which vj is an arbitrary eigenvector of In. Upon selecting wk = zk and vj = yj, the Kronecker sum has eigenvalues αj + βk and eigenvectors yj ⊗ zk .
2.6.4 KRONECKER FORM
OF
QUADRATIC PRODUCTS
Let R be a second-order n × n tensor. The quadratic product a Rb is easily derived: if r = VEC(R), then T
a T Rb = tr[ba T R] = VEC T ([ba T ]T )VEC(R) = VEC T (ab T )VEC(R) = b T ⊗ a T r.
(2.60)
2.6.5 KRONECKER PRODUCT OPERATORS FOR FOURTH-ORDER TENSORS Let A and B be second-order n × n tensors, and let C be a fourth-order n × n × n × n tensor. Suppose that A = CB, which is equivalent to aij = cijklbkl in which the range of i, j, k, and l is (1, n). The TEN22 operator is introduced implicitly using VEC(A) = TEN 22(C)VEC(B).
(2.61)
Note that TEN 22(ACB)VEC(D) = VEC(ACBD) = I n ⊗ A VEC(CBD) = I n ⊗ A TEN 22(C) VEC(BD) = I n ⊗ A TEN 22(C)I n ⊗ B VEC(D),
© 2003 by CRC CRC Press LLC
(2.62)
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Mathematical Foundations: Tensors
37 −1
hence, TEN22(ACB) = In ⊗ ATEN22(C)In ⊗ B. Upon writing B = C A, it is obvious −1 that VEC(B) = TEN22(C )VEC(A). However, TEN22(C)VEC(B) = VEC(A), thus −1 −1 −1 VEC(B) = [TEN22(C)] VEC(A). We conclude that TEN22(C ) = TEN22 (C). T T ˆ Furthermore, by writing A = CB , it is also obvious that Un a = TEN22(C)Unb, ˆ ) = U TEN22(C) U . The inverse of the TEN22 operator is introduced thus TEN22(C n2 n2 using the relation ITEN22(TEN22(C)) = C.
2.6.6 TRANSFORMATION PROPERTIES
OF
VEC
AND
TEN22
Suppose that A and B are true second-order n × n tensors and C is a fourth-order n × n × n × n tensor such that A = CB. All are referred to a coordinate system denoted as Y. Let the unitary matrix (tensor) Qn represent a rotation that gives rise to a coordinate system Y′. Let A′, B′, and C′ denote the counterparts of A, B, and C. T Now, since A′ = Qn AQ n, VEC(A ′) = Q ⊗ QVEC(A). −1
(2.63) −1
−1
However, note that (Q ⊗ Q) = Q ⊗ Q = Q ⊗ Q = (Q ⊗ Q) . Hence, 2 Q ⊗ Q is a unitary matrix (tensor) in an n vector space. However, not all rotations 2 in n –dimensional space can be expressed in the form Q ⊗ Q. It follows that VEC(A) 2 transforms as an n × 1 vector under rotations of the form Q ⊗ Q. Now write A′ = C′ B′, from which T
T
T
Q ⊗ Q VEC(A) = TEN 22(C′)Q ⊗ QVEC(B).
(2.64a)
TEN 22(C′) = Q ⊗ Q TEN 22(C)(Q ⊗ Q) T ,
(2.64b)
It follows that
thus TEN22(C) transforms a second-order n × n tensor under rotations of the form Q ⊗ Q. Finally, letting Ca and Cb denote third-order n × n × n tensors, respectively, thereby satisfying relations of the form A = Cab and b = Cb A, it is readily shown that TEN21(Ca) and TEN12(Cb) satisfy 2
2
TEN 21(C′a ) = Q ⊗ QTEN 21(C a )Q T
n2 × n
TEN12(C′b ) = QTEN12(C b )Q T ⊗ Q T
n × n2 ,
which we call tensors of order (2,1) and (1,2), respectively.
© 2003 by CRC CRC Press LLC
(2.65)
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Finite Element Analysis: Thermomechanics of Solids
2.6.7 KRONECKER PRODUCT FUNCTIONS FOR TENSOR OUTER PRODUCTS Tensor outer products are commonly used in continuum mechanics. For example, Hooke’s Law in isotropic linear elasticity with coefficients µˆ and λˆ can be written as ˆ δ E Tij = µˆ (δ ikδ jl + δ ilδ jk ) Ekl + λδ ij kl kl
i, j = 1, 2, 3,
(2.66)
in which Tij and Eij are entries of the (small-deformation) stress and strain tensors denoted by T and E. Here, δij denotes the substitution (Kronecker) tensor. Equation 2.66 exhibits three tensor outer products of the identity (Kronecker) tensor I: δik δjl, δil δjk, and δijδkl. In general, let A and B be two nonsingular n × n second-order tensors with entries a ij and bij; let a = VEC(A) and b = VEC(B). There are 24 permutations of the indices ijkl corresponding to outer products of tensors A and B. Recalling the definitions of the Kronecker product, we introduce three basic Kronecker-product functions: C1 (A, B) = ab T
C 2 (A, B) = A ⊗ B
C 3 (A, B) = A ⊗ BUn2 .
(2.67)
Twenty-four outer product-Kronecker product pairs are obtained as follows: TEN 22( aij bkl ) = C1 (A, B)
TEN 22(bij akl ) = C1 (B, A)
TEN 22( a ji bkl ) = C1 (A T , B)
TEN 22(bji akl ) = C1 (BT , A)
TEN 22( aij blk ) = C1 (A, BT )
TEN 22(bij alk ) = C1 (B, A T )
TEN 22( a ji blk ) = C1 (A T , BT )
TEN 22(bji alk ) = C1 (BT , A T )
TEN 22( aik bjl ) = C2 (B, A)
TEN 22(bik a jl ) = C2 (A, B)
TEN 22( aki bjl ) = C2 (B, A T )
TEN 22(bki a jl ) = C2 (A, BT )
TEN 22( aik blj ) = C2 (BT , A)
TEN 22(bik alj ) = C2 (A T , B)
TEN 22( aki blj ) = C2 (BT , A T )
TEN 22(bki alj ) = C2 (A T , BT )
TEN 22( ail bjk ) = C3 (B, A)
TEN 22(bil a jk ) = C3 (A, B)
TEN 22( ali bjk ) = C3 (B, A T )
TEN 22(bli a jk ) = C3 (A, BT )
TEN 22( ail bkj ) = C3 (BT , A)
TEN 22(bil akj ) = C3 (A T , B)
TEN 22( ali bkj ) = C3 (BT , A T )
TEN 22(bli akj ) = C3 (A T , BT )
© 2003 by CRC CRC Press LLC
,
(2.68)
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Mathematical Foundations: Tensors
39
With t = VEC(T) and e = VEC(E), and noting that U9e = e (since E is symmetric), we now restate Equation 2.66 as t = [ µˆ [C2 (I, I) + C3 ( I, I)] + λˆC1 (I, I)]e
(2.69)
The proof is presented for several of the relations in Equation 2.68. We introduce tensors R and S with entries rij and sij. Also, let s = VEC(S) and r = VEC(R). a.) Suppose that sij = aij bkl rkl. However, aij bkl rkl = aij b lk rkl , in which case b kl th T T is the kl entry of B . It follows that S = tr(B R)A. Hence, T
T
s = VEC T [(BT ) T ]VEC(R)a = VEC T (B)VEC(R)a = ab T r = C1 (A, B)r.
(2.70)
Since s = TEN22(aij bkl)r, it follows that TEN 22( aij bkl ) = C1 (A, B), as shown in Equation 2.68. T T b.) Suppose that sij = aik bjl rkl. However, aik bjl rkl = aik rkl b lj, thus S = ARB . Now s = VEC(ARBT ) = I ⊗ AVEC(RBT ) = I ⊗ AB ⊗ Ir = B ⊗ Ar = C 2 (B, A)r,
(2.71)
as shown in Equation 2.68. T T T c.) Suppose sij = ail bjk rkl. However, ail bjk rkl = ail r lk bkj, thus S = AR B . Now s = VEC(AR T BT ) = I ⊗ AB ⊗ IVEC(R T ) = B ⊗ AU 9 r = C3 (B, A)r, as shown in Equation 2.68.
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Finite Element Analysis: Thermomechanics of Solids
2.6.8 KRONECKER EXPRESSIONS FOR SYMMETRY CLASSES IN FOURTH-ORDER TENSORS Let C denote a fourth-order tensor with entries cijkl. If the entries observe cijkl = c jikl
( a)
cijkl = cijlk
(b) ,
cijkl = cklij
(c )
(2.73)
then C is said to be totally symmetric. A fourth-order tensor C satisfying Equation 2.73a but not 2.73b or c is called symmetric. Kronecker-product conditions for symmetry are now stated. The fourth-order tensor C is totally symmetric if and only if TEN22(C) = TEN22T (C)
( a)
Un2 TEN22(C) = TEN22(C)
(b) .
TEN22(C)Un2 = TEN22(C)
(c )
(2.74)
Equation 2.74a is equivalent to symmetry with respect to exchange of ij and kl in C. Total symmetry also implies that, for any second-order n × n tensor B, the corresponding tensor A = CB is symmetric. Thus, if a = VEC(A) and b = VEC(B), then a = TEN22(C)b. However, U 2 a = TEN 22(C)b . Multiplying through the later n expression with Un2 implies Equation 2.74b. For any n × n tensor A, the tensor B = −1 −1 −1 C A is symmetric. It follows that b = TEN22(C )a = TEN22 (C)a, and Un2 b = −1 −1 −1 −1 TEN22 Ca. Thus, TEN22(C ) = Un2 TEN22 (C). Also, TEN22(C) = [ Un2TEN22 −1 (C)] = TEN22(C) Un2 . We now draw the immediate conclusion that Un2 TEN22(C) Un2 = TEN22(C) if C is totally symmetric. We next prove the following: C
−1
is totally symmetric if C is totally symmetric.
(2.75)
−1 −1 Note that TEN22(C) Un2 = TEN22(C) implies that Un2 TEN22(C ) = TEN22(C ), −1 −1 while Un2 TEN22(C) = TEN22(C) implies that TEN22(C ) Un2 = TEN22(C ). Finally, we prove the following: for a nonsingular n × n tensor G, T
GCG is totally symmetric if C is totally symmetric. −T
(2.76)
Equation 2.76 implies that TEN22(GCG ) = I ⊗ G TEN22(C)I ⊗ G , so that T TEN22(GCG ) is certainly symmetric. Next, consider whether A′ given by A ′ = GCG T B′
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T
(2.77)
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41
is symmetric, in which case B′ is a second-order, nonsingular n × n tensor. However, we can write G −1 A ′G − T = CG −1 B′ G − T . −1
(2.78)
−T
Now G A′G is symmetric since C is totally symmetric, and therefore A′ is symmetric. Next, consider whether B′ given by the following is symmetric: B′ = G − T C −1G −1 A ′
(2.79)
G T B′ G = C −1G −1 A ′G.
(2.80)
However, we can write
−1
T
Since C is totally symmetric, it follows that G B′G is symmetric, and hence T B′ is symmetric. We conclude that GCG is totally symmetric.
2.6.9 DIFFERENTIALS
TENSOR INVARIANTS
OF
Let A be a symmetric 3 × 3 tensor, with invariants I1(A), I2(A), and I3(A). For a scalar-valued function f(A), df (A) =
∂f ∂f da = tr dA , ∂A ∂aij ij
∂f ∂f . = ∂A ij ∂aij
(2.81)
However, with a = VEC(A), we can also write ∂f T df (A) = VEC T VEC( dA) ∂A =
∂f da. ∂a
(2.82)
Taking this further, ∂I1 ∂ T (i a ) = i T = ∂a ∂a ∂I2 ∂ 1 T 2 (i a) − a T a = ∂a ∂a 2 = I1i T − a T
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Finite Element Analysis: Thermomechanics of Solids
and dI3 = tr(A 2 dA) − I1tr(A dA) + I2 dA = tr(A −1dA / I3 )
(2.84)
so that ∂I3 = VEC(A −1 ) / I3 ∂a
(2.85)
2.7 EXERCISES 1. Given a symmetric n × n tensor σ, prove that tr(σ − tr(σ ) In /n) = 0. 2. Prove that if σ is a symmetric tensor with entries σij,
ε ijk σ jk = 0,
i = 1, 2, 3.
3. Verify using 2 × 2 tensors that tr(AB) = tr(BA) . 4. Express I3 as a function of I1 and I2. 5. Using 2 × 2 tensors and 2 × 1 vectors, verify the six relations given for Kronecker products. 6. Write out the 9 × 9 quantity TEN22(C) in Equation 2.66. 7. Using a 2 × 2 tensor A, write out the differential of ln(A).
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3
Introduction to Variational and Numerical Methods
3.1 INTRODUCTION TO VARIATIONAL METHODS Let u(x) be a vector-valued function of position vector x, and consider a vectorvalued function F(u(x), u′(x),x), in which u′(x) = ∂u/∂x. Furthermore, let v(x) be a function such that v(x) = 0 when u(x) = 0 and v′(x) = 0 when u′(x) = 0, but which is otherwise arbitrary. The differential d F measures how much F changes if x changes. The variation δ F measures how much F changes if u and u′ change at fixed x. Following Ewing, we introduce the vector-valued function φ (e: F) as follows (Ewing, 1985): Φ (e:F) = F(u( x ) + ev( x ), u′( x ) + ev′( x ), x ) − F(u( x ), u′( x ), x )
(3.1)
The variation δ F is defined by dΦ δ F = e , de e=0
(3.2)
with x fixed. Elementary manipulation demonstrates that
δF = ∂F
∂F ∂F ev + tr ev ′ , ∂u ′ ∂u
(3.3)
∂F
in which ∂u ′ ev ′ = ∂u ′ ev ′ij . If F = u, then δ F = δ u = ev. If F = u ′, then δ F = δ u′ = ij ev′. This suggests the form
δF=
∂F ∂F δ u + tr δ u′ . ∂u′ ∂u
(3.4)
The variational operator exhibits five important properties: 1. δ (.) commutes with linear differential operators and integrals. For example, if S denotes a prescribed contour of integration:
∫ δ ( )dS = δ ∫ ( )dS.
(3.5)
43
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Finite Element Analysis: Thermomechanics of Solids
2. δ ( f ) vanishes when its argument f is prescribed. 3. δ (.) satisfies the same operational rules as d(.). For example, if the scalars q and r are both subject to variation, then
δ (qr ) = qδ (r ) + δ (q)r.
(3.6)
4. If f is a prescribed function of (scalar) x, and if u (x) is subject to variation, then
δ ( fu) = fδu.
(3.7)
5. Other than for number 2, the variation is arbitrary. For example, for two T vectors v and w, v d w = 0 implies that v and w are orthogonal to each T other. However, v δ w implies that v = 0, since only the zero vector can be orthogonal to an arbitrary vector. As a simple example, Figure 3.1 depicts a rod of length L, cross-sectional area A, and elastic modulus E. At x = 0, the rod is built in, while at x = L, the tensile force P is applied. Inertia is neglected. The governing equations are in terms of displacement u, stress S, and (linear) strain E: du dx
strain-displacement
E=
stress-strain
S = EE
equilibrium
dσ =0 dx
(3.8)
Combining the equations furnishes EA
d 2u = 0. dx 2
(3.9)
The following steps serve to derive a variational equation that is equivalent to the differential equation and endpoint conditions (boundary conditions and constraints).
E,A P L
FIGURE 3.1 Rod under uniaxial tension.
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Introduction to Variational and Numerical Methods
45
Step 1: Multiply by the variation of the variable to be determined (u) and integrate over the domain.
∫
L
δ u EA
0
d 2u Adx = 0. dx 2
(3.10)
Differential equations to be satisfied at every point in the domain are 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 positive definite.
∫
L
0
d dx
δuEA du − dδu EA du dx = 0 dx dx dx
(3.11)
However, the first term is the integral of a derivative, so that
∫
L
0
dδu du du dx EA dx dx = δuEA dx . 0 L
(3.12)
Step 3: Identify the primary and secondary variables. The primary variable is present in the endpoint terms (rhs) under the variational symbol, l, and is u. The conjugate secondary variable is EA du . dx Step 4: Satisfy the constraints and boundary conditions. At x = 0, u is prescribed, thus δu = 0. At x = L, the load P = EA du is prescribed. Also, note dx du = δ ( 1 EA( du )2 ) . that ( du ) E A 2 dx dx dx Step 5: Form the variational equation; the equations and boundary conditions are consolidated into one integral equation, δF = 0, where˙ F=
∫
L
0
2
1 du EA dx − Pu( L). 2 dx
(3.13)
th
The j variation of a vector-valued quantity F is defined by d jΦ δ jF = e j j . de e=0
(3.14)
It follows that δ u = 0 and δ u′ = 0. By restricting F to a scalar-valued function F and x to reduce to x, we obtain 2
δ 2 F = {δ u T
δu δ u ′ T}H , δ u ′
2
∂ T ∂ F ∂u ∂u H= T ∂ ∂ ∂u ∂u F ′
and H is known as the Hessian matrix.
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T ∂ ∂ F ∂u ∂u ′ , T ∂ ∂ F ∂u ′ ∂u ′
(3.15)
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Finite Element Analysis: Thermomechanics of Solids
Now consider G given by G=
∫ F(x, u(x), u′(x))dV + ∫ h (x)u(x) dS, T
(3.16)
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 concentrate on a three-dimensional, rectangular coordinate system and suppose that δ G = 0, as in the Principle of Stationary Potential Energy in elasticity. Note that ∂ ∂F ∂F ∂ ∂ δ u = tr δ u ′ + F δ u. ∂u ′ ∂x ∂u ′ ∂x ∂u ′
(3.17)
The first and last terms in Equation 3.17 can be recognized as divergences of vectors. We now invoke the divergence theorem to obtain 0 = δG ∂F
∂F
=
∫ ∂u δ u + tr ∂u′ δ u′ dV + ∫ h (x)δu(x) dS
=
∫ ∂u − ∂x ∂u′ δ u dV + ∫ n
∂F
∂ ∂F
T
T
∂F δ udS + h T ( x )δ u( x ) dS. ∂u ′
∫
(3.18)
For suitable continuity properties of u, arbitrariness of δ u implies that δ G = 0 is equivalent to the following Euler equation, boundary conditions, and constraints (the latter two are not uniquely determined by the variational principle): ∂F ∂ ∂F − = 0 T. ∂u ∂x ∂u ′ u( x ) prescribed T ∂F T T n ∂u ′ + h1 ( x ) = 0 on S
(3.19) x on S1 x on S − S1
Let D > 0 denote a second-order tensor, and let π denote a vector that is a nonlinear function of a second vector u, which is subject to variation. The function F = 12 π T Dπ satisfies
δ F = δ uT
∂π T Dπ ∂u T
δ 2 F = δ uT
∂ ∂π T ∂π T ∂π Dπ δ u + δ u T T δ uDπ . T ∂u ∂u ∂u ∂u (3.20)
Despite the fact that D > 0, in the current nonlinear example, the specific vector ∗ u satisfying δ F = 0 may correspond to a stationary point rather than a minimum.
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47
3.2 NEWTON ITERATION AND ARC-LENGTH METHODS 3.2.1 NEWTON ITERATION Letting f and x denote scalars, consider the nonlinear algebraic equation f(x; λ) = 0, in which λ is a parameter we will call the load intensity. Such equations are often solved numerically by a two-track process: the load intensity λ is increased progressively using small increments. At each increment, the unknown x is computed using th an iteration procedure. Suppose that at the n increment of λ, an accurate solution is achieved as xn. Further suppose for simplicity’s sake that xn is “close” to the actual st (0) solution xn+1 for the (n + 1) increment of λ. Using x = xn as the starting value, Newton iteration provides iterates according to the scheme −1
df x ( j +1) = x ( j ) − f ( x ( j ) ). dx x ( j )
(3.21)
Let ∆n+1,j denote the increment xn( +j +11) − xn( +j )1. Then, to first-order in the Taylor series −1 df −1 df ∆ n +1,j − ∆ n +1,j−1 = − f(x(j) ) − f(x(j−1) ) dx x(j−1) dx x(j) −1
[
]
df ≈ − f(x(j) ) − f(x(j−1) ) dx x(j) −1
df df ≈ − ∆ n +1,j−1 + 0 2 (j) dx dx x x(j) ≈ ∆ n +1,j−1 + 0 2
(3.22)
in which 0 refers to second-order terms in increments. It follows that ∆n+1,j ≈ 0 . For this reason, Newton iteration is said to converge quadratically (presumably to the correct solution if the initial iterate is “sufficiently close”). When the iteration scheme converges to the solution, the load intensity is incremented again. Consider f (x) = (0) (0) (x − 1)2. If x = 1/2, the iterates are 1/2, 3/4, 7/8, and 15/16. If x = 2, the iterates are 3/2, 5/4, 9/8, and 17/16. In both cases, the error is halved in each iteration. The nonlinear, finite element poses nonlinear, algebraic equations of the form 2
2
δ u T [ϕ(u) − λv] = 0,
(3.23)
in which u and ϕ are n × 1 vectors, v is a constant n × 1 unit vector, and λ represents st “load intensity.” The Newton iteration scheme provides the ( j + 1) iterate for un+1 as ∂ϕ ∆ n +1,j = − ∂u u(nj+)1
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−1
[ϕ ( u ) − λ v ] , ( j) n +1
j+1
u(jn ++11) − u(nj+)1 = ∆ n +1,j .
(3.24)
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Finite Element Analysis: Thermomechanics of Solids
in which, for example, the initial iterate is un. One can avoid the use of an explicit matrix inverse by solving the linear system ∂ϕ ( j) ∂u ( j ) ∆ n+1, j = ϕ u n+1 − v j +1 un +1
(
3.2.2 CRITICAL POINTS
)
AND THE
u (nj++11) = u (nj+)1 + ∆ n+1, j .
(3.25)
ARC-LENGTH METHOD ∂ϕ
∗
A point λ at which the Jacobian matrix J = ∂u is singular is called a critical point, and corresponds to important phenomena such as buckling. There often is good reason to attempt to continue calculations through critical points, such as to compute a postbuckled configuration. Arc-length methods are suitable for doing so. Here, we present a version with a simple eigenstructure. Suppose that the change in load intensity is regarded as a variable. Introduce th the “constraint” on the size of the increment for the n load step:
ψ (u, λ ) = β 2 [λ − λ n ] + v T (u − u n ) − Σ 2 = 0,
(3.26)
in which Σ is interpreted as the arc length in n + 1 dimensional space of u and λ. Also, β > 0. Now, 2
ϕ (u) − λv 0 = . ψ 0
(3.27)
Newton iteration now is expressed as
( ) (
ϕ u ( j ) − λ( j ) v u (nj++11) − u (nj+)1 n +1 n +1 , J ′ = − ( j) λ( j +1) − λ( j ) ψ λ u , n+1 n +1 n +1 n
)
J J′ = v T
−v . β 2
(3.28)
An advantage is gained if J′ can be made nonsingular even though J is singular. Τ 2 β > 0. Then Suppose that J is symmetric and we can choose β such that J + v v /β J′ admits the “triangularization” J + v v T / β2 J′ = 0T
− v / β I T β v /β
0 . β
(3.29)
Τ 2 2 β ). Ideally, β is chosen to maximize The determinant of J′ is now β det(J + v v /β det (J′).
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49
3.3 EXERCISES 1. Directly apply variational calculus to F, given by F=
∫
L
0
2
1 du EA dx − Pu( L ) dx 2
to verify that δ F = 0 gives rise to the Euler equation EA
d 2u =0 dx 2
What endpoint conditions (not unique) are compatible with δ F = 0? 2. The governing equation for an Euler-Bernoulli beam in Figure 3.2 is
EI
d 4w =0 dx 4
in which w is the vertical displacement of the neutral (centroidal) axis. The shear force V and the bending moment M satisfy M = − EI
d 2w dx 2
V = EI
d 3w dx 3
Using integration by parts twice, obtain the function F such that δ F = 0 is equivalent to the foregoing differential equation together with the boundary conditions for a cantilevered beam of length L: w(0) = w ′(0) = 0 Z
M ( L) = 0, V( L) = V0 . neutral axis v0
V E,I
x L FIGURE 3.2 Cantilevered beam.
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4
Kinematics of Deformation
The current chapter provides a review of the mathematics for describing deformation of continua. A more complete account is given, for example, in Chandrasekharaiah and Debnath (1994).
4.1 KINEMATICS 4.1.1 DISPLACEMENT In finite-element analysis 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 1.1 in Chapter 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 that 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.
deformed
x e2 X
undeformed
e1 FIGURE 4.1 Position vectors in deformed and undeformed configurations.
51
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Finite Element Analysis: Thermomechanics of Solids
Q' ds
P'
Q * P *
dS
FIGURE 4.2 Deformed and undeformed distances between adjacent points.
4.1.2 DISPLACEMENT VECTOR The vector u(X) represents the displacement from position X to x: u(X, t ) = x − X.
(4.1)
Now consider two close points, P and Q, in the undeformed configuration. The vector difference XP − XQ is represented as a differential d X with squared length 2 T dS = d X d X. The corresponding quantity in the deformed configuration is dx, with 2 T dS = dx dx.
4.1.3 DEFORMATION GRADIENT TENSOR The deformation gradient tensor F is introduced as dx = Fd X
F=
∂x ∂X
(4.2)
F satisfies the polar-decomposition theorem: F = UΣV T ,
(4.3)
in which U and V are orthogonal and Σ is a positive definite diagonal tensor whose
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Kinematics of Deformation
53
entries λj, the singular values of F, are called the principal stretches. λ1 Σ = 0 0
0
λ2 0
0 0 λ3
(4.4)
Based on Equation 4.3, F can be visualized as representing a rotation, followed by a stretch, followed by a second rotation.
4.2 STRAIN The deformation-induced change in squared length is given by ds2 − dS2 = d X T 2 Ed X
1 E = [FT F − I], 2
(4.5)
in which E denotes the Lagrangian strain tensor. Also of interest is the Right CauchyT Green strain C = F F = 2E + I. Note that F = I + ∂ u /∂ X. If quadratic terms in ∂ u /∂ X are neglected, the linear-strain tensor EL is recovered as EL =
T 1 ∂u ∂u + . 2 ∂X ∂X
(4.6)
Upon application of Equation 4.3, E is rewritten as 1 E = V ( Σ 2 − I)V T 2
(4.7)
1
Under pure rotation x = QX, F = Q and E = 2 [Q Q − I] = 0. The case of pure rotation in small strain is considered in a subsequent section.
4.2.1 F, E, EL
AND
u
IN
T
ORTHOGONAL COORDINATES
Let Y1, Y2, and Y3 be orthogonal coordinates of a point in an undeformed configuration, with y1, y2, y3 orthogonal coordinates in the deformed configuration. The Γ2,Γ Γ3 and γ1,γγ2,γγ3. corresponding orthonormal base vectors are Γ1,Γ 4.2.1.1 Deformation Gradient and Lagrangian Strain Tensors Recalling relations introduced in Chapter 1 for orthogonal coordinates, the differential position vectors are expressed as dR =
∑ dY H Γ α
α
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α
α
dr =
∑ dy h γ β β
β
β
(4.8)
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Finite Element Analysis: Thermomechanics of Solids
dr =
∑ dy h γ α α
α
α
=
∑ ∑ dy h (γ α α
β
=
∑ ∑ dy h q α α
β
=
⋅ Γ β )Γ β
T Γ βα β
α
∑∑∑q
T βα
hα dyα H dY Γ Hζ dYζ ζ ζ β
∑∑∑q
T βα
hα dyα (Γ ∧ Γ ζ ) Hζ dYζ Γ ζ , Hζ dYζ β
ζ
=
α
α
β
ζ
β
α
α
(4.9)
in which Γβ denotes the base vector in the curvilinear system used for the undeformed configuration. This can be written as dr = Q T F ′dR,
T [Q T ]βα = qβα
[F ′]αζ =
hα ∂yα , Hζ ∂Yζ
(4.10)
in which Q is the orthogonal tensor representing transformation from the undeformed to the deformed coordinate system. It follows that 1 1 E = [ F T F − I ] = [ F′ T F′ − I ], 2 2 from which [ E]ij =
1 2
∑ β
h 2β ∂y β ∂y β − δ H H ∂Y ∂Y ij . i j i j
(4.11)
Displacement Vector The position vectors can be written in the form R = ZiΓi, r = z jγj. The displacement vector referred to the undeformed base vectors is u = [z j q ji − Z i ]Γ i ,
q ji = γ j ⋅ Γ i .
(4.12)
Cylindrical Coordinates In cylindrical coordinates, u = [r cos(θ − Θ) − R] e R + r sin(θ − Θ)eθ + ( z − Z ) e Z .
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(4.13)
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55 2
We now apply the chain rule to ds in cylindrical coordinates: ds 2 = dr ⋅ dr = dr 2 + r 2 dθ 2 + dz 2 dr dr dθ r dθ dθ 1 dr = dR + RdΘ + dZ + r dR + RdΘ + r dZ R dΘ dZ dR R dΘ dZ dR 2
dz 1 dz dz RdΘ + dZ + dR + R dΘ dZ dR
RdΘ
= {dR
2
2
dR dZ} C RdΘ , dZ
(4.14)
in which dr dθ dz cRR = + r + dR dR dR 2
2
2
eRR =
1 (c − 1) 2 RR
eΘΘ =
1 (c − 1) 2 ΘΘ
eZZ =
1 (c − 1) 2 ZZ
dr 1 dr dθ r dθ dz 1 dz + r + cRΘ = dR R dΘ dR R dΘ dR R dΘ
eRΘ =
1 c 2 RΘ
1 dr dr r dθ dθ 1 dz dz + + cΘZ = r R dΘ dZ R dΘ dZ R dΘ dZ
eΘZ =
1 c 2 ΘZ
dr dr dθ dθ dz dz cZR = + r r + dZ dR dZ dR dZ dR
eZR =
1 c . 2 ZR
1 dr r dθ 1 dz + cΘΘ = + R dΘ R d Θ R d Θ 2
2
dr dθ dz cZZ = + r + dZ dZ dZ 2
2
2
2
(4.15) 4.2.1.2 Linear-Strain Tensor in Cylindrical Coordinates If quadratic terms in the displacements and their derivatives are neglected, then uR ≈ r − R du dr ≈1+ R dR dR 1 dr 1 du R ≈ R dΘ R d Θ dr du R ≈ dZ dZ
uΘ ≈θ −Θ R
uZ ≈ z − Z ,
u dθ d u du ≈ R Θ = Θ − Θ dR dR R dR R R + uR u 1 duΘ 1 duΘ r dθ = 1+ ≈1+ R + R dΘ R R dΘ R R dΘ dθ duΘ ≈ r dZ dZ r
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(4.16) dz duZ ≈ dR dR 1 dz 1 duZ ≈ R dΘ R dΘ du dz ≈1+ Z dZ dZ
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Finite Element Analysis: Thermomechanics of Solids
giving rise to the linear-strain tensor du R dR du u 1 1 ∂u R EL = Θ − Θ + 2 dR R R ∂R du 1 du Z + R dZ 2 dR
1 2
duΘ − dR uR
uΘ
+
+
1 ∂u R
R R ∂R 1 duΘ
R R dΘ 1 du Z 1 duΘ
+ 2 dZ
R
1 du Z
2 dR 1 duΘ 2
dΘ
+
dZ +
du R
dZ 1 du Z
. (4.17)
R dΘ
du Z dZ
The divergence of u in cylindrical coordinates is given by ∇ ⋅ u = trace ddur = trace(EL), from which ∇⋅u =
duR uR 1 duΘ duZ + + + , dR R R dΘ dZ
(4.18)
which agrees with the expression given in Schey (1973).
4.2.2 VELOCITY-GRADIENT TENSOR, DEFORMATION-RATE TENSOR, AND SPIN TENSOR We now introduce the particle velocity v = ∂ x/∂ t and assume that it is an explicit function of x(t) and t. The velocity-gradient tensor L is introduced using dv = Ldx, from which L= =
dv dx dv dX dX dx
˙ −1 . = FF
(4.19)
Its symmetric part, called the deformation-rate tensor, 1 D = [L + LT ], 2
(4.20)
can 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: 1 E˙ = [F T F˙ + F˙ T F] 2 1 ˙ −1 = F T [FF + F − T F˙ T ]F 2 = F T DF.
© 2003 by CRC CRC Press LLC
(4.21)
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The antisymmetric portion of L is called the spin tensor W: 1 W = [L − LT ]. 2
(4.22)
Suppose the deformation consists only of a time-dependent, rigid-body motion: x(t ) = Q(t )X + b(t ),
Q T (t )Q(t ) = I.
(4.23)
Clearly, F = Q and E = 0. Furthermore, ˙ T, L = QQ
(4.24)
˙ T + QQ ˙ T = QQ ˙ T + (QQ ˙ T )T . which is antisymmetric since 0 = I˙ = [QQ T ]• = QQ T ˙ Hence, D = 0 and W = QQ , thus explaining the name of W. 4.2.2.1 v, L, D, and W in Orthogonal Coordinates The velocity v(y, t) in orthogonal coordinates is given by v(y, t ) =
dr = dt
∑v γ α
α,
vα = hα
α
dyα . dt
(4.25)
Based on what we learned in Chapter 1, with v denoting the velocity vector in orthogonal coordinates, dv 1 ∂v j [L]βα = = δ jβ + v j c jβα . dr βα hα ∂yα c αjβ = α
α 1 (1 − δ αβ ) hα
β j
hβ
(4.26)
β
∂ 2 xα ∂yβ = hβ ∂yk ∂y j ∂x k
j 1
1
2
2
Of course, [D]βα = [[L]βα + [L]αβ ], [W]βα = [[L]βα − [L]αβ ]. 4.2.2.2 Cylindrical Coordinates The velocity vector in cylindrical coordinates is v=
dr dθ dz e +r e + e dt r dt θ dt z
= vr e r + vθ eθ + vz e z .
© 2003 by CRC CRC Press LLC
(4.27)
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Finite Element Analysis: Thermomechanics of Solids
Observe that dv = dvr e r + dvθ eθ + dvz e z + vr de r + vθ deθ = [dvr − vθ dθ ]e r + [dvθ + vr dθ ]eθ + dvz e z .
(4.28)
Converting to matrix-vector notation, we get v dv 1 dv r dvr dr + rdθ + r dz − θ rdθ dr r dθ dz r dv dv dv v 1 dv = θ dr + θ rdθ + θ dz + r rdθ r dθ dz r dr dv z dv z 1 dv z dr + rdθ + dz dr r dθ dz
=
dr L rdθ , dz
dvr dr dv L= θ dr dv z dr
1 dv r r dθ 1 dvθ
− +
vθ r vr
r dθ r 1 dv z r dθ
dz dvθ . dz dv z dz dv r
(4.29)
4.2.2.3 Spherical Coordinates Now v=
dr dθ dφ e r + r cos φ eθ + r e dt φ dt dt
= vr e r + vθ eθ + vφ eφ e r = cos φ (cos θ e1 + sin θ e 2 ) + sin φ e 3
(4.30)
eθ = − sin θ e1 + cos θ e 2 eφ = − sin φ (cos θ e1 + sin θ e 2 ) + cos φ e 3 . Next,
v′ = Qv,
vr v = vθ , v φ
cos φ cos θ Q = − sin θ − sin φ cos θ
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v′ referred to e1, e2 , e3 (4.31) cos φ sin θ cos θ − sin φ sin θ
sin φ 0 cos φ
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59
dv = QQT dv * + dQQT v dvr dv* = dvθ referred to e r , eθ , eφ dv φ
= dv * + dQQT v,
(4.32)
Recall that 0 dQ(θ ) T Q (θ ) = − cos φ dt 0
cos φ
0 0 dθ sin φ + 0 dt −1 0
0 − sin φ
0 0 0
1 dφ . 0 dt 0
(4.33)
Thus, it follows that 0 1 − cos φ dQQ T = r cos φ 0
cos φ
0 0 1 sin φ r cos φ dθ + 0 r −1 0
0 − sin φ
0 0 0
1 0 rdφ 0 (4.34)
and cos φvθ dθ + vφ dφ dQQ T v = − cos φvr dθ + sin φvφ dθ − sin φvθ dθ − vr dφ 0 1 0 = r 0
vθ − vr + tan φvφ − tan φvθ
vφ dr 0 r cos φ dθ . − vr r dφ
(4.35)
dr dφ dvθ r cos φ dθ dφ dvφ rdφ dφ
(4.36)
Finally, dvr dvr dr dv dv* = dvθ = θ dr dv dvφ φ dr
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1
dv r
1 dv r
r cos φ dθ dvθ 1
r 1
r cos φ dθ dvφ 1
r 1
r cos φ dθ
r
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Finite Element Analysis: Thermomechanics of Solids
and dvr dr dv L= θ dr dv φ dr dvr dr dv = θ dr dv φ dr
1
dv r
r cos φ dθ dvθ 1
r 1
r cos φ dθ dvφ 1
r 1
r cos φ dθ
r
1 1
0 dφ dvθ 1 + 0 dφ r dvφ 0 dφ
1 dv r
dv r
+
vθ
−
r cos φ dθ
−
− vr + tan φvφ − tan φvθ dφ r 1 dvθ . r dφ dvφ vr − dφ r
1 dv r
r cos φ dθ r v r − tan φvφ dvθ
r cos φ dθ dvφ 1
vθ
r
r tan φvθ
1
r
r
+
vφ 0 − vr
vφ
(4.37)
The divergence of v is given by trace(L), thus, ∇⋅v =
dvr 1 dvθ vr − tan φvφ 1 dvφ + − + , dr r cos φ dθ r r dφ
(4.38)
which is again in agreement with Schey (1973).
4.3 DIFFERENTIAL VOLUME ELEMENT The volume spanned by the differential-position vector dR is given by the vector triple-product dV0 = d X1 ⋅ d X 2 × d X 3 = d X1d X 2 d X 3 d X1 = dX1e1
d X 2 = dX2 e 2
The vectors dXi deform into dxj = verified to be
dx j dx i
d X 3 = dX3 e 3 .
(4.39)
ej dXi. The deformed volume is now readily
dV = dx1 ⋅ dx 2 × dx 3 1
= JdV0 ,
J = det (F) = det 2 (C),
(4.40)
and J is called the Jacobian. To obtain J for small strain, we invoke invariance and J = det(C ) to find 1
1
det 2 (C) = det 2 [I + 2 E] = (1 + 2 EI )(1 + 2 EII )(1 + 2 EIII ),
© 2003 by CRC CRC Press LLC
(4.41)
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61
in which EI, EII, EIII are the eigenvalues of EL, assumed to be much less that unity. The linear-volume strain follows as 1
evol = det 2 (C) − 1 = 1 + 2( EI + EII + EIII ) + quadratic terms − 1 ≈ tr( EL ),
(4.42)
using the approximation 1 + x ≈ 1 + x/2 if x << 1. The time derivative of J is prominent in incremental formulations in continuum mechanics. Recalling that J = I3 , d 1 dI3 J= dt 2 J dt =
J ˙) tr(C −1C 2
=
J tr(F −1 F − T [F T F˙ + F˙ T F]) 2
=
J tr(F −1 F˙ + F −1 F − T F˙ T F) 2
=
J [tr(F −1 F˙ ) + tr(F −1 F − T F˙ T F)] 2
F −1 F˙ + F − T F˙ T = J tr 2 = J tr(D).
(4.43)
4.4 DIFFERENTIAL SURFACE ELEMENT Let dS denote a surface element in the deformed configuration, with exterior unit normal n, as illustrated in Figure 4.3. The corresponding quantities from the reference configuration are dS0 and n0. A surface element dS obeys the transformation (Chandrashekharaiah and Debnath, 1994): n dS = JF − T n 0 dS0 ,
(4.44)
from which we conclude that dS = J n T0 C −1 n 0 dS0
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n=
F −Tn0 n T0 C −1 n 0
.
(4.45)
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n
n0
dx2
dS0
dX2
dS dX1 dx1 FIGURE 4.3 Surface patches in undeformed and deformed configurations.
During deformation, the surface normal changes direction, a fact which is important, for example, in contact problems. In incremental variational methods, we consider the differential dn and d(ndS): dt
d dJ − T d F −T [n dS] = F n 0 dS0 + J n 0 dS0 . dt dt dt
(4.46)
However, recalling Equation 4.43, dJ = J tr(D) dt dJ − T F n 0 dS0 = tr(D) JF − T n 0 dS0 dt = tr(D)n dS.
(4.47)
−T
Also, since d(F F ) = 0, then T
dF − T dF T − T F = −F −T dt dt = − LT F − T .
(4.48)
d[ndS] = [tr(D)I − LT ]ndS. dt
(4.49)
Finally, we have
Next, we find with some effort that −1
−T
dn dF = dt dt
n0 n T0 C −1 n 0
= [(n T Dn)I − LT ]n.
© 2003 by CRC CRC Press LLC
−
−T
F n0
n T0
dC dt
T 0
n0
−1
n T0 C −1 n 0 n C n 0 (4.50)
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63
4.5 ROTATION TENSOR Elementary manipulation furnishes du =
du dX dX
= [ EL + ω ]d X,
(4.51)
in which the antisymmetric rotation tensor ω appears: 1 ∂u ∂u − 2 ∂X ∂X
ω=
T
.
(4.52)
Consider pure rotation with small angle θ : cos θ x = − sin θ 0 θ2 1 − 2 ≈ −θ 0
sin θ cos θ 0
θ 1−
θ
2
2
0
0 0X 1 0 0 X, 1
(4.53)
thus, θ2 − 2 u= 0 0 θ2 − 2 EL = 0 0
0 θ
−
2
0
0 −
2
θ
2
2
0
0 0 0 X + −θ 0 0 0 0 , 0
θ 0 0
0 ω = −θ 0
0 0X 0
θ 0 0
0 0 . 0
(4.54)
(4.55)
Evidently, the normal strains do not vanish, but are second-order in θ, while ω is first-order in θ. Under the assumption of small deformation, nonlinear terms are neglected so that EL is regarded as vanishing.
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Finite Element Analysis: Thermomechanics of Solids
Y
L θ L
L
θ
X
L FIGURE 4.4 Element in undeformed and deformed configurations.
Consider as a second example the following derivation in which the sides of the unit square rotate toward each other by 2θ. The deformation can be described by the relations θ2 x = (1 − cos θ ) X + sin θY ≈ 1 − X + θY 2
(4.56)
θ2 y = sin θX + (1 − cos θ )Y ≈ θX + 1 − Y , 2 from which we conclude that θ2 − 2 EL = θ 0
θ −
θ
2
2
0
0 0 . 0
(4.57)
Upon neglecting quadratic terms, we conclude that linear-shear strain is a measure of how much the axes rotate toward each other, while the rotation tensor is a measure of how much the axes rotate in the same sense.
4.6 COMPATIBILITY CONDITIONS FOR EL AND D The following paragraphs present the compatibility equations for the linear-strain tensor, allowing EL to be integrated to produce a displacement field u(X), which is unique to within a rigid-body motion. It should be evident that a parallel argument produces exactly the same result for the velocity vector v(x) starting from a given deformation-rate tensor D(x).
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65
There are two major “paths” to solutions of problems in continuum mechanics. For example, Assume/approximate u. Apply the strain-displacement relations to estimate E, apply the stress-strain relations to estimate S, and then apply equilibrium relations to obtain the field equation. Solve the field equation applying boundary conditions and constraints. Assume/approximate S, and then estimate E using the strain-displacement relations. Apply compatibility relations to obtain the field equation, and then solve the field equation and boundary conditions and constraints. To understand the second solution path, we focus on the following compatibility relation. Suppose the linear strains are known as functions of the position X. Under what circumstances can the strain-displacement relations be integrated to produce a displacement field that is unique to within a rigid-body displacement? Recall that du dX dX
du =
= [ EL + ω ]d X
(4.58)
Evidently, EL is only part of the derivative of u. In 2-D rectilinear coordinates, we will see that the compatibility equation, guaranteeing unique u to within a rigidbody motion, is given by: ∂2 E12 ∂2 E22 1 ∂2 E = 211 + ∂x1∂x2 2 ∂x2 ∂x12
(4.59)
The proof of the compatibility equation is as follows. We consider a path in the physical (X1, X2, X3) space, along which the arc length is denoted by λ. Accordingly, the position vector X is a function of λ. The integral of du, taken from λ = 0 to λ* is u(X(λ *)) = u(0) +
∫
X ( λ *)
∫
X ( λ *)
0
= u(0) +
du dX dX E(X(λ *))dX +
0
∫
X ( λλ *)
ω(X)dX.
(4.60)
0
The term u(0) can be interpreted as the rigid-body translation. The first integral can be evaluated since E(X) and X(λ) are given functions. The second integral can be rewritten using an elementary transformation as
∫
X ( λ *)
ω(X)dX = −
0
© 2003 by CRC CRC Press LLC
∫
0
X ( λ *)
ω(X * − X)d(X * − X).
(4.61)
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Finite Element Analysis: Thermomechanics of Solids
Note the following: d[ω(X)(X * − X)] = ω(X)d (X * − X) + [(X * − X)T dω T (X * − X)]T = ω(X)d (X * − X) − [(X * − X)T dω(X * − X)]T T
dω(X * − X) d ( X * − X ) . = ω(X)d (X * − X) − (X * − X)T d (X * − X) (4.62) Consequently, integration by parts furnishes −
∫
0
X ( λ *)
ω(X)d (X * − X) = ω(X(0))(X(λ *) − X(0)) +
∫
T
0
X ( λ* )− X ( 0 )
(X * − X) T
dω(X * − X) d (X * − X) . (4.63) d (X * − X)
The term ω (X(0))(X(λ*) − X(0)) can be identified as a rigid-body rotation since the rotation tensor ω (X(0)) is independent of position. We now focus on the second term. Now that rigid-body translation and rotation have been accommodated, the integral should give a unique value regardless of the path. Consider two paths, path 1 and path 2, from λ = 0 to λ*. The integral over these paths should produce the same unique values. Therefore, the closed path 0 to λ* along path 1 followed by λ* to 0 in a negative sense along path 2 should vanish. Furthermore, path 1 and path 2 are arbitrary, except for terminating at the same points. It follows that along any closed path
∫ (X * − X)
T
dω(X * − X) d (X * − X) = 0 T . d (X * − X)
(4.64)
ω 1(X) ω 2(X) ω 3(X)}, in which ω i denotes Let us now write ω (X) as ω (X) = {ω th the vector corresponding to the i column of ω. The compatibility condition can now be written as
∫ (X * − X)
T
dω j (X * − X) d (X * − X)
d (X * − X) = 0,
j = 1, 2, 3.
(4.65)
Ψ ⋅ dX = 0 over For a suitable differential function Ψ(X), the condition for Ψ an arbitrary contour is that the curl of Ψ vanishes, from which we express the compatibility conditions as dω j (X) ∇ × (X * − X)T = 0, ( X * − X) d
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j = 1, 2, 3.
(4.66)
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67
This can be expanded at considerable effort to derive 81 compatibility equations, including Equation 4.59.
4.7 SAMPLE PROBLEMS 1. Referring to Figure 4.5, determine u, F, J, and E as functions of X and Y. Use H = 1.0, W = 1.0, a = 0.1, b = 0.1, c = 0.3, d = 0.2, e = 0.2, f = 0.1. Assume a unit thickness in the Z-direction in both the deformed and undeformed configurations. Solution: Since straight sides are deformed into straight sides, the deformation pattern can be assumed in the form x = αX + βY + γXY
y = δX + εY + ζXY ,
and it is necessary to determine α, β, γ, δ, ε, and ζ from the coordinates of the vertices in the deformed configuration. At ( X , Y ) = (W , 0) : W + a = αW
b = δW
e = βH
H + f = εH
At ( X , Y ) = (0, H ) :
At ( X , Y ) = (W , H ) : W + c = αH + βH + γ WH
H + d = δW + εH + ζWH
After elementary manipulation, a = 1.1 W f ε = 1+ = 1.2 H
α = 1+
e = 0.2 H W + c − (W + a ) − e γ = = 0.4 WH
β=
Y
b = 0.1 W H + d − b − (H + f ) ζ= = 0. WH
δ=
Y
c e H
d
H f b W undeformed plate
X
W a deformed plate
FIGURE 4.5 Plate elements in undeformed and deformed states.
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X
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Finite Element Analysis: Thermomechanics of Solids
The 2 × 2 deformation gradient tensor F and its determinant are now:
β + γX 1.1 + 0.4Y = ε + ζX 0.1
α + γY F= δ + ζY
0.2 + 0.4 X 1.2 .
J = 1.3 + 0.48Y − 0.04 X The displacement vector is x − X (α − 1) X + βY + γXY 0.1X + 0.2Y + 0.4 XY u= = . = 0.1X + 0.2Y y − Y δX + (ε − 1)Y + ζXY The Lagrangian strain E =
E=
1 1.1 + 0.4Y 2 0.2 + 0.4 X
1 [FTF 2
− I] is
0.1 1.1 + 0.4Y 1.2 0.1
0.11 + 0.44Y + 0.08Y 2 = 0.17 + 0.22 X + 0.04Y + 0.08 XY
0.2 + 0.4 X 1 1 − 1.2 2 0
0 1
0.17 + 0.22 X + 0.04Y + 0.08 XY 0.24 + 0.08 X + 0.08 X 2
2. Figure 4.6 shows a square element at time t and at t + dt. Estimate L, D, and W at time t. Use a = 0.1dt, b = 1 + 0.2dt, c = 0.2dt, d = 1 + 0.4dt, e = 0.05dt, f = 0.1dt, g = 1 − 0.1dt, h = 1 + 0.5dt. Assume a unit thickness in the Z-direction in both the deformed and undeformed configurations. Y
Y
d c
g h 1
f
e X 1 element at time t
X b a element at time t+dt
FIGURE 4.6 Element experiencing rigid body motion and deformation.
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69
Solution: First, represent the deformed position vectors in terms of the undeformed position vectors using eight coefficients determined using the given geometry. In particular, x = α + βX + γY + δXY
y = ε + ζX + ηY + θXY .
Following procedures analogous to Problem 1, we find:
α = 0.1dt
ε = 0.05dt
β = 1 + 0.2 dt
ζ = 0.1dt
γ = 0.2 dt
η = 1 − 0.1dt
δ = −0.5dt
θ = −0.05dt.
The velocities can be estimated using vx ≈
x−X dt
and vy ≈
y−Y , dt
from which
v x = 0.1 + 0.2 X + 0.2Y − 0.5 XY v y = 0.05 + 0.1X − 0.1Y − 0.05 XY . The tensors L, D, and W are now readily found as: 0.2 − 0.5Y L= 0.1 − 0.05Y
0.2 − 0.5 X −0.1 − 0.05 X
0.2 − 0.5Y D= 0.15 − 0.25 X − 0.025Y 0 W= 0.4 + 0.25 X − 0.025Y
0.15 − 0.25 X − 0.025Y −0.1 − 0.05 X −0.4 − 0.25 X + 0.025Y . 0
4.8 EXERCISES 1. Consider a one-dimensional deformation in which x = (1 + λ)X. What value of λ is the linear-strain εL error by 5% relative to the Lagrangian strain E? Use the error measure error =
E − EL . E
2. A 1 × 1 square plate has a constant (2 × 2) Lagrangian-strain tensor E. What is the deformed length of the diagonal? What is the volume change?
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Finite Element Analysis: Thermomechanics of Solids
If the linear strain EL is now approximated as E, what is the diagonal and what is the volume change? Take 0.1 E= 0.05
0.05 . 0.1
3. Verify that the expressions in the text for F and E in cylindrical coordinates are consistent with the equation F = Q F ′,
T [Q T ]βα = qβα
[F ′]αζ =
hα ∂yα , Hζ ∂Yζ
in which Q is the orthogonal tensor representing transformation from the undeformed to the deformed coordinate system, and also with
[ E]ij =
1 2
∑ β
hβ2 ∂yβ ∂yβ δ − H H ∂Y ∂Y ij . i j i j
4. Obtain expressions for u, F, E, and EL in spherical coordinates. 5. For cylindrical coordinates, determine the Lagrangian strain in the following two cases: ( a) Pure radial expansion:
r = λR, θ = Θ, z = Z
(b) Torsion:
r = R, θ = Θ + λZ , z = Z
6. In spherical coordinates, determine the Lagrangian strain for pure radial expansion. r = λR, θ = Θ, φ = Φ 7. Obtain v, L, D, and W in spherical coordinates. 8. For cylindrical coordinates, find L, D, and W for the following flows: (a) pure radial flow:
vr = f (r ),
vθ = 0,
vz = 0
(b) pipe flow:
vr = 0,
vθ = 0,
vz = f (r )
(c) cylindrical flow:
vr = 0,
vθ = f (r ),
vz = 0
(d) torsional flow:
vr = 0,
vθ = f ( z ),
vz = 0
9. In spherical coordinates, find L, D, and W for pure radial expansion: vr = f (r ),
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vθ = 0,
vφ = 0
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71
10. For linear strain in rectilinear coordinates and 2-D, the compatibility relation is ∂2 E12 ∂2 E22 1 ∂2 E = 211 + . ∂x1∂x2 2 ∂x2 ∂x12 Find the implications of this relation for a linear-strain field assumed to be given by E11 = a1 X 2 + a2 XY + a3Y 2 E22 = b1 X 2 + b2 XY + b3Y 2 E12 = c1 X 2 + c2 XY + c3Y 2 . 11. Consider a square L × L plate (undeformed configuration) with linear strains E11 = a1 + a2 X + a3Y E22 = b1 + b2 X + b3Y E12 = c1 + c2 X + c3Y . Assuming that the origin does not move, find the deformed position of (X,Y) = (L,L).
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5
Mechanical Equilibrium and the Principle of Virtual Work
5.1 TRACTION AND STRESS 5.1.1 CAUCHY STRESS Consider a differential tetrahedron enclosing the point x in the deformed configuration. The area of the inclined face is dS, and dSi is the area of the face whose exterior normal vector is −ei. Simple vector analysis serves to derive that ni = dSi/dS (see Exercise 1 in Chapter 1). Next, let dP denote the force on a surface element dS, and (i ) let dP denote the force on area dSi. The traction vector is introduced by τ = dP/dS. As the tetrahedron shrinks to a point, the contribution of volume forces, such as inertia, decays faster than surface forces. Balance of forces on the tetrahedron now requires that dPj =
∑ dP . i j
(5.1)
i
The traction vector acting on the inclined face is defined by
τ=
dP , dS
∑
dPj( i )
∑
dPj( i ) dSi dSi dS
(5.2)
from which
τj =
dS
i
=
i
= Tij ni .
(5.3)
73
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3
dF n
dF
(2)
dS
(2)
dS dS
dF
(1)
dF (1) dS
2
(3)
(3)
1 FIGURE 5.1 Equilibrium of a tetrahedron.
with Tij =
dPj( i ) dSi
.
(5.4)
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 the th ij entry of the Cauchy stress T. In matrix-vector notation, the stress-traction relation is written as τ = T T n.
(5.5)
The next section will show that T is symmetric by virtue of the balance of angular T momentum. Equation 5.5 implies that T is a tensor, thus, it follows that T is a tensor. To visualize T, consider a differential cube. Positive stresses are shown on faces pointing in positive directions, as shown in Figure 5.2. 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 bar in Figure 5.3. The traction vector te1 acts at x = L, while the traction vector −te1 acts at x = 0. At x = L, the corresponding stress is t11 = te1 ⋅ e1 = t. At x = 0, the stress is given by (−te1) ⋅ (e1) = t. Clearly, the stress at both ends, and in fact throughout the bar, is positive (tensile). We will see later that the stress tensor is symmetric by virtue of the balance of angular momentum.
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3
T33 T32
T31
T23
T13 T11
T21
T22
T12
2
1 FIGURE 5.2 Illustration of the stress tensor.
Y
t
t
X
Z FIGURE 5.3 Tractions on a bar experiencing uniaxial tension.
5.1.2 1ST PIOLA-KIRCHHOFF STRESS The transformation to undeformed coordinates is now considered. In Chapter 3, we saw that τdS = T Τ ndS = T Τ JF − T n 0 dS0 = S T n 0 dS0 ,
S = JF −1T.
(5.6)
S is known as the 1st Piola-Kirchhoff stress tensor, and it is not symmetric.
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5.1.3 2ND PIOLA-KIRCHHOFF STRESS We next derive the stress tensor that is conjugate to the Lagrangian-strain rate, i.e., corresponds to the correct amount of work per unit undeformed volume. At a segment dS at x on the deformed boundary, assuming static conditions, the rate of work dW˙ of the tractions is dW˙ = dP T u˙ = τ T u˙ dS.
(5.7)
Over the surface S, shifting to tensor-indicial notation and invoking the divergence theorem, we find W˙ = = =
∫ τ u˙ dS T
∫ T n u˙ dS ij i
∂u˙ j
∫ ∂x
j
∫
Tij dV + u˙ j
i
∂Tij ∂xi
We will shortly see that static equilibrium implies that us to conclude that
W˙ =
∂u˙ j
∫ ∂x
∂Tij ∂x i
Tij dV .
= 0, which enables
(5.9)
i
Returning to matrix-vector notation, since
˙ = W
(5.8)
dV .
∂u˙ j ∂x i
= [L]ji, then
∫ tr(TL )dV T
=
∫ tr[T(D − W)]dV ,
=
∫ tr[TD]dV
(5.10)
since the trace vanishes for the product of a symmetric and antisymmetric tensor.
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To convert to undeformed coordinates, ˙ = W
∫ tr(TD)dV
=
∫ tr( JTF
=
∫
=
∫ tr(SE˙ )dV,
−1
E˙ F − T )dV0
−T
(5.11) −1
tr( JF TF E˙ )dV0 S = JF − T TF −1 .
The tensor S is called the 2nd Piola-Kirchhoff stress tensor. It is symmetric if T is symmetric. ˙ −1 ) = J tr(F −1TF˙ ) = tr( S F˙ ), so that the 1st PiolaNote that J tr(TL) = J tr(TFF Kirchhoff stress is said to be conjugate to the deformation-gradient tensor F.
5.2 STRESS FLUX Consider two deformations, x1 and x2, differing only by a rigid body motion: x 2 = V(t )x 1 + b(t ) ,
(5.12)
in which V(t) is orthogonal. A tensor A(x) is objective (see Eringen) if A( x 2 ) = VA( x1 )V T .
(5.13)
It turns out that the matrix of time derivatives of the Cauchy stress, T˙ , is not objective, while the deformation rate tensor, D, is. Thus, if ξ is a fourth-order tensor, ˙ = ξD would be senseless. Instead, the time a constitutive equation of the form T derivative is replaced with an objective stress flux, as explained in the following. First, note that F2 = VF1
(5.14)
L 2 = F˙ 2 F2−1 ˙ T VF + VF˙ ]F −1 V T = [VV 1 1 1 = Ω + VL1 V T ,
˙ T. Ω = VV
T The tensor Ω is antisymmetric since dl/dt = 0 = Ω + Ω .
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We seek a stress flux affording the simplest conversion from deformed to undeformed coordinates. Note that dS d = [JF − T TF −1 ] dt dt = Jtr(D)F − T TF −1 + JF˙ − T TF −1 + JF − T T˙F −1 + JF − T TF˙ −1 .
(5.15)
˙ −1 and F˙ − T = − F − T F˙ T F − T . Now, However, (F −1 F). = 0 so that F˙ −1 = − F −1 FF dS ° = JF − T T F −1 , dt
(5.16a)
o T = T˙ + tr( D)T − LT − TLT ,
(5.16b)
in which
o
is known as the Truesdell stress flux. Under pure rotation, F = Q and S˙ = JQ T Q T . o To prove the objectivity of T , note that o
T 2 = T˙2 + T2 tr(D 2 ) − L 2 T2 − T2 LT2
[
]
= [VT1 V T ]• + VT V T tr(D1 ) − [VL1 V T + Ω]VT1V T − VT1T V T VLT1 V T − Ω = VT˙1V T + ΩVT1V T − VT1V T Ω + VT1V T tr(D1 ) − VL1T1 V T − VT1LT1 V T − ΩVT1 V T + VT1 V T Ω o
= V T 1 VT .
(5.16c)
The choice of stress flux is not unique. For example, the stress flux given by
o
T = T − Ttr(D) = T˙ − LT − T LT
(5.17)
is also objective, as is the widely used Jaumann stress flux: ∆
T = T˙ + WT − T W,
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W=
1 (L − LT ). 2
(5.18)
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5.3 BALANCE OF MASS, LINEAR MOMENTUM, AND ANGULAR MOMENTUM 5.3.1 BALANCE
OF
MASS
Balance of mass requires that the total mass of an isolated body not change. For example, d dt
∫ ρ dV = 0,
(5.19)
in which ρ( x, t ) is the mass density. Since dV = JdV0 , it follows that ρJ = ρ 0 . In addition, 0= =
d ( ρJ ) dt ∂ ∂ ( ρJ ) + ( ρJ )v( x, t ), ∂t ∂x
v ( x, t ) =
∂x . ∂t
(5.20)
5.3.2 RAYLEIGH TRANSPORT THEOREM Let w( x, t ) denote a vector-valued function. Conversion of the volume integral from deformed to undeformed coordinates is achieved as
∫ ρw(x, t)dV = ∫ V
V0
ρ 0 w( x, t )dV0 .
(5.21)
The Rayleigh Transport Theorem is d dt
5.3.3 BALANCE
OF
d
∫ ρw(x, t)dV = ∫ ρ dt w(x, t)dV = 0 .
(5.22)
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: P=
∫ τ dS = dt ∫ ρ dt dV . d
du
(5.23)
Invoking Equation 5.22 yields
∫ © 2003 by CRC CRC Press LLC
τ dS =
∫
ρ
d 2u dV . dt 2
(5.24)
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In current coordinates, application of the divergence theorem furnishes the equilibrium equation
∫ τ dS = ∫ T n dS T
∫
= [∇ T T T ]T dV .
(5.25)
Equation 5.24 now becomes
∫
T T d 2uT T dV = 0 T . ∇ − ρ dt 2
(5.26)
Since this equation applies not only to the whole body, but also to arbitrary subdomains of the body, the argument of the integral in Equation 5.26 must vanish pointwise: ∇ TT T = ρ
d 2 uT . dt 2
(5.27)
To convert to undeformed coordinates, the 1st Piola-Kirchhoff stress is invoked to furnish
∫
S T n 0 dS0 =
∫
ρ0
d 2uT dV0 , dt 2
(5.28)
and the divergence theorem furnishes ∇ T0 S T = ρ0
d 2uT , dt 2
(5.29)
in which ∇ο denotes the divergence operator referred to undeformed coordinates. This equation will later be the starting point in the formulation of incremental variational principles.
5.3.4 BALANCE
OF
ANGULAR MOMENTUM
Assuming that only surface forces are present, relative to the origin, the total moment of the traction is equal to the rate of change of angular momentum:
∫ x × τ dS = dt ∫ x × ρ x˙ dV. d
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(5.30)
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To examine this principle further, it is convenient to use tensor-indicial notation. First, note using the divergence theorem that
∫ x × τ dS = ∫ ε
ijk
x j τk dS
=
∫ε
ijk
x j Tlk nl dS
=
∫ ∂x (ε
∂
ijk
x j Tlk )dV
l
=
∫ε
ijk
δ jl Tlk dV + ε ijk x j
∫
=
∫ε
ijk
Tjk dV + ε ijk x j
∫
∂ T dV ∂xl lk
∂ T dV . ∂xl lk
(5.31)
Continuing, d dt
∫ x × ρ x˙ dV = ∫ x × ρ ˙˙x dV + ∫ x˙ × ρ x˙ dV =
∫ x × ρ ˙˙x dV
=
∫ε
ijk
x j ρ˙˙ x k dV .
(5.32)
Balance of angular momentum can thus be restated as
0=
∫ε
ijk
∂ Tjk dV + ε ijk x j Tlk − ρ˙˙ x k dV . ∂xl
∫
(5.33)
The second term vanishes by virtue of balance of linear momentum (see Equation 5.27), leaving
ε ijk Tjk = 0 ,
(5.34)
which implies that T is symmetric: T = T (see exercises in Chapter 1). It follows from Equation 5.6 and Equation 5.11 that S is also symmetric, but that S is not symmetric. T
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5.4 PRINCIPLE OF VIRTUAL WORK The balance of linear and angular momentum leads to auxiliary variational principles that are fundamental to the finite-element method. Variational methods were introduced in Chapter 3. We recall the balance of linear momentum in rectilinear coordinates as ∂ T − ρu˙˙k = 0 , ∂xl kl
(5.35)
and Tlk = Tkl by virtue of the balance of angular momentum. We have tacitly assumed that x˙˙ = u˙˙k , 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 ∂
∫ δu ∂x T k
kl
l
− ρu˙˙k dV = 0 ,
(5.36)
in which δu k is an admissible variation of uk . Consider the spatial dependence of uk to be subject to variation, but not the temporal dependence. For example, if uk can be represented, at least locally, as u = N T ( x)γ ( t ) ,
(5.37a)
δu k = [N T ( x )]kl δγ l ( t ).
(5.37b)
then
The second term in the variational equation remains as − ∫ δuk ρu˙˙k dV . The first term is integrated by parts only once for reasons that will be identified shortly: it becomes ∫ ∂ [δu T ]dV − ∫ ∂δu k T dV . From the divergence theorem, ∂x l
∂x l
k lk
lk
∂
∫ ∂x [δu T ]dV = ∫ n [δu T ]dS k lk
l
k lk
l
∫
= δuk τ k dS,
(5.38)
which can be interpreted as the virtual work of the tractions on the exterior boundary. Next, since T is symmetric,
∫
∂δuk T dV = δ 3kl Tlk dV ∂xl lk
∫
∂u 1 ∂u 3kl = k + l , 2 ∂xl ∂x k
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(5.39)
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83
and we call 3kl the Eulerian strain. The term ∫δ 3kl Tlk dV can be called the virtual work of the stresses. Next, to evaluate ∫ δuk τ k dS, we suppose that the exterior boundary consists of three zones: S = S1 + S2 + S3 . On S1, the displacement uk is prescribed, causing the integral over S1 to vanish. On S2, suppose that the traction is prescribed as τ k (s). The contribution is ∫ S δukτ k (s)dS . Finally, on S3, suppose that tk = 2 τ k (s) − [A(s)]kl ul (s), thus furnishing ∫ S3 δukτ k (s)dS − ∫ S3 δuk [A(s)]kl ul (s)dS. The various contributions are consolidated into the Principle of Virtual Work (Zienkiewicz and Taylor, 1989) as
∫ δ 3 T dV + ∫ δu ρu˙˙ dV = ∫ δu τ (s)dS + ∫ δu τ (s)dS − ∫ δu [ A(s)] u (s)dS kl lk
k
k
k k
S2
s3
k k
s3
k
kl
l
(5.40) Now consider the case in which, as in classical elasticity, Tlk = d lkmn 3mn
T = D3,
(5.41)
in which dlkmn are the entries of a fourth-order, constant, positive-definite tensor D. The first term in Equation 5.40 now becomes δ ∫ 1 3kl d lkmn 3mn dV , with a positive2 definite integrand. Achieving this outcome is the motivation behind integrating by parts just once. In the language of Chapter 3, uk is the primary variable and τk is the secondary variable: on any boundary point, either uk or τk should be specified, or, more generally, τk should be specified as a function of uk. Finally, application of T the interpolation model (see Equation 5.37) and cancellation of δ γ furnishes the ordinary differential equation M˙˙ γ + [K + H]γ = f (t ) ,
(5.42)
in which
M=
∫ ρN(x)N (x)dV
H=
∫ N(x)AN (x)dV
T
T
S3
K= f (t ) =
∫
∫ BDB dV
S2 + S3
T
(5.43)
N(s) τ k (s)dS
and in which VEC(T ) = χVEC(3) , χ = TEN22(D). In addition, VEC(3) = BT ( x )γ . B is derived from the strain-displacement relations, and 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 a consistent force vector. These notions will be addressed in greater detail in subsequent chapters.
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To convert the Principle of Virtual Work to undeformed coordinates,
∫ δu ρu˙˙ dV → ∫ δu ρ u˙˙ dV k
∫
S2
δukτ k (s)dS +
∫
S3
k
0 k
k
δukτ k (s)dS →
∫
S20
0
δukτ k0 (s0 )dS0 +
∫
S30
(5.44)
δukτ k0 (s0 )dS0 ,
using, from Chapter 3,
µ = J n T0 C −1 n 0 .
τ k0 = µτ k
(5.45)
Next,
∫ δu [A(s)] u (s)dS → ∫ δu [µA(s)] u (s )dS . S3
k
kl
l
k
S3
kl l
0
0
(5.46)
Some manipulation is required to convert the virtual work of the stresses. Observe that δ3 =
1 ∂δ u ∂δ u + 2 ∂ x ∂x
T
=
T 1 ∂δ u ∂X ∂δ u ∂X + 2 ∂X ∂x ∂X ∂x
=
1 [δFF −1 + F − T δF T ] 2
1 = F − T [F T δ F + δF T F] F −1 2 = F − T δ EF −1 .
(5.47)
Now
∫ δ3
T dV =
kl lk
∫ tr(δ 3T)dV
=
∫ tr(F
=
∫ tr(δ EJ F
=
∫ tr(δ ES)dV
−T
δ E F −1 T) JdV0 −1
T F − T )dV0
0
∫
= δE ji Sij dV0 .
© 2003 by CRC CRC Press LLC
(5.48)
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Consolidating the terms, the Principle of Virtual Work in undeformed coordinates is
∫ δE S dV + ∫ δu ρ u˙˙ dV = ∫ ji ij
0
k
0 k
0
S20
−
δuk τ k0 (s0 )dS0 +
∫
S30
∫
S30
δuk τ k0 (s0 )dS0
δuk [ µA(s 0 )]kl ul (s0 )dS0 .
(5.49)
5.5 SAMPLE PROBLEMS 1. The following plate is subject to the tractions shown. At B, there is a frictionless roller support. Find the reactions at A and B. Assume that the plate has thickness W. y
τ = τ0[ ex + ey ]
x L
H x A
B L
FIGURE 5.4 Plate under traction.
Solution: The reactions are denoted as RAx , RAy , RBx , and RBy and are all assumed to be positive. Clearly, RBx = 0. Force equilibrium requires that RAx − tW RAy + RBy − tW
∫
L
∫
L
0
0
x tW dx = 0 ⇒ RAx = 2L L x tW . dx = 0 ⇒ RAy + RBy = 2L L
Balance of moments about A requires that LRBy + tW Accordingly, RAy =
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5 tW 6 L
∫
L
0
.
x2 tW . dx = 0 ⇒ RBy = − 3L L
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2. The figure shows a square plate of uniform thickness W. It experiences Cauchy stresses given by T11 T = T21 0
T12 0 0
0 0 . 0
On diagonal AB, find the total transverse force and moment about A. y B
x A
L
FIGURE 5.5 Plate element under stress.
Solution: The axes are rotated through 45 degrees so that the x-axis coincides with the diagonal (x′-axis). Now, Tx ′x ′ =
Txx + T xy 2
Ty′y′ =
Txx − T xy 2
The transverse force is T Fy′ = xx − T xy 2 WL. 2 The moment about A is M=
© 2003 by CRC CRC Press LLC
2 LFy′ . 2
Tx ′y′ = −
Txx . 2
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3. The plate illustrated has thickness W, width dx, and height dy. The stresses Txz, Tyz, and Tzz vanish. Find the equations resulting from balancing forces and moments. In particular, prove that Txy = Tyx . Tyy + dTyy
Tyx + dTyx
Txy + dTxy Txx + dTxx
dy
Txx Txy
dx Tyx
Tyy FIGURE 5.6 Moment balance on a plate element under nonuniform stress.
Solution: For the x-direction forces, dTyx dTyx dT dT W Txx + xx dx − Txx dy + W Tyx + dy − Tyx dx = 0 ⇒ xx + =0. dx dy dx dy For y-direction forces, dTxy dTyy dTxy dTyy W Txy + dx − Txy dy + W Tyy + dy − Tyy dx = 0 ⇒ + = 0. dx dy dx dy For moments about A, dTyx dT dy −W Txx + xx dx − Txx dy − W Tyx + dy dxdy dx dy 2 dTxy dTyy dx dx dydx + W Tyy + dy − Tyy dx = 0. + W Txy + dx dy 2
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Finite Element Analysis: Thermomechanics of Solids
Consequently, neglecting terms of third order in differentials, −WTyx dxdy + WTxy dydx = 0 ⇒ Txy = Tyx . 4. At point (0,0,0), the tractions τ1, τ 2, τ 3 act on planes with normal vectors n1, n2, and n3. Find the Cauchy stress tensor T given n1 =
1 [e + e 2 + e 3 ] 3 1
τ1 =
1 [6e 1 + 9e 2 + 15e 3 ] 3
n2 =
1 [e + e 2 − e 3 ] 3 1
τ2 =
1 [0e 1 + 1e 2 + 3e 3 ] 3
n3 =
1 [e − e 2 − e 3 ] 3 1
τ3 = −
1 [4e 1 + 5e 2 + 7e 3 ]. 3
Solution: This problem requires application of the stress-traction relation. Now 6 T11 1 9 = T 3 21 12 T31 ⇒ T11 + T12 + T13 = 6
T12 T22 T32
T13 1 1 T23 1 3 T33 1
(1) T21 + T22 + T23 = 9
(2) T31 + T32 + T33 = 1 (3) .
Similarly,
1 2 ⇒ T11 + T12 = 3
3 T11 5 = T21 7 T31
( 4)
(7)
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T22 T32
T21 + T22 = 5
−4 T11 1 −5 = T 3 21 −6 T31 ⇒ T11 − T12 − T13 = −4
T12
T12 T22 T32
T13 1 1 T23 1 2 T33 0 T31 + T32 = 7
(5)
(6)
T13 1 1 T23 −1 3 T33 −1
T21 − T22 − T23 = −5
(8)
T31 − T32 − T33 = −6
(9)
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The solution is straightforward. For example, adding (1) and (7) furnishes T11 = 1, from which (4) implies T12 = 2. Solving the equations furnishes us with 1 T = 2 3
3 4 . 5
2 3 4
5. Given the Cauchy stress T, find S and S if x(t) = Q(t)X, Q(t)Q(t) = I. T
−T
Solution: Clearly F = Q, thus J = 1. Also, F T Q(t)T and S = Q(t)TQ(t) .
= Q. It is immediate that S =
5.6 EXERCISES 1. In classical linear elasticity, introduce the isotropic stress and isotropic linear strain as s = tr( S)
e = tr( EL ),
and introduce the deviatoric stress and strain using 1 sd = s − si 3
1 ed = e − esi. 3
Verify that, if s˙ = zµe˙ + λtr(ε )i i Tsd = 0
iTed = 0
sd = 2 µed s = (2 µ + 3λ )e . 2. Find the reaction forces in the following problem. There is a frictionless roller support at B.
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y
τ = τ0[–ex + ey ][1 –y/H]
H x A
B L
FIGURE 5.7 Plate element under traction.
3. Find the net force and moment about A (thickness = W) on the diagonal in the following plate, given the Cauchy stresses x T = y 0
y xy 0
0 0 . 0
y B
H
x
A L FIGURE 5.8 Plate element experiencing stress.
4. At point (0,0,0), the tractions τ1, τ 2, τ 3 act on planes with normal vectors n1, n2, and n3. Find the Cauchy stress tensor T, given, n1 =
1 [e + e 2 + e 3 ] 3 1
τ1 =
n2 =
1 [e + e 2 ] 2 1
τ2 =
n3 =
1 [e − e 2 − e 3 ] 3 1
τ3 = −
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1 [2e 1 − 5e 2 + 6e 3 ] 3 1 [e − e 2 + e 3 ] 2 1 1 [6e 1 − 1e 2 + 2e 3 ]. 3
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nd
5. Given the Cauchy stress tensor T, find the 1 and 2 Piola-Kirchhoff stress ∆(t)X, in which tensors if x(t) = Q(t)∆ 1 + at ∆ (t ) = 0 0
0 1 + bt 0
0 0 . 1 + ct
6. The Cauchy stress tensor is given by T11 T = T12 0
T12 T22 0
0 0 . 0
Now suppose the 1-2 axes are rotated through +θ degrees around the z-axis, to furnish 1′ − 2′ axes. Prove that T1′ 1′ =
T11 + T22 T11 − T22 + cos 2θ + T12 sin 2θ 2 2
T2′2′ =
T11 + T22 T11 − T22 − cos 2θ − T12 sin 2θ 2 2
T1′ 2′ = −
T11 − T22 sin 2θ + T12 cos 2θ 2
7. Verify that Eij( L ) =
1 λ Tij − Tkk δ ij 2µ 2 µ + 3λ
is the inverse of Tij = 2 µEij( L ) + λEkk( L )δ ij . 8. In undeformed coordinates and Kronecker Product (VEC) notation, the 2nd Piola-Kirchhoff stress for incompressible hyperelastic materials can be written as s=
∂w ′ = 2ϕ1′n1′ + 2ϕ 2′ n ′2 − pI3 n 3 . ∂e
Find the corresponding expression in deformed coordinates. Derive ψ 1′ and ψ 2′ , in which direct transformation furnishes t = 2ψ 1′m1′ + 2ψ 2′ m ′2 − pi .
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9. Prove that, under uniaxial tension in an isotropic, linearly elastic material, E22 = E33. 10. Obtain λ in terms of E and ν. (See Problem 7.) ( L) 11. The bulk modulus κ is defined by tkk = 3κekk . Obtain κ as a function of E and ν. (See Problem 7.) 12. The 2" × 2" × 2" cube shown in Figure 5.9 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 7 100 lbf. Take E = 10 psi and ν = 1/3. Find all nonzero stresses and strains. What is the volume change? What are the principal stresses and strains? What is the maximum shear stress?
y
F
x z
FIGURE 5.9 Strain in a constrained cube.
13. Write out the proof of the Principle of Virtual Work in deformed coordinates. 14. Write out the conversion of the Principle of Virtual Work to undeformed coordinates. 15. Write out the steps whereby the Principle of Virtual Work leads to M˙˙ γ + [K + H]γ = f (t ) . 16. Suppose that u = N (x)γγ (t) and assume two dimensions. In 2-D, find B T in terms of the derivatives of N (x), in which T
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T
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Mechanical Equilibrium and the Principle of Virtual Work
311 321 T = B ( x )γ (t ), 312 322 1 N T (x) = 0
u1 T = N ( x ) γ (t ) u2
x
y
0
0
0
0
1
x
0 Φ y 0
Φ is a 1 × n constant matrix (row vector); γ is n × 1.
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93
0 . Φ
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Stress-Strain Relation and the Tangent-Modulus Tensor
6.1 STRESS-STRAIN BEHAVIOR: CLASSICAL LINEAR ELASTICITY Under the assumption of linear strain, the distinction between the Cauchy and PiolaKirchhoff stresses vanishes. The stress is assumed to be given as a linear function of linear strain by the relation Tij = cijkl Ekl( L ) ,
(6.1)
in which cijkl are constants and are the entries of a 3 × 3 × 3 × 3 fourth-order tensor, C. If T and EL were not symmetrical, C might have as many as 81 distinct entries. However, due to the symmetry of T and EL there are no more than 36 distinct entries. Thermodynamic arguments in subsequent sections will provide a rationale for the Maxwell relations: ∂Tij ∂Ekl
=
∂Tkl . ∂Eij
(6.2)
It follows that cijkl = cklij, which implies that there are, at most, 21 distinct coefficients. There are no further arguments from general principles for fewer coefficients. Instead, the number of distinct coefficients is specific to a material, and reflects the degree of symmetry in the material. The smallest number of distinct coefficients is achieved in the case of isotropy, which can be explained physically as follows. Suppose a thin plate of elastic material is tested such that thin strips are removed at several angles and then subjected to uniaxial tension. If the measured stress-strain curves are the same and independent of the orientation at which they are cut, the material is isotropic. Otherwise, it exhibits anisotropy, but may still exhibit limited types of symmetry, such as transverse isotropy or orthotropy. The notion of isotropy is illustrated in Figure 6.1. In isotropic, linear-elastic materials (which implies linear strain), the number of distinct coefficients can be reduced to two, m and l, as illustrated by Lame’s equation, Tij = 2 µ Eij( L ) + λ Ekk( L )δ ij .
(6.3) 95
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Finite Element Analysis: Thermomechanics of Solids
T 34 1
2
1
2 3 4
E FIGURE 6.1 Illustration of isotropy.
This can be inverted to furnish Eij( L ) =
1 λ T − T δ . 2 µ ij 2 µ + 3λ kk ij
(6.4)
The classical elastic modulus E0 and Poisson’s ratio n represent response under uniaxial tension only, provided that T11 = T, Tij = 0. Otherwise, E=
T11 E11( L )
ν=−
( L) E( L) E22 = − 33 . ( L) E11 E11( L )
(6.5)
1 ν λ = , E 2 µ 2 µ + 3λ
(6.6)
It is readily verified that 1 1 λ = 1− E 2 µ 2 µ + 3λ from which it is immediate that 1 1+ν = , 2µ E
(6.7)
Leaving the case of uniaxial tension for the normal (diagonal) stresses and strains, we can write E11( L ) = =
1 1 λ λ 1− T − (T + T ) 2 µ 2 µ + 3λ 11 2 µ 2 µ + 3λ 22 33 1 [T − ν (T22 + T33 )], E 11 ( L) E22 =
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1 [T − ν (T33 + T11 )], E 22
(6.8) (6.9)
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97
and 1 [T − ν (T11 + T22 )], E 33
( L) E33 =
(6.10)
and the off-diagonal terms satisfy E12( L ) =
1+ν T E 12
( L) E23 =
1+ν T E 23
( L) E31 =
1+ν T . E 31
(6.11)
6.2 ISOTHERMAL TANGENT-MODULUS TENSOR 6.2.1 CLASSICAL ELASTICITY Under small deformation, the fourth-order tangent-modulus tensor D in linear elasticity is defined by dT = DdEL .
(6.12)
In linear isotropic elasticity, the stress-strain relations are written in the Lame’s form as T = 2 µEL + λ tr( EL )I.
(6.13)
Using Kronecker Product notation from Chapter 2, this can be rewritten as VEC(T ) = [2 µI ⊗ I + λiiT ]VEC( EL ),
(6.14)
from which we conclude that D = ITEN22(2 µI ⊗ I + λiiT ).
(6.15)
6.2.2 COMPRESSIBLE HYPERELASTIC MATERIALS In isotropic hyperelasticity, which is descriptive of compressible rubber elasticity, nd the 2 Piola-Kirchhoff stress is taken to be derivable from a strain-energy function that depends on the principal invariants I1, I2, I3 of the Right Cauchy-Green strain tensor: S=
dw dw =2 dE dC
s = VEC( S)
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sT =
dw dw =2 de dc
e = VEC( E)
(a) (6.16) c = VEC( C ).
(b)
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Finite Element Analysis: Thermomechanics of Solids
Now, s = 2φi ni ,
φi =
∂w , ∂I i
niT =
∂I i . ∂c
(6.17)
From Chapter 2, n1 = i
n2 = I1i − c
n3 = VEC(C−1 ) I3 .
(6.18)
The tangent-modulus tensor Do, referred to the undeformed configuration, is given by dS = DodE
ds = TEN 22( Do )dE,
(6.19)
and now TEN 22( Do ) = 4φij ni nTj + 4φi A i ,
Ai =
dni . dc
(6.20)
Finally, A1 =
di = 0T dc
A2 =
d I dc 2
=
d [ I i − c] dc 1
= iiT − I 9 A3 = =
(a)
(b)
(6.21)
d [VEC(C −1 ) I3 ] dc d [ I i − I1c + VEC(C 2 )] dc 2
= inT2 − ciT + C ⊕ C = I1[iiT − I 9 ] − [ic T + ciT ] + C ⊕ C
(c)
In deriving A3 we have taken advantage of the Cayley-Hamilton theorem (see Chapter 2).
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99
6.3 INCOMPRESSIBLE AND NEAR-INCOMPRESSIBLE HYPERELASTIC MATERIALS Polymeric materials, such as natural rubber, are often nearly incompressible. For some applications, they can be idealized as incompressible. However, for applications involving confinement, such as in the corners of seal wells, it may be necessary to accommodate the small degree of incompressibility to achieve high accuracy. Incompressibility and near-incompressibility represent internal constraints. The principal (e.g., Lagrangian) strains are not independent, and the stresses are not determined completely by the strains. Instead, differences in the principal stresses are determined by differences in principal strains (Oden, 1972). An additional field must be introduced to enforce the internal constraint, and we will see that this internal field can be taken as the hydrostatic pressure (referred to the current configuration).
6.3.1 INCOMPRESSIBILITY The constraint of incompressibility is expressed by the relation J = 1. Now, J = det F = det 2 (F) = det(F) det(F T ) = det 2 (F T F) = I3 ,
(6.22)
and consequently, I3 = 1.
(6.23)
The constraint of incompressibility can be enforced using a Lagrange multiplier (see 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, w′, similar to w ′ = w( I1′, I2′ ) −
1 p( I3 − 1), 2
I1′ =
I1 , I31/ 3
I2′ =
I2 , I32 / 3
(6.24)
in which w is interpreted as the conventional strain-energy function, but with dependence on I3 (=1) removed. I1′ and I2′ are called the deviatoric invariants. For reasons to be explained in a later chapter presenting variational principles, this form serves
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Finite Element Analysis: Thermomechanics of Solids
to enforce incompressibility, with S now given by s=
∂w ′ ∂e
= 2φ1′n1′ + 2φ 2′ n ′2 − pI 3 n 3
φ1′ =
∂w ∂I1′
φ 2′ =
∂w ∂I2′
(6.25) ∂I ′ n1′ = 1 ∂c
T
T
∂I ′ n ′2 = 2 . ∂c −1
−T
To convert to deformed coordinates, recall that S = JF T F . It is left to the reader in an exercise to derive ψ 1′ and ψ 2′ in which t = VEC(T ) = 2ψ 1′m1′ + 2ψ 2′ m ′2 − pi
(6.26)
in which iTm1′ = 0 and iTm′2 = 0. It follows that p = −tr(T)/3 since tr(T ) = i T t = 2ψ 1′i T m 1′ + 2ψ 2′ i T m ′2 − pi T i = − i T ip = −3 p.
(6.27)
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 subsequent chapters that it should be defined ∗ as D using ds de T TEN 22( D ∗ ) = − ds dp
ds dp 0
ds = 4[φ1′A1′ + φ 2′ A ′2 + φ11 ′ n1′ (n1′ ) T + φ12 ′ n1′ (n ′2 ) T de
(6.28)
+ φ 21 ′ n ′2 (n1′ ) T + φ 22 ′ n ′2 (n ′2 ) T ] ds = − I3 n3 . dp Example: Uniaxial Tension Consider the Neo-Hookean elastomer satisfying w = α [I1 − 3]
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and subject to
I3 = 1.
(6.29)
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101
We seek the relation between s1 and e1, which will be obtained twice: once by enforcing the incompressibility constraint a priori, and the second by enforcing the constraint a posteriori. a priori: Assume for the sake of brevity that e2 = e3. I3 = 1 implies that c2 = 1/ c1 . The strain-energy function now is w = α [c1 + 2c − 3]. The stress, s1, is 1 now found as
s1 = 2
dw 1 = 2α 1 − 3/2 . dc1 c1
(6.30)
a posteriori: Use the augmented function w ′ = α [ I1 − 3] −
p [ I − 1]. 2 3
(6.31)
Now
s1 = 2
dw ′ = 2α − p/c1 dc1
0 = s2 = 2
dw ′ = 2α − p/c2 dc 2 (6.32)
dw ′ = 2α − p/c3 0 = s3 = 2 dc3 0=
dw ′ = 0 → I 3 = 1. dp
Thus, it follows that c2 = c3 = 1/ c1 and p/c2 = 2α . We conclude that s1 = 2α − p / c1 = 2α −
c2 p / c2 c1
c = 2α 1 − 2 c1 1 = 2α 1 − 3/ 2 . c1
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(6.33)
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Finite Element Analysis: Thermomechanics of Solids
a posteriori with deviatoric invariants: Consider the augmented function with deviatoric invariants:
[
]
w ′ = α I1 I31/ 3 − 3 − s1 = 2
p [ I − 1] 2 3
(a)
1 I3 dw ′ 1 I1 I3 = 2α 1 3 − − p 43 dc1 I 3 I c c 3 1 1 3
(b) (6.34)
1 I3 dw ′ 1 I1 I3 0 = s2 = 2 = 2α 1 3 − − p 43 dc2 I 3 I c c 3 2 2 3
(c)
1 I3 dw ′ 1 I1 I3 = 2α 1 3 − − p 43 dc3 3 I3 c3 c3 I3
(d)
0 = s3 = 2
Hence, c2 = c3. Furthermore, dw ′ = 0 → I3 = 1. dp
(6.34e)
Equation 6.34 now implies that 2α c2 − c1 p = , 3 c2 c2
(6.35)
and substitution into Equation 6.34b furnishes 1 s1 = 2α 1 − 3/ 2 , c1
(6.36)
as in the a priori case and in the first a posteriori case. Now, the Lagrange multiplier p can be interpreted as the pressure referred to current coordinates.
6.3.2 NEAR-INCOMPRESSIBILITY As will be seen in Chapter 18, the augmented strain-energy function w ′′ = w( I1′, I2′ ) −
1 p2 p[ I3 − 1] − 2 2κ
(6.37)
[
(6.38)
serves to enforce the constraint
]
p = −κ I3 − 1 .
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103
Here, k is the bulk modulus, and it is assumed to be quite large compared to, for example, the small strain-shear modulus. The tangent-modulus tensor is now ds de T = − ds dp
( )
TEN 22 D *
ds dp . 1 κ
(6.39)
6.4 NONLINEAR MATERIALS AT LARGE DEFORMATION Suppose that the constitutive relations are measured at a constant temperature in the current configuration as o
T = DD,
(6.40)
in which the fourth-order tangent-modulus tensor D can, in general, be a function of stress, strain, temperature, and internal-state variables (discussed in subsequent o chapters). This form is attractive since T and D are both objective. Conversion to undeformed coordinates is realized by S˙ = JF −1 DDF − T −1
(6.41)
−T
= JF DF E˙ F −1F − T . If s = VEC(S) and e = VEC(E), then s˙ = JI ⊗ F −1VEC( DF − T E˙ F −1F − T ) = JI ⊗ F −1F −1 ⊗ ITEN 22( D)VEC(F − T E˙ F −1 ) = JF −1 ⊗ F −1TEN 22( D)(F − T ⊗ I)(I ⊗ F − T )VEC( E˙ ) = JF −1 ⊗ F −1TEN 22( D)F − T ⊗ F − T e˙ = TEN 22( Do )e˙ .
(6.42)
Recalling Chapter 2, it follows that S˙ = Do E˙ ,
(6.43)
Do = ITEN 22( JF −1 ⊗ F −1TEN 22( D)F − T ⊗ F − T )
(6.44)
in which
is the tangent-modulus tensor Do referred to undeformed coordinates.
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Finite Element Analysis: Thermomechanics of Solids
Suppose instead that the Jaumann stress flux is used and that ∆
T = DD.
(6.45)
Now ∆ S˙ = JF − T T + Ttr( D) − (L + W)T − T (LT − W) F −1
[
]
= JF − T DF − T E˙ F −1 + Ttr(F − T E˙ F −1 ) − (L + W)T − T (LT − W) F −1.
(6.46)
For this flux, there does not appear to be any way that S˙ can be written in the form of Equation 6.42, i.e., is determined by E˙ . To see this, consider VEC( S˙ ) = TEN 22( D ′)VEC( E˙ ) − TEN 22( D ′′)VEC(L + W)
(a)
TEN 22( D ′) = [I ⊗ (F − T ) 2 TEN 22( JF − T DF − T ) + VEC(T )VEC T (FF − T )]
(b)
TEN 22( D ′′) = J[I ⊗ F −1 ]2 [T ⊗ I − I ⊗ TU 9 ]
(c) (6.47)
In the second term in Equation 6.47a, VEC(L + W) cannot be eliminated in favor of VEC(D), and hence in favor of VEC( E˙ ) . Note that VEC(D) =
1 1 VEC(L + LT ) = (I + U 9 )VEC(L). 2 2
(6.48)
I + U9 is singular, as seen in the following argument. Recall that U9 is symmetric and thus has real eigenvalues. However, U 29 = I and thus the eigenvalues of U9 are either 1 or −1. Some of the eigenvalues must be −1. Otherwise, U9 would be the identity matrix, in which case it would not, in general, have the permutation property identified in Chapter 2. Thus, some of the eigenvalues of I + U9 vanish. Instead, we write ˙ ) − D ′′VEC(D) − D ′′VEC( D + 2 W), VEC( S˙ ) = D ′VEC(E
(6.49)
and we see that if the Jaumann stress flux is used, S˙ is determined by E˙ and the spin W. Recall that the spin does not vanish under rigid-body rotation.
6.5 EXERCISES 1. In classical linear elasticity, introduce the isotropic stress and isotropic linear strain as s = tr( S)
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e = tr( EL ),
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105
and introduce the deviatoric stress and strain using 1 sd = s − si 3
1 ed = e − ei. 3
Verify that iTsd = 0
iTed = 0
sd = 2 µed s = (2 µ + 3λ )e. 2. Verify that Equation 6.3 can be inverted to furnish e=
1 λ t− (i T t )i, 2 µ (2 µ + 3λ )
t = VEC(T )
3. In classical linear elasticity, under uniaxial stress, the elastic modulus E and Poisson’s ratio ν are defined by E=
σ 11 ε11
ν=−
ε ε 22 = − 33 , ε11 ε11
σ 11 ≠ 0 σ ij = 0
ij ≠ 11 .
Prove that 1 1 λ = 1− E 2 µ (2 µ + 3λ )
ν=−
λ 2( µ + λ )
µ=
E . 2(1 + ν )
4. Obtain v, L, D, and W in spherical coordinates. 5. For cylindrical coordinates, find L, D, and W for the following flows: (a) pure radial flow:
vr = f (r ),
vθ = 0,
vz = 0
(b) pipe flow:
vr = 0,
vθ = 0,
vz = f (r )
(c) cylindrical flow:
vr = 0,
vθ = f (r ),
vz = 0
(d) torsional flow:
vr = 0,
vθ = f ( z ),
vz = 0
nd
6. In undeformed coordinates, the 2 Piola-Kirchhoff stress for an incompressible, hyperelastic material is given by s=
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∂w ′ = 2ϕ1′n1′ + 2ϕ 2′ n′2 − pI3n3 . ∂e
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Finite Element Analysis: Thermomechanics of Solids
Find the corresponding expression in deformed coordinates: derive ψ 1′ and ψ 2′ , in which direct transformation furnishes t = 2ψ 1′m1′ + 2ψ 2′ m′2 − pi. 7. Prove that under uniaxial tension in an isotropic linearly elastic material e22 = e33. (Problem 9, Chapter 5.) 8. Obtain λ in terms of E and ν. (Problem 10, Chapter 5.) ( L) 9. The bulk modulus K is defined by tkk = 3Κekk . Obtain K as a function of E and ν. (Problem 11, Chapter 5.) 10. The 2" × 2" × 2" shown in Figure 6.2 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 lbf. Take E = 10 ^ 7 psi and ν = 1/3. Find all nonzero stresses and strains. What is the volume change? What are the principal stresses and strains? What is the maximum shear stress? (Problem 12, Chapter 5.) v E
x z
FIGURE 6.2 Strain in a constrained plate.
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7
Thermal and Thermomechanical Response
7.1 BALANCE OF ENERGY AND PRODUCTION OF ENTROPY 7.1.1 BALANCE
OF
ENERGY
The total energy increase in a body, including internal energy and kinetic energy, is equal to the corresponding work done on the body and the heat added to the body. In rate form, K˙ + Ξ˙ = W˙ + Q˙ ,
(7.1)
in which: Ξ is the internal energy with density ξ Ξ˙ =
∫ ρξ˙ dV;
(7.2a)
W˙ is the rate of mechanical work, satisfying W˙ =
∫ u˙ τ dS, T
(7.2b)
Q˙ is the rate of heat input, with heat production h and heat flux q, satisfying Q˙ =
∫ ρhdV − ∫ n q dS, and T
(7.2c)
K˙ is the rate of increase in the kinetic energy, K˙ =
∫
ρ u˙ T
du˙ dV . dt
(7.2d)
It has been tacitly assumed that all work is done on S, and that body forces do no work. 107
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Finite Element Analysis: Thermomechanics of Solids
Invoking the divergence theorem and balance of linear momentum furnishes
du˙
∫ ρξ˙ + u˙ ρ dt − ∇ T − tr(T D) − ρh + ∇ qdV = 0. T
T
T
(7.3)
The inner 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 undeformed coordinates, is obtained as
ρξ˙ = tr(TD) − ∇ T q + ρh.
(7.4)
To convert to undeformed coordinates, note that
∫ n qdS = ∫ q J F T
T
=
−T
n 0 dS0
∫ q n dS T 0
0
q 0 = J F −1q.
0
(7.5)
In undeformed coordinates, Equation 7.3 is rewritten as
∫ [ρ ξ˙ − tr(SE)˙ − ρ h + ∇ q ] dV 0
0
T 0
0
0
= 0,
(7.6a)
furnishing the local form
ρ 0ξ˙ − tr( SE˙ ) − ρ 0 h + ∇ T0 q 0 = 0.
(7.6b)
7.1.2 ENTROPY PRODUCTION INEQUALITY Following the thermodynamics of ideal and non-ideal gases, the entropy production inequality is introduced as follows (see Callen, 1985): H˙ =
∫
ρη˙ dV ≥
∫
h dV − T
∫
nTq dS, T
(7.7a)
in which H is the total entropy, η is the specific entropy per unit mass, and T is the absolute temperature. This relation provides a “framework” for describing the irreversible nature of dissipative processes, such as heat flow and plastic deformation. We apply the divergence theorem to the surface integral and obtain the local form of the entropy production inequality:
ρTη˙ ≥ h − ∇ T q + q T ∇T/T.
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(7.7b)
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109
The corresponding relation in undeformed coordinates is
ρ 0 Tη˙ ≥ h − ∇ T0 q 0 + q T0 ∇ 0 T/ T.
(7.7c)
7.1.3 THERMODYNAMIC POTENTIALS The Balance of Energy introduces the internal energy Ξ, which is an extensive variable—its value accumulates over the domain. The mass and volume averages of extensive variables are also 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 heat flux is completely converted into entropy according to Q˙ = TH˙ .
(7.8)
(We shall consider several irreversible effects, such as plasticity, viscosity, and heat conduction.) In undeformed coordinates, the balance of energy for reversible processes can be written as ˙ + ρ Tη˙ . ρ 0ξ˙ = tr( SE) 0
(7.9)
We call this equation the thermal equilibrium equation. It is assumed to be integrable, so that the internal energy is dependent on the current state represented by the current values of the state variables E and η. For the sake of understanding, we can think of T as a thermal stress and η as a thermal strain. Clearly, η˙ = 0 if there is no heat input across the surface or generated in the volume. Consequently, the entropy is a convenient 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 that maximize the internal energy. Other thermodynamic potentials, depending on alternate state variables, can be introduced by a Lorentz transformation, as illustrated in the following equation. 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:
ρ 0 g = ρ 0ξ − tr( SE) − ρ 0 Tη,
(7.10a)
ρ 0 g˙ = −tr( ES˙ ) − ρ 0ηT˙ .
(7.10b)
from which
Stability of thermodynamic equilibrium requires that S and T assume values that minimize g under suitable conditions. This potential is of interest in fluids experiencing adiabatic conditions since the pressure (stress) is accessible to measurement using, for example, pitot tubes.
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Finite Element Analysis: Thermomechanics of Solids
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
ρ 0 f = ρ 0ξ − ρ 0 Tη,
(7.11a)
ρ0 f˙ = tr( SE˙ ) − ρ0ηT˙ .
(7.11b)
furnishing
It is evident that f is a function of both an intensive and an extensive variable. At thermodynamic equilibrium, 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 ρ0h = ρ0x − tr(SE), and
ρ 0 h˙ = −tr( ES˙ ) + ρ 0 Tη˙ .
(7.12)
The enthalpy also 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.
7.2 CLASSICAL COUPLED LINEAR THERMOELASTICITY The classical theory of coupled thermoelasticity in isotropic media corresponds to the restriction to the linear-strain tensor, E ≈ E˙ L and to the stress-strain temperature relation T = 2 µE + λ[tr( E) − α (T − T0 )]I.
(7.13)
Here, α is the volumetric coefficient of thermal expansion, typically a small number in metals. If the temperature increases without stress being applied, the strain increases according to evol = tr(E) = α(T − T0). Thermoelastic processes are assumed to be reversible, in which case, −∇ ⋅ q + h = ρTη˙ . It is also assumed that the specific heat at constant strain, ce, given by ce = T
∂η , ∂T
(7.14)
is constant. The balance of energy is restated as
ρ 0ξ˙ = tr(TE˙ ) − ρTη˙
(7.15)
Recalling that ξ is a function of the extensive variables E and η, to convert to E and T as state variables, which are accessible to measurement, we recall the
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Helmholtz Free Energy f = e + Tη. Since f˙ is an exact differential, to ensure path independence, we infer the Maxwell relation:
−ρ
∂η ∂T . = ∂E T ∂T E˙
(7.16)
Returning to the energy-balance equation, ∂ξ ∂ξ ˙ E˙ + ρ ρξ˙ = tr ρ η ∂η E ∂E η ∂ξ ∂ξ ∂η ∂ξ ∂η ˙ ˙ = tr ρ +ρ E + ρ T. ∂η E ∂E T ∂η E ∂T E ∂E η
(7.17)
Also, note that
T=
∂ξ ∂E η
T=
∂ξ , ∂η E
(7.18a)
thus ∂ξ ∂ξ ∂η ∂ξ ∂η ˙ ˙ ρξ˙ = tr ρ +ρ E + ρ T ∂η E ∂E T ∂η E ∂T E ∂E η ∂T ˙ ∂η ˙ = tr T − T T. E + ρT ∂T E ∂T E
(7.18b)
We previously identified the coefficient of specific heat, assumed constant, as ce = T ∂∂Tη , so that E
∂T ˙ ˙ ρξ˙ = tr T − T E + ρce T. ∂T
(7.19)
From Equation 7.13, upon approximating T as T0, T0 ∂∂TT = −αλT0 I. From Fourier’s Law, q = −k∇T. Thus, the thermal-field equation now can be written as ˙ ) + ρc T˙ . k∇2 T = αλT0tr(E e
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(7.20)
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The balance of linear momentum, together with the stress-strain and straindisplacement relations of linear isotropic thermoelasticity, imply that ∂ ∂x j
1 ∂u ∂u ∂u ∂u 2 µ i + j + λ k − α (T − T0 )δ ij = ρ 2i , ∂ x ∂ x x ∂ t 2 ∂ j i k
(7.21)
from which we obtain the mechanical-field equation (Navier’s Equation for Thermoelasticity):
µ∇ 2 u + (λ + µ )∇tr( E) − αλ∇T = ρ
∂u . ∂t 2
(7.22)
The thermal-field equation depends on the mechanical field through the term αλT0 tr( E˙ ). Consequently, if E is static, there is no coupling. Similarly, the mechanical field depends on the thermal field through αλ∇T, which often is quite small in, for example, metals, if the assumption of reversibility is a reasonable approximation. We next derive the entropy. Since ce = T ∂∂Τη is constant, we conclude that it E has the form
ρη = ρη0 + ρc e ln(T/T0 ) + ρη * ( E),
(7.23)
where η*(E) remains to be determined. We take η0 to vanish. However, ρ ∂∂Eη = ρ ∂∂ηE* = T − ∂∂TT E = αλI, implying that
ρη = ρη0 + ρce ln(T/T0 ) + αλtr( E).
(7.24)
Now consider f, for which the fundamental relation in Equation 7.11b implies ∂f = −η ∂T E
ρ
∂f = T. ∂E T
(7.25)
Integrating the entropy,
ρf = ρf0 − ρce [T ln(T/T0 ) − 1] − αλtr(E)T + ρf * ( E),
(7.26)
*
in which f (E), remains to be determined. Integrating the stress,
ρf = µ tr( E 2 ) +
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λ 2 tr (E) − αλ tr( E)(T − T0 ) + ρf ** (T), 2
(7.27)
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**
in which f (T) also remains to be determined. However, reconciling the two forms furnishes (taking f0 = 0)
ρf = µ tr(E2 ) +
λ 2 tr (E) − αλ tr(E)T − ρce [T ln(T/T0 ) − 1]. 2
(7.28)
7.3 THERMAL AND THERMOMECHANICAL ANALOGS OF THE PRINCIPLE OF VIRTUAL WORK 7.3.1 CONDUCTIVE HEAT TRANSFER For a linear, isotropic, thermoelastic medium, the Fourier Heat Conduction Law becomes q = −kT∇T, in which kT is the thermal conductance, assumed positive. Neglecting coupling to the mechanical field, the thermal-field equation in an isotropic medium experiencing small deformation can be written as −∇ T kT ∇T + ρce T˙ = 0.
(7.29)
We now construct a thermal counterpart of the principle of virtual work. Multiplying by δ T and using integration by parts, we obtain
∫ δ ∇ Tk ∇TdV + ∫ δ Tρc T˙ dV = ∫ δ Tn qdS T
T
T
e
(7.30)
Clearly, T is regarded as the primary variable, and the associated secondary variable is q. Suppose that the boundary is decomposed into three segments: S = SI + SII + SIII. On SI, the temperature T is prescribed as, for example, T0. It follows that δ T = 0 on SI. On SII, the heat flux q is prescribed as q0. Consequently, δ TnTq → δ TnTq0. On SIII, the heat flux is dependent on the surface temperature through a heat-transfer vector: h: q = q0 − h(T − T0). The right side of Equation 7.30 now becomes
∫ δ Tn qdS = ∫ T
S
SII
δ Tn T q 0 dS +
∫
SIII
δ Tn T q 0 dS −
∫
SIII
δ Tn T h(T − T0 )dS.
(7.31)
We now suppose that T is approximated using an interpolation function of the form T − T0 ~ N TT ( x )θ (t )
δ T ~ N TT ( x )δ θ (t ),
(7.32)
δ ∇T ~ BTT ( x )δ θ (t ),
(7.33)
from which we obtain ∇T ~ BTT ( x )θ(t )
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in which BT is the thermal analog of the strain-displacement matrix. Upon substitution of the interpolation models, the thermal-field equation now reduces to the system of ordinary differential equations: K T θ (t ) + M T θ˙ (t ) = fT ,
(7.34)
in which K T = K T1 + K T 2
Thermal Stiffness Matrix
K T1 =
∫ B (x)k B (x)dV
Conductance Matrix
K T2 =
∫
Surface Conductance Matrix
MT = fT =
T
SIII
T T
T
N T ( x )n T hN TT ( x )dS
∫ N (x)ρc N (x)dV T
∫
SII
e
Thermal Mass Matrix; Capacitance Matrix
T T
N T ( x )n T q 0 dS +
∫
SIII
N T ( x )n T q 0 dS
Consistent Thermal Force; Consistent Heat Flux
7.3.2 COUPLED LINEAR ISOTROPIC THERMOELASTICITY The thermal-field equation is repeated as − kT ∇ 2 T = αλT0 tr( E˙ ) + ρc e T˙ .
(7.35)
Following the same steps used for conductive heat transfer furnishes the variational principle
∫ δ ∇ Tk ∇TdV + ∫ δ Tρc T˙ dV + ∫ δ TαλT tr( E˙ )dV = ∫ δ Tn qdS. T
T
T
e
(7.36)
0
The principle of virtual work for the mechanical field is recalled as
∫ tr(δET )dV + ∫ δu ρu˙˙dV = ∫ δu τ(s)dS. T
T
(7.37)
S
Also, recall that T = 2µE + λ[tr(E) − α(T − T0)]I. Consequently,
∫ tr(δ E[2µE + λ tr(E)I])dV − ∫ tr(δ Eλα (T − T )I)dV + ∫ δu ρu˙˙dV = ∫ δu τ(s)dS. T
0
T
S
(7.38)
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Then introduce the interpolation models, u( x ) = N T ( x )γ (t ),
(7.39)
from which we can derive B(x) and b(x), thus satisfying VEC( E) = BT ( x )γ (t )
tr( E) = b T ( x )γ (t ).
(7.40)
It follows that Kγ (t ) + M˙˙ γ (t ) − Ω T θ(t ) = f .
(7.41)
We assume that the traction τ(S) is specified everywhere as τ0(S) on S. Here, K=
∫ B (x)DB(x)dV
Stiffness Matrix
M=
∫ N(x)ρN (x)dV
Mass Matrix
T
T
∫
(7.42)
Ω = αλb ( x )N T ( x )dV Thermoelastic Matrix f=
T
∫ N(x)τdS
Consistent Force Vector .
For the thermal field, assuming that the heat flux q is specified as q0 on the surface, variational methods, together with the interpolation models, furnish the equation 1 1 1 K T θ(t ) + M T θ˙ (t ) + Ωγ˙ (t ) = f , T0 T0 T0 T
fT =
∫ N (x)n q dS. T
T
0
(7.43)
The combined equations for a thermoelastic medium are now written in state (first-order) form as γ˙ (t ) γ˙ (t ) f (t ) d W1 γ (t ) + W2 γ (t ) = 0 dt θ(t ) θ(t ) T10 fT (t ) M W1 = 0 0
0 K 0
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0 0 , 1 MT T0
0 W2 = − K Ω
K 0 0
−Ω T 0 . 1 MT T0
(7.44)
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Note that W1 is positive-definite and symmetric, while 12 (W2 + W2T ) is positivesemidefinite, implying that coupled linear thermoelasticity is at least marginally stable, whereas a strictly elastic system is strictly marginally stable. Thus, thermal conduction has a stabilizing effect, which can be shown to be analogous to viscous dissipation.
7.4 EXERCISES 1. Express the thermal equilibrium equation in: (a) cylindrical coordinates (b) spherical coordinates 2. Derive the specific heat at constant stress, rather than at constant strain. 3. Write down the coupled thermal and elastic equations for a one-dimensional member.
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8
Introduction to the FiniteElement Method
8.1 INTRODUCTION In thermomechanical members and structures, finite-element analysis (FEA) is typically invoked to compute displacement and temperature fields from known applied loads and heat fluxes. FEA has emerged in recent years as an essential resource for 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 acceptance has benefited from rapid progress in related computer hardware and software, especially computer-aided design (CAD) systems. Today, a number of highly developed, user-friendly finite-element codes are available commercially. The purpose of this chapter is to introduce finite-element theory and practice. The next three chapters focus on linear elasticity and thermal response, both static and dynamic, of basic structural members. After that, nonlinear thermomechanical response is considered. In FEA practice, a design file developed using CAD is often “imported” into finiteelement codes, from which point little or no additional effort is required to develop the finite-element model and perform sophisticated thermomechanical analysis and simulation. CAD integrated with an analysis tool, such as FEA, is an example of computer-aided engineering (CAE). CAE is a powerful resource with the potential of identifying design problems much more efficiently and rapidly than by “trial and error.” A major FEM 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 be diagnostic, supporting interpretation of product-failure data. Analysis also can be used to assess performance, for example, by determining whether the design-stiffness coefficient for a rubber spring is attained.
8.2 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 the design of a mechanical system can 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), 117
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3.
4.
5. 6. 7.
8.
in terms of their nodal values, which now become the unknowns in the finite-element model. 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. 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. Equilibrium principles in variational form are applied using the various approximations within each element, leading to element equilibrium equations. The element equilibrium equations are assembled to provide a global equilibrium equation. 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. 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 can revise the finite-element model and repeat the computations. When the model is validated, postprocessing, with heavy reliance on graphics, then serves to interpret the results, for example, determining whether the underlying design is satisfactory. If problems with the design are identified, the analyst can then choose to revise the design. The revised design is modeled, and the process of validation and interpretation is repeated.
8.3 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 and “raw” output files are generated. The postprocessor module typically contains (interfaces to) graphical utilities, thus 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.
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119
The analyst can use benchmark solutions, special cases, or experimental data to validate the analysis. With validation, the analyst gains confidence in, for example, the mesh. He or 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 as follows: 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 analogs 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. 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, for example, five. 9. Symmetric configurations should have symmetric meshes. 10. The density of elements should be greater in domains with higher gradients. 11. Interior angles in elements should not be excessively acute or obtuse, for example, less than 45° or greater than 135°. 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.
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There are two classes of errors in finite-element analysis: 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 round-off. 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, 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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9
Element Fields in Linear Problems
This chapter presents interpolation models in physical coordinates for the most part, for the sake of simplicity and brevity. However, in finite-element codes, the physical coordinates are replaced by natural coordinates using relations similar to interpolation models. Natural coordinates allow use of Gaussian quadrature for integration and, to some extent, reduce the sensitivity of the elements to geometric details in the physical mesh. Several examples of the use of natural coordinates are given.
9.1 INTERPOLATION MODELS 9.1.1 ONE-DIMENSIONAL MEMBERS 9.1.1.1 Rods The governing equation for the displacements in rods (also bars, tendons, and shafts) is
EA
∂2u ∂2u = ρ A , ∂x 2 ∂t 2
(9.1)
in which u(x, t) denotes the radial displacement, E, Α and ρ are constants, x is the spatial coordinate, and t denotes time. Since the displacement is governed by a second-order 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 xe+1, as shown in Figure 9.1. The lowest-order interpolation model consistent with two integration constants is linear, in the form
u( x, t ) = ϕ Tm1 ( x )Φm1γ m1 (t ),
ue (t ) γ m1 (t ) = , ue+1 (t )
ϕ Tm1 ( x ) = (1
x ).
(9.2)
We seek to identify Φm1 in terms of the nodal values of u. Letting ue = u(xe) and ue+1 = u(xe+1), furnishes ue (t ) = (1
xe )Φm1γ m1 (t ),
ue+1 (t ) = (1
xe+1 )Φm1γ m1 (t ).
(9.3)
121
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ue
ue+1 x
xe
xe+1
FIGURE 9.1 Rod element.
wet
wet +1
we
we+1 x
xe
xe+1
FIGURE 9.2 Beam element.
However, from the meaning of γm1(t), we conclude that 1 Φm1 = 1
−1
− xe , 1
xe 1 xe+1 = le −1 xe+1
le = xe+1 − xe .
(9.4)
9.1.1.2 Beams The equation for a beam, following Euler-Bernoulli theory, is:
EI
∂4w ∂2u + ρA 2 = 0, 4 ∂x ∂t
(9.5)
in which w(x, t) denotes the transverse displacement of the beam’s neutral axis, and I is a constant. In the spatial domain, there are four constants of integration. In an element, they can be supplied implicitly by the values of w and w′ = ∂w/∂x at each of the two element nodes. Referring to Figure 9.2, we introduce the interpolation model for w(x, t):
w( x, t ) = ϕ Τb1 ( x )Φb1γ b1 (t ),
© 2003 by CRC CRC Press LLC
(
ϕ Τb1 ( x ) = 1
x
x2
)
x3 ,
we − we′ γ m1 (t ) = . (9.6) we+1 − we′+1
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123
Enforcing this model at xe and at xe+1 furnishes 1 0 Φb1 = 1 0
−1
xe
x e2
−1
−2 x e
x e+1
x e2+1
−1
−2 x e+1
x e3 −3 x e2 . x e3+1 2 −3 x e+1
(9.7)
9.1.1.3 Beam Columns Beam columns are of interest, among other reasons, in predicting buckling according to the Euler criterion. The z–displacement w of the neutral axis is assumed to depend only on x and the x–displacement. Also, u is modeled as u( x, z ) = u0 ( x ) − z
∂w( x ) , ∂x
(9.8)
in which u0(x) represents the stretching of the neutral axis. It is necessary to know u0(x), w(x) and ∂w∂(xx ) at xe and xe+1. The interpolation model is now u( x, z, t ) = (1
x )Φm1γ m1 − z(1
ue (t ) γ ml = , ue+1 (t )
x
x 3 )Φb1γ b1,
x2
(9.9)
we − we′ γ bl = , we+1 − we′+1
and
Φm1 =
1 le
x e+1 −1
1 0 Φ b1= 1 0
− xe , 1
xe
x e2
−1
−2 x e
x e+1
x e2+1
−1
−2 x e+1
−1
x e3 −3 x e2 . x e3+1 −3 x e2+1
9.1.1.4 Temperature Model: One Dimension The temperature variable to be determined is T − T0, in which T0 is a reference temperature assumed to be independent of x. The governing equation for a onedimensional conductor is kA
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∂2 T ∂T = ρAce . 2 ∂x ∂t
(9.10)
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z Xe+2,Ye+2 Y Xe+1,Ye+1
Xe,Ye
X
middle surface FIGURE 9.3 Triangular plate element.
This equation is formally the same as for a rod equation (see Equation 9.1), furnishing the interpolation model for the element as T( x, t ) − T0 = ϕ Tm1 ( x )Φm1θ1 (t ),
Te (t ) − T0 θ1 (t ) = . Te+1 (t ) − T0
(9.11)
9.1.2 INTERPOLATION MODELS: TWO DIMENSIONS 9.1.2.1 Membrane Plate Now suppose that the displacements u(x, y, t) and v(x, y, t) are modeled on the triangular plate element in Figure 9.3, using the values ue(t), ve(t), ue+1(t), ve+1(t), ue+2(t), and ve+2(t). This element arises in plane stress and plane strain, and is called a membrane plate element. A linear model suffices for each quantity because it provides three coefficients to match three nodal values. The interpolation model now is Τ u( x, y, t ) ϕ m 2 = v( x, y, t ) 0 T
ue (t ) γ u (t ) = ue+1 (t ) , ue+2 (t )
0 T ΦmΤ2 ϕ Τm 2 0 T
ve (t ) γ v (t ) = ve+1 (t ) , ve+2 (t )
ϕm2
1 = x , y
0 T γ u (t ) ΦmΤ2 γ v (t )
ΦmΤ 2
1 = 1 1
(9.12)
xe xe+1 xe + 2
−1
ye ye+1 . ye+2
9.1.2.2 Plate with Bending Stresses In a plate element experiencing bending only, the in-plane displacements, u and v, are expressed by
u( x, y, z, t ) = − z
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∂w ∂w , v( x, y, z, t ) = − z , ∂x ∂y
(9.13)
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125
in which z = 0 at the middle plane. The out-of-plane displacement, w, is assumed to be a function of x and y only. An example of an interpolation model is introduced as follows to express w(x, y) throughout the element in terms of the nodal values of w, ∂∂wx and ∂∂wy : w( x, y, t ) = ϕ Tb 2 ( x, y)Φb 2 γ b 2 (t ) ϕ Tb 2 ( x, y) = 1
x
γ Tb 2 (t ) = we
1 0 0 1 Φb−21 = 0 0 1 0 0
x2
y
( )
− ∂∂w x
− ∂∂w y
e +1
y2
xy
e
− ∂∂w y w e+2
(9.14) 1 2 ( x y + y2 x) 2
x3
we +1
e
( )
− ∂∂w x
e+2
− ∂∂w x
( )
e +1
− ∂∂w y
e+2
xe ye ( xe + ye )
xe
ye
xe2
xe ye
ye2
xe3
−1
0
−2 xe
− ye
0
−3 xe2
−( xe ye + 12 ye2 )
0
−1
0
− xe
−2 xe
0
−( 12 xe2 + xe ye )
xe +1
ye +1
xe2+1
xe +1 ye +1
ye2+1
xe3+1
−1
0
−2 xe +1
− ye +1
0
0
−1
0
− xe +1
−2 ye +1
0
xe + 2
ye + 2
xe2+ 2
xe + 2 ye + 2
ye2+ 2
xe3+ 2
−1
0
−2 xe + 2
− ye + 2
0
0
−1
0
− xe + 2
−2 ye + 2
−3 x
−3 x
1 2
xe +1 ye +1 ( xe +1 + ye +1 ) −( xe +1 ye +1 + 12 ye2+1 )
2 e +1
2 e+2
0
1 2
−( 12 xe2+1 + xe +1 ye +1 ) 1 2
y3
xe + 2 ye + 2 ( xe + 2 + ye + 2 ) −( xe + 2 ye + 2 + 12 ye2+ 2 ) −( 12 xe2+ 2 + xe + 2 ye + 2 )
ye3 0 −3ye2 ye3+1 0 . −3ye2+1 ye3+ 2 0 −3ye2+ 2
It follows that − z ∂ϕΤb 2 u(x, y, z, t) ∂x Τ v(x, y, z, t) = − z ∂ϕ b 2 Φ γ ( t ). b2 b2 ∂y w(x, y, z, t) ϕ Τb 2
(9.15)
9.1.2.3 Plate with Stretching and Bending Finally, for a plate experiencing both stretching and bending, the displacements are assumed to satisfy u( x, y, z, t ) = u0 ( x, y, z, t ) − z
© 2003 by CRC CRC Press LLC
∂w , ∂x
v( x, y, z, t ) = v0 ( x, y, z, t ) − z
∂w ∂y
(9.16)
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Finite Element Analysis: Thermomechanics of Solids
and w is a function only of x, y, and t(not z). Here, z = 0 at the middle surface, while u0 and v0 represent the in-plane displacements. Using the nodal values of u0, v0, and w0, a combined interpolation model is obtained as − z ∂ϕΤb 2 u( x, y, z, t ) u0 ( x, y, t ) ∂x Τ v( x, y, z, t ) = v ( x, y, t ) + − z ∂ϕb 2 Φ γ (t ) 0 b2 b2 ∂ y w( x, y, z, t ) w0 ( x, y, t ) ϕ Τb 2 ϕ Τ m2 = 0T T 0
∂ϕΤ
− z ∂xb 2 Φm 2 ∂ϕΤb 2 − z ∂y 0 Τ ϕ b 2 0
0T ϕ Τm 2 0T
0 Φm 2 0
0 γ u (t ) 0 γ v (t ) . Φb 2 γ w (t )
(9.17)
9.1.2.4 Temperature Field in Two Dimensions In the two-dimensional, triangular element illustrated in Figure 9.3, the linear interpolation model for the temperature is T − T0 = ϕ Τm 2 Φm 2 θ 2 ,
θ Τ2 = (Te − T0
Te+1 − T0
Te+2 − T0 ).
(9.18)
9.1.2.5 Axisymmetric Elements An axisymmetric element is displayed in Figure 9.4. It is applicable to bodies that are axisymmetric and are submitted to axisymmetric loads, such as all-around pressure. The radial displacement is denoted by u, and the axial displacement is denoted by w. The tangential displacement v vanishes, while radial and axial displacements are independent of θ. Now u and w depend on r, z, and t. There are two distinct situations that require distinct interpolation models. In the first case, none of the nodes are on the axis of revolution (r = 0), while in the second case, one or two nodes are, in fact, on the axis. In the first case, the linear e+2 e+1 r θ FIGURE 9.4 Axisymmetric element.
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e
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127
interpolation model is given by 0 T ΦaT1 ϕ Ta1 0 T
T u(r, z, t ) ϕ a1 = w(r, z, t ) 0 T
ϕ Ta1 = (1
r
z ),
1 Φa1 = 1 1
0 T γ ua1 (t ) . ΦaT1 γ wa1 (t )
−1
ze ze+1 , ze+2
re re+1 re+2
γ ua1
ue = ue+1 , ue+2
(9.19)
γ wa1
we = we+1 we + 2
Now suppose that there are nodes on the axis, and note that the radial displacements are constrained to vanish on the axis. For reasons shown later, it is necessary to enforce the symmetry constraints a priori in the formulation of the displacement model. In particular, suppose that node e is on the axis, with nodes e + 1 and e + 2 defined counterclockwise at the other vertices. The linear interpolation model is now 0 T ΦaT2 ϕ Ta 2 0 T
T u(r, z, t ) ϕ a 2 = w(r, z, t ) 0 T
ϕ
T a2
= (r
0 T γ ua 2 (t ) ΦaT2 γ wa 2 (t )
re+1 Φa 2 = re+2
z − ze ),
(9.20)
−1
ze+1 − ze . ze+2 − ze
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.
9.1.3 INTERPOLATION MODELS: THREE DIMENSIONS We next consider the tetrahedron illustrated in Figure 9.5. A linear interpolation model for the temperature can be expressed as T − T0 = ϕ 3TT Φ3T θ3 ϕ 3TT = (1 1 1 Φ3T = 1 1
x
z)
xe
ye
xe+1
ye+1
xe + 2
ye+2
xe + 3
ye+3
θ3TT = {Te − To
© 2003 by CRC CRC Press LLC
y
(9.21)
Te+1 − To
ze ze+1 ze+2 ze+3
−1
Te+2 − To
Te+3 − To}
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Finite Element Analysis: Thermomechanics of Solids
z e+3
0
y
e
x
e+2
e+1
FIGURE 9.5 Tetrahedral element.
For elasticity with displacements u, v, and w, the corresponding interpolation model is T u( x, y, z, t ) ϕ 3 T v( x, y, z, t ) = 0 w( x, y, z, t ) 0T
ue (t ) ue+1 (t ) γ u (t ) = , ue+2 (t ) ue+3 (t )
0T ϕ 3T 0T
0T Φ3 0T 0 ϕ 3T 0
ve ( t ) ve+1 (t ) γ v (t ) = , ve+2 (t ) ve+3 (t )
0 Φ3 0
0 γ u (t ) 0 γ v (t ) Φ3 γ w (t )
(9.22)
we (t ) we+1 (t ) γ w (t ) = we+2 (t ) we+3 (t )
9.2 STRAIN-DISPLACEMENT RELATIONS AND THERMAL ANALOGS 9.2.1 STRAIN-DISPLACEMENT RELATIONS: ONE DIMENSION For the rod, the strain is given by ε = E11 = ∂∂ux . An estimate for ε implied by the interpolation model in Equation 9.3 has the form
ε ( x, t ) = β Tm1 ( x )Φm1 γ m1 ( t ),
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(9.23)
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129
βTm1 =
dϕ Tm1 = (0 dx
1)
For the beam, the corresponding relation is
ε ( x , z, t ) = − z
∂w , ∂x
(9.24)
from which the consistent approximation is obtained:
ε ( x, z, t ) = − zβ Tb1 ( x )Φb1 γ b1 ( t ), βTb1 =
dϕ Tb1 = (0 dx
1
(9.25)
3x 2 )
2x
For the beam column, the strain is given by
ε ( x, z, t ) = − zβ Tmb1 ( x )Φmb1 γ mb1 ( t ),
(
β Tmb1 = β Tm1
Φm1 − zβ Tb1 0
)
(9.26)
0 γ m1 (t ) Φmb1 γ b1 (t )
9.2.2 STRAIN-DISPLACEMENT RELATIONS: TWO DIMENSIONS In two dimensions, the (linear) strain tensor is given by Exx E( x, y) = Exy
Exy = Eyy 12
(
∂u ∂x ∂u + ∂v ∂y ∂x
)
(
1 ∂u + ∂v 2 ∂y ∂x ∂v ∂y
).
(9.27)
We will see later 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. In traditional finite-element notation, we obtain [cf. (Zienkiewicz and Taylor, 1989)] Exx γ u2 ˆ ε ′ = Eyy = βTm 2 Φ m2 γ v2 E xy
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(9.28)
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Finite Element Analysis: Thermomechanics of Solids
βΤm 2
∂ϕΤm 2 , ∂y 1 ∂ϕΤm 2 2 ∂x
∂ϕΤm 2 ∂x = 0Τ 1 ∂ϕΤ ∂ym 2 2
0Τ
Φ ˆ = m2 Φ m2 0
0 Φm 2
The prime in e′ is introduced temporarily to call attention to the fact that it does not equal VEC(EL). Hereafter, the prime will not be displayed. For a plate with bending stresses only, ∂2w ∂x 2 2 e ′( x, y, z, t ) = − z ∂ w2 , ∂y ∂2w ∂x∂y
(9.29)
from which
e ′( x, y, z, t ) = − zβ Tb 2 ( x )Φb 2 γ b 2 (t ),
βb2
∂2ϕTb 2 ∂x 2 ∂2ϕT b2 = . 2 ∂y ∂2ϕTb 2 ∂x∂y
(9.30)
For a plate experiencing both membrane and bending stresses, the relations can be combined to furnish e ′( x, y, z, t ) = β Tmb 2 ( x, y, z )Φmb 2 γ mb 2 (t )
(
βTmb 2 = βTm 2
Φm 2 − zβTb 2 , Φmb 2 = 0
)
9.2.3 AXISYMMETRIC ELEMENT
ON
AXIS
OF
(9.31)
0 γ m 2 ( t ) , γ mb 2 ( t ) = Φb 2 γ b2 (t )
REVOLUTION
For the 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 e(r, z, t ) = 1 2
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∂u ∂r u r ∂w ∂z
Φa1 = β Ta1 0
( ∂∂uz + ∂∂wr )
0 γ ua1 (t ) Φa1 γ wa1 (t )
(9.32)
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0 12 β Ta1 = 0 0
131
1
0
0
0
1
z r
0
0
0
0
0
0
0
1 2
0
1 2
0 0 1 0
in which the prime is no longer displayed. If element e is now located on the axis of revolution, we obtain Φa 2 e(r, z, t ) = baT2 0
β Ta 2
9.2.4 THERMAL ANALOG
1 1 = 0 0
IN
0 γ ua 2 (t ) Φa1 γ wa1 (t )
0
0
0
z − ze r
0
0
0
0
0
1 2
0
1 2
(9.33)
0 0 1 0
TWO DIMENSIONS
The thermal analog of the strain is the temperature gradient satisfying 0 β TT2 = 0
∇T = β TT 2 ΦT 2 θ 2 ,
1 0
0 . 1
(9.34)
9.2.5 THREE-DIMENSIONAL ELEMENTS Recalling the tetrahedral element in the previous section, the strain relation can be written as Exx Eyy Ezz e= = Exy 12 Eyz 12 E 1 zx 2
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∂u ∂x ∂v ∂y
( (
∂w ∂z ∂u + ∂v ∂y ∂x ∂v + ∂w ∂z ∂y
Φ3 = β 3T 0 0
) )
( ∂∂wx + ∂∂uz )
0 Φ3 0
0 γ u 3 0 γ v 3 Φ3 γ w 3
(9.35)
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Finite Element Analysis: Thermomechanics of Solids
9.2.6 THERMAL ANALOG
IN
THREE DIMENSIONS
Again referring to the tetrahedral element, the relation for the temperature gradient is
∇T = β 3TT Φ 3T θ 3 ,
β 3TT
0 = 0 0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0 0 0 . 0 1 (9.36) 0
9.3 STRESS-STRAIN-TEMPERATURE RELATIONS IN LINEAR THERMOELASTICITY 9.3.1 OVERVIEW 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 Lame’s form by S = 2 µEL + λtr( EL )I,
(9.37)
in which I is the identity tensor. The Lame’s coefficients are denoted by λ and µ, and are given in terms of the familiar elastic modulus E and Poisson ratio ν as
µ=
E , 2(1 + ν )
λ=
νE . (1 − 2ν )(1 + ν )
(9.38)
Letting s = VEC(S) and e = VEC(EL), the stress-strain relations are written using Kronecker product operators as s = De,
D = 2 µI 9 + λii T ,
(9.39)
and D is the tangent-modulus tensor introduced in the previous chapters.
9.3.2 ONE-DIMENSIONAL MEMBERS For a beam column, recalling the strain-displacement model, S11 = σ( x, z, t ) = E ε = − z Eβ Tmb1 ( x )Φmb1 γ mb1 (t ).
(9.40)
The cases of a rod and a beam are recovered by setting γ m1 or γ b1 equal to zero vectors, respectively.
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133
9.3.3 TWO-DIMENSIONAL ELEMENTS 9.3.3.1 Membrane Response In two-dimensional elements, several cases can be distinguished. We first consider elements in plane stress. It is convenient to use Hooke’s Law in the form Exx =
1 [ S − ν ( Syy + Szz )] E xx
Eyy =
1 [ S − ν ( Syy + Szz )] E yy
Ezz =
1 [ S − ν ( Sxx + Syy )] E zz
(9.41)
1+ν Exy = S E xy Eyz =
1+ν S E yz
Ezx =
1+ν S E zx
Under plane stress, Sxz = Syz = Szz = 0. Now, Ezz ≠ 0, but we will see that it is not of present interest since it does not influence the solution process. Later on, for reasons such as tolerances, we may wish to calculate it, but this is a postprocessing issue. Consequently, in plane stress, the stress-strain relations reduce to Exx =
1 ( S − νSyy ) E xx
Eyy =
1 ( S − νSxx ) E yy
Exy =
1+ν S E xy
(9.42)
In traditional finite-element notation, this can be written as Sxx Exx Syy = Dm 21 Eyy . S E xy xy
(9.43)
in which
Dm21
1 = E −ν 0
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−ν 1 0
0 0 1 + ν
−1
1 E ν = 1−ν2 0
ν 1 0
0 0 1 + ν
(9.44)
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Finite Element Analysis: Thermomechanics of Solids
can 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). Under plane strain, Exz = Eyz = Ezz = 0. Now, Szz ≠ 0. However, this stress is not of great interest since its value is not needed for attaining a solution. The quantity may be of interest later, e.g., to check whether yield conditions in plasticity are met, but it can be obtained at that point by a postprocessing procedure. The equations of plane strain are Sxx E xx Syy = Dm22 E yy , S E xy xy
Dm22 =
*
E 1 − ν *2
1 ν * 0
ν* 1 0
(9.45)
0 E ν 0 , E * = , ν* = 2 1 − ν 1 − ν 1 + ν *
and Dm22 is the tangent-modulus tensor in plane strain. However, in the finite-element method, we will see that we are interested in a slightly different quantity from Dm21 or Dm22 , as follows. The Principle of Virtual Work uses the strain-energy density given by 12 Sij Eij. However, elementary manipulation serves to prove that
1 1 Sij Eij = ( Sxx 2 2
Syy
1 Szz )0 0
0 1 0
0 Exx 0 Eyy . 2 Exy
(9.46)
In the Principle of Virtual Work, the tangent-modulus tensor in plane stress is replaced by
Dm21 ′
1 E ν = 1−ν2 0
ν 1 0
0 2(1 + ν ) 0
(9.47)
and similarly for plane strain. (This peculiarity is an artifact of traditional finiteelement notation and does not appear if VEC notation is used.) The stresses are now given in terms of nodal displacements by Sxx (x, y, t) Exx γ u2 ˆ Syy (x, y, t) = Dm 2 i Eyy = Dm′ 2 i β Tm 2 Φ , m2 γ ν2 S (x, y, t) E xy xy
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1 i= 2
plane stress (9.48) plane strain
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135
9.3.3.2 Two-Dimensional Members: Bending Response Thin plates experiencing bending only are assumed to be in a state of plane stress. The tangent-modulus matrix is given in Equation 9.47, and an approximation for the strain is obtained as
Sxx Exx ˆ γ (t ). Syy = Dm 21 Eyy − z Dm 21β Tb 2 Φ b2 b2 S E xy xy
(9.49)
9.3.4 ELEMENT FOR PLATE WITH MEMBRANE AND BENDING RESPONSE Plane stress is also applicable, consequently:
Sxx Exx ˆ γ (t ). Syy = Dm 21 Eyy = Dm 21βTmb 2 ( x, y, z )Φ mb 2 mb 2 S E xy xy
(9.50)
9.3.5 AXISYMMETRIC ELEMENT It is sufficient to consider the case in which none of the nodes of the element are located on the axis of revolution.
Srr Err ˆ Φ Sθθ Eθθ a1 T = Da = Daβ a1 0 Szz Ezz S E rz rz 1 −ν Da = E −ν 0
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−ν
−ν
1
−ν
−ν
1
0
0
0 γ ua1 (t ) , ˆ γ wa1 (t ) Φ a1
0 0 0 1 + ν
−1
(9.51)
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Finite Element Analysis: Thermomechanics of Solids
For use in the Principle of Virtual Work, Da is modified to furnish Da′ , given by 1 0 Da′ = E 0 0
0
0
1
0
0
1
0
0
0 1 0 −ν 0 −ν 2 0
−ν
−ν
1
−ν
−ν
1
0
0
0 0 0 1 + ν
−1
1 0 0 0
0
0
1
0
0
1
0
0
0 0 . (9.52) 0 2
9.3.6 THREE-DIMENSIONAL ELEMENT All six stresses and strains are now present. Using traditional notation, we write Sxx Exx Syy Eyy Φ3 Szz Ezz T = D3 = D3β3 0 Sxy Exy 0 Syz Eyz S E zx zx 1 −ν −ν D3 = E 0 0 0
0
0 γ u 3 0 γ v 3 Φ3 γ w 3
Φ3 0
−ν
−ν
0
0
1
−ν
0
0
−ν
1
0
0
0
0
1+ν
0
0
0
0
1+ν
0
0
0
0
−ν
−ν
0
0
1
−ν
0
0
−ν
1
0
0
0
0
1+ν
0
0
0
0
1+ν
0
0
0
0
0 0 0 0 0 1 + ν
(9.53)
−1
and for the Principle of Virtual Work, 1 −ν −ν D3′ = EJ 6 0 0 0
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−1
0 0 0 J6 0 0 1 + ν
(9.54)
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Element Fields in Linear Problems
1 0 0 J6 = 0 0 0
9.3.7 ELEMENTS
FOR
137
0
0
0
0
1
0
0
0
0
1
0
0
0
0
2
0
0
0
0
2
0
0
0
0
0 0 0 0 0 2
CONDUCTIVE HEAT TRANSFER
Assuming the isotropic version of Fourier’s Law, the heat-flux vector, which can be considered the thermal analog of the stress, satisfies β ΤT 1ΦT 1θ1 q = − k β ΤT 2 ΦT 2 θ 2 Τ β T 3 ΦT 3θ 3
1− D 2−D
(9.55)
3 − D.
9.4 EXERCISES 1. Modify the rod element to replace the physical coordinate x with the “natural coordinate” ξ in which
ξ = ax + b,
2.
3. 4. 5. 6.
ξ ( x e ) = −1,
ξ ( x e+1 ) = +1.
Rewrite the interpolation model using natural coordinates, and perform the inverse to obtain Φ m1. 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 Φ b1. Verify that the inverse given in Equation 9.44 is correct. Invert Equation 9.39 to express E as a function of s. Find the correct −1 expression for D . −1 Write out the 9 × 9 tensors D and D that correspond to Lame’s equation. For the axisymmetric triangular element, find the strain-displacement T matrix β in the cases in which two of the nodes are located on the axis of revolution. Note that for axisymmetric problems, ur = 0 at r = 0.
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Finite Element Analysis: Thermomechanics of Solids
7. The stress-strain relations in two-dimensional plane stress can be written as Exx 1 1 Eyy = −ν E E 0 xy
−ν 1 0
0 Sxx 0 Syy 1 + ν Sxy
Prove that in plane strain Exx 1 1 Eyy = * −ν * E E 0 xy
−ν * 1 0
0 Sxx 0 Syy , 1 + ν * Sxy
where E* = E /(1 − ν 2 ), ν * = ν /(1 − ν ) . 8. If Exx 1 1 − ν 2 = Eyy E −ν (1 + ν )
−ν (1 + ν ) Sxx , 1 − ν 2 Syy
find the matrix Dˆ such that Sxx Exx E ˆ . D = Syy (1 + ν )(1 − 2ν ) Eyy What happens if ν → 1/2?
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10
Element and Global Stiffness and Mass Matrices
10.1 APPLICATION OF THE PRINCIPLE OF VIRTUAL WORK Elements of variational calculus were discussed in Chapter 3, and the Principle of Virtual Work was introduced in Chapter 5. Under static conditions, the principle is repeated here as
∫ δ E S dV + ∫ δu ρu˙˙ dV = ∫ δu τ dS. ij ij
i
i
(10.1)
i i
As before, δ represents the variational operator. We assume for our purposes that the displacement, the strain, and the stress satisfy representations of the form u( x, t ) = ϕ T ( x )Φ γ ( t ),
E = β T ( x )Φ γ ( t ),
S = DE,
(10.2)
in which E and S are written as one-dimensional arrays in accordance with traditional finite-element notation. For use in the Principle of Virtual Work, we need D′, which introduces the factor 2 into the entries corresponding to shear. We suppose that the boundary is decomposed into four segments: S = SI + SII + SIII + SIV . On SI, u is prescribed, in which event δ u vanishes. On SII , the traction τ is prescribed as τ 0. On SIII, there is an elastic foundation described by τ = τ 0 − 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 τ = τ 0 − Bü. The term on the right now becomes
∫ δu τ dS = δ γ ∫ T
i i
SII +SIII +SIV
− δγ T
Φ T ϕ( x )τ 0 dS
∫
Φ T ϕ( x )Aϕ T ( x )ΦdSγ (t )
∫
Φ T ϕ( x )Bϕ T ( x )ΦdS˙˙ γ (t ).
SIII
− δγ T
(10.3)
SIV
139
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Finite Element Analysis: Thermomechanics of Solids
The term on the left in Equation 10.1 becomes
∫ δE S dV = δ γ Kγ (t),
K=
∫
M=
T
ij ij
δ ui ρ˙˙u i dV = δ γ M˙˙γ ( t ), T
∫ Φ β(x)D′β (x)Φ dV T
T
(10.4)
∫ ρΦ ϕ(x)ϕ (x)Φ dV, T
T
in which K is called the stiffness matrix and M is called the mass matrix. Canceling the arbitrary variation and bringing terms with unknowns to the left side furnishes the equation as follows: (K + K S )γ (t ) + (M + M S )˙˙ γ (t ) = f f=
∫
KS =
∫
Φ ϕ( x )Aϕ ( x )ΦdS
MS =
∫
Φ T ϕ( x )Bϕ T ( x )ΦdS.
SII +SIII
Φ T ϕ( x )τ 0 dS
T
(10.5) T
SIII
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. Sample Problem 1: One element rod Consider a rod with modulus E, mass density ρ, area A, and length L. It is built in at x = 0. At x = L, there is a concentrated mass m to which is attached a spring of stiffness k, as illustrated in Figure 10.1. The stiffness and mass matrices, from the domain, reduce to the scalar values K → EA/L, M → ρAL/3, MS → m, KS → k. ρAL + k )γ + ( 3 + m)γ˙˙ = f . The governing equation is ( EA L Sample Problem 2: Beam element Consider a one-element model of a cantilevered beam to which a solid disk is welded at x = L. Attached at L is a linear spring and a torsional spring, the latter having the property that the moment developed is proportional to the slope of the beam. P
E,A,L,ρ
m
FIGURE 10.1 Rod with inertial and compliant boundary conditions.
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k
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141
z y
V0 M0
r
m x
E,I,A,L,ρ
kT k
FIGURE 10.2 Beam with translational and rotational inertial and compliant boundary conditions.
The shear force V0 and the moment M0 act at L. The interpolation model, incorporating the constraints w(0, t) = −w′ (0, t) = 0 a priori, is L2 w( x , t ) = ( x x ) −2 L 2 3
−1
L3 w( L, t ) . −3 L2 − w ′( L, t )
(10.6)
The stiffness and mass matrices, due to the domain, are readily shown to be K=
EI 12 L3 6 L
6L , 4 L2
13 35 M = ρAL 11 210 L
L . 2 1 105 L 11 210
(10.7)
The stiffness and mass contributions from the boundary conditions are k KS = 0
0 , kT
m MS = 0
0 . mr 2 2
(10.8)
The governing equation is now ˙˙( L, t ) w w( L, t ) V0 (M + M S ) + (K + K S ) = . ˙˙ ′( L, t ) − w ′( L, t ) M0 −w
(10.9)
10.2 THERMAL COUNTERPART OF THE PRINCIPLE OF VIRTUAL WORK For our purposes, we focus on the equation of conductive heat transfer as k∇ 2 T = ρce
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∂T . ∂t
(10.10)
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Multiplying by the variation of T − T0, integrating by parts, and applying the divergence theorem furnishes
∫ δ∇ Tk∇TdV + ∫ δ Tρc T
e
∂T dV = δ Tn T qdS. ∂t
∫
(10.11)
Now suppose that the interpolation models for temperature in the current element furnish a relation of the form T − T0 = ϕ TT ( x )ΦT θ(t ),
∇T = β TT ( x )ΦT θ(t ),
q = − kTT ( x )ΦT θ(t ) .
(10.12)
The terms on the left in Equation 10.11 can now be written as
∫ δ∇ Tk∇TdV → δθ (t)K θ(t), T
T
T
∫
∂T δ Tρce dV → δ θT (t )MT θ˙ (t ), ∂t
∫
K T = kΦTTβT βTTΦT dV (10.13)
∫ ρc Φ ϕ ϕ Φ dV.
MT =
e
T T
T
T T
T
KT and MT can be called the thermal stiffness (or conductance) matrix and thermal mass (or capacitance) matrix, respectively. Suppose that the boundary S has four zones: S = SI + SII + SIII + SIV . On SI, the temperature is prescribed as T1, from which we conclude that δ T = 0. On SII, the T T T heat flux is prescribed as n q1. On SIII, the heat flux satisfies n q = n q1 − h1 (T − T T T0), while on SIV , n q = n q1 − h2 d T/dt. The governing finite-element equation is now [M T + M TS ]θ˙ (t ) + [K T + K TS ]θ( t) = fT (t ) M TS =
∫
fT (t ) =
∫ Φ ϕ n q dS,
SIV
Ω
ΦTT ϕ T h2 ϕ TT ΦT dS, T T
T
T
1
K TS =
∫
SIII
(10.14)
ΦTT ϕ T h1ϕ TT ΦT dS
Ω = SII + SIII + SIV
10.3 ASSEMBLAGE AND IMPOSITION OF CONSTRAINTS 10.3.1 RODS Consider the assemblage consisting of two rod elements, denoted as e and e + 1 [see Figure 10.3(a)]. There are three nodes, numbered n, n + 1, and n + 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 + 1, the force of element e + 1 on element e is equal to and opposite the force of element e on element e + 1. It is helpful to carefully define global (assemblage
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Pn+2
Pn
e
e+1
n
n+1 a. forces in global system
P1(e)
P1(e+1)
e
e+1 P2 (e+1)
P2 (e)
b. forces in local system
FIGURE 10.3 Assembly of rod elements.
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, P1( e ) − P2( e+1) = 0 (e) 2
since no external load is applied. Clearly, P The elements satisfy 1 k (e) −1 1 k ( e+1) −1 and in this case, k = k equations: (e)
(e+1)
(10.15) ( e +1) 1
= Pn and P
−1 un − P2( e ) = 1 un+1 P1( e ) −1 un+1 − P2( e+1) = 1 un+2 P1( e+1)
(10.16)
= EA/L. These relations can be written as four separate k ( e ) un − k ( e )un+1 = − P2( e )
(i)
− k ( e ) un + k ( e )un+1 = P1( e )
(ii)
k ( e+1) un+1 − k ( e+1)un+2 = − P2( e+1)
(iii)
− k ( e+1) un+1 + k ( e+1)un+2 = P1( e+1)
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= Pn+2.
(10.17)
(iv)
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Add (ii) and (iii) and apply Equation 10.15 to obtain k ( e ) un − k ( e )un+1 = − P2( e )
(i)
− k ( e ) un + [k ( e ) + k ( e+1) ]un+1 − k ( e+1)un+2 = 0
(ii + iii)
− k ( e+1) un+1 − k ( e+1)un+2 = P1( e+1)
(10.18)
(iv)
and in matrix form k (e) − k ( e ) 0
−k (e) k ( e ) + k ( e+1) − k ( e+1)
un − Pn − k ( e+1) un+1 = 0 k ( e+1) un+2 Pn+1 0
(10.19)
The assembled stiffness matrix shown in Equation 10.19 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 entry on the right side of the upper matrix and the highest entry on the left side of the lowest matrix. The overlay structure is depicted in Figure 10.4. K(1) K(2) K(3) K(4) K(5) (4)
(5)
k22 +k11
K(N–2) K(N–1) K(N)
FIGURE 10.4 Assembled beam stiffness matrix.
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Now, the equations of the individual elements are written in the global system as ˜ ( e ) , K ( e+1) → K ˜ ( e+1) K (e) → K
˜ (e) K
1 = k ( e ) −1 0
−1
0 0 , 0
1 0
˜ ( e+1) K
0 = k ( e+1) 0 0
0 1 −1
0 −1 . 1
(10.20)
(g)
The global stiffness matrix (the assembled stiffness matrix K of the two-element ˜ (e) + K ˜ ( e+1) . member) is the direct sum of the element stiffness matrices: K ( g ) = K (g) ( e ) ˜ . In this notation, the strain energy in the two elements Generally, K = ∑e K can be written in the form V( e ) = V( e+1) =
1 ( e )Τ ˜ ( e ) ( e ) γ K γ 2 1 ( e )Τ ˜ ( e+1) ( e ) γ K γ 2
T
γ ( e ) = (un
un+1
(10.21)
un+2 ).
The total strain energy of the two elements is 12 γ T K ( g ) γ . (g) Finally, notice that K is singular: the sum of the rows is the zero vector, as is the sum of the columns. In this form, an attempt to solve the system will give rise to “rigid-body motion.” To illustrate this reasoning, suppose, for simplicity’s sake, (e) (e+1) that k = k , in which case equilibrium requires that Pn = Pn+2. If computations were performed with perfect accuracy, the equation would pose no difficulty. However, in performing computations, errors arise. For example, Pn is computed as Pˆn = Pn + ε n and Pˆn+1 = Pn+1 + ε n+1 , Pn+1 = Pn = P . Computationally, there is now an unbalanced force, ε n+1 − ε n. In the absence of mass, this, in principle, implies infinite accelerations. In the finite-element method, the problem of rigid-body motion can be detected if the output exhibits large deformation. The problem is easily suppressed using constraints. In particular, symmetry implies that un+1 = 0. Recalling Equation 10.19, we now have 1 EA −1 L 0
−1 2 −1
0 −1 1
un − Pn 0 = R , un+2 Pn+1
(10.22)
in which R is a reaction force that 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 un+2 since it introduces the
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new unknown R. The first and third equations are now rewritten as 0 un − P + ε n , = 1 un+1 P + ε n+1
EA 1 L 0
(10.23)
with the solution EA , L
un = [ − P + ε n ]
un+2 = [ P + ε n+1 ]
EA . L
(10.24)
To preserve symmetry, it is necessary for un + un+1 = 0. However, the sum is computed as un + un+1 =
ε n + ε n+1
( ELA )
.
(10.25)
The reaction force is given by R = −[εn + εn+1], and, in this case, it can be considered as a measure of computational error. Note that Equation 10.23 can be obtained from Equation 10.22 by simply “striking out” the second row of the matrices and vectors and the second column of the matrix. The same assembly arguments apply to the inertial forces as to the elastic forces. Omitting the details, the kinetic energies T of the two elements are T=
1 ˙ ( e )T ˜ ( e ) ˜ ( e+1) ˙ ( e ) γ [M + M ]γ 2
1 ˜ ( e ) = m( e ) M 1/ 2 1 m( e ) = ρAl 3 T
γ˙ ( e ) = {u˙e
(10.26)
1/ 2 1
(e)
u˙e+1
u˙e+2}
10.3.2 BEAMS A similar argument applies for beams. The potential energy and stiffness matrix of th the e element can be written as V (e) =
K
(e)
1 T (e) γ K γb 2 b
(e) K11 = ( e )T K 21
(e) K12 (e) K 22
γ Tb = {we − we′ we+1 − we′+1}
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(10.27)
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In a two-element beam model analogous to the previous rod model, V (e+1) V , implying that
K
(g)
(e) K 11 ( e )T = K 21 0
0 K . K
(e) K 12 ( e +1) K + K 11 K (21e+1)T
( e +1) 12 ( e +1) 22
(e) 22
(g)
Generally, in a global coordinate system, K
(g)
=V
(e)
+
(10.28)
(e) = ∑e K .
10.3.3 TWO-DIMENSIONAL ELEMENTS We next consider assembly in 2-D. Consider the model depicted in Figure 10.5 consisting of four rectangular elements, denoted as element e, e + 1, e + 2, and e + 3. The nodes are also numbered in the global system. Locally, the nodes in an element are numbered in a counterclockwise fashion. Suppose there is one degree of freedom per node (e.g., x-displacement) and one corresponding force. e+3,4
e+3,3
e+2,4
e+3
e+3,1
e+2
e+3,2
7
8
e+2,2
e+2
6
4
5 e
1
e+2,1
9
e+3
e,4
e+2,3
e+1
2
3
e,3
e+1,3 e+1,4
e
e,1
e+1
e,2
FIGURE 10.5 2-D assembly process.
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e+1,1
e+1,2
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In the local systems, the force on the center node induces displacements according to fe,3 = k3(,e1)ue,1 + k3(,e2)ue,2 + k3(,e3)ue,3 + k3(,e4)ue,4 fe+1,4 = k4( e,1+1)ue+1,1 + k4( e,2+1)ue+1,2 + k4( e,3+1)ue+1,3 + k4( e,4+1)ue+1,4 fe+2,1 = k1(,e1+2 )ue+2,1 + k1(,e2+2 )ue+2,2 + k1(,e3+2 )ue+2,3 + k1(,e4+2 )ue+2,4
(10.29)
fe+3,2 = k2( e,1+3)ue+3,1 + k2( e,2+3)ue+3,2 + k2( e,3+3)ue+3,3 + k2( e,4+3)ue+3,4 Globally, ue,1 → u1
ue,2 → u2
ue,3 → u5
ue,4 → u6
ue+1,1 → u2
ue+1,2 → u3
ue+1,3 → u4
ue+1,4 → u5
ue+2,1 → u5
ue+2,2 → u4
ue+2,3 → u9
ue+2,4 → u8
ue+3,1 → u6
ue+3,2 → u5
ue+3,3 → u8
ue+3,4 → u7
(10.30)
Adding the forces of the elements on the center node gives
[
]
[ ]u + [k
]
f5 = k3(,e1)u1 + k3(,e2) + k4( e,1+1) u2 + k4( e,2+1)u3 + k4( e,3+1) + k1(,e2+2 ) u4
[
+ k3(,e3) + k4( e,4+1) + k1(,e1+2 ) + k2( e,2+3)
[
(e) 3,4
5
]
+ k2( e,1+3) u6
]
+ k2( e,4+3)u7 + k1(,e4+2 ) + k2( e,3+3) u8 + k1(,e3+2 )u9
(10.31)
Taking advantage of the symmetry of the stiffness matrix, this implies that the fifth row of the stiffness matrix is
{
κ 5T = k3(,e1)
[k
(e) 3,3
[k
(e) 3,2
+ k4( e,1+1)
]
k4( e,2+1)
[k
]
( e +1) 4 ,3
+ k1(,e2+2 )
}
+ k4( e,4+1) + k1(,e1+2 ) + k2( e,2+3) K symmetry K
] (10.32)
Finally, for later use, we consider damping, which generates a stress proportional to the strain rate. In linear problems, it leads to a vector-matrix equation of the form M˙˙ γ + Dγ˙ + Kγ = f (t ).
(10.33)
At the element level, the counterpart of the kinetic energy and the strain energy (e) is the Rayleigh Damping Function, D , given by D ( e ) = 12 γ˙ ( e ) T D ( e ) γ˙ ( e ) , and the conth sistent damping force on the e element is fd( e ) (t ) = ∂˙ D( e ) = D( e ) γ˙ ( e ) . ∂γ
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(10.34)
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The Rayleigh Damping Function is additive over the elements. Accordingly, if ˆ ( e ) is the damping matrix of the eth element referred to the global system, the D assembled damping matrix is given by D( g ) =
∑ Dˆ
(e)
(10.35)
.
e
It should be evident that the global stiffness, mass, and damping matrices have the same bandwidth; the force on one given node depends on the displacements (velocities and accelerations) of the nodes of the elements connected at the given node, thus determining the bandwidth.
10.3 EXERCISES 1. The equation of static equilibrium in the presence of body forces, such as gravity, is expressed by ∂Sij ∂X j
= − bi .
Without the body forces, the dynamic Principle of Virtual Work is derived as
∫
∫
δ Eij Sij dV + δ ui ρ
∂ 2ui dV = δ uiτ i dS. ∂t 2
∫
How should the second equation be modified to include body forces? 2. Consider a one-dimensional system described by a sixth-order differential equation:
Q
d 6q = 0, dx 6
Q a constant.
Consider an element from xe to xe+1. Using the natural coordinate ξ = −1 when x = xe′ = +1 when x = xe+1, for an interpolation model with the minimum order that is meaningful, obtain expressions for ϕ(ξ), Φ, and γ, serving to express q as q(ξ ) = ϕ T (ξ )Φ γ . 3. Write down the assembled mass and stiffness matrices of the following three-element configuration (using rod elements). The elastic modulus is E, the mass density is ρ, and the cross-sectional area is A.
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Finite Element Analysis: Thermomechanics of Solids
y
L
L/2
L
4. Assemble the stiffness coefficients associated with node n, shown in the following figure, assuming plane-stress elements. The modulus is E, and (1) (2) (3) the Poisson’s ratio is ν . K , K , and K denote the stiffness matrices of the elements.
K (3) K (1)
n
K (2)
5. Suppose that a rod satisfies δ Ψ = 0, in which Ψ is given by Ψ=
∫
L
0
2
1 du Pu( L) . dx − E 2 dx
Use the interpolation model u( x ) = {1
u( x e , t ) x}Φ . u( xe+1, t )
For an element xe < x < xe+1, find the matrix Ke such that u( xe , t ) fe Ke = . u( xe+1, t ) fe+1 6. Redo this derivation for Ke in the previous exercise, using the natural coordinate ξ, whereby ξ = ax + b, in which a and b are such that −1 = axe + b, +1 = axe+1 + b. 7. Next, regard the nodal-displacement vector as a function of t. Find the matrix Me such that ••
u( x e , t ) u( xe , t ) fe + Me K = , e u( xe+1, t ) u( xe+1, t ) fe+1
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x
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151
in which ρ is the mass density. Derive Me using both physical and natural coordinates. 8. Show that, for the rod under gravity, a two-element model gives the exact answer at x = 1, as well as a much better approximation to the exact displacement distribution.
E A ρ
g L
9. Apply the method of the previous exercise to consider a stepped rod, as shown in the figure, with each segment modeled as one element. Is the displacement at x = 2L still exact?
E1
A1
L1
ρ1 g
E2
A2 ρ2
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L2
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11
Solution Methods for Linear Problems
11.1 NUMERICAL METHODS IN FEA 11.1.1 SOLVING THE FINITE-ELEMENT EQUATIONS: STATIC PROBLEMS Consider the numerical solution of the linear system Kγγ = f, in which K is the positive-definite and symmetric stiffness matrix. In many problems, it has a large T dimension, but is also banded. The matrix can be “triangularized”: K = LL , in which L is a lower triangular, nonsingular matrix (zeroes in all entries above the T diagonal). We can introduce z = L γ and obtain z by solving Lz = f. Next, γ can be T computed by solving L γ = z. Now Lz = f can be conveniently solved by forward substitution. In particular, Lz = f can be expanded as l11 l21 l31 . . l n1
0
.
.
.
l22
.
.
.
l32
l33
.
.
.
.
.
.
.
.
.
.
ln 2
.
.
.
0 z1 f1 . z 2 f2 . z3 f3 = . . . . 0 . . lnn zn fn
(11.1)
Assuming that the diagonal entries are not too small, this equation can be solved, starting from the upper-left entry, using simple arithmetic: z1 = f1/l11, z2 = [ f2 − l21z1]/ l22, z3 = [ f3 − l31z1 − l32z2]/l33,… . T Next, the equation L γ = z can be solved using backward substitution. The equation is expanded as l11 0 0 . . 0
l12
.
.
.
l22
.
.
.
0
.
.
.
.
.
ln−2,n−2
ln−2,n−1
.
.
0
ln−1,n−1
0
.
0
0
l1n γ 1 f1 . γ 2 . . γ 3 . = . ln−2,n . fn−2 ln−1,n . fn−1 lnn γ n fn
(11.2)
153
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Starting from the lower-right entry, solution can be achieved using simple arithmetic as γn = fn/lnn,
γ n−1 = [ fn−1 − ln−1,1γ n ]/ln−1,n−1 , γ n−2 = [ fn−2 − ln−2,nγ n − ln−2,n−1γ n−1 ]/ln−2,n−2 , K . In both procedures, only one unknown is encountered in each step (row).
11.1.2 MATRIX TRIANGULARIZATION OF LINEAR SYSTEMS
SOLUTION
AND
We next consider how to triangularize K. Suppose that the upper-left ( j − 1) × ( j − 1) block Kj−1 has been triangularized: K j −1 = L j −1LTj −1 . In determining whether the j × j block Kj can be triangularized, we consider K j −1 Kj = T k j
k j L j −1 = k jj λ Tj
0 LTj −1 l jj 0 T
λj , l jj
(11.3)
in which kj is a ( j − 1) × 1 array of the first j − 1 entries of the j column of Kj. Simple manipulation suffices to furnish kj and ljj. th
k j = L j −1λ j (11.4)
l jj = k jj − λ Tj λ j
Note that λj can be conveniently computed using forward substitution. Also, note that ljj = k jj − kTj K −j −11k j . The fact that Kj > 0 implies that ljj is real. Obviously, st the triangularization process proceeds to the ( j + 1) block and on to the complete stiffness matrix. As an example, consider 1 A 3 = 12 13
1 2 1 3 1 4
1 4 . 1 5
1 3
(11.5)
Clearly, L1 = LT1 → 1. For the second block, 1 λ 2
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0 1 l22 0
λ 2 1 = l22 12
, 1 3 1 2
(11.6)
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155
from which λ 2 = 1/2 and l22 = 1/3 − (1/2) 2 = 1/ 12 . Now 1 L2 = 1/2
0 . 1/ 12
(11.7)
We now proceed to the full matrix: 1 12 13
1 2 1 3 1 4
T 1 4 = L3L3 1 5
1 3
1 = 12 l31
0 1 0 0 l33 0
0 1/ 12 l32
1 = 12 l31
1 2
1 2
1/ 12 0
l31 l32 l33
l31 / 2 + l32 / 12 2 2 2 l31 + l32 + l33 l31
1 3
l31 / 2 + l32 / 12
2 We conclude that l31 = 1/3, l32 = 1/ 12 , l33 = 1/5 − 1/9 − 1/12 =
11.1.3 TRIANGULARIZATION
(11.8)
1 180
.
ASYMMETRIC MATRICES
OF
Asymmetric stiffness matrices arise in a number of finite-element problems, including problems with unsteady rotation and thermomechanical coupling. If the matrix is still nonsingular, it can be decomposed into the product of a lower-triangular and an upper-triangular matrix: K = LU.
(11.9)
th
Now, the j block of the stiffness matrix admits the decomposition K j −1 Kj = T k2 j
k1 j L j −1 = k jj λ Tj
L j −1U j −1 = T λ j U j −1
0 U j −1 l jj 0 T
λ Tj u j + u jj l jj
uj u jj
L j −1u j
(11.10)
in which it is assumed that the ( j − 1) block has been decomposed in the previous step. Now, uj is obtained by forward substitution using Lj−1uj = k1j, and λj can be obtained by forward substitution using U Tj −1λ j = k 2 j . Finally, u jj l jj = k jj − λ Tj u j , for th
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which purpose ujj can be arbitrarily set to unity. An equation of the form Kx = f can now be solved by forward substitution applied to Lz = f, followed by backward substitution applied to Ux = z.
11.2 TIME INTEGRATION: STABILITY AND ACCURACY Much insight can be gained from considering the model equation: dy = − λy, dt
(11.11)
in which λ is complex. If Re(λ) > 0, for the initial value y(0) = y0, y(t) = y0 exp(−λt), then clearly y(t) → 0. The system is called asymptotically stable in this case. We now consider whether numerical-integration schemes to integrate Equation 11.11 have stability properties corresponding to asymptotic stability. For this purpose, we apply the trapezoidal rule, the properties of which will be discussed in a subsequent section. Consider time steps of duration h, and suppose that the solution th has been calculated through the n time step, and we seek to compute the solution at st the (n + 1) time step. The trapezoidal rule is given by dy yn+1 − yn ≈ , dt h
− λy ≈ −
yn+1 =
1 − λh/2 y 1 + λh/2 n
λ [ y + yn ]. 2 n+1
(11.12)
Consequently,
n
1 − λh/2 = y0 1 + λh/2
(11.13)
Clearly, yn+1 → 0 if 11 −+ λλhh//22 < 1, and yn+1 → ∞ if 11 −+ λλhh//22 > 1, in which |·| implies the magnitude. If the first inequality is satisfied, the numerical method is called A-stable (see Dahlquist and Bjork, 1974). We next write λ = λr + iλi, and now A-stability requires that λih 2
2
λih 2
2
1 − λr h 2
2
+
1 + λr h 2
2
+
< 1.
(11.14)
A-stability implies that λr > 0, which is precisely the condition for asymptotic stability. Consider the matrix-vector system arising in the finite-element method: M˙˙ γ + Dγ˙ + Kγ = 0,
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γ (0) = γ 0 ,
γ˙ (0) = γ˙ 0 ,
(11.15)
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in which M, D, and K are positive-definite. Elementary manipulation serves to derive that d dt
1 γ˙ T Mγ˙ + 1 γ T Kγ = − γ˙ T Dγ˙ < 0. 2 2
(11.16)
It follows that γ˙ → 0 and γ → 0. We conclude that the system is asymptotically stable. Introducing the vector p = γ˙ , the n-dimensional, second-order system is written in state form as the (2n)-dimensional, first-order system of ordinary differential equations: M 0
•
0 p D + I γ − I
K p f = . 0 γ 0
(11.17)
We next apply the trapezoidal rule to the system: M 0
1 0 h (p n+1 − p n ) D + I 1 ( γ n+1 − γ n ) − I h
1 1 K 2 (p n+1 + p n ) = 2 (fn+1 + fn ) . 0 1 ( γ n+1 + γ n ) 0 2
(11.18)
From the equation in the lower row, pn+1 = 2h [γγn+1 − γn] − pn. Eliminating pn+1 in the upper row furnishes a formula underlying the classical Newmark method: h h2 K D = M + D + K 2 4
K D γ n+1 = rn+1 ,
rn+1
(11.19)
h h2 h h h2 (f + f ) = M + D − K yn + M + D p n + 2 4 2 2 4 n+1 n
and KD can be called the dynamic stiffness matrix. Equation 11.19 can be solved by triangularization of KD, followed by forward and backward substitution.
11.3 NEWMARK’S METHOD To fix the important notions, consider the model equation dy = f ( y). dx
(11.20)
Suppose this equation is modeled as
α yn+1 + β yn + h[γ fn+1 + δ fn ] = 0.
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(11.21)
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We now use the Taylor series to express yn+1 and fn+1 in terms of yn and fn. Noting that yn′ = fn and yn′′ = fn′, we obtain 0 = α [ yn + yn′ h + yn′′h 2 /2] + βyn + hγ [ yn′ + yn′′h] + hδyn′ .
(11.22)
2
For exact agreement through h , the coefficients must satisfy
α +β =0
α +γ +δ = 0
α /2 + γ = 0.
(11.23)
We also introduce the convenient normalization γ + δ = 1. Simple manipulation serves to derive that α = −1, β = 1, γ = 1/2, δ = 1/2, thus furnishing yn+1 − yn 1 = [ f ( yn+1 ) + f ( yn )], 2 h
(11.24)
which can be recognized as the trapezoidal rule. The trapezoidal rule is unique and optimal in having the following three characteristics: It is a “one-step method” using only the values at the beginning of the current time step. 2 It is second-order-accurate; it agrees exactly with the Taylor series through h . Applied to dy/dt + λy = 0, with initial condition y(0) = y0, it is A-stable whenever a system described by the equation is asymptotically stable.
11.4 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 that is optimal in an important sense, called Gaussian quadrature, has long been known. It is based on converting physical coordinates to natural coordinates. Consider ∫ ba f ( x )dx. Let ξ = b −1 a [2x − (a + b)]. Clearly, ξ maps the interval [a, b] into the interval [−1,1]. The integral now becomes 1 1 Now consider the power series b − a ∫ −1 f (ξ )dξ. f (ξ ) = α 0 + α 1ξ + α 2ξ 2 + α 3ξ 3 + α 4ξ 4 + α 5ξ 5 + L,
(11.25)
from which
∫
1
−1
f (ξ )dξ = 2α 1 + 0 +
2 2 α + 0 + α 5 + 0 + L. 3 3 5
(11.26)
The advantages illustrated for integration on a symmetric interval demonstrate that, st with n function evaluations, an integral can be evaluated exactly through (2n − 1) order.
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Consider the first 2n − 1 terms in a power-series representation for a function: g(ξ ) = α 1 + α 2ξ + L + α 2 nξ 2 n−1 .
(11.27)
Assume that n integration (Gauss) points ξi and n weights are used as follows:
∫
−1
g(ξ )dξ =
n
n
n
1
n
∑ g(ξ )w = α ∑ w + α ∑ w α + L + α ∑ w ξ i
i
i
1
i =1
i
2
i =1
2 n −1 i i
2n
i
i =1
(11.28)
.
i =1
Comparison with Equation 11.26 implies that
∑
n
n
n
wi = 1,
i =1
∑ i =1
wiξ = 0,
∑
n
wiξi2 = 2/3, K,
i =1
∑
wiξi2 n−2 =
i =1
2 , 2n − 1
n
∑w ξ
2 n −1 i i
= 0.
i =1
(11.29) It is necessary to solve for n integration points, ξi, and n weights, wi. These are 2n−i universal quantities. To integrate a given function, g(ξ), exactly through ξ , it is necessary to perform n function evaluations, namely to compute g(ξi). As an example, we seek two Gauss points and two weights. For n = 2, w1 + w2 = 2
(i),
w1ξ1 + w2ξ2 = 0
(ii)
2 3
(iii),
w1ξ13 + w2ξ23 = 0
(iv)
w1ξ12 + w2ξ22 =
(11.30)
From (ii) and (iv), w1ξ1[ξ12 − ξ22 ] = 0, leading to ξ2 = −ξ1. From (i) and (iii), it follows that −ξ2 = ξ1 = 1/ 3. The normalization w1 = 1 implies that w2 = 1.
11.5 MODAL ANALYSIS BY FEA 11.5.1 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˙˙ γ + Kγ = 0,
γ (0) = γ 0 ,
γ˙ (0) = γ˙ 0 .
(11.31)
Assume a solution of the form γ = γˆ exp(λt ), which furnishes upon substitution [K + λ2 M]γˆ = 0.
(11.32)
The j eigenvalue, λj, is obtained by solving det(K + λ2j M) = 0, and a corresponding eigenvector vector, γj, can also be computed (see Sample Problem 2). th
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For the sake of generality, suppose that λj and γj are complex. Let γ Hj denote the complex conjugate (Hermitian) transpose of γj. Now, λ2j satisfies
λ2j = −
γ Hj Kγ j γ Hj Mγ j
.
Since M and K are real and positive-definite, it follows that λj is pure imaginary: λj = iωj. Without loss of generality, we can take γj to be real and orthonormal. Sample Problem 1 As an example, consider k12 . k22
(11.33)
(λ2 )2 + [k11 + k22 ]λ2 + [k11k22 + k122 ] = 0,
(11.34)
1 M= 0
0 1
k11 K= k12
Now det[K + λ I] = 0 reduces to 2
with the roots
[ 1 = [−[k 2
λ2+,− =
1 −[k11 + k22 ] ± [k11 + k22 ]2 − 4[k11k22 + k122 ] 2 11
+ k22 ] ± [k11 − k22 ]2 − 4 k122
]
] (11.35)
so that both λ2+ and λ2− are negative (since k11 and k22 are positive). th th We now consider eigenvectors. The eigenvalue equations for the i and j eigenvectors are written as [K + ω 2j M]g j = 0
[K + ω k2 M]g k = 0
(11.36)
It is easily seen that the eigenvectors have arbitrary magnitudes, and for convenience, we assume that they have unit magnitude γ Tj γ j = 1. Simple manipulation furnishes that γ Tk Kγ j − γ Tj Kγ k − [ω 2j γ Tk Mγ j − ω k2 γ Tj Mγ k ] = 0.
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(11.37)
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Symmetry of K and M implies that γ Tk Kγ j − γ Tj Kγ k = 0,
[ω 2j γ Tk Mγ j − ω k2 γ Tj Mγ k ] = (ω 2j − ω k2 )γ Tk Mγ j = 0. (11.38)
Assuming for convenience that the eigenvalues are all distinct, it follows that γ Tj Mγ k = 0,
γ Tj Kγ k = 0,
j ≠ k.
(11.39)
The eigenvectors are thus said to be orthogonal with respect to M and K. The th th quantities µ j = γ Tj Mγ j and κ j = γ Tj Kγ j are called the ( j ) modal mass and ( j ) modal stiffness. Sample Problem 2 Consider 0 γ 1 1 γ 2
2 0
••
+
k 2 m −1
−1 γ 1 0 = . 1 γ 2 0
(11.40)
2 2 2 2 Let ζ = ω j /ω 0 , ω 0 = k/m. For the determinant to vanish, 1 − ζ ±2 = ±1/ 2 . Using 1 − ζ 2 = 1/ 2 , the first eigenvector satisfies ±
2 −1
(1) −1 γ 1 0 = , (1) 1/ 2 γ 2 0
[γ ] + [γ ] (1) 2 1
(1) 2 2
= 1,
(11.41)
implying that γ 1(1) = 1/ 3 , γ 2(1) = 2 / 3 . The corresponding procedures for the second eigenvalue furnish that γ 1( 2 ) = 1/ 3 , γ 2( 2 ) = − 2 / 3 . It is readily verified that T
T
µ1 = 4/3,
T
T
κ1 =
γ ( 2 ) Mγ (1) = γ (1) Mγ ( 2 ) = 0, γ ( 2 ) Kγ (1) = γ (1) Kγ ( 2 ) = 0,
µ 2 = 4/3,
4 [1 − 1/ 2 ], 3
κ2 =
4 [1 + 1/ 2 ] 3
(11.42)
The modal matrix X is now defined as X = [γ 1
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γ2
γ3
L
γ n ].
(11.43)
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Finite Element Analysis: Thermomechanics of Solids th
T
T
Since the jk entries of X MX and X KX are γ Tj Mγ k and γ Tj Kγ k , respectively, it follows that µ1 0 X T MX = . . .
0
.
.
µ2
.
.
.
.
.
.
.
.
.
.
.
. . . . µ n
κ 1 0 X T KX = . . .
0
.
.
κ2
.
.
.
.
.
.
.
.
.
.
.
. . . . (11.44) . κ n
The modal matrix is said to be orthogonal with respect to M and K, but it is −1 T not purely orthogonal since X ≠ X . The governing equation is now rewritten as ξ + X T KXξ = g, X T MX˙˙
ξ = X −1 γ ,
g = XTf,
(11.45)
implying the uncoupled modes
µ j ξ˙˙j + κ j ξ j = g j (t ).
(11.46)
Suppose that gj(t) = gj0 sin (ω t). Neglecting transients, the steady-state solution th for the j mode is
ξj =
gj0
κ j − ω 2µ j
sin(ω t ).
(11.47)
It is evident that if ω 2 ~ ω 2j = κ j /µ j (resonance), the response amplitude for the j 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. th
11.5.2 COMPUTATION
OF
EIGENVECTORS
AND
EIGENVALUES
Consider Kγ j = ω 2j Mγ j , with γ Tj γ j = 1. Many methods have been proposed to compute the eigenvalues and eigenvectors of a large system. Here, we describe a method that is easy to visualize, which we call the hypercircle method. The vectors Kγγj and Mγγj must be parallel to each other. Furthermore, the vectors Kγ j / γ Tj K 2γ j and Mγ j / γ Tj M2γ j must terminate at the same point in a hypersphere in nth dimensional space. Suppose that γ (jν ) is the ν iterate and that the two vectors
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163
do not coincide in direction. Another iterate can be attained by an interval-halving method:
Mγˆ (jν +1) =
Mγ (ν ) Kγ (ν ) 1 j j + T 2 ( ν ) 2 γ (ν ) T M 2 γ (ν ) γ K γ (ν ) j j j j
,
γ (jν +1) = γˆ (jν +1)
T γˆ (jν +1) M2 γˆ (jν +1) .
(11.48) Alternatively, note that C(γγ j ) =
γ Tj K
Mγ j
γ Tj K 2 γ j
γ Tj M 2 γ j
(11.49)
is the cosine of the angle between two unit vectors, and as such it assumes the maximum value of unity when the vectors coincide. A search can be executed on the hypersphere in the vicinity of the current point, seeking the path along which C(γγj) increases until the maximum is attained. Once the eigenvector γj is found, the corresponding eigenvalue is found from 2 ω j = ( γ Tj Kγ j )/( γ Tj Mγ j ). Now, an efficient scheme is needed to “deflate” the system so that ω12 ,and γ1 no longer are part of the eigenstructure in order to ensure that the solution scheme does not converge to values that have already been calculated. Given γ1 and ω12 , we can construct a vector p2 that is M-orthogonal to γ1 by using an intermediate vector pˆ 2 : p2 = pˆ 2 − γ 1T Mpˆ 2 γ 1. Clearly, γ 1T Mp2 = γ 1T Mpˆ 2 − γ 1T Mpˆ 2 γ 1T Mγ 1 = 0 assuming µ1 = 1. However, p2 is also clearly orthogonal to Kγγ1 since it is collinear with Mγγ1. A similar procedure leads to vectors pj, which are orthogonal to each other and to Mγγ1 and Kγγ1. For example, with p2 set to unit magnitude, p 3 = pˆ 3 − p T2 pˆ 3p 2 − γ 1T Mpˆ 3 γ 1 . Now, p T2 p 3 = 0 and γ 1T Mpˆ 3 = 0. th Introduce the matrix X1 as follows: X1 = [γγ1 p2 p3 … pn]. For the k eigenvalue, we can write
[X KX T 1
1
]
− ω 2k X1TMX1 X1−1 γ k = 0,
(11.50)
which decomposes to ω12 0T
0 1 − ω k2 ˜ 0 T K n −1
0 0 0 . = ˜ η 0 M n −1 n −1
(11.51)
This implies the “deflated” eigenvalue problem ˜ −ω M ˜ ]η = 0. [K n −1 n −1 n −1 k
(11.52)
The eigenvalues of the deflated system are also eigenvalues of the original system. The eigenvector ηn−1 can be used to compute the eigenvectors of the original system
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using transformations involving the matrix X1. The eigenvalues and eigenvectors of the deflated system can be computed by, for example, the hypercircle method described previously.
11.6 EXERCISES 1. Verify that the triangular factors L3 and LT3 for A3 in Equation 11.5 are correct. 2. Using A3 in Equation 11.5, use forward substitution followed by backward substitution to solve 1 A 3 γ = 1 . 1 3. Triangularize the matrix 36 K = 30 18
30 41 23
18 23. 14
4. For the model equation dy/dx = f(y), develop a two-step numerical-integration model: (αyn+1 + βyn + γ yn−1 ) + h[δ f ( yn+1 ) + ε f ( yn ) + ζ f ( yn−1 )] = 0. What is the order of the integration method (highest power in h with exact agreement with the Taylor series)? 5. Find the integration (Gauss) points and weights for n = 3. 6. In the damped linear-mechanical system M˙˙ γ + Dγ˙ + Kγ = f (t ), th suppose that γ(t) = γn at the n time step. Derive KD and rn+1 such that γ st at the (n + 1) time step satisfies
K D γ n+1 = rn+1 . 7. For the linear system 36 30 24
30 41 32
24 γ 1 1 32 γ 2 = 2 27 γ 3 3
triangularize the matrix and solve for γ1, γ2, γ3.
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165
8. (a) Find the modal masses µ1, µ2 and the modal stiffnesses κ1 and κ2 of the system 1 3 0
••
0 γ 1 1 + 27 2 γ 2 −1
−1 γ 1 10 = sin(10t ). 2 γ 2 20
(b) Determine the steady-state response of the system (i.e., particular solution to the equation). 9. Triangularize 2 µ A/ L 0 A
0 4 µ AL/Y 2 2 µ A/ L
−A
−2 µ A/ L . 0
10. Put the following equations in state form, apply the trapezoidal rule, and triangularize the ensuing dynamic stiffness matrix: M˙˙ γ + Kγ − Σπ = f ,
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Σ T γ = 0.
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12
Rotating and Unrestrained Elastic Bodies
12.1 FINITE ELEMENTS IN ROTATION We first consider rotation about a fixed axis. The coordinate system is embedded in the fixed point and rotates. The undeformed position vector, X′ in the rotated system is related to its counterpart, X, in the unrotated system by X ′ = Q(t )X. The counterpart for the deformed position is x ′ = Q(t )x. The displacement also satisfies u ′ = Q(t )u. ˙ (t )Q T (t )(*). The rotation is represented by the “axial” vector, ω, satisfying ω × (*) = Q T ˙ (t )Q (t ) is antisymmetric). The time derivatives in rotating coordinates (Recall that Q satisfy du′ ∂u′ = + ω × u′ dt ∂t d 2 u′ ∂ 2 u′ = 2 + 2ω × u ′ + ω × ω × u + α × u dt 2 ∂t
(12.1)
∂(.) where α = ∂∂ωt and ∂ t imply differentiation with the coordinate system instantaneously fixed. The rightmost four terms in (12.1) are called the translational, Coriolis, centrifugal, and angular accelerations, respectively. 2 Applied to the Principle of Virtual Work, the inertial term becomes ∫ δ u′ T ρ d 2 dt [u′ + X′]dV . Assuming that u′ = ϕ T (X′)Φγ (t ),
∫
δ u′ T ρ
d 2γ dγ d 2 u′ T dV = δ γ Μ + G1 + (G 2 + A ) γ 2 2 dt dt dt
∫
G1 = Φ T ρ ϕ Ω ϕ T d VΦ,
∫
Α = Φ T ρ ϕΑϕ T d VΦ.
M = Φ T ρ ϕ ϕ T d VΦ, G2 = Φ T ρ ϕ Ω2ϕ T d VΦ,
(12.2)
∫
∫
The matrix M is the conventional positive-definite and symmetric mass matrix; the Coriolis matrix G1 is antisymmetric; the centrifugal matrix G2 is negative-definite; and ˙ = QQ ˙ T, A = Ω ˙. the angular acceleration matrix A is antisymmetric. Also, Ω 167
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z ω
y
r E,A,L,ρ
θ x
f0 sin ωt
FIGURE 12.1 Elastic rod on rotating rigid shaft.
There is also a rigid-body force term:
∫
δ u′ T ρ
d2 X′d V = −δγ T frot , dt 2
∫
frot = Φ T ρϕ [Ω2 + A]X′d V.
(12.3)
The governing equation is now M
d2γ dγ + G1 + [K + G 2 + A]γ = f − frot . dt 2 dt
(12.4)
Consider a rod attached to a thin shaft rotating steadily at angular velocity ω (see Figure 12.1), with f0 = 0. If r is the undeformed position along the shaft, the governing equation is EA
d 2u = −ω 2 [r + u]. dr 2
(12.5)
Assuming a one-element model with u(r, t ) = ru( L, t ) L , we obtain EA ρω 2 AL 2 L − u( L, t ) = − ρω A 3 =− ω=
∫
L
0
ρω 2 AL2 3
r rdr L (12.6)
Clearly, u(L, t) becomes unbounded if ω becomes equal to the natural frequency ρAL , in which case, ω is called a critical speed. EA 3 3
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Rotating and Unrestrained Elastic Bodies
12.2
169
FINITE-ELEMENT ANALYSIS FOR UNCONSTRAINED ELASTIC BODIES
Consider the response of an elastic body that has no fixed points, as in spacecraft. It is assumed that the tractions are prescribed on the undeformed surface. The response is referred to the “body axes” corresponding to axes embedded in the corresponding rigid body, for example, the principal axes of the moment of inertia tensor. The position vector r of a point in the body can be decomposed as follows: r = rc + ξ + u ,
(12.7)
in which rc is the position vector to the center of mass; ξ is the relative positionvector from the center of the mass to the undeformed position of the current point, referred to rotating axes; and u is the displacement from the undeformed to the deformed position in the rotating system (see Figure 12.2). The balance of linear momentum becomes ∂Sij ∂ξ j
= ρ[˙˙ rci + ξ˙˙i + u˙˙i ].
(12.8)
Recall that ξ˙ = ω × ξ, and the corresponding variational relation is δ ξ = δ u × ξ, ω = θ˙ . It follows that δr = δrc + δ θ˙ × (ξ + u) + δ u′, in which δ u′ is the variation of u with the axes instantaneously held fixed. The quantities rc, θ, and u′ can be varied independently since there is no constraint relating them. For δ rc, we have ∂Sij
∫ δr ∂ξ ci
j
rci + ξ˙˙i + u˙˙i ] dV = 0. − ρ[˙˙
(12.9)
z u(ξξ,t) ξ(t)
rc(t)
y x FIGURE 12.2 3-D unconstrained 3-D element.
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Now ∫ δ rci ρξ˙˙i dV = δ rci ∫ ρξ˙˙i dV = 0 by virtue of the definition of the center of mass. Next ∫ δ rci ρu˙˙i dV = 0 on the assumption, for which a strong argument can be made, that deformation does not affect the position of the center of mass. Then, ∫ δ rci ρ˙˙ rci dV = δ rci ˙˙ rci ∫ ρdV = δ rci mr˙˙ci . Finally, ∂Sij
∫δ r
∂ξ j
ci
dV = δ rci
∂Sij
∫ ∂ξ
dV
j
∫
= δ rci Sij n j dS
∫
= δ rci τ i dS = δ rci Pi
(12.10)
Consequently P = mr··c, as expected. θ×ξ Consider the variation with respect to ξ : d ξ = dθ
∫
∂S δξi ij − ρ[˙˙ rci + ξ˙˙i + u˙˙i ] dV = 0. ∂ξ j
(12.11)
First,
∫ δξ ρ˙˙r dV = ˙˙r ∫ δξ ρdV, i
ci
ci
i
(12.12)
and δ [ ∫ ρξi dV ] = 0, Recall that ξ˙ = ω × ξ and the corresponding variational relation is δ ξ = δθ × ξ, ω = θ˙. Now,
∫ δ ξ ρu˙˙ d V = ∫ [δθ × ξ] ρu˙˙ dV = δθ ∫ ξ × ρu˙˙dV. T
i
i
i
i
(12.13)
It is common to assume that ∫ ξ × uρdV = 0, which also implies that ∫ ξ × ρu˙˙i dV = 0. This assumption implies that the body axes in the deformed configuration are not affected by deformation and are obtained by rotating the undeformed axes according to rigid-body relations. Next, consider δξi:
∫δξ ρξ˙˙ dV = δ θ ∫ ρξ × [ω × ω × ξ + α × ξ]ρdV T
i
i
= δ θT − ρ[ω × ξ × ξ × ω + ξ × ξ × α ]ρdV
∫
= δ θT [ω × Jω + Jα ]
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(12.14)
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171
in which J = − ∫ ρ « 2 dV, and « (.) = ξ × (.). Clearly, as the matrix corresponding to a 2 vector product, « is antisymmetric. We determined in an earlier discussion that « is negative-definite, thus, the moment-of-inertia tensor J is positive-definite. Finally, ∂Sij
∫δξ ∂ξ i
dV =
j
∂ξ i
∂
∫ ∂ξ (δξ S )dV − ∫ δ ∂ξ S dV i ij
ij
j
j
∫
= (δξi Sij ni )dS
∫
= δξiti dS = δ θ T m,
∫
m = ξ × t dS
(12.15)
The vector-valued quantity m is recognized as the moment of the tractions referred to the center of mass. We thus obtain the Euler equation of a rigid body as m = ω × Jω + Jα .
(12.16)
Equations 12.10 and 12.16 can be solved separately from the finite-element equation for the displacements, to be presented next, to determine the origin and orientation of the “body axes.” Now consider the variation δ u′: ∂σ ij
∫ δu′ ∂ξ i
j
− ρ[˙˙ rci + ξ˙˙i + u˙˙i ] dV = 0.
(12.17)
First, note that ∫ δui′ρdVr˙˙ci = 0 since ∫ δui′ρdV = 0. The remaining terms are exactly the same as for a body with a fixed point, and consequently it reduces to M
d2γ dγ + G1 + [K + G 2 + A] γ = f − frot dt 2 dt
∫
G2 = Φ T ρ ϕΩ2ϕ T dVΦ,
∫
frot = Φ T ρ ϕ[Ω2 + A]X′dV
G1 = Φ T ρ ϕΩϕ T dVΦ, A = Φ T ρ ϕAϕ T dVΦ,
(12.18)
∫
∫
12.3 EXERCISES 1. Consider a one-element model for a steadily rotating rod (see Figure 12.1) to which is applied an oscillatory force, as shown. Find the steady-state solution. 2. Find the exact solution for u(r) in Exercise 1. Does it agree with the oneelement solution? Try two elements. Is the solution better? By how much?
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3. Prove that the inertia tensor J is positive-definite. Find J for the brick “element” with density ρ.
c
b a
4. Write down the proof that for an unrestrained elastic body, if
∫ ρudV = 0 and ∫ ρξ × udV = 0,
then F = m˙˙r0
M = Jα + ω × Jω.
5. 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 this situation couples extension and bending.
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13
Thermal, Thermoelastic, and Incompressible Media
13.1 TRANSIENT CONDUCTIVE-HEAT TRANSFER 13.1.1 FINITE-ELEMENT EQUATION The governing equation for conductive-heat transfer without heat sources, assuming an isotropic medium, is k∇ 2 T = ρce T˙ .
(13.1)
With the interpolation model T(t) − T0 = ϕ TT (x)ΦTθ( t ) and ∇T = β TT ΦT θ (t ), the finite-element equation assumes the form K Tθ + MT θ˙ = −q(t )
∫
K T = ΦTT β T kβ TT ΦT dV ,
∫
MT = ΦTT ϕ T ρceϕTT ΦTT dV
(13.2)
T
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 time rates) solid-mechanics equations, in which information propagates into the medium as finite velocity waves, and in which oscillatory response occurs in response to a perturbation.
13.1.2 DIRECT INTEGRATION
BY THE
TRAPEZOIDAL RULE
Equation 13.1 is already in state form since it is first-order, and the trapezoidal rule can be applied directly: MT
q + qn θ n+1 − θ n θ + θn + K T n+1 = − n+1 , h 2 2
(13.3)
from which K DT θn+1 = rn+1 K DT = MT +
(13.4)
h h h K r = MT θn − K T θn − (q n+1 + q n ) 2 T n+1 2 2 173
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For the assumed conditions, the dynamic thermal stiffness matrix is positivedefinite, and for the current time step, the equation can be solved in the same manner as in the static counterpart, namely forward substitution followed by backward substitution.
13.1.3 MODAL ANALYSIS Modes are not of much interest in thermal problems since the modes are not oscillatory or useful to visualize. However, the 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 T vanishes and that θ at t = 0 does not vanish. Multiplying the equation by θ and elementary manipulation furnishes that d θT MT θ T = −θ K T θ < 0. dt 2
(13.5)
T Clearly, the product θ MT θ decreases continuously. However, it only vanishes if θ vanishes. To examine the modes, assume a solution of the form θ(t) = θ 0j exp(λjt). The eigenvectors θ 0j satisfy
µ j =k , θT0 j MT θ0 k = Tj j≠k 0
κ j = k θT0 j K T θ0 k = Tj j≠k 0
(13.6)
and we call µ Tj and κ Tj the j modal thermal mass and j modal thermal stiffness, θ01 … θ0n ], and again respectively. We can also form the modal matrix Θ = [θ th
µ Tj 0 ΘTMT Θ = . . 0
th
0
.
.
µ Tj
.
.
.
.
.
.
.
.
.
.
.
0 . . , . .
κ Tj 0 ΘTK T Θ = . . 0
0
.
.
κ Tj
.
.
.
.
.
.
.
.
.
.
.
0 . . . . . (13.7)
−1 T T Let ξ = Θ θ and g(t) = Θ q(t). Pre- and postmultiplying Equation 13.2 with Θ and Θ, respectively, furnishes the decoupled equation
µTjξ˙ j + κ Tjξ j = g j .
© 2003 by CRC CRC Press LLC
(13.8)
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175
Suppose, for convenience, that gj is a constant. Then, the general solution is of the form κ ξ j = ξ j 0 exp − Tj µTj
t +
κ exp − Tj (t − T ) gj dτ , 0 µTj
∫
t
(13.9)
illustrating the monotonically decreasing nature of the response. Now there are n uncoupled single degrees of freedom.
13.2 COUPLED LINEAR THERMOELASTICITY 13.2.1 FINITE-ELEMENT EQUATION The classical theory of coupled thermoelasticity accommodates the fact that the thermal and mechanical fields interact. For isotropic materials, assuming that temperature only affects the volume of an element, the stress-strain relation is Sij = 2 µEij + λ ( Ekk − α (T − T0 ))δ ij ,
(13.10)
in which α denotes the volumetric thermal-expansion coefficient. The equilibrium ∂S equation is repeated as ∂xij = ρu˙˙i . The Principle of Virtual Work implies that j
∫ δE [2µE + λE δ ] dV + ∫ δu ρü dV − αλ ∫ δE (T − T )δ ij
ij
kk ij
o
i
i
o
ij
0
ij
∫
dVo = δu j t j dSo . (13.11)
Now consider the interpolation models u = N T ( x )γ (t ),
E ij → E = BT ( x )γ (t ),
T − T0 = v T ( x )θ (t ), ∇T = BTT ( x )θ (t ), (13.12)
in which E is the strain written as a column vector in conventional finite-element notation. The usual procedures furnish the finite-element equation M˙˙ γ (t ) + Kγ (t ) − Σθ(t ) = f (t ),
∫
Σ = αλ Bν T dVo .
(13.13)
The quantity Σ is the thermomechanical stiffness matrix. If there are nm displacement degrees of freedom and nt thermal degrees of freedom, the quantities appearing in the equation are M, K: nm × nm ,
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γ (t ), f (t ): nm × 1,
Σ: nm × nt ,
θ (t ): nt × 1.
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We next address the thermal field. The energy-balance equation (from Equation 7.35), including mechanical effects, is given by k∇ 2 T = ρce T˙ + αλT0 tr( E˙ ).
(13.14)
Application of the usual variational methods imply that
K T θ(t ) + MT θ˙ (t ) + T0 Σ T γ˙ (t ) = −q,
q=
∫ ν n ⋅ qdS.
(13.15)
Case 1: Suppose that T is constant. At the global level, θ (t ) = − T0 K T−1 Σ T γ˙ (t ) + T0 K T−1 q. Thus, the thermal field is eliminated at the global level, giving the new governing equation as γ (t ) + T0 ΣK T−1 Σ T γ˙ (t ) + Kγ (t ) = f (t ). M˙˙
(13.16)
Conductive-heat transfer is analogous to damping. The mechanical system is now asymptotically stable rather than asymptotically marginally stable. We next put the global equations in state form: Q1z˙ + Q 2 z = f M Q1 = 0 0 0 Q 2 = − K ΣT
0 K 0 K 0 0
0 M T / T0 0
−Σ 0 K T / T0
(13.17) γ˙ z = γ θ f f = 0 −q / T0
Clearly, Equation 13.17 can be integrated numerically using the trapezoidal rule: Q + h Q z = Q − h Q z + h [f + f ]. 1 2 2 n+1 1 2 2 n 2 n+1 n
© 2003 by CRC CRC Press LLC
(13.18)
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177
Now, consider asymptotic stability, for which purpose it is sufficient to take f = 0, T z(0) = z0. Upon premultiplying Equation 13.17 by z , we obtain d 1 T z Q1z = − z T Q 2 z dt 2 = −z T
[
]
1 Q + Q 2T z 2 2
= −θθT K T θ
(13.19)
and z must be real. Assuming that θ ≠ 0, it follows that z ↓ 0, and hence the system is asymptotically stable.
13.2.2 THERMOELASTICITY
IN A
ROD
Consider a rod that is built into a large, rigid, nonconducting temperature reservoir at x = 0. The force, f0, and heat flux, −q0, are prescribed at x = L. A single element models the rod. Now, u( x, t ) = xγ (t )/ L,
E( x, t ) = γ (t )/ L,
T − T0 = xθ (t )/ L,
dT = θ (t )/ L. dx
(13.20)
The thermoelastic stiffness matrix becomes Σ = αλ ∫Bν dV → Σ = αλ A/2. The governing equations are now T
1 ρAL ˙˙ EA γ+ γ − α λ Aθ = f0 2 3 L 1 1 ρce AL ˙ 1 kA θ+ θ + α λ Aγ˙ = − q 0 2 T0 3 T0 L
(13.21)
13.3 COMPRESSIBLE ELASTIC MEDIA For a compressible elastic material, the isotropic stress Skk and the dilatational strain Ekk are related by Skk = 3κEkk, in which the bulk modulus κ satisfies κ = E/[3(1 − 2ν)]. Clearly, as ν → 1/2, the pressure, p = −Skk /3, needed to attain a finite compressive volume strain (Ekk < 0) becomes infinite. At the limit ν = 1/2, the material is said to satisfy the internal constraint of incompressibility. Consider the case of plane strain, in which Ezz = 0. The tangent modulus matrix D is readily found from 1−ν2 Sxx E −ν (1 + ν ) Syy = (1 + ν )(1 − 2ν ) 0 Szz
© 2003 by CRC CRC Press LLC
−ν (1 + ν ) 1−ν2 0
0 Exx 0 Eyy . 1 − 2ν Ezz
(13.22)
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Clearly, D becomes unbounded as ν → 1/2. Furthermore, suppose that for a material to be nearly incompressible, ν is estimated as .495, while the correct value is .49. It might be supposed that the estimated value is a good approximation for −1 the correct value. However, for the correct value, (1 − 2ν) = 50. For the estimated −1 value, (1 − 2ν) = 100, implying 100 percent error!
13.4 INCOMPRESSIBLE ELASTIC MEDIA In an incompressible material, a pressure field arises that serves to enforce the 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. The Principle of Virtual Work is now expressed in terms of the displacements and pressure, and an adjoining equation is introduced to enforce the constraint a posteriori. The pressure can be shown to serve as a Lagrange multiplier, and the displacement vector and the pressure are varied independently. In incompressible materials, to preserve finite stresses, we suppose that the second Lame coefficient satisfies λ → ∞ as tr(E) → 0 in such a way that the product is an indeterminate quantity denoted by p:
λtr( E) → − p.
(13.23)
The Lame form of the constitutive relations becomes Sij = 2 µEij − pδ ij ,
(13.24)
together with the incompressibility constraint Eijδij = 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 = ∂Eij , and the domain term in the Principle of Virtual Work can be rewritten as ∫δEijSijdV = ∫δ wdV. The elastic-strain energy is given by w = µ Ei j Ei j + λ2 Ekk2. For reasons explained shortly, we introduce the augmented strain-energy function w ′ = µEij Eij − pEkk
(13.25)
and assume the variational principle
∫ δw′dV + ∫ δ u ρ u˙˙ dV = ∫ δ u τ dS . T
o
© 2003 by CRC CRC Press LLC
T
o
o
o
(13.26)
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179
Now, considering u and p to vary independently, the integrand of the first term becomes δw′ = δEij[2µEij − pδij ] − δ pEkk , furnishing two variational relations:
∫ δE S dV + ∫ δu ρ u˙˙ dV = ∫ δ u τ dS T
ij ij
o
T
o
o
(a)
o
(13.27)
∫ δpE dV = 0 kk
(b)
0
The first relation is recognized as the Principle of Virtual Work, and the second equation serves to enforce the internal constraint of incompressibility. We now introduce the interpolation models: u = N T ( x ) g (t ) Ekk = bT ( x ) g(t )
e = BT ( x ) g ( t ) p( x ) = x T ( x ) p(t )
(13.28)
Substitution serves to derive that Mγ˙˙(t ) + Kγ (t ) − Σπ(t ) = f
∫
Σ = bξ T dVo ,
(13.29)
Σ T γ (t ) = 0
Assuming that these equations apply at the global level, use of state form furnishes M 0 0 T
0 K 0T
0 γ˙ (t ) 0 d 0 γ˙ (t ) + − K dt 0 π(t ) Σ T
K 0 0T
− Σ γ˙ (t ) f (t ) 0 γ (t ) = 0 . 0 π(t ) 0
(13.30)
The second matrix is antisymmetric. Furthermore, the system exhibits marginal asymptotic stability; namely, if f(t) = 0 while γ˙ (0), γ (0), and π (0) do not all vanish, then d 1 T ( γ˙ (t ) dt 2
© 2003 by CRC CRC Press LLC
γ T (t )
M π T (t )) 0 0 T
0 K 0T
0 γ˙ (t ) 0 γ (t ) = 0 0 π(t )
(13.31)
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Finite Element Analysis: Thermomechanics of Solids
13.5 EXERCISES 1. Find the exact solution for a circular rod of length L, radius r, mass density ρ, specific heat ce , conductivity k, and cross-sectional area A = πr2. The initial temperature is T0, and the rod is built into a large wall at fixed temperature T0 (see figure below). However, at time t = 0, the temperature T1 is imposed at x = L. Compare the exact solution to the one- and twoelement solutions. Note that for a one-element model,
ρc AL kA θ ( L, t ) + e θ˙( L, t ) = − q( L). 3 L L r
T0
T1,t>0
2. State the equations of a thermoelastic rod, and put the equations for the thermoelastic behavior of a rod in state form. 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. γ + Kγ − Σπ = f, M˙˙
Σ T γ = 0.
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 µA / L 0 A
0 4 µAL / Y 2 2 AL / Y
−A
u( L) f −2 AL / Y v(Y ) = 0 0 p 0
and that this implies that 3µ u( L) = f (which can also be shown by an a L priori argument).
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14
Torsion and Buckling
14.1 TORSION OF PRISMATIC BARS Figure 14.1 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 θ(z), and at height L, it is θ0. Ordinarily, in the 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 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 stresses 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. The compatibility condition (see Chapter 4) is enforced, furnishing
z T θ0 r0 section after twist
section before twist L
θ
x z y φ
x FIGURE 14.1 Twist of a prismatic rod. 181
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Finite Element Analysis: Thermomechanics of Solids
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 φ and at radial position r. Clearly, X = r cos φ and Y = r sin φ. Twist induces a rotation through angle θ (z), but it does not affect the radial position. Now, x = r cos(φ + θ), y = r sin(φ + θ). Use of double-angle formulae furnishes the displacements, and restriction to small angles θ furnishes, to first order, u = −Yθ ,
v = Xθ .
(14.1)
It is also assumed that torsion does not increase the length of the member, which is attained by requiring that axial displacement w only depends on X and Y. The quantity w(X, Y) is called the warping function. It is readily verified that all strains vanish except Exz and Eyz, for which Exz =
1 ∂w ∂θ − y , 2 ∂x ∂z
1 ∂w ∂θ + x . 2 ∂y ∂z
Eyz =
(14.2)
Equilibrium requires that ∂Sxz ∂Syz + = 0. ∂x ∂y
(14.3)
The equilibrium relation can be identically satisfied by a potential function y for which Sxz =
∂ψ , ∂y
Syz = −
∂ψ . ∂x
(14.4)
We must satisfy the compatibility condition to ensure that the strain field arises from a displacement field that 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 =
1 1 ∂ψ Sxz = , 2µ 2 µ ∂y
Eyz =
1 1 ∂ψ Syz = − . 2µ 2 µ ∂x
Compatibility (integrability) now requires that −
∂2 w ∂x∂y
=
∂2 w , ∂y∂x
furnishing
∂ 1 ∂ψ 1 dθ ∂ 1 ∂ψ 1 dθ + y + − − x = 0, ∂y 2 µ ∂y 2 dz ∂x 2 µ ∂x 2 dz
© 2003 by CRC CRC Press LLC
(14.5)
(14.6)
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183
ψ
n ds
dy
dψ z
n x = cos χ = dy/ds n y = sin χ = dx/ds
–dx
ψ FIGURE 14.2 Illustration of geometric relation.
which in turn furnishes Poisson’s Equation for the potential function y : ∂ 2ψ ∂ 2ψ dθ + 2 = −2 µ . ∂x 2 ∂y dz
(14.7)
For boundary conditions, assume that the lateral boundaries of the member are unloaded. The stress-traction relation already implies that τx = 0 and τy = 0 on the lateral boundary S. For traction τz to vanish, we require that
τ z = n x Sxz + ny Syz = 0 on S.
(14.8)
Upon examining Figure 14.2, it can be seen that nx = dy/ds and ny = −dx/ds, in which s is the arc length along the boundary at z. Consequently,
τz =
dy dx S − S ds xz ds yz
=
dy ∂ψ dx ∂ψ + ds ∂y ds ∂x
=
dψ ds
(14.9)
Now, on S, ddsψ = 0, and therefore ψ is a constant, which can, in general, be taken as zero. We next consider the total torque on the member. Figure 14.3 depicts the cross section at z. The torque on the element at x and y is given by dT = xSyz dxdy − ySxz dxdy dψ dψ = − x −y dxdy dx dy
© 2003 by CRC CRC Press LLC
(14.10)
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Finite Element Analysis: Thermomechanics of Solids
z
dy dx
sxz
syz
y x
y
x FIGURE 14.3 Evaluation of twisting moment.
Integration furnishes dψ dψ T = − x dxdy +y dy dx
∫
d ( xψ ) d ( yψ ) dx dy = − + − ψ dx + dy dxdy dx dy
∫
xψ = − ∇ ⋅ dxdy + 2 ψ dxdy yψ
∫
∫
(14.11)
Application of the divergence theorem to the first term leads to ∫ψ [xnx + yny]ds, which vanishes since y vanishes on S. Finally,
∫
T = 2 ψ dxdy.
(14.12)
We apply variational methods to the Poisson Equation, considering the stresspotential function y to be the unknown. Now,
∫ δψ [∇ ⋅ ∇ψ + 2µθ ′]dxdy = 0. © 2003 by CRC CRC Press LLC
(14.13)
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185
Integration by parts, use of the divergence theorem, and imposition of the “constraint” y = 0 on S furnishes
∫ (∇δψ ) ⋅ ∇ψ dxdy = ∫ δψ 2µθ ′ dxdy.
(14.14) th
The integrals are evaluated over a set of small elements. In the e element, approximate ψ as ν TT ( x, y)ψ e ηe , in which νT is a vector with dimension (number of rows) equal to the number of nodal values of ψ. The gradient ∇ψ has a corresponding interpolation model ∇ψ = β TT ( x, y)ψ e ηe , in which βT is a matrix. The finite-element counterpart of the Poisson Equation at the element level is written as K (Te ) ηe = 2 µθ ′ fT( e )
∫
K (Te ) = ψ eT β T β TT dxdyψ e
(14.15)
∫
fT( e ) = ψ eT ν T ( x, y)dxdy and the stiffness matrix should be nonsingular, since the constraint y = 0 on S has −1 already been used. It follows that, globally, ηg = 2 µθ ′ K (Tg ) fT( g ) . The torque satisfies
∫
T = 2ψ dxdy = 2 ηTg fT( g ) −1
= 4 µθ ′ fT( g ) K(Tg ) fT( g ) T
(14.16)
In the theory of torsion, it is common to introduce the torsional constant J, for T −1 which T = 2µJθ ′. It follows that J = 2 fT( g ) K(Tg ) fT( g ) .
14.2 BUCKLING OF BEAMS AND PLATES 14.2.1 EULER BUCKLING
OF
BEAM COLUMNS
14.2.1.1 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 chapters. For now, we will focus on beams and plates, using classical equations in which, by retaining lowest-order corrections for geometric nonlinearity, in-plane compressive forces appear.
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z
y
Q0
E,I,A,L,ρ
M0
P
x
FIGURE 14.4 Euler buckling of a beam column.
For the beam shown in Figure 14.4, the classical Euler buckling equation is ˙˙ = 0, EIw iv + Pw ′′ + ρAw
(14.17)
and P is the axial compressive force. The interpolation model for w(x) is recalled T Φ γ. Following the usual variational procedures (integration by parts) as w(x) = ϕ (x)Φ furnishes
∫ δ wρAw˙˙ dx → δ γ M˙˙γ , T
∫ δ w[EIw
iv
∫
M = Φ T ϕ ( x )ρAϕ T ( x )dVΦ
∫
(14.18)
∫
+ Pw ′′]dx = δ w ′′EIw ′′dx − δ w ′Pw ′dx − [(δ w )( − Pw ′ − EIw ′′′)]0L − [( −δ w ′)( − EIw ′′)]0L
At x = 0, both δ w and −δ w′ vanish, while the shear force V and the bending moment M are identified as V = −EIw′′′ and M = −EIw′′. The “effective shear force” Q is defined as Q = −Pw′ − EIw′′′. For the specific case illustrated in Figure 14.3, for a one-element model, we can use the interpolation formula
w( x ) = ( x
2
L2 x ) −2 L 3
−1
L3 γ (t ), −3 L2
w( L ) γ (t ) = . − w ′( L)
(14.19)
The mass matrix is shown, after some algebra, to be
M = ρALK 0 ,
© 2003 by CRC CRC Press LLC
13 35 K0 = 11 210 L
L . 1 L2 105 11 210
(14.20)
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187
Similarly,
∫
δ w ′Pw ′dx = δ γ T
∫ δ w′′EIw′′dx = δ γ
T
P K γ, L 1
65 K1 = 1 10 L
EI K γ, L3 2
12 K2 = 6 L
L 2 2 15 L 1 10
(14.21)
6L 4 L2
The governing equation is written in finite-element form as EI K − P K γ + ρALK ˙˙ 0γ = f, L3 2 L 1
Q0 f = . M0
(14.22)
In a static problem, ˙˙ γ = 0, the solution has the form 2 PL cof K2 − K1 EI f, γ= 2 PL det K2 − K1 EI
in which cof denotes the cofactor, and γ → ∞ for values of 2 det(K2 − PL EI K1 ) = 0.
(14.23)
PL2 EI
which render
14.2.1.2 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) = f0 exp(iωt), in which f0 is a known vector. The displacement function satisfies γ(t) = γ0 exp(iωt), in which the amplitude vector γ0 satisfies EI K − P K − ω 2 ρALK γ = f . 0 0 0 L3 2 L 1
(14.24)
Resonance occurs at a frequency ω0, for which EI P det 3 K 2 − K1 − ω 02 ρALK 0 = 0. L L
(14.25)
1 Clearly, ω 02 is an eigenvalue of the matrix ρAL K 0−1 / 2 [ EL3I K 2 − PL K1 ]K 0−1 / 2 . The 2 resonance frequency ω 0 is reduced by the presence of P and vanishes precisely at the critical value of P.
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V0
P M0 L
L
FIGURE 14.5 Buckling of a clamped-clamped beam.
14.2.1.3 Sample Problem: Interpretation of Buckling Modes Consider static buckling of a clamped-clamped beam, as shown in Figure 14.5. This configuration can be replaced with two beams of length L, for which the right beam experiences shear force V1 and bending moment M1, while the left beam experiences shear force V0 − V1 and bending moment M0 − M1. The beam on the right is governed by
∫
65 K1 = 1 10 L
P δ w ′Pw ′dx = δ γ K γ, L 1 T
∫ δ w′′EIw′′dx = δ γ
T
EI K γ, L3 2
12 K2 = 6 L
L 2 L2 15 1 10
(14.26)
6L 4 L2
The governing equation is written in finite-element form as 12 EI 3 L 6 L
6 L P 65 − 1 L 4 L2 L 10
V1 f = , M1
L γ = f 2 2 15 L 1 10
(14.27)
w( L ) γ = − w ′( L)
Consider the symmetric case in which M0 = 0, with the implication that w′(L) = 0. The equation reduces to 12 EI − 6 P w( L) = V , 1 L3 5 L
(14.28)
from which we obtain the critical buckling load given by P1 = 10EI/L . 2
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189
Next, consider the antisymmetric case in which V0 = 0 and w(L) = 0. The counterpart of Equation 14.28 is now 4 EI − 2 PL ( − w ′( L)) = M , 1 L 15
(14.29)
and P2 = 30EI/L . If neither the constraint of symmetry nor axisymmetry is applicable, there are two critical buckling loads, to be obtained in Exercise 2 as 27.8 EI2 and 8.9 EI2 . These values L L are close enough to the symmetric and antisymmetric cases to suggest an interpretation of the two buckling loads as corresponding to the two “pure” buckling modes. Compare the obtained values with the exact solution, assuming static conditions. Consider the symmetric case. Let w(x) = wc(x) + 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 demonstrated in Equation 14.17, wc(x) has a general solution of the form wc(x) = α + β x + γ cos κ x + δ sin κ x, in which κ = PL2 / EI . Now, w = −w′ = 0 at x = 0, −w′(L) = 0, and EIw′′′(L) = V1, expressed as the conditions 2
1α + 0 β + 1γ + 0δ = − w p (0) 0α + 1β + 0γ + κδ = − w ′p (0) (14.30)
0α + Lβ − γκ sin κL + δκ cosκL = − w ′p ( L) 0α + 0 β + γκ 3 sin κL − δκ 3 cosκL = − EIw ′′′ p ( L ) + V1 or otherwise stated − w p (0) − w ′p (0) Bz = , − w ′p ( L) − EIw ′′′( L) + V p 1
1 0 B= 0 0
0
1
1
0
L
−κ sin κL
0
κ 3 sin κL
κ , κ cosκL −κ 3 cosκL 0
α β z= γ δ (14.31)
For the solution to “blow up,” it is necessary for the matrix B to be singular, which it is if the corresponding homogeneous problem has a solution. Accordingly, we seek conditions under which there exists a nonvanishing vector z, for which Bz = 0. Direct elimination of α and β furnishes α = −γ and β = −κδ. The remaining coefficients must satisfy − sin κL sin κL
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cosκL − κL γ 0 = . cosκL δ 0
(14.32)
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A nonvanishing solution is possible only if the determinant vanishes, which reduces to sin κL = 0. This equation has many solutions for kL, including kL = 0. 2 2 2 The lowest nontrivial solution is kL = p, from which Pcrit = π EI/L = 9.87 EI/L . 2 Clearly, the symmetric solution in the previous two-element model (Pcrit = 10 EI/L ) gives an accurate result. For the antisymmetric case, the corresponding result is that tan κ L = κ L. The lowest meaningful root of this equation is kL = 4.49 (see Brush and Almroth, 1975), 2 giving Pcrit = 20.19 EI/L . Clearly, the axisymmetric part of the two-element model is not as accurate, unlike the symmetric part. This issue is addressed further in the subsequent exercises. Up 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. Consider a known initial distribution w0(x). The governing equation is d2 d2 d2 EI 2 ( w − w0 ) + P 2 ( w − w0 ) = 0, 2 dx dx dx
(14.33)
d2 d2 d2 d2 d2 d2 w . I w P w I w P E + = E + 0 dx 2 dx 2 0 dx 2 dx 2 dx 2 dx 2
(14.34)
or equivalently,
The crookedness is modeled as a perturbation. Similarly, if the cross-sectional properties of the beam column exhibit a small amount of variation, for example, EI(x) = EI0[1 + ϑ sin(π x/L)], the imperfection can also be modeled as a perturbation.
14.2.2 EULER BUCKLING
OF
PLATES
The governing equation for a plate element subject to in-plane loads is Eh 2 ∂2 w ∂2 w ∂2 w ∇ 4 w + Px 2 + Py 2 + Pxy =0 2 12(1 − ν ) ∂x ∂y ∂x∂y
(14.35)
(see Wang 1953), in which the loads are illustrated as shown in Figure 14.6. The usual z
y Pxy Pyx
x
h
Px FIGURE 14.6 Plate element with in-plane compressive loads.
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Py
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191
variational methods furnish, with some effort,
∫ δ w∇ w dA = ∫ δw(n ⋅ ∇)∇ w dS − ∫ δ ∇w ⋅ (nP ⋅ ∇)∇w dS + ∫ tr(δ WW)dA, 4
2
(14.36) in which W = ∇∇ w (a matrix!). In addition, T
∫
∂2 w ∂2 w ∂2 w δw Px 2 + Py 2 + Pxy dA = dw(nx ∂y ∂x∂y ∂x
∫
ny )p dS
− {δ wx
wx δ wy}P dA, wy
∫
[ [
P ∂w + 1 P x ∂x 2 xy p= 1 P ∂w + P y 2 xy ∂x
∂w ∂y ∂w ∂y
] , ]
Px P= 12 Pxy
1 2
(14.37)
Pxy Py
For simplicity’s sake, assume that w( x, y) = ϕ Tb 2 Φb 2 γ b 2 from which we can obtain the form wx ∇w = = β1Tb 2 Φb 2 γ b 2 , wy
VEC(W) = βT2 b 2 Φb 2 γ b 2 .
(14.38)
We also assume that the secondary variables (n ⋅ ∇)∇ w, (n ⋅ ∇)∇w, and also [ P ∂ w + 1 P ∂w ] x ∂x 2 xy ∂y are prescribed on S. ∂w ∂ w 1 [ Pxy + P ] y ∂y 2 ∂x These conditions serve to obtain 2
[K b 21 − K b 22 ]γ b 2 = f K b 21 =
(14.39)
Eh 2 Φ T β βT dAΦb 2 12(1 − ν 2 ) b 2 2 b 2 2 b 2
∫
∫
K b 22 = ΦbT2 β1b 2 Pβ1Tb 2 dVΦb 2 and f reflects the quantities prescribed on S. As illustrated in Figure 14.7, we now consider a three-dimensional loading space in which Px, Py , and Pxy correspond to the axes, and seek to determine a surface in the space of critical values at which buckling occurs. In this space, a straight line
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Pxy
λ φ θ
Py
Px FIGURE 14.7 Loading space for plate buckling.
emanating from the origin represents a proportional loading path. Let the load intensity, λ, denote the distance to a given point on this line. By analogy with spherical coordinates, there exist two angles, θ and φ, such that Px = λ cos θ cos φ ,
Py = λ sin θ cos φ ,
Pxy = λ sin φ
(14.40)
Now, ˆ (θ , ϕ ) K b 22 = λK b 22
∫
ˆ (θ , ϕ ) = Φ T β Pˆ (θ , ϕ )βT dVΦ K b 22 b2 b2 1b 2 1b 2 cos θ cosϕ Pˆ (θ , φ ) = sin ϕ
(14.41)
sin ϕ
sin θ cosϕ
For each pair (q, f), buckling occurs at a critical load intensity, λcrit(θ, φ), satisfying ˆ ] = 0. det[K b 21 − λ crit (θ , φ )K b 22
(14.42)
A surface of critical load intensities, λcrit(θ, φ), can be drawn in the loading space shown in Figure 14.7 by evaluating λcrit(θ, φ) over all values of (q, f) and discarding values that are negative.
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14.3 EXERCISES 1. Consider the triangular member shown to be modeled as one finite element. Assume that
ψ ( x, y) = (1
1 y)1 1
x
x1 x2 x3
y1 y2 y3
−1
ψ1 ψ 2 . ψ 3
Find KT , f T , and the torsional constant T. y
(2,3) (3,2) (1,1)
x
Triangular shaft cross section.
2. Find the torsional constant for a unit square cross section using two triangular elements. 3. Derive the matrices K0, K1, and K2 in Equations 14.20 and 14.21. 4. Compute the two critical values in Equation 14.23. 5. Use the four-element model shown in Figure 14.5, and determine how much improvement, if any, occurs in the symmetric and antisymmetric cases. 6. Consider a two-element model and a four-element model of the simplesimple case shown in the following figure. Compare Pcrit in the symmetric and antisymmetric cases with exact values. V0
P M0 L
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L
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7. Consider a cantilevered beam with a compressive load P, as shown in the following figure. The equation is
EI
d 4w d 2w + P 2 = 0. 4 dx dx
The primary variables at x = L are w(L) and −w′(L), and
∫
L
0
δ w ′′EIw ′′dx = δ γ T
EI 12 L3 6 L
6L γ , 4 L2
∫
L
0
δ w ′Pw ′dx = δ γ T
P 12 L 6 L
6L γ . 4 L2
Find the critical buckling load(s). V0 M0
L
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P
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15
Introduction to Contact Problems
15.1 INTRODUCTION: THE 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 so clear. This is especially true if loads are transmitted to the member along an interface with a second member. This class of problems is called contact problems, and they are arguably the most common boundary conditions encountered in practical problems. The finite-element community has devoted, and continues to devote, a great deal of effort to this complex problem, leading to gap and interface elements for contact. Here, we introduce gap elements. First, consider the three-spring configuration in Figure 15.1. 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
δ
2 k kc = 3k
(15.1)
δ ≥ g.
From the viewpoint of the finite-element method, Figure 15.1 poses the following difficulty. If a node is set at the lowest point on spring B and at the point directly P
contactor
δ
A k
B
C
k
k g
target
FIGURE 15.1 Simple contact problem.
195
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P
contactor
δ
A k
B
C
k
k kg
target
FIGURE 15.2 Spring representing contact element.
below it on the target, these nodes are not initially connected, but are later connected in the physical problem. Furthermore, it is necessary to satisfy the nonpenetration constraint whereby the middle spring does not move through the target. 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 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 15.2. The stiffness of the middle spring (B in series with the contact spring) is now denoted as km, and −1
1 1 km = + . k kg
(15.2)
It is desirable for the middle spring to be soft when the gap is open (g > δ ) and to be stiff when the gap is closed (g ≤ δ ): k /100 g > δ kg = 100 k g ≤ δ .
(15.3)
Elementary algebra serves to demonstrate that 2 k + 0.01k kc ≈ 2 k + 0.99k
g>δ g ≤ δ.
(15.4)
Consequently, the model with the contact is too stiff by 0.5% when the gap is open, and too soft by 0.33% when the gap is closed (contact). One conclusion that can be drawn from this example is that the stiffness of the gap element should be related to the stiffnesses of the contactor and target in the vicinity of the contact point.
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197
15.2 POINT-TO-POINT CONTACT Generally, it is not known what points will come into contact, and there is no guarantee that target nodes will come into contact with foundation nodes. The gap elements can be used to account for the unknown contact area, as follows. Figure 15.3 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. th The gap (i.e., the distance in the undeformed configuration) from the i node of th the contactor to the j node of the target is denoted by gij. (For the purpose of this th discussion, the gap is constant, i.e., not updated.) In point-to-point contact, the i 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 it does not occur at a node.) It follows that each node of the target is connected by a spring to each of the nodes on the contactor. The angle between the spring and the normal at the contactor node is αij, while the angle between the spring and the th normal to the target is αji. Under load, the i contactor node experiences displacement th th uij in the direction of the j target node, and the j target node experiences displaceth th ment uji. For example, the spring connecting the i contactor node with the j target node has stiffness kij, given by kijlower kij = kijupper
δ ij < gij
(15.5)
δ ij ≥ gij ,
in which δij = uij + uji is the relative displacement. The force in the spring connecting th th the i contactor node and j target node is fij = kij(gij)δij. The total normal force th experienced by the i contactor node is fi = ∑j fij cos(αij).
candidate contactor contact surface dSc c1
c2
c3 α31 k(g31)
t1
t2 dSt
candidate target contact surface FIGURE 15.3 Point-to-point contact.
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As an example of how the spring stiffness might depend upon the gap, consider the function kij ( gij − δ ij ) = k0 ε + (1 − 2ε ) 2 tan −1 α ( gij − δ ij − γ ) 2 − ( gij − δ ij − γ ) , 2 π
(15.6)
where γ, α, and ε are positive parameters selected as follows. When gij − δij − γ > 0, kij attains the lower-shelf value, k0ε, and we assume that ε << 1. If gij − δij − γ < 0, kij approaches the upper-shelf value, k0 (1 − ε). We choose γ to be a small value to attain a narrow transition range from the lower- to the upper-shelf values. In the range 0 < gij − δij < γ, there is a rapid but continuous transition from the lower-shelf (soft) value to the upper-shelf (stiff) value. If we now choose α such that αγ = 1, kij becomes k0 /2, when the gap closes (gij = δij). The spring characteristic is illustrated in Figure 15.4. The total normal force on a contactor node is the sum of the individual contactelement forces, namely Nc
ftj =
∑ k (g ij
ij
− δ ij )δ ij cos(α ij ).
(15.7)
i
Clearly, significant forces are exerted only by the contact elements that are “closed.”
k ij εk 0
k0
k 0 /2
εk 0
FIGURE 15.4 Illustration of a gap-stiffness function.
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δij –gij
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199
candidate target contact surface dSt t1
t2
t3
element connecting node t3 with nodes c1 and c2
c2 dSc
c1
c3
candidate contactor contact surface FIGURE 15.5 Element for point-to-surface contact.
15.3 POINT-TO-SURFACE CONTACT We now briefly consider point-to-surface contact, illustrated in Figure 15.5 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 is written as k([g1 − δ1], ˆ , in which g1 − δ1 is the gap between nodes t1 and c1, and K ˆ is the [g2 − δ2]) K geometric part of the stiffness matrix of a triangular elastic element. The stiffness matrix of the element can 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 on the candidate contactor nodes. In some finite-element codes, the foregoing scheme is 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 proportional to the normal traction through a friction coefficient. This model does not appear to consider sliding and can be considered a bonded contact. Advanced models address sliding contact and incorporate friction laws not based on the Coulomb model.
15.4 EXERCISES 1. Consider a finite-element model for a set of springs, illustrated in the following figure. A load moves the plate on the left toward the fixed plate on the right. What is the load-deflection curve of the configuration? For a finite-element model, an additional bilinear spring is supplied, as shown. What is the load-deflection curve of the finite-element model? Identify a kg value for which the load-deflection behavior of the finiteelement model is close to the actual configuration. Why is the new spring needed in the finite-element model?
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F
k
k k
k k H L
2. Suppose a contact element is added in the previous problem, in which the stiffness (spring rate) satisfies k ( gij − δ ij ) = k0 ε + (1 − 2ε ) 2 tan −1 α ( gij − δ ij − γ )2 − ( gij − δ ij − γ ) . 2 π Suppose αγ = 1, kL = k/100, and ku = 100k. Compute the stiffness k for the configuration as a function of the deflection δ.
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16
Introduction to Nonlinear FEA
16.1 OVERVIEW The previous section addressed 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 are nonlinear. 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, can 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 can be strongly dependent on normal pressures. Fortunately, much of the linear finite-element method can be adapted to nonlinear problems, as explained in this chapter. The next chapter focuses on isothermal problems. The extension to thermomechanical problems will be presented in a subsequent chapter.
16.2 TYPES OF NONLINEARITY There are three major types of nonlinearity in thermomechanical boundary-value problems: material nonlinearity, geometric nonlinearity, and boundary-condition nonlinearity. Nonlinearities can also be present if the formulation is referred to deformed coordinates, possibly introducing stress fluxes and converted coordinates. Material nonlinearity can occur through nonlinear dependence of the stress on the strain or temperature, including temperature dependence of the tangent modulus tensor. Metals undergoing plastic flow exhibit nonlinear material behavior. Geometric nonlinearity occurs because of large deformation, especially in problems referred to undeformed coordinates. Rubber components typically exhibit large deformation and require nonlinear kinematic descriptions. In this situation, a strain measure needs to be chosen, as does the stress conjugate to it. Boundary-condition nonlinearity occurs because of nonlinear supports on the boundary and contact. An example of a nonlinear support is a rubber pad placed under a machine to absorb vibrations. For finite-element methods for nonlinear problems, the loads or time steps are applied in increments. Then, incremental variational principles, together with interpolation models, furnish algebraic (static) or ordinary differential equations (dynamic) 201
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in terms of vector-valued incremental displacements or incremental temperatures. For mechanical systems, a typical equation is K( γ )∆γ + Μ( γ )∆˙˙ γ = ∆f,
(16.1)
in which ∆γγ is the incremental displacement vector, ∆f is the incremental force vector, K(γγ) is the tangent stiffness matrix, and M(γγ) is the tangent mass matrix. It will be seen that this type of equation is a realization of the optimal Newton iteration method for nonlinear equations.
16.3 COMBINED INCREMENTAL AND ITERATIVE METHODS: A SIMPLE EXAMPLE Consider a one-dimensional rod of nonlinear material under small deformation in which the elastic modulus is a function of strain: E = E0(1 + αε), ε = E11. Under static loading, the equilibrium equation is AE 0 γ 1 + α γ = P . L L
(16.2)
Suppose the load is applied in increments, ∆ jP, and that the load at the n load th step is Pn. Suppose further that the solution γ n is known at the n load. We now consider the steps necessary to determine the solution γ n+1 for Pn+1. We introduce the th increment ∆nγ = g n+1 − γ n. Subtracting Equation 16.2 at the n step from the equation st th at the (n + 1) step, the incremental equilibrium equation for the n increment is now th
AE 0 γ + ∆ nγ 1 + 2α n+1 ∆ γ = ∆nP . n L L
(16.3)
This equation is quadratic in the increments. In fact, geometric nonlinearity generally leads to a quadratic function of increments. The error of neglecting the nonlinearity may be small if the load increment is sufficiently small. However, we will retain the nonlinear term for the sake of illustrating the use of iterative procedures. In particular, the previous equation can be written with obvious identifications in the form g( x ) = βx + ζ x 2 − η = 0.
(16.4)
Newton iteration (discussed in Chapter 3) furnishes the optimal iteration scheme th for the ν iterate: [βx ν + ζ x ν − η] , β + 2ζ x ν 2
x ν +1 = x ν −
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x 0 = η / β.
(16.5)
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203
SAMPLE PROBLEM 1 The efficiency of this scheme is addressed as follows. Consider ζ = 1, β = 12, and η = 1. The correct solution is x = 0.0828. Starting with the initial value x = 0, the 1 first two iterates are, approximately, 121 = 0.833, and 121 (1 − 144 ) = 0.0828 . This illustrates the rapid quadratic convergence of Newton iteration.
16.4 FINITE STRETCHING OF A RUBBER ROD UNDER GRAVITY: A SIMPLE EXAMPLE Figure 16.1 shows a rubber rod under gravity. It is assumed to attain finite strain. The rod is also assumed to experience uniaxial tension. The figure refers to the undeformed configuration, with the element occupying the interval (Xe, Xe+1). Its length is le, its cross-sectional area is Ae, and its mass density is ρ. It is composed of rubber and is stretched axially by the loads Pe and Pe+1. 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 ) = x(X ) − X.
16.4.1 NONLINEAR STRAIN-DISPLACEMENT RELATIONS The element is assumed short enough that a satisfactory approximation for the displacement u(X ) is provided by the linear interpolation model u( X ) = ue + ( X − Xe )(ue+1 − ue ) le = NTae ,
NT =
1 {x − x le e
x − x e},
(16.6)
in which a eT = {ue ue+1}. Now, u(Xe) = ue and u(Xe+1) = ue+1 are viewed as the unknowns to be determined using the finite-element method. The Lagrangian strain
Pe xe,ue g
xe+1,ue+1 Pe+1
FIGURE 16.1 Rubber rod under load.
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E = Exx is given by E=
∂u 1 ∂u + ∂x 2 ∂x
2
=
ue+1 − ue 1 ue+1 − ue + 2 le le
=
1 {−1 le
1} a e +
2
1 1 T 1 ae 2 2 le −1
−1 ae . 1
(16.7)
Note that dE = BT da e , BL =
1 −1 le 1
B = B L + B NL a e B NL =
1 le2
1 −1
(16.8)
−1 . 1
The vector BL and the matrix BNL are the strain-displacement matrices. The following sections illustrate two different approaches to formulating a finiteelement model for the rubber rod. One is based on satisfying the incompressibility constraint a priori. The second is based on satisfying the constraint a posteriori, which is typical of finite-element code practice. In addition, we also include a slight variant involving a near-incompressibility constraint in an a posteriori manner.
16.4.2 STRESS
AND
TANGENT MODULUS RELATIONS
The Neohookean strain-energy density function, w, accommodating incompressibility, is expressed in terms of the eigenvalues c1, c2, and c3 of C = 2E + I, as follows: w=
D (c + c + c3 − 3), 2 1 2
subject to c1c2 c3 − 1 = 0 ,
(16.9)
in which D is the (small-strain) elastic modulus. We can show, with some effort, that under uniaxial tension, c2 = c3. We first enforce the incompressibility constraint a priori by using the substitution c2 = c3 =
1 . c1
(16.10)
After elementary manipulations, w=
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D [2 E + 1 + 2 /(2 E + 1) −1/ 2 − 3] . 2
(16.11)
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205
nd
The (2 Piola-Kirchhoff) stress, S, defined in Chapter 5, is now obtained as S= =
∂w ∂E 1 D[1 − (2 E + 1)−3/ 2 ]. 3
(16.12)
The tangent modulus, D, is also required: ∂S ∂E
DT =
= D(2 E + 1)−5/ 2 .
(16.13)
If the strain E is small compared to unity, DT reduces to D. We next satisfy the incompressibility constraint a posteriori. An augmented strain-energy function w* is introduced by w* =
D p [c + c + c3 − 3] − (c1c2 c3 − 1) , 2 1 2 2
(16.14)
in which the Lagrange multiplier, p, is the (true) hydrostatic pressure. The augmented energy is stationary with respect to p as well as c1, c2, and c3, from which it follows that c1c2c3 − 1 = 0 (incompressibility). The stresses satisfy S1 =
∂w* ∂w* D p c1c2c3 =2 = − 2 2 c1 ∂E1 ∂c1
S2 =
∂w* ∂w* D p c1c2c3 =2 = − =0 2 2 c2 ∂E2 ∂c2
S3 =
∂w * ∂w * D p c1c2c3 =2 = − = 0. 2 2 c3 ∂E3 ∂c3
(16.15)
It is readily shown that c2 = c3. Enforcement of the stationary condition for p (c1c2c3 − 1 = 0), with S2 = 0, now furnishes that p = D / c1 . It follows that S1 = D[1 − c1−3/ 2 ]/ 2 , in agreement with Equation 16.12.
16.4.3 INCREMENTAL EQUILIBRIUM RELATION The Principle of Virtual Work states the condition for static equilibrium as
∫
Xe +1
Xe
BSe AdX − P e − ρg
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∫
Xe +1
Xe
N T AdX a e = 0,
Pee Pe = , Pe e+1
(16.16)
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in which the third term represents the weight of the element, while Pe represents the forces from the adjacent elements. In an incremental formulation, we can replace the loads and displacements by their differential forms. In particular,
d Pe =
∫
Xe +1
BdSAe dX +
Xe
∫
Xe +1
Xe
d BSAe dX
= K e da e = {K1e + K 2 e + K 3e + K 4 e} da e
(16.17)
in which K1e = Ae
=
∫
Xe +1
Xe
Xe
B L DT BTL dX −1 1
1 −1
DT Ae le
=
K 2 e = Ae
∫
Xe +1
(
)
DT B L a eTB NL + B NL a e BTL dX −1 1
DT Ae [ue+1 − ue ] 1 2 le le −1
K 3e = Ae
=
∫
(16.18a)
Xe +1
Xe
B NL a ea eTBTNL dX −1 1
2 DT Ae ue+1 − ue 1 le le −1
K 4 e = Ae
=
(16.18b)
∫
Xe +1
Xe
(16.18c)
B L SdX
SAe 1 le −1
−1 1
(16.18d)
and DT =
© 2003 by CRC CRC Press LLC
1 le
∫
Xe +1
Xe
DT dX ,
S=
1 le
∫
Xe +1
Xe
SdX.
(16.19)
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207
Combining Equations 16.18 and 16.19 produces the simple relation 1 Ke = κ −1
−1 , 1
κ=
DT Ae le
2 ue+1 − ue ue+1 − ue SAe 1 2 + + l + l . le e e
(16.20)
Suppose (ue+1 − ue)/le is small compared to unity, with the consequence that E is also small compared to unity. It follows in this case that DT ≈ D, S ≈ S. As a result, Ke reduces to the stiffness matrix for a rod element of linearly elastic material experiencing small strain: Ke =
EAe le
1 −1
−1 . 1
(16.21)
Several special cases illuminate additional aspects of finite-element modeling. Sample Problem 1: Single Element Built in at One End Figure 16.2 depicts a single-element model of a rod that is built in at X = 0: Xe = X0 = 0. At the opposite end, Xe+1 = X1 = 1. The rod is submitted to the load P. The displacement at X = 0 is subject to the constraint u(0) = u0 = 0, so that Equation 16.20 becomes 1 κ −1
−1 0 − dPr = , 1 du1 dP
(16.22)
in which dPr is an incremental reaction force that is not known a priori (of course, from equilibrium, dPr = dP). For the only unknown reaction, this last equation “condenses” to kdu1 = dP .
x1
A,ρ
X
u1
P
FIGURE 16.2 Rubber rod element under load: built in at top.
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(16.23)
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For later use, the current shape function degenerates to the expression N → N = X/X1. The (Lagrangian) strain is given by 2
E(u1 ) =
u1 1 u1 + , X1 2 X1
(16.24)
and the strain-displacement matrices reduce to B→ B=
u 1 + 12 . X1 X1
(16.25)
nd
The (2 Piola-Kirchhoff) stress S is now a function of u1: 3 − 2 2 1 1 u u S(u1 ) = D1 − 2 1 + 1 − 1 . 3 X1 2 X1
16.4.4 NUMERICAL SOLUTION
BY
(16.26)
NEWTON ITERATION
Equation 16.22 can be solved by Newton iteration, sometimes called “load balancing,” the reason for which is explained shortly. We introduce the “residual” function φ (u1) = 0 as follows. The Principle of Virtual Work implies that if gravity is considered in Figure 16.2,
φ (u1 ) =
∫
X1
0
BSA dX − P − ρg
∫
X1
0
N T A dX a e
3 − 2 2 1 u1 1 u1 1 u1 X = + 2 AX1 D1 − 2 + − 1 − P − ρgA 1 u1 2 le le 3 X1 2 X1
= 0.
(16.27) th
when u1 is the solution. Consider an iteration process in which the j iterate u1j has been determined. Newton iteration determines the next iterate, u1j +1, using
κ (u1j )∆ j u1 = −φ (u1j ) u1j +1 = u1j + ∆ j u1.
(16.28)
Convergence of this scheme to the correct value u1 is usually rapid, provided that the initial iterate u10 is sufficiently close to u1. A satisfactory initial iterate is
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209
often obtained using an incremental procedure. Suppose that the solution has been obtained at previous load steps. A starting iterate for the current load step is obtained by extrapolating from the previous solutions. As an example, suppose the solution is known at the load P = k∆P, in which ∆P is a load increment. The load is now incremented to produce P = (k + 1)∆P, and the solution u1,k+1 at this load is determined using Newton iteration. It is frequently satisfactory to start the iteration using u10,k +1 = u1. This numerical procedure is illustrated more extensively at the end of this chapter. Sample Problem 2: Assembled Stiffness Matrix for a Two-Element Model Assemblage procedures are used to combine the element-equilibrium relations to obtain the global equilibrium relation for an assemblage of elements. For elements ‘e’ and ‘e + 1’, the element-equilibrium relations are expanded as k11e due + k12e due+1 = dPee
(i)
e e k21 due + k22 due+1 = dPee+1
(ii) (16.29)
k11e+1due+1 + k12e+1due+2 = dPee++11
(iii)
e +1 e +1 k21 due+1 + k22 due+2 = dPee++21
(iv).
The superscript indicates the element index, and the subscript indicates the node index. If no external force is applied at xe+1, the interelement force balance is expressed as dPee+1 + dPee++11 = 0 .
(16.30)
Adding Equation 16.29(ii) and Equation 16.29(iii) furnishes
(
)
e e k21 due + k22 + k11e+1 due+1 + k12e+1due+2 = 0 .
(16.31)
Equations 16.29, 16.30, and 16.31 are expressed in matrix form as k11e e k21 0
k12e e k22 + k11e+1 e +1 k21
0 due dPee k12e+1 due+1 = 0 . e+1 e +1 k22 due+2 dPe+2
(16.32)
Equation 16.32 illustrates that the (incremental) global stiffness matrix is formed by “overlaying” Ke and Ke+1, with the entries added in the intersection.
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Pee
xe,ue
P,Ae
e Pe+1 e+1 Pe+1
xe+1,ue+1
P,Ae+1
xe+2,ue+2
ρe+1 e+2
x FIGURE 16.3 Two rubber rods under load.
Sample Problem 3: Two Identical Elements under Gravity under Equal End-Loads If Ke = Ke+1 and dPee = − dPee++11 = − dP , overlaying the element matrices leads to the global (two-element) relation 1 κ −1 0
−1 2 −1
0 du1 − dP −1 du2 = 0 . 1 du3 dP
(16.33)
Note that Equation 16.33 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, due to numerical errors in the load increments, the condition for static equilibrium is not satisfied numerically, and therefore that the body is predicted to accelerate indefinitely (undergo rigid-body motion). However, we also know that the configuration under incremental loads is symmetric, implying a constraint du2 = 0. This constraint permits “condensation,” that is, reducing Equation 16.33 to a system with two unknowns by eliminating rows and columns associated with the middle incremental displacement: 1 κ −1
−1 du1 − dP = . 1 du3 dP
(16.34)
The condensed matrix is now proportional to the identity matrix, and the system has a solution. In general, stiffness matrices can easily be singular or nearly singular
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211
(with a large condition number) unless constraints are used to suppress “rigid-body modes.”
16.5 ILLUSTRATION OF NEWTON ITERATION Unfortunately, since elastomers or metals experiencing plasticity are typically quite compliant, it is common to encounter convergence problems, for which there are four major approaches: 1. 2. 3. 4.
Increasing stiffness, such as by introducing additional constraints if available Reducing load-step sizes and reforming the stiffness matrix after each iteration Switching to displacement control rather than load control Using an arc-length method (described in Chapter 3)
For an equation of the form
ψ ( x ) = 0,
(16.35)
Newton iteration seeks a solution through an iterative process given by −1
dψ ( x j ) x j +1 = x j − ψ (x j ) . dx
(16.36)
Clearly, using two sequential iterates, −1
−1
dψ ( x j ) dψ ( x j −1 ) x j +1 − x j = x j − x j −1 − ψ (xj ) − ψ ( x j −1 ) dx dx −1 dψ ( x *) d 2ψ ( x *) ( *) = x j − x j −1 − + x − x j dx 2 dx
× ψ ( x *) +
dψ ( x *) d 2ψ ( x *) dψ ( x *) ( x j −1 − x*) ( x j − x*) − + dx 2 dx dx
× ψ ( x *) +
dψ ( x *) ( x j −1 − x*) + L dx
−1 dψ ( x *) dψ ( x *) ( x − x j −1 ) + O([ x j − x j−1 ]2 ) = 1 − dx j dx
for some x* between x j and x j−1.
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−1
(16.37)
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encounter difficulties with convergence Y 1
X FIGURE 16.4 Illustration of the inverse tangent function.
16.5.1 EXAMPLE The performance of Newton iteration procedure is illustrated using a simple example showing convergence problems. Consider
ϕ ( x) =
2 tan −1 ( x ) − y = 0. π
(16.38)
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 = 1. The curve appears as shown in Figure 16.4. 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 is far from the initial iterate. th Suppose that y is incremented such that the n value of y is yn = n∆y.
(16.39)
To obtain the solution of the (n + 1) step, the Newton iteration procedure st th generates the (ν + 1) iterate from the ν iterate, as follows: st
x nν++11
=
x nν+1
( )
−1
dφ x nν+1 − φ x nν+1 , dx
( )
(16.40)
ν in which x n+1 is the ν iterate for the solution xn+1. 0 Of course, a starting iterate, x n+1 , is needed. An attractive candidate is xn+1. However, this may not be good enough when convergence difficulties appear. Newton iteration was implemented for this example in a simple problem using increments of 0.001. A stopping criterion of 10 iterations was used. Convergence was found to fail near y = 0.999. th
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16.6 EXERCISES 1. Doing the algebra, make judicious choices of starting iterates, and develop the first few iterates for the roots of the equations x 2 + 3x + 2 = 0 x2 − 1 = 0 ( x − 1)2 = 0 2. Write a simple program implementing Newton iteration, and apply it to the examples in Exercise 1. 3. Using Example 5.1, write a simple program using Newton iteration to compute X(Y ) using increments ∆y = 0.001. Set the stopping criterion as the first of: (a) 10 iterates or (b) an absolute difference of less than 0.001 in the successive values of x. Note the value of x at which convergence fails, and compare the values of x with accurate values available from the formulae and tables in Abramawitz and Stegun (1997).
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17
Incremental Principle of Virtual Work
17.1 INCREMENTAL KINEMATICS Recall that the displacement vector u(X) is assumed to admit a satisfactory approxT Φ γ (t). Also recall that the imation at the element level in the form u(X) = ϕ (X)Φ deformation-gradient tensor is given by F = ∂∂xx . Suppose that the body under study is subjected to a load vector, P, which is applied incrementally via load increments, th ∆jP. The load at the n load step is denoted as Pn. The solution, Pn, is known, and the solution of the increments of the displacements is sought. Let ∆nu = un+1 − un, T Φ∆nγ. By suitably arranging the derivatives of ∆nu with respect so that ∆nu = ϕ (X)Φ to X, a matrix, M(X), can easily be determined for which VEC(∆nF) = M(X)∆nγ. T We next consider the Lagrangian strain, E(X) = 12 (F F − I). Using Kronecker Product algebra from Chapter 2, we readily find that, to first order in increments, ∆ n e = VEC( ∆ n E) 1 = VEC [FT ∆ n F + ∆ n FT F] 2 1 = [I ⊗ FT + FT ⊗ IU]VEC( ∆ n F) 2 = G∆ n g,
1 GT = [I ⊗ FT + FT ⊗ IU]M (X)∆ n g. 2
(17.1)
This form shows the advantages of Kronecker Product notation. Namely, it enables moving the incremental displacement vector to the end of the expression outside of domain integrals, which we will encounter subsequently. Alternatively, for the current configuration, a suitable strain measure is the −T −1 Eulerian strain, « = 12 (I − F F ), which refers to deformed coordinates. Note that −1 −1 −1 −1 −T −T T −T since ∆n(FF ) = 0, ∆nF = −F ∆nFF . Similarly, ∆nF = −F ∆nF F . Simple manipulation furnishes that 1 VEC( ∆ n« ) = [F − T F −1 ⊗ F − T U + F −1 ⊗ F − T F −1 ]M∆ n g. 2
(17.2)
There also are geometric changes for which an incremental representation is −1 useful. For example, since the Jacobian J = det(F) satisfies dJ = Jtr(F dF), we obtain
215
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Finite Element Analysis: Thermomechanics of Solids
the approximate formula ∆J = Jtr(F −1∆ n F) = JVEC T (F −1 )VEC( ∆ n F) T
= VEC T (F − T ) JM∆ n γ .
(17.3)
Also of interest are d[n dS] = [tr(D)I − LT ]n dS dt dn = [(nT Dn)I − LT ]n dt
(17.4)
d dS = [tr(D) − nT Dn]dS dt Using Equation 17.4, we obtain the incremental forms ∆ n [ndS] = dS[nVEC T (F − T ) − nT ⊗ F − T U]M∆ n γ ∆ n n = [n(nT F − T ) ⊗ nT − nT ⊗ F − T U]M∆ n γ
(17.5)
∆ n dS = dS[nVEC T (F − T ) − n(nT F − T ) ⊗ nT ]M∆ n γ .
17.2 INCREMENTAL STRESSES For the purposes of deriving an incremental variational principle, we shall see that st the incremental 1 Piola-Kirchhoff stress, ∆ n S , is the starting point. However, to o formulate mechanical properties, the objective increment of the Cauchy stress, ∆ n T, is the starting point. Furthermore, in the resulting variational statement, which we called the Incremental Principle of Virtual Work, we find that the quantity that nd appears is the increment of the 2 Piola-Kirchhoff stress, ∆nS. From Chapter 5, we learned that S = SF T , from which, to first order, ∆ n S = ∆ n SF T + S∆ n F T .
(17.6)
For the Cauchy stress, the increment must take into account the rotation of the underlying coordinate system and thereby be objective. We recall the objective o Truesdell stress flux, ∂ T / ∂t, introduced in Chapter 5: o
∂ T/∂t = ∂T/∂t + Ttr(D) − LT − TLT .
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(17.7)
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217
Among the possible stress fluxes, it is unique in that it is proportional to the nd rate of the 2 Piola-Kirchhoff stress, namely o
∂S/∂t = J F −1∂ T/∂tF − T .
(17.8)
An objective Truesdell stress increment is readily obtained as o VEC( ∆ n T ) = 1 F T ⊗ FVEC( ∆ n S).
J
(17.9)
o
Furthermore, once VEC( ∆ n T ) has been determined, the (nonobjective) increment of the Cauchy stress can be computed using o
∆ n T = ∆ n T + Ttr( ∆ n FF −1 ) − ∆ n FF −1T − T F − T ∆ n F T ,
(17.10)
from which o
VEC( ∆ nT ) = VEC( ∆ n T ) + [TVEC T (F − T ) − (T F − T ) ⊗ I − I ⊗ (T F − T )]M∆ n γ . (17.11)
17.3 INCREMENTAL EQUILIBRIUM EQUATION We now express the incremental equation of nonlinear solid mechanics (assuming that there is no net rigid-body motion). In the deformed (Eulerian) configuration, equilibrium at tn requires
∫ T n dS = ∫ ρu˙˙ dV. T
(17.12)
Referred to the undeformed (Lagrangian) configuration, this equation becomes
∫ S n dS = ∫ ρ u˙˙ dV , T
0
0
0
(17.13)
0
st
in which, as indicated before, S is the 1 Piola-Kirchhoff stress, S denotes the surface (boundary) in the deformed configuration, and n0 is the surface normal vector in the undeformed configuration. Suppose the solution for S is known as Sn at time tn and is sought at tn+1. We introduce the increment ∆ n S to denote Sn+1 − Sn . A similar definition is introduced for the increment of the displacements. Now, equilibrium applied to Sn+1 and Sn implies
∫∆ S n T
n
© 2003 by CRC CRC Press LLC
0
dS0 =
∫ ρ ∆ u˙˙ dV . 0
n
0
(17.14)
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Finite Element Analysis: Thermomechanics of Solids
Application of the divergence theorem furnishes the differential equation ˙˙ T . ∇ T ∆ n S T = ρ0 ∆ n u
(17.15)
17.4 INCREMENTAL PRINCIPLE OF VIRTUAL WORK To derive a variational principle for the current formulation, the quantity to be varied is the incremental displacement vector since it is now the unknown. Following Chapter 5, Equation 17.15 is multiplied by (δ∆nu) . Integration is performed over the domain. The Gauss divergence theorem is invoked once. Terms appearing on the boundary are identified as primary and secondary variables. Boundary conditions and constraints are applied. T
The reasoning process is similar to that in the derivation of the Principle of Virtual Work in finite deformation in which u is the unknown, and furnishes
∫ tr(δ∆ E ∆ S)dV + ∫ δ∆ FS∆ F dV + ∫ δ∆ u ρ ∆ u˙˙ dV = ∫ δ∆ u ∆ τ dS , T
n
T
n
0
n
n
T
0
T
n
n
0
0
n
n 0
0
(17.16)
in which τ 0 is the traction experienced by dS0. The fourth term describes the virtual external work of the incremental tractions. The first term describes the virtual internal work of the incremental stresses. The third term describes the virtual internal work of the incremental inertial forces. The second term has no counterpart in the previously formulated Principle of Virtual Work in Chapter 5, and arises because of geometric nonlinearity. We simply call it the geometric stiffness integral. Due to the importance of this relation, Equation 17.16 is derived in detail in the equations that follow. It is convenient to perform the derivation using tensor-indicial notation: ∂
∂
∂
∫ δ∆ u ∂X (∆ S )dV = ∫ ∂X [δ∆ u (∆ S )]dV − ∫ ∂X [δ∆ u ]∆ S dV n i
n ij
0
n i
j
n ij
0
n i
j
n ij
0
j
∫
(17.17)
= δ∆ nui ρ0 ∆ nu˙˙i dV0 .
The first term on the right is converted using the divergence theorem to ∂
∫ ∂X [δ∆ u (∆ S )]dV = ∫ δ∆ u (n ∆ S )dS n i
n ij
0
n i
j
n ij
0
j
∫
= δ∆ n uiτ 0 j dS0 which is recognized as the fourth term in Equation 17.16.
© 2003 by CRC CRC Press LLC
(17.18)
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219
To first-order in increments, the second term on the right is written, using tensor notation, as ∂
∫ ∂X [δ∆ u ]∆ S dV = ∫ tr(δ∆ F∆ S )dV n i
n ij
n
0
n
0
j
∫
= tr(δ∆ n F[ ∆ n SFT + S∆ n FT ])dV0
∫
∫
= tr(FTδ∆ n F∆ n S)dV0 + tr(δ∆ n FS∆ n FT )dV0
(17.19)
The second term is recognized as the second term in Equation 17.16. The first term now becomes 1
∫ tr(F δ∆ F∆ S)dV = ∫ tr 2 [F δ∆ F + δ∆ F F]∆ S dV T
n
n
0
T
T
n
n
n
0
∫
= tr(δ∆ nΕ T ∆ nS)dV0
(17.20)
which is recognized as the first term in Equation 17.16.
17.5 INCREMENTAL FINITE-ELEMENT EQUATION For present purposes, let us suppose constitutive relations in the form ∆ n S = D( X, γ n )∆ n E,
(17.21)
in which D(X, γ ) is the fourth-order tangent modulus tensor. It is rewritten as ∆ n s = χ( X , γ n ) ∆ n e s = VEC( S),
e = VEC( E),
(17.22)
χ = TEN22( D).
Also for present purposes, we assume that ∆ττ 0 is prescribed on the boundary S0, a common but frequently unrealistic assumption that is addressed in a subsequent section. In VEC notation, and using the interpolation models, Equation 17.16 becomes
δ∆ n γ T [(K T + K G )∆ n γ + M∆ n ˙˙γ − ∆ n f ] = 0
(17.23)
KT =
∫ M GχG M dV
KG =
∫ M S ⊗ IM dV
M=
∫ ρ Φ ϕϕ Φ dV
∆n f =
∫ ρ Φ ϕ∆τ dS
T
T
0
T
0
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T
0
T
0
T
0
0
0
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Finite Element Analysis: Thermomechanics of Solids
KT is now called the tangent modulus matrix, KG is the geometric stiffness matrix, M is the (incremental) mass matrix, and ∆nf is the incremental force vector.
17.6 INCREMENTAL CONTRIBUTIONS FROM NONLINEAR BOUNDARY CONDITIONS Again, let Ii denote the principal invariants of C, and let i = VEC(I), c2 = VEC(C ), n iT = ∂Ii /∂c, and Ai = ∂ni /∂c. Recall from Chapter 2 that 2
n1 = i
n2 = I1i − c
n3 = I2 i − I1c + c2
I9 = I ⊗ I
A1 = 0
A2 = ii T − I9
A3 = I ⊗ C + C ⊗ I − (ic T + ciT ) + I1 (iiT − I 9 ).
(17.24)
Equation 17.23 is complete if increments of tractions are prescribed on the undeformed surface S0. We now consider the more complex situation in which τ is referred to the deformed surface S, on which they are prescribed functions of u. From Chandrasekharaiah and Debnath (1994), conversion is obtained using τ dS = τ 0 dS0
n T q dS = n T0 q 0 dS0
dS = µdS0
−1
µ = J n C n0 = T 0
(17.25) n ⊗ n n3 T 0
T 0
and from Nicholson and Lin (1997b) ∆µ ≈ dµ = m T dc ≈ m T ∆c,
m T = n T0 ⊗ n T0 A 3 / 2 µ.
(17.26)
Suppose that ∆ττ is expressed on S as follows: ∆τ = ∆ τ − A TM ∆u.
(17.27)
Here, dττ is prescribed, while AM is a known function of u. Also, S0 is the undeformed counterpart of S. These relations are capable of modeling boundary conditions, such as support by a nonlinear elastic foundation. From the fact that τ dS = τ0 dS0 = µτ0 dS, we conclude that τ = µ τ0. It follows that
∆τ0 = =
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1 τ ∆τ − 2 ∆µ µ µ 1 τ ( ∆ τ − ATM ∆u) − 2 mT ∆c µ µ
(17.28)
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Incremental Principle of Virtual Work
221
From the Incremental Principle of Virtual Work, the rhs term is written as 1
τ
∫ δ∆u ∆τ dS = ∫ δ∆u µ (∆ τ − A ∆u) − µ T
T
0
0
T M
2
mT ∆c dS0 .
(17.29)
Recalling the interpolation models for the increments, we obtain an incremental force vector plus two boundary contribution to the stiffness terms. In particular,
∫ δ∆u ∆τ dS = δ∆γ ∆f − δ∆γ [ K T
T
0
∆f =
∫ µ ∆τ dS , 1
0
T
0
∫
K BF = NATM N T dS0 ,
BF
+ KBN ]∆γ
K BN =
τ
∫µ
2
(17.30)
mT G dS0
The first boundary contribution is from the nonlinear elastic foundation coupling the traction and displacement increments on the boundary. The second arises from geometric nonlinearity when the traction increment is prescribed on the current configuration.
17.7 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, gear teeth. The extent of the contact zone is an unknown to be determined as part of the solution process. Solution of contact problems, introduced in Chapter 15, 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 for one particular formulation, which is mostly of interest for explicitly addressing the effect of large deformation. Figure 17.1 shows a contactor moving into contact with a foundation that is assumed to be rigid. We seek to follow the development of the contact area and the tractions arising throughout it. From Chapter 15, we recall that 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 comes into contact with the contactor point at x. We define a gap function, g, using y(x) = x + gn. Let m be the surface normalvector to the target at y(x). Let Sc be the candidate contact surface on the contactor, whose undeformed counterpart is S0c. There also is a candidate contact surface Sf on the foundation. We limit our attention to bonded contact, in which particles coming into contact with each other remain in contact. Algorithms for sliding contact with and without friction are available. For simplicity’s sake, we also assume that shear tractions, in
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contactor
n g m y(x)
foundation
FIGURE 17.1 Contact.
the osculating plane of point of interest, are negligible. Suppose that the interface can be represented by an elastic foundation satisfying the incremental relation ∆τ n = − k ( g)∆un .
(17.31)
Here, τn = n τ and un = n u are the normal components of the traction and displacement vectors. Since the only traction to consider is the normal traction (to the contactor surface), the transverse components of ∆u 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, T
k ( g) =
T
kH π
π − arctan(α g − ε ) + k , r L k 2
k H /k L >> 1.
(17.32)
As in Chapter 15, when g is positive, the gap is open and k approaches kL, which should be chosen as a small number, theoretically zero. When g becomes negative, the gap is closed and k approaches kH, which should be chosen as a large number, theoretically infinity to prevent penetration of the rigid body). Under the assumption that only the normal traction on the contactor surface is important, it follows that τ = tnn, from which ∆τ = ∆τ n n + τ n ∆ n.
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The contact model contributes the matrix Kc to the stiffness matrix as follows (see Nicholson and Lin, 1997b):
∫ δ∆u ∆τµ dS = ∫ δ∆u T
0
T
n
∆τ n µ dS0
= −δ∆γ T K c ∆γ
∫
(17.34)
∫
∫
K c = −2 Nnτ n mT βT dSc 0 + kc ( g)NnnT N T µ dSc 0 + Nτ n µhT dSc 0 . To update the gap, use the following relations proved in Nicholson and Lin (1997-b). T The differential vector, dy, is tangent to the foundation surface, hence, m dy = 0. It follows that 0 = mT du + gmT d n + m Tn dg ∆g = −
(17.35)
mT ∆u + gmT ∆ n . mTn
Using Equation 17.5, we may derive, with some effort, that ∆g = Γ T ∆γ ,
ΓT = −
mT N + ghT , mTn
hT = [n[(nT F − T ) ⊗ nT ] − nT ⊗ F − T ]MT . (17.36)
17.8 INTERPRETATION AS NEWTON ITERATION The (nonincremental) Principle of Virtual Work can be restated in the undeformed configuration as
∫ tr(δ ES)dV + ∫ δ u ρu˙˙ dV = ∫ δ u τ dS . T
T
o
o
o
(17.37)
We assume for convenience that τ is prescribed on So. The interpolation model satisfies the form
[
]
δe = BTL + BTNL ( γ ) δ γ BTL = BTNL = Fu =
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1 (I ⊗ I + I ⊗ IU)M(X) 2
(
)
1 I ⊗ FuT + Fu ⊗ IU M(X) 2 ∂u . ∂X
(17.38)
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Clearly, Fu and BNL are linear in γ. Τ Upon cancellation of the variation d γ , an algebraic equation is obtained as
∫
∫
∫
˙˙ dVo , Φ( γ , f ) = [(B L + B NL ( γ )]s dVo + N T ρu
f = N T τ o dSo .
(17.39)
At the load step, Newton iteration is expressed as
(
∂Φ (ν) J= γ n+1, fn+1 ∂γ
)
(
γ (nv++11) = γ (nν+)1 − J −1Φ γ (nν+)1,fn+1 ,
)
−1
(17.40)
or as a linear system
(
)
(
)
J γ (nv++11) − γ (nv+)1 = Φ γ (nv+)1, fn+1 ,
[
]
(17.41)
γ (nv++11) = γ (nv+)1 + γ (vn++11) − γ (nv+)1 . If the load increments are small enough, the starting iterate can be estimated as the th solution from the n 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 = KT + KG ,
(17.42)
so that the incremental stiffness matrix is the same as the Jacobian matrix in Newton iteration. This, of course, is a satisfying result. The Jacobian matrix can 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 solution can be viewed as the first iterate in a Newton iteration scheme. The one-time incremental solution can potentially be improved by additional iterations, as shown in Equation 17.41, but at the cost of computing the “residual” Φ at each load step.
17.9 BUCKLING Finite-element equations based on classical buckling equations for beams and plates were addressed in Chapter 14. In the classical equations, geometrically nonlinear terms appear through a linear correction term, thereby furnishing linear equations. Here, in the absence of inertia and nonlinearity in the boundary conditions, we briefly present a general viewpoint based on the incremental equilibrium equation (K T + K G )∆γ n+1 = ∆fn+1 .
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This solution predicts a large incremental displacement if the stiffness matrix KT + KG is ill-conditioned or outright singular. Of course, in elastic media, KT is positive-definite. However, in the presence of in-plane compression, KG may have a negative eigenvalue whose magnitude is comparable to the smallest positive eigenvalue of KT . To see this recall that KT =
∫ M GχG M dV T
T
0
KG =
∫ M S ⊗ IM dV . T
0
(17.44)
We suppose that the element in question is thin in a local z (out-of-plane direction). This suggests the assumption of plane stress. Now, in plate-and-shell theory, it is necessary 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 z-direction is out of the plane, in the geometric stiffness term, S11I S ⊗ I → S12 I 0 I
S12 I S22 I 0I
0I 0I . 0 I
(17.45)
In classical buckling, it is assumed that loads applied proportionately induce proportionate in-plane stresses. Thus, 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. Instead, if l denotes the distance along the line followed by the load point in proportional loading, the stresses become numerical functions of l. As a simple alternative to the general case, we consider buckling of a single element and suppose that the stresses appearing in Equation 17.46 are applied in a compressive sense along the faces of the element in a proportional manner whereby ( − Sˆ11 )I S ⊗ I → λ ( − Sˆ12 )I 0I
( − Sˆ12 )I ( − Sˆ22 )I 0I
0I 0 I , 0 I
(17.46)
in which the circumflex implies a reference value along the stress path at which l = 1, and the negative signs on the stresses are present since buckling is associated with compressive stresses. At the element level, the equation now becomes ˆ )∆γ = ∆f . (K T − λK G n +1 n +1
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At a given load increment, the critical buckling load for the current path, as a function of two angles determining the path in the stress space illustrated in ˆ ) singular. Chapter 14, is obtained by computing the l value rendering (K T − λK G
17.10 EXERCISES 1. Assuming linear interpolation models for u,v in a plane triangular membrane element with vertices (0,0),(1,0),(0,1), obtain the matrix M, G, BL, and BNL. 2. Repeat Exercise 1 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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Tangent-Modulus Tensors for Thermomechanical Response of Elastomers
18.1 INTRODUCTION 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 = − trace(ττ)/3 in incompressible or near-incompressible materials): u( X , t ) = NT ( X )γ (t ),
T( X , t ) − T0 = ν T ( X )θ (t ),
p = ξ T ( X )ψ (t ),
(18.1)
in which T0 is the temperature in the reference configuration, assumed constant. Here, N, ν, and ξ are shape functions and γ, θ, and ψ are vectors of nodal values. Application of the strain-displacement relations and their thermal analogs furnishes f1 = VEC(F − I) = M1 g = U T M2 g, f2 = VEC(F T − I) = M2 g, δe = b Tδ g, b = M2G, GT =
(18.2)
1 T (F ⊗ I + I ⊗ F T U), ∇T = bTT q 2
in which U is a 9 × 9 universal permutation tensor such that VEC(A ) = UVEC(A), and e = VEC(E) is the Lagrangian strain vector. Also, ∇ is the gradient operator referred to the deformed configuration. The matrix β and the vector βT are typically expressed in terms of isoparametric coordinates. T
18.2 COMPRESSIBLE ELASTOMERS The Helmholtz potential was introduced in Chapter 7 and shown to underlie the relations of classical coupled thermoelasticity. The thermohyperelastic properties of compressible elastomers are also derived from the Helmholtz free-energy density φ (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) = ρ0φ (T, E) (T constant), in which ρ0 is the density in the undeformed configuration. Typically, the elastomer is assumed to be isotropic, in which case φ can be expressed as a function of T, I1, I2, and I3. Alternatively, it may be expressed as a function of T and the stretch ratios λ1, λ2, and λ3. 227
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With φ known as a function of T, I1, I2, and I3, the entropy density η per unit mass and the specific heat ce at constant strain are obtained as ce = T
∂η ∂T E
η=−
∂φ . ∂T
(18.3)
The 2 Piola-Kirchhoff stress, s = VEC(S), is obtained from nd
s T = ρ0
∂φ ∂e
=2 T
∑ρ φ n , 0 i
i
φi =
i
∂φ . ∂Ii
(18.4)
Also of importance is the (isothermal) tangent-modulus matrix DT =
∂s =4 ∂e T
∑ ∑ ρ φ n n + 4∑ ρ φ A , 0 ij
i
i
T j
j
0 i
i
i
φij =
∂2φ . ∂Ii ∂I j
(18.5)
An expression for DT has been derived by Nicholson and Lin (1997c) for compressible, incompressible, and near-incompressible elastomers described by strain-energy functions (Helmholtz free-energy functions) and based on the use of stretch ratios (singular values of F) rather than invariants.
18.3 INCOMPRESSIBLE AND NEAR-INCOMPRESSIBLE ELASTOMERS When the temperature T is held constant, elastomers often satisfy the constraint of incompressibility or near-incompressibility. The constraint is accommodated by augmenting φ with terms involving a new parameter similar to a Lagrange multiplier. Typically, this new parameter is related to the pressure p. The thermohyperelastic properties of incompressible and near-incompressible elastomers can be 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 p. If the elastomer is incompressible at a constant temperature, the augmented Helmholtz function, φ, can be written as
φ = φd ( J1, J2 , T) − λξ ( J , T)/ρ0 ,
J1 = I1/ I31/ 3 ,
J2 = I2 / I32 / 3 ,
(18.6)
where ξ is a material function satisfying the constraint ξ(J, T ) = 0 and J = I31 2 = det(F). It is easily shown that φd depends on the deviatoric Lagrangian strain Ed, due to the introduction of the deviatoric invariants J2 and J3. The Lagrange multiplier λ is, in fact, the (true) pressure p: p = −trace(T )/3 =
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∂ξ . ∂J T
(18.7)
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For an elastomer that is near-incompressible at a constant temperature, φ can be written as
ρ0φ = ρ0φd ( J1, J2 , T) − pξ ( J , T) − p2 /2κ ,
(18.8)
in which κ0 is a constant. The near-incompressibility constraint is expressed by ∂φ/∂p = 0, which implies p = −κξ ( J , T).
(18.9)
The bulk modulus κ is given by
κ =−
∂p ∂ξ . =κ0 ∂J T ∂J T
(18.10)
Chen et al. (1997) presented sufficient conditions under which near-incompressible models reduce to the incompressible case as κ → ∞. Nicholson and Lin (1996) formulated the relations
ξ ( J , T) = f 3 (T) J − 1, φd = φ 1( J1 , J 2 ) + φ 2 (T), φ 2 (T) = ce T(1 − ln(T/T0 )), (18.11) with the consequence that p = −κ 0 ( f 3 (T) J − 1),
κ = f 3 (T)κ 0 .
(18.12)
Equation 18.12 provides a linear pressure-volume relation in which thermomechanical effects are confined to thermal expansion expressed using a constant-volume coefficient α. If the constraint is assumed to be satisfied a priori, the Helmholtz free energy is recovered as
φ ( I1, I2 , I3 ) = φd ( J1, J2 , T) + κ 0 ( f 3 (T) − 1)2 /2 ρ0 .
(18.13)
Alternatively, the latter term results from retaining the lowest nonvanishing term 3 in a Taylor-series representation of φ about f (T)J − 1. Given Equation 18.13, the entropy now includes a term involving p:
η=−
∂φd + πα f 4 (T)/ρ0 , ∂T
π = p/ f 3 (T).
(18.14)
The stress and the tangent-modulus matrices are correspondingly modified: s T = ρ0
∂φd ∂e
− π f 3 (T)n3T / J T,π
(18.15) T
∂s ∂ ∂φd DTP = = ρ0 − π f 3 (T) 2A3 / J − n3n3T / J 3 ∂e ∂e ∂e T,π
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]
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18.3.1 SPECIFIC EXPRESSIONS
FOR THE
HELMHOLTZ POTENTIAL
There are two broad approaches to the formulation of Helmholtz potential: To express φ as a function of I1, I2, and I3, and T (and p) To express φ as a function of the principal stretches λ1, λ2, and λ3, and T (and p). The latter approach is thought to possess the convenient feature of allowing direct use of test data, for example, from uniaxial tension. We will now examine several cases. 18.3.1.1 Invariant-Based Incompressible Models: Isothermal Problems The strain-energy function depends only on I1, I2, and incompressibility is expressed by the constraint I3 = 1, assumed to be satisfied a priori. In this category, the most widely used models include the Neo-Hookean material:
φ = C1 ( I1 − 3),
I3 = 1
(18.16)
and the (two-term) Mooney-Rivlin material:
φ = C1 ( I1 − 3) + C2 ( I2 − 3),
I3 = 1,
(18.17)
in which C1 and C2 are material constants. Most finite-element codes with hyperelastic elements support the Mooney-Rivlin model. In principle, Mooney-Rivlin coefficients C1 and C2 can be determined independently by “fitting” suitable loaddeflection curves, for example, uniaxial tension. Values for several different rubber compounds are listed in Nicholson and Nelson (1990). 18.3.1.2 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 v0 the Poisson’s ratio, referred to the undeformed configuration. The two models are: ν0 1 − 2ν 0 1 + ν0 1 1− 2ν 0 ρ 0φ1 = G0 I1 + I3 − I3 − ν0 ν0 2
(18.18)
1 I ρ 0φ 2 = G0 2 + 2 I3 − 5 2 I3 Let w denote the Helmholtz free energy evaluated at a constant temperature, in which case it is the strain energy. We note a general expression for w which is implemented
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231
in several commercial finite-element codes (e.g., ANSYS, 2000): w( J1, J2 , J ) =
∑ ∑ C ( J − 3) ( J − 3) + ∑ ( J − 1) /D , i
ij
i
j
1
k
2
r
j
k
Jr = J /(1 + Eth ),
k
(18.19) in which Eth is called the thermal expansion strain, while Cij and Dk are material constants. Several codes also provide software for estimating the model coefficients from user-supplied data. Several authors have attempted to uncouple the response into isochoric (incompressible) and volumetric parts even in the compressible range, giving rise to functions of the form φ = φ1(J1, J2) + φ2(J). A number of proposed forms for φ2 are discussed in Holzappel (1996). 18.3.1.3 Thermomechanical Behavior under Nonisothermal Conditions Now we come to the accommodation of coupled thermomechanical effects. Simple extensions of, for example, 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,
ρ0φ = C1 ( J1 − 2) + C2 ( J2 − 3) + ρ0ce T(1 − ln(T/To ) − κ ( f 3 (T) J − 1)π − π 2 /2κ , (18.20) in which π = p/f (T ). As previously mentioned, a model similar to Nicholson and Lin (1996) has been proposed by Holzappel and Simo (1996) for compressible elastomers described using stretch ratios. 3
18.4 STRETCH RATIO-BASED MODELS: ISOTHERMAL CONDITIONS For compressible elastomers, Valanis and Landel (1967) proposed a strain-energy function based on the decomposition
φ (λ1 , λ 2 , λ3 , T) = φ (λ1 , T) + φ (λ 2 , T) + φ (λ3 , T),
T fixed.
(18.21)
Ogden (1986) has proposed the form N
ρ 0φ (λ , T) =
∑ µ (λ
αp
p
1
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− 1),
T fixed.
(18.22)
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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 and 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 × 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) =
∑φ A . j
j
(18.23)
0
Here, φj 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 = VEC(F) and a = VEC(A), df ( a) =
1 T F′ ⊕ F′da + Wdω 2 ∝
F ′( A ) =
∑ jφ A j
0
j −1
df dF = ITEN 22 dA da
(18.24)
1 W = −( F − AF′/2)T ( F − AF′/2) + ( AT ⊗ F′ − F′ T ⊗ A) 2 ω = VEC(dΩ Ω), in which dΩ Ω is an antisymmetric tensor representing the Also, dω rate of rotation of the principal directions. The critical step is to determine a matrix J ω = −Jda. It is shown in Dahlquist and Bjork that J = −[AT A]−1W, in such that Wdω T I which [A A] is the Morse-Penrose inverse [(Dahlquist and Bjork(1974)]. Thus, df /da = F′ T ⊕ F′/2 − [AT A]′ W.
(18.25)
We now apply the tensor derivative to elastomers modeled using stretch ratios, especially in the model presented by Ogden (1986). In particular, a strain-energy function, w, was proposed, which, for compressible elastomers and isothermal response, is equivalent to the form w = tr
∑ ξ [C
ζi
i
i
− I ] ,
(18.26)
in which ci are the eigenvalues of C, and ξi, ζi are material properties. The tangentmodulus tensor χ appearing in Chapter 17 for the incremental form of the Principle
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of Virtual Work is obtained as
χ=4
∑ζ ξ (ζ − 1)C
ζi −2
i i
i
⊕ Cζ i −2 /2 + 4
i
Wi = −
3 − ζi 2
ζ
∑ ζ ξ [A i i
T
A]′ Wi
(18.27)
i
ζ
C i ⊗C i +
ζ i − 1 ζi −2 ζ −2 [C ⊗ C − C ⊗ C i ]. 2
(18.28)
18.5 EXTENSION TO THERMOHYPERELASTIC MATERIALS Equations 18.27 and 18.28 can be extended to thermohyperelastic behavior as follows, based on 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 can be described as mechanically nonlinear but thermally linear. Due to the role of thermal expansion, it is desirable to uncouple dilatational and deviatoric effects as much as possible. To this end, we introduce the deviatoric Cauchyˆ = C / I 1/ 3 in which I3 is the third principal invariant of C. Now, modGreen strain C 3 ifying w and expanding it in J − 1, ( J = I31/ 2 ) and retaining lowest-order terms gives w = tr
∑ ξ [Cˆ
ζi
i
i
1 − I] + κ ( J − 1)2 , 2
(18.29)
in which κ is the bulk modulus. The expression for χ in Equation 18.27 is affected by these modifications. To accommodate thermal effects, it is necessary to recognize that w is simply the Helmholtz free-energy density ρoφ under isothermal conditions, in which ρo is the mass density in the undeformed configuration. It is assumed that φ = 0 in the undeformed configuration. As for invariant-based models, we can obtain a function φ with three terms: a purely mechanical term φM, a purely thermal term φT , and a mixed term φTM. Now, with entropy, η, φ satisfies the relations s T = ρo
∂ϕ ∂e T
η=−
∂ϕ . ∂T e
(18.30)
Following conventional practice, the specific heat at constant strain, ce = T∂η /∂Te, is assumed to be constant, from which we obtain
φT = ce T [ln(T/T0 ) + 1].
(18.31)
On the assumption that thermal effects in shear (i.e., deviatoric effects) can be
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neglected relative to thermal effects in dilatation, the purely mechanical effect is equated with the deviatoric term in Equation 18.29: φM = tr
∑ ξ [Cˆ
ζi
i
i
− Ι ] .
(18.32)
Of greatest interest is φTM. The development of Nicholson and Lin (1996) furnishes
φ TM =
κ [β 3 (T) J − 1]2 2ρ
β (T) = (1 + α (T/T0 )/3) −1 .
(18.33)
The tangent-modulus tensor χ′ = ∂ s/∂ e now has two parts: XM + XTM, in which XM is recognized as χ, derived in Equation 18.29. Without providing the details, Kronecker Product algebra furnishes the following:
χTM =
n nT κ 3 β3 β 2 n3n3T + ( β 3 J − 1)A3 − 3 2 3 . Jρ J J
(18.34)
The foregoing discussion of stretch-based 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, 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, Equations 18.33 and 18.34 can be extended to incorporate the constraints of incompressibility and near-incompressibility. The tangent-modulus tensor presented here only addresses the differential of stress with respect to strain. However, if coupled heat transfer (conduction and radiation) is considered, a 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).
18.6 THERMOMECHANICS OF DAMPED ELASTOMERS Thermoviscohyperelasticity is a topic central to important applications, such as rubber mounts in hot engines. The current section introduces a thermoviscohyperelastic constitutive model thought to be suitable for near-incompressible elastomers exhibiting modest levels of viscous damping following a Voigt 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), models viscous
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dissipation and arises directly from the entropy-production inequality. It provides a consistent thermodynamic framework for describing damping in terms of a viscosity tensor that depends 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 and from the Principle of Virtual Work. Several quantities, such as internal energy density, χ, have reversible and irreversible portions, indicated by the subscripts r and i: χ = χr + χi. The thermodynamic formulation in the succeeding paragraphs is referred to undeformed coordinates. There are several types of viscoelastic behaviors 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 that 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. This type of damping is viewed as arising in small motions superimposed on the large strains, which already reflect long-term viscoelastic effects.
18.6.1 BALANCE
OF
ENERGY
The conventional equation for the balance of energy is expressed as
ρ0 χ˙ = s T e˙ − ∇0T q 0 + ρ0 h = srT e˙ + siT e˙ − ∇0T q 0 + ρ0 h
(18.35)
where s = VEC(S) and e = VEC(E). Here, χ is the internal energy per unit mass, q0 is the heat-flux vector, ∇0 is the divergence operator referred to undeformed coordinates, and h is the heat input per unit mass, for simplicity’s sake, assumed independent of temperature. The state variables are thus e and T. The Helmholtz free energy, φr per unit mass, and the entropy, η per unit mass, are introduced using
φr = χ − Tη.
(18.36)
∇T0 q 0 − ρ0 h = sTr e˙ + sTi e˙ − ρ0 Tη˙ − ρ0ηT˙ − ρ0 φ˙ r .
(18.37)
Now,
18.6.2 ENTROPY PRODUCTION INEQUALITY The entropy-production inequality is stated as
ρ0 Tη˙ ≥ −∇T0 q 0 + ρ0 h + q0T ∇T/T ≥ ρ0 φ˙r − sTr e˙ − siT e˙ + ρ0 Tη˙ + ρ0ηT˙ + q T0 ∇T/T
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(18.38)
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The Helmholtz potential is assumed to represent reversible thermohyperelastic effects. We decompose η into reversible and irreversible portions: η = ηr + ηi. Now, φr , ηr , and ηi are assumed to be differentiable functions of E and T. Furthermore, we suppose that ηi = ηi1 + ηi2 and
[
]
ρ 0 Tη˙ i 2 = −∇ T0 q0 + ρ 0 h i .
(18.39)
This allows us to say that the viscous dissipation is “absorbed” as heat. We also suppose that reversible effects are “absorbed” as a portion of the heat input as follows:
[
]
ρ0 Tη˙ r = −∇T0 q0 + ρ0 h r .
(18.40)
In addition, from conventional arguments,
ρ∂φr /∂e = sTr
∂φr /∂T = −ηr ,
(18.41)
it follows that sTi e˙ − q T0 ∇0 T/T ≥ − ρ0ηi1T˙ .
(18.42)
Inequality as shown in Equation 18.42 can be satisfied if ρ0ηi T˙ ≥ 0 and sTi e˙ ≥ 0
(a)
− q T0 ∇T/T ≥ 0
(18.43)
(b).
Inequality as shown in Equation 18.43b is conventionally assumed to express the fact that heat flows irreversibly from cold to hot zones. Inequality as shown in Equation 18.43a requires that viscous effects be dissipative.
18.6.3 DISSIPATION POTENTIAL Following Ziegler and Wehrli (1987), the specific dissipation potential Ψ(q0 , e˙ , e, T) = − ρ0ηi T˙ is introduced, for which siT = ρ0 Λ i ∂Ψ/∂e˙
(a).
− ∇T0 T/T = Λ t ρ0 ∂Ψ/∂q 0 .
(b).
(18.44)
The function Ψ is selected such that Λi and Λt are positive scalars, in which case the inequalities in Equations 18.44a and 18.44b require that (∂Ψ/∂e˙ ) e˙ ≥ 0
(∂Ψ/∂q 0 )q 0 ≥ 0.
(18.45)
This can be interpreted as indicating the convexity of a dissipation surface in (e˙ , q0 ) space. Clearly, to state the constitutive relations, it is sufficient to specify φr and Ψ.
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A simple illustration is now provided showing how the dissipation potential in Equation 18.45 provides a “framework” for describing dissipative effects. On the expectation that properties governing heat transfer are not affected by strain, we introduce the decomposition Ψ = Ψi + Ψt
1
ρ0 Ψt = [q0T q0 ]2 [ Λ t /2].
(18.46)
Now, Ψt represents thermal effects, and we assume for simplicity’s sake that Λt is a material constant. Inequality in Equation 18.46 implies that −∇0 T/T = q 0 /Λ t .
(18.47)
This is essentially the conventional Fourier law of heat conduction, with Λt recognized as the thermal conductivity. As an elementary example of viscous dissipation, suppose that Ψi = µ (T, J1, J2 )e˙ Te˙ / 2
∆ i = 1,
(18.48)
in which µ(T, J1, J2) is the viscosity. Hence, si = µ (T, J1, J2 )e˙ ,
(18.49)
and Equation 18.44a requires that µ be positive.
18.6.4 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 18.40 is repeated as ˙ ρ0 φ˙r = sTr e˙ − ρ0ηr T.
(18.50)
The ensuing Maxwell relation is ∂srT /∂T = − ρ0 ∂ηr /∂e.
(18.51)
Conventional operations furnish the reversible part of the equation of thermal equilibrium (balance of energy): [−∇T0 q0 + ρ0 h]r = − T(∂sTr /∂T)e˙ + ρ0ce T˙ ,
ce = T∂ηr /∂T.
(18.52)
For the irreversible part, we recall the relations
(
)
− ∇T0 q 0 − ρ0 h i = − sTi e˙ + ρ0ci T˙
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(18.53)
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and ∂η ∂η ρ0 Tη˙i 2 = ρ0 T i 2 e˙ + ρ0 T i 2 T˙ ∂e ∂T
(18.54)
and − siT = ρ0 T
∂ηi 2 , ∂e
ci = T
∂ηi 2 . ∂T
(18.55)
Upon adding the relations, we obtain the thermal-field equation −∇T0 q 0 + ρ0 h = − T ∂sTr /∂Te˙ − siTe˙ + ρ0 (ce + ci )T˙ .
(18.56)
It is easily seen that Equation 18.55 directly reduces to a well-known expression in classical linear thermoelasticity. In addition, under adiabatic conditions in which −∇T0 q 0 + ρ0 h = 0, most of the “viscous” work, s iT e˙ , is “absorbed” as a temperature increase controlled by ρ0(ce + ci), while a smaller portion is “absorbed” into the elastic strain-energy field.
18.7 CONSTITUTIVE MODEL: POTENTIAL FUNCTIONS 18.7.1 HELMHOLTZ FREE-ENERGY DENSITY In the moderately damped, thermohyperelastic material, the elastic (reversible) stress is assumed to satisfy a thermohyperelastic constitutive relation suitable for nearincompressible elastomers. In particular,
φr = φrm ( I1 , I2 , I3 ) + φrt (T) + φrtm (T, I3 ) + φro .
(18.57)
Here, φrm 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. Again, I1, I2, I3 are the principal invariants of the (right) Cauchy-Green strain tensor. The formulation can easily be adapted to stretch ratio-based models, such as the Ogden (1986) model. The function φrt(T) represents the purely thermal portion of the Helmholtz free-energy density. Finally, φrtm(T, I3) represents thermomechanical effects, again based on the assumption that the primary coupling is through volumetric expansion. The quantity φro represents the Helmholtz free energy in the reference state, and, for simplicity’s sake, is assumed to vanish. The forms of φrt and φrtm are introduced in the current presentation:
φrt (T) = ce T[1 − ln(T/T0 )] φrtm (T, I3 ) =
κ [ f 3 (T) J − 1]2 2 ρ0
α J = I31/ 2 = det( F) f (T) = 1 + ( T − T0 ) 3
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(18.58) (18.59) −1
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and α is the volumetric coefficient of thermal expansion. For the sake of illustration, for Φrm(I1, I2, I3), we display the classical two-term Mooney-Rivlin model:
φrm ( I1, I2 , I3 ) = C1 ( J1 − 1) + C2 ( J2 − 1),
I3 = 1,
(18.60)
in which J1 = I1/ I31/ 3 and J2 = I2 / I32 / 3 are the “deviatoric invariants” of C. The reversible stress is obtained as sr = 2φ j n j n1 = i,
i = VEC(I)
φ j = ∂φr /∂I j
n1 = I1i − c,
c = VEC(C)
n3 = I3VEC(C−1 )
(18.61)
18.7.2 SPECIFIC DISSIPATION POTENTIAL Fourier’s law of conduction is obtained from:
ρΨt = [q0T q0 ]1/ 2 [kt /2].
(18.62)
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 Ψi, of which an example was already given in Section 6.3. Here, taking a more general viewpoint, we seek expressions of the form Ψi = 12 e˙ T Dv (e, T)e˙ , in which Dv is called the viscosity tensor; it is symmetric and positive-definite. (Of course, the correct expression is determined by experiments.) The simplest example was furnished in Section 18.6.3. As a second example, to ensure isotropy, suppose that Ψi is a function of J˙1 and J˙2 : Ψi ( J˙1 , J˙2 ), and note that I J˙1 = m1T e˙, m1T = 2 iT − 1 I1 n3T I31 3 3 3
I J˙2 = mT2 e˙, mT2 = 2 nT2 − 23 I2 n3T I32 3 . 3 (18.63)
For an expression reminiscent of the two-term Mooney-Rivlin strain-energy function, let us consider the specific form
[
]
Ψi = µ1 (T) C1v J˙12 /2 + C2 v J˙2 2 ,
(18.64)
in which C1v and C2v are positive material coefficients. We obtain the viscosity tensor,
[
]
Dv = µ (T) C1v m1m1T + C2 v m2 mT2 .
(18.65)
Unfortunately, this tensor is only positive-semidefinite. As a second example, suppose that the dissipation potential is expressed in terms of the deformation rate
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tensor D, in particular, Ψi = µ(T)tr(D )/2, which has the advantage in that the deformation rate tensor D is in the observed (current) configuration. With d = VEC(D), 2
d = F − T ⊗ F − T e˙
Dv = µ (T)(2ε + I )−1 ⊗ (2ε + I )−1,
(18.66)
which is positive-definite.
18.8 VARIATIONAL PRINCIPLES 18.8.1 MECHANICAL EQUILIBRIUM In this section, we present one of several possible formulations for the finite-element equations of interest, neglecting inertia. Application of variational methods to the mechanical field furnishes the Principle of Virtual Work in the form
∫ tr(δES )dV = ∫ δu τ dS − ∫ tr(δES )dV , T
0
i
0
0
0
r
(18.67)
in which τ0 denotes the traction vector on the undeformed surface S0. As illustrated in the examples in the previous section, we expect that the dissipation potential has the form Ψi = 12 e˙ T Dv(e, T) e˙ , from which si = Dv e˙ ,
(18.68)
in which Dv is again the viscosity tensor, and it will be taken as symmetric and positive-definite. (It is positive-definite since siTe˙ ≥ 0 for all e˙ .) Equation 18.67 is thus rewritten as
∫ δe D e˙dV = ∫ δu t dS − ∫ δe s dV . T
T
i
0
T
0
r
0
0
(18.69)
18.8.2 THERMAL EQUILIBRIUM The equation for thermal equilibrium is rewritten as
[
]
ρ0 (η˙ r + η˙i 2 ) = −∇T0 q0 + ρ0 h / T,
(18.70)
and note that it is in rate form, in contrast to the equation of mechanical equilibrium (see Equation 18.69). For the sake of a unified rate formulation, we first introduce the integrated form of this relation. The current value of ηr + ηi2, assuming that the initial values of the entropies vanish, is now given by
ρ0 (ηr + ηi 2 ) =
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∫ [[−∇ q + ρ h]/ T]dt. T 0 0
0
(18.71)
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241
The corresponding variational principle is stated as
∫ δ Tρ (η + η 0
r
i2
∫ ∫ [[−∇ q + ρ h]/ T]dt dV .
)dV0 = δ T
T 0 0
0
0
(18.72)
With some effort, a rate (incremental) variational principle can be obtained in the form
∫ δ T[−T∂s /∂Te˙ − s e˙ + ρ (c + c )T˙ ]/ TdV = ∫ δ T[[−∇ q T r
T i
0
e
i
T 0
0
0
] ]
+ ρ0 h / T dV0 .
(18.73)
(The motivation behind writing Equations 18.71 and 18.72 is simply to establish how a rate principle is obtained in the thermal field as the counterpart of the rate (incremental) principle for the mechanical field.) Upon approximating T in the denominator by T0 and letting k denote the thermal conductivity, we can obtain the thermal-equilibrium equation in the form
∫ δ T[−∂s /∂Te − s e˙ − ρ (c + c )T˙ ]/ T dV T r
T i
∫
0
e
i
∫
0
0
∫
+ [k (∇0δ T)T ∇0δ T/T0 ]dV0 = δ Tρ0 h/T0 dV0 − nT (δ Tq/T0 )dS0
(18.74)
Using interpolation models for displacement and temperature, Equations 18.69 and 18.74 reduce directly into finite-element equations for the mechanical and thermal fields.
18.9 EXERCISES 1. Derive explicit forms of the stress and tangent-modulus tensors using the Helmoltz potential in Equation 18.20. 2. Derive the quantities m1 and m2 in Equation 18.63. 3. Verify Equation 18.58 by recovering ce upon differentiating twice with respect to T. 4. Substitute the required interpolation models in Equations 18.69 and 18.74 to obtain an element-level finite-element equation for the mechanical and thermal fields.
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19
Inelastic and Thermoinelastic Materials
19.1 PLASTICITY Plasticity and thermoplasticity are topics central to the analysis of important applications, such as metal forming, ballistics, and welding. The main goal of this section is to present a model of plasticity and thermoplasticity, along with variational and finite-element statements, accommodating the challenging problems of finite strain and kinematic hardening.
19.1.1 KINEMATICS Elastic and plastic deformation satisfies the additive decomposition D = D r + Di ,
(19.1)
from which we can formally introduce strains: ∃=
∫ Ddt
∃r =
∫ D dt r
∃i =
∫ D dt. i
(19.2)
The Lagrangian strain E satisfies the decomposition E˙ = F T DF
∫
Er = F T D r Fdt
∫
Ei = F T D i Fdt.
(19.3)
Typically, plastic strain is viewed as permanent strain. As illustrated in Figure 19.1, in a uniaxial tensile specimen, the stress, S11, can 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 equal to zero, there still is a residual strain, Ei, which is identified as the inelastic strain. However, if instead the stress was increased to point C, it would encounter reversed loading at point D, which reflects the fact that the elastic region need not include the zerostress value.
19.1.2 PLASTICITY We will present a constitutive equation for plasticity to illustrate how the tangent modulus is stated. The ideas leading to the equation will be presented subsequently in the section on thermoplasticity. With χ e = ITEN22(De) and De denoting the 243
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C S11
A E E D
E B Ei
F
E11
FIGURE 19.1 Illustration of inelastic strain.
tangent-modulus tensor relating the elastic-strain rate to the stress rate (assuming a linear relation), the constitutive equation of interest is e˙ i = Ci s˙,
e˙ e = Ces˙, T
∂Ψ ∂Ψi Ci = i H, ∂s ∂s
−1
Ce = χ e ,
k˙ = Ke˙ i
∂Ψ ∂Ψ ∂Ψ T H = − i + i K i . ∂k ∂s ∂ei
(19.4)
In Equation 19.4, Ψi is the yield function. Ψi = 0 determines a closed convex surface in stress space called the yield surface. (We will see later that Ψi also serves as a [complementary] dissipation potential.) The stress point remains on the yield surface during plastic flow, and is moving toward its exterior. The plastic strain rate, expressed as a vector, is typically assumed to be normal to the yield surface at the stress point. If the stress point is interior to, or moving tangentially on, the yield surface, only elastic deformation occurs. On all interior paths, for example, due to unloading, the response is only elastic. Plastic deformation induces “hardening,” corresponding to a nonvanishing value of Ci. Finally, k is a history-parameter vector, introduced to represent dependence on the history of plastic strain, for example, through the amount of plastic work. The yield surface is distorted and moved by plastic strain. In Figure 19.2(a), the conventional model of isotropic hardening is illustrated in which the yield surface expands as a result of plastic deformation. This model is unrealistic in predicting a growing elastic region. Reversed plastic loading is encountered at much higher stresses than isotropic hardening predicts. An alternative is kinematic hardening (see Figure 19.2[b]), in which the yield surface moves with the stress point. Within a few percentage points of plastic strain, the yield surface may cease to encircle the origin.
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Sll
path of stress point
Sl
Slll
principal stresses Sl Sll Slll
FIGURE 19.2(a) Illustration of yield-surface expansion under isotropic hardening.
Sll
path of stress point
Sl
Slll FIGURE 19.2(b) Illustration of yield-surface motion under kinematic hardening.
A reference point interior to the yield surface, sometimes called the back stress, must be identified to serve as the point at which the elastic strain vanishes. Combined isotropic and kinematic hardening are shown in Figure 19.2(c). However, the yield surface contracts, which is closer to actual observations (e.g., Ellyin [1997]).
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Sll
path of stress point
Sl
Slll FIGURE 19.2(c) Illustration of combined kinematic and isotropic hardening.
The rate of movement must exceed the rate of contraction for the material to remain stable with a positive tangent modulus. Combining the elastic and inelastic portions furnishes the tangent-modulus tensor:
[
χ = χe−1 + Ci
]
−1
= [I + χeCi ]−1 χe .
(19.5)
Suppose that in uniaxial tension, the elastic modulus is Ee and the inelastic modulus-relating stress and inelastic strain increments are Ei, and Ei << Ee. The total Ei uniaxial modulus is then . (1 + E i / E e )
19.2 THERMOPLASTICITY As in Chapter 18, two potential functions are introduced to provide a systematic way to describe irreversible and dissipative effects. The first is interpreted as the Helmholtz free-energy density, and the second is for dissipative effects. To accommodate kinematic hardening, we also assume an extension of the Green and Naghdi (G-N) (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.
19.2.1 BALANCE
OF
ENERGY
The conventional equation for energy balance is augmented using a vector-valued, workless internal variable, α 0, regarded as representing “microstructural rearrangements”:
ρo χ˙ 0 = sT e˙ r + sT e˙ i − ∇T0 q0 + ρo h + βT0 α˙ 0 ,
(19.6)
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where χo is the internal energy per unit mass in the undeformed configuration and s = VEC(S), e = VEC(E), and β0 is the flux per unit mass associated with α0. However, βr. Also, c is the internal energy per unit mass, q0 is the note that β = 0, thus βi = −β heat-flux vector referred to undeformed coordinates, and h is the heat input per unit mass, for simplicity’s sake, assumed independent of temperature. The state variables are Er , Ei , T, and α 0. The next few paragraphs will go over some of the same ground as for damped elastomers in Chapter 18, except for two major points. In that chapter, the stress was assumed to decompose into reversible and irreversible portions in the spirit of elementary Voigt models. In the current context, the strain shows the decomposition in the spirit of the classical Maxwell model. In addition, as seen in the following, it proves beneficial to introduce a workless internal variable to give the model the flexibility to accommodate phenomena such as kinematic hardening. The Helmholtz free energy, φ , per unit mass and the entropy, η, per unit mass are introduced using
φ = χ − Tη.
(19.7)
∇T0 q0 − ρ0 h = sT e˙ r + sT e˙ i − ρ0 Tη˙ − ρ0ηT˙ − ρ0φ˙ + βT0 α˙ 0 .
(19.8)
Now,
19.2.2 ENTROPY-PRODUCTION INEQUALITY The entropy now satisfies
ρ0 Tη˙ ≥ −∇T0 q0 + ρ0 h + q0T ∇T/ T ≥ ρ0φ˙ − sT e˙ r − sT e˙ i + ρ0 Tη˙ + ρ0ηT˙ + q0T∇T/ T − βT0 α˙ 0 .
(19.9)
Viewing φr as a differentiable function of er , T, and α 0, we conclude that ∂φr / ∂e r = sT
∂φr / ∂T = − ρ0ηr
∂φr / ∂α 0 = βT0 r .
(19.10)
Extending the G-N formulation, let s* = ρ0∂φi /∂ei and assume that ηi = −∂φi /∂T α 0 = βT0i . Now, and ρ0∂φi /∂α T
s*T = ρ0 ∂φi / ∂ei
ηi = − ∂φi / ∂T
ρ0 ∂φi / ∂ α 0 = βT0 i .
(19.11)
The entropy-production inequality (see Equation 19.9) is now restated as (sT − s*T )e˙ i − q0T ∇0 T/ T ≥ 0.
(19.12)
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The inequality shown in Equation 19.12 can be satisfied if (sT − s*T )e˙ i ≥ 0
(a)
− q0T ∇0 T/ T ≥ 0
(b).
(19.13)
The first inequality involves a quantity, s* = VEC(S*), with dimensions of stress. In the subsequent sections, s* will be viewed as a reference stress, often called the back stress, which is interior to a yield surface and can be used to characterize the motion of the yield surface in stress space. 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 to regard e˙ r as vanishing at the origin. Instead, e˙ r = 0 is now associated with a moving-reference stress interior to the yield surface, identified here as the back stress s*.
19.2.3 DISSIPATION POTENTIAL As in Chapter 18, we introduce a specific dissipation potential, Ψi, for which e˙ iT = ρ0 Λ i ∂Ψi / ∂ ∋
(i)
− ∇T0 T / T = Λ t ρ0 ∂Ψi / ∂q 0
(ii)
∋ = s − s* (iii), (19.14a)
from which, with Λi > 0 and Λt > 0,
ρ0 Λ i (∂Ψ / ∂∋) ∋ + ρ0 Λ t (∂Ψ / ∂q 0 )q 0 > 0.
(19.14b)
On the expectation that properties governing heat transfer are not affected by strain, we introduce the decomposition into inelastic and thermal portions:
Ψ = Ψi + Ψt
ρ0 Ψt =
Λt q T0 q 0 , 2
(19.15a)
where Ψi represents mechanical effects and is identified in the subsequent sections. The thermal constitutive relation derived from the dissipation potential implies Fourier’s law: −∇0 T / T = q 0 / Λ t . The inelastic portion is discussed in the following section.
(19.15b)
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19.3 THERMOINELASTIC TANGENT-MODULUS TENSOR The elastic strain rate satisfies a thermohypoelastic constitutive relation: e˙ r = Cr (s − s*)• + ar T˙ .
(19.16)
Cr is a 9 × 9 second-order, elastic compliance tensor, and ar is the 9 × 1 thermoelastic expansion vector, with both presumed to be known from measurements. Analogously, for rate-independent thermoplasticity, we seek tensors Ci and ai , depending on ∋ , ei , and T such that e˙ i = Ci (s − s*)• + ai T˙
(19.17a)
˙ e˙ = [Cr + Ci ](s − s*)• + ( ar + ai )T.
(19.17b)
During thermoplastic deformation, the stress and temperature satisfy a thermoplastic yield condition of the form Πi (∋, ei , k, T, ηi 2 ) = 0,
(19.18)
and Πi is called the yield function. Here, the vector k is introduced to represent the effect of the history of inelastic strain, e˙ i , such as work hardening. It is assumed to be given by a relation of the form k˙ = K (ei , k, T)e˙ i .
(19.19)
˙ = 0 during thermoplastic flow, from The “consistency condition” requires that Π i which d Πi d Πi d Πi ˙ d Πi ˙ d Πi ∋˙ + e˙ + k+ T+ η˙ = 0. d∋ dei i dk dT dηi i
(19.20)
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, dΠ i e˙ i = Λ i d∋
T
(a)
d Πi η˙ i = Λ i dηi
(b).
(19.21)
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Equation 19.14a suggests that the yield function may be identified as the dissipation potential: Πi = ρ0Ψi. Standard manipulation furnishes e˙ i = Ci ∋˙ + ai T˙ T
∂Ψ ∂Ψi Ci = i H ∂∋ ∂∋ ∂Ψ ci = i ∂T
2
H
η˙i 2 = biT ∋˙ + ci T˙ T
∂Ψ ∂Ψi ai = bi = i H ∂∋ ∂T
(19.22)
∂Ψ ∂Ψ ∂Ψ T ∂Ψ ∂Ψ i i H = − i + i K i + ∂k ∂∋ ∂ηi 2 ∂T ∂ei
and H must be positive for Λi to be positive. Note that, in the current formulation, the dependence of the yield function on temperature accounts for ci. The thermodynamic inequality shown in Equation 19.13a is now satisfied if H > 0. T Next, note that s* depends on ei , T, and α 0 since s* = ρ0∂φi /∂ei. For simplicity’s sake, we neglect dependence on α 0 and assume that a relation of the following form can be measured for s*: T
s˙* = Γe˙ i + ϑ T˙
Γ=
∂ ∂ Ψ ∂ei ∂ei I
ϑ T = ∂2 Ψi / ∂ei ∂T.
(19.23)
From Equations 19.16 and 19.17, the thermoinelastic tangent-modulus tensor and thermal thermomechanical vector are obtained as e˙ = Cs˙ + aT˙ C = (C r + Ci )[I − (I + ΓCi ) −1 ΓCi ] a = [ ar + ai + (C r + Ci )Γai − (C r + Ci )(I + ΓCi ) −1ϑ ]
(19.24a)
(19.24b)
If appropriate, the foregoing formulation can be augmented to accommodate plastic incompressibility.
19.3.1 EXAMPLE We now provide a simple example using the Helmholtz free-energy density function and the dissipation-potential function to derive constitutive relations. The following expression involves a Von Mises yield function, linear kinematic hardening, linear work hardening, and linear thermal softening.
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i. Helmholtz free-energy density:
φ = φr + φi
ρ0φi = k3eiT ei
ρ0φr = eTr Cr−1e r / 2 − a T Cr−1 (T − T0 )e r + ρ0 c′r T(1 − ln(T / T0 ))
(19.25)
in which c′r is a known constant. From Equation 19.28, ∋ = ρ0 (∂φr / ∂e r )T = Cr−1[e r − ar ( T − T0 )].
(19.26)
Finally, cr = −T
∂ 2φ r = c′ ∂T 2
(19.27)
ii. Dissipation potential: Ψ = Ψi + Ψt
ρ0 Ψt =
Λt q T0 q 0 2
Ψi = ∋T ∋ − [k0 + k1k − k2 (T − T0 )] = 0
(19.28)
k˙ =∋T e˙ i
(19.29)
Straightforward manipulations serve to derive H = k1 ∋ T ∋ = k1[k0 + k1k − k2 (T − T0 )] Ci =
∋∋T H ∋T ∋
ci = k22 H
ai = bi =
k2 ∋ ∋T ∋
(19.30)
H
(19.31)
Consider a two-stage thermomechanical loading, as illustrated schematically in nd Figure 19.3. Let SI , SII , SIII denote the principal values of the 2 Piola-Kirchhoff stress, and suppose that SIII = 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 (S1, S2) 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 and T3. The plastic strain must increase, thus, the center of the yield surface moves. In addition, strain hardening tends to cause the yield surface to expand, while the 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.
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Sll T3 T1
T2
T0
T4
T0 T0 T0
T0
Sl T0
Slll FIGURE 19.3 Effect of load and temperature on yield surface.
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 occur within elements, essentially reducing the continuity order of the fields (discontinuity in strain gradients). Special procedures have been developed in some codes to address this difficulty. The shrinkage of the yield surface with temperature provides an element of the explanation of the phenomenon of adiabatic shear banding, which is commonly encountered in some materials during impact or metal forming. In rapid processes, plastic work is mostly converted into heat and on into high temperatures. There is not enough time for the heat to flow away from the spot experiencing high deformation. However, the process is unstable while the stress level is maintained. Namely, as the material gets hotter, the rate of plastic work accelerates, thanks to the softening evident in Figure 19.3. The instability is manifested in small, periodically spaced bands, in the center of which the material is melted and resolidified, usually in a much more brittle form than before. These bands can nucleate brittle failure.
19.4 TANGENT-MODULUS TENSOR IN VISCOPLASTICITY The thermodynamic discussion in the previous section 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 case viscoplasticity and thermoviscoplasticity are attractive models. An example
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253
of a constitutive model, for example, following Perzyna (1971), is given in undeformed coordinates as
µω e˙i =< 1 −
k > Ψi
∂Ψi ∂∋ ∂Ψi ∂∋
T
∂Ψi ∂∋
T
,
<1−
1 − k , k >= Ψi Ψi 0,
k ≥0 Ψi otherwise, if
1−
(19.32) and Ψi(∋, ei, k, T, ηi) is a loading surface function. The elastic response is still considered linear in the form ˙ e˙ r = χ r−1s˙ + α r T.
(19.33)
Recall from thermoplasticity that T
∂ ∂ Γ= Ψ ∂ei ∂ei i
s˙* = Γe˙ i + ϑT˙
ϑ T = ∂2 Ψi /∂ei ∂T.
(19.34)
Corresponding to ∋, there is a reference stress s′ and a corresponding vector ∋′ = s′ − s* such that Ψi( ∋′, ei, k, T, ηi) = k(ei, k, T, ηi) determines a quasi-static, referenceyield surface. The vectors ∋ and ∋′ have the same origin and direction, but the latter terminates at the reference surface, while the latter terminates outside the reference surface if inelastic flow is occurring. Interior to the surface, no inelastic flow occurs. If exterior to the surface, inelastic flow occurs at a rate dependent on the distance to the exterior of the reference surface. This situation is illustrated in Figure 19.4. Sll
S
∗
stress point Sl
Slll FIGURE 19.4 Illustration of reference surface in viscoplasticity.
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It should be evident that viscoplasticity and thermoviscoplasticity can be formulated to accommodate phenomena such as kinematic hardening and thermal shrinkage of the reference-yield surface. The tangent-modulus matrix now reduces to elastic relations, and viscoplastic effects can be treated as an initial force (after canceling the variation) since
χ k ⋅s˙ = χ r e˙ + χ r α r T˙ − Γ − r < 1 − > Ψi µv
∂Ψi ∂∋ ∂Ψi ∂∋
T
∂Ψi ∂∋
T
.
(19.35)
In particular, the Incremental Principle of Virtual Work is now stated, to first order, as
∫ δ∆e χ ∆edV + ∫ δ∆e χ α ∆TdV + ∫ δ∆u ρ ∆u˙˙dV T
T
r
o
T
r
r
o
o
χ k = δ∆uT ∆τ dSo + δ∆eT Γ − r < 1 − > µv Ψi
∫
∫
o
∂Ψi ∂∋ ∂Ψi ∂∋
T
∂Ψi ∂∋
T
dVo
(19.36)
19.5 CONTINUUM DAMAGE MECHANICS Ductile fracture occurs by processes associated with the notion of damage. An internal-damage variable is introduced that accumulates with plastic deformation. It also manifests itself in reductions in properties, such as the experimental values of the elastic modulus and yield stress. When the damage level in a given element reaches a known or assumed critical value, the element is considered to have failed. It is then removed from the mesh (considered to be no longer supporting the load). The displacement and temperature fields are recalculated to reflect 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 Thomasson (1990), considers damage to occur by a specific mechanism occurring in a threestage process: nucleation of voids, their subsequent growth, and their coalescence to form a macroscopic defect. The coalescence event is 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˙ = Ξ( f , ei , T ),
(19.37)
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255
S11
Ep1
Sy1
Ep3
Ep2
Sy3
Sy2 E2
E3
E1
E11 FIGURE 19.5 Illustration of effect of damage on elastic-plastic properties.
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 empirical in nature and is not dependent on a specific mechanism. It uses the parameter D, which is interpreted as the fraction of damaged area Ad to total area Ao that the stress (traction) acts on. Consider a uniaxial tensile specimen with damage, but experiencing elastic behavior. Suppose that damaged area Ad can no longer support a load (is damaged). For a given load P, the true stress at a point in the undamaged zone is S = A −P A = 1 −1 D AP = 1 −1 D S ′. Here, S′ o D o is a nominal stress, but is also the measured stress. If E is the elastic modulus measured in an undamaged specimen, the modulus measured in the current specimen will be E′ = E(1 − D), demonstrating that damage is manifest in small changes in properties. As an illustration of damage, suppose specimens are loaded into the plastic range, unloaded, and then loaded again. Without damage, the stress-strain curve should return to its original path. However, due to the damage, there are slight changes in the elastic slope, in the yield stress, and in the slope after yield (exaggerated in Figure 19.5). From the standpoint of thermodynamics, damage is a dissipative internal variable. In reality, the amount of mechanical or thermal energy absorbed by damage is probably small, so that its role in the energy-balance equation can be neglected. At the risk of being slightly conservative, in dynamic (adiabatic) problems, the plastic work can be assumed to be completely converted into heat. Even so, for the sake of a consistent framework for treating dissipation, a dissipation potential, Ψd, can be introduced for damage, as has been done, for example, by Bonora (1997). The contribution to the irreversible entropy production can be introduced in the form D D˙ ≥ 0, in which D is the “force” associated with flux D˙ . Positive dissipation is assured if we assume
D
=
∂Ψd , ∂D˙
Ψd =
1 Λ (e , T, k ) D˙ 2 , 2 d i
Λ d (ei , T, k ) > 0.
(19.38)
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An example of a satisfactory function is Λd (ei , T, k) = Λdo ∫(s − s*) e˙ i dt, Λdo a positive constant, showing damage to depend on plastic work. Specific examples of constitutive relations for damage are given, for example, in Bonora (1997). At the current values of the damage parameter, the finite-element equations are solved for the nodal displacements, from which the inelastic strains can be computed. This information can then be used to update the damage-parameter values. Upon doing so, the damage-parameter values are compared to critical values. As stated previously, if the critical value is obtained, the element is deleted. The path of deleted elements can be viewed as a crack. The code LS-DYNA Ver. 9.5, (2000) incorporates a material model that includes viscoplasticity and damage mechanics. It can easily be upgraded to include thermal effects, assuming that 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 (see Moraes [2002]).
19.6 EXERCISES 1. In isothermal plasticity, assuming the following yield function, find the uniaxial stress-strain curve: Ψi = (s − k1ei )T (s − k1ei ) − ko + k2
∫
t
0
e˙ i T e˙ i dt .
Assume small strain and that the plastic strain is incompressible: tr(Ei ) = 0. 2. Regard the expression in Exercise 1 as defining the reference-yield surface in viscoplasticity, with viscosity hv. Find the stress-strain curve under uniaxial tension if a constant strain rate is imposed.
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Advanced Numerical Methods
In nonlinear finite-element analysis, solutions are typically sought using Newton iteration, either in classical form or augmented as an arc-length method to bypass critical points in the load-deflection behavior. Here, two additional topics of interest are briefly presented.
20.1 ITERATIVE TRIANGULARIZATION OF PERTURBED MATRICES 20.1.1 INTRODUCTION In solving large linear systems, it is often attractive to use Cholesky triangularization followed by forward and backward substitutions. In computational problems, such as in the nonlinear finite-element method, solutions are attained incrementally, with the stiffness matrix slightly modified whenever it is updated. The goal here is to introduce and demonstrate an iterative method of determining the changes in the triangular factors ensuing from modifying the stiffness matrix. A heuristic convergence argument is given, as well as a simple example indicating rapid convergence. Apparently, no efficient iterative method for matrix triangularization has previously been established. The finite-element method often is applied to problems requiring solution of large linear systems of the form K0γ0 = f0, in which the stiffness matrix K0 is positivedefinite, symmetric, and may be banded. As discussed in a previous chapter, an attractive method of solution is based on Cholesky decomposition (triangularization), T in which K0 = L0 L 0 and L0 is lower-triangular, and it is also banded if K0 is banded. The decomposition enables an efficient solution process consisting of forward substitution followed by backward substitution. Often, however, the stiffness matrix is updated during the solution process, leading to a slightly different (perturbed) matrix, K = K0 + ∆K, in which ∆K is small when compared to K0. For example, this situation may occur in modeling nonlinear problems using an updated Lagrangian scheme and load incrementation. Given the fact that triangular factors are available for K0, it would appear to be attractive to use an iteration scheme for the perturbed matrix K, in which the initial iterate is L0. The iteration scheme should not involve solving intermediate linear systems except by using current triangular factors. A scheme is introduced in the following section and produces, in a simple example, good estimates within a few iterations. The solution of perturbed linear systems has been the subject of many investigations. Schemes based on explicit matrix inversion include the Sherman-MorrisonWoodbury formulae (see Golub and Van Loan [1996]). An alternate method is to carry bothersome terms to the right side and iterate. For example, the perturbed linear 257
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system can be written as K 0 ∆γ = ∆f − ∆Kγ 0 − ∆K∆γ ,
(20.1)
and an iterative-solution procedure, assuming convergence, can be employed as K 0 ∆γ ( j +1) = ∆f − ∆Kγ 0 − ∆K∆γ ( j ) .
(20.2)
Unfortunately, in a typical nonlinear problem involving incremental loading, especially in systems with decreasing stiffness, it will eventually be necessary to update the triangular factors frequently.
20.1.2 NOTATION
AND
BACKGROUND
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 can be decomposed as A = A l + diag(A) + A u,
(20.3)
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. For later use, we introduce the matrix functions: lower(A) = Al +
1 diag(A), 2
upper(A) = Au +
1 diag(A). 2
(20.4)
Note that: (a) the product of two lower-triangular matrices is also lower-triangular, and (b) the inverse of a nonsingular, lower-triangular matrix is also lower-triangular. Likewise, the product of two upper-triangular matrices is upper-triangular, and the inverse of a nonsingular, upper-triangular matrix is upper-triangular. In proof of (a), (1) (2) th let L and L be two n × n lower-triangular matrices. The ij entry of the product (1) (1) ( 2 ) matrix is given by ∑ k =1,n lik lkj . Since L is lower-triangular, lik(1) vanishes unless k ≤ i. Similarly, lkj( 2 ) vanishes unless k ≥ j. Clearly, all entries of ∑ k =1,n lik(1)lkj( 2 ) vanish (1) (2) unless i ≥ j, which is to say that L L is lower-triangular. In proof of (b), let A denote the inverse of a lower-triangular matrix L. We multiply th T the j column of A by L and set it equal to the vector ej (ej = {0….. 1…..0} with th unity in the j position): now, l11a1 j
=0
l21a1 j + l22 a2 j
=0
l31a1 j + l32 a2 j + l33a3 j
=0
M l j1a1 j + l j 2 a2 j + l j 3a3 j + L + l jj a jj = 1
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(20.5)
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259 −1
−1
Forward substitution establishes that akj = 0, if k < j, and ajj = l jj , thus, A = L is lower-triangular.
20.1.3 ITERATION SCHEME Let K0 denote a symmetric, positive-definite matrix, for which the unique triangular T factors are L0 and L 0. 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 can be written as [K 0 + ∆K] = [L 0 + ∆L][LT0 + ∆LT ].
(20.6)
We can rewrite Equation 20.6 as [I + L−01 ∆L][I + ∆LT L−0T ] = L−01 [K 0 + ∆K]L−0T ,
(20.7)
L−01 ∆L + ∆LT L−0T = L−01 ∆KL−0T − L−01 ∆L∆LT L−0T .
(20.8)
from which
−1
Note that L0 ∆L is lower-triangular. It follows that
(
)
∆L = L 0 lower L−01 ∆KL−0T − L−01 ∆L∆LT L−0T .
(20.9)
The factor of 1/2 in the definition of the lower and upper matrix functions is motivated by the fact that the diagonal entries of L−01 ∆L and ∆LT L−0T are the same. Furthermore, for banded matrices, if ∆L and L0 have the same semibandwidth, b, it follows that, for the correct value of ∆L, L−01 ∆KL−0T − L−01 ∆L∆LT L−0T is also banded, with a bandwidth no greater than b. Unfortunately, it is not yet clear how to take advantage of this behavior. An iteration scheme based on Equation 20.9 is introduced as
(
∆L( j +1) = L 0 lower L−01 ∆KL−0T − L−01 ∆L( j ) ∆L( j ) L−0T
(
∆L(1) = L 0 lower L−01 ∆KL−0T
T
)
)
(20.10)
Explicit formation of the fully populated inverses L−01 and L−0T can be avoided by using forward and backward substitution. In particular, L−01 ∆K = [L−01 ∆k1 L−01 ∆k 2 K L−01 ∆k n ], where ∆k1 is the first column of ∆K. We can now solve for b j = L−01∆k j by solving the system L0 bj = ∆kj.
20.1.4 HEURISTIC CONVERGENCE ARGUMENT For an approximate convergence argument, we use the similar relation ∆ A = ∆K − 2A −1 ( ∆A) ∞ ( ∆A),
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(20.11)
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in which (∆A)∞ is the solution (converged iterate) for ∆A. Consider the iteration scheme ∆A ( j +1) = ∆K − 2A −1 ( ∆A) ∞ ( ∆A)( j ) .
(20.12)
Subtraction of two successive iterates and application of matrix-norm inequalities furnish ∆A ( j +2 ) − ∆A ( j +1) = 2A −1 ∆A ∞ [ ∆A ( j +1) − ∆A ( j ) ].
(20.13)
Convergence is assured in this example if σ (A −1 ∆ A ∝ ) < 12 , in which s denotes the spectral radius (see Dahlquist and Bjork [1974]). An approximate convergence criterion is obtained as max λ j ( ∆A ∝ ) < j
1 min λ (A), 2 k k
(20.14)
in which λj(∆A) denotes the j eigenvalue of the n × n matrix ∆A. Clearly, convergence is expected if the perturbation matrix has a sufficiently small norm. Applied to the current problem, we also expect convergence will occur if max j | λ j ( ∆L∞ ) |< 12 mink | λk (L) |. th
20.1.5 SAMPLE PROBLEM Let L0 and K0 be given by a L0 = b
0 , c
ab . b 2 + c 2
a2 K0 = ab
(20.15)
Now suppose that the matrices are perturbed according to a L= b
0 , c + d
a2 K0 = ab
b 2 + (c + d )2 ab
(20.16)
so that 0 ∆K = 0
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. d (2c + d ) 0
(20.17)
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We are interested in the case in which d/c << 1, for example, d/c = 0.1, ensuring that the perturbation is small. We also use the fact that L−01 =
1 c ac − b
0 . a
(20.18)
The correct answer, which should emerge from the iteration scheme, is 0 ∆L∝ = 0
0 . d
(20.19)
The initial iterate is found from straightforward manipulation as 1 d . d 1+ 2 c
0 ∆L = 0
0
(1)
(20.20)
The ratio of the norms of the error is error =
norm( ∆L(1) ) − norm( ∆L∝ ) norm( ∆L∝ )
1 d d 1 + −d 2 c = d
= 5%
(20.21)
Letting ∆ = d (1 + 12 dc ), the second iterate is found, after straightforward manipulation, as 0 ∆L( 2 ) = 0
0 1 ∆
2
∆−
0 = 0
2 c
0 d 1 −
2
d 2 c
− 1 d3 8 c 3
(20.22)
The relative error is now error =
d2 c2
1 + 1 8
d c
1 = 0.011 + 80 Clearly, this is a significant improvement over the initial iterate.
© 2003 by CRC CRC Press LLC
(20.23)
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20.2 OZAWA’S METHOD FOR INCOMPRESSIBLE MATERIALS In this section, thermal and inertial effects are neglected and the traction is assumed to be prescribed on the undeformed exterior boundary. Using a two-field formulation for an incompressible elastomer leads to an incremental relation, in global form, as follows (see Nicholson, 1995): K MM T − K MP
K MP dγ dfM = = . 0 dψ 0
(20.24)
If KMM is singular, it can be replaced with K′MM = KMM + χKMP K MP , and χ can be chosen to render K′MM positive-definite (see Zienkiewicz, 1989). The presence of zeroes on the diagonal poses computational difficulties, which have received considerable attention. Here, we discuss a modification of the Ozawa method discussed by Zienkiewicz and Taylor (1989). In particular, Equation 20.24 is replaced with the iteration scheme T
K ′MM − K TMP
K MP dγ I / ρa dψ
j +1
dfM = j , dψ / ρa
(20.25)
th
in which the superscript j denotes the j iterate and ra is an acceleration parameter. This scheme converges rapidly for suitable choices of ra. If the assumed pressure fields are discontinuous at the element boundaries, this method can be used at the element level to eliminate pressure variables (see Hughes, 1987). In this event, the global equilibrium equation only involves displacement degrees-of-freedom. For each iteration, it is necessary to solve a linear system. Computation can be expedited using a convenient version of the LU decomposition. Let L1 and L2 denote lower-triangular matrices arising in the following Cholesky decompositions: K′MM = L1L1
T
I/ra + K MPL1L1 KMP = L2L2. T
T
T
(20.26)
Then, a triangularization is attained as K ′MM − K TMP
K MP L1 = I / ρa − K TMP L−1T
0 LT1 L2 0T
L−11K MP . LT2
(20.27)
Forward and backward substitution can now be exploited to solve the linear system arising in the incremental finite-element method.
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20.3 EXERCISES 1. Examine the first two iterates for the matrices a K = b d a K + ∆K = b d a2 = ab ad
0 a 0 0 f 0
0 c e 0 c e+g
d a2 e = ab f ad
b c 0
0 a 0 0 f 0 ab
ab b2 + c2 bd + ce
bd + ce d 2 + e 2 + f 2 ad
d e + g f
b c 0
bd + c(e + g) . d 2 + (e + g) 2 + f 2 ad
b2 + c2 bd + c(e + g)
2. Verify that the product and inverse of lower-triangular matrices are lowertriangular using a L1 = b
0 , c
d L2 = e
0 . f
3. Verify the triangularization scheme in the matrix 2 A = −1 −1
−1 2 1
1 −1. 0
Use the triangular factors to solve the equation 2 Ab = −1. 1
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Monographs and Texts Abramawitz, M. and Stegun, I., Eds, Handbook of Mathematical Functions, with Formulas, Graphs and Mathematical Tables, Dover Publications, Inc., New York, 1997. Belytschko, T., Liu, W.K., and Moran, B., Nonlinear Finite Elements for Continua and Structures, John Wiley and Sons, New York, 2000. Bonet, J. and Wood, R., Nonlinear Continuum Mechanics for Finite Element Analysis, Cambridge University Press, Cambridge, 1997. Brush, D. and Almroth, B., Buckling of Bars, Plates and Shells, McGraw-Hill Book Company, New York, 1975. Callen, H.B., Thermodynamics and an Introduction to Thermostatistics, 2nd ed., John Wiley and Sons, New York, 1985. Chandrasekharaiah, D. and Debnath, L., Continuum Mechanics, Academic Press, San Diego, 1994. Chung, T.J., Continuum Mechanics, Prentice-Halls, Englewood Cliffs, NJ, 1988. Dahlquist, G. and Bjork, A., Numerical Methods, Prentice-Hall, Englewood Cliffs, NJ, 1974. Ellyin, F., Fatigue Damage, Crack Growth and Life Prediction, Chapman & Hall, FL, 1997. Eringen, A.C., Nonlinear Theory of Continuous Media, McGraw-Hill, New York, 1962. Ewing, G.M., Calculus of Variations with Applications, Dover Publications, Mineola, N.Y., 1985. Gent, A.N., Ed, Engineering with Rubber: How to Design Rubber Components, Hanser, New York, 1992. Golub, G. and Van Loan, C., Matrix Computations, Johns Hopkins University Press, Baltimore, MD, 1996. Graham, A., Kronecker Products and Matrix Calculus with Applications, Ellis Horwood, Ltd., Chichester, 1981. Hughes, T., The Finite Element Method: Linear Static and Dynamic Analysis, Dover Publishers Inc., New York, 2000. Kleiber, M., Incremental Finite Element Modeling in Non-Linear Solid Mechanics, Ellis Horwood, Ltd., Chichester, 1989. Oden, J.T., Finite Elements of Nonlinear Continua, McGraw-Hill Book Company, New York, 1972. Schey, H.M., DIV, GRAD, CURL and All That, Norton, New York, 1973. Thomason, P.F., Ductile Fracture of Metals, Pergamon Press, Oxford, 1990. Wang, C-T., Applied Elasticity, The Maple Press Company, York, PA, 1953. Zienkiewicz, O.C. and Taylor, R.L., The Finite Element Method. Volume 2, 4th ed, McGrawHill Book Company, London, 1989.
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Articles and Other Sources ANSYS User Manual, ver 6.0, Swanson Analysis Systems, 2000. Blatz, P.J. and Ko, W.L., Application of finite elastic theory to the deformation of rubbery materials, Trans. Soc. Rheol., 6, 223, 1962. Bonora, N., A nonlinear CDM model for ductile failure, Engineering Fracture Mechanics, 58, 1/2, 11, 1997. Chen, J., Wan, W., Wu, C.T., and Duan, W., On the perturbed Lagrangian formulation for nearly incompressible and incompressible hyperelasticity, Comp. Methods Appl. Mech. Eng., 142, 335, 1997. Dillon, O.W., A nonlinear thermoelasticity theory, J. Mech. Phys. Solids, 10, 123, 1962. Green, A. and Naghdi, P., A general theory of an elastic-plastic continuum, Arch. Rat. Mech. Anal., 18/19, 1965. Gurson, A.L., Continuum theory of ductile rupture by void nucleation and growth, part 1: yield criteria and flow rules for porous ductile media, J. Eng. Mat’s Technol., 99, 2, 1977. Holzappel, G., On large strain viscoelasticity: continuum applications and finite element applications to elastomeric structures, Int. J. Numer. Meth. Eng., 39, 3903, 1996. Holzappel, G. and Simo, J., Entropy elasticity of isotropic rubber-like solids at finite strain, Comp. Methods Appl. Mech. Eng., 132, 17, 1996. LS-DYNA, ver. 95, Livermore Software Technology Corporation, Livermore, CA, 2000. Moraes, R. and Nicholson D.W., Local damage criterion for ductile fracture with application to welds under dynamic loads, in Advances in Fracture and Damage Mechanics, Guagliano, M. and Aliabadi, M.H., Eds., 2nd ed. Hoggar Press, Geneva, 277, 2002. Nicholson, D.W. and Nelson, N., Finite element analysis in design with rubber (Rubber Reviews), Rubber Chem. Tech., 63, 638, 1990. Nicholson, D.W., Tangent modulus matrix for finite element analysis of hyperelastic materials, Acta Mech., 112, 187, 1995. Nicholson, D.W. and Lin, B., Theory of thermohyperelasticity for near-incompressible elastomers, Acta Mech., 116, 15, 1996. Nicholson, D.W. and Lin, B., Finite element method for thermomechanical response of nearincompressible elastomers, Acta Mech., 124, 181, 1997a. Nicholson, D.W. and Lin, B., Incremental finite element equations for thermomechanical of elastomers: effect of boundary conditions including contact, Acta Mech., 128, 1–2, 81, 1997b. Nicholson, D.W. and Lin, B., On the tangent modulus tensor in hyperelasticty, Acta Mech., 131, 1997c. Nicholson, D.W., Nelson, N., Lin, B., and Farinella, A., Finite element analysis of hyperelastic components, Appl. Mech. Rev., 51, 5, 303, 1999. Ogden, R.W., Recent advances in the phenomenological theory of rubber elasticity, Rubber Chem. Tech., 59, 361, 1986. Perzyna, P., Thermodynamic theory of viscoplasticity, Adv. Appl. Mech., 11, 1971. Valanis, K. and Landel, R.F., The strain energy function of a hyperelastic material in terms of the extension ratios, J. Appl. Physics, 38, 2997, 1967.
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Tvergaard, V., Influence of voids on shear band instabilities under plane strain conditions, Int. J. Fracture, 17, 389–407, 1981. Ziegler, H. and Wehrli, C., The derivation of constitutive relations from the free energy and the dissipation function, Adv. Appl. Mech., 25, 187, 1987.
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