An Introduction to Continuum Mechanics

An Introduction to Continuum Mechanics

This is Volume 158 in MATHEMATICS IN SCIENCE AND ENGINEERING A Series of Monographs and Textbooks Edited by RICHARD BELLMAN, University of Southern California The complete listing of books in this series is available from the Publisher upon request.

An Introduction to Continuum Mechanics

MORTON E. GURTIN Department of Mathematics Carnegie-Mellon University Pittsburgh. Pennsylvania

1981

ACADEMIC PRESS A Subsidiary of Harcourt Brace Jovanovich. Publishers

New York London

Toronto Sydney San Francisco

For Amy and Bill

eOPYRIGHT © 1981, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHER.

ACADEMIC PRESS, INC. III Fifth Avenue, New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. 24/28 Oval Road, London NWI 7DX

library of Congress Cataloging in PUblication Data Gurtin, Morton E. An introduction to continuum mechanics. (Mathematics in science and engineering) Bibliography: p. Includes index. 1. Continuum mechanics. I. Title. II. Series. QA80B.2.GB6 531 BO-2335 ISBN 0-12-309750-9 AACR2 AMS (MOS) 1970 Subject Classification: 73899

PRINTED IN THE UNITED STATES OF AMERICA

81 82 83 84

9 8 7 6 S 4 3 2 1

Contents

ix

Preface Acknowledgments

xi

Chapter I Tensor Algebra 1. Points. Vectors. Tensors 2. Spectral Theorem. Cayley-Hamilton Theorem. Polar Decomposition Theorem Selected References

Chapter IT

11

17

Tensor Analysis

3. Differentiation 4. Gradient. Divergence. Curl 5. The Divergence Theorem. Stokes' Theorem Selected References

19

29 37 40

Chapter ill Kinematics 41 54

6. Bodies. Deformations. Strain 7. Small Deformations v

vi 8. 9. 10. 11.

CONTENTS

Motions Types of Motions. Spin. Rate of Stretching Transport Theorems. Volume. Isochoric Motions Spin. Circulation. Vorticity Selected References

Chapter IV 12. 13.

58

67 77 80 85

Mass. Momentum

Conservation of Mass Linear and Angular Momentum. Center of Mass Selected Reference

87 92 95

Chapter V Force 14. 15.

Force. Stress. Balance of Momentum Consequences of Momentum Balance Selected References

Chapter VI 16. 17. 18. 19.

20. 21.

115 119 122 130 137

Change in Observer. Invariance of Material Response

Change in Observer Invariance under a Change in Observer Selected References

Chapter VIII 22. 23. 24.

Constitutive Assumptions. Inviscid Fluids

Constitutive Assumptions Ideal Fluids Steady, Plane, Irrotational Flow of an Ideal Fluid Elastic Fluids Selected References

Chapter VII

97 108 113

139 143 145

Newtonian Fluids. The Navier-Stokes Equations

Newtonian Fluids Some Simple Solutions for Plane Steady Flow Uniqueness and Stability Selected References

147 156 159 164

CONTENTS

ChapterIX 25. 26. 27. 28. 29.

Vll

Finite Elasticity

Elastic Bodies Simple Shear of a Homogeneous and Isotropic Elastic Body The Piola- Kirchhoff Stress Hyperelastic Bodies The Elasticity Tensor Selected References

165 175 178 184 194 198

Chapter X Linear Elasticity 30. 31. 32. 33. 34. 35.

Derivation of the Linear Theory Some Simple Solutions Linear Elastostatics Bending and Torsion Linear Elastodynarnics Progressive Waves Selected References

199 201 205 214 220 223 226

Appendix 36. The Exponential Function 37. Isotropic Functions 38. General Scheme of Notation

227 229 239

References

243

Hints for Selected Exercises

247

Index

261

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Preface

This book presents an introduction to the classical theories of continuum mechanics; in particular, to the theories of ideal, compressible, and viscous fluids, and to the linear and nonlinear theories of elasticity. These theories are important, not only because they are applicable to a majority of the problems in continuum mechanics arising in practice, but because they form a solid base upon which one can readily construct more complex theories of material behavior. Further, although attention is limited to the classical theories, the treatment is modem with a major emphasis on foundations and structure. I have used direct-as opposed to component-notation throughout. While engineers and physicists might at first fmd this a bit difficult, I believe that the additional effort required is more than compensated for by the resulting gain in clarity and insight. For those not familiar with direct notation, and to make the book reasonably self-contained, I have included two lengthy chapters on tensor algebra and analysis. The book is designed to form a one- or two-semester course in continuum mechanics at the first-year graduate level and is based on courses I have taught overthe past fifteen years to mathematicians, engineers, and physicists at Brown University and at Carnegie-Mellon University. With the exception of a list of general references at the end of each chapter, I have omitted almost all reference to the literature. Those interested in questions of priority or history are referred to the encyclopedia articles of Truesdell and Toupin [1], Truesdell and Noll [1], Serrin, [1], and Gurtin [1]. ix

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Acknowledgments

lowe so much to so many people. To begin, this book would never have been written without the inspiration and influence of Eli Sternberg, Walter Noll, and Clifford Truesdell, who, through discussions, courses, and writings, both published and unpublished, have endowed me with the background necessary for a project of this type. I want to thank D. Carlson, L. Martins, I. Murdoch, P. Podio-Guidugli, E. Sternberg, and W. Williams for their detailed criticism of the manuscript, and for their many valuable discussions concerning the organization and presentation of the material. I would also like to thank J. Anderson, J. Marsden, L. Murphy, D. Owen, D. Reynolds, R. Sampaio, K. Spear, S. Spector, and R. Temam for valuable comments. The proof (p. 23) that the square-root function (on tensors) is smooth comes from unpublished lecture notes of Noll; the proof of the smoothness lemma (p. 65) is due to Martins; Section 33, concerning bending and torsion of linear elastic bodies, is based on unpublished lecture notes of Sternberg; the first paragraph of Section 21 , concerning invariance under a change in observer, is a paraphrasing of a remark by Truesdell [2]. Because this book has been written over a period of fifteen years, and because of a "fading memory", I have probably neglected to mention many other sources from which I have profited; if so, I apologize. Finally, I am extremely grateful to Nancy Colmer for her careful and accurate typing (and retyping!) of the manuscript.

xi

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CHAPTER

I Tensor Algebra

1.

POINTS. VECTORS. TENSORS

The space under consideration will always be a three-dimensional euclidean point space $. The term point will be reserved for elements of $, the term vector for elements of the associated vector space "Y. The difference

v=y-x of two points is a vector (Fig. 1); the sum

y=x+v ofa point x and a vector v is a point. The sum of two points is not a meaningful concept. y

Figure 1

x

The inner product of two vectors u and v will be designated by u . v, and we define u 2 = u· u.

I.

2

TENSOR ALGEBRA

We use the symbol ~ for the reals, ~+ for the strictly positive reals. Representation Theorem for Linear Forms. 1 there exists a unique vector a such that

Let l/J: "1/ ~ ~ be linear. Then

l/J(v) = a' v

for every vector v.

A cartesian coordinate frame consists of an orthonormal basis {e;} = {el , e2, e3} together with a point 0 called the origin. We assume once and for all that a single, fixed cartesian coordinate frame is given. The (cartesian) components of a vector U are given by Ui =

u'e i ,

so that U·V=LUiV i. i

Similarly, the coordinates of a point x are Xi

= (x - o)·e i •

The span sp{ u, v, ... , w} of a set {u, v, ... , w} of vectors is the subspace of "1/consisting of all linear combinations of these vectors: sp{u, v, ... , w} = {oeu

+ {3v + ... + yWIIX, {3, ... , y E ~}.

(We will also use this notation for vector spaces other than "1/.) Given a vector v, we write

{v}l- = {ulu- v = O} for the subspace of "1/ consisting of all vectors perpendicular to v. We use the term tensor as a synonym for "linear transformation from "1into "1/." Thus a tensor S is a linear map that assigns to each vector u a vector v = Su. The set of all tensors forms a vector space if addition and scalar multiplication are defined pointwise; that is, S + T and IXS (IX E ~) are the tensors defined by (S + T)v = Sv + Tv, (IXS)V

=

IX(SV).

The zero element in this space is the zero tensor 0 which maps every vector v into the zero vector:

Ov = O. 1

Cf., e.g., Halmos [I, §67].

1. POINTS. VECTORS. TENSORS

3

Another important tensor is the identity I defined by

Iv = v for every vector v. The product ST of two tensors is the tensor ST = SoT; that is, (ST)v = S(Tv)

for all v. We use the standard notation S2 = SS, etc. Generally, ST #- TS. If ST = TS, we say that Sand T commute. We write ST for the transpose of S; ST is the unique tensor with the property

for all vectors u and v. It then follows that (S

+ T)T = ST + TT, (ST) T = TTST,

(1)

(ST)T = S. A tensor S is symmetric if

skew if

Every tensor S can be expressed uniquely as the sum of a symmetric tensor E and a skew tensor W: S = E + W; in fact, E =!(S

+ ST),

W = t(S - ST). We call E the symmetric part of S, W the skew part of S.

I.

4

TENSOR ALGEBRA

The tensorproduct a (8) b of two vectors a and b is the tensor that assigns to each vector v the vector (b· v)a: (a (8) b)v = (b· v)a. Then (a (8) b)T = (b (8) a), (a (8) b)(c (8) d)

=

(b· c)a (8) d,

(e i (8) ei)(ej (8) e)

=

O, { e, (8) e.,

L e, (8) e, =

ii'j

i =

i.

(2)

I.

i

Let e be a unit vector. Then e (8) e applied to a vector v gives (v· e)e,

which is the projection of v in the direction of e, while I - e (8) e applied to v gives v - (v· e)e, which is the projection of v onto the plane perpendicular to e (Fig. 2).

Figure 2

The components Sij of a tensor S are defined by Sij = ei . Se j. With this definition v = Su is equivalent to V;

=

L

SijU j•

j

Further,

S and

=

L Sije; (8) ej i.j

(3)

5

1. POINTS. VECTORS. TENSORS

We write [S] for the matrix

It then follows that

CST] = [SF, CST] = [S] [T], and

1 0 0] [

[1]=010. 001

The trace is the linear operation that assigns to each tensor S a scalar tr S and satisfies

=

tr(u ® v)

u· v

for all vectors u and v. By (3) and the linearity of tr, tr S =

tr(~

Sije; ® ej) = l,)

~

Sij tree; ® e)

I,)

=

I Sije;· ej = I i,j

s..

i

Thus the trace is well defined: trS

=

IS;;. i

This operation has the following properties: tr ST = tr S, tr(ST) = tr(TS). The space of all tensors has a natural inner product

which in components has the form S·T = IS;J'ij' i,j

(4)

6

I.

TENSOR ALGEBRA

Then loS = tr S, R (ST) = (STR) T = (RT T) S, 0

0

U

0

0

(5)

Sv = So (u ® v),

(a ® b) (u ® v) 0

=

(a u)(b v). 0

0

More important is the following Proposition (a)

IfS is symmetric,

SoT = SoTT = SO H(T + TT)}. (b)

IfW is skew,

WoT = -WoT T = WO {~T (c)

- TT)}.

IfS is symmetric and W skew,

SoW=O. (d) (e) (f)

1fT S = Ofor every tensor S, then T = O. 1fT S = Ofor every symmetric S, then T is skew. 1fT W = 0 for every skew W, then T is symmetric. 0

0

0

We define the determinant of a tensor S to be the determinant of the matrix [S]: det S = det[S]. This definition is independent of our choice of basis {e.}. A tensor S is invertible if there exists a tensor S - 1, called the inverse of S, such that SS-l = S-lS = I. It follows that S is invertible if and only if det S =1= O.

The identities det(ST) = (det S)(det T), det ST = det S, det(S-l) = (det S)- \ (ST)-l = T- 1S-t, (S-l)T

= (ST)-l

(6)

1.

POINTS. VECTORS. TENSORS

7

will be useful. For convenience, we use the abbreviation

A tensor Q is orthogonal if it preserves inner products:

QU'Qv = u-v for all vectors u and v. A necessary and sufficient condition that Q be orthogonal is that

or equivalently,

An orthogonal tensor with positive determinant is called a rotation. (Rotations are sometimes called proper orthogonal tensors.) Every orthogonal tensor is either a rotation or the product of a rotation with - I. If R =I I is a rotation, then the set of all vectors v such that

Rv = v forms a one-dimensional subspace of 1/ called the axis of R. A tensor S is positive definite provided v'Sv > 0 for all vectors v =I O. Throughout this book we will use the following notation: Lin Lin + Sym Skw Psym Orth Orth +

=

the set of all tensors;

= the set of all tensors S with det S > 0; the the = the = the = the = =

set of all symmetric tensors; set of all skew tensors; set of all symmetric, positive definite tensors; set of all orthogonal tensors; set of all rotations.

The sets Lin +, Orth, and Orth + are groups under multiplication; in fact, Orth + is a subgroup of both Orth and Lin ". Orth is the orthogonal group; Orth + is the rotation group (proper orthogonal group). On any three-dimensional vector space there are exactly two cross products, and one is the negative of the other. We assume that one such cross product, written u x v

8

I.

TENSOR ALGEBRA

for all u and v, has been singled out. Intuitively, u x v will represent the righthanded cross product of u and v; thus if

then the basis {e;} is right handed and the components of u x v relative to {e;} are

Further, u x v = -v x u, u x u = 0,

u . (v x w) = w' (u x v) = v· (w xu). When u, v, and ware linearly independent, the magnitude of the scalar u· (v x w)

represents the volume of the parallelepiped 9 determined by u, v,w. Further, I det S

=

Su' (Sv x Sw)

u-fv x w) and hence

_ vol(S(9» Id et S I - vol(9) , which gives a geometrical interpretation of the determinant (Fig. 3). Here S(9) is the image of 9 under S, and vol designates the volume. There is a one-to-one correspondence between vectors and skew tensors: given any skew tensor W there exists a unique vector w such that Wv

=wx

v

for every v, and conversely; indeed,

[W] = [

~

-(3 corresponds to WI

1

= oc,

Cf., e.g., Nickerson, Spencer, and Steenrod [I, §S.2).

(7)

1. POINTS. VECTORS. TENSORS

9

Figure 3

We call w the axial vector corresponding to W. It follows from (7) that (for W =1= 0) the null space of W, that is the set of all v such that Wv= 0, is equal to the one-dimensional subspace spanned by w. This subspace iscalled axis ofW. We will frequently use the facts that "Y and Lin are normed vector spaces and that the standard operations of tensor analysis are continuous. In particular, on "Y and Lin, the sum, inner product, and scalar product are continuous, as are the tensor product on "Y and the product, trace, transpose, and determinant on suitable subsets of Lin.

EXERCISES

1. Choose a E "Y and let 1jJ: "Y ...... IR be defined by ljJ(v) = a' v. Show that

a 2.

= Li ljJ(ei)ei'

Prove the representation theorem for linear forms (page I).

3. Show that the sum S 4.

+ T and product ST are tensors.

Establish the existence and uniqueness of the transpose ST of S.

5. Show that the tensor product a ® b is a tensor. 6.

Prove that (a) (b) (c)

7.

Sea ® b) = (Sa) ® b, (a ® b)S = a ® (STb). Li (Sei) ® e, = S.

Establish (1), (2), (4), and (5).

I.

10

TENSOR ALGEBRA

8.

Show that

9.

(a) v = Su is equivalent to Vi = Lj SijUj, (b) (ST)ij = Sji' (c) (ST)ij = 1kj , (d) (a ® b)ij = a.b], (e) S . T = Li,j Sij Iij . Prove that the operation S· T is indeed an inner product; that is, show that (a) S· T = T . S, (b) S . T is linear in T for S fixed, (c) S·S ~ 0, (d) S· S = 0 only when S = O.

Lk s.,

10. Establish the proposition on p. 6. 11. Show that the trace of a tensor equals the trace of its symmetric part, so that, in particular, the trace of a skew tensor is zero. 12. Prove that Q is orthogonal if and only if QTQ = I. 13. Show that Q is orthogonal if and only if H = Q - I satisfies HH T = HTH. H + H T + HH T = 0, 14.

Let ({J: iI x iI x iI -+ IR be trilinear and skew symmetric; that is, ({J is linear in each argument and ({J(u, v, w)

= - ({J(v, U, w) = - ({J(u, w, v) = - ({J(w, v, u)

for all u, v, WE iI. Let SELin. Show that ({J(Se 1, e2, e 3)

+ ({J(et> Se2' e 3) + ({J(el, e 2, Se3)

= (tr S)({J(e 1, e2' e 3)·

15. Let Q be an orthogonal tensor, and let e be a vector with Qe = e. (a)

Show that QTe

(b)

= e.

Let W be the axial vector corresponding to the skew part of Q. Show that w is parallel to e.

16. Show that if w is the axial vector of W E Skw, then [w] 17.

1

= j2,WI.

Let DE Psym, Q E Orth. Show that QDQT E Psym.

2. SPECTRAL, CAYLEY-HAMILTON, AND OTHER THEOREMS

11

2. SPECTRAL THEOREM. CAYLEY-HAMILTON THEOREM. POLAR DECOMPOSITION THEOREM A scalar W is an eigenvalue of a tensor S if there exists a unit vector e such that Se = we, in which case e is an eigenvector. The characteristic space for S corresponding to W is the subspace of 1/ consisting of all vectors v that satisfy the equation Sv = wv. If this space has dimension n, then w is said to have multiplicity n. The spectrum ofS is the list (Wt, W2, ..•), where Wt :5; W2 :5; •.. are the eigenvalues ofS with each eigenvalue repeated a number of times equal to its multiplicity. Proposition (a) (b)

The eigenvalues of a positive definite tensor are strictly positive. The characteristic spaces ofasymmetric tensor are mutually orthoqonal.

Proof Let W be an eigenvalue of a positive definite tensor S, and let e be a corresponding eigenvector. Then, since Se = we and [e] = 1, W = e-Se > O. To prove (b) let wand A. be distinct eigenvalues of a symmetric tensor S, and let

Su = wu,

Sv

= A.V,

so that u belongs to the characteristic space of w, v to the characteristic space of A. Then wu • v = v· Su = u . Sv = AU' v, and, since ca

=f;

A,

U •

v = O.

0

The next result! is one of the central theorems oflinear algebra. Spectral Theorem. Let S be symmetric. Then there is an orthonormal basis for 1/ consisting entirely of eigenvectors of S. Moreover, for any such basis e., e2' e3 the corresponding eigenvalues Wt, W2' W3' when ordered, form the entire spectrum ofS and S

=

LWjej (8)ej.

(1)

j

I Cf., e.g., Bowen and Wang [I, §27]; Halmos [1, §79]; Stewart [I, §37]. The first two references state the theorem in terms of perpendicular projections onto characteristic spaces (cf. Exercise 4).

I.

12

TENSOR ALGEBRA

Conversely, ifS has the form (1) with {e.} orthonormal, then WI' w z , eigenvalues ofS with e., e 2, e3 corresponding eigenvectors. Further:

W3

are

(a) S has exactly three distinct eigenvalues ifand only if the characteristic spaces ofS are three mutually perpendicular lines through o. (b) S has exactly two distinct eigenvalues if and only if S admits the representation (2)

In this case WI and W2 are the two distinct eigenvalues and the corresponding characteristic spaces are sp{e} and {e}', respectively. Conversely, if sp{e} and {elL (lei = 1) are the characteristic spaces for S, then S must have the form (2). (c) S has exactly one eigenvalue ifand only if

S = wI.

(3)

In this case W is the eigenvalue and "1/ the corresponding characteristic space. Conversely, if"l/ is a characteristic space for S, then S has the form (3).

The relation (1) is called a spectral decomposition of S. Note that, by (1), the matrix ofS relative to a basis {eJ of eigenvectors is diagonal:

Further, in view of (a)-(c), the characteristic spaces of a symmetric tensor S are three mutually perpendicular lines through 0; or a line 1through 0 and the plane through 0 perpendicular to I; or S has only one characteristic space, "1/ itself. Thus if Olt~ (01: = 1, ... , n :::;; 3) denote the characteristic spaces ofS, then any vector v can be written in the form

v=

L v~,

(4)

~

Commutation Theorem.

Suppose that two tensors Sand T commute. Then T leaves each characteristic space ofS invariant; that is, ifv belongs to a characteristic space ofS, then Tv belongs to the same characteristic space. Conversely, if T leaves each characteristic space ofa symmetric tensor S invariant, then Sand T commute. Proof

Let Sand T commute. Suppose that Sv = wv. Then S(Tv)

= T(Sv) = w(Tv),

so that Tv belongs to the same characteristic space as v.

2. SPECTRAL, CAYLEY-HAMILTON, AND OTHER THEOREMS

13

To prove the converse assertion choose a vector v and decompose v as in (4). 1fT leaves each characteristic space 0/./,% ofS invariant, then Tv, E IJIJ,% and S(Tv,%) = w,%(Tv,%) = T(w,%v,%) = T(Sv,%), where w'% is the eigenvalue corresponding to 0/./,%' We therefore conclude, with the aid of (4), that

L STv,% =

STy =

'%

LTSv,% = TSv. '%

Thus, since v was chosen arbitrarily, ST = TS.

0

There is only one subspace of "Y that every rotation leaves invariant; namely "Y itself. We therefore have the following corollary of the spectral theorem.

Corollary.

A symmetric tensor S commutes with every rotation

S

=

if and only if

wI.

Crucial to our development of continuum mechanics is the polar decomposition theorem; our proof of this theorem is based on the

Square-Root Theorem. Let C be symmetric and positive definite. Then there is a unique positive definite, symmetric tensor U such that We write

U 2 = C.

.JC for U.

Proof

(Existence)

Let C = LWiei ®ei i

be a spectral decomposition of C E Psym and define U E Psym by

U=L~ei®ei' i

(Since co, > 0, this definition makes sense.) Then U 2 = C is a direct consequence of (1.2h.

(Uniqueness')

Suppose

U 2 = V2 = C with U, V E Psym. Let e be an eigenvector ofC with W > 0 the corresponding eigenvalue. Then, letting A = JOj,

o= I

Stephenson (1).

(U 2

-

wI)e = (U

+ AI)(U - AI)e.

I.

14

TENSOR ALGEBRA

Let v = (U - ..1.I)e. Then Uv = -..1.V and v must vanish, for otherwise -A. would be an eigenvalue of U, an impossibility since U is positive definite and A. > O. Hence Ue = ..1.e. Similarly, Ve

=

..1.e,

and Ue = Ve for every eigenvector e of C. Since we can form a basis of eigenvectors of C (cf. the spectral theorem), U and V must coincide. 0 Polar Decomposition Theorem. Let F E Lin +. Then there exist positive definite, symmetric tensors D, V and a rotation R such that (5)

F = RU = YR.

Moreover, each of these decompositions is unique; infact, U

= JFTF,

We call the representation F polar decomposition of F.

V

= JFF T •

(6)

= RU (resp. F = VR) the right (resp. left)

Proof. Our first step will be to show that FTF and FF T belong to Psym. Both tensors are clearly symmetric. Moreover, v • FTFv

= (Fv) • (Fv)

~

0,

and this inner product can equal zero only ifFv = 0, or equivalently, since F is invertible, only if v = o. Thus FTF E Psym. Similarly, FF T E Psym. (Uniqueness) Let F = RU be a right polar decomposition of F. Then since R is a rotation, FTF

= URTRU = U 2 •

But by the square-root theorem there can be at most one U E Psym whose square is FTF. Thus (6)1 holds and U is unique; since R = FU-l, R is also unique. On the other hand, if F = VR is a left polar decomposition, then FF T

= V2,

and V is determined uniquely by (6)z, R by the relation R = V - 1 F.

2. SPECTRAL, CAYLEY-HAMILTON, AND OTHER THEOREMS

(Existence)

15

Define U E Psym by (6)1 and let R

= FU- 1 .

To verify F = RU is a right polar decomposition we have only to show that R E Orth + . Since det F > 0 and det U > 0 (the latter because all eigenvalues ofU are strictly positive), det R > o. Thus we have only to show that R E Orth. But this follows from the calculation RTR = U- 1FTFU- t = U- 1U2U- 1 = I. Finally, define

v = RUR T. Then V E Psym (Exercise 1.17) and VR = RURTR = RU = F.

D

Given a tensor S, the determinant of S - wI admits the representation det(S - wI) = _w 3 + 11(S)W 2 - 12(S)W + 13(S) , (7) for every wEIR, where = tr S,

11 (S)

12(S) = t[(tr S)2 - tr(S2)], liS) = det

(8)

s.

We call (9)

the list of principal invariants' of S. When S is symmetric J s is completely characterized by the spectrum (w 1 , W2' w 3 ) ofS. Indeed, a simple calculation shows that 11 (S)

=

W1

+

12 (S) = W tW 2

liS) =

W2

+

W3'

+ W 2W3 + W tW3,

(10)

W tW 2W3·

Moreover, the above characterization is a one-to-one correspondence. To see this note first that WEIR is an eigenvalue of a tensor S if and only if W satisfies the characteristic equation det(S - wI) = 0, [ liS) are called invariants of S because of the way they transform under the orthogonal group: lk(QSQT) = liS) for all Q E Orth [cf. (37.3)].

I.

16

TENSOR ALGEBRA

or equivalently, (11)

Further, when S is symmetric the multiplicity of an eigenvalue w is equal to its multiplicity as a root of (11).1 Thus we have the following Proposition.

Let Sand T be symmetric and suppose that

f s

= fT'

Then Sand T have the same spectrum.

More important is the Cayley-Hamilton Theorem." equation:

Every tensor S satisfies its own characteristic (12)

EXERCISES

1. Determine the spectrum, the characteristic spaces, and a spectral de-

composition for each of the following tensors: A =

B

IXI

+ pm @ m,

= m@n+n@m.

Here IX and p are scalars, while m and n are orthogonal unit vectors. 2. Granted the validity of the portion of the spectral theorem ending with (1), prove the remainder of the theorem. 3. Let D E Sym, Q E Orth. Show that the spectrum ofD equals the spectrum of QDQT. Show further that if e is an eigenvector of D, then Qe is an eigenvector of QDQT corresponding to the same eigenvalue. 4. A tensor P is a perpendicular projection if P is symmetric and p 2 = P. (a) Let n be a unit vector. Show that each of the following tensors is a perpendicular projection: I,

(b)

0,

n@n,

1- n@n.

(13)

Show that, conversely, if P is a perpendicular projection, then P admits one of the representations (13).

1

Cf., e.g., Bowen and Wang [1, Theorem 27.4); Halmos [I, §78, Theorem 6).

2

cr. e.g., Bowen and Wang [1, Theorem 26.1); Halmos[1, §58).

SELECTED REFERENCES

17

5. Show that a (not necessarily symmetric) tensor S commutes with every skew tensor W if and only if S = wI.

(14)

6. Let F = RU and F = VR denote the right and left polar decompositions of FE Lin+. (a) (b)

Show that U and V have the same spectrum (WI' W2' w 3 ) . Show that F and R admit the representations F

=

L Wi f i e e., i

where e, and f i are, respectively, the eigenvectors of U and V corresponding to Wi' 7. Let R be the rotation corresponding to the polar decomposition of F E Lin +. Show that R is the closest rotation to F in the sense that, IF -

RI < IF - QI

(15)

for all rotations Q =I' R. The result (15) suggests that an alternate proof" of the polar decomposition can be based on the following variational problem: Find a rotation R that minimizes IF - Q lover all rotations Q. SELECTED REFERENCES

Bowen and Wang [1]. Brinkman and Klotz [1]. Chadwick [1, Chapter 1]. Halmos [1]. Martins and Podio-Guidugli [1]. Nickerson, Spencer, and Steenrod [1, Chapters 1-5]. Stewart [1].

1

Such a proof is given by Martins and Podio-Guidugli [1].

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CHAPTER

II Tensor Analysis

3. DIFFERENTIATION

In this section we introduce a notion of differentiation sufficientlygeneral to include scalar, point, vector, or tensor functions whose arguments are scalars, points, vectors, or tensors. To accomplish this we use the fact that IR, "Y, and Lin are normed vector spaces, and where necessary phrase our definitions in terms of such spaces. Let fJII and "If'" denote normed vector spaces, and let f be defined in a neighborhood of zero in fJII and have values in "If'". We say that f(u) approaches zerofaster than u, and write f(u)

=

as u

o(u)

or, more simply, f(u) = o(u),

if lim Ilf(u)II = • ->0

."'0

[u] 19

o.

-+

0,

II.

20

TENSOR ANALYSIS

[Here 11£(0)11 is the norm of £(0) on 11', while 11011 is the norm of 0 on 0lI.] Similarly, £(0)

=

g(o)

+ 0(0)

signifies that £(0) - g(o) = 0(0). Note that this latter definition also has meaning when f and g have values in C, since in this instance f - g has values in the vector space "Y. As an example consider the function cp(t) =

r, Then

cp(t) = o(t)

if and only if IX > 1. Let g be a function whose values are scalars, vectors, tensors, or points, and whose domain is an open interval f0 of IJl The derivative g(t) of gat t, if it exists, is defined by g(t) = dd g(t) = lim! [g(t + IX) - g(t)]. t ~-->O IX

(1)

If the values of g are points, then g(t + IX) - g(t) is a point difference and hence a vector. Thus the derivative of a point function is a vector. Similarly, the derivative of a vector function is a vector, and the derivative of a tensor function is a tensor. We say that g is smooth if g(t) exists at each t E f0, and if the function g is continuous on f0. Let g be differentiable at t. Then (1) implies that

lim! [g(t ~-o

IX

+ IX)

- g(t) - IXg(t)] = 0,

or equivalently, g(t + IX) = g(t) + IXg(t) + p(IX).

(2)

Clearly,

is linear in IX; thus g(t

+ IX)

- g(t)

is equal to a term linear in IX plus a term that approaches zero faster than IX. In dealing with functions whose domains lie in spaces of dimension greater than one the most useful definition of a derivative is based on this result; we define

3.

21

DIFFERENTIATION

the derivative to be the linear map which approximates get + a) - get) for small a. Thus let Ol/ and "fII be finite-dimensional normed vector spaces, let q; be an open subset of Ol/, and let

g: q;

"fII.

----+

We say that g is differentiable at x E q; if the difference g(x

+ u)

- g(x)

is equal to a linear function of u plus a term that approaches zero faster than u. More precisely, g is differentiable at x if there exists a linear transformation (3)

Dg(x): Ol/ ----+ if"

such that g(x

+ u) = g(x) + Dg(x) [u] + o(u)

(4)

as u ----+ O. If Dg(x) exists, it is unique; in fact, for each u, Dg(x) [u]

= lim! [g(x + au) cx-+O cxelRl

a

g(x)]

= dd

a

g(x

+ auHx=o-

We call Dg(x) the derivative of g at x, Since any two norms on a finitedimensional vector space are equivalent, Dg(x) is independent ofthe choice of norms on iJIt and "fII. If g is differentiable at each x E q;, then Dg denotes the map x f-+ Dg(x) whose domain is q; and whose codomain is the space of linear transformations from iJIt into if". This space is finite dimensional and can be normed in a natural manner; thus it makes sense to talk about the continuity and differentiability of Dg. In particular, we say that g is of class C 1 , or smooth, if g is differentiable at each point of q; and Dg is continuous. Continuing in this manner, we say g is of class C2 if g is of class C 1 and Dg is smooth, and so forth. Of course, the three-dimensional euclidean space is is not a normed vector space. However, when the domain q; of g is contained in is the above definition remains valid provided we replace iJIt in (3) by the vector space 1associated with is. Similarly, when g has values in is we need only replace if' in (3) by "1"'. Note that by (2), when the domain q; of g is contained in ~, Dg(t) [a]

for every a E

~.

= ag(t)

(5)

22

II.

TENSOR ANALYSIS

Smooth-Inverse Theorem. 1 Let ~ be an open subset of a finite-dimensional normed vector space 0/1. Further, let g: ~ -+ 0/1 be a one-to-one function ofclass C" (n ~ 1) and assume that the linear transformation Dg(x): 0/1 -+ 0/1 is invertible at each x E~. Then g-l is of class C".

Often, the easiest method of computing derivatives is to appeal directly to the definition. We demonstrate this procedure with some examples. Consider first the function tp: "Y" -+ ~ defined by qJ(V) = v· v.

Then qJ(V

+ u)

= v· v

+ 2v' u + u· u =

qJ(v)

+ 2v' U + o(u),

so that DqJ(v) [u] = 2v' u.

Similarly, for G: Lin

-+

Lin defined by G(A) = A 2

we have G(A

+ V) =

A2

+ AV +

VA

+V2 =

G(A)

+ AV + VA + o(V),

so that DG(A) [V] = AV

Next, let G: Lin

-+

+ VA.

(6)

Lin be defined by G(A) = A 3 •

Then G(A

+ V)

+ V)3 = A 3 + A 2V + AVA + VA 2 + o(V) = G(A) + A 2V + AVA + VA 2 + o(V),

= (A

and hence

As our last example, let L: 0/1 L(x

-+

"If'" be linear. Then

+ u) =

L(x)

+ L(u),

1 This result is a direct consequence of the inverse function theorem (cf. Dieudonne [I, Theorem 10.2.5]).

3.

23

DIFFERENTIATION

so that, trivially, DL(x) = L.

(7)

The next two results involve far less trivial computations.

Theorem (Derivative of the determinant).

Let cp be defined on the set ofall

invertible tensors A by cp(A)

= det A.

Then cp is smooth. Infact, Dcp(A) [V]

= (det A) tr(VA -1)

(8)

for every tensor V. Proof

By (2.7) with co = -1, det(I + 8) = I + tr 8 + 0(8)

as 8

--+

O. Thus, for A invertible and V

E

Lin,

det(A + V) = det[(I + VA -1)A] = (det A) det(I + VA -1) = (det A) [1 + tr(VA -1) + o(V)] = det A + (det A) tr(VA -1) + o(V) as V

--+

O. Therefore, since the map

V

f-+

(det A) tr(VA -1)

is linear, (8) must be valid. The proof that Dcp is continuous follows from the continuity of the determinant, trace, and inverse operations. 0

Theorem (Derivative of the square root).

Thefunction H: Psym

--+

Psym

defined by

H(C)

=.jC

is smooth. Proof By the square-root theorem (page 13), the function H is one-toone with inverse G: Psym --+ Psym defined by

G(A) = A 2 •

Clearly, G is smooth with derivative DG(A): Sym --+ Sym given by (6). Thus, in view of the smooth-inverse theorem, to complete the proof it suffices to show that DG(A) is invertible at each A E Psym, or equivalently, that DG(A) [V] = 0

implies

V = O.

II.

24

TENSOR ANALYSIS

Assume the former holds, so that, by (6), AU

+ UA = O.

Let A be an eigenvalue of A with e a corresponding eigenvector. Then AUe

+ UAe = AUe + AUe =

0

and A(Ue)

=

-

A(Ue).

If Ue # 0, then - A is an eigenvalue of A. But A is positive definite, so that A and - A cannot both be eigenvalues. Thus Ue = 0, and this must hold for every eigenvector e of A. By the spectral theorem, there is a basis for 1/ of eigenvectors of A. Thus U = O. D It will frequently be necessary to compute the derivative of a product n(f, g) of two functions f and g. In tensor analysis there are many different products available; for example, the product of a scalar cp and a vector v

n(cp, v)

=

cpv,

the inner product of two vectors n(u, v) = u • v,

the tensor product of two vectors n(u, v)

= u ® v,

the action of a tensor S on a vector v n(S, v)

= Sv,

and so forth. The above operations have one property in common, bilinearity. Therefore, in order to establish a product rule valid in all cases of interest, we consider the general product operation

n: ff x

is smooth on fYI if et> is smooth on Ii, and ifet> and Vet> have continuous extensions to all of fYI; in this case we also write et> and Vet> for the extended functions. An analogous interpretation applies to the statement" et> is of class eN on fYI"; note that (for fYI open or closed) this will be true if and only if the components of et> have continuous partial derivatives of all orders 5. N on fYI. A vector field of the form v = Vcp satisfies curl v = 0 (Exercise 5a). Our next theorem, which we state without proof, gives the converse of this result. Potential Theorem.' Let v be a smooth point field on an open or closed, simply connected region fYI, and assume that curl v Then there is a class

e

2

= O.

scalar field cp on fYI such that

v = Vcp. We close this section by proving that vector fields with constant gradients are affine. 1

Cf., e.g., Fleming [I, Corollary 2, p. 279].

II.

36

TENSOR ANALYSIS

Proposition. Let f be a smooth point- or vector-valuedfield on an open or closed region !71, and assume that F = Vf is constant on !71. Then f(x) = f(y) + F(x - y)

(9)

for all x, y E !71. Proof Choose x, y E !71. Since !71 is connected there is a curve c in !71 from y to x. Thus

f(x) - f(y) =

i I 0

=F

d

=

da f(c(a)) do

f

Ii 0

Vf(c(a))c(a) da

c(a) do = F(x - y).

D

By (9) any vector field fwith constant gradient F can be written in the form f(x) = a

+ F(x

- x o)

with a E 1/ and Xo E $. Moreover, the point X o can be arbitrarily chosen: (Of course, a depends on the choice of xo.)

EXERCISES

1.

Let 0(, 0) is the closed ball of radius 8 centered at

Xo'

Therefore,

for every closed ball 0 c fJIl, then

Proof

Va

I(xo) - ~va ino dV I,

= vol(Oa)' Then

t, ::;; ~ Va

i

1(xo) - (x)j dVx

no

which tends to zero as 8 1

o.

Let

t, = where

=

::;;

sup I(x o) - (x) I

xen o -+

0, since is continuous.

See, e.g., Kellogg [1, Chapter 4].

0

if

5, THE DIVERGENCE THEOREM. STOKES' THEOREM

39

Note that the localization theorem and the divergence theorem together yield the following interesting interpretation of the divergence: div v(x) div S(x)

= lim

~-o vo

= lim

~-o vo

l~n)

l~n)

~ ~

r v . dA, r So dA. Jilfi

Jilfi

0

6

6

We will make use of Stokes' theorem, but only in the following very special form. Stokes' Theorem.' Let v be a smooth vector field on an open set [j£ in Iff. Further, let n be a disc in [j£ (Fig. 3), let 0 be a unit normal to n, and let the bounding circle c(O"), 0 ~ 0" ~ 1, be oriented so that [c(O) x c(O")] • 0 > 0, Then

In(curl v) .

0

dA =

i

V

0 O. (b) Compute f(f!l) and use this result to demonstrate that f is not a deformation. (c) Show that (5)2 is not satisfied. 7. SMALL DEFORMATIONS

We now study the behavior of the various kinematical fields when the displacement gradient Vu is small. Since f(p)

+ u(p),

=

p

=

I + Vu;

it follows that F

(1)

hence the Cauchy-Green strain tensors C and B, defined by (6.6), obey the relations C = I + Vu + VuT + VuT Vu,

B

=

I

+ Vu + VuT + Vu VuT •

(2)

7.

55

SMALL DEFORMA nONS

When the deformation is rigid, C = B = I and

Vu + VuT + VuVuT

=

O.

(3)

Moreover, in this case Vu is constant, because F is. The tensor field (4)

is called the infinitesimal strain; clearly

C = I + 2E + VuT Vu,

Proposition.

(5)

+ 2E + Vu VuT .

B= I

Let f. (0 <s < eo) be a one-parameter family of deformations

with

IVu.1 = e. Then

+ o(e)

2E. = C. - I as s

~

O. Further,

=

B. - I

+ o(e)

(6)

if each f. is rigid, then VUE =

Vui'

-

+ o(e).

(7)

Proof The result (6) is a trivial consequence of (2), while (7) follows from (3). 0 This proposition asserts that to within an error of order o(e) the tensors

2E., C. - I, and B. - I coincide. It asserts, in addition, that to within the same error the displacement gradient corresponding to a rigid deformation is skew. The above discussion should motivate the following definition: An infinitesimal rigid displacement of f!4 is a vector field u on f!4 with Vu constant and skew; or equivalently, a vector field u that admits the representation u(p)

=

u(q)

+ W(p

- q)

(8)

for all p, q E f!4, where W is skew (cf. the proposition on page 36). Of course, using the relation between skew tensors and vectors, we can also write u in the form u(p) = u(q)

+ co x

(p - q)

with cothe axial vector corresponding to W.

III.

56

KINEMATICS

Theorem (Characterization of infinitesimal rigid displacements). a smooth vector field on f!4. Then the following are equivalent:

(a) (b)

u is an infinitesimal rigid displacement. u has the projection property: for all p, q

E

f!4,

(p - q) . [u(p) - u(q)] = (c) (d)

Let u be

o.

Vu(p) is skew at each p E f!4. The infinitesimal strain E(p) = 0 at each p E f!4.

Proof

(a)

=>

(b).

Let u be rigid. Then (8) implies

(p - q) • [u(p) - u(q)J = (p - q) . W(p - q) = 0,

since W is skew. (b) => (a). If we differentiate the expression in (b) with respect to p, we arrive at u(p) - u(q)

+ VU(p)T(p

- q) = 0,

and this result, when differentiated with respect to q, yields - Vu(q) - Vu(p? = O.

(9)

Taking p = q tells us that Vu(p) is skew; hence (9) implies that Vu(p) = Vu(q)

for all p, q E 14, and Vu is constant. Thus (a) holds. (a) ¢> (c). Trivially, (a) implies (c). To prove the converse assertion assume that H(p) = Vu(p) is skew at each p E 14. We must show that H is constant. Let n be an open ball in f!4. Choose p, q E n and let c(a) = q

+ a(p

- q),

0

~ a ~

1,

so that c describes the straight line from q to p. Then u(p) - u(q) =

so that

i

Vu(x) dx

=

f f

(p - q) • [u(p) - u(q)] =

H(c(a»c(a) do =

f

H(c(a»(p - q) do,

(p - q) . H(c(a»(p - q) do = 0,

since H is skew. Thus u has the projection property on n, and the argument given previously [in the proof of the assertion (b) => (a)] tells us that H is constant on n. But n is an arbitrary open ball in f!4; thus H is constant on f!4. (c) ¢> (d). This is a trivial consequence of (4). 0

7.

SMALL DEFORMA nONS

57

EXERCISES

1. Under the hypotheses of the proposition containing (6) show that

E. = U. - I + 0(8) = V. - I + 0(8), det F. - 1 = div U.

+ 0(8).

Give a physical interpretation of det F. - 1 in terms of the volume change in the deformation f e : 2. Let u and v be smooth vector fields on rJI and suppose that u and v correspond to the same infinitesimal strain. Show that u - v is an infinitesimal rigid displacement. For the remaining exercises u is a smooth vector field on rJI and E is the corresponding infinitesimal strain. Also, in 3 and 5, rJI is bounded.

3. Define the mean strain E by vol(rJI)E =

L

E dV.

Show that vol(rJI)E

=

"21 Jr

oal

(u

® n + n ® u) dA,

so that E depends only on the boundary values of u. 4. Let

w

=

t(Vu - VuT ) .

Show that

5. (Korn's inequality)

IEI 2

+ IWl 2 =

IEl 2

-

IWI 2 = Vu' VuT •

Let u be of class C 2 and suppose that

u=O Show that

IVuI2 ,

on

i}rJI.

III.

58

KINEMATICS

6. Consider the deformation defined in cylindrical coordinates Xl

= r cos e,

PI = R cos

X2

= r sin e,

P2

e, = R sin e,

P3

=

Z,

by

e= e

r = R,

+

rLZ,

z

z.

=

A deformation of this type is called a pure torsion; it describes a situation in which a cylinder with generators parallel to the Z-axis is twisted uniformly along its length with cross sections remaining parallel and plane. The constant rL represents the angle of twist per unit length. Show that the corresponding displacement is given by

/3 P2(COS /3 -

uI(p) = PI(COS

uip)

=

/3, 1) + PI sin /3, 1) - P2 sin

/3

= rLP3'

U3(P) = O.

Show further that both Vu and u approach zero as a fact,

-+

0, and that, in

UI(P) = -rLP2P3 + o(rL),

(10)

uip) = rLPIP3 + o(rL)

as

rL -+

O.

8. MOTIONS Let fJd be a body. A motion of fJd is a class C 3 function x:fJd x

~-+g

with x(', t), for each fixed t, a deformation of fJd (Fig. 6). Thus a motion is a smooth one-parameter family of deformations, the time t being the parameter. We refer to ' x = x(p, t) as the place occupied by the material point p at time t, and write fJd t

= x(fJd, t)

1 We carefully distinguish between the motion x and its values x, and between the reference map p and material points p.

8.

59

MOTIONS

• x = X(P. t)

Figure 6

for the region of space occupied by the body at t. It is often more convenient to work with places and times rather than with material points and times, and for this reason we introduce the trajectory

At each t, x(', t) is a one-to-one mapping of flI onto inverse p(', t): flit

flit;

hence it has an

flI

~

such that x(p(x, r), t)

Given (x, t)

E

= x,

p(x(p, r), t) = p.

!Y,

p = p(x, t) is the material point that occupies the place x at time t. The map p: !Y ~ flI so defined is called the reference map of the motion. We call

.

a x(p, t)

x(p, t)

= at

x(p, t)

= :t22

the velocity and x(p, t)

60

III.

KINEMATICS

the acceleration. Using the reference map p we can describe the velocity x(p, r) as a function vex, t) of the place x and time t. Specifically,

v: ,'Y -+

'f~

is defined by

vex, t)

=

x(p(x, t), t)

and is called the spatial description of the velocity. The vector vex, t) is the velocity of the material point which at time t occupies the place x. More generally, any field associated with the motion can be expressed as a function of the material point and time with domain fJ4 x IR, or as a function of the place and time with domain ,'Y. We therefore introduce the following terminology: a material field is a function with domain fJ4 x IR; a spatial field is a function with domain :Y. The field i is material, the field v is spatial. It is a simple matter to transform a material field into a spatial field, and vice versa. We define the spatial description Domain Arguments Gradient with respect to first argument Derivative with respect to second argument (time) Divergence Curl

~x~

Material point p and time t

Spatial field n

.r Place x and time t grad

n

0 smooth. (Here {ei } is an orthonormal basis.) Compute p, v, and L, and determine the streamlines. 6. Define the spatial gradient and spatial time derivative of a material field and show that X'

7. Consider a surface

[I'

= 0,

grad x = I.

in fJI of the form [I'

= {p E .@lqJ(p) = O},

where .@ is an open subset of fJI and qJ is a smooth scalar field on .@ with VqJ never zero on [1'. Let x be a motion of fJI. Then, at time t, [I' occupies the surface [1', =

{x E .@,Il/J(x, t) = O},

where .@, = x(.@, t) and l/J(x, t)

=

qJ(p(x, t».

Show that: (a) VqJ(p) is normal to [I' at p E [1'; (b) grad l/J(x, t) is normal to [1', at x E [1',; (c) VqJ = F T (grad l/J)..., and hence grad l/J(x, t) never vanishes on Y't; (d) IVqJ 12 = (grad l/J)... • B(grad l/J)_, where B = FFT is the leftCauchyGreen strain tensor; (e) l/J' = - v • grad l/J. 9. TYPES OF MOTIONS. SPIN. RATE OF STRETCHING A motion x is steady if for all time t and v' = 0

everywhere on the trajectory ff. Note that !!I = fJlo x IR, because the body occupies the same region fJlo for all time. Also, since the velocity field v is independent of time, we may consider v as a function xr->v(x) on fJlo . Thus in a steady motion the particles that cross a given place x all cross x with the same velocity v(x). Of course, for a given material point p the velocity will generally change with time, since x(p, t) = v(x(p, t». Consider now a (not necessarily steady) motion x and choose a point p with x(p, T) E ofJI, at some time T. Then (6.5)2 with f = x(', T) implies that

68

III.

KINEMATICS

P E iJ!!J and a second application of (6.5h, this time with f = x(', r), tells us that x(p, t) E iJ!!Jt for all t. Thus a material point once on the boundary lies on the boundary for all time. If the boundary is independent of time (iJ!!J t = iJ!!J o for all r), as is the case in a steady motion, then x(p, t), as a function of t, describes a curve on iJ!!J o , and x(p, t) is tangent to iJ!!J o . Thus we have the following Proposition. In a steady motion the velocity field is tangent to the boundary; i.e., v(x) is tangent to iJ!!J o at each x E iJ!!J o .

In a steady motion path lines and streamlines satisfy the same autonomous differential equation s(t)

= v(s(t».

Thus, as a consequence of the uniqueness theorem for ordinary differential equations, we have the following Proposition. In a steady motion every path line is a streamline and every streamline is a path line.

Let be a smooth field on the trajectory of a steady motion. Then is steady if '

=

0

(1)

[in which case we consider as a function x ....... (x) on !!JoJ. Proposition. Let qJ be a smooth, steady scalar field on the trajectory of a steady motion. Then the following are equivalent: (a)

qJ is constant on streamlines; that is, given any streamline s,

d

dt qJ(s(t» = 0 for all t.

(b) q, = O. (c) v· grad qJ = O. Proof

Note first that, by (8.4)1 and (1),

q, =

v • grad qJ;

thus (b) and (c) are equivalent. Next, for any streamline s, :t qJ(s(t»

=

s(t) • grad qJ(s(t»

=

v(s(t» • grad qJ(s(t»,

(2)

so that (c) implies (a). On the other hand, since every point of !!J o has a streamline passing through it, (a) and (2) imply (c). D

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

69

x(q, t)

x(a, t)

Figure 7

x(P,

t)

The number bet) = Ix(p, t) - x(q, t) I

(3)

represents the distance at time t between the material points p and q. Similarly, the angle OCt) at time t subtended by the material points a, p, q is the angle between the vectors x(a, t) - x(p, t) and x(q, t) - x(p, t). (See Fig. 7.) A motion x is rigid if

a

at Ix(p, t) - x(q, t)! = 0

(4)

for all materials points p and q and each time t. Thus a motion is rigid if the distance between any two material points remains constant in time. Theorem (Characterization of rigid motions). Let x be a motion, and let v be the corresponding velocity field. Then the following are equivalent:

(a) x is rigid. (b) At each time t, v(', t) has the form ofan infinitesimal rigid displacement of fJI,; that is, v(', t) admits the representation

Vex, t)

=

v(y, t) + W(t)(x - y)

for all x, y E fJI where Wet) is a skew tensor. " (c) The velocity gradient L(x, t) is skew at each (x, t) Proof.

(5) E

.'!I.

If we use (3) to differentiate b(t)2, we find that bet) J(t) = [x(p, t) - x(q, t)] • [x(p, t) - x(q, t)],

or equivalently, letting x and y denote the places occupied by p and q at time t, bet) J(t) = (x - y) • [vex, t) - v(y, t)J.

(6)

By (6), (4), and the fact that bet) 'I- 0 for p 'I- q, x is rigid if and only if v(-, t) has the projection property at each time t. The equivalence of (a), (b), and (c) is therefore a direct consequence of the theorem characterizing infinitesimal rigid displacements (page 56). D

70

III.

KINEMATICS

Let ro(t) be the axial vector corresponding to W(t); then (5) becomes v(x, t) = v(y, t)

+ ro(t)

x (x - y),

which is the classical formula for the velocity field of a rigid motion. The vector function ro is called the angular velocity. Note that curl v

= 2ro,

which gives a physical interpretation of curl v, at least for rigid motions. For convenience, we suppress the argument t and write v(x)

v(y)

=

+ ro x (x - y).

Assume ro ¥- O. Then for fixed y the velocity field XHro x (x - y) vanishes for x on the line {y + O(rolO( E

~}

and represents a rigid rotation about this line. Thus given any fixed y, v is the sum of a uniform velocity field with constant value v(y) and a rigid rotation about the line through y spanned by roo For this reason we calli = sp{ro} the spin axis. For future use, we note that 1= l(t) can also be specified as the set of all vectors e such that We=O. As we have seen, a rigid motion is characterized (at each time) by a velocity gradient which is both constant and skew. We now study the case in which the gradient is still constant, but is symmetric rather than skew. Thus consider a velocity field of the form v(x)

= D(x - y)

with D a symmetric tensor. By the spectral theorem D is the sum of three tensors of the form O(e ® e,

lei = 1,

with corresponding e's mutually orthogonal. It therefore suffices to limit our discussion to the velocity field

= O(e ® e)(x - y). Relative to a coordinate frame with e = e., v has components v(x)

(Vl' 0, 0),

(7)

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

---

-----

---

---- ----

•y Figure 8

71

--- ---

and is described in Fig. 8. What we have shown is that every velocity field with gradient symmetric and constant is (modulo a constant field) the sum of three velocity fields of the form (7) with" axes" e mutually perpendicular. Now consider a general velocity field v. Since L = grad v, it follows that v(x)

=

v(y) + L(y)(x - y) + o(x - y)

as x --+ y, where y is a given point, and where we have suppressed the argument t. Let D and W, respectively, denote the symmetric and skew parts ofL: D

= !

+ div v) dV, (5)

' dV

+

f

JO~t

v • n dA.

10. TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS

Proof d dt

79

In view of (2),

i

+ so that in an isochoric motion with v = 0 on of!Jt ,

i

WvdV = O.

91,

3.

Derive (5)2 for . Show that

v = grad ( q>' + v;). so that the acceleration is also the gradient of a potential. SELECTED REFERENCES

Chadwick [1, Chapter 2]. Eringen [1, Chapter 2]. Germain [1, Chapters 1, 5]. Serrin [1, §§1l, 17,21-23,25-29]. Truesdell [1, Chapter 2]. Truesdell and Noll [1, §§21-25]. Truesdell and Toupin [1, §§13-149].

(10)

This page intentionally left blank

CHAPTER

IV Mass. Momentum

12. CONSERVATION OF MASS One of the most important properties of bodies is that they possess mass. We here consider bodies whose mass is distributed continuously. No matter how severely such a body is deformed, its mass is the integral of a density field; that is, given any deformation f there is a density field Pr on f( fJi) such that the mass m(&l) of any part &l is given by

f.

m(&l) =

Pr dV.

r(&')

Since the mass of a part cannot be altered by deforming the part, m(&l) is independent of the deformation f. We now make the above ideas precise. A mass distribution for fJi is a family of smooth density fields

Pr: f(fJi)

-+

IR+,

one for each deformation f, such that

f.

r(&')

Pr dV =

f.

PI dV == m(&l)

1(&')

87

(1)

88

IV.

MASS. MOMENTUM

for any part f!/' and all deformations f and g. The number Pr(x) represents the density at the place x E f(~) in the deformation f, and equation (I) expresses conservation ofmass. We denote by Po the density field Pr when f(p) = p for all p E~. Thus Po(p) gives the density at p when the body is in the reference position. Note that, by the localization theorem, m(nel) . Po(p) = lim l(n )' s

el-O VO

where n el is the ball of radius /j centered at p. As our next result shows, the reference density Po determines the density in all deformations. Proposition.

Let f be a deformation of ~ and let F

= Vf. Then

Pr(x) det F(p) = Po(p) provided x Proof

(2)

= f(p). By (1) and the definition of Po,

r

Pr(x) dY.,

=

Jr(9'l

f

Po(p) dVp '

9'

On the other hand, if we change the variable of integration on the left side from x to p, we arrive at Lpr(f(p» det F(p) dv.,. Thus

L

[p,(f(p» det F(p) - Po(p)] dVp = 0

for every part f!/', and (2) follows from the localization theorem.

0

Given a motion x of ~ we will always write p(x, t) for the density at the place x E ~t in the deformation x(., t). Thus p::r

-+

IR+

is defined by p(x, r) = PX(.,t)(x); we will refer to P as the density in the motion x. In view of (1), m(f!/') =

f

9',

p(x, r) dV"

==

f

9',

P dV,

89

12. CONSERVATION OF MASS

and we have the following

Theorem.

F or every part f!J' and time t, d dt

f

=

p dV

O.

(3)

~t

If F = Vx is the deformation gradient in the motion, then (2) takes the form

p(x, t) det F(p, t)

=

Po(p)

(4)

provided x = x(p, t). Thus p is the spatial description of po/det F, and we conclude from the smoothness lemma that p is smooth on !Y.

Theorem (Local conservation of mass)

p + p div v = 0, p' + div(pv) = O. Proof

(5)

By (4),

+ p( det

p( det F):

F). = 0,

which with (10.2) yields (5k Next, by (8.4)1'

P=

p'

+ v· grad p,

and (5)1' when combined with this relation, implies (5h.

D

Since a motion is isochoric if and only if div v = 0, (5) has the following

Corollary.

A motion is isochoric

if and only if p = O.

Equation (3) expresses conservation of mass for a region f!jJt that moves with the body. It is often more convenient, however, to work with afixed region &/ called a control volume. Of course, material will generally flow into and out of fit, a fact which will become clear in the derivation of the resulting conservation law. By a control volume at time t we mean a bounded regular region &/ with &/

!fl t

c

for all r in some neighborhood of t, Thus for J sufficiently small we have the situation shown in Fig. 1. Let n denote the outward unit normal to 8&/. By the divergence theorem,

rdiv(pv) dV Jr pv : n dA.

J9I

=

091

90

IV.

MASS. MOMENTUM

Figure 1

Further, Lp'(X, t) dVx = L :t p(x, t) dv., = :t Lp(X, t) dVx '

where the last step uses the fact that .o/l is independent of t. Conservation of mass (5h, when combined with the relations above, yields the following

Theorem (Conservation of mass for a control volume). volume at time t. Then d dt

f ~

f

p(x, t) dVx = -

Let ~ be a control

p(x, t)v(x, t)· n{'x) dA x •

(6)

o~

Since n is the outward unit normal to a~, pv • n represents the mass flow, per unit area, out of ~ across its boundary. Thus (6) asserts that the rate of increase of mass in ~ is equal to the mass flow into ~ across its boundary. Lemma.

Let 0 be a smooth scalar field on f!It, let A. be a scalar field on iJf!It, let d be the set of all smooth vector fields v on f!It that satisfy div(pv) = 0

in

f!It,

v • n = A. on

and define the kinetic energy % on d by

%{v}

=

L~

pdV.

Show that if v E d has the form v

= grad cp,

iJf!It,

92

IV.

MASS. MOMENTUM

then

X{v}

X{g}

~

for all g E d, with equality holding only when v = g.

13. LINEAR AND ANGULAR MOMENTUM. CENTER OF MASS Let x be a motion of fll. Given a part 9 of fll, the linear momentum 1(9, t) and the angular momentum a(9, t) (about the origin 0) of 9 at time tare defined by 1(9, t) =

f f

vp dV,

@,

a(9, t) =

(1)

r x vp dV,

@,

where r: Iff ~

r

is the position vector

=x -

r(x)

Proposition.

(2)

o.

For every part 9 and time t,

=

i(9, t)

f f

vp dV,

@,

a(9, t) =

(3)

r x vp dV.

@,

Proof

The identity (12.8) yields (3)1 and a(9, t) =

f

(r x v)"p dV.

@,

But by (8.6),

(r x v)"

=r x

and, since v x v = 0, (3}z follows.

v+ v x

v,

0

Assume now that fJ8 is bounded, so that m(fll) is finite. Then the center of mass e(r) at time t is the point of space defined by e(r) -

0 =

m(~)

L,r p dV.

(4)

13. LINEAR AND ANGULAR MOMENTUM. CENTER OF MASS

93

(It is not difficult to show that this definition is independent of the choice of origin 0.) If we differentiate (4) with respect to t and use (12.8) and (8.6), we

find that

m(~) LtVP dV,

ci(t) =

so that ci represents the average velocity of the body. This result and (1)1 imply that l(gjI, t) = m(gjI)ci(t).

(5)

Thus the linear momentum of a body gjI is the same as that of a particle of mass m( gjI) attached to the center ofmass of gjI.

EXERCISES

In these exercises :14 is bounded. 1.

Show that IX(t) - z = _1_ m(gjI)

f

(x - z)p(x, t) dV,.,

(6)

ga,

for any point z, so that IX(t) is independent of the choice of origin. 2. Other types of momenta of interest are the angular momentum a.(t) relative to a moving point z(t) and the spin angular momentum aspin(t):

asPin(t)

=

f

(7)

r" x V"P dV.

gat

.

Here rz(x, t) = x - z(t)

(8)

is the position vector from z(t), r,,(x, t) is the position vector from IX(t), and V" =

i" =

V -

ci

is the velocity relative to IX. For convenience, let I(t) = l(gjI, t) and a(t) = a(gjI, t). Show that a z = aspin a

=

aspin

+ (IX

- z) x I,

+ (IX - 0) x i.

94

IV.

MASS. MOMENTUM

The term «x - z) x I is usually referred to as the orbital angular momentum about z; it represents the angular momentum /JI would have if all of its mass were concentrated at the center of mass. 3. Consider a rigid motion. By (9.17),

Xo(y, t)

=

q(t)

+ Q(t)(y

- z)

(9)

for all y E /JI o and all t. (Here X o is x, at r = 0; i.e., X o is the motion taking the configuration at time r = 0 as reference.) (a)

Show that

= q(t) + Q(t)[(X(O) - z]

(X(t)

(10)

and hence [noting that xo(y, t) is well defined for all y E C]

!X(t) = xo((X(O), t). What is the meaning of this result? (b)

Show that the angular velocity co(t) is the axial vector ofQ(t)Q(t)T and that

v.. = co x r... (c)

A vector function k on

~

(11)

rotates with the body if k(t) = Q(t)k(O)

(12)

for all t. [Note that Q(O) = I, since xo(y, 0) = y for all y.] Show that k rotates with the body if and only if

k= (d)

co x k.

(13)

Use the identity

f x (d x f)

=

(f 21 - f ® f)d

to show that

as p in

=

Jco,

where J(t) =

(e)

fBI,(r;1 -

r.. ® r..)p dV

(14)

is the inertia tensor of /JI t relative to the center of mass. Show that J(t) = Q(t)J(O)Q(t)T,

(15)

and use this fact to prove that the matrix [J(t)] of J(t)-relative to any orthonormal basis {e;(t)} that rotates with the body-is independent of t.

95

SELECTED REFERENCE

(f)

Construct an orthonormal basis {ei(t)} that rotates with the body and has J1

[J] = [

0

o

J2

0

0] O.

0

J3

In this case {ei(t)} is called a principal basis and the corresponding numbers J; are called moments of inertia. Let w;(t) denote the components of ro(t) with respect to {e;(t)}. Show that the components of aspin(t) relative to this basis are given by

+ (J 3 J 2 W2 + (J 1 -

(aspin)! = J l w!

J 2 )W 2 W 3 ,

(a.Pin) 2 =

J 3)WI W3,

(a.Pinh

= J 3 W3 +

(J 2

-

J l )W I W 2 '

4. Define the kinetic energy S(t) and the relative kinetic energy S ,,(t) by

S ,,(t)

=

t Jr ,v; DI

(a)

Prove Konig's theorem: S = SOl

(b)

+ tm(81)ci2.

Consider a rigid motion. Use the identity (d x £)2

= d· (£21 - f

(8) £)d

to prove that

5. Show that

J = (tr M)I - M, where M(t) =

JrDI, f" (8) f"P dV

is called the Euler tensor.

SELECTED REFERENCE

Truesdell and Toupin [1, §§155-171].

p dV.

This page intentionally left blank

CHAPTER

v Force

14. FORCE. STRESS. BALANCE OF MOMENTUM During a motion mechanical interactions between parts of a body or between a body and its environment are described by forces. Here we shall be concerned with three types offorces: (i) contact forces between separate parts of a body; (ii) contact forces exerted on the boundary of a body by its environment; (iii) body forces exerted on the interior points of a body by the environment. One of the most important and far reaching axioms in continuum mechanics is Cauchy's hypothesis concerning the form of the contact forces. Cauchy assumed 1 the existence of a surface force density s(n, x, t) defined for each unit vector n and every (x, t) in the trajectory ff of the motion (Fig. 1). This field has the following property: Let!/' be an oriented surface in !!4t with positive unit normal n at x. Then s(n, x, t) is the force, per unit area, exerted across!/' upon the material on the negative side of!/' by the material on the positive side. Cauchy's hypothesis is quite strong; indeed, if ~ is an oriented surface tangent to !/' at x and having the same positive unit normal there, then the force per unit area at x is the same on ~ as on !/' (Fig. 2). 1 Noll [3) has shown that Cauchy's hypothesis actually follows from balance of linear momentum under very general assumptions concerning the form of the surface force s. (Cf. also Truesdell [I, p. 136); Gurtin and Williams [1).)

97

98

V.

FORCE

s(n, x, I)

+

Figure 2

Figure I

To determine the contact force between two separate parts rY' and!!} at time t (Fig. 3) one simply integrates s over the surface of contact

!/t = rY't n !!}t; thus

f

s(o",

X,

t) dA"

==

Sf',

f

s(o) dA

Sf',

gives the force exerted on rY' by!!} at time t. Here 0" is the outward unit normal to orY't at x. For points on the boundary of !!It, s(o, X, t)-with n the outward unit normal to o!!lt at x-gives the surface force, per unit area, applied to the body by the environment. This force is usually referred to as the surface traction. In any case, given a part rY',

i

s(o) dA

iJ~,

represents! the total contact force exerted on rY' at time t (Fig. 4). The environment can also exert forces on interior points of 14, a classical example being the force field due to gravity. Such forces are determined by a vector field b on .r; b(x, t) gives the force, per unit volume, exerted by the environment on x. Thus for any part rY' the integral

f

J~,

b(x, t) dV" ==

i

~,

b dV

gives that part of the environmental force on rY' not due to contact. 1 Here it is tacit that, given any part [7; and time t, 5(n., x, r) is an integrable function ofx on of!l'" This assumption of integrability actually follows from the momentum balance laws (I) and our hypotheses (i) and (ii) concerning force systems, Indeed, (i) trivially implies integrability on surfaces of polyhedra, and this is all that is needed to establish the existence of the stress T; (9) and (11) then yield integrability on of!l', for any part [7;,

14.

FORCE. STRESS. 8ALANCE OF MOMENTUM

Figure 3

99

Figure 4

The above discussion should motivate the following definitions. Let AI be the set of all unit vectors. By a system of forces for 9B during a motion (with trajectory ff) we mean a pair (s, b) of functions

b: ff

s: AI x ff ~ "f/,

with (i) (ii)

"//,

~

s(n, x, t), for each n E .K and t, a smooth function of x on 9B 1 ; b(x, t), for each t, a continuous function of x on 9B t •

We call s the surface force and b the body force; and we define the force f(&>, t) and moment m(&>, t) (about 0) on a part &> at time t by f(&>, t) =

i

s(n) dA +

i

r x s(n) dA +

ofJ',

m(&>, t) =

f

b dV,

fJ',

o!¥.,

f

r x b dV.

fJ',

Here n is the outward unit normal to iJ&>t and r is the position vector (13.2). The basic axioms connecting motion and force are the momentum balance laws. These laws assert that for every part &>and time t, f(&>, t)

=

l(&>, t),

(1)

m(&>, t) = a(&>, t); they express, respectively, balance of linear momentum and balance of angular momentum [cf. (13.1)J. Tacit in the statement of axiom (1) is the existence of an observer (frame of reference) relative to which the motion and forces are measured. The existence of such an observer is nontrivial, since the momentum balance laws are generally not invariant under changes in observer (cf. Exercise 21.3). Observers relative to which (1) hold are called inertial; in practice the fixed stars are often used to define the class of inertial observers.

v.

100

FORCE

An obvious consequence of (1)1 and (13.5) is that f(PA, t) = m(PA)iX(t),

provided PA is bounded. Thus the totalforce on afinite body is equal to the mass of the body times the acceleration of its mass center. By virtue of (13.3) the laws of momentum balance can be written as follows:

r

Jo&'t

sen) dA

f

+

fP,

f =f

b dV =

r r x sen) dA + ffP,r x b dV J0&',

vp dV,

e,

fP,

(2)

r x vp dV.

If we introduce the total body force

b* = b - pv,

(3)

which includes the inertial body force - pv, and define

f*(&" t) = m*(&', t) =

r

J ofP,

r

JofP,

sen) dA

+

f

r x sen) dA

fP,

b* dV,

+

f

fP,

(4)

r x b* dV,

then (1) takes the simple form

f*(&" t) = 0,

(5)

Our next result gives a far less trivial characterization of the momentum balance laws. Recall that an infinltesimal rigid displacement (ofS) is a mapping w: S -+ 1/ of the form w(x) = wo

+ W(x

- 0)

(6)

with Wo a vector and Wa skew tensor [cf. (7.8)]. Theorem of Virtual Work. Let (s, b) be a system offorces for PA during a motion. Then a necessary and sufficient condition that the momentum balance laws be satisfied is that given any part &' and time t,

r

s(n)· w dA +

J OfPt

f

fP,

b*· w dV = 0

(7)

for every infinitesimal rigid displacement w. Proof

Note first that by (13.2) we can write (6) in the form w

=

Wo

+ ro x r,

(8)

14.

101

FORCE. STRESS. BALANCE OF MOMENTUM

with co the axial vector corresponding to W. Let cp(wo, co) denote the left side of (7) with w given by (8). Then cp(wo, co) involves integrals of the fields s • w = s· W o

+ s . (co

x r),

where we have written s for s(n), and since

k·(co x r)

=

co·(r x k),

it follows that

= h, . w = s· w

Wo • S + m- (r x s), Wo • b; + co· (r x b*).

Thus by (4), cp(wo, co) =

t) + co· m*(&', t),

W o • f*(&"

and cp( Wo, co) = 0 for all vectors

Wo

and co if and only if (5) hold.

0

The next theorem is one of the central results of continuum mechanics. Its main assertion is that s(n) is linear in n. Cauchy's Theorem (Existence of stress). Let (s, b) be a system offorcesfor f1l during a motion. Then a necessary and sufficient condition that the momentum balance laws be satisfied is that there exist a spatial tensor field T (called the Cauchy stress) such that (a) for each unit vector n,

s(n) = Tn; (b) (c)

(9)

T is symmetric; T satisfies the equation of motion

+b=

div T

Proof

pt.

(10)

In stating condition (c) we do not assume explicitly that T(x, t) is smooth in x,

(11)

so that, a priori, it is not clear whether or not (c) makes sense. We therefore begin our proof by showing that (11) is a direct consequence of (9). Indeed, by (9), for an orthonormal basis {eJ,

L s(ei) e e, = L (Tei) ® ei; i

i

thus, in view of Exercise 1.6c, T(x, t) =

L s(e;, x, r) ® e.,

and (11) follows from (12) and property (i) of force systems.

(12)

v.

102

FORCE

We are now in a position to establish the necessity and sufficiency of (a)-(c). N ecessit y. Assume that (1) are satisfied. For convenience, we fix the time t and suppress it as an argument in most of what follows. The proof will proceed in a number of steps. Assertion I. Given any x unit vector k with

E [fBI>

any orthonormal basis {e.], and any (i = 1,2,3),

(13)

if follows that

s(k, x) = -

L (k· ei)s( -ei' x).

(14)

i

Proof First let x belong to the interior of [fBr' Let b > 0 and consider the tetrahedron '§s with the following properties: the faces of '§~ are 9' ~, 9' 1s» 9' 2~, and 9' 3~' where k and - e, are the outward unit normals to (J'§s on 9's and 9'i~' respectively; the vertex opposite to 9'~ is x; the distance from x to 9'~ is b (Fig. 5). Clearly, '§~ is contained in the interior of [fBI for all sufficiently small say 0 s 00 . Next, b*(x, t), defined by (3), is continuous in x, since b(x, r), p(x, t), and v(x, t) have this property; hence b*(·, t) is bounded on '§~o (for t fixed). Thus if we apply (5) 1 to the part f!jJ which occupies the region '§s at time t, and use (4) 1, we conclude that

s,

IL,,,s(n) dA I :s; "vol(,§~) for all 0 :s; 00 , where" is independent of

o.

~----~~.e2

Figure 5

(15)

14.

FORCE. STRESS. BALANCE OF MOMENTUM

103

Let A(15) denote the area of [1'6' Since A(15) is a constant times 15 2 , while vol(~6) is a constant times 15 3 , we may conclude from (15) that

A~15) Lf~(O) as 15

-+

dA

-+

0

s(k) dA +

L:

f

O. But

f

JiJ~6

f9'6

s(o) dA =

i

s( -ei) dA,

9"6

and, since s(o, x) is continuous in x for each fixed

0,

A~15) f9'6S(k) dA -+ s(k, x), A~15)

L'6

S(

-e;) dA

-+

(k· ei)s( -ei> x),

where we have used the fact that the area of [l'i6 is A(15)k • ei' Combining the above relations we conclude that (14) holds as long as x lies in the interior of fIl t • But x 1-+ s(o, x) is continuous on f!4 t • Thus, by continuity, (14) must hold everywhere on f!4 t • Assertion 2 (Newton's law of action and reaction). map 01-+ s(o, x) is continuous on .AI and satisfies

=

s(o, x)

For each x E f!4 t the

-s( -0, x),

(16)

Proof Since the right side of(14) is a continuous function ofk, s(k, x) is continuous on the set of all unit vectors k consistent with (13). Thus, since this is true for every choice of orthonormal basis {e.},s(k, x) must be a continuous function ofk everywhere on JV: Therefore, ifin (14) we let k -+ e 1, we arrive at s(e 1, x) = -s( -e 1 , x), and again, since the basis rei} is arbitrary, (16) must hold. Assertion 3.

Given any x E f!4 t and any orthonormal basis {eJ, s(k, x)

=

L: (k· ei)s(ei, x)

(17)

i

for all unit vectors k. Proof Choose an orthonormal basis {e.} and a unit vector k that does not lie in a coordinate plane (that is, in a plane spanned by two e.), Then there is no i such that k· e, = 0, and we can define a new orthonormal basis rei} by ei

=

[sgn(k· ei)]ei'

104

V.

Then k ei > 0 for i the basis {eJ yields 0

s(k, x) = -

I

=

FORCE

1,2,3, and Assertion 1 together with (16) applied to

I

(k i\)s( -ei, x) =. 0

i

(k ei)s(ei' x) = 0

i

I

(k e;)s(ei , x). 0

i

Thus (17) holds as long as k does not lie in a coordinate plane. But the map Ol--+s(o, x) is continuous on A: Thus (17) holds for all unit vectors k, and Assertion 3 is proved. Choose an orthonormal basis {ei} and let T(x, t) be defined by (12). Then T(x, t) is smooth in x, and, in view of (17), (9) holds. By (9), balance of linear momentum (2)1 takes the form

i

To dA +

a~

f

=

b dV

~

f vp

dV,

~

or equivalently, using the divergence theorem,

f

(divT

+b-

pv)dV = O.

~,

By the localization theorem, this relation can hold for every part & and time t only if the equation of motion (10) is satisfied. To complete the proof of necessity we have only to establish the symmetry ofT. Let w be any smooth vector field on !!Jr' We then conclude, with the aid of the divergence theorem and (4.2)5' that

i

ToowdA =

iJ~t

i f

f

(TTw)oodA =

iJ~t

=

div(TTw)dV

~t

(w div T 0

+ To grad w) dV.

~t

Thus, by (3), (9), and (10),

i

s(o) w dA 0

iJ~t

+

f

b* w dV 0

~t

=

f

To grad w dV.

(18)

e,

In particular, if we take w equal to the infinitesimal rigid displacement (6), then grad w = Wand (7) implies

f

ToW dV = 0

~t

for every part &, so that ToW

= O.

Since this result must hold for every skew tensor W, (f) of the proposition on page 6 yields the symmetry of T.

14.

FORCE. STRESS. BALANCE OF MOMENTUM

105

Tn

Figure 6

Sufficiency. Assume that there exists a symmetric spatial tensor field T consistent with (9) and (10). Clearly, (18) holds in the present circumstances. Since T is symmetric, and since the gradient of an infinitesimal rigid displacement is skew, when w is such a field the right side of(18) vanishes. We therefore conclude from the theorem of virtual work that the momentum balance laws hold. 0

Actually, one can show that (a) and (c) are equivalent to balance oflinear momentum (1)1; and that granted (1)1' the symmetry of T is equivalent to balance of angular momentum (1)2. Let T = T(x, t) be the stress at a particular place and time. If Tn

=

[n] = 1,

ern,

then a is a principal stress and n is a principal direction, so that principal stresses and principal directions are eigenvalues and eigenvectors ofT. Since T is symmetric, there exist three mutually perpendicular principal directions and three corresponding principal stresses. Consider an arbitrary oriented plane surface with positive unit normal n at x (Fig. 6). Then the surface force Tn can be decomposed into the sum of a normal force (0 • Tn)n

=

(n ® o)Tn

and a shearing force (I - n ® n)Tn, and it follows that n is a principal direction if and only if the corresponding shearing force vanishes. A fluid at rest is incapable of exerting shearing forces. In this instance Tn is parallel to n for each unit vector n, and every such vector is an eigenvector ofT. In view of the discussion given in the paragraph following the spectral theorem, T has only one characteristic space, 1/ itself, and (c) of the spectral theorem implies that

T =-nI

v.

106

FORCE

10I

t

I

I I

Pressure

Pure tension

t

t

-----Pure shear

t

Figure 7

with n a scalar. n is called the pressure of the fluid. Note that in this case the force (per unit area) on any surface in the fluid is -no. Two other important states of stress are: (a) Pure tension (or compression) with tensile stress a in the direction e, where lei = 1: T

=

rr(e ® e).

(b) Pure shear with shear stress, relative to the direction pair (k, 0), where k and 0 are orthogonal unit vectors: T = -r(k ®

0

+ 0 ® k),

The surface force fields corresponding to the above examples (with T constant) are shown in Fig. 7. EXERCISES

In Exercises 1, 4, and 8, fll is bounded. 1. The moment mit) about a moving point z(t) is defined by mz(t) =

r r, x s(o) dA + f r, x b dV,

JiJ9I,

91,

where r z is the position vector from z [cf. (13.8)]. (a) Let f(t) = f(fll, t). Show that, for y: ~ -+ $,

m, = Illy + (y - z) x f. (b) Let l(t) = l(fll, t). Show that [cf. (13.7)]

2. Prove that the momentum balance laws (1) hold for every choice of origin (constant in time) if they hold for one such choice.

14. FORCE. STRESS. BALANCE OF MOMENTUM

107

3. Show that the Cauchy stress is uniquely determined by the force system. 4. Consider a rigid motion with angular velocity roo Let {ej(t)} denote a principal basis of inertia and let J, denote the corresponding moments of inertia (relative to e) (cf. Exercise 13.3f). Further, let wlt) and mj(t) denote the components of ro(t) and m..(t) relative to {ei(t)}. Derive Euler's equations

m 1 = J 1Wl mz

=

m3 =

+ (J 3 -

JZ)WZ W3'

Jzwz + (J 1 - J 3)WIW3,

+

J 3 W3

(Jz -

Jd W I W Z '

These relations supplemented by f

m(91)li

=

constitute the basic equations of rigid body mechanics. When f and m.. are known they provide a system of nonlinear ordinary differential equations for ro and ~. 5. Let T be the stress at a particular place and time. Suppose that the corresponding surface force on a given plane !/ is perpendicular to g, while the surface force on any plane perpendicular to g vanishes. Show that T is a pure tension. 6. Let T be a pure shear. Compute the corresponding principal stresses and directions. 7. Prove directly that s(o, x) = -s( -0, x) by applying (2)1 to the part [JJ that occupies the region fYt o at time t. Here fYt o is the rectangular region which is centered at x, which has dimensions (j x (j X (jz, and which has n normal to the (j x (j faces (Fig. 8).

Figure 8

Ii

8. Prove Da Silva's theorem: At a given time the total moment on a body can be made to vanish by subjecting the surface and body forces to a suitable rotation; that is, there exists an orthogonal tensor Q such that

i

O~t

r x Qs(o) dA

+

f

~t

r x Qb dV = O.

V.

108

FORCE

Show in addition that

r

(QTr) x s(n) dA +

JoflB•

i

(QTr) x b dV = O.

flB,

What is the meaning of this relation? 9.

Suppose that at time t the surface traction vanishes on the boundary ofJB t • Show that at any x E afJBt the stress vector on any plane perpendicular to afJB t is tangent to the boundary.

15. CONSEQUENCES OF MOMENTUM BALANCE By Cauchy's theorem (and Exercise 14.3), to each force system consistent with the momentum balance laws there corresponds exactly one symmetric tensor field T consistent with (14.9) and (14.10). Conversely, the force system (s, b) is completely determined by the stress T and the motion x. Indeed, the surface force s is given by (14.9), while the body force b is easily computed using the equation of motion (14.10) [with p computed using (12.4)]: s(n) = Tn,

b=

pv -

div T.

This observation motivates the following definition. By a dynamical process we mean a pair (x, T) with . (a) (b) (c)

x a motion, T a symmetric tensor field on the trajectory ff of x, and T(x, t) a smooth function of x on fJB t •

Further, ifv and p are the velocity field and density corresponding to x, then the list (v, p, T) is called a flow. In view of the above discussion, to each force system consistent with the momentum balance laws there corresponds exactly one dynamical process (or equivalently, exactly one flow), and conversely. For the remainder ofthis section (v, p, T) is a flow and (s, b) is the associated force system. Theorem (Balance of momentum for a control volume). volume at time t. Then at that time

Let 9i be a control

r s(n) dA + Jutrb dV = dd Jutrvp dV + Jrout(pv)v' n dA,

Jout

t

(1)

r r x s(n) dA + Jutrr x b dV = ddt Jutrr x vp dV + Jfoutr x (pv)v' n dA.

Jout

109

15. CONSEQUENCES OF MOMENTUM BALANCE

Figure 9

We will prove only (1)\. First of all, by (4.2)4' (8.5), and (12.5h,

Proof

pv

pv' + p(grad v)v = (pv)' - p'v + (grad v)(pv) = (pv)' + v div(pv) + (grad v)(pv) = (pV)' + div(v ® pv).

=

Thus, since

L

div T dV = 191Tn dA = 191 s(n) dA,

L(pv)'

f

91

div(v ® pv) dV =

dV =

:t Lpv

dV,

i~

..

!::f; -,

r (v ® pv)n dA = JriJ91(pv)v· n dA,

JiJ91

if we integrate the equation of motion (14.10) over f7l we arrive at (1k

D

Equation (1)\ asserts that the totalforce on the control volume f7l is equal to the rate at which the linear momentum ofthe material in f7l is increasing plus the rate ofoutflow ofmomentum across of7l. Equation (1h has an analogous interpretation. A flow (v, p, T) is steady if fJl1 = fJlo for all t and Vi

= 0,

o' = 0,

T' = O.

In this case we call fJI0 the flowregion. Note that our definitions are consistent: in a steady flow the underlying motion is a steady motion (cf. Section 9). As an example of the last theorem consider the steady flow of a fluid through a curved pipe (Fig. 9), and let f7l be the control volume bounded by the pipe walls and the cross sections marking the ends ofthe pipe. Assume that the stress is a pressure and that the velocity, density, and pressure at the entrance and exit have the constant values and respectively, with e\ perpendicular to the entrance cross section and ez perpendicular to the exit cross section. Since the flow is steady,

~dt

f

91

vpdV = O.

v.

110

FORCE

Further, v- n must vanish at the pipe walls; therefore

i

(pv)v-ndA =

P2v~A2e2

- PlviA1e l,

iJEJt

where AI and A 2 are the areas of the entrance and exit cross sections, respectively. Let f denote the total force exerted by the fluid on the pipe walls. Then the total force on the control volume 91 is

and (1)1 yields

= (1t 1 + Plvi)A1e1

f

- (1t2

+ P2v~)A2e2.

A similar analysis, based on conservation of mass (12.6), tells us that P 1v1A 1 = P2V2A2 == em,

and hence f = (1tlAI

+ emv 1)e1

(1t2A2 + mv 2)e2.

-

Thus we have a simple expression for the force on the pipe walls in terms of conditions at the entrance and exit. Although the assumptions underlying this example are restrictive, they are, in fact, good approximations for a large class of physical situations.

Theorem of Power Expended. For every part

i

sen) - v dA

o<

+

f

b - v dV =

&It

f

(!J

and time t,

T - D dV + :

&It

t

f~

P dV,

(2)

&It

where

D

=

!(grad v + grad vT)

is the stretching. Proof

Since T is symmetric, (a) of the proposition on page 6 implies T - grad v = T - D.

Also, by (12.8),

so that (14.3) yields

f

&It

b. - v dV

=

f

&It

b - v dV -

~ dt

f

&It

2

v P dV.

2

15.

CONSEQUENCES OF MOMENTUM BALANCE

Equation (14.18) (with w identity. D

111

v) and the above relations yield the desired

=

The terms

f

~

p

dV

fT. D dV

and

~.

~.

are called, respectively, the kineticenergy and the stress power of fJ/ at time t. The theorem of power expended asserts that the power expended on fJ/ by the surface and body forces is equal to the stress power plus the rate of change of kinetic energy. A flow is potential if the velocity is the gradient of a potential: v = grad (fJ. Note that (fJ is a spatial field of class C 2 (because v is C 2 ) ; hence curl v = curl grad tp = 0 and potential flows are irrotational. Conversely, by the potential theorem (page 35), if a flow is irrotational, and if {fit is simply connected at some (and hence every) time t, then the flow is potential. For a body force b, the field b/p represents the body force per unit mass. We say that the body force is conservative with potential [3 if

b/p = -grad

[3.

(3)

If the flow is also steady the equation of motion (14.10) implies that b' = 0; in this case we will require that

[3' =

O.

Bernoulli's Theorem. Consider afiow (v, p, T) with stress a pressure -nI and body force conservative with potential [3. (a)

Ifthefiow is potential,

grad ((fJ'

(b)

grad n = O.

(4)

Ifthefiow is steady,

v. (c)

+ ~2 + [3) + ~

grad(~

+ [3) + ~

v . grad n

=

O.

(5)

Ifthefiow is steady and irrotational,

grad(~

+

[3) + ~

grad n = O.

(6)

112

V. FORCE

By hypothesis,

Proof

T = -nI, so that div T = -grad n; thus the equation of motion (14.10) takes the form - grad n + b = pv,

(7)

or equivalently, by (3),

v = - .!. grad n -

grad p.

p

(8)

On the other hand, by (9.10)1' for a potential (and consequently irrotational) flow

v = v' + ! gradtv")

=

grad ( q/ +

~2);

(9)

for a steady flow v•v= v•

grad(~)

(10)

(where we have used the fact that v· Wv = 0, since W is skew); for a steady, irrotationa1 flow

v=

grad(~}

(11)

The relations (9)-(11), when combined with (8), yield the desired results (4)-(6).

0 EXERCISES

1. Consider a statical situation in which a (bounded) body occupies the region fJlo for all time. Let b: fJlo -+ 1/ and T: fJlo -+ Sym with T smooth

satisfy divT + b = O. Define the mean stress T through vol(fJloyr =

r T dV.

J£to

113

SELECTED REFERENCES

(a)

(Signorini's theorem) Show that T is completely determined by the surface traction Tn and the body force b as follows:

vol(86'o)T (b)

=

io~o

(Tn ® r) dA +

r b ® r dV.

J~o

Assume that b = 0 and that iJ86'o consists of two closed surfaces Yo and Y 1 with Y 1 enclosing Yo (Fig. 10). Assume further that Yo and Y 1 are acted on by uniform pressures no and n1' so that sen) = -non

on

Yo,

sen) = -n1n

on

Y

1,

Figure 10

with no and n 1 constants. Show that T is a pressure of amount

where 00 and 01 are, respectively, the volumes enclosed by Yo and Y 1• 2. Prove (l)z. 3. Derive the following counterpart of (2) for a control volume 9l:

i

o~

s(n)· v dA

=

I

~

+

r b· v dV

J~

d T . D dV + -d

I

2

-v P dV t ~ 2

+

I

SELECTED REFERENCES

Gurtin and Martins [1]. Noll [3, 4]. Truesdell [1, Chapter 3]. Truesdell and Toupin [1, §§195-238].

o~

2

-pv

2

(v . n) dA.

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CHAPTER

VI Constitutive Assumptions. Inviscid Fluids

16. CONSTITUTIVE ASSUMPTIONS The axioms laid down for force-namely the laws ofmomentum balanceare common to most bodies in nature. These laws, however, are insufficient to fully characterize the behavior of bodies because they do not distinguish between different types of materials. Physical experience has shown that two bodies of the same size and shape subjected to the same motion will generally not have the same resulting force distribution. For example, two thin wires of the same length and diameter, one of steel and one of copper, will require different forces to produce the same elongation. We therefore introduce additional hypotheses, called constitutive assumptions, which serve to distinguish different types of material behavior. Here we will consider three types of constitutive assumptions. (i) Constraints on the possible deformations the body may undergo. The simplest constraint is that only rigid motions be possible, a constraint which forms the basis of rigid body mechanics. Another example is the assumption of incompressibility, under which only isochoric deformations are permissible. This assumption is realistic for liquids such as water under normal flow 115

116

VI.

CONSTITUTIVE ASSUMPTIONS. IN VISCID FLUIDS

conditions and even describes the behavior of gases like air provided the velocities are not too large. (ii) Assumptions on the form of the stress tensor. The most widely used assumption of this type is that the stress be a pressure, an assumption valid for most fluids when viscous effects are negligible. (iii) Constitutive equations relating the stress to the motion. Here a classical example is the equation of state of a gas giving the pressure as a function of the density. We now make these ideas precise. A material body is a body f!J together with a mass distribution and a family C(J of dynamical processes. C(J is called the constitutive class of the body; it consists of those dynamical processes consistent with the constitutive assumptions of the body. A dynamical process (x, T) is isochoric if x(', t)

is an isochoric deformation

(1)

at each t; a material body is incompressible if each (x, T) E C(J is isochoric. By (1), given any part f!JJ,

vol(f!JJt ) = vol(f!JJ) for all t, which implies (10.4). Thus every motion of an incompressible body is isochoric. The restriction (1), however, implies more: not only does each motion preserve volume, but the volume of any part throughout the motion must be the same as its volume in the reference configuration. Further, in view of (6.17) and (c) of the theorem characterizing isochoric motions (page 78), every flow (v, p, T) of an incompressible body must have det F

= 1,

div v =

o.

(2)

Also, by (2)1 and (12.4), p(x, t) = Po(p) for x = x(p, t). Thus, in particular, when Po is constant,

in this case we refer to Po as the density of the body. A dynamical process (x, T) is Eulerian if the stress T is a pressure; that is, if

T = -1tI with n a scalar field on the trajectory of x.

16. CONSTITUTIVE ASSUMPTIONS

117

Figure 1

An ideal fluid is a material body consistent with the following two constitutive assumptions: (a) The constitutive class is the set of all isochoric, Eulerian dynamical processes. (b) The density Po is constant. Thus an ideal fluid is an incompressible material body for which the stress is a pressure in every flow. Note that for an ideal fluid the pressure is not determined uniquely by the motion: an infinite number of pressure fields correspond to each motion. This degree of flexibility in the pressure is common to all incompressible materials. That such a property is physically reasonable can be inferred from the following example. Consider a ball composed of an ideal fluid under a time-independent uniform pressure 11: (Fig. 0, and assume, for the moment, that all body forces, including gravity, are absent. Then the ball should remain in the same configuration for all time. Moreover, since the ball is incompressible, an increase or decrease in pressure should not result in a deformation. Thus the same "motion" should correspond to all uniform pressure fields. This argument is easily extended to nonuniform pressure fields and arbitrary motions, but for consistency the requirement of null body forces must be dropped to insure satisfaction of the equation of motion. We next introduce a material body, called an elastic fluid, in which compressibility effects are not ignored, and for which the pressure is completely specified by the motion. Here the constitutive class is defined by a smooth response function

giving the pressure when the density is known: 11:

= ft(p).

(3)

That is, the constitutive class is the set of all Eulerian dynamical processes (x, - nI) which obey the constitutive equation (3). We will consistently write (v, p, 11:) in place of (v, p, - nI)

118

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

for the flow of an ideal fluid or of an elastic fluid. (Of course, P = Po in the former case.) An inviscid fluid is often characterized by the requirement that it be incapable of exerting shearing forces; in this sense both ideal and elastic fluids are inviscid.

EXERCISE

1. Consider a material body f:!4 with constitutive class C(j. A simple constraint for f:!4 is a function y: Lin" -+ ~

such that each dynamical process (x, T) E C(j satisfies y(F) = O.

(4)

For such a material one generally lays down the following constraint axiom: the stress is determined by the motion only to within a stress N that does no work in any motion consistent with the constraint. [The rate at which a stress N does work is given by the stress power, per unit volume, N'D (cf. page 111), where D is the stretching.] We now make this idea precise. Let PlJ be the set of all possible stretching tensors; that is, PlJ is the set of all tensors D with the following property: for some C 2 function F: ~ -+ Lin + consistent with (4), D is the symmetric part of L = F(t)F(t)-l

at some fixed time t. Let fJt = PlJ.1;

that is, fJt

= {NESymIN'D = 0 forall DEPlJ}.

Then the constraint axiom can be stated as follows: If (x, T) belongs to C(j, then so also does every dynamical process of the form (x, T + N) with N(x, t) E fJt for all (x, t) in the trajectory of x. We call fJt the reaction space.

17.

119

IDEAL FLUIDS

An incompressible material can be defined by the simple constraint y(F) = det F - 1.

(5)

Show that this constraint satisfies the constraint axiom if and only if the corresponding reaction space is the set of all tensors of the form - nI, n E IR; i.e., if and only if the stress is determined by the motion at most to within an arbitrary pressure field. Show further that the constitutive assumptions of an ideal fluid are consistent with the constraint axiom.

17. IDEAL FLUIDS Consider an ideal fluid with density Po. The basic equations are the equation of motion (15.7) and the constraint equation (16.2)z: -grad n

+b=

Pov,

(1)

div v = O. When the body force is conservative with potential

v = -grad(~

+

p, (15.8) implies

p),

and the acceleration is the gradient of a potential. We therefore have the following corollary of the results established in Section 11.

Theorem (Properties of ideal fluid motion).

The flow of an ideal fluid under a conservative body force has the following properties: (a) (b) (c)

If the flow is irrotational at one time, it is irrotational at all times. The flow preserves circulation. Vortex lines are transported with the fluid.

As we shall see, the equations of motion of a fluid are greatly simplified when the flow is assumed to be irrotational. The above theorem implies that the motion of an ideal fluid under a conservative body force is irrotational if the flow starts from a state of rest. This furnishes the usual justification for the assumption ofirrotationality. For a plane flow in an infinite region a much stronger assertion can be made. Indeed, if the flow is steady and in a uniform state at infinity [i.e., if grad vex) ~ 0 as x ~ 00], and if each streamline begins or ends at infinity, then, since W is constant on streamlines (cf. the proposition on page 81), the flow must necessarily be irrotational.

VI.

120

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

Bernoulli's Theorem for IdealFluids, Let (v, Po, n) be a fiow ofan idea/fluid under a conservative body force with potential 13. (a)

If the flow is potential (v = grad cp),

+ ~ + ~ + fJ)

grad ( cp' (b)

=

o.

If the flow is steady, 2

+ ~ + 13)' =

(v

2

Po

0,

(2)

so that (v 212) + (nl Po) + fJ is constant on streamlines. (c) If the flow is steady and irrotational,

v2

n

2

Po

- + - + 13 = const

(3)

everywhere. Proof. Conclusion (a) follows from (15.4) (with P = Po). To establish (b) and (c) let

v2 2

n Po

1]' =

0,

v' grad

I'[ =

= - + - + 13·

I'[

For a steady flow

and by (15.5),

0;

hence ~ = 1]'

+ v ' grad I]

= 0,

which is (2). Finally, for a steady, irrotational flow (15.6) yields grad which, with

1]' =

0, implies (3).

I] =

0,

D

By Bernoulli's theorem, for a steady, irrotational flow under a conservative body force equations (1) reduce to

div v = 0, 2

v

-2

n

curl v = 0,

+ - + 13 = Po

(4)

const.

17.

121

IDEAL FLUIDS

These equations are also sufficient to characterize this type of flow. Indeed, suppose that (v, n) is a steady solution of (4) on a region of space flI o . Then, since W = 0, (9.10)1 implies

v=

v'

+ grad(~)

grad(~)

=

=

-grad(~

+ p),

and the basic equations (1) (with b/po = -grad p) are satisfied. In a steady motion the velocity is tangent to the boundary (cf. the proposition on page 68); thus (4) should be supplemented by the boundary condition

v'n=O

on

ofll o .

(5)

When the flow is nonsteady we are left with the system (1) to solve. The basic nonlinearity of these equations is obscured by the material time derivative in (1)1. Indeed, in terms of spatial operators these equations take the form v'

+ (grad v)v = div v

=

- grad n

+ b,

(6)

0,

where, for convenience, we have written n for n/po and b for b/po. Equations (6) are usually referred to as Euler's equations. EXERCISES

1.

Show that in the flow of an ideal fluid the stress power is zero (cf. page 111 ).

2. Consider the flow of a bounded ideal fluid under a conservative body force, and suppose that for each t and each x E ofllt> v(x, t) is tangent to ofll r • Show that

d -d t

f

v2 dV

= 0,

111,

so that the kinetic energy is constant. 3. Consider a homogeneous motion of the form

x(p, t)

= Po +

F(t) [p - Po]'

Show that if x represents a motion of an ideal fluid under conservative body forces, then F(t)F(t) -1 must be symmetric at each t. 4. Consider a steady, irrotational flow of an ideal fluid over an obstacle fJl (Fig. 2); that is, ~ is a bounded regular region whose interior lies outside the flow region flI o and whose boundary is a subset of ofll o . Assume that

122

VI.

CONSTITUTIVE ASSUMPTIONS. IN VISCID FLUIDS

Figure 2

the body force is zero. Show that the total force exerted on fJi by the fluid is equal to

18. STEADY, PLANE, IRROTATIONAL FLOW OF AN IDEAL FLUID For a steady, plane flow the velocity field has the form vex) =

Vt(Xt,

+ V2(Xt, x2)e2,

x2)e t

(1)

and the flow region can be identified with a region fJi in ~2. For convenience, we identify v and x with vectors in ~2 and write, in place of (1), vex) = (Vt(x),

vix»,

Then the basic equations (17.4) reduce to oV t oX t

+

oV2 _ OX2 -

0 ,

(2) 1t

!(vf + vD + - = const Po

with 1t also a field on fJi. Here, for convenience, we have assumed that the body force is zero. For the moment let {J

=

-V2'

18. STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID

123

Then the first two relations in (2) are the Cauchy-Riemann equations:

orx Op OX2 = - OX 1; these are necessary and sufficient conditions that

g(z) = rx(X1' X2)

+ ip(X1' x 2)

be an analytic function of the complex variable

on r7l (considered as a region in the complex plane C). Thus we have the following

Consider a steady, plane, irrotational flow of an ideal fluid. Let C be defined by

Theorem.

g: ~

--+

(3)

Then g is an analytic function. Conversely, any analyticfunction g generates, in the sense of(3), a solution of(2)1.2' The function g is called the complex velocity. Let c = (C1' C2) be a curve in~. Then c can also be considered as a curve

in C. Given any complex function h on

i

h(z) dz

=

f

~

we write

h(e(u»c(u) do ;

in particular, we define r(c) = ig(Z) dz.

(4)

To interpret r(c) note first that k(u) = (C2(U),

-c 1(u»

is normal to c at c(u). Therefore when c is the boundary curve of quirement that v· n = 0 on o~ [cf. (17.5)] becomes

(5) ~

the re(6)

Thus in this instance

ig(Z)dZ = f(V 1 - iV2)(C1 + iC2)du = f(V 1C1 + v2c2)du,

124

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

-ll1I

Figure 3

and

r(c) =

i

V(X) • dx

is the circulation around c. Next,

Ikl =

Jci + c~,

so that Ik(O") I do is the element of arc length along c. The vector f(c) = -

f

n(c(O"»k(O") da

(7)

therefore represents the integral along c of the surface force - no (0 = k/ Ik I) with respect to arc length. In fact, for a simple closed curve oriented counterclockwise as shown in Fig. 3, f(c) gives the total force (per unit length in the X3direction) on the material inside c. Blasius-Kutta-Joukowski Theorem. Consider a steady, plane, irrotational flow of an ideal fluid in a region qj whose boundary is a simple closed curve c. Let g be the complex velocity ofthe flow. Then

f

ipo "g(z) 2 dz. fl(C) - ific) = 2

(8)

If in addition qj is the region exterior to c and the velocity is uniform at infinity in the sense that g(z)

-+

a

as

z-+oo

(9)

with a real, then (10)

fl(c) = 0,

Proof

Since c is closed,

fkdO" = 0,

18. STEADY, PLANE, IRROTATJONAL FLOW OF IDEAL FLUID

125

and we conclude from (7) and the Bernoulli equation (2h that

= P°

f(e)

2

f

v2 k da.

Thus 1 fl(e) - if2(e) = Po I V2(C2 + ic 1) do = ipo I\2(C 1 - ic 2) do, 2 0 2 0

On the other hand, 192 dz =

f

(Vl - iV2)2(l\

+ ic2) da,

and a simple calculation using (6) yields (vi + V~)(Cl

- ic 2) = (Vl - iV2)2(C1 + ic 2)·

Thus (8) holds. Assume now that fJl is exterior to e and that (9) holds. Assume further that the origin 0 lies inside e. Then, since g is analytic and (9) holds, the Laurent expansion of g about the origin has the form g(z) =

U

+ -IXZ1 + -1XZ22 + ...

for any z E fJl. Hence (4) and Cauchy's theorem of residues imply F(c) = 2ni1X 1.

Next, since g is analytic, we can compute g2 by termwise multiplication. Thus 2 2 2UIX1 ( ) =u +--+ gz z

lXi + 2UIX2 Z2

+"',

and a second application of Cauchy's theorem tells us that 192 dz

= 4niuIX 1 = 2ur(e). 0

Therefore, in view of (8), (10) holds.

If fJl is simply connected, then g can be written as the derivative of an analytic function w: g

=

dw dz

126

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

(This result extends to multiply connected regions provided one is willing to admit multivalued functions.) The function w: ~ -+ C is called the complex potential; the real and imaginary parts of w, denoted qJ and t/J, respectively, generate the velocity v through

Let s

=

(SI' S2) be a streamline so that

s(t)

= v(s(t».

(11)

Then

and t/J is constant on streamlines. For this reason t/J is called the stream function. Let c be a curve. Then c is essentially a segment of a streamline s if there is a smooth one-to-one mapping r of [0, IJ onto a closed interval of IR such that c(a) = s(-r(a»,

O~a~1.

(13)

Proposition. Consider a flow with a complex potential. Let c be a curve in the flow region and assume that v is nowhere zero on c. Then the following are equivalent: (a) (b) (c) (d)

v l(c(a»cia) = vic(a»c l(a)for 0 ~ a ~ 1. v(c(a» is parallel to c(a)for 0 ~ a ~ 1. c is essentially a segment of a streamline. t/J is constant on c; that is,for 0 ~ a ~ 1,

d da t/J(c(a» = O. Proof

We will show that (d) - (a) - (b) - (c).

(d) - (a).

This follows from the identity [cf. (12)J dd t/J(c(a» = -vic(a»c l(a) a

+ v l(c(a»cia).

(a) - (b). Here we use the fact that (a) and (b) are each equivalent to the assertion that k(a)· v(c(a» = 0 for 0 ~ a ~ 1, where k, defined by (5), is normal to the curve c.

18. STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID

127

(c) => (b). If c is essentially a segment of a streamline s, then (11), (13), and the chain rule yield c(o) = s(r(u»t(u) = v(s("C(u»)t(u) = i(u)v(c(u»,

(14)

and (b) is satisfied. (b) => (c). Assume that (b) holds. Then, since v(c(u» and c(u) never vanish, c(u)

= tX(u)v(c(u»,

o s o s 1,

with tX a continuous function on [0, 1] which never vanishes. Let "C(u) =

J:

tX().) d)',

let s be the streamline which passes through c(O) at time t = 0, so that s satisfies (11) and s(O) = c(O); and let d be the curve in d(u)

[R2 defined

=

by

s("C(u»,

0 :s; a :s; 1.

(15)

Then d(O) = c(O) and, using the argument (14), ci(u) = tX(u)v(d(u»,

O:S;u:S;1.

Thus c and d satisfy the same differential equation and the same initial condition. We therefore conclude from the uniqueness theorem for ordinary differential equations that c = d, so that, by (15),c is essentially a segment of a streamline. 0 For problems involving flows about a stationary region one usually assumes, as the boundary condition, that the boundary of the region be a streamline. In view ofthe last proposition, this insures that the velocity field be tangent to the boundary, or equivalently, that the boundary be impenetrable to the fluid (cf. the proposition on page 68). As an example consider the region exterior to the unit circle and assume that the velocity is uniform and in the xrdirection at infinity, so that (9) is satisfied. Consider the flow generated by the complex potential

128

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

Figure 4

In terms of cylindrical coordinates z = re" the stream function tf; has the form

tf;(r, e) = a(r - ~)

sin

e,

so that tf; has the constant value zero on the unit circle. Further, the complex velocity

satisfies (9). The streamlines are shown in Fig. 4. For this flow F(unit circle) = 0, and by the Blasius-Kutta-Joukowski theorem there is no net force on the boundary. Another flow with the desired properties is generated by w(z)

= a(z + ~ + iy log z).

This potential has a stream function

tf;(r, e) = a[(r - ~)

sin e

+ y log rJ'

which again vanishes on the circle r = 1. Further, w generates a complex velocity g(z)

= a

1 ( 1 - Z2

+ ~iY) ,

which satisfies (9). A point at which g(z) = 0 is called a stagnation point. Points z on the unit circle are given by the equation z = e". A necessary and sufficient condition that there exist a stagnation point at z = ei9 is that .

II

sm o = -

Y

2'

and hence that - 2 ~ Y ~ 2. Assume Y ~ O. Then: (a) for 0 ~ Y < 2 there are two stagnation points on the cylinder; (b) for Y = 2 there is one stagnation

18.

STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID

129

(a)

Figure 5

point on the cylinder; (c) for y > 2 there are no stagnation points on the cylinder. These three cases are shown in Fig. 5 with the stagnation points denoted by S. For these flows r(unit circle) = - 2n«y provided the unit circle has a counter-clockwise orientation; thus there is no drag (force in the xI-direction), but there is a lift (force in the x2-direction) equal to 2npo«2 y. The problem of flow about an airfoil is far more complicated than the simple flow presented in Fig. 5,but the results are qualitatively the same. There the circulation is adjusted as shown in Fig. 6 to insure two stagnation points with the aft stagnation point coincident with the (sharp) trailing edge (KuttaJoukowski condition).

Figure 6

130

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS EXERCISES

1.

Show that the complex potential w(z) = _Ce-inpzP+l

represents the flow past a wedge (Fig. 7) of angle 21X, where C and {3 ({3 > -1) are real constants and {3n

IX

= 1 + {3'

Determine the stagnation points (if any).

Figure 7

2. Sketch the streamlines of the complex potential w(z)

=

Cz 2 ,

where C is a real constant, and show that the absolute value of the velocity is proportional to the distance from the origin. 3. Consider a steady, plane, irrotational flow of an ideal fluid. Let c be a curve in the flow region with

Let no be the pressure at a point at which the velocity is Vo' Show that the force (8) on c is given by .

fl(C) - if2(C) = -

Po (no + -2-o2) [(YA V

. YB) + I(XA - x B)]

IPO . +2

f.

cg

2

dz.

19. ELASTIC FLUIDS The basic equations for the flow of an elastic fluid are the equation of motion -grad n

+b=

pt,

(1)

ELASTIC FLUIDS

131

p' + div(pv) = 0,

(2)

19.

conservation of mass and the constitutive equation n

= ft(p).

(3)

We assume that ft has a strictly positive derivative, and we define functions " > 0 and e on ~ + by 2( ) _ dft(p) p - ----;[P'

tc

e(p)

=

r

,,2(0

p.

(4)

e:

(

where p* is an arbitrarily chosen value of the density. The function ,,(p) is called the sound speed; the reason for this terminology will become apparent when we discuss the acoustic equations. In a flow e(p) may be interpreted as a spatial field. We may therefore conclude from the chain rule that ,,2(p) 1 dft(p) 1 grad e(p) = - - grad p = - -d- grad p = - grad n. p

p

p

P

(5)

Thus when the body force is conservative with potential 13, (1) takes the form

v = - grad(e(p) + 13), and the acceleration is the gradient of a potential. We therefore have the following corollary of the results established in Section 11.

Theorem (Properties of elastic fluid motion).

The flow of an elastic fluid under a conservative body force has the following properties: (a) (b) (c)

If the flow is irrotational at one time, it is irrotational at all times. The flow preserves circulation. Vortex lines are transported with the fluid.

Thus, in particular, the flow of an elastic fluid under a conservative body force is irrotational provided the flow starts from rest.

Bernoulli's Theorem for Elastic Fluids. Let (v, p, n) be a flow of an elastic fluid under a conservative body force with potential 13. Let e be defined by (4). (a)

If the flow is potential,

grad(llJ'

+ ~ + e(p) + 13) = o.

132

VI. (b)

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

If the flow is steady,

(~

+ e(p) +

p)" = 0,

(6)

so that (v 2 j2) + e(p) + p is constant on streamlines. (c) If the flow is steady and irrotational,

v2

2 + e(p) + p =

const

(7)

everywhere. Proof

Conclusion (a) follows from (15.4) and (5). Let '1 =

v2

2 + e(p) + p.

For a steady flow '1' = 0,

v • grad '1 = 0,

where we have used (15.5) and (5); hence ~ =

'1' + v • grad '1 = 0,

which is (6). Finally, for a steady, irrotational flow, (5) and (15.6) yield grad '1 = 0, which with '1' = implies (7). 0

°

If we define

then (1)-(3) take the form v'

+ (grad v)v + rt.(p) grad p = p' + div(pv) =

bjp,

(8)

0.

Equations (8) constitute a nonlinear system for p and v and are generally quite difficult to solve. Further, for many problems of interest the solution will not be smooth because of the appearance of shock waves (surfaces across which the velocity suffers a jump discontinuity), and the notion of a weak solution must be introduced. A careful discussion ofthese matters, however, is beyond the scope of this book. We now consider flows which are close to a given rest state with constant density Po. We therefore assume that Ip - PoI,Igrad p I, lvi, and Igrad vi are small.' Let [) be an upper bound for these fields. Then, since rt.(p) grad p = rt.(Po) grad p

+ [rt.(p)

- rt.(Po)] grad p,

I Since our interest lies only in deriving the asymptotic form of the field equations in the limit as (j ..... 0, it is not necessary to work with dimensionless quantities.

133

19. ELASTIC FLUIDS

and since rx. is continuous, we have, formally, rx.(p) grad p = rx.(Po) grad p

as

~

+ 0( 0, (4h implies that dejdp > 0, so that e(p) is an invertible function of p ; we may therefore write p = e- 1 ( C

_Igra~

C(12)

and use this relation to express K(p) as a function

A(gradtp) K

=

K

(-l(C s -

Igrad2 CPI2))

(15)

of grad cpo Of Course C and hence R will generally vary from flow to flow. EXERCISES

1.

An ideal gas is an elastic fluid defined by a constitutive equation of the form (16)

137

SELECTED REFERENCES

where x > 0 and y > 1 are strictly positive constants. [Note that an ideal gas is not an ideal fluid. Actually, (16) represents the isentropic behavior of an ideal gas with constant specific heats.] (a)

Show jhat the sound speed K = K(p) is given by K

(b)

2

= yn/p.

Show that the relation (15) is given by

y- 1 j(2(grad qJ) = -2- (v 2 with v a constant. 2. Consider the ideal gas (16) in equilibrium (v body force

-

[grad qJ1 2 )

=

0) under the gravitational

b = -pge3' where g is the gravitational constant. Assume that n(x) = no = constant at

X3

= O.

Determine the pressure distribution as a function of height x 3 • 3. Show that for an elastic fluid the stress power of a part &> at time t is

f

T'DdV

=

~,

-f

ttbp di/

~,

(cr. page 111), where 1

v=P

is the specific volume (i.e., the volume per unit mass). SELECTED REFERENCES

Courant and Friedrichs [1]. Hughes and Marsden [1, §§9-11, 13, 14]. Lamb [1, Chapters 1-10]. Meyer [1, §6]. Serrin [1, §§15-18, 35,45,46].

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CHAPTER

VII Change in Observer. Invariance of Material Response

20. CHANGE IN OBSERVER Two observers viewing a moving body will generally record different motions for the body; in fact, the two recorded motions will differ by the rigid motion which represents the movement of one observer relative to the other. We now make this idea precise. Let x and x* be motions of fJI. Then x and x* are related by a change in observer if x*(p, t) = q(t) + Q(t) [x(p, r) - 0]

(1)

for every material point p and time t, where q(t) is a point of space and Q(t) is a rotation. That is, writing f(x, t)

= q(t) + Q(t)(x -

0),

(2)

then at each time t the deformation x*(., t) is simply the deformation x(', t) followed by the rigid deformation f~·, t): x*(', t) = f(·, t) x(', t). 0

139

(3)

140

VII.

OBSERVER CHANGE. INVARIANCE OF RESPONSE

o

X(o,'>j

f(o, I)

,,*(,/)

• Figure I

For convenience, we write !!Ai for the region of space occupied by !!A at time t in the motion x*: !!Ai = x*(!!A, t);

then the relation between x and x* is illustrated by Fig. 1. We now determine the manner.in which the various kinematical quantities transform under a change in observer. Letting

F

=

Vx,

F* = \lx*

and differentiating (1) with respect to p, we arrive at F*(p, t) = Q(t)F(p, r),

(4)

which is the transformation law for the deformation gradient. Note that, since det Q = 1, (4) implies det F*

=

det F.

(5)

Next, let F = RU = VR,

F* = R*U* = V*R*

be the polar decompositions ofF and F*. Then (4) implies F* = R*U* = QRU, and, since QR is a rotation, we conclude from the uniqueness of the polar decomposition that R* = QR, Also,

U* =

u.

141

20. CHANGE IN OBSERVER ~

91,

:/~I

Figure 2

and therefore, since Y

•

fJI~

f( -, I)

= RUR T , = QYQT.

y*

From these relations it follows that the Cauchy-Green strain tensors C = U 2, B = v-, C* = U*2, B* = y*2 transform according to C*

=

C,

Intuitively it is clear that if x E !!AI and x* E x*

=

!!A~

are related through

f(x, t),

(6)

then x and x* must correspond to the same material point. To verify that this is indeed the case, note that, by (3), x(-, t)

= f(·, t) - 1 x*(-, t), 0

which is easily inverted to give pC t)

=

p*C t) f(·, t), 0

where p and p* are the reference maps corresponding to the motions x and x*, respectively (Fig. 2). Thus p(x, t)

= p*(f(x, t), t),

or equivalently, by (6), p(x, t) = p*(x*, t),

(7)

so that x and x* correspond to the same material point. As a consequence of (1), x*(p, t) = q(t)

+ Q(t)x(p, t) + Q(t) [x(p, t)

- o].

If we take p = p*(x*, t) in the left side and p = p(x, t) in the right side of this relation-a substitution justified by (7) provided x* and x are related by (6)-we conclude that

v*(x*, t) = cj' 0)

154

VIII. NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

so that the kinetic energy decreases with time. This result represents a type of stability inherent in viscous fluids; in Section 24 we will establish a stronger form of stability: we will show that under these conditions the kinetic energy actually decreases exponentially with time. It is instructive to write the Navier-Stokes equations in dimensionless form. Consider a solution of (15) with

b

O.

=

Let I and v be numbers with I a typical length (such as the diameter of a body) and v a typical velocity (such as a mainstream velocity). Further, identify points x with their position vectors x - 0 from a given origin 0, and define the dimensionless position vector

the dimensionless time tv

l

=1'

the dimensionless velocity

1 v(x, l) = - v(x, t), v

and the dimensionless pressure

Then grad

v=

I

-

v

[where grad v(x, l)

grad v,

I

Y'=2 V' , v

grad ito

=

I

2 v

.grad no

= V;;y(x, l), etc.], so that (15) reduces to v'

+ (grad v)V =

1 Re L\v - grad ito, (20)

div v = 0, where

Iv Re=v

is a dimensionless quantity called the Reynolds number of the flow.

155

22. NEWTONIAN FLUIDS

Equations (20) show that a solution of the Navier-Stokes equations with a given Reynolds number can be used to generate solutions which have different length and velocity scales, but the same Reynolds number. This fact allows one to model a given flow situation in the laboratory by adjusting the length and velocity scales and the viscosity to give an experimentally tractable problem with the same Reynolds number. Some typical values for the kinematical viscosity are! :

water: v = 1.004 air: v

=

15.05

X

10- 2

cm 2jsec,

X

10- 2

cm 2jsec.

EXERCISES

1. A Reiner-Rivlin jluid 2 is defined by the constitutive assumptions of the Newtonian fluid with the assumption of linearity removed. Use (37.15) to show that the response of a Reiner-Rivlin fluid is independent of the observer if and only if the constitutive equation has the form T = -nI

+ (l(o(.J"o)D + (l(l(.J"o)D 2

with (l(o(.J"o) and (l(l(.J"O) scalar functions of the list.J"o = (0, liD), of principal invariants of D.

2. Consider a Newtonian fluid in a fixed, bounded region assume that

v=o

on

§I

13 (D ))

of space and

G.ry(

for all time. (a)

Show that the rate at which the fluid dissipates energy, i.e.,

can be written in the alternative forms

indicating that all of the e,nergy dissipation is due to spin. At 20°C and atmospheric pressure. Cf., e.g., Bird, Stewart, and Lightfoot [I, p. 8]. There are better models of non-Newtonian fluids (cf. Truesdell and Noll [I, Chapter E] and Coleman, Markovitz, and Noll [I]). 1

2

156 (b)

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

The surface force exerted on expression

of7t by the fluid is given

s = no - 2JlWo = no

by the simple

+ JlO x curl v,

where 0 is the outward unit normal to for a plane portion of the boundary.

of7t.

Establish this relation

23. SOME SIMPLE SOLUTIONS FOR PLANE STEADY FLOW For steady flow in the absence ofbody forces the Navier-Stokes equations (22.15) reduce to po(grad v)v = Jl .1v - grad n, div v

= o.

(I)

Consider the plane velocity field v(x)

= Vl(X l, X 2)el'

(2)

Then (1)2 implies that

so that V l = grad v is

V 1(X2)

and v is a simple shear (see Fig. I). Further, the matrix of

(3)

and div v = 0,

(grad v)v =

o.

Therefore (I) reduces to

02V 1

on

ox~

OXl'

Jl-=-

(4)

23. SOME SIMPLE SOLUTIONS FOR PLANE STEADY FLOW

157

X2

Figure 1

We now consider two specific problems consistent with the above flow. Problem 1 (Flow between two plates). Consider the flow between two infinite flat plates, one at x 2 = 0 and one at Xz = h, with the bottom plate held stationary and the top plate moving in the x 1 direction with velocity v (Fig. 2). Then the boundary conditions are

v 1 (0) = 0,

v 1 (h) = v.

(5)

We assume in addition that there is no pressure drop in the xcdirection, so that 11: == const. Then (4)1 yields Vi

=

(X

+ 1h z ,

and this relation satisfies the boundary conditions (5) if and only if (X = 0 and

P = v/h. Thus the solution of our problem is Vi

= ox-fh.

(6)

In view of (22.13), (3), and (6), the stress T is the constant field [T]

=

-1 :[~ o ~ ~] 0

+ fl;'

1

[~ ~ ~];

0 0 0

v

Figure 2

158

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

that is,

TI 2

1l'U

= T2 1 = h'

Thus the force per unit area exerted by the fluid on the top plate has a normal component tt and a tangential (shearing) component - 1l'U/h. This fact furnishes a method of determining the viscosity of the fluid. Problem 2 (Flow between two fixed plates under a pressure gradient). Again we consider the above configuration, but now we hold the two plates fixed in space (Fig. 3). Then the relevant boundary conditions are (7) If the pressure were constant, then the solution of the first problem would tell us that the velocity is identically zero. We therefore allow a pressure drop; in particular, since v = V(X 2), (4)1 implies that

- - --u 2

°s A.
such that

°

(7)

163

24. UNIQUENESS AND STABILITY

Proof

In view of (6), (v, n) satisfies the system v'

+ (grad v)v = v dv - grad IX, div v = 0,

on

v = 0

v(x, 0)

(8)

i}fJl x [0, (0),

= vo(x),

where IX = n + {3. If we take the inner product of (8)1 with v and use the identities (4) 1, 2, we arrive at the relation t(v 2 )'

+ v· gradt« + tv 2 ) =

v div{(grad

V)T V} -

v Igrad v1 2 .

If we integrate this relation over fJl, we conclude, with the aid of the divergence theorem, (8h, 3, and the lemma, that

d

dt IIvl1

2

+ 2vllgrad vl1 2

where [grad vl1

2

=

= 0,

L

Igrad vl 2 dV.

By the Poincare inequality (Exercise 1) there exists a constant that

..1. 0

> 0 such

Thus

where A.

= VA. o , and hence

which integrates to give or equivalently, Ilvll(t) s Ilvll(O)e- l ' . In view of (8)4' this inequality clearly implies (7).

0

Remark. It should be emphasized that the preceding two theorems require that the solution be smooth for all time and therefore may be misleading, since there is no global regularity theorem for the Navier-Stokes equations. 1 1

cr. Ladyzhenskaya [I, §6]; Temam [I, Chapter III].

VIII.

164

NEWTONIAN FLUIDS. NAVIER~STOKES

EQUATIONS

EXERCISES

l.

Supply the details of the following proof of the Poincare inequality. Let be a smooth scalar field on fJf with cp = 0 on a~. Since fJf is bounded, cp can be extended to a box ~ (with sides parallel to coordinate planes) by defining cp = 0 on ~ - fJf. Suppose that for x E ~,x\ varies between o: and fJ. Define cp\ = acpjax\. Then tp

cp2(X)

= 2 flcp(e,X2,X3)cp\(e,X2,X3)de,

and hence

r cp2 dV = 2 J~ J",('cp(e, x 2, x 3)cp\(e, x 2, x 3) de dV

x

J!R

~

2

X

s

(L f (L f

cp2(e, x 2, x 3) de dVx ) \/2 CPT(e, x 2, X3) de dVx ) 1/2

2(fJ - rx)(L cp2 dV yl2 (L [grad cp

2 1

dV yl2

which implies the Poincare inequality

L cp2 dV

~

4(fJ - rx)2 L1grad cpl2 dV.

Show that this implies an analogous result for a smooth vector field v on fJf with v = 0 on afJf. 2. Establish (4). SELECTED REFERENCES

Bird, Stewart, and Lightfoot [l, Part I]. Hughes and Marsden [I, §§ 16, 17]. Ladyzhenskaya [I]. Lamb [I, Chapter I]. Noll [I]. Serrin [l, Chapter G]. Temam [I].

CHAPTER

IX Finite Elasticity

25. ELASTIC BODIES

In classical mechanics the force on an elastic spring depends only on the change in length of the spring; this force is independent of the past history of the length as well as the rate at which the length is changing with time. For a body the deformation gradient F measures local distance changes, and it therefore seems natural to define an elastic body as one whose constitutive equation gives the stress T(x, t) at x = x(p, r) when the deformation gradient F(p, t) is known: T(x, t)

= t(F(p, r), p).

(1)

Formally, then, an elastic body is a material body whose constitutive class C(j is defined by a smooth response function

t: Lin" x

fJ4 --+ Sym

as follows: C(j is the set of all dynamical processes (x, T) consistent with (1). Note that for an elastic body the function t is completely determined by the response of the material to time-independent homogeneous motions of the form x(p, r)

= q + F(p - Po);

that is, to homogeneous deformations. 165

(2)

IX.

166

FINITE ELASTICITY

When convenient, and when there is no danger of confusion, we will write T(F) in place of T(F, p). Proposition. A necessary and sufficient condition that the response of an elastic body be independent of the observer is that the response function t satisfy

QT(F)QT = T(QF)

(3)

for every F E Lin + and every Q E Orth + • Proof Choose FE Lin + arbitrarily, let x be the motion (2), and let T be defined by (1), so that (x, T) E ~, the constitutive class of the body. Assume that the response is independent of the observer. Then every (x*, T*) related to (x, T) by a change in observer must also belong to ~. By (20.4) and (21.1), this can happen only if, given any rotation Q,

QTQT

=

T(QF).

Since T = T(F), this clearly implies (3). Conversely, if (3) is satisfied, then, in view of (20.4) and (21.1), the response is independent of the observer. D We assume henceforth that the response is independent of the observer, so that (3) holds.

The importance of the strain tensors U and C is brought out by the next result, which gives alternative forms for the constitutive equation (1). Corollary (Reduced constitutive equations). The response function T is completely determined by its restriction to Psym; infact,

T(F) = RT(U)R T for every F E Lin +, where R E Orth + is the rotation tensor and U right stretch tensor corresponding to F; that is, F = RU is the decomposition ofF. Further, there exist smooth response functions from Psym into Sym such that

(4)

E Psym the right polar * T, - and T T,

T(F) = Ft(U)FT, T(F) = RT(C)R T ,

(5)

T(F) = FT(C)FT , with C

U2

FTF the right Cauchy-Green strain tensor corresponding to F. Proof To derive (4) we simply choose Q in (3) equal to R T. Next, since F = RU, (4) can be written in the form =

=

T(F) = FU- 1T(U)U- 1FT.

25. ELASTIC BODIES

167

T(U) = U- 1T(U)U- 1,

(6)

Thus if we define we are led to (5h. On the other hand, if we let T(C) = T(C 1 / 2 ) , T(C)

= T(C 1 / 2 ) ,

(7)

for C E Psym, then (4) and (5)1 imply (5h,3' Finally, by (6),.T is smooth, while (7) and the smoothness ofthe square root (page 23) yield the smoothness ofT and T. 0 The converse is also true: each of the constitutive equations in (4) and (5) is independent of the observer. We leave the proof of this assertion as an exercise. * T, - and -T depend also It is important to note that the response functions T, on the material point p under consideration. Suppose that we rotate a specimen of material and then perform an experiment upon it. If the outcome is the same as ifthe specimen had not been rotated, then the rotation is called a symmetry transformation. To fix ideas consider the two-dimensional example shown in Fig. 1; there an elastic ring is connected by identical mutually perpendicular elastic springs which meet at the center ofthe ring. The forces required to produce any given deformation are the same for any prerotation of the ring by a multiple of 90°, so that such rotations are symmetry transformations. Moreover, they are the only symmetry transformations, since any other prerotation can be detected by some subsequent deformation. We now apply these ideas to the general elastic body 114. Choose a point p of 114 and let fr denote the homogeneous deformation from p with deformation gradient F: f,(q) = p

+ F(q

- p)

for every q E 114. Consider two experiments: (1) Deform 114 with the homogeneous deformation f r . (2) Rotate 114 with the rotation fQ (Q E Orth +) about p and then deform the rotated body with the homogeneous deformation fF (Fig. 2). In this case the total deformation is fr ° fQ and the deformation gradient is FQ.

Figure 1

IX.

168

FINITE ELASTICITY

Figure 2

The stress at p is

T(F, p) in Experiment 1 and

T(FQ, p) in Experiment 2. If for Q fixed these two stress tensors coincide for each F, then Q is called a symmetry transformation; Q has the property that the response of the material at p is the same before and after the rotation f Q • Thus, more succinctly, a symmetry transformation at p is a tensor Q E Orth + such that

T(F, p)

=

T(FQ, p)

(8)

for every F E Lin ", We write 0 and Q E Orth ". An incompressible elastic body is an incompressible material body defined by a constitutive equation of the form T

=

-nI

+ t(F),

(22)

or, more precisely, for x = x(p, t), T(x, t)

= -1t(p, t)I +

t(F(p, t), p)

with

t: Unim x

f!J -+

Sym

smooth. (As in the case of an incompressible fluid the pressure 1t is not uniquely determined by the motion.) For such a body: (a) Determine necessary and sufficient conditions that the response be independent of the observer. (b) Define the notions of material symmetry and isotropy. (c) Show that in the case of isotropy the constitutive equation can be written in the form (23) with fa = (ll(D), 12 (D), 1). [The quantities 1tappearing in (22) and (23) are not necessarily the same.]

26. SIMPLE SHEAR

175

Two important examples of (23) are the Mooney-Rivlin material for which T = -nI + PoB + PIB- I with

Po and PI constants, and the neo-Hookean material for which T

with

= -nI + PoB

Po constant. 26. SIMPLE SHEAR OF A HOMOGENEOUS AND ISOTROPIC ELASTIC BODY

Let f1l be a homogeneous, isotropic body in the shape of a cube. Consider the deformation x = x(p) defined (in cartesian components) by Xl

= PI + ypz,

where

Y = tan 8 is the shearing strain (Fig. 4). Since this deformation is homogeneous, the corresponding stress T is constant. Thus (x, T) represents a solution of the basic equations with body force b = o. (Cf. the discussion given at the end of the last section.) The matrix corresponding to the deformation gradient F is given by I

Y 0]

[F]=OIO [ 001

x

• e2

e2

0

0

e,

Figure 4

x(~)

e.

IX.

176

FINITE ELASTICITY

and the matrix for the left Cauchy-Green strain tensor B = FFT is

IH

[B]~[l~y' A simple calculation shows that

[Br' det{B - wI)

=

~ [~Y ~YY' H 1

_w 3

+ {3 +

y2)W 2 -

{3 +

y2)W

+ 1,

and hence the list of principal invariants of B is §B

=

(3

+ yl,3 + y2,

1).

We therefore conclude from (25.12) that T = Po{y2)I

+ Pl{y2)B + pz< y2)B - l ,

Hence Tl 3 = T23 = 0 and (l)

where (2) is the generalized shear modulus. The result (1) asserts that the shear stress Tl2 is an odd function of the shear strain y. In linear elasticity theory (cf. page 202) the normal stresses Tl l , T2 2 , and T3 3 in simple shear are zero. Here Tll

= y2 Pl + r

T22 = y2P2 + r

(3)

26. SIMPLE SHEAR

177

Thus for the normal stresses to vanish both /31 and and this, in turn, would imply that J1 = 0. Thus if for

/32 would have to be zero,

y #- 0,

(4)

which is a reasonable assumption, then it is impossible to produce a simple shear by applying shear stresses alone. Further (2)-(4) imply that, for y #- 0,

which is the Poynting effect. In fact, (1)-(3) yield the important result

This relation is independent of the material properties of the body; it is satisfied by every isotropic elastic body in simple shear.

EXERCISES

1.

Show that the components of the unit normals on the slanted faces of the deformed cube are given by nl

= ±(I + y2)-1/2

n2

= +: y(I + y2)-1/2,

n3

=

0,

and use this fact to show that T 22

_

yT12 and 1 + y2

TI 2 1 + y2

represent, respectively, normal and tangential components of the surface force Tn on these faces. 2. Consider the uniform extension defined by

x3 =

WP3'

Show that the corresponding stress T is a pure tension in the direction e.; i.e.,

IX.

178

FINITE ELASTICITY

if and only if (J

=

Po + Pl;.2 + P2;.-2,

0= Po + PlQi + P2 W - 2, where the scalar functions Po, Pl' and (through the invariants J B) .

P2 are each functions of;.2 and w 2

27. THE PIOLA-KIRCHHOFF STRESS The Cauchy stress T measures the contact force per unit area in the deformed configuration. In many problems of interest-especially those involving solids-it is not convenient to work with T, since the deformed configuration is not known in advance. For this reason we introduce a stress tensor which gives the force measured per unit area in the reference configuration. Let (x, T) be a dynamical process. Then given a part 9, we can use (6.18h to write the total surface force on 9 at time t (see Fig. 5) as an integral over 09; the result is

r

Jot?',

Tm dA

=

r

Jot?'

(det F)T",F-Tn dA,

where m and n, respectively, are the outward unit normal fields on 09, and 09, while T", is the material description of T. Thus if we let (1)

then

r

Jot?',

TmdA

=

S: PA x

~

r SndA.

Jot?'

(2)

We call the field -. Lin

defined by (1) the Piola-Kirchhoff stress.' By (2), Sn is the surface force measured per unit area in the reference configuration. Similarly, if b is the body force corresponding to (x, T), then

r b dV = Jt?'r b",(det F) dV = Jt?'r bo dV,

Jt?'t 1

In the literature S is often referred to as thefirst Piola-Kirchhoffstress.

(3)

27.

THE PIOLA-KIRCHHOFF STRESS

179

~-+.D(P)

Figure S

where bo = (det F)b....

(4)

We call bo the reference body force; bo gives the body force measured per unit volume in the reference configuration.

Proposition.

The Piola-Kirchhoffstress S satisfies the balance equations

f

J8~

f

J8~

Sn dA

(x - 0) X Sn dA

+

+ f (x J~

f

bo dV =

J~

f ~

- 0) X bo dV

xPo dV,

=

f ~

(5) (x - 0) X xpo dV

for every part 9. Proof.

By (12.7),

f

tp dV

=

~,

f

xPo dV.

~

This relation, (2), and (3), when combined with balance of linear momentum (14.2)1' imply (5)1. Similarly, (6.18h, (12.7), (14.2h, and an analogous argumentyields (5)2. 0

Proposition. S satisfies the field equations

= Po X, SF T = FS T •

Div S + bo

Proof.

(6)

By (5)1 and the divergence theorem,

L

(Div S

+ bo -

Po x) dV

=

0,

180

IX.

FINITE ELASTICITY

and, since this relation must be satisfied for every part &>, (6)1 must hold. Further, (1) implies T", = (det F)-ISFT ,

0

and (6)z follows from the symmetry of T.

It is important to note that, by (6)z, S generally is not symmetric.

Next, by the symmetry ofT, (a) of the proposition on page 6, (8.8)1' and (1.5)z,

and therefore

f

T . D dV

~.

=

f

(det F)(T",F- T )

•

F dV =

f.s- F

dV.

Thus the stress power of &> at time t is given by

LF S·

dV.

Also, (1), (4), (6.18)1' and (6.14)1 imply

f Tm. v dA = r (Tv)· m dA = f Sn· idA, lB., lB., lB.

f.,

b· v dV

=

f.

(7)

bo • i dV,

while (12.7) yields

We therefore have the following alternative version of (15.2). Theorem of Power Expended.

Given any part &>,

r Sn· i dA + f• bo • it dV = f•S • F dV + ddt .I.r i2 Po dV. lB. 2

(8)

The results established thus far are consequences of balance of momentum and are independent of the particular constitution of the body. Assume now that the body is elastic, so that by (1) and (25.1),S is given by a constitutive equation of the form

S = S(F)

(9)

27. THE PIOLA-l(IRCHHOFF STRESS

181

S(F) = (det F)T(F)F- T.

(10)

with (As before, we do not mention the explicit dependence of S on the material point p.) Choose Q E Orth ". Then det(QF) = det F, (QF)-T = QF- T, so that (25.3) and (10) imply S(QF)

=

det(QF)T(QF)(QF)-T

=

(det F)QT(F)QTQF- T = QS(F).

Thus S(QF) = QS(F)

(11)

for every F E Lin + and Q E Orth ". This relation expresses the requirement that the response be independent of the observer; it can be used to deduce constitutive equations for S analogous to (25.5). It is easier, however, to proceed directly from (25.5h. Indeed, (25.5h, (9), and (10) imply that

S

=

(det F)FT(C),

(12)

and if we define seC) = Jdet C fCc),

(13)

S = FS(C).

(14)

then, since

(12) reduces to We leave it as an exercise to show that this constitutive equation is independent of the observer. Note that, by (13) and the smoothness off (cf. the corollary on page 166), S is a smooth mapping of Psym into Sym. In view of (o), and (14), we can rewrite the basic system offield equations (25.13)-(25.15) in the form

S = FS(C), C = FTF,

Div S

F

+ bo =

=

Vx

(15)

Po X.

The relation (6h follows automatically from the fact that S has symmetric values. Also, since the density enters (15) only through its reference value Po, which is assumed known a priori, balance of mass (25.15) need not be included in the list of field equations.

IX.

182

FINITE ELASTICITY

All of the fields in (15) have fJI x IR as their domain, and the operator Div is with respect to the material point p in fJI. In contrast, some of the fields in (25.13)-(25.15) are defined on the trajectory ff, and, more importantly, the operator div is with respect to the place x in the current configuration. For this reason the formulation (15) is more convenient than (25.13)-(25.15) in problems for which the trajectory is not known in advance. The boundary-value problems of finite elasticity are obtained by adjoining to (15) suitable initial and boundary conditions. As initial conditions one usually specifies the initial motion and velocity: x(p, 0)

=

x(p, 0)

xo(p),

=

(16)

vo(p)

with Xo and Vo prescribed functions on fJI. To specify the boundary conditions one considers complementary regular 1 subsets [/1 and [/2 of ofJI (so that

o

where [/IX is the relative interior of [/IX) and then prescribes the motion on [/1' the surface traction on [/2: x =

x

on

[/1

x [0, (0),

on

[/2

x [0, (0)

(17)

with x and s prescribed vector fields on [/1 X [0, (0) and [/2 X [0, (0), respectively. Another form of boundary condition arises when one specifies the surface traction Tm on the deformed surface X([/2' t). A simple example of this type of condition arises when one considers the effects of a uniform pressure no. Here T(x, t)m(x)

=

-nom(x)

for all x E X([/2' t) and all t ~ 0. We can also write this condition as a restriction on the Piela-Kirchhoff stress S; the result is (cf. Exercise 6) on

[/2

x [0, (0).

(18)

Clearly, this is not a special case of (17) because F(p, t) is not known in advance. We can, of course, generalize (17) to include this case by allowing s to be a function s(F, p, t) ofF, p, and t, rather than a function ofp and t only. [Actually, one can show that the combination (det F)F-Tn depends only on the tangential gradient of x on [/2'] 1 Cf. Kellogg [I]; Gurtin [I, p. 14]. Roughly speaking, each' 5/'. is a relatively closed, piecewise smooth surface with boundary a piecewise smooth closed curve.

183

27. THE PIOLA-KIRCHHOFF STRESS

In the statical theory all of the fields are independent of time and the underlying boundary-value problem consists in finding a deformation f that satisfies the field equations S

= FS(C),

C = FTF,

(19)

F = Vf,

Div S + b o = 0, and the boundary conditions on

So

9'1'

=

s

on

9'2

(20)

with f and s prescribed functions on 9'1 and 9'2' respectively. When tractions are prescribed over the entire boundary, that is, when (20) has the form So

=

s

on

(21)

i}[fI,

(5)1 implies that

Jr s dA + iJl1I

i

111

bo dV = 0.

This relation involves only the data and furnishes a necessary condition for the existence of a solution. On the other hand, (5)z yields

r (f -

J

iJl1I

0) x

s dA

+

i

111

(f - 0) x b o dV = 0,

and, because of the presence of the deformation f, is not a restriction on the data, but rather a compatibility condition which will automatically be satisfied by any solution of the boundary-value problem (19), (21). This difference between force and moment balance is illustrated in Fig. 6. As long

-

----~ Figure 6

184

IX.

FINITE ELASTICITY

as the forces acting on f!4 obey balance of forces, we expect a solution, even though moments are not balanced in the reference configuration; the body will simply deform to insure balance of moments in the deformed configuration. EXERCISES

1.

Give the details of the proof of (5h.

2.

Show that

S(FQ)

3.

S(F)Q

=

(22)

for every Q in the symmetry group f§ and every F that Sand S are invariant under f§. Use (11) to show that

S(Q)·

Q=

E

Lin +. Show further

0

(23)

for any smooth function (of time) Q with values in Orth + . 4.

Let f!4 be bounded. Consider a dynamical process and define qJ =

~

L

2

u Po dV,

where u(p, t) = x(p, t) - P is the displacement. Show that

iP =

f ~

iJ2 po dV -

f

~

S· Vu dV +

r

Jo~

u· So dA.

5. Show that the constitutive equation (14) is independent of the observer. 6.

Equation (6.18h holds for any sufficiently regular subsurface Y' of of!):

i

T(x)m(x) dA; =

f(9')

f

T(f(p»G(p)o(p) dA p •

9'

Use this fact to establish (18).

28. HVPERELASTIC BODIES Thus far the only general restriction we have placed on constitutive equations is the requirement that material response be independent of the observer. Thermodynamics also serves to constrain constitutive behavior, and a standard thermodynamic axiom (consistent with a purely mechanical theory) is the requirement of nonnegative work in closed processes. We now study the implications of this axiom for elastic bodies.

28.

185

HYPERELASTIC BODIES

Let (x, T) be a dynamical process, and let b be the corresponding body force. Then given any part f!I', the work on f!I' during a time interval [to, t 1] is given by

f' {r to

Tn' v dA

J{j~t

+

f

+

Lbo'

b· v dV} dt,

~t

or equivalently, in view of (27.7), by

I' {fa~

Sn • i dA

(1)

i dV} dt

with S the Piola-Kirchhoff stress and bo the reference body force. Let us agree to call (x, T) closed during [to, t 1 ] provided (2)

x(p, to) = x(p, t 1),

for all p E 911. If we integrate (27.8) between to and t 1 and use (2h, we see that for processes of this type (1) reduces to

ff tl

to

S . F dV dt.

~

We consider now an elastic body and restrict our attention to dynamical processes belonging to the constitutive class 0 let p~: IR --+ IR be a Coo function with .the following properties:

0;

(i)

p~

(ii)

p~(t)

= 0 whenever It I ~ fJ;

(iii)

f~oo

pit) dt = l.

~

An example of such a family of functions is furnished by

{

pit) =

, ~ exp ( -

fJ2 1 -

t

2)'

It I
0, there must exist a 150 > such that det Ftl > 0 on IR for 0 < 15 < bo . For this range of 15, F, can be considered as a mapping of IR into Lin +, This completes the proof of Assertion 2. Our last step is

°

Assertion 3. The body is hyperelastic. Proof Let F be an arbitrary piecewise smooth, closed curve in Lin + , and let F tl (0 < 15 < 15 0 ) be the family established in Assertion 2. By (a) of Assertion 2,

190

IX.

FINITE ELASTICITY

as long as C> < 1. Let C>I = min(1, C>o). Then for 0 < C> < C>I, each F" satisfies the hypotheses of Assertion 1 (with to = -1 and t l = 2); hence

r

I

({J,,(t) dt

(16)

0,

=

where (17)

By (a) of Assertion 2 the integral from - 1to we can therefore rewrite (16) as

II

({J" dt

o

+ fO

({J" dt

+

C> and

fl+"

-"

from 1 +

({J" dt

=

o.

C>

to 2 vanishes;

(18)

I

By (b) and (c) of Assertion 2 in conjunction with Lebesgue's dominated convergence theorem I (recall that F has at most a finite number of discontinuities),

f

({J" dt -.

f

S(F) • F dt.

On the other hand, (b) of Assertion 2 and (17) imply that the remaining integrals in (18) converge to zero as C> -. O. Thus

f

S(F) • F dt

= O.

What we have shown is that the integral of S over any piecewise smooth, closed curve in Lin + vanishes. Since Lin + is an open, connected subset of (the vector space) Lin, a standard theorem- in vector analysis tells us that S is the derivative of a smooth scalar function b on Lin +. Thus (5) holds and the proof is complete. 0 Hyperelastic materials have several interesting properties; an example is the following direct consequence of (7).

Proposition.

For a hyperelastic material the work is zero in closed processes.

If we write IT(p, t)

=

b(F(p, r), t)

for the values of b in a process, then (6) implies 0I

2

= S· F,

Cf., e.g., Natanson [I, p. 161]. cr., e.g., Nickerson, Spencer, and Steenrod [I, Theorem 8.4].

28.

191

HYPERELASTIC BODIES

and the theorem of power expended (27.8) has the following important corollary. Theorem (Balance of energy for hyperelastic materials). Each dynamical process for a hyperelastic body satisfies the energy equation

L~

SO • xdA

+

Lbo· x

dV

= :t

L

(a

+ Po ~)

dV

for each part &>. Here S is the Piola-Kirchhoffstress, Po is the reference density, and bo is the reference body force (27.4).

The term Ladv

represents the strain energy of &>. The energy equation asserts that the power expended on &> must equal the rate at which the total energy of &> is changing. As a direct consequence of the above theorem we have the following important Corollary (Conservation of energy). Assume that the body is finite and hyperelastic. Consider a dynamical process for the body corresponding to body force b = 0, and suppose that So·

x=

0

afJI

on

for all time. Then the total energy is constant:

L(a

+ Po ~) dV

=

const.

EXERCISES

1. Consider a hyperelastic body with strain-energy density fJ. (For con-

venience, we suppress dependence on the material point p.) (a) Use (27.23) and (5) to show that fJ(Q) = fJ(I)

(19)

for every Q E Orth + . (Here you may use the connectivity of Orth + to insure the existence of a smooth curve R: [0, 1] -+ Orth + with R(O) = I and R(l) = Q.) (b) Use (27.11) and (5) to show that fJ( QF) = fJ(F)

for every Q E Orth + and F

E

Lin + .

IX.

192 (c)

FINITE ELASTICITY

Use (27.22) and (5) to show that &(FQ)

(d)

= &(F)

for every Q in the symmetry group and every .F E Lin + . Show that &(F) = &(U) and that there exists a function iJ such that &(F) = iJ(C).

(e)

Here U is the right stretch tensor, C the right Cauchy-Green strain tensor. Show that

S=

2DiJ,

where DiJ(C) E Sym with components

o~ .. iJ(C) 1)

(f)

has an interpretation analogous to D&(F). Assume that the material at p is isotropic. Show that iJ(C)

= iJ(B)

with B the left Cauchy-Green strain tensor, and that

[cf. (37.4)J. Let

Show that the Piela-Kirchhoff and Cauchy stress tensors are given by S =2{iJ 1F

+ iJ2 [ (tr B)I -

BJF + (det B)iJ3F- T } ,

T = 2(det F)-1{(det B)iJ3 1 + [iJ1

+ (tr B)iJ2JB -

iJ2B 2 } ,

where we have omitted the arguments of iJk' Extend the results of this section to incompressible elastic bodies (cf. Exercise 25.6). 3. Consider a Newtonian fluid. Show that the work is nonnegative in closed processes if and only if the viscosity u is nonnegative.

2.

28. 4.

193

HYPERELASTIC BODIES

Consider the statical (time-independent) behavior of a homogeneous hyperelastic body without body forces. Show that Div[a(F)I

-

FTS] = 0

and hence that Ia9[6(F)n

5.

-

FTSn]d A

0

=

for every part 9of 9.(This result is of importance in fracture mechanics.) (Principle of stationary potential energy) For a hyperelastic body the (statical) mixed problem discussed in Section 27 can be stated as follows: find a C2 deformation f such that S

=

F

S(F) = Db(F), Div S

+ b,

=

Vf,

=0

on 9 and

f=P

on

Y,,

Sn=B

on Y,.

(20)

Let us agree to call a C2 function f : 9 4 8 with det Vf > 0 kinematically admissible i f f satisfies the boundary condition (20),. Assume that 9 is bounded. Define the potential energy @ on the set of kinematically admissible functions by

-

@{f} = Ia8(Vf) dV - IBb, u dV -

L-

I u dA,

where u(p) = f(p) - p, We say that the variation of @ is zero at f, and write

6@{f} = 0, if

d -

da

@{f

+ ag} la=,

=

0

for every g with f + ag in the domain of @ for all sufficiently small a (that is, for every C 2 function g : 9d -+ Y with g = 0 on Yl). Show that

b@{f) = 0 provided f is a solution of the mixed problem.

IX.

194

FINITE ELASTICITY

29. THE ELASTICITY TENSOR The behavior of the constitutive equation S = S(F) [cr. (27.9)] near F = I is governed by the linear transformation C: Lin

--+ Lin

defined by C

=

(1)

DS(I).

(As before we have suppressed mention of the dependence on the material point p.) C is called the elasticity tensor for the material point p; TOughly speaking, C is the derivative of the Piela-Kirchhoff stress with respect to F at F = I. The importance of this tensor will become apparent in the next section, where we deduce the linearized theory appropriate to small deformations from the reference configuration. For convenience, we assume throughout this section that the residual stress vanishes:

S(I)

= t(I) = O.

(2)

Our first result shows that, because of (2), C could also have been defined as the derivative of r, the response function for the Cauchy stress. Proposition C Proof

=

(3)

Dt(I).

By (27.10), S(F)FT = qJ(F)t(F),

where qJ(F) = det F. If we differentiate this relation with respect to F, using the product rule, we find that S(F)HT

+ DS(F)[H]FT =

qJ(F) Dt(F) [H]

+ DqJ(F) [H]t(F)

for every HELin. Evaluating this expression at F = I, we conclude, with the aid of (2), that DS(I) [H] = Dt(I) [H], which implies (3). 0

29. Proposition

195

THE ELASTICITY TENSOR

(Properties of the elasticity tensor)

(a) C[H] E Symfor every HELin; (b) C[W] = Ofor every W E Skw. Proof.

Assertion (a) follows from the relation C[H]

=

d drx 1'(1 + rxHH.=o

and the fact that l' has symmetric values. To prove (b) choose W E Skw and take Q(t)

= eW t ,

so that Q(t) E Orth + (cf. Section 36). Then (25.3) with F = I and (2) imply t(Q(t» = 0,

and this relation, when differentiated with respect to t, yields Dt(Q(t»

toon = O.

Since Q(O) = I and Q(O) = W, if we evaluate this expression at t = 0 and use (3), we are led to (b). 0 Let

denote the symmetric part of HELin. Then H = E + W, with W skew, and (b) and the linearity of C imply that C[H]

=

C[E];

(4)

hence C is completely determined by its restriction to Sym. Our next step is to establish the invariance properties of C. Proposition.

C is invariant under the symmetry group

'§ for the material at

p.

Proof. In view of (25.9)1' the last theorem in Section 37, and (3), for HELin and Q E e,

QC[H]QT = QDt(l) [H]QT = Dt(QIQT) [QHQT] = Dt(l) [QHQ~] = C[QHQT], and C is invariant under '§.

0

IX.

196

FINITE ELASTICITY

Since C has values in Sym [cf. (a) above], this proposition and (37.22) have the following obvious but important consequence. Theorem. Assume that the material at p is isotropic. Then there exist scalars Jl and A such that

C[E] = 2JlE + A(tr E)I

(5)

for every symmetric tensor E. The scalars Jl = Jl(p) and A = A(p) are called the Lame moduli at p.

We say that C is symmetric if H'C[G] = G' C[H] for all tensors Hand G. Since C[W] = 0 for all skew W, C can never be positive definite in the usual sense. There are, however, important situations in which C restricted to Sym has this property. Thus let us agree to call C positive definite if

E' C[E] > 0 for all symmetric tensors E #- O. Proposition. Assume that the material at p is isotropic. Then C is symmetric. Moreover, C is positive definite if and only if the Lame moduli obey the inequalities: 2Jl + 3A > O.

u > 0,

(6)

Proof Let Hand G be tensors with symmetric parts Us and G s ' respectively. Then, by (a) and (b) of the proposition on page 195, and since I· H, = tr H, (5) implies

H • C[G] = H s ' C[Gs] = 2JlHs ' G s

+ A(tr H)(tr G) =

G . C[H],

so that C is symmetric. Next, choose a symmetric tensor H and let. a=!trH,

H o = H - aI

so that

H

=

H,

+ aI,

tr H o

= O.

Then, since I • H, = 0, H • C[H] = 2Jl(1H o 12

+ 3ct 2 ) + 9A.a 2

= 2Jll H, 12

+ 3a2(2ti + 3A).

Trivially, (6) implies that C is positive definite. Conversely, if C is positive definite, then by choosing H = aI we conclude that 2Jl + 3A > 0, and by choosing H with tr H = 0 we see that Jl > O. 0

197

29. THE ELASTICITY TENSOR

Assume that the material at p is hyperelastic. Then C is sym-

Proposition. metric.

Proof

Since

[ef. (28.5)], the proposition follows from the symmetry of the second derivative. To see this note that

o

of3 ~(I + txH + f3G)

=

+ txH + f3G) . G,

SCI

and hence

02 Otx of3 ~(I

+ txH + f3 G) !a=P=o = DS(I)[H]· G

=

C[H] . G;

by switching the order of differentiation we see that this expression equals

02

of3 ctx ~(I + txH + f3 G ) la=P=o = and the symmetry of C follows.

C[G]' H,

D EXERCISES

1. Show that, as a consequence of (27.11), DS(F) [WF]

for all F

E

Lin + and W

E

= WS(F)

(7)

Skw.

2. Relate C = DS(I) and L = Dt(I) for the case in which (2) is not satisfied. 3. C is strongly elliptic if

A· C[A] > 0 whenever A has the form A = a ® e, a (a) (b)

=1=

0, C

=1=

O.

Show that if C is positive definite, then C is strongly elliptic. Show that for an isotropic material C is strongly elliptic if and only if

2Jl.

+ A. >

O.

198

IX.

FINITE ELASTICITY

SELECTED ~ERENCES

Coleman and Noll [1]. Green and Zerna [1]. Sternberg and Knowles [1]. Truesdell and Noll [1, Chapter OJ. Wang and Truesdell [1].

CHAPTER

x Linear Elasticity

30. DERIVATION OF THE LINEAR THEORY We now deduce the linearized theory appropriate to situations in which the displacement gradient Vu is small. The crucial step is the linearization of the general constitutive equation (1)

for the Piela-Kirchhoff stress [cf. (27.9)]. In order to discuss the behavior of this equation as 0= Vu tends to zero, we consider S(F) as a function of H using the relation

F

=

I

+0

[cf. (7.1)]. Theorem (Asymptotic form of the constitutive relation). residual stress vanishes. Then

S(F) = C[E] as H

-+

+ 0(0)

Assume that the (2)

0, where C is the elasticity tensor (29.1) and

E = 1(H + H T ) is the infinitesimal strain (7.4). 199

(3)

x.

200

LINEAR ELASTICITY

Proof Since the residual stress vanishes, we may conclude from (29.1), (29.2), and (29.4) that

S(F) = S(I

+ H)

+ DS(I) [HJ + o(H) = C[HJ + o(H) = C[EJ + o(H). D = S(I)

Using (2) we can write the asymptotic form of the constitutive equation (1) as

S = C[EJ

+ o(Vu).

(4)

Therefore, if the residual stress in the reference configuration vanishes, then to within terms ofo(Vu) as Vu -. 0 the stress S is a linearfunction ofthe infinitesimal strain E. Also, since C has symmetric values [cf. (a) of the proposition on page 195J, to within the same error S is symmetric. The linear theory of elasticity is based on the stress-strain law (4) with the terms of order o(Vu) neglected, the strain-displacement relation (3), and the equation of motion (27.6)1: S = C[EJ, E

=

t(Vu

Div S

+ VuT) ,

+ bo =

(5)

Poii.

Note that these equations are expressed in terms of the displacement

u(p, t)

=

x(p, t) - p,

rather than the motion X. It is important to emphasize that the formal derivation of the linearized constitutive equation (5)1 was based on the following two assumptions: (a) (b)

The residual stress in the reference configuration vanishes. The displacement gradient is small.

Note that, by (5)1 and the theorem on page 56, E = S = 0 in an infinitesimal rigid displacement. This is an important property of the linearized theory. Given C, Po, and bo , (5) is a linear system of partial differential equations for the fields u, E, and S. By (29.5), when the body is isotropic (5)1 may be replaced by S = 2J-lE + A.(tr E)I. Moreover, when the body is homogeneous, Po, p; and A. are constants.

(6)

201

31. SOME SIMPLE SOLUTIONS

Assume now that f!l is homogeneous and isotropic. Then, since Div(Vu

+ VU T) = L\u + V Div u

and tr E = Div u, the equations (5)2.3 and (6) are easily combined to give the displacement equation of motion Ji. L\u

+ (A + Ji.)V Div u + bo = Poii.

In the statical theory ii = 0 and we have the displacement equation of equilibrium Ji. L\u

+ (A + Ji.)V Div u + bo =

O.

(8)

EXERCISES

1.

Let u be a C 4 solution of (8) with bo = O. Show that Div u and Curl u are harmonic functions. Show further that u is biharmonic; that is,

= O. with b o =

L\L\u 4

Let u be a C solution of (7) O. Show that Div u and Curl u satisfy wave equations. 3. (Boussinesq-Papkovitch-Neuber solution) Let qJ and g be harmonic fields on f!l (with qJ scalar valued and g vector valued) and define

2.

u = g - exV(r . g

+ qJ),

Ji.+A ex = 2(2Ji. + A)'

where r(p) = p - o. Show that u is a solution of (8) with bo = O. 4. Consider the case in which the residual stress S(I) =F O. Show that S(F)

=

8(1)

+

W8(1)

+ C[E] + o(H).

31. SOME SIMPLE SOLUTIONS Assume now that the body is homogeneous and isotropic. Then any statical (i.e., time-independent) displacement field u with E constant generates a solution of the field equations (30.5h. 3 and (30.6) with S constant and

x.

202

LINEAR ELASTICITY

L

Figure I

I

bo = O. We now discuss some particular solutions of this type, using cartesian coordinates where convenient. (a)

Pure shear.

Let u(p)

= YP2 el

(Fig. I), so that the matrices of E and S are

[E]

=

0] 21[0Y 0Y 0,

~ ~], o

000

0

with 't

= p.y.

Thus p. determines the response of the body in shear, at least within the linear theory, and for this reason is called the shearmodulus. Note that, in contrast to the general nonlinear theory (cf. Section 26), the normal stresses are all zero; the only stress present is the shear stress 'to (b) Uniform compression or expansion. Here u(p) = B(p - 0).

Then E

= BI,

S = -nl,

n = -3KB, where

is the modulus of compression. For our third solution it is simpler to work with the stress-strain law (30.6) inverted to give E as a function of S. This inversion is easily accomplished upon noting that tr S

=

(2p.

+ 3A.) tr E,

31. SOME SIMPLE SOLUTIONS

203

1 E = 2 [S A. A. (tr S)IJ. J1. 2J1. + 3

(1)

and hence

(c)

Pure tension.

Here we want the stress to have the form (1

0 0]

[S] = 0 0 O. [ 000 The corresponding strain tensor is then given by

0]

oI

0

o

I

1=

-VG,

with 1

G

= E (1,

and E

=

J1.(2J1. + 3..1.) J1. + A. '

Note that the strain E corresponds to a displacement field of the form u(p) = GP1e 1 + lPzez + lp3e3.

The modulus E is obtained by dividing the tensile stress (1 by the longitudinal strain G produced by it. It is known as Young's modulus. The modulus v is the ratio of the lateral contraction to the longitudinal strain of a bar under pure tension. It is known as Poisson's ratio. If we write Eo and So for the traceless parts of E and S, that is, Eo = E - i{tr E)I,

So = S - :ktr S)I,

(2)

then the isotropic constitutive relation (30.6) is equivalent to the following pair of relations: So = 2J1.Eo , tr S = 3K tr E. Another important form for this stress-strain relation is the one taken by the inverted relation (1) when Young's modulus and Poisson's ratio are used: E =

1

E [(1 + v)S -

v(tr S)I].

(3)

X.

204

LINEAR ELASTICITY

Since an elastic solid should increase its length when pulled, should decrease its volume when acted on by a pure pressure, and should respond to a positive shearing strain by a positive shearing stress, we would expect that E > 0,

Il > 0.

K> 0,

Also, a pure tensile stress should produce a contraction in the direction perpendicular to it; thus v> O. Even though these inequalities are physically well motivated, we will not assume that they hold, for in many circumstances other (somewhat weaker) assumptions are more natural. In particular, we will usually assume that C is positive definite. A simple computation, based on (29.6), shows that this restriction is equivalent to either of the following two sets of inequalities: (i) (ii)

Il > 0, E > 0,

> 0; - 1< v
0, which contradicts (4). Thus r = 00 and (5) implies (2). 0 Proposition.

Proof

Let W be a skew tensor. Then eW t is a rotation for each t ~ O.

Let X(t) = eW t for t ~ O. Then, by definition, X=WX,

X(O)

=

I.

Let

Then, since W is skew, Z = XX T + XX T = WXX T + XXTW T = WXX T - XXTW,

and Z satisfies Z=WZ-ZW,

Z(O) = I.

This initial-value problem has the unique solution Z(t) = I for all t

~

O. Thus

X(t)X(t)T = I

and X(t) is orthogonal. But from the preceding proposition X(t) X(t) is a rotation. 0

E

Lin +. Thus

229

37. ISOTROPIC FUNCTIONS EXERCISE

1. Let A E Lin and Vo E "Y. Show that the function

=

eA1v o

Av(t),

t

vet) satisfies the initial-value problem vet)

=

v(O)

= Vo '

> 0,

SELECTED REFERENCE

Hirsch and Smale [1].

37. ISOTROPIC FUNCTIONS Let ~ c Orth. A set d c Lin is invariant under ~ if QAQTEd whenever AEdandQE~.

Proposition.

Thefollowing sets are invariant under Orth: Lin,

Lin +,

Orth,

Orth +,

Sym,

Skw,

Psym.

Proof We will give the proof only for Lin +. Choose A E Lin and Q E Orth. Then det(QAQT) = (det A)(det Q)2 = det A, since [det QI = 1. Thus A E Lin" implies QAQT E Lin+.

(1)

0

Let d c Lin. A scalar function qJ: d --+ IR is invariant under ~ if d is invariant under ~ and qJ(A)

=

qJ(QAQT)

(2)

for every A E d and Q E ~. Similarly, a tensor function G: d --+ Lin is invariant under ~ if d is invariant under ~ and

for every A E d and Q E ~. An isotropic function is a function invariant under Orth.

230

APPENDIX

Proposition. Let 0 mass of &' subscript indicating material description, also mach number outward unit normal to boundary symbol for "small order u" origin set of all orthogonal tensors set of all rotations (proper orthogonal tensors) material point reference map part of :J4 set of all symmetric, positive definite tensors orthogonal tensor rotation tensor position vector

Place of definition or first occurrence

92 41 58 98 179 46 46 34 194 32,61 30,61 61 21 71 6 203 55 I 42 42 123 168 199 15 3 220 63 92 7 7 87 60,133 37 19 2 7 7 41 59 41 7 7 46 92

240

Symhol IR IR+

S s Sym Skw ,;

sp tr t T

or u 'f/{E} U

V v v

r vol W w

x

x or.

p

cp

{,;} liS) K(p) Ak ). tt

Po P 8(F) Jl. v

ro V ~

Q9

x

ep cp'

APPENDIX

Name

reals strictly-positive reals Piola-Kirchhoff stress stress vector set of all symmetric tensors set of all skew tensors subscript indicating spatial description, also kinematically admissible state span trace time Cauchy stress trajectory displacement strain energy right stretch tensor left stretch tensor spatial description of the velocity speed vector space associated with r! volume spin complex potential place motion center of mass potential of body force potential of flow potential energy principal invariants of a tensor S sound speed principal stretches Lame modulus pressure reference density density in motion strain-energy density viscosity, also shear modulus kinematic viscosity, also Poisson's ratio angular velocity gradient, material gradient laplacian tensor product cross product material time derivative of cp spatial time derivative of cp

Place of definition or first occurrence I I

178 97 7 7 60,208 2 5 58 101 59 42 205 46 46 60 133 I

37 71 126 58 58 92 III III

208 15 131 45 196 106 88 88 186 149,196 151.203 70 29,60 32 4 7 61 61

38.

241

GENERAL SCHEME OF NOTATION GENERAL NOTATION

Expression

~ agp ~

gp U ,F gp n .17 gpc:F XEgp

j: gp .......'1" x ....... j(x)

jog

Meaning

interior of gp boundary of gp closure of gp union of gp and .F intersection of gp and .F gp is a subset of .F x is an element of the set gp j maps the set gp into the set .'1"; gp is the domain, :J7 the codomain the mapping that carries x into j(x); e.g., x ....... Xl is the mapping that carries every real number x into its square composition of the mappingsj andg; that is,(f 0 g) (x) =

{xIR(x) holds}

j(g(x»

the set of all x such that R(x) holds; e.g.,{x10 ~ x is the interval [0, 1]

~

I}

This page intentionally left blank

References

R. G. Bartle [I] The Elements of Real Analysis, 2nd ed. New York: Wiley (1976). R. B. Bird, W. E. Stewart, and E. N. Lightfoot [I] Transport Phenomena. New York: Wiley (1960). R. M. Bowen and C. C. Wang [1] Introduction to Vectors and Tensors, Vol. 1. New York: Plenum (1976). H. W. Brinkmann and E. A. Klotz [I] Linear Alqebra and Analytic Geometry. Reading: Addison-Wesley (1971). P. Chadwick [1] Continuum Mechanics. New York: Wiley (1976). B. D. Coleman, H. Markovitz, and W. Noll [I] Viscometric flows of non-Newtonian fluids. Springer Tracts in Natural Philosophy 5. New York: Springer-Verlag (1966). B. D. Coleman and W. Noll [I] Material symmetry and thermostatic inequalities in finite elastic deformations. Arch. Rational Mech. Anal. 15,87-111 (1964). R. Courant and K. O. Friedrichs [1] Supersonic Flow and Shock Waves. New York: Wiley (Interscience) (1948). J. Dieudonne [1] Foundations of Modern Analysis. New York: Academic Press (1960). A. C. Eringen [I] Nonlinear Theory of Continuous Media. New York: McGraw-Hili (1962). G. Fichera [1] Existence theorems in elasticity. Handbuch der Physik 692' Berlin: Springer-Verlag (1972). 243

244

REFERENCES

W. Fleming [I] Functions of Several Variables. Reading: Addison-Wesley (1965). P. Germain [I] Cours de Mecanique des Milieux Continus, Vol. 1. Paris: Masson (1973). A. E. Green and W. Zerna [I] Theoretical Continuum Mechanics. London and New York: Oxford Univ. Press (1954). M. E. Gurtin [I] The linear theory ofelasticity. Handbuch der Physik 6a 2 . Berlin: Springer-Verlag (1972). [2] A short proof of the representation theorem for isotropic, linear stress-strain relations. J. Elasticity 4, 243-245 (1974). M. E. Gurtin and L. C. Martins [I] Cauchy's theorem in classical physics. Arch. Rational Mech. Anal. 60,305-324 (1976). M. E. Gurtin and W. O. Williams [I] An axiomatic foundation for continuum thermodynamics. Arch. Rational Mech. Anal. 26, 83-117 (1967). P. R. Halmos [I] Finite-Dimensional Vector Spaces, 2nd ed. New York: Van Nostrand-Reinhold (1958). M. W. Hirsch and S. Smale [I] Differential Equations, Dynamical Systems and Linear Algebra. New York: Academic Press (1974). T. J. R. Hughes and J. E. Marsden [I] A Short Course in Fluid Mechanics. Boston: Publish or Perish (1976). O. D. Kellogg [I] Foundations of Potential Theory. Berlin and New York: Springer-Verlag (1929). Reprinted New York: Dover (1953). O. A. Ladyzhenskaya [I] The Mathematical Theory of Viscous Incompressible Flow. New York: Gordon & Breach (1963). H. Lamb [I] Hydrodynamics, 6th ed. London and New York: Cambridge Univ. Press (1932). Reprinted New York: Dover (1945). A. E. H. Love [I] A Treatise on the Mathematical Theory of Elasticity, 4th ed. London and New York: Cambridge Univ, Press (1927). Reprinted New York: Dover (1944). L. C. Martins and P. Podio-Guidugli [I] A variational approach to the polar decomposition theorem. Rend. Accad. Naz. Lincei (Classe Sci. Fis., Mat. Natur.) 66(6), 487-493 (1979). [2] A new proof of the representation theorem for isotropic, linear constitutive relations. J. Elasticity 8, 319-322 (1978). R. E. Meyer [I] Introduction to Mathematical Fluid Dynamics. New York: Wiley (lnterscience) (1971). I. P. Natanson [I] Theory of Functions ofa Real Variable. New York: Ungar (1955). W. Noll [I] On the continuity of the solid and fluid states. J. Rational Mech. Anal. 4,3-81 (1955). [2] A mathematical theory of the mechanical behavior of continuous media. Arch. Rational Mech. Anal. 2, 197-226 (1958). Reprinted in W. Noll, The Foundations of Mechanics and Thermodynamics. Berlin: Springer-Verlag (1974). [3] The foundations of classical mechanics in the light of recent advances in continuum mechanics. The Axiomatic Method, with Special Reference to Geometry and Physics.

REFERENCES

245

Amsterdam: North-Holland Pub!. (1959). Reprinted in W. Noll, The Foundations of Mechanics and Thermodynamics. Berlin: Springer-Verlag (1974). [4] La rnecanique classique, basee sur un axiome d'objectivite. La Methode Axiomatique dans les Mecaniques Classiques el Nouvelles. Paris: Gauthier-Villars (1963). Reprinted in W. Noll, The Foundations of Mechanics and Thermodynamics. Berlin: Springer-Verlag (1974). H. K. Nickerson, D. C. Spencer, and N. E. Steenrod [I] Advanced Calculus. New York: V a =

°

and (14) holds trivially. We therefore have the desired result for S E Sym and S E Skw. Complete the proof by showing that if S commutes with every WE Skw, then so also do the symmetric and skew parts of S. 7. Choose Q E Orth +, Q #- R. Use the identity IF - Q 12

= IF 12

-

2F • Q

+3

SECTION

5

249

to show that F - Q 12

1

where F

IF - R 12

-

=

2U . (I - Qo),

= RU is the right polar decomposition of F and Qo = RTQ =F I.

Next, establish the identity 2U • (I - Qo) = U· (Qo - I)(Qo - I)T = tr{(Qo - I)TU(Qo - I)}, and show that U •(I - Qo) > 0 by showing that (Qo - I?U(Qo - I) E Psym. SECTION 3 2. Differentiate AA - 1 = I. 3. Use the chain rule. 6. Differentiate QQT = I. 7. Take Q(t) = eW t• 8c. Show that Dcp(A) [U]

= U·

f

{S(A o

+ IXA) + IX DS(A o + IXA) [A]} da,

and show further that the integrand is equal to d dlX [IXS(A o

+ IXA)].

SECTION 4 2. Show that div(STa) is linear in a and use the representation theorem for linear forms. SECTION 5 l a.

Use the fact that, given any vector a,

(1

o~

v ® n dA)a =

1

(v

o~

® n)a dA =

1

(v

o~

® a)n dA.

250

HINTS FOR SELECTED EXERCISES

SECTION 6 6. 9.

Use (2.7) and (2.10) to show that .fc = (3, 3, 1) if and only if the spectrum of Cis (1, 1, 1). Let and be right polar decompositions of F 1 = Vf 1 and F 2 = Vf2 , and note that U 1(p) = U 2 (p) for all P E .sf. Define g = f 2 fi 1. Choose q E f 1(.sf) arbitrarily and let P = fi 1(q). Show that 0

Vg(q)

=

R 2(p)R 1(p?

and appeal to the characterization theorem for rigid deformations. 10.

To establish (18)1 let v = ~>iei'

r

write

L: ei' r

v . m dA =

J

J

i

ol(go)

Vim

dA,

ol(go)

and use (14.2). For (18)z let a be a vector, write a'

f

Tm dA =

J

ol(go)

r

(TTa)· m dA,

J

ol(go)

and use (18)1' For (18h use the identity a . (r x Tm) = (Tm) . (a x r) = m . TT(a x r)

[r(x) = x - 0]

to write a'

r

J

r x Tm dA

=

ol(go)

r

J

m . TT(a x r) dA

iJf(go)

and appeal to (18)1 again. 12. To verify (5)1 use the continuity of f and f- 1 together with the fact that for a continuous function the inverse image of an open set is open. To prove (5)2 note that, since fJI is closed, f(fJI) = f(~)

u f(J.sf),

and since f(fJI) is closed, f(fJI) = f(fJlt u Jf(.sf).

13b.

Use these identities with (5)1 to verify (5)z. Show that f(fJI) is not closed.

SECTION

251

8

SECTION 7 I.

Consider C and U as functions of H = Vu: C

=

U

t(H),

Use the chain rule and the identity Dt(O) [S]

t

= O(H).

=

0 2 to conclude that

= 2 DO(O) [S]

for all tensors S, and then conclude from (5)1 that O(H) = I + E + o(H).

5. Use Exercise 4 in conjunction with Exercise 4.9b. 6. Use the trigonometric identities cos(E> sin(E>

+ {3) = + {3) =

cos E> cos {3 - sin E> sin {3, sin E> cos {3

+ cos E> sin {3,

together with the estimates

= 1 + 0({3), sin {3 = {3 + 0({3).

cos {3

[It is a simple matter to verify that Vu -+ 0 as a -+ 0 without explicitly computing the gradient. Indeed, one simply notes that u, as a function of a and p, is smooth for (a, p) E ~ X 8; one then obtains the above limit by first evaluating u at a = 0 and then taking its gradient.]

SECTION 8 3. Compute div v using (5) and Exercise 4.9a. 7a. Choose PEg and let c(o), 0::;; c(O) = p. Use the relation

0" ::;;

1, be a smooth curve in g with

d dO" cp(c(O"» = 0 to show that Vcp(p) is normal to c(O).

252

HINTS FOR SELECTED EXERCISES

SECTION 9 Id,

Show that the two terms in the right side of (15) are symmetric and skew, respectively, and use the uniqueness of the expansion L = D + W of L into symmetric and skew parts.

3. Show that IxiY, t) - xiz, t)1 = Iy - z],

so that 4c.

xl,

t) is a rigid deformation.

By (19), (n + I)

(!!1 •

T

T

•

C = F An + 1 F = (C) = (F An F) .

SECTION 10 2.

Use (4.2)4 and (9.11).

SECTION 11 1. Use (9.12), (10.2), and (2h.

2. Take the curl of (9.lOh and use the identity curl(w x v) = (grad w)v - (grad v)w + (div v)w - (div w)v and (10.2). 3.

Fix Yand r, and let g(t)

=

v(xt(y, t), t)w(xiY, t), r).

It suffices to show that

g(t)

=

Fly, t)g('t).

By (9), g satisfies a certain ordinary differential equation. Show that f(t) = Fly, t)g('t) satisfies the same differential equation and f('t) = g('t). Appeal to the uniqueness theorem for ordinary differential equations. 5.

Use (9.10).

SECTION

13

253

SECTION 12 3. Show that p(y, r) = p(x, t) det Fly, t)

and then proceed as in the proof of (7). 4. Let f = g - v and use the divergence theorem to prove that f{v} + f{f}.

:f( {g}

=

SECTION 13 3a. Use (9) and (12.7) to convert (4) to an integral over PA o . 3b. Use (10) to eliminate q(t) from (9) and derive xo(y, t)

= a(t) +

Q(t)[y - a(O)].

By (9.13), v(x o(y, t), t) = xo(y, t) = !i(t)

+ Q(t)[y

- a(O)].

Choose x E PAt arbitrarily, and let y be such that x = xo(y, t), i.e., y - a(O) = Q(t)T[X - a(t)].

Then v(x, t) = !i(t)

+ Q(t)Q(t?[x - a(t)].

Compare the gradients (with respect to x) of the above equation and (9.7).

3c. (If) Define g(t) = Q(t)k(O) and show that g and k satisfy the same differential equation and g(O) = k(O). Appeal to the uniqueness theorem for ordinary differential equations. 3e. To establish (15) transform (14) to an integral over PA o using the same steps as in (3a). Then use (12) (with k = e) and (15) to show that ei(t) • J(t)eit) = ej(O) • J(O)ej(O).

3f.

Choose {ei(O)} to be an orthonormal basis of eigenvectors. Since a.pin(t) =

L Jijwit)ej(t), i.]

where Jij are the components of J(t) relative to {ej(t)}, we may conclude, with the aid of (13), that

a. pin = L [JijWjei + JijWj(co i.j

x ej)].

254

HINTS FOR SELECTED EXERCISES

Show that

4a.

i

v,,-pdV

=0

iJIJ,

and then compute Use (11).

4b.

.Y(

with v = v"- + lX.

SECTION 14 8. Prove that it suffices to find a tensor Q such that G(Q) =

r

Jo

r ® Qs(n) dA

+

i

r ® Qb dV

iJIJ,

@,

is symmetric. Show that G(Q) = G(I)QT, take the left polar decomposition of G(I), and choose QT to make G(I)QT symmetric. Here you will need the following extended version of the polar decomposition theorem 1: given F E Lin there exist unique positive semidefinite, symmetric tensors D, V and a (not necessarily unique) orthogonal tensor R such that F = RD = VR. 9. We must show that n· T(x, t)k = 0, whenever 0 is normal to-and k tangent to-ofJl t at x. Use the symmetry of T and the fact that ofJlt is traction-free.

SECTION 15 la. Use Exercise 5.lb. lb. Use Exercise 5.la. 2. Use components. I

cr., e.g.,

Halmos [I, §83].

SECTION

22

255

SECTION 16 1. Show that q) =

Given any tensor N

E

{D

E

Syml tr D = O}.

Sym, write N

=

-1tI

+ No,

where 1t = - -! tr N, so that tr No = O. Show that N . D = 0 for every DE q) if and only if No = O. SECTION 17

2. Use the divergence theorem to express the term involving b in (15.2) as an integral over iJtJ4 t • 3. Compute grad v. 4. Use Bernoulli's theorem. SECTION 20

2a. Use Exercise 2.5. SECTION 21

2. Use (20.1), (6.18)2' and (1) to show that

f

T*n* dA = Q(t)

iJ&t

f

Tn dA.

iJ&,

Apply the divergence theorem to both sides and then use (6.14)1 to convert the integral over fY'~ to an integral over fY'. SECTION 22

2a. Use Exercises 4.9b and 7.4. 2b. Use coordinates with, say, e3 = n together with the relation div v = 0 to show that (grad V)T n = O.

256

HINTS FOR SELECTED EXERCISES

SECTION 25 For HE Yep with det H # 1 take F, = H" and Gn = H- n (or vice versa). 4e. Choose FE Lin + arbitrarily and take H = (det F)1/3F- 1 to reduce T to a function of the density. Show next that T is isotropic and use this isotropy to reduce the stress to a pressure (cf. the corollary on page 13). 5b. Choose H E Yep. By (16) and (19), Tg(FG- 1) = T(F) = T(FH) = Tg(FHG- 1 ) 4c.

for every F

E

Lin +, so that

fg(F) = Tg(FGHG- 1) for every F E Lin ". Thus H E Yep implies GHG - 1E Ye p(g) and GYep G - 1 C Ye p(g). Similarly, Ye p(g) c GYe p G -I. Use (17) and (20). 5d. Let G = RU be a right polar decomposition of G. Then SC.

Yep(g)

= RUYep U-1R- 1•

Choose Q E Yep. Then Q

= RUQU- 1R- 1

belongs to Ye p(g). Thus, since Q, (QR)U

QE Orth +

(why?),

= (RQ)(QTUQ)

(0() RQRT.

represent polar decompositions ofthe same tensor; hence Q = Thus Q E Yep implies RQR T E Yep(g), and RYepR T c Yep(g). Similarly, Yep(g) c RYepRT. Equation (0() also implies that U = QTUQ. When p is isotropic this result and the corollary on page 13 tell us that U = AI, so that G = AR. 6a. Without loss in generality add the constraint for all F 6c. Define

E

tr T(F) = 0 Unim (cf. the remarks on page 148).

f(F) = T«det F) - 1/3 F)

({3)

for all FE Lin +. Then f(F) = T(F) for all FE Unim, so that f provides an extension of T to Lin +. Show that f(F) must reduce to an isotropic function of B = FFT and use (37.21).

SECTION

29

257

SECTION 27 6. Transform

f

(T +

1t o I)m

dA

.(9',t)

to an integral over!/' ( c !/'2) and use the fact that!/' is arbitrary.

SECTION 28 1a. Integrate u(R)o along a path from R = I to R = Q. lb. Substitute (5) into (27.11) and use the chain rule to verify that DF[u(QF) - u(F)] = O.

Ie,

Integrate this from F = I [using (19)]. Use the chain rule to differentiate u(F) = u(FTF)

with respect to F and appeal to (5) and (27.14). 2. Use the incompressibility of the body to show that (nI)· D

= 0,

so that the term involving the pressure 11: does not enter the expression for the work. Then use the extension (p) to reduce the problem to the one studied in this section. 3. Show that the work done on a part

I

t

to

{!IJ

during [to, t 1] is given by

1

fT. D dV dt iP,

provided the process is closed during this time interval. 4. Use components.

SECTION 29 1. Extend the proof of (b) of the theorem on page 195.

258

HINTS FOR SELECTED EXERCISES

SECTION 30 4.

Use (29.7) with F = I.

SECTION 32 2. Use Exercise 15.1. 9. The following theorem is needed: if w, are class C" (N ~ 1) fields on []It with

then there exists a class C" + 1 field

t/J

on

[]It

such that

The compatibility equation in the form (E ll , 2

-

E 12 , l ) ' 2 = (E 12 , 2

establishes the existence of a field E ll , 2 = (E 12

and hence fields

Ua

t/J

-

E 22 , l ) ' l

such that

+ t/J),l,

such that

Ell

E2 2

= U 1,l' = U2,2'

E 12

+ t/J = U 1,2'

E 12

-

t/J

= U 2,l'

11. Equation (13h, written in the form

implies the existence of fields

and there exists a

qJ such

that

t/J

and () such that

SECTION

34

259

SECTION 33 4. Assume that the boundary of the cross section is parametrized by PI(a) and pz«(1) with (1 the arc lerigth. Then the vector with components

is normal to this boundary at each that

(1.

Use this fact, (7), and (8) to show

5c. Show that

and use the divergence theorem on verify that

g>0

(as a region in

f

S13dA

= O.

f

SZ3dA

= O.

[RZ) and

(11) to

Yo

Similarly,

Yo

SECTION 34 2. Show that

L{

{It(t - T)' C[E(t

+ T)J - E(t

L

{P{Ii(t - T)' ii(t

+ T) -

+ T) . C[E(t -

o(t

+ T) • o(t

T)]} dt dV = 2'¥t{E}(t),

- r)} dt dV = -2.Jt'"{u}(t)

(where we have suppressed the argument p), and that the left sides sum toct>(t). To derive these identities write the integrands on the left sides as derivatives with respect to r.

260

HINTS FOR SELECTED EXERCISES

5. Show that the convolution of the left side of (12) with A(t) = t is equal to

A*

LS * E dV + LpU * ii dV,

where the convolution of two tensor fields is defined using the inner product of tensors in the integrand in (11). Show further, using the commutativity of the convolution operation and the symmetry of C, that

s*E=s*E. In view of the above remarks we have the identity obtained by taking the convolution of (12) with I, and if we differentiate this relation twice with respect to time, we arrive at (12).

SECTION 37

2. Choose U and v with IU I = Iv I, let Q be the orthogonal tensor carrying u into v, and show that qJ(u) = qJ(v). 3. Choose v and let Q be any rotation about v. Show that for this choice of Q, Qq(v) = q(v), and use this fact to prove that q(v) is parallel to v. Thus q(v) = qJ(v)v. Show that qJ is isotropic and use Exercise 1. 4. Use induction.

Index

A

Acceleration, 60 Acoustic equations, 133 Acoustic tensor, 224 Airy stress function, 214 Angle of twist, 58 Angular velocity, 70 Anisotropic material point, 169 Axial vector, 9 B

Balance of energy for hypere1astic material, 191 for linear elastic material, 220 for viscous fluid, 153 Bending of bar, 214-217 Bernoulli's theorem, 111 for elastic fluid, 131 for ideal fluid, 120 Betti's reciprocal theorem, 206 Blasius-Kutta-Joukowski theorem, 124 Body, 41 Body force, 99 Boundary, 35 Boussinesq-Papkovich-Neuber solution, 201

C Cartesian coordinate frame, 2 Cauchy stress, see Stress, Cauchy Cauchy's theorem, 101 Cayley-Hamilton theorem, 16 Center of mass, 92 Chain rule, 26 Characteristic space, 11 Circulation, 82 Closed region, 35 0',21 Commutation of tensors, 3 Commutation theorem, 12 Compatibility equation, 213 Complex potential, 126 Complex velocity, 123 Connected set, 35 Conservation of mass, 88 for control volume, 90 local, 89 Conservative body force, 111 Constitutive assumption, 115-119 Control volume, 89 Cross product, 7 Curl, 32 material, 61 spatial, 61 261

262

INDEX

Curve, 34 c1osed,34 length of, 34

G

D

Da Silva's theorem, 107 Deformation, 42 Deformation gradient, 42 Density, 88 Derivative, 20-21 invariance of, 237 Determinant, 6 derivative of, 23 Differentiable, 21 Displacement, 42 Displacement equation of equilibrium, 201 Displacement equation of motion, 201 Displacement problem of elastostatics, 207 Divergence material, 61 spatial, 61 of tensor field, 30 of vector field, 30 Divergence theorem, 37 Dynamical process, 108 E

Eigenvalue, 11 Eigenvector, 11 Elastic body, 165 incompressible, 174 Elastic fluid, 117, 130-137 Elasticity tensor, 194-197 Elastic process, 220 Elastic state, 205 Equation of motion, 101 Euclidean point space, 1 Eulerian process, 116 Euler's equations for ideal fluids, 121 Euler tensor, 95 Exponential function, 227-229 Extended symmetry group, 172 Extension, 44 Extra stress, 148 F

Finite elasticity, 165-198 Flow, 108 Flow region, 109

Gradient material, 60 of point field, 30 of scalar field, 29 spatial, 61 of vector field, 30 Graffi's reciprocal theorem, 223 H

Harmonic field, 32 Hellinger-Prange-Reissner principle, 212 Homogeneous body, 171 Homogeneous deformation, 42 Hu-Washizu principle, 212 Hyperelastic body, 186 I

Ideal fluid, 117, 119-130 Incompressible body, 116 Inertia tensor, 94 Infinitesimal rigid displacement, 55 characterization of, 56 Infinitesimal strain, 55 Inner product of tensors, 5 of vectors, 1 Integral of vector field around curve, 34 Interior of set, 35 Invariance of function, 229 Invariance of set, 229 Invariance under change in observer, 143145 for elastic body, 166 for Newtonian fluid, 149 Inverse of tensor, 6 Irrotational motion, 81 Isochoric deformation, 51 Isochoric dynamical process, 116 Isochoric motion, 78 Isotropic function definition, 229 representation theorem for linear tensor functions, 235, 236 representation theorem for scalar functions,230 representation theorem for tensor functions, 233, 235 Isotropic material point, 169

263

INDEX

K

Kelvin's theorem, 83 Kinematically admissible state, 208 Kinematical viscosity, 151 Kinetic energy, lIl, 220 Konig's theorem, 95 Kom's inequality, 57 L

Lagrange-Cauchy theorem, 81 Lame moduli, 196 Laplacian, 32 Left Cauchy-Green strain tensor, 46 Left stretch tensor, 46 Lin, 7 Lint, 7 Linear elasticity, 199-226 Linear form, representation theorem, 1 Localization theorem, 38 Longitudinal sound speed, 225 M

Mach number, 133 Mass, 87 Mass distribution, 87 Material body, 116 Material curve, 82 Material description, 60 Material field, 60 Material point, 41 Material time derivative, 60, 61 Mean strain, 57 Minimum complementary energy, principle of,211 Minimum potential energy, principle of, 193,208 Mixed problem of elastodynamics, 221 Mixed problem of elastostatics, 207 Modulus of compression, 202 Momentum angular, 92 balance for control volume, 108 balance of, 99 linear, 92 spin angular, 93 Motion, 58 Motion relative to time T, 75 Multiplicity of eigenvalue, 11

N

Navier-Stokes equations, 151 Newtonian fluid, 148 Normal force, 105

o Observer change, 139 Open region, 35 Origin, 2 Orth,7 Orth", 7 Orthogonal tensor, 7 p

Part of body, 41 Path line, 63 Piola-Kirchhoff stress, see Stress, PiolaKirchhoff Place, 58 Plane motion, 74 poincare inequality ,I 163, 164 Point, 1 Poisson's ratio, 203 Polar decomposition of tensor, 14 Polar decomposition theorem, 14 Position vector, 62 Positive definite tensor, 7 Potential flow, 111 Potential theorem, 35 Power expended, theorem of, 110, 180 Pressure, 106 Principal invariants, 15 list of, 15 Principal stress, see Stress, principal Principal stretch, 45 Progressive wave, 223-226 Psym, 7 Pure torsion, 58 R

Reduced constitutive equation, 166 Reference body force, 179 Reference configuration, 41 Reference density, 88 Regular region, 37 Reiner-Rivlin fluid, 155 Reynolds number, 154 Reynolds' transport theorem, 78

264

INDEX

Right Cauchy-Green strain tensor, 46 Right stretch tensor, 46 Rigid deformation, 48 characterization of, 49 Rigid motion, 69 Rivlin-Ericksen tensors, 76 Rotation, 7 about point, 43 tensor, 46

s Shearing force, 105 Shear modulus, 202 Signorini's theorem, 113 Simple shear of elastic body, 175-178 of Newtonian fluid, 156-159 Simply connected set, 35 Skew tensor, 3 Skw, 7 Small deformation, 54-58 Smooth, 20 Smooth-inverse theorem, 22 Smoothness lemma, 60 Sonic flow, 133 Sound speed, 131 Span, 2 Spatial description, 60 Spatial description of velocity, 60 Spatial field, 60 Spatial time derivative, 61 Spectral decomposition, 12 Spectral theorem, 11 Spectrum, 11 Spin, 71 axis, 71 transport of, 80 Square root of tensor, 13 derivative of, 23 Square-root theorem, 13 Stability theorem for Navier-Stokes equations, 162 Stagnation point, 128 Stationary potential energy, principle of, 193 Steady flow, 109 Steady motion, 67 Stiffness matrix, 210 Stokes flow, 152 Stokes' theorem, 39 Strain energy, 205

Strain-energy density, 186 Stream function, 126 Streamline, 64 Stress Cauchy, 101 Piola-Kirchhoff, 178 principal, 105 shear, 106 tensile, 106 Stress power, III Stretch, 44 Stretching, 71 Subsonic flow, 133 Supersonic flow, 133 Surface force, 99 Surface traction, 98 Sym, 7 Symmetric tensor, 3 Symmetry group, 168 Symmetry transformation, 168 T Tensor, 2 skew part of, 3 symmetric part of, 3 Tensor product, 4 Time, 58 Torsion of circular cylinder, 217-219 Torsional rigidity, 219 Total body force, 100 Trace, 5 Traction problem of elastostatics, 208 Trajectory of motion, 59 Transfer theorem, 231 Transformation law for Cauchy-Green strain tensors, 141 for Cauchy stress, 144 for deformation gradient, 140 for rotation tensor, 140 for spin, 142 for stretch tensors, 140, 141 for stretching, 142 for velocity gradient, 142 Transverse sound speed, 225

u Unimodular group, 172 Uniqueness for linear elastodynamics, 222

265

INDEX

for linear elastostatics, 207 for Navier-Stokes equations, 160

v Vector, 1 Velocity gradient, 63 Virtual work, theorem of, 100 Viscosity, 149 Viscous flow problem, 160 Volume of set, 37

Vortex line, 83 W

Wang's lemma, 232 Wave equation, 133 Work in closed processes, 185 Y

Young's modulus, 203