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Finite Element Analysis Thermomechanics of Solids
David W...

Author:
David W. Nicholson

0749_Frame_FM Page iii Wednesday, February 19, 2003 4:53 PM

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

© 2003 by CRC CRC Press LLC

(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

© 2003 by CRC CRC Press LLC

(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 =

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

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 ′ ) =

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

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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Mathematical Foundations: Tensors

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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Mathematical Foundations: Tensors

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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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

© 2003 by CRC CRC Press LLC

−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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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 .

© 2003 by CRC CRC Press LLC

(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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57

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

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

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 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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Finite Element Analysis: Thermomechanics of Solids

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,

© 2003 by CRC CRC Press LLC

γ (0) = γ 0 ,

γ˙ (0) = γ˙ 0 ,

(11.15)

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157

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.

© 2003 by CRC CRC Press LLC

(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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159

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.

© 2003 by CRC CRC Press LLC

(11.37)

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161

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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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 ,

© 2003 by CRC CRC Press LLC

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

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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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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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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 ,

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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

© 2003 by CRC CRC Press LLC

cosκL − κL γ 0 = . cosκL δ 0

(14.32)

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Finite Element Analysis: Thermomechanics of Solids

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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Finite Element Analysis: Thermomechanics of Solids

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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193

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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Finite Element Analysis: Thermomechanics of Solids

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

© 2003 by CRC CRC Press LLC

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 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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Introduction to Contact Problems

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.

© 2003 by CRC CRC Press LLC

t3

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Finite Element Analysis: Thermomechanics of Solids

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.

(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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229

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

© 2003 by CRC CRC Press LLC

− 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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233

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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245

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 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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251

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

,

= 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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Inelastic and Thermoinelastic Materials

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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20

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),

© 2003 by CRC CRC Press LLC

(20.11)

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Finite Element Analysis: Thermomechanics of Solids

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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Advanced Numerical Methods

261

We are interested in the case in which d/c

Finite Element Analysis Thermomechanics of Solids

David W. Nicholson

CRC PR E S S Boca Raton London New York Washington, D.C.

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0749_Frame_FM Page iv Wednesday, February 19, 2003 4:53 PM

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

© 2003 by CRC CRC Press LLC

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)

© 2003 by CRC CRC Press LLC

v ⋅ w → vTw

(1.13)

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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

© 2003 by CRC CRC Press LLC

∇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

© 2003 by CRC CRC Press LLC

(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

© 2003 by CRC CRC Press LLC

(1.42)

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Mathematical Foundations: Vectors and Matrices

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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Mathematical Foundations: Vectors and Matrices

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 =

© 2003 by CRC CRC Press LLC

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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Mathematical Foundations: Vectors and Matrices

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

© 2003 by CRC CRC Press LLC

cos θ − sin θ 0

0 dθ 0 dt 1

(1.67)

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14

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

© 2003 by CRC CRC Press LLC

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

=

© 2003 by CRC CRC Press LLC

∂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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Mathematical Foundations: Vectors and Matrices

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 ′ ) =

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

cαjβ =

α 1 (1 − δ αβ ) hα

j

β hβ

(a.4)

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Mathematical Foundations: Vectors and Matrices

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

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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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20

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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Mathematical Foundations: Vectors and Matrices

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

© 2003 by CRC CRC Press LLC

sin θ cos θ

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22

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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Mathematical Foundations: Vectors and Matrices

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.

© 2003 by CRC CRC Press LLC

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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

© 2003 by CRC CRC Press LLC

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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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Mathematical Foundations: Tensors

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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Mathematical Foundations: Tensors

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.

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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).

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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

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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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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.

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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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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.

© 2003 by CRC CRC Press LLC

(2.72)

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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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(2.83)

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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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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.

© 2003 by CRC CRC Press LLC

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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Introduction to Variational and Numerical Methods

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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Introduction to Variational and Numerical Methods

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 .

© 2003 by CRC CRC Press LLC

(4.13)

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Kinematics of Deformation

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

© 2003 by CRC CRC Press LLC

(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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Kinematics of Deformation

57

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

© 2003 by CRC CRC Press LLC

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

© 2003 by CRC CRC Press LLC

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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Kinematics of Deformation

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 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

© 2003 by CRC CRC Press LLC

0 1 l22 0

λ 2 1 = l22 12

, 1 3 1 2

(11.6)

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Solution Methods for Linear Problems

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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Finite Element Analysis: Thermomechanics of Solids

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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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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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 ]

© 2003 by CRC CRC Press LLC

(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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Finite Element Analysis: Thermomechanics of Solids

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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Thermal, Thermoelastic, and Incompressible Media

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 ,

© 2003 by CRC CRC Press LLC

γ (t ), f (t ): nm × 1,

Σ: nm × nt ,

θ (t ): nt × 1.

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Finite Element Analysis: Thermomechanics of Solids

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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Finite Element Analysis: Thermomechanics of Solids

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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Thermal, Thermoelastic, and Incompressible Media

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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Torsion and Buckling

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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Torsion and Buckling

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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Finite Element Analysis: Thermomechanics of Solids

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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Torsion and Buckling

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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Finite Element Analysis: Thermomechanics of Solids

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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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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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 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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Finite Element Analysis: Thermomechanics of Solids

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.

(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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(17.33)

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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 =

© 2003 by CRC CRC Press LLC

1 (I ⊗ I + I ⊗ IU)M(X) 2

(

)

1 I ⊗ FuT + Fu ⊗ IU M(X) 2 ∂u . ∂X

(17.38)

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Finite Element Analysis: Thermomechanics of Solids

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 .

© 2003 by CRC CRC Press LLC

(17.43)

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225

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

© 2003 by CRC CRC Press LLC

(17.47)

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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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18

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 =

© 2003 by CRC CRC Press LLC

∂ξ . ∂J T

(18.7)

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229

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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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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Finite Element Analysis: Thermomechanics of Solids

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 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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251

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

,

= 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

© 2003 by CRC CRC Press LLC

(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),

© 2003 by CRC CRC Press LLC

(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

© 2003 by CRC CRC Press LLC

. d (2c + d ) 0

(20.17)

0749_Frame_C20 Page 261 Wednesday, February 19, 2003 5:35 PM

Advanced Numerical Methods

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We are interested in the case in which d/c

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