Tensor Analysis and Continuum Mechanics 9781402010552, 1402010559

Citation preview

Tensor Analysis and Continuum Mechanics

Tensor Analysis and Continuum Mechanics by

Yves R. Talpaert Faculties of Science and Schools of Engineering at Algiers University, Algeria; Brussels University, Belgium; Bujumbura University, Burundi; Libreville University, Gabon; Lome University, Togo; Lubumbashi University, Zaire and Ouagadougou University, Burkina Faso

Springer-Science+Business Media, B.Y.

A C.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN 978-90-481-6190-4 ISBN 978-94-015-9988-7 (eBook) DOI 10.1007/978-94-015-9988-7

Printed on acid-free paper

All Rights Reserved © 2002 Springer Science+Business Media Dordrecht Originally published by Kluwer Academic Publishers in 2002. Softcover reprint of the hardcover 1st edition 2002

No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the Publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work.

CONTENTS

PREFACE ........................................................... xv

Chapter 1.

TENSORS

1

1.

FIRST STEPS WITH TENSORS ......................... .

1.1

Multilinear forms .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear mapping .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multilinear form ...........................................

1.2

Dual space, vectors and covectors . . . . . . . . . . . . . . . . . . . . . . . . . . . Dual space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Expression of a covector .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Einstein summation convention. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Change of basis and cobasis '" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.3

Tensors and tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

I 2 2 3 3 3 5

7

to

Tensor product of multilinear forms. . . . . . . . . . . . . . . . . . . . . . . . . . to Tensor of type (?) ......................................... II Tensor of type (~) ......................................... 12 Tensor of type Tensor of type

(g) .........................................

14

m......................................... 16

Tensor of type (:) ......................................... 18 Tensor of type

2. 2.1

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

22

OPERATIONS ON TENSORS ........ . . . . . . . . . . . .. .. . . . . .

25

Tensor algebra ........................................... 25 Addition of tensors ....................................... , Multiplication of a tensor by a scalar ..... . . . . . . . . . . . . . . . . . . . .. Tensor multiplication ......................................

2.2

20

Symmetric and antisymmetric tensors

25 25 26

Contraction and tensor criteria ..... . . . . . . . . . . . . . . . . . . . . . . . .. 27 Contraction ............................................... 27 Tensor criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 31 vii

viii

Contents 3.

EUCLIDEAN VECTOR SPACE ............................ 33

3.1

Pre-Euclidean vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 33 Scalar multiplication and pre-Euclidean space ................... 33 Fundamental tensor .. , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 34

3.2

Canonical isomorphism and conjugate tensor . . . . . . . . . . . . . . . .. 35 35

Canonical isomorphism ..................................... Conjugate tensor and reciprocal basis .... . . . . . . . . . . . . . . . . . . . . .. Covariant and contravariant representations of vectors ............ Representation of tensors of order 2 and contracted products .......

37 40

42

3.3

Euclidean vector spaces ............................... . . . . . 45

4.

EXTERIOR ALGEBRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 49

4.1

p-forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Definition of a p-form ...................................... Exterior product of I-forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Expression of a p-form ..................................... Exterior product of p-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Exterior algebra. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

49 49 50 52 56

57

4.2

q-vectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 59

5.

POINT SPACES ......................................... 63

5.1

Point space and natural frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 63 Point space ............................................... 63 Coordinate system and frame of reference ...................... 64 Natural frame ............................................. 66

5.2

Tensor fields and metric element .. . . . . . . . . . . . . . . . . . . . . . . . . .. 69 Transformations of curvilinear coordinates ..................... 69 Tensor fields .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Metric element .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

5.3

73 74

Christoffel symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 75 Definition of Christoffel symbols. . . . . . . . . . . . . . . . . . . . . . . . . . . .. 75 Ricci identities and Christoffel formulae ........................ 78

5.4

Absolute differential, Covariant derivative, Geodesic .. . . . . . . . . Absolute differential of a vector, covariant derivatives ............ Absolute differential of a tensor, covariant derivatives . . . . . . . . . . .. Geodesic and Euler's equations ..................... . . . . . . . . Absolute derivative of a vector (along a curve) ..................

5.5

5.6

Volume form and adjoint. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80 80 83 85 86

88 88

Volume form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Adjoint. .. .. . .. .... . .. .. . . ... . . .. .. . .. . ... . .. ..... .. .. .

91

Differential operators ....................................

92

Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Divergence .............................................. 99 Curl. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 101

Contents Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

ix 103

EXERCISES .................................................. 106

Chapter 2

LAGRANGIAN AND EULERIAN DESCRIPTIONS

147

1.

LAGRANGIAN DESCRIPTION ..........................

147

1.1

Configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

147

1.2

Deformation and Lagrangian Description ...................

148

1.3

Flow and hypotheses of continuity .... . . . . . . . . . . . . . . . . . . . . .

152

1.4

Trajectories. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

153

1.5

Streakline. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 154

1.6

Velocity and acceleration of a particle . . . . . . . . . . . . . . . . . . . . . .. 155

1.7

Abstract configuration .................................... 156

2.

EULERIAN DESCRIPTION ... . . . . . . . . . . . . . . . . . . . . . . . . . .. 157

2.1

Definition; Comparison between L- and E-descriptions ..... . .. 157

2.2

Trajectory and velocity ................... . . . . . . . . . . . . . . .. 158

2.3

Streamline .............................................. 161

2.4

Steady motion ........................................... 163

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 165

Chapter 3

DEFORMATIONS

171

1.

HOMOGENEOUS TRANSFORMATION .......... . . . . . . ..

172

1.1

Definition of homogeneous transformations ...... . . . . . . . . . ..

172

1.2

Convective transport .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Convective transport of a vector ............................ Convective transport of a volume . . . . . . . . . . . . . . . . . . . . . . . . . . .. Simple shear .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

174 174 174 176

1.3

Cauchy-Green deformation tensor and stretch ............... (Right) Cauchy-Green deformation tensor. . . . . . . . . . . . . . . . . . . . Stretch ................................................. Shear angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

177 177 180 181

x

Contents Principal stretches ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

182

1.4

Finite strain tensor .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

184

1.5

Polar decomposition ..................................... 186 186 Pure stretch and rotation .................................. Euler-Almansi strain tensor .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 190

1.6

Rigid body transformation ................................ 191

2.

TANGENTIAL HOMOGENEOUS TRANSFORMATION. . . .

193

2.1

Deformation gradient ....................................

193

2.2

Homogeneous transformations of elements .................. 196 Transport of vectors, volume deformation, and area deformation ... 196 Stretches ........ ....................................... 199 Strain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 201

2.3

Displacement and gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Material displacement gradient .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. Spatial displacement gradient .... . . . . . . . . . . . . . . . . . . . . . . . . . .. Curvilinear coordinate system ..............................

3.

INFINITESIMAL TRANSFORMATION . . . . . . . . . . . . . . . . . .. 2 I 0

3.1

Tensor notions relating to infinitesimal transformations ........ 2 I I

203 204 206 208

3.2

Compatibility conditions .................................. 216

3.3

Rigid body transformation ....... ......................... 220

EXERCISES ................................................... 222

Chapter 4

KINEMATICS OF CONTINUA

263

1.

LAGRANGIAN KINEMATICS

263

1.1

Homogeneous transformation motion ......................

264

1.2

General motion and gradient

266

2.

EULERIAN KINEMATICS ............................... 268

2.1

Homogeneous transformation motion ....................... 268 Velocity field ............................................. 268 Material derivative of a vector .... . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Material derivative of a volume .............................. 269 Eulerian rates ......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 271

2.2

General motion and velocity gradient . . . . . . . . . . . . . . . . . . . . . .. 274 Velocity gradient tensor and Eulerian rates ......... . . . . . . . . . .. 274

Contents

xi

Lagrangian and Eulerian strain tensors ... . . . . . . . . . . . . . . . . . . . .. 279 Rate of rotation .......................................... 280 Decomposition of motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 285 2.3

Rigid body motion ....................................... 286

3.

MATERIAL DERIVATIVES OF CmCULATION, FLUX, AND VOLUME .............................................. 287

3.1

About the particle derivative ... . . . . . . . . . . . . . . . . . . . . . . . . . .. Physical quantity ......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vector field ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Tensor field ...... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

3.2

Material derivative of circulation ........................... 293

3.3

Material derivative of flux .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

3.4

Material derivative of volume integral ..... . . . . . . . . . . . . . . . .. 299 Lagrangian and Eulerian approaches ......... . . . . . . . . . . . . . . . .. 299 Proper motion case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 303

287 287 290 292

296

EXERCISES .................................................. 305

Chapter 5

FUNDAMENTAL LAWS; PRINCIPLE OF VIRTUAL WORK

315

1.

CONSERVATION OF MASS AND CONTINUITY EQUATION 315

1.1

Axiom of mass conservation ..............................

315

1.2

Continuity equation ...................................... Continuity equation in the Lagrangian description ........ . . . . . .. Continuity equation in the Eulerian description ..... . . . . . . . . . . .. Mass flow rate ...........................................

316 316 317 319

1.3

Material derivative of integral of mass density . . . . . . . . . . . . . ..

320

1.4

Isochoric motion, steady and irrotational flows .............. Isochoric motion ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Steady flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Steady isochoric flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Irrotational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Isochoric irrotational flow ....... . . . . . . . . . . . . . . . . . . . . . . . . . ..

322 322 326 327 328 329

2.

FUNDAMENTAL LAWS OF DYNAMICS .................

330

2.1

Body forces and surface forces .. . . . . . . . . . . . . . . . . . . . . . . . . . .

330

2.2

Principles of linear momentum and moment of momentum. . . .. 333

2.3

Cauchy's stress tensor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 337

xii

Contents 2.4

Cauchy's stress tensor and principles of dynamics ............ Linear momentum principle and equilibrium equations .. . . . . . . . .. Moment of momentum principle ............................ The generalized Cauchy's theorem. . . . . . . . . . . . . . . . . . . . . . . . .. Poisson's theorem ...... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

342 342 344 346 347

3.

THEOREM OF KINETIC ENERGY ....................... 348

3.1

Theorem of kinetic energy in the Eulerian description ......... 348

3.2

Theorem of kinetic energy in the Lagrangian description

350

4.

STUDY OF STRESSES..................................

356

4.1

Reciprocity of stresses .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 356

4.2

Principal stresses ......................................... 358

4.3

Stress invariants; deviator .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

364

4.4

Stress quadric of Cauchy and Lame stress ellipsoid

370

4.5

Geometrical constructions and Mohr's circles ................ (Mohr's) stress plane ...................................... Stress vector and plane of Mohr . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Description of Mohr's circles ............................... Particular stresses .........................................

374 374 375 378 380

5.

PRINCIPLE OF VIRTUAL WORK ........................ 384

5.1

Preliminary recalls ....................................... 384

5.2

Rigid body motion ....................................... 386

5.3

Expressions of virtual power (and virtual work) .............. 389

5.4

Principle of virtual work ................................. , 391

6.

THERMOMECHANICS AND BALANCE EQUATIONS ....

393

6.1

Balance equation .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proper motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Material domain .......................................... Fixed domain ............................................

393 393 398 401

6.2

First principle of thermodynamics .......................... Principle ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Balance equations and local forms . . . . . . . . . . . . . . . . . . . . . . . . . .. Potential energy of body forces . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Internal energy and balance equation .........................

402 402 404 406 407

6.3

Second principle of thermodynamics ........... . . . . . . . . . . .. 409 Principle ............................................... 409 Clausius-Duhem inequality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 411 Dissipation and reversibility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 412

Contents 6.4

xiii

Conclusion and constitutive equations ....................... 414

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 417

Chapter 6

LINEAR ELASTICITY

455

1.

ELASTICITY AND TESTS ............................... 455

2.

GENERALIZED HOOKE'S LAW IN LINEAR ELASTICITY . 458

2.1

Generalized Hooke's law ............ . . . . . . . . . . . . . . . . . . . .. 458

2.2

Quadratic forms and strain energy function .................. 459

2.3

Isotropic material and Lame coefficients ..................... 463 Constitutive equations ...................................... 463 Young's modulus and Poisson's ratio ......................... 466 Bulk modulus ........................................... 469 Shear modulus ........................................... 471 Hooke's law's expression in a general coordinate system .......... 472 Navier's equations of motion ................................ 475

3.

EQUATIONS AND PRINCIPLES IN ELASTOSTATICS ..... 477

3.1

Navier's equation; the Beltrami equations of compatibility. . . .

478

3.2

Principle ofsuperposition .............. . . . . . . . . . . . . . . . . ..

481

3.3

Saint-Venant's principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 484

4.

CLASSICAL PROBLEMS .......... .. . . .. . .. . .. .. . .. . . ... 485

4.1

Plane problems .......................................... 485 Plane stress problems ................ . . . . . . . . . . . . . . . . . . . . .. 485 Plane strain problems ...................................... 488

4.2

Classical problems in elastostatics .......................... 491 Uniaxial stresses .......................................... 491 Torsion of a circular cylinder body ............................ 493 Torsion of cylindrical shafts ................................. 496

EXERCISES ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 501

SUMMARY OF FORMULAE .............................................

541

xiv

Contents

BIBLIOGRAPHY ......................................................... 575 GLOSSARY OF SYMBOLS ............................................... 577 INDEX .................................................................... 585

PREFACE

This book is designed for students in engineering, physics and mathematics. The material can be taught from the beginning of the third academic year. It could also be used for selfstudy, given its pedagogical structure and the numerous solved problems which prepare for modem physics and technology. One of the original aspects of this work is the development together of the basic theory of tensors and the foundations of continuum mechanics. Why two books in one? Firstly, Tensor Analysis provides a thorough introduction of intrinsic mathematical entities, called tensors, which is essential for continuum mechanics. This way of proceeding greatly unifies the various subjects. Only some basic knowledge of linear algebra is necessary to start out on the topic of tensors. The essence of the mathematical foundations is introduced in a practical way. Tensor developments are often too abstract, since they are either aimed at algebraists only, or too quickly applied to physicists and engineers. Here a good balance has been found which allows these extremes to be brought closer together. Though the exposition of tensor theory forms a subject in itself, it is viewed not only as an autonomous mathematical discipline, but as a preparation for theories of physics and engineering. More specifically, because this part of the work deals with tensors in general coordinates and not solely in Cartesian coordinates, it will greatly help with many different disciplines such as differential geometry, analytical mechanics, continuum mechanics, special relativity, general relativity, cosmology, electromagnetism, quantum mechanics, etc .. Secondly, the foundations of Continuum Mechanics constitute the most important part of this work. It involves chapters on the Lagrangian and Eulerian descriptions, deformations, kinematics of continua, fundamental laws, the principle of virtual work, and linear elasticity. These chapters lay the groundwork for other or more technical subjects as fluid mechanics, strength of materials, plasticity and viscoelasticity, thermoelasticity and thermodynamics, nonlinear continuum mechanics, finite element methods in continuum mechanics, etc .. xv

xvi

Preface

The reader will quickly discover the importance of these chapters and, given their controlled and logical progression as well as their role of introducing the above mentioned disciplines, will eagerly take up the challenge. Unlike other authors who denote tensors by (several) lines above and below bold letters, I simply represent both general tensors and vectors by bold letters. This simplification is possible since the reader gradually gains experience in dealing with tensors and because tensors of orders higher than 2 are not frequently used in continuum mechanics. In addition, the difference in level of operation dot '.' or '.' has no mathematical meaning, it has only a pedagogical value, namely: bringing down dots specifies that the corresponding operation results are real numbers. This stipulated simplification leads the reader to think carefully about operations between tensors and about the types of different tensors. Of course, scalars are not designated by bold letters. According to usage the vertical brackets completely enclose the elements of matrices, whereas they partly enclose the normal mathematical expressions. Terms of the continuum mechanics terminology are sometimes translated into French for French speaking readers. The important propositions and the formulae to be framed are shown by

W

and~.

The summary of formulae and glossary of symbols should make the assimilation of notions easier. All the proofs and the 95 solved exercises are described in detail.

Acknowledgements. Many thanks are due to my former students who let me expound a part of the material that resulted in this book.

I wish to express my gratitude to Kluwer Academic Publishers for their cooperation. I would particularly appreciate it if readers would let me know of any errors or further suggestions. Any faculty of science or engineering school, interested in my analysis of continuum mechanics and in the way differential geometry and theoretical mechanics have been developed in my previous books, can contact me for a possible collaboration.

Yves R. Talpaert [email protected]

CHAPTER 1

TENSORS

The idea of tensor took form at the end of the 19th century when it became necessary to express pressure forces in continua. But the first important developments of the notion of tensor date back to the very beginning of the 20th century; they are generally owed to Ricci, Levi-Civita, E. Cartan, .... The name tensor, introduced by the physicist Voigt, is reminiscent of tension in fluids, elastic solids, .... This terminology designates intrinsic mathematical entities which are suitable for the expression of the laws of mechanics regardless of the choice of coordinate system. The reader has already encountered tensors, since vectors and linear forms are examples of them. It is pointless insisting on the considerable importance that tensors have gained through

the developments of exact and applied sciences in the 20th century, more especially in continuum mechanics, special relativity, general relativity, quantum mechanics, differential geometry, Riemannian geometry, analytical mechanics, fluid dynamics, cosmology, electromagnetism and so on I.

1. FIRST STEPS WITH TENSORS 1.1

MUL TILINEAR FORMS

Let E and F be finite-dimensional real vector spaces.

I This chapter. is based on our books: Mecanique Analytique vol.2 (1982), Differential Geometry with Applications to Mechanics and Physics (2000), where tensors are considered in the manifold context, and Mechanics, Tensors and Virtual Works (2002), where tensors are applied to theoretical mechanics.

Y. R. Talpaert, Tensor Analysis and Continuum Mechanics © Springer Science+Business Media Dordrecht 2002

2

Chapter 1

1.1.1

Linear Mapping

D

A mapping g: E ---+ F: x is linear if Vx,y

E

E, Vk

H

g(x)

R:

E

g(x + y) = g(x) + g(y) ,

g(kx)

= k g(x).

Let us denote by L( E; F) the set of (continuous) linear mappings of E to F. D

The addition in L(E; F) is the mapping L(E;F)x L(E;F) ---+ L(E; F): (g,h)

H

g +h

such that the sum g + h is the linear mapping defined by E ---+ F : x D

H

(g + h)(x) = g(x) + h(x).

The multiplication of a linear mapping g of E into F by a scalar k is the mapping Rx L(E;F) ---+ L(E;F): (k,g)

H

kg

such that the product kg (of g by k) is the linear mapping defined by E ---+ F : (k, g)(x) = kg(x). We know that L(E; F) provided with the two previous laws of addition and multiplication has the structure of a vector space.

1.1.2

Multilinear Form

In mechanics we particularize F by choosing this vector space to be R. So we will consider the vector space L( E; R) later on. D

A linear form on E is a mapping f : E ---+ R: x such that Vx,y

E

H

f(x)

E, Vk E R:

f(x + y) = f(x) + f(y),

f(kx)

A linear form on E is also called a one-jorm or covector. Let

E(I) , ... , E(p)

be p vector spaces.

= kf(x).

3

Tensors D

A p-linear form defined on the Cartesian product of p spaces E(l) x ... x E(p) is a

mapping which is linear with respect to each vector, that is, \fx(l)'Y(l) EE(l), ... , \fx(p),Y(p) EE(p), \fkER: f(x(l)

+ Y(l),X(2)""'x(p») =

f(X(l),X(2)""'x(p»)

+ f(Y(l),X(2)""'x(p»)

Provided with laws of addition and multiplication by a scalar defined as before, the space L p (E; R) of p-linear forms on E has the structure of a vector space.

1.2

DUAL SPACE, VECTORS AND COVECTORS

1.2.1

Dual Space

D

The vector space of linear forms defined on E is called the dual space of E.

It is denoted by E* .

So, the dual space is a vector space the elements of which, called covectors, are linear functions E ~ R . It is a space of functions. Example 1. The row vectors are covectors (or l-forms). With the multiplication of matrices, a row vector (linearly) associates a real to each column vector. For instance:

Example 2. In quantum mechanics the I-forms called bras and denoted associate complexes

1.2.2

(¢, 'If)

to vectors called kets and denoted I'If) .

Expression of a Covector Let E be a real vector space of dimension n.

(¢ I

linearly

4

Chapter 1

A covector on E is a linear mapping f: E vector xEE.

~

R which associates a real f(x) to each

We denote by xl, ... ,xn the components ofx with respect to a basis (el, ... ,en ) of E. The real f(x) is written: f(x) = f(xle l + ... + xne n ) = xlf(e l ) + ... + xn f(e n ).

By letting we have n

f(x) =

IJ; xi. i=l

We mention that the image of x under f is sometimes called the value of the form (for x). We are now going to express the covector f with respect to the dual basis. The dual basis (e*l, ... ,e*n) of the basis (el, ... ,en ) is such that: (1-1) where 15~ is the Kronecker delta l and the n linear forms making up the dual basis are e*i :E~R:xHe*i(x)=xi.

Thus

\;:fx E E : n

f(x)

= IJ; e*i (x), i=l

which leads to the expression of the covector n

f

=

IJ; e*i.

(1-2)

i=l

Remark. The reader will compare the previous expression with that of a vector: n

x=Ixie i · i=l

I

The Kronecker delta is the symbol

I

15; = 15ij = 15ij = { 0

if

if

i = j,

i *- j.

5

Tensors

So according to usage the components of vectors show an upper index and the components of covectors a lower index. Notation. Generally we will represent the covectors (or linear forms) by Greek characters, and since they are the elements of a vector space, namely E* , we have decided to write them in bold characters.

1.2.3

Einstein Summation Convention

The Einstein summation convention consists in removing the summation sign more precisely:

~,

Summation is implied when an index is repeated on upper and lower levels 1• For example, we denote:

n

aUkxiyizk

n

n

= I I I aljkx'ylzk i=l i=l k=l

On the one hand, any repeated index of summation is called a dummy index because it does not matter what the letter is; for instance:

brsbllb12 b ijX i x i= rsX X = llX X + 12X X

+

In particular, the Kronecker symbol is such that

0/1

=

011 +

° + ... 22

=

1+ 1+ ...

Attention! In the summation convention we emphasize that an index of summation shall never be repeated more than once.

1 The Einstein convention will also be used with indices at the same height in the frame of the usual Euclidean space when considering orthononnal bases.

6

Chapter 1

So there is no question of writing: j aJbx , J biiCjX Y i

(never!), J

(never!).

In the same manner, given

Q = aijxiyj with yj =b(x',

if we want to express Q as a function of the coordinates Xi only, then there is no question of writing: (never!), but we must firstly change the dummy index in yJ , for instance: yj = b£x k in order to obtain the following right expression: j i k Q -_ aij bkX x .

In the same manner, if we want to multiply aix i and biyi we must first rename the dummy index of one term of the product and write aib jX i yj . On the other hand, there is another type of index. An index which appears once in each expression is called afree index. So, for instance, the equation i = 1,2; j = 1,2,3,4

represents the following system of equations: Y

I

= blk X k ,

y2 = b;x'

that is, explicitly: l l Y I = bI XI + b2X2 + bl3 X3 + bl4X4 , y2 =b12XI +bix 2 +bix 3 +b1x4.

In this example the index i is free and the indexj is dummy. Remark 1. The reader must take care to write the equations correctly. Any free introduced index must appear in every term.

So the following equations are

7

Tensors meaningful:

Meaningless:

(never!)

Zj=Xj+Yj'

(never!)

a pq

-

bpC;. ' q

Remark 2. Factoring is possible with the Kronecker symbol. For instance, given an orthonormal basis of the Euclidean space, by viewing n; = liijnj (with summation over}) the equation is written:

1.2.4

Change of Basis and Cobasis

Let (e 1 ,e2) be a basis of a 2-dimensional vector space called the unprimed basis and (e;, e;) be another basis called the primed basis to make matters simpler. The following change of basis e; == a~e1 +a 12e 2 e; == a;e 1 + a~e2

is written in a condensed manner:

where

a = (a~) and {3 = ({3,k ) are inverse matrices and thus a~{3; = Ii;.

The equalities

imply by comparison: I

I,]

2

2,]

X ==alx

I

+a 2x

X ==alx

/2

,

2 /2

+a 2 x

In the same manner the equalities X==X

,]

/

/2

/

I

2

e l +x e 2 =X e l +X e 2

= X 1({3I' lei

2({31' {32/) + {32/) I e2 + x 2e l + 2 e 2

8

Chapter 1

imply by comparison

We generalize, and in an n-dimensional space E we say: PRI

The matrix associated with the expression of unprimed components as functions of those primed is the transpose of the matrix associated with the expression of primed basis vectors as functions of the unprimed.

Proof The matrices

(al) and (p~) being inverse, we denote (1-3) (1-4)

The equations imply (1-5) By comparing the following explicit expressions I

I,]

1,2

X =alx +a 2x

+ ... ,

and so on, the proposition is thus proved. PR2

The matrix associated with the expression of primed components as a function of the unprimed components is the transpose of the matrix associated with the expression of unprimed basis vectors as functions of the primed. It is the inverse and transpose of the matrix associated with the expression of primed basis vectors as functions of the unprimed.

Proof Since

the equalities x=x'Je }'. =x'e.1 =xiRJe'. I-'I}

imply

x'} = Plx ' . By comparing the following explicit expressions X

,]

el

1/1

= 1-'1 X 1/ 1 ,

= 1-'1 e l

I

1/1 2 X + ... , + 1-'2

1/ 2 , + 1-'1 e 2 + ...

(1-6)

Tensors

9

and so on, the proposition is thus proved. Weare now going to show the formulae of the change of dual bases (a dual basis is also called a cobasis). From every

because e'*J(ek)=oj we deduce:

= x'J = f3! Xi = f3( e*i (x)

e'*J (x)

which implies e'*J =f3!e*i

(1-7)

and (1-8) The reader will easily say the propositions which refer to (1-7) and (1-8). Example. Primed axes being obtained from a 90° direct rotation! about the e 2 -axis, express the vector 4e 3 with respect to the primed basis ( e;). The 'primed' basis vectors e~

= a~el are explicitly

and the 'unprimed' vectors e i

= f3!e~

[e l

::

]

[

=

0

0

are explicitly:

1][e; 1

~ 1 ~ ~ :~

.

From PR2 we deduce:

1

A direct rotation is also called a counterclockwise rotation.

10

Chapter 1

The given vector (such that

x3

= 4)

This result is obvious because e3

1.3

is written - 4ei since xrl

= -4,

x,2

= X,3 = o.

= -ei.

TENSORS AND TENSOR PRODUCT Let E(I)' ... ,E(p), ... , E(p+q) (or simply E) be finite-dimensional vector spaces.

1.3.1

Tensor Product of Multilinear Forms

Let / be a p-linear form defined by E(I)x ... x

E(p) ~ R: (x(l)' ... 'x(P»

H

f(x(I)' ... 'x(P»'

let h be a q-linear form defined by

D

r:F

The tensor product of a p-linear form/ and a q-linear form h is the (p+q)-linear form denoted / ® h : E(I) x ... x E(p+q) ~ R : (x(l) ,... , x(p+q»

H

/

® h (x(l) ,... , x(p+q»

such that

Example 1. If we consider two linear forms/and h defined on E, namely: /:E ~ R:XH lex), h :E

~

R : y H hey),

then the tensor product of linear forms/and h is the bilinear form /®h:ExE ~R:(X,Y)H /®h(x,y) =/(x) hey).

Given a basis (e;) of E, the linear forms/and h have the respective values: lex) = J; Xi = J; e Oi (X), hey) = hj yj = hj e Oj (y)

and the corresponding value of the tensor product is / ® h(x,y) = lex) hey) = J; hj Xi yj .

\1

Tensors Example 2. The tensor product of two bilinear forms f : E(l)x E(2) ~

R: (X(l)'X(2)) H

h: E(3) X E(4) ~

R : (X(3),X(4)) H

f(X(I),X(2)) ' h(X(3)'X(4))

is the following quadrilinear form

1.3.2

Tensor of Type (~)

D

A tensor of type (~) or covector is a linear form defined on E,

It is an element of the vector space E' ,

According to usage the covectors are generally denoted by Greek letters; for instance: wEE', Thus the definition of a covector or tensor of type (~) is expressed as follows: w: E

with \::fa,b

E

R, \::fx,y

E

~ R:XHW(X)

E:

w(ax + by)

= aw(x) + bw(y) ,

The co vector expressed as (1-2) is written: where

The image of any vector x under w is the real w(x) = OJie*i (xie i) = OJixie*i (e i)

this value being also denoted by (w,X)

= w(X) ,

(1-10)

12

Chapter 1

Change of basis We recall that a linear form m behaves towards any vector x in the following way: w(x) = w(x' e,) = w(x'j

(distributivity) (mixed associativity) x=O.

The real x.y is called the scalar product of x andy. A vector space E provided with the scalar multiplication is called a pre-Euclidean vector space.

D

We note that the first three properties show a symmetric bilinear form since it is linear with respect to every element of any pair of vectors and is commutative. The fourth property shows this form is nondegenerate.

3.1.2

Fundamental Tensor

PRIO The scalar multiplication on E defines the nondegenerate symmetric bilinear form g: ExE

that is, a tensor of type D

e j(X'Yb) = UiW)XiYj = (ub,X)(W'Yb) = (U b ·x)(w· Yb)' Xb . 13 , Y

= 13 (Xb,y) = U 18> Wb(Xb,Y) = uiwje i 18> e*j (Xb,y) =UiWjXiy j =(u,Xb)(Wb'Y)=(U'Xb)(W b .y),

Xb ·/4' Yb = 14(X b'Yb) = ul8> W(Xb'Yb) = uiwje i l8>e j (x b'Yb) = uiw j XiYj = (U'X b)(W'Yb) = (u· Xb)(W' Yb)'

Since we have u b . x = U· Xb and w b ' Y = W· Yb' the four previous reals are equal and we denote the common value by t(x,y)=x·t·y,

that is, (1-49) where u b and x are tensors of order 1 of opposite variances, obviously, and so are Wb and

y. If we now consider a tensor of type (:), for example U 18> WbEE 18> E* , we have Vy E E :

(I-50) sInce

[likewise for u b ® W E E* ® E]. From the definition of the contracted multiplication and if we consider vectors and co vectors a, b, c, d such that the following operations are possible, then the reader will immediately prove: (a ® b) . (c 18> d) = (b.c) a 18> d

(I-51 )

and the double contraction: (a ® b) : (c ® d)

= (b.c)(a.d) .

(I-52)

44

Chapter 1

(ii) Second, we view the general tensors of order 2 by considering the canonical isomorphism between E and E*. The generalized expressions of the previous tensors I\, 12, 13 and 14 are obviously: I I -- t ije *i!O. '

3 0 0] [

S=020,

001

001

[0-I 0]

R = FS-l = I 0 0 , 001

The vectors of the principal basis are

By considering C (= S2) we deduce that the squared principal stretches are (~F

=9,

Of course, these values correspond to the following

Thus the principal stretches are ~=3,

The principal strains Li

= teA;

~=2,

A:J=1.

-I) are the following eigenvalues of L

= t(C - /):

229

Deformations

The principal unit elongations are

Exercise 3.

Let us consider the transformation defined by XI

=X2

X 2 =-XI ,

,

X3

=2X3

.

(i) Calculate the deformed volume of a cube with unit edges (10-2 m) in the first octant. (ii) Find the deformed area vectors of the left, front and lower faces. (iii) Express and characterize the right pure stretch tensor S and the rotation tensor R

which decompose F. Answer. (i) From

[ 0 1 OJ

F= -1 0 0 ,

o

0 2

we deduce that J=detF=2.

Thus the deformed volume is equal to J.1

= 2J.1o = 2

( 10-6 m3 ).

(ii) The unit vector corresponding to the face of the coordinate plane

OXI X3

is -e2 and

SInce

[0-1 0:

F-I= 1 0

o

0

,

0 112

then by recalling (3-58) the area vector of this face is expressed as

a = a (-e 2 ) =

2[~1 ~ ~]A [~11 -2e =

o

0

V2

l (l0-4

m2),

0

since J = 2 and A =1. Likewise we find that the area vector of the front face (corresponding to el ) is a = -2 e2 and that of the lower face (corresponding to -e3 ) is a = -e3 .

230

Chapter 3

(iii) By calculating

[100]

S2=C= IF-F= 0 1 0 ,

o0

4

we remark that there is a pure stretch Il.:J =2 in the direction of e3 • The tensor R=F-S-I

[-10010]0

=

o0

1

is associated with a rotation of 90° about e3 •

Exercise 4.

We consider the following transformation between two configurations Co and CI referred to the same frame of orthonormal basis (ei ):

where the dimensions of the constants k and / are respectively [L-1r l ] and [L]. (i) Find the positions of particles initially located at a o (0,0,0), bo (0,0,-/), Co (0,0,-1/2)

and do (0, I ,-1/2), at given time t. Determine the positions of the straight segments Clabo and codo at time t. (U) Let us consider a pair of neighboring points (Po,qo) such that Po is (0,X2,X3) and qo

is

(0,X2+dX 2,X3+dX3).

In

Co this pair of points defines the vector

dX = dX 2 e 2 + dX 3 e3. Calculate the transported vector dx at time t and deduce the

squared norm ds 2 • Verify this last result by using the Cauchy-Green deformation tensor. (iii) In a configuration Cr let us denote by

qi the coordinates of a point of

coordinates Xi in the initial configuration Co. Express the coordinates of this point at time t by referring to the configuration Cr.

231

Deformations

Answer. (i) The positions of the particles at Go and bo remain unaltered while the particle located at Co occupies the position c(O,kI 2t/4,-1/2) at time t and the particle at do occupies the position d(O,l + kI 2t/4,-1/2). The material segment Gobo is the set of particles whose Lagrangian variables are

We know that the material coordinates Xi of a particle remain unchanged (unlike the spatial coordinates x} Since

the segment Gb, which is the transformed segment of Gobo at time t, is then defined by

V x3 E [-1,0]:

XI=O,

x 2 +ktx3(x 3+l)=0.

Thus the segment Gobo becomes at time t this delimited parabola in the plane

Xl

=0.

Likewise the material segment codo is the set of particles whose Lagrangian variables are

By introducing the spatial coordinates as previously and since 0 S;X2 S; 1is equivalent to

kl 2t/4 S; x 2S; 1+k12 t/4, then the segment cd, which is the transformed segment of codo at time t, is defined by

Thus the segment codo becomes at time t the straight segment cd of unit length along the straight of equations

Xl

=0, x3 =-t; this is a translation.

(ii) The deformation gradient F is defined by

o o

232

Chapter 3

The components of the transported vector dx = F ·dX are expressed as dx ' that is:

[~ Thus the squared norm of this vector is

The Cauchy-Green deformation tensor C = IFF is explicitly

The squared norm of dx is 2

.

ds = dx.dx = Cij dX ' dX J = dX.C ·dX , that is explicitly:

which is actually the obtained value of ds 2 . (iii) From

we deduce: and thus Xl

=X1 =';1,

X2

= X2 - ktX\X 3 +/) =

X3

= X3 =';3.

e +k(T - t).; \.;

3

+/),

= F; dX J ,

233

Deformations

Exercise 5.

Given an orthononnal Cartesian basis let us consider the following transfonnation

between two configurations Co and CI and where the dimension of a ER+ is [L-Il(i) Depict the defonnation_ (ii) Find the stretches in the directions of coordinate axes. (iii) Find the infinitesimal strain tensor

6 and the unit elongations for the elements which were initially in the directions of basis vectors.

(iv) Give an estimate of the shear angle between orthogonal directions defined by e2 and

e3' at point (0,0,2) originally, if a =10- 5 • Answer. (i) The displacement components are

There is no displacement in the directions of the axes oXI and oX3 . Every point of the axis oX2 is fixed, but any material straight interval of the axis oX3 transfonned into a parabolic line segment, that is, { XI = 0, X 2= a (X3)2 }. Also, a material line {XI = 0, X 2 = a} becomes the parabola {XI =0, X 2= a+a (X3)2 }.

'>-_---_...1.-_ _ _ _ _

el

Xl Fig. 21

e2

X2

IS

234

Chapter 3

(ii) The displacement gradient is

vu=[~

0

2U:' 1

0 0

and, from F= 1+ V U , we deduce: 0

dx, =FdX,

=f:

2~X' J[~' J= e, , &,

0

0

dx,

dx,

=FdX, =f:

l:

0

2a:,lE}&,e"

0

2a:,

=FdX, =

0

Hl, 1=(m

x,

Thus the stretches are

(iii) The infinitesimal strain tensor is

The unit elongation for the element which was in the direction of e] is

and likewise

e, +e,)&,.

235

Deformations

Of course, for very small deformations we have o(e3) == 2a 2(X3)2

which is confirmed by

(iv) The required shear angle is () == 2e(e2 ,e3 )

= 2623 = 2a X 3 •

For instance, if a=10-5 and X3 =2 then we obtain () == 4xlO-5 •

Exercise 6. Given an orthonormal Cartesian basis let us consider the following transformation

between two configurations Co and Ct. (i) Express the tensor F of this homogeneous transformation. (ii) Determine the transported vector of any vector, the transport of a volume, and the

transport of an area vector. (iii) Find the principal directions of the Cauchy-Green deformation tensor and the

principal stretches. (iv) Find the principal directions of the Green-Lagrange strain tensor and the principal strains. (v) Give the condition of existence of infinitesimal transformations and express the

infinitesimal strain tensor. Answer. (i) The tensor F is expressed as follows: F = (1 + kt)e 1 ®e1 +e 2 ®e 2 +e 3 ®e 3 • (ii) The transported vector of V =

Vi e) + V2 e2+ VJ e3 l+kt

v=F·V=

f

o

~

o

is

236

Chapter 3

The convective transport of a volume Po is PI

=(l+kt)po

0

The transported vector of the unit area vector IN = Nl e 1 + N2 e 2+ N 3 e 3 is the vector

o o

~lEl

which leads to

Since the norm of the transported area vector n is the previous square root, we deduce that every area A is transformed into the following

a=A~ NI2+(l+kt)2(N;+N;)

0

(iii) The Cauchy-Green deformation tensor is expressed as follows: C= IF of= (l+kt)2e) ®e, +e 2 ®e 2 +e3®e 3

0

The eigenvalues of

c=

0 01

(l+ktf 0 1 0

r

o

0 1

An eigenvector W corresponding to the first eigenvalue is such that

C oW=(l+kt)2W,

thus a principal direction is defined bye)

0

For the double eigenvalue we recall that every direction perpendicular to e1 is principal. From

we deduce that the principal stretches are

237

Deformations (iv) The Green-Lagrange strain tensor is expressed as follows: L

1

et

2

2

2

= -(C -J) = (kt+-)e ®e j

j

•

The principal directions of L are those of C. The principal strains are

~] =kt+ k2t2 2

[ that is, ~] =

t (C

ll

-1) ]

and

(v) The required condition

[[vult «1

is equivalent to the following

[kt[« 1 and thus, by considering L, we obtain:

e=kte1 ®e1

•

Exercise 7. In the classic example of the simple shear deformation between two configurations

Co and C2 defined by [see (3-4)] we know that the deformation gradient is such that:

12k 0 F= [ 0 I 0 ,

1

o

0 I

also written:

F= e j 0e 1 + 2ke 1 ®e 2 + e 2 ®e 2 + e3 ®e 3 • We also know that the volume per unit original volume is J =detF=l. (i)

Is the simple shear a homogeneous transformation?

(ii)

Find the transported vectors ofthe vectors of the orthonormal Cartesian basis (e;). Show in a figure the transformation of material lines along the respective vectors e] and e2 whose origin Po belongs to the plane ox] x 2 •

238

Chapter 3

(iii)

Determine the Cauchy-Green deformation tensor C.

(iv)

What are the stretches which were in the directions of basis vectors and in the direction of the bisector of the angle X I OX2 • Determine the corresponding unit elongations.

(v)

Calculate the shear angle for initial orthogonal directions defined by el and e2'

(vi)

Find the Green-Lagrange strain tensor L and the strains along the directions of the basis vectors.

(vii) Find the principal stretches, the principal strains, and the transported vectors of the

eigenvectors E, of L. Determine the principal basis (if;) and the direct principal basis

(i().

(viii) From the right pure stretch tensor S with respect to the principal basis (if;) express

the rotation tensor R with respect to (e i ). If k = 1 calculate the rotation tensor and the transported vectors of the direct principal basis (X;) . Characterize the polar decomposition. (ix)

Calculate the left Cauchy-Green deformation tensor eLand the Euler-Almansi strain tensor e. How does e22 (for instance) characterize a material line segment?

(x)

In the case of very small deformations, that is, for k 0):

In,,;e,). 3

tjit)

=

l=!

Thus the stress vector t(ei ) as follows:

t(it)

= t(n,e,)

t(it)

=

on the inclined face is expressed from the stress vectors

3

I

nj

(5-46)

t(e;) .

j~l

We set t(e')=a .. Jl

where

aJi

e j'

(5-47)

represents the jth component of the stress vector on the face whose outward

normal vector is e j , that is: t j(e i ) =a ji'

(5-48)

Thus (5-46) is written:

with summation over i and j. Thus the components

0 the flux is outward, which leads to a decrease of Q.

Third, besides the convection and diffusion, a change of Q can be caused by a source of Q within jj( which can produce or remove a part of Q with regard to jj( . The gained or lost part of Q due to the source is denoted by Qs =

1- pq s df1.

(5-149)

D,

In conclusion, this primary analysis leads to the next balance equation: (5-150a)

that is

~ r

d t !s,

r_

r

pqdf1=-r_ pqVr.darpd.da+ pq s df1. JoD, ,l,Dt !s,

(5-150b)

Remark. If we put together the convection and diffusion terms to define the total flux vector rp=rpc+rpd'

then the balance equation is written:

~ r

d t !s,

I

In French

pqdf1 =-

r_ rp.da + !s,r pq s df1.

JoD,

(5-151)

d is called the flux de diffusion and rpd is the vecteur densite surfacique de flux diffusif.

396

Chapter 5

Example. In the particular case in which the quantity Q is the mass M of expressed as follows:

D"

it is

1, pqdJl,

M =

which means that p q is the element of mass per unit volume or density p, thus q = 1 . The mass conservation law expresses that the mass of proper motion and there is no source within

qs

=

15"

15,

remains constant during the

that is:

O.

In consequence the balance equation (5-150b) is reduced to Eq. (5-14):

~ r pdJl+ JaD, r_ pVr.da=O,

dt 1,

which proves that the exchanges of mass through the boundary mass flow rate (PR5).

a15t

are opposed to the

(ii) Vector quantity

We assume there exists a vector quantity associated with by Q(t)

=

15,

and defined, at time t,

1 p(x,t)q(x,t) dJi. I[),

First, the change of Q can be caused by the motion of the domain material medium.

15,

with respect to the

In this case of convection we say: D

The convection jlux of Q through the boundary

a15t

is the vector surface integral (5-152)

where 'Pc = pq ® Vr is said to be the convectionjlux tensor. This convection flux tensor is also written: (5-153) Thus the rate of change of Q owed to convection is written:

397

Fundamental Laws; Principle of Virtual Work

~ .b

dt

pq dp = -cI> c

I

=-

r _ pq (V,. da) .

.l3D,

Second, the change of Q can be caused by particles passing through the boundary on both sides independently of the motion of D[. In this case of diffusion we say: D

aD[

The diffusion flux of Q through the boundary

is the vector surface integral (5-154)

where fPd(x,t) is said to be the diffusionflux tensor. Third, besides the convection and diffusion a change of Q can be caused by a source of Q within Dt which can produce or remove a part of Q with regard to D[. The gained or lost part due to the source is denoted by (5-155)

In conclusion, this primary analysis leads to the next balance equation: d

=-Q = -cI>c -cI>d + Qs,

(5-156a)

dt

that is

~ .v,r pq dp = - .l3D, r_ pq(V,. da) - .l3D, r_ fPd' da + ob,r pqs dp.

dt

(5-156b)

Remark. If we put together the convection and diffusion terms to define the total flux tensor fP = fPc +fPd'

then the balance equation is written:

~ r

dt .fB,

pqdp=-

r_

.l3D,

fP.da+

r

.fB,

pqsdp.

(5-157)

398

Chapter 5

Example. In the particular case in which the quantity Q associated with 15(, it is expressed as follows:

IS

the linear momentum

en=1, pVdp,

which means that the vector quantity q is V. Thus the convection flux of the linear momentum en is

(5-158)

6.1.2

Material Domain In the case of a material domain D( we know that Vr = 0 (V = W) and the

· · =d·IS the matena ·ld · · -d . denvatlve envatlve dt dt (i) Scalar quantity

From above we deduce that the balance equation (5-150b) becomes in the case of a material domain: (5-159a) It is also [because (5-19b)]:

1,pddqt dp =-1

D,

'Pd· da +

1 pqs dp ,

(5-159b)

and [because (4-67)]:

i , p(aatq + V.grad q)dp = - ./aD"f 'Pd· da + 1 pqs dp

(5-159c)

f (!!...-(pq)+pqdivV)dp =- f 'Pd. da + f pqs dp JaD, 1, 1, dt

(5-159d)

and [because (4-101 )]:

and [because (4-105)]:

f

1,

~(pq)dp+ f pqV.da=_f 'Pd. da + f pqsdp. at ./aD, ./aD, 1,

(5-15ge)

Fundamental Laws; Principle of Virtual Work Remark. The second tenn

1aD, pqV.da

399

does not represent the convection flux of Q

through the boundary since Vr = 0 implies that the convection flux vanishes; but we can say this second tenn represents the convection flux through the boundary of a fixed domain D which coincides with D t at time t. Example 1. If Q is the mass of a material domain we find again the law of mass conservation whose expression is a balance equation where q = I,

qJd

= 0,

qs

=

0,

that is: dfl=o, !!....lp dt ,

or

r

1,

(d p + pdivV) dfl = 0, dt

or

Example 2. We recall that the theorem of kinetic energy has the following expression: d

V2

-dt ob,r p2- dfl = JD,r f. V dfl

+

r

JaD,

(17' V).n da -

r

b,

17: d dfl.

In the particular case in which Q is the kinetic energy T, the previous equation is actually a balance equation of type (S-IS9a), where 1

qs =-(f.V-u:d), p but we note that the source proceeds from internal forces and body forces too. (ii) Vector quantity

For a material domain Dt , the (vector) balance equation (5-IS6b) becomes

!!.... J p q d fl = - JoD,r qJd • da dt If),

+

r

.b,

p qs d fl ,

(S-160a)

or (5-160b)

400

Chapter 5

or (5-l60c) or

r (~(pq) + pqdivV) df-L = - LD" 'I'd' da + 1 pq, df-L,

(5-160d)

r ~(pq)df-L+ r pq(V.n)da =- JiJD, r 'I'd ·da + Jo,r at lJD,

(5-l60e)

Jo, dt or

Jo,

pq,df-L.

Example. We recall that the equation of the linear momentum principle is

~ 1pVdf-L= dt ,

L u·nda+ 1fdf-L. D,

,

In the particular case in which the vector quantity Q is the linear momentum ~, q is equal to V and the previous equation is really a balance equation of type (5-l60a) where

q,

q=V,

f

=-.

p

(iii) Local formulation

We can formulate a balance equation locally in various ways. Indeed, from various Eqs. (5-159) and by converting the surface integrals into volume integrals for any material domain, we obtain: dq . p - = - dlVrpd + P q s ' dt

(5-l60a)

p(a q + V.gradq) = - divrpd + pqs'

(5-l60b)

~(pq)+pqdivV=-divrpd+

(5-l60c)

at

dt

~(pq)+div(pqV) =

at

pqs'

-divrpd + pqs'

(5-l60d)

Likewise, from various Eqs. (5-160), we obtain: dq d' Pdi=- IVrpd+Pqs'

(5-l6la)

401

Fundamental Laws; Principle of Virtual Work

p(a q + gradq ·V) = -div'Pd + pqs' at

(5-161b)

~(pq) + pqdivV = -div'Pd + pqs '

(5-161c)

~(pq)+div(pq®V) = -div'Pd + pqs.

(5-161d)

dt

at

6.1.3

Fixed Domain For any fixed domain D we know that W = 0 [See Section 3.4.2 of Chapter 4].

(i) Scalar quantity Since the domain is independent of time and if the quantities under the integral signs fulfill conditions which allow the permutation of derivative and integral signs, the balance equation (5-150b) becomes:

1:/pq)dP

= - LDpqV.da - LD 'Pd· da+

1pqs dp.

Unlike the case of any material domain, for a fixed domain the term

r

JeD

(5-162)

pqV.da really

represents the convection flux through aD. (ii) Vector quantity

In the same manner, given a fixed domain D the vector balance equation (5-156b) becomes: f ~(pq)dp = - r pq(V.da) - r 'Pd ·da+ r pqsdp, .b at JeD JeD JD

(5-163a)

or equivalently (5-163b)

Example. The mass conservation, the linear momentum principle, and the theorem of kinetic energy can be expressed from a balance equation given a fixed clomain. With regard to the mass conservation, the balance equation (5-162), such that q = 1,

'Pd = 0,

qs = 0,

Chapter 5

402

becomes rapdfl+r

.In

at

JaD

pV.da=O,

which actually expresses the mass conservation for a fixed domain D and where the second integral is the flow rate through aD . As regards the linear momentum principle, the balance equation (5-163a) such that q=V,

'Pd

= -rr,

becomes

r ~(pV)dfl=- ~ r pV(Vfz)da+ ~ r rr·fzda+ .Inr fdfl.

~&

This is actually the Euler formulation (5-39) of the linear momentum principle. Finally, with regards to the theorem of kinetic energy, the balance equation (5-162), such that 'Pd = -rr' V,

1 q, = -(f. V - rr : d),

p

becomes

a v2 v2 r -(p-)dfl + r (rr· V).da + r f. V dfl - r rr: d dfl . at 2 = - JraD p-V.da 2 JiJD JD JD

.In

6.2 FIRST PRINCIPLE OF THERMODYNAMICS Until now only mechanical quantities have been considered. But we know that mechanical changes bring about thermal changes and conversely (chemical and electromagnetic energies are not considered in this book). In other words, the theorem of kinetic energy involves mechanical quantities, but we know that non-mechanical energies must be considered. More precisely, we are going to obtain a balance equation for a continuum studied in thermo mechanics .

Let D t be a material domain included in a material system. 6.2.1

Principle

The first principle of thermodynamics postulates the convertibility between mechanics and thermal energies, more precisely it brings into play:

Fundamental Laws; Principle of Virtual Work (i)

The kinetic energy of the material inside D[:

T= (ii)

V2

r p-dJl, JD, 2

(5-76)

The (mechanical) power of externalforces exerted on D[: 9>(e)

=

1 f.V dJl+ L t . V da, (n)

(5-74)

D,

I

(iii)

403

The rate of increase of total heat l into D[ :

Q=-

r

!3D,

q.da+

r

ob,

rdJl,

(5-164)

(the minus sign because;' is outward). This so called rate is more exactly the rate of change of heat into D t

•

The first term making up Q is the part of the rate owed to surface actions. It is defined by q , which is called the heat flux vector per unit area (at t) by conduction following from diffusion processes through aD[ or simply: D

The vector q is said to be the heat flux vecto?

The quantity - q . Ii under the integral sign is the rate of increase of heat per unit area, at t, by conduction (short range actions) The second term making up Q is due to long range actions gained by D[ , it is defined by r. D

The quantity r is called the radiant heat constant per unit volume (at t).3

A last term must be considered in the first principle. Indeed, besides external supplies which may alter the kinetic energy (as the motion of particles of a boiling liquid), there are those which can modify the energy of the material system without changing the kinetic energy. Thus we introduce:

(iv)

The internal energy of D[: E(i)

= =

i

pu dJl

(5-165a)

I

L

udm.

(5-165b)

'MDI

I

Called in French: 'Taux de chaleur reryue " more exactly: 'Taux de chaleur echangee'.

2

Called in French: 'Vecteur densite surfacique du taux de chaleur reryue' or 'Vecteur courant de chaleur'. Called in French: 'Densite volumique de taux de chaleur fournie a D, par des sources exterieures '.

3

Chapter 5

404

D

The internal energy per unit volume pu is called the strain energy function l • The internal energy per unit mass u is called the specific internal energy or the strain energy per unit mas;'.

The balance equation of energy or first principle of thermodynamics is the following PR46 At every time, the material derivative of the (kinetic and internal) energy for the material domain Dr included in a material system is the sum of the power of external forces exerted on Dr and the rate of increase of total heat:

d -(T+ E(i)) = ~e) +Q.

(5-166)

dt

Remark. Before continuing, let us give dimensions and units: [Q]

=[ML2r3]

[E(il]

=[Mer2]

(W),

[q]=[Mr 3 ] (Wm-2 ),

[r]=[ML- l r

(J),

[u]=[er 2 ] (Jkg- l ),

[pu] = [ML- l r

3] 2]

(Wm-3 ), (Jm-3 ).

By considering Eq. (5-164), that is Q=

1, (r- divq) dj.J,

we say: D

The evolution of Dt is adiabatic if there is no heat loss or gain for Dt , that is, if Q=O.

In particular, an evolution such that q

6.2.2

=0

and r

=0

is adiabatic.

Balance Equations and Local Forms The first principle of thermodynamics is expressed as

8V'

I

2

d V2 -1 p(-+u)dj.J=-L dt, 2

D,

q.da+ L t(nl.Vda+

Called in French: 'Energie interne volumique '. Called in French: 'Energie interne massique '.

D,

1f. Vdj.J+ 1rdj.J. ,

,

(5-167)

Fundamental Laws; Principle of Virtual Work

405

Of course, it is a balance equation such that

V2 q=-+u 2

since, given the symmetry of

0"

1

qs =-(f. V +r ),

'Pd =q-O"'V,

'

p

we recall that

(0"' n).v = (0"' V).n

The time rate in the previous principle equation can be obviously written: d V2 .b, p-(-+u)dp dt 2

or

av

v

2

2

f p(- ( - + u) + V.grad(- + u)) dp .h, at 2 2 or

a v v f p(-+u)V.da. J~ -(p(-+u))dp+ & 2 ~ 2 2

2

Remark. As before, if we consider any domain immediately obtain:

J V2 -=- f_ p(_+ u) dJ1 dt Jb, 2

= -

15

1

moving with a proper motion we

V2 f _ p(-+ u)Vr.da - f _ (q-O"' V).da + f_ (f.V + r) dJ1. 2 JaD, Jb,

JaD,

In the same manner we obtain for a fixed domain: d V2 V2 f p(-+u)dJ1=_f p(-+u)V.da- f (q-a·V).da+ f (f.V+r)df.1. dt JJ 2 JaD 2 JaD .b

-

Local forms The various local formulations of the balance equation (5-160) lead to d V2 P - ( - + u) = dive 0" . V - q) + f. V + r , dt 2

a v2 V2 p(-(-+u) + V.grad(-+u))

at

2

2

d V2 V2 -(p(-+u)) +p(-+u) divV dt 2 2

(5-168a)

=

div(O"· V -q)+ f. V +r,

(5-168b)

=

div(O"· V -q)+ I.V +r,

(5-168c)

406

ChapterS

a at

V2

V2

-(p(-+u)) + div(p(-+u) V) = div(u· V -q)+ f. V +r, 2 2

(5-168d)

Training. The reader will express every local fonn given an orthononnal Cartesian basis, for instance:

6.2.3 Potential Energy of Body Forces

V: x

H

In the case in which the body force f is conservative there is a potential function V(x) such that

f

=

-gradV.

The power developed by f is

r f.V dJ.1 = - 1,r gradY. V dJ.1 = - 1,r

1, By letting CV

dV dJ.1

dt

= -~

r V dJ.1.

dt 1,

1

= , V(x) dJi ,

the power developed by f is written:

1, f. V dJi = _ dCV . dt In this special case the first principle of thennodynamics is expressed as

~(T+E(i)+CV)= r ~

~

(u·n.V-q.n)da+

r rdJl.

~

(5-169)

It is a balance equation for the energy T + E(i) + CV such that

V2

q=-+u+V, 2

IJ'd=q-u·V,

qs

r p

=-.

Remark. In the case of an adiabatic evolution of a material domain Dt , if the surface forces exerted on aDt are negligible then the first principle of thennodynamics expresses a special conservation law, namely:

d -(T+E(jJ +CV)=O.

dt

(5-170)

Fundamental Laws; Principle of Virtual Work 6.2.4

407

Internal Energy and Balance Equation

By subtracting the fundamental equations of first principle of thermodynamics and theorem of kinetic energy: d -(T + E(i» = 9{e) +0, dt

dT =

dt

gJ(e)

+ gJ(i) ,

we obtain the important equation dE(i)

(5-171)

Tt=O-gJ(i)'

which shows that the power

gJ(i)

of internal forces has opposite effects on the time rates of

T and E(i) respectively; for instance, a positive

gJ(i)

results in an increase of T and a

decrease ofE(i) . Eq. (5-171) is written:

.!!.r dt ob,

pu dp =

r

ob,

(J': d dp -

r

JoD,

q.da +

r rdp.

ob,

(5-172)

It is a balance equation (where the quantity Q is E(i» such that

q=u,

'Pd=q,

I qs=-(O':d+r). p

Training. The reader will immediately express this equation in the cases of proper motions and fixed domains. Local forms. By taking the last results into account, the various local forms (5-160) of the balance equation lead to the following expressions of the energy equation: du . p - = (J':d -d,vq+r, dt p(au + V.gradu) at

(5-173a)

= (J': d -divq +r,

(5-173b)

- divq + r ,

(5-173c)

.!!.-(pu) + pu divV = dt

0' : d

~(pu)+ div(puV) = (J': d -divq+ r.

at

(5-173d)

408

Chapter 5

Eq. (l73b) is interesting since it holds the different terms due to heat, namely those of - heat convection V. grad u , - heat conduction - div q, - heat radiation r.

In addition this equation shows a non-steady term of internal energy p au and a term of

at

power per unit volume owed to (mechanical) internal forces (Y : d . Let us specify that this power per unit volume owed to internal forces is put in the second member of different equations, though this term does not derive from external forces. Eq. (1 73a) is written as follows: W'

du 1 1 . -=-(Y:d+-(r-dlvq). dt p p

(5-174)

This equation is rendered into the following proposition: PR47 The rate of change of specific internal energy is the sum of the specific power (that is, per unit mass) of internal forces plus the specific rate of increase of total heat. Given an orthonormal Cartesian basis, this is expressed as du 1 1 = -(Yydij +-(r -q;J. dt p p ,

-

(5-175)

Remark. The first principle of thermodynamics is not concerned with the reversible or irreversible character of the process of energy conversion. We simply mention that the stress tensor can be expressed as a sum of a reversible part (YR and an irreversible part (YI : (Y = (YR + (YI .

By recalling Section 6.2.3 we can also view the part of reversible power

~~

=-

i (YR: d dp I

to be R

~(i)

dZ

=-di'

where Z plays the role of a deformation energy. Thus Eq. (5-171) leads to the following time rate d -(E(j)-Z)=Q+ dt

i (Y:d dp. I

I

409

Fundamental Laws; Principle of Virtual Work 6.3 SECOND PRINCIPLE OF THERMODYNAMICS

Let us consider the second principle of thermodynamics which deals with the type of evolution of the material domain D t included in a material system. Let us study this evolution by introducing two essential notions developed in thermodynamics. First, the absolute temperature T is a positive function of the empirical temperature and is defined at every point of the material domain and at every time. Second, the entropy is a quantity which estimates the 'energy degradation' and shows the 'disorderly state' of the material system. It is denoted by S=

1, ps dJ.1,

(5-176)

where s is the specific entropy (per unit mass) 1• We recall that the entropy in any material system is the sum of the entropies of its parts.

6.3.1

Principle

By referring to the rate of increase of total heat, the second principle of thermodynamics introduces the following notion. D

The external entropy supply rate 2 is composed of a part of external heat supplied to D t through aDt by conduction and a part gained by D t owed to external body sources, namely:

/(e)=-1

D,

~q.da+l~dJ.1. T ,T

(5-177)

Let us state the second principle of thermodynamics: PR48 The material derivative of the entropy associated with a material domain Dt never less than the external entropy supply rate:

IS

dS dt

-:::::r(e) ,

that is: dS > r ~ dJ.1 dt - JD, T

I

2

f

aD,

~ q. da . T

Called in French: 'Entropie massique'. Called in French: 'Taux d'apport exterieur d'entropie'.

(5-178)

410

ChapterS

We note that

We mention that the equality in the principle expression holds for reversible processes and the inequality for irreversible processes. For the general irreversible case (in which the reversible hypothesis cannot be considered), the time rate of change of entropy dS is strictly greater than /(e) . dt The difference between this time rate of change of entropy and /(e) characterizes the internal production of entropy and we say: D

The internal entropy production rate l is the non-negative quantity dS

/(i) == di -/(e) .

(5-179)

In this manner dS == /(e) + /(i) can be viewed as a 'generalized balance equation' where dt the quantity Q is the entropy S, namely:

!!.-1 ps d,u == - L ~q.da + 1~d,u + /(i), dtt DtT tT

(5-180)

and such that q

==s,

q

qJd

But there is a complementary term

==

T'

I(i) ~

r q, == pT .

0.

We mention that

Remark. Let us observe that for an adiabatic evolution characterized by q == 0 and r == 0, the external entropy supply rate vanishes, and thus the following consequence dS

di== /(i)

~O

means that the entropy cannot decrease, even without any external supply.

1

Called in French: 'Taux de production interne d'entropie '.

Fundamental Laws; Principle of Virtual Work 6.3.2

411

Clausius-Duhem Inequality The inequality of the second principle has the following local forms:

ds . q r p dt :2: -dzv T + T '

(5-181a)

as q r p(8i+V.grads):2:-diVT+f'

(5-181b)

:r (ps) +ps divV :2: - div ~ +; ,

(5-181c)

~(ps)+div(psV) :2:-div~+~

(5-181d)

at

T

T

.

Eq. (5-18Ia) is also written:

ds I. 1 r p-:2: --dzvq-q.grad-+dt T T T or equivalently [since T > 0 and grad ~ =

T

-~ gradT]: T

ds 1 :2: - div q + - q. grad T + r . dt T

PT -

By recalling Eq. (5-173a), that is:

· d zvq-r = u: d - pdu - , dt the last inequality becomes the Clausius-Duhem inequality:

ds du 1 u: d + peT - - - ) - -q.gradT :2: O. dt dt T

(5-182)

By introducing the function If/ = u- Ts,

the Clausius-Duhem inequality is written:

dlf/ dt

dT dt

1

u: d - p(-+ s - ) - -q.gradT :2: O.

T

(5-183)

412

6.3.3

Chapter 5 Dissipation and Reversibility Let us introduce the following natural notion.

D

The internal entropy production rate per unit volume associated with

f(i) =

rei)

1 rei) dJl.

IS

the density (5-184)

I

The first local form of the 'generalized balance equation' (5-180) becomes

ds

. q

r

(5-185)

rei) = p dt +dlVy- T .

Thus the local form (5-181a) of the second principle of thermodynamics is simply written: (5-186) that is:

ds dt

.

1

rei) T = pT -+ dlvq --q.gradT -r ~ o. T

(5-187)

The internal entropy production is connected to the dissipation so defined: D

The dissipation per unit volume is the non-negative quantity rp

=

r(i)T,

(5-188)

the dissipation is the integral (/J

=

1rp dJl . I

We emphasize that the second principle of thermodynamics expresses the condition rp~O

everywhere in D t and at every time. It is profitable to split rp into two parts according to rp = rp) + rp2 ,

whose definitions are the following.

413

Fundamental Laws; Principle of Virtual Work D

The intrinsic dissipation per unit volume is the quantity qJ] =

P T -ds + d·Ivq dt

(5-189)

r ,

the intrinsic dissipation is (/J]

.6

= , qJ] dJ.l .

This intrinsic dissipation results from the dissipation of mechanical energy into heat. In other words,

qJ]

such that P T -ds dt

= - d·Ivq + r + qJl

may be considered as an internal heat source. We immediately have qJl

= a:d

ds du + peT - - - ) dt dt

dlf

dT

=a:d - p ( - + s - ) . dt dt

D

The thermal dissipation per unit volume is the quantity qJ2

=-

1 T q.gradT,

(5-190)

the thermal dissipation is (/J2 =

.rb,

qJ2

dJ.l .

The thermal dissipation results from heat exchanges by conduction.

Remark. The thermal dissipation per unit volume vanishes if the evolution of Dt adiabatic, but also if its evolution is isothermal, since in this case gradT = 0 . D

The evolution of a domain Dt is said to be non-dissipative if the dissipation per unit volume

D

IS

qJ

is negligible everywhere in D t and at any time.

The evolution of a domain Dt is said to be reversible if D t and at any time.

It is said to be irreversible if qJ is positive.

qJl = qJ2 =

0 everywhere in

414

Chapter 5

The case of (ideal) reversible evolution corresponds to the equality in the expression of the second principle of thermodynamics. PR49 For every reversible evolution there is no internal entropy production rate, unlike irreversible evolutions. Proof Since any reversible evolution is non-dissipative, that is: (jJ

Y(i) = T =0,

we conclude that the internal entropy production rate

I(i)

vanishes.

To end this paragraph let us compare the various balance equations. QorQ

Mass Linear momentum

q orq

or

qs or qs

(jJd

0

0

m ~

(jJd

V

f

-(1'

P

V2

-(1'·V

1

-(f.v - (1': d) P

Kinetic energy

T

Internal energy

E(i)

u

q

1 -((1' : d + r)

Energy

T+E(i)

-+u

q-(1"V

-(f.v +r)

Entropy

S

s

2

V2

p

1

p

2

q

r

T

pT

6.4 CONCLUSIONS AND CONSTITUTIVE EQUATIONS Given a continuum in which only the mechanical and thermal phenomena are considered, let us draw up the list of independent balance equations and corresponding unknown functions of time and positions. We recall that the body forces f supposed to be known.

(per unit volume) and the radiant heat source r are

Fundamental Laws; Principle of Virtual Work

415

(i) The equation of continuity:

ap + div(pV) == 0, at

that is: I scalar equation and 4 unknowns p, Vi' (ii) The equation of motion (balance equation oflinear momentum):

dV f + d·Iva, p--== dt that is: 3 scalar equations and 9 unknowns a ij

.

But the moment of momentum principle implies the symmetry of a, and thus there are only 6 independent unknowns a ij . Thus the total comes to 1 + 3 == 4 equations and 4 + 9 - 3 == 10 unknowns. (iii) The energy equation

d V2 P - ( - + u) == dive a . V - q) + f.v +r , dt 2

that is: 1 scalar equation and 4 unknowns u, qi' Thus, finally, there are 5 equations and 14 unknowns which are the density p, the three velocity components Vi' the six independent stress components a ij , the three components of the heat flux vector qi' and the specific internal energy u. (iv) Last but not least, the inequation of the entropy ds dt

q T

r T

p-~-div-+-

introduces two unknowns: the entropy s and the absolute temperature T . There are 16 unknowns and 11 equations are missing! We must find eleven equations in order to hope the system of partial differential equations be solved ('to hope' because boundary and initial conditions must be considered in addition!). Unlike the five previous equations which are not dependent on the medium (and thus called 'universal'), six ofthe additional equations will characterize the physical properties of the continuum studied. These supplementary equations are called the constitutive equations. The remaining additional equations are three temperature-heat conduction relations and two thermodynamics equations of state.

416

Chapter 5

Remark 1. Constitutive equations cannot take the complex properties of any particular material into account, but they are used for defining 'ideal materials' as shown in the next chapter which deals with ideal elastic solids. Remark 2. For numerous problems of continuum mechanics the 'uncoupled thermoelastic theory' can be considered; that is, the interaction between mechanical and thermal processes can be neglected. It is the case for particular evolutions such as isothermal or adiabatic evolutions, but also if the process nature allows the above mentioned approximation; for example, when the heat production leads to slight variations of temperature. In the 'uncoupled thermoelastic theory' it is assumed that the temperature is known or determined from non-mechanical information.

Of course, the temperature field influences p and various physical constants such as elastic or viscosity coefficients. In the above mentioned approximation the mechanical processes are only governed by the continuity equation and the motion equations, while there are ten unknowns p, Vi and a ij , therefore six constitutive equations are required. We are going to have a closer look at the previous considerations in the next chapter.

Fundamental Laws; Principle of Virtual Work

417

EXERCISES Exercise 1.

The velocity field of a gas moving through a 5 meter pipe of x 2 -axis is defined by

When entering (x 2 = 0) the gas has a (mass) density p equal to 1.5 kg/m 3 . Find p at the end (X2 = 5) in the case of a steady motion. Answer. The continuity equation (5-13) is simply

a

-(pV2 )=0.

&2

Since pV2 does not explicitly depend on

X2'

we immediately have

(pV2 )1 (X],X2,X,) = (pV2 )1 (X],O,X,) = 0.075 (kgm-2 S-l).

From _ 0.075 (kglm 3 ) p - 0.02X2 + 0.05 ' we deduce the following (mass) density for x 2 = 5 (m):

pIX2~S =0.5 (kglm3 ).

Exercise 2. An infinitesimal deformation is specified by the displacement field of components u 1 =A(x3 -!)X1X 2 +Bx2' u2 = u3

C( x 3 -l)xi + DX2'

= Ex~ +FX2'

where the constants A, B, C, D, E, F and I are sufficiently small and have the required dimensions. (i) Find the infinitesimal strain tensor. (ii) From the continuity equation, determine the ratio of the (mass) density in the initial

configuration to that after deformation. (iii) Find relationships between constants if the continuum is incompressible.

418

Chapter 5

Answer. (i) Since

we know that e is the symmetric part of [

~; ].

Thus

.] [A(X 3 -1)x2 0 [ au, = ax ] 0

A(X3 -1)Xl + B 2C(x3 -1)x2 + D 3Ex~ +F

leads to

A (x -l)x + B 2 3 I 2

A(X3 -1)x2 e

=

A B -(X3 -1)Xl +-

2

2

A

2C(x3 -1)x2 + D 3E+C 2 F - - - x +222

2 X1X2

A

2 X1X2 3E+C 2 F - - - x +2 2 2 o

(ii) The continuity equation in Lagrangian description, namely:

P(X,t) J(X,t)

=

Po(X,O),

is briefly written:

Po =J. P

But for infinitesimal deformations we know that J == 1 + tre

and thus

Po

- == 1+(A +2C)(X3 -1)x2 + D. P

(iii) In the case of an incompressible medium, since J = 1 the infinitesimal deformation is such that tre = 0,

that is

(A +2C)(x3 -1)x2 + D = 0 and thus

D=O,

A+2C =0.

Fundamental Laws; Principle of Virtual Work

419

Exercise 3.

In spherical coordinates r, B, rp we recall that every axisymmetric motion is such that the components of the velocity field are only meridian and independent of the position of the meridian plane; that is: V=Vr (r,B)1r +Ve(r,B)Ie · If the motion is steady determine the components of the velocity field as well as the corresponding vector expression as functions of a stream function.

Answer. The continuity equation for a steady axisymmetric motion is written: .

d/v(pV)

1 8

1

8

.

= 2-(pr 2Vr ) + - .--(pVosmB) = o. r 8r

rsmB 8B

We know that a differential form is exact simply connected domain D is zero.

iff its integral along any closed curve r of a

By multiplying the previous expression by r2 sinB and considering the Stokes' theorem (Green-Riemann formula), we obtain: Vr ) + ~(prVo sin B)] dr /\ dB II [~(pr2sinB 8r 8B = H) [d(pr 2sinB Vr ) /\ dB - d(prVo sinB) /\ dr] 'lJ

=

1r (pr 2sinB Vr dB - prVo sinB dr) = o.

There is a so called stream function ljI(r, B) of dimension [L3T- 1] such that pr2 sinB Vr dB - prVo sinB dr

= dljl.

We deduce that pV =_~81j1 o rsinB 8r '

where Pc is a constant of (mass) density dimension. Since in this problem we simply have gradljl

81j1 8r

181j1

= - I r + --Ie, rae

we immediately deduce the corresponding vector expression pV =

~ gradljl /\ I¢. r

smB

420

Chapter 5

Exercise 4. Given a frame of reference {o;e l ,e 2 ,e3 }, the edges of length 0.4 (m) of an elastic cube are parallel to the Cartesian coordinates axes passing through the center 0 of the cube. The forces to be considered are the gravitational attraction and surface forces whose stress field is given by the tensor 0.5xj + 1.3 0'

=

10

5 X

[

-0.25x2

-0.25(Xj +X 3)]

0.25x2

1.1

- 0.85x2

- 0.25(xj + x 3)

- 0.85x2

1.884x3 +1.2

-

(Pa).

We assume that the (mass) density of the elastic cube is constant and equal to 8xl0 3 (kg m ~3) and the gravitational acceleration equals 9.8 (m s ~2 ). (i)

By considering the previous given forces (per unit volume and per unit area) prove that the equilibrium equations are verified.

(ii) Determine the stress vector on the plane at the center 1

0

whose unit normal vector is

Jj(e l +e2 +e3)'

(iii) What are the resultant and the sum of all moments about

the face in the plane

Xl

= -0.2

0

of the stresses acting on

(m).

Answer. (i) The force of gravity per unit volume has the following components

/3 The equilibrium equations

O'ik,k

a

+ It

=

=-pg=-0.784xl0 5

0 are the following identities expressed in 10 5 Pa:

a

a

-(-0.25(XI +X3» + -(-0.85X2) + -(1. 884x3 + 1.2) - 0.784 = O. Oxl oX 2 oX 3 (ii) The stress vector in) =

0' .

n, at 0, is explicitly:

421

Fundamental Laws; Principle of Virtual Work (iii) For the plane with equation xI

= -0.2 the outward unit vector is -

el ,

and thus

n l =-1.

The stress vector at any point (-0.2, x 2' x 3 ) of this face is explicitly:

1.3-0.1 -0.2SX2

-0.2SX2

0.OS-0.2Sx3

- 0.8SX2

tl] [ [t2 = t3

=[

1.1

0.OS-0.2SX 3] [-1] -0.8SX2 . 0 l. 884x3 + 1.2

0

-1.2 ] 0.2Sx 2 (bars). 0.2Sx 3 - O.OS

The resultant of the stresses exerted on the domain D of the face of equation the following vector:

Xl

=

-0.2 is

- 1.2 10.21°2 dx 2 dx 3

0.25

0.2 0.2 0. 2 1°·2 dx3 0.25 1 x 2 dx 2 0.2 0.2 2 1°·2 X3 dx, - 0.05 1°·21°·2 dx 2 dx 3 0..2 dx 2 0.2' 0.2 0.2

L

-19200: =

[

0

(N).

-800

The sum of all moments of the stresses about 0 is M

= o

r

JD

x

1\ t(iI)

da

'

where x denotes the position vector of any point of the face in the plane of equation XI = -0.2. This sum of moments is explicitly:

422

Chapter 5

£

0.21°. 2

0.2

(-0.05X2 + 1.2x 2) dx 2 dx 3

0.2

Exercise 5. At a given point we consider the stress tensor

0'

[1 1 2]1 (MPa).

== 1 0

2 (i)

1

1

Determine the stress vector on the plane through the given point and of equation

x + y + 2z - 4 == 0 and find its normal component. (ii) Given e; ==

~ (e l +e2 + e3 )

and e; ==

Js

(2e 2 + e3 ), express the shear component 0';3'

Answer. (i) Since the unit normal vector to the plane is , 1 n== j6(e, +e 2 +2e3)'

the stress vector is

and

Fundamental Laws; Principle of Virtual Work

423

(ii) We have I

_

0"13 -

t,(e;) 1

-

e'

I'

t,(ej)

that is, 0";3 =

e;.(O" ·e;) =

~ '" 2.06559

-vIS

(MPa).

Exercise 6. A fluid moves around a sphere of radius R. The components of the velocity field are, in spherical coordinates:

=~ sinO

v

r2

T

¢

where T is a positive constant of dimension [T] . The stress tensor is 0

0

0

0

_ 3cR sin¢ T r3

0

0"=

3cR 3 sin¢ T r3 0

-----

3

0

where c is a positive constant of dimension [ML-1T- 1]. The body forces are neglected. (i) Find the trajectories of particles.

(ii) Give the Lagrangian description of the motion. (iii) Is the motion isochoric? (iv) Determine the dynam about the origin of surface forces exerted on the sphere. (v) In the Eulerian description express the time rate of change of the distance between the particle at point (r, 0, ¢) at time t and the fixed plane of equation ¢ = 0 . (vi) Given two very small colored lines in the respective directions 1, and 1,p' find the

rate of shear (or shearing), and do the fibers move closer or away from each other (angular velocity!)? Answer. (i) It is a steady motion, the trajectories and streamlines coincide.

Their equations are

424

Chapter 5 dr

dB

d¢

Vr

Vo

R3 sinB T r2

They are circles defined by constant values of the radial distance r and colatitude B_ (ii) Let us choose such constants respectively equal to ro and Bo-

Since the radius of the circular trajectory is ro sinBo, then the velocity of the corresponding particle is the constant V¢ = (ro sinBo h.} and thus V¢ t R3 t ¢=----+¢o =--1 +¢orosmBo Tro The Lagrangian description of the motion is obtained by referring to the initial constants ro, Bo and ¢o, that is:

(iii) The motion is isochoric because divV = 8Vr +~ 8Vo +_1_8V¢ + cotBVo 8r r 8B rsinB 8¢ r

+~Vr = 0_ r

(iv) Given the basis (l,,1e,1¢)' the stress vector t=u-;' at any point of the sphere is explicitly written:

-

=

o o0 o ~ sinB 0

~

[1]l

0SinB] -0=

0

0

00

- 3 ; sinB

-3~ sinB 1¢, T

because;' = lr The moment of t about 0, namely M 0 = R 1r lr R

10 0

1¢ 0

0

0

-3~sinB

Thus the required dynam is

T

/\

t , is the following vector

=3~R sinBlo_ T

1

425

Fundamental Laws; Principle of Virtual Work

(v) The distance between the point (r, 0, ¢) and the plane of equation ¢

y

= r sinO sin¢ .

The time rate of change of y is

y= ay +V.grady, at

where the gradient of y is explicitly

grad y =

ay ar 1 ay r

ao

ay r sinO a¢ 1

Thus the distance sought is

y = Vr sinOsin¢ + VecosO sin¢ + V¢ cos¢ R3 sinO cos¢ T r2 (vi) In spherical coordinates we have the following tensor

grad V =

o

o

o

o o

and since

2d(x,t) = grad V + tgradV, we deduce:

= 0 is

426

Chapter 5

2d(x,t)

=

o

0

-3!f sinB

0

0

0

0

0

-3!f sinB T

. d rate 0 f shear IS . Thus the reqUire

-

r3

T

r3

R3- 3 sinB 3-'

T r

Except for B = 0 and e = ff, this rate is always negative. Thus the angular velocity between the fibers is positive and the fibers move away from each other.

Exercise 7. We consider a continuum for which every particle is viewed as a microstructure. The motion of the system is defined by a vector field V which is the velocity of the mass center of the microstructure and by the angular velocity vector m of the microstructure (at x, at time t). For a mass element dm we introduce the spherical inertia tensor I dm , where k is the radius of gyration of the considered microstructure.

e

In this problem the couple C of body forces per unit volume and the couple-stress vector M per unit area are taken into consideration. Furthermore, it can be proved that there is a tensor m for every x and every time t such that at a point of a surface we have M = m . where is the unit normal to the surface.

n,

n

State the equations of the moment of momentum principle in integral and local forms by using the rigid body theory.

Answer. Since the moment of momentum associated with a mass element dm of a material domain D t is (x 1\ V + em)dm,

then the equation of the moment of momentum principle is

~! (xl\pV+pk2m)dfl= r (xl\j+C)dfl+ r dt 'D, obi kD,

(xl\u·n+M)da

or, equivalently, by using Eq. (5-64):

~l(XI\Pv+pem)dfl= r (xl\j+C+xl\divu+f1:u+divm)dfl. dt 1, I

We deduce the local form:

Fundamental Laws; Principle of Virtual Work X /\

r

P + pe dO) = x /\ f + C + dt

X /\

427

diva + p : a + div m ,

where we recall that p is the orientation tensor (of order 3). It is equivalent to X /\

(p r

- f - diva) + pe

dO) dt

=

C + p:a + div m

and, given the equation of the linear momentum principle, this is reduced to pk 2 dO) _- C + p:a + d·IV m. dt

Thus the vector p:a is different from zero in general and the Cauchy stress tensor is not symmetric any more. Weare going to make p: a anti symmetric parts:

explicit by decomposing a

into symmetric and

First, we know that the adjoint of the anti symmetric tensor of components a~k is the covector with components

(*a A)i -_ -1 Pyk a AJk 2

_

-

1

'd="= a AJk •

-liiJk'. The principal direction associated with the following equations

O"IIl'

determined by (n;n,nfn,n~I)' is obtained from

aR2 sinO n~I + aR2 sinO n~I

= 0,

aR2 sinO n~I + aR2 sinO n~I = 0, aR2 sinO n~I = 0,

whose solution is and thus the corresponding unit vector of the principal basis, such that this basis is direct, IS

(iv) The maximum shear stress value is equal to aR 2 sinO.

(v)

For R = 1 and 0 = 7r/4, the principal stress values are O"m

J2

=-a. 2

For R = 1 and 0

= 7r, the principal stress values are

The ensuing Mohr's circles are schematized as follows:

-2Q

Fig. 50

454

ChapterS

(vi) Since the mean normal stress exerted on the body surface is US

I = -tru = 2aR 2 cosO,

3

the deviator stress tensor is explicitly

u

D

"U-U'

"lDR'~ine

°

°

The connection between the principal stress values of u D and u leads to the following principal stress values u D : UID

= aR2 (2cosO + sinO) -

2aR 2cosO

= aR 2sinO,

ug =2aR 2cosO-2aR 2cosO=0, UI~

= aR2 (2cosO -

sinO) - 2aR2 cosO = -aR2 sinO.

These values are really the eigenvalues of u u(u 2 _a 2R 4 sin 20)

D,

whose characteristic equation is

= 0.

Of course, the principal directions of u D are those of the stress tensor u. (vii) The invariants of u Dare

II =tru O =0, 100

DO

I

DO

DO 12

12 =-(uiiu jj -uij u j , ) =-(-OI2U21 -U 21 ( 2 2 D 13 = detu = 0.

Thus the characteristic equation of u D is actually u

3 _

(a 2 R 4 sin 2 0) u

= 0.

24'2 )=-a R Sill 0

CHAPTER 6

LINEAR ELASTICITY

We know that the principles of conservation of mass, conservation of linear momentum, conservation of moment of momentum, and of conservation of energy hold for every continuum, whatever material we consider in this study. However, the notions introduced until now, notably these principles, do not let us determine the behavior of a specific material. On the one hand, experience shows that the responses of different materials under the same loading conditions are extremely varied; on the other hand, it shows that the behavior of a given material changes greatly under various loads. Of course, the introduction of constitutive equations cannot allow us to describe all the observed phenomena, but these supplementary equations are useful for the definition of idealized continua.

1. ELASTICITY AND TESTS In particular, the deformation of certain materials completely vanishes when the loads are removed and it is said that the material has an 'elastic' behavior. The initial 'natural' state remains in the elastic material memory. In other words, an elastic material 'remembers' its sole unstressed state but 'has forgotten' all the intermediate states. Thus, more precisely, we say: D

A material or a body is elastic if the stress tensor u only depends on the deformation between the initial unstressed state and the final stressed state.

455

Y. R. Talpaert, Tensor Analysis and Continuum Mechanics © Springer Science+Business Media Dordrecht 2002

456

Chapter 6

So, the elastic behavior is independent of the 'motion' between the two states mentioned, only the strain effect is to be considered. On the contrary, beyond a given level of loading, continua may permanently be deformed and fracture can even occur. To know the mechanical behavior of materials it is useful to test specimens of those whose shapes are very simple as slender circular cylinders, slender right angled parallelepipeds for instance.

, .... F

- ....

....

F

.,

Fig. 51 The most frequent test consists in applying tensile forces axially to the material!. The connection between the magnitude of the force and the elongation of the body is illustrated in the following figure.

Ci~_

. . . _ . . . ________

~_~_~

_______

L..-_ _ _ _ _ _ _ _ _ _ _ _ _

E

Fig. 52 Let us consider the initial length Zof the cylinder. The stress a = FIA (N m -2), where F is the axial tensile force exerted at the extremities of the specimen and A the initial area of the cross-section, is represented as a function of the axial strain E: = LiZ I Z, where LiZ is the elongation. By choosing the previous ratios referring to the cross-section and length of the test body we have obtained a representation that does not depend on specimen dimensions.

I

In French a traction test is called: 'Essai de traction '.

Linear elasticity

457

For values corresponding to the line oa the continuum has an elastic behavior and reverts to its initial state as soon as the load is removed. For higher tensile forces leading to stress values greater than an elastic limit a e there is a permanent deformation, which means that ,11 (and thus [;) is different from zero, whereas the load vanishes. Finally, for increasing tensile forces the material can break. The slope of the curve oa is a fundamental constant in linear elasticity called Young's modulus E = a / [;. It characterizes each material such that F / A < a e and will later be intensively used. For example, the Young's modulus ofa steel specimen is approximately equal to 2xl0 5 (MPa), and by considering a cylinder of radius equal to 0.03 (m) which carries a load of 5 xl 0 5 (N) we obtain:

and thus

,11 "",0.883x1O-3 • [

So the axial strain [;, that is, the (dimensionless) relative elongation ,1[/[ , is of the order of 10-3 . Given this smallness, the infinitesimal strain theory is applicable to this steel deformation. Of course, the axial strain 1 F

6=--

EA

is all the more large since the force F is large. When the material elongates it shrinks perpendicularly to the elongation direction. This lateral contraction is proportional to the increase of the tensile force. By considering conditions such that an initial circular cross-section of diameter d remains circular after deformation, the relative change of diameter is then denoted by the following lateral strain 6

, = -M -. d

The ratio between lateral and axial strains is another important constant characterizing the material if the strains are small. It is called Poisson's ratio and is defined in this particular case as follows: 6'

v=-_· 6

Experience reveals that this real number is always positive but smaller than 0.5. For steel the value of v is of the order ofO.3 and in general between 0.2 and 0.3 for metals.

458

Chapter 6

In the following sections, unless otherwise specified, the thermal effects will not be taken into account, isothermal and adiabatic processes will not be considered. In other words, the temperature is constant and no heat variation is viewed. The isothermal hypothesis is valid for very slow evolutions of materials which conduct heat quickly or for equilibrium states in the cases of uniform and constant temperatures. The adiabatic hypothesis fits vibrations or waves which have no time for heat exchanges. After introducing linear elasticity in general, we will study the properties of classical elasticity, namely those of homogeneous and isotropic materials. In brief, a homogeneous continuum is such that its material properties do not differ at different neighborhoods and an isotropic continuum is such that its material properties do not depend on direction.

2. THE GENERALIZED HOOKE'S LAW IN LINEAR ELASTICITY 2.1 THE GENERALIZED HOOKE'S LAW

We assume that the displacement gradients are 'sufficiently small' and displacements too, so that the Lagrangian and Eulerian descriptions of vectors and tensors can be considered as indistinguishable. Thus at any point and every time the material behavior is known from u in function of by knowing that the 'natural' state is described as 'no strain-no stress'.

6,

From the previous experience and ensuing comments we state that the stress tensor is expressed as a function u of the infinitesimal strain tensor 6 such that u(O)=O,

that is, explicitly: urs(O) =0.

Let us assume these functions 6, that is, explicitly: urs

eyrs

can be expanded in Taylor's series of components of

= Ersy 6 JI.. + ... - ErsIl

-

6

11

+ E rs12

6

21

+ E rs21 "12 co + ...

,

where the coefficients Ersil of dimension [ML-IT-2 ] are to be specified. A priori these coefficients are functions of position vectors, but we will assume these are constants; that is, the material is said to be homogeneous.

Linear elasticity

459

In linear elasticity we assume that the various ars are linearly depending on

8ij'

Thus

given 'sufficiently small' deformations, the constitutive equations for a linear elastic body are the nine equations (6-1)

r,s,i,j E {l,2,3}, which relate the stress tensor u and the infinitesimal strain tensor

&.

The equivalent tensor relation (6-2)

a=E:&

is called the generalized Hooke's law. It expresses the stress tensor as the specified double contraction.

D

The tensor

if of 81 constant components

ErSij

is called the elasticity tensor.

It is really a tensor of fourth order since the double contraction between infinitesimal strain tensor & leads to the stress tensor a .

if and the

The symmetry of both stress and infinitesimal strain tensors implies there are at most 36 independent components of if . Indeed, the symmetry of

&

implies that

and thus These 27 equalities reduce the number of independent elastic components to 54. By considering the symmetry of u for a linear elastic body, it then follows that we have 18 relations of symmetry, namely:

and the number of independent elastic components is 36 at the most.

2.2 QUADRATIC FORMS AND STRAIN ENERGY FUNCTION

To make progress let us introduce an important positive-definite quadratic form l which characterizes linear elasticity, namely:

I We recal\ that U can only vanish if all the strain components are zero. The quadratic fonn U is a convex function (positive second derivatives of U), this is equivalent to stating that the quadratic fonn is positivedefmite.

460

Chapter 6 1 U(E) = - (E: E: E) 2

(6-3)

1 U(E) = - (E : u).

(6-4)

or 2

Given an orthonormal basis it is explicitly written: 1 - .. U(E) = - [; E rslJ [; . 2 sr jI

or

PRI

(6-5)

The positive definite quadratic form U is symmetric in pairs of indices of the independent components of E are at most 21 in number.

Proof From the expression of the quadratic form U we deduce the differential

By changing the dummy indices this expression is written:

Thus by comparison we conclude that £,ijkl = £,klij.

Since £,ijkl = £, jikl = £, jllk = £,ijlk

we conclude that the various independent components of

E

are

£,1111, £,1112, £,1ll3, £,1122, £,1123, £,ll33, £,1212, £,1213, £,1222, £,1223, £,1233, £,1313. £,1322, £,1323, £,1333, £,2222, £,2223, £,2233, £,2323, £,2333, £,3333.

E and

Linear elasticity

PR2

The partial derivative of the quadratic fonn U with respect to a.

461

&

is the stress tensor

Proof We have

au a 1 1 -- = - ( - & : E :&)=-(E:& + &:E) = E:& a& a& 2 2 =0'.

We can also prove this proposition in explicit fonn. Indeed, we have dU =

~(Eklij & d& + Eijkl& de) 2 M, M,

= Eijkl&kl d&y =0' ijd&ij

and thus, by considering the various &ij as independent variables, we have actually

au

.

--=0".

a&ij

PR3

In the infinitesimal context the quadratic fonn U is the strain energy function.

Proof The energy equation in local form du p-= a:d-divq+r,

dt

which gives the rate of change of internal energy, becomes du p-=a:d,

dt

when the thermal effects are ignored. This particular balance equation is explicitly written: P il=aiJd

,

According to usage we denote this last relation by P du

= a' d&,

and we have obtained the following result: ij

au

a =p--' a&ij

(6-6)

462

Chapter 6

The quadratic form is really the strain energy function U

= pu.

We are now going to introduce the dual quadratic form U' as follows. The quadratic form 1 1 U(c) =-(c: E :c) = -csra rs 2 2

is also written: 1

rs

-a c sr ' 2

Since U is a positive-definite form its discriminant is nonzero and Eq. (6-1) can be inverted to obtain the strain components C if in terms of the stress components a rs , namely: [c=C:a]

(6-7)

and U(c) is successively written: _ 1 rs _ 1 rs qp _ 1 . C- . U(c) - -a c sr - -a C srpq a - -a. .a. 2 2 2

Thus if we introduce the following quadratic form 1 U'(a)=-a:C :a,

(6-8)

2

we have obtained the duality equation, namely: U(c)

PR4

= U*(a).

(6-9)

The partial derivative of the dual quadratic form U* with respect to a is the infinitesimal strain tensor c .

Proof We have

au' a 1 1 -- = - ( - a : C :a)=-(C :a+a:C)=C:a 2 2

aa aa =c.

Thus we have obtained in explicit form:

au'

aa if

=

c if '

which is the dual expression ofEq. (6-6).

(6-10)

463

Linear elasticity 2.3 ISOTROPIC MATERIAL AND LAME COEFFICIENTS

We know that the number of elastic coefficients strain energy function exists.

ltijkl

is reduced to 21 when a

The more a material has material symmetries, the less it has elastic coefficients in its behavior constitutive equations. We are going to consider the simple case of isotropic material. D

CJr

A material is said to be isotropic if its mechanical properties are independent of direction. On the contrary, it is said to be anisotropic if its mechanical properties are described with reference to direction.

2.3.1 Constitutive Equations

Since every isotropic material is equivalent in all directions, the corresponding definite quadratic form (6-3) is a quadratic invariant of the infinitesimal strain tensor &. By denoting

invariants of &, the quadratic form is necessarily a linear combination of second order invariants 1? = E:iiE: jj and 12 . Thus we explicitly denote: (6-11)

that is:

By considering double contractions, we equivalently denote: A

U(&) = - (I: &)2 + Jl &: & 2

(6-11 ')

A

U(&) = - (tr&)2 + Jl & : & . 2

At this stage it is interesting to consider the sum of the spherical strain tensor &s and the deviator strain tensor 8 D :

464

Chapter 6

We recall that every strain deviator tensor is associated with a shear deformation for which the dilatation vanishes (trc D = 0); on the other hand, every spherical strain tensor represents a uniform deformation. In addition we note that spaces of shear and uniform deformations are supplementary. To make progress we observe that CUC jj

= (buC s + Ci~)(bjjCS + cf) = 9(C S )2,

because

We also have

Thus the quadratic form (6-11) is expressed as U(c)

= -1 (9,1,(C S )2 + 6j.J(C S )2 + 2j.Jcffc~) 2

and, by letting 2

K=,1,+-j.J, 3

it is written: (6-12) Since it is a sum of squared terms the quadratic form is positive-definite iff 3K

D

= 3,1, + 2j.J > 0,

The constants A, and j.J are called the Lame coefficients.

Their dimensions are those of

PR5

j.J>o.

It, namely:

[ML-IT-2 ].

Hooke's law's equation u = It : C is expressed as a fimction of Lame coefficients as follows: (6-13) u=,1,(trc)I+2j.Je.

au

Proof We recall that the components of the stress tensor u = -

ac

are

Linear elasticity

465

.. = __ au

aU

a£ij ,

and, given an orthononnal basis, they are expressed as (6-14)

au = At5ij£kk + 2,u£ji'

In conclusion, the constitutive equations for a linear isotropic elastic material are respectively all =A(£ll +£22 +£33)+2,u£1l '

a22 = A(£" + £22 + £33) + 2,u£22 , a 33 =A(£11 +£22 +£33)+2,u£33 '

(6-15)

= 2,u£12 ' a l3 = 2,u£l3 , a 23 = 2,u£23 .

aI2

Each Eq. (6-14) can be inverted to express the strain components in function of the stress components as follows: 1 £u =-(aij -At5ij£kk)' 2,u

Since the addition of first three Eqs. (6-15) is a kk

= 3A£kk + 2,u£kk = (3,1 + 2,u)£kk'

(6-16)

we deduce the following strain components

A

1

= -(a -

t5a kk)

(6-17)

Ei=-(a(tru) 1). 2,u 3A + 2,u

(6-18)

£ .. U

2,u

U

3,1 + 2,u

U

and thus 1

PR6

A

The stress and strain tensors are spherical together; that is:

£ff = O. Proof The comparison between dU

= a ij dEi ij = 3a s dEi s +

aff dEiff

and the differential of the expression (6-12) of U, that is, dU

= 9K£s dEi s

+ 2,uEiff d&ff,

466

Chapter 6

implies the following relationship between spherical tensors as well as deviator tensors: (6-19) D

D

(6-20)

aij =2peij.

This proves the proposition. Remark 1. For a uniform compressive stress as < 0 there is a uniform compressive strain (or uniform contraction) e S < 0 since K is positive. Remark 2. We emphasize that Eq. (6-16) gives the sum of the first three Eqs. (6-15), that is: (6-21) tra = (3A.+ 2p)ekk = 3Ktre. This result also follows from as = 3K e S since we successively have tra

2.3.2

= tra S = 3a s = 9K e S = 3Ktre.

Young's Modulus and Poisson's Ratio

We recall that a stress is said to be uniaxial if there is one and only one nonzero normal stress component. The uniaxial stress type approximates the stress in the cylindrical bar considered in the introductory test specimen. For instance, let us choose the principal stress in axial direction. Thus Eq. (6-17) gives

A

1 ell

A+p

= 2p (all - 3A. + 2p all) = p(3A. + 2p) all' A -----a 2p(3A. + 2p)

(6-22) (6-23)

II'

all the other strain components are zero. We note that (6-24)

Two important constants are to be introduced.

467

Linear elasticity D

CF

The constant

a" /

£11

of dimension [ML-'T-2 ] is called the Young's modulus.

It is denoted by (6-25)

D

The real number -

£22

=-

£11

£33

is called Poisson's ratio.

£11

It is denoted by

(6-26)

v=---

2(1 + ji)

Of course, we obtain a relation between E, and (6-26), namely: E ji= . 2(1 + v)

ji,

and v by eliminating 1 between Eqs. (6-25) (6-27)

Since

the constitutive equation (6-18) is written: £

I+v =-a

E

v - - (tra)]

E

(6-28)

'

that is, explicitly: (6-29)

Of course, we can obtain 1 in function of v and E by eliminating and (6-26), namely: vE 1=----(1- 2v)(! + v)

ji

between Eqs. (6-25) (6-30)

Thus Eq. (6-13) is written: E

vE

a = -- e + (tre)] , I+v (l-2v)(l+v)

(6-31 )

that is explicitly: (6-32)

468

Chapter 6

We also have

A

E

2(A+.u)=-=---v (1-2v)(J+v)

(6-33)

(l-v)E = --'-------'---

(6-34)

and 1

2

'" + .u

(I - 2v)(I + v)

Now let us make explicit the constitutive equations in terms of v and E for a linear isotropic elastic material. So Eq. (6-29) leads to 1

cll =-(O'll

E

c22

-v(0'22 +0'33»'

1

= -(an E

v(0'33 + all»,

1

c33 =-(0'33 -V(O'II +0'22», E 1+ V

(6-35)

cl2 =£0'12'

c[3

1+ V

= --0'[3 E

,

I +V c23 = --0'23 ' E

and Eq. (6-32) leads to E vE + (cll +c22 +c33), l+v (I-2v)(J+v) E vE 0'22 = --c22 + (cll + c22 + C33)' l+v (l-2v)(l+v) E vE 0'33 = --c33 + (cll + c22 + c33)' l+v (1-2v)(l+v) E 0'12 = --c12 ' l+v all = - - c l l

E

0'[3 = --c[3' I +v E

0'23 = --c23 . l+v

(6-36)

469

Linear Elasticity 2.3.3

Bulk Modulus

Another stress state corresponding to a uniform (normal) tension is defined by the following stress tensor u=uI. It is sometimes called a hydrostatic stress state.

Eq. (6-16) becomes 3u

= (3,1, + 2J1.) trs.

Thus we have 2 u K=A+-J1.=- . 3 tre

D

The ratio K of the hydrostatic stress u to the dilatation modulus l .

(6-37)

IS

called the bulk

Of course, given a positive value of u, K is all the more great as the dilatation is small. Let us express K otherwise From (6-25) we know that A+J1. 3K =3A+2J1. = - - E J1. A =(-+l)E. J1.

(6-38)

Let us express AI J1. as a function of v . Since the Lame parameter J1. is such that

E

-=2J1. l+v

(6-30) may be written: ,1,= 2J1.v . 1-2v

Thus by replacing AI J1. with 2v/(1- 2v) in Eq. (6-38) we obtain: K=

1

E 3(1- 2v)

Called in French: 'Module de rigidite ala dilatation·.

(6-39)

470

Chapter 6

Since K and p are positive we necessarily have

E>O,

1

-l

(isochoric motion),

P(X,t) = Po (X,O)

,

( irrotational flow),

(isochoric irrotational flow).

Section 2. ten)

lim LJF

=

Lla--.O

,1a '

t=1:+(1',

~= R=

r

V dm

JM D,

r

JM Dt

0

dt ob,

r

hi

pV dll ,

r

py dll,

ydm=

[MR] = [R(e)] M(e) ~r

= ob, r

r

r

ob,

.In,

~ r dt ob,

fdll+

f dll +

r ~(pV)dll+ r

~&

~

x

1\

r

Mo=

JM D,

x

1\ V

dm

= ob, r

xl\ydm=

r

.vI

0'

pVdll=

p y dll =

h,

= JrM0,

'»lto

pV dll =

r

r

JaD,

.faD,

t(ir)da,

ten) da,

p(V.lnJVda=

r

ob,

x

1\

f dll +

r

r

~

JaD,

fdll+ x

1\

r

~

ten) da ,

t(n)da,

x 1\ pV dll , xl\py dll.

564

r

Jot

Summary of Formulae x /\ P

r dp = .br x /\ f

dp +

l

r

JaD,

x /\ t(Ii) da,

r x /\ ~(pV)dp + r x /\ p(V.1n)V da = r x /\ f ot JaD obi

obi

I

~ lqdpt dt

t(Ii)

=

I

1(q+qdivV)dp, ,

=u.1

(n) _ - Up

tJ

n'

ni ,

tee,) =

U

J

ten)

=" 3

~

n.t(e,)

=(j.

I

)l

)i'

ne I J'

i=1

1 (p r D,

- f - divu) dp

=

0,

p r=f + divu,

=f

p (oV + gradV. V)

ot

+divu,

oV V2 P ( - + (cur/V) /\ V + grad - ) = f + divu ,

ot

2

~(p V) + div(pV @ V) =

ot

f + divu,

1

PYr =fr +orurr +-oeure +ozu rz + r PYo

1

urr-u ee , r

2

= fo + 0ruOr + -oou oe + ozuoz +-UOr ' r r 1

1

r

r

PYz =fz +oruzr +-oeuze +ozu zz +-uzr'

dp +

r

JaD I

x

/\ten)

da,

Summary of Formulae 1 1 1 P Yr = Ir + 8 rrr rr + -8 err rB + -.-8¢rrr¢ + -(2rrrr - rree - a ¢¢ + rrrO cot B) ,

r

rsmB

1

r

1

1

P Ye = Ie + 8 rrrer + -8 erree + - . - 8 ¢ae¢ + -(3a re + (rree - rr¢¢)cotB), r rsmB r 1 1 1 P Y¢ = I¢ + 8 rrr¢r + -8 orr¢o + - . - 8 ¢a¢¢ + -(3a r¢ + 2a B¢ cotB),

r

;; + rrik,k

rsmB

r

= 0,

r x/\pydp=~r x/\pVdp= r x/\fdp+ r x/\rr']n da , dt JD, JD, JoD,

JLJ,

div(x /\ rr ·1n ) x /\ P Y =

= x /\ divrr + p: rr

X /\

f

+

X /\

,

divrr + p : rr (+ C)

(if couple C per unit volume).

Section 3. 1 dV 2 - p-2 dt

= f. V + dive rr . V) - rr: d ,

-i f. V dp + i

~e)

-

Dt

Dt

t(n), V da ,

~i) = - JLJ, r rr: d d p ,

r 1 2 T = Ji - pV dp, D, 2 dT dt

-=~e)+~i) ,

1f(X,t)

= J(X,t)F-1(X,t) .rr(x). tF-1(X,t),

rr:d= (F-1·rr·tF-'):L

= ~1f:L, J

~i) = - JDo r 1f(X,t): L(X,t) dpo '

rrr -:d dm

J,

M D,

df

P

= rr . da ,

=

r

J,

MDo

1f.

- : L dm,

Po

1f(X,t)· dA

= F-' (X,t)· df ,

~ = F-1 . rr . tF-', Po P

565

566

Summary of Formulae

B(X,t)

= J(X,t) a(x) . tF-'(X,t),

B(X,!)

= F(X,t) '1r(X,t),

1r(X,t) = F-' (X,t)· B(X,t),

dV dflo = 1 J I·V dflo + 1 1 -Po-dt 2

I

Do

Do2

dV 2

I

- Po - 2 dt

-

= Po I· V

r

Po

Po

r = Po 1 + div B ,

JDo

r

dflo =

r

JDo

t B·V.

Do

•

t

+ dlv( B . V) -

J I dflo + f

JaDo

dA -

l 'L Do

1r:

dflo,

.

1r : L

B. dA ,

Section 4. ( ')' , , ann=t (Ii), .n=a·n.n=n.a·n,

det( a ij - a 0 ij ) = 0 , 3 2 I a -ail a +-(aiiall-aijajl)a-det(aij) =0, 2 I, =aji = tra,

12

= 7.I (a;p jj

-

a;pji)

= 7.I((tra) 2 -tra 2) ,

13 = deta,

aD

=a

- as,

a=a S +a D ,

tru D = 0,

2

a 3 - Il a +I2a-I3=0,

Summary of Formulae

567

(principal values),

Section 5.

ar

n

.

or=L-joqJ, aq

j=1

V* = L. ~

~ q *j , aqJ

j=1

i

=

Or(e)

t

1t

f·or dl1 +

aD,

(Ii)

.or da,

.b (f + divu). V* dl1 + .b u: d' dl1 + .b u: Q' dl1 , ~~) = -.b u: d* dl1 , ~~) = - 1, r u d; dl1, ~:)

=

t

t

t

If

t