finite element aanalysis er. gourab's notes on engineering

CRC PR ESSBoca Raton London New York Washington, D.C.

FiniteElementAnalysisThermomechanics of Solids

David W. Nicholson

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

© 2003 by CRC CRC Press LLC

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No claim to original U.S. Government worksInternational Standard Book Number 0-8493-0749-X

Library of Congress Card Number 2002041419Printed in the United States of America 1 2 3 4 5 6 7 8 9 0

Printed on acid-free paper

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 2003620.1

′

121—dc21 2002041419

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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 thermome-chanical and nonlinear applications. Most academic departments offering advanceddegrees in mechanical engineering, civil engineering, and aerospace engineering offera

first-level course

in the finite-element method, and by now, almost all undergraduatesof such programs have some exposure to the finite-element method. A number ofdepartments offer a second-level course. It is hoped that this text will appeal toinstructors of such courses. Of course, it hopefully will also be helpful to engineersengaged in self-study on nonlinear and thermomechanical finite-element analysis.

The principles of the finite-element method are presented for application to themechanical, thermal, and thermomechanical response, both static and dynamic, oflinear and nonlinear solids. It provides an integrated treatment of:

• Basic principles, material models, and contact models (for example, linearelasticity, hyperelasticity, and thermohyperelasticity).

• Computational, numerical, and software-design aspects (such as finite-element data structures).

• Modeling principles and strategies (including mesh design).

The text is designed

for a second-level course, as a reference work, or for selfstudy

. Familiarity is assumed with the finite-element method at the level of a first-level graduate or advanced undergraduate course.

A first-level course in the finite-element method, for which many excellent booksare 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 second-level course. Nor is there much emphasis on

coupled

thermomechanical problems.However, many engineers could benefit from a text covering nonlinear problemsand the associated continuum thermomechanics. Such a text could be used in aformal class or for self-study. Many important applications have significant nonlin-earity, making nonlinear finite-element modeling necessary. As a few examples, wemention polymer processing; metal forming; rubber components, such as tires andseals; biomechanics; and crashworthiness. Many applications combine thermal andmechanical response, such as rubber seals in hot engines. Engineers coping withsuch applications have access to powerful finite-element codes and computers.However, they often lack and urgently need an in-depth but compact exposition ofthe finite-element method, which provides a foundation for addressing future prob-lems. It is hoped this text also fills this need.

Of necessity, a selection of topics has been made, and topics are given coverageproportional 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 stillcompact 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 towhich the author is indebted are:

1. Zienkiewicz, O.C. and Taylor R.L.,

The Finite Element Method

, Vols. 1and 2, McGraw Hill, London, 1989.

2. Kleiber, M.,

Incremental Finite Element Modeling in Nonlinear SolidMechanics

, Chichester, Ellis Horwood, Ltd., 1989.3. Bonet, J. and Wood, R.D.,

Nonlinear Continuum Mechanics for FiniteElement Analysis

, Cambridge, Cambridge University Press, 1997.4. Belytschko, T., Lui, W.K, and Moran, B.,

Nonlinear Finite Elements forContinua 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

withthermal effects.

4. Inclusion of nonlinear boundary conditions, including contact, in an inte-grated

incremental variational

formulation.

Kronecker Product algebra (KPA) has been widely used in control theory for manyyears (Graham, 1982). It is highly compact and satisfies simple rules: for example,the inverse of a Kronecker Product of two nonsingular matrices is the KroneckerProduct of the inverses. Recently, a number of extensions of KPA have been intro-duced and shown to permit compact expressions for otherwise elaborate quantitiesin continuum and computational mechanics. Examples include:

1. Compact expressions for the tangent-modulus tensors in hyperelasticity(invariant-based and stretch-based; compressible, incompressible, and near-incompressible), thermohyperelasticity, and finite-strain plasticity.

2. A general, compact expression for the tangent stiffness matrix in nonlinearFEA, including nonlinear boundary conditions, such as contact.

KPA with recent extensions can completely replace other notations in most casesof interest here. In the author’s experience, students experience little difficulty ingaining a command of it.

The first three chapters concern mathematical foundations, and Kronecker Productnotation for tensors is introduced. The next four chapters cover relevant linear andnonlinear continuum thermomechanics to enable a unified account of the finite-elementmethod. Chapters 8 through 15 represent a compact presentation of the finite-elementmethod in linear elastic, thermal, and thermomechanical media, including solutionmethods. The final five chapters address nonlinear problems based on a unified set ofincremental variational principles. Material nonlinearity is treated also, as is geometricnonlinearity and nonlinearity due to boundary conditions. Several numerical issues innonlinear 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 manyyears of teaching and research, made possible through the love and patience of mywife Linda and son Michael, to whom this is dedicated. To my profound love, Icheerfully add my immense gratitude. Much of my understanding of the advancedtopics addressed in this book arose through the highly successful research of mythen-doctoral student, Dr. Baojiu Lin. He continued his invaluable support by pro-viding comments and corrections on the manuscript. Also deserving of acknowledg-ment is Ashok Balasubranamian, whose careful attention to the manuscript hasgreatly helped to make it achieve my objectives. Thanks are due to hundreds ofgraduate students in my courses in continuum mechanics and finite elements overthe years. I tested and refined the materials through the courses and benefitedimmensely 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 especiallyfor 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 Depart-ment of Mechanical, Materials, and Aerospace Engineer-ing at the University of Central Florida, Orlando, Florida,where he has served as the Chair (1990–1995) and InterimChair (2000–2002). He earned an S.B. degree at MIT in1966 and a doctorate at Yale in 1971. He has 31 years ofexperience since the doctorate, including research inindustry (Goodyear, 7 years) and government (the NavalSurface Weapons Center, 6 years), and faculty positionsat Stevens Institute of Technology and UCF. He hasauthored over 140 articles on the finite-element method,continuum mechanics, fracture mechanics, and dynamics.He has served as a technical editor for

Applied MechanicsReview

,

as an associate editor of

Tire Science and Technology

,

and as a member ofscientific advisory committees for a number of international conferences. He hasbeen the instructor in over 20 short courses and workshops. Recently, he served asExecutive Chair of the XXIst

Southeastern Conference on Theoretical and AppliedMechanics and as co-editor of

Developments in Theoretical and Applied Mechanics

,

Volume XXI

.

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Table of Contents

Chapter 1

Mathematical Foundations: Vectors and Matrices...............................1

1.1 Introduction....................................................................................................11.1.1 Range and Summation Convention ................................................11.1.2 Substitution Operator ......................................................................1

1.2 Vectors............................................................................................................21.2.1 Notation ...........................................................................................21.2.2 Gradient, Divergence, and Curl ......................................................4

1.3 Matrices..........................................................................................................51.3.1 Eigenvalues and Eigenvectors.........................................................81.3.2 Coordinate Transformations............................................................91.3.3 Transformations of Vectors .............................................................91.3.4 Orthogonal Curvilinear Coordinates.............................................111.3.5 Gradient Operator..........................................................................161.3.6 Divergence and Curl of Vectors ....................................................17Appendix I: Divergence and Curl of Vectors in Orthogonal Curvilinear Coordinates................................................................................18

Derivatives of Base Vectors ..........................................................18Divergence.....................................................................................19Curl ................................................................................................20

1.4 Exercises ......................................................................................................20

Chapter 2

Mathematical Foundations: Tensors ..................................................25

2.1 Tensors .........................................................................................................252.2 Divergence, Curl, and Laplacian of a Tensor .............................................27

2.2.1 Divergence.....................................................................................272.2.2 Curl and Laplacian ........................................................................28

2.3 Invariants......................................................................................................292.4 Positive Definiteness....................................................................................302.5 Polar Decomposition Theorem....................................................................312.6 Kronecker Products on Tensors...................................................................32

2.6.1

VEC

Operator and the Kronecker Product ...................................322.6.2 Fundamental Relations for Kronecker Products...........................332.6.3 Eigenstructures of Kronecker Products ........................................352.6.4 Kronecker Form of Quadratic Products........................................362.6.5 Kronecker Product Operators for Fourth-Order Tensors..............362.6.6 Transformation Properties of

VEC

and

TEN

22 ............................372.6.7 Kronecker Product Functions for Tensor Outer Products ............38

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2.6.8 Kronecker Expressions for Symmetry Classes in Fourth-Order Tensors ................................................................40

2.6.9 Differentials of Tensor Invariants .................................................412.7 Exercises ......................................................................................................42

Chapter 3

Introduction to Variational and Numerical Methods.........................43

3.1 Introduction to Variational Methods............................................................433.2 Newton Iteration and Arc-Length Methods ................................................47

3.2.1 Newton Iteration............................................................................473.2.2 Critical Points and the Arc-Length Method .................................48

3.3 Exercises ......................................................................................................49

Chapter 4

Kinematics of Deformation ...............................................................51

4.1 Kinematics ...................................................................................................514.1.1 Displacement .................................................................................514.1.2 Displacement Vector......................................................................524.1.3 Deformation Gradient Tensor .......................................................52

4.2 Strain ............................................................................................................534.2.1 F, E, E

L

and u in Orthogonal Coordinates....................................534.2.2 Velocity-Gradient Tensor, Deformation-Rate Tensor,

and Spin Tensor.............................................................................564.3 Differential Volume Element .......................................................................604.4 Differential Surface Element .......................................................................614.5 Rotation Tensor............................................................................................634.6 Compatibility Conditions For

E

L

and D .....................................................644.7 Sample Problems .........................................................................................674.8 Exercises ......................................................................................................69

Chapter 5

Mechanical Equilibrium and the Principle of Virtual Work .............73

5.1 Traction and Stress ......................................................................................735.1.1 Cauchy Stress ................................................................................735.1.2 1st Piola-Kirchhoff Stress .............................................................755.1.3 2nd Piola-Kirchhoff Stress............................................................76

5.2 Stress Flux ...................................................................................................775.3 Balance of Mass, Linear Momentum, and Angular Momentum................79

5.3.1 Balance of Mass ............................................................................795.3.2 Rayleigh Transport Theorem ........................................................795.3.3 Balance of Linear Momentum ......................................................795.3.4 Balance of Angular Momentum....................................................80

5.4 Principle of Virtual Work ............................................................................825.5 Sample Problems .........................................................................................855.6 Exercises ......................................................................................................89

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Chapter 6

Stress-Strain Relation and the Tangent-Modulus Tensor ..................95

6.1 Stress-Strain Behavior: Classical Linear Elasticity ....................................956.2 Isothermal Tangent-Modulus Tensor...........................................................97

6.2.1 Classical Elasticity ........................................................................976.2.2 Compressible Hyperelastic Materials ...........................................97

6.3 Incompressible and Near-Incompressible Hyperelastic Materials..............996.3.1 Incompressibility ...........................................................................996.3.2 Near-Incompressibility ................................................................102

6.4 Nonlinear Materials at Large Deformation...............................................1036.5 Exercises ....................................................................................................104

Chapter 7

Thermal and Thermomechanical Response.....................................107

7.1 Balance of Energy and Production of Entropy.........................................1077.1.1 Balance of Energy.......................................................................1077.1.2 Entropy Production Inequality ....................................................1087.1.3 Thermodynamic Potentials..........................................................109

7.2 Classical Coupled Linear Thermoelasticity ..............................................1107.3 Thermal and Thermomechanical Analogs of the Principle

of Virtual Work ..........................................................................................1137.3.1 Conductive Heat Transfer ...........................................................1137.3.2 Coupled Linear Isotropic Thermoelasticity ................................114

7.4 Exercises ....................................................................................................116

Chapter 8

Introduction to the Finite-Element Method.....................................117

8.1 Introduction................................................................................................1178.2 Overview of the Finite-Element Method ..................................................1178.3 Mesh Development ....................................................................................118

Chapter 9

Element Fields in Linear Problems .................................................121

9.1 Interpolation Models..................................................................................1219.1.1 One-Dimensional Members ........................................................1219.1.2 Interpolation Models: Two Dimensions......................................1249.1.3 Interpolation Models: Three Dimensions ...................................127

9.2 Strain-Displacement Relations and Thermal Analogs ..............................1289.2.1 Strain-Displacement Relations: One Dimension ........................1289.2.2 Strain-Displacement Relations: Two Dimensions ......................1299.2.3 Axisymmetric Element on Axis of Revolution ..........................1309.2.4 Thermal Analog in Two Dimensions..........................................1319.2.5 Three-Dimensional Elements......................................................1319.2.6 Thermal Analog in Three Dimensions........................................132

9.3 Stress-Strain-Temperature Relations in Linear Thermoelasticity .............1329.3.1 Overview .....................................................................................1329.3.2 One-Dimensional Members ........................................................1329.3.3 Two-Dimensional Elements ........................................................133

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9.3.4 Element for Plate with Membrane and Bending Response .......1359.3.5 Axisymmetric Element................................................................1359.3.6 Three-Dimensional Element .......................................................1369.3.7 Elements for Conductive Heat Transfer .....................................137

9.4 Exercises ....................................................................................................137

Chapter 10

Element and Global Stiffness and Mass Matrices ..........................139

10.1 Application of the Principle of Virtual Work............................................13910.2 Thermal Counterpart of the Principle of Virtual Work.............................14110.3 Assemblage and Imposition of Constraints ..............................................142

10.3.1 Rods.............................................................................................14210.3.2 Beams ..........................................................................................14610.3.3 Two-Dimensional Elements ........................................................147

10.3 Exercises ....................................................................................................149

Chapter 11

Solution Methods for Linear Problems ...........................................153

11.1 Numerical Methods in FEA ......................................................................15311.1.1 Solving the Finite-Element Equations: Static Problems ............15311.1.2 Matrix Triangularization and Solution of Linear Systems.........15411.1.3 Triangularization of Asymmetric Matrices.................................155

11.2 Time Integration: Stability and Accuracy .................................................15611.3 Newmark’s Method ...................................................................................15711.4 Integral Evaluation by Gaussian Quadrature ............................................15811.5 Modal Analysis by FEA............................................................................159

11.5.1 Modal Decomposition .................................................................15911.5.2 Computation of Eigenvectors and Eigenvalues ..........................162

11.6 Exercises ....................................................................................................164

Chapter 12

Rotating and Unrestrained Elastic Bodies.......................................167

12.1 Finite Elements in Rotation.......................................................................16712.2 Finite-Element Analysis for Unconstrained Elastic Bodies......................16912.3 Exercises ....................................................................................................171

Chapter 13

Thermal, Thermoelastic, and Incompressible Media ......................173

13.1 Transient Conductive-Heat Transfer..........................................................17313.1.1 Finite-Element Equation .............................................................17313.1.2 Direct Integration by the Trapezoidal Rule ................................17313.1.3 Modal Analysis............................................................................174

13.2 Coupled Linear Thermoelasticity ..............................................................17513.2.1 Finite-Element Equation .............................................................17513.2.2 Thermoelasticity in a Rod...........................................................177

13.3 Compressible Elastic Media ......................................................................17713.4 Incompressible Elastic Media ...................................................................17813.5 Exercises ....................................................................................................180

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Chapter 14

Torsion and Buckling.......................................................................181

14.1 Torsion of Prismatic Bars..........................................................................18114.2 Buckling of Beams and Plates ..................................................................185

14.2.1 Euler Buckling of Beam Columns..............................................18514.2.2 Euler Buckling of Plates .............................................................190

14.3 Exercises ....................................................................................................193

Chapter 15

Introduction to Contact Problems....................................................195

15.1 Introduction: the Gap.................................................................................19515.2 Point-to-Point Contact ...............................................................................19715.3 Point-to-Surface Contact ...........................................................................19915.4 Exercises ....................................................................................................199

Chapter 16

Introduction to Nonlinear FEA........................................................201

16.1 Overview....................................................................................................20116.2 Types of Nonlinearity ................................................................................20116.3 Combined Incremental and Iterative Methods: a Simple Example..........20216.4 Finite Stretching of a Rubber Rod under Gravity: a Simple Example ....203

16.4.1 Nonlinear Strain-Displacement Relations...................................20316.4.2 Stress and Tangent Modulus Relations.......................................20416.4.3 Incremental Equilibrium Relation...............................................20516.4.4 Numerical Solution by Newton Iteration....................................208

16.5 Illustration of Newton Iteration.................................................................21116.5.1 Example.......................................................................................212

16.6 Exercises ....................................................................................................213

Chapter 17

Incremental Principle of Virtual Work ............................................215

17.1 Incremental Kinematics .............................................................................21517.2 Incremental Stresses ..................................................................................21617.3 Incremental Equilibrium Equation ............................................................21717.4 Incremental Principle of Virtual Work ......................................................21817.5 Incremental Finite-Element Equation .......................................................21917.6 Incremental Contributions from Nonlinear Boundary Conditions ...........22017.7 Effect of Variable Contact .........................................................................22117.8 Interpretation as Newton Iteration.............................................................22317.9 Buckling.....................................................................................................22417.10 Exercises ....................................................................................................226

Chapter 18

Tangent-Modulus Tensors for Thermomechanical Response of Elastomers ...................................................................227

18.1 Introduction................................................................................................22718.2 Compressible Elastomers ..........................................................................22718.3 Incompressible and Near-Incompressible Elastomers ..............................228

18.3.1 Specific Expressions for the Helmholtz Potential ......................230

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18.4 Stretch Ratio-Based Models: Isothermal Conditions................................23118.5 Extension to Thermohyperelastic Materials..............................................23318.6 Thermomechanics of Damped Elastomers................................................234

18.6.1 Balance of Energy.......................................................................23518.6.2 Entropy Production Inequality ....................................................23518.6.3 Dissipation Potential ...................................................................23618.6.4 Thermal-Field Equation for Damped Elastomers.......................237

18.7 Constitutive Model: Potential Functions...................................................23818.7.1 Helmholtz Free-Energy Density .................................................23818.7.2 Specific Dissipation Potential .....................................................239

18.8 Variational Principles.................................................................................24018.8.1 Mechanical Equilibrium..............................................................24018.8.2 Thermal Equilibrium...................................................................240

18.9 Exercises ....................................................................................................241

Chapter 19

Inelastic and Thermoinelastic Materials..........................................243

19.1 Plasticity.....................................................................................................24319.1.1 Kinematics...................................................................................24319.1.2 Plasticity ......................................................................................243

19.2 Thermoplasticity ........................................................................................24619.2.1 Balance of Energy.......................................................................24619.2.2 Entropy-Production Inequality ....................................................24719.2.3 Dissipation Potential ...................................................................248

19.3 Thermoinelastic Tangent-Modulus Tensor ................................................24919.3.1 Example.......................................................................................250

19.4 Tangent-Modulus Tensor in Viscoplasticity ..............................................25219.5 Continuum Damage Mechanics ................................................................25419.6 Exercises ....................................................................................................256

Chapter 20

Advanced Numerical Methods ........................................................257

20.1 Iterative Triangularization of Perturbed Matrices .....................................25720.1.1 Introduction .................................................................................25720.1.2 Notation and Background ...........................................................25820.1.3 Iteration Scheme..........................................................................25920.1.4 Heuristic Convergence Argument ...............................................25920.1.5 Sample Problem ..........................................................................260

20.2 Ozawa’s Method for Incompressible Materials ........................................26220.3 Exercises ....................................................................................................263

Monographs and Texts

........................................................................................265

Articles and Other Sources

.................................................................................267

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


1

Mathematical Foundations: Vectorsand Matrices

1.1 INTRODUCTION

This chapter provides an overview of mathematical relations, which will prove usefulin the subsequent chapters. Chandrashekharaiah and Debnath (1994) provide a morecomplete discussion of the concepts introduced here.

1.1.1 R

ANGE

AND

S

UMMATION

C

ONVENTION

Unless otherwise noted, repeated Latin indices imply summation over the range1 to 3. For example:

(1.1)

(1.2)

The repeated index is “summed out” and, therefore, dummy. The quantity

a

ij

b

jk

inEquation (1.2) has two free indices,

i

and

k

(and later will be shown to be the

ik

th

entry of a second-order tensor). Note that Greek indices do not imply summation.Thus,

a

α

b

α

=

a

1

b

1

if

α

=

1.

1.1.2 S

UBSTITUTION

O

PERATOR

The quantity,

δ

ij

, later to be called the Kronecker tensor, has the property that

(1.3)

For example,

δ

ij

v

j

=

1

×

v

i

, thus illustrating the substitution property.

1

a b a b a b a b a bi i i i

i

= = + +=

∑1

3

1 1 2 2 3 3

a b a b a b a bij jk i k i k i k= + +1 1 2 2 3 3

δ ij

i j

i j=

=≠

1

0

0749_Frame_C01 Page 1 Wednesday, February 19, 2003 4:55 PM


2


1.2 VECTORS

1.2.1 N

OTATION

Throughout this and the following chapters, orthogonal coordinate systems will beused. Figure 1.1 shows such a system, with base vectors

e

1

,

e

2

, and

e

3

. The scalarproduct of vector analysis satisfies

(1.4)

The vector product satisfies

(1.5)

It is an obvious step to introduce the alternating operator,

ε

ijk

, also known as the

ijk

th

entry of the permutation tensor:

(1.6)

FIGURE 1.1

Rectilinear coordinate system.

3

2

1v1

v2

v

e1

e2

e3

v3

e ei j ij⋅ = δ

e e

e

e

0

i j

k

k

i j ijk

i j ijk

i j

× =

≠

− ≠

=

and in right-handed order

and not in right-handed order

ε ijk i j k

ijk

ijk

ijk

= × ⋅

= −

[ ]e e e

1

1

0

distinct and in right-handed order

distinct but not in right-handed order

not distinct



Mathematical Foundations: Vectors and Matrices

3

Consider two vectors,

v

and

w

. It is convenient to use two different types ofnotation. In

tensor indicial notation

, denoted by (*T),

v

and

w

are represented as

*T) (1.7)

Occasionally, base vectors are not displayed, so that

v

is denoted by

v

i

. Bydisplaying base vectors, tensor indicial notation is explicit and minimizes confusionand ambiguity. However, it is also cumbersome.

In this text, the “default” is

matrix-vector

(*M) notation, illustrated by

*M) (1.8)

It is compact, but also risks confusion by not displaying the underlying basevectors. In *M notation, the transposes

v

T

and

w

T

are also introduced; they aredisplayed as “row vectors”:

*M) (1.9)

The scalar product of

v

and

w

is written as

*T)

(1.10)

The magnitude of

v

is defined by

*T) (1.11)

The scalar product of

v

and

w

satisfies

*T) (1.12)

in which

θ

vw

is the angle between the vectors

v

and

w

. The scalar, or dot, product is

*M) (1.13)

v e w e= =v wi i i i

v w=

=

v

v

v

w

w

w

1

2

3

1

2

3

v wT T= = v v v w w w1 2 3 1 2 3

v w e e

e e

⋅ = ⋅

= ⋅

=

=

( ) ( )v w

v w

v w

v w

i i j j

i j i j

i j ij

i i

δ

v v v= ⋅

v w v w⋅ = cosθvw

v w v wT⋅ →



4


The vector, or cross, product is written as

*T)(1.14)

Additional results on vector notation are presented in the next section, whichintroduces matrix notation. Finally, the vector product satisfies

*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

1.2.2 G

RADIENT

, D

IVERGENCE

,

AND

C

URL

The derivative,

d

φ

/

dx

, of a scalar

φ

with respect to a vector

x

is defined implicitly by

*M) (1.16)

and it is a row vector whose

i

th

entry is

d

φ

/

dx

i

. In three-dimensional rectangularcoordinates, the gradient and divergence operators are defined by

*M) (1.17)

and clearly,

*M) (1.18)

The gradient of a scalar function

φ

satisfies the following integral relation:

(1.19)

The expression will be seen to be a tensor (see Chapter 2). Clearly,

(1.20)

v w e e

e

× = ×

=

v w

v w

i j i j

ijk i j kε

v w v w× = sinθvw

12 v w× .

ddφ φ=d

dx

x

∇ =

∂∂∂∂∂

∂

( )

( )

( )

( )

x

y

z

d

dx

T

= ∇( ) ( )

∇ = ∫∫ φ φdV dSn

∇vT

∇ = ∇ ∇ ∇vT [ ]v v v1 2 3



Mathematical Foundations: Vectors and Matrices

5

from which we obtain the integral relation

(1.21)

Another important relation is the divergence theorem. Let

V

denote the volumeof 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 withinthe body. The divergence of

v

satisfies

*M) (1.22)

The curl of vector

v

,

∇

×

v

, is expressed by

(1.23)

which is the conventional cross-product, except that the divergence operator replacesthe first vector. The curl satisfies the curl theorem, analogous to the divergencetheorem (Schey, 1973):

(1.24)

Finally, the reader may verify, with some effort that, for a vector

v

(

X

) and apath

X

(

S

) in which

S

is the length along the path,

. (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 isarbitrary, thus giving the condition for

v

to have a path-independent integral as

. (1.26)

1.3 MATRICES

An

n

×

n

matrix is simply an array of numbers arranged in rows and columns, alsoknown as a second-order array. For the matrix

A

, the entry

a

ij

occupies the intersectionof the

i

th

row and the

j

th

column. We may also introduce the

n

×

1 first-order array a,in which ai denotes the ith entry. We likewise refer to the 1 × n array, aT, as first-order.

∇ = ∫∫ v nvT TdV dS

d

ddV dS

vx

n vT= ∫∫

( )∇ × = ∂∂

v i ijkj

kxvε

∇ × = ×∫ ∫v n vdV dS

v X n v⋅ = ⋅ ∇ ×∫ ∫d S dS( )

∇ × =v 0



6 Finite Element Analysis: Thermomechanics of Solids

In the current context, a first-order array is not a vector unless it is associated witha coordinate system and certain transformation properties, to be introduced shortly.In the following, all matrices are real unless otherwise noted. Several properties offirst- 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.The transpose of a matrix, A, denoted AT, is a matrix in which A is

called symmetric if A = AT, and it is called antisymmetric if A = −AT.The product of two matrices, A and B, is the matrix, C, for which

*T) (1.27)

Consider the following to visualize matrix multiplication. Let the first-orderarray denote the ith row of A, while the first-order array bj denotes the jth columnof B. Then cij can be written as

*T) (1.28)

The product of a matrix A and a first-order array c is the first-order array din which the ith entry is di = aijcj.

The ijth entry of the identity matrix I is δij. Thus, it exhibits ones on the diagonalpositions (i = j) and zeroes off-diagonal (i ≠ j). Thus, I is the matrixcounterpart of the substitution operator.

The determinant of A is given by

*T) (1.29)

Suppose a and b are two non-zero, first-order n × 1 arrays. If det(A) = 0, thematrix A is singular, in which case there is no solution to equations of the formAa = b. However, if b = 0, there may be multiple solutions. If det(A) ≠ 0, then thereis a unique, nontrivial solution a.

Let A and B be n × n nonsingular matrices. The determinant has the followinguseful properties:

*M) (1.30)

a aij jiT =

c a bij ik kj=

a iT

cij i j= a bT

det( )A = 16

ε εijk pqr ip jq kra a a

det( ) det( )det( )

det( ) det( )

det( )

AB A B

A AT

=

=

=I 1



Mathematical Foundations: Vectors and Matrices 7

If det(A) ≠ 0, then A is nonsingular and there exists an inverse matrix, A−1,for which

*M) (1.31)

The transpose of a matrix product satisfies

*M) (1.32)

The inverse of a matrix product satisfies

*M) (1.33)

If c and d are two 3 × 1 vectors, the vector product c × d generates the vectorc × d = Cd, in which C is an antisymmetric matrix given by

*M) (1.34)

Recalling that c × d = εikjckdj, and noting that εikjck denotes the (ij)th componentof an antisymmetric tensor, it is immediate that [C]ij = εikjck.

If c and d are two vectors, the outer product cdT generates the matrix C given by

*M) (1.35)

We will see later that C is a second-order tensor if c and d have the transformationproperties of vectors.

An n × n matrix A can be decomposed into symmetric and antisymmetricmatrices using

(1.36)

AA A A I− −= =1 1

( )AB B AT T T=

( )AB B A− − −=1 1 1

C =

−

−

−

0

0

0

3 2

3 1

2 1

c c

c c

c c

C =

c d c d c d

c d c d c d

c d c d c d

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

A A A A A A A A AT T= + = + = −s a s a, [ ], [ ]12

12




1.3.1 EIGENVALUES AND EIGENVECTORS

In this case, A is again an n × n tensor. The eigenvalue equation is

(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

(1.38)

The equation det(A − λjI) = 0 is expanded as (2 − λj)2 − 1, with roots λ1,2 = 1, 3, and

(1.39)

Note that in each case, the rows are multiples of each other, so that only one rowis independent. We next determine the eigenvectors. It is easily seen that magnitudesof the eigenvectors are arbitrary. For example, if x1 is an eigenvector, so is 10x1.Accordingly, the magnitudes are arbitrarily set to unity. For x1 = x11 x12

T,

(1.40)

from which we conclude that A parallel argument furnishes

If A is symmetric, the eigenvalues and eigenvectors are real and the eigenvectorsare orthogonal to each other: The eigenvalue equations can be “stackedup,” as follows.

(1.41)

With obvious identifications,

(1.42)

( )A I x 0− =λ j j

A =

2 1

1 2

A I A I− =

− =−

−

λ λ1 2

1 1

1 1

1 1

1 1

x x x x11 12 112

1220 1+ = + =

x T1 1 1 2= − / .

x T2 1 1 2= / .

x xTi j ij= δ .

A x x x ] x x x[ : : [ : : ]

. . .

. . .

. . . . .

. . .

. . .

1 2 1 2

1

2

1

0

0

0

0

K Kn n

n

n

=

−

λ

λ

λ

λ

AX X= ΛΛ




and X is the modal matrix. Let yij represent the ijth entry of Y = XTX

(1.43)

so that Y = I. We can conclude that X is an orthogonal tensor: XT = X−1. Further,

(1.44)

and X can be interpreted as representing a rotation from the reference axes to theprincipal axes.

1.3.2 COORDINATE TRANSFORMATIONS

Suppose that the vectors v and w are depicted in a second-coordinate system whosebase vectors are denoted by Now, can be represented as a linear sum of thebase vectors ei:

*T) (1.45)

But then It follows that δij = = (qikek) ⋅ (qjlel) =qikqjlδkl, so that

*T)

In *M) notation, this is written as

*M) (1.46)

in which case the matrix Q is called orthogonal. An analogous argument proves thatQTQ = I. From Equation (1.30), 1 = det(QQT) = det(Q)det(QT) = det2(Q). Right-handed 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 while v isreferred to ei. Now

*T)

(1.47)

yij i j ij= =x xT δ

X AX A X XT T= =ΛΛ ΛΛ

′e j . ′e j

′ =e ej ji iq

e ei j ij ijq⋅ ′ = = ′cos( ).θ ′ ⋅ ′e ei j

q q q qik jk ik kj

ij

=

=

T

δ

QQ IT =

′e j ,

′ = ′ ′

= ′

=

v e

e

e

v

v q

v

j j

j ji i

i i




It follows that and hence

*M) (1.48)

in which qij is the jith entry of QT.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 vectorsei. It is a vector if and only if, upon a rotation of the coordinate system to basevectors v′ transforms according to Equation (1.48).

Since is likewise equal to dφ,

*M) (1.49)

for which reason dφ/dx is called a contravariant vector, while v is properly called acovariant vector.

Finally, to display the base vectors to which the tensor A is referred (i.e., intensor-indicial notation), we introduce the outer product

(1.50)

with the matrix-vector counterpart Now

(1.51)

Note the useful result that

In this notation, given a vector b = bkek,

(1.52)

as expected.

v v qi j ji= ′ ,

v Q v v QvT= ′ ′ = ( ) ( )a b

′e j ,( )d

d dφx x′ ′

d

d

d

d

φ φx x

QT

′=

e ei j∧

e eTi j .

A e e= ∧aij i j

e e e ei j k i jk∧ ⋅ = δ

Ab e e e

e e e

e

e

= ∧ ⋅

= ∧ ⋅

=

=

a b

a b

a b

a b

ij i j k k

ij k i j k

ij k i jk

ij j i

δ




1.3.4 ORTHOGONAL CURVILINEAR COORDINATES

The position vector of a point, P, referring to a three-dimensional, rectilinear, coor-dinate system is expressed in tensor-indicial notation as RP = xiei. The position vectorconnecting two “sufficiently close” points P and Q is given by

(1.53)

where

(1.54)

with arc length

(1.55)

Suppose now that the coordinates are transformed to yj coordinates: xi = xi(yj).The same position vector, now referred to the transformed system, is

(1.56)

in which hα is called the scale factor. Recall that the use of Greek letters for indicesimplies no summation. Clearly, γγγγα is a unit vector. Conversely, if the transformationis reversed,

(1.57)

∆∆R R R dR= − ≈P Q x

dR ex i idx=

dS dx dxx i i=

dR g

g

y

j j

dy

h

hdx

dy

dx

dy

=

=

=

∑ α α

α α α

αα α

1

3

γγ

γγα

α α

α

α α

=

=

dxdy

dxdy

dxdy

i

ii

i

j j

h

dx

dy

e

e1

dR g

g

y i i

i

ji j

dy

dy

dxdx

=

=




then the consequence is that

(1.58)

We restrict attention to orthogonal coordinate systems yj, with the property that

(1.59)

The length of the vector dRy is now

(1.60)

Under restriction to orthogonal coordinate systems, the initial base vectors ei can beexpressed in terms of γγγγα using

(1.61)

and furnishing

(1.62)

Also of interest is the volume element; the volume determined by the vectordRy is given by the vector triple product

(1.63)

and h1h2h3 is known as the Jacobian of the transformation. For cylindrical coordinatesusing r, θ, and z, as shown in Figure 1.2, x1 = rcosθ, x2 = rsinθ, and x3 = z. Simplemanipulation furnishes that hr = 1, hθ = r, hz = 1, and

(1.64)

which, of course, are orthonormal vectors. Also of interest are the relations der =eθ dθ and deθ = −er dθ.

e gji

ji

j

dy

dx

dy

dxh= = ∑ α

α αα

γγ

γγ γγα β αβδT =

dS h dy dyy i i= α

e e

e

Ti j i j

i

i

jj

i j

i

j

k

jk

h

x

y

h h

x

y

x

y

= ( )=

∂∂

=∂∂

∂∂

γγ γγ

γγ1

1

1h h

x

y

x

yi j

i

j

k

jik

∂∂

∂∂

= δ

dV h dy h dy h dy

h h h dy dy dy

y = ⋅ ×

=

( ) [ ]1 1 1 2 2 2 3 3 3

1 2 3 1 2 3

γγ γγ γγ

e e e e e e e er z= + = − + =cos sin sin cosθ θ θ θθ1 2 1 2 3




Transformation of the coordinate system from rectilinear to cylindrical coordi-nates can be viewed as a rotation of the coordinate system through θ. Thus, if thevector v is referred to the reference rectilinear system and v′ is the same vectorreferred to a cylindrical coordinate system, then in two dimensions,

(1.65)

If v′ is differentiated, for example, with respect to time t, there is a contributionfrom the rotation of the coordinate system: for example, if v and θ are functions oftime t,

(1.66)

where the partial derivative implies differentiation with θ instantaneously held fixedand

(1.67)

FIGURE 1.2 Cylindrical coordinate system.

x1

x3

x2

er

ez

eθ

r

θ

′ = = −

v Q v Q( ) ( )

cos sin

sin cosθ θ

θ θ

θ θ

0

0

0 0 1

d

dt

d

dt

d

dt

t

d

dt

′ = +

= ∂∂

′ + ′

v Q vQ

v

vQ

Q vT

( )( )

( )( )

θ θ

θ θ

d

dt

d

dt

Q( )sin cos

cos sinθ

θ θ

θ θ θ=

−

− −

0

0

0 0 1




Now is an antisymmetric matrix ΩΩΩΩ (to be identified later as a tensor)since

(1.68)

In fact,

(1.69)

It follows that

(1.70)

in which ωωωω is the axial vector of ΩΩΩΩ.Referring to Figure 1.3, spherical coordinates r, θ, and φ are introduced by the

transformation

(1.71)

FIGURE 1.3 Spherical coordinate system.

x3eφ

eθ

x2

x1

er

π φφ

φ

θ

θ

ddt

Q TQ( ) ( )θ θ

0 Q QQ

QQ

QT T TT

= = +

d

dt

d

dt

d

dt( ( ) ( ))

( )( )

( )( )θ θ θ θ θ θ

d

dt

d

dt

QQT( )

( )θ θ θ= −

0 1 0

1 0 0

0 0 0

d

dt t′ = ∂

∂′ + × ′v v vωω

x r x r x r1 2 3= = =cos cos sin cos sinθ φ θ φ φ




The position vector is given by

(1.72)

Now er has the same direction as the position vector: r = rer. Thus, it follows that

(1.73)

Following the general procedure in the preceding paragraphs,

(1.74)

The differential of the position vector furnishes

(1.75)

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:

(1.76)

Eliminating e1, e2, e3 in favor of er, eθ, eφ and using *M notation permits writing

(1.77)

r e e e

e e e

= + +

= + +

x x x

r r r

1 1 2 2 3 3

1 2 3cos cos sin cos sinθ φ θ φ φ

e e e er = + +cos cos sin cos sinθ φ θ φ φ1 2 3

∂∂

=∂∂

= −∂∂

= −

∂∂

=∂∂

=∂∂

= −

∂∂

=∂∂

=∂∂

=

x

r

xr

xr

x

r

xr

xr

x

r

x xr

1 1 1

2 2 2

3 3 30

cos cos sin cos cos sin

sin cos cos cos sin sin

sin cos

θ φθ

θ φφ

θ φ

θ φθ

θ φφ

θ φ

φθ φ

φ

d dr r d rdrr e e e= + +cosφ θ φθ φ

e e e e

e e e

e e e e

e e e e

e e e e

e e e

r r

r

r

= + +

= − +

= − + +

= − −

= + −

= +

cos cos sin cos sin

sin cos

sin [cos sin ] cos

cos cos sin sin cos

sin cos cos sin sin

sin cos

θ φ θ φ φ

θ θ

φ θ θ φ

θ φ θ φ θ

θ φ θ φ θ

φ φ

θ

φ

θ φ

θ φ

φ

1 2 3

1 2

1 2 3

1

2

3 .

v e e e v e e e= + + ′ = + +v v v v v vr r1 1 2 2 3 3 .θ θ φ φ

′ = = −

− −

v Q v Q( , ) , ( , )

cos cos sin cos sin

sin cos

sin cos sin sin cos

.θ φ θ φ

θ φ θ φ φ

θ θ

φ θ φ θ φ

0




Suppose now that v(t), θ, and φ are functions of time. As in cylindrical coordinates,

(1.78)

where ωωωω is the axial vector of After some manipulation,

(1.79)

1.3.5 GRADIENT OPERATOR

In rectilinear coordinates, let ψ be a scalar-valued function of x: ψ(x), starting withthe chain rule

*T)

(1.80)

Clearly, dψ is a scalar and is unaffected by a coordinate transformation. Supposethat x = x(y): dr′ = gidyi. Observe that

(1.81)

d

dt t′ = ∂

∂′ + × ′v v vωω

ddt

Q TQ( ) ( ).θ θ

ddt

ddt

ddt

ddt

Q( )sin cos cos cos

cos sin

sin sin cos sin

cos sin sin sin cos

cos cos sin cos sin

( )( )

cos

cos

θθ φ θ φ

θ θ

θ φ θ φ

θ

θ φ θ φ φ

θ φ θ φ φ

φ

θ θ

φ

φ

=

−

− −

−

+

− −

− − −

= −

0

0

0

0 0 0

0 0

0

Q QT sinsin

sin

φ

φ

θ φ

0 0

0 0 1

0 0 0

1 0 0−

+

−

ddt

ddt

dx

dx

d d dxx

ii

i i ii

ψ ψ

ψ ψ ψ

= ∂∂

= ∇ ⋅ = ∇ = ∂∂

[ ] ,r r e e

dx

dx

h yh dy

h yh dy

iiψ ψ

ψ

ψ

α αα α

α

α

α ααβ β β

β

= ∂∂

= ∂∂

= ∂∂

⋅

∑

∑ ∑

1

γγγγ




implying the identification

(1.82)

For cylindrical coordinates in tensor-indicial notation with er = γγγγr, eθ = γγγγθ, ez = γγγγz,

(1.83)

and in spherical coordinates

(1.84)

1.3.6 DIVERGENCE AND CURL OF VECTORS

Under orthogonal transformations, the divergence and curl operators are invariantand satisfy the divergence and curl theorems, respectively. Unfortunately, the trans-formation properties of the divergence and curl operators are elaborate. The readeris referred to texts in continuum mechanics, such as Chung (1988). The developmentis 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 vectorreferred to orthogonal coordinates. The divergence and curl satisfy

(1.85)

and

(1.86)

and in cylindrical coordinates:

(1.87)

( )∇ ′ = ∂∂∑ψ ψα

α αα

γγh y

∇ = ∂∂

+ ∂∂

+ ∂∂

ψ ψ ψθ

ψθee

er zr r z

∇ = ∂∂

+ ∂∂

+ ∂∂

ψ ψφ

ψθ

ψφ

θee e

rz

r r rcos

( ) ( ) ( ) ( )∇ ⋅ ′ = ∂∂

′ + ∂∂

′ + ∂∂

′

v

1

1 2 3 12 3 1

23 1 2

31 2 3h h h y

h h vy

h h vy

h h v

( ) ( ) ( )

( ) ( )

( ) ( )

∇ × ′ = ∂∂

′ − ∂∂

′

− ∂∂

′ − ∂∂

′

+ ∂∂

′ − ∂∂

′

v1

1 2 31

23 3

32 2 1

21

3 33

1 1 2

31

2 22

1 1

h h hh

yh v

yh v

hy

h vy

h v

hy

h vy

h v

γγ

γγ

γγ 3

( )( )

∇ ⋅ ′ =∂

∂+

∂∂

+∂∂

v1 1r

rv

r r

v v

zr zθ

θ




and

(1.88)

APPENDIX I: DIVERGENCE AND CURL OF VECTORSIN ORTHOGONAL CURVILINEAR COORDINATES

DERIVATIVES OF BASE VECTORS

In tensor-indicial notation, a vector v can be represented in rectilinear coordinatesas v = vkek. In orthogonal curvilinear coordinates, it is written as

A line segment dr = dxiei transforms to dr′ = dykgk. Recall that

(a.1)

From Equation (a.1),

(a.2)

The bracketed quantities are known as Cristoffel symbols. From Equations (a.1 and a.2),

(a.3)

Continuing,

(a.4)

( )( ) ( )

∇ × ′ =∂∂

−∂

∂

−∂∂

−∂∂

+∂

∂−

∂∂

v e e e

1r

v rv

z

v

r

v

zr

rv

r

vzr

z r rzθ θ

θθ

θ

′ = ∑ ′ =vα α αv γγ

α ααα

∑ ′v hg .

e g

g e g g

kl

kl

k

kk

y

xh

y

x

x

yh

=∂∂

=∂∂

=∂∂

= ⋅

∑ ββ

ββ

αα

α α α

γγ

∂∂

=∂

∂ ∂

=

=∂

∂ ∂∂∂∑

geα α

ββ

β βα β

α β α β

y

x

y y

jh

jh

x

y y

y

x

j k jk

k j k

2

2

γγ ,

dh

dy

d

dy

jh

j j

αα

α

α

α α

= ⋅

=

γγg

∂∂

=∂∂

−∂∂

= = −

∑

γγ γγ

γγ

α

α

α α

α

α

α β ββ

α βα

αβ βδα β

y h y h

h

y

c ch j

h

j j j

j j

1

11

g

, ( )




DIVERGENCE

The development that follows is based on the fact that

(a.5)

The differential of v′ is readily seen to be

(a.6)

First, note that

(a.7)

Similarly,

(a.8)

∇ ⋅ = ′∇ ⋅ ′ = ′′

=

v v

vr

vr

trd

dtr

d

d

d dv v dj j j j′ = +v γγ γγ

dvv

ydy

h

v

yh dy

h

v

yh dy

h

v

yd

j jj

kj k

jj

jj

jj

γγ γγ

γγ

γγ γγ γγ

γγ γγ

=∂∂

=∂∂

=∂∂

∧

⋅

=∂∂

∧

⋅ ′

∑

∑ ∑

∑

1

1

1

α αα α

α

α αα

αβ β β

β

α αα

α

( )

( )

r

v d vy

dy

v

h yh dy

v

h yh dy

v

h yd

v

hc

j j jj

kk

j j

j j

j j

jj

γγγγ

γγ

γγγγ γγ

γγγγ

γγ γγ

=∂∂

=∂∂

=∂∂

∧

⋅

=∂∂

∧

⋅ ′

= ∧

∑

∑ ∑

∑

∑

α αα α

α

α αα

αβ β β

β

α αα

α

ααβ β α

β

( )

( )

r

⋅ ′∑α

dr




Consequently,

(a.9)

CURL

In rectilinear coordinates, the individual entries of the curl can be expressed as adivergence, as follows. For the ith entry,

(a.10)

Consequently, the curl of v can be written as

(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 areasof 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

2. Prove that if σσσσ is a symmetric tensor with entries σij, that

3. If v and w are n × 1 vectors, prove that v × w can be written as

dvdr h

v

yv c

h

v

yv cj

j j j j j

=∂∂

+

∇ ⋅ =

∂∂

+

∑

βα α αβ βα

α

α

ααα

α

δ1 1, v

[ ]

( ) ( )

( )

∇ × =∂∂

= ∂∂

=

= ∇ ⋅

v

w

i ijkk

j

jji

ji

jki k

i

v

x

xw w v

ε

ε

∇ × =

∇ ⋅

∇ ⋅

∇ ⋅

v

w

w

w

( )

( )

( )

1

2

3

n e e e= + +A

A

A

A

A

A1

12

23

3

ε σijk jk i= =0 1 2 3, , , .

v w Vw× =




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

(a) Verify that AB ≠ BA.(b) Verify that (AB)T = BTAT.

5. Consider a matrix C given by

Verify that its inverse is given by

6. For the matrices in Exercise 4, find the inverses and verify that

7. Consider the matrix

FIGURE 1.4 Geometry of a tetrahedron.

e1

e2

A1

e3

A3

A2

3

2

A

n

1

A B=−

−

=

1 1 2

1 3 1 4

1 1 3

1 2 1 4

/

/ /

/

/ /

C =

a b

c d

C− =−

−

−

1 1ad bc

d b

c a

( )AB B A1 1 1− − −=

Q =−

cos sin

sin cos

θ θ

θ θ




Verify that(a) QQT = QTQ(b) QT = Q−1

(c) For any 2 × 1 vector 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 (one-

dimensional arrays)

verify that

9. Verify the divergence theorem using the following block, where

10. For the vector and geometry of Exercise 9, verify that

FIGURE 1.5 Test figure for the divergence theorem.

Qa a=

a b=

=

q

r

s

t

a Cb b C aT T T=

v =−

+

x y

x y

Y

X1

1

n v v× = ∇ ×∫ ∫dS dV




11. Using the geometry of Exercise 9, verify that

using

12. Obtain the expressions for the gradient, divergence, and curl in sphericalcoordinates.

n A AT× = ∇ ×∫ ∫dS dV

a x y x y

a x y x y

a x y x y

a x y x y

112 2

122 2

212 2

222 2

= + + +

= + + −

= + − −

= − − −



25

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 thisrelation is preserved under rotation. In particular, simple manipulation furnishes that

∗

)

(2.1)

The square matrix

A

is now called a second-order tensor if and only if

A

′

=

QAQ

T

.Let

A

and

B

be second-order

n

×

n

tensors. The manipulations that followdemonstrate that

A

T

, (

A

+

B

),

AB

, and

A

−−−−

1

are also tensors.

(2.2)

(2.3)

(2.4)

(2.5)

2

′ =

=

=

= ′

v Qv

QAw

QAQ Qw

QAQ w

T

T .

( ) ( )A QAQ

Q A Q

T T T

T T T

′ =

=T

′ ′ =

=

=

A B QAQ QBQ

QA QQ BQ

QABQ

T T

T T

T

( )( )

( )

( )A B A B

QAQ QBQ

Q A B Q

T T

T

+ ′ = ′ + ′

= +

= +( )

′ =

=

=

− −

− − −

−

A QAQ

Q A Q

QA Q

1 T 1

T 1 1 1

1 T

( )

.



26


Let

x

denote an

n

×

1 vector. The outer product,

xx

T

, is a second-order tensor since

(2.6)

Next,

(2.7)

However,

(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, from which

(2.9)

and also

(2.10)

We conclude that

(2.11)

Furthermore, if

d

u

′

is a vector generated from

d

u

by rotation in the oppositesense from the coordinate axes, then

d

u

′ =

Q

d

u

and

d

x

=

Q

d

x

′

. Hence,

Q

is a tensor.Also, since , it is apparent that

(2.12)

from which we conclude that is a tensor. We can similarly show that

I

and

0

are tensors.

( )

( )( )

( )

xx x x

Qx Qx

Q xx Q

T T

T

T T

′ = ′ ′

=

=

d d dd

d

d

d2φ φ= =

x H x H

x xT

T

.

d d d d

d d

′ ′ ′ = ′

= ′

x H x Q x H Q x

x Q H Q x

T T

T T

( )

( ) ,

d du xux

= ∂∂

,

d du xux

T TT

= ∂∂

d du xx

uT TT

T= ∂∂

.

∂∂

= ∂∂

ux

ux

T T

T .

d d′ = ′∂ ′∂ ′u xux

∂ ′∂ ′

= ∂∂

ux

Qux

QT,

∂∂ux




27

2.2 DIVERGENCE, CURL, AND LAPLACIAN OF A TENSOR

Suppose

A

is a tensor and

b

is an arbitrary, spatially constant vector of compatibledimension. The divergence and curl of a vector

have already been defined. For laterpurposes, we need to extend the definition of the divergence and the curl to

A

.

2.2.1 D

IVERGENCE

Recall the divergence theorem Let , in which

b

is anarbitrary constant vector. Now

(2.13)

Consequently, we must define the divergence of

A

such that

*

) (2.14)

In tensor-indicial notation,

(2.15)

Application of the divergence theorem to the vector

c

j

=

b

i

a

ij

furnishes

(2.16)

Since

b

is arbitrary, we conclude that

(2.17)

Thus, if we are to write as a (column) vector, mixing tensor- and matrix-vector notation,

(2.18)

∫ = ∫ ∇c n cT TdS dV. c A bT=

b An A b

A b

b A

T T T

T T

T T T T

dS dV

dV

dV

∫ ∫∫

∫

= ∇

= ∇

= ∇

( )

[ ] .

An A 0T T TdS dV− ∇ =∫∫ [ ] .

b a n dS b dVi ij j i i∫ ∫− ∇[ ] =[ ] .T T TA 0

bx

a dVij

ij i

∂∂

− ∇[ ]

=∫ [ ] .T T TA 0

[ ] .∇ = ∂∂

= ∂∂

T T TA ij

ijj

jixa

xa

∇ ⋅ A

∇ ⋅ = ∇A AT T[ ]T .



28


It should be evident that ( ) has different meanings when applied to a tensoras opposed to a vector.

Suppose

A

is written in the form

(2.19)

in which corresponds to the

i

th

row of It is easily seen that

(2.20)

2.2.2 C

URL

AND

L

APLACIAN

The curl of vector

c

satisfies the curl theorem Using tensor-indicial notation,

(2.21)

From the divergence theorem applied to the tensor

c

ij

=

ε

ijk

a

kl

b

l

,

(2.22)

Let denote the row vector (array) corresponding to the

l

th

row of

A

:

=

a

lk

. It follows that

(2.23)

∇ ⋅

A1T

2T

3T

=

αααααα

,

ααiT A T:[ ] .ααi j ija=

∇ = ∇ ∇ ∇T T T1

T TA ( ).αα αα αα2 3

∫ ∇ × = ∫ ×c n cdV dS.

n c× =

=

= =

∫ ∫∫∫

dS n a b dS

n a dS b

c n dSb c a b

ijk j kl l

ijk j kl l

ij j l ij ijk kl l

ε

ε

ε, .

n Ab

A b AT

×

= ∂∂

=∂∂

= ∇ ×

∇ × = ∂∂

∫ ∫

∫

∫

dSx

a b dV

a

xdV b

dVx

a

i jijk kl l

ijkkl

jl

iil ijk

jkl

( )

, [ .

ε

ε

ε if ]

ααlT [ ]ααl k

T

∇ × = ∇ × ∇ × ∇ ×[ ]A 1 2 3αα αα αα .




29

If ββββ

I

is the array for the

I

th

column of

A

, then

(2.24)

The Laplacian applied to

A

is defined by

(2.25)

It follows, therefore, that

(2.26)

The vectors ββββ

i

satisfy the Helmholtz decomposition

(2.27)

Observe from the following results that

(2.28)

An integral theorem for the Laplacian of a tensor is now found as

(2.29)

2.3 INVARIANTS

Letting

A

denote a nonsingular, symmetric, 3

×

3 tensor, the equation det(

A

−

λ

l

)

=

0 can be expanded as

(2.30

)

in which

(2.31)

Here,

tr

(

A

)

=

δ

ij

a

ij

denotes the

trace

of

A

. Equation 2.30 also implies the Cayley-Hamilton theorem:

(2.32)

∇ × = ∇ × ∇ × ∇ ×[ ]AT1 2 3ββ ββ ββ .

[ ] .∇ = ∇2 2A ij ija

∇ = ∇ ∇ ∇2 21

22

23A [ ].ββ ββ ββ

∇ = ∇ ∇ ⋅ − ∇ × ∇ ×2ββ ββ ββi i i( ) .

∇ = ∇ ∇ ⋅ − ∇ × ∇ ×2 T T TA A A( ) [ ] .

∇ = ∇ − × ∇ ×∫ ∫ ∫2 T T TA n A n AdV ) dS dS( [ ] .

λ λ λ31

22 3 0− + − =I I I ,

I tr I tr tr I1 22 2

3

12

= = − =( ) [ ( ) ( ) det( ).A A A A ]

A A A I31

22 3− + − =I I I 0,



30


from which

(2.33)

The trace of any

n

×

n

symmetric tensor

B

is invariant under orthogonal trans-formations (rotations), such as

tr

(B′) = tr(B), since

(2.34)

Likewise, tr(A2) and tr(A3) are invariant since A, A2, and A3 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 ispositive definiteness, defined as follows. The symmetric n × n tensor A is positive-definite, written A > 0, if, for all nonvanishing n × 1 vectors x, the quadratic productq(A, x) = xTAx > 0. The importance of this property is shown in the followingexample. Let Π = xTAx − xTf, in which f is known and A > 0. After some simplemanipulation,

(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 sakeof demonstration,

(2.36)

I tr I tr I tr

I I I

33

12

2

31 2

1 2

13

= − +

= − +− −

[ ( ) ( ) ( )]

[ ]

A A A

A A A1 I

′ =

=

=

a q q a

a q q

a

pq pq pr qs rs pq

rs pr qs

rs rs

δ δ

δ .

12

d dd

d

d

dd

d d

2Π Π=

=

xx x

x

x A x

TT

T .

x Ax x X x

y y y X x

T T T

T T

=

= =

= ∑

ΛΛΛΛ

ΛΛ , ( )

.λi i

i

y2



Mathematical Foundations: Tensors 31

The last expression can be positive for arbitrary y (arbitrary x) only if λi > 0, i =1, 2,…, n. The matrix A is semidefinite if xTAx ≥ 0, and negative-definite (writtenA < 0), if xTAx < 0. If B is a nonsingular tensor, then BTB > 0, since q(BTB, x) =xTBTBx = yT y > 0 (in which y = Bx and Ω denotes the quadratic product). If B issingular, for example if B = yyΤΤΤΤ where y is an n × 1 vector, BΤΤΤΤB is positive-semidefinitesince a nonzero eigenvector x of B can be found for which the quadratic productq(BTB, x) vanishes.

Now suppose that B is a nonsingular, antisymmetric tensor. Multiplying throughBxj = λ j xj with BT

furnishes

(2.37)

Since BΤΤΤΤB is positive-definite, it follows that Thus, λ j is imaginary: using . Consequently, , demonstrating that B2

is negative-definite.

2.5 POLAR DECOMPOSITION THEOREM

For an n × n matrix B, BTB > 0. If the modal matrix of B is denoted by Xb, we canwrite

(2.38a)

in which Y is an (unknown) orthogonal tensor. In general, we can write

. (2.38b)

To “justify” Equation 2.38b, we introduce the square root using

(2.38c)

B Bx B x

Bx

x

T Tj j j

j j

j j

=

= −

= −

λ

λ

λ2 .

− >λ j2 0.

λ µj ji= i = −1 B x x x2 2 2j j j j j= = −λ µ

B B X X

X YY X

X Y X Y

T T

T12 T

12

T12 T

12

T

=

=

=

b b b

b b b b

b b b b

∆∆

∆∆ ∆∆

∆∆ ∆∆

( ) ( )

( ) ( ) ,

B Y XT12= ( )∆∆b b

B BT

B B XT T= =

b b b b

n

∆∆ ∆∆1

2

1

2

1

2

0

0

0

0

X ,

. . .

. . .

. . . . .

. . . .

. . .

,

λ

λ

λ




in which the positive square roots are used. It is easy to verify that andthat . Note that

(2.38d)

Thus, is an orthogonal tensor, called, for example, Z, and hence wecan write

(2.38e)

Finally, noting that , we make the iden-tification in Equation 2.38b. Equation 2.38 plays a major role in theinterpretation 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.

(2.39)

The inverse VEC operator, IVEC, is introduced by the obvious relationIVEC(VEC(A)) = A. The Kronecker product of an n × m matrix A and an r × smatrix B generates an nr × ms matrix, as follows.

(2.40)

If m, n, r, and s are equal to n, and if A and B are tensors, then A ⊗ Btransforms as a second-order n2 × n2 tensor in a sense that is explained subsequently.

( )B B BT 2 =B BT > 0

B B B B B B B B B B B

I

T T T T T( )

( )

= ( )

( )

=

− − − −1

2

1

2

1

2

1

2 [ ] B

T

.

B B BT( )−1/2

B Z B B

ZX X

T

T

=

= b b b∆∆1

2 .

( )( ) ( )ZX ZX Z X X Z ZZ IT T TbT

bT T

b b = = =Y ZXT T= b

VEC a a a a an n nn( ) .,A T= −11 21 31 1L

A B

B B B

B

B B

⊗ =

a a a

a

a a

m

n nm

11 12 1

21

1

. .

. . . .

. . . . .

. . . . .

. . .

.




Equation 2.40 implies that the n2 × 1 Kronecker product of two n × 1 vectors aand b is written as

(2.41)

2.6.2 FUNDAMENTAL RELATIONS FOR KRONECKER PRODUCTS

Six basic relations are introduced, followed by a number of subsidiary relations.The proofs of the first five relations are based on Graham (1981).

Relation 1: Let A denote an n × m real matrix, with entry aij in the i th row andj th column. Let I = (j − 1)n + i and J = (i − 1)m + j. Let Unm denote the nm × nmmatrix, independent of A, satisfying

. (2.42)

Then,

(2.43)

Note that uJK = uJI = 1 and uIK = uIJ = 1, with all other entries vanishing. Henceif 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

(2.44)

Relation 3: If In denotes the n × n identity matrix, and if B denotes an n × ntensor, then

(2.45)

Relation 4: Let A, B, C, and D, respectively, denote m × n, r × s, n × p, and s ×q matrices. Then,

(2.46)

a b

b

b

b

⊗ =

a

a

an

1

2

.

.

.

uK I

K Iu

K J

K JJK IK=

=

≠

=

=

≠

1

0

1

0

,

,

,

,

VEC VECnm( ) ( ).A U AT =

tr VEC VEC( ) ( ) ( ).AB A BT T=

I B I BT Tn n⊗ = ⊗( ) .

( ) ( ) .A B C D AC BD⊗ ⊗ = ⊗




Relation 5: If A, B, and C are n × m, m × r, and r × s matrices, then

(2.47)

Relation 6: If a and b are n × 1 vectors, then

. (2.48)

As proof of Relation 6, if I = ( j − 1)n + i, then the I th entry of VEC(baT) isbiaj. It is also the I th entry of a ⊗ b. Hence, a ⊗ b = VEC(baT) = VEC([abT]T).

Symmetry of Unn was established in Relation 1. Note that VEC(A) = Unn VEC(AT) =U2

nnVEC(A) if A is n × n, and hence the matrix Unn satisfies

. (2.49)

Unn is hereafter called the permutation tensor for n × n matrices. If A is symmetric, thenVEC(A) = 0. If A is antisymmetric, then (Unn + Inn)VEC(A) = 0.

If A and B are second-order n × n tensors, then

(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

VEC(A) = VEC(AIn). If Inn is the identity tensor in n2-dimensional space, then In ⊗ In =Inn since = In ⊗ InVEC(In). Now, in = Ini, thus In ⊗ In = .

If A, B, and C denote n × n tensors, then

(2.51)

However, by a parallel argument,

(2.52)

VEC VEC( ) ( ).ACB B A CT= ⊗

a b abT T⊗ = VEC([ ] )

U U U UT 1nn n nn nn nn2

2= = = −I

( )U Inn n− 2

tr VEC VEC

VEC VEC

VEC VEC

VEC VEC

tr

T T

Tnn

nnT

T T

( ) ( ) ( )

( ) ( )

[ ( )] ( )

( ) ( )

( ),

AB B A

B U A

U B A

B A

BA

=

=

=

=

=

VEC VECn n n( ) ( )I I I= In2

VEC VEC

VEC

VEC

n

n n

( ) ( )

( )( ) ( )

( ).

ACB I A CB

I A B I C

B A C

T T= ⊗

= ⊗ ⊗

= ⊗

VEC VEC

VEC

VEC

T T T T

T

n

[( ) ] ( )

( )

( ).

ACB BC A

A B C

A BU C

=

= ⊗

= ⊗ 2




The permutation tensor arises in the relation

(2.53)

Consequently, if C is arbitrary,

(2.54)

and, upon using the relation , we obtain an important result:

(2.55)

If A and B are nonsingular n × n tensors, then

(2.56)

The Kronecker sum and difference appear frequently (for example, in controltheory) and are defined as follows:

(2.57)

The Kronecker sum and difference of two n × n tensors are n2 × n2 tensors, asexplained in the following section.

2.6.3 EIGENSTRUCTURES OF KRONECKER PRODUCTS

Let αj and βk denote the eigenvalues of A and B, and let yj and zk denote thecorresponding eigenvectors. The Kronecker product, sum, and difference have thefollowing eigenstructures:

(2.58)

A B α j − βk yj ⊗ zk

VEC VECT T

n

T[( ) ] ( ).ACB U ACB= 2

U B A C A BU Cn n

VEC VEC2 2⊗ = ⊗( ) ( ),

U U 1

n n2 2= −

B A U A BU⊗ = ⊗n n2 2 .

( )( )

.

A B A B AA BB

I I

I

1 1 1 1⊗ ⊗ = ⊗

= ⊗

=

− − − −

n n

n2

A B A I I B A B A I I B⊕ = ⊗ + ⊗ = ⊗ − ⊗n n n n .

expression eigenvalue eigenvectorjk jkth th

j k j k

j k j k

A B y z

A B y z

⊗ ⊗

⊕ + ⊗

α β

α β




As proof,

(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 correspondingquantities for In ⊗ B are βk × 1 and vj × zk , in which vj is an arbitrary eigenvectorof 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 aTRb is easily derived:if r = VEC(R), then

(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 × ntensor. Suppose that A = CB, which is equivalent to aij = cijklbkl in which the rangeof i, j, k, and l is (1, n). The TEN22 operator is introduced implicitly using

(2.61)

Note that

(2.62)

α β α βj j k k j k j k

j k

j k

y z y z

Ay Bz

A B y z

⊗ = ⊗

= ⊗

= ⊗ ⊗( )( ).

a Rb ba R

ba R

ab R

b a r

T T

T T T

T T

T T

=

=

=

= ⊗

tr

VEC VEC

VEC VEC

[ ]

([ ] ) ( )

( ) ( )

.

VEC TEN VEC( ) ( ) ( ).A C B= 22

TEN VEC VEC

VEC

TEN VEC

TEN VEC

n

n

n n

22

22

22

( ) ( ) ( )

( )

( ) ( )

( ) ( ),

ACB D ACBD

I A CBD

I A C BD

I A C I B D

=

= ⊗

= ⊗

= ⊗ ⊗




hence, TEN22(ACB) = In ⊗ ATEN22(C)In ⊗ B. Upon writing B = C−1A, it is obviousthat VEC(B) = TEN22(C−1)VEC(A). However, TEN22(C)VEC(B) = VEC(A), thusVEC(B) = [TEN22(C)]−1VEC(A). We conclude that TEN22(C−1) = TEN22−1(C).Furthermore, by writing AT = BT, it is also obvious that Un a = TEN22(C)Unb,thus TEN22( ) = TEN22(C) . The inverse of the TEN22 operator is introducedusing 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-ordern × n × n × n tensor such that A = CB. All are referred to a coordinate systemdenoted as Y. Let the unitary matrix (tensor) Qn represent a rotation that gives riseto a coordinate system Y′. Let A′, B′, and C′ denote the counterparts of A, B, and C.Now, since A′ = Qn AQT

n,

(2.63)

However, note that (Q ⊗ Q)T = QT ⊗ QT = Q−−−−1 ⊗ Q−−−−1 = (Q ⊗ Q)−−−−1. Hence,Q ⊗ Q is a unitary matrix (tensor) in an n2 vector space. However, not all rotationsin n2–dimensional space can be expressed in the form Q ⊗ Q. It follows that VEC(A)transforms as an n2 × 1 vector under rotations of the form Q ⊗ Q.

Now write A′ = C′ B′, from which

(2.64a)

It follows that

, (2.64b)

thus TEN22(C) transforms a second-order n2 × n2 tensor under rotations of the formQ ⊗ Q.

Finally, letting Ca and Cb denote third-order n × n × n tensors, respectively,thereby satisfying relations of the form A = Cab and b = CbA, it is readily shownthat TEN21(Ca) and TEN12(Cb) satisfy

(2.65)

which we call tensors of order (2,1) and (1,2), respectively.

CC U

n2 Un2

VEC VEC( ) ( ).′ = ⊗A Q Q A

Q Q A C Q Q B⊗ = ′ ⊗VEC TEN VEC( ) ( ) ( ).22

TEN TEN22 22( ) ( )( )′ = ⊗ ⊗C Q Q C Q Q T

TEN TEN n n

TEN TEN n n

a a

b b

21 21

12 12

2

2

( ) ( )

( ) ( ) ,

′ = ⊗ ×

′ = ⊗ ×

C Q Q C Q

C Q C Q Q

T

T T




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 can be written as

(2.66)

in which Tij and Eij are entries of the (small-deformation) stress and strain tensorsdenoted by T and E. Here, δij denotes the substitution (Kronecker) tensor. Equation 2.66exhibits three tensor outer products of the identity (Kronecker) tensor I: δikδjl, δilδjk,and δijδkl. In general, let A and B be two nonsingular n × n second-order tensorswith entries aij and bij; let a = VEC(A) and b = VEC(B). There are 24 permutationsof the indices ijkl corresponding to outer products of tensors A and B. Recalling thedefinitions of the Kronecker product, we introduce three basic Kronecker-productfunctions:

(2.67)

Twenty-four outer product-Kronecker product pairs are obtained as follows:

(2.68)

ˆ ˆµ λ and

T E E i jij ik jl il jk kl ij kl kl = + + =ˆ ( ) ˆ , , , ,µ δ δ δ δ λδ δ 1 2 3

C A B ab C A B A B C A B A BU1T

2 3( , ) ( , ) ( , ) .= = ⊗ = ⊗ n2

TEN a b TEN b a

TEN a b TEN b a

TEN a b TEN b a

TEN a b

ij kl ij kl

ji kl ji kl

ij lk ij lk

ji lk

22 22

22 22

22 22

22

( ) ( , ) ( ) ( , )

( ) ( , ) ( ) ( , )

( ) ( , ) ( ) ( , )

( ) ( ,

= =

= =

= =

=

C A B C B A

C A B C B A

C A B C B A

C A

1 1

1T

1T

1T

1T

1T BB C B A

C B A C A B

C B A C A B

C B A C

T1

T T

2 2

2T

2T

2T

) ( ) ( , )

( ) ( , ) ( ) ( , )

( ) ( , ) ( ) ( , )

( ) ( , ) ( )

TEN b a

TEN a b TEN b a

TEN a b TEN b a

TEN a b TEN b a

ji lk

ik jl ik jl

ki jl ki jl

ik lj ik lj

22

22 22

22 22

22 22

=

= =

= =

= = 22T

2T T

2T T

3 3

3T

3T

A B

C B A C A B

C B A C A B

C B A C A B

( , )

( ) ( , ) ( ) ( , )

( ) ( , ) ( ) ( , )

( ) ( , ) ( ) ( , )

(

TEN a b TEN b a

TEN a b TEN b a

TEN a b TEN b a

TEN a

ki lj ki lj

il jk il jk

li jk li jk

il

22 22

22 22

22 22

22

= =

= =

= =

bb TEN b a

TEN a b TEN b a

kj il kj

li kj li kj

) ( ) ( ) ( , )

( ) ( , ) ( ) ( , )

,

= =

= =

C B ,A C A B

C B A C A B

3T

3T

3T T

3T T

22

22 22




With t = VEC(T) and e = VEC(E), and noting that U9e = e (since E is symmetric),we now restate Equation 2.66 as

(2.69)

The proof is presented for several of the relations in Equation 2.68. We introducetensors R and S with entries rij and sij. Also, let s = VEC(S) and r = VEC(R).

a.) Suppose that sij = aijbklrkl. However, aijbklrkl = aijbTlkrkl, in which case bT

kl

is the kl th entry of BT. It follows that S = tr(BTR)A. Hence,

(2.70)

Since s = TEN22(aij bkl)r, it follows that , asshown in Equation 2.68.

b.) Suppose that sij = aikbjlrkl. However, aikbjlrkl = aikrklbTlj, thus S = ARBT. Now

(2.71)

as shown in Equation 2.68.c.) Suppose sij = ailbjkrkl. However, ailbjkrkl = ailr

Tlkbkj, thus S = ARTBT. Now

(2.72)

as shown in Equation 2.68.

t e= + +[ ˆ[ ( , ) ( , )] ˆ ( , )]µ λC I I C I I C I I2 3 1

s B R a

B R a

ab r

C A B r

T T T

T

T

1

=

=

=

=

VEC VEC

VEC VEC

[( ) ] ( )

( ) ( )

( , ) .

TEN a bij kl22( ) ( , )= C A B1

s ARB

I A RB

I AB Ir

B Ar

C B A r

T

T

=

= ⊗

= ⊗ ⊗

= ⊗

=

VEC

VEC

( )

( )

( , ) ,2

s AR B

I AB I R

B AU r

C B A r

T T

T

=

= ⊗ ⊗

= ⊗

=

VEC

VEC

( )

( )

( , ) ,

9

3




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

, (2.73)

then C is said to be totally symmetric. A fourth-order tensor C satisfying Equation2.73a but not 2.73b or c is called symmetric.

Kronecker-product conditions for symmetry are now stated. The fourth-ordertensor C is totally symmetric if and only if

. (2.74)

Equation 2.74a is equivalent to symmetry with respect to exchange of ij and klin C. Total symmetry also implies that, for any second-order n × n tensor B, thecorresponding tensor A = CB is symmetric. Thus, if a = VEC(A) and b = VEC(B),then a = TEN22(C)b. However, . Multiplying through the laterexpression with implies Equation 2.74b. For any n × n tensor A, the tensor B =C−1A is symmetric. It follows that b = TEN22(C−1)a = TEN22−1(C)a, and b =TEN22−1Ca. Thus, TEN22(C−1) = TEN22−1(C). Also, TEN22(C) = [ TEN22−1

(C)]−1 = TEN22(C) . We now draw the immediate conclusion that TEN22(C) = TEN22(C) if C is totally symmetric.We next prove the following:

C−1 is totally symmetric if C is totally symmetric. (2.75)

Note that TEN22(C) = TEN22(C) implies that TEN22(C−1) = TEN22(C−1),while TEN22(C) = TEN22(C) implies that TEN22(C−1) = TEN22(C−1).

Finally, we prove the following: for a nonsingular n × n tensor G,

GCGT is totally symmetric if C is totally symmetric. (2.76)

Equation 2.76 implies that TEN22(GCG−T) = I ⊗ G TEN22(C)I ⊗ GT, so thatTEN22(GCGT) is certainly symmetric. Next, consider whether A′ given by

(2.77)

c c a

c c b

c c c

ijkl jikl

ijkl ijlk

ijkl klij

=

=

=

)

)

)

(

(

(

TEN22 ) TEN22

TEN22( ) TEN22

TEN22 ) TEN22

T( ( ) (

( ) (

( ( ) (

C C

U C C

C U C

=

=

=

a

b

c

n

n

)

)

)

2

2

U a C bn

TEN2 22= ( )U

n2

Un2

Un2 U

n2

Un2 U

n2

Un2

Un2 U

n2

Un2 U

n2

′ = ′A BTGCG




is symmetric, in which case B′ is a second-order, nonsingular n × n tensor. However,we can write

(2.78)

Now G−−−−1A′G−T is symmetric since C is totally symmetric, and therefore A′ issymmetric. Next, consider whether B′ given by the following is symmetric:

(2.79)

However, we can write

(2.80)

Since C−−−−1 is totally symmetric, it follows that GTB′G is symmetric, and henceB′ is symmetric. We conclude that GCGT is totally symmetric.

2.6.9 DIFFERENTIALS OF TENSOR INVARIANTS

Let A be a symmetric 3 × 3 tensor, with invariants I1(A), I2(A), and I3(A). For ascalar-valued function f(A),

(2.81)

However, with a = VEC(A), we can also write

(2.82)

Taking this further,

(2.83)

G A G G G1 T 1 T− − − −′ = ′C B .

′ = ′− − −B G G AT 1 1C

G B G G GT 1 1′ = ′− −C A .

dff

ada tr

fd

f f

aijij

ij ij

( ) , .AA

AA

= ∂∂

= ∂∂

∂∂

= ∂∂

df VECf

VEC d

fd

( ) ( )

.

AA

A

aa

TT

= ∂∂

= ∂∂

∂∂

= ∂∂

=

∂∂

= ∂∂

−

= −

I

I

I

1

2 2

1

12

a ai a i

a ai a a a

i a

T T

T T

T T

( )

( )




and

(2.84)

so that

(2.85)

2.7 EXERCISES

1. Given a symmetric n × n tensor σσσσ, prove that

.

2. Prove that if σσσσ is a symmetric tensor with entries σij,

3. Verify using 2 × 2 tensors that

.

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).

dI tr d I tr d I d

tr d I

32

1 2

13

= − +

= −

( ) ( )

( / )

A A A A A

A A

∂∂

= −IVEC I3 1

3aA( ) /

tr tr I nn( ( ) / )σσ σσ− = 0

ε σijk jk i= =0 1 2 3, , , .

tr tr( ) ( )AB BA=



43

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 vector-valued function

F

(

u

(

x

)

,

u

′(

x

)

,

x

)

, in which

u

′(

x

)

=

∂

u

/∂

x

. Furthermore, let

v

(

x

)

be afunction such that

v

(

x

)

=

0

when

u

(

x

)

=

0

and

v

′(

x

)

=

0

when

u

′(

x

)

=

0

, but whichis 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 atfixed

x

. Following Ewing, we introduce the vector-valued function φφφφ

(

e

:

F

)

as follows(Ewing, 1985):

(3.1)

The variation

δ

F

is defined by

(3.2)

with

x

fixed. Elementary manipulation demonstrates that

(3.3)

in which

.

If

If

then

δ

F

=

δ

u

′

=

e

v

′

.

This suggests the form

(3.4)

The variational operator exhibits five important properties:

1.

δ

(

.

)

commutes with linear differential operators and integrals. For exam-ple, if

S

denotes a prescribed contour of integration:

(3.5)

3

Φ ( : ) ( ( ) ( ), ( ) ( ), ) ( ( ), ( ), )e e eF F u x v x u x v x x F u x u x x= + ′ + ′ − ′

δ F =

=

ee

,e

d

d

ΦΦ0

δ FFu

vFu

v= ∂∂

+ ∂∂ ′

′

e e ,tr

∂∂ ′

∂∂ ′′ = ′F

uFv ve e

ijiju

F u F u v= = =, .then δ δ e F u= ′,

δ δ δFFu

uFu

u= ∂∂

+ ∂∂ ′

′

tr .

δ δ( ( .) )dS dS∫ ∫=

0749_Frmae_C03 Page 43 Wednesday, February 19, 2003 5:01 PM


44


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

(3.6)

4. If

f

is a prescribed function of (scalar)

x

, and if

u

(

x

)

is subject to variation,then

(3.7)

5. Other than for number 2, the variation is arbitrary. For example, for twovectors

v

and

w

,

v

T

d

w

=

0 implies that

v

and

w

are orthogonal to eachother. However,

v

T

δ

w

implies that

v

=

0

, since only the zero vector canbe orthogonal to an arbitrary vector.

As a simple example, Figure 3.1 depicts a rod of length L, cross-sectional areaA, and elastic modulus E. At

x

=

0, the rod is built in, while at

x

=

L

, the tensileforce P is applied. Inertia is neglected. The governing equations are in terms ofdisplacement

u, stress S, and (linear) strain E:

strain-displacement

stress-strain

equilibrium (3.8)

Combining the equations furnishes

(3.9)

The following steps serve to derive a variational equation that is equivalent to thedifferential equation and endpoint conditions (boundary conditions and constraints).

FIGURE 3.1 Rod under uniaxial tension.

δ δ δ( ) ( ) ( )qr q r q r= + .

δ δ( )fu f u= .

Edudx

=

S E= E

ddxσ = 0

EAd u

dx

2

2 0= .

E,A

L

P



Introduction to Variational and Numerical Methods 45

Step 1: Multiply by the variation of the variable to be determined (u) andintegrate over the domain.

(3.10)

Differential equations to be satisfied at every point in the domain are replacedwith an integral equation whose integrand includes an arbitrary function.

Step 2: Integrate by parts, as needed, to render the argument in the domainintegral positive definite.

(3.11)

However, the first term is the integral of a derivative, so that

. (3.12)

Step 3: Identify the primary and secondary variables. The primary variable ispresent in the endpoint terms (rhs) under the variational symbol, l, and isu. The conjugate secondary variable is

Step 4: Satisfy the constraints and boundary conditions. At x = 0, u is pre-scribed, thus δu = 0. At x = L, the load is prescribed. Also, notethat .

Step 5: Form the variational equation; the equations and boundary conditionsare consolidated into one integral equation, δF = 0, where˙

(3.13)

The jth variation of a vector-valued quantity F is defined by

(3.14)

It follows that δ 2u = 0 and δ2u′ = 0. By restricting F to a scalar-valued functionF and x to reduce to x, we obtain

(3.15)

and H is known as the Hessian matrix.

δu Ad udx

AdxL

E2

20

0∫ = .

d

dxu A

du

dx

d u

dxA

du

dxdx

L

δ δE

−

=∫ E

00

d udx

Adudx

dx u Adudx

L Lδ δ

=∫ E E0 0

EA dudx

.

P A dudx= E

( ) ( ( ) )dudx

dudx

dudxA AE E= δ 1

22

F Adu

dxdx Pu L

L

=

−∫ 1

20

2

E ( ).

δ j jj

je

edde

F =

=

Φ

0

.

δ δ δδ

δ2F

F F

F F

= ′′

=

∂∂

∂∂

∂∂

∂∂ ′

∂∂ ′

∂∂

∂∂ ′

∂∂ ′

, ,u u Hu

uH u u u u

u u u u

T T

T T

T T




Now consider G given by

(3.16)

in which V again denotes the volume of a domain and S denotes its surface area. Inaddition, h is a prescribed (known) function on S. G is called a functional since itgenerates a number for every function u(x). We first concentrate on a three-dimensional,rectangular coordinate system and suppose that δG = 0, as in the Principle ofStationary Potential Energy in elasticity. Note that

(3.17)

The first and last terms in Equation 3.17 can be recognized as divergences ofvectors. We now invoke the divergence theorem to obtain

(3.18)

For suitable continuity properties of u, arbitrariness of δu implies that δG = 0is equivalent to the following Euler equation, boundary conditions, and constraints(the latter two are not uniquely determined by the variational principle):

(3.19)

Let D > 0 denote a second-order tensor, and let ππππ denote a vector that is anonlinear function of a second vector u, which is subject to variation. The function

satisfies

(3.20)

Despite the fact that D > 0, in the current nonlinear example, the specific vectoru∗ satisfying δF = 0 may correspond to a stationary point rather than a minimum.

G F dV dS= ′ +∫ ∫( , ( ), ( )) ( ) ( )x u x u x h x u xT ,

∂∂

∂∂ ′

= ∂∂ ′

′

+ ∂∂

∂∂ ′

x u

uu

ux u

uF F

F .δ δ δtr

0 =

= ∂∂

+ ∂∂ ′

′

+

= ∂∂

− ∂∂

∂∂ ′

+ ∂

∂ ′+

∫ ∫

∫ ∫ ∫

δ

δ δ δ

δ δ δ

G

F FdV dS

F FdV

FdS dS.

uu

uu h x u x

u x uu n

uu h x u x

T

T T

tr ( ) ( )

( ) ( )

∂∂

− ∂∂

∂∂ ′

=F F

u x u0T.

u x

nu

h x 0T1T T

( )

( )

prescribed x S

S x S S

on F

on on

1

1

∂∂ ′

+ = −

F = 12 ππ ππTD

δ δ δ δ δ δ δF F .= ∂∂

= ∂∂

∂∂

+ ∂∂

∂∂

u

uD u

uD

uu u

u uuDT

T

TT

T

TT

T

T

ππ ππ ππ ππ ππ ππ ππ2




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 oftensolved numerically by a two-track process: the load intensity λ is increased progres-sively using small increments. At each increment, the unknown x is computed usingan iteration procedure. Suppose that at the nth increment of λ, an accurate solutionis achieved as xn. Further suppose for simplicity’s sake that xn is “close” to the actualsolution xn+1 for the (n + 1)st increment of λ. Using x(0) = xn as the starting value,Newton iteration provides iterates according to the scheme

(3.21)

Let ∆n+1,j denote the increment Then, to first-order in the Taylor series

(3.22)

in which 02 refers to second-order terms in increments. It follows that ∆n+1,j ≈ 02. Forthis reason, Newton iteration is said to converge quadratically (presumably to thecorrect solution if the initial iterate is “sufficiently close”). When the iteration schemeconverges to the solution, the load intensity is incremented again. Consider f (x) =(x − 1)2. If x(0) = 1/2, the iterates are 1/2, 3/4, 7/8, and 15/16. If x(0) = 2, the iteratesare 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

(3.23)

in which u and ϕϕϕϕ are n × 1 vectors, v is a constant n × 1 unit vector, and λ represents“load intensity.” The Newton iteration scheme provides the ( j + 1)st iterate for un+1 as

(3.24)

x xdf

dxf xj j

x

j

j

( ) ( ) ( )

( )( ) .+

−

= −

11

x xnj

nj

++

+−11

1( ) ( ) .

∆ ∆

∆

n 1,j n 1,j 1x

1(j)

x

1(j 1)

x

(j) (j 1)

x xn 1,j 1

dfdx

f(x )dfdx

f(x

dfdx

f(x ) f(x

dfdx

dfdx

(j) (j )

(j)

(j) (j)

+ + −

− −−

−−

−

+ −

− = −

−

≈ −

−[ ]

≈ −

−1

1

1

)

)

++

≈ ++ −

0

0

2

n 1,j 1∆ 2

δ λu u vT [ ( ) ] ,ϕϕ − = 0

∆ ∆n 1,j n 1j

j 1 n(j 1)

nj

n 1,j

n 1j

.+

−

+ + ++

+ += − ∂∂

( ) −[ ] − =+

ϕϕ ϕϕu

u v u uu( )

( ) ( ),

1

1 1λ




in which, for example, the initial iterate is un. One can avoid the use of an explicitmatrix inverse by solving the linear system

(3.25)

3.2.2 CRITICAL POINTS AND THE ARC-LENGTH METHOD

A point λ∗ at which the Jacobian matrix is singular is called a critical point,and corresponds to important phenomena such as buckling. There often is goodreason to attempt to continue calculations through critical points, such as to computea postbuckled configuration. Arc-length methods are suitable for doing so. Here, wepresent a version with a simple eigenstructure.

Suppose that the change in load intensity is regarded as a variable. Introducethe “constraint” on the size of the increment for the nth load step:

(3.26)

in which Σ2 is interpreted as the arc length in n + 1 dimensional space of u and λ.Also, β > 0. Now,

(3.27)

Newton iteration now is expressed as

(3.28)

An advantage is gained if J′ can be made nonsingular even though J is singular.Suppose that J is symmetric and we can choose β such that J + v vΤΤΤΤ/ββββ2 > 0. ThenJ′ admits the “triangularization”

(3.29)

The determinant of J′ is now ββββ2det(J + v vΤΤΤΤ/ββββ2). Ideally, β is chosen to maximizedet (J′).

∂∂

= ( ) − = ++

+ + + ++

+ +ϕu

u v u uun

jn j n

jj n

jn

jn j

1

1 1 1 11

1 1( )

,( ) ( ) ( )

, .∆ ∆ϕϕ

J = ∂∂ϕϕu

ψ λ β λ λu v u uT, [ ] ( ) ,( ) = − + − − =2 2 0n n Σ

ϕϕ( ).

u v 0

0

−

=

λ

ψ

′−

−

= −

( ) −

( )

′ =−

++( )

+( )

++( ) ( )

+( )

+( )

+( )

+

Ju u u v

uJ

J v

vT

nj

nj

nj

nj

nj

nj

nj

n

11

1

11

1 1

1 12λ λ

λ

ψ λ β

ϕϕ

,, .

′ =+ −

J

J v v v

0

I 0

v

T

T T

/ /

/.

ββ ββ

ββ ββ ββ

2




3.3 EXERCISES

1. Directly apply variational calculus to F, given by

to verify that δF = 0 gives rise to the Euler equation

What endpoint conditions (not unique) are compatible with δF = 0?2. The governing equation for an Euler-Bernoulli beam in Figure 3.2 is

in which w is the vertical displacement of the neutral (centroidal) axis.The shear force V and the bending moment M satisfy

Using integration by parts twice, obtain the function F such that δF = 0is equivalent to the foregoing differential equation together with theboundary conditions for a cantilevered beam of length L:

FIGURE 3.2 Cantilevered beam.

F EA P LL

=

−∫ 1

2

2

0

du

dxdx u( )

EAd udx

2

2 0=

EId wdx

4

4 0=

M EI V EI= − =d w

dx

d w

dx

2

2

3

3

w w V( ( ( , ( .0) 0) M L) V L) 0= ′ = = =0 0

Z

V

L

E,I

neutral axis

x

v0



51

Kinematicsof Deformation

The current chapter provides a review of the mathematics for describing deformationof continua. A more complete account is given, for example, in Chandrasekharaiahand Debnath (1994).

4.1 KINEMATICS

4.1.1 D

ISPLACEMENT

In finite-element analysis for finite deformation, it is necessary to carefully distin-guish between the current (or “deformed”) configuration (i.e., at the current time orload step) and a reference configuration, which is usually considered strain-free.Here, both configurations are referred to the same orthogonal coordinate systemcharacterized by the base vectors

e

1

,

e

2

,

e

3

(see Figure 1.1 in Chapter 1). Consider abody with volume

V

and surface

S

in the current configuration. The particle Poccupies a position represented by the position vector

x

, and experiences (empirical)temperature

T

. In the corresponding undeformed configuration, the position of P isdescribed by

X

, and the temperature has the value

T

0

independent of

X

. It is nowassumed that

x

is a function of

X

and

t

and that

T

is also a function of

X

and

t

. Therelations are written as

x

(

X

,

t

) and

T

(

X

,

t

), and it is assumed that

x

and

T

arecontinuously differentiable in

X

and

t

through whatever order needed in the subse-quent development.

FIGURE 4.1

Position vectors in deformed and undeformed configurations.

4

e2

e1

X

x

undeformed

deformed



52


4.1.2 D

ISPLACEMENT

V

ECTOR

The vector

u

(

X

) represents the displacement from position

X

to

x

:

(4.1)

Now consider two close points, P and Q, in the undeformed configuration. Thevector difference

X

P

−

X

Q

is represented as a differential

d

X

with squared length

dS

2

=

d

X

T

d

X

. The corresponding quantity in the deformed configuration is

d

x

, with

dS

2

=

d

x

T

d

x

.

4.1.3 D

EFORMATION

G

RADIENT

T

ENSOR

The deformation gradient tensor

F

is introduced as

(4.2)

F

satisfies the polar-decomposition theorem:

(4.3)

in which

U

and

V

are orthogonal and ΣΣΣΣ

is a positive definite diagonal tensor whose

FIGURE 4.2

Deformed and undeformed distances between adjacent points.

ds

Q'

P'

Q *

P *dS

u( ) .X x X, t = −

d dx F X FxX

= = ∂∂

F U V= Σ T ,



Kinematics of Deformation

53

entries

λ

j

, the singular values of

F

, are called the principal stretches.

(4.4)

Based on Equation 4.3,

F

can be visualized as representing a rotation, followedby a stretch, followed by a second rotation.

4.2 STRAIN

The deformation-induced change in squared length is given by

(4.5)

in which

E

denotes the

Lagrangian strain tensor

. Also of interest is the

Right Cauchy-Green strain

C

=

F

T

F

=

2

E

+

I

. Note that

F

=

I

+

∂

u

/

∂

X

. If quadratic terms in

∂

u

/

∂

X

are neglected, the linear-strain tensor

E

L

is recovered as

(4.6)

Upon application of Equation 4.3,

E

is rewritten as

(4.7)

Under pure rotation

x

=

QX

,

F

=

Q

and

E

=

[

Q

T

Q

−

I

]

=

0

. The case of purerotation in small strain is considered in a subsequent section.

4.2.1 F

,

E

, EL AND u IN ORTHOGONAL COORDINATES

Let Y1, Y2, and Y3 be orthogonal coordinates of a point in an undeformed configura-tion, with y1, y2, y3 orthogonal coordinates in the deformed configuration. Thecorresponding orthonormal base vectors are ΓΓΓΓ1,ΓΓΓΓ2,ΓΓΓΓ3 and γγγγ1,γγγγ2,γγγγ3.

4.2.1.1 Deformation Gradient and Lagrangian Strain Tensors

Recalling relations introduced in Chapter 1 for orthogonal coordinates, the differentialposition vectors are expressed as

(4.8)

Σ =

λ

λ

λ

1

2

3

0 0

0 0

0 0

ds dS d d 2 2 212

− = = −X F F ITT XE E [ ],

EL = ∂∂

+ ∂∂

12

uX

uX

T

.

E = −

V I VT12

2( )ΣΣ

12

d dY H d dy hR r= =∑ ∑α α αα

β β ββ

ΓΓ γγ




(4.9)

in which ΓΓΓΓβ denotes the base vector in the curvilinear system used for the undeformedconfiguration.

This can be written as

(4.10)

in which Q is the orthogonal tensor representing transformation from the undeformedto the deformed coordinate system. It follows that

from which

(4.11)

Displacement Vector

The position vectors can be written in the form R = ZiΓΓΓΓi, r = z jγγγγj. The displacementvector referred to the undeformed base vectors is

(4.12)

Cylindrical Coordinates

In cylindrical coordinates,

(4.13)

d dy h

dy h

dy h q

qh

H

dy

dYH dY

qh

H

dy

dYH dY

r

T

T

T

=

= ⋅

=

=

= ∧

∑

∑∑

∑∑

∑∑∑

∑∑∑

α α αα

α α α β βαβ

α α βα βαβ

βαα

ζ

α

ζζ ζ β

αβζ

βαα

ζ

α

ζβ ζ ζ ζ ζ

αβζ

γ

γγ

ΓΓ ΓΓ

ΓΓ

ΓΓ

ΓΓ ΓΓ )) ΓΓ

( )

( ,

d d qh

H

y

Yr Q F R Q FT T T= ′ = ′ =

∂∂

, [ [ ] , ] βα βα αζα

ζ

α

ζ

E F F I F F I= − = ′ ′ −12

12

[ ] [ ],T T

[ ] .E iji j i j

ij

h

H H

y

Y

y

Y=

∂∂

∂∂

−

∑1

2

2β β β

β

δ

u = − = ⋅[ ] , .z Z j ji i i ji j iq qΓΓ γγ ΓΓ

u e e e= − − + − + −[ cos( ) ] sin( ) ( ) .r R r z ZR Zθ θ θΘ Θ



Kinematics of Deformation 55

We now apply the chain rule to ds2 in cylindrical coordinates:

(4.14)

in which

(4.15)

4.2.1.2 Linear-Strain Tensor in Cylindrical Coordinates

If quadratic terms in the displacements and their derivatives are neglected, then

(4.16)

ds d d dr r d dz

drdR

dRR

drd

RddrdZ

dZ rddR

dRrR

dd

Rd rddZ

dZ

dzdR

dRR

dzd

RddzdZ

dZ

dR Rd dZ

dR

Rd

dZ

2 2 2 2 2

2 2

2

1

1

= ⋅ = + +

= + +

+ + +

+ + +

=

r r

C

θ

θ θ θΘ

ΘΘ

Θ

ΘΘ

Θ Θ

,

cdrdR

rddR

dzdR

e c

cR

drd

rR

dd R

dzd

e c

cdrdZ

rddZ

dzdZ

e

RR RR RR

ZZ ZZ

=

+

+

= −

=

+

+

= −

=

+

+

2 2 2

2 2 2

2 2 2

12

1

1 1 12

1

θ

θ

θ

( )

( )ΘΘ ΘΘ ΘΘΘ Θ Θ

== −

=

+

+

=

=

+

+

12

1

1 1 12

1 1

( )c

cdrdR R

drd

rddR

rR

dd

dzdR R

dzd

e c

cR

drd

drdZ

rR

dd

rddZ R

dzd

ZZ

R R R

Z

Θ Θ Θ

Θ

Θ Θ Θ

Θ Θ

θ θ

θ θΘΘ Θ Θ

=

=

+

+

=

dzdZ

e c

cdrdZ

drdR

rddZ

rddR

dzdZ

dzdR

e c

Z Z

ZR ZR ZR

12

12

θ θ.

u r Ru

Ru z ZR Z≈ − ≈ − ≈ − Θ Θθ ,

dr

dR

du

dRr

d

dRR

d

dR

u

R

du

dR

u

R

dz

dR

du

dR

R

dr

d R

du

d

r

R

d

d

R u

R R

du

d

u

R R

du

d R

dz

d R

du

ddr

dZ

du

dZr

d

dZ

du

dZ

R Z

R R R Z

R

≈ + ≈

= − ≈

≈ =+

+

≈ + + ≈

≈ ≈

1

1 11

1 1 1 1

θ

θ

θ

Θ Θ Θ

Θ Θ

Θ

Θ Θ Θ Θ Θ Θ Θ1

dzdz

dZ

du

dZZ≈ +1




giving rise to the linear-strain tensor

(4.17)

The divergence of u in cylindrical coordinates is given by ∇ ⋅ u = trace =trace(EL), from which

(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 explicitfunction of x(t) and t. The velocity-gradient tensor L is introduced using dv = Ldx,from which

(4.19)

Its symmetric part, called the deformation-rate tensor,

(4.20)

can be regarded as a strain rate referred to the current configuration. The correspond-ing strain rate referred to the undeformed configuration is the Lagrangian strain rate:

(4.21)

EL

R R

R

R

du

dR

du

dR

u

R R

u

R

du

dR

du

dZ

du

dR

u

R R

u

R

u

R R

du

d

du

dZ R

du

d

du

dR

du

dZ

du

dZ

R Z

R Z

Z

=

− +

+

− +

+ +

+

+

∂

∂∂

∂

1

2

1 1

2

1

2

1 1 1

2

1

1

2

1

2

Θ Θ

Θ Θ Θ Θ

Θ

Θ Θ11

R

du

d

du

dZ

Z Z

Θ

.

ddur

∇⋅ = + + +udu

dR

u

R R

du

d

du

dZR R Z1 Θ

Θ,

Lvx

vX

Xx

FF 1

=

=

= −

d

d

d

d

d

d

˙ .

D L LT= +12

[ ],

˙ [ ˙ ˙ ]

[ ˙ ˙ ]

.

E = +

= +

=

− −

12

12

F F F F

F FF F F F

F DF

T T

T 1 T T

T




The antisymmetric portion of L is called the spin tensor W:

(4.22)

Suppose the deformation consists only of a time-dependent, rigid-body motion:

(4.23)

Clearly, F = Q and E = 0. Furthermore,

(4.24)

which is antisymmetric since .Hence, D = 0 and , 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

(4.25)

Based on what we learned in Chapter 1, with v denoting the velocity vector inorthogonal coordinates,

(4.26)

Of course,

4.2.2.2 Cylindrical Coordinates

The velocity vector in cylindrical coordinates is

(4.27)

W L LT= −12

[ ].

x Q X b Q Q IT( ) ( ) ( ), ( ) ( ) .t t t t t= + =

L QQT= ˙ ,

0 I QQ QQ QQ QQ QQT T T T T T= = = + = +˙ [ ] ˙ ˙ ˙ ( ˙ )•

W QQT= ˙

v yr

( , ) , .td

dtv v h

dy

dt= = =∑ α α

αα α

αγγ

[ ] .L βαβα α α

β βαδ=

=∂∂

+

d

d h

v

yvj

j j j

vr

1c

cα βα

αβ β

βα β

δα β

α β

j

k j k

hh

hx

y y

y

x

j

j

= −

=∂

∂ ∂∂∂

11

2

( )

[ ] [[ ] [ ] ] [ ] [[ ] [ ] ].,D L L W L Lβα βα αβ βα βα αβ= + = −1

2

1

2

v e e e

e e e

= + +

= + +

dr

dtr

d

dt

dz

dt

v v v

r z

r r z z

θθ

θ θ .




Observe that

(4.28)

Converting to matrix-vector notation, we get

(4.29)

4.2.2.3 Spherical Coordinates

Now

(4.30)

Next,

(4.31)

d dv dv dv v d v d

dv v d dv v d dv

r r z z r r

r r r z z

v e e e e e

e e e

= + + + +

= − + + +

θ θ θ θ

θ θ θθ θ[ ] [ ] .

d

dv

dr r

dv

d

dv

dz

v

rdv

dr r

dv

d

dv

dz

v

rdv

dr r

dv

d

dv

dz

dv

dr r

dv

r r r

z z

r r

dr rd dz rd

dr rd dz rd

dr rd dz

dr

rd

dz

r

z

v

L L

= +

= =

+ + −

+ +

+ +

1

1

1

1

θ

θ

θ

θ θ

θ θ

θ

θ

θ

θ θ θ

,

dd

v

r

dv

dzdv

dr r

dv

d

v

r

dv

dzdv

dr r

dv

d

dv

dz

r

z z z

r

θ

θ

θ

θ

θ θ θ

−

+

1

1

.

v e e e

e e e

e e e e

e e e

e e e e

= + +

= + +

= + +

= − +

= − + +

dr

dtr

d

dtr

d

dt

v v v

r

r r

r

cos

cos (cos sin ) sin

sin cos

sin (cos sin ) cos .

φ θ φ

φ θ θ φ

θ θ

φ θ θ φ

θ φ

θ θ φ φ

θ

φ

1 2 3

1 2

1 2 3

′ = =

′

= −

− −

v Qv v v e e e

Q

, , , ,

cos cos cos sin sin

sin cos

sin cos sin sin cos

referred to

v

v

v

r

θ

φ

φ θ φ θ φ

θ θ

φ θ φ θ φ

1 2 3

0




(4.32)

Recall that

(4.33)

Thus, it follows that

(4.34)

and

(4.35)

Finally,

(4.36)

d d d

d d d

dv

dv

dv

r

r

v QQ v * QQ v

v * QQ v v* e e e

T T

T

= +

= + =

referred to , , ,θ

φ

θ φ

d

dt

d

dt

d

dt

QQ

( )( )

cos

cos sin

sin

.θ θ

φ

φ φ

φ

θ φT = −

−

+

−

0 0

0

0 0

0 0 1

0 0 0

1 0 0

dr

r dr

rdQQT = −

−

+

−

10 0

0

0 0

10 0 1

0 0 0

1 0 0cos

cos

cos sin

sin

cosφ

φ

φ φ

φ

φ θ φ

d

v d v d

v d v d

v d v d

r

v v

v v

v v

dr

r d

r d

r

r

r

r

QQ vT =

+

− +

− −

= − +

− −

cos

cos sin

sin

tan

tan

cos .

φ θ φ

φ θ φ θ

φ θ φ

φ

φ

φ θ

φ

θ φ

φ

θ

θ φ

φ

θ

1

0

0 0

0

dv

dv

dv

dv

dr

r d

rd

r

dv

dr r

dv

d r

dv

ddv

dr r

dv

d r

dv

ddv

dr r

dv

d r

dv

d

r r r

* =

=

θ

φ

φ θ φ

φ θ φ

φ θ φ

θ θ θ

φ φ φ

φ θ

φ

1 1

1 1

1 1

cos

cos

cos

cos




and

(4.37)

The divergence of v is given by trace(L), thus,

(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 vectortriple-product

(4.39)

The vectors dXi deform into dxj = ejdXi. The deformed volume is now readilyverified to be

(4.40)

and J is called the Jacobian. To obtain J for small strain, we invoke invariance and to find

(4.41)

L =

+ − +

− −

=

dv

dr r

dv

d r

dv

ddv

dr r

dv

d r

dv

ddv

dr r

dv

d r

dv

d

dv

dr

r r r

r

r

v v

v v

v v

r

r

1 1

1 1

1 1

1

0

0 0

0

cos

cos

cos

tan

tan

φ θ φ

φ θ φ

φ θ φ

θ θ θ

φ φ φ

θ φ

φ

θ

φ

φ

11 1

1 1

1 1

r

dv

d

v

r r

dv

d

v

rdv

dr r

dv

d

v v

r r

dv

ddv

dr r

dv

d

v

r r

dv

d

v

r

r r

r

r

cos

cos

tan

cos

tan

.

φ θ φ

φ θ

φ

φ

φ θ

φ

φ

θ φ

θ θ φ θ

φ φ θ φ

+ +

−

− −

−

∇⋅ = + −−

+vdv

dr r

dv

d

v v

r r

dv

dr r1 1

cos

tan,

φ θφ

φθ φ φ

dV d d d d d d

d dX d dX d dX

0 1 2 3 1 2 3

1 1 1 2 2 2 3 3 3

= ⋅ × =

= = =

X X X X X X

X e X e X e .

dx

dj

ix

dV d d d

JdV J

= ⋅ ×

= = =

x x x

F C

1 2 3

0

12, ( ) ( ), det det

J = det( )C

det det12

12 2

1 2 1 2 1 2

( ) [ ]

( )( )( ),

C I= +

= + + +

E

E E EI II III




in which EI, EII, EIII are the eigenvalues of EL, assumed to be much less that unity.The linear-volume strain follows as

(4.42)

using the approximation 1 + x/2 if x << 1.The time derivative of J is prominent in incremental formulations in continuum

mechanics. Recalling that ,

(4.43)

4.4 DIFFERENTIAL SURFACE ELEMENT

Let dS denote a surface element in the deformed configuration, with exterior unitnormal n, as illustrated in Figure 4.3. The corresponding quantities from the refer-ence configuration are dS0 and n0. A surface element dS obeys the transformation(Chandrashekharaiah and Debnath, 1994):

(4.44)

from which we conclude that

(4.45)

e det

E E E

tr

vol

I II III

L

= −

= + + + + −

≈

12 1

1 2 1

( )

( )

( ),

C

quadratic terms

E

1+ ≈x

J I= 3

d

dtJ

J

dI

dt

Jtr

Jtr

Jtr

Jtr tr

J tr

J tr

=

=

= +

= +

= +

= +

=

−

− −

− − −

− − −

− −

12

2

2

2

2

2

3

( ˙ )

( [ ˙ ˙ ])

( ˙ ˙ )

[ ( ˙ ) ( ˙ )]

˙ ˙

C C

F F F F F F

F F F F F F

F F F F F F

F F F F

1

1 T T T

1 1 T T

1 1 T T

1 T T

(( ).D

n F nTdS J dS= −0 0 ,

dS J dS= =−−

−n C n n

F n

n C nT 1

T

T 10 0 00

0 0

.




During deformation, the surface normal changes direction, a fact which is impor-tant, for example, in contact problems. In incremental variational methods, weconsider the differential and d(ndS):

(4.46)

However, recalling Equation 4.43,

(4.47)

Also, since d(FTF−T) = 0, then

(4.48)

Finally, we have

(4.49)

Next, we find with some effort that

(4.50)

FIGURE 4.3 Surface patches in undeformed and deformed configurations.

dX1dx1

dS0

n0

n

dx2

dS

dX2

ddtn

d

dtdS

dJ

dtdS J

d

dtdS[ ] .n F n

FnT

T

= +−−

0 0 0 0

dJ

dtJ tr

dJ

dtdS tr J dS

tr dS

=

=

=

− −

( )

( )

( ) .

D

F n D F n

D n

T T0 0 0 0

d

dt

d

dt

TT

TT

T T

FF

FF

L F

−− −

−

= −

= − .

ddt

[ ][ ( ) ] .

nD I L n

dStr dST= −

ddt

ddt

d

dtn F n

n C n

F n

n C n

n n

n C n

n Dn I L n

T

T

T

T

T

T 1

T T

C

= −

= −

−

−

−

− −

−

0

0 0

0

0 0

0 0

0 01 1

1

[( ) ] .




4.5 ROTATION TENSOR

Elementary manipulation furnishes

(4.51)

in which the antisymmetric rotation tensor ωωωω appears:

. (4.52)

Consider pure rotation with small angle θ :

(4.53)

thus,

(4.54)

(4.55)

Evidently, the normal strains do not vanish, but are second-order in θ, while ωωωωis first-order in θ. Under the assumption of small deformation, nonlinear terms areneglected so that EL is regarded as vanishing.

ddd

d

dL

uuX

X

X

=

= +[ ] ,E ωω

ωω = ∂∂

− ∂∂

12

uX

uX

T

x X

X

= −

≈

−

− −

cos sin

sin cos

,

θ θ

θ θ

θ

θ

θ

θ

0

0

0 0 1

1 0

1 0

0 0 1

2

22

2

u X X=

−

−

+ −

θ

θ

θ

θ

2

22

2

0 0

0 0

0 0 0

0

0 0

0 0 0

0

EL =

−

−

= −

θ

θ

θ

θ

2

22

2

0 0

0 0

0 0 0

0

0 0

0 0 0

, .

0

ωω




Consider as a second example the following derivation in which the sides of theunit square rotate toward each other by 2θ.

The deformation can be described by the relations

(4.56)


(4.57)

Upon neglecting quadratic terms, we conclude that linear-shear strain is a mea-sure of how much the axes rotate toward each other, while the rotation tensor is ameasure of how much the axes rotate in the same sense.

4.6 COMPATIBILITY CONDITIONS FOR EL AND D

The following paragraphs present the compatibility equations for the linear-straintensor, allowing EL to be integrated to produce a displacement field u(X), which isunique to within a rigid-body motion. It should be evident that a parallel argumentproduces exactly the same result for the velocity vector v(x) starting from a givendeformation-rate tensor D(x).

FIGURE 4.4 Element in undeformed and deformed configurations.

Y

LL

L

L

X

θ

θ

x X Y X Y

y X Y X Y

= − + ≈ −

+

= + − ≈ + −

( cos ) sin

sin ( cos ) ,

1 12

1 12

2

2

θ θ θ θ

θ θ θ θ

EL =

−

−

θ

θ

θ

θ

2

22

2

0

0

0 0 0

.




There are two major “paths” to solutions of problems in continuum mechanics.For example,

Assume/approximate u. Apply the strain-displacement relations to estimate E,apply the stress-strain relations to estimate S, and then apply equilibriumrelations to obtain the field equation. Solve the field equation applyingboundary conditions and constraints.

Assume/approximate S, and then estimate E using the strain-displacementrelations. Apply compatibility relations to obtain the field equation, andthen solve the field equation and boundary conditions and constraints.

To understand the second solution path, we focus on the following compatibilityrelation. Suppose the linear strains are known as functions of the position X. Underwhat circumstances can the strain-displacement relations be integrated to producea displacement field that is unique to within a rigid-body displacement? Recall that

(4.58)

Evidently, EL is only part of the derivative of u. In 2-D rectilinear coordinates,we will see that the compatibility equation, guaranteeing unique u to within a rigid-body motion, is given by:

(4.59)

The proof of the compatibility equation is as follows. We consider a path in thephysical (X1, X2, X3) space, along which the arc length is denoted by λ. Accordingly,the position vector X is a function of λ. The integral of du, taken from λ = 0 to λ* is

(4.60)

The term u(0) can be interpreted as the rigid-body translation. The first integralcan be evaluated since E(X) and X(λ) are given functions. The second integral canbe rewritten using an elementary transformation as

(4.61)

ddd

d

dL

uuX

X

X

=

= +[ ]E ωω

∂∂ ∂

= ∂∂

+ ∂∂

212

1 2

211

22

222

12

12

E

x x

E

x

E

x

u X uuX

X

u X X X X

X

XX

( ( *)) ( )

( ) ( ( *)) ( ) .

( *)

( *)( *)

λ

λ

λ

λλλ

= +

= + +

∫

∫∫

0

0

0

00

dd

d

d dE ωω

ωω ωω( ) ( * ) ( * ).( *)

( *)X X X X X X

X

Xd d

0

0λ

λ∫ ∫= − − −




Note the following:

(4.62)

Consequently, integration by parts furnishes

(4.63)

The term ωωωω(X(0))(X(λ*) − X(0)) can be identified as a rigid-body rotation sincethe rotation tensor ωωωω(X(0)) is independent of position. We now focus on the secondterm. Now that rigid-body translation and rotation have been accommodated, theintegral should give a unique value regardless of the path. Consider two paths, path1 and path 2, from λ = 0 to λ*. The integral over these paths should produce the sameunique values. Therefore, the closed path 0 to λ* along path 1 followed by λ* to 0 ina negative sense along path 2 should vanish. Furthermore, path 1 and path 2 arearbitrary, except for terminating at the same points. It follows that along any closed path

(4.64)

Let us now write ωωωω(X) as ωωωω(X) = ωωωω1(X) ωωωω2(X) ωωωω3(X), in which ωωωω i denotesthe vector corresponding to the i th column of ωωωω. The compatibility condition cannow be written as

(4.65)

For a suitable differential function Ψ(X), the condition for ΨΨΨΨ ⋅ dX = 0 overan arbitrary contour is that the curl of Ψ vanishes, from which we express thecompatibility conditions as

(4.66)

d d d

d d

ddd

d

T T T

T T

T

[ ( )( * )] ( ) ( * ) [( * ) ( * )]

( ) ( * ) [( * ) ( * )]

( ) ( * ) ( * )( * )

( * )( * )

ωω ωω ωω

ωω ωω

ωω ωω

X X X X X X X X X X

X X X X X X X

X X X X XX X

X XX X

− = − + − −

= − − − −

= − − − −−

−

T

.

− − = −

+ − −−

−

∫

∫ −

ωω ωω

ωω

( ) ( ) ( ( ))( ( *) ( ))

( * )( * )

( * )( * ) .

( *)

( ) ( )*

X X * X X X X

X XX X

X XX X

X

X X

d

d

ddT

T

λ

λ

λ0

0

0

0 0

( * )( * )

( * )( * ) .X X

X XX X

X X 0− −−

− =∫ T Tdd

dωω

( * )( * )

( * )( * ) , , , .X X

X X

X XX X−

−−

− = =∫ T jd

dd j

ωω0 1 2 3

∇ × −−

= =( * )

( )

( * ), , , .X X

X

X XT d

djjωω

0 1 2 3




This can be expanded at considerable effort to derive 81 compatibility equations,including Equation 4.59.

4.7 SAMPLE PROBLEMS

1. Referring to Figure 4.5, determine u, F, J, and E as functions of X and Y.Use H = 1.0, W = 1.0, a = 0.1, b = 0.1, c = 0.3, d = 0.2, e = 0.2, f = 0.1.Assume a unit thickness in the Z-direction in both the deformed andundeformed configurations.

Solution: Since straight sides are deformed into straight sides, the deformationpattern can be assumed in the form

and it is necessary to determine α, β, γ, δ, ε, and ζ from the coordinatesof the vertices in the deformed configuration.

After elementary manipulation,

FIGURE 4.5 Plate elements in undeformed and deformed states.

x X Y XY y X Y XY= + + = + +α β γ δ ε ζ ,

At

At

At

( , ) ( , ) :

( , ) ( , ) :

( , ) ( , ) :

X Y W W a W b W

X Y H e H H f H

X Y W H W c H H WH H d W H WH

= + = =

= = + =

= + = + + + = + +

0

0

α δ

β ε

α β γ δ ε ζ

α β δ

ε γ ζ

= + = = = = =

= + = = + − + − = = + − − + =

1 1 1 0 2 0 1

1 1 2 0 4 0

a

W

e

H

b

Wf

H

W c W a e

WH

H d b H f

WH

. . .

.( )

.( )

.

Y Y

W W

H

X X

undeformed plate deformed plate

H

a

b

d

c

e

f




The 2 × 2 deformation gradient tensor F and its determinant are now:

.

The displacement vector is

The Lagrangian strain E = [FTF − I] is

2. Figure 4.6 shows a square element at time t and at t + dt. Estimate L, D,and W at time t. Use a = 0.1dt, b = 1 + 0.2dt, c = 0.2dt, d = 1 + 0.4dt,e = 0.05dt, f = 0.1dt, g = 1 − 0.1dt, h = 1 + 0.5dt. Assume a unit thicknessin the Z-direction in both the deformed and undeformed configurations.

FIGURE 4.6 Element experiencing rigid body motion and deformation.

F =+ +

+ +

=+ +

= + −

α γ β γ

δ ζ ε ζ

Y X

Y X

Y X

J Y X

1 1 0 4 0 2 0 4

0 1 1 2

1 3 0 48 0 04

. . . .

. .

. . .

u =−

−

=

−( ) + +

+ −( ) +

=

+ +

+

x X

y Y

X Y XY

X Y XY

X Y XY

X Y

α β γ

δ ε ζ

1

1

0 1 0 2 0 4

0 1 0 2

. . .

. ..

12

E =+

+

+ +

−

=+ + + + +

+

12

1 1 0 4 0 1

0 2 0 4 1 2

1 1 0 4 0 2 0 4

0 1 1 2

12

1 0

0 1

0 11 0 44 0 08 0 17 0 22 0 04 0 08

0 17 0

2

. . .

. . .

. . . .

. .

. . . . . . .

.

Y

X

Y X

Y Y X Y XY

.. . . . . .22 0 04 0 08 0 24 0 08 0 08 2X Y XY X X+ + + +

Y Y

X X

1

1a

e

g

c

d

h

f

b

element at time t element at time t+dt




Solution: First, represent the deformed position vectors in terms of the unde-formed position vectors using eight coefficients determined using the givengeometry. In particular,

Following procedures analogous to Problem 1, we find:

The velocities can be estimated using vx ≈ and vy ≈ , from which

The tensors L, D, and W are now readily found as:

4.8 EXERCISES

1. Consider a one-dimensional deformation in which x = (1 + λ)X. Whatvalue of λ is the linear-strain εL error by 5% relative to the Lagrangianstrain E? Use the error measure

2. A 1 × 1 square plate has a constant (2 × 2) Lagrangian-strain tensor E.What is the deformed length of the diagonal? What is the volume change?

x X Y XY y X Y XY= + + + = + + +α β γ δ ε ζ η θ .

α ε

β ζ

γ η

δ θ

= =

= + =

= = −

= − = −

0 1 0 05

1 0 2 0 1

0 2 1 0 1

0 5 0 05

. .

. .

. .

. . .

dt dt

dt dt

dt dt

dt dt

x Xdt− y Y

dt−

v X Y XY

v X Y XY

x

y

= + + −

= + − −

0 1 0 2 0 2 0 5

0 05 0 1 0 1 0 05

. . . .

. . . . .

L

D

W

=− −

− − −

=− − −

− − − −

=− − +

0 2 0 5 0 2 0 5

0 1 0 05 0 1 0 05

0 2 0 5 0 15 0 25 0 025

0 15 0 25 0 025 0 1 0 05

0 0 4 0 25 0 025

0 4

. . . .

. . . .

. . . . .

. . . . .

. . .

.

Y X

Y X

Y X Y

X Y X

X Y

++ −

0 25 0 025 0. ..

X Y

error L=−E E

E.




If the linear strain EL is now approximated as E, what is the diagonal andwhat is the volume change? Take

3. Verify that the expressions in the text for F and E in cylindrical coordinatesare consistent with the equation

in which Q is the orthogonal tensor representing transformation from theundeformed to the deformed coordinate system, and also with

4. Obtain expressions for u, F, E, and EL in spherical coordinates.5. For cylindrical coordinates, determine the Lagrangian strain in the fol-

lowing two cases:

6. In spherical coordinates, determine the Lagrangian strain for pure radialexpansion.

7. Obtain v, L, D, and W in spherical coordinates.8. For cylindrical coordinates, find L, D, and W for the following flows:

9. In spherical coordinates, find L, D, and W for pure radial expansion:

E =

0 1 0 05

0 05 0 1

. .

. ..

F Q F Q FT T= ′ = ′ =∂∂

, [ [ , ] ]βα βα αζα

ζ

α

ζ

qh

H

y

Y

[ ] .E iji j i j

ij

h

H H

y

Y

y

Y=

∂∂

∂∂

−

∑1

2

2β β β

β

δ

( ) Pure radial expansion:

( ) Torsion:

a r R z Z

b r R Z z Z

= = =

= = + =

λ θ

θ λ

, ,

, ,

Θ

Θ

r R= = =λ θ φ, , Θ Φ

(a) pure radial flow:

(b) pipe flow:

(c) cylindrical flow:

(d) torsional flow:

v f r v v

v v v f r

v v f r v

v v f z v

r z

r z

r z

r z

= = =

= = =

= = =

= = =

( ), ,

, , ( )

, ( ),

, ( ),

θ

θ

θ

θ

0 0

0 0

0 0

0 0

v f r v vr = = =( ), , θ φ0 0




10. For linear strain in rectilinear coordinates and 2-D, the compatibilityrelation is

Find the implications of this relation for a linear-strain field assumed tobe given by

11. Consider a square L × L plate (undeformed configuration) with linearstrains

Assuming that the origin does not move, find the deformed position of(X,Y) = (L,L).

∂∂ ∂

= ∂∂

+ ∂∂

212

1 2

211

22

222

12

12

E

x x

E

x

E

x.

E a X a XY a Y

E b X b XY b Y

E c X c XY c Y

11 12

2 32

22 12

2 32

12 12

2 32

= + +

= + +

= + + .

E a a X a Y

E b b X b Y

E c c X c Y

11 1 2 3

22 1 2 3

12 1 2 3

= + +

= + +

= + + .



73

Mechanical Equilibrium and the Principleof Virtual Work

5.1 TRACTION AND STRESS

5.1.1 C

AUCHY

S

TRESS

Consider a differential tetrahedron enclosing the point

x

in the deformed configura-tion. The area of the inclined face is

dS

, and

dS

i

is the area of the face whose exteriornormal vector is

−

e

i

. Simple vector analysis serves to derive that

n

i

=

dS

i

/

dS

(seeExercise 1 in Chapter 1). Next, let

d

P

denote the force on a surface element

dS

, andlet

d

P

(

i

)

denote the force on area

dS

i

. The traction vector is introduced by ττττ

=

d

P

/

dS

.As the tetrahedron shrinks to a point, the contribution of volume forces, such asinertia, decays faster than surface forces. Balance of forces on the tetrahedron nowrequires that

. (5.1)

The traction vector acting on the inclined face is defined by

, (5.2)

from which

(5.3)

5

dP dPj ji

i

= ∑

ττ = ddSP

τ jj

i

i

ji

i

i

i

ij i

dP

dS

dP

dS

dS

dS

T n

=

=

=

∑

∑

( )

( )

.



74


with

. (5.4)

It is readily seen that

T

ij

can be interpreted as the intensity of the force actingin the

j

direction on the facet pointing in the

−

i

direction, and is recognized as the

ij

th

entry of the Cauchy stress

T

. In matrix-vector notation, the stress-traction relationis written as

. (5.5)

The next section will show that

T

is symmetric by virtue of the balance of angularmomentum. Equation 5.5 implies that

T

T

is a tensor, thus, it follows that

T

is atensor. To visualize

T

, consider a differential cube. Positive stresses are shown onfaces pointing in positive directions, as shown in Figure 5.2.

In traditional depictions, the stresses on the back faces are represented by arrowspointing in negative directions. However, this depiction can be confusing—thearrows actually represent the directions of the traction components. Consider theone-dimensional bar in Figure 5.3. The traction vector

t

e

1

acts at

x

=

L

, while thetraction vector

−

t

e

1

acts at

x

=

0. At

x

=

L

, the corresponding stress is

t

11

=

t

e

1

⋅

e

1

=

t

.At

x

=

0, the stress is given by (

−

t

e

1

)

⋅

(

e

1

) =

t

. Clearly, the stress at both ends, andin fact throughout the bar, is positive (tensile).

We will see later that the stress tensor is symmetric by virtue of the balance ofangular momentum.

FIGURE 5.1

Equilibrium of a tetrahedron.

1

3

dF (3)

dF (2)

dF

n

dF (1)

dS(3)

dS(2) dSdS(1)

2

TdP

dSijj

i

i

=( )

ττ = T Tn



Mechanical Equilibrium and the Principle of Virtual Work

75

5.1.2 1

ST

P

IOLA

-K

IRCHHOFF

S

TRESS

The transformation to undeformed coordinates is now considered. In Chapter 3, wesaw that

(5.6)

is known as the 1st Piola-Kirchhoff stress tensor, and it is not symmetric.

FIGURE 5.2

Illustration of the stress tensor.

FIGURE 5.3

Tractions on a bar experiencing uniaxial tension.

1

2

3

T33

T32

T23

T21T12T11

T13

T31

T22

Y

Z

t t X

ττ ΤΤ

ΤΤ

dS dS

dS

dS

0 0

=

=

= =

−

−

T

T

S S T

n

JF n

n JF

T

T0 0

1, .

S



76


5.1.3 2

ND

P

IOLA

-K

IRCHHOFF

S

TRESS

We next derive the stress tensor that is conjugate to the Lagrangian-strain rate, i.e.,corresponds to the correct amount of work per unit undeformed volume. At a segment

dS

at

x

on the deformed boundary, assuming static conditions, the rate of work of the tractions is

(5.7)

Over the surface

S

, shifting to tensor-indicial notation and invoking the divergencetheorem, we find

(5.8)

We will shortly see that static equilibrium implies that = 0, which enablesus to conclude that

. (5.9)

Returning to matrix-vector notation, since

=

[

L

]

ji

, then

,

(5.10)

since the trace vanishes for the product of a symmetric and antisymmetric tensor.

dW

dW d

dS

T

T

˙ ˙

˙ .

=

=

P u

uττ

˙ ˙

˙

˙˙ .

W dS

T n u dS

u

xT dV u

T

xdV

ij i j

j

iij j

ij

i

=

=

=∂∂

+∂∂

∫∫

∫ ∫

ττTu

∂∂T

xij

i

˙˙

Wu

xT dVj

iij=

∂∂∫

∂∂u

xj

i

˙ ( )

[ ( )]

[ ]

W dV

dV

dV

=

= −

=

∫∫∫

tr

tr

tr

T

T

T

L

D W

D

T




77

To convert to undeformed coordinates,

(5.11)

The tensor

S

is called the 2nd Piola-Kirchhoff stress tensor. It is symmetric if

T

is symmetric. Note that so that the 1st Piola-

Kirchhoff stress is said to be conjugate to the deformation-gradient tensor

F

.

5.2 STRESS FLUX

Consider two deformations,

x

1

and

x

2

, differing only by a rigid body motion:

, (5.12)

in which

V

(

t

) is orthogonal. A tensor

A

(

x

) is objective (see Eringen) if

. (5.13)

It turns out that the matrix of time derivatives of the Cauchy stress, , is notobjective, while the deformation rate tensor,

D

, is. Thus, if ξξξξ

is a fourth-order tensor,a constitutive equation of the form

=

ξξξξ

D

would be senseless. Instead, the timederivative is replaced with an objective stress flux, as explained in the following.First, note that

(5.14)

The tensor ΩΩΩΩ

is antisymmetric since

d

l

/

dt

=

0

=

ΩΩΩΩ

+

ΩΩΩΩ

T

.

˙ ( )

( ˙ )

( ˙ )

( ˙ ) , .

W dV

dV

dV

dV J

=

=

=

= =

∫∫∫∫

− −

− −

− −

tr

tr

tr

tr

T

T E

T E

SE T

D

F F

F F

S F F

1 T

T 1

T 1

J

J

0

0

J J Jtr tr tr tr( ) ( ˙ ) ( ˙ ) ( ˙ ),T T T SL FF F F F1 1= = =− −

x V x b2 1= +( ) ( )t t

A x VA x VT( ) ( )2 1=

T

T

F VF2 1=

L F F

VV VF VF F V

VL V VV

2 2 21

T1 1 1

1 T

1T T

=

= +

= + =

−

−

˙

[ ˙ ˙ ]

, ˙ .ΩΩ ΩΩ



78


We seek a stress flux affording the simplest conversion from deformed to unde-formed coordinates. Note that

(5.15)

However, . Now,

, (5.16a)

in which

(5.16b)

is known as the

Truesdell stress flux

. Under pure rotation, .To prove the objectivity of , note that

(5.16c)

The choice of stress flux is not unique. For example, the stress flux given by

(5.17)

is also objective, as is the widely used Jaumann stress flux:

(5.18)

ddt

ddt

J

J J J J

ST

T T T T

=

= + + +

− −

− − − − − − − −

[ ]

( ) ˙ ˙ ˙ .

F F

D F F F F F F F F

T 1

T 1 T 1 T 1 T 1tr

( ) ˙ ˙ ˙ ˙F F 0 F F FF F F F F1 1 1 1 T T T T− − − − − − −= = − = −. so that and

ddt

JS

T= − ° −F FT 1

T T D T T To

= + − −˙ ( ) ,tr TL L

F Q S JQ QT= = and ˙ To

To

T T T T T

T T T

T T T T

T

o2 2 2 2 2 2 2 2

1 1 1

1 1 1 1 1

1

= + − −

= + − + − −[ ]= + − +

− −

•

˙ ( )

[ ] ( ) [ ]

˙ ( )

tr

tr T

tr

T

T

D L L

V V V V D VL V V V V V VL V

V V V V V V V V D

VL V V

1T T

1T T T T

1T

T T T T

1T

ΩΩ ΩΩ

ΩΩ ΩΩ

TT T T

T

1 1L V V V V V

V V

T1

T1

T

1T

T − +

=

ΩΩ ΩΩ

o.

T T

T T

= −

= − −

oTtr

T T

( )

˙

D

L L

T T T T∆

= + − = −˙ , ( ).W W W L LT 12




79

5.3 BALANCE OF MASS, LINEAR MOMENTUM, AND ANGULAR MOMENTUM

5.3.1 B

ALANCE

OF

MASS

Balance of mass requires that the total mass of an isolated body not change. Forexample,

(5.19)

in which is the mass density. Since , it follows that . Inaddition,

(5.20)

5.3.2 RAYLEIGH TRANSPORT THEOREM

Let denote a vector-valued function. Conversion of the volume integral fromdeformed to undeformed coordinates is achieved as

. (5.21)

The Rayleigh Transport Theorem is

. (5.22)

5.3.3 BALANCE OF LINEAR MOMENTUM

In a fixed-coordinate system, balance of linear momentum requires that the totalforce on a body with volume V and surface S be equal to the rate of change of linearmomentum:

. (5.23)

Invoking Equation 5.22 yields

. (5.24)

d

dtdVρ∫ = 0,

ρ( , )x t dV JdV= 0 ρ ρJ = 0

0 =

= ∂∂

+ ∂∂

= ∂∂

ddt

J

tJ J t t

t

( )

( ) ( ) ( , ), ( , ) .

ρ

ρ ρx

v x v xx

w x( , )t

ρ ρw x w x( , ) ( , )t dV t dVV V∫ ∫= 0 0

0

d

dtt dV

d

dtt dVρ ρw x w x 0( , ) ( , )∫ ∫=

=

Pu= =∫ ∫ττ dS

d

dt

d

dtdVρ

ττ dSd

dtdV∫ ∫= ρ

2

2

u




In current coordinates, application of the divergence theorem furnishes the equi-librium equation

(5.25)

Equation 5.24 now becomes

. (5.26)

Since this equation applies not only to the whole body, but also to arbitrarysubdomains of the body, the argument of the integral in Equation 5.26 must vanishpointwise:

(5.27)

To convert to undeformed coordinates, the 1st Piola-Kirchhoff stress is invokedto furnish

(5.28)

and the divergence theorem furnishes

(5.29)

in which ∇ο denotes the divergence operator referred to undeformed coordinates.This equation will later be the starting point in the formulation of incrementalvariational principles.

5.3.4 BALANCE OF ANGULAR MOMENTUM

Assuming that only surface forces are present, relative to the origin, the total momentof the traction is equal to the rate of change of angular momentum:

(5.30)

ττd d

d

S S

VT

∫ ∫∫

=

= ∇

T

T

T

T T

n

[ ] .

∇ −

=∫ T T Tu

0T ρ ddt

dT2

2 V

∇ =T T uT ρ d

dt

2

2

T

.

S T0

T

nu

dddt

dS V0 0

2

2 0∫ ∫= ρ ,

∇ =0 0

2

2T T

TuS ρ d

dt,

x x x× = ×∫ ∫ττ dSd

dtdVρ ˙ .



Mechanical Equilibrium and the Principle of Virtual Work 81

To examine this principle further, it is convenient to use tensor-indicial notation.First, note using the divergence theorem that

(5.31)

Continuing,

(5.32)

Balance of angular momentum can thus be restated as

. (5.33)

The second term vanishes by virtue of balance of linear momentum (see Equation5.27), leaving

, (5.34)

which implies that T is symmetric: T = TT (see exercises in Chapter 1). It followsfrom Equation 5.6 and Equation 5.11 that S is also symmetric, but that is notsymmetric.

x × =

=

=

= + ∂∂

= + ∂∂

∫ ∫∫∫

∫ ∫

∫ ∫

ττ ττdS x dS

x T n dS

xx T dV

T dV xx

T dV

T dV xx

T dV

ijk j k

ijk j lk l

lijk j lk

ijk jl lk ijk jl

lk

ijk jk ijk jl

lk

ε

ε

ε

ε δ ε

ε ε

∂∂

( )

.

d

dtdV dV dV

dV

x x dVijk j k

x x x x x x

x x

× = × + ×

= ×

=

∫ ∫ ∫∫∫

ρ ρ ρ

ρ

˙ ˙ ˙ ˙

˙

˙ .ε ρ

0 = + ∂∂

−

∫ ∫ε ε ρijk jk ijk j

llk kT dV x

xT x dV˙

ε ijk jkT = 0

S




5.4 PRINCIPLE OF VIRTUAL WORK

The balance of linear and angular momentum leads to auxiliary variational principlesthat are fundamental to the finite-element method. Variational methods were intro-duced in Chapter 3. We recall the balance of linear momentum in rectilinear coor-dinates as

, (5.35)

and by virtue of the balance of angular momentum. We have tacitly assumedthat , which is to say that deformed positions are referred to a coordinatesystem that does not translate or rotate. A variational principle is sought from

, (5.36)

in which is an admissible variation of . Consider the spatial dependence of to be subject to variation, but not the temporal dependence. For example, if

can be represented, at least locally, as

, (5.37a)

then

(5.37b)

The second term in the variational equation remains as . The firstterm is integrated by parts only once for reasons that will be identified shortly: itbecomes . From the divergence theorem,

(5.38)

which can be interpreted as the virtual work of the tractions on the exterior boundary.Next, since T is symmetric,

(5.39)

∂∂

− =x

T ul

kl kρ˙ 0

T Tlk kl=˙ ˙x uk=

δux

T u dVkl

kl k

∂∂

−

=∫ ρ˙ 0

δuk uk

uk uk

uT= N ( ) ( )x tγγ

δ δγu tk kl l= [ ( )] ( ).N xT

− ∫ δ ρu u dVk k˙

∫ − ∫∂∂ ∂

∂

x xlk lk

llku T dV T dV

uk[ ]δ

δ

∂∂

=

=

∫ ∫

∫x

u T dV n u T dS

u dS

lk lk l k lk

k k

[ ] [ ]

,

δ δ

δ τ

∂∂

=

=∂∂

+∂∂

∫ ∫δδ

u

xT dV T dV

u

x

u

x

k

llk kl lk

klk

l

l

k

3

312

,




and we call 3kl the Eulerian strain. The term ∫δ 3klTlkdV can be called the virtualwork of the stresses. Next, to evaluate , we suppose that the exteriorboundary consists of three zones: . On S1, the displacement uk isprescribed, causing the integral over S1 to vanish. On S2, suppose that the traction isprescribed as . The contribution is . Finally, on S3, suppose that tk =

, thus furnishing The var-ious contributions are consolidated into the Principle of Virtual Work (Zienkiewiczand Taylor, 1989) as

(5.40)

Now consider the case in which, as in classical elasticity,

(5.41)

in which dlkmn are the entries of a fourth-order, constant, positive-definite tensor D.The first term in Equation 5.40 now becomes , with a positive-definite integrand. Achieving this outcome is the motivation behind integrating byparts just once. In the language of Chapter 3, uk is the primary variable and τk isthe secondary variable: on any boundary point, either uk or τk should be specified,or, more generally, τk should be specified as a function of uk. Finally, application ofthe interpolation model (see Equation 5.37) and cancellation of δ γγγγT furnishes theordinary differential equation

, (5.42)

in which

(5.43)

and in which , χ = TEN22(D). In addition, .B is derived from the strain-displacement relations, and M is the positive-definitemass matrix. K is the positive-definite matrix representing the domain contributionto the stiffness matrix. H is the boundary contribution to the stiffness matrix, and fis a consistent force vector. These notions will be addressed in greater detail insubsequent chapters.

∫ δ τu dSk k

S S S S= + +1 2 3

τ k ( )s ∫S k ku dS2δ τ ( )s

τ k kl lu−( ) [ ( )] ( )s A s s ∫ − ∫S k k S k kl lu dS u u dS3 3δ τ δ( ) [ ( )] ( ) .s A s s

δ δ ρ δ τ δ τ δ3332

kl lk k k k k k k k kl lS

T dV u u dV u dS u dS u A u dS∫ ∫∫∫∫+ = + −˙ ( ) ( ) [ ( )] ( )s s s sss

Tlk lkmn mn= =d ,3 T D3

δ ∫ 12

3 3kl lkmn mn Vd d

M K H f˙ [ ] ( )γγ γγ+ + = t

M N x N x K B B

H N x AN x f N s s

T T

T

= =

= =

∫ ∫∫ ∫ +

ρ ( ) ( )

( ) ( ) ( ) ( ) ( )

dV dV

dV t dSS

kS S

D

3 2 3

ττ

VEC VEC( ) ( )T = χ 3 VEC( ) ( )3 B xT= γγ




To convert the Principle of Virtual Work to undeformed coordinates,

(5.44)

using, from Chapter 3,

. (5.45)

Next,

(5.46)

Some manipulation is required to convert the virtual work of the stresses.Observe that

(5.47)

Now

(5.48)

δ ρ δ ρ

δ τ δ τ δ τ δ τ

u u dV u u dV

u dS u dS u dS u dS

k k k k

k kS

k kS

k kS

k kS

˙ ˙

( ) ( ) ( ) ( ) ,

∫ ∫∫ ∫ ∫ ∫

→

+ → +

0 0

00 0

00 0

2 3 20 30

s s s s

τ µτ µk k J0 = = − n C n0T 1

0

δ δ µu u dS u u dSk kl lS

k kl lS

[ ( )] ( ) [ ( )] ( ) .A s s A s s3 3

0 0∫ ∫→

δ δ δ

δ δ

δ δ

δ δ

δ

3ux

ux

uX

Xx

uX

Xx

FF F F

F F F F F F

F F

T

T

1 T T

T T T 1

T

= ∂∂

+ ∂∂

= ∂∂

∂∂

+ ∂∂

∂∂

= +

= +

=

− −

− −

−

12

12

12

12

[ ]

[ ]

E −−1.

δ δ

δ

δ

δ

δ

3

0

0

0

0

kl lk

ji ij

T dV tr dV

tr JdV

tr J dV

tr dV

E S dV

∫ ∫∫∫∫∫

=

=

=

=

=

− −

− −

( )

( )

( )

( )

.

3T

F F T

E F F

T 1

1 T

E

T

ES




Consolidating the terms, the Principle of Virtual Work in undeformed coordinatesis

(5.49)

5.5 SAMPLE PROBLEMS

1. The following plate is subject to the tractions shown. At B, there is africtionless roller support. Find the reactions at A and B. Assume that theplate has thickness W.

Solution: The reactions are denoted as and are allassumed to be positive. Clearly, RBx = 0. Force equilibrium requires that

Balance of moments about A requires that

.

Accordingly, RAy = .

FIGURE 5.4 Plate under traction.

δ δ ρ δ τ δ τ

δ µ

E S dV u u dV u dS u dS

u u dS

ji ij k k k kS

k kS

k kl lS

0 0 00

0 00

0 0

0 0 0

20 30

30

∫ ∫ ∫ ∫

∫

+ = +

−

˙ ( ) ( )

[ ( )] ( ) .

s s

A s s

y

H

A B

L

x

τ = τ0[ ex + ey ] xL

R R R RAx Ay Bx By, , , and

R tWx

Ldx R

tW

L

R R tWx

Ldx R R

tW

L

Ax

L

Ax

Ay By

L

Ay By

− = ⇒ =

+ − = ⇒ + =

∫

∫0

0

02

02

.

LR tWx

Ldx R

tW

LBy

L

By+ = ⇒ = −∫2

00

3

5

6

tW

L




2. The figure shows a square plate of uniform thickness W. It experiencesCauchy stresses given by

.

On diagonal AB, find the total transverse force and moment about A.

Solution: The axes are rotated through 45 degrees so that the x-axis coincideswith the diagonal (x′-axis). Now,

The transverse force is

The moment about A is

.

FIGURE 5.5 Plate element under stress.

T =

T T

T

11 12

21

0

0 0

0 0 0

y

x

A

B

L

TT

T TT

T TT

x xxx

xy y yxx

xy x yxx

′ ′ ′ ′ ′ ′= + = − = −2 2 2

.

FT

T WLyxx

xy′ = −

2

2 .

M LFy= ′2

2




3. The plate illustrated has thickness W, width dx, and height dy. The stressesTxz, Tyz, and Tzz vanish. Find the equations resulting from balancing forcesand moments. In particular, prove that

.

Solution: For the x-direction forces,

.

For y-direction forces,

.

For moments about A,

FIGURE 5.6 Moment balance on a plate element under nonuniform stress.

T Txy yx=

Txx

Txy

Tyx + dTyx

Txy + dTxy

Txx + dTxx

Tyy + dTyy

Tyx

Tyy

dx

dy

W TdT

dxdx T dy W T

dT

dydy T dx

dT

dx

dT

dyxxxx

xx yxyx

yxxx yx+ −

+ + −

= ⇒ + =0 0

W TdT

dxdx T dy W T

dT

dydy T dx

dT

dx

dT

dyxyxy

xy yyyy

yyxy yy+ −

+ + −

= ⇒ + =0 0

− + −

− +

+ +

+ + −

=

W TdT

dxdx T dy

dyW T

dT

dydy dxdy

W TdT

dxdx dydx W T

dT

dydy T dx

dx

xxxx

xx yxyx

xyxy

yyyy

yy

2

20 .




Consequently, neglecting terms of third order in differentials,

.

4. At point (0,0,0), the tractions ττττ1, ττττ 2, ττττ 3 act on planes with normal vectorsn1, n2, and n3. Find the Cauchy stress tensor T given

Solution: This problem requires application of the stress-traction relation.Now

.

Similarly,

− + = ⇒ =WT dxdy WT dydx T Tyx xy xy yx0

n e e e e e e

n e e e e e e

n e e e e e e

1 1 1 1

1 1

1 1

= + + = + +

= + − = + +

= − − = − + +

1

3

1

36 9 15

1

3

1

30 1 3

1

3

1

34 5 7

2 3 2 3

2 2 3 2 2 3

3 2 3 3 2 3

[ [ ]

[ [ ]

[ [ ].

]

]

]

ττ

ττ

ττ

1

3

6

9

12

1

3

1

1

1

11 12 13

21 22 23

31 32 33

=

T T T

T T T

T T T

⇒ + + = + + = + + =T T T T T T T T T11 12 13 21 22 23 31 32 336 9 1 1) 2) 3)( ( (

1

2

3

5

7

1

2

1

1

0

11 12 13

21 22 23

31 32 33

=

T T T

T T T

T T T

⇒ + = + = + =T T T T T T11 12 21 22 31 323 5 7 4) 5) 6)( ( (

1

3

4

5

6

1

3

1

1

1

11 12 13

21 22 23

31 32 33

−

−

−

=

−

−

T T T

T T T

T T T

⇒ − − = − − − = − − − = −T T T T T T T T T11 12 13 21 22 23 31 32 334 5 6 7) 8) 9)( ( (




The solution is straightforward. For example, adding (1) and (7) furnishes T11 =1, from which (4) implies T12 = 2.

Solving the equations furnishes us with

.

5. Given the Cauchy stress T, find and S if x(t) = Q(t)X, Q(t)Q(t)T = I.

Solution: Clearly F = Q, thus J = 1. Also, F−T = Q. It is immediate that =Q(t)T and S = Q(t)TQ(t)T.

5.6 EXERCISES

1. In classical linear elasticity, introduce the isotropic stress and isotropiclinear strain as

and introduce the deviatoric stress and strain using

Verify that, if

.

2. Find the reaction forces in the following problem. There is a frictionlessroller support at B.

T =

1 2 3

2 3 4

3 4 5

S

S

s e= =tr tr( ) ( ),S E L

s s e ed di i= − = −13

13

s es .

˙ ˙ ( )s e i= +z trµ λ εε

i iTd

Tds = =0 0 e

s ed d= 2µ

s = +( )2 3µ λ e




3. Find the net force and moment about A (thickness = W) on the diagonalin the following plate, given the Cauchy stresses

.

4. At point (0,0,0), the tractions ττττ1, ττττ 2, ττττ 3 act on planes with normal vectorsn1, n2, and n3. Find the Cauchy stress tensor T, given,

FIGURE 5.7 Plate element under traction.

FIGURE 5.8 Plate element experiencing stress.

y

H

A B

L

x

τ = τ0[–ex + ey ][1 –y/H]

T

x y

y xy=

0

0

0 0 0

y

xA

L

H

B

n e e e e e e

n e e e e e

n e e e e e e

1 1

1 1

1 1

1 2 3 1 2 3

2 2 2 2 3

3 2 3 3 2 3

1

3

1

32 5 6

1

2

1

2

1

3

1

36 1 2

= + + = − +

= + = − +

= − − = − − +

[ [ ]

[ [ ]

[ [ ].

]

]

]

ττ

ττ

ττ




5. Given the Cauchy stress tensor T, find the 1st and 2nd Piola-Kirchhoff stresstensors if x(t) = Q(t)∆∆∆∆(t)X, in which

.

6. The Cauchy stress tensor is given by

.

Now suppose the 1-2 axes are rotated through +θ degrees around thez-axis, to furnish 1′ − 2′ axes. Prove that

7. Verify that

is the inverse of .8. In undeformed coordinates and Kronecker Product (VEC) notation, the

2nd Piola-Kirchhoff stress for incompressible hyperelastic materials canbe written as

.

Find the corresponding expression in deformed coordinates. Derive, in which direct transformation furnishes

.

∆∆( )t

at

bt

ct

=

+

+

+

1 0 0

0 1 0

0 0 1

T =

T T

T T

11 12

12 22

0

0

0 0 0

TT T T T

T

TT T T T

T

TT T

T

′ ′

′ ′

′ ′

=+

+−

+

=+

−−

−

= −−

+

1 111 22 11 22

12

2 211 22 11 22

12

1 211 22

12

2 22 2

2 22 2

22 2

cos sin

cos sin

sin cos

θ θ

θ θ

θ θ

E T TijL

ij kk ij( ) = −

+

12 2 3µ

λµ λ

δ

T E Eij ijL

kkL

ij= +2µ λ δ( ) ( )

se

n n n= ∂ ′∂

= ′ ′ + ′ ′ −wpI2 21 1 2 2 3 3ϕ ϕ

′ ′ψ ψ1 2 and

t m m i= ′ ′ + ′ ′ −2 21 1 2 2ψ ψ p




9. Prove that, under uniaxial tension in an isotropic, linearly elastic material,E22 = E33.

10. Obtain λ in terms of E and ν. (See Problem 7.)11. The bulk modulus κ is defined by . Obtain κ as a function of

E and ν. (See Problem 7.)12. The 2" × 2" × 2" cube shown in Figure 5.9 is confined on its sides facing

the ± x faces by rigid, frictionless walls. The sides facing the ±z faces arefree. The top and bottom faces are subjected to a compressive force of100 lbf. Take E = 107 psi and ν = 1/3. Find all nonzero stresses and strains.What is the volume change? What are the principal stresses and strains?What is the maximum shear stress?

13. Write out the proof of the Principle of Virtual Work in deformed coordi-nates.

14. Write out the conversion of the Principle of Virtual Work to undeformedcoordinates.

15. Write out the steps whereby the Principle of Virtual Work leads to

.

16. Suppose that u = NT(x)γγγγ(t) and assume two dimensions. In 2-D, find BT

in terms of the derivatives of NT(x), in which

FIGURE 5.9 Strain in a constrained cube.

t ekk kkL= 3κ ( )

y

F

x

z

M K H f˙ [ ] ( )γγ γγ+ + = t




ΦΦΦΦ is a 1 × n constant matrix (row vector); γγγγ is n × 1.

3

3

3

3

1 0 0 0

0 0 0 1

11

21

12

22

1

2

= ( )

=

=

B x N x

N x0

0

T T

T

( ) , ( ) ( )

( ) .

γγ γγ

ΦΦ

ΦΦ

tu

ut

x y

x y



95

Stress-Strain Relation and the Tangent-Modulus Tensor

6.1 STRESS-STRAIN BEHAVIOR: CLASSICAL LINEAR ELASTICITY

Under the assumption of linear strain, the distinction between the Cauchy and Piola-Kirchhoff stresses vanishes. The stress is assumed to be given as a linear functionof linear strain by the relation

(6.1)

in which

c

ijkl

are constants and are the entries of a 3

×

3

×

3

×

3 fourth-order tensor,

C

. If

T

and

E

L

were not symmetrical,

C

might have as many as 81 distinct entries.However, due to the symmetry of

T

and

E

L

there are no more than 36 distinct entries.Thermodynamic arguments in subsequent sections will provide a rationale for the

Maxwell

relations:

(6.2)

It follows that

c

ijkl

=

c

klij

, which implies that there are, at most, 21 distinct coeffi-cients. There are no further arguments from general principles for fewer coefficients.Instead, the number of distinct coefficients is specific to a material, and reflects thedegree of symmetry in the material. The smallest number of distinct coefficients isachieved in the case of isotropy, which can be explained physically as follows.Suppose a thin plate of elastic material is tested such that thin strips are removedat several angles and then subjected to uniaxial tension. If the measured stress-straincurves are the same and independent of the orientation at which they are cut, thematerial is isotropic. Otherwise, it exhibits anisotropy, but may still exhibit limitedtypes of symmetry, such as transverse isotropy or orthotropy. The notion of isotropyis illustrated in Figure 6.1.

In isotropic, linear-elastic materials (which implies linear strain), the number ofdistinct coefficients can be reduced to two,

m

and

l

, as illustrated by Lame’s equation,

(6.3)

6

T c Eij ijkl klL= ( ),

∂∂

=∂∂

T

E

T

Eij

kl

kl

ij

.

T E Eij ijL

kkL

ij= +2µ λ δ( ) ( ) .



96


This can be inverted to furnish

(6.4)

The classical elastic modulus E

0

and Poisson’s ratio

n

represent response

underuniaxial tension only

, provided that

T

11

=

T

,

T

ij

= 0. Otherwise,

(6.5)

It is readily verified that

(6.6)

from which it is immediate that

(6.7)

Leaving the case of uniaxial tension for the normal (diagonal) stresses andstrains, we can write

(6.8)

(6.9)

FIGURE 6.1

Illustration of isotropy.

1 23

4

T

2 1

34

E

E T TijL

ij kk ij( ) .= −

+

12 2 3µ

λµ λ

δ

E = = − = −T

E

E

E

E

EL

L

L

L

L11

11

22

11

33

11( )

( )

( )

( )

( ) .ν

1 12

12 3

12 2 3E

E

= −+

=

+µλ

µ λν

µλ

µ λ,

12

1µ

ν= +E

,

E T T T

T T T

L11 11 22 33

11 22 33

12

12 3

12 2 3

1

( ) ( )

[ ( )],

= −+

−

++

= − +

µλ

µ λ µλ

µ λ

νE

E T T TL22 22 33 11

1( ) [ ( )],= − +E

ν




97

and

(6.10)

and the off-diagonal terms satisfy

(6.11)

6.2 ISOTHERMAL TANGENT-MODULUS TENSOR

6.2.1 C

LASSICAL

E

LASTICITY

Under small deformation, the fourth-order tangent-modulus tensor

D

in linear elas-ticity is defined by

(6.12)

In linear isotropic elasticity, the stress-strain relations are written in the Lame’sform as

(6.13)

Using Kronecker Product notation from Chapter 2, this can be rewritten as

(6.14)


(6.15)

6.2.2 C

OMPRESSIBLE

H

YPERELASTIC

M

ATERIALS

In isotropic hyperelasticity, which is descriptive of compressible rubber elasticity,the 2

nd

Piola-Kirchhoff stress is taken to be derivable from a strain-energy functionthat depends on the principal invariants

I

1

,

I

2

,

I

3

of the Right Cauchy-Green straintensor:

(6.16)

E T T TL33 33 11 22

1( ) [ ( )],= − +E

ν

E T E T E TL L L12 12 23 23 31 31

1 1 1( ) ( ) ( ) .= + = + = +ν ν νE

E

E

d dT D E= L .

T E E= +2µ λL L Itr( ) .

VEC VEC( )T E= ⊗ +[ ] ( ),2µ λI I iiTL

D = ⊗ +ITEN22 2( ).µ λI I iiT

SE C

se

s S e E c C

= = = =

= = =

dd

dd

dd

dd

(a)

(b)

w w w w

VEC VEC VEC

2 2T

c

( ( () ) ).



98


Now,

(6.17)

From Chapter 2,

(6.18)

The tangent-modulus tensor

D

o

, referred to the undeformed configuration, isgiven by

(6.19)

and now

(6.20)

Finally,

(6.21)

In deriving

A

3

we have taken advantage of the Cayley-Hamilton theorem (seeChapter 2).

s n nT= = ∂∂

= ∂∂

2φ φi i ii c

, , .w

ii

I

I

n n i n 11 2 1 3 3= = − = −i c CI VEC I( ) .

dS D E s D E= =o oTENd d )d22( ,

TEN o22( )d

dD

c= + =4 4φ φij i j i i i

in n A AnT , .

A

A

A

1

2 2

1

9

31

3

2 12

2

1

= =

=

= −

= −

=

= − +

= − + ⊕

= − − + + ⊕

−

dd

(a)

dd

dd

[

(b)

dd

dd

(c)

i0

i

ii I

c

i

in i

ii I i i

T

T

T T

T9

T T

c

c

cc

C

cc C

c C C

c c C C

I

I

VEC I

I I VEC

I

]

[ ( ) ]

[ ( )]

[ ] [ ]




99

6.3 INCOMPRESSIBLE AND NEAR-INCOMPRESSIBLE HYPERELASTIC MATERIALS

Polymeric materials, such as natural rubber, are often nearly incompressible. Forsome applications, they can be idealized as incompressible. However, for applica-tions involving confinement, such as in the corners of seal wells, it may be necessaryto accommodate the small degree of incompressibility to achieve high accuracy.Incompressibility and near-incompressibility represent

internal constraints

. Theprincipal (e.g., Lagrangian) strains are not independent, and the stresses are notdetermined completely by the strains. Instead, differences in the principal stressesare determined by differences in principal strains (Oden, 1972). An additional fieldmust be introduced to enforce the internal constraint, and we will see that this internalfield can be taken as the hydrostatic pressure (referred to the current configuration).

6.3.1 I

NCOMPRESSIBILITY

The constraint of incompressibility is expressed by the relation

J

=

1. Now,

(6.22)

and consequently,

(6.23)

The constraint of incompressibility can be enforced using a Lagrange multiplier(see Oden, 1972), denoted here as

p

. The multiplier depends on

X

and is, in fact,the additional field just mentioned. Oden (1972) proposed introducing an augmentedstrain-energy function,

w

′

, similar to

(6.24)

in which

w

is interpreted as the conventional strain-energy function, but with depen-dence on

I

3

(

=

1) removed. are called the deviatoric invariants. For reasonsto be explained in a later chapter presenting variational principles, this form serves

J

I

=

=

=

=

=

det

det ( )

det( )det( )

det ( )

,

F

F

F F

F F

2

2

3

T

T

I3 1= .

′ = ′ ′ − − ′ = ′ =w w I I p I II

II

I

I( ) ( ) ,

/ /1 2 3 11

221

21

3

1 3

3

2 3, , ,

′ ′I I1 2 and



100


to enforce incompressibility, with

S

now given by

(6.25)

To convert to deformed coordinates, recall that

S

= J

F

−

1

T

F

−

T

. It is left to thereader in an exercise to derive in which

(6.26)

in which It follows that

p

=

−

tr

(

T

)

/

3 since

(6.27)

Evidently, the Lagrange multiplier enforcing incompressibility is the “true”hydrostatic pressure.

Finally, the tangent-modulus tensor is somewhat more complicated because d

S

depends on d

E

and d

p

. We will see in subsequent chapters that it should be definedas

D

∗

using

(6.28)

Example: Uniaxial Tension

Consider the Neo-Hookean elastomer satisfying

(6.29)

se

c c

= ∂ ′∂

= ′ ′ + ′ ′ −

′ = ∂∂ ′

′ = ∂∂ ′

′ =∂ ′∂

′ =∂ ′∂

w

p

w

I

w

I

I I

2 21 2 2 3 3

11

22

11

22

φ φ

φ φ

1n n n

n nT T

I

.

′ ′ψ ψ1 2 and

t m m= = ′ ′ + ′ ′ −VEC p( )T 2 21 1 2 2ψ ψ i

i m i mT T′ = ′ =1 20 0 and .

tr

p

p

p

( )

.

T =

= ′ ′ + ′ ′ −

= −

= −

i t

i m i m i i

i i

T

T1

T T

T

2 2

3

1 2 2ψ ψ

TENp

p

22( )D

se

s

s∗ =

−

dd

dd

dd

T

0

dd

dd

se

s

= ′ ′ + ′ ′ + ′ ′ ′ + ′ ′ ′

+ ′ ′ ′ + ′ ′ ′

= −

4 1 1 2 2 11 1 1 12 1 2

21 2 1 22 2 2

3 3

[ ( ) ( )

( ) ( ) ]

.

φ φ φ φ

φ φ

A A n n n n

n n n n

T T

T T

pI n

w I= −[ ] =α I and subject to1 33 1.




101

We seek the relation between

s

1

and

e

1

, which will be obtained twice: once byenforcing the incompressibility constraint

a priori

, and the second by enforcing theconstraint

a posteriori

.

a priori

: Assume for the sake of brevity that

e

2

=

e

3

.

I

3

=

1 implies that. The strain-energy function now is The stress,

s

1

, isnow found as

(6.30)

a posteriori

: Use the augmented function

(6.31)

Now

(6.32)

Thus, it follows that We conclude that

(6.33)

c c2 11= / w c c= + −α[ ].11

2 3

swc c1

1 13 22 2 11= = −

dd

α / .

′ = − − −w Ip

Iα[ ] [ ].1 332

1

sw

cp c

sw

cp c

sw

cp c

w

pI

11

1

22

2

33

3

3

2 2

0 2 2

0 2 2

0 0 1

= ′ = −

= = ′ = −

= = ′ = −

= ′ = → =

dd

dd

dd

dd

α

α

α

/

/

/

.

c c c p c2 3 1 21 2= = =/ and / α.

s p c

c

cp c

c

c

c

1 1

2

12

2

1

13 2

2

2

2 1

2 11

= −

= −

= −

= −

α

α

α

α

/

/

./



102


a posteriori with deviatoric invariants

: Consider the augmented function withdeviatoric invariants:

(6.34)

Hence,

c

2

=

c

3

. Furthermore,

(6.34e)

Equation 6.34 now implies that

(6.35)

and substitution into Equation 6.34b furnishes

(6.36)

as in the

a priori

case and in the first

a posteriori

case. Now, the Lagrange multiplier

p

can be interpreted as the pressure referred to current coordinates.

6.3.2 N

EAR

-I

NCOMPRESSIBILITY

As will be seen in Chapter 18, the augmented strain-energy function

(6.37)

serves to enforce the constraint

(6.38)

′ = −[ ] − −

= ′ = −

−

= = ′ = −

−

=

w I Ip

I

sw

c I

I

I

I

cp

I

c

sw

c I

I

I

I

cp

I

c

α

α

α

1 31 3

3

11 3

1 31

34 3

3

1

3

1

22 3

1 31

34 3

3

2

3

2

32

1

2 21 1

3

0 2 21 1

3

0

/ [ ] (a)

dd

(b)

dd

(c)

ssw

c I

I

I

I

cp

I

c33 3

1 31

34 3

3

3

3

3

2 21 1

3= ′ = −

−d

d(d)α

dd

′ = → =wp

I0 13 .

23

2 1

2 2

α c cc

pc

−

= ,

sc1

13 22 11= −

α / ,

′′ = ′ ′ − − −w w I I p Ip

( , ) [ ]1 2 3

212

12κ

p I= − −[ ]κ 3 1 .




103

Here,

k

is the bulk modulus, and it is assumed to be quite large compared to,for example, the small strain-shear modulus. The tangent-modulus tensor is now

(6.39)

6.4 NONLINEAR MATERIALS AT LARGE DEFORMATION

Suppose that the constitutive relations are measured at a constant temperature in thecurrent configuration as

(6.40)

in which the fourth-order tangent-modulus tensor

D

can, in general, be a functionof stress, strain, temperature, and internal-state variables (discussed in subsequentchapters). This form is attractive since and

D

are both objective. Conversion toundeformed coordinates is realized by

(6.41)

If

s

=

VEC

(

S

) and

e

=

VEC

(

E), then

(6.42)

Recalling Chapter 2, it follows that

(6.43)

in which

(6.44)

is the tangent-modulus tensor Do referred to undeformed coordinates.

TENp

p

221

D

se

s

s* .( ) =

−

dd

dd

dd

T

κ

To

= DD,

To

˙

˙ .

S D

D E

=

=

− −

− − − −

J

J

F DF

F F F F

T

T T

1

1 1

˙ ( ˙ )

( ) ( ˙ )

( )( )( ) ˙

( ) ˙

( )˙.

s = ⊗

= ⊗ ⊗

= ⊗ ⊗ ⊗

= ⊗ ⊗

=

− − − −

− − − −

− − − −

− − − −

J

J

J ( )

J

I F F F F

I F F I F F

F F F I I F

F F F F e

e

T T

T

T T

T T

1 1

1 1 1

1 1

1 1

22

22

22

22

VEC

TEN VEC

TEN VEC

TEN

TEN o

D E

D

D E

D

D

E

˙ ˙ ,S D E= o

D Do ITEN TEN= ⊗ ⊗− − − −22 221 1( ( ) )JF F F FT T




Suppose instead that the Jaumann stress flux is used and that

(6.45)

Now

(6.46)

For this flux, there does not appear to be any way that can be written in theform of Equation 6.42, i.e., is determined by . To see this, consider

(6.47)

In the second term in Equation 6.47a, VEC(L + W) cannot be eliminated in favorof VEC(D), and hence in favor of . Note that

(6.48)

I + U9 is singular, as seen in the following argument. Recall that U9 is symmetricand thus has real eigenvalues. However, and thus the eigenvalues of U9 areeither 1 or −1. Some of the eigenvalues must be −1. Otherwise, U9 would be theidentity matrix, in which case it would not, in general, have the permutation propertyidentified in Chapter 2. Thus, some of the eigenvalues of I + U9 vanish. Instead, wewrite

(6.49)

and we see that if the Jaumann stress flux is used, is determined by and thespin W. Recall that the spin does not vanish under rigid-body rotation.

6.5 EXERCISES

1. In classical linear elasticity, introduce the isotropic stress and isotropiclinear strain as

T D∆

= D.

˙ ( ) ( )

˙ ( ˙ ) ( ) ( ) .

S T D T T

D E T E T T

= + − + − −

= + − + − −[ ]

− −

− − − − − −

J ( )

J

F L W L W F

F F F F F L W L W F

T T

T T T T

T∆

tr

tr

1

1 1 1

SE

VEC TEN VEC TEN VEC

TEN TEN VEC VEC

TEN

( ˙ ( ) ( ˙ ( ) ( )

( ) [ ( ) ( ) ( ) ( )]

( ) [ [ ]

S D E D

D D T

D T T

) )

J

J ]

= ′ − ′′ +

′ = ⊗ +

′′ = ⊗ ⊗ − ⊗

− − − −

−

22 22

22 22

22

2

1 29

L W

I F F F FF

I F I I U

T T T T T

(a)

(b)

(c)

VEC( )E

VEC VEC VEC( ) (D L L I U LT) ( ( ) ).= + = +12

12 9

U I92 =

VEC VEC VEC VEC( ) ( ) ( ) ( ),˙ ˙S D D D D= ′ − ′′ − ′′ +E D W2

S E

s e L= =tr tr( ) ( ,S E )



Stress-Strain Relation and the Tangent-Modulus Tensor 105

and introduce the deviatoric stress and strain using

Verify that

2. Verify that Equation 6.3 can be inverted to furnish

3. In classical linear elasticity, under uniaxial stress, the elastic modulus Eand Poisson’s ratio ν are defined by

Prove that

4. Obtain v, L, D, and W in spherical coordinates.5. For cylindrical coordinates, find L, D, and W for the following flows:

6. In undeformed coordinates, the 2nd Piola-Kirchhoff stress for an incom-pressible, hyperelastic material is given by

s s i e e id d= − = −13

13

s e .

i s i e

s e

Td

Td

d d

= =

=

= +

0 0

2

2 3

µ

µ λs e( ) .

e t i t i tT= −+

=1

2 2 3µλ

µ λ( )( ) , ( )VEC T

Eijij

= = − = −≠

= ≠

σε

νεε

εε

σ

σ11

11

22

11

33

11

11 0

0 11 ,

.

1 12

12 3 2 2 1E

E= −+

= −

+=

+µλ

µ λν λ

µ λµ

ν( ) ( ) ( ).

(a) pure radial flow:

(b) pipe flow:

(c) cylindrical flow:

(d) torsional flow:

v f r v v

v v v f r

v v f r v

v v f z v

r z

r z

r z

r z

= ( ) = =

= = = ( )

= = ( ) =

= = ( ) =

, ,

, ,

, ,

, ,

θ

θ

θ

θ

0 0

0 0

0 0

0 0

se

n n n= ∂ ′∂

= ′ ′ + ′ ′ −wpI2 21 1 2 2 3 3ϕ ϕ .




Find the corresponding expression in deformed coordinates: derive in which direct transformation furnishes

7. Prove that under uniaxial tension in an isotropic linearly elastic materiale22 = e33. (Problem 9, Chapter 5.)

8. Obtain λ in terms of E and ν. (Problem 10, Chapter 5.)9. The bulk modulus K is defined by Obtain K as a function

of E and ν. (Problem 11, Chapter 5.)10. The 2" × 2" × 2" shown in Figure 6.2 is confined on its sides facing the

± x faces by rigid, frictionless walls. The sides facing the ± z faces arefree. The top and bottom faces are subjected to a compressive force of100 lbf. Take E = 10 7 psi and ν = 1/3. Find all nonzero stresses andstrains. What is the volume change? What are the principal stresses andstrains? What is the maximum shear stress? (Problem 12, Chapter 5.)

FIGURE 6.2 Strain in a constrained plate.

′ ′ψ ψ1 2 and ,

t m m i= ′ ′ + ′ ′ −2 21 1 2 2ψ ψ p .

t ekk kkL= 3Κ ( ).

z

v

x

E



107

Thermal and Thermomechanical Response

7.1 BALANCE OF ENERGY AND PRODUCTIONOF ENTROPY

7.1.1 B

ALANCE

OF

E

NERGY

The total energy increase in a body, including internal energy and kinetic energy, isequal to the corresponding work done on the body and the heat added to the body.In rate form,

(7.1)

in which:

Ξ

is the internal energy with density

ξ

(7.2a)

is the rate of mechanical work, satisfying

(7.2b)

is the rate of heat input, with heat production

h

and heat flux

q

, satisfying

(7.2c)

is the rate of increase in the kinetic energy,

(7.2d)

It has been tacitly assumed that all work is done on

S

, and that body forces dono work.

7

˙ ˙ ˙ ˙,K + = +Ξ W Q

˙ ˙ ;Ξ = ∫ ρξdV

W

˙ ˙ ,W dS= ∫ uTττ

Q

˙ ,Q hdV dS= −∫ ∫ρ n qT and

K

˙ ˙˙

.K = ∫ ρuuT d

dtdV



108


Invoking the divergence theorem and balance of linear momentum furnishes

(7.3)

The inner bracketed term inside the integrand vanishes by virtue of the balanceof linear momentum. The relation holds for arbitrary volumes, from

which the localform of balance of energy, referred to undeformed coordinates, is obtained as

(7.4)

To convert to undeformed coordinates, note that

(7.5)

In undeformed coordinates, Equation 7.3 is rewritten as

(7.6a)

furnishing the

local form

(7.6b)

7.1.2 E

NTROPY

P

RODUCTION

I

NEQUALITY

Following the thermodynamics of ideal and non-ideal gases, the entropy productioninequality is introduced as follows (see Callen, 1985):

(7.7a)

in which

H

is the total entropy,

η

is the specific entropy per unit mass, and T is theabsolute temperature. This relation provides a “framework” for describing the irre-versible nature of dissipative processes, such as heat flow and plastic deformation.We apply the divergence theorem to the surface integral and obtain the local formof the entropy production inequality:

(7.7b)

ρξ ρ ρ˙ ˙˙

( ) .+ − ∇

− − + ∇

=∫ uu

D qT T Td

dttr h dVT T 0

ρξ ρ˙ ( ) .= − ∇ +tr TD qT h

n q q F n

q n q F q

T T T

T

dS J dS

dS J

∫ ∫∫

=

= =

−

−

0 0

0 0 0 01 .

ρ ξ ρ0 0 0 0 0 0˙ ( ˙ ,− − + ∇[ ] =∫ tr dVSE) h Tq

ρ ξ ρ0 0 0 0 0˙ ( ˙ ) .− − + ∇ =tr SE h Tq

˙ ˙ ,H = ≥ −∫ ∫ ∫ρηdVhT

dVT

dSn qT

ρ ηT h T/T˙ .≥ − ∇ + ∇T Tq q




109

The corresponding relation in undeformed coordinates is

(7.7c)

7.1.3 T

HERMODYNAMIC

P

OTENTIALS

The Balance of Energy introduces the internal energy

Ξ

, which is an extensivevariable—its value accumulates over the domain. The mass and volume averagesof extensive variables are also referred to as extensive variables. This contrasts withintensive, or pointwise, variables, such as the stresses and the temperature. Anotherextensive variable is the entropy

H

. In reversible elastic systems, the heat flux iscompletely converted into entropy according to

(7.8)

(We shall consider several irreversible effects, such as plasticity, viscosity, and heatconduction.) In undeformed coordinates, the balance of energy for reversible pro-cesses can be written as

(7.9)

We call this equation the thermal equilibrium equation. It is assumed to beintegrable, so that the internal energy is dependent on the current state representedby the current values of the state variables

E

and

η

. For the sake of understanding,we can think of

T

as a thermal stress and

η

as a thermal strain. Clearly, ifthere is no heat input across the surface or generated in the volume. Consequently,the entropy is a convenient state variable for representing adiabatic processes.

In Callen (1985), a development is given for the stability of thermodynamicequilibrium, according to which, under suitable conditions, the strain and the entropydensity assume values that maximize the internal energy. Other thermodynamicpotentials, depending on alternate state variables, can be introduced by a Lorentztransformation, as illustrated in the following equation. Doing so is attractive if thenew state variables are accessible to measurement. For example, the Gibbs FreeEnergy (density) is a function of the intensive variables

S

and T:

(7.10a)

from which

(7.10b)

Stability of thermodynamic equilibrium requires that

S

and T assume values thatminimize

g

under suitable conditions. This potential is of interest in fluids experi-encing adiabatic conditions since the pressure (stress) is accessible to measurementusing, for example, pitot tubes.

ρ η0 0 0 0 0T h T T˙ / .≥ − ∇ + ∇T Tq q

˙ ˙ .Q H= T

ρ ξ ρ η0 0˙ ( ˙ ˙.= +tr SE) T

η = 0

ρ ρ ξ ρ η0 0g tr= − −( ) ,SE 0T

ρ ρ η0 0 T˙ ( ˙) ˙ .g tr= − −ES



110


In solid continua, the stress is often more difficult to measure than the strain.Accordingly, for solids, the Helmholtz Free Energy (density)

f

is introduced using

(7.11a)

furnishing

(7.11b)

It is evident that

f

is a function of both an intensive and an extensive variable.At thermodynamic equilibrium, it exhibits a (stationary) saddle point rather than amaximum or a minimum. Finally, for the sake of completeness, we mention a fourthpotential, known as the enthalpy

ρ

0

h

=

ρ

0

x

−

tr

(

SE

), and

(7.12)

The enthalpy also is a function of an extensive variable and an intensive variableand exhibits a saddle point at equilibrium. It is attractive in fluids under adiabaticconditions.

7.2 CLASSICAL COUPLED LINEAR THERMOELASTICITY

The classical theory of coupled thermoelasticity in isotropic media corresponds tothe restriction to the linear-strain tensor, and to the stress-strain temperaturerelation

(7.13)

Here,

α

is the volumetric coefficient of thermal expansion, typically a smallnumber in metals. If the temperature increases without stress being applied, thestrain increases according to

e

vol

=

tr

(

E

)

=

α

(

T

−

T

0

). Thermoelastic processes areassumed to be reversible, in which case, It is also assumed thatthe specific heat at constant strain,

c

e

, given by

(7.14)

is constant. The balance of energy is restated as

(7.15)

Recalling that

ξ

is a function of the extensive variables

E

and

η

, to convert to

E

and T as state variables, which are accessible to measurement, we recall the

ρ ρ ξ ρ η0 0 0f = − T ,

ρ ρ η0 0˙ ( ˙ ) ˙ .f tr= −SE T

ρ ρ η0 0˙ ( ˙ ˙.h ) T= − +tr ES

E E≈ ˙L

T E E= + − −2 0µ αλ[ ( ) ( )] .tr T T I

−∇ ⋅ + =q h Tρ η.

c TTe = ∂

∂η

,

ρ ξ ρ η0˙ ( ˙ ) ˙= −tr TE T




111

Helmholtz Free Energy

f

=

e

+

T

η

. Since is an exact differential, to ensure pathindependence, we infer the Maxwell relation:

(7.16)

Returning to the energy-balance equation,

(7.17)

Also, note that

(7.18a)

thus

(7.18b)

We previously identified the coefficient of specific heat, assumed constant, as

c

e

=

so that

(7.19)

From Equation 7.13, upon approximating T as T

0

, From Fou-rier’s Law,

q

=

−

k

∇

T

. Thus, the thermal-field equation now can be written as

(7.20)

f

− ∂∂

= ∂∂

ρ ηE

T

T ET ˙.

ρξ ρ ξ ρ ξη

η

ρ ξ ρ ξη

η ρ ξη

η

η

η

˙ ˙ ˙

˙ ˙

= ∂∂

+ ∂

∂

= ∂∂

+ ∂∂

∂∂

+ ∂

∂∂∂

tr

tr

EE

E EE

E

E T E ETT.

T = ∂∂

= ∂∂

ξ ξηηE

TE

,

ρξ ρ ξ ρ ξη

η ρ ξη

η

ρ η

η

˙ ˙ ˙

˙ ˙ .

= ∂∂

+ ∂∂

∂∂

+ ∂

∂∂∂

= − ∂∂

+ ∂

∂

tr

trE

E E

TT

E

E T E E

E

ET

T

TTT

TT

T ∂∂

ηT E

,

ρξ ρ˙ ˙ ˙= − ∂∂

+tr ceTT

TT

T.E

T .T0 0∂∂ = −T αλT I

k tr∇ = +20T T c Teαλ ρ( ˙ ) ˙E .



112


The balance of linear momentum, together with the stress-strain and strain-displacement relations of linear isotropic thermoelasticity, imply that

(7.21)

from which we obtain the mechanical-field equation (Navier’s Equation forThermoelasticity):

(7.22)

The thermal-field equation depends on the mechanical field through the term Consequently, if

E

is static, there is no coupling. Similarly, the mechan-ical field depends on the thermal field through

αλ

∇

T

, which often is quite small in,for example, metals, if the assumption of reversibility is a reasonable approximation.

We next derive the entropy. Since is constant, we conclude that ithas the form

(7.23)

where

η

*(

E

) remains to be determined. We take

η

0

to vanish. However, implying that

(7.24)

Now consider

f

, for which the fundamental relation in Equation 7.11b implies

(7.25)

Integrating the entropy,

(7.26)

in which

f

*

(

E

), remains to be determined. Integrating the stress,

(7.27)

∂∂

∂∂

+∂∂

+∂∂

− −

=∂∂x

u

x

u

x

u

x

u

tj

i

j

j

i

k

kij

i212 0 2µ λ α δ ρ( ) ,T T

µ λ µ αλ ρ∇ + + ∇ − ∇ = ∂∂

22uu

( ) ( ) .tr E Tt

αλT0tr( ˙ ).E

ce = ∂∂T ηΤ E

ρη ρη ρ ρη= + +0 0c T Te ln( / ) ( ),* E

ρ ρη η∂∂

∂∂= =E ET

*

− =∂∂T

ET αλI,

ρη ρη ρ αλ= + +0 0c T Te ln( / ) ( ).tr E

∂∂

= − ∂∂

=f f

ET η ρ

ET

T

.

ρ ρ ρ αλ ρf f tr fe= − − − +0 0 1c T T T T[ ln( / ) ] ( ) ( ),*E E

ρ µ λ αλ ρf tr tr tr f= + − − +( ) ( ) ( )( ) ( ),**E E2 202

E T T T




113

in which

f

**

(T) also remains to be determined. However, reconciling the two formsfurnishes (taking

f

0

=

0)

(7.28)

7.3 THERMAL AND THERMOMECHANICAL ANALOGS OF THE PRINCIPLE OF VIRTUAL WORK

7.3.1 C

ONDUCTIVE

H

EAT

T

RANSFER

For a linear, isotropic, thermoelastic medium, the Fourier Heat Conduction Lawbecomes

q

=

−

k

T

∇

T, in which

k

T

is the thermal conductance, assumed positive.Neglecting coupling to the mechanical field, the thermal-field equation in an isotropicmedium experiencing small deformation can be written as

(7.29)

We now construct a thermal counterpart of the principle of virtual work. Mul-tiplying by

δ

T and using integration by parts, we obtain

(7.30)

Clearly, T is regarded as the primary variable, and the associated secondaryvariable is

q

. Suppose that the boundary is decomposed into three segments:

S

=

S

I

+

S

II

+

S

III

. On

S

I

, the temperature T is prescribed as, for example, T

0

. It follows that

δ

T

=

0 on

S

I

. On

S

II

, the heat flux

q

is prescribed as

q

0. Consequently, δ TnTq →δ TnTq0. On SIII, the heat flux is dependent on the surface temperature through aheat-transfer vector: h: q = q0 − h(T − T0). The right side of Equation 7.30 now becomes

(7.31)

We now suppose that T is approximated using an interpolation function of theform

(7.32)

from which we obtain

(7.33)

ρ µ λ αλ ρf tr tr tr= + − − −( ) ) ) [ ln( / ) ].E (E (E2 202

1T c T T Te

−∇ ∇ + =Tk cT eT Tρ ˙ .0

δ δ ρ δ∇ ∇ + =∫ ∫ ∫T Tn qT T T T Tk dV c dV dST e˙

δ δ δ δT T T T T Tn q n q n q nT T T TdS dS dS dSS S S SII III III∫ ∫ ∫ ∫= + − −0 0 0h( .)

T T T− 0 ~ ( ) ( ) ~ ( ) ( ),N x N xTT

TTθθ θθt tδ δ

∇ ∇T T~ ( ) ( ) ~ ( ) ( ),B x B xTT

TTt tθθ θθδ δ




in which BT is the thermal analog of the strain-displacement matrix. Upon substitutionof the interpolation models, the thermal-field equation now reduces to the systemof ordinary differential equations:

(7.34)

in which

7.3.2 COUPLED LINEAR ISOTROPIC THERMOELASTICITY

The thermal-field equation is repeated as

(7.35)

Following the same steps used for conductive heat transfer furnishes the varia-tional principle

(7.36)

The principle of virtual work for the mechanical field is recalled as

(7.37)

Also, recall that T = 2µE + λ[tr(E) − α(T − T0)]I. Consequently,

(7.38)

Thermal Stiffness Matrix

Conductance Matrix

Surface Conductance Matrix

Thermal Mass Matrix; Capacitance Matrix

Consistent Thermal Force;Consistent Heat Flux

K M fT T Tθθ )) θθ ))( ˙ ( ,t t+ =

K K KT T T= +1 2

K B x B xT T TT

1 = ∫ ( ) ( )k dVT

K N x n N xT TT

TT

2 = ∫ ( ) ( )h dSSIII

M N x N xT T TT= ∫ ( ) ( )ρce dV

f N x n q N x n qT TT

TT= +∫ ∫( ) ( )0 0dS dS

S SII III

− ∇ = +k trT2

0T T c Teαλ ρ( ˙ ) ˙ .E

δ δ ρ δ αλ δ∇ ∇ + + =∫ ∫ ∫ ∫T Tn qT T T c T T T TT ek dV dV tr dV dS˙ ( ˙ ) .0 E

tr dV dV dSS

( ) ˙ ( ) .δ δ ρ δET∫ ∫ ∫+ =u u u sT Tττ

tr tr dV tr dV dV dST T

S( [ ( ) ]) ( ( ) ) ˙ ( ) .δ µ λ δ λα δ ρ δE E E I E I u u u s2 0+ − − + =∫ ∫ ∫ ∫T T ττ



Thermal and Thermomechanical Response 115

Then introduce the interpolation models,

(7.39)

from which we can derive B(x) and b(x), thus satisfying

(7.40)

It follows that

(7.41)

We assume that the traction ττττ(S) is specified everywhere as ττττ0(S) on S. Here,

(7.42)

For the thermal field, assuming that the heat flux q is specified as q0 on thesurface, variational methods, together with the interpolation models, furnish theequation

(7.43)

The combined equations for a thermoelastic medium are now written in state(first-order) form as

(7.44)

u x N xT( ) ( ) ( ),= γγ t

VEC t tr t( ) ( ) ( ) ( ) ( ) ( ).E E= =B x b xT Tγγ γγ

K M fTγγ γγ ΩΩ θθ( ) ˙ ( ) ( ) .t t t+ − =

K B x B x

M N x N x

b x N x

f N x

T

T

TT

=

=

=

=

∫∫∫∫

( ) ( )

( ) ( )

( ) ( )

( ) .

D dV Stiffness Matrix

dV Mass Matrix

dV Thermoelastic Matrix

dS Consistent Force Vector

ρ

αλΩΩ

ττ

1 1 1

0 0 0T T T K M f f N x n qT T T T T

T0θθ θθ ΩΩγγ( ) ˙( ) ˙ ( ) , ( ) .t t t dS+ + = = ∫

W W

f

0

f

1 2

T

d

dt

t

t

t

t

t

t

t

t

˙ ( )

( )

( )

˙ ( )

( )

( )

( )

( )

γγ

γγ

θθ

γγ

γγ

θθ

+

=

1

0T

W

M 0 0

0 K 0

0 0 M

W

0 K

K 0 0

0 M

1

T

2

T

T

=

=

−

−

1 1

0 0T T

, .

ΩΩ

ΩΩ




Note that W1 is positive-definite and symmetric, while is positive-semidefinite, implying that coupled linear thermoelasticity is at least marginallystable, whereas a strictly elastic system is strictly marginally stable. Thus, thermalconduction has a stabilizing effect, which can be shown to be analogous to viscousdissipation.

7.4 EXERCISES

1. Express the thermal equilibrium equation in:(a) cylindrical coordinates(b) spherical coordinates

2. Derive the specific heat at constant stress, rather than at constant strain.3. Write down the coupled thermal and elastic equations for a one-dimensional

member.

12 ( )W WT

2 2+



117

Introduction to the Finite-Element Method

8.1 INTRODUCTION

In thermomechanical members and structures, finite-element analysis (FEA) is typi-cally invoked to compute displacement and temperature fields from known appliedloads and heat fluxes. FEA has emerged in recent years as an essential resource formechanical and structural designers. Its use is often mandated by standards such asthe ASME Pressure Vessel Code, by insurance requirements, and even by law. Itsacceptance has benefited from rapid progress in related computer hardware and soft-ware, especially computer-aided design (CAD) systems. Today, a number of highlydeveloped, user-friendly finite-element codes are available commercially. The purposeof this chapter is to introduce finite-element theory and practice. The next three chaptersfocus on linear elasticity and thermal response, both static and dynamic, of basicstructural members. After that, nonlinear thermomechanical response is considered.

In FEA practice, a design file developed using CAD is often “imported” into finite-element codes, from which point little or no additional effort is required to developthe finite-element model and perform sophisticated thermomechanical analysis andsimulation. CAD integrated with an analysis tool, such as FEA, is an example ofcomputer-aided engineering (CAE). CAE is a powerful resource with the potential ofidentifying design problems much more efficiently and rapidly than by “trial and error.”

A major FEM application is the determination of stresses and temperatures ina component or member in locations where failure is thought most likely. If thestresses or temperatures exceed allowable or safe values, the product can be rede-signed and then reanalyzed. Analysis can be diagnostic, supporting interpretation ofproduct-failure data. Analysis also can be used to assess performance, for example,by determining whether the design-stiffness coefficient for a rubber spring is attained.

8.2 OVERVIEW OF THE FINITE-ELEMENT METHOD

Consider a thermoelastic body with force and heat applied to its exterior boundary.The finite-element method serves to determine the displacement vector

u

(

X

,

t

) andthe temperature T(

X

,

t

) as functions of the undeformed position

X

and time

t

. Theprocess of creating a finite-element model to support the design of a mechanicalsystem can be viewed as having (at least) eight steps:

1.

The body is first discretized, i.e., it is modeled as a mesh of finite elementsconnected at nodes.

2. Within each element, interpolation models are introduced to provideapproximate expressions for the unknowns, typically

u

(

X

,

t

) and T(

X

,

t

),

8



118


in terms of their nodal values, which now become the unknowns in thefinite-element model.

3. The strain-displacement relation and its thermal analog are applied to theapproximations for

u

and T to furnish approximations for the (Lagrangian)strain and the thermal gradient.

4. The stress-strain relation and its thermal analog (Fourier’s Law) areapplied to obtain approximations to stress

S

and heat flux

q

in terms ofthe nodal values of

u

and T.5. Equilibrium principles in variational form are applied using the various approx-

imations within each element, leading to

element equilibrium equations

.6.

The element equilibrium equations are assembled to provide a

globalequilibrium equation

.7. Prescribed kinematic and temperature conditions on the boundaries (

con-straints

) are applied to the global equilibrium equations, thereby reducingthe number of degrees of freedom and eliminating “rigid-body” modes.

8. The resulting global equilibrium equations are then solved using computeralgorithms.

The output is postprocessed. Initially, the output should be compared to dataor benchmarks, or otherwise validated, to establish that the model correctly repre-sents the underlying mechanical system. If not satisfied, the analyst can revise thefinite-element model and repeat the computations. When the model is validated,postprocessing, with heavy reliance on graphics, then serves to interpret the results,for example, determining whether the underlying design is satisfactory. If problemswith the design are identified, the analyst can then choose to revise the design. Therevised design is modeled, and the process of validation and interpretation is repeated.

8.3 MESH DEVELOPMENT

Finite-element simulation has classically been viewed as having three stages:

pre-processing, analysis

, and

postprocessing

. The input file developed at the preprocess-ing stage consists of several elements:

1. control information (type of analysis, etc.)2. material properties (e.g., elastic modulus)3. mesh (element types, nodal coordinates, connectivities)4. applied force and heat flux data5. supports and constraints (e.g., prescribed displacements)6. initial conditions (dynamic problems)

In problems without severe stress concentrations, much of the mesh data can bedeveloped conveniently using automatic-mesh generation. With the input file devel-oped, the analysis processor is activated and “raw” output files are generated. Thepostprocessor module typically contains (interfaces to) graphical utilities, thus facil-itating display of output in the form chosen by the analyst, for example, contours ofthe Von Mises stress. Two problems arise at this stage:

validation

and

interpretation

.



Introduction to the Finite-Element Method

119

The analyst can use benchmark solutions, special cases, or experimental data tovalidate the analysis. With validation, the analyst gains confidence in, for example,the mesh. He or she still may face problems of interpretation, particularly if theoutput is voluminous. Fortunately, current graphical-display systems make interpre-tation easier and more reliable, such as by displaying high stress regions in vividcolors. Postprocessors often allow the analyst to “zoom in” on regions of highinterest, for example, where rubber is highly confined. More recent methods basedon virtual-reality technology enable the analyst to fly through and otherwise becomeimmersed in the model.

The goal of mesh design is to select the number and location of finite-elementnodes and element types so that the associated analyses are sufficiently accurate.Several methods include automatic-mesh generation with adaptive capabilities,which serve to produce and iteratively refine the mesh based on a user-selected errortolerance. Even so, satisfactory meshes are not necessarily obtained, so that modelediting by the analyst may be necessary. Several practical rules are as follows:

1. Nodes should be located where concentrated loads and heat fluxes areapplied.

2. Nodes should be located where displacements and temperatures are con-strained or prescribed in a concentrated manner, for example, where “pins”prevent movement.

3. Nodes should be located where concentrated springs and masses and theirthermal analogs are present.

4. Nodes should be located along lines and surface patches, over whichpressures, shear stresses, compliant foundations, distributed heat fluxes,and surface convection are applied.

5. Nodes should be located at boundary points where the applied tractionsand heat fluxes experience discontinuities.

6. Nodes should be located along lines of symmetry.7. Nodes should be located along interfaces between different materials or

components.8. Element-aspect ratios (ratio of largest to smallest element dimensions)

should be no greater than, for example, five.9. Symmetric configurations should have symmetric meshes.

10. The density of elements should be greater in domains with higher gradi-ents.

11. Interior angles in elements should not be excessively acute or obtuse, forexample, less than 45

°

or greater than 135

°

.12. Element-density variations should be gradual rather than abrupt.13. Meshes should be uniform in subdomains with low gradients.14. Element orientations should be staggered to prevent “bias.”

In modeling a configuration, a good practice is initially to develop the mesh locallyin domains expected to have high gradients, and thereafter to develop the meshin the intervening low-gradient domains, thereby “reconciling” the high-gradientdomains.



120


There are two classes of errors in finite-element analysis:

Modeling error

ensues from inaccuracies in such input data as the materialproperties, boundary conditions, and initial values. In addition, there often arecompromises in the mesh, for example, modeling sharp corners as rounded.

Numerical error

is primarily due to truncation and round-off. As a practicalmatter, error in a finite-element simulation is often assessed by comparingsolutions from two meshes, the second of which is a refinement of the first.

The sensitivity of finite-element computations to error is to some extent con-trollable. If the condition number of the stiffness matrix (the ratio of the maximumto the minimum eigenvalue) is modest, sensitivity is reduced. Typically, the conditionnumber increases rapidly as the number of nodes in a system grows. In addition,highly irregular meshes tend to produce high-condition numbers. Models mixingsoft components, for example, rubber, with stiff components, such as steel plates,are also likely to have high-condition numbers. Where possible, the model shouldbe designed to reduce the condition number.



121

Element Fields in Linear Problems

This chapter presents interpolation models in physical coordinates for the most part,for the sake of simplicity and brevity. However, in finite-element codes, the physicalcoordinates are replaced by natural coordinates using relations similar to interpola-tion models. Natural coordinates allow use of Gaussian quadrature for integrationand, to some extent, reduce the sensitivity of the elements to geometric details inthe physical mesh. Several examples of the use of natural coordinates are given.

9.1 INTERPOLATION MODELS

9.1.1 O

NE

-D

IMENSIONAL

M

EMBERS

9.1.1.1 Rods

The governing equation for the displacements in rods (also bars, tendons, andshafts) is

(9.1)

in which

u

(

x

,

t

) denotes the radial displacement, E,

Α

and

ρ

are constants,

x

is thespatial coordinate, and

t

denotes time. Since the displacement is governed by asecond-order differential equation, in the spatial domain, it requires two (time-dependent) constants of integration. Applied to an element, the two constants canbe supplied implicitly using two nodal displacements as functions of time. We nowapproximate

u

(

x

,

t

) using its values at

x

e

and

x

e

+

1

, as shown in Figure 9.1.The lowest-order interpolation model consistent with two integration constants

is linear, in the form

(9.2)

We seek to identify ΦΦΦΦ

m1

in terms of the nodal values of

u

. Letting

u

e

=

u

(

x

e

) and

u

e

+

1

=

u

(

x

e

+

1

), furnishes

(9.3)

9

EAu

xA

u

t

∂∂

= ∂∂

2

2

2

2ρ ,

u x t x t tu t

u tx xm

Tm m m

e

e

mT( , ) ( ) ( ), ( )

( )

( )( ) ( ).= =

=

+

ϕϕ ΦΦ γγ γγ ϕϕ1 1 1 1

1

1 1,

u t x t u t x te e m m e e m m( ) ( ) ( ), ( ) ( ).= = ( )+ +1 11 1 1 1 1 1ΦΦ γγ ΦΦ γγ



122


However, from the meaning of γγγγ

m

1

(

t

), we conclude that

(9.4)

9.1.1.2 Beams

The equation for a beam, following Euler-Bernoulli theory, is:

(9.5)

in which

w

(

x

,

t

) denotes the transverse displacement of the beam’s neutral axis, and

I

is a constant. In the spatial domain, there are four constants of integration. In anelement, they can be supplied implicitly by the values of

w

and

w

′

=

∂

w

/∂

x

at eachof the two element nodes. Referring to Figure 9.2, we introduce the interpolationmodel for

w

(

x

,

t

):

(9.6)

FIGURE 9.1

Rod element.

FIGURE 9.2

Beam element.

ue

xe xe+1

ue+1x

wet

we

wet+1

we+1

xe xe+1

x

ΦΦm

e

e e

e e

e e e

x

x l

x xl x x1

1

1

1

1

1

1

1

1 1=

=−

−

= −+

−+

+, .

EIw

xA

u

t

∂∂

+ ∂∂

=4

4

2

2 0ρ ,

w x t x t x x x x t

w

w

w

w

b b b b m

e

e

e

e

( , ) ( ) ( ), ( ) , ( ) .= = ( ) =− ′

− ′

+

+

ϕϕ ΦΦ γγ ϕϕ γγΤΤ ΤΤ1 1 1 1

2 31

1

1

1



Element Fields in Linear Problems

123

Enforcing this model at

x

e

and at

x

e

+

1

furnishes

(9.7)

9.1.1.3 Beam Columns

Beam columns are of interest, among other reasons, in predicting buckling accordingto the Euler criterion. The

z

–displacement

w

of the neutral axis is assumed to dependonly on

x

and the

x

–displacement. Also,

u

is modeled as

(9.8)

in which

u

0

(

x

) represents the stretching of the neutral axis. It is necessary to know

u

0

(

x

),

w

(

x

) and at

x

e

and

x

e

+

1

. The interpolation model is now

(9.9)

and

9.1.1.4 Temperature Model: One Dimension

The temperature variable to be determined is T

−

T

0

, in which T

0

is a referencetemperature assumed to be independent of

x

. The governing equation for a one-dimensional conductor is

(9.10)

ΦΦb1

2 3

2

1 12

13

1 12

11

0 1 2 3

1

0 1 2 3

=− − −

− − −

+ + +

+ +

−x x x

x x

x x x

x x

e e e

e e

e e e

e e

.

u x z u x zw x

x( , ) ( )

( ),= − ∂

∂0

∂∂w x

x( )

u x z t x z x x xm m b b( , , ) ( ) ( ) ,= −1 11 12 3

1 1ΦΦ γγ ΦΦ γγ

γγ γγm

e

e

b

e

e

e

e

u t

u t

w

w

w

w

l l , =

=

− ′

− ′

+ +

+

( )

( ),

1 1

1

ΦΦ ΦΦm b 1

1

1

2 3

2

1 12

13

1 12

1

1

1 1

1

0 1 2 3

1

0 1 2 3

=−

−

=− − −

− − −

+

+ + +

+ +

−

l

x x

x x x

x x

x x x

x x

e

e e

e e e

e e

e e e

e e

, .

kAx

Acte

∂∂

= ∂∂

2

2

T Tρ .



124


This equation is formally the same as for a rod equation (see Equation 9.1),furnishing the interpolation model for the element as

(9.11)

9.1.2 I

NTERPOLATION

M

ODELS

: T

WO

D

IMENSIONS

9.1.2.1 Membrane Plate

Now suppose that the displacements

u

(

x

,

y

,

t

) and

v

(

x

,

y

,

t

) are modeled on thetriangular plate element in Figure 9.3, using the values

u

e

(

t

),

v

e

(

t

),

u

e

+

1

(t), ve+1(t), ue+2(t),and ve+2(t). This element arises in plane stress and plane strain, and is called amembrane plate element. A linear model suffices for each quantity because it providesthree coefficients to match three nodal values. The interpolation model now is

(9.12)

.

9.1.2.2 Plate with Bending Stresses

In a plate element experiencing bending only, the in-plane displacements, u and v,are expressed by

(9.13)

FIGURE 9.3 Triangular plate element.

z

Y

middle surface

Xe+1,Ye+1

Xe+2,Ye+2

Xe,Ye X

T T T T

T T( , ) ( ) ( ), ( )

( )

( ).x t x t t

t

tm m

e

e

− = =−

−

+0 1 1 1 1

0

1 0

ϕϕ ΦΦ θθ θθT

u x y t

v x y t

t

t

m

m

m

m

u

v

( , , )

( , , )

( )

( )

=

ϕϕ

ϕϕ

ΦΦ

ΦΦ

γγ

γγ

ΤΤ

ΤΤ

ΤΤ

ΤΤ

2

2

2

2

0

0

0

0

T

T

T

T

γγ γγ ϕϕ ΦΦΤΤu v m m ( )

( )

( )

( )

, ( )

( )

( )

( )

, ,t

u t

u t

u t

t

v t

v t

v t

x

y

x y

x y

e

e

e

e

e

e

e e

e e=

=

=

=+

+

+

+

+ +1

2

1

2

2 2 1 1

1 1

1

11 2 2

1

x ye e+ +

−

u x y z t zw

xv x y z t z

w

y( , , , ) , ( , , , ) ,= − ∂

∂= − ∂

∂



Element Fields in Linear Problems 125

in which z = 0 at the middle plane. The out-of-plane displacement, w, is assumedto be a function of x and y only.

An example of an interpolation model is introduced as follows to express w(x, y)throughout the element in terms of the nodal values of :

(9.14)

.

It follows that

(9.15)

9.1.2.3 Plate with Stretching and Bending

Finally, for a plate experiencing both stretching and bending, the displacements areassumed to satisfy

(9.16)

w wx

wy, and ∂

∂∂∂

w x y t x y tb b b( , , ) ( , ) ( )= ϕϕ ΦΦ γγ2 2 2T

ϕϕb x y x y x xy y x x y y x y22 2 3 2 2 31

12

T ( , ) ( )= +

γγb e e ee e

ee e e

wx

wy

wx

wy

wx

wyw

2 1 1

12 2 2

T ( )

t w w= −( ) −

−( )

−

−( ) −

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+ +

++ + +

ΦΦb

e e e e e e e e e e e e

e e e e e e

e e e e e e

e e e

x y x x y y x x y x y y

x y x x y y

x x x x y y

x y x x

21

2 2 3 12

3

2 12

2

12

2 2

1 1 12

1

0 1 0 2 0 3 0

0 0 1 0 2 0 3

1

−

+ + +

=

+

− − − − − +

− − − − + −

( )

( )

( )

ee e e e e e e e e

e e e e e e

e e e e

y y x x y x y y

x y x x y y

x y x x

+ + + + + + + + +

+ + + + + +

+ + + +

+

− − − − − +

− − − − +

1 1 12

13 1

2 1 1 1 1 13

1 1 12

1 112 1

2

1 112 1

2

0 1 0 2 0 3 0

0 0 1 0 2 0

( )

( )

( 11 1 12

2 2 22

2 2 22

23 1

2 2 2 2 2 23

2 2 22

2 212 2

2

3

1

0 1 0 2 0 3 0

0 0

y y

x y x x y y x x y x y y

x y x x y y

e e

e e e e e e e e e e e e

e e e e e e

+ +

+ + + + + + + + + + + +

+ + + + + +

−

+

− − − − − +

)

( )

( )

−− − − − + −

+ + + + + +1 0 2 0 32 212 2

22 2 2

2x y x x y ye e e e e e( )

u(x, y, z, t)

v(x, y, z, t)

w(x, y, z, t)

z

z t

b

b

b

b b

x

y

=

−

−

∂∂

∂∂

ϕϕ

ϕϕ

ΤΤ

ΤΤ

ΤΤϕϕ

ΦΦ γγ

2

2

2

2 2 ( ).

u x y z t u x y z t zwx

v x y z t v x y z t zwy

( , , , ) ( , , , ) , ( , , , ) ( , , , )= − ∂∂

= − ∂∂0 0




and w is a function only of x, y, and t(not z). Here, z = 0 at the middle surface, whileu0 and v0 represent the in-plane displacements. Using the nodal values of u0, v0, andw0, a combined interpolation model is obtained as

(9.17)

9.1.2.4 Temperature Field in Two Dimensions

In the two-dimensional, triangular element illustrated in Figure 9.3, the linear inter-polation model for the temperature is

(9.18)

9.1.2.5 Axisymmetric Elements

An axisymmetric element is displayed in Figure 9.4. It is applicable to bodies thatare axisymmetric and are submitted to axisymmetric loads, such as all-aroundpressure. The radial displacement is denoted by u, and the axial displacement isdenoted by w. The tangential displacement v vanishes, while radial and axial dis-placements are independent of θ. Now u and w depend on r, z, and t.

There are two distinct situations that require distinct interpolation models. Inthe first case, none of the nodes are on the axis of revolution (r = 0), while in thesecond case, one or two nodes are, in fact, on the axis. In the first case, the linear

FIGURE 9.4 Axisymmetric element.

u x y z t

v x y z t

w x y z t

u x y t

v x y t

w x y t

z

z

b

b

b

x

y

( , , , )

( , , , )

( , , , )

( , , )

( , , )

( , , )

=

+

−

−

∂∂

∂∂

0

0

0

2

2

2

ϕϕ

ϕϕ

ΤΤ

ΤΤ

ΤΤϕϕ

=

−

−

∂∂

∂∂

ΦΦ γγ

ϕϕ

ϕϕ

ϕϕ

ΦΦ

ΦΦ

ΦΦ

γγ

γγ

γγ

ΤΤΤΤ

ΤΤΤΤ

ΤΤ

ϕϕ

ϕϕ

b b

mb

mb

b

m

m

b

u

v

w

t

z

z

t

t

t

x

y

2 2

22

22

2

2

2

2

( )

( )

( )

( )

0

0

0 0

0 0

0 0

0 0

T

T

T T

.

T T T T T T T T− = = − − −+ +0 2 2 2 2 0 1 0 2 0ϕϕ ΦΦ θθ θθΤΤ ΤΤm m e e e, ( ).

r

e+2

e

e+1

θ




interpolation model is given by

(9.19)

Now suppose that there are nodes on the axis, and note that the radial displace-ments are constrained to vanish on the axis. For reasons shown later, it is necessaryto enforce the symmetry constraints a priori in the formulation of the displacementmodel. In particular, suppose that node e is on the axis, with nodes e + 1 and e + 2defined counterclockwise at the other vertices. The linear interpolation model is now

(9.20)

A similar formulation can be used if two nodes are on the axis of symmetry sothat the u displacement in the element is modeled using only one nodal displacement,with a coefficient vanishing at each of the nodes on the axis of revolution.

9.1.3 INTERPOLATION MODELS: THREE DIMENSIONS

We next consider the tetrahedron illustrated in Figure 9.5.A linear interpolation model for the temperature can be expressed as

(9.21)

u r z t

w r z t

t

t

a

a

a

a

ua

wa

( , , )

( , , )

( )

( ).

=

ϕϕ

ϕϕ

ΦΦ

ΦΦ

γγ

γγ1

1

1

1

1

1

T T

T T

T T

T T

0

0

0

0

ϕϕ ΦΦ γγ γγa a

e e

e e

e e

ua

e

e

e

wa

e

e

e

r z

r z

r z

r z

u

u

u

w

w

w

1 1 1 1

2 2

1

1 1

2

1 1

2

1

1

1

1

T = =

=

=

+ +

+ +

−

+

+

+

+

( ), , ,

u r z t

w r z t

t

t

a

a

a

a

ua

wa

( , , )

( , , )

( )

( )

=

ϕϕ 00

00 ϕϕ

ΦΦ 00

00 ΦΦ

γγ

γγ2

2

2

2

2

2

T T

T T

T T

T T

ϕϕ ΦΦa e a

e e e

e e e

r z zr z z

r z z2 2

1 1

2 2

1

T = − =−

−

+ +

+ +

−

( ), .

T T− =0 3 3 3ϕϕ ΦΦ θθT TT

ϕϕ

ΦΦ

3

3

1 1 1

2 2 2

3 3 3

1

3 1 2 3

1

1

1

1

1

T

T

TT T T T T T T T T

T =

=

= − − − −

+ + +

+ + +

+ + +

−

+ + +

( )

x y z

x y z

x y z

x y z

x y z

e e e

e e e

e e e

e e e

e o e o e o e oθ




For elasticity with displacements u, v, and w, the corresponding interpolationmodel is

(9.22)

9.2 STRAIN-DISPLACEMENT RELATIONS AND THERMAL ANALOGS

9.2.1 STRAIN-DISPLACEMENT RELATIONS: ONE DIMENSION

For the rod, the strain is given by . An estimate for ε implied by theinterpolation model in Equation 9.3 has the form

(9.23)

FIGURE 9.5 Tetrahedral element.

x e+1

e+2

e+3

y

e

z

0

u x y z t

v x y z t

w x y z t

t

t

t

u

v

w

( , , , )

( , , , )

( , , , )

( )

( )

( )

=

ϕϕ 00 00

00 ϕϕ 00

00 00 ϕϕ

ΦΦ 00 00

00 ΦΦ 00

00 00 ΦΦ

γγ

γγ

γγ

3

3

3

3

3

3

T T T

T T T

T T T

γγ γγ γγu

e

e

e

e

v

e

e

e

e

w

e

et

u t

u t

u t

u t

t

v t

v t

v t

v t

t

w t

w t( )

( )

( )

( )

( )

, ( )

( )

( )

( )

( )

, ( )

( )

(=

=

=+

+

+

+

+

+

+1

2

3

1

2

3

1

))

( )

( )

w t

w t

e

e

+

+

2

3

ε = = ∂∂E ux11

ε( , ) ( ) ( ),x t x= ββ ΦΦ γγm m m t1 1 1T




For the beam, the corresponding relation is

(9.24)

from which the consistent approximation is obtained:

(9.25)

For the beam column, the strain is given by

(9.26)

9.2.2 STRAIN-DISPLACEMENT RELATIONS: TWO DIMENSIONS

In two dimensions, the (linear) strain tensor is given by

(9.27)

We will see later the two important cases of plane stress and plane strain. In thelatter case, Ezz vanishes and szz is not needed to achieve a solution. In the formercase, szz vanishes and Ezz is not needed for solution.

In traditional finite-element notation, we obtain [cf. (Zienkiewicz and Taylor, 1989)]

(9.28)

ββ ϕϕm

md

dx11 0 1T

T

= = ( )

ε( , , ) ,x z t zw

x= − ∂

∂

ε( , , ) ( ) ( ),x z t z xb b b= − ββ ΦΦ γγ1 1 1T t

ββϕϕ

bbd

dxx x1

1 20 1 2 3TT

= = ( )

ε( , , ) ( ) ( ),x z t z xb b b= − ββ ΦΦ γγm m m t1 1 1T

ββ ββ ββΦΦ 00

00 ΦΦ

γγ

γγmb m b

m

mb

m

b

zt

t1 1 1

1

1

1

1

T T T= −( )

( )

( )

E( ) .x yE E

E E

xx xy

xy yy

ux

uy

vx

uy

vx

vy

, =

=( )

( )

∂∂

∂∂ + ∂

∂∂∂ + ∂

∂∂∂

12

12

′ =

=

εε ββ ΦΦ

γγ

γγ

E

E

E

xx

yy

xy

mT

m

u

v

2 2

2

2

ˆ




The prime in e′ is introduced temporarily to call attention to the fact that it doesnot equal VEC(EL). Hereafter, the prime will not be displayed.

For a plate with bending stresses only,

(9.29)

from which

. (9.30)

For a plate experiencing both membrane and bending stresses, the relations canbe combined to furnish

(9.31)

9.2.3 AXISYMMETRIC ELEMENT ON AXIS OF REVOLUTION

For the toroidal element with a triangular cross section, it is necessary to considertwo cases. If there are no nodes on the axis of revolution, then

(9.32)

ββ

00

00 ΦΦΦΦ 00

00 ΦΦΤΤ

ΤΤΤΤ

ΤΤΤΤ

ΤΤ ΤΤ

ϕϕ

ϕϕ

ϕϕ ϕϕm

m

m

m m

m

m

m

x

y

y x

2

2

2

2 2

2

2

212

12

=

=

∂∂

∂∂

∂∂

∂∂

, ˆ

′ = −

∂∂∂∂∂∂ ∂

e ( , , , ) ,x y z t z

wx

wy

wx y

2

2

2

2

2

′ = − =

∂∂

∂∂

∂∂ ∂

e ( , , , ) ( ) ( ),x y z t z x t

x

y

x y

bT

b b b

bT

bT

bT

ββ ΦΦ γγ ββ

ϕϕ

ϕϕ

ϕϕ2 2 2 2

22

2

22

22

2

′ =e ( , , , ) ( , , ) ( )x y z t x y z tmb mb mbββ ΦΦ γγ2 2 2T

ββ ββ ββ ΦΦΦΦ 00

00 ΦΦγγ

γγ

γγmb m b mb

m

b

mb

m

b

tt

t2 2 2 2

2

2

2

2

2

T T T= −( ) =

=

z , , ( )

( )

( )

e( , , )( )

( )r z t

t

t

ur

ur

wz

uz

wr

a

a

a

ua

wa

=

( )

=

∂∂

∂∂

∂∂ + ∂

∂12

1

1

1

1

1

ββΦΦ

ΦΦ

γγ

γγT

0

0




in which the prime is no longer displayed. If element e is now located on the axisof revolution, we obtain

(9.33)

9.2.4 THERMAL ANALOG IN TWO DIMENSIONS

The thermal analog of the strain is the temperature gradient satisfying

(9.34)

9.2.5 THREE-DIMENSIONAL ELEMENTS

Recalling the tetrahedral element in the previous section, the strain relation can bewritten as

(9.35)

ββa

zr

1

0 1 0 0 0 0

1 0 0 0

0 0 0 0 0 1

0 0 0 0

12

12

12

T =

e b( , , )( )

( )r z t

t

ta

a

a

ua

wa

=

2

2

1

2

1

T0

0

ΦΦ

ΦΦ

γγ

γγ

ββa

z zer

2

1 0 0 0 0

1 0 0 0

0 0 0 0 1

0 0 012

12

T =

−

∇ = =

T ββ ΦΦ θθ ββTT

T T2 2 2 2

0 1 0

0 0 1, .T

e =

= ( )( )( )

=

∂∂∂∂∂∂

∂∂ + ∂

∂∂∂ + ∂

∂∂∂ + ∂

∂

E

E

E

E

E

E

xx

yy

zz

xy

yz

zx

ux

vy

wz

uy

vx

vz

wy

wx

uz

12

12

12

3ββTT

0 0

0 0

0 0

ΦΦ

ΦΦ

ΦΦ

γγ

γγ

γγ

3

3

3

3

3

3

u

v

w




9.2.6 THERMAL ANALOG IN THREE DIMENSIONS

Again referring to the tetrahedral element, the relation for the temperature gradient is

(9.36)

9.3 STRESS-STRAIN-TEMPERATURE RELATIONSIN LINEAR THERMOELASTICITY

9.3.1 OVERVIEW

If S is the stress tensor under small deformation, the stress-strain relation for alinearly elastic, isotropic solid under small strain is given in Lame’s form by

(9.37)

in which I is the identity tensor. The Lame’s coefficients are denoted by λ and µ,and are given in terms of the familiar elastic modulus E and Poisson ratio ν as

(9.38)

Letting s = VEC(S) and e = VEC(EL), the stress-strain relations are written usingKronecker product operators as

(9.39)

and D is the tangent-modulus tensor introduced in the previous chapters.

9.3.2 ONE-DIMENSIONAL MEMBERS

For a beam column, recalling the strain-displacement model,

(9.40)

The cases of a rod and a beam are recovered by setting γγγγm1 or γγγγb1 equal to zerovectors, respectively.

∇ = =

T ββ θθ ββ3 3 3 3

0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 1

T T TT TΦ , .

S E E= +2µ λL Ltr( ) ,I

µν

λ νν ν

=+

=− +

E

E2 1 1 2 1( )

,( )( )

.

s e= = +D D I iiT, , 2 9µ λ

S x z t z x t11 1 1 1= = = −σ( , , ) ( ) ( ).E E mb mb mbε ββ ΦΦ γγT




9.3.3 TWO-DIMENSIONAL ELEMENTS

9.3.3.1 Membrane Response

In two-dimensional elements, several cases can be distinguished. We first considerelements in plane stress. It is convenient to use Hooke’s Law in the form

(9.41)

Under plane stress, Sxz = Syz = Szz = 0. Now, Ezz ≠ 0, but we will see that it isnot of present interest since it does not influence the solution process. Later on, forreasons such as tolerances, we may wish to calculate it, but this is a postprocessingissue. Consequently, in plane stress, the stress-strain relations reduce to

(9.42)

In traditional finite-element notation, this can be written as

(9.43)

in which

(9.44)

E S S S

E S S S

E S S S

E S

E S

E S

xx xx yy zz

yy yy yy zz

zz zz xx yy

xy xy

yz yz

zx zx

= − +

= − +

= − +

= +

= +

= +

1

1

1

1

1

1

E

E

E

E

E

E

[ ( )]

[ ( )]

[ ( )]

ν

ν

ν

ν

ν

ν

E S S

E S S

E S

xx xx yy

yy yy xx

xy xy

= −

= −

= +

1

1

1

E

E

E

( )

( )

ν

ν

ν

S

S

S

E

E

E

xx

yy

xy

m

xx

yy

xy

=

D 21 .

Dm21

1

2

1 0

1 0

0 0 11

1 0

1 0

0 0 1

=

−

−

+

=−

+

−

EE

ν

ν

νν

ν

ν

ν




can be called the tangent-modulus matrix under plane stress (note that, unlike theprevious definition, it is not based on the VEC operator and is not a tensor).

Under plane strain, Exz = Eyz = Ezz = 0. Now, Szz ≠ 0. However, this stress is notof great interest since its value is not needed for attaining a solution. The quantitymay be of interest later, e.g., to check whether yield conditions in plasticity are met,but it can be obtained at that point by a postprocessing procedure. The equations ofplane strain are

(9.45)

and Dm22 is the tangent-modulus tensor in plane strain.However, in the finite-element method, we will see that we are interested in a

slightly different quantity from Dm21 or Dm22, as follows. The Principle of VirtualWork uses the strain-energy density given by . However, elementary manip-ulation serves to prove that

(9.46)

In the Principle of Virtual Work, the tangent-modulus tensor in plane stress isreplaced by

(9.47)

and similarly for plane strain. (This peculiarity is an artifact of traditional finite-element notation and does not appear if VEC notation is used.) The stresses are nowgiven in terms of nodal displacements by

(9.48)

S

S

S

E

E

E

xx

yy

xy

xx

yy

xy

=

Dm22 ,

Dm22 2 21

1 0

1 0

0 0 11 1

=−

+

=−

=−

E, E

E,

*

*

*

*

*

* *

ν

ν

ν

νν

ν νν

12 S Eij ij

12

12

1 0 0

0 1 0

0 0 2

S E S S S

E

E

E

ij ij xx yy zz

xx

yy

xy

=

( ) .

′ =−

+

Dm21 21

1 0

1 0

0 0 2 1

Eν

ν

ν

ν( )

S

S

S

E

E

E

xx

yy

xy

xx

yy

xy

(x, y, t)

(x, y, t)

(x, y, t)

plane stress

plane strain

=

= ′

=

D Dm i m i m m

ui2 2 2 2

2

2

1

2ββ ΦΦ

γγ

γγT ˆ ,

ν




9.3.3.2 Two-Dimensional Members: Bending Response

Thin plates experiencing bending only are assumed to be in a state of plane stress.The tangent-modulus matrix is given in Equation 9.47, and an approximation forthe strain is obtained as

(9.49)

9.3.4 ELEMENT FOR PLATE WITH MEMBRANE AND BENDING RESPONSE

Plane stress is also applicable, consequently:

(9.50)

9.3.5 AXISYMMETRIC ELEMENT

It is sufficient to consider the case in which none of the nodes of the element arelocated on the axis of revolution.

(9.51)

S

S

S

Exx

yy

xy

xx

yy

xy

=

−D Dm m b b bE

E

z t21 21 2 2 2ββ ΦΦ γγT ˆ ( ).

S

S

S

E

E

E

x y z t

xx

yy

xy

m

xx

yy

xy

m mb mb mb

=

=D D21 21 2 2 2ββ ΦΦ γγT ( ) ˆ ( )., ,

S

S

S

S

E

E

E

E

t

t

rr

zz

rz

a

rr

zz

rz

a a

a

a

ua

wa

θθ θθ

=

=

D D ββ

ΦΦ

ΦΦ

γγ

γγ1

1

1

1

1

T0

0

ˆ

ˆ

( )

( ),

Da =

− −

− −

− −

+

−

E

1 0

1 0

1 0

0 0 0 1

1ν ν

ν ν

ν ν

ν




For use in the Principle of Virtual Work, Da is modified to furnish , given by

(9.52)

9.3.6 THREE-DIMENSIONAL ELEMENT

All six stresses and strains are now present. Using traditional notation, we write

(9.53)

and for the Principle of Virtual Work,

(9.54)

′Da

′ =

− −

− −

− −

+

−

Da E

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 2

1 0

1 0

1 0

0 0 0 1

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 2

1ν ν

ν ν

ν ν

ν

.

S

S

S

S

S

S

E

E

E

E

E

E

xx

yy

zz

xy

yz

zx

xx

yy

zz

xy

yz

zx

u

v

w

=

=

D D3 3 3

3

3

3

3

3

3

ββ

ΦΦ 00 00

00 ΦΦ 00

00 00 ΦΦ

γγ

γγ

γγ

T

D3

11 0 0 0

1 0 0 0

1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

=

− −

− −

− −

+

+

+

−

E

ν ν

ν ν

ν ν

ν

ν

ν

′ =

− −

− −

− −

+

+

+

−

D3 6

1

6

1 0 0 0

1 0 0 0

1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

EJ J

ν ν

ν ν

ν ν

ν

ν

ν




9.3.7 ELEMENTS FOR CONDUCTIVE HEAT TRANSFER

Assuming the isotropic version of Fourier’s Law, the heat-flux vector, which can beconsidered the thermal analog of the stress, satisfies

(9.55)

9.4 EXERCISES

1. Modify the rod element to replace the physical coordinate x with the“natural coordinate” ξ in which

Rewrite the interpolation model using natural coordinates, and performthe inverse to obtain ΦΦΦΦm1.

2. Rewrite the Euler-Bernoulli equation for the beam using the previoustransformation. Rewrite the interpolation model using natural coordinates,and perform the inverse to obtain ΦΦΦΦb1.

3. Verify that the inverse given in Equation 9.44 is correct.4. Invert Equation 9.39 to express E as a function of s. Find the correct

expression for D−1.5. Write out the 9 × 9 tensors D and D−1 that correspond to Lame’s equation.6. For the axisymmetric triangular element, find the strain-displacement

matrix ββββT in the cases in which two of the nodes are located on the axisof revolution. Note that for axisymmetric problems, ur = 0 at r = 0.

J6

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 2 0 0

0 0 0 0 2 0

0 0 0 0 0 2

=

q = −

−

−

−

k

D

D

D

T T

T T

T T

ββ ΦΦ θθ

ββ ΦΦ θθ

ββ ΦΦ θθ

ΤΤ

ΤΤ

ΤΤ

1 1 1

2 2 2

3 3 3

1

2

3 .

ξ ξ ξ= + = − = ++ax b x xe e, ( ) , ( ) . 1 11




7. The stress-strain relations in two-dimensional plane stress can be writtenas

Prove that in plane strain

,

where

.

8. If

,

find the matrix such that

.

What happens if ν → 1/2?

E

E

E

S

S

S

xx

yy

xy

xx

yy

xy

=

−

−

+

11 0

1 0

0 0 1E

ν

ν

ν

E

E

E

S

S

S

xx

yy

xy

xx

yy

xy

=

−

−

+

11 0

1 0

0 0 1E

*

*

*

*

ν

ν

ν

E E ** /( ), /( )= − = −1 12ν ν ν ν

E

E

S

S

xx

yy

xx

yy

=

− − +

− + −

1 1 1

1 1

2

2E

ν ν ν

ν ν ν

( )

( )

D

S

S

E E

E

xx

yy

xx

yy

=

+( ) −( )

1 1 2ν νD



139

Element and Global Stiffness and Mass Matrices

10.1 APPLICATION OF THE PRINCIPLEOF VIRTUAL WORK

Elements of variational calculus were discussed in Chapter 3, and the Principle ofVirtual Work was introduced in Chapter 5. Under static conditions, the principle isrepeated here as

(10.1)

As before,

δ

represents the variational operator. We assume for our purposesthat the displacement, the strain, and the stress satisfy representations of the form

(10.2)

in which

E

and

S

are written as one-dimensional arrays in accordance with traditionalfinite-element notation. For use in the Principle of Virtual Work, we need

D

′

, whichintroduces the factor 2 into the entries corresponding to shear. We suppose that theboundary is decomposed into four segments: S

=

S

I

+

S

II

+

S

III

+

S

IV

. On S

I

,

u

isprescribed, in which event

δ

u

vanishes. On S

II

, the traction ττττ

is prescribed as ττττ

0

.On S

III

, there is an elastic foundation described by ττττ

=

ττττ

0

−

A

(

x

)

u

, in which

A

(

x

) isa known matrix function of

x

. On S

IV

, there are inertial boundary conditions, byvirtue of which ττττ

=

ττττ

0

−

Bü

. The term on the right now becomes

(10.3)

10

δ δ ρ δ τE S dV u u dV u dSij ij i i i i∫ ∫ ∫+ =˙ .

u x ) x T( ( ) ( ) ( ) ( ), t t , t , ,= = =ϕϕ ΦΦγγ ββ ΦΦγγE S ET x D

δ τ δ

δ

δ

u dS dS

dS t

dS t

i i∫ ∫

∫

∫

=

−

−

+ +γγ ΦΦ ϕϕ ττ

γγ ΦΦ ϕϕ ϕϕ ΦΦ γγ

γγ ΦΦ ϕϕ ϕϕ ΦΦ γγ

T T0

T T T

T T T

x

x A x

x B x

( )

( ) ( ) ( )

( ) ( ) ˙ ( ).

S S S

S

S

II III IV

III

IV



140


The term on the left in Equation 10.1 becomes

(10.4)

in which

K

is called the stiffness matrix and

M

is called the mass matrix. Cancelingthe arbitrary variation and bringing terms with unknowns to the left side furnishesthe equation as follows:

(10.5)

Clearly, elastic supports on S

III

furnish a boundary contribution to the stiffnessmatrix, while mass on the boundary segment S

IV

furnishes a contribution to the massmatrix.

Sample Problem 1: One element rod

Consider a rod with modulus E, mass density

ρ

, area

A

, and length

L

. It is built inat

x

=

0. At

x

=

L

, there is a concentrated mass

m

to which is attached a spring ofstiffness

k

, as illustrated in Figure 10.1. The stiffness and mass matrices, from thedomain, reduce to the scalar values

K

→

EA

/

L

,

M

→

ρ

AL

/

3,

M

S

→

m

,

K

S

→

k

.The governing equation is .

Sample Problem 2: Beam element

Consider a one-element model of a cantilevered beam to which a solid disk is weldedat

x

=

L

. Attached at

L

is a linear spring and a torsional spring, the latter havingthe property that the moment developed is proportional to the slope of the beam.

FIGURE 10.1

Rod with inertial and compliant boundary conditions.

δ

δ ρ

E S dV

u dV

ij ij

i i

∫ ∫∫ ∫

= = ′

= =

δδγγ γγ ΦΦ ββ ββ ΦΦ

δδγγ γγ ρρΦΦ ϕϕ ϕϕ ΦΦ

T T T

T T T

K K x D x

M M x x

( ), ( ) ( )

˙ ˙ ( ), ( ) ( ) ,

t dV

u t dV

( ) ( ) ( )˙K K M M f+ + + =S St tγγ γγ( )

f x

K x A x

M x B x

T0

T T

T T

=

=

=

+∫

∫

∫

ΦΦ ϕϕ ττ

ΦΦ ϕϕ ϕϕ ΦΦ

ΦΦ ϕϕ ϕϕ ΦΦ

( )

( ) ( )

( ) ( ) .

dS

dS

dS

S

S

S S

S

S

II III

III

IV

( ) ( )˙EAL

ALk m f+ + + =γ γρ

3

E,A,L,ρ

P

m k



Element and Global Stiffness and Mass Matrices

141

The shear force

V

0

and the moment

M

0

act at

L

. The interpolation model, incorpo-rating the constraints

w

(0,

t

)

=

−

w

′

(0,

t

)

=

0

a priori

, is

(10.6)

The stiffness and mass matrices, due to the domain, are readily shown to be

(10.7)

The stiffness and mass contributions from the boundary conditions are

(10.8)

The governing equation is now

(10.9)

10.2 THERMAL COUNTERPART OF THE PRINCIPLEOF VIRTUAL WORK

For our purposes, we focus on the equation of conductive heat transfer as

. (10.10)

FIGURE 10.2

Beam with translational and rotational inertial and compliant boundary conditions.

z

y

E,I,A,L,ρ

M0

kT

k

x

V0

mr

w x t x xL L

L L

w L t

w L t( , ) ( )

( , )

( , ).=

− −

− ′

−

2 3

2 3

2

1

2 3

K M=

=

EI

L

L

L LAL

L

L L3 2

1335

11210

11210

1105

2

12 6

6 4, . ρ

K MS

T

S mr

k

k

m=

=

0

0

0

02

2

, .

( )˙ ( , )

˙ ( , )( )

( , )

( , ).M M K K+

− ′

+ +

− ′

=

S S

w L t

w L t

w L t

w L t

V

M

0

0

k cte∇ = ∂

∂2T

Tρ



142


Multiplying by the variation of

T

−

T

0

, integrating by parts, and applying thedivergence theorem furnishes

(10.11)

Now suppose that the interpolation models for temperature in the current elementfurnish a relation of the form

. (10.12)

The terms on the left in Equation 10.11 can now be written as

(10.13)

K

T

and

M

T

can be called the thermal stiffness (or conductance) matrix and thermalmass (or capacitance) matrix, respectively.

Suppose that the boundary

S

has four zones: S

=

S

I

+

S

II

+

S

III

+

S

IV

. On S

I

, thetemperature is prescribed as T

1

, from which we conclude that

δ

T

=

0. On S

II

, theheat flux is prescribed as

n

T

q

1

. On S

III

,

the heat flux satisfies

n

T

q

=

n

T

q

1

−

h1 (T −T0), while on SIV, nTq = nTq1 − h2 d T/dt. The governing finite-element equation isnow

(10.14)

10.3 ASSEMBLAGE AND IMPOSITION OF CONSTRAINTS

10.3.1 RODS

Consider the assemblage consisting of two rod elements, denoted as e and e + 1[see Figure 10.3(a)]. There are three nodes, numbered n, n + 1, and n + 2. We firstconsider assemblage of the stiffness matrices, based on two principles: (a) the forcesat the nodes are in equilibrium, and (b) the displacements at the nodes are continuous.Principle (a) implies that, in the absence of forces applied externally to the node, atnode n + 1, the force of element e + 1 on element e is equal to and opposite the forceof element e on element e + 1. It is helpful to carefully define global (assemblage

δ δ ρ δ∇ ∇ + ∂∂

=∫ ∫ ∫T Tn qT T TT

Tk dV ct

dV dSe .

T T T − = ∇ = = −0 ϕϕ ΦΦ θθ ββ ΦΦ θθ ΦΦ θθT T T T T Tt t k tT T Tx x q x( ) ( ), ( ) ( ), ( ) ( )

δ δ

δ ρ δ ρ

∇ ∇ → =

∂∂

→ =

∫ ∫∫ ∫

TT T T T T T

e T T e T T T T

k dV t t k dV

ct

dV t t c dV

T T

TT

θθ θθ ΦΦ ββ ββ ΦΦ

θθ θθ ΦΦ ϕϕ ϕϕ ΦΦ

T T T

T T T

K K

M M

( ) ( ),

( ) ˙( ), .

[ ]˙( ) [ ] ( ) ( )M M K K fT TS T TS Tt t t+ + + =θθ θθ

M K

f n q

T TTS T

TT T

TT TS T T T T

T TT

TT

h dS h dS

t dS

= =

= = + +

∫ ∫

∫

ΦΦ ϕϕ ϕϕ ΦΦ ΦΦ ϕϕ ϕϕ ΦΦ

ΦΦ ϕϕ

2 1

1

S S

II III IV

IV III

S S S

,

( ) ,Ω

Ω



Element and Global Stiffness and Mass Matrices 143

level) and local (element level) systems of notation. The global system of forces isshown in (a), while the local system is shown in (b). At the center node,

(10.15)

since no external load is applied. Clearly, .The elements satisfy

(10.16)

and in this case, k(e) = k(e+1) = EA/L. These relations can be written as four separateequations:

(10.17)

FIGURE 10.3 Assembly of rod elements.

Pn Pn+2

P1(e+1)

P2 (e+1)

P1(e)

P2(e)

n n+1

e+1e

e e+1

a. forces in global system

b. forces in local system

P Pe e1 2

1 0( ) ( )− =+

P P P Pen

en2 1

12

( ) ( )= =++ and

ku

u

P

P

ku

u

P

P

e n

n

e

e

e n

n

e

e

( )

( )

( )

( )

( )

( )

1 1

1 1

1 1

1 1

1

2

1

1 1

2

21

11

−

−

=

−

−

−

=

−

+

+ +

+

+

+

k u k u P

k u k u P

k u k u P

k u k u P

en

en

e

en

en

e

en

en

e

en

en

e

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

(

(

(

− = −

− + =

− = −

− + =

+

+

++

++

+

++

++

+

1 2

1 1

11

12 2

1

11

12 1

1

i)

ii)

iii)

iv)(




Add (ii) and (iii) and apply Equation 10.15 to obtain

(10.18)

and in matrix form

(10.19)

The assembled stiffness matrix shown in Equation 10.19 can be visualized asan overlay of two element stiffness matrices, referred to global indices, in whichthere is an intersection of the overlay. The intersection contains the sum of the lowestentry on the right side of the upper matrix and the highest entry on the left side ofthe lowest matrix. The overlay structure is depicted in Figure 10.4.

FIGURE 10.4 Assembled beam stiffness matrix.

k u k u P

k u k k u k u

k u k u P

en

en

e

en

e en

en

en

en

e

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

[ ]

(

(

(

− = −

− + + − =

− − =

+

+

++

++

++

++

+

1 2

11

12

11

12 1

1

0

i)

ii iii)

iv)

k k

k k k k

k k

u

u

u

P

P

e e

e e e e

e e

n

n

n

n

n

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

−

− + −

−

=

−

+ +

+ +

+

+ +

0

0

01 1

1 1

1

2 1

K(1)

K(2)

K(3)

K(4)

k22+k11

K(5)

K(N–2)

K(N–1)

K(N)

(4) (5)




Now, the equations of the individual elements are written in the global system as

(10.20)

The global stiffness matrix (the assembled stiffness matrix K(g) of the two-elementmember) is the direct sum of the element stiffness matrices: .

Generally, K(g) = ∑e . In this notation, the strain energy in the two elementscan be written in the form

(10.21)

The total strain energy of the two elements is Finally, notice that K(g) is singular: the sum of the rows is the zero vector, as is

the sum of the columns. In this form, an attempt to solve the system will give riseto “rigid-body motion.” To illustrate this reasoning, suppose, for simplicity’s sake,that k(e) = k(e+1), in which case equilibrium requires that Pn = Pn+2. If computationswere performed with perfect accuracy, the equation would pose no difficulty. How-ever, in performing computations, errors arise. For example, Pn is computed as

and . Computationally, there is now anunbalanced force, . In the absence of mass, this, in principle, implies infiniteaccelerations. In the finite-element method, the problem of rigid-body motion canbe detected if the output exhibits large deformation.

The problem is easily suppressed using constraints. In particular, symmetryimplies that un+1 = 0. Recalling Equation 10.19, we now have

, (10.22)

in which R is a reaction force that arises to enforce physical symmetry in the presenceof numerically generated asymmetry. The equation corresponding to the secondequation is useless in predicting the unknowns un and un+2 since it introduces the

K K K K

K K

e e e e

e e

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

˜ , ˜

˜ , ˜ .

→ →

=

−

−

= −

−

+ +

+ +

1 1

1 1

1 1 0

1 1 0

0 0 0

0 0 0

0 1 1

0 1 1

k ke e

K K Kg e e( ) ( ) ( )˜ ˜= + +1

˜ ( )K e

V K

V K

e e e e

e e e e

e T

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

˜

˜

( ).

=

=

=

+ +

+ +

12

12

1 1

1 2

γγ γγ

γγ γγ

γγ

ΤΤ

ΤΤ

u u un n n

12 γγ γγT gK( ) .

P Pn n n= + ε ˆ ,P P P P Pn n n n n+ + + += + = =1 1 1 1ε ε εn n+ −1

EA

L

u

u

P

R

P

n

n

n

n

1 1 0

1 2 1

0 1 1

0

2 1

−

− −

−

=

−

+ +




new unknown R. The first and third equations are now rewritten as

, (10.23)

with the solution

. (10.24)

To preserve symmetry, it is necessary for un + un+1 = 0. However, the sum iscomputed as

(10.25)

The reaction force is given by R = −[εn + εn+1], and, in this case, it can be consideredas a measure of computational error.

Note that Equation 10.23 can be obtained from Equation 10.22 by simply “strikingout” the second row of the matrices and vectors and the second column of the matrix.

The same assembly arguments apply to the inertial forces as to the elastic forces.Omitting the details, the kinetic energies T of the two elements are

(10.26)

10.3.2 BEAMS

A similar argument applies for beams. The potential energy and stiffness matrix ofthe eth element can be written as

(10.27)

EAL

u

u

P

P

n

n

n

n

1 0

0 1 1 1

=

− +

+

+ +

ε

ε

u PA

Lu P

A

Ln n n n= − + = ++ +[ ] [ ]ε εE,

E2 1

u un nn n

AL

+ =+

( )++

11ε ε

E.

T = + +12

1˙ [ ˜ ˜ ˙( ) ( ) ( ) ( )γγ ]]γγe e e eTM M

˜/

/

˙ ˙ ˙ ˙

( ) ( )

( )( )

( )

M e e

ee

e T

=

=

= + +

m

m Al

u u ue e e

1 1 2

1 2 1

13

1 2

ρ

γγ

V

w w w w

b b

b e e e e

( ) ( )

( )

( ) ( )

( ) ( )

e e

e11e

12e

21e T

22e

T

K

KK K

K K

=

=

= − ′ − ′+ +

12

1 1

γγ γγ

γγ

T




In a two-element beam model analogous to the previous rod model, V (g) = V (e) +V (e+1), implying that

. (10.28)

Generally, in a global coordinate system, K(g) = ∑∑∑∑e K(e).

10.3.3 TWO-DIMENSIONAL ELEMENTS

We next consider assembly in 2-D. Consider the model depicted in Figure 10.5consisting of four rectangular elements, denoted as element e, e + 1, e + 2, and e + 3.The nodes are also numbered in the global system. Locally, the nodes in an elementare numbered in a counterclockwise fashion. Suppose there is one degree of freedomper node (e.g., x-displacement) and one corresponding force.

FIGURE 10.5 2-D assembly process.

K

K K 0

K K K K

0 K K

g11e

12e

21e T

22e

11e

12e

21e T

22e

( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

= +

+ +

+ +

1 1

1 1

e+3,4

e+3,1 e+3,2 e+2,1 e+2,2

e+1,3

e+1,2e+1,1e,2e,1

e,4 e,3

e+2,4

e+1,4

e+2,3

e+3

e+3

e+2

e+2

e

e

e+1

e+1

e+3,3

7 8 9

1 2 3

65

4




In the local systems, the force on the center node induces displacements according to

(10.29)

Globally,

(10.30)

Adding the forces of the elements on the center node gives

(10.31)

Taking advantage of the symmetry of the stiffness matrix, this implies that thefifth row of the stiffness matrix is

(10.32)

Finally, for later use, we consider damping, which generates a stress proportionalto the strain rate. In linear problems, it leads to a vector-matrix equation of the form

. (10.33)

At the element level, the counterpart of the kinetic energy and the strain energyis the Rayleigh Damping Function, D(e), given by , and the con-sistent damping force on the eth element is

(10.34)

f k u k u k u k u

f k u k u k u k u

ee

ee

ee

ee

e

ee

ee

ee

ee

,( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 3 1 1 3 2 2 3 3 3 3 4 4

1 4 4 11

1 1 4 21

1 2 4 31

1 3 4 41

= + + +

= + + +++

++

++

++

, , , , , , , ,

, , , , , , , , ee

e ,e

ee

ee

ee

e

ee

ee

e

f k u k u k u k u

f k u k u k

+

++

++

++

++

+

++

++

+

= + + +

= + +

1 4

2 1 1 12

2 1 1 22

2 2 1 32

2 3 1 42

2 4

3 2 2 13

3 1 2 23

3 2 2

,

, , , , , , , ,

, , , , ,

( ) ( ) ( ) ( )

( ) ( ),, , , ,3

33 3 2 4

33 4

( ) ( )ee

eeu k u+

++

++

u u u u u u u u

u u u u u u u u

u u u u u u u u

u u u

e e e e

e e e e

e e e e

e e

, , , ,

, , , ,

, , , ,

, ,

1 1 2 2 3 5 4 6

1 1 2 1 2 3 1 3 4 1 4 5

2 1 5 2 2 4 2 3 9 2 4 8

3 1 6 3 2

→ → → →

→ → → →

→ → → →

→ →

+ + + +

+ + + +

+ + uu u u u ue e5 3 3 8 3 4 7+ +→ →, ,

f k u k k u k u k k u

k k k k u k

e e e e e e

e e e e

5 3 1 1 3 2 4 11

2 4 21

3 4 31

1 22

4

3 3 4 41

1 12

2 23

5 3

= + +[ ] + + +[ ]+ + + +[ ] +

+ + + +

+ + +

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,,( )

,( )

,( )

,( )

,( )

,( )

4 2 13

6

2 43

7 1 42

2 33

8 1 32

9

e e

e e e e

k u

k u k k u k u

+[ ]+ + +[ ] +

+

+ + + +

κ 5 3 1 3 2 4 11

4 21

4 31

1 22

3 3 4 41

1 12

2 23

T = +[ ] +[ ]+ + +[ ]

+ + + +

+ + +

k k k k k k

k k k k symmetry

e e e e e e

e e e e

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,( )

,( ) K K

M D K f˙ ˙ ( )γγ γγ γγ+ + = t

D( ) ( ) ( ) ( )˙ ˙e e T e eD= 12 γγ γγ

fd t( ) ( ) ( ) ( )( ) ˙˙ .e e e eD= ∂

∂ =γγ γγD




The Rayleigh Damping Function is additive over the elements. Accordingly, if is the damping matrix of the eth element referred to the global system, the

assembled damping matrix is given by

(10.35)

It should be evident that the global stiffness, mass, and damping matrices havethe same bandwidth; the force on one given node depends on the displacements(velocities and accelerations) of the nodes of the elements connected at the givennode, thus determining the bandwidth.

10.3 EXERCISES

1. The equation of static equilibrium in the presence of body forces, suchas gravity, is expressed by

Without the body forces, the dynamic Principle of Virtual Work is derived as

How should the second equation be modified to include body forces?2. Consider a one-dimensional system described by a sixth-order differential

equation:

Consider an element from xe to xe+1. Using the natural coordinate ξ = −1when x = xe′ = +1 when x = xe+1, for an interpolation model with theminimum order that is meaningful, obtain expressions for ϕϕϕϕ(ξ), ΦΦΦΦ, and γγγγ,serving to express q as

.

3. Write down the assembled mass and stiffness matrices of the followingthree-element configuration (using rod elements). The elastic modulus isE, the mass density is ρ, and the cross-sectional area is A.

ˆ ( )D e

D Dg e

e

( ) ( )ˆ .= ∑

∂∂

= −S

Xbij

ji.

δ δ ρ δ τE S dV uu

tdV dSij ij i

2i

2 i iu∫ ∫ ∫+∂∂

= .

Qd q

dxQ

6

6 0= , a constant.

q T( ) ( )ξ ξ= ϕϕ ΦΦγγ




4. Assemble the stiffness coefficients associated with node n, shown in thefollowing figure, assuming plane-stress elements. The modulus is E, andthe Poisson’s ratio is . K(1), K(2), and K(3) denote the stiffness matricesof the elements.

5. Suppose that a rod satisfies δΨ = 0, in which Ψ is given by

Use the interpolation model

For an element xe < x < xe+1, find the matrix Ke such that

6. Redo this derivation for Ke in the previous exercise, using the naturalcoordinate ξ, whereby ξ = ax + b, in which a and b are such that −1 =axe + b, +1 = axe+1 + b.

7. Next, regard the nodal-displacement vector as a function of t. Find thematrix Me such that

y

L LL/2 x

ν

K (1)

K (3)

K (2)

n

Ψ =

−∫ 1

2

2

0

dudx

dxPu LL ( )

.E

u x xu x t

u x t

e

e

( ) ( , )

( , ).=

+

11

ΦΦ

Ke

e

e

e

e

u x t

u x t

f

f

( , )

( , ).

+ +

=

1 1

M Ke

e

e

e

e

e

e

e

u x t

u x t

u x t

u x t

f

f

( , )

( , )

( , )

( , ),

+

••

+ +

+

=

1 1 1




in which ρ is the mass density. Derive Me using both physical and naturalcoordinates.

8. Show that, for the rod under gravity, a two-element model gives the exactanswer at x = 1, as well as a much better approximation to the exact dis-placement distribution.

9. Apply the method of the previous exercise to consider a stepped rod, asshown in the figure, with each segment modeled as one element. Is thedisplacement at x = 2L still exact?

EAρ

Lg

g

E1

A1 L1

L2

ρ1

E2

A2

ρ2



153

Solution Methodsfor Linear Problems

11.1 NUMERICAL METHODS IN FEA

11.1.1 S

OLVING

THE

F

INITE

-E

LEMENT

E

QUATIONS

: S

TATIC

P

ROBLEMS

Consider the numerical solution of the linear system

Kγγγγ

=

f

, in which

K

is thepositive-definite and symmetric stiffness matrix. In many problems, it has a largedimension, but is also banded. The matrix can be “triangularized”:

K

=

LL

T

, inwhich

L

is a lower triangular, nonsingular matrix (zeroes in all entries above thediagonal). We can introduce

z

=

L

Tγγγγ

and obtain

z

by solving

Lz

=

f

. Next, γγγγ

can becomputed by solving

L

Tγγγγ

=

z

. Now

Lz

=

f

can be conveniently solved by forwardsubstitution. In particular,

Lz

=

f

can be expanded as

(11.1)

Assuming that the diagonal entries are not too small, this equation can be solved,starting from the upper-left entry, using simple arithmetic:

z

1

=

f

1

/

l

11

,

z

2

=

[

f

2

−

l

21

z

1

]

/

l

22

,

z

3

=

[

f

3

−

l

31

z

1

−

l

32

z

2

]

/

l

33

,

…

.Next, the equation

L

Tγγγγ

=

z

can be solved using backward substitution. Theequation is expanded as

(11.2)

11

l

l l

l l l

l l l

z

z

z

z

f

f

f

fn n nn n n

11

21 22

31 32 33

1 2

1

2

3

1

2

3

0 0

0

. . .

. . . .

. . .

. . . . . .

. . . . .

. . .

.

.

.

.

=

.

l l l

l

l l l

l l

l

n

n n n n n n

n n n n

nn n

11 12 1

22

2 2 2 1 2

1 1 1

1

2

3

0

0 0

0

0 0 0 0

. . .

. . . .

. . . .

. . .

. . .

.

.

.

, , ,

, ,

− − − − −

− − −

γ

γ

γ

γ

=

−

−

f

f

f

f

n

n

n

1

2

1

.

..



154


Starting from the lower-right entry, solution can be achieved using simple arith-metic as

γ

n

=

f

n

/

l

nn

,

In both procedures, only one unknown is encountered in each step (row).

11.1.2 M

ATRIX

T

RIANGULARIZATION

AND

S

OLUTION

OF

L

INEAR

S

YSTEMS

We next consider how to triangularize

K

. Suppose that the upper-left (

j

−

1)

×

(

j

−

1) block

K

j

−

1

has been triangularized: In determining whether the

j

×

j

block

K

j

can be triangularized, we consider

(11.3)

in which

k

j

is a (

j

−

1)

×

1 array of the first

j

−

1 entries of the

j

th

column of

K

j

.Simple manipulation suffices to furnish

k

j

and

l

jj

.

(11.4)

Note that λλλλ

j

can be conveniently computed using forward substitution. Also,note that

l

jj

=

The fact that

K

j

>

0 implies that

l

jj

is real. Obviously,the triangularization process proceeds to the (

j

+

1)

st

block and on to the completestiffness matrix.

As an example, consider

(11.5)

Clearly, For the second block,

(11.6)

γ γ γ γ γn n n n n n n n n n n n n n n nf l l f l l l− − − − − − − − − − − − −= − = − −1 1 1 1 1 1 2 2 2 2 1 1 2 2[ ]/ , [ ]/ , ., , , , , K

K L LTj j j− − −=1 1 1.

KK k

k k

L 0 L

0T T

T

Tj

j j

j jj

j

j jj

j j

jjl l

=

=

− − −1 1 1

λλ

λλ,

k j j j

jj jj j jl k

=

= −

−L

T

1λλ

λλ λλ

kjj j j j− −−k kT 1K 1

.

A3 =

1 12

13

12

13

14

13

14

15

.

L L1 1T= → 1.

1 0 1

0

1

2 22

2

22

12

12

13λ

λ

l l

=

,



Solution Methods for Linear Problems

155

from which

λ

2

=

1/2 and Now

(11.7)

We now proceed to the full matrix:

(11.8)

We conclude that

l

31

=

1/3, l32 = = 1/5 − 1/9 − 1/12 =

11.1.3 TRIANGULARIZATION OF ASYMMETRIC MATRICES

Asymmetric stiffness matrices arise in a number of finite-element problems, includ-ing problems with unsteady rotation and thermomechanical coupling. If the matrixis still nonsingular, it can be decomposed into the product of a lower-triangular andan upper-triangular matrix:

(11.9)

Now, the jth block of the stiffness matrix admits the decomposition

(11.10)

in which it is assumed that the ( j − 1)th block has been decomposed in the previousstep. Now, uj is obtained by forward substitution using Lj−1uj = k1j, and λλλλj can beobtained by forward substitution using Finally, for

l2221 3 1 2 1 12= − =/ ( / ) / .

L2

1 0

1 2 1 12=

/ /.

1

1 0 0

1 12 0

1

0 1 12

0 0

1

2 12

2

12

13

12

13

14

13

14

15

3 3

12

31 32 33

12 31

32

33

12 31

12

13 31 32

31 31

=

=

= +

+

L LT

/ /

/ /

/

l l l

l

l

l

l

l l

l l ll l l l32 312

322

33212/ + +

1 12/ , l332 1

180 .

K LU= .

KK k

k

L 0 U

0

L U L

U

T T T

T T

j

j j

j jj

j

j jj

j j

jj

j j j j

j j j j jj jj

k l

l

=

=

=+

− − −

− − −

−

1 1

2

1 1

1 1 1

1

λλ

λλ λλ

u

u

u

u u

U kTj j j− =1 2λλ . u ujj jj jj j jl k= − λλT ,




which purpose ujj can be arbitrarily set to unity. An equation of the form Kx = f cannow be solved by forward substitution applied to Lz = f, followed by backwardsubstitution applied to Ux = z.

11.2 TIME INTEGRATION: STABILITY AND ACCURACY

Much insight can be gained from considering the model equation:

(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 Equation11.11 have stability properties corresponding to asymptotic stability. For this pur-pose, we apply the trapezoidal rule, the properties of which will be discussed in asubsequent section. Consider time steps of duration h, and suppose that the solutionhas been calculated through the nth time step, and we seek to compute the solution atthe (n + 1)st time step. The trapezoidal rule is given by

(11.12)

Consequently,

(11.13)

Clearly, yn+1 → 0 if < 1, and yn+1 → ∞ if > 1, in which | · | impliesthe 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-stabilityrequires that

(11.14)

A-stability implies that λr > 0, which is precisely the condition for asymptoticstability.

Consider the matrix-vector system arising in the finite-element method:

(11.15)

dy

dty= −λ ,

dy

dt

y y

hy y yn n

n n≈−

− ≈ − +++

112

, [ ]. λ λ

yh

hy

h

hy

n n

n

+ = −+

= −+

1

0

1 21 2

1 21 2

λλ

λλ

//

//

1 21 2

−+

λλ

hh//

1 21 2

−+

λλ

hh//

1

11

2 2

2 2

2 2

2 2

− +

+ +<

λ λ

λ λ

rh ih

rh ih.

M D K 0˙ ˙ , ( ˙ ( ˙ ,γγ γγ γγ γγ γγ γγ γγ+ + = = = 0 , 0) )0 0



Solution Methods for Linear Problems 157

in which M, D, and K are positive-definite. Elementary manipulation serves to derivethat

(11.16)

It follows that and γγγγ → 0. We conclude that the system is asymptoticallystable.

Introducing the vector the n-dimensional, second-order system is writtenin state form as the (2n)-dimensional, first-order system of ordinary differentialequations:

(11.17)

We next apply the trapezoidal rule to the system:

(11.18)

From the equation in the lower row, pn+1 = [γγγγn+1 − γγγγn] − pn. Eliminating pn+1

in the upper row furnishes a formula underlying the classical Newmark method:

(11.19)

and KD can be called the dynamic stiffness matrix. Equation 11.19 can be solvedby triangularization of KD, followed by forward and backward substitution.

11.3 NEWMARK’S METHOD

To fix the important notions, consider the model equation

(11.20)

Suppose this equation is modeled as

(11.21)

d

dt

12

12

˙ ˙ ˙ ˙ .γγ γγ γγ γγ γγ γγT T TM K D 0+

= − <

γγ → 0

p = ˙,γγ

M 0

0 I

p D K

I 0

p f

0

+

−

=

•

γγ γγ.

M 0

0 I

p p D K

I 0

p p f f

0

−

−

+−

+

+

=+

+

+

+

+

+

1

1

12

12

12

1

1

1

1

1h

h

n n

n n

n n

n n

n n( )

( )

( )

( )

( ).

γγ γγ γγ γγ

2h

K r K M D K

r M D K M D p f f

D Dγγ n n

n n n n n

h h

h hy

h h h

+ +

+ +

= = + +

= + −

+ +

+ +

1 1

2

1

2 2

1

2 4

2 4 2 2 4

,

( )

dy

dxf y= ( ).

α β γ δy y h f fn n n n+ ++ + + =1 1 0[ ] .




We now use the Taylor series to express yn+1 and fn+1 in terms of yn and fn. Notingthat and we obtain

(11.22)

For exact agreement through h2, the coefficients must satisfy

(11.23)

We also introduce the convenient normalization γ + δ = 1. Simple manipulationserves to derive that α = −1, β = 1, γ = 1/2, δ = 1/2, thus furnishing

(11.24)

which can be recognized as the trapezoidal rule. The trapezoidal rule is unique and optimal in having the following three char-

acteristics:

It is a “one-step method” using only the values at the beginning of the currenttime step.

It is second-order-accurate; it agrees exactly with the Taylor series through h2.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 efficiencyof 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 physicalcoordinates to natural coordinates. Consider Let ξ = [2x − (a + b)].Clearly, ξ maps the interval [a, b] into the interval [−1,1]. The integral now becomes

Now consider the power series

(11.25)

from which

(11.26)

The advantages illustrated for integration on a symmetric interval demonstrate that,with n function evaluations, an integral can be evaluated exactly through (2n − 1)st order.

′ =y fn n ′′ = ′y fn n ,

0 22= + ′ + ′′ + + ′ + ′′ + ′α β γ δ[ / ] [ ] .y y h y h y h y y h h yn n n n n n n

α β α γ δ α γ+ = + + = + =0 0 0 2/ .

y y

hf y f yn n

n n+

+−

= +11

12

[ ( ) ( )],

∫ ab f x dx( ) . 1

b a−

11

1b a f d− ∫ − ( ) .ξ ξ

f ( ) 0 1 22

33

44 5ξ α α ξ α ξ α ξ α ξ α ξ= + + + + + +5 L,

f d( .ξ ξ α α α) 2 0

23

025

01

1

1 3 5−∫ = + + + + + +L




Consider the first 2n − 1 terms in a power-series representation for a function:

(11.27)

Assume that n integration (Gauss) points ξi and n weights are used as follows:

(11.28)

Comparison with Equation 11.26 implies that

(11.29)

It is necessary to solve for n integration points, ξi, and n weights, wi. These areuniversal quantities. To integrate a given function, g(ξ), exactly through ξ2n−i, it isnecessary to perform n function evaluations, namely to compute g(ξi).

As an example, we seek two Gauss points and two weights. For n = 2,

(11.30)

From (ii) and (iv), leading to ξ2 = −ξ1. From (i) and (iii), itfollows that −ξ2 = ξ1 = 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 mechanicalsystem, which is unforced but has nonzero initial values, is described by

. (11.31)

Assume a solution of the form which furnishes upon substitution

(11.32)

The jth eigenvalue, λj, is obtained by solving and a corre-sponding eigenvector vector, γγγγj, can also be computed (see Sample Problem 2).

g nn( .ξ α α ξ α ξ) 1 2 2

2 1= + + + −L

g d g w w w wi i

i

n

i

i

n

i i

i

n

n i in

i

n

( ( .ξ ξ ξ α α α α ξ) )1

1

1

1 2

1

2

1− = = =

−

=∫ ∑ ∑ ∑ ∑= = + + +

1

2 1L

w w w wn

wi

i

n

i

i

n

i i

i

n

i in

i

n

i in

i

n

= = =

−

=

−

=∑ ∑ ∑ ∑ ∑= = = =

−=

1 1

2

1

2 2

1

2 1

1

1 0 2 32

2 10, , / , , , . ξ ξ ξ ξK

w w w w

w w w w

1 2 1 1 2 2

1 12

2 22

1 13

2 23

2 0

23

0

+ = + =

+ = + =

(i), (ii)

(iii), (iv)

ξ ξ

ξ ξ ξ ξ

w1 1 12

22 0ξ ξ ξ[ ] ,− =

1/ 3.

M K 0˙ , ( , ˙ ( ˙γγ γγ γγ γγ γγ γγ+ = = = 0 00 0) )

γγ γγ = ˆ ( ),exp tλ

[ ]ˆ .K M 0+ =λ2 γγ

det( ) ,K M+ =λ j2 0




For the sake of generality, suppose that λj and γγγγj are complex. Let denote thecomplex conjugate (Hermitian) transpose of γγγγj. Now, satisfies

.

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 γγγγj to be real and orthonormal.

Sample Problem 1

As an example, consider

(11.33)

Now det[K + λ2I] = 0 reduces to

(11.34)

with the roots

(11.35)

so that both are negative (since k11 and k22 are positive).We now consider eigenvectors. The eigenvalue equations for the ith and jth

eigenvectors are written as

(11.36)

It is easily seen that the eigenvectors have arbitrary magnitudes, and for conve-nience, we assume that they have unit magnitude Simple manipulationfurnishes that

(11.37)

γγ jH

λ j2

λ jj j

j j

2 = −γγ γγγγ γγ

H

H

K

M

M K=

=

1 0

0 1

11 12

12 22

k k

k k.

( ) [ ] [ ] ,λ λ2 211 22

211 22 12

2 0+ + + + =k k k k k

λ+ − = − + ± + − +[ ]= − + ± − −[ ]

, [ ] [ ] [ ]

[ ] [ ]

211 22 11 22

211 22 12

2

11 22 11 222

122

12

4

12

4

k k k k k k k

k k k k k

λ λ+ −2 2 and

[ [K M 0 K M 0+ = + =ω ωj j k k2 2]g g ]

γγ γγj jT = 1.

γγ γγ γγ γγ γγ γγ γγ γγk j j k j k j k j kT T T TK K M M− − − =[ ] .ω ω2 2 0




Symmetry of K and M implies that

(11.38)

Assuming for convenience that the eigenvalues are all distinct, it follows that

(11.39)

The eigenvectors are thus said to be orthogonal with respect to M and K. Thequantities and are called the ( jth) modal mass and ( jth)modal stiffness.

Sample Problem 2

Consider

(11.40)

Let For the determinant to vanish, Using the first eigenvector satisfies

(11.41)

implying that The corresponding procedures for the sec-ond eigenvalue furnish that It is readily verified that

(11.42)

The modal matrix X is now defined as

(11.43)

γγ γγ γγ γγ γγ γγ γγ γγ γγ γγk j j k j k j k j k j k k jT T T T TK K M M M− = − = − =0 02 2 2 2, [ ( ) . ]ω ω ω ω

γγ γγ γγ γγj k j k j kT TM K= = ≠0 0, , .

µ j j j= γγ γγTM κ j j j= γγ γγTK

2 0

0 1

2 1

1 1

0

0

1

2

1

2

+

−

−

=

••γ

γ

γ

γk

m.

ζ ω ω ω2 202

02= =j k m/ , / . 1 1 22− = ±±ζ / .

1 1 22− =±ζ / ,

2 1

1 1 2

0

01

11

21 1

1 2

21 2−

−

=

[ ] + [ ] =

/, ,

( )

( )

( ) ( )γ

γγ γ

γ γ11

211 3 2 3( ) ( )/ , / .= =

γ γ12

221 3 2 3( ) ( )/ , / .= = −

γγ γγ γγ γγ

γγ γγ γγ γγ

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

, / , / ,

, [ / ], [ / ]

2 1 1 21 2

2 1 1 21 2

0 4 3 4 3

043

1 1 243

1 1 2

T T

T T

M M

K K

= = = =

= = = − = +

µ µ

κ κ

X = [ ].γγ γγ γγ γγ1 2 3 L n




Since the jkth entries of XTMX and XTKX are and respectively,it follows that

(11.44)

The modal matrix is said to be orthogonal with respect to M and K, but it isnot purely orthogonal since X−1 ≠ XT.

The governing equation is now rewritten as

(11.45)

implying the uncoupled modes

(11.46)

Suppose that gj(t) = gj0 sin (ω t). Neglecting transients, the steady-state solutionfor the jth mode is

(11.47)

It is evident that if (resonance), the response amplitude for thejth mode is much greater than for the other modes, so that the structural motion underthis excitation frequency illustrates the mode. For this reason, the modes can easilybe animated.

11.5.2 COMPUTATION OF EIGENVECTORS AND EIGENVALUES

Consider with Many methods have been proposed tocompute the eigenvalues and eigenvectors of a large system. Here, we describe amethod that is easy to visualize, which we call the hypercircle method. The vectorsKγγγγj and Mγγγγj must be parallel to each other. Furthermore, the vectors and must terminate at the same point in a hypersphere in n-dimensional space. Suppose that is the ν th iterate and that the two vectors

γγ γγj kTM γγ γγj k

TK ,

X MX X KXT T=

=

µ

µ

µ

κ

κ

κ

1

2

1

2

0

0

0

0

. . .

. . .

. . . . .

. . . . .

. . . .

. . .

. . .

. . . . .

. . . . .

. . . .

.

n n

X MX X KX g X g X fT T 1 T˙ , , ,ξξ ξξ ξξ γγ+ = = =−

µ ξ κ ξj j j j jg t˙ ( ).+ =

ξκ ω µ

ωjj

j j

gt=

−0

2 sin( ).

ω ω κ µ2 2~ /j j j=

K Mγγ γγj j j= ω 2 , γγ γγj jT = 1.

K KTγ γ γj j j/ 2

M MTγ γ γj j j/ 2

γγ j( )ν




do not coincide in direction. Another iterate can be attained by an interval-halvingmethod:

(11.48)

Alternatively, note that

(11.49)

is the cosine of the angle between two unit vectors, and as such it assumes themaximum value of unity when the vectors coincide. A search can be executed onthe hypersphere in the vicinity of the current point, seeking the path along which C(γγγγj)increases until the maximum is attained.

Once the eigenvector γγγγj is found, the corresponding eigenvalue is found from Now, an efficient scheme is needed to “deflate” the system

so that and γγγγ1 no longer are part of the eigenstructure in order to ensure that thesolution scheme does not converge to values that have already been calculated.

Given γγγγ1 and we can construct a vector p2 that is M-orthogonal to γγγγ1 byusing an intermediate vector Clearly,

However, p2 is also clearly orthogonal to Kγγγγ1

since it is collinear with Mγγγγ1. A similar procedure leads to vectors pj, which are orthog-onal to each other and to Mγγγγ1 and Kγγγγ1. For example, with p2 set to unit magnitude,

Now, Introduce the matrix X1 as follows: X1 = [γγγγ1 p2 p3

… pn]. For the kth eigen-value, we can write

(11.50)

which decomposes to

(11.51)

This implies the “deflated” eigenvalue problem

(11.52)

The eigenvalues of the deflated system are also eigenvalues of the original system.The eigenvector ηηηηn−1 can be used to compute the eigenvectors of the original system

MM

M

K

KM

Tˆ , ˆ ˆ ˆ .( )( )

( ) ( )

( )

( ) ( )

( ) ( ) ( ) ( )γγγγ

γγ γγ

γγ

γγγγ γγ γγ γγ

γγj

j

j j

j

j j

j j j jT T ν

ν

ν ν

ν

ν ν

ν ν ν ν+ + + + += +

=1

2 2

1 1 1 2 112

C(γγγγ

γγ γγ

γγ

γγ γγjj

j j

j

j j

) =T

T T

K

K

M

M2 2

ω j2 = ( )/( ).γγ γγ γγ γγj j j j

T TK Mω1

2,

ω12,

ˆ : ˆ ˆ .p p p MpT2 2 2 1 2 1 = − γγ γγ γγ γγ1 2 1 2

T TMp Mp= −ˆγγ γγ γγ1 2 1 1 10 1T TMp Mˆ .= =assuming µ

p3 = ˆ ˆ ˆ .p p p p MpT T3 2 3 2 1 3 1− − γγ γγ p p MpT T

2 3 1 30 0= = and γγ ˆ .

X KX X MX X 0T T 11 1

21 1 1−[ ] =−ωω γγk k ,

ωω

12

1

2

1 1

1 0 00

0 K

0

0 M 0T T˜ ˜.

n

k

n n− − −

−

=

ηη

[ ˜ ˜ ] .K M 0n k n n− − −− =1 1 1ω ηη




using transformations involving the matrix X1. The eigenvalues and eigenvectors ofthe deflated system can be computed by, for example, the hypercircle methoddescribed previously.

11.6 EXERCISES

1. Verify that the triangular factors L3 and for A3 in Equation 11.5 are correct.2. Using A3 in Equation 11.5, use forward substitution followed by backward

substitution to solve

3. Triangularize the matrix

4. For the model equation dy/dx = f(y), develop a two-step numerical-integrationmodel:

What is the order of the integration method (highest power in h with exactagreement with the Taylor series)?

5. Find the integration (Gauss) points and weights for n = 3.6. In the damped linear-mechanical system

suppose that γγγγ(t) = γγγγn at the nth time step. Derive KD and rn+1 such that γγγγat the (n + 1)st time step satisfies

7. For the linear system

triangularize the matrix and solve for γ1, γ2, γ3.

LT3

A3γγ =

1

1

1

.

K =

36 30 18

30 41 23

18 23 14

.

( ) [ ( ) ( ) ( )] .α β γ δ ε ζy y y h f y f y f yn n n n n n+ − + −+ + + + + =1 1 1 1 0

M D K f˙ ˙ ( ),γγ γγ γγ+ + = t

K rDγγ n n+ +=1 1.

36 30 24

30 41 32

24 32 27

1

2

3

1

2

3

=

γ

γ

γ




8. (a) Find the modal masses µ1, µ2 and the modal stiffnesses κ1 and κ2 ofthe system

(b) Determine the steady-state response of the system (i.e., particularsolution to the equation).

9. Triangularize

10. Put the following equations in state form, apply the trapezoidal rule, andtriangularize the ensuing dynamic stiffness matrix:

31 0

0 227

1 1

1 2

10

2010

1

2

1

2

+

−

−

=

( )

••γ

γ

γ

γsin .t

2 0

0 4 2

2 0

2

µ

µ µ

µ

A L A

AL Y A L

A A L

/

/ /

/

.

−

−

M K f 0T˙ , .γγ γγ ΣΣππ ΣΣ γγ+ − = =



167

Rotating and Unrestrained Elastic Bodies

12.1 FINITE ELEMENTS IN ROTATION

We first consider rotation about a fixed axis. The coordinate system is embedded inthe fixed point and rotates. The undeformed position vector,

X

′

in the rotated systemis related to its counterpart,

X

, in the unrotated system by The counterpartfor the deformed position is The displacement also satisfies The rotation is represented by the “axial” vector, ωωωω

, satisfying (Recall that is antisymmetric). The time derivatives in rotating coordinatessatisfy

(12.1)

where imply differentiation with the coordinate system instanta-neously fixed. The rightmost four terms in (12.1) are called the translational, Coriolis,centrifugal, and angular accelerations, respectively.

Applied to the Principle of Virtual Work, the inertial term becomes Assuming that

(12.2)

The matrix

M

is the conventional positive-definite and symmetric mass matrix; theCoriolis matrix

G

1

is antisymmetric; the centrifugal matrix

G

2

is negative-definite; andthe angular acceleration matrix

A

is antisymmetric. Also,

12

′ =X Q X( ) .t′ =x Q x( ) .t ′ =u Q u( ) .t

ωω × =(*) ˙ ( ) ( )(*).Q QTt t˙ ( ) ( )Q QTt t

ddt t

ddt t

′ = ∂ ′∂

+ × ′

′ = ∂ ′∂

+ × ′ + × × + ×

u uu

u uu u u

ωω

ωω ωω ωω αα2

2

2

2 2

αα ωω= ∂∂

∂∂t t and (.)

∫ ′δ ρu T ddt

2

2

[ ′ + ′u X ] .dV ′ = ′u XTϕϕ ΦΦγγ( ) ( )t ,

δ ρ δ′ ′ = + + +

∫ u u G G AT Td

dtdV

ddt

ddt

2

2

2

2 1 2γγ ΜΜ γγ γγ γγ( )

M G

G

T T T T

T T T T

= =

= =

∫ ∫∫ ∫

ΦΦ ϕϕϕϕ ΦΦ ΦΦ ϕϕΩΩϕϕ ΦΦ

ΦΦ ϕϕΩΩ ϕϕ ΦΦ ΑΑ ΦΦ ϕϕ ϕϕ ΦΦ

ρ ρ

ρ ρ

d d

d d

V V

V V

, ,

, .

1

22 ΑΑ

˙ ˙ , ˙ .ΩΩ ΩΩ= =QQT A



168

Finite Element of Analysis: Thermomechanics of Solids

There is also a rigid-body force term:

(12.3)

The governing equation is now

(12.4)

Consider a rod attached to a thin shaft rotating steadily at angular velocity

ω

(see Figure 12.1), with

f

0

=

0. If

r

is the undeformed position along the shaft, the governing equation is

(12.5)

Assuming a one-element model with we obtain

(12.6)

Clearly,

u

(

L

,

t

) becomes unbounded if

ω

becomes equal to the natural frequency in which case,

ω

is called a critical speed.

FIGURE 12.1

Elastic rod on rotating rigid shaft.

z

ω

θ

x

ry

E,A,L,ρ

f0 sin ωt

δ ρ δ ρ′ ′ = − = + ′∫ ∫u X f f A XT Trot rot

Tddt

d d2

22V , V.γγ ΦΦ ϕϕ ΩΩ[ ]

M G K G A f fd

dt

d

dt rot

2

2 1 2

γγ γγ γγ+ + + + = −[ ] .

EAd u

drr u

2

22= − +ω [ ].

u r t ru L t L( , ) ( , )= ,

EA

L

ALu L t A

r

Lrdr

AL

L

−

= −

= −

∫ρω ρω

ρω

22

0

2 2

3

3

( , )

ω ρ= EA AL3 3

,



Rotating and Unrestrained Elastic Bodies

169

12.2 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. Theresponse is referred to the “body axes” corresponding to axes embedded in thecorresponding rigid body, for example, the principal axes of the moment of inertiatensor. The position vector

r

of a point in the body can be decomposed as follows:

, (12.7)

in which

r

c

is the position vector to the center of mass; is the relative position-vector 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 thedeformed position in the rotating system (see Figure 12.2).

The balance of linear momentum becomes

(12.8)

Recall that and the corresponding variational relation is It follows that in which

δ

u

′

is the variation of

u

with the axes instantaneously held fixed. The quantities

r

c

, θθθθ

, and

u

′

can be variedindependently since there is no constraint relating them. For

δ

r

c

, we have

(12.9)

FIGURE 12.2

3-D unconstrained 3-D element.

r r uc= + +ξξ

ξξ

∂∂

= + +S

r uij

jci i iξ

ρ ξ[˙ ˙ ˙ ].

ξξ ωω ξξ= × , δ δξξ ξξ= ×u ,ω = θθ. δ δ δ δr r u uc= + × + + ′˙ ( )θθ ξξ ,

δξ

ρ ξrS

r u dVciij

jci i i

∂∂

− + +

=∫ [˙ ˙ ˙ ] .0

z

x

y

u(ξξ,t)

ξ(t)

rc(t)



170

Finite Element of Analysis: Thermomechanics of Solids

Now by virtue of the definition of the center of mass.Next on the assumption, for which a strong argument can be made,that deformation does not affect the position of the center of mass. Then,

Finally,

(12.10)

Consequently

P

=

m

r

··

c

, as expected.Consider the variation with respect to ξξξξ

:

dξξξξ

=

dθθθθ

×

ξξξξ

(12.11)

First,

(12.12)

and Recall that and the corresponding variational relationis Now,

(12.13)

It is common to assume that which also implies that This assumption implies that the body axes in the deformed configuration are notaffected by deformation and are obtained by rotating the undeformed axes accordingto rigid-body relations.

Next, consider

δξ

i

:

(12.14)

∫ = ∫ =δ ρξ δ ρξr dV r dVci i ci i˙ ˙ 0

∫ =δ ρr u dVci i˙ 0

∫ = ∫ =δ ρ δ ρ δr r dV r r dV r mrci ci ci ci ci ci˙ ˙ ˙ .

δξ

δξ

δ

δ τ

δ

rS

dV rS

dV

r S n dS

r dS

r P

ciij

jci

ij

j

ci ij j

ci i

ci i

∂∂

=∂∂

=

=

=

∫ ∫

∫∫

δξξ

ρ ξiij

jci i i

Sr u dV

∂∂

− + +

=∫ [˙ ˙ ˙ ] .0

δξ ρ δξ ρi ci ci ir dV r dV˙ ˙ ,∫ ∫=

δ ρξ[ ]∫ =idV 0, ξξ ωω ξξ= ×δ δξξ ξξ= ×θθ , ω = ˙.θθ

δ ρ δ ρ δ ρξξ ξξ ξξi i i iTu d u dV dV˙ [ ] ˙ ˙ .V∫ ∫ ∫= × = ×θθ θθ u

∫ × =ξξ uρdV 0, ∫ × =ξξ ρ˙ .u dVi 0

δξ ρξ δ ρ ρ

δ ρ ρ

δ

i idV dV

dV

˙ [ ]

[ ]

[ ]

∫ ∫∫

= × × × + ×

= − × × × + × ×

= × +

θθ

θθ

θθ

T

T

T J J

ξξ ωω ωω ξξ αα ξξ

ωω ξξ ξξ ωω ξξ ξξ αα

ωω ωω αα



Rotating and Unrestrained Elastic Bodies 171

in which Clearly, as the matrix corresponding to avector product, « is antisymmetric. We determined in an earlier discussion that «2 isnegative-definite, thus, the moment-of-inertia tensor J is positive-definite. Finally,

(12.15)

The vector-valued quantity m is recognized as the moment of the tractionsreferred to the center of mass. We thus obtain the Euler equation of a rigid body as

. (12.16)

Equations 12.10 and 12.16 can be solved separately from the finite-elementequation for the displacements, to be presented next, to determine the origin andorientation of the “body axes.”

Now consider the variation δu′:

(12.17)

First, note that since The remaining terms are exactlythe same as for a body with a fixed point, and consequently it reduces to

(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-statesolution.

2. Find the exact solution for u(r) in Exercise 1. Does it agree with the one-element solution? Try two elements. Is the solution better? By how much?

J = − ∫ = ×ρρ .. ξξ ..2dV, ( ) ( ). and « «

δξξ ξ

δξ δ ξξ

δξ

δξ

δ

iij

j ji ij

i

jij

i ij i

i i

T

SdV S dV dV

S n dS

t dS

dS

∂∂

= ∂∂

− ∂∂

=

=

= = ×

∫ ∫ ∫

∫∫

∫

( )

( )

,

S

θθ ξξm m t

m J J= × +ωω ωω αα

δσξ

ρ ξ′∂∂

− + +

=∫ u r u dViij

jci i i[˙ ˙ ˙ ] .0

∫ ′ =δ ρu dVri ci˙ 0 ∫ ′ =δ ρu dVi 0.

M G K G A f fd

dt

d

dt rot

2

2 1 2

γγ γγ γγ+ + + + = −[ ]

G G

A f X

T

T

1 2

2

= =

= = + ′

∫ ∫∫ ∫

ΦΦ ϕϕΩΩϕϕ ΦΦ ΦΦ ϕϕΩΩ ϕϕ ΦΦ

ΦΦ ϕϕ ϕϕ ΦΦ ΦΦ ϕϕ ΩΩ

ρ ρ

ρ ρ

T T T

Trot

T

dV dV

dV dV

, ,

,

2

[ ]A A



172 Finite Element of Analysis: Thermomechanics of Solids

3. Prove that the inertia tensor J is positive-definite.Find J for the brick “element” with density ρ.

4. Write down the proof that for an unrestrained elastic body,

if

then

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, givingrise to Coriolis effects in bending. Derive the ensuing one-element model.Note that this situation couples extension and bending.

ab

c

ρ ρu u∫ ∫= × =dV dV0 0 and ξξ ,

F r M= = + ×m˙0 .J Jαα ωω ωω



173

Thermal, Thermoelastic,and Incompressible Media

13.1 TRANSIENT CONDUCTIVE-HEAT TRANSFER

13.1.1 F

INITE

-E

LEMENT

E

QUATION

The governing equation for conductive-heat transfer without heat sources, assumingan isotropic medium, is

(13.1)

With the interpolation model and thefinite-element equation assumes the form

(13.2)

This equation is parabolic (first-order in the time rates), and implies that thetemperature changes occur immediately at all points in the domain, but at smallerinitial rates away from where the heat is added. This contrasts with the hyperbolic(second-order time rates) solid-mechanics equations, in which information propa-gates into the medium as finite velocity waves, and in which oscillatory responseoccurs in response to a perturbation.

13.1.2 D

IRECT

I

NTEGRATION

BY

THE

T

RAPEZOIDAL

R

ULE

Equation 13.1 is already in state form since it is first-order, and the trapezoidal rulecan be applied directly:

(13.3)

from which

(13.4)

13

k ce∇ =2T Tρ ˙ .

T(t) T x t0 TT

T− = ϕϕ ( ) ( )ΦΦ θθ ∇ =T ββ θθTT

T tΦΦ ( ),

K M

K M

T

TT

θθ θθ

ββ ββ ϕϕ ϕϕ

+ = −

= =∫ ∫T

TT

T TT

T T TT

T e T TT

t

k dV dVT

˙ ( )

,

q

ΦΦ ΦΦ ΦΦ ΦΦρc

M Kq q

Tn n

Tn n n n

h

θθ θθ θθ θθ+ + +−+

+= −

+1 1 1

2 2,

K rDT n nθθ + +=1 1

K M K r M K q qDT T T n T n T n n n

h h h= + = − − ++ +2 2 21 1θθ θθ ( )



174


For the assumed conditions, the dynamic thermal stiffness matrix is positive-definite, and for the current time step, the equation can be solved in the same manneras in the static counterpart, namely forward substitution followed by backwardsubstitution.

13.1.3 M

ODAL

A

NALYSIS

Modes are not of much interest in thermal problems since the modes are notoscillatory or useful to visualize. However, the equation can still be decomposedinto independent single degree of freedom systems. First, we note that the thermalsystem is asymptotically stable. In particular, suppose the inhomogeneous termvanishes and that θθθθ

at

t

=

0 does not vanish. Multiplying the equation by θθθθ

T

andelementary manipulation furnishes that

(13.5)

Clearly, the product θθθθ

T

M

Tθθθθ

decreases continuously. However, it only vanishes if θθθθ

vanishes.To examine the modes, assume a solution of the form θθθθ

(

t

)

=

θθθθ

0

j

exp(

λ

j

t

). Theeigenvectors θθθθ

0

j

satisfy

(13.6)

and we call

µ

Tj

and

κ

Tj

the

j

th

modal thermal mass and

j

th

modal thermal stiffness,respectively. We can also form the modal matrix ΘΘΘΘ

= [θθθθ

01

…

θθθθ

0

n

], and again

(13.7)

Let ξξξξ

=

ΘΘΘΘ

−

1θθθθ

and g(

t

)

=

ΘΘΘΘ

T

q

(

t

). Pre- and postmultiplying Equation 13.2 with ΘΘΘΘ

T

and ΘΘΘΘ

, respectively, furnishes the decoupled equation

(13.8)

ddt

TT

θθ θθ θθ θθT

TMK

20

= − < .

θθ θθ θθ θθ0 0 0 00 0j T kTj

j T kTjj k

j k

j k

j kT TM K=

=≠

==≠

µ κ

,

ΘΘ ΘΘ ΘΘ ΘΘT TM

. .

. . .

. . . . .

. . . . .

. . . .

K

. .

. . .

. . . . .

. . . . .

. . . .

T

Tj

Tj

T

Tj

Tj

=

=

µ

µ

κ

κ

0 0

0

0

0 0

0

0

, .

µ ξ κ ξTj j Tj j jg˙ .+ =



Thermal, Thermoelastic, and Incompressible Media

175

Suppose, for convenience, that

g

j

is a constant. Then, the general solution is ofthe form

(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 F

INITE

-E

LEMENT

E

QUATION

The classical theory of coupled thermoelasticity accommodates the fact that thethermal and mechanical fields interact. For isotropic materials, assuming that tem-perature only affects the volume of an element, the stress-strain relation is

(13.10)

in which

α

denotes the volumetric thermal-expansion coefficient. The equilibriumequation is repeated as . The Principle of Virtual Work implies that

(13.11)

Now consider the interpolation models

(13.12)

in which

E

is the strain written as a column vector in conventional finite-elementnotation. The usual procedures furnish the finite-element equation

(13.13)

The quantity ΣΣΣΣ

is the

thermomechanical stiffness

matrix

. If there are

n

m

displace-ment degrees of freedom and

n

t

thermal degrees of freedom, the quantities appearingin the equation are

ξ ξκµ

κµ

τj jTj

Tj

Tj

Tj

t

t t T g= −

+ − −

∫0

0exp exp ( ) ,j d

S E Eij ij kk ij= + − −2 0µ λ α δ( ( )) ,T T

∂∂ =S

xij

j

uiρ˙

δ µ λ δ δ ρ αλ δ δ δE E E dV u ü dV E dV u t dSij ij kk ij o i i o ij ij o j j o[ ] ( .2 0+ + − − =∫ ∫ ∫ ∫T T )

u N E B BT T= → = − = ∇ =T x x x x( ) ( ), ( ) ( ), ( ) ( ) ( ) ( ),γγ γγ ,, t t t tij TT E T T T0 v θθ θθ

M K f˙ ( ) ( ) ( ) ( ), .γγ γγ ΣΣθθt t t t dVo+ − = = ∫ Σ αλ B Tνν

M K f, : , ( ), ( ): , : , ( ): . n n t t n n n t nm m m m t t× × × ×γγ ΣΣ θθ1 1



176


We next address the thermal field. The energy-balance equation (from Equation7.35), including mechanical effects, is given by

(13.14)

Application of the usual variational methods imply that

(13.15)

Case 1:

Suppose that T is constant. At the global level, Thus,the thermal field is eliminated at the global level, giving the new governing equationas

(13.16)

Conductive-heat transfer is analogous to damping. The mechanical system isnow asymptotically stable rather than asymptotically marginally stable.

We next put the global equations in

state form

:

(13.17)

Clearly, Equation 13.17 can be integrated numerically using the trapezoidal rule:

(13.18)

k c tre∇ = +20T T Tρ αλ˙ ( ˙ ).E

K M q q n qT TTt t t dSθθ θθ ΣΣ γγ( ) ˙( ) ˙ ( ) , .+ + = − = ⋅∫T 0 νν

θθ ΣΣ γγ( ) ˙ ( ) .t tT T= − +− −T T0 0K K q1 T 1

M K K f1 T˙ ( ) ˙ ( ) ( ) ( ).γγ ΣΣ ΣΣ γγ γγt t t tT+ + =−T0

Q z Q z f1 2˙ + =

Q

M 0 0

0 K 0

0 0 M

z

Q

0 K

K 0 0

0 K

f

f

0

qT

1

0

2

0 0

=

=

=

−

−

=

−

T

T

T

T

T

/

˙

/ /

γγ

γγ

θθ

ΣΣ

ΣΣ

Q Q z Q Q z f f1 2 1 1 2 12 2 2+

= −

+ ++ +h h h

n n n n[ ].



Thermal, Thermoelastic, and Incompressible Media 177

Now, consider asymptotic stability, for which purpose it is sufficient to take f = 0,z(0) = z0. Upon premultiplying Equation 13.17 by zT, we obtain

(13.19)

and z must be real. Assuming that θθθθ ≠ 0, it follows that z ↓ 0, and hence the systemis asymptotically stable.

13.2.2 THERMOELASTICITY IN A ROD

Consider a rod that is built into a large, rigid, nonconducting temperature reservoirat x = 0. The force, f0, and heat flux, −q0, are prescribed at x = L. A single elementmodels the rod. Now,

(13.20)

The thermoelastic stiffness matrix becomes ΣΣΣΣ = αλ∫BνTdV → Σ = αλA/2. Thegoverning equations are now

(13.21)

13.3 COMPRESSIBLE ELASTIC MEDIA

For a compressible elastic material, the isotropic stress Skk and the dilatational strainEkk 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 compressivevolume strain (Ekk < 0) becomes infinite. At the limit ν = 1/2, the material is saidto satisfy the internal constraint of incompressibility.

Consider the case of plane strain, in which Ezz = 0. The tangent modulus matrixD is readily found from

(13.22)

d

dt

T

12

12

1 2

2 2

z Q z z Q z

z Q Q z

K

T T

T T

T

= −

= − +[ ]= −θθ θθ

u x t x t L E x t t L x t Ld

dxt L( , ) ( )/ , ( , ) ( )/ , ( )/ , ( )/ .= = − = =γ γ θ θ T T

T0

ρ γ γ α λ θ

ρ θ θ α λ γ

AL AL

A

1 c AL kAL

Ae

3

31 1

20 0

˙

˙ ˙

+ − =

+ + = −

E 12

f

T Tq

0

0

S

S

S

E

E

E

xx

yy

zz

xx

yy

zz

=+ −

− − +( )− +( ) −

−

E1 1 0

1 1 0

0 0 1 2

2

2

( )( ).

1 1 2ν ν

ν ν ν

ν ν ν

ν




Clearly, D becomes unbounded as ν → 1/2. Furthermore, suppose that for amaterial to be nearly incompressible, ν is estimated as .495, while the correct valueis .49. It might be supposed that the estimated value is a good approximation forthe correct value. However, for the correct value, (1 − 2ν)−1 = 50. For the estimatedvalue, (1 − 2ν)−1 = 100, implying 100 percent error!

13.4 INCOMPRESSIBLE ELASTIC MEDIA

In an incompressible material, a pressure field arises that serves to enforce theconstraint. Since the trace of the strains vanishes everywhere, the strains are notsufficient to determine the stresses. However, the strains together with the pressureare sufficient. In FEA, a general interpolation model is used at the outset for thedisplacement field. The Principle of Virtual Work is now expressed in terms ofthe displacements and pressure, and an adjoining equation is introduced to enforcethe constraint a posteriori. The pressure can be shown to serve as a Lagrangemultiplier, and the displacement vector and the pressure are varied independently.

In incompressible materials, to preserve finite stresses, we suppose that thesecond Lame coefficient satisfies λ → ∞ as tr(E) → 0 in such a way that the productis an indeterminate quantity denoted by p:

(13.23)

The Lame form of the constitutive relations becomes

(13.24)

together with the incompressibility constraint Eijδi j = 0. There now are two inde-pendent principal strains and the pressure with which to determine the three principalstresses.

In a compressible elastic material, the strain-energy function w satisfies Sij = ,and the domain term in the Principle of Virtual Work can be rewritten as ∫δEijSijdV =∫δwdV. The elastic-strain energy is given by w = µEijEij + . For reasonsexplained shortly, we introduce the augmented strain-energy function

(13.25)

and assume the variational principle

(13.26)

λtr p( ) .E → −

S E pij ij ij= −2µ δ ,

∂∂

wEij

λ2 Ekk

2

′ = −w E E pEij ij kkµ

δ δ ρ δ′ + =∫ ∫ ∫w dV dV dSoT

o oT

ou u u˙ .ττ



Thermal, Thermoelastic, and Incompressible Media 179

Now, considering u and p to vary independently, the integrand of the first termbecomes δw′ = δEij[2µEij − pδij] − δpEkk , furnishing two variational relations:

(13.27)

The first relation is recognized as the Principle of Virtual Work, and the secondequation serves to enforce the internal constraint of incompressibility.

We now introduce the interpolation models:

(13.28)

Substitution serves to derive that

(13.29)

Assuming that these equations apply at the global level, use of state formfurnishes

(13.30)

The second matrix is antisymmetric. Furthermore, the system exhibits marginalasymptotic stability; namely, if f(t) = 0 while (0), γγγγ (0), and ππππ (0) do not all vanish,then

(13.31)

δ δ ρ δ

δ

E S dV u u dV dS

pE dV

ij ij oT

o oT

o

kk

∫ ∫ ∫∫

+ =

=

˙ u ττ (a)

(b)0 0

u N B x

b x x

= =

= =

T T

kkT T

t t

E x t p t

( (

( ( ) ( )

x g g

g x p

) ( ) ) ( )

) ( ) ( )

e

M K f˙ ( ) ( ) ( )

, ( )

γ t t t

dV to

+ − =

= =∫γγ ΣΣππ

ΣΣ ξξ ΣΣ γγb T T 0

M 0 0

0 K 0

0 0 0

0 K

K 0 0

0 0

f

0

0T T T T

+

−

−

=

d

dt

t

t

t

t

t

t

t˙ ( )

˙ ( )

( )

˙ ( )

( )

( )

( )γγ

γγ

ππ

ΣΣ

ΣΣ

γγ

γγ

ππ

.

γγ

d

dtt t t

t

t

t

12

0( ˙ ( ) ( ) ( ))

˙ ( )

( )

( )

γγ γγ ππ

γγ

γγ

ππ

T T T

T T

M 0 0

0 K 0

0 0 0

=




13.5 EXERCISES

1. Find the exact solution for a circular rod of length L, radius r, mass densityρ, specific heat ce , conductivity k, and cross-sectional area A = πr2. Theinitial temperature is T0, and the rod is built into a large wall at fixedtemperature T0 (see figure below). However, at time t = 0, the temperatureT1 is imposed at x = L. Compare the exact solution to the one- and two-element solutions. Note that for a one-element model,

2. State the equations of a thermoelastic rod, and put the equations for thethermoelastic behavior of a rod in state form.

3. Put the following equations in state form, apply the trapezoidal rule, andtriangularize the ensuing dynamic stiffness matrix, assuming that the tri-angular factors of M and K are known.

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 lengthis L, the lateral dimension is Y, and the interpolation models are linear forthe displacements (u linear in x, with v,w linear in y) and constant for thepressure. Show that the finite-element equation assumes the form

and that this implies that 3µ = f (which can also be shown by an apriori argument).

kAL

L tc AL

L t q Leθ ρ θ( , ) ˙( , ) ( ).+ = −3

T0 T1,t>0

r

L

M K f, 0T˙ .γγ γγ ΣΣππ ΣΣ γγ+ − = =

2 0

0 4 2

2 0

0

0

2

µ

µ

A L A

AL Y AL Y

A AL Y

u L

v Y

p

f/

/ /

/

−

−

=

( )

( )

u LL( )



181

Torsion and Buckling

14.1 TORSION OF PRISMATIC BARS

Figure 14.1 illustrates a member experiencing torsion. The member in this case iscylindrical with length

L

and radius

r

0

. The base is fixed, and a torque is applied atthe top surface, which causes the member to twist. The twist at height

z

is

θ

(z), andat height

L

, it is

θ

0

.Ordinarily, in the finite-element problems so far considered, the displacement is

the basic unknown. It is approximated by an interpolation model, from which anapproximation for the strain tensor is obtained. Then, an approximation for the stresstensor is obtained using the stress-strain relations. The nodal displacements aresolved by an equilibrium principle, in the form of the Principle of Virtual Work. Inthe current problem, an alternative path is followed in which stresses or, moreprecisely, a stress potential, is the unknown. The strains are determined from thestresses. However, for arbitrary stresses satisfying equilibrium, the strain field maynot be compatible. The compatibility condition (see Chapter 4) is enforced, furnishing

FIGURE 14.1

Twist of a prismatic rod.

14

section before twist

section after twist

zT

θ0

r0

θ

φ

x

z

y

L

x



182


a partial differential equation known as the Poisson Equation. A variational argumentis applied to furnish a finite-element expression for the torsional constant of thesection.

For the member before twist, consider points

X

and

Y

at angle

φ

and at radialposition

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 smallangles

θ

furnishes, to first order,

(14.1)

It is also assumed that torsion does not increase the length of the member, whichis attained by requiring that axial displacement

w

only depends on

X

and

Y

. Thequantity

w

(

X

,

Y

) is called the

warping function

. It is readily verified that all strains vanish except

E

xz

and

E

yz

, for which

(14.2)

Equilibrium requires that

(14.3)

The equilibrium relation can be identically satisfied by a potential function

y

for which

(14.4)

We must satisfy the compatibility condition to ensure that the strain field arisesfrom a displacement field that is unique to within a rigid-body translation androtation. (Compatibility is automatically satisfied if the displacements are consideredthe unknowns and are approximated by a continuous interpolation model. Here, thestresses are the unknowns.) From the stress-strain relation,

(14.5)

Compatibility (integrability) now requires that , furnishing

(14.6)

u Y v X= − =θ θ, .

Ewx

yz

Ewy

xzxz yz= ∂

∂− ∂

∂

= ∂∂

+ ∂∂

12

12

θ θ, .

∂∂

+∂∂

=S

x

S

yxz yz 0.

Sy

Sxxz yz= ∂

∂= − ∂

∂ψ ψ

, .

E Sy

E Sxxz xz yz yz= = ∂

∂= = − ∂

∂1

21

21

21

2µ µψ

µ µψ

, .

∂∂ ∂

∂∂ ∂=2 2w

x yw

y x

− ∂∂

∂∂

+

+ ∂

∂− ∂

∂−

=

y yy

d

dz x xx

d

dz

12

12

12

12

0µ

ψ θµ

ψ θ,




183

which in turn furnishes Poisson’s Equation for the potential function

y

:

(14.7)

For boundary conditions, assume that the lateral boundaries of the member areunloaded. The stress-traction relation already implies that

τ

x

=

0 and

τ

y

=

0 on thelateral boundary

S

. For traction

τ

z

to vanish, we require that

(14.8)

Upon examining Figure 14.2, it can be seen that

n

x

=

dy

/

ds

and

n

y

=

−

dx

/

ds

, inwhich

s

is the arc length along the boundary at

z

. Consequently,

(14.9)

Now, on and therefore

ψ

is a constant, which can, in general, be takenas zero.

We next consider the total torque on the member. Figure 14.3 depicts the crosssection at

z

. The torque on the element at

x

and

y

is given by

(14.10)

FIGURE 14.2

Illustration of geometric relation.

ψ

ψ

dψ z –dx

dy dsn

nx = cos χ = dy/dsny = sin χ = dx/ds

∂∂

+ ∂∂

= −2

2

2

2 2ψ ψ µ θx y

d

dz.

τ z x xz y yzn S n S S= + = 0 on .

τ

ψ ψ

ψ

z xz yz

dy

dsS

dx

dsS

dy

ds y

dx

ds x

d

ds

= −

= ∂∂

+ ∂∂

=

S dds, ψ = 0,

dT xS dxdy yS dxdy

xd

dxy

d

dydxdy

yz xz= −

= − −

ψ ψ



184


Integration furnishes

(14.11)

Application of the divergence theorem to the first term leads to

∫

ψ

[

xn

x

+

yn

y

]

ds

,which vanishes since

y

vanishes on

S

. Finally,

(14.12)

We apply variational methods to the Poisson Equation, considering the stress-potential function

y

to be the unknown. Now,

(14.13)

FIGURE 14.3

Evaluation of twisting moment.

z

x

xy

y

dx syz

sxz

dy

T xddx

yddy

dxdy

d xdx

d ydy

dxdx

dydy

dxdy

x

ydxdy dxdy

= − +

= − +

− +

= − ∇⋅

+

∫

∫

∫ ∫

ψ ψ

ψ ψ ψ

ψ

ψψ

( ) ( )

2

T dxdy= ∫2 ψ .

δψ ψ µθ[ ] .∇ ⋅ ∇ + ′ =∫ 2 0dxdy




185

Integration by parts, use of the divergence theorem, and imposition of the“constraint”

y

=

0 on

S

furnishes

(14.14)

The integrals are evaluated over a set of small elements. In the

e

th

element,approximate

ψ

as in which

ν

T

is a vector with dimension (number ofrows) equal to the number of nodal values of

ψ

. The gradient

∇

ψ

has a correspondinginterpolation model in which ββββ

T

is a matrix. The finite-elementcounterpart of the Poisson Equation at the element level is written as

(14.15)

and the stiffness matrix should be nonsingular, since the constraint

y

=

0 on

S

hasalready been used. It follows that, globally, The torque satisfies

(14.16)

In the theory of torsion, it is common to introduce the torsional constant

J,

forwhich

T

=

2

µ

J

θ

′

. It follows that

14.2 BUCKLING OF BEAMS AND PLATES

14.2.1 E

ULER

B

UCKLING

OF

B

EAM

C

OLUMNS

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 equilibriumstates that the member potentially can sustain: compression only, or compressionwith bending. The member will “snap” to the second state if it involves less “potentialenergy” than the first state. The notions explaining buckling are addressed in detail insubsequent chapters. For now, we will focus on beams and plates, using classicalequations in which, by retaining lowest-order corrections for geometric nonlinearity,in-plane compressive forces appear.

( ) .∇ ⋅∇ = ′∫ ∫δψ ψ δψ µθdxdy dxdy2

νT e ex yT ( , ) ,ψψ ηη

∇ =ψ ββ ψψ ηηT e ex yT ( , ) ,

K f

K

f

T T

T

Te

e Te

Te

e T T e

Te

e T

dxdy

x y dxdy

( ) ( )

( )

( ) ( , )

ηη

ψψ ββ ββ ψψ

ψψ

= ′

=

=

∫∫

2µθ

ν

ηηgg 1 gK f= ′ −

2µθ T T( ) ( ) .

T dxdy

T

T T T

=

=

= ′

∫( )

( ) ( )− ( )

2

2

4

ψ

µθ

ηηgT g

g T g 1 g

f

f K f

J Tg

Tg

TgT

=−

21

f K f( ) ( ) ( ).



186


For the beam shown in Figure 14.4, the classical Euler buckling equation is

(14.17)

and

P

is the axial compressive force. The interpolation model for

w(x) is recalledas w(x) = ϕϕϕϕT(x)ΦΦΦΦγγγγ. Following the usual variational procedures (integration by parts)furnishes

(14.18)

At x = 0, both δ w and −δw′ vanish, while the shear force V and the bendingmoment 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 canuse the interpolation formula

(14.19)

The mass matrix is shown, after some algebra, to be

(14.20)

FIGURE 14.4 Euler buckling of a beam column.

z

E,I,A,L,ρ

y

x

Q0

M0P

EIw Pw Awiv + ′′ + =ρ ˙ ,0

δ ρ δ ρw Aw dx x A x dV˙ ˙ , ( ) ( )∫ ∫→ =γγ γγ ΦΦ ϕϕ ϕϕ ΦΦT T TM M

δ δ δ

δ δ

w Iw Pw dx w Iw dx w Pw dx

w Pw Iw w Iw

iv

L L

[ ]

[( )( )] [( )( )]

E E

E E

+ ′′ = ′′ ′′ − ′ ′

− − ′ − ′′′ − − ′ − ′′

∫ ∫ ∫0 0

w x x xL L

L Lt t

w L

w L( ) ( ) ( ) ( )

( )

( ).=

− −

=− ′

−

2 3

2 3

2

1

2 3γγ γγ,

M K K= =

ρALL

L L0 0 2

1335

11210

11210

1105

, .



Torsion and Buckling 187

Similarly,

(14.21)

The governing equation is written in finite-element form as

(14.22)

In a static problem, the solution has the form

(14.23)

in which cof denotes the cofactor, and γγγγ → ∞ for values of which renderdet(K2

14.2.1.2 Dynamic Buckling

In a dynamic problem, it may be of interest to determine the effect of P on theresonance frequency. Suppose that f(t) = f0 exp(iωt), in which f0 is a known vector.The displacement function satisfies γγγγ(t) = γγγγ0 exp(iωt), in which the amplitude vectorγγγγ0 satisfies

(14.24)

Resonance occurs at a frequency ω0, for which

(14.25)

Clearly, is an eigenvalue of the matrix Theresonance frequency is reduced by the presence of P and vanishes precisely atthe critical value of P.

δ δ

δ δ

′ ′ = =

′′ ′′ = =

∫

∫

w Pw dxPL

L

L L

w Iw dxI

L

L

L L

T

T

γγ γγ

γγ γγ

K K

K K

1 1 2

3 2 2 2

65

110

110

215

12 6

6 4

,

EE

,

E,

IL

PL

ALQ

M3 2 1 0

0

0

K K K f f−

+ = =

γγ γγρ ˙ .

˙ ,γγ = 0

γγ =−

−

cof

det

,

K K

K Kf

2 1

2 1

PL

I

PL

I

2

2

E

E

PLI2

E− =PL

I2

1 0E K ) .

EIL

PL

AL3 2 12

0 0 0K K K f− −

=ω ρ γγ .

det .EIL

PL

AL3 2 1 02

0 0K K K− −

=ω ρ

ω02 1

01 2

2 1 01 2

3ρALEK K K K− −−/ /[ ] .IL

PL

ω02




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 beamexperiences shear force V0 − V1 and bending moment M0 − M1. The beam on theright is governed by

(14.26)

The governing equation is written in finite-element form as

(14.27)

Consider the symmetric case in which M0 = 0, with the implication that w′(L) = 0.The equation reduces to

(14.28)

from which we obtain the critical buckling load given by P1 = 10EI/L2.

FIGURE 14.5 Buckling of a clamped-clamped beam.

L L

M0

V0

P

δ δ

δ δ

′ ′ = =

′′ ′′ = =

∫

∫

w Pw dxPL

L

L L

w Iw dxI

L

L

L L

T

T

γγ γγ

γγ γγ

K K

K K

1 1 2

3 2 2 2

65

110

110

215

12 6

6 4

,

EE

,

EIL

L

L L

PL

L

L L3 2 2

12 6

6 4

65

110

110

215

−

=γγ f

f =

=

− ′

V

M

w L

w L

1

1

,( )

( ) γγ

12653 1

EI

L

P

Lw L V−

=( ) ,




Next, consider the antisymmetric case in which V0 = 0 and w(L) = 0. The coun-terpart of Equation 14.28 is now

(14.29)

and P2 = 30EI/L2.If neither the constraint of symmetry nor axisymmetry is applicable, there are two

critical buckling loads, to be obtained in Exercise 2 as and These valuesare close enough to the symmetric and antisymmetric cases to suggest an interpretationof the two buckling loads as corresponding to the two “pure” buckling modes.

Compare the obtained values with the exact solution, assuming static condi-tions. Consider the symmetric case. Let w(x) = wc(x) + wp(x), in which wc(x) isthe characteristic solution and wp(x) is the particular solution reflecting the per-turbation. From the Euler buckling equation demonstrated in Equation 14.17, wc(x)has a general solution of the form wc(x) = α + βx + γ cosκx + δ sinκx, in whichκ = Now, w = −w′ = 0 at x = 0, −w′(L) = 0, and EIw′′′(L) = V1, expressedas the conditions

(14.30)

or otherwise stated

(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 remainingcoefficients must satisfy

(14.32)

42

15 1

EI

LPL w L M−

− ′ =( ( )) ,

27 8 2. EIL 8 9 2. .EI

L

PL I2 / .E

1 0 1 0 0

0 1 0 0

0

0 0 3 31

α β γ δ

α β γ κδ

α β γκ κ δκ κ

α β γκ κ δκ κ

+ + + = −

+ + + = − ′

+ − + = − ′

+ + − = − ′′′ +

w

w

L L L w L

L L EIw L V

p

p

p

p

( )

( )

sin cos ( )

sin cos ( )

Bz B z=

−

− ′

− ′

− ′′′ +

=−

−

=

w

w

w L

Iw L V

L L L

L L

p

p

p

p

( )

( )

( )

( )

sin cos

sin cos

,

0

0

1 0 1 0

0 1 0

0

0 013 3E

, κ

κ κ κ κ

κ κ κ κ

α

β

γ

δ

− −

=

sin cos

sin cos.

κ κ κ

κ κ

γ

δ

L L L

L L

0

0




A nonvanishing solution is possible only if the determinant vanishes, whichreduces to sinκL = 0. This equation has many solutions for kL, including kL = 0.The lowest nontrivial solution is kL = p, from which Pcrit = π2EI/L2 = 9.87 EI/L2.Clearly, the symmetric solution in the previous two-element model (Pcrit = 10 EI/L2)gives an accurate result.

For the antisymmetric case, the corresponding result is that tan κL = κL. Thelowest meaningful root of this equation is kL = 4.49 (see Brush and Almroth, 1975),giving Pcrit = 20.19 EI/L2. Clearly, the axisymmetric part of the two-element model isnot as accurate, unlike the symmetric part. This issue is addressed further in thesubsequent exercises.

Up to this point, it has been implicitly assumed that the beam column is initiallyperfectly straight. This assumption can lead to overestimates of the critical bucklingload. Consider a known initial distribution w0(x). The governing equation is

(14.33)

or equivalently,

(14.34)

The crookedness is modeled as a perturbation. Similarly, if the cross-sectionalproperties of the beam column exhibit a small amount of variation, for example,EI(x) = EI0[1 + ϑ sin(πx/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

(14.35)

(see Wang 1953), in which the loads are illustrated as shown in Figure 14.6. The usual

FIGURE 14.6 Plate element with in-plane compressive loads.

d

dxI

d

dxw w P

d

dxw w

2

2

2

2 0

2

2 0 0E ( ) ( ) ,− + − =

d

dxI

d

dxw P

d

dxw

d

dxI

d

dxw P

d

dxw

2

2

2

2

2

2

2

2

2

2 0

2

2 0E E+ = + .

Ehw P

w

xP

w

yP

w

x yx y xy

2

24

2

2

2

2

2

12 10

( )−∇ + ∂

∂+ ∂

∂+ ∂

∂ ∂=

ν

z

xPx

Pyx

PyPxy

h

y




variational methods furnish, with some effort,

(14.36)

in which W = ∇∇Tw (a matrix!). In addition,

(14.37)

For simplicity’s sake, assume that from which we can obtainthe form

(14.38)

We also assume that the secondary variables (n ⋅ ∇)∇2w, (n ⋅ ∇)∇w, and also

are prescribed on S.

These conditions serve to obtain

(14.39)

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 inthe space of critical values at which buckling occurs. In this space, a straight line

δ δ δ δw w dA w w dS w w dS tr dA∇ = ⋅∇ ∇ − ∇ ⋅ ⋅∇ ∇ +∫ ∫ ∫ ∫4 2( ) ( ) ( ) ,n nP WW

δ

δ δ

w Pw

xP

wy

Pw

x ydA dw n n dS

w ww

wdA

x y

x y

x

y

x y xy

∂∂

∂∂

∂∂ ∂

2

2

2

2

2

+ +

=

−

∫ ∫

∫

( )

,

p

P

p P=+[ ]

+[ ]

=

∂∂

∂∂

∂∂

∂∂

P P

P P

P P

P P

x xy

xy y

x xy

xy y

wx

wy

wx

wy

12

12

12

12

,

w x y b b b( , ) = ϕϕ ΦΦ γγ2 2 2T

∇ =

= ( ) =w

w

wVEC

x

y

b b b b b bββ ΦΦ γγ ββ ΦΦ γγ1 2 2 2 2 2 2 2T TW, .

[ ]

[ ]

P P

P P

x xy

xy y

wx

wy

wx

wy

∂∂

∂∂

∂∂

∂∂

+

+

12

12

[ ]K K fb b b21 22 2− =γγ

K

K

T T

T T

b b b b b

b b b b b

Ed

d

21

2

2 2 2 2 2 2 2

22 2 1 2 1 2 2

12 1=

−

=

∫

∫

hA

( )νΦΦ ββ ββ ΦΦ

ΦΦ ββ ββ ΦΦP V




emanating from the origin represents a proportional loading path. Let the loadintensity, λ, denote the distance to a given point on this line. By analogy withspherical coordinates, there exist two angles, θ and φ, such that

(14.40)

Now,

(14.41)

For each pair (q, f), buckling occurs at a critical load intensity, λcrit(θ, φ),satisfying

(14.42)

A surface of critical load intensities, λcrit(θ, φ), can be drawn in the loading spaceshown in Figure 14.7 by evaluating λcrit(θ, φ) over all values of (q, f) and discardingvalues that are negative.

FIGURE 14.7 Loading space for plate buckling.

Pxy

Py

Px

λφ

θ

P P Px y xy= = =λ θ φ λ θ φ λ φcos cos , sin cos , sin

K K

K P

P

T T

b b

b b b b bdV

22 22

22 2 1 2 1 2 2

=

=

=

∫λ θ ϕ

θ ϕ θ ϕ

θ φθ ϕ ϕ

ϕ θ ϕ

ˆ ( , )

ˆ ( , ) ˆ ( , )

ˆ ( , )cos sin

sin sin

ΦΦ ββ ββ ΦΦ

cos

cos

det[ ( , ) ˆ ] .K Kb crit b21 22 0− =λ θ φ




14.3 EXERCISES

1. Consider the triangular member shown to be modeled as one finite ele-ment. Assume that

Find KT , fT , and the torsional constant T.

2. Find the torsional constant for a unit square cross section using twotriangular 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 simple-

simple case shown in the following figure. Compare Pcrit in the symmetricand antisymmetric cases with exact values.

Triangular shaft cross section.

ψ

ψ

ψ

ψ

( , ) ( ) .x y x y

x y

x y

x y

=

−

1

1

1

1

1 1

2 2

3 3

1

1

2

3

y

x

(2,3)

(3,2)

(1,1)

V0

M0

L L

P




7. Consider a cantilevered beam with a compressive load P, as shown in thefollowing figure. The equation is

The primary variables at x = L are w(L) and −w′(L), and

Find the critical buckling load(s).

EId wdx

Pd wdx

4

4

2

2 0+ = .

δ δ δ δ′′ ′′ =

′ ′ =

∫ ∫w Iw dx

IL

L

L Lw Pw dx

PL

L

L L

L

EE

0

L

03 2 2

12 6

6 4

12 6

6 4γγ γγ γγ γγT T, .

L

V0

M0P



195

Introduction to Contact Problems

15.1 INTRODUCTION: THE GAP

In many practical problems, the information required to develop a finite-elementmodel, for example, the geometry of a member and the properties of its constituentmaterials, can be determined with little uncertainty or ambiguity. However, oftenthe loads experienced by the member are not so clear. This is especially true if loadsare transmitted to the member along an interface with a second member. This classof problems is called contact problems, and they are arguably the most commonboundary conditions encountered in practical problems. The finite-element commu-nity has devoted, and continues to devote, a great deal of effort to this complexproblem, leading to gap and interface elements for contact. Here, we introduce gapelements.

First, consider the three-spring configuration in Figure 15.1. All springs are ofstiffness

k

. Springs

A

and

C

extend from the top plate, called the

contactor

, to thebottom plate, called the

target

. The bottom of spring

B

is initially remote from thetarget by a gap

g

. The exact stiffness of this configuration is

(15.1)

From the viewpoint of the finite-element method, Figure 15.1 poses the followingdifficulty. If a node is set at the lowest point on spring

B

and at the point directly

FIGURE 15.1

Simple contact problem.

15

kk g

k gc =

<

≥

2

3

δ

δ .

contactorP

A B C

δ

k k k

gtarget



196


below it on the target, these nodes are not initially connected, but are later connectedin the physical problem. Furthermore, it is necessary to satisfy the

nonpenetrationconstraint

whereby the middle spring does not move through the target. If the nodesare considered unconnected in the finite-element model, there is nothing to enforcethe nonpenetration constraint. If, however, the nodes are considered connected, thestiffness is artificially high.

This difficulty is overcome in an approximate sense by a bilinear contact element.In particular, we introduce a new spring,

k

g

, as shown in Figure 15.2.The stiffness of the middle spring (

B

in series with the contact spring) is nowdenoted as

k

m

, and

(15.2)

It is desirable for the middle spring to be soft when the gap is open (

g

>

δ

) andto be stiff when the gap is closed (

g

≤

δ

):

(15.3)

Elementary algebra serves to demonstrate that

(15.4)

Consequently, the model with the contact is too stiff by 0.5% when the gapis open, and too soft by 0.33% when the gap is closed (contact). One conclusionthat can be drawn from this example is that the stiffness of the gap element shouldbe related to the stiffnesses of the contactor and target in the vicinity of the contactpoint.

FIGURE 15.2

Spring representing contact element.

contactor

target

P

A B C

δ

k k k

kg

kk km

g

= +

−1 1

1

.

kk g

k gg =

>

≤

/100

δ

δ100 .

kk k g

k k gc ≈

+ >

+ ≤

2 0 01

2 0 99

.

. .

δ

δ



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 noguarantee that target nodes will come into contact with foundation nodes. The gapelements can be used to account for the unknown contact area, as follows. Figure 15.3shows a contactor and a target, on which are indicated candidate contact areas, d

S

c

and d

S

t

, containing nodes c1, c2,…..,cn, t1, t2,…..,tn. The candidate contact areasmust contain all points for which there is a possibility of establishing contact.

The gap (i.e., the distance in the undeformed configuration) from the

i

th

node ofthe contactor to the

j

th

node of the target is denoted by

g

ij

. (For the purpose of thisdiscussion, the gap is constant, i.e., not updated.) In point-to-point contact, the

i

th

node on the contactor is connected to each node of the target by a spring with abilinear stiffness. (Clearly, this element may miss the edge of the contact zone whenit does not occur at a node.) It follows that each node of the target is connected bya spring to each of the nodes on the contactor. The angle between the spring andthe normal at the contactor node is

α

ij

, while the angle between the spring and thenormal to the target is

α

ji

. Under load, the

i

th

contactor node experiences displacement

u

ij

in the direction of the

j

th

target node, and the

j

th

target node experiences displace-ment

u

ji

. For example, the spring connecting the

i

th

contactor node with the

j

th

targetnode has stiffness

k

ij

, given by

(15.5)

in which

δ

ij

=

u

ij

+

u

ji

is the relative displacement. The force in the spring connectingthe

i

th

contactor node and

j

th

target node is

f

ij

=

k

ij

(

g

ij

)

δ

ij

. The total normal forceexperienced by the

i

th

contactor node is

f

i

=

∑

j

f

ij

cos(

α

ij

).

FIGURE 15.3

Point-to-point contact.

kk g

k gij

ijlower ij ij

ijupper ij ij

=<

≥

δ

δ ,

candidate contactor contact surface

candidate target contact surface

dSc

dSt

c1

t1 t2 t3

c2 c3

α31

k(g31)



198


As an example of how the spring stiffness might depend upon the gap, considerthe function

(15.6)

where

γ

,

α

, and

ε

are positive parameters selected as follows. When

g

ij

−

δ

ij

−

γ

>

0,

k

ij

attains the lower-shelf value,

k

0

ε

,

and we assume that

ε

<<

1. If

g

ij

−

δ

ij

−

γ

<

0,

k

ij

approaches the upper-shelf value,

k

0

(1

−

ε

). We choose

γ

to be a small value toattain a narrow transition range from the lower- to the upper-shelf values. In therange 0

<

g

ij

−

δ

ij

<

γ

, there is a rapid but continuous transition from the lower-shelf(soft) value to the upper-shelf (stiff) value. If we now choose

α

such that

αγ

=

1,

k

ij

becomes

k

0

/2,

when the gap closes (

g

ij

=

δ

ij

). The spring characteristic is illustratedin Figure 15.4.

The total normal force on a contactor node is the sum of the individual contact-element forces, namely

(15.7)

Clearly, significant forces are exerted only by the contact elements that are“closed.”

FIGURE 15.4

Illustration of a gap-stiffness function.

k g

k g g

ij ij ij

ij ij ij ij

( )

( ) tan ( ) ( ) ,

−

= + − − − − − −

−

δ

ε ε πα δ γ δ γ0

1 21 2 22

f k gtj ij ij ij ij ij

i

Nc

= −∑ ( ) cos( ).δ δ α

kij

k0 /2

k0εk0

εk0δij –gij



Introduction to Contact Problems 199

15.3 POINT-TO-SURFACE CONTACT

We now briefly consider point-to-surface contact, illustrated in Figure 15.5 using atriangular element. Here, target node t3 is connected via a triangular element tocontactor nodes c1 and c2. The stiffness matrix of the element is written as k([g1 − δ1],[g2 − δ2]) , in which g1 − δ1 is the gap between nodes t1 and c1, and is thegeometric part of the stiffness matrix of a triangular elastic element. The stiffnessmatrix of the element can be made a function of both gaps. Total force normal tothe target node is the sum of the forces exerted by the contact elements on thecandidate contactor nodes.

In some finite-element codes, the foregoing scheme is used to approximate thetangential force in the case of friction. Namely, an “elastic-friction” force is assumedin which the tangential tractions are assumed proportional to the normal tractionthrough a friction coefficient. This model does not appear to consider sliding andcan be considered a bonded contact. Advanced models address sliding contact andincorporate friction laws not based on the Coulomb model.

15.4 EXERCISES

1. Consider a finite-element model for a set of springs, illustrated in thefollowing figure. A load moves the plate on the left toward the fixed plateon the right.What is the load-deflection curve of the configuration? For a finite-element model, an additional bilinear spring is supplied, asshown. What is the load-deflection curve of the finite-element model? Identify a kg value for which the load-deflection behavior of the finite-element model is close to the actual configuration.Why is the new spring needed in the finite-element model?

FIGURE 15.5 Element for point-to-surface contact.

candidate target contact surface

candidate contactor contact surface

element connecting nodet3 with nodes c1 and c2

dSt

dSc

t2t1

c1 c2 c3

t3

K K




2. Suppose a contact element is added in the previous problem, in which thestiffness (spring rate) satisfies

Suppose αγ = 1, kL = k/100, and ku = 100k. Compute the stiffness k forthe configuration as a function of the deflection δ.

F

HL

k

k

kk

k

k g

k g g

ij ij

ij ij ij ij

( )

( ) tan ( ) ( ) .

−

= + − − − − − −

−

δ

ε ε πα δ γ δ γ0

1 21 2 22



201

Introduction to Nonlinear FEA

16.1 OVERVIEW

The previous section addressed finite-element methods for linear problems. Appli-cations that linear methods serve to analyze include structures under mild loads,disks and rotors spinning at modest angular velocities, and heated plates. However,a large number of problems are nonlinear. Plasticity is a nonlinear materials theorysuited for metals in metal forming, vehicle crash, and ballistics applications. Inproblems with high levels of heat input, mechanical properties, such as the elasticmodulus, and thermal properties, such as the coefficient of specific heat, can bestrongly temperature-dependent. Rubber seals and gaskets commonly experiencestrains exceeding 50%. Soft biological tissues typically are modeled as rubberlike.Many problems involve variable contact, for example, meshing gear teeth. Heatconducted across electrical contacts can be strongly dependent on normal pressures.Fortunately, much of the linear finite-element method can be adapted to nonlinearproblems, as explained in this chapter. The next chapter focuses on isothermalproblems. The extension to thermomechanical problems will be presented in asubsequent chapter.

16.2 TYPES OF NONLINEARITY

There are three major types of nonlinearity in thermomechanical boundary-valueproblems:

material nonlinearity,

geometric nonlinearity

, and

boundary-conditionnonlinearity

. Nonlinearities can also be present if the formulation is referred todeformed coordinates, possibly introducing stress fluxes and converted coordinates.

Material nonlinearity can occur through nonlinear dependence of the stress onthe strain or temperature, including temperature dependence of the tangent modulustensor. Metals undergoing plastic flow exhibit nonlinear material behavior.

Geometric nonlinearity occurs because of large deformation, especially in prob-lems referred to undeformed coordinates. Rubber components typically exhibit largedeformation and require nonlinear kinematic descriptions. In this situation, a strainmeasure needs to be chosen, as does the stress conjugate to it. Boundary-conditionnonlinearity occurs because of nonlinear supports on the boundary and contact. Anexample of a nonlinear support is a rubber pad placed under a machine to absorbvibrations.

For finite-element methods for nonlinear problems, the loads or time steps areapplied in increments. Then, incremental variational principles, together with inter-polation models, furnish algebraic (static) or ordinary differential equations (dynamic)

16



202


in terms of vector-valued incremental displacements or incremental temperatures.For mechanical systems, a typical equation is

(16.1)

in which

∆γγγγ

is the

incremental displacement vector

,

∆

f

is the

incremental forcevector

,

K

(γγγγ

) is the

tangent stiffness matrix

, and

M

(γγγγ

) is the

tangent mass matrix

. Itwill be seen that this type of equation is a realization of the optimal Newton iterationmethod for nonlinear equations.

16.3 COMBINED INCREMENTAL AND ITERATIVE METHODS: A SIMPLE EXAMPLE

Consider a one-dimensional rod of nonlinear material under small deformation in whichthe elastic modulus is a function of strain: E

=

E

0

(1

+

αε

),

ε

=

E

11

. Under static loading,the equilibrium equation is

. (16.2)

Suppose the load is applied in increments,

∆

j

P

, and that the load at the

n

th

loadstep is

P

n

. Suppose further that the solution

γ

n

is known at the

n

th

load. We nowconsider the steps necessary to determine the solution

γ

n

+

1

for

P

n

+

1

. We introduce theincrement

∆

n

γ

=

g

n

+

1

−

γ

n

. Subtracting Equation 16.2 at the

n

th

step from the equationat the (

n

+

1

)

st

step, the incremental equilibrium equation for the

n

th

increment is now

. (16.3)

This equation is quadratic in the increments. In fact, geometric nonlinearitygenerally leads to a quadratic function of increments. The error of neglecting thenonlinearity may be small if the load increment is sufficiently small. However, wewill retain the nonlinear term for the sake of illustrating the use of iterative procedures.In particular, the previous equation can be written with obvious identifications inthe form

. (16.4)

Newton iteration (discussed in Chapter 3) furnishes the optimal iteration schemefor the

ν

th

iterate:

(16.5)

K( ) ( ) ˙γγ γγ ΜΜ γγ γγ∆ ∆ ∆+ = f,

A

L LP

E0 1+

=α γ γ

A

L LPn n

n n

E0 11 2+ +

=+α γ γ γ∆ ∆ ∆

g x x x( ) = + − =β ζ η2 0

x xx x

xxν ν

ν ν

νβ ζ η

β ζη β+ = − + −

+=1

20

2[ ]

, / .



Introduction to Nonlinear FEA

203

S

AMPLE

P

ROBLEM

1

The efficiency of this scheme is addressed as follows. Consider

ζ

=

1,

β

=

12, and

η

=

1. The correct solution is

x

=

0.0828. Starting with the initial value

x

=

0, thefirst two iterates are, approximately, . This illus-trates the rapid

quadratic

convergence of Newton iteration.

16.4 FINITE STRETCHING OF A RUBBER ROD UNDER GRAVITY: A SIMPLE EXAMPLE

Figure 16.1 shows a rubber rod under gravity. It is assumed to attain finite strain.The rod is also assumed to experience uniaxial tension. The figure refers to theundeformed configuration, with the element occupying the interval (

X

e

,

X

e

+

1

). Itslength is

l

e

, its cross-sectional area is

A

e

, and its mass density is

ρ

. It is composedof rubber and is stretched axially by the loads

P

e

and

P

e

+

1

. Prior to stretching, a givenmaterial particle is located at

X

. After deformation, it is located at

x

(

X

), and thedisplacement

u

(

X

) is given by

u

(

X

)

=

x

(

X

)

−

X

.

16.4.1 N

ONLINEAR

S

TRAIN

-D

ISPLACEMENT

R

ELATIONS

The element is assumed short enough that a satisfactory approximation for thedisplacement

u

(

X

) is provided by the linear interpolation model

(16.6)

in which . Now,

u

(

X

e

)

=

u

e

and

u

(

X

e

+

1

)

=

u

e

+

1

are viewed as theunknowns to be determined using the finite-element method. The Lagrangian strain

FIGURE 16.1

Rubber rod under load.

112

112

11440 833 1 0 0828= − =. , ( ) .and

u X u X X u u l

lx x x x

e e e e e

ee

e e

( ) ( )( )

, ,

= + − −

= = − −

+1

1N a NT T

aTe e eu u= + 1

Pe

Pe+1

xe,ue

xe+1,ue+1

g




E = Exx is given by

(16.7)

Note that

(16.8)

The vector BL and the matrix BNL are the strain-displacement matrices. The following sections illustrate two different approaches to formulating a finite-

element model for the rubber rod. One is based on satisfying the incompressibilityconstraint a priori. The second is based on satisfying the constraint a posteriori,which is typical of finite-element code practice. In addition, we also include a slightvariant involving a near-incompressibility constraint in an a posteriori manner.

16.4.2 STRESS AND TANGENT MODULUS RELATIONS

The Neohookean strain-energy density function, w, accommodating incompressibil-ity, is expressed in terms of the eigenvalues c1, c2, and c3 of C = 2E + I, as follows:

, (16.9)

in which D is the (small-strain) elastic modulus. We can show, with some effort,that under uniaxial tension, c2 = c3.

We first enforce the incompressibility constraint a priori by using the substitution

. (16.10)

After elementary manipulations,

. (16.11)

Eux

ux

u u

l

u u

l

l l

e e

e

e e

e

ee e

ee

= ∂∂

+ ∂∂

=−

+−

= − +−

−

+ +

12

12

11 1

12

11 1

1 1

2

1 1

2

2 .a a aT

dE d e L NL e= = +B a B B aT , B

B BLe

NLel l

=−

=

−

−

1 1

1

1 1 1

1 12 .

wD

c c c c c c= + + − − =2

3 1 01 2 3 1 2 3( ), subject to

c cc2 3

1

1= =

wD

E E= + + + −−

22 1 2 2 1 31 2[ /( ) ]/



Introduction to Nonlinear FEA 205

The (2nd Piola-Kirchhoff) stress, S, defined in Chapter 5, is now obtained as

(16.12)

The tangent modulus, D, is also required:

(16.13)

If the strain E is small compared to unity, DT reduces to D. We next satisfy the incompressibility constraint a posteriori. An augmented

strain-energy function w* is introduced by

, (16.14)

in which the Lagrange multiplier, p, is the (true) hydrostatic pressure. The augmentedenergy is stationary with respect to p as well as c1, c2, and c3, from which it followsthat c1c2c3 − 1 = 0 (incompressibility).

The stresses satisfy

(16.15)

It is readily shown that c2 = c3. Enforcement of the stationary condition forp (c1c2c3 − 1 = 0), with S2 = 0, now furnishes that . It follows that

, in agreement with Equation 16.12.

16.4.3 INCREMENTAL EQUILIBRIUM RELATION

The Principle of Virtual Work states the condition for static equilibrium as

(16.16)

SwE

D E

= ∂∂

= − + −13

1 2 1 3 2[ ( ) ]./

DSE

D E

T = ∂∂

= + −( ) ./2 1 5 2

wD

c c cp

c c c* [ ] ( )= + + − − −2

32

11 2 3 1 2 3

SwE

wc

D p c c c

c

SwE

wc

D p c c c

c

SwE

wc

D p c c c

c

11 1

1 2 3

1

22 2

1 2 3

2

33 3

1 2 3

3

22 2

22 2

0

22 2

0

= ∂∂

= ∂∂

= −

= ∂∂

= ∂∂

= − =

= ∂∂

= ∂∂

= − =

* *

* *

* *.

p D c= / 1

S D c1 13 21 2= − −[ ]//

B P N a 0 PTS AdX g AdXP

Pe

X

Xe

X

X

ee e

e

ee

e

e

e

e+ +

∫ ∫− −

= =

+

1 1

1

ρ , ,




in which the third term represents the weight of the element, while Pe represents theforces from the adjacent elements. In an incremental formulation, we can replacethe loads and displacements by their differential forms. In particular,

(16.17)

in which

(16.18a)

(16.18b)

(16.18c)

(16.18d)

and

(16.19)

d dSA dX d SA dX

d

d

ee

X

X

eX

X

e e

e e e e e

e

e

e

e

P B B

K a

K K K K a

= +

=

= + + +

+ +

∫ ∫1 1

1 2 3 4

K B BT1

1

1 1

1 1

e e L T LX

X

T e

e

A D dX

D A

l

e

e

=

=−

−

+

∫

K B a B B a BT T2

1

1

21 1

1 1

e e T L e NL NL e LX

X

T e

e

e e

e

A D dX

D A

l

u u

l

e

e

= +( )

=− −

−

+

∫+[ ]

K B a a BT T3

1

2

1

1 1

1 1

e e NL e e NLX

X

T e

e

e e

e

A dX

D A

l

u u

l

e

e

=

=−

−

−

+

∫+

K B4

1

1 1

1 1

e e LX

X

e

e

A SdX

SA

l

e

e

=

=−

−

+

∫

Dl

D dX Sl

SdXTe

TX

X

e X

X

e

e

e

e

= =+ +

∫ ∫1 11 1

, .




Combining Equations 16.18 and 16.19 produces the simple relation

. (16.20)

Suppose (ue+1 − ue)/le is small compared to unity, with the consequence that Eis also small compared to unity. It follows in this case that . As aresult, Ke reduces to the stiffness matrix for a rod element of linearly elastic materialexperiencing small strain:

. (16.21)

Several special cases illuminate additional aspects of finite-element modeling.

Sample Problem 1: Single Element Built in at One End

Figure 16.2 depicts a single-element model of a rod that is built in at X = 0: Xe =X0 = 0. At the opposite end, Xe+1 = X1 = 1. The rod is submitted to the load P. Thedisplacement at X = 0 is subject to the constraint u(0) = u0 = 0, so that Equation16.20 becomes

, (16.22)

in which dPr is an incremental reaction force that is not known a priori (of course,from equilibrium, dPr = dP). For the only unknown reaction, this last equation“condenses” to

. (16.23)

FIGURE 16.2 Rubber rod element under load: built in at top.

KeT e

e

e e

e

e e

e

e

e

D A

l

u u

l

u u

l

SA

l=

−

−

= +−

+−

++ +κ κ

1 1

1 11 2 1 1

2

,

D D S ST ≈ ≈,

Kee

e

A

l=

−

−

E 1 1

1 1

κ1 1

1 1

0

1

−

−

=

−

du

dP

dP

r

kdu dP1 =

x1

u1

A,ρ

PX




For later use, the current shape function degenerates to the expression N → N =X/X1.

The (Lagrangian) strain is given by

, (16.24)

and the strain-displacement matrices reduce to

. (16.25)

The (2nd Piola-Kirchhoff) stress S is now a function of u1:

. (16.26)

16.4.4 NUMERICAL SOLUTION BY NEWTON ITERATION

Equation 16.22 can be solved by Newton iteration, sometimes called “load balanc-ing,” the reason for which is explained shortly. We introduce the “residual” functionφ(u1) = 0 as follows. The Principle of Virtual Work implies that if gravity is con-sidered in Figure 16.2,

(16.27)

when u1 is the solution. Consider an iteration process in which the jth iterate hasbeen determined. Newton iteration determines the next iterate, , using

(16.28)

Convergence of this scheme to the correct value u1 is usually rapid, providedthat the initial iterate is sufficiently close to u1. A satisfactory initial iterate is

E uu

X

u

X( )1

1

1

1

1

212

= +

B → = +BX

u

X

1

1

1

12

S u Du

X

u

X( )1

1

1

1

1

23

213

1 212

1= − +

−

−

φ ρ( )u BSA dX P g A dX

l

u

lAX D

u

X

u

X

X X

e

e e

10 0

12 1

1

1

1

1

232

1 1

1 13

1 212

1

= − −

= +

− +

−

∫ ∫−

N aT

− −

=

P gAX

uρ 112

0.

u j1

u j1

1+

κ φu u u

u u u

jj

j

j jj

1 1 1

11

1 1

( ) = − ( )= ++

∆

∆ .

u10




often obtained using an incremental procedure. Suppose that the solution has beenobtained at previous load steps. A starting iterate for the current load step is obtainedby extrapolating from the previous solutions. As an example, suppose the solutionis known at the load P = k∆P, in which ∆P is a load increment. The load is nowincremented to produce P = (k + 1)∆P, and the solution u1,k+1 at this load is determinedusing Newton iteration. It is frequently satisfactory to start the iteration using

. This numerical procedure is illustrated more extensively at the end ofthis chapter.

Sample Problem 2: Assembled Stiffness Matrix for a Two-Element Model

Assemblage procedures are used to combine the element-equilibrium relations toobtain the global equilibrium relation for an assemblage of elements. For elements‘e’ and ‘e + 1’, the element-equilibrium relations are expanded as

(16.29)

The superscript indicates the element index, and the subscript indicates the nodeindex. If no external force is applied at xe+1, the interelement force balance isexpressed as

. (16.30)

Adding Equation 16.29(ii) and Equation 16.29(iii) furnishes

. (16.31)

Equations 16.29, 16.30, and 16.31 are expressed in matrix form as

. (16.32)

Equation 16.32 illustrates that the (incremental) global stiffness matrix is formedby “overlaying” Ke and Ke+1, with the entries added in the intersection.

u uk1 10

1, + =

k du k du dP

k du k du dP

k du k du dP

k du k du dP

ee

ee e

e

ee

ee e

e

ee

ee e

e

ee

ee e

e

11 12 1

21 22 1 1

111

1 121

2 11

211

1 221

2 21

+ =

+ =

+ =

+ =

+

+ +

++

++ +

+

++

++ +

+

(i)

(ii)

(iii)

(iv).

dP dPee

ee

+ +++ =1 1

1 0

k du k k du k duee

e ee

ee21 22 11

11 12

12 0+ +( ) + =+

++

+

k k

k k k k

k k

du

du

du

dP

dP

e e

e e e e

e e

e

e

e

ee

ee

11 12

21 22 111

121

211

221

1

2 21

0

0

0+

=

+ +

+ +

+

+ ++




Sample Problem 3: Two Identical Elements under Gravity under Equal End-Loads

If Ke = Ke+1 and , overlaying the element matrices leads to theglobal (two-element) relation

. (16.33)

Note that Equation 16.33 has no solution since the global stiffness matrix hasno inverse: the second row is the negative of the sum of the first and last rows. Thissuggests that, due to numerical errors in the load increments, the condition for staticequilibrium is not satisfied numerically, and therefore that the body is predicted toaccelerate indefinitely (undergo rigid-body motion). However, we also know that theconfiguration under incremental loads is symmetric, implying a constraint du2 = 0.This constraint permits “condensation,” that is, reducing Equation 16.33 to a systemwith two unknowns by eliminating rows and columns associated with the middleincremental displacement:

. (16.34)

The condensed matrix is now proportional to the identity matrix, and the systemhas a solution. In general, stiffness matrices can easily be singular or nearly singular

FIGURE 16.3 Two rubber rods under load.

xe,ue

ρe+1

Pe+1xe+1,ue+1

xe+2,ue+2

xe+2

e+1

Pe

P,Ae

P,Ae+1

Pee

e+1

dP dP dPee

ee= − = −++11

κ

1 1 0

1 2 1

0 1 1

0

1

2

3

−

− −

−

=

−

du

du

du

dP

dP

κ1 1

1 1

1

3

−

−

=

−

du

du

dP

dP




(with a large condition number) unless constraints are used to suppress “rigid-bodymodes.”

16.5 ILLUSTRATION OF NEWTON ITERATION

Unfortunately, since elastomers or metals experiencing plasticity are typically quitecompliant, it is common to encounter convergence problems, for which there arefour major approaches:

1. Increasing stiffness, such as by introducing additional constraints if available 2. Reducing load-step sizes and reforming the stiffness matrix after each iteration3. Switching to displacement control rather than load control4. Using an arc-length method (described in Chapter 3)

For an equation of the form

, (16.35)

Newton iteration seeks a solution through an iterative process given by

. (16.36)

Clearly, using two sequential iterates,

(16.37)

for some x* between xj and xj−1.

ψ ( )x = 0

x xd x

dxxj j

jj+

−

= −

1

1ψψ

( )( )

x x x xd x

dxx

d x

dxx

x xd x

dxd x

dxx x

xd x

dxx

j j j jj

jj

j

j j 2 j

+ −

−−

−

−

−

−

− = − −

−

= − − + −

× +

1 1

11

1

1

1

1

ψψ

ψψ

ψ ψ

ψ ψ

( )( )

( )( )

( *) ( *)( *)

( *)( *)

(

2

jj j

j

j j j

x*d x

dxd x

dxx x*

xd x

dxx x*

d xdx

d xdx

x x O x

−

− + −

× + −

+

= −

− + −

−

−

−

−

−

)( *) ( *)

( )

( *)( *)

( )

( *) ( *)( ) ([

ψ ψ

ψ ψ

ψ ψ

2

2 1

1

1

1

11

L

xxj−12] )




16.5.1 EXAMPLE

The performance of Newton iteration procedure is illustrated using a simple exampleshowing convergence problems. Consider

. (16.38)

The goal is to find the solution of x as y is incremented in the range (0,1). Clearly,x approaches infinity as y approaches unity, so that the goal is to generate the x(y)relationship accurately as close as possible to y = 1. The curve appears as shown inFigure 16.4. As y is incremented by small amounts just below unity, the differencesin x due to the increment are large, so that the solution value is far from the initialiterate.

Suppose that y is incremented such that the nth value of y is

. (16.39)

To obtain the solution of the (n + 1)st step, the Newton iteration proceduregenerates the (ν + 1)st iterate from the νth iterate, as follows:

, (16.40)

in which is the ν th iterate for the solution xn+1. Of course, a starting iterate, , is needed. An attractive candidate is xn+1.

However, this may not be good enough when convergence difficulties appear.Newton iteration was implemented for this example in a simple problem using

increments of 0.001. A stopping criterion of 10 iterations was used. Convergencewas found to fail near y = 0.999.

FIGURE 16.4 Illustration of the inverse tangent function.

Y

1

X

encounter difficulties with convergence

ϕπ

( ) tan ( )x x y= − =−201

y n yn = ∆

x xd x

dxxn n

nn+

++

+

−

+= −( )

( )11

11

1

1ν ν

ννφ

φ

xn+1ν

xn+10




16.6 EXERCISES

1. Doing the algebra, make judicious choices of starting iterates, and developthe first few iterates for the roots of the equations

2. Write a simple program implementing Newton iteration, and apply it tothe examples in Exercise 1.

3. Using Example 5.1, write a simple program using Newton iteration tocompute X(Y ) using increments ∆y = 0.001. Set the stopping criterion asthe first of: (a) 10 iterates or (b) an absolute difference of less than 0.001in the successive values of x. Note the value of x at which convergencefails, and compare the values of x with accurate values available from theformulae and tables in Abramawitz and Stegun (1997).

x x

x

x

2

2

2

3 2 0

1 0

1 0

+ + =

− =

− =

( )



215

Incremental Principleof Virtual Work

17.1 INCREMENTAL KINEMATICS

Recall that the displacement vector

u

(

X

) is assumed to admit a satisfactory approx-imation at the element level in the form

u

(

X

)

=

ϕϕϕϕ

T

(

X

)ΦΦΦΦ γγγγ

(

t

). Also recall that thedeformation-gradient tensor is given by

F

=

Suppose that the body under studyis subjected to a load vector,

P

, which is applied incrementally via load increments,

∆

j

P

. The load at the

n

th

load step is denoted as

P

n

. The solution,

P

n

, is known, andthe solution of the increments of the displacements is sought. Let

∆

n

u

=

u

n

+

1

−

u

n

,so that

∆

n

u

=

ϕϕϕϕ

T

(

X

)ΦΦΦΦ

∆

nγγγγ

. By suitably arranging the derivatives of

∆

n

u

with respectto

X

, a matrix,

M

(

X

), can easily be determined for which

VEC

(

∆

n

F

)

=

M

(

X

)

∆

nγγγγ

.We next consider the Lagrangian strain,

E

(

X

)

=

(

F

T

F

−

I

). Using KroneckerProduct algebra from Chapter 2, we readily find that, to first order in increments,

(17.1)

This form shows the advantages of Kronecker Product notation. Namely, itenables moving the incremental displacement vector to the end of the expressionoutside of domain integrals, which we will encounter subsequently.

Alternatively, for the current configuration, a suitable strain measure is theEulerian strain,

«

=

(

I

−

F

−

T

F

−−−−

1

), which refers to deformed coordinates. Note thatsince

∆

n

(

FF

−−−−

1

)

=

0

,

∆

n

F

−−−−

1

=

−

F

−−−−

1

∆

n

FF

−−−−

1

. Similarly,

∆

n

F

−−−−

T

=

−

F

−

T

∆

n

F

T

F

−

T

. Simplemanipulation furnishes that

(17.2)

There also are geometric changes for which an incremental representation isuseful. For example, since the Jacobian

J

=

det(

F

) satisfies

dJ

=

Jtr

(

F

−

1

d

F

), we obtain

17

∂∂

xx .

12

∆ ∆

∆ ∆

∆

∆ ∆

n n

n nT

n

n n

VEC

VEC

VEC

e

G G

T

T T

T T T

=

= +

= ⊗ + ⊗

= = ⊗ + ⊗

( )

[ ]

[ ] ( )

, [ ] ( )

E

M

12

12

12

F F F F

I F F IU F

g I F F IU X g .

12

VEC nT T

n( ) [ ] .∆ ∆= ⊗ + ⊗− − − − − −12

1 1F F F U F F F M gT 1«



216


the approximate formula

(17.3)

Also of interest are

(17.4)

Using Equation 17.4, we obtain the incremental forms

(17.5)

17.2 INCREMENTAL STRESSES

For the purposes of deriving an incremental variational principle, we shall see thatthe incremental 1

st

Piola-Kirchhoff stress, , is the starting point. However, toformulate mechanical properties, the objective increment of the Cauchy stress, is the starting point. Furthermore, in the resulting variational statement, which wecalled the

Incremental Principle of Virtual Work

, we find that the quantity thatappears is the increment of the 2

nd

Piola-Kirchhoff stress,

∆

n

S

.From Chapter 5, we learned that

, from which, to first order,

(17.6)

For the Cauchy stress, the increment must take into account the rotation of theunderlying coordinate system and thereby be objective. We recall the objectiveTruesdell stress flux, , introduced in Chapter 5:

(17.7)

∆ ∆

∆

∆

J Jtr

JVEC VEC

VEC J

n

n

Tn

T

=

=

=

−

−

−

( )

( ) ( )

( )

F F

F F

F

1

1T

T M γγ .

d dSdt

tr dS

ddt

ddt

dS tr dS

T

T T

T

[ ][ ( ) ]

[( ) ]

[ ( ) ]

nD n

nn Dn n

D

= −

= −

= −

I L

I L

n Dn

∆∆ ∆∆ γγ

γγ

γγ

nT T T T

n

nT T T T T

n

nT T T T T

n

dS dS VEC

dS dS VEC

[ ] [ ( ) ]

[ ( ) ]

[ ( ) ( ) ] .

n n F n F U

n n n F n n F U

n F n n F n

= − ⊗

= ⊗ − ⊗

= − ⊗

− −

− −

− −

M

M

M

∆ ∆

∆ ∆

∆nS∆o n T,

S S= FT

∆ ∆ ∆n nT

nTS S S= +F F .

∂ ∂To

/ t

∂ ∂ = ∂ ∂ + − −T T T To

/ / ( ) .t t tr TD L LT




217

Among the possible stress fluxes, it is unique in that it is proportional to therate of the 2

nd

Piola-Kirchhoff stress, namely

(17.8)

An objective Truesdell stress increment is readily obtained as

(17.9)

Furthermore, once has been determined, the (nonobjective) incre-ment of the Cauchy stress can be computed using

(17.10)

from which

(17.11)

17.3 INCREMENTAL EQUILIBRIUM EQUATION

We now express the incremental equation of nonlinear solid mechanics (assumingthat there is no net rigid-body motion). In the deformed (Eulerian) configuration,equilibrium at

t

n

requires

(17.12)

Referred to the undeformed (Lagrangian) configuration, this equation becomes

(17.13)

in which, as indicated before, is the 1

st


S

denotes the surface(boundary) in the deformed configuration, and

n

0

is the surface normal vector in theundeformed configuration. Suppose the solution for is known as at time

t

n

andis sought at

t

n

+

1

. We introduce the increment to denote A similardefinition is introduced for the increment of the displacements. Now, equilibriumapplied to and implies

(17.14)

∂ ∂ ∂ ∂S T/ / .t J t T= − −F F1o

VEC VECn

TnJ( ) ( ).∆ ∆

oT S= ⊗1 F F

VEC( )∆o

n T

∆ ∆ ∆ ∆ ∆o

n n n nT

nTtrT T T T T= + − −− − −( ) ,FF FF F F1 1

VEC VEC VECn nT T T T

n( ) ( ) [ ( ) ( ) ( )] .∆ ∆ ∆T T T T T M= + − ⊗ − ⊗− − −oF F I I F γγ

TT dS dVn u∫ ∫= ρ˙ .

S T dS dVn u0 0 0 0∫ ∫= ρ ˙ ,

S

S Sn

∆nS S Sn n+ −1 .

Sn+1 Sn

∆ ∆nT

ndS dVS n u0 0 0 0∫ ∫= ρ ˙ .



218


Application of the divergence theorem furnishes the differential equation

(17.15)

17.4 INCREMENTAL PRINCIPLE OF VIRTUAL WORK

To derive a variational principle for the current formulation, the quantity to be varied isthe incremental displacement vector since it is now the unknown. Following Chapter 5,

Equation 17.15 is multiplied by (

δ

∆

n

u

)

T

.Integration is performed over the domain.The Gauss divergence theorem is invoked once.Terms appearing on the boundary are identified as primary and secondary

variables.Boundary conditions and constraints are applied.

The reasoning process is similar to that in the derivation of the Principle ofVirtual Work in finite deformation in which

u

is the unknown, and furnishes

(17.16)

in which ττττ

0

is the traction experienced by

dS

0

. The fourth term describes the virtualexternal work of the incremental tractions. The first term describes the virtual internalwork of the incremental stresses. The third term describes the virtual internal workof the incremental inertial forces. The second term has no counterpart in the previ-ously formulated Principle of Virtual Work in Chapter 5, and arises because ofgeometric nonlinearity. We simply call it the geometric stiffness integral. Due to theimportance of this relation, Equation 17.16 is derived in detail in the equations thatfollow. It is convenient to perform the derivation using tensor-indicial notation:

(17.17)

The first term on the right is converted using the divergence theorem to

(17.18)

which is recognized as the fourth term in Equation 17.16.

∇T∆ ∆nT

nTS = ρ0

˙ .u

tr dV dV dV dSn n n n nT

n nT

n( ) ˙ ,δ δ δ ρ δ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E S ST TF F u u u0 0 0 0 0 0∫ ∫ ∫ ∫+ + = ττ

δ δ δ

δ ρ

∆ ∆ ∆ ∆ ∆ ∆

∆ ∆

n ij

n ijj

n i n ijj

n i n ij

n i n i

uX

S dVX

u S dVX

u S dV

u u dV

∂∂

∂∂

∂∂

( ) [ ( )] [ ]

˙ .

0 0 0

0 0

∫ ∫ ∫

∫

= −

=

∂∂X

u S dV u n S dS

u dS

n i n ij n i j n ij

n i j

j

[ ( )] ( )δ δ

δ τ

∆ ∆ ∆ ∆

∆

0 0

0 0

∫ ∫

∫

=

=




219

To first-order in increments, the second term on the right is written, using tensornotation, as

(17.19)

The second term is recognized as the second term in Equation 17.16.The first term now becomes

(17.20)

which is recognized as the first term in Equation 17.16.

17.5 INCREMENTAL FINITE-ELEMENT EQUATION

For present purposes, let us suppose constitutive relations in the form

(17.21)

in which D(X, γ) is the fourth-order tangent modulus tensor. It is rewritten as

(17.22)

Also for present purposes, we assume that ∆ττττ0 is prescribed on the boundary S0, acommon but frequently unrealistic assumption that is addressed in a subsequent section.

In VEC notation, and using the interpolation models, Equation 17.16 becomes

(17.23)

∂∂X

u S dV tr dV

tr dV

tr dV tr dV

jn i n ij n n

n nT

nT

Tn n n n

T

[ ] ( )

( [ ])

( ) ( )

δ δ

δ

δ δ

∆ ∆ ∆ ∆

∆ ∆ ∆

∆ ∆ ∆ ∆

0 0

0

0 0

∫ ∫

∫∫ ∫

=

= +

= +

F

F F F

F F F F

S

S S

S S

tr dV tr dV

tr dV

Tn n

Tn n

Tn

nT

n

( ) [ ]

( )

F F F F F Fδ δ δ

δ

∆ ∆ ∆ ∆ ∆

∆ ∆

S S

S

0 0

0

12∫ ∫

∫

= +

= Ε

∆ ∆n n nS D E= ( , ) ,X γ

∆ ∆n n ns X= χχ( , )γ e

s S e D= = =VEC VEC TEN22( ), ( ), ( ). E χ

δ∆ ∆ ∆ ∆n T G n n nγγ γγ γγT K K M f[( ) ˙ ]+ + − = 0

K M M K M S M

M

TT T

GT

T Tn

T

dV dV

dV dS

= = ⊗

= =

∫ ∫∫ ∫

G Gχχ 0 0

0 0 0 0 0

I

fρ ρΦΦ ϕϕϕϕ ΦΦ ΦΦ ϕϕ ττ∆ ∆




KT is now called the tangent modulus matrix, KG is the geometric stiffness matrix,M is the (incremental) mass matrix, and ∆nf is the incremental force vector.

17.6 INCREMENTAL CONTRIBUTIONS FROM NONLINEAR BOUNDARY CONDITIONS

Again, let Ii denote the principal invariants of C, and let i = VEC(I), c2 = VEC(C2),, and Ai = ∂ni /∂c. Recall from Chapter 2 that

(17.24)

Equation 17.23 is complete if increments of tractions are prescribed on theundeformed surface S0. We now consider the more complex situation in which ττττ isreferred to the deformed surface S, on which they are prescribed functions of u.From Chandrasekharaiah and Debnath (1994), conversion is obtained using

(17.25)

and from Nicholson and Lin (1997b)

(17.26)

Suppose that ∆ττττ is expressed on S as follows:

(17.27)

Here, is prescribed, while AM is a known function of u. Also, S0 is theundeformed counterpart of S. These relations are capable of modeling boundaryconditions, such as support by a nonlinear elastic foundation.

From the fact that ττττ dS = ττττ0 dS0 = µττττ0 dS, we conclude that ττττ = µττττ0. It follows that

(17.28)

n cTi iI= ∂ ∂/

n n c

A A I

1 2 1 3 2 1 2 9

1 2 9 3 1 9

= = − = − + = ⊗

= = − = ⊗ + ⊗ − + + −

i i c c

ii

I I I

IT T T T

n i I I I

0 A I C C I ic ci ii I( ) ( ).

ττ ττdS dS dS dS

dS dS J

T T

T T T

= =

= = = ⊗−

0 0 0 0 0

0 0 0 0 0 3

n q n q

n C n n n n1µ µ

∆ ∆µ µ µ≈ = ≈ = ⊗d dm c m c m n n AT T T T T, / . 0 0 3 2

∆ ∆ ∆ττ ττ= − A uMT .

dττ

∆ ∆ ∆

∆ ∆ ∆

ττ ττ ττ

ττ ττ

0 2

2

1

1

= −

= − −

µ µµ

µ µ( )A mM

T Tu c



Incremental Principle of Virtual Work 221

From the Incremental Principle of Virtual Work, the rhs term is written as

(17.29)

Recalling the interpolation models for the increments, we obtain an incrementalforce vector plus two boundary contribution to the stiffness terms. In particular,

(17.30)

The first boundary contribution is from the nonlinear elastic foundation couplingthe traction and displacement increments on the boundary. The second arises fromgeometric nonlinearity when the traction increment is prescribed on the currentconfiguration.

17.7 EFFECT OF VARIABLE CONTACT

In many, if not most, “real-world” problems, loads are transmitted to the member ofinterest via contact with other members, for example, gear teeth. The extent of thecontact zone is an unknown to be determined as part of the solution process. Solutionof contact problems, introduced in Chapter 15, is a difficult problem that has absorbedthe attention of many investigators. Some algorithms are suited primarily for linearkinematics. Here, a development is given for one particular formulation, which ismostly of interest for explicitly addressing the effect of large deformation.

Figure 17.1 shows a contactor moving into contact with a foundation that isassumed to be rigid. We seek to follow the development of the contact area and thetractions arising throughout it. From Chapter 15, we recall that corresponding to apoint x on the contactor surface there is a target point y(x) on the foundation towhich the normal n(x) at x points. As the contactor starts to deform, n(x) rotatesand points toward a new value, y(x). As the point x approaches contact, the pointy(x) approaches the foundation point, which comes into contact with the contactorpoint at x.

We define a gap function, g, using y(x) = x + gn. Let m be the surface normal-vector to the target at y(x). Let Sc be the candidate contact surface on the contactor,whose undeformed counterpart is S0c. There also is a candidate contact surface Sf

on the foundation. We limit our attention to bonded contact, in which particles coming into contact

with each other remain in contact. Algorithms for sliding contact with and withoutfriction are available. For simplicity’s sake, we also assume that shear tractions, in

δ δµ µ

∆ ∆ ∆ ∆ ∆ ∆u u A u cT TMT TdS dSττ ττ ττ

0 0 2 0

1∫ ∫= − −

( ) .m

δ δ δ∆ ∆ ∆ ∆ ∆ ∆uT T TBF BNdSττ γγ γγ γγ0 0∫ = − +f K K[ ]

∆ ∆f K NA N K m= = =∫ ∫ ∫10 0 0µ µ

ττ ττdS dS dSBF M

T TBN 2

T, , G




the osculating plane of point of interest, are negligible. Suppose that the interfacecan be represented by an elastic foundation satisfying the incremental relation

(17.31)

Here, τn = nTτ and un = nT u are the normal components of the traction anddisplacement vectors. Since the only traction to consider is the normal traction (tothe contactor surface), the transverse components of ∆u are not needed (do not resultfrom work). Also, k(g) is a nonlinear stiffness function given in terms of the gap by,for example,

(17.32)

As in Chapter 15, when g is positive, the gap is open and k approaches kL, whichshould 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 isimportant, it follows that ττττ = tnn, from which

(17.33)

FIGURE 17.1 Contact.

foundation

y(x)

contactor

gn

m

∆ ∆τn nk g u= − ( ) .

k gk

g k k kHk r L H L( ) ( ) , / .= − −

+ >>π

π α ε2

1arctan

∆ ∆ ∆ττ = +τ τn nn n.




The contact model contributes the matrix Kc to the stiffness matrix as follows(see Nicholson and Lin, 1997b):

(17.34)

To update the gap, use the following relations proved in Nicholson and Lin (1997-b).The differential vector, dy, is tangent to the foundation surface, hence, mTdy = 0. Itfollows that

(17.35)

Using Equation 17.5, we may derive, with some effort, that

(17.36)

17.8 INTERPRETATION AS NEWTON ITERATION

The (nonincremental) Principle of Virtual Work can be restated in the undeformedconfiguration as

(17.37)

We assume for convenience that ττττ is prescribed on So. The interpolation modelsatisfies the form

(17.38)

δ µ δ τ µ

δ

∆ ∆ ∆ ∆

∆ ∆

u

K

T ττ

γγ γγ

dS u dSnT

n

Tc

0 0∫ ∫=

= −

K Nn m Nnn N N hc nT T

c cT T

c nT

cdS k g dS dS= − + +∫ ∫ ∫2 0 0 0τ µ τ µββ ( ) .

0 = + +

= − +

m u m n m n

m u m nm n

T T T

T T

T

d g d dg

gg∆ ∆ ∆

.

∆ ∆gg= = − + = ⊗ − ⊗− −ΓΓ γγ ΓΓT T

T T

TT T T T T T TN h

nh n n F n n F M, , [ [( ) ] ] .

mm

tr dV dV dSoT

oT

o( ) ˙ .δ δ ρ δES u u u∫ ∫ ∫+ = ττ

( )

δ δe B B

B I I I IU X

B I F F IU X

FuX

= +[ ]= ⊗ + ⊗

= ⊗ + ⊗( )

= ∂∂

LT

NLT

LT

NLT

uT

u

u

γγ γγ

12

12

( ) ( )

( )

.

M

M




Clearly, Fu and BNL are linear in γγγγ.Upon cancellation of the variation d γγγγ Τ, an algebraic equation is obtained as

(17.39)

At the load step, Newton iteration is expressed as

(17.40)

or as a linear system

(17.41)

If the load increments are small enough, the starting iterate can be estimated as thesolution from the nth load step. Also, a stopping (convergence) criterion is neededto determine when the effort to generate additional iterates is not rewarded byincreased accuracy.

Careful examination of the relations from this and the incremental formulationsuncovers that

(17.42)

so that the incremental stiffness matrix is the same as the Jacobian matrix in Newtoniteration. This, of course, is a satisfying result. The Jacobian matrix can be calculatedby conventional finite-element procedures at the element level followed by conven-tional assembly procedures. If the incremental equation is only solved once at eachload increment, the solution can be viewed as the first iterate in a Newton iterationscheme. The one-time incremental solution can potentially be improved by additionaliterations, 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 plateswere addressed in Chapter 14. In the classical equations, geometrically nonlinearterms appear through a linear correction term, thereby furnishing linear equations.Here, in the absence of inertia and nonlinearity in the boundary conditions, we brieflypresent a general viewpoint based on the incremental equilibrium equation

(17.43)

ΦΦ γγ γγ ττ( , ) [( ( ] ˙ , .f B B ) s u f N= + + =∫ ∫ ∫L NL oT

oT

o odV dV dSN ρ

γγ γγ ΦΦ γγ ΦΦ γγnv )

n)

n)

n n)

n,++

+−

+ + + +

−

= − ( ) = ∂∂ ( )

1

11

11 1 1 1

1

( ( ( (, ,ν ν ν

γJ f J f

J fγγ γγ ΦΦ γγ

γγ γγ γγ γγ

nv )

nv

nv

n

nv )

nv

n(v )

nv

++

+ + +

++

+ ++

+

−( ) = ( )= + −[ ]

11

1 1 1

11

1 11

1

( ( ) ( )

( ( ) ( )

, ,

.

J K K= +T G ,

( ) .K K fT G n n+ =+ +∆ ∆γγ 1 1




This solution predicts a large incremental displacement if the stiffness matrixKT + KG is ill-conditioned or outright singular. Of course, in elastic media, KT ispositive-definite. However, in the presence of in-plane compression, KG may havea negative eigenvalue whose magnitude is comparable to the smallest positive eigen-value of KT . To see this recall that

(17.44)

We suppose that the element in question is thin in a local z (out-of-planedirection). This suggests the assumption of plane stress. Now, in plate-and-shelltheory, it is necessary to add a transverse shear stress on the element boundaries toallow the element to support transverse loads. We assume that the transverse shearstresses only appear in the incremental force term and the tangent stiffness term,and that the geometric stiffness term strictly satisfies the plane-stress assumption. Itfollows that if the z-direction is out of the plane, in the geometric stiffness term,

(17.45)

In classical buckling, it is assumed that loads applied proportionately induceproportionate 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-planestresses, arises in the eigenvalue problem for the critical buckling load. In nonlinearproblems, 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 proportionalloading, the stresses become numerical functions of l.

As a simple alternative to the general case, we consider buckling of a singleelement and suppose that the stresses appearing in Equation 17.46 are applied in acompressive sense along the faces of the element in a proportional manner whereby

(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 withcompressive stresses. At the element level, the equation now becomes

(17.47)

K G G K ITT T

GTdV dV= = ⊗∫ ∫M M M Mχ 0 0 S .

S I

I I I

I I I

I I I

⊗ →

S S

S S

11 12

12 22

0

0

0 0 0

.

S I⊗ →

− −

− −

λ

( ˆ ) ( ˆ )

( ˆ ) ( ˆ ) ,

S S

S S

11 12

12 22

0

0

0 0 0

I I I

I I I

I I I

( ˆ ) .K K fT G n n− =+ +λ ∆ ∆γγ 1 1




At a given load increment, the critical buckling load for the current path, as afunction of two angles determining the path in the stress space illustrated inChapter 14, is obtained by computing the l value rendering singular.

17.10 EXERCISES

1. Assuming linear interpolation models for u,v in a plane triangularmembrane 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 atetrahedral element with vertices (0,0,0),(1,0,0),(0,1,0),(0,0,1).

( ˆ )K KT G− λ



227

Tangent-Modulus Tensors for Thermomechanical Response of Elastomers

18.1 INTRODUCTION

Within an element, the finite-element method makes use of interpolation models forthe displacement vector

u

(

X

,

t

) and temperature T(

X

,

t

) (and pressure

p

=

−

trace

(ττττ

)/3in incompressible or near-incompressible materials):

(18.1)

in which T

0

is the temperature in the reference configuration, assumed constant.Here,

N

,

ν

, and ξξξξ

are shape functions and γγγγ

, θθθθ

, and ψψψψ

are vectors of nodal values.Application of the strain-displacement relations and their thermal analogs furnishes

(18.2)

in which

U

is a 9

×

9 universal permutation tensor such that

VEC

(

A

T

)

=

U

VEC

(

A

),and

e

=

VEC

(

E

) is the Lagrangian strain vector. Also,

∇

is the gradient operatorreferred to the deformed configuration. The matrix ββββ

and the vector ββββ

T

are typicallyexpressed in terms of isoparametric coordinates.

18.2 COMPRESSIBLE ELASTOMERS

The Helmholtz potential was introduced in Chapter 7 and shown to underlie therelations of classical coupled thermoelasticity. The thermohyperelastic properties ofcompressible 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 isconventional to introduce the strain energy density

w

(

E

)

=

ρ

0

φ

(T,

E

) (T constant),in which

ρ

0

is the density in the undeformed configuration. Typically, the elastomeris assumed to be isotropic, in which case

φ

can be expressed as a function of

T

,

I

1

,

I

2

, and

I

3

. Alternatively, it may be expressed as a function of T and the stretch ratios

λ

1

,

λ

2

, and

λ

3

.

18

u N( , ) ( ) ( ), ( , ) ( ) ( ), ( ) ( ),X X X X Xt t t t p t= − = =T T Tγ ν θ ξ ψ T T 0

f F I U f F I e

F I I F U

T T T

T T T T

1 1 2 2 2

2

12

= − = = = − = =

= = ⊗ + ⊗ ∇ =

VEC VEC

T

( ) , ( ) ,

, ( ),

M g M g M g, b g

b M b q

δ δ

G G T



228


With

φ

known as a function of T,

I

1

,

I

2

, and

I

3

, the entropy density

η

per unitmass and the specific heat

c

e

at constant strain are obtained as

(18.3)

The 2

nd


s

=

VEC

(

S

), is obtained from

(18.4)

Also of importance is the (isothermal) tangent-modulus matrix

(18.5)

An expression for

D

T

has been derived by Nicholson and Lin (1997c) forcompressible, incompressible, and near-incompressible elastomers described bystrain-energy functions (Helmholtz free-energy functions) and based on the use ofstretch 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 ofincompressibility or near-incompressibility. The constraint is accommodated byaugmenting

φ

with terms involving a new parameter similar to a Lagrange multiplier.Typically, this new parameter is related to the pressure

p

. The thermohyperelasticproperties of incompressible and near-incompressible elastomers can be derivedfrom the augmented Helmholtz free energy, which is a function of

E

,

T

, and

p

. Theconstraint introduces additional terms into the governing finite-element equationsand requires an interpolation model for

p

.If the elastomer is incompressible at a constant temperature, the augmented

Helmholtz function,

φ

, can be written as

(18.6)

where

ξ

is a material function satisfying the constraint

ξ

(

J

,

T

)

=

0 and It is easily shown that

φ

d

depends on the deviatoric Lagrangian strain

E

d

, due to theintroduction of the deviatoric invariants

J

2

and

J

3

. The Lagrange multiplier

λ

is, infact, the (true) pressure

p

:

(18.7)

ceE

= ∂∂

= − ∂∂

TT

T

η η φ.

s nT

Ti i

i

iiI

= ∂∂

= = ∂∂∑ρ φ ρ φ φ φ

0 02e

, .

Dse

AT ij i j

ji

i i

i

iji jI I

= ∂∂

= + = ∂∂ ∂∑∑ ∑

T

4 40 0

2

ρ φ ρ φ φ φn nT , .

φ φ= − = =d J J J J I I J I I( , , ( , / , / ,/ /1 2 0 1 1 3

1 32 2 3

2 3T) T)/ , λξ ρ

J I= =31 2 det( ).F

p traceJ

= − = ∂∂

( )/ .T 3ξ

T



Tangent-Modulus Tensors for Thermomechanical Response of Elastomers

229

For an elastomer that is near-incompressible at a constant temperature,

φ

can bewritten as

(18.8)

in which

κ

0

is a constant. The near-incompressibility constraint is expressed by

∂φ

/

∂

p

=

0, which implies

(18.9)

The bulk modulus

κ

is given by

(18.10)

Chen et al. (1997) presented sufficient conditions under which near-incompress-ible models reduce to the incompressible case as

κ

→

∞

. Nicholson and Lin (1996)formulated the relations

(18.11)

with the consequence that

(18.12)

Equation 18.12 provides a linear pressure-volume relation in which thermome-chanical effects are confined to thermal expansion expressed using a constant-volumecoefficient

α

. If the constraint is assumed to be satisfied

a priori

, the Helmholtz freeenergy is recovered as

(18.13)

Alternatively, the latter term results from retaining the lowest nonvanishing termin a Taylor-series representation of

φ

about

f

3

(T)

J

−

1. Given Equation 18.13, the entropy now includes a term involving

p

:

(18.14)

The stress and the tangent-modulus matrices are correspondingly modified:

(18.15)

ρ ρ ξ κ0 0 1 22 2φ φ= − −d J J p J p( , , ) ( , ) / ,T T

p J= −κξ( , ).T

κ κ ξ= − ∂∂

= ∂∂

pJ JT T

0 .

ξ( , ( ) , ( , ) ( , ) ( ( )J f J J J c lnd eT) T T) (T T T/T ),= − = + = −31 1 2 2 2 01 1φ φ φ φ

p f J f= − − =κ κ κ03 3

01( ( ), ) .T) (T

φ φ( , , ) ( , , ( ) ) / .I I I J J fd1 2 3 1 2 03 2

01 2= + −T) (Tκ ρ

η πα ρ π= −∂∂

+ =φd f p fT

(T (T40

3)/ , / ).

s n

A n n

T d Tf J

f J J

=∂∂

−

= ∂∂

= ∂∂

∂∂

− −[ ]

ρ φ π

ρ φ π

π

π

0T,

33

TPT,

0

Td 3

3 3 3T

(T

T) 2

e

se e e

) /

( / /D 3



230


18.3.1 S

PECIFIC

E

XPRESSIONS

FOR

THE

H

ELMHOLTZ

P

OTENTIAL

There are two broad approaches to the formulation of Helmholtz potential:

To express

φ

as a function of

I

1

, 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 directuse of test data, for example, from uniaxial tension. We will now examine severalcases.

18.3.1.1 Invariant-Based Incompressible Models: Isothermal Problems

The strain-energy function depends only on I1, I2, and incompressibility is expressedby the constraint I3 = 1, assumed to be satisfied a priori. In this category, the mostwidely used models include the Neo-Hookean material:

(18.16)

and the (two-term) Mooney-Rivlin material:

(18.17)

in which C1 and C2 are material constants. Most finite-element codes with hyper-elastic elements support the Mooney-Rivlin model. In principle, Mooney-Rivlincoefficients C1 and C2 can be determined independently by “fitting” suitable load-deflection curves, for example, uniaxial tension. Values for several different rubbercompounds 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 config-uration. The two models are:

(18.18)

Let w denote the Helmholtz free energy evaluated at a constant temperature, in whichcase it is the strain energy. We note a general expression for w which is implemented

φ = − =C I I1 1 33 1( ),

φ = − + − =C I C I I1 1 2 2 33 3 1( ) ( ), ,

ρ φν

νν

ν

ρ φ

νν

0 1 0 10

03 3

1 2 0

0

0 2 02

33

12

1 2 1

12

2 5

0

0= +−

− −+

= + −

−G I I I

GI

II



Tangent-Modulus Tensors for Thermomechanical Response of Elastomers 231

in several commercial finite-element codes (e.g., ANSYS, 2000):

(18.19)

in which Eth is called the thermal expansion strain, while Cij and Dk are materialconstants. Several codes also provide software for estimating the model coefficientsfrom user-supplied data.

Several authors have attempted to uncouple the response into isochoric (incom-pressible) and volumetric parts even in the compressible range, giving rise to func-tions of the form φ = φ1(J1, J2) + φ2(J). A number of proposed forms for φ2 arediscussed in Holzappel (1996).

18.3.1.3 Thermomechanical Behavior under Nonisothermal Conditions

Now we come to the accommodation of coupled thermomechanical effects. Simpleextensions of, for example, the Mooney-Rivlin material have been proposed byDillon (1962), Nicholson and Nelson (1990), and Nicholson (1995) for compressibleelastomers, and in Nicholson and Lin (1996) for incompressible and near-incom-pressible elastomers. From the latter,

(18.20)

in which π = p/f 3(T ). As previously mentioned, a model similar to Nicholson andLin (1996) has been proposed by Holzappel and Simo (1996) for compressibleelastomers described using stretch ratios.

18.4 STRETCH RATIO-BASED MODELS: ISOTHERMAL CONDITIONS

For compressible elastomers, Valanis and Landel (1967) proposed a strain-energyfunction based on the decomposition

(18.21)

Ogden (1986) has proposed the form

(18.22)

w J J J C J J J D J J Eiji j

ji

rk

k

k

r th( , , ) ( ) ( ) ( ) / , /( ),1 2 1 23 3 1 1= − − + − = +∑∑ ∑

ρ ρ κ π π κ0 1 1 2 2 03 22 3 1 1 2φ = − + − + − − − −C J C J c ln f Je o( ) ( ) ( ( / ) ( ( ) ) / ,T T T T

φ φ φ φ( , , , ( , ( , ( , , .λ λ λ λ λ λ1 2 3 1 2 3T) T) T) T) T = + + fixed

ρ λ µ λα0

1

1φ( , ( ), .T) T = −∑ pp

N

fixed




In principle, in incompressible isotropic elastomers, stretch ratio-based modelshave 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 thermohyper-elastic materials under isothermal conditions and accommodate thermal effects.From Nicholson and Lin (1997c), we invoke the expression for the differential of atensor-valued isotropic function of a tensor. Namely, let A denote a nonsingularn × n tensor with distinct eigenvalues, and let F(A) be a tensor-valued isotropicfunction of A, admitting representation as a convergent polynomial:

(18.23)

Here, φj are constants. A compact expression for the differential dF(A) is pre-sented using Kronecker Product notation.

The reader is referred to Nicholson and Lin (1997c) for the derivation of thefollowing expression. With f = VEC(F) and a = VEC(A),

(18.24)

Also, dωωωω = VEC(dΩΩΩΩ), in which dΩΩΩΩ is an antisymmetric tensor representing therate of rotation of the principal directions. The critical step is to determine a matrix Jsuch that Wdωωωω = −Jda. It is shown in Dahlquist and Bjork that J = −[AT A]−1W, inwhich [AT A]I is the Morse-Penrose inverse [(Dahlquist and Bjork(1974)]. Thus,

(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-energyfunction, w, was proposed, which, for compressible elastomers and isothermalresponse, is equivalent to the form

(18.26)

in which ci are the eigenvalues of C, and ξi, ζi are material properties. The tangent-modulus tensor χχχχ appearing in Chapter 17 for the incremental form of the Principle

F A A( ) .=∝

∑φ jj

0

d d d

jdd

ITENddj

j

f a F F a

F AF F AF A F F A

T

T T T

( )

( )

( / ) ( / ) ( )

= ′ ⊕ ′ +

′ = =

= − − ′ − ′ + ⊗ ′ − ′ ⊗

−∝

∑

12

22

2 212

1

0

W

ωω

F A AFA

fa

φ

W

d df a F F A A W/ / [ ] .= ′ ⊕ ′ − ′T T2

w tr i

i

i= −

∑ξ ζ[ ] ,C I




of Virtual Work is obtained as

(18.27)

(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 expan-sion, that volume changes are small, and that volume changes depend linearly ontemperature. Thus, materials of present interest can be described as mechanicallynonlinear but thermally linear.

Due to the role of thermal expansion, it is desirable to uncouple dilatational anddeviatoric effects as much as possible. To this end, we introduce the deviatoric Cauchy-Green strain in which I3 is the third principal invariant of C. Now, mod-ifying w and expanding it in J − 1, and retaining lowest-order terms gives

(18.29)

in which κ is the bulk modulus. The expression for χχχχ in Equation 18.27 is affectedby these modifications.

To accommodate thermal effects, it is necessary to recognize that w is simplythe Helmholtz free-energy density ρoφ under isothermal conditions, in which ρo isthe mass density in the undeformed configuration. It is assumed that φ = 0 in theundeformed configuration. As for invariant-based models, we can obtain a functionφ with three terms: a purely mechanical term φM, a purely thermal term φT , and amixed term φTM. Now, with entropy, η, φ satisfies the relations

(18.30)

Following conventional practice, the specific heat at constant strain, ce = T∂η/∂Te,is assumed to be constant, from which we obtain

(18.31)

On the assumption that thermal effects in shear (i.e., deviatoric effects) can be

χχ = − ⊕ + ′− −∑ ∑4 1 2 42 2ζ ξ ζ ζ ξζ ζi i i

i

i i

i

Ti i( ) / [ ]C C A WA i

Wii i i i i i= − ⊗ + ⊗ − ⊗− − − −3

2

1

22 2ζ ζζ ζ ζ ζC C C C C C[ ].

ˆ / /C C= I31 3

( /J I= 31 2 )

w tr Ji

i

i= −

+ −∑ξ κζ[ ˆ ] ( ) ,C I

12

1 2

sTo= ∂

∂= − ∂

∂ρ ϕ η ϕ

eT e

T

.

φT ec ln= +T T/T ]0[ ( ) .1




neglected relative to thermal effects in dilatation, the purely mechanical effect isequated with the deviatoric term in Equation 18.29:

(18.32)

Of greatest interest is φTM. The development of Nicholson and Lin (1996) furnishes

(18.33)

The tangent-modulus tensor χχχχ′ = ∂s/∂e now has two parts: XM + XTM, in whichXM is recognized as χχχχ, derived in Equation 18.29. Without providing the details,Kronecker Product algebra furnishes the following:

(18.34)

The foregoing discussion of stretch-based thermohyperelastic models has beenlimited to compressible elastomers. However, many elastomers used in applications,such as seals, are incompressible or near-incompressible. For such applications, aswe have seen, an additional field variable is introduced, namely, the hydrostaticpressure (referred to deformed coordinates). It serves as a Lagrange multiplierenforcing the incompressibility and near-incompressibility constraints. Followingthe approach for invariant-based models, Equations 18.33 and 18.34 can be extendedto incorporate the constraints of incompressibility and near-incompressibility.

The tangent-modulus tensor presented here only addresses the differential ofstress with respect to strain. However, if coupled heat transfer (conduction andradiation) is considered, a general expression for the tangent-modulus tensor isrequired, expressing increments of stress and entropy in terms of increments of strainand temperature. A development accommodating heat transfer for invariant-basedelastomers is given in Nicholson and Lin (1997a).

18.6 THERMOMECHANICS OF DAMPED ELASTOMERS

Thermoviscohyperelasticity is a topic central to important applications, such asrubber mounts in hot engines. The current section introduces a thermoviscohy-perelastic constitutive model thought to be suitable for near-incompressible elas-tomers exhibiting modest levels of viscous damping following a Voigt model. Twopotential functions are used to provide a systematic treatment of reversible andirreversible effects. One is the familiar Helmholtz free energy in terms of the strainand the temperature; it describes reversible, thermohyperelastic effects. The secondpotential function, based on the model of Ziegler and Wehrli (1987), models viscous

φ ΙM i

i

tr i= −

∑ξ ζ[ ˆ ] .C

φ TM

J= − = + −κ βρ

β α[ ( ]( ) ( ( )/ ) .

3 2

011

21 3

T) T T/T

χχTM J JJ

J= + − −

κρ

β β β33

2 3 33

33 3

21n n An nT

T

( ) .




dissipation and arises directly from the entropy-production inequality. It provides aconsistent thermodynamic framework for describing damping in terms of a viscositytensor that depends on strain and temperature.

The formulation leads to a simple energy-balance equation, which is used toderive a rate-variational principle. Together with the Principle of Virtual Work,variational equations governing coupled thermal and mechanical effects are pre-sented. Finite-element equations are derived from the thermal-equilibrium equationand from the Principle of Virtual Work. Several quantities, such as internal energydensity, χ, have reversible and irreversible portions, indicated by the subscripts rand i: χ = χr + χi. The thermodynamic formulation in the succeeding paragraphs isreferred to undeformed coordinates.

There are several types of viscoelastic behaviors in elastomers, especially if theycontain fillers such as carbon black. For example, under load, elastomers experiencestress softening and compression set, which are long-term viscoelastic phenomena.Of interest here is the type of damping that is usually assumed in vibration isolationin which the stresses have an elastic and a viscous portion reminiscent of the classicalVoigt model, and the viscous portion is proportional to strain rates. The time con-stants are small. This type of damping is viewed as arising in small motions super-imposed 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

(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 coor-dinates, and h is the heat input per unit mass, for simplicity’s sake, assumed inde-pendent of temperature. The state variables are thus e and T. The Helmholtz freeenergy, φr per unit mass, and the entropy, η per unit mass, are introduced using

(18.36)

Now,

(18.37)

18.6.2 ENTROPY PRODUCTION INEQUALITY

The entropy-production inequality is stated as

(18.38)

ρ χ ρ

ρ

0 0 0 0

0 0 0

˙ ˙

˙ ˙

= − ∇ +

= + − ∇ +

s e q

s e s e q

T T

T T T

h

hr i

φr = −χ ηT .

∇ − = + − − −0 0 0 0 0 0T

rhq ρ ρ η ρ η ρs e s eT Tr i T T˙ ˙ ˙ ˙ ˙ .φφ

ρ η ρ

ρ ρ η ρ η

0 0 0 0 0

0 0 0 0

T T/T

T T T/T

˙

˙ ˙ ˙ ˙ ˙

≥ −∇ + + ∇

≥ − − + + + ∇

T T

r rT

iT T

hq

s e s e q

q

φ




The Helmholtz potential is assumed to represent reversible thermohyperelasticeffects. 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

(18.39)

This allows us to say that the viscous dissipation is “absorbed” as heat. We alsosuppose that reversible effects are “absorbed” as a portion of the heat input as follows:

(18.40)

In addition, from conventional arguments,

(18.41)

it follows that

(18.42)

Inequality as shown in Equation 18.42 can be satisfied if and

(18.43)

Inequality as shown in Equation 18.43b is conventionally assumed to expressthe fact that heat flows irreversibly from cold to hot zones. Inequality as shown inEquation 18.43a requires that viscous effects be dissipative.

18.6.3 DISSIPATION POTENTIAL

Following Ziegler and Wehrli (1987), the specific dissipation potential = − ρ0ηi is introduced, for which

(18.44)

The function Ψ is selected such that Λi and Λt are positive scalars, in which casethe inequalities in Equations 18.44a and 18.44b require that

(18.45)

This can be interpreted as indicating the convexity of a dissipation surface in space. Clearly, to state the constitutive relations, it is sufficient to specify φr and Ψ.

ρ η ρ0 2 0 0 0T ˙ .i ih= −∇ +[ ]Tq

ρ η ρ0 0 0 0T ˙ .r rh= −∇ +[ ]Tq

ρ η∂ ∂ = ∂ ∂ = −φ φr rT

r r/ / ,e s T

s eT Ti˙ ˙.− ∇ ≥ −q0 0 0 1T/T ρ ηi T

ρ η0 0iT ≥

s eTi (a) T/T (b).˙ ≥ − ∇ ≥0 00qT

Ψ( , ˙, ,q0 e e T)T

s ei iT

tT = ∂ ∂ − ∇ = ∂ ∂ρ ρ0 0 0 0Λ Ψ Λ Ψ/ ˙ / .(a). T/T (b).q

( / ˙ ˙ ( / .∂ ∂ ≥ ∂ ∂ ≥Ψ Ψe e) )0 00 0q q

(˙, )e q0




A simple illustration is now provided showing how the dissipation potential inEquation 18.45 provides a “framework” for describing dissipative effects. On theexpectation that properties governing heat transfer are not affected by strain, weintroduce the decomposition

(18.46)

Now, Ψt represents thermal effects, and we assume for simplicity’s sake that Λt isa material constant. Inequality in Equation 18.46 implies that

(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 dissi-pation, suppose that

(18.48)

in which µ(T, J1, J2) is the viscosity. Hence,

(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 thestress, entropy, and internal energy. Equation 18.40 is repeated as

(18.50)

The ensuing Maxwell relation is

(18.51)

Conventional operations furnish the reversible part of the equation of thermalequilibrium (balance of energy):

(18.52)

For the irreversible part, we recall the relations

(18.53)

Ψ Ψ Ψ Ψ Λ= + =i t t tρ0 0 0

12 2[ ] [ / ].q qT

−∇ =0 0T/T q / .Λt

Ψ ∆i iJ J= =µ( , , )˙ ˙ ,T /1 2 2 1e eT

s ei J J= µ( , , )˙,T 1 2

ρ φ ρ η0 0˙ ˙ ˙

r r= −s eTr T.

∂ ∂ = − ∂ ∂srT

r/ / .T ρ η0 e

[ ] / ˙ ˙ , /−∇ + = − ∂ ∂ + = ∂ ∂0 0 0 0Tq ρ ρ ηh c cr r

Te e rT( T) T T T.s e

− ∇ −( ) = − +0 0 0 0Tq ρ ρh c

i is eTi T˙ ˙




and

(18.54)

and

(18.55)

Upon adding the relations, we obtain the thermal-field equation

(18.56)

It is easily seen that Equation 18.55 directly reduces to a well-known expressionin classical linear thermoelasticity. In addition, under adiabatic conditions in which

most of the “viscous” work, , is “absorbed” as a temperatureincrease controlled by ρ0(ce + ci), while a smaller portion is “absorbed” into theelastic 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) stressis assumed to satisfy a thermohyperelastic constitutive relation suitable for near-incompressible elastomers. In particular,

(18.57)

Here, φrm represents the purely mechanical response and can be identified as theconventional, 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 stretchratio-based models, such as the Ogden (1986) model. The function φrt(T) representsthe purely thermal portion of the Helmholtz free-energy density. Finally, φrtm(T, I3)represents thermomechanical effects, again based on the assumption that the primarycoupling is through volumetric expansion. The quantity φro represents the Helmholtzfree 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:

(18.58)

(18.59)

ρ η ρ η ρ η0 2 0

20

2T T TT

T˙ ˙ ˙i

i i= ∂∂

+ ∂∂e

e

− = ∂∂

= ∂∂

se

Ti

ii

icρ η η0

2 2T TT

, .

−∇ + = − ∂ ∂ − + +0 0 0 0T T Tq s e s eρ ρh c cr i e iT T T/ ˙ ˙ ( ) ˙ .

−∇ + =0 0 0 0Tq ρ h , s eTi

˙

φ φ φ φ φr rm rt rtm roI I I I= + + +( , , ) ( ) ( , ) .1 2 3 3T T

φrt ec ln( [ ( / )T) T T T ]= −1 0

φrtm I f J( , ) [ ( ]T T)30

3 2

21= −κ

ρ

J I f= = = + −

−

31 2

0

1

13

/ ( ) ( ( )det T) T TFα




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:

(18.60)

in which are the “deviatoric invariants” of C. The revers-ible stress is obtained as

(18.61)

18.7.2 SPECIFIC DISSIPATION POTENTIAL

Fourier’s law of conduction is obtained from:

(18.62)

The viscous stress si depends on the shear part of the strain rate as well as thetemperature. However, since the elastomers of interest are nearly incompressible, togood 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 anexample was already given in Section 6.3. Here, taking a more general viewpoint,we seek expressions of the form , in which Dv is called theviscosity tensor; it is symmetric and positive-definite. (Of course, the correct expres-sion is determined by experiments.) The simplest example was furnished in Section18.6.3. As a second example, to ensure isotropy, suppose that Ψi is a function of and and note that

(18.63)

For an expression reminiscent of the two-term Mooney-Rivlin strain-energy func-tion, let us consider the specific form

(18.64)

in which C1v and C2v are positive material coefficients. We obtain the viscosity tensor,

(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

φrm I I I C J C J I( , , ) ( ) ( ), ,1 2 3 1 1 2 2 31 1 1= − + − =

J I I J I I1 1 31 3

2 2 32 3= =/ // / and

s nr j j j r jI

VEC I VEC I VEC

= = ∂ ∂

= = = − = = −

2

1 1 1 3 3

φ φ φ

/

, ( ) , ( ) ( ) n i i I n i c c C n C 1

ρΨt tk= [ ] [ / ]./q q0 01 2 2T

ΨiT

v= 12 ˙ ( , )˙e e eD T

J1˙ : ( ˙ , ˙ ),J J Ji2 1 2Ψ

˙ ˙, ˙ ˙, .J II I J I

I I1 1 13

3 31 3

2 2 2 2 3 32 32 1

3 2 23

1 2

3= = −

= = −

m m i n m m n nT T T T T T T Te e

Ψi v vC J C J= +[ ]µ1 1 12

2 222(T ,) ˙ / ˙

Dv v vC C= +[ ]µ(T .) 1 1 1 2 2 2m m m mT T




tensor D, in particular, Ψi = µ(T)tr(D2)/2, which has the advantage in that thedeformation rate tensor D is in the observed (current) configuration. With d =VEC(D),

(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-elementequations of interest, neglecting inertia. Application of variational methods to themechanical field furnishes the Principle of Virtual Work in the form

(18.67)

in which ττττ0 denotes the traction vector on the undeformed surface S0. As illustratedin the examples in the previous section, we expect that the dissipation potential hasthe form Ψi = Dv(e, T) from which

(18.68)

in which Dv is again the viscosity tensor, and it will be taken as symmetric andpositive-definite. (It is positive-definite since ) Equation 18.67 isthus rewritten as

(18.69)

18.8.2 THERMAL EQUILIBRIUM

The equation for thermal equilibrium is rewritten as

(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 introducethe integrated form of this relation. The current value of ηr + ηi2, assuming that theinitial values of the entropies vanish, is now given by

(18.71)

d F F e= ⊗ = + ⊗ +− − − −T Tv

˙ )( ) ( )D µ ε ε(T ,2 21 1I I

tr dV dS tr dVi r( ) ( ) ,δ δ δES u EST0 0 0 0∫ ∫ ∫= −ττ

12 eT e,

s ei v= D ˙,

s e eTi

˙ ˙≥ 0 for all .

δ δ δe D e u t eTi

T TrdV dS dV˙ .0 0 0 0∫ ∫ ∫= − s

ρ η η ρ0 2 0 0 0( ˙ ˙ ) / ,r iT h+ = −∇ +[ ]q T

ρ η η ρ0 2 0 0 0( ) / .r i h dt+ = −∇ +[ ][ ]∫ Tq T




The corresponding variational principle is stated as

(18.72)

With some effort, a rate (incremental) variational principle can be obtained inthe form

(18.73)

(The motivation behind writing Equations 18.71 and 18.72 is simply to establishhow 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 thedenominator by T0 and letting k denote the thermal conductivity, we can obtain thethermal-equilibrium equation in the form

(18.74)

Using interpolation models for displacement and temperature, Equations 18.69and 18.74 reduce directly into finite-element equations for the mechanical andthermal fields.

18.9 EXERCISES

1. Derive explicit forms of the stress and tangent-modulus tensors using theHelmoltz 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 andthermal fields.

δ ρ η η δ ρT T T0 2 0 0 0 0 0( ) / .r i dV h dt dV+ = −∇ +[ ][ ]

∫ ∫∫ Tq

δ ρ δ ρT T T )T T T Ti− ∂ ∂ − + +[ ] = −∇ +[ ][ ]∫ ∫s e s e qTrT

e ic c dV h dV/ ˙ ˙ ( ˙ / / .0 0 0 0 0 0T

δ ρ

δ δ δ ρ δ

T T c c )T T

T T/T T T T T

i−∂ ∂ − − +[ ]+ ∇ ∇ = −

∫∫ ∫ ∫

s e s eTr e i

T T

dV

k dV h dV dS

T / ˙ ( ˙ /

[ ( ) ] / ( / )

0 0 0

0 0 0 0 0 0 0 0 0n q



243

Inelastic and Thermoinelastic Materials

19.1 PLASTICITY

Plasticity and thermoplasticity are topics central to the analysis of important appli-cations, such as metal forming, ballistics, and welding. The main goal of this sectionis to present a model of plasticity and thermoplasticity, along with variational andfinite-element statements, accommodating the challenging problems of finite strainand kinematic hardening.

19.1.1 K

INEMATICS

Elastic and plastic deformation satisfies the additive decomposition

(19.1)

from which we can formally introduce strains:

(19.2)

The Lagrangian strain

E

satisfies the decomposition

(19.3)

Typically, plastic strain is viewed as permanent strain. As illustrated in Figure 19.1,in a uniaxial tensile specimen, the stress,

S

11

, can be increased to the point A, andthen unloaded along the path AB. The slope of the unloading portion is E, the sameas that of the initial elastic portion. When the stress becomes equal to zero, therestill is a residual strain,

E

i

, which is identified as the inelastic strain. However, ifinstead the stress was increased to point C, it would encounter reversed loading atpoint D, which reflects the fact that the elastic region need not include the zero-stress value.

19.1.2 P

LASTICITY

We will present a constitutive equation for plasticity to illustrate how the tangentmodulus is stated. The ideas leading to the equation will be presented subsequentlyin the section on thermoplasticity. With χχχχ

e

=

ITEN

22(

D

e

) and

D

e

denoting the

19

D D D= +r i ,

∃ = ∃ = ∃ =∫ ∫ ∫D D Ddt dt dtr r i i .

˙ .E E E= = =∫ ∫F DF F D F F D FT Tr

Ti r idt dt



244


tangent-modulus tensor relating the elastic-strain rate to the stress rate (assuminga linear relation), the constitutive equation of interest is

(19.4)

In Equation 19.4, ΨΨΨΨ

i

is the yield function. ΨΨΨΨ

i

=

0 determines a closed convexsurface in stress space called the yield surface. (We will see later that ΨΨΨΨ

i

also servesas a [complementary] dissipation potential.) The stress point remains on the yieldsurface 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 thestress point. If the stress point is interior to, or moving tangentially on, the yieldsurface, only elastic deformation occurs. On all interior paths, for example, due tounloading, the response is only elastic. Plastic deformation induces “hardening,”corresponding to a nonvanishing value of

C

i

.

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), theconventional model of isotropic hardening is illustrated in which the yield surfaceexpands as a result of plastic deformation. This model is unrealistic in predicting agrowing elastic region. Reversed plastic loading is encountered at much higherstresses than isotropic hardening predicts. An alternative is kinematic hardening (seeFigure 19.2[b]), in which the yield surface moves with the stress point. Within a fewpercentage points of plastic strain, the yield surface may cease to encircle the origin.

FIGURE 19.1

Illustration of inelastic strain.

S11

E11

AC

E

E

E

FEi

B

D

˙ ˙ ˙ ˙, ˙ ˙

.

e C s e C s C K

C

ei i e e e i

ii i i

i

i iH H

= = = =

= ∂∂

∂∂

= − ∂∂

+ ∂∂

∂∂

−, ,

,

χχ

ΨΨ ΨΨ ΨΨ ΨΨ ΨΨ

1

k

s s e kK

s

T T

e



245

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. Com-bined isotropic and kinematic hardening are shown in Figure 19.2(c). However, theyield surface contracts, which is closer to actual observations (e.g., Ellyin [1997]).

FIGURE 19.2(a)

Illustration of yield-surface expansion under isotropic hardening.

FIGURE 19.2(b)

Illustration of yield-surface motion under kinematic hardening.

Sll

Sl

Slll principal stresses Sl Sll Slll

path of stress point

Slll

Sll

Sl



246


The rate of movement must exceed the rate of contraction for the material to remainstable with a positive tangent modulus.

Combining the elastic and inelastic portions furnishes the tangent-modulus tensor:

(19.5)

Suppose that in uniaxial tension, the elastic modulus is E

e

and the inelasticmodulus-relating stress and inelastic strain increments are E

i

,

and E

i

<< E

e

. The totaluniaxial modulus is then

19.2 THERMOPLASTICITY

As in Chapter 18, two potential functions are introduced to provide a systematicway to describe irreversible and dissipative effects. The first is interpreted as theHelmholtz free-energy density, and the second is for dissipative effects. To accom-modate kinematic hardening, we also assume an extension of the Green and Naghdi(G-N) (1965) formulation, in which the Helmholtz free energy decomposes intoreversible and irreversible parts, with the irreversible part depending on the “plasticstrain.” Here, it also depends on the temperature and a workless

internal state variable

.

19.2.1 B

ALANCE

OF

E

NERGY

The conventional equation for energy balance is augmented using a vector-valued, work-less internal variable, αααα

0

, regarded as representing “microstructural rearrangements”:

(19.6)

FIGURE 19.2(c)

Illustration of combined kinematic and isotropic hardening.

Slll

Sll

Sl


χχ χχ χχ χχ= +[ ] = +− − −e i e i e

1 1 1C C[ ] .I

E

E Ei

i e( / ).

1 +

ρ ρo r i oh˙ ˙ ˙ ˙ ,χ0 0 0 0 0= + − ∇ + +s e eT T T Tqs ββ αα



247

where

χ

o

is the internal energy per unit mass in the undeformed configuration and

s

=

VEC

(

S

),

e

=

VEC

(

E

), and ββββ

0

is the flux per unit mass associated with αααα

0

. However,note that ββββ

=

0

, thus ββββ

i

=

−ββββ

r

. Also,

c

is the internal energy per unit mass,

q

0

is theheat-flux vector referred to undeformed coordinates, and

h

is the heat input per unitmass, for simplicity’s sake, assumed independent of temperature. The state variablesare

E

r

,

E

i

, T, and αααα

0

.The next few paragraphs will go over some of the same ground as for damped

elastomers in Chapter 18, except for two major points. In that chapter, the stresswas assumed to decompose into reversible and irreversible portions in the spirit ofelementary Voigt models. In the current context, the strain shows the decompositionin the spirit of the classical Maxwell model. In addition, as seen in the following,it proves beneficial to introduce a workless internal variable to give the model theflexibility to accommodate phenomena such as kinematic hardening.

The Helmholtz free energy,

φ

, per unit mass and the entropy,

η

, per unit massare introduced using

(19.7)

Now,

(19.8)

19.2.2 E

NTROPY

-P

RODUCTION

I

NEQUALITY

The entropy now satisfies

(19.9)

Viewing

φ

r

as a differentiable function of

e

r

, T, and αααα

0

, we conclude that

(19.10)

Extending the G-N formulation, let

s

*

T

=

ρ

0

∂

φ

i

/

∂

e

i

and assume that

η

i

=

−∂

φ

i

/

∂

Tand

ρ

0

∂

φ

i

/

∂αααα

0

=

. Now,

(19.11)

The entropy-production inequality (see Equation 19.9) is now restated as

(19.12)

φ = −χ ηT .

∇0 0 0 0 0T Tq − = + − − − +ρ ρ η ρ η ρ φ0 0 0T Th T

rT

is s˙ ˙ ˙ ˙ ˙ ˙ .e e ββ αα

ρ η ρ

ρ φ ρ η ρ η

0 0 0 0 0

0 0 0 0 0 0

T T/T

T T T/T

˙

˙ ˙ ˙ ˙ ˙ ˙ .

≥ − + +

≥ − − + + + −

∇ ∇

∇

T T

Tr i

hq q

qT Ts e s e T ββ αα

∂ ∂ ∂ ∂ ∂ ∂φ φ ρ η φr rT

r r r rT/ / / .e s= = − = T 0 0 0αα ββ

ββ0iT

s e*Ti i i i i i

T= = − =ρ φ η φ ρ φ0 0 0 0∂ ∂ ∂ ∂ ∂ ∂/ / / . T αα ββ

( )˙ .s s eT Ti*− − ≥qT

0 0 0∇ T/T



The inequality shown in Equation 19.12 can be satisfied if

(19.13)

The first inequality involves a quantity, s* = VEC(S*), with dimensions of stress.In the subsequent sections, s* will be viewed as a reference stress, often called theback stress, which is interior to a yield surface and can be used to characterize themotion of the yield surface in stress space. In classical kinematic hardening in whichthe hyperspherical yield surface does not change size or shape but just moves, thereference stress is simply the geometric center. If kinematic hardening occurs, asstated before, the yield surface need not include the origin even with small amountsof plastic deformation. Thus, there is no reason to regard as vanishing at theorigin. Instead, = 0 is now associated with a moving-reference stress interior tothe yield surface, identified here as the back stress s*.

19.2.3 DISSIPATION POTENTIAL

As in Chapter 18, we introduce a specific dissipation potential, Ψi, for which

(19.14a)

from which, with Λi > 0 and Λt > 0,

(19.14b)

On the expectation that properties governing heat transfer are not affected bystrain, we introduce the decomposition into inelastic and thermal portions:

(19.15a)

where Ψi represents mechanical effects and is identified in the subsequent sections.The thermal constitutive relation derived from the dissipation potential impliesFourier’s law:

(19.15b)

The inelastic portion is discussed in the following section.

( )˙ s s eT Ti*− ≥ − ∇ ≥0 00 0 (a) T/T (b).qT

er

er

˙ / /e s si iiT

i t *= ∋ − ∇ = ∂ ∋ = −ρ ρ0 0 0 0Λ Λ (i) T / T (ii) (iii),∂ ∂ ∂Ψ ΨT q

ρ ρ0 0 0 0Λ Λi t( / ( / ) .∂ ∂ ∂ >Ψ ∋) ∋ + ∂Ψ 0q q

Ψ Ψ Ψ Ψ= + =i t tt ρ0 0 02

Λq qT ,

−∇ =0 0T / T q / .Λt


Inelastic and Thermoinelastic Materials 249

19.3 THERMOINELASTIC TANGENT-MODULUS TENSOR

The elastic strain rate satisfies a thermohypoelastic constitutive relation:

(19.16)

Cr is a 9 × 9 second-order, elastic compliance tensor, and ar is the 9 × 1thermoelastic expansion vector, with both presumed to be known from measure-ments. Analogously, for rate-independent thermoplasticity, we seek tensors Ci andai , depending on , ei, and T such that

(19.17a)

(19.17b)

During thermoplastic deformation, the stress and temperature satisfy a thermo-plastic yield condition of the form

(19.18)

and Πi is called the yield function. Here, the vector k is introduced to represent theeffect of the history of inelastic strain, , such as work hardening. It is assumed tobe given by a relation of the form

(19.19)

The “consistency condition” requires that = 0 during thermoplastic flow, fromwhich

(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 currentstress point,

(19.21)

˙ ( ) ˙ .e s sr r r*= − +•C Ta

∋

˙ ( ) ˙e C s si i i*= − +• a T

˙ [ ]( ) ( ) ˙e s s= + − + +•C Cr i r i* a a T.

Πi i i( , , , , ) ,∋ =e k T η 2 0

ei

˙ ( , , )˙ .k K k= e ei iT

Πi

d

d

d

d

d

d

d

d

d

di i

ii

i i ii

Π Π Π Π Π∋

∋ ˙ ˙ ˙ ˙ .+ + + + =e

ek

kT

Tiη

η 0

˙ ˙ei ii

T

iid

d

d

d=

=Λ Π Λ Π∋

(a) (b).ii

ηη



Equation 19.14a suggests that the yield function may be identified as the dissi-pation potential: Πi = ρ0Ψi. Standard manipulation furnishes

(19.22)

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 thermody-namic inequality shown in Equation 19.13a is now satisfied if H > 0.

Next, note that s* depends on ei , T, and αααα 0 since s*T = ρ0∂φi /∂ei. For simplicity’ssake, we neglect dependence on αααα 0 and assume that a relation of the following formcan be measured for s*:

(19.23)

From Equations 19.16 and 19.17, the thermoinelastic tangent-modulus tensorand thermal thermomechanical vector are obtained as

(19.24a)

(19.24b)

If appropriate, the foregoing formulation can be augmented to accommodateplastic incompressibility.

19.3.1 EXAMPLE

We now provide a simple example using the Helmholtz free-energy density functionand the dissipation-potential function to derive constitutive relations. The followingexpression involves a Von Mises yield function, linear kinematic hardening, linearwork hardening, and linear thermal softening.

˙ ˙ ˙ ˙ ˙ ˙ei i i i iT

i

ii i

i ii

T

i

ii i

i

i

H H

TH H

= ∋ + = ∋ +

= ∂∂∋

∂∂∋

= = ∂∂∋

∂∂

= ∂∂

= − ∂∂

+ ∂∂

∂

C T c T

T

c

a b

a b

e kK

T

η 2

2

CΨ Ψ Ψ Ψ

Ψ Ψ Ψ ΨΨ Ψ Ψi

T

i

i

i

∂∋

+ ∂∂

∂∂

η 2 T

˙ ˙ ˙ /s ee e

e* ii i

T

Ti i= + =

= ∂ ∂ ∂ΓΓ ΓΓϑ ϑT T.I

∂∂

∂∂

Ψ Ψ2

˙ ˙ ˙e s= +C aT

C C C I I C C

C C C C I C

= + − +

= + + + − + +

−

−

( )[ ( ) ]

[ ( ) ( )( ) ]

r i i i

r i r i i r i i

ΓΓ ΓΓ

ΓΓ ΓΓ

1

1a a a a ϑ



i. Helmholtz free-energy density:

(19.25)

in which is a known constant. From Equation 19.28,

(19.26)

Finally,

(19.27)

ii. Dissipation potential:

(19.28)

(19.29)

Straightforward manipulations serve to derive

(19.30)

(19.31)

Consider a two-stage thermomechanical loading, as illustrated schematically inFigure 19.3. Let SI, SII, SIII denote the principal values of the 2nd Piola-Kirchhoffstress, and suppose that SIII = 0. In the first stage, with the temperature held fixedat T0, the stresses are applied proportionally well into the plastic range. The centerof the yield surface moves along a line in the (S1, S2) plane, and the yield surfaceexpands as it moves. In the second stage, suppose that the stresses S1 and S2 arefixed, but that the temperature increases to T1 and then to T2 and T3. The plasticstrain must increase, thus, the center of the yield surface moves. In addition, strainhardening tends to cause the yield surface to expand, while the increased temperaturetends to make it contract. However, in this case, thermal softening must dominatestrain hardening, and contraction must occur since the center of the yield surfacemust move further along the path shown even as the yield surface continues to “kiss”the fixed stresses SI and SII.

φ φ φ φ

φ

= + =

= − − + ′ −− −

r i i i i

r r r r r r r

k

ln

T T c T (T / T )

ρ

ρ ρ

0 3

01 1

0 0 02 1

e e

C C

T

T Tae e e/ ( ) ( )

′cr

∋ = ∂ ∂ = − −−ρ φ01

0( / ) [ ( )].r r r r re eT C a T T

c TT

crr= −

∂∂

= ′2

2

φ

Ψ Ψ Ψ Ψ= + =i t tt T ρ0 0 02

Λq q

ΨiT

ik k k k k= ∋ ∋ − + − − = =∋T [ ( )] ˙ ˙0 1 2 0 0T T e

H k k k k k k= ∋ ∋ = + − −1 1 0 1 2 0T [ ( )]T T

C a bT

T Ti i i iH k Hk

H= ∋∋∋ ∋

= = = ∋∋ ∋

c 22 2



Unfortunately, accurate finite-element computations in plasticity and thermo-plasticity often require close attention to the location of the front of the yieldedzone. This front will occur within elements, essentially reducing the continuity orderof the fields (discontinuity in strain gradients). Special procedures have been devel-oped in some codes to address this difficulty.

The shrinkage of the yield surface with temperature provides an element ofthe explanation of the phenomenon of adiabatic shear banding, which is commonlyencountered in some materials during impact or metal forming. In rapid processes,plastic work is mostly converted into heat and on into high temperatures. Thereis not enough time for the heat to flow away from the spot experiencing highdeformation. However, the process is unstable while the stress level is maintained.Namely, as the material gets hotter, the rate of plastic work accelerates, thanks tothe softening evident in Figure 19.3. The instability is manifested in small, peri-odically 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 brittlefailure.

19.4 TANGENT-MODULUS TENSORIN VISCOPLASTICITY

The thermodynamic discussion in the previous section applies to thermoinelasticdeformation, for which the first example given concerned quasi-static plasticity andthermoplasticity. However, it is equally applicable when rate sensitivity is present, inwhich case viscoplasticity and thermoviscoplasticity are attractive models. An example

FIGURE 19.3 Effect of load and temperature on yield surface.

Slll

Sll

SlT0

T0

T0

T0

T0

T0

T1T2

T3

T4



of a constitutive model, for example, following Perzyna (1971), is given in undeformedcoordinates as

(19.32)

and Ψi(∋, ei, k, T, ηi) is a loading surface function. The elastic response is stillconsidered linear in the form

(19.33)

Recall from thermoplasticity that

(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, reference-yield surface. The vectors and ′ have the same origin and direction, but the latterterminates at the reference surface, while the latter terminates outside the referencesurface 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 tothe exterior of the reference surface. This situation is illustrated in Figure 19.4.

FIGURE 19.4 Illustration of reference surface in viscoplasticity.

µω =< − >

< − >=− − ≥

∂

∂ ∋

∂

∂∋

∂

∂∋

˙ ,ei

T

T

k kk k

1 11 1 0

Ψ ΨΨ Ψ

Ψ

Ψ Ψi

i

i i

, if

0, otherwise, ii i

˙ ˙ ˙e sr r r= +−χχ αα1 T.

˙ ˙ ˙ .s ee e

e* ii i

T

iT

i i= + =

= ∂ΓΓ ΓΓϑ ϑT T∂

∂∂

∂Ψ ∂ Ψ /∂2

∋ ∋∋

∋ ∋

Slll

Sll

Sl

S∗stress point



It should be evident that viscoplasticity and thermoviscoplasticity can be for-mulated to accommodate phenomena such as kinematic hardening and thermalshrinkage of the reference-yield surface.

The tangent-modulus matrix now reduces to elastic relations, and viscoplasticeffects can be treated as an initial force (after canceling the variation) since

(19.35)

In particular, the Incremental Principle of Virtual Work is now stated, to first order, as

(19.36)

19.5 CONTINUUM DAMAGE MECHANICS

Ductile fracture occurs by processes associated with the notion of damage. Aninternal-damage variable is introduced that accumulates with plastic deformation.It also manifests itself in reductions in properties, such as the experimental valuesof the elastic modulus and yield stress. When the damage level in a given elementreaches 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 elementdeletion.

There are two different schools of thought on the suitable notion of a damageparameter. One, associated with Gurson (1977), Tvergaard (1981), and Thomasson(1990), considers damage to occur by a specific mechanism occurring in a three-stage process: nucleation of voids, their subsequent growth, and their coalescenceto form a macroscopic defect. The coalescence event is used as a criterion for elementfailure. The parameter used to measure damage is the void-volume fraction f. Modelsand criteria for the three processes have been formulated. For both nucleation andgrowth, evolution of f is governed by a constitutive equation of the form

(19.37)

⋅ = + − −

< − >

∂

∂∋

∂

∂∋

∂

∂∋

˙ ˙ ˙ .s eχχ χχ αα ΓΓ χχΨΨr r r

r

v

T

T

kT

i

i

i i

µ1

Ψ

Ψ Ψ

δ δ δ ρ

δ δµ

∆ ∆ ∆ ∆ ∆ ∆

∆ ∆ ∆

e e e u u

u e

Tr o

Tr r o

To o

To

T r

v

T

T o

dV dV dV

dSk

dV

i

∫ ∫ ∫

∫ ∫

+ +

+ −

< − >

∂

∂∋

∂

∂∋

∂

∂∋

χχ χχ αα

ττ ΓΓ χχ

T

= i

i i

˙

1Ψ

Ψ

Ψ Ψ

˙ ( , , ),f f Ti= ΞΞ e



for which several specific forms have been proposed. To this point, a nominal stressis used in the sense that the reduced ability of material to support stress is notaccommodated.

The second school of thought is more empirical in nature and is not dependenton a specific mechanism. It uses the parameter D, which is interpreted as the fractionof damaged area Ad to total area Ao that the stress (traction) acts on. Consider a uniaxialtensile specimen with damage, but experiencing elastic behavior. Suppose that dam-aged area Ad can no longer support a load (is damaged). For a given load P, the truestress at a point in the undamaged zone is Here, S′is a nominal stress, but is also the measured stress. If E is the elastic modulusmeasured in an undamaged specimen, the modulus measured in the current specimenwill be E′ = E(1 − D), demonstrating that damage is manifest in small changes inproperties.

As an illustration of damage, suppose specimens are loaded into the plasticrange, unloaded, and then loaded again. Without damage, the stress-strain curveshould return to its original path. However, due to the damage, there are slightchanges in the elastic slope, in the yield stress, and in the slope after yield (exag-gerated 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 probablysmall, so that its role in the energy-balance equation can be neglected. At the risk ofbeing slightly conservative, in dynamic (adiabatic) problems, the plastic work can beassumed to be completely converted into heat. Even so, for the sake of a consistentframework for treating dissipation, a dissipation potential, ΨΨΨΨd, can be introduced fordamage, as has been done, for example, by Bonora (1997). The contribution to theirreversible entropy production can be introduced in the form ≥ 0, in which is the “force” associated with flux . Positive dissipation is assured if we assume

(19.38)

FIGURE 19.5 Illustration of effect of damage on elastic-plastic properties.

S11

E11

Sy1

Sy2

Ep1 Ep2

E1

E2

Sy3

Ep3

E3

S SD D= = = ′− − −P

A A1

1P

A1

1o D o.

D D DD

D = ∂∂

= >Ψ

Ψdd i d iD

D˙ ( ˙ , (, ,T, ) ,T, ) 0.d

12

2Λ Λe ek k



An example of a satisfactory function is Λd(ei, T, k) = Λdo ∫(s − s*) dt, Λdo apositive 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 aresolved for the nodal displacements, from which the inelastic strains can be computed.This information can then be used to update the damage-parameter values. Upondoing so, the damage-parameter values are compared to critical values. As statedpreviously, if the critical value is obtained, the element is deleted. The path of deletedelements can be viewed as a crack.

The code LS-DYNA Ver. 9.5, (2000) incorporates a material model that includesviscoplasticity and damage mechanics. It can easily be upgraded to include thermaleffects, assuming that all viscoplastic work is turned into heat. Such a model hasbeen shown to reproduce the location and path of a crack in a dynamically loadedwelded structure (see Moraes [2002]).

19.6 EXERCISES

1. In isothermal plasticity, assuming the following yield function, find theuniaxial stress-strain curve:

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 underuniaxial tension if a constant strain rate is imposed.

ei

Ψi 0

.= − − − +

∫( ) ( ) ˙ ˙s e s e e ek k k k dti

Ti o i

Ti

t

1 1 2


© 2003 by CRC CRC Press LLC © 2003 by CRC CRC Press LLC

257

Advanced Numerical Methods

In nonlinear finite-element analysis, solutions are typically sought using Newtoniteration, either in classical form or augmented as an arc-length method to bypasscritical points in the load-deflection behavior. Here, two additional topics of interestare briefly presented.

20.1 ITERATIVE TRIANGULARIZATIONOF PERTURBED MATRICES

20.1.1 I

NTRODUCTION

In solving large linear systems, it is often attractive to use Cholesky triangularizationfollowed by forward and backward substitutions. In computational problems, such asin the nonlinear finite-element method, solutions are attained incrementally, with thestiffness matrix slightly modified whenever it is updated. The goal here is to introduceand demonstrate an iterative method of determining the changes in the triangularfactors ensuing from modifying the stiffness matrix. A heuristic convergence argu-ment is given, as well as a simple example indicating rapid convergence. Apparently, noefficient iterative method for matrix triangularization has previously been established.

The finite-element method often is applied to problems requiring solution oflarge linear systems of the form

K

0γγγγ

0

=

f

0

, in which the stiffness matrix

K

0

is positive-definite, symmetric, and may be banded. As discussed in a previous chapter, anattractive method of solution is based on Cholesky decomposition (triangularization),in which

K

0

=

L

0

L

T

0

and

L

0

is lower-triangular, and it is also banded if

K

0

is banded.The decomposition enables an efficient solution process consisting of forward sub-stitution followed by backward substitution. Often, however, the stiffness matrix isupdated during the solution process, leading to a slightly different (perturbed) matrix,

K

=

K

0

+

∆

K

, in which

∆

K

is small when compared to

K

0

. For example, this situationmay occur in modeling nonlinear problems using an updated Lagrangian schemeand load incrementation. Given the fact that triangular factors are available for

K

0

,it would appear to be attractive to use an iteration scheme for the perturbed matrix

K

, in which the initial iterate is

L

0

. The iteration scheme should not involve solvingintermediate linear systems except by using current triangular factors. A scheme isintroduced in the following section and produces, in a simple example, good estimateswithin a few iterations.

The solution of perturbed linear systems has been the subject of many investiga-tions. Schemes based on explicit matrix inversion include the Sherman-Morrison-Woodbury formulae (see Golub and Van Loan [1996]). An alternate method is tocarry bothersome terms to the right side and iterate. For example, the perturbed linear

20



258


system can be written as

(20.1)

and an iterative-solution procedure, assuming convergence, can be employed as

(20.2)

Unfortunately, in a typical nonlinear problem involving incremental loading, espe-cially in systems with decreasing stiffness, it will eventually be necessary to updatethe triangular factors frequently.

20.1.2 N

OTATION

AND

B

ACKGROUND

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 entriesvanish. Consider a nonsingular real matrix

A

. It can be decomposed as

, (20.3)

in which

diag

(

A

) consists of the diagonal entries of

A

, with zeroes elsewhere;

A

l

coincides with

A

below the diagonal with all other entries set to zero; and

A

u

coincideswith

A

above the diagonal, with all other entries set to zero. For later use, weintroduce the matrix functions:

(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 theinverse of a nonsingular, upper-triangular matrix is upper-triangular. In proof of (a),let

L

(1)

and

L

(2)

be two

n

×

n

lower-triangular matrices. The

ij

th

entry of the productmatrix is given by Since

L

(1)

is lower-triangular, vanishes unless

k

≤

i

. Similarly, vanishes unless

k

≥

j

. Clearly, all entries of vanishunless

i

≥

j

, which is to say that

L

(1)

L

(2)

is lower-triangular.In proof of (b), let

A

denote the inverse of a lower-triangular matrix

L

. We multiplythe

j

th

column of

A

by

L

and set it equal to the vector

e

j

(

e

j

T

=

0

…..

1

…..

0 withunity in the

j

th

position): now,

(20.5)

K f K K0 0∆ ∆ ∆ ∆ ∆γγ γγ γγ= − − ,

K f K K01

0∆ ∆ ∆ ∆ ∆γγ γγ γγ( ) ( ).j j+ = − −

A A A A= + +l udiag( )

lower diag upper diagl u( ) ( ), ( ) ( ).A A A A A A= + = +12

12

∑ =k n ik kjl l11 2

,( ) ( ) . lik

( )1

lkj( )2 ∑ =k n ik kjl l1

1 2,

( ) ( )

l a

l a l a

l a l a l a

l a l a l a l a

j

j j

j j j

j j j j j j jj jj

11 1

21 1 22 2

31 1 32 2 33 3

1 1 2 2 3 3

0

0

0

1

=

+ =

+ + =

+ + + + =

M

L



Advanced Numerical Methods

259

Forward substitution establishes that

a

kj

=

0, if

k

<

j

, and

a

jj

=

l

−

1

jj

, thus,

A

=

L

−

1

islower-triangular.

20.1.3 I

TERATION

S

CHEME

Let

K

0

denote a symmetric, positive-definite matrix, for which the unique triangularfactors are

L

0

and

L

T

0

. If

K

0

is banded, the maximum width of its rows (the bandwidth)equals 2

b

−

1, in which

b

is the bandwidth of

L

0

. The factors of the perturbed matrix

K

can be written as

(20.6)

We can rewrite Equation 20.6 as

(20.7)

from which

(20.8)

Note that

L

0

−−−−

1

∆

L

is lower-triangular. It follows that

(20.9)

The factor of 1/2 in the definition of the

lower

and

upper

matrix functions ismotivated by the fact that the diagonal entries of and are the same.

Furthermore, for banded matrices, if

∆

L

and

L

0

have the same semibandwidth,

b

, it follows that, for the correct value of is alsobanded, with a bandwidth no greater than

b

. Unfortunately, it is not yet clear howto take advantage of this behavior.

An iteration scheme based on Equation 20.9 is introduced as

(20.10)

Explicit formation of the fully populated inverses can be avoidedby using forward and backward substitution. In particular,

, where

∆

k

1

is the first column of

∆

K

. We can now solve for by solving the system

L

0

b

j

= ∆kj.

20.1.4 HEURISTIC CONVERGENCE ARGUMENT

For an approximate convergence argument, we use the similar relation

(20.11)

[ ] [ ][ ].K K L L L LT T0 0 0+ = + +∆ ∆ ∆

[ ][ ] [ ] ,I L L I L L L K K L1 T T 1 T+ + = +− − − −0 0 0 0 0∆ ∆ ∆

L L L L L KL L L L L1 T T 1 T 1 T T0 0 0 0 0 0− − − − − −+ = −∆ ∆ ∆ ∆ ∆ .

∆ ∆ ∆ ∆L L L KL L L L L1 T 1 T T= −( )− − − −0 0 0 0 0lower .

L L10− ∆ ∆L LT T

0−

∆ ∆ ∆ ∆L L KL L L L L1 T 1 T T, 0 0 0 0− − − −−

∆ ∆ ∆ ∆

∆ ∆

L L L KL L L L L

L L L KL

1 T 1 T T

1 T

j j jlower

lower

+( ) − − − −

( ) − −

= −( )= ( )

10 0 0 0 0

10 0 0

( ) ( )

L L1 T0 0− − and

L K L k01

01

1− −=∆ ∆[

L k L k01

2 01− −∆ ∆K n ]

b L kj j= −01∆

∆ ∆ ∆ ∆A K A A A= − −∞2 1( ) ( ),




in which (∆A)∞ is the solution (converged iterate) for ∆A. Consider the iterationscheme

(20.12)

Subtraction of two successive iterates and application of matrix-norm inequali-ties furnish

(20.13)

Convergence is assured in this example if , in which s denotes thespectral radius (see Dahlquist and Bjork [1974]). An approximate convergencecriterion is obtained as

(20.14)

in which λj(∆A) denotes the jth eigenvalue of the n × n matrix ∆A. Clearly, convergenceis expected if the perturbation matrix has a sufficiently small norm. Applied to the currentproblem, we also expect convergence will occur if .

20.1.5 SAMPLE PROBLEM

Let L0 and K0 be given by

(20.15)

Now suppose that the matrices are perturbed according to

(20.16)

so that

(20.17)

∆ ∆ ∆ ∆A K A A A( ) ( )( ) ( ) .j j+ −∞= −1 12

∆ ∆ ∆ ∆ ∆A A A A A A( ) ( ) ( ) ( )[ ].j j j j+ + −∞

+− = −2 1 1 12

σ ( A )A 1−∝ <∆ 1

2

max minj j k

kλ λ( ) ( ),∆A A∝ < 1

2

max minj j k k| ( ) | | ( ) |λ λ∆L L∞ < 12

L K0 0

2

2 2

0=

=+

a

b c

a ab

ab b c, .

L K=+

=+ +

a

b c d

a ab

ab b c d

00

2

2 2,

( )

∆K =+

0 0

0 2d c d( ).



Advanced Numerical Methods 261

We are interested in the case in which d/c << 1, for example, d/c = 0.1, ensuringthat the perturbation is small. We also use the fact that

(20.18)

The correct answer, which should emerge from the iteration scheme, is

(20.19)

The initial iterate is found from straightforward manipulation as

(20.20)

The ratio of the norms of the error is

(20.21)

Letting , the second iterate is found, after straightforward manip-ulation, as

(20.22)

The relative error is now

(20.23)

Clearly, this is a significant improvement over the initial iterate.

L 10

1 0− =

−

ac

c

b a.

∆L∝ =

0 0

0 d.

∆L1

0 0

0 11

2

( ) = +

d

d

c

.

errornorm norm

norm

d d

d

d

c

= −

=+

−

=

∝

∝

( ) ( )

( )

%

( )∆ ∆∆

L LL

1

1

5

1

2

∆ = +d dc( )1 1

2

∆∆ ∆L( )2

0 0

0

0 0

0 1

1

2

18

2

2

2

3

3

=−

=− −

c

dc

dc

d

errordc

dc

= +

= +

2

2 118

0 01 11

80.




20.2 OZAWA’S METHOD FOR INCOMPRESSIBLE MATERIALS

In this section, thermal and inertial effects are neglected and the traction is assumedto be prescribed on the undeformed exterior boundary. Using a two-field formulationfor an incompressible elastomer leads to an incremental relation, in global form, asfollows (see Nicholson, 1995):

. (20.24)

If KMM is singular, it can be replaced with K′MM = KMM + χKMP KTMP, and χ can be

chosen to render K′MM positive-definite (see Zienkiewicz, 1989).The presence of zeroes on the diagonal poses computational difficulties, which

have received considerable attention. Here, we discuss a modification of the Ozawamethod discussed by Zienkiewicz and Taylor (1989). In particular, Equation 20.24is replaced with the iteration scheme

(20.25)

in which the superscript j denotes the jth iterate and ra is an acceleration parameter.This scheme converges rapidly for suitable choices of ra.

If the assumed pressure fields are discontinuous at the element boundaries, thismethod can be used at the element level to eliminate pressure variables (see Hughes,1987). In this event, the global equilibrium equation only involves displacementdegrees-of-freedom.

For each iteration, it is necessary to solve a linear system. Computation can beexpedited using a convenient version of the LU decomposition. Let L1 and L2 denotelower-triangular matrices arising in the following Cholesky decompositions:

K′MM = L1L1T I/ra + KT

MPL1L1TKMP = L2L2

T. (20.26)

Then, a triangularization is attained as

. (20.27)

Forward and backward substitution can now be exploited to solve the linearsystem arising in the incremental finite-element method.

K K

K

f

0

MM MP

MPT

M

−

=

=

0

d

d

dγγ

ψ

′

−

=

+K K

K I

fMM MP

MPT

a

jM

ja

d

d

d

d/ /,

ρ ρ

γγ

ψψ ψψ

1

′

−

=−

−

−K K

K I

L 0

K L L

L L K

0 L

MM MP

MPT

a MPT T

TMP

T T/ρ

1

1 2

1 11

2



Advanced Numerical Methods 263

20.3 EXERCISES

1. Examine the first two iterates for the matrices

2. Verify that the product and inverse of lower-triangular matrices are lower-triangular using

3. Verify the triangularization scheme in the matrix

Use the triangular factors to solve the equation

K

K K

=

= + +

+ + +

+ =

+

+

a

b c

d e f

a b d

c e

f

a ab ad

ab b c bd ce

ad bd ce d e f

a

b c

d e g f

a b d

c e

0 0

0 0

0 0

0 0

0 0

2

2 2

2 2 2

∆ gg

f

a ab ad

ab b c bd c e g

ad bd c e g d e g f

0 0

2

2 2

2 2 2

= + + +

+ + + + +

( )

( ) ( )

.

L L1 2

0 0=

=

a

b c

d

e f, .

A =

−

− −

−

2 1 1

1 2 1

1 1 0

.

Ab = −

2

1

1

.



265

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