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Theory of Elasticity 

Second Revised and Enlarged Edition 



Course of Theoretical Physics 

Volume 7 



. D. Landau and E. M. Lifshi 

Institute of Physical Problems 
USSR Academy of Sciences 






Pergamon Press 



Course of Theoretical Physics 
Volume 7 

THEORY OF ELASTICITY 
Second Revised and Enlarged Edition 

L D. LANDAU and E. M. UFSHITZ 

Institute of Physical Problems, USSR Academy 
of Sciences 

This is a comprehensive textbook covering 
not only the ordinary theory of the 
deformation of solids, but also some topics 
not usually found in textbooks on the 
subject, such as thermal conduction and 
viscosity in solids, and various problems 
in the theory of elastic vibrations and 
waves. The authors have discussed only 
briefly certain special matters, such as 
complex mathematical methods in the 
theory of elasticity and the theory of 
shells, which are outside the scope of this 
book. 

As well as many minor corrections and 
additions, a chapter on the macroscopic 
theory of dislocations has been added in 
this edition, while some further errors have 
been eliminated. 




COURSE OF THEORETICAL PHYSICS 
Volume 7 

THEORY OF ELASTICITY 



OTHER TITLES IN THE SERIES 

Vol. 1. MECHANICS 

Vol. 2. THE CLASSICAL THEORY OF FIELDS 

Vol. 3. QUANTUM MECHANICS— NON-RELATIVISTIC THEORY 

Vol. 4. RELATIVISTIC QUANTUM THEORY 

Vol. 5. STATISTICAL PHYSICS 

Vol. 6. FLUID MECHANICS 

Vol. 8. ELECTRODYNAMICS OF CONTINUOUS MEDIA 

Vol. 9. PHYSICAL KINETICS 



THEORY OF ELASTICITY 



by 
L. D. LANDAU and E. M. LIFSHITZ 

INSTITUTE OF PHYSICAL PROBLEMS, U.S.S.R. ACADEMY OF SCIENCES 



Volume 7 of Course of Theoretical Physics 

Translated from the Russian by 
J. B. SYKES and W. H. REID 



SECOND ENGLISH EDITION, REVISED AND ENLARGED 



PERGAMON PRESS 

OXFORD • LONDON • EDINBURGH • NEW YORK 
TORONTO • SYDNEY • PARIS • BRAUNSCHWEIG 



Pergamon Press Ltd., Headington Hill Hall, Oxford 

4 & 5 Fitzroy Square, London W.l 

Pergamon Press (Scotland) Ltd., 2 & 3 Teviot Place, Edinburgh 1 

Pergamon Press Inc., Maxwell House, Fairview Park, Elmsford, New York 10523 

Pergamon of Canada Ltd., 207 Queen's Quay West, Toronto 1 

Pergamon Press (Aust.) Pty. Ltd., 19a Boundary Street, Rushcutters Bay, 
N.S.W. 2011, Australia 

Pergamon Press S.A.R.L., 24 rue des Ecoles, Paris 5° 

Vieweg & Sohn GmbH, Burgplatz 1, Braunschweig 



Copyright © 1959 and 1970 
Pergamon Press Ltd. 



All Rights Reserved. No part of this publication" may be reproduced, 

stored in a retrieval system, or transmitted, in any form or by any 

means, electronic, mechanical, photocopying, recording or otherwise, 

without the prior permission of Pergamon Press Limited. 



First English edition 1959 
Second (revised) edition 1970 



Library of Congress Catalog Card No. 77-91701 



Printed in Great Britain by J. W. Arrowsmith Ltd., Bristol 
08 006465 5 



CONTENTS 

Page 
Prefaces to the English editions vii 

Notation viii 



I. FUNDAMENTAL EQUATIONS 

§1. The strain tensor 1 

§2. The stress tensor 4 

§3. The thermodynamics of deformation 8 

§4. Hooke's law 10 

§5. Homogeneous deformations 13 

§6. Deformations with change of temperature 15 

§7. The equations of equilibrium for isotropic bodies 17 

§8. Equilibrium of an elastic medium bounded by a plane 25 

§9. Solid bodies in contact 30 

§10. The elastic properties of crystals 37 



II. THE EQUILIBRIUM OF RODS AND PLATES 

§11. The energy of a bent plate 44 

§12. The equation of equilibrium for a plate 46 

§13. Longitudinal deformations of plates 53 

§14. Large deflections of plates 58 

§15. Deformations of shells 62 

§16. Torsion of rods 68 

§17. Bending of rods 75 

§18. The energy of a deformed rod 78 

§19. The equations of equilibrium of rods 82 

§20. Small deflections of rods 89 

§21. The stability of elastic systems 97 



III. ELASTIC WAVES 

§22. Elastic waves in an isotropic medium 101 

§23. Elastic waves in crystals 106 

§24. Surface waves 109 

§25. Vibration of rods and plates 113 

§26. Anharmonic vibrations 118 



IV. DISLOCATIONS 

§27. Elastic deformations in the presence of a dislocation 123 

§28. The action of a stress field on a dislocation 131 

§29. A continuous distribution of dislocations 134 

§30. Distribution of interacting dislocations 139 

§31. Equilibrium of a crack in an elastic medium 144 



vi Contents 



V. THERMAL CONDUCTION AND VISCOSITY IN SOLIDS 



Page 



§32. The equation of thermal conduction in solids 150 

§33. Thermal conduction in crystals 152 

§34. Viscosity of solids 153 

§35. The absorption of sound in solids 155 

§36. Highly viscous fluids 161 

Index 163 



PREFACE TO THE FIRST ENGLISH EDITION 

The present volume of our Theoretical Physics deals with the theory of 
elasticity. 

Being written by physicists, and primarily for physicists, it naturally 
includes not only the ordinary theory of the deformation of solids, but also 
some topics not usually found in textbooks on the subject, such as thermal 
conduction and viscosity in solids, and various problems in the theory of 
elastic vibrations and waves. On the other hand, we have discussed only 
very briefly certain special matters, such as complex mathematical methods 
in the theory of elasticity and the theory of shells, which are outside the scope 
of this book. 

Our thanks are due to Dr. Sykes and Dr. Reid for their excellent trans- 
lation of the book. 

Moscow L. D. Landau 

E. M. Lifshitz 



PREFACE TO THE SECOND ENGLISH EDITION 

As well as some minor corrections and additions, a chapter on the macro- 
scopic theory of dislocations has been added in this edition. The chapter has 
been written jointly by myself and A. M. Kosevich. 

A number of useful comments have been made by G. I. Barenblatt, V. L. 
Ginzburg, M. A. Isakovich, I. M. Lifshitz and I. M. Shmushkevich for the 
Russian edition, while the vigilance of Dr. Sykes and Dr. Reid has made it 
possible to eliminate some further errors from the English translation. 

I should like to express here my sincere gratitude to all the above-named. 

Moscow E. M. Lifshitz 



NOTATION 
p density of matter 
u displacement vector 

1 / dut duiA 
uuc = -I 1 1 strain tensor 



2\dxjc dxi 



aoe stress tensor 

K modulus of compression 

/x modulus of rigidity 

E Young's modulus 

a Poisson's ratio 

c% longitudinal velocity of sound 

ct transverse velocity of sound 

Ci and Ct are expressed in terms of K, ju, or of E, a by formulae given in 

§22. 
The quantities K, p, E and a are related by 

E = 9KfMl(3K+fi) 

a = (3K-2fM)l2(3K+tJL) 

K= EI3(l-2o) 

fi = £"/2(l+a) 



CHAPTER I 

FUNDAMENTAL EQUATIONS 

§1. The strain tensor 

The mechanics of solid bodies, regarded as continuous media, forms the 
content of the theory of elasticity.] 

Under the action of applied forces, solid bodies exhibit deformation to 
some extent, i.e. they change in shape and volume. The deformation of a 
body is described mathematically in the following way. The position of any 
point in the body is defined by its radius vector r (with components x\ = x, 
x 2 = y, xz = z) in some co-ordinate system. When the body is deformed, 
every point in it is in general displaced. Let us consider some particular 
point; let its radius vector before the deformation be r, and after the deforma- 
tion have a different value r' (with components x'i). The displacement of 
this point due to the deformation is then given by the vector r' - r, which we 
shall denote by u: 

Ui = x'i — Xi. (1-1) 

The vector u is called the displacement vector. The co-ordinates x'i of the 
displaced point are, of course, functions of the co-ordinates Xi of the point 
before displacement. The displacement vector Ui is therefore also a function 
of the co-ordinates Xi. If the vector u is given as a function of Xi, the defor- 
mation of the body is entirely determined. 

When a body is deformed, the distances between its points change. Let 
us consider two points very close together. If the radius vector joining them 
before the deformation is dxt, the radius vector joining the same two points 
in the deformed body is dx'i = dxt + dm. The distance between the points 
is d/ = -v/(d*i 2 + d#2 2 + dff 3 2 ) before the deformation, and dl' = \/(dx'i 2 + 
+ dx'2 2 +dx's 2 ) after it. Using the general summation rule, J we can write 
d/ 2 = dxi 2 , dl' 2 = dx'i 2 = (dxi + dui) 2 . Substituting dm = (dmldx k )dx k ,we 

can write 

dm . . dm dm , 

dl' 2 = d/ 2 + 2 dxt dx k + dx k dxi. 

dx k dx k dxi 

Since the summation is taken over both suffixes i and k in the second term 
on the right, we can put (dmldx k )dxidx k = {du k jdxi)dxidx k . In the third 

t The basic equations of elasticity theory were established in the 1 820's by Cauchy and by Poisson. 

| In accordance with the usual rule, we omit the sign of summation over vector and tensor suffixes. 
Summation over the values 1 , 2, 3 is understood with respect to all suffixes which appear twice in a 
given term. 

I* 1 



2 Fundamental Equations §1 

term, we interchange the suffixes i and /. Then dZ' 2 takes the final form 

d/'2 = dP + 2u ik dxt dx k , (1.2) 

where the tensor ut k is defined as 

1 / dui dujc dui dui \ 
2\dx k dxt dxt dxjc) 
These expressions give the change in an element of length when the body is 
deformed. 

The tensor u\ k is called the strain tensor. We see from its definition that 
it is symmetrical, i.e. 

Uik = u ki . (1.4) 

This result has been obtained by writing the term 2(dui/dx k )dxi dx k in d/' 2 
in the explicitly symmetrical form 

/ But du k \ 
\ dx k oxi I 

Like any symmetrical tensor, Uf k can be diagonalised at any given point. 
This means that, at any given point, we can choose co-ordinate axes (the 
principal axes of the tensor) in such a way that only the diagonal components 
«n, "22, «33 of the tensor u\ k are different from zero. These components, the 
principal values of the strain tensor, will be denoted by m (1) , m (2) , m (3) . It should 
be remembered, of course, that, if the tensor ui k is diagonalised at any point 
in the body, it will not in general be diagonal at any other point. 

If the strain tensor is diagonalised at a given point, the element of length 
(1.2) near it becomes 

dZ' 2 = (Sac + 2uuc) dxt dx k 
= (1 + 2 M <i>) d*i 2 + (1 + 2w< 2 >) d*2 2 + (1 + 2m«») dx 3 2 . 
We see that the expression is the sum of three independent terms. This 
means that the strain in any volume element may be regarded as composed 
of independent strains in three mutually perpendicular directions, namely 
those of the principal axes of the strain tensor. Each of these strains is a 
simple extension (or compression) in the corresponding direction : the length 
d#i along the first principal axis becomes dx\ = -\/(l+2u a) ) dx\, and simi- 
larly for the other two axes. The quantity \/{\ + 2u (i) ) — 1 is consequently 
equal to the relative extension (dx'i — dxi)[dxt along the ith principal axis. 

In almost all cases occurring in practice, the strains are small. This means 
that the change in any distance in the body is small compared with the 
distance itself. In other words, the relative extensions are small compared 
with unity. In what follows we shall suppose that all strains are small. 

If a body is subjected to a small deformation, all the components of the 
strain tensor are small, since they give, as we have seen, the relative changes 
in lengths in the body. The displacement vector Ui, however, may 
sometimes be large, even for small strains. For example, let us consider a 
long thin rod. Even for a large deflection, in which the ends of the rod move 



§1 The strain tensor 3 

a considerable distance, the extensions and compressions in the rod itself 
will be small. 

Except in such special cases, f the displacement vector for a small defor- 
mation is itself small. For it is evident that a three-dimensional body (i.e. 
one whose dimension in no direction is small) cannot be deformed in such a 
way that parts of it move a considerable distance without the occurrence of 
considerable extensions and compressions in the body. 

Thin rods will be discussed in Chapter II. In other cases m is small for 
small deformations, and we can therefore neglect the last term in the general 
expression (1.3), as being of the second order of smallness. Thus, for small 
deformations, the strain tensor is given by 

_ 1 tM duk\ 
2\dxic dxil 
The relative extensions of the elements of length along the principal axes of 
the strain tensor (at a given point) are, to within higher-order quantities, 
VXl +2u (i) )— 1 & u (i) , i.e. they are the principal values of the tensor tin. 

Let us consider an infinitesimal volume element dV, and find its volume 
dV after the deformation. To do so, we take the principal axes of the strain 
tensor, at the point considered, as the co-ordinate axes. Then the elements of 
length d^i, d*2, dx$ along these axes become, after the deformation, d#'i 
= (l+w (1) ) d#i, etc. The volume dV is the product d#i da^ d#3, while dV' 
is dx\ dx' 2 dx' z . Thus dV = dV{\ + u<-»)(l + w (2) )(l + m (3) ). Neglecting higher- 
order terms, we therefore have dV = dF(l+« (1) + M (2) + M (3) ). The sum 
M (D -|- w (2) + M (3) of the principal values of a tensor is well known to be invariant, 
and is equal to the sum of the diagonal components uu =Mn + M22+«33 in 
any co-ordinate system. Thus 

dV = dV(l + u u ). (1.6) 

We see that the sum of the diagonal components of the strain tensor is the 
relative volume change {dV — dV)jdV. 

It is often convenient to use the components of the strain tensor in spherical 
or cylindrical co-ordinates. We give here, for reference, the corresponding 
formulae, which express the components in terms of the derivatives of the 
components of the displacement vector in the same co-ordinates. In spherical 
co-ordinates r, 6, (f>, we have 

du r 1 du e u r 1 du* u e u r 

u r r = —, u ee = -—-+—-, u H = —^ -— - + — cot0 + — , 
dr r do r r sin o of r r 



Y r\dd Y J r si] 



dUg BUg Ug 1 Bu r > (1.7) 

2u r e = — +-■ 



sin0 9^' dr r r dd 

1 du r dUs, Ux 

2ufr = H -• 

r sin 9 d<f> dr r 



t Which include, besides deformations of thin rods, those of thin plates to form cylindrical surfaces. 



4 Fundamental Equations §2 

In cylindrical co-ordinates r, (f>, z, 

du r 1 duf u r du z 

u rr = -r— , «^ = - —7- H , u zz = — — , 

or yr r d<f> r dz 

1 du z du A du r du z 

2«„ = -—+—*, 2u rz = -^ + -^- f } (1.8) 



r 30 d# 3s- 3r 

aw, W, 1 du r 

or r r dtp 

1 

§2. The stress tensor 

In a body that is not deformed, the arrangement of the molecules corre- 
sponds to a state of thermal equilibrium. All parts of the body are in mechani- 
cal equilibrium. This means that, if some portion of the body is considered, 
the resultant of the forces on that portion is zero. 

When a deformation occurs, the arrangement of the molecules is changed, 
and the body ceases to be in its original state of equilibrium. Forces there- 
fore arise which tend to return the body to equilibrium. These internal 
forces which occur when a body is deformed are called internal stresses. If 
no deformation occurs, there are no internal stresses. 

The internal stresses are due to molecular forces, i.e. the forces of inter- 
action between the molecules. An important fact in the theory of elasticity is 
that the molecular forces have a very short range of action. Their effect 
extends only to the neighbourhood of the molecule exerting them, over a 
distance of the same order as that between the molecules, whereas in the 
theory of elasticity, which is a macroscopic theory, the only distances con- 
sidered are those large compared with the distances between the molecules. 
The range of action of the molecular forces should therefore be taken as zero 
in the theory of elasticity. We can say that the forces which cause the internal 
stresses are, as regards the theory of elasticity, "near-action" forces, which act 
from any point only to neighbouring points. Hence it follows that the forces 
exerted on any part of the body by surrounding parts act only on the surface 
of that part. 

The following reservation should be made here. The above asserioon is 
not valid in cases where the deformation of the body results in macroscopic 
electric fields in it (pyroelectric and piezoelectric bodies). We shall not discuss 
such bodies in this book, however. 

Let us consider the total force on some portion of the body. Firstly, this 
total force is equal to the sum of all the forces on all the volume elements in 
that portion of the body, i.e. it can be written as the volume integral jTdV, 
where F is the force per unit volume and FdFthe force on the volume element 
dV. Secondly, the forces with which various parts of the portion considered 
act on one another cannot give anything but zero in the total resultant force, 
since they cancel by Newton's third law. The required total force can there- 
fore be regarded as the sum of the forces exerted on the given portion of the 



§2 The stress tensor 5 

body by the portions surrounding it. From above, however, these forces act 
on the surface of that portion, and so the resultant force can be represented 
as the sum of forces acting on all the surface elements, i.e. as an integral 
over the surface. 

Thus, for any portion of the body, each of the three components j FfdV 
of the resultant of all the internal stresses can be transformed into an integral 
over the surface. As we know from vector analysis, the integral of a scalar 
over an arbitrary volume can be transformed into an integral over the surface 
if the scalar is the divergence of a vector. In the present case we have the 
integral of a vector, and not of a scalar. Hence the vector Ft must be the 
divergence of a tensor of rank two, i.e. be of the form 

Ft = doac/dxjc. (2.1) 

Then the force on any volume can be written as an integral over the closed 
surface bounding that volume :f 

IV, dV = f ^ dV = L ik d/fc, (2.2) 

where d/< are the components of the surface element vector df, directed (as 
usual) along the outward normal. J 

The tensor aw is called the stress tensor. As we see from (2.2), aacdfjc is the 
tth component of the force on the surface element df. By taking elements 
of area in the planes of xy, yz, zx, we find that the component am of the stress 
tensor is the ith. component of the force on unit area perpendicular to the 
x*-axis. For instance, the force on unit area perpendicular to the ar-axis, 
normal to the area (i.e. along the #-axis) is <t xx , and the tangential forces 
(along the y and z axes) are a yx and a zx . 

The following remark should be made concerning the sign of the force 
oikdfk. The surface integral in (2.2) is the force exerted on the volume 
enclosed by the surface by the surrounding parts of the body. The force 
which this volume exerts on the surface surrounding it is the same with the 
opposite sign. Hence, for example, the force exerted by the internal stresses 
on the surface of the body itself is —fontdfk, where the integral is taken over 
the surface of the body and df is along the outward normal. 

Let us determine the moment of the forces on a portion of the body. The 
moment of the force F can be written as an antisymmetrical tensor of rank 
two, whose components are FiXk — FjcXi, where xi are the co-ordinates of the 



t The integral over a closed surface is transformed into one over the volume enclosed by the 
surface by replacing the surface element d/ 4 by the operator dVd/dxi. 

J Strictly speaking, to determine the total force on a deformed portion of the body we should 
integrate, not over the old co-ordinates x t , but over the co-ordinates x' t of the points of the deformed 
body. The derivatives (2.1) should therefore be taken with respect to x' t . However, in view of the 
smallness of the deformation, the derivatives with respect to x ( and x\ differ only by higher-order 
quantities, and so the derivatives can be taken with respect to the co-ordinates x t . 



6 Fundamental Equations §2 

point where the force is applied.f Hence the moment of the forces on the 
volume element dV is (FiX k — F k Xi)dV, and the moment of the forces on the 
whole volume is Mi k = J(FiX k — F k Xi)dV. Like the total force on any volume, 
this moment can be expressed as an integral over the surface bounding the 
volume. Substituting the expression (2.1) for Fi, we find 

M ik = (— — x k — xA dV 

J \ dxi oxi ] 

rd(auXjc-o k iXi) C( dx k dx t \ 

= ; dv ~ au i> ° kl v~ dv - 

J oxi J \ dxi dxi ! 

In the second term we use the fact that the derivative of a co-ordinate with 
respect to itself is unity, and with respect to another co-ordinate is zero 
(since the three co-ordinates are independent variables). Thus dx k jdxi = 8 k i, 
where Sfcz is the unit tensor; the multiplication gives CT^Sfc/ = at k , a k ion = <?ki- 
In the first term, the integrand is the divergence of a tensor; the 
integral can be transformed into one over the surface. The result is 
Mac = §(cTux k - ajciXi)dfi+ S(crjci - o ik )dV. If M ik is to be an integral over the 
surface only, the second term must vanish identically, i.e. we must have 

&ik = <*ki- (2.3) 

Thus we reach the important result that the stress tensor is symmetrical. 
The moment of the forces on a portion of the body can then be written 
simply as 

M ik = j(FiX k - F k x t ) d V = j((T U x k - a k iXi) dfi . (2.4) 

It is easy to find the stress tensor for a body undergoing uniform com- 
pression from all sides {hydrostatic compression). In this case a pressure of 
the same magnitude acts on every unit area on the surface of the body, and its 
direction is along the inward normal. If this pressure is denoted by />, a force 
—pdfi acts on the surface element dfi. This force, in terms of the stress 
tensor, must be cxi k df k . Writing — pdfi = —p8i k df k , we see that the stress 
tensor in hydrostatic compression is 

f^ik = -poik- (2.5) 

Its non-zero components are simply equal to the pressure. 

In the general case of an arbitrary deformation, the non-diagonal com- 
ponents of the stress tensor are also non-zero. This means that not only a 
normal force but also tangential (shearing) stresses act on each surface 
element. These latter stresses tend to move the surface elements relative to 
each other. 



f The moment of the force F is denned as the vector product FXr, and we know from vector 
analysis that the components of a vector product form an antisymmetrical tensor of rank two as written 
here. 



§2 The stress tensor 1 

In equilibrium the internal stresses in every volume element must balance, 
i.e. we must have Ft = 0. Thus the equations of equilibrium for a deformed 
body are 

dcTiicjdxjc = 0. (2.6) 

If the body is in a gravitational field, the sum F + />g of the internal stresses 
and the force of gravity (pg per unit volume) must vanish ; p is the density f and 
g the gravitational acceleration vector, directed vertically downwards. In 
this case the equations of equilibrium are 

daikldx k +pgi = 0. (2.7) 

The external forces applied to the surface of the body (which are the usual 
cause of deformation) appear in the boundary conditions on the equations of 
equilibrium. Let P be the external force on unit area of the surface of the 
body, so that a force P d/ acts on a surface element d/. In equilibrium, this 
must be balanced by the force — aw d/& of the internal stresses acting on that 
element. Thus we must have Pi df— a i1c df k = 0. Writing d/fc = tik df, 
where n is a unit vector along the outward normal to the surface, we find 

(Wit = Pi. (2.8) 

This is the condition which must be satisfied at every point on the surface of 
a body in equilibrium. 

We shall derive also a formula giving the mean value of the stress tensor 
in a deformed body. To do so, we multiply equation (2.6) by xjc and integrate 
over the whole volume: 

——xjc dV = -\ - dV- an—- dV = 0. 

J oxi J oxi J oxi 

The first integral on the right is transformed into a surface integral; in the 
second integral we put dxjc/dxi = 8m. The result is §auxjc d/i — jaw dV = 0. 
Substituting (2.8) in the first integral, we find §PiXjc df = JV^ dV = Vdnc, 
where V is the volume of the body and d^ the mean value of the stress tensor. 
Since a^ = ajd, this formula can be written in the symmetrical form 

a m = (1/2F) j (PiXk + P k xi) df. (2. 9) 

Thus the mean value of the stress tensor can be found immediately from the 
external forces acting on the body, without solving the equations of equili- 
brium. 



t Strictly speaking, the density of a body changes when it is deformed. An allowance for this 
change, however, involves higher-order quantities in the case of small deformations, and is therefore 
unimportant. 



8 Fundamental Equations §3 

§3. The thermodynamics of deformation 

Let us consider some deformed body, and suppose that the deformation 
is changed in such a way that the displacement vector m changes by a small 
amount dm; and let us determine the work done by the internal stresses in 
this change. Multiplying the force Fi = dcrikldxjc by the displacement Sw$ and 
integrating over the volume of the body, we have J8R dV = ftdaikjdxkjSui dV, 
where 8R denotes the work done by the internal stresses per unit volume. 
We integrate by parts, obtaining 

8RdV = koikhuidfjc— <*ik— — dV. 

By considering an infinite medium which is not deformed at infinity, we 
make the surface of integration in the first integral tend to infinity; then 
aijc = on the surface, and the integral is zero. The second integral can, 
by virtue of the symmetry of the tensor cm, be written 



C 1 r I dhui dhujA 

J Z.J \ UXfc UA% J 

- _if hi— dUk \ dV 

2 J \dxjc dxi/ 
= — vik&Uik dV. 



Thus we find 



SR = —oikhuoc. (3.1) 

This formula gives the work SR in terms of the change in the strain tensor. 

If the deformation of the body is fairly small, it returns to its original 
undeformed state when the external forces causing the deformation cease 
to act. Such deformations are said to be elastic. For large deformations, the 
removal of the external forces does not result in the total disappearance of the 
deformation ; a residual deformation remains, so that the state of the body is 
not that which existed before the forces were applied. Such deformations 
are said to be plastic. In what follows we shall consider only elastic defor- 
mations. 

We shall also suppose that the process of deformation occurs so slowly 
that thermodynamic equilibrium is established in the body at every instant, 
in accordance with the external conditions. This assumption is almost always 
justified in practice. The process will then be thermodynamically reversible. 

In what follows we shall take all such thermodynamic quantities as the 
entropy S, the internal energy $, etc., relative to unit volume of the body,f 

f The following remark should be made here. Strictly speaking, the unit volumes before and after 
the deformation should be distinguished, since they in general contain different amounts of matter. 
We shall always relate the thermodynamic quantities to unit volume of the undeformed body, i.e. 
to the amount of matter therein, which may occupy a different volume after the deformation. Accord- 
ingly, the total energy of the body, for example, is obtained by integrating £ over the volume of the 
undeformed body. 



§3 The thermodynamics of deformation 9 

and not relative to unit mass as in fluid mechanics, and denote them by the 
corresponding capital letters. 

An infinitesimal change d& in the internal energy is equal to the difference 
between the heat acquired by the unit volume considered and the work dR 
done by the internal stresses. The amount of heat is, for a reversible process, 
TdS, where T is the temperature. Thus d£ = TdS-dR; with dR given 
by (3.1), we obtain 

d£ = TdS+a ik du ik . (3.2) 

This is the fundamental thermodynamic relation for deformed bodies. 

In hydrostatic compression, the stress tensor is a ik = —phk (2.5). Then 
one duac = -phk dune = -p duu. We have seen, however (cf. (1.6)), that the 
sum uu is the relative volume change due to the deformation. If we consider 
unit volume, therefore, uu is simply the change in that volume, and dun is 
the volume element dV. The thermodynamic relation then takes its usual form 

te = TdS-pdV. 

Introducing the free energy of the body, F = S— TS, we find the form 

dF = -SdT+a ik du ik (3.3) 

of the relation (3.2). Finally, the thermodynamic potential $ is defined as 

<J> = £-TS-o ik u ik = F-a ik u ik . (3.4) 

This is a generalisation of the usual expression O = <f — TS+pV.-f Substi- 
tuting (3.4) in (3.3), we find 

dd> = -SdT-u ik d(j ik . (3.5) 

The independent variables in (3.2) and (3.3) are respectively S, u ik and 
T, Ui k . The components of the stress tensor can be obtained by differentiating 
S or F with respect to the components of the strain tensor, for constant 
entropy S or temperature T respectively: 

a ik = (d<?ldu ik )s = {dFjdu ik ) T - (3.6) 

Similarly, by differentiating <1> with respect to the components <j ik , we can 
obtain the components u\ k : 

u ik = -(d^lda ik ) T . (3.7) 



t For hydrostatic compression, the expression (3.4) becomes <b= F+ pu it = F + p(V —V ), 
where V —V is the volume change resulting from the deformation. Hence we see that the definition 
of used here differs by a term — pV from the usual definition O = F+ pV. 



10 Fundamental Equations §4 

§4. Hooke's law 

In order to be able to apply the general formulae of thermodynamics to 
any particular case, we must know the free energy F of the body as a function 
of the strain tensor. This expression is easily obtained by using the fact that 
the deformation is small and expanding the free energy in powers of w^Jfc- We 
shall at present consider only isotropic bodies. The corresponding results 
for crystals will be obtained in §10. 

In considering a deformed body at some temperature (constant throughout 
the body), we shall take the undeformed state to be the state of the body in the 
absence of external forces and at the same temperature ; this last condition is 
necessary on account of the thermal expansion (see §6). Then, for ua = 0, 
the internal stresses are zero also, i.e. a^ = 0. Since ow = dFjdutk, it 
follows that there is no linear term in the expansion of F in powers of uac. 

Next, since the free energy is a scalar, each term in the expansion of F 
must be a scalar also. Two independent scalars of the second degree can be 
formed from the components of the symmetrical tensor m^: they can be 
taken as the squared sum of the diagonal components (uu 2 ) and the sum of 
the squares of all the components (uac 2 ). Expanding F in powers of «^» we 
therefore have as far as terms of the second order 

F = Fo+%\u u 2 + iJLU ik 2 . (4.1) 

This is the general expression for the free energy of a deformed isotropic 
body. The quantities A and [m are called Lame coefficients. 

We have seen in §1 that the change in volume in the deformation is given 
by the sum uu. If this sum is zero, then the volume of the body is unchanged 
by the deformation, only its shape being altered. Such a deformation is 
called a pure shear. 

The opposite case is that of a deformation which causes a change in the 
volume of the body but no change in its shape. Each volume element of the 
body retains its shape also. We have seen in §1 that the tensor of such a 
deformation is uac = constant x S^. Such a deformation is called a hydro- 
static compression. 

Any deformation can be represented as the sum of a pure shear and a 
hydrostatic compression. To do so, we need only use the identity 

mk = (uuc-$8ikUii) + %8 ik uu. (4.2) 

The first term on the right is evidently a pure shear, since the sum of its 
diagonal terms is zero (8u = 3). The second term is a hydrostatic compres- 
sion. 

As a general expression for the free energy of a deformed isotropic body, 
it is convenient to replace (4.1) by another formula, using this decomposition 
of an arbitrary deformation into a pure shear and a hydrostatic compression. 
We take as the two independent scalars of the second degree the sums of the 



§4 Hooke's law 1 1 

squared components of the two terms in (4.2). Then F becomesf 

F = 11(11* - \hmif + Wm*. (4.3) 

The quantities K and ju. are called respectively the bulk modulus or modulus of 
hydrostatic compression (or simply the modulus of compression) and the shear 
modulus or modulus of rigidity. K is related to the Lame coefficients by 

K = A + foi. (4.4) 

In a state of thermodynamic equilibrium, the free energy is a minimum. 
If no external forces act on the body, then F as a function of uw must have a 
minimum for uac - 0. This means that the quadratic form (4.3) must be 
positive. If the tensor u ilc is such that uu = 0, only the first term remains 
in (4.3) ; if, on the other hand, the tensor is of the form u* = constant x S^, 
then only the second term remains. Hence it follows that a necessary (and 
evidently sufficient) condition for the form (4.3) to be positive is that each 
of the coefficients K and ju, is positive. Thus we conclude that the moduli of 
compression and rigidity are always positive: 

K > 0, ft > 0. (4.5) 

We now use the general thermodynamic relation (3.6) to determine the 
stress tensor. To calculate the derivatives dFjduijc, we write the total differ- 
ential dF (for constant temperature) : 

dF = Kuu dun + 2p{uui - \uiihoc) d(uijc - Iuu^m). 

In the second term, multiplication of the first parenthesis by oik gives zero, 
leaving dF = Ku u duii + 2[4iHk-%uu8ik) &Uik, or writing dun = 8 ik duac, 

dF = [Kuuhuc + l^uuc-^uiihijc)] du ik . 

Hence the stress tensor is 

G ilc = Kuuhi]c + 2^{uiic-\8iicUii). (4.6) 

This expression determines the stress tensor in terms of the strain tensor for 
an isotropic body. It shows, in particular, that, if the deformation is a pure 
shear or a pure hydrostatic compression, the relation between a* and uac is 
determined only by the modulus of rigidity or of hydrostatic compression 
respectively. 

It is not difficult to obtain the converse formula which expresses uac in 
terms of 0%. To do so, we find the sum an of the diagonal terms. Since this 
sum is zero for the second term of (4.6), we have an = 3Kuu, or 

uu = oiifiK. (4.7) 



f The constant term F is the free energy of the undeformed body, and is of no further interest. 
We shall therefore omit it, for brevity, taking F to be only the free energy of the deformation (the 
elastic free energy, as it is called). 



12 Fundamental Equations §4 

Substituting this expression in (4.6) and so determining u ik , we find 

utjc = S ik aiij9K+(a ik -l8 ik aii)/2fi y (4.8) 

which gives the strain tensor in terms of the stress tensor. 

Equation (4.7) shows that the relative change in volume (uu) in any 
deformation of an isotropic body depends only on the sum a u of the diagonal 
components of the stress tensor, and the relation between uu and <r u is 
determined only by the modulus of hydrostatic compression. In hydrostatic 
compression of a body, the stress tensor is a ik = -p8 ik . Hence we have 
in this case, from (4.7), 

uu = -p/K. (4.9) 

Since the deformations are small, uu and p are small quantities, and we can 
write the ratio uu/p of the relative volume change to the pressure in the 
differential form (l/V)(dVfdp) T . Thus 

JL_ 1 ( dV \ 
~K~ ~ V\dp) t 

The quantity \\K is called the coefficient of hydrostatic compression (or simply 
the coefficient of compression). 

We see from (4.8) that the strain tensor Ui k is a linear function of the stress 
tensor Oi k . That is, the deformation is proportional to the applied forces. 
This law, valid for small deformations, is called Hooke's law.f 

We may give also a useful form of the expression for the free energy of a 
deformed body, which is obtained immediately from the fact that F is quad- 
ratic in the strain tensor. According to Euler's theorem, Ui k dFjdui k — 2F, 
whence, since dF\dui k = oi ki we have 

F = \o ik u i1c . (4.10) 

If we substitute in this formula the ua as linear combinations of the 
components oi ky the elastic energy will be represented as a quadratic function 
of the ai k . Again applying Euler's theorem, we obtain cr ik dF/da ik = 2F, and 
a comparison with (4.10) shows that 

u ik = dFjda ik . (4.11) 

It should be emphasised, however, that, whereas the formula o% = dF/duac 
is a general relation of thermodynamics, the inverse formula (4.11) is applic- 
able only if Hooke's law is valid. 



f Hooke's law is actually applicable to almost all elastic deformation. The reason is that deforma- 
tions usually cease to be elastic when they are still so small that Hooke's law is a good approximation. 
Substances such as rubber form an exception. 



§5 Homogeneous deformations 13 

§5. Homogeneous deformations 

Let us consider some simple cases of what are called homogeneous deforma- 
tions, i.e. those in which the strain tensor is constant throughout the volume 
of the body. For example, the hydrostatic compression already considered 
is a homogeneous deformation. 

We first consider a simple extension (or compression) of a rod. Let the 
rod be along the #-axis, and let forces be applied to its ends which stretch it 
in both directions. These forces act uniformly over the end surfaces of the 
rod ; let the force on unit area be p. 

Since the deformation is homogeneous, i.e. uuc is constant through the 
body, the stress tensor o% is also constant, and so it can be determined at once 
from the boundary conditions (2.8). There is no external force on the sides 
of the rod, and therefore o-^« & = 0. Since the unit vector n on the side of the 
rod is perpendicular to the #-axis, i.e. n z = 0, it follows that all the com- 
ponents aw except o zz are zero. On the end surface we have a z tni = p, or 
ozz = P- 

From the general expression (4.8) which relates the components of the 
strain and stress tensors, we see that all the components uuc with i ^ k are 
zero. For the remaining components we find 

1/1 1 \ 1/ 1 1\ 

u xx = u yy - --(_ - _jp, u zz = 3(3^ + -)* (5-1) 

The component u zz gives the relative lengthening of the rod. The coeffi- 
cient of p is called the coefficient of extension, and its reciprocal is the modulus 
of extension or Young's modulus, E: 

u zz = pJE, (5.2) 

where 

E = 9Kfjil(3K+fji). (5.3) 

The components u X x and u yy give the relative compression of the rod in the 
transverse direction. The ratio of the transverse compression to the longi- 
tudinal extension is called Poisson's ratio, cr:j- 

u xx = -<yu zz , (5.4) 

where 

o = %(3K-2 H ,)/(3K+ H ,). (5.5) 



f The use of a to denote Poisson's ratio and a^ to denote the components of the stress tensor can- 
not lead to ambiguity, since the latter always have suffixes. 



14 



Fundamental Equations 



§5 



Since K and p are always positive, Poisson's ratio can vary between - 1 
(for K = 0) and J (for p = 0). Thus f 

- 1 < <t < i. (5.6) 

Finally, the relative increase in the volume of the rod is 

m = pllK. (5.7) 

The free energy of a stretched rod can be obtained immediately from formula 
(4.10). Since only the component a zz is not zero, we have F = \a zz u zz , 
whence 

F = p2/2E. (5.8) 

In what follows we shall, as is customary, use E and a instead of K and p. 
Inverting formulae (5.3) and (5.5), we havej 

p = E/2(l + cx), K = 5/3(1 -2a). (5.9) 

We shall write out here the general formulae of §4, with the coefficients 
expressed in terms of E and a. The free energy is 

y -^ + i^)' (5 - 10) 

The stress tensor is given in terms of the strain tensor by 

E t a 



Vik 



Conversely, 



1 + a 



Uik + 



l-2a 



mS 



ik\ 



Uik = [( 1 + a ) aw - oaiiSuc] IE. 



(5.11) 



(5.12) 



Since formulae (5.11) and (5.12) are in frequent use, we shall give them also 
in component form : 

E \ 

&xx = tt - — rz — ^r-z[(l- a ) u xx + o(uyy + u zz )], 



f yy 



&ZZ 



o X y 



(l+a)(l-2a) u 

E 

(l+a)(l-2a) 

E 

(l+a)(l-2a) 
E 

Uxy, &xz = 



[(1 - a)u yy + a(u xx + U zz )], 

[(1 - a)U ZZ + 0(U XX + Uyy)], 



E 



E 



1+CT 



1 + a 



u xtt a yz 



1+a 



Uy Z , 



(5.13) 



t In practice, PoiSSON's ratio varies only between and i. There are no substances known for 
which a < 0, i.e. which would expand transversely when stretched longitudinally. It may be men- 
tioned that the inequality a > corresponds to A > 0, where A is the Lame coefficient appearing 
in (4.1); in other words, both terms in (4.1), as well as in (4.3), are always positive in practice, although 
this is not thermodynamically necessary. Values of a close to i (e.g. for rubber) correspond to a 
modulus of rigidity which is small compared with the modulus of compression. 

t The second Lame coefficient is A = Eaj{\ — 2a)(14-a). 



6 Deformations with change of temperature 

and conversely 

1 
U X z = — \?xx— o - (o'j/2/ + o'zz)]. 



u vv — ~^;\. a yy~ < T ( cr zx+ &zz)]> 
hi 



15 



U zz = —[cr Z z — cr(oxx+Vyy)]> 

hi 



( (5-14) 



U X y 



1 + a 



E 



-a X y, U X z — 



1 + a 



E 



-<7xZt Uy Z — 



1 + a 

~E 



~ a yz' 



Let us now consider the compression of a rod whose sides are fixed in 
such a way that they cannot move. The external forces which cause the 
compression of the rod are applied to its ends and act along its length, which 
we again take to be along the #-axis. Such a deformation is called a 
unilateral compression. Since the rod is deformed only in the ^-direction, 
only the component u zz of m^ is not zero. Then we have from (5.11) 



E 



o X x — a yy — 



(l + a)(l-2cr) 



U ZZ> &ZZ — 



27(1 -Or) 



(l + o)(l-2o) 



u zz . 



Again denoting the compressing force by p (a zz = p, which is negative for 
a compression), we have 



Uzz = />(1 + a)(l - 2a)lE(l - a). 



(5.15) 



The coefficient of p is called the coefficient of unilateral compression. For the 
transverse stresses we have 



°xx = Oyy = />a/(l — o). 
Finally, the free energy of the rod is 

F = />2(1 + CT )(l _ 2a)/2E(l - a). 



(5.16) 
(5.17) 



§6. Deformations with change of temperature 

Let us now consider deformations which are accompanied by a change in 
the temperature of the body; this can occur either as a result of the deforma- 
tion process itself, or from external causes. 

We shall regard as the undeformed state the state of the body in the absence 
of external forces at some given temperature To. If the body is at a tempera- 
ture T different from To, then, even if there are no external forces, it will in 
general be deformed, on account of thermal expansion. In the expansion of 
the free energy F(T), there will therefore be terms linear, as well as quadratic, 
in the strain tensor. From the components of the tensor uac, of rank two, 



16 Fundamental Equations §6 

we can form only one linear scalar quantity, the sum uu of its diagonal com- 
ponents. We shall also assume that the temperature change T—Tq which 
accompanies the deformation is small. We can then suppose that the coeffi- 
cient of uu in the expansion of F (which must vanish for T = To) is simply 
proportional to the difference T— To. Thus we find the free energy to be 
(instead of (4.3)) 

F(T) = Fo{T)-K*{T- Toyn+niuijc-^ncUnY + Wm 2 , (6.1) 

where the coefficient of T—Tq has been written as — Ktx.. The quantities 
/Lt, K and a can here be supposed constant; an allowance for their tempera- 
ture dependence would lead to terms of higher order. 

Differentiating F with respect to um, we obtain the stress tensor: 

a ik = -KatT-To^iK + KunSiit + lrfuik-iSiKUH). (6.2) 

The first term gives the additional stresses caused by the change in tempera- 
ture. In free thermal expansion of the body (external forces being absent), 
there can be no internal stresses. Equating a^ to zero, we find that ui k is of 
the form constant x §*&, and 

uu = <x(T— T ). (6.3) 

But uu is the relative change in volume caused by the deformation. Thus a 
is just the thermal expansion coefficient of the body. 

Among the various (thermodynamic) types of deformation, isothermal and 
adiabatic deformations are of importance. In isothermal deformations, the 
temperature of the body does not change. Accordingly, we must put T = To 
in (6.1), returning to the usual formulae; the coefficients K and //, may there- 
fore be called isothermal moduli. 

A deformation is adiabatic if there is no exchange of heat between the 
various parts of the body (or, of course, between the body and the surround- 
ing medium). The entropy S remains constant. It is the derivative — dFjdT 
of the free energy with respect to temperature. Differentiating the expression 
(6.1), we have as far as terms of the first order in utk 

S(T) = S {T) + Kauu. (6.4) 

Putting S constant, we can determine the change of temperature T—T due 
to the deformation, which is therefore proportional to uu. Substituting this 
expression for T- T in (6.2), we obtain for a ik an expression of the usual 
kind, 

oik = Ka,aUiiSik + 2[j,(uik-^SikUii) t (6.5) 

with the same modulus of rigidity \x. but a different modulus of compression 
Kad. The relation between the adiabatic modulus K &A and the ordinary 
isothermal modulus K can also be found directly from the thermodynamic 
formula 

/ dV\ _ (W\ T(dVldT)j? 



\ dp i s \ dp / t C 



p 



§7 The equations of equilibrium for isotropic bodies 17 

where C p is the specific heat per unit volume at constant pressure. If V is 
taken to be the volume occupied by matter which before the deformation 
occupied unit volume, the derivatives dVjdT and dVjdp give the relative 
volume changes in heating and compression respectively. That is, 

(dV/dT) p = a, (dVldp) s = -1/JT.d, (dVldp) T = -1(K. 
Thus we find the relation between the adiabatic and isothermal moduli to be 
l/JSTad = 1/K- TaPICp, juad = H- (6-6) 

For the adiabatic Young's modulus and Poisson's ratio we easily obtain 

E o + ET**I9C p 

E& * = \-ET^\9C p Cad " 1-ET«*I9C P - {bJ) 

In practice, ETa. 2 jC p is usually small, and it is therefore sufficiently accurate 
to put 

£ad = E+ E2T<x?j9C p , cr ad = a + (1 + o)ETa?l9C p . (6.8) 

In isothermal deformation, the stress tensor is given in terms of the 
derivatives of the free energy : 

aik = (dF/duacJT' 

For constant entropy, on the other hand, we have (see (3.6)) 

aw = {d$lduac)s, 

where & is the internal energy. Accordingly, the expression analogous to 
(4.3) determines, for adiabatic deformations, not the free energy but the in- 
ternal energy per unit volume : 

£ = \K & mi 2 + K U M ~ \mhkf. (6.9) 

§7. The equations of equilibrium for isotropic bodies 

Let us now derive the equations of equilibrium for isotropic solid bodies. 
To do so, we substitute in the general equations (2.7) 

doijc/dxjc + pgi = 
the expression (5.11) for the stress tensor. We have 

daw Ea dun E duik 

+ ■ 



dxic (1 + ct)(1-2ct) dxi 1 + ct dxic 
Substituting 

l / dut du]c\ 
2\dx]c dx t i 



18 Fundamental Equations §7 

we obtain the equations of equilibrium in the form 
E dZ Ui E d*ui 



2(1 + a) dx k * 2(1 + ct)(1 - 2a) dx t dx t 

These equations can be conveniently rewritten in vector notation. The 
quantities dhii/dxjc 2 are components of the vector A u > and dui/dxi = div u. 
Thus the equations of equilibrium become 

1 J jm 2(1 + a) 

AU + -T- grad div u = -p g . (7.2) 

1 — La h. 

It is sometimes useful to transform this equation by using the vector identity 
grad div u = A u+ curl curl u. Then (7.2) becomes 

1-2(7 

grad div u curl curl u 

2(1 -a) 

(l + c,)(l-2a) 
= " PS E(l-«) ■ (7 ' 3) 

We have written the equations of equilibrium for a uniform gravitational 
field, since this is the body force most usually encountered in the theory of 
elasticity. If there are other body forces, the vector pg on the right-hand 
side of the equation must be replaced accordingly. 

A very important case is that where the deformation of the body is caused, 
not by body forces, but by forces applied to its surface. The equation of 
equilibrium then becomes 

(1 - 2a) A u + grad div u = (7.4) 

or 

2(1 - a) grad div u- (1 - 2a) curl curl u = 0. (7.5) 

The external forces appear in the solution only through the boundary con- 
ditions. 
Taking the divergence of equation (7.4) and using the identity 

div grad = A> 
we find 

A div u = 0, (7.6) 

i.e. div u (which determines the volume change due to the deformation) is a 
harmonic function. Taking the Laplacian of equation (7.4), we then obtain 

AAu = 0, (7.7) 

i.e. in equilibrium the displacement vector satisfies the biharmonic equation. 
These results remain valid in a uniform gravitational field (since the right- 
hand side of equation (7.2) gives zero on differentiation), but not in the 
general case of external forces which vary through the body. 



§7 The equations of equilibrium for isotropic bodies 19 

The fact that the displacement vector satisfies the biharmonic equation 
does not, of course, mean that the general integral of the equations of equili- 
brium (in the absence of body forces) is an arbitrary biharmonic vector; it 
must be remembered that the function u(x, y, z) also satisfies the lower- 
order differential equation (7.4). It is possible, however, to express the general 
integral of the equations of equilibrium in terms of the derivatives of an 
arbitrary biharmonic vector (see Problem 10). 

If the body is non-uniformly heated, an additional term appears in the 
equation of equilibrium. The stress tensor must include the term 

-Ka(T-T )S ik 

(see (6.2)), and daijcjdxjc accordingly contains a term 

-KoidT/dxi^ -[E*l3(l-2<j)]dTldxi. 

The equation of equilibrium thus takes the form 

— - grad div u curl curl u = a grad T. (7.8) 

1 + a 2(1 + a) 

Let us consider the particular case of a plane deformation, in which one 
component of the displacement vector (u z ) is zero throughout the body, 
while the components u x , u y depend only on x and y. The components 
u ZZ) Uxz, u yz of the strain tensor then vanish identically, and therefore so do 
the components cr xz , o yz of the stress tensor (but not the longitudinal stress 
<T ZZ , the existence of which is implied by the constancy of the length of the 
body in the jsr-direction). f 

Since all quantities are independent of the co-ordinate z, the equations of 
equilibrium (in the absence of external body forces) doikjdxjc = reduce in 
this case to two equations : 

do X z daxy _ dvyx doyy _ ,_ ~. 

dx dy dx dy 

The most general functions a xx , a xy , a yy satisfying these equations are of 
the form 

°zx = &x\ty\ °xy = -&xjdxdy, a yy = d 2 x/dx 2 , (7.10) 

where x is an arbitrary function of x and y. It is easy to obtain an equation 
which must be satisfied by this function. Such an equation must exist, since the 
three quantities a xx , a xy , a yy can be expressed in terms of the two quantities 
u x , u y , and are therefore not independent. Using formulae (5.13), we find, 
for a plane deformation, 

o xx + o yy = E(u xx + U yy )l(l + cr)(l - 2a). 



t The use of the theory of functions of a complex variable provides very powerful methods of 
solving plane problems in the theory of elasticity. See N. I. Muskhelishvili, Some Basic Problems 
of the Mathematical Theory of Elasticity, 2nd English ed., P. Noordhoff, Groningen 1963. 



20 Fundamental Equations §7 

But 

du x duy 

Vxx+Vyy = AX> U xx + Uyy = — 1 = dlVU, 

ox By 

and, since by (7.6) div u is harmonic, we conclude that the function x satisfies 
the equation 

A Ax = 0, (7.11) 

i.e. it is biharmonic. This function is called the stress function. When the 
plane problem has been solved and the function x is known, the longitudinal 
stress o zz is determined at once from the formula 

<*zz = oE(u xx + u yy )l(l + cr)(\-2a) = a(a xx + a yy ), 
or 

<y zz =°Ax- (7.12) 

PROBLEMS 

Problem 1. Determine the deformation of a long rod (of length /) standing vertically in a 
gravitational field. 

Solution. We take the 2-axis along the axis of the rod, and the xy-plane in the plane of 
its lower end. The equations of equilibrium are doxi/dxt = doytjdxi = 0, da zi /dx { = pg. 
On the sides of the rod all the components a t u except a zz must vanish, and on the upper 
end (z = I) a xz = a yz = a lz = 0. The solution of the equations of equilibrium satisfying 
these conditions is a zz = — pg(l-z), with all other am zero. From one we find «<» to be 
u X x = u yy = opg(l—z)jE, u zz = —pg(l—z)/E, u X y — u xt = u yt = 0, and hence by inte- 
gration we have the components of the displacement vector, u x — opg(l—z)x/E, Uy == 
o P g(l-z)y/E, u z = -(pg/2E){l 2 -(l-z) 2 -o(x 2 +y*)}. The expression for u z satisfies the 
boundary condition u z = only at one point on the lower end of the rod. Hence the solution 
obtained is not valid near the lower end. 

Problem 2. Determine the deformation of a hollow sphere (of external and internal radii 
R 2 and i? 2 ) with a pressure p x inside and p 2 outside. 

Solution. We use spherical co-ordinates, with the origin at the centre of the sphere. 
The displacement vector u is everywhere radial, and is a function of r alone. Hence curl u=0, 
and equation (7.5) becomes grad div u = 0. Hence 

1 d(r*u) 

div u = — = constant = 3a, 

r 2 dr 

or u = ar+bjr*. The components of the strain tensor are (see formulae (1.7)) u„ == a—2b/r 3 , 
uee = «00 = a+b/r*. The radial stress is 

E E 2E b 

&rr = — — —A{\ — a)Urr + 2aU ee } = —a 



(l + cr)(l-2(7) lv ' w l-2a l + o-r3 

The constants a and b are determined from the boundary conditions: a„ = — p x at r — R u 
and o„ — —pi at r — i?g. Hence we find 

- P1 R 1*-P2 R # 1 ~ 2a h _ Rl 3R 2 3 (pl-p2) 1 + ct 

R 2 3 -Ri 3 ' E ' " i^-fli 3 '~2E' 



§7 The equations of equilibrium for isotropic bodies 21 

For example, the stress distribution in a spherical shell with a pressure pi=* p inside and 
p 2 = outside is given by 

pR?l Ra 3 \ pR$ I Rs? \ 

For a thin spherical shell of thickness h — R 2 —Rj, <^ R we have approximately 

u = pR 2 (l - a)l2Eh, a ee = o H = IpRjh, o rr = \p, 

where a„ is the mean value of the radial stress over the thickness of the shell. 

The stress distribution in an infinite elastic medium with a spherical cavity (of radius R) 
subjected to hydrostatic compression is obtained by putting i? x = R, i^ = oo, p x = 0, 
Ps = p: 

/i RZ \ I R3 \ 

°rr = -P\\ ~ — j, o„ = o„ = -p\\+—y 

At the surface of the cavity the tangential stresses a ee = a^ = —3p/2, i.e. they exceed the 
pressure at infinity. 

Problem 3. Determine the deformation of a solid sphere (of radius R) in its own gravi- 
tational field. 

Solution. The force of gravity on unit mass in a spherical body is —gt/R. Substituting 
this expression in place of g in equation (7.3), we obtain the following equation for the radial 
displacement: 

E{\-a) d/ld(r2«) 



- a) d / 1 d(r%) \ r 

-2a)M72 dr / = P8 R' 



(l + <r)(l-2a) 

The solution finite for r — which satisfies the condition a rr = for r — R is 

#>fl(l-2a)(l + <r) /3-a r2 

u = r 



105(1 -a) 



IS- a r* \ 



It should be noticed that the substance is compressed (u rr < 0) inside a spherical surface of 
radius i?\/{(3— <0/3(l +a)} and stretched outside it (u rr > 0). The pressure at the centre of 
the sphere is (3 —a)g pRj 10(1 —o). 

Problem 4. Determine the deformation of a cylindrical pipe (of external and internal radii 
jR 8 and i?i), with a pressure p inside and no pressure outside.f 

Solution. We use cylindrical co-ordinates, with the ar-axis along the axis of the pipe. 
When the pressure is uniform along the pipe, the deformation is a purely radial displacement 
w r = u(r). Similarly to Problem 2, we have 

1 d(m) 

div u = — = constant = 2a. 

r dr 

Hence u = ar+bjr. The non-zero components of the strain tensor are (see formulae (1.8)) 
Mr, = dtt/dr = a—b/r 2 , u^ = u/r = a+b/r*. From the conditions a„ = at r = R t , 
and o„ = —p at r = R lt we find 

pR? (l + g)(l-2or) pR^RJ l + o- 

a ~R 2 2 -Ri*' E ' ~ R 2 *-R!* '~E~' 



t In Problems 4, 5 and 7 it is assumed that the length of the cylinder is maintained constant, so 
that there is no longitudinal deformation. 



22 Fundamental Equations §7 

The stress distribution is given by the formulae 

a zz = 2paR^/(R 2 z -Ri z ). 

Problem 5. Determine the deformation of a cylinder rotating uniformly about its axis. 

Solution. Replacing the gravitational force in (7.3) by the centrifugal force pftV (where 
ft is the angular velocity), we have in cylindrical co-ordinates the following equation for the 
displacement u f = u(r): 



? (l- g ) £/ljM\ _ 

a)(l-2a)dr\r dr J H 



E(l-a) d/1 d(ru) 
(1 + o-)(1-2(t) dr\r dr , 

The solution which is finite for r = and satisfies the condition a„ = for r = R is 



" = —sm^r r[{)l 

Problem 6. Determine the deformation of a non-uniformly heated sphere with a spherically 
symmetrical temperature distribution. 

Solution. In spherical co-ordinates, equation (7.8) for a purely radial deformation is 

d/1 d(r^u)\ l + o- dr 






dr\r 2 dr / 3(1 -a) dr 

The solution which is finite for r = and satisfies the condition a„ — for r = R is 

v ' 

The temperature T(r) is measured from the value for which the sphere, if uniformly heated, 
is regarded as undeformed. In the above formula the temperature in question is taken as that 
of the outer surface of the sphere, so that T(R) = 0. 

Problem 7. The same as Problem 6, but for a non-uniformly heated cylinder with an 
axially symmetrical temperature distribution. 

Solution. We similarly have in cylindrical co-ordinates 

v ' o 

Problem 8. Determine the deformation of an infinite elastic medium with a given tempera- 
ture distribution T(x, y, z) which is such that the temperature tends to a constant value To 
at infinity, there being no deformation there. 

Solution. Equation (7.8) has an obvious solution for which curl u = and 

div u = <x(l + o)[T(x, y, z) - T J/3(1 - a). 



§7 The equations of equilibrium for isotropic bodies 23 

The vector u, whose divergence is a given function defined in all space and vanishing at 
infinity, and whose curl is zero identically, can be written, as we know from vector analysis, 
in the form 

u(x, y, z) = - — grad dV , 

4n J r 

where 

r = ^{( X - X >)2 + (y-y') 2 + (z-z') 2 }. 
We therefore obtain the general solution of the problem in the form 

a(l+a) ^CT'-Tq 



u = — 



127T(1-Cr) 



-a) CI -To 

_jL grad ° dV ', (1) 

— cr) J r 



wHere T' == T(x', y\ z'). 

If a finite quantity of heat q is evolved in a very small volume at the origin, the temperature 
distribution can be written T—T = (qlC)S(x)S(y)8(z), where C is the specific heat of the 
medium. The integral in (1) is then qlCr, and the deformation is given by 

a(l + a)q T 
u = 



12tt(1 -a)C r3 

Problem 9. Derive the equations of equilibrium for an isotropic body (in the absence of 
body forces) in terms of the components of the stress tensor. 

Solution. The required system of equations contains the three equations 

daaddXk = (1) 

and also the equations resulting from the fact that the six different components of k<* are 
not independent quantities. To derive these equations, we first write down the system of 
differential relations satisfied by the components of the tensor «<*. It is easy to see that the 
quantities 

1 / dUi dujc\ 

2\dxjc dxil 

satisfy identii ally the relations 



d 2 Uik d 2 ui m d 2 uu d 2 uje T 

•+ . . = — — + ■ 



dxidx m dxfdxjc dxjcdx m dx%dxi 

Here there are only six essentially different relations, namely those corresponding to the fol- 
lowing values of i, k, I, m : 1122, 1133,2233, 1123, 2213,3312. All these are retained if the above 
tensor equation is contracted with respect to / and m : 

d 2 U tt _ d 2 Uu d 2 Ukl 

dxidxjc dxjcdxi dxidxi 

Substituting here unc in terms of o,a: according to (5.12) and using (1), we obtain the re- 
quired equations: 

(l + o)AcTi k +—^- = 0. (3) 

OXfOXjc 
These equations remain valid in the presence of external forces constant throughout the body. 



24 Fundamental Equations §7 

Contracting equation (3) with respect to the suffixes i and k, we find that A<*ii = 0, i.e. 
<tji is a harmonic function. Taking the Laplacian of equation (3), we then find that AA"<* = 0, 
i.e. the components ant are biharmonic functions. These results follow also from (7.6) 
and (7.7), since aw and unc are linearly related. 

Problem 10. Express the general integral of the equations of equilibrium (in the absence 
of body forces) in terms of an arbitrary biharmonic vector (B. G. Galerkin 1930). 
Solution. It is natural to seek a solution of equation (7.4) in the form 

u = Af +A grad div f. 

Hence div u = (1 +A) div Af. Substituting in (7.4), we obtain 

(l-2a)AAf+[2(l-(r)^ + l] grad div Af= 0. 

From this we see that, if f is an arbitrary biharmonic vector (A Af = 0)> then 

u = Af - — r grad divf. 

2(1 - a) 

Problem 1 1 . Express the stresses a„, a^, a r $ for a plane deformation (in polar co-ordinates 
r, <j>) as derivatives of the stress function. 

Solution. Since the required expressions cannot depend on the choice of the initial line 
of <f>, they do not contain <f> explicitly. Hence we can proceed as follows : we transform the 
Cartesian derivatives (7.10) into derivatives with respect to r, <f>, and use the results that 
a„ = (axs^.o, o H = (a tfy )^_ , <^ = (<^i/)0=o, the angle <f> being measured from the *-axis. 
Thus 

arr = ~rJr + ^ W ° H ~ dr2 ' ^ ~ dr \ r d V 

Problem 12. Determine the stress distribution in an infinite elastic medium containing 
a spherical cavity and subjected to a homogeneous deformation at infinity. 

Solution. A general homogeneous deformation can be represented as a combination of a 
homogeneous hydrostatic extension (or compression) and a homogeneous shear. The former 
has been considered in Problem 2, so that we need only consider a homogeneous shear. 

Let <7ij; (0) be the homogeneous stress field which would be found in all space if the cavity 
were absent: in a pure shear a«<°> = 0. The corresponding displacement vector is denoted 
by u<°>, and we seek the required solution in the form u = uC'+u' 1 ', where the function u<*> 
arising from the presence of the cavity is zero at infinity. 

Any solution of the biharmonic equation can be written as a linear combination of centrally 
symmetrical solutions and their spatial derivatives of various orders. The functions r\ 
r, 1, 1/r are independent centrally symmetrical solutions. Hence the most general form of a 
biharmonic vector u (1> , depending only on the components of the constant tensor ane w 
as parameters and vanishing at infinity, is 

Ui a) = A(m «»—[-)+Bo kl «» - )+Ca k P r. (1) 

dx k \r) dxidx k dxi\rj dxidx k dxi 

Substituting this expression in equation (7.4), we obtain 

{1 _2a)^+4-^ L = [2(l-2cr)C+(A + 2C)]o kl « » l\ I = 0, 
v dx? dx t dxi dxidx k dxi r 

whence A = — 4C(1 — a). Two further relations between the constants A, B, C are obtained 
from the condition at the surface of the cavity: (<7,-fc< > + <r<*: (1) )rt* = for r = R (R being the 



§8 Equilibrium of an elastic medium bounded by a plane 25 

radius of the cavity, the origin at its centre, and n a unit vector parallel to r). A somewhat 
lengthy calculation, using (1), gives the following values: 

B = Ci? 2 /5, C = 5#3(1 + a )j2E(7 - 5 a). 

The final expression for the stress distribution is 

5(l-2a)/ J R\s 3 jr^ 



I 5(\-2a)/R\ A 3 (R\*\ 

l 1+J ^r(7) + 7^l7) ) + 

15 /R\ 3 ( /R\ 2 \ 
+ —-—( — ) CT -(~ ) \(oii (0) nkni+o k Pnini) + 



15 /R\*l /R\ 2 \ , ns 

+ 2(7^)(T) 3 ( 1 - 2CT -(7) 2 | 8 " < ' iAB - 

In order to obtain the stress distribution for arbitrary one*- ) (not a pure shear), ao: (0) in 
this expression must be replaced by ct,* (0) — £§<& ff» (0) , and the expression 

r R3 l 

corresponding to a deformation homogeneous at infinity (cf. Problem 2) must be added. We 
may give here the general formula for the stresses at the surface of the cavity: 

15 l 
a ik = - — — {(1 - (y){(Tik i0) - crtPmnjc- a k i i0) nini) + 
7 — 5cr l 

5CT-1 ) 

+ vim m nin m nin k - crcW *^^ * + — — <Jii {0) (8ik - nm) \. 



Near the cavity, the stresses considerably exceed the stresses at infinity, but this extends 
over only a short distance (the concentration of stresses). For example, if the medium is 
subjected to a homogeneous extension (only o lz W different from zero), the greatest stress 
occurs on the equator of the cavity, where 

27- 15a 

§8. Equilibrium of an elastic medium bounded by a plane 

Let us consider an elastic medium occupying a half-space, i.e. bounded 
on one side by an infinite plane, and determine the deformation of the 



26 Fundamental Equations §8 

medium caused by forces applied to its free surf ace. f The distribution of 
these forces need satisfy only one condition: they must vanish at infinity in 
such a way that there is no deformation at infinity. In such a case the equa- 
tions of equilibrium can be integrated in a general form. 

The equation of equilibrium (7.4) holds throughout the space occupied 
by the medium: 

graddivu + (l-2cr)Au = 0. (8.1) 

We seek a solution of this equation in the form 

u = f+grad<£, (8.2) 

where <£ is some scalar and the vector f satisfies Laplace's equation: 

Af=0. (8.3) 

Substituting (8.2) in (8.1), we then obtain the following equation for <f>: 

2(l-a)A<£ = -divf. (8.4) 

We take the free surface of the elastic medium as the #y-plane; the medium 
is in z > 0. We write the functions f x and f y as the ^-derivatives of some 
functions g x and g y : 

fx = dgxfiz* fy = dgy/dz- (8-5) 

Since f x and f y are harmonic functions, we can always choose the functions 
g x and g y so as to satisfy Laplace's equation : 

Ag* = 0, Agy = 0. (8.6) 

Equation (8.4) then becomes 

2(1-.)A^= --(— +—+/,). 

Since g x , gy and f z are harmonic functions, we easily see that a function 
<j> which satisfies this equation can be written as 

where i/j is again a harmonic function: 

A«A = o. ( 8 - 8 ) 



| The most direct and regular method of solving this problem is to use Fourier's method on 
equation (8.1). In that case, however, some fairly complicated integrals have to be calculated. The 
method given below is based on a number of artificial devices, but the calculations are simpler. 



z-0 



= -2(1 + a)P z [E. (8.10) 

z = 



§8 Equilibrium of an elastic medium bounded by a plane 27 

Thus the problem of determining the displacement u reduces to that 
of finding the functions g Xy g yy f z , iff, all of which satisfy Laplace's equation. 

We shall now write out the boundary conditions which must be satisfied at 
the free surface of the medium (the plane z = 0). Since the unit outward 
normal vector n is in the negative ^-direction, it follows from the general 
formula (2.8) that a iz = -P t . Using for a ik the general expression (5.11) and 
expressing the components of the vector u in terms of the auxiliary quantities 
g x> g y , f z and «/r, we obtain after a simple calculation the boundary conditions 

r d 2 gx -\ Vdl l-2ar _ 1 /dgz [ dg v \ +2 #j- 

L 5* 2 J z« o + ldx[ 2(1 - of* 2(1 - a) \ dx dy) dz). 

= -2(1 +*)P X /E, (8.9) 

La^ 2 Jz-o Uvl2(l-ar 2(l-ff)U* 3y/ aWJz=o 
= -2(l + a)P 2 ,/£ , , 

[K'-(t4') rf 3 

The components P x , P y , P z of the external forces applied to the surface are 
given functions of the co-ordinates x and y, and vanish at infinity. 

The formulae by which the auxiliary quantities g X} g y ,f z and \jj were defined 
do not determine them uniquely. We can therefore impose an arbitrary 
additional condition on these quantities, and it is convenient to make the 
quantity in the braces in equations (8.9) vanish :f 

dy 
Then the conditions (8.9) become simply 

raagfc-i _ 2(l + o) ra^-l _ 2(l + a) p 

La^Lo * *' L"^"J z =o~ £ ^ 

Equations (8.l0)-(8.l2) suffice to determine completely the harmonic 
functions g x , g y , f z and ifj. 

For simplicity, we shall consider the case where the free surface of an 
elastic half-space is subjected to a concentrated force F, i.e. one which is 
applied to an area so small that it can be regarded as a point. The effect of 
this force is the same as that of surface forces given by P = FS(x)8(y), the 
origin being at the point of application of the force. If we know the solution 
for a concentrated force, we can immediately find the solution for any force 
distribution P(#, y). For, if 

Ui = G ik (x,y, z)F k (8.13) 

f We shall not prove here that this condition can in fact be imposed; this follows from the absence 
of contradiction in the result. 



( i - 2 ^-(S + l ! ) +4(1 - <7) S= o - (8 - n) 



(8.12) 



28 Fundamental Equations §8 

is the displacement due to the action of a concentrated force F applied at 
the origin, then the displacement caused by forces T*{x,y) is given by the 
integralf 

Ui = f J G«(*-*\ y-y\ z)P k (x', y') dx' d/. (8.14) 

We know from potential theory that a harmonic function / which is zero 
at infinity and has a given normal derivative df/dz on the plane z = is 
given by the formula 

1 rr\df(x\y\z)l dx'dy' 

f(x, y, z) = , 

2ttJ J L dz Jz = o r 
where 

r = \/{(x — x') 2 + (y—y') 2 + z 2 }. 

Since the quantities dg x /dz, dgyjdz and that in the braces in equation (8.10) 
satisfy Laplace's equation, while equations (8.10) and (8.12) determine the 
values of their normal derivatives on the plane z = 0, we have 

/,_ &t + ^L\ J± . i±f f [**<>?> u d/ 

\ 8* 9y / dz ttE J J r 



\ + a F t 



tE r 



dgx 


l + o F x 


dgy 1+ff Fy 


dz 


ttE ' r' 


dz ttE r' 



(8.15) 

(8.16) 
r oz ttCj r 

where now r = ^(x 2 +y 2 + z 2 ). 

The expressions for the components of the required vector u involve the 
derivatives of g Xi g y with respect to x, y, z, but not g x , g y themselves. To 
calculate dg x jdx, dgyjdy, we differentiate equations (8.16) with respect to 
x and y respectively : 

d 2 g x _ _l + a F^ d 2 gy = 1 + or F y y 

dxdz ttE r 3 ' dydz ttE r 3 

Now, integrating over z from oo to z, we obtain 

dg x 1+ a F x x 



dx ttE ' r(r + z)' 
dgy 1 + a F p y 



(8.17) 



dy ttE r{r + z) 

We shall not pause to complete the remaining calculations, which are 
elementary but laborious. We determine f z and diff/dz from equations (8.11), 

t In mathematical terms, Gw is the Green's tensor for the equations of equilibrium of a semi-infinite 
medium. 



§8 



Equilibrium of an elastic medium bounded by a plane 



29 



(8.15) and (8.17). Knowing di/jjdz, it is easy to calculate dtfjjdx and difjjdy by 
integrating with respect to z and then differentiating with respect to x and y. 
We thus obtain all the quantities needed to calculate the displacement vector 
from (8.2), (8.5) and (8.7). The following are the final formulae: 



l + <r( 



xz (l-2,r)*-| 2(l-a)r+z 

—\r z -\ v x -r 

r 3 r(r + z) J r(r + z) 

\2r(ar + z) + z 2 ]x ) 

r 6 {r + z)* ) 



Uy _ Ltfff^ - (1-2^1 2(1- CT ), + , I 



+ 



(r + #) J " r(r + z) 

[2r(ar + z) + z 2 ]y 



r 3 (r + z) 2 



(xF x +yF y )\, 



1 + erf r2(l — CT ) z 2 i r 1-2ct zi ) 

In particular, the displacement of points on the surface of the medium is 
given by putting z = 0: 



1 + cr 1 



u x 



Uu = 



(l-2o> 



: lax \ 

-F z + 2(l-a)F x +-—(xF x +yFy)\, 
2ttE r\r r 2 ) 

1 + alf (1-2ct)v lay \ 

— -.- - -^ ^F z + 2(l-a)F y + -/(xF x +yF y )\, 

2ttE r{ r r 2 j 



(8.19) 



»* = "T^-- k 1 " "X^ + O ~ 2ct ) " («^r+^y) 



PROBLEM 

Determine the deformation of an infinite elastic medium when a force F is applied to a 
small region in it. t 

Solution. If we consider the deformation at distances r which are large compared with 
the dimension of the region where the force is applied, we can suppose that the force is 
applied at a point. The equation of equilibrium is (cf. (7.2)) 



Au + 



1 



l-2o- 



grad div u = 



2(1 + a) 
E 



Fo(r), 



(1) 



where 8(r) = S(x)8(y)8(z), the origin being at the point where the force is applied. We seek 
the solution in the form u = Uo+Ux, where u„ satisfies the Poisson-type equation 



A«o = - 2(1 J~ g) Fo(r). 
E 



(2) 



■f The corresponding problem for an arbitrary infinite anisotropic medium has been solved by 
I. M. Lifshitz and L. N. RozentsveIg (Zhttrnal experimental' not i teoreticheskoi fiziki 17, 783, 1947). 



30 Fundamental Equations §9 

We then have for ui the equation 

grad div ui + (l — 2o)/\vl\ = — grad div un. (3) 

The solution of equation (2) which vanishes at infinity is uo = (1 + o)F/2nEr. Taking the 
curl of equation (3), we have A curl ui = 0. At infinity we must have curl ui = 0. But a 
function harmonic in all space and zero at infinity must be zero identically. Thus curl ui = 0, 
and we can therefore write ui = grad^. From (3) we obtain grad {2(1 — a) A^ + div uo} = 0. 
Hence it follows that the quantity in braces is a constant, and it must be zero at infinity; we 
therefore have in all space 

dlVUo 1 + ct /1\ 

£</, = = F- grad (-). 

r 2(1 -a) 4ttE(1-ct) & \rf 

If i/i is a solution of the equation A«A = l/ r > then 

1 + CT 

6 = F« grad ib. 

Y 4ttE(1-o) 

Taking the solution iff = \r, which has no singularities, we obtain 

1 + ct (F.n)n-F 

ui = grad </> = — — , 

57r.c(l — ct) r 

where n is a unit vector parallel to the radius vector r. The final result is 

1 + CT (3-4<r)F+n(n.F) 



$7rE(l-a) r 

On putting this formula into the form (8.13) we obtain the Green's tensor for the equa- 
tions of equilibrium of an infinite isotropic medium:! 

1 + CT 1 

87tE(1 — ct) r 



1 



Hk 



1 d*r 



47Tfil r 4(1 — ct) dxfdxjc 



§9. Solid bodies in contact 

Let two solid bodies be in contact at a point which is not a singular point 
on either surface. Fig. la shows a cross-section of the two surfaces near 
the point of contact O. The surfaces have a common tangent plane at O, 
which we take as the xy-plane. We regard the positive ^-direction as being 
into either body (i.e. in opposite directions for the two bodies) and denote 
the corresponding co-ordinates by z and z'. 



f The fact that the components of the tensor Gxk are first-order homogeneous functions of the co- 
ordinates x, y, z is evident from arguments of homogeneity applied to the form of equation (1), where 
the left-hand side is a linear combination of the second derivatives of the components of the vector u, 
and the right-hand side is a third-order homogeneous function (S(ar) = a _3 S(r)). 

This property remains valid in the general case of an arbitrary anisotropic medium. 



§9 



Solid bodies in contact 



31 



Near a point of ordinary contact with the acy-plane, the equation of the 
surface can be written 

z = K afi x^ fii (9.1) 

where summation is understood over the values 1, 2, of the repeated suffixes 
a, j8 (xi = x, x 2 = y), and K afi is a symmetrical tensor of rank two, which 
characterises the curvature of the surface: the principal values of the tensor 
K ap are l/2Ri and 1/2^2, where Ri and R 2 are the principal radii of curvature 
of the surface at the point of contact. A similar relation for the surface of 
the other body near the point of contact can be written 

Z' = K'apXJCfi. (9-2) 

Let us now assume that the two bodies are pressed together by applied 
forces, and approach a short distance A.f Then a deformation occurs near 
the original point of contact, and the two bodies will be in contact over a 
small but finite portion of their surfaces. Let u z and u' z be the components 
(along the * and z' axes respectively) of the corresponding displacement 
vectors for points on the surfaces of the two bodies. The broken lines 




Fig. 1 

in Fig. lb show the surfaces as they would be in the absence of any deforma- 
tion, while the continuous lines show the surfaces of the deformed bodies; the 
letters z and z' denote the distances given by equations (9.1) and (9.2). It 
is seen at once from the figure that the equation 

(z + u z ) + (z' + u' z ) = h, 



or 



(*«*+*'«*)*«**+«« + «'« = n > 



(9.3) 



f This contact problem in the theory of elasticity was first solved by H. Hertz. 



32 Fundamental Equations §9 

holds everywhere in the region of contact. At points outside the region of 
contact, we have 

z + z' + u z +u' z < h. 

We choose the x andj> axes to be the principal axes of the tensor K afi + K afi . 
Denoting the principal values of this tensor by A and B,-\ we can rewrite 
equation (9.3) as 

Ax* + By 2 + u z + u' z = h. (9.4) 

We denote by P z {x, y) the pressure between the two deformed bodies at 
points in the region of contact; outside this region, of course P z = 0. To 
determine the relation between P z and the displacements u z> u' z , we can 
with sufficient accuracy regard the surfaces as plane and use the formulae 
obtained in §8. According to the third of formulae (8.19) and (8.14), the 
displacement u z under the action of normal forces P z (x, y) is given by 



~ ttE J J" 



u z = -^-\\-^-dx'dy, 
Mz = -^-JJ-T— dxdy > 

where a, a' and E, E' are the Poisson's ratios and the Young's moduli of the 
two bodies. Since P z = outside the region of contact, the integration ex- 
tends only over this region. It may be noted that, from these formulae, the 
ratio u z \u' z is constant: 

u z \u' z = (1 - CT 2)£'/(1 - CT '2)£. (9.6) 

The relations (9.4) and (9.6) together give the displacements u Zi u' z at every 
point of the region of contact (although (9.5) and (9.6), of course, relate to 
points outside that region also). 

Substituting the expressions (9.5) in (9.4), we obtain 

l/l-o* l-<x'2\ r rP z (x',y') 

-I— — + — jp— ) j j r dx' d/ = h-Ax*-By2. (9.7) 



t The quantities A and B are relaced to the radii of curvature i? 1( R 2 and R\, R' t by 

2(A + B) = — +— +—+—, 
V ' R! R 2 R\ R' 2 

/ 1 1 \ 2 / 1 1 \2 

T \Ri i? 2 /U'i rJ' 

where <f> is the angle between the normal sections whose radii of curvature are R t and R\. 

The radii of curvature are regarded as positive if the centre of curvature lies within the body con- 
cerned, and negative in the contrary case. 



§9 Solid bodies in contact 33 

This integral equation determines the distribution of the pressure P z over 
the region of contact. Its solution can be found by analogy with the following 
results of potential theory. The idea of using this analogy arises as follows: 
firstly, the integral on the left-hand side of equation (9.7) is of a type com- 
monly found in potential theory, where such integrals give the potential of a 
charge distribution; secondly, the potential inside a uniformly charged 
ellipsoid is a quadratic function of the co-ordinates. 

If the ellipsoid x 2 la 2 +y 2 jb 2 + z 2 lc 2 = 1 is uniformly charged (with volume 
charge density p), the potential in the ellipsoid is given by 

<K x > y> z ) 

00 

n x 2 y 2 z 2 \ d£ 

= irpabcj |1 - — - -^j - J^r] V{{a 2 + i){b 2 + £ ){c 2 + t )} ' 
o 

In the limiting case of an ellipsoid which is very much flattened in the 
z- direction (c -> 0), we have 

00 

Hx,y) = «pabcj{l -JL-- _^j ; 



in passing to the limit c -> we must, of course, put z = for points inside 
the ellipsoid. The potential <j>{x, y, z) can also be written as 

,_ c c c p dx ' d y' dz ' 

mytZ) ~ J J J v{(^-^) 2 +(^-y) 2 +(^-^) 2 } , 

where the integration is over the volume of the ellipsoid. In passing to the 
limit c ->0, we must put z = z' = in the radicand; integrating over z' 
between the limits 

±c^{\-(x' 2 \a 2 )-{y' 2 \b 2 )\ 

we obtain 

r rdx' dy' 1/ x' 2 y' 2 \ 

where 

r= V{{x-x') 2 + {y-y'Y}, 

and the integration is over the area inside the ellipse 

x '2j a 2 +y '2j b 2 = 1. 



34 Fundamental Equations §9 

Equating the two expressions for <f>(x, v), we obtain the identity 

r rdx' dy' If x' 2 y' 2 \ 

J J ~~r V \ ~a~ 2 W) 

Comparing this relation with equation (9.7), we see that the right-hand 
sides are quadratic functions of x and y of the same form, and the left-hand 
sides are integrals of the same form. We can therefore deduce immediately 
that the region of contact (i.e. the region of integration in (9.7)) is bounded 
by an ellipse of the form 

x 2 y 2 

and that the function P z (x, y) must be of the form 

Pfay) = constant x ^ (l - *- - £). 

Taking the constant such that the integral jjP z dx dy over the region of 
contact is equal to the given total force F which moves the bodies together, 
we obtain 

'^-zWf 1 -*-£)• (9 ' 10) 

This formula gives the distribution of pressure over the area of the region of 
contact. It may be pointed out that the pressure at the centre of this region 
is f times the mean pressure FjTrab. 

Substituting (9.10) in equation (9.7) and replacing the resulting integral 
in accordance with (9.8), we obtain 

00 

FDr/ x 2 y 2 \ 

-ft 1 -*rr£i) "fMc+flc+flo 



= h-Ax 2 -By\ 
where 

4\ E + E' )' 
This equation must hold identically for all values of x and y inside the 



§9 Solid bodies in contact 35 

ellipse (9.9) ; the coefficients of x and y and the free terms must therefore be 
respectively equal on each side. Hence we find 

FDr d| 

1 g (9.11) 



tu r 

TT J 




V{(<* 2 + Z)(b 2 + m' 



FDr dg 



bD r 
A = — - 

77 J (d 



B 



FDr d£ 



bu r 

IT J I 



(b 2 + i)V{(a 2 + ZW + m 

Equations (9.12) determine the semi-axes a and b of the region of contact 
from the given force F {A and B being known for given bodies). The 
relation (9.11) then gives the distance of approach h as a function of the force 
F. The right-hand sides of these equations involve elliptic integrals. 

Thus the problem of bodies in contact can be regarded as completely 
solved. The form of the surfaces (i.e. the displacements u z , u' z ) outside the 
region of contact is determined by the same formulae (9.5) and (9.10); the 
values of the integrals can be found immediately from the analogy with the 
potential outside a charged ellipsoid. Finally, the formulae of §8 enable us to 
find also the deformation at various points in the bodies (but only, of course, 
at distances small compared with the dimensions of the bodies). 

Let us apply these formulae to the case of contact between two spheres of 
radii R and R'. Here A = B = 1/2R+1/2R'. It is clear from symmetry 
that a = b, i.e. the region of contact is a circle. From (9.12) we find the 
radius a of this circle to be 

a = Fii3{DRR'l(R + R')}V3. (9.13) 

h is in this case the difference between the sum R + R' and the distance be- 
tween the centres of the spheres. From (9.10) we obtain the following 
relation between F and h : 

h _ F^Jz^l+l)] 1 ' 3 . (9.14) 

It should be noticed that h is proportional to F m ; conversely, the force F 
varies as h 3/2 . We can write down also the potential energy U of the spheres 
in contact. Since —F= —dU/dh, we have 



2 / RR' 
U = #5/2 — 
SD 



I RR' 

JrTW < 9 - 15 > 



Finally, it may be mentioned that a relation of the form h = constant x F 2/3 , 
or F = constant x A 3/2 , holds not only for spheres but also for other finite 



36 Fundamental Equations §9 

bodies in contact. This is easily seen from similarity arguments. If we make 
the substitution 

a 2 -> oca 2 , b 2 -> <xb 2 , F -> a 3/2 F, 

where a is an arbitrary constant, equations (9.12) remain unchanged. In 
equation (9.11), the right-hand side is multiplied by a, and so h must be 
replaced by aJi if this equation is to remain unchanged. Hence it follows 
that F must be proportional to h 3/2 . 

PROBLEMS 

Problem 1. Determine the time for which two colliding elastic spheres remain in contact. 

Solution. In a system of co-ordinates in which the centre of mass of the two spheres is 
at rest, the energy before the collision is equal to the kinetic energy of the relative motion 
£/«>*> where v is the relative velocity of the colliding spheres and /x = mjW^/Cwii+wia) their 
reduced mass. During the collision, the total energy is the sum of the kinetic energy, which 
may be written £/Jt 8 , and the potential energy (9.15). By the law of conservation of energy 
we have 

(dh\ 2 .„,„ , 4 / RR' 



lahY 4 / 

J— +kh 5/2 = u.v 2 , k = — A 
r \dt) r 5ZW- 



R + R' 

The maximum approach h of the spheres corresponds to the time when their relative velocity 
h = 0, and is h = {nlk)*'W*. 

The time t during which the collision takes place (i.e. h varies from to h and back) is 
h„ i 

dx 



T 



Jv(» 2 -^ /2 //x) Uv J 



V(*> 2 - Mfi'*/p) \ k 2 v] J V(i - * 2/5 ) ' 

or 

51X9/10) IW l#W ' 

By using the statical formulae obtained in the text to solve this problem, we have neglected 
elastic oscillations of the spheres resulting from the collision. If this is legitimate, the velocity 
v must be small compared with the velocity of sound. In practice, however, the validity of 
the theory is limited by the still more stringent requirement that the resulting deformations 
should not exceed the elastic limit of the substance. 

Problem 2. Determine the dimensions of the region of contact and the pressure distri- 
bution when two cylinders are pressed together along a generator. 

Solution. In this case the region of contact is a narrow strip along the length of the 
cylinders. Its width 2a and the pressure distribution across it can be found from the formulae 
in the text by going to the limit bja -*■ oo. The pressure distribution will be of the form 
Pz(x) = constant X \/(l —x 3 la 2 ), where x is the co-ordinate across the strip; normalising 
the pressure to give a force F per unit length, we obtain 

IF 



IF II x*\ 



Substituting this expression in (9.7) and effecting the integration by means of (9.8), we have 

4DF f df 8DF 



A = 



i 



3tt J (a 2 + £) 3/2 | 3-rra 2 



§10 The elastic properties of crystals 37 

One of the radii of curvature of a cylindrical surface is infinite, and the other is the radius 
of the cylinder; in this case, therefore, A = 1/2R+112R', B = 0. We have finally for the 
width of the region of contact 

16DF RR' 



I/16DF RR' \ 
a= VI 3tt ' R + R') 



§10. The elastic properties of crystals 

The change in the free energy in isothermal compression of a crystal is, as 
with isotropic bodies, a quadratic function of the strain tensor. Unlike what 
happens for isotropic bodies, however, this function contains not just two 
coefficients, but a larger number of them. The general form of the free energy 
of a deformed crystal is 

F = ^iklmUikUim, (10.1) 

where X iJc i m is a tensor of rank four, called the elastic modulus tensor. Since 
the strain tensor is symmetrical, the product uacUim is unchanged when the 
suffixes i, k, or /, m, or /, / and k, m, are interchanged. Hence we see that the 
tensor Xmm can be defined so that it has the same symmetry properties: 

Xiklm = Xmm = ^ikml — hmik> (10.2) 

A simple calculation shows that the number of different components of a 
tensor of rank four having these symmetry properties is in general 21. 

In accordance with the expression (10.1) for the free energy, the stress 
tensor for a crystal is given in terms of the strain tensor by 

one = dFjdUije — XikimUim't (10.3) 

cf. also the last footnote to this section. 

If the crystal possesses symmetry, relations exist between the various 
components of the tensor X ik i m , so that the number of independent com- 
ponents is less than 21. 

We shall discuss these relations for each possible type of macroscopic 
symmetry of crystals, i.e. for each of the crystal classes, dividing these into the 
corresponding crystal systems. 

(1) Triclinic system. Triclinic symmetry (classes C\ and C*) does not place 
any restrictions on the components of the tensor Xikim, and the system of co- 
ordinates may be chosen arbitrarily as regards the symmetry. All the 21 
moduli of elasticity are non-zero and independent. However, the arbitrariness 
of the choice of co-ordinate system enables us to impose additional conditions 
on the components of the tensor Xikim- Since the orientation of the co-ordinate 
system relative to the body is defined by three quantities (angles of rotation), 
there can be three such conditions; for example, three of the components may 



38 Fundamental Equations §10 

be taken as zero. Then the independent quantities which describe the elastic 
properties of the crystal will be 18 non-zero moduli and 3 angles defining the 
orientation of the axes in the crystal. 

(2) Monoclinic system. Let us consider the class C s ; we take a co-ordinate 
system with the *y-plane as the plane of symmetry. On reflection in this 
plane, the co-ordinates undergo the transformation x -> x, y -+y, z -> — z. 
The components of a tensor are transformed as the products of the corres- 
ponding co-ordinates. It is therefore clear that, in the transformation men- 
tioned, all components X ik i m whose suffixes include z an odd number of 
times (1 or 3) will change sign, while the other components will remain un- 
changed. By the symmetry of the crystal, however, all quantities characterising 
its properties (including all components Xiicim) must remain unchanged on 
reflection in the plane of symmetry. Hence it is evident that all components 
with an odd number of suffixes z must be zero. Accordingly, the general 
expression for the elastic free energy of a crystal belonging to the monoclinic 
system is 

* — iX xxxx u xx +2^yyyy u yy + 2^zzzzUzz 2 + XxxyyUxxUyy + X X x ZZ UxxUzz + 

+ AyyzzUyyU zz + 2X X yxyU X y 2 + 2X xzxz U xz 2 + 2Xy zyz Uy z 2 + 2\ X xxyUxxU xy + 

+ 2Xyyy x UyyUy x + 2Xxy ZZ U X yU zz + 4X ZZ yzUx Z Uy z . ( 1 0.4) 

This contains 13 independent coefficients. A similar expression is obtained 
for the class C2, and also for the class C^n, which contains both symmetry 
elements (C2 and an). In the argument given, however, the direction of only 
one co-ordinate axis (that of z) is fixed ; those of x and y can have arbitrary 
directions in the perpendicular plane. This arbitrariness can be used to make 
one coefficient, say X xyzz , vanish by a suitable choice of axes. Then the 13 
quantities which describe the elastic properties of the crystal will be 12 non- 
zero moduli and one angle defining the orientation of the axes in the xy-plane. 

(3) Orthorhombic system. In all the classes of this system (Cz v , D 2 , £>2ft) the 
choice of co-ordinate axes is determined by the symmetry, and the expression 
obtained for the free energy is the same for each class. 

Let us consider, for example, the class D^n', we take the three planes of 
symmetry as the co-ordinate planes. Reflections in each of these planes are 
transformations in which one co-ordinate changes sign and the other two 
remain unchanged. It is evident therefore that the only non-zero components 
X-ikim are those whose suffixes contain each of x, y, z an even number of times ; 
the other components would have to change sign on reflection in some plane 
of symmetry. Thus the general expression for the free energy in the ortho- 
rhombic system is 

F — \XxxxxUxz +2*yyyyUyy +^X zzzz U zz 2 + X X xyyU X xUyy + XxxzzUxxUzz + 

+ Xyy ZZ UyyU ZZ + 2X X y x yU X y 2 + 2X XZXZ U XZ % + 2Xy Z y Z Uy z 2 . (10.5) 

It contains nine moduli of elasticity. 



§10 The elastic properties of crystals 39 

(4) Tetragonal system. Let us consider the class C 4t ,; we take the axis C 4 
as the #-axis, and the x and y axes perpendicular to two of the vertical 
planes of symmetry. Reflections in these two planes signify transformations 

x -> — x, y -+y, z ->z 

and 

x -> x, y -> —y, z-+z; 

all components X ik im with an odd number of like suffixes therefore vanish. 
Furthermore, a rotation through an angle £tt about the axis C 4 is the trans- 
formation 

x -+y, v -> —x, z -+z. 
Hence we have 

Xxxxx = "yyyyy X X xzz — Ayyzz> X X zxz = Ayzyz- 

The remaining transformations in the class C iv do not give any further 
conditions. Thus the free energy of crystals in the tetragonal system is 

F =» %\xxxx(Uxx 2 + Uyy 2 ) + &zzzzUzz 2 + hzxzz(UxzUzz + UyyU zz ) + 

+ ^xxyyUxxUyy + 2X X yxyU X y 2 + 1\ X zxz{Uxz 2 + %z 2 )- (10.6) 

It contains six moduli of elasticity. 

A similar result is obtained for those other classes of the tetragonal system 
where the natural choice of the co-ordinate axes is determined by symmetry 
(I>2d> £>4, D^ h ). In the classes C 4 , S 4 , C ih , on the other hand, only the choice 
of the s-axis is unique (along the axis C 4 or 5 4 ). The requirements of symmetry 
then allow a further component X XX xy = -X yy yx in addition to those which 
appear in (10.6). These components may be made to vanish by suitably 
choosing the directions of the x and y axes, and F then reduces to the form 
(10.6). 

(5) Rhotnbohedral system. Let us consider the class C 3v ; we take the third- 
order axis as the sr-axis, and the j-axis perpendicular to one of the vertical 
planes of symmetry. In order to find the restrictions imposed on the com- 
ponents of the tensor X ilc im by the presence of the axis C 3 , it is convenient 
to make a formal transformation using the complex co-ordinates £ = x + iy, 
r\ = x — iy, the z co-ordinate remaining unchanged. We transform the 
tensor Xium to the new co-ordinate system also, so that its suffixes take the 
values I, 77, z. It is easy to see that, in a rotation through 2^/3 about the 
axis C3, the new co-ordinates undergo the transformation £ -> £e 2nt/3 , 
rj _> ^ e -2^f/3 ? z ->z. By symmetry only those components Xucim which are 
unchanged by this transformation can be different from zero. These com- 
ponents are evidently the ones whose suffixes contain £ three times, or r\ 
three times (since (^2^/3)3 = e 2ni = i) ? G r £ and r\ the same number of times 
(since e 2^/3 e -2rf/3 = j^ i >e . X ZZZZ} X g7liv ^uw *£?**» ^£*vz> ^Hl*> ^w«- 



40 Fundamental Equations §10 

Furthermore, a reflection in the symmetry plane perpendicular to the jy-axis 
gives the transformation x -> x, y -> -y, z -> z, or f -> rj, 77 -> £. Since 
*££*z becomes A^z in this transformation, these two components must be 
equal. Thus crystals of the rhombohedral system have only six moduli of 
elasticity. In order to obtain an expression for the free energy, we must form 
the sum \,\iicimUiicUim, in which the suffixes take the values £, t], z; since F 
is to be expressed in terms of the components of the strain tensor in 
the co-ordinates x, y, z, we must express in terms of these the components 
in the co-ordinates £, rj, z. This is easily done by using the fact that the 
components of the tensor u ik transform as the products of the corresponding 
co-ordinates. For example, since 

£ 2 = (x+iy) 2 = x 2 —y 2 + 2ixy, 
it follows that 

u ii = u %% — u yy "f" 2iu X y. 

Consequently, the expression for F is found to be 

F = &ZZZzU Z Z 2 + 2A iviv (u zx +U 1/V ) 2 + A^ vv {(u Xa; -Uyy) 2 + 4u X y 2 } + 
+ 2X^zz(U zz + Uyy)u ZZ + A\ Zrl z{u X z 2 + Uy z 2 ) + ^^ z {(u xx - Uyy)u xz - 2u X yUy z }. 

(10.7) 
This contains 6 independent coefficients. A similar result is obtained for the 
classes D 3 and D M , but in the classes C3 and ^6, where the choice of the x and 3/ 
axes remains arbitrary, requirements of symmetry allow also a non-zero value 
of the difference X i ^ z -X 7j1j7jZ . This, however, can be made to vanish by a 
suitable choice of the x and y axes. 

(6) Hexagonal system. Let us consider the class Cq; we take the sixth- 
order axis as the sr-axis, and again use the co-ordinates $ = x+iy> r\ = x — iy. 
In a rotation through an angle \n about the #-axis, the co-ordinates £, r\ 
undergo the transformation £ -> ge^' 3 , r\ -> rje-^ 3 . Hence we see that only 
those components Amm are non-zero which contain the same number of 
suffixes i and r\. These are A zzzz , \ gv g v A £f „, X gr)ZZ , X gZrjZ . Other symmetry 
elements in the hexagonal system give no further restrictions. There are 
therefore only five moduli of elasticity. The free energy is 

F = %X zzzz U ZZ 2 + 2X Ev £ v (U XX + Uyy) 2 + X iivv [(u XX -Uyy) 2 + 4u X y 2 ] + 

+ 2X^ZZU ZZ (U XX + Uyy) + iX iZvZ (u xz 2 + U yz 2 ). (10.8) 

It should be noticed that a deformation in the ry-plane (for which u xx , 
Uyy and u xy are non-zero) is determined by only two moduli of elasticity, 
as for an isotropic body; that is, the elastic properties of a hexagonal crystal 
are isotropic in the plane perpendicular to the sixth-order axis. 

For this reason the choice of axis directions in this plane is unimportant and 
does not affect the form of F. The expression (10.8) therefore applies to all 
classes of the hexagonal system. 



§10 The elastic properties of crystals 41 

(7) Cubic system. We take the axes along the three fourth-order axes of 
the cubic system. Since there is tetragonal symmetry (with the fourth-order 
axis in the ^-direction), the number of different components of the tensor 
hklm is limited to at most the following six: X xxxx , ^zzzz, ^xxzz, ^xxyy, ^xyxy, 
^xzxz' Rotations through far about the x and y axes give respectively the 
transformations x -> x, y -> — z, z ->y, and x -> z, y -»■ y, z -> — x. The 
components listed are therefore equal in successive pairs. Thus there remain 
only three different moduli of elasticity. The free energy of crystals of the 
cubic system is 

■** = 2^XXXX\ U XX +Uyy + Uzz ) + A XX yy[U xx Uyy-\- U XX U ZZ~\~ Uyytlzz) + 

+ 2\ z y X y(ll x y 2 + U XZ 2 + Uy Z 2 ). (10.9) 

We may recapitulate the number of independent parameters (elastic moduli 
or angles defining the orientation of axes in the crystal) for the classes of the 
various systems : 



Triclinic 


21 


Monoclinic 


13 


Orthorhombic 


9 


Tetragonal (C4, S4, Cm) 


7 


Tetragonal (C 4v , D 2 d, #4, D 4h ) 


6 


Rhombohedral (C3, 56) 


7 


Rhombohedral (C^v, D3, Dsa) 


6 


Hexagonal 


5 


Cubic 


3 



The least number of non-zero moduli that is possible by suitable choice of 
the co-ordinate axes is the same for all the classes in each system : 

Triclinic 18 

Monoclinic 12 

Orthorhombic 9 

Tetragonal 6 

Rhombohedral 6 

Hexagonal 5 

Cubic 3 

All the above discussion relates, of course, to single crystals. Polycrystalline 
bodies whose component crystallites are sufficiently small may be regarded 
as isotropic bodies (since we are concerned with deformations in regions 
large compared with the dimensions of the crystallites). Like any isotropic 
body, a polycrystal has only two moduli of elasticity. It might be thought at 
first sight that these moduli could be obtained from those of the individual 
crystallites by simple averaging. This is not so, however. If we regard the 
deformation of a polycrystal as the result of a deformation of its component 
crystallites, it would in principle be necessary to solve the equations of 



42 Fundamental Equations §10 

equilibrium for every crystallite, taking into account the appropriate boun- 
dary conditions at their surfaces of separation. Hence we see that the relation 
between the elastic properties of the whole crystal and those of its component 
crystallites depends on the actual form of the latter and the amount of correla- 
tion of their mutual orientations. There is therefore no general relation 
between the moduli of elasticity of a polycrystal and those of a single crystal 
of the same substance. 

The moduli of an isotropic polycrystal can be calculated with fair accuracy 
from those of a single crystal only when the elastic properties of the single 
crystal are nearly isotropic.f In a first approximation, the moduli of elasticity 
of the polycrystal can then simply be put equal to the "isotropic part" of the 
moduli of the single crystal. In the next approximation, terms appear which 
are quadratic in the small "anisotropic part" of these moduli. It is found % 
that these correction terms are independent of the shape of the crystallites 
and of the correlation of their orientations, and can be calculated in a general 
form. 

Finally, let us consider the thermal expansion of crystals. In isotropic 
bodies, the thermal expansion is the same in every direction, so that the 
strain tensor in free thermal expansion is (see §6) um = $<x.(T— Tb)S$fc, where 
a is the thermal expansion coefficient. In crystals, however, we must put 

u ik = \*i k {T-T ) t (10.10) 

where a^ is a tensor of rank two, symmetrical in the suffixes i and k. Let us 
calculate the number of independent components of this tensor in crystals 
of the various systems. The simplest way of doing this is to use the result of 
tensor algebra that to every symmetrical tensor of rank two there corresponds 
a tensor ellipsoid.^ It follows at once from considerations of symmetry that, 
for triclinic, monoclinic and orthorhombic symmetry, the tensor ellipsoid has 
three axes of different length. For tetragonal, rhombohedral and hexagonal 
symmetry, on the other hand, we have an ellipsoid of revolution (with its 
axis of symmetry along the axes C4, C3 and C§ respectively). Finally, for cubic 
symmetry the ellipsoid becomes a sphere. An ellipsoid of three axes is 
determined by three quantities, an ellipsoid of revolution by two, and a 
sphere by one (the radius). Thus the number of independent components 
of the tensor a^ in crystals of the various systems is as follows: triclinic, 
monoclinic and orthorhombic, 3 ; tetragonal, rhombohedral and hexagonal, 2 ; 
cubic, 1. 

Crystals of the first three systems are said to be biaxial, and those of the 
second three systems uniaxial. It should be noticed that the thermal expan- 
sion of crystals of the cubic system is determined by one quantity only, i.e. 
they behave in this respect as isotropic bodies. 



t For a "nearly isotropic" cubic crystal (e.g.), the difference X X xxx~ X X xyy— 2X X yxv must be small. 
t I. M. Lifshitz and L. N. RozentsveIg, Zhurnal eksperimental'noii teoreticheskoi fiziki 16, 967, 1946. 
§ Determined by the equation «ikX{Xic — 1. 



§10 The elastic properties of crystals 43 

PROBLEM 
Determine the Young's modulus of a cubic crystal as a function of direction. 

Solution. We take the axes of co-ordinates along the three axes of the fourth order. Let 
the axis of a rod cut from the crystal be in the direction of the unit vector n. The stress 
tensor one in the extended rod must satisfy the following conditions: when one is multiplied 
by n { , the resulting extension force must be parallel to n (condition at the ends of the rod) ; 
when it is multiplied by a vector perpendicular to n, the result must be zero (condition on the 
sides of the rod). Such a tensor must be of the form one = />«<«*, where p is the extension 
force per unit area of the ends of the rod. Calculating the components one by means of the 
expression (10.9) for the free energy! and comparing them with the formulae one = ptiitik, 
we find the components of the strain tensor to be 

(ci + 2c 2 )n x 2 -c 2 

u xx = p- r— , u xy = pn x n y l2cz, 

{ci-C2)(ci + 2c 2 ) 

and similarly for the remaining components. Here we have put X xxxx = c lt Xxxyy = c t , 
Axyxy — Cg. 

The relative longitudinal extension of the rod is u = (dl'—dl)ldl, where dl' is given by 
formula (1.2) and dxjdl — n t . For small deformations this gives u — Uijcmnjc. The Young's 
modulus is determined by the coefficient of proportionality in p = Eu, and is 

*- fcrSfcrfi - ^((w+^+w)]" 1 . 

E has extremum values in the directions of the edges (i.e. of the co-ordinate axes) and of the 
spatial diagonals of the cube. 



f In calculating <r,fc, the following fact must be borne in mind. If we effect the calculation, not 
directly from the formulae auc = KkimUm, but by differentiation of the expression for the free energy 
with respect to the components of the tensor unc, the derivatives with respect to mk with i ^ k give 
twice the values of the corresponding components o.-fc. This is because the expressions aa = dFjduuc 
are meaningful only as indicating that dF = one dw.fc; in the sum <r<k du<jfc, however, the term in the 
differential dune for each component with i ^ k of the symmetrical tensor «<& appears twice. 



CHAPTER II 

THE EQUILIBRIUM OF RODS AND PLATES 

§11. The energy of a bent plate 

In this chapter we shall study some particular cases of the equilibrium of 
deformed bodies, and we begin with that of thin deformed plates. When we 
speak of a thin plate, we mean that its thickness is small compared with its 
dimensions in the other two directions. The deformations themselves are 
supposed small, as before. In the present case the deformation is small if the 
displacements of points in the plate are small compared with its thickness. 

The general equations of equilibrium are considerably simplified when 
applied to thin plates. It is more convenient, however, not to derive these 
simplified equations directly from the general ones, but to calculate afresh 
the free energy of a bent plate and then vary that energy. 

When a plate is bent, it is stretched at some points and compressed at 
others: on the convex side there is evidently an extension, which decreases 
as we penetrate into the plate, finally becoming zero, after which a gradually 
increasing compression is found. The plate therefore contains a neutral 
surface, on which there is no extension or compression, and on opposite sides 
of which the deformation has opposite signs. The neutral surface clearly 
lies midway through the plate. 




Fig. 2 

We take a co-ordinate system with the origin on the neutral surface and the 
sr-axis normal to the surface. The xy-plane is that of the undeformed plate. 
We denote by £ the vertical displacement of a point on the neutral surface, 
i.e. its z co-ordinate (Fig. 2). The components of its displacement in the 
xy-plane are evidently of the second order of smallness relative to £, and can 
therefore be put equal to zero. Thus the displacement vector for points on the 
neutral surface is 

u x ®> = uy^ = 0, « z <°> = i{x,y). (11.1) 

44 



§1 1 The energy of a bent plate 45 

For further calculations it is necessary to note the following property of 
the stresses in a deformed plate. Since the plate is thin, comparatively small 
forces on its surface are needed to bend it. These forces are always consider- 
ably less than the internal stresses caused in the deformed plate by the ex- 
tension and compression of its parts. We can therefore neglect the forces P< 
in the boundary condition (2.8), leaving a^n^ = 0. Since the plate is only 
slightly bent, we can suppose that the normal vector n is along the #-axis. 
Thus we must have on both surfaces of the plate a xz = a yz = a zz = 0. Since 
the plate is thin, however, these quantities must be small within the plate 
if they are zero on each surface. We therefore conclude that the components 
a xz , a yz , a zz are small compared with the remaining components of the stress 
tensor everywhere in the plate. We can therefore equate them to zero and 
use this condition to determine the components of the strain tensor. 

By the general formulae (5.13), we have 



E E 

o Z x = ~ u zxy a zy = - u zyi 

l + cr l + o 

E 

° zz = Ti — ^ — ^rrft 1 ~ CT )"zz+ a ( u *x+ u vv)}- 

{l + a)[l — Za) 



(11.2) 



Equating these expressions to zero, we obtain 8u x /8z — — 8u z /8x, 
duyfdz = —8u z /dy, u zz = — o(u X x + u y y)j(\ — o). In the first two of these 
equations u z can, with sufficient accuracy, be replaced by £(x, y):du x /dz = 

— dt,/ dx, duyjdz = — dt,jdy> whence 

u x = —zd^/dx, u y = — zdt,ldy. (11.3) 

The constants of integration are put equal to zero in order to make 

u x = u y = for z = 0. 

Knowing u x and u y , we can determine all the components of the strain 
tensor : 

u xx = —zd^/dx 2 , Uyy — —zd 2 Hdy 2 , u xy = — zd^jdxdy^ 

a (d% &l\ (11.4) 

u xz = uy z = 0, lta ._^_ + _j. 

We can now calculate the free energy F per unit volume of the plate, using 
the general formula (5.10). A simple calculation gives the expression 



E 

F= z* — 

1 + 



4 w^ + £?£) 2 + r^r-^i). (11 . 5) 

crl2(l-<7)\a*2 dy*J [\dxdyl dx*dy*\l V ' 



The total free energy of the plate is obtained by integrating over the volume. 
The integration over z is from — \h to + \h t where h is the thickness of the 



46 The Equilibrium of Rods and Plates §12 



plate, and that over x, y is over the surface of the plate. The result is that 
the total free energy F pl = J*F dV of a deformed plate is 

Pl 24(l-o2)j)l\dx2 + dy2) + 

+2(, -' ) (® , -55)] d '*" (n - 6) 

the element of area can with sufficient accuracy be written as dx dy simply, 
since the deformation is small. 

Having obtained the expression for the free energy, we can regard the plate 
as being of infinitesimal thickness, i.e. as being a geometrical surface, since 
we are interested only in the form which it takes under the action of the 
applied forces, and not in the distribution of deformations inside it. The 
quantity £ is then the displacement of points on the plate, regarded as a surface, 
when it is bent. 

§12. The equation of equilibrium for a plate 

The equation of equilibrium for a plate can be derived from the condition 
that its free energy is a minimum. To do so, we must calculate the variation 
of the expression (11.6). 

We divide the integral in (11.6) into two, and vary the two parts separately. 
The first integral can be written in the form J(A £) 2 d/, where d/ = da? dy 
is a surface element and A = d 2 Jdx 2 + d 2 jdy 2 is here (and in §§13, 14) the 
two-dimensional Laplacian. Varying this integral, we have 

8hj(A0*df= JAZA^df 

= J*A£ div graded/ 

= Jdiv ( A £ grad S£) d/- Jgrad S£ • grad A £ d/. 

All the vector operators, of course, relate to the two-dimensional co-ordinate 
system (x, y). The first integral on the right can be transformed into an 
integral along a closed contour enclosing the plate :f 

Jdiv( AC gradSQ d/ = j> A£(n . gradS£) 61 

r 2S£ 

= <fA£-^d/, 

J on 

where djdn denotes differentiation along the outward normal to the contour. 

t The transformation formula for two-dimensional integrals is exactly analogous to the one for three 
dimensions. The volume element dV is replaced by the surface element d/ (a scalar), and the surface 
element df is replaced by a contour element dl multiplied by the vector n along the outward normal to 
the contour. The integral over df is converted into one over dl by replacing the operator dfd/dxt by 
fit dl. For instance, if ^ is a scalar, we have J grad <t> df = $ <£n dl. 



§12 The equation of equilibrium for a plate 

In the second integral we use the same transformation to obtain 

Jgrad8£ . grad A£ d/ = jdiv(S£ grad AC) d/- JS£A 2 £ d/ 



47 



= fa(n • grad A£) d/- J8£A 2 £ d/ 

?A^ 
dn 



JB^dl-faAHdf 



Substituting these results, we find that 



^J(A0 2 d/= JHAHdf-j 8^dl+j>A^-dl. (12.1) 

The transformation of the variation of the second integral in (11.6) is 
somewhat more lengthy. This transformation is conveniently effected in 



^U". 



Fig. 3 



components, and not in vector form. We have 

J Wdxdy/ dx 2 dy 2 ) J 



It 



= 2 



d% d 2 bt, d% d 2 S£ d 2 8£ dH 

dx 2 dy 2 



dxdy dxdy dx 2 dy 2 
The integrand can be written 



d/. 



a /dS£ dH 



B8CdH\ 
dx dy 2 / 



d /3S£ d 2 C 



a§£ dH\ 



dx\ dy dxdy dx dy 2 } ' dy\ dx dxdy dy dx 2 )' 

i.e. as the (two-dimensional) divergence of a certain vector. The variation 
can therefore be written as a contour integral: 



J Wdxdy / dx 2 dy 2 ) J \dx dxdy dy dx 2 ) 

J \ dy dxdy dx dy 2 ) 



+ 



(12.2) 



dy dxdy 
where 6 is the angle between the #-axis and the normal to the contour (Fig. 3). 



48 The Equilibrium of Rods and Plates §12 

The derivatives of S£ with respect to x and y are expressed in terms of 
its derivatives along the normal n and the tangent 1 to the contour : 



B 


a 


a 


— = 


cos — 


— sin — 


Bx 


Bn 


dl 



a . a a 

— = sin0 l-cos0 — . 

By Bn dl 

Then formula (12.2) becomes 

J WdxdyJ dx 2 By 2 ) 

r B8C( B% B 2 l B%\ 

= Ad/— 2 sin cos — - -sin 2 0— - -cos 2 0— - + 

J Bn { BxBy Bx 2 By 2 ) 

r B8U I B% BH\ BH ) 

+ <hdl— - sin0cos0 — + (co S 20 - sin20)— - . 

J Bl { \ By 2 Bx 2 / BxBy) 

The second integral may be integrated by parts. Since it is taken along a 
closed contour, the limits of integration are the same point, and we have 
simply 

r 9 ( / b 2 i a 2 £\ a 2 £ \ 

-^ 8i ;-|sin*cos*(- - _) + (c o*-.uW>— }. 

Collecting all the above expressions and multiplying by the coefficients 
shown in formula (11.6), we obtain the following final expression for the 
variation of the free energy: 

Eh 3 I r 
r [BAl Bl [B 2 l B 2 t\ 



B 2 t\ 

+ (cos 2 0-sin 2 0) 

BxBy) . 



+ 



|_Jldz(A£ + (l-<7)(2sin0cos0 

J Bn { \ BxBy 



BxBy 

B 2 £\\\ 
— sin 2 cos 2 



" -«))). (12.3) 



Bx 2 By 2 ) 

In order to derive from this the equation of equilibrium for the plate, we 
must equate to zero the sum of the variation 8F and the variation 8U of the 
potential energy of the plate due to the external forces acting on it. This 
latter variation is minus the work done by the external forces in deforming the 



§12 The equation of equilibrium for a plate 49 

plate. Let P be the external force acting on the plate, per unit areaf and 
normal to the surface. Then the work done by the external forces when the 
points on the plate are displaced a distance S£ is JPS£ d/. Thus the condition 
for the total free energy of the plate to be a minimum is 

SF pl -J>S£d/=0. (12.4) 

On the left-hand side of this equation we have both surface and contour 
integrals. The surface integral is 

The variation 8£ in this integral is arbitrary. The integral can therefore 
vanish only if the coefficient of S£ is zero, i.e. 

Eh* 

AH-P = 0. (12.5) 



12(1 -a 2 ) 



This is the equation of equilibrium for a plate bent by external forces acting 
on it J. 

The boundary conditions for this equation are obtained by equating to 
zero the contour integrals in (12.3). Here various particular cases have to be 
considered. Let us suppose that part of the edge of the plate is free, i.e. no 
external forces act on it. Then the variations S£ and hdljdn on this part of 
the edge are arbitrary, and their coefficients in the contour integrals must be 
zero. This gives the equations 

H(1-ct) — COS0SU10 + 

dn dl{ \ dx 2 dy 2 / 

+ (sin 2 0-cos 2 0) — - = 0, (12.6) 

dxdy) 

( d 2 l d 2 l d 2 t\ 

A£+(l-cr) 2 sin cos 0-^- -sin 2 0— - -cos 2 0— - = 0, (12.7) 
I dxdy dx 2 dy 2 ) 

which must hold at all free points on the edge of the plate. 

The boundary conditions (12.6) and (12.7) are very complex. Considerable 
simplifications occur when the edge of the plate is clamped or supported. If 
it is clamped (Fig. 4a), no vertical displacement is possible, and moreover no 



f The force P may be the result of body forces (e.g. the force of gravity), and is then equal to the 
integral of the body force over the thickness of the plate. 

X The coefficient D = Eh 3 jl2(l — a 9 ) in this equation is called the fiexural rigidity or cylindrical 
rigidity of the plate. 



50 The Equilibrium of Rods and Plates §12 

bending is possible at the edge. The angle through which a given part of the 
edge turns from its initial position is (for small displacements £) the derivative 
dt,Jdn. Thus the variations S£ and Sd£/dn must be zero at clamped edges, so 
that the contour integrals in (12.3) are zero identically. The boundary con- 
ditions have in this case the simple form 

I = 0, dijdn = 0. (12.8) 

The first of these expresses the fact that the edge of the plate undergoes no 
vertical displacement in the deformation, and the second that it remains 
horizontal. 



m 



(b) 

Fig. 4 



It is easy to determine the reaction forces on a plate at a point where it 
is clamped. These are equal and opposite to the forces exerted by the plate 
on its support. As we know from mechanics, the force in any direction is 
equal to the space derivative, in that direction, of the energy. In particular, 
the force exerted by the plate on its support is given by minus the derivative 
of the energy with respect to the displacement £ of the edge of the plate, and 
the reaction force by this derivative itself. The derivative in question, how- 
ever, is just the coefficient of S£ in the second integral in (12.3). Thus the 
reaction force per unit length is equal to the expression on the left of equation 
(12.6) (which, of course, is not now zero), multiplied by Eh z jl2(l — a 2 ). 

Similarly, the moment of the reaction forces is given by the expression on 
the left of equation (12.7), multiplied by the same factor. This follows at 
once from the result of mechanics that the moment of the force is equal to 
the derivative of the energy with respect to the angle through which the 
body turns. This angle is dt,/dn, so that the corresponding moment is given 
by the coefficient of BSl/dn in the third integral in (12.3). Both these expres- 
sions (that for the force and that for the moment) can be very much simplified 
by virtue of the conditions (12.8). Since £ and dt,jdn are zero everywhere on 
the edge of the plate, their tangential derivatives of all orders are zero also. 
Using this and converting the derivatives with respect to x andjy in (12.6) 
and (12.7) into those in the directions of n and 1, we obtain the following 
simple expressions for the reaction force F and the reaction moment M: 

Eh^ [dK dd a 2 n 

F= — - + , (12.9) 

12(l- CT 2)La«3 d/ drfil 

Eh* 3 2 £ 

M = 1. (12.10) 

12(l-cr 2 )d«2 v 



§12 The equation of equilibrium for a plate 51 

Another important case is that where the plate is supported (Fig. 4b), 
i.e. the edge rests on a fixed support, but is not clamped to it. In this case 
there is again no vertical displacement at the edge of the plate (i.e. on the 
line where it rests on the support), but its direction can vary. Accordingly, 
we have in (12.3) S£ = in the contour integral, but dS£/d» # 0. Hence 
only the condition (12.7) remains valid, and not (12.6). The expression on the 
left of (12.6) gives as before the reaction force at the points where the plate is 
supported ; the moment of this force is zero in equilibrium. The boundary 
condition (12.7) can be simplified by converting to the derivatives in the direc- 
tion of n and 1 and using the fact that, since £ = everywhere on the edge, the 
derivatives dljdl and d 2 £/d/ 2 are also zero. We then have the boundary 
conditions in the form 

dK dd dl 

£ = , — + a - = 0. (12.11) 

dn 2 dl dn 



PROBLEMS 

Problem 1 . Determine the deflection of a circular plate (of radius R) with clamped edges, 
placed horizontally in a gravitational field. 

Solution. We take polar co-ordinates, with the origin at the centre of the plate. The force 
on unit area of the surface of the plate is P = phg. Equation (12.5) becomes A a £ = 64j3, 
where j8 = 3pg(l—o*)J16h t E; positive values of £ correspond to displacements downward. 
Since £ is a function of r only, we can put A = r~ l d(rd/dr)/dr. The general integral is 
£ = j3r*+ar 3 +b+cr*log(r/R)+dlog(rlR). In the case in question we must put d — 0, 
since log(r[R) becomes infinite at r — 0, and c = 0, since this term gives a singularity in 
A £ at r — (corresponding to a force applied at the centre of the plate; see Problem 3). The 
constants a and b are determined from the boundary conditions £ = 0, d£/dr = for r — R. 
The result is £ = j3(.R»-r a )*. 

Problem 2. The same as Problem 1, but for a plate with supported edges. 
Solution. The boundary conditions (12.11) for a circular plate are 

d% o-d£ 
£ = 0, — -+ — - = 0. 
dr 2 r dr 

The solution is similar to that of Problem 1, and the result is 

'5 + a 



/5 + cr \ 

£ = j8(#2_ r 2)| R2-r*\. 



Problem 3. Determine the deflection of a circular plate with clamped edges when a force 
/ is applied to its centre. 

Solution. We have A a £ = everywhere except at the origin. Integration gives 
£ = ar2 + b + cr*log(rlR), 

the log r term again being omitted. The total force on the plate is equal to the force / at its 



52 The Equilibrium of Rods and Plates §12 

centre. The integral of A 2 £ over the surface of the plate must therefore be 

12(1 -cr2) 



„ f 12(1-0^) 

2tt r A2£ dr = — i if. 

J Eh? J 

o 



Hence c = 3(1— o^f/lirEh 3 . The constants a and b are determined from the boundary 
conditions. The result is 

3/(1 - ct2) 

Problem 4. The same as Problem 3, but for a plate with supported edges. 
Solution. 



£ = 



3/(l-<T2 )f-3 + o-_ o _ o , # 

4ttM3 



r3 + o- in 

(R*-r*)-2r*log — \. 

Ll + o- r J 



Problem 5. Determine the deflection of a circular plate suspended by its centre and in a 
gravitational field. 

Solution. The equation for £ and its general solution are the same as in Problem 1. 
Since the displacement at the centre is { = 0, we have c = 0. The constants a and b are 
determined from the boundary conditions (12.6) and (12.7), which are, for circular symmetry, 

dA£ d/d2£ ld£\ n d2£ ad£ 

= — — + -— = 0, —-+-— = 0. 



r dr/ ' dr 2 r dr 



dr dr\dr 2 

The result is 

R 3 + ct 



r R 3 + a! 

I = 0r2 r2 + 8#2 1og— +2ZP . 

L r 1 + o-J 



Problem 6. A thin layer (of thickness h) is torn off a body by external forces acting against 
surface tension forces at the surface of separation. With given external forces, equilibrium is 
established for a definite area of the surface separated and a definite shape of the layer 
removed (Fig. 5). Derive a formula relating the surface tension to the shape of the 
layer removed. f 



Fig. 5 

Solution. The layer removed can be regarded as a plate with one edge (the line of separa- 
tion) clamped. The bending moment on the layer is given by formula (12.10). The work 
done by this moment when the length of the separated surface increases by 8* is 

MdSC/dx = MSxdH/dx 2 

(the work of the bending force F itself is a second-order quantity). The equilibrium condition 
is that this work should be equal to the change in the surface energy, i.e. to 2a.hx, where a is 



| This problem was discussed by I. V. Obreimov (1930) in connection with a method which he 
developed for measuring the surface tension of mica. The measurements which he made by this 
method were the first direct measurements of the surface tension of solids. 



§13 Longitudinal deformations of plates 53 

the surface-tension coefficient, the factor 2 allowing for the creation of two free surfaces by 
the separation. Thus 

Eh* / 3 2 £\ 2 



o 8 )!" 



24(1 - a 2 ) \dxV 

§13. Longitudinal deformations of plates 

Longitudinal deformations occurring in the plane of the plate, and not 
resulting in any bending, form a special case of deformations of thin plates. 
Let us derive the equations of equilibrium for such deformations. 

If the plate is sufficiently thin, the deformation may be regarded as uniform 
over its thickness. The strain tensor is then a function of x and y only (the 
;cy-plane being that of the plate) and is independent of z. Longitudinal 
deformations of a plate are usually caused either by forces applied to its edges 
or by body forces in its plane. The boundary conditions on both surfaces of 
the plate are then atknjc = 0, or, since the normal vector is parallel to the 
ar-axis, at z = 0, i.e. a xz = a yz = a zz = 0. It should be noticed, however, 
that in the approximate theory given below these conditions continue to 
hold even when the external tension forces are applied to the surfaces of the 
plate, since these forces are still small compared with the resulting longi- 
tudinal internal stresses (<j xx , &yy, <*xy) in the plate. Since they are zero at 
both surfaces, the quantities a xz , a yz , a zz must be small throughout the 
thickness of the plate, and we can therefore take them as approximately zero 
everywhere in the plate. 

Equating to zero the expressions (11.2), we obtain the relations 

U ZZ = - 0(U ZX + Uyy)l{\- a), U XZ = Uy Z = 0. (13-1) 

Substituting in the general formulae (5.13), we obtain for the non-zero com- 
ponents of the stress tensor 

E " 

a xx — ~ 7\ u xx + aU yy)t 
1 — CT Z 

E 

a VV = Z -AUyy+OUxx), > (13.2) 

1— O* 

E 

°xy = ~ u xy- 

1 + or 

It should be noticed that the formal transformation 

E -> E/(l - a 2 ), a -> a/(l - a) (13.3) 

converts these expressions into those which give the relation between the 
stresses a xx , a xy , a yy and the strains u xx , u yy , u zz for a plane deformation 
(formulae (5.13) with u zz = 0). 

Having thus eliminated the displacement u Zy we can regard the plate as a 
two-dimensional medium (an "elastic plane"), of zero thickness, and take 



54 The Equilibrium of Rods and Plates §13 

the displacement vector u to be a two-dimensional vector with components 
u x and Up. If P x and P y are the components of the external body force per 
unit area of the plate, the general equations of equilibrium are 

\ ox ay J 

(da yx do yy \ 

\ ox oy I 

Substituting the expressions (13.2), we obtain the equations of equilibrium in 
the form 



J 1 d 2 u x 1 8 2 u x 1 d 2 Uy\ 

Ehl h H ) + P x = 0, 

ll-cx 2 dx 2 2(1 + a) dy 2 2(1 -a) dxdyj 

I 1 d 2 u v 1 d 2 u v 1 d 2 u x \ 

Ehl - + — + -\ + Py = 0. 

ll-a 2 By 2 2(1 + a) dx 2 2(1 -a) dxdyj 



(13.4) 



These equations can be written in the two-dimensional vector form 

grad divu-^1 - <*) curl curlu = - (1 - o 2 )F/Eh, (13.5) 

where all the vector operators are two-dimensional. 

In particular, the equation of equilibrium in the absence of body forces is 

grad divu— 1(1 — a) curl curlu = 0. (13.6) 

It differs from the equation of equilibrium for a plane deformation of a body 
infinite in the ^-direction (§7) only by the sign of the coefficient (in accordance 
with (13.3)).f As for a plane deformation, we can introduce the stress function 
defined by 

<? xx = d 2 X /dy 2 , o xy = - d 2 x/dxdy, a yy = d 2 X /dx 2 , (13.7) 

whereby we automatically satisfy the equations of equilibrium in the form 

da xx du xy da yx da yy 

1 = (J, 1 = u. 

dx dy dx dy 

The stress function, as before, satisfies the biharmonic equation, since for 
Ax we have 

A% = oxx+o-yy = E(u xx + Uyy)l(l - a) = {£/(l-cr)}divu; 

this differs only by a factor from the result for a plane deformation. 

It may be pointed out that the stress distribution in a plate deformed by 
given forces applied to its edges is independent of the elastic constants of the 



f A deformation homogeneous in the ^-direction for which a zx = a zv = a zz = everywhere is 
sometimes called a state of plane stress, as distinct from a plane deformation, for which u zx = u zy = 
u tz = everywhere. 



§13 Longitudinal deformations of plates 55 

material. For these constants appear neither in the biharmonic equation 
satisfied by the stress function, nor in the formulae (13.7) which determine 
the components o% from that function (nor, therefore, in the boundary 
conditions at the edges of the plate). 

PROBLEMS 

Problem 1 . Determine the deformation of a plane disc rotating uniformly about an axis 
through its centre perpendicular to its plane. 

Solution. The required solution differs only in the constant coefficients from the solution 
obtained in §7, Problem 5, for the plane deformation of a rotating cylinder. The radial 
displacement u r = u(r) is given by the formula 



u = r 

8E 



/3 + o- \ 

(tt/ 2 -4 



This is the expression which gives that of §7, Problem 5, if the substitution (13.3) is made. 

Problem 2. Determine the deformation of a semi-infinite plate (with a straight edge) 
under the action of a concentrated force in its plane, applied to a point on the edge. 






-. .^- \ 



Fig. 6 

Solution. We take polar co-ordinates, with the angle <f> measured from the direction of 
the applied force; it takes values from — (Jw+a) to frr— a, where a is the angle between the 
direction of the force and the normal to the edge of the plate (Fig. 6). At every point of the 
edge except that where the force is applied (the origin) we must have a^ = <r r a = 0. Using 
the expressions for a^ and a r ^ obtained in §7, Problem 11, we find that the stress function 
must therefore satisfy the conditions 

dx 1 dx 

—— = constant, --— - = constant, for <f> = -(£?! + a), (£77- -a). 
or r d(f) 

Both conditions are satisfied if x = rf(<f>). With this substitution, the biharmonic equation 

fl d / d\ a 2 \2 
[rlh-X ~dr) + ~d^) X = 

gives solutions for f(<f>) of the forms sin 4>, cos <f>, <f> sin <f>, <f> cos <j>. The first two of these lead to 
stresses which are zero identically. The solution which gives the correct value for the force 
applied at the origin is 

X = -(Flir)rtf, sincf>, a rr = -{IFIttt) cos«£, ct^ = a H = 0, (1) 

where F is the force per unit thickness of the plate. For, projecting the internal stresses on 
directions parallel and perpendicular to the force F, and integrating over a small semicircle 



56 



The Equilibrium of Rods and Plates 



§13 



centred at the origin (whose radius then tends to zero), we obtain 

J u rr r COS (j> d<f> = —F, 

a rr r sin <j> d<f> = 0, 

i.e. the values required to balance the external force applied at the origin. 

Formulae (1) determine the required stress distribution. It is purely radial: only a radial 
compression force acts on any area perpendicular to the radius. The lines of equal stress are 
the circles r = d cos <f>, which pass through the origin and whose centres lie on the line of 
action of the force F (Fig. 6). 

The components of the strain tensor are u„ = o„\E, u^ = — aa rr /E, u r ^ = 0. From these 
we find by integration (using the expressions (1.8) for the components mic in polar co- 
ordinates) the displacement vector : 

IF (l-o)F 
u r = \og(r/a) cos j> (f> sin <f>, 

7tE 7tE 

2oF 2F t , x . # (l-a)F 
u A = — —sin <f) H log(r/«) sin <j> -\ (sin ^ — <f> cos <j>). 



rE 



TT. 



E 



tE 



Here the constants of integration have been chosen so as to give zero displacement (trans- 
lation and rotation) of the plate as a whole : an arbitrarily chosen point at a distance a from the 
origin on the line of action of the force is assumed to remain fixed. 

Using the solution obtained above, we can obtain the solution for any distribution of forces 
acting on the edge of the plate (cf. §8). It is, of course, inapplicable in the immediate neigh- 
bourhood of the origin. 




Fig. 7 



Problem 3. Determine the deformation of an infinite wedge-shaped plate (of angle 2a) 
due to a force applied at its apex. 

Solution. The stress distribution is given by formulae which differ from those of Problem 
2 only in their normalisation. If the force acts along the mid-line of the wedge (Fi in Fig. 7), 
we have a„ = —(Fi cos <£)/K«+i s i n 2a), a r ^ = a^ = 0. If, on the other hand, the force 
acts perpendicular to this direction (F t in Fig. 7), then 

a rr = — (Fz cos 4>)jr{a — \ sin 2a). 

In each case the angle <f> is measured from the direction of the force. 

Problem 4. Determine the deformation of a circular disc (of radius R) compressed by two 
equal and opposite forces Fh applied at the ends of a diameter (Fig. 8). 
Solution. The solution is obtained by superposing three internal stress distributions. 



§13 



Longitudinal deformations of plates 



Two of these are 



oVr* = 



= -(IFIirrdcosfa, cr a V^i = o {1) Mi = °> 



o® W . = -(2F/77T2) COS 02, aP>r* = * (2 W, = 0, 

where n, & and r 2 , <f> 2 are the polar co-ordinates of an arbitrary point P with origins at A 
and B respectively. These are the stresses due to a normal force F applied to a point on the 
edge of a half-plane; see Problem 2. The third distribution, o(*Uk = (F/wi?)8<*, is a uniform 
extension of definite intensity. For, if the point P is on the edge of the disc, we have 
r x — 2R cos 4>i, *t = 2R cos ^, so that c^V/, = <* w r t r t = —F/irR. Since the directions of 
r x and r 2 at this point are perpendicular, we see that the first two stress distributions give 
a uniform compression on the edge of the disc. These forces can be just balanced by the 
uniform tension given by the third system, so that the edge of the disc is free from stress, as it 
should be. 




Fig. 8 

Problem 5. Determine the stress distribution in an infinite sheet with a circular aperture 
(of radius R) under uniform tension. 

Solution. The uniform tension of a continuous sheet corresponds to stresses o^xx = T, 
ff(0) v» = <* w xy = 0, where T is the tension force. These in turn correspond to the stress 
function x (0) — \Ty* = %Tr* sin 2 ^ = J7V*(1 —cos 2<f>). When there is a circular aperture 
(with the centre as the origin of polar co-ordinates r, <f>), we seek the stress function in the 
form x = X (0) +X (I) , X (1) =f(.i r )+F(r) cos 2<f>. The integral of the biharmonic equation which 
is independent of <f> is of the form /(r) = ar* log r+br'+c log r, and in the integral pro- 
portional to cos 2<f> we have F(r) = d^+ei^+glr 2 . The constants are determined by the 
conditions o^Uk = for r = oo and a rr = c f ^ = for r = R. The result is 

x (i) = iT R^-logr+ (l - |L) cos2^J, 



and the stress distribution is given by 



3i?2x 

— J cos 20}, 



o*-»r{i+--(i+— Jco.2*J, 

0^.-17(1+—-—) sin 2f 



In particular, at the edge of the aperture we have <* K 

o^q = 3T, i.e. three times the stress at infinity (cf. §?, Problem 12) 



"ft" 



T(l —2 cos 24), and for <f> = ±frr, 



58 The Equilibrium of Rods and Plates §14 

§14. Large deflections of plates 

The theory of the bending of thin plates given in §§11-13 is applicable only 
to fairly small deflections. Anticipating the result given below, it may be 
mentioned here that the condition for that theory to be applicable is that the 
deflection £ is small compared with the thickness h of the plate. Let us now 
derive the equations of equilibrium for a plate undergoing large deflections. 
The deflection £ is not now supposed small compared with h. It should be 
emphasised, however, that the deformation itself must still be small, in the 
sense that the components of the strain tensor must be small. In practice, 
this usually implies the condition £ <^ /, i.e. the deflection must be small 
compared with the dimension / of the plate. 

The bending of a plate in general involves a stretching of it.f For small 
deflections this stretching can be neglected. For large deflections, however, 
this is not possible ; there is therefore no neutral surface in a plate undergoing 
large deflections. The existence of a stretching which accompanies the 
bending is peculiar to plates, and distinguishes them from thin rods, which 
can undergo large deflections without any general stretching. This property 
of plates is a purely geometrical one. For example, let a flat circular plate be 
bent into a segment of a spherical surface. If the bending is such that the 
circumference of the plate remains constant, its diameter must increase. If the 
diameter is constant, on the other hand, the circumference must be reduced. 

The energy (11.6), which may be called the pure bending energy, is only 
the part of the total energy which arises from the non-uniformity of the 
tension and compression through the thickness of the plate, in the absence 
of any general stretching. The total energy includes also a part due to this 
general stretching; this may be called the stretching energy. 

Deformations consisting of pure bending and pure stretching have been 
considered in §§11-13. We can therefore use the results obtained in these 
sections. It is not necessary to consider the structure of the plate across its 
thickness, and we can regard it as a two-dimensional surface of negligible 
thickness. 

We first derive an expression for the strain tensor pertaining to the stretch- 
ing of a plate (regarded as a surface) which is simultaneously bent and 
stretched in its plane. Let u be the two-dimensional displacement vector 
(with components u x , u y ) for pure stretching; £, as before, denotes the trans- 
verse displacement in bending. Then the element of length d/ = y^dx 2 + dy 2 ) 
of the undeformed plate is transformed by the deformation into an element 
dl', whose square is given by d/' 2 = (dx + du x ) 2 + (dy + du y ) 2 + dl 2 . Putting 
here du x = (du x /dx) dx + (du x /dy) dy, and similarly for du y and d£, we 
obtain to within higher-order terms d/' 2 = dl 2 + 2u afi dx 0l dx fi , where the 
two-dimensional strain tensor is defined as 

i/du a du B \ i dt, dt 

— ( — -+ — -1+ • (14.1) 

2\dx s dxj 2dx a dx# 



u, 



*B 



f An exception is, for instance, the bending of a flat plate into a cylindrical surface. 



§14 Large deflections of plates 59 

(In this and the following sections, Greek suffixes take the two values x and_y ; 
as usual, summation over repeated suffixes is understood.) The terms quad- 
ratic in the derivatives of u a are here omitted; the same cannot, of course, be 
done with the derivatives of £, since there are no corresponding first-order 
terms. 

The stress tensor <r a/? due to the stretching of the plate is given by formula 
(13.2), in which u afi must be replaced by the total strain tensor given by 
formula (14.1). The pure bending energy is given by formula (11.6), and can 
be written J Yi(£) dx dy, where *Fi(£) denotes the integrand in (11.6). The 
stretching energy per unit volume of the plate is, by the general formulae, 
\u afi a ap . The energy per unit surface area is obtained by multiplying by h, 
so that the total stretching energy can be written J T 2 (« a/f ) d/, where 

T 2 = \hu^a^. (14.2) 

Thus the total free energy of a plate undergoing large deflections is 

F P i= j{Ti(0 + T 2 ( Wa/? )}d/. (14.3) 

Before deriving the equations of equilibrium, let us estimate the relative 
magnitude of the two parts of the energy. The first derivatives of £ are of 
the order of £//, where / is the dimension of the plate, and the second deriva- 
tives are of the order of £/Z 2 . Hence we see from (11.6) that Ti ~ M 3 £ 2 // 4 . 
The order of magnitude of the tensor components u afi is £ 2 // 2 , and so 
T 2 ~ £A£ 4 // 4 . A comparison shows that the neglect of T 2 in the approximate 
theory of the bending of plates is valid only if £ 2 <^ A 2 . 

The condition of minimum energy is 8F+ SU = 0, where U is the poten- 
tial energy in the field of the external forces. We shall suppose that the 
external stretching forces, if any, can be neglected in comparison with the 
bending forces. (This is always valid unless the stretching forces are very 
large, since a thin plate is much more easily bent than stretched.) Then we 
have for 8U the same expression as in §12: 8U = -J"PS£ d/, where P is the 
external force per unit area of the plate. The variation of the integral J Ti d/ 
has already been calculated in §12, and is 

r ew r 

The contour integrals in (12.3) are omitted, since they give only the boundary 
conditions on the equation of equilibrium, and not that equation itself, which 
is of interest here. 

Finally, let us calculate the variation of the integral J Y 2 d/. The variation 
must be taken both with respect to the components of the vector u and with 
respect to £. We have 

r raT 2 

SPF 2 d/= — Su^df. 
J J ou a 

The derivatives of the free energy per unit volume with respect to u xp are 



60 The Equilibrium of Rods and Plates §14 

a a/} ; hence d^l^Uafi = ha afi . Substituting also for u a/} the expression (14.1), 
we obtain 

8JV 2 df=hj* a/i 8u a0 df 

r (B8u a dhu B dt, 3S£ 3S£ d£\ 

= pUJ — -+ — -+——+— —4/; 

J l dxo dxg, dx a dxo dx a dxj 



or, by the symmetry of a a/? , 

(d8u a dSC 3£ 

Cg 0X0 

Integrating by parts, we obtain 

da^ . 3/ 3£ 

*0 



8 [V 2 d/ = h LJ^l + ^L ■£-} d/. 
J J I oa^ 3*^ 3#J 

ts, we obtain 
8 |V 2 d/ = -h f |^8« a + J-( ff JL\ 8 (;\ d /. 



The contour integrals along the circumference of the plate are again omitted. 
Collecting the above results, we have 



8F vl + 8U = (U AH-h — (o aS -^\-p)8t,-h-^8u} d/ = 0. 

P JLll2(l-o*) dxA^dxJ I dx B a l J 



In order that this relation should be satisfied identically, the coefficients of 
81 and 8u x must each be zero. Thus we obtain the equations 



M3 3 / 3£ 

AH-h— ' 

12(1 -a 2 ) dx 



do^dx, = 0. (14.5) 

The unknown functions here are the two components u Xl u y of the vector 
u and the transverse displacement £. The solution of the equations gives both 
the form of the bent plate (i.e. the function £(#, y)) and the extension resulting 
from the bending. Equations (14.4) and (14.5) can be somewhat simplified 
by introducing the function x related to a afi by (13.7). Equation (14.4) then 
becomes 



/3 2 v 3 2 £ 3 2 v 3 2 £ 3 2 v 3 2 £ \ 

-hi— £— + ——- 2— *-— I =P. (14.6) 

\ 3V 2 dx 2 dx 2 dy 2 dxdy dxdyf 



EW> (d 2 x d 2 £ d 2 X dH &X d 2 £ 

12(1 — a 2 ) \ dy 2 dx 2 dx 2 dy 2 dxdy dxdy/ 



Equations (14.5) are satisfied automatically by the expressions (13.7). Hence 
another equation is needed ; this can be obtained by eliminating u a from the 
relations (13.7) and (13.2). 

To do this, we proceed as follows. We express w a/ff in terms of a a/? , obtaining 
from (13.2) 

U X X = (VXX— 0-Oyy)IE, Uyy — {Oyy— OO XX )/E, U X y = (1 + <j)(J X y/E. 



§14 Large deflections of plates 61 

Substituting here the expression (14.1) for u a/s , and (13.7) for o a p, we find 
the equations 

dx + 2\dx) ~ E\dy 2 ° dx 2 )' 

dy + 2\dy) E\dx 2 ° dy 2 )' 
du x du y d^dt, _ 2(1 + a) d 2 x 



dy dx dx dy E dxdy 

We take d 2 jdy 2 of the first, d 2 /dx 2 of the second, — d 2 /dxdy of the third, and 
add. The terms in u x and u y then cancel, and we have the equation 

2 jdH d% i d 2 £\ 2 \ _ 

\ dx 2 dy 2 \dxdyl ) 

Equations (14.6) and (14.7) form a complete system of equations for large 
deflections of thin plates (A. Foppl 1907). These equations are very compli- 
cated, and cannot be solved exactly, even in very simple cases. It should be 
noticed that they are non-linear. 

We may mention briefly a particular case of deformations of thin plates, 
that of membranes. A membrane is a thin plate subject to large external 
stretching forces applied at its circumference. In this case we can neglect 
the additional longitudinal stresses caused by bending of the plate, and 
therefore suppose that the components of the tensor cr a/? are simply equal to 
the constant external stretching forces. In equation (14.4) we can then 
neglect the first term in comparison with the second, and we obtain the 
equation of equilibrium 

hcr a ,~?^-+P=0 t (14.8) 

with the boundary condition that £ = at the edge of the membrane. This 
is a linear equation. The case of isotropic stretching, when the extension of 
the membrane is the same in all directions, is particularly simple. Let T be 
the absolute magnitude of the stretching force per unit length of the edge of 
the membrane. Then ha afi = T8 afi , and we obtain the equation of equili- 
brium in the form 

TAt + P = 0. (14.9) 

PROBLEMS 
Problem 1. Determine the deflection of a plate as a function of the force on it when 

Solution. An estimate of the terms in equation (14.7) shows that x ~ EC*. For { ^> h, 
the first term in (14.6) is small compared with the second, which is of the order of magnitude 
htxll* ~ EH* 1 1* (I being the dimension of the plate). If this is comparable with the external 



62 The Equilibrium of Rods and Plates §15 

force P, we have £ ~ (l l P{Eh)l. Hence, in particular, we see that £ is proportional to the 
cube root of the force. 

Problem 2. Determine the deformation of a circular membrane (of radius R) placed 
horizontally in a gravitational field. 

Solution. We have P = pgh; in polar co-ordinates, (14.9) becomes 



r dr \ or! 



1 d / d£\ P gh_ 

T' 



The solution finite for r — and zero for r = R is £ = pgh(R 2 —r 2 )/4T. 

§15. Deformations of shells 

In discussing hitherto the deformations of thin plates, we have always 
assumed that the plate is flat in its undeformed state. However, deformations 
of plates which are curved in the undeformed state (called shells) have 
properties which are fundamentally different from those of the deformations 
of flat plates. 

The stretching which accompanies the bending of a flat plate is a second- 
order effect in comparison with the bending deflection itself. This is seen, 
for example, from the fact that the strain tensor (14.1), which gives this 
stretching, is quadratic in £. The situation is entirely different in the defor- 
mation of shells : here the stretching is a first-order effect, and therefore is 
important even for small bending deflections. This property is most easily 
seen from a simple example, that of the uniform stretching of a spherical 
shell. If every point undergoes the same radial displacement £, the length 
of the equator increases by Inl,. The relative extension is IttX^IttR — t,jR f 
and hence the strain tensor also is proportional to the first power of £. This 
effect tends to zero as R -> oo, i.e. as the curvature tends to zero, and is 
therefore due to the curvature of the shell. 

Let i? be the order of magnitude of the radius of curvature of the shell, 
which is usually of the same order as its dimension. Then the strain tensor 
for the stretching which accompanies the bending is of the order of £jR, 
the corresponding stress tensor is ~ E£jR, and the deformation energy per 
unit area is, by (14.2), of the order of Eh{ljRf. The pure bending energy, on 
the other hand, is of the order of Eh^jR^, as before. We see that the ratio of 
the two is of the order of (R/h) 2 , i.e. it is very large. It should be emphasised 
that this is true whatever the ratio of the bending deflection £ to the thickness 
h, whereas in the bending of flat plates the stretching was important only 
for £ £ h. 

In some cases there may be a special type of bending of the shell in 
which no stretching occurs. For example, a cylindrical shell (open at both 
ends) can be deformed without stretching if all the generators remain parallel 
(i.e. if the shell is, as it were, compressed along some generator). Such 
deformations without stretching are geometrically possible if the shell has 
free edges (i.e. is not closed) or if it is closed but its curvature has opposite 



§15 Deformations of shells 63 

signs at different points. For example, a closed spherical shell cannot be 
bent without being stretched, but if a hole is cut in it (the edge of the hole 
not being fixed), then such a deformation becomes possible. Since the pure 
bending energy is small compared with the stretching energy, it is clear that, 
if any given shell permits deformation without stretching, then such defor- 
mations will, in general, actually occur when arbitrary external forces act on 
the shell. The requirement that the bending is unaccompanied by stretching 
places considerable restrictions on the possible displacements u u . These 
restrictions are purely geometrical, and can be expressed as differential 
equations, which must be contained in the complete system of equilibrium 
equations for such deformations. We shall not pause to discuss this question 
further. 

If, however, the deformation of the shell involves stretching, then the 
tensile stresses are in general large compared with the bending stresses, 
which may be neglected. Shells for which this is done are called membranes. 

The stretching energy of a shell can be calculated as the integral 

F P i = ^hju xfi a afi df (15.1) 

taken over the surface. Here w a/ff (oc, /? = 1, 2) is the two-dimensional strain 
tensor in the appropriate curvilinear co-ordinates, and the stress tensor a a p 
is related to u a/3 by formulae (13.2), which can be written, in two-dimensional 
tensor notation, as 

(V = E[(l - a)u ap + o8 a/) u Yy ]l(l - CT 2). (15.2) 

A case requiring special consideration is that where the shell is subjected 
to the action of forces applied to points or lines on the surface and directed 
through the shell. These may be, in particular, the reaction forces on the shell 
at points (or lines) where it is fixed. The concentrated forces result in a 
bending of the shell in small regions near the points where they are applied; 
let d be the dimension of such a region for a force /applied at a point (so that 
its area is of the order of d 2 ). Since the deflection £ varies considerably over a 
distance d, the bending energy per unit area is of the order of Eh z l 2 /d' i , and the 
total bending energy (over an area ~ d 2 ) is of the order oiEtfitpjd 2 . The strain 
tensor for the stretching is again ~ £/i?, and the total stretching energy due to 
the concentrated forces is ~ Eh£, 2 d 2 /R 2 . Since the bending energy increases 
and the stretching energy decreases with decreasing d, it is clear that both 
energies must be taken into account in determining the deformation near the 
point of application of the forces. The size d of the region of bending is given 
in order of magnitude by the condition that the sum of these energies is a 
minimum, whence 

d ~ y/(hR). (15.3) 



64 



The Equilibrium of Rods and Plates 



§15 



The energy ~ Eh 2 £, 2 /R. Varying this with respect to £ and equating the result 
to the work done by the force/, we find the deflection £ ~ fR/Eh 2 . 

However, if the forces acting on the shell are sufficiently large, the shape of 
the shell may be considerably changed by bulges which form in it. The 
determination of the deformation as a function of the applied loads requires 
special investigation in this unusual case.f 

Let a convex shell (with edges fixed in such a way that it is geometrically 
rigid) be subjected to the action of a large concentrated force / along the in- 
ward normal. For simplicity we shall assume that the shell is part of a sphere 
of radius R. The region of the bulge will be a spherical cap which is almost a 
mirror image of its original shape (Fig. 9 shows a meridional section of the 
shell). The problem is to determine the size of the bulge as a function of the 
force. 

The major part of the elastic energy is concentrated in a narrow strip near 
the edge of the bulge, where the bending of the shell is relatively large ; we 
shall call this the bending strip and denote its width by d. This energy may be 
estimated, assuming that the radius r of the bulge region is much less than R, 
so that the angle a <^ 1 (Fig. 9). Then r — R sin a ~ Ra, and the depth of the 




Fig. 9 



bulge H = 2R(1 — cos a) ~ jRa 2 . Let £ denote the displacement of points on 
the shell in the bending strip. Just as previously, we find that the energies of 
bending along the meridian and of stretching along the circle of latitude:}: per 
unit surface area are respectively, in order of magnitude, Eh 3 t, 2 ld 4 and 



t The results given below are due to A. V. Pogorelov (1960). A more precise analysis of the problem 
together with some similar ones is given in his book Teoriya obolochek pri zakriticheskikh deformatsiyakh 
{Theory of Shells at Supercritical Deformations), Moscow 1965. 

% The curvature of the shell does not affect the bending along the meridian in the first approximation, 
so that this bending occurs without any general stretching along the meridian, as in the cylindrical 
bending of a flat plate. 



§15 Deformations of shells 65 

Ehl 2 /R 2 . The order of magnitude of the displacement £ is in this case deter- 
mined geometrically: the direction of the meridian changes by an angle ~a 
over the width d, and so £ ~ <xd ~ rd\R. Multiplying by the area of the bend- 
ing strip (~n/), we obtain the energies Eh^jR 2 d and EhdW/R*. The condi- 
tion for their sum to be a minimum again gives d~-\/{hR), and the total 
elastic energy is then ~ Er 3 (h/R) 5/2 , orf 

constant x Eh™. H^ l2 jR. (15.4) 

In this derivation it has been assumed that d<^r; formula (15.4) is therefore 
valid if the condition^ 

Rhjr*<^\ (15.5) 

holds. 

The required relation between the depth of the bulge H and the applied 
force / is obtained by equating/ to the derivative of the energy (15.4) with 
respect to H. Thus we find 

■ H~f 2 R 2 \E 2 hs. ■ . . -, , : ; . (15.6) 

It should be noticed that this relation is non-linear. 

Finally, let the deformation (bulge) of the shell occur under a uniform 
external pressure^). In this case the work done ispAV, where AV~ Hr 2 ~ H 2 R 
is the change in the volume within the shell when the bulge occurs. Equating 
to zero the derivative with respect to H of the total free energy (the difference 
between the elastic energy (15.4) and this work), we obtain 

H~h*E 2 lR*p 2 . (15.7) 

The inverse variation (H increasing when p decreases) shows that in this case 
the bulge is unstable. The value of H given by formula (15.7) corresponds to 
unstable equilibrium for a given p: bulges with larger values of H grow of 
their own accord, while smaller ones shrink (it is easy to verify that (15.7) 
corresponds to a maximum and not a minimum of the total free energy). 
There is a critical value p cr of the external load beyond which even small 
changes in the shape of the shell increases in size spontaneously. This value 
may be defined as that which gives H~h in (15.7): 

p CT ~Eh 2 IR 2 . (15.8) 

We shall add to the above brief account of shell theory only a few simple 
examples in the following Problems. 



t A more accurate calculation shows that the constant coefficient is 1.2 (1— <r 2 ) _3 ' 4 . 

t When a bulge is formed, the outer layers of a spherical segment become the inner ones and are 
therefore compressed, while the inner layers become the outer ones and are stretched. The relative 
extension (or compression) ~ hjR, and so the corresponding total energy in the region of the bulge 
~ E(h/R) 2 kr*. With the condition (15.5) it is in fact small in comparison with the energy (15.4) in 
the bending strip. 

-5* 



66 



The Equilibrium of Rods and Plates 



§15 



PROBLEMS 
Problem 1 . Derive the equations of equilibrium for a spherical shell (of radius R) deformed 
symmetrically about an axis through its centre. 

Solution. We take as two-dimensional co-ordinates on the surface of the shell the angles 
0, <f> in a system of spherical polar co-ordinates, whose origin is at the centre of the sphere and 
polar axis along the axis of symmetry of the deformed shell. 

Let P r be the external radial force per unit surface area. This force must be balanced by a 
radial resultant of internal stresses acting tangentially on an element of the shell. The con- 
dition is 



h(a,,+ a e0 )IR = P r 



(1) 



This equation is exactly analogous to Laplace's equation for the pressure difference between 
two media caused by surface tension at the surface of separation. 

Next, let Qz(0) be the resultant of all external forces on the part of the shell lying above the 
co-latitude 9; this resultant is along the polar axis. The force Qz(9) must be balanced by the 
projection on the polar axis of the stresses lirRhagg sin 9 acting on the cross-section 2irRh sin 9 
of the shell at that latitude. Hence 



2TrRha ee sin 2 = Q z (6). 



(2) 



Equations (1) and (2) determine the stress distribution, and the strain tensor is then given 
by the formulae 



Finally, the displacement vector is obtained from the equations 



= 0. 



u ee 



1 / du e 

— I Yu r 

R\dd 



)■ 



1 
U d( p — — (u d cotd + u r ). 
R 



(3) 



(4) 



Problem 2. Determine the deformation under its own weight of a hemispherical shell 
convex upwards, the edge of which moves freely on a horizontal support (Fig. 10). 




Fig. 10 

Solution. We have P r = — pgh cos 9, Q z = —2iTR 2 pgh(l —cos 9); Q z is the total weight 
of the shell above the circle of co-latitude 9. From (1) and (2) of Problem 1 we find 



a eo - - 



Rpg 



1 + COS 0' 



a rr 



= Rpg\ cos 01. 

^\1+COS0 / 



From (3) we calculate u^ and ugg, and then obtain ug and u r from (4); the constant in the 
integration of the first equation (4) is chosen so that for 9 = \n we have Ug — 0. The result 



Ug = 



R 2 pg(l + a){ COS0 



E 



1 + cos 6 



+ log(l + cos0) sin0, 



U r = (1 



2+. 



/ i 

-cos 0— cos log(l + COS 0) J. 



E { 1 + ff 

The value of u T for = \ti gives the horizontal displacement of the support. 



§15 



Deformations of Shells 



67 



Problem 3. Determine the deformation of a hemispherical shell with clamped edges, 
convex downwards and filled with liquid (Fig. 11); the weight of the shell itself can be 
neglected in comparison with that of the liquid. 




Fig. 11 



Solution. We have 

P r = P ogR cos d, P* = 0, 

f 2 

Q z = 2ttR? P r cos sin d0 = -nR?pog(l - cos30), 



where p is the density of the liquid. We find from (1) and (2) of Problem 1 



a ee 



R 2 pog 1 — cos 3 
~3h sin 2 ' 

The displacements are 



R 2 Pog -l + 3cos0-2cos 3 



2h 



sin 2 



R?pog(l + a) . r cos0 

u d = : sin 



U r = 



3Eh 

R3f >0g(l + cr) 

3Eh 



r cos0 I 

— -+log(l + cos0) , 

LI + cos 6 J 



r 3 cos 1 
cos01og(l + cos0)-l + . 



For 9 — $n, u r is not zero as it should be. This means that the shell is actually so severely bent 
near the clamped edge that the above solution is invalid. 

Problem 4. A shell in the form of a spherical cap rests on a fixed support (Fig. 12). Deter- 
mine the bending resulting from the weight Q of the shell. 




Fig. 12 



68 The Equilibrium of Rods and Plates §16 

Solution. The main deformation occurs near the edge, which is bent as shown by the 
dashed line in Fig. 12. The displacement ug is small compared with the radial displacement 
M r = £• Since £ decreases rapidly as we move away from the supported edge, the deformation 
can be regarded as that of a long flat plate (of length 2irR sin a). This deformation is composed 
of a bending and a stretching of the plate. The relative extension at each point is Z/R (R 
being the radius of the shell), and therefore the stretching energy is Et?l2R* per unit volume. 
Using as the independent variable the distance x from the line of support, we have for the 
total stretching energy 

Eh 



eh r 
jPipi = 2ttR sin a—— £ 2 dx, 



The bending energy is 



Effi f/d2A2 

i<2 P i = 2ttR sin a — I dx. 

24(1-0*) J \dW 

Varying the sum F pl = F lp i +F 2 p i with respect to £, we obtain 

d*£ 12(1-0*) 

— -+— - X = 0. 

d*4 h*R* 

For x -*■ oo, £ must tend to zero, and for * = we must have the boundary conditions of 
zero moment of the forces (£" = 0) and equality of the normal force and the corresponding 
component of the force of gravity: 

27ri?sina £'" = Qcosa. 

12(1 - a*) * 

The solution which satisfies these conditions is £ = Ae~ KX cos kx, where 

_ r3(l-o2)-|i/4 _ 0cot a r3i?2(i-o-2)-|i/4 
L A 2 #2 J ' A " Eh [ 8W J ' 

The bending of the shell is 

d = £(0) cos a = A cos a. 

§16. Torsion of rods 

Let us now consider the deformation of thin rods. This differs from all 
the cases hitherto considered, in that the displacement vector u may be large 
even for small strains, i.e. when the tensor u^ is small.f For example, when 
a long thin rod is slightly bent, its ends may move a considerable distance, 
even though the relative displacements of neighbouring points in the rod 
are small. 

There are two types of deformation of a rod which may be accompanied by 
a large displacement of certain parts of it. One of these consists in bending 



f The only exception is a simple extension of a rod without change of shape, in which case the 
vector u is always small if the tensor wjfc is small, i.e. if the extension is small. 



§16 Torsion of rods 69 

the rod, and the other in twisting it. We shall begin by considering the latter 
case. 

A torsional deformation is one in which, although the rod remains straight, 
each transverse section is rotated through some angle relative to those below 
it. If the rod is long, even a slight torsion causes sufficiently distant cross- 
sections to turn through large angles. The generators on the sides of the rod, 
which are parallel to its axis, become helical in form under torsion. 

Let us consider a thin straight rod of arbitrary cross-section. We take a 
co-ordinate system with the sr-axis along the axis of the rod and the origin 
somewhere inside the rod. We use also the torsion angle t, which is the angle 
of rotation per unit length of the rod. This means that two neighbouring 
cross-sections at a distance dsr will rotate through a relative angle d^ = t dz 
(so that t = d<f>(dz). The torsional deformation itself, i.e. the relative dis- 
placement of adjoining parts of the rod, is assumed small. The condition 
for this to be so is that the relative angle turned through by cross-sections 
of the rod at a distance apart of the order of its transverse dimension R is 
small, i.e. 

rR < 1. (16.1) 

Let us examine a small portion of the length of the rod near the origin, and 
determine the displacements u of the points of the rod in that portion. As 
the undisplaced cross-section we take that given by the «y-plane. When a 
radius vector r turns through a small angle 8<f>, the displacement of its end 
is given by 

8r = 5«J>Xr, (16.2) 

where S<|> is a vector whose magnitude is the angle of rotation and whose 
direction is that of the axis of rotation. In the present case, the rotation is 
about the #-axis, and for points of co-ordinate z the angle of rotation relative 
to the ary-plane is rz (since t can be regarded as a constant in some region 
near the origin). Then formula (16.2) gives for the components u x , u y of the 
displacement vector 

u x = -Tzy, u y = tzx. (16.3) 

When the rod is twisted, the points in it in general undergo a displacement 
along the sr-axis also. Since for t = this displacement is zero, it may be 
supposed proportional to t when t is small. Thus 

u z = nls( X) y), (16.4) 

where i/r(#, y) is some function of x and y, called the torsion function. As a 
result of the deformation described by formulae (16.3) and (16.4), each cross- 
section of the rod rotates about the #-axis, and also becomes curved instead 
of plane. It should be noted that, by taking the origin at a particular point in 
the #y-plane, we "fix" a certain point in the cross-section of the rod in such a 



70 The Equilibrium of Rods and Plates §16 

way that it cannot move in that plane (but it can move in the ^-direction). 
A different choice of origin would not, of course, affect the torsional deforma- 
tion itself, but would give only an unimportant displacement of the rod as a 
whole. 

Knowing u, we can find the components of the strain tensor. Since u is 
small in the region under consideration, we can use the formula 



uoc = \{duijdxk+du]cjdxi). 



The result is 



Uxx — u yy — u xy — u zz — 0, 
Uxz = $t(— - yj, u yz = \r\— + x\. (16.5) 

It should be noticed that uu = ; in other words, torsion does not result in 
a change in volume, i.e. it is a pure shear deformation. 
For the components of the stress tensor we find 

&xx — a yy = a zz = a xy = 0, 

a xz = 2fiu zz = firi- yu o yz = 2[xu yz = /*rl— + *J. (16.6) 

Here it is more convenient to use the modulus of rigidity /x in place of E and 
a. Since only a xz and a yz are different from zero, the general equations 
of equilibrium dancjdxjc = reduce to 

i^+^fl = 0. (16.7) 

dx dy 

Substituting (16.6), we find that the torsion function must satisfy the equation 

A«A = 0, (16.8) 

where A is the two-dimensional Laplacian. 

It is rather more convenient, however, to use a different auxiliary function 
X ( x > y)> defined by 

a xz = 2firdxldy, Oyz = -2firdx/dx; (16.9) 

this function satisfies more convenient boundary conditions on the circum- 
ference of the rod (see below). Comparing (16.9) and (16.6), we obtain 

*± = y + 2 d I, d l=- x -2% (16.10) 

dx dy dy dx 

Differentiating the first of these with respect to y, the second with respect to 



§16 Torsion of rods 71 

x, and subtracting, we obtain for the function x the equation 

AX=-1- (16.11) 

To determine the boundary conditions on the surface of the rod, we note 
that, since the rod is thin, the external forces on its sides must be small com- 
pared with the internal stresses in the rod, and can therefore be put equal to 
zero in seeking the boundary conditions. This fact is exactly analogous to 
what we found in discussing the bending of thin plates. Thus we must have 
oiknjc = on the sides of the rod; since the ^-direction is along the axis, 
rig = 0, and this equation becomes 

aZxHx + Ozytly = 0. 

Substituting (16.9), we obtain 

d X d X ~ 

— n x n y — "• 

dy dx 

The components of the vector normal to a plane contour (the circumference 
of the rod) are n x — — dyjdl, n y = dxjdl, where x and y are co-ordinates 
of points on the contour and d/ is an element of arc. Thus we have 

dx ^x 

— dx H dy = dx = 0, 

dx dy 

whence x — constant, i.e. x ls constant on the circumference. Since only 
the derivatives of the function x appear in the definitions (16.9), it is clear 
that any constant may be added to x- If the cross-section is singly connected, 
we can therefore use, without loss of generality, the boundary condition 

X = (16.12) 

on equation (16.1 l).f 




Fig. 13 
For a multiply connected cross-section, however, x v*d\ have different 
constant values on each of the closed curves bounding the cross-section. 



f The problem of determining the torsion deformation from equation (16.11) with the boundary 
condition (16.12) is formally identical with that of determining the bending of a uniformly loaded 
plane membrane from equation (14.9). 

It is useful to note also an analogy with fluid mechanics: an equation of the form (16.11) determines 
the velocity distribution v(x, y) for a viscous fluid in a pipe, and the boundary condition (16.12) 
corresponds to the condition v = at the fixed walls of the pipe (see Fluid Mechanics, §17). 



72 The Equilibrium of Rods and Plates §16 

Hence we can put x = on only one of these curves, for instance the outer- 
most (Co in Fig. 13). The values of x on the remaining bounding curves are 
found from conditions which are a consequence of the one-valuedness of the 
displacement u z = rifj(x,y) as a function of the co-ordinates. For, since the 
torsion function tfj(x, y) is one- valued, the integral of its differential d^r round 
a closed contour must be zero. Using the relations (16.10), we therefore 
have 



j**-§Q**+%*y) 



= — 2 (1) I — dy dx\ — 2 (p (# dy—y dx) 

= 0, 



or 



— d/= -S, (16.13) 

dn 

where dx/dn is the derivative of the function x along the outward normal 
to the curve, and S the area enclosed by the curve. Applying (16.13) to each 
of the closed curves C\, C2, ••• , we obtain the required conditions. 

Let us determine the free energy of a rod under torsion. The energy per 
unit volume is 

F = \0UcUilc = O xz U xz + OyzUy Z = ( O xz 2 + Oy z 2 )/2fl 

or, substituting (16.9), 

where grad denotes the two-dimensional gradient. The torsional energy 
per unit length of the rod is obtained by integrating over the cross-section 
of the rod, i.e. it is £CV 2 , where the constant C = 4/x J (grad x) 2 d/, and is 
called the torsional rigidity of the rod. The total elastic energy of the rod is 
equal to the integral 

F rod = i J Ct2 d*, (16.14) 

taken along its length. 

Putting 

(gradx) 2 = div(xgradx)-xAx = divfo grad*) + x 

and transforming the integral of the first term into one along the circumference 
of the rod, we obtain 

C = 4/x^x^d/+4^Jxd/. (16.15) 



§16 Torsion of rods 73 

If the cross-section is singly connected, the first term vanishes by the 
boundary condition x = 0> leaving 

C = Aii J" x dx d y- (16.16) 

For a multiply connected cross-section (Fig. 13), we put x = on the outer 
boundary C and denote by xfr the constant values of x on the inner boun- 
daries Ck, obtaining by (16.13) 

c = 4/*2x*s*+4/* jx <** 4y; ( 16 - 17 ) 

* 

it should be remembered that, in integrating in the first term in (16.15), we go 
anti-clockwise round the contour Co and clockwise round all the others. 

Let us consider now a more usual case of torsion, where one of the ends of 
the rod is held fixed and the external forces are applied only to the other end. 
These forces are such that they cause only a twisting of the rod, and no other 
deformation such as bending. In other words, they form a couple which twists 
the rod about its axis. The moment of this couple will be denoted by M. 

We should expect that, in such a case, the torsion angle t is constant 
along the rod. This can be seen, for example, from the condition that the free 
energy of the rod is a minimum in equilibrium. The total energy of a de- 
formed rod is equal to the sum F T0 <i+ U, where U is the potential energy 
due to the action of the external forces. Substituting in (16.14) r = df/dz 
and varying with respect to the angle <f>, we find 

or, integrating by parts, 

_ f C-^8<f>dz + 8U+[Cr8<f>] = 0. 
J dsr 

The last term on the left is the difference of the values at the limits of inte- 
gration, i.e. at the ends of the rod. One of these ends, say the lower one, is 
fixed, so that 8cf> = there. The variation 8 U of the potential energy is 
minus the work done by the external forces in rotation through an angle 8(f). 
As we know from mechanics, the work done by a couple in such a rotation 
is equal to the product M8<f> of the angle of rotation and the moment of the 
couple. Since there are no other external forces, 8U = -M8<f>, and we 
have 

C^8<f>dz+[8<f>(-M+Cr)] = 0. (16.18) 



i< 



dz 
The second term on the left has its value at the upper end of the rod. In the 



74 The Equilibrium of Rods and Plates §16 

integral over z, the variation 8<f> is arbitrary, and so we must have 

Cdr/dz = 0, 
i.e. 

t = constant. (16.19) 

Thus the torsion angle is constant along the rod. The total angle of rotation 
of the upper end of the rod relative to the lower end is rl, where / is the length 
of the rod. 

In equation (16.18), the second term also must be zero, and we obtain the 
following expression for the constant torsion angle : 

r = M/C. (16.20) 

PROBLEMS 

Problem 1. Determine the torsional rigidity of a rod whose cross-section is a circle of 
radius R. 

Solution. The solutions of Problems 1-4 are formally identical with those of problems of 
the motion of a viscous fluid in a pipe of corresponding cross-section (see the last footnote 
to this section). The discharge Q is here represented by C. 

For a rod of circular cross-section we have, taking the origin at the centre of the circle, 
X = i(R 2 — x 2 — V 2 ), and the torsional rigidity is C = £/«r.R 4 . For the function tf> we have, 
from (16.10), i// = constant. A constant xjt, however, corresponds by (16.4) to a simple dis- 
placement of the whole rod along the s-axis, and so we can suppose that tfi = 0. Thus the 
transverse sections of a circular rod undergoing torsion remain plane. 

Problem 2. The same as Problem 1, but for an elliptical cross-section of semi-axes a 
and b. 

Solution. The torsional rigidity is C = 7r/xa s 6 s /(a s +6 2 ). The distribution of longitudinal 
displacements is given by the torsion function ifi = (b* — a 2 )*y/(6*-f-a 2 ), where the co-ordinate 
axes coincide with those of the ellipse. 

Problem 3. The same as Problem 1, but for an equilateral triangular cross-section of 
side a. 

Solution. The torsional rigidity is C = \/3 /jui*J80. The torsion function is 

i/j = y{xy/Z +y){x-y/3 —y)]6a 

the origin being at the centre of the triangle and the x-axis along an altitude. 

Problem 4. The same as Problem 1, but for a rod in the form of a long thin plate (of 
width d and thickness h<^d). 

Solution. The problem is equivalent to that of viscous fluid flow between plane parallel 
walls. The result is that C = ^fidh s . 

Problem 5. The same as Problem 1, but for a cylindrical pipe of internal and external 
radii i? x and R 2 respectively. 

Solution. The function x = i(-^2 a — r 2 ) (in polar co-ordinates) satisfies the condition 
(16.13) at both boundaries of the annular cross-section of the pipe. From formula (16.17) 
we then find C = i/wKiV — i?i 4 ). 

Problem 6. The same as Problem 1, but for a thin- walled pipe of arbitrary cross-section. 

Solution. Since the walls are thin, we can assume that x varies through the wall thickness 

h, from zero on one side to Xi on the other, according to the linear law x = XiVlh (y being a 



§17 Bending of rods 75 

co-ordinate measured through the wall). Then the condition (16.13) gives XiUh = S, 
where L is the perimeter of the pipe cross-section and S the area which it encloses. The 
second term in the expression (16.17) is small compared with the first, and we obtain 
C = 4hS 2 ^IL. If the pipe is cut longitudinally along a generator, the torsional rigidity falls 
sharply, becoming (by the result of Problem 4) C = \ixLtf. 

§17. Bending of rods 

A bent rod is stretched at some points and compressed at others. Lines on 
the convex side of the bent rod are extended, and those on the concave side 
are compressed. As with plates, there is a neutral surface in the rod, which 
undergoes neither extension nor compression. It separates the region of 
compression from the region of extension. 

Let us begin by investigating a bending deformation in a small portion of 
the length of the rod, where the bending may be supposed slight; by this we 
here mean that not only the strain tensor but also the magnitudes of the dis- 
placements of points in the rod are small. We take a co-ordinate system with 
the origin on the neutral surface in the portion considered, and the sr-axis 
parallel to the axis of the undeformed rod. Let the bending occur in the 
•sw-plane.f 

As in the bending of plates and the twisting of rods, the external forces on 
the sides of a thin bent rod are small compared with the internal stresses, and 
can be taken as zero in determining the boundary conditions at the sides of the 
rod. Thus we have everywhere on the sides of the rod a^njc = 0, or, since 
n z = 0, a xx n x + a xy n y = 0, and similarly for i = y, z. We take a point on 
the circumference of a cross-section for which the normal n is parallel to the 
ar-axis. There will be another such point somewhere on the opposite side 
of the rod. At both these points % = 0, and the above equation gives 
a xx = 0. Since the rod is thin, however, a xx must be small everywhere in the 
cross-section if it vanishes on either side. We can therefore put a xx = 
everywhere in the rod. In a similar manner, it can be seen that all the com- 
ponents of the stress tensor except a zz must be zero. That is, in the bending 
of a thin rod only the extension (or compression) component of the internal 
stress tensor is large. A deformation in which only the component a zz of 
the stress tensor is non-zero is just a simple extension or compression (§5). 
Thus there is a simple extension or compression in every volume element of 
a bent rod. The amount of this varies, of course, from point to point in every 
cross-section, and so the whole rod is bent. 

It is easy to determine the relative extension at any point in the rod. Let 
us consider an element of length dz parallel to the axis of the rod and near 
the origin. On bending, the length of this element becomes dz'. The only 
elements which remain unchanged are those which lie in the neutral surface. 
Let R be the radius of curvature of the neutral surface near the origin. The 

f In a rod undergoing only small deflections we can suppose that the bending occurs in a single 
plane. This follows from the result of differential geometry that the deviation of a slightly bent curve 
from a plane (its torsion) is of a higher order of smallness than its curvature. 



76 The Equilibrium of Rods and Plates §17 

lengths dz and dz' can be regarded as elements of arcs of circles whose radii 
are respectively R and R + x, x being the co-ordinate of the point where 
dz f lies. Hence 

, , R + x , / x\ 

The relative extension is therefore (dz' — dz)/dz = xjR. 

The relative extension of the element dz, however, is equal to the com- 
ponent u zz of the strain tensor. Thus 

u zz = x/R. (17.1) 

We can now find a zz by using the relation a zz - Eu zz which holds for a 
simple extension. This gives 

a zz = ExjR. (17.2) 

The position of the neutral surface in a bent rod has now to be determined. 
This can be done from the condition that the deformation considered must 
be pure bending, with no general extension or compression of the rod. The 
total internal stress force on a cross-section of the rod must therefore be 
zero, i.e. the integral J a zz df, taken over a cross-section, must vanish. Using 
the expression (17.2) for a zz , we obtain the condition 

jxdf=0. (17.3) 

We can now bring in the centre of mass of the cross-section, which is that 
of a uniform flat disc of the same shape. The co-ordinates of the centre of 
mass are, as we know, given by the integrals J x d/// df, $y d// / df. Thus the 
condition (17.3) signifies that, in a co-ordinate system with the origin in the 
neutral surface, the x co-ordinate of the centre of mass of any cross-section 
is zero. The neutral surface therefore passes through the centres of mass 
of the cross-sections of the rod. 

Two components of the strain tensor besides u zz are non-zero, since for a 
simple extension we have u X x = tt yy = —au zz . Knowing the strain tensor, 
we can easily find the displacement also : 

u zz — du z Jdz = xjR, duzjdx = du y /dy = — ax/R, 

du z du x du x du v du u du z 

— -+ — - = 0, +^=0, — - + — ~ = 0. 

dx dz By dx dz dy 

Integration of these equations gives the following expressions for the com- 
ponents of the displacement : 

"^"a^ 2 "^ 2- ^' (17.4) 

u y = — axyJR, u z = xz/R. 

The constants of integration have been put equal to zero; this means that 
we "fix" the origin. 



§17 



Bending of rods 



77 



It is seen from formulae (17.4) that the points initially on a cross-section 
z — constant = %o will be found, after the deformation, on the surface 
z = zo+u z = zo(l+xlR). We see that, in the approximation used, the 
cross-sections remain plane but are turned through an angle relative to their 
initial positions. The shape of the cross-section changes, however; for 
example, when a rod of rectangular cross-section (sides a, b) is bent, the sides 
y = + \b of the cross-section become y = ±\b + u y = ±\b{\ — ax\R\ i.e. 
no longer parallel but still straight. The sides x = ±\a> however, are bent 
into the parabolic curves 



(Fig. 14). 



x = ±\a + u x = ±\a- — [*o 2 + <Ki« 2 -:>' 2 )] 

Lis. 










— j— b — •- 


n 


a 

\ 


1 

/ 
1 
1 


i 
\ 








\ 



Fig. 14 



The free energy per unit volume of the rod is 
\<JikUQz = \o lz u zz = %Ex 2 /R 2 . 
Integrating over the cross-section of the rod, we have 



K£/* 2 )J** 2 d/. 



(17.5) 



This is the free energy per unit length of a bent rod. The radius of curvature 
R is that of the neutral surface. However, since the rod is thin, JR can here 
be regarded, to the same approximation, as the radius of curvature of the 
bent rod itself, regarded as a line (often called an "elastic line"). 

In the expression (17.5) it is convenient to introduce the moment of 
inertia of the cross-section. The moment of inertia about they-axis in its plane 
is defined as 



/„« J>d/, 



(17.6) 



analogously to the ordinary moment of inertia, but with the surface element 
d/ instead of the mass element. Then the free energy per unit length of the 
rod can be written 



\EI y IR\ 



(17.7) 



78 The Equilibrium of Rods and Plates §18 

We can also determine the moment of the internal stress forces on a given 
cross-section of the rod (the bending moment). A force a zz d/ = (xE/R) d/ 
acts in the z-direction on the surface element d/ of the cross-section. Its 
moment about the y-axis is xa zz d/. Hence the total moment of the forces 
about this axis is 

My = (E/R) jx*df= ElyfR. (17.8) 

Thus the curvature \jR of the elastic line is proportional to the bending 
moment on the cross-section concerned. 

The magnitude of I y depends on the direction of the jy-axis in the cross- 
sectional plane. It is convenient to express I y in terms of the principal 
moments of inertia. If 6 is the angle between the jy-axis and one of the 
principal axes of inertia in the cross-section, we know from mechanics that 

I y = h cos 2 d+I 2 sin 2 0, (17.9) 

where h and 7 2 are the principal moments of inertia. The planes through 
the #-axis and the principal axes of inertia are called the principal planes of 
bending. 

If, for example, the cross-section is rectangular (with sides a, b), its centre 
of mass is at the centre of the rectangle, and the principal axes of inertia 
are parallel to the sides. The principal moments of inertia are 

h = a*b/12, h = 0& 3 /12. (17.10) 

For a circular cross-section of radius R, the centre of mass is at the centre 
of the circle, and the principal axes are arbitrary. The moment of inertia 
about any axis lying in the cross-section and passing through the centre is 

/ = IttR\ (17.11) 

§18. The energy of a deformed rod 

In §17 we have discussed only a small portion of the length of a bent rod. 
In going on to investigate the deformation throughout the rod, we must 
begin by finding a suitable method of describing this deformation. It is 
important to note that, when a rod undergoes large bending deflections,-)- 
there is in general a twisting of it as well, so that the resulting deformation 
is a combination of pure bending and torsion. 

To describe the deformation, it is convenient to proceed as follows. We 
divide the rod into infinitesimal elements, each of which is bounded by two 
adjacent cross-sections. For each such element we use a co-ordinate system 
f , rj, £, so chosen that all the systems are parallel in the undeformed state, 
and their £-axes are parallel to the axis of the rod. When the rod is bent, the 



t By this, it should be remembered, we mean that the vector u is not small, but the strain tensor 
is still small. 



§18 The energy of a deformed rod 79 

co-ordinate system in each element is rotated, and in general differently in 
different elements. Any two adjacent systems are rotated through an infini- 
tesimal relative angle. 

Let d<|> be the vector of the angle of relative rotation of two systems at a 
distance d/ apart along the rod (we know that an infinitesimal angle of rotation 
can be regarded as a vector parallel to the axis of rotation ; its components 
are the angles of rotation about each of the three axes of co-ordinates). 

To describe the deformation, we use the vector 

SI = d*/d/, (18.1) 

which gives the "rate" of rotation of the co-ordinate axes along the rod. If 
the deformation is a pure torsion, the co-ordinate system rotates only about 
the axis of the rod, i.e. about the £-axis. In this case, therefore, the vector 
SI is parallel to the axis of the rod, and is just the torsion angle r used in §16. 
Correspondingly, in the general case of an arbitrary deformation we can call 
the component Q^ of the vector SI the torsion angle. For a pure bending of the 
rod in a single plane, on the other hand, the vector SI has no component Q^, 
i.e. it lies in the l^-plane at each point. If we take the plane of bending as the 
££-plane, then the rotation is about the ry-axis at every point, i.e. SI is parallel 
to the 77-axis. 

We take a unit vector t tangential to the rod (regarded as an elastic line). 
The derivative dt/d/ is the curvature vector of the line; its magnitude is 
1/R, where R is the radius of curvature, f and its direction is that of the 
principal normal to the curve. The change in a vector due to an infinitesimal 
rotation is equal to the vector product of the rotation vector and the vector 
itself. Hence the change in the vector t between two neighbouring points of 
the elastic line is given by dt = d<t> X t, or, dividing by d/, 

dt/d/= Slxt. (18.2) 

Multiplying this equation vectorially by t, we have 

SI = txdt/d/+t(t- a). (18.3) 

The direction of the tangent vector at any point is the same as that of the 
£-axis at that point. Hence t • SI = Q f . Using the unit vector n along the 
principal normal (n = R dt/d/), we can therefore put 

SI = tXnIR + tfy. (18.4) 

The first term on the right is a vector with two components Q g , £l r 
The unit vector t Xn is the binormal unit vector. Thus the components Q £ , 
Q v form a vector along the binormal to the rod, whose magnitude equals the 
curvature 1/R. 



f It may be recalled that any curve in space is characterised at each point by a curvature and a 
torsion. This torsion (which we shall not use) should not be confused with the torsional deformation, 
which is a twisting of a rod about its axis. 



80 The Equilibrium of Rods and Plates §18 

By using the vector SI to characterise the deformation and ascertaining 
its properties, we can derive an expression for the elastic free energy of a 
bent rod. The elastic energy per unit length of the rod is a quadratic function 
of the deformation, i.e., in this case, a quadratic function of the components 
of the vector SI. It is easy to see that there can be no terms in this quadratic 
form proportional to QgQg and Q^Q^. For, since the rod is uniform along 
its length, all quantities, and in particular the energy, must remain constant 
when the direction of the positive £-axis is reversed, i.e. when £ is replaced 
by — £, whereas the products mentioned change sign. 

For Qg = Q = we have a pure torsion, and the expression for the 
energy must be that obtained in §16. Thus the term in Q^ 2 in the free 
energy is \C£l^. 

Finally, the terms quadratic in Q.g and Q. v can be obtained by starting from 
the expression (17.7) for the energy of a slightly bent short section of the rod. 
Let us suppose that the rod is only slightly bent. We take the ££-plane as 
the plane of bending, so that the component Qg is zero ; there is also no torsion 
in a slight bending. The expression for the energy must then be that given 
by (17.7), i.e. %EI v jR 2 . We have seen, however, that 1/JR 2 is the square of the 
two-dimensional vector (Q.%, Q. v ). Hence the energy must be of the foi'm 
^ElyQ,^. For an arbitrary choice of the £ and v\ axes this expression becomes, 
as we know from mechanics, 

i£(/„Q f » + 2Z,gQ ? Q £ +/ K £y), 

where I V7]y / g, 1^ are the components of the inertia tensor for the cross- 
section of the rod. It is convenient to take the £ and 17 axes to coincide with 
the principal axes of inertia. We then have simply ^E(IiQ^ 2 + l2Sl v 2 ) } where 
7i, I2 are the principal moments of inertia. Since the coefficients of £2g 2 and 
Q, v 2 are constants, the resulting expression must be valid for large deflections 
also. 

Finally, integrating over the length of the rod, we obtain the following 
expression for the elastic free energy of a bent rod : 

Frod = J* {ihEQf + lhEnz + iCn&dl. (18.5) 

Next, we can express in terms of SI the moment of the forces acting on 
a cross-section of the rod. This is easily done by again using the results 
previously obtained for pure torsion and pure bending. In pure torsion, the 
moment of the forces about the axis of the rod is Cr. Hence we conclude 
that, in the general case, the moment M^ about the £-axis must be CD.^. 
Next, in a slight deflection in the f£-plane, the moment about the 77-axis is 
EhjR. In such a bending, however, the vector SI is along the 17-axis, so that 
1/R is just the magnitude of SI, and EI 2 /R = EI 2 Q. Hence we conclude 
that, in the general case, we must have M g — EIi£l g , M v = EI 2 & V (the $ and 
r\ axes being along the principal axes of inertia in the cross-section). Thus 



Frod = J"| 



§18 The energy of a deformed rod 81 

the components of the moment vector M are 

M { = Em gy M v = Eh& r M c = CQ C . (18.6) 

The elastic energy (18.5), expressed in terms of the moment of the forces, is 

Mr 2 M 2 M? \ 

i^!_ + i!!i_ + ^i- d/. (18.7) 

ZhE 2hE 2C J 

An important case of the bending of rods is that of a slight bending, in 
which the deviation from the initial position is everywhere small compared 
with the length of the rod. In this case torsion can be supposed absent, and 
we can put Q £ = 0, so that (18.4) gives simply 

Sl = txn//? = txdt/d/. (18.8) 

We take a co-ordinate system x, y, z fixed in space, with the 2-axis along the 
axis of the undeformed rod (instead of the system £, rj, £ for each point in the 
rod), and denote by X, Y the co-ordinates x, y for points on the elastic line; 
X and Y give the displacement of points on the line from their positions 
before the deformation. 

Since the bending is only slight, the tangent vector t is almost parallel 
to the .sr-axis, and the difference in direction can be approximately neglected. 
The unit tangent vector is the derivative t = dr/d/ of the radius vector r 
of a point on the curve with respect to its length. Hence 

dt/d/= d 2 r/d/ 2 ^d 2 r/d* 2 ; 

the derivative with respect to the length can be approximately replaced by 
the derivative with respect to z. In particular, the x and y components of 
this vector are respectively d 2 Xjdz 2 and d 2 Y/d;s 2 . The components ft £ , Cl v 
are, to the same accuracy, equal to Q xt Q v , and we have from (18.8) 

Q g = -d 2 Y/d* 2 , Q v = d 2 X/d* 2 . (18.9) 

Substituting these expressions in (18.5), we obtain the elastic energy of a 
slightly bent rod in the form 

ri /d 2 Y\ 2 /d*X\ 2 \ 

F <«^ E iH-w) +h (-^)h (18 - 10) 

Here h and h are the moments of inertia about the axes of x and y respectively, 
which are the principal axes of inertia. 

In particular, for a rod of circular cross-section, h = h = A a nd the 
integrand is just the sum of the squared second derivatives, which in the 
approximation considered is the square of the curvature : 



/d*x\ 2 /d 2 yy 
\ow + \d^7 



/d*X\ 2 /d 2 Y\ 2 1 



82 The Equilibrium of Rods and Plates §19 

Hence formula (18.10) can be plausibly generalised to the case of slight 
bending of a circular rod having any shape (not necessarily straight) in its 
undeformed state. To do so, we must write the bending energy as 

Frod = \EI J(— - — J d*, (18.11) 

where R is the radius of curvature at any point of the undeformed rod. This 
expression has a minimum, as it should, in the undeformed state (R = R ), 
and for R -> co it becomes formula (18.10). 

§19. The equations of equilibrium of rods 

We can now derive the equations of equilibrium for a bent rod. We again 
consider an infinitesimal element bounded by two adjoining cross-sections 
of the rod, and calculate the total force acting on it. We denote by F the 
resultant internal stress on a cross-section.f The components of this vector 
are the integrals of a^ over the cross-section : 

Fi^ja^df. (19.1) 

If we regard the two adjoining cross-sections as the ends of the element, a 
force F+dF acts on the upper end, and -F on the lower end; the sum of 
these is the differential dF. Next, let K be the external force on the rod per 
unit length. Then an external force K dl acts on the element of length d/. 
The resultant of the forces on the element is therefore dF+K d/. This must 
be zero in equilibrium. Thus we have 

dF/dZ = - K. (19.2) 

A second equation is obtained from the condition that the total moment of 
the forces on the element is zero. Let M be the moment of the internal 
stresses on the cross-section. This is the moment about a point (the origin) 
which lies in the plane of the cross-section; its components are given by 
formulae (18.6). We shall calculate the total moment, on the element con- 
sidered, about a point O lying in the plane of its upper end. Then the 
internal stresses on this end give a moment M+dM. The moment about O 
of the internal stresses on the lower end of the element is composed of the 
moment — M of those forces about the origin O' in the plane of the lower 
end and the moment about O of the total force — F on that end. This latter 
moment is — dl X — F, where dl is the vector of the element of length of the 
rod between O' and O. The moment due to the external forces K is of a 
higher order of smallness. Thus the total moment acting on the element 
considered is dM + dlxF. In equilibrium, this must be zero: 

dM + dlxF = 0. 



| This notation will not lead to any confusion with the free energy, which does not appear in 
§§19-21. 



§19 The equations of equilibrium of rods 83 

Dividing this equation by d/ and using the fact that dl/d/ = t is the unit 
vector tangential to the rod (regarded as a line), we have 

dM/d/=Fxt. (19.3) 

Equations (19.2) and (19.3) form a complete set of equilibrium equations 
for a rod bent in any manner. 

If the external forces on the rod are concentrated, i.e. applied only at 
isolated points of the rod, the equilibrium equations at all other points are 
much simplified. For K = we have from (19.2) 

F = constant, (19.4) 

i.e. the stress resultant is constant along any portion of the rod between 
points where forces are applied. The values of the constant are found from 
the fact that the difference F 2 -Fi of the forces at two points 1 and 2 is 

F 2 -Fi= -SK, (19.5) 

where the sum is over all forces applied to the segment of the rod between 
the two points. It should be noticed that, in the difference F2-F1, the 
point 2 is further from the point from which / is measured than is the point 1 ; 
this is important in determining the signs in equation (19.5). In particular, if 
only one concentrated force f acts on the rod, and is applied at its free end, 
then F = constant = f at all points of the rod. 

The second equilibrium equation (19.3) is also simplified. Putting 
t = dl/d/ = dr/d/ (where r is the radius vector from any fixed point to the 
point considered) and integrating, we obtain 

M = FXr+ constant, (19.6) 

since F is constant. 

If concentrated forces also are absent, and the rod is bent by the application 
of concentrated moments, i.e. of concentrated couples, then F = constant 
at all points of the rod, while M is discontinuous at points where couples 
are applied, the discontinuity being equal to the moment of the couple. 

Let us consider also the boundary conditions at the ends of a bent rod. 
Various cases are possible. 

The end of the rod is said to be clamped (Fig. 4a, §12) if it cannot move 
either longitudinally or transversely, and moreover its direction (i.e. the direc- 
tion of the tangent to the rod) cannot change. In this case the boundary 
conditions are that the co-ordinates of the end of the rod and the unit tangen- 
tial vector t there are given. The reaction force and moment exerted on the 
rod by the clamp are determined by solving the equations. 

The opposite case is that of a free end, whose position and direction are 
arbitrary. In this case the boundary conditions are that the force F and 
moment M must be zero at the end of the rod.f 



t If a concentrated force f is applied to the free end of the rod, the boundary condition is F = f 
not F = 0. 



84 The Equilibrium of Rods and Plates §19 

If the end of the rod is fixed to a hinge, it cannot be displaced, but its 
direction can vary. In this case the moment of the forces on the freely turning 
end must be zero. 

Finally, if the rod is supported (Fig. 4b), it can slide at the point of support 
but cannot undergo transverse displacements. In this case the direction t of 
the rod at the support and the point on the rod at which it is supported are 
unknown. The moment of the forces at the point of support must be zero, 
since the rod can turn freely, and the force F at that point must be perpen- 
dicular to the rod; a longitudinal force would cause a further sliding of the 
rod at this point. 

The boundary conditions for other modes of fixing the rod can easily be 
established in a similar manner. We shall not pause to add to the typical 
examples already given. 

It was mentioned at the beginning of §18 that a rod of arbitrary cross- 
section undergoing large deflections is in general twisted also, even if no 
external twisting moment is applied to the rod. An exception occurs when a 
rod is bent in one of its principal planes, in which case there is no torsion. 
For a rod of circular cross-section no torsion results for any bending (if there 
is no external twisting moment, of course). This can be seen as follows. The 
twisting is given by the component Q% = SI • t of the vector SI. Let us 
calculate the derivative of this along the rod. To do so, we use the fact that 
Q c = MJC: 

d „, x ~ d ^ dM dt 

— (M-t) = C — *- = t+M — . 

d/ v ' 61 61 61 

Substituting (19.3), we see that the first term is zero, so that 

C6SiJ61= M-dt/d/. 

For a rod of circular cross-section, h = I 2 = I; by (18.3) and (18.6), we can 
therefore write M in the form 

M = EItx6t/6l+tCQ c . (19.7) 

Multiplying by dt/d/, we have zero on the right-hand side, so that 

6QJ61 = 0, 
whence 

D s = constant, (19.8) 

i.e. the torsion angle is constant along the rod. If no twisting moments are 
applied to the ends of the rod, then Q,^ is zero at the ends, and there is no 
torsion anywhere in the rod. 

For a rod of circular cross-section, we can therefore put for pure bending 

dr d 2 r 

M = EIX Xdt/61 = EI— x— . (19.9) 

61 d/ 2 



§19 The equations of equilibrium of rods 85 

Substituting this in (19.3), we obtain the equation for pure bending of a 
circular rod: 

<fr d 3 r dr ,« rt < A v 

£/ d7*d^ = Fx d? (mo) 

PROBLEMS 

Problem 1. Reduce to quadratures the problem of determining the shape of a rod of 
circular cross-section bent in one plane by concentrated forces. 

Solution. Let us consider a portion of the rod lying between points where the forces 
are applied ; on such a portion F is constant. We take the plane of the bent rod as the xy- 
plane, with the y-axis parallel to the force F, and introduce the angle 8 between the tangent 
to the rod and the y-axis. Then dxjdl = sin 8, dyjdl = cos 8, where x, y are the co-ordinates 
of a point on the rod. Expanding the vector products in (19.10), we obtain the following 
equation for 8 as a function of the arc length I: £7d 2 0/d/ a — Fsin 8 = 0. A first integration 
gives lEI(dOld[)*+Fcos 8 = c lt and 

/ = ± VdEI) f ,. Z ^+C2. (1) 

J v(ci — Fcosv) 

The function 8(1) can be obtained in terms of elliptic functions. The co-ordinates 

x = J sin 6 dl t y = j cos 6 d/ 

are 

x = ± y/[2EI(ci - F cos 6)1 F 2 ] + constant, 

r cos0d0 (2) 

y = +W(iEI) + constant. 

The moment M (19.9) is parallel to the ar-axis, and its magnitude is M = EId8/dl. 

Problem 2. Determine the shape of a bent rod with one end clamped and the other under 
a force f perpendicular to the original direction of the rod (Fig. 15). 




Fig. 15 

Solution. We have F = constant = f everywhere on the rod. At the clamped end 
(/ = 0), 8 = \it, and at the free end (Z = L, the length of the rod) M = 0, i.e. 8' = 0. Putting 
8(L) = 8 , we have in (1), Problem 1, Ci — /cos 8 and 



= VQEiiflj 



dO 



-v/(cos do — cos 8) 



86 The Equilibrium of Rods and Plates §19 

Hence we obtain the equation for 6 : 



= V(iEIlf)j 



dd 



\/(cos #o — cos 6) 



The shape of the rod is given by 

x = ^/ (2EI jf)[^/ (cos 6 ) -V (cos do -cos 6) \, 

in 



y = V(EI/2f)j 



cosddd 



yYcos do — cos 6) 
e 

Problem 3. The same as Problem 2, but for a force f parallel to the original direction of 
the rod. 



Fig. 16 

Solution. We have F = — f; the co-ordinate axes are taken as shown in Fig. 16. The 
boundary conditions are 8 = for I = 0, 0' = for / = L. Then 



vwwj 



ie 



yYcos 6 — cos #o) 
o 



where 

d = 6(L) 

is given by 

0. 



- vmif)j 



dd 



\/(cOS0— COS0o) 


For * and y we obtain 

x = V(2W)[V(l~ cos 0o)- V(cos0-cos0 o )], 
e 



y=V(EI/2f)j 



cos 6 dd 



-\/(cos 6 — cos do) 



§19 



The equations of equilibrium of rods 



87 



For a small deflection, O <^ 1, and we can write 

L 



v(£///) !v(^)= w(£///) ' 



i.e. 0q does not appear. This shows that, in accordance with the result of §21, Problem 3, 
the solution in question exists only for / 5s ir 2 EI14L 2 , i.e. when the rectilinear shape ceases 
to be stable. 




Problem 4. The same as Problem 2, but for the case where both ends of the rod are sup- 
ported and a force f is applied at its centre. The distance between the supports is L . 

Solution. We take the co-ordinate axes as shown in Fig. 17. The force F is constant on 
each of the segments AB and BC, and on each is perpendicular to the direction of the rod at 
the point of support A or C. The difference between the values of F on AB and BC is f, 
and so we conclude that, on AB, F sin 8 = — i/, where o is the angle between the y-axis 
and the line AC. At the point A (/ = 0) we have the conditions = \tt and M = 0, i.e. 
0' = 0, so that on AB 



I EI sin O r 

= V ? Ji 



/ 



■\/cos8 



x = 2 



EI sin #o cos 6 

1 ' 



y = 



/EI sin 6 f 



cos0d0. 



The angle 8 is determined from the condition that the projection of AB on the straight line 
AC must be iL , whence 



\u 



J EI sin O fi 

= v7 J" 



EI sin O fcos(0-0 o ) 
<\/sin0 



d0. 



*• 



For some value O lying between and Jw the derivative dfld0 o (/being regarded as a function 
of 6 ) passes through zero to positive values. A further decrease in O > i-e. increase in the 
deflection, would mean a decrease in /. This means that the solution found here becomes 
unstable, the rod collapsing between the supports. 

Problem 5. Reduce to quadratures the problem of three-dimensional bending of a rod 
under the action of concentrated forces. 



88 The Equilibrium of Rods and Plates §19 

Solution. Let us consider a segment of the rod between points where forces are applied, 
on which F = constant. Integrating (19.10), we obtain 

dr d2r 

the constant of integration has been written as a vector cF parallel to F, since, by appro- 
priately choosing the origin, i.e. by adding a constant vector to r, we can eliminate any vector 
perpendicular to F. Multiplying (1) scalar ly and vectorially by r' (the prime denoting 
differentiation with respect to I), and using the fact that r'«r" = (since r' 2 = 1), we obtain 
FTXr'+cF^r' = 0, EIv" = (Fxr)Xr'+cFxr'. In components (with the ar-axis parallel 
to F) we obtain (xy' — yx') + cz' = 0, EIz" = —F(xx'+yy'). Using cylindrical polar co- 
ordinates r, <f>, z, we have 

r*<f>' + cz' = 0, EIz" = -Err'. (2) 

The second of these gives 

z' = F{A-r*)\2Eh (3) 

where A is a constant. Combining (2) and (3) with the identity r' 2 +r*^'* -J-*'* = 1, we find 

r dr 



d/ = 



VI> 2 - (r 2 + c z )(A - r2)2F2/4£2/2j' 

and then (2) and (3) give 

{A-r*)rdr 



z = 



/ 

c 




2EI J y/[r* - F 2 (r* + c*)(A - r^ffim*] 

(A-r*)dr 
V[r 2 - F 2 (r 2 + c 2 )(A - r2)*/4£ 2 / 2 ] 

which gives the shape of the bent rod. 

Problem 6. A rod of circular cross-section is subjected to torsion (with torsion angle r) 
and twisted into a spiral. Determine the force and moment which must be applied to the 
ends of the rod to keep it in this state. 

Solution. Let R be the radius of the cylinder on whose surface the spiral lies (and along 
whose axis we take the ^-direction) and a the angle between the tangent to the spiral and a 
plane perpendicular to the sr-axis ; the pitch h of the spiral is related to a and R by h — 2nR tan a. 
The equation of the spiral is x — R cos <$>, y = R sin <f>, z = <j>R tan a, where <f> is the angle 
of rotation about the »-axis. The element of length is 61 — (Rjcos a)d<£. Substituting these 
expressions in (19.7), we calculate the components of the vector M, and then the force F 
from formula (19.3); F is constant everywhere on the rod. The result is that the force F is 
parallel to the z-axis and its magnitude is F = F z = (O/ R) sin a— (EI/R 3 ) cos 2 a sin a. 
The moment M has a ^-component M t — Cr sin a-\-(EI/R) cos 3 a and a ^-component, along 
the tangent to the cross-section of the cylinder, M$ = FR. 

Problem 7. Determine the form of a flexible wire (whose resistance to bending can be 
neglected in comparison with its resistance to stretching) suspended at two points and in a 
gravitational field. 

Solution. We take the plane of the wire as the jry-plane, with the y-axis vertically down- 
wards. In equation (19.3) we can neglect the term dM/dZ, since M is proportional to EI. 
Then Fxt = 0, i.e. F is parallel to t at every point, and we can put F = Ft. Equation (19.2) 
then gives 

d / dx\ n d ( n dy\ 

d7 



A dl! dl\ 61 J H 



§20 Small deflections of rods 89 

where q is the weight of the wire per unit length; hence F 6x1 61 = c, Fdyldl — ql, and so 
F = V(«"+flV). so that dxldl = A/V(A 2 +P), dyjdl = UV(A*+P), where ^4 = elq. 
Integration gives x = A sivh-\llA), y — ^(A*+P), whence y = A cosh (x/A), i.e. the 
wire takes the form of a catenary. The choice of origin and the constant A are determined 
by the fact that the curve must pass through the two given points and have a given length. 

§20. Small deflections of rods 

The equations of equilibrium are considerably simplified in the important 
case of small deflections of rods. This case holds if the direction of the 
vector t tangential to the rod varies only slowly along its length, i.e. the deriva- 
tive dt/d/ is small. In other words, the radius of curvature of the bent rod is 
everywhere large compared with the length of the rod. In practice, this 
condition amounts to requiring that the transverse deflection of the rod is 
small compared with its length. It should be emphasised that the deflection 
need not be small compared with the thickness of the rod, as it had to be in 
the approximate theory of small deflections of plates given in §§ll-12.f 

Differentiating (19.3) with respect to the length, we have 

d2M dF dt ^^ 

= Xt+Fx— . (20.1) 

d/2 61 dl K ' 

The second term contains the small quantity dt/d/, and so can usually be 
neglected (some exceptional cases are discussed below). Substituting in the 
first term dF/d/ = — K, we obtain the equation of equilibrium in the form 

d 2 M/d/2 = txK. (20.2) 

We write this equation in components, substituting in it from (18.6) and 
(18.9) 

M z = -EhY", My = EIzX", M z = 0, (20.3) 

where the prime denotes differentiation with respect to z. The unit vector t 
may be supposed to be parallel to the #-axis. Then (20.2) gives 

EI 2 X™-K X = 0, EhY^-Ky = 0. (20.4) 

These equations give the deflections X and Y as functions of z, i.e. the shape 
of a slightly bent rod. 

The stress resultant F on a cross-section of the rod can also be expressed in 
terms of the derivatives of X and Y. Substituting (20.3) in (19.3), we obtain 

F x = -EhX"\ F y = -EhY"'. (20.5) 

We see that the second derivatives give the moment of the internal stresses, 
while the third derivatives give the stress resultant. The force (20.5) is 
called the shearing force. If the bending is due to concentrated forces, the 
shearing force is constant along each segment of the rod between points 



f We shall not give the complex theory of the bending of rods which are not straight when un- 
deformed, but only consider one simple example (see Problems 8 and 9). 



90 The Equilibrium of Rods and Plates §20 

where forces are applied, and has a discontinuity at each of these points 
equal to the force applied there. 

The quantities EI% and EI\ are called the flexural rigidities of the rod in 
the xz and yz planes respectively, f 

If the external forces applied to the rod act in one plane, the bending takes 
place in one plane, though not in general the same plane. The angle between 
the two planes is easily found. If a is the angle between the plane of action of 
the forces and the first principal plane of bending (the a:2-plane), the equa- 
tions of equilibrium become X* iv > = (K/I 2 E) cos a, Y< lv > = (KlhE) sin a. 
The two equations differ only in the coefficient of K. Hence X and Y are 
proportional, and Y = (Xh/Ii) tan a. The angle 6 between the plane of 
bending and the xsr-plane is given by 

tan0 = (7 2 //i)tana. (20.6) 

For a rod of circular cross-section h = h and <x = 6, i.e. the bending occurs 
in the plane of action of the forces. The same is true for a rod of any cross- 
section when a = 0, i.e. when the forces act in a principal plane. The magni- 
tude of the deflection £ = V( X2 + Y2 ) satisfies the equation 

£/£(iv) = k, I = IihlVVi 2 cos2 a + / 2 2 sin2 a ). (20.7) 

The shearing force F is in the same plane as K, and its magnitude is 

F = -Ell'". (20.8) 

Here I is the "effective" moment of inertia of the cross-section of the rod. 

We can write down explicitly the boundary conditions on the equations of 
equilibrium for a slightly bent rod. If the end of the rod is clamped, we must 
have X = Y = there, and also X' = Y' = 0, since its direction cannot 
change. Thus the conditions at a clamped end are 

X = Y = 0, X' = y = 0. (20.9) 

The reaction force and moment at the point of support are determined from 
the known solution by formulae (20.3) and (20.5). 

When the bending is sufficiently slight, the hinging and supporting of a 
point on the rod are equivalent as regards the boundary conditions. The 
reason is that, in the latter case, the longitudinal displacement of the rod at 
its point of support is of the second order of smallness compared with the 



f An equation of the form 

DXW-K X = (20.4a) 

also describes the bending of a thin plate in certain limiting cases. Let a rectangular plate (with 
sides a, b and thickness h) be fixed along its sides a (parallel to the y-axis) and bent along its sides b 
(parallel to the 2-axis) by a load uniform in the y-direction. In the general case of arbitrary a and b, 
the two-dimensional equation (12.5), with the appropriate boundary conditions at the fixed and free 
edges, must be used to determine the bending. In the limiting case a g> b, however, the deformation 
may be regarded as uniform in the y-direction, and then the two-dimensional equilibrium equation 
becomes of the form (20.4a), with the flexural rigidity replaced by D = E/i 3 a/12(l -o 2 ). Equation 
(20.4a) is also applicable to the opposite limiting case a <g b, when the plate can be regarded as a 
rod of length b with a narrow rectangular cross-section (a rectangle of sides a and h); in this case, 
however, the flexural rigidity is D = EI2 = Eh 3 ajl2. 



§20 Small deflections of rods 91 

transverse deflection, and can therefore be neglected. The boundary con- 
ditions of zero transverse displacement and moment give 

X = Y = 0, X" = Y" = 0. (20.10) 

The direction of the end of the rod and the reaction force at the point of 
support are obtained by solving the equations. 

Finally, at a free end, the force F and moment M must be zero. According 
to (20.3) and (20.5), this gives the conditions 

X" = Y" = 0, X'" = Y'" = 0. (20.11) 

If a concentrated force is applied at the free end, then F must be equal to this 
force, and not to zero. 

It is not difficult to generalise equations (20.4) to the case of a rod of 
variable cross-section. For such a rod the moments of inertia I\ and 1% are 
functions of z. Formulae (20.3), which determine the moment at any cross- 
section, are still valid. Substitution in (20.2) now gives 

d 2 / d 2 Y\ d 2 / d 2 ^\ 

E — (h = K y , E—(h — A = K *> ( 20A2 ) 

d*2\ dW d* 2 \ <W 

in which Ij and h must be differentiated. The shearing force is 

d / d 2 X\ d / d 2 Y\ 

' — \h , F y = -E — (h . (20.13) 

d*\ d* 2 / V d*\ d* 2 / v ' 

Let us return to equations (20.1). Our neglect of the second term on the 
right-hand side may in some cases be illegitimate, even if the bending is 
slight. The cases involved are those in which a large internal stress resultant 
acts along the rod, i.e. F z is very large. Such a force is usually caused by a 
strong tension of the rod by external stretching forces applied to its ends. 
We denote by T the constant lengthwise stress F z . If the rod is strongly 
compressed instead of being extended, T will be negative. In expanding the 
vector product F Xdt/d/ we must now retain the terms in T, but those in F x 
and F y can again be neglected. Substituting X", Y", 1 for the components 
of the vector dt/d/, we obtain the equations of equilibrium in the form 

hEX^)-TX"-K x = 0, 

(20.14) 
I 1 EYM-TY"-K v = 0. v ' 

The expressions (20.5) for the shearing force will now contain additional 
terms giving the projections of the force T (along the vector t) on the x and 
y axes : 

F x = -EI 2 X'" + TX', F y = -EhY'" + TY'. (20.15) 

These formulae can also, of course, be obtained directly from (19.3). 

In some cases a large force T can result from the bending itself, even if 
no stretching forces are applied. Let us consider a rod with both ends 



F x = -E 



92 The Equilibrium of Rods and Plates §20 

clamped or hinged to fixed supports, so that no longitudinal displacement is 
possible. Then the bending of the rod must result in an extension of it, 
which leads to a force T in the rod. It is easy to estimate the magnitude of the 
deflection for which this force becomes important. The length L + AL of the 
bent rod is given by 

L 

L+AL = JV(l + *' 2 + Y' 2 ) d*, 
o 

taken along the straight line joining the points of support. For slight bending 
the square root can be expanded in series, and we find 

L 


The stress force in simple stretching is equal to the relative extension multi- 
plied by Young's modulus and by the area S of the cross-section of the rod. 
Thus the force T is 

E'er L 

T = — J(Z' 2 + Y'2) d*. (20.16) 

2L 

If 8 is the order of magnitude of the transverse bending, the derivatives 
X' and Y' are of the order of 8jL, so that the integral in (20.16) is of the 
order of S 2 /L, and T ~ ES(8/L) 2 . The orders of magnitude of the first and 
second terms in (20.14) are respectively EI8/L* and T8/L 2 ~ ES8 3 /L*. The 
moment of inertia / is of the order of A 4 , and S ~ h 2 , where h is the thickness 
of the rod. Substituting, we easily find that the first and second terms in 
(20.14) are comparable in magnitude if 8 ~ h. Thus, when a rod with fixed 
ends is bent, the equations of equilibrium can be used in the form (20.4) only 
if the deflection is small in comparison with the thickness of the rod. If 8 
is not small compared with h (but still, of course, small compared with L), 
equations (20.14) must be used. The force T in these equations is not known 
a priori. It must first be regarded as a parameter in the solution, and then 
determined by formula (20.16) from the solution obtained; this gives the 
relation between T and the bending forces applied to the rod. 

The opposite limiting case is that where the resistance of the rod to bending 
is small compared with its resistance to stretching, so that the first terms in 
equations (20.14) can be neglected in comparison with the second terms. 
Physically this case can be realized either by a very strong tension force T or 
by a small value of EI, which can result from a small thickness h. Rods under 
strong tension are called strings. In such cases the equations of equilibrium 
are 

TX" + K X = 0, TY" + K y = 0. (20.17) 



§20 Small deflections of rods 93 

The ends of the string are fixed, in the sense that their co-ordinates are given, 



i.e. 



X = Y = 0. (20.18) 

The direction of the ends cannot be decided arbitrarily, but is given by the 
solution of the equations. 

In conclusion, we may show how the equations of equilibrium of a slightly 
bent rod may be obtained from the variational principle, using the expression 
(18.10) for the elastic energy: 

F to a = lEJ{hY"2 + l2X"2}dz. 

In equilibrium the sum of this energy and the potential energy due to the 
external forces K acting on the rod must be a minimum, i.e. we must have 
8F TO &-$(K x 8X+K y 8Y)<\z = 0, where the second term is the work done 
by the external forces in an infinitesimal displacement of the rod. In varying 
■Prod, we effect a repeated integration by parts: 

l8JX"*dz = jX"8X"dz 

= [X"8X']- jX'"8X'dz 

= [X ,, 8X']-[X ,,, 8X]+ jX™8Xdz, 
and similarly for the integral of Y" 2 . Collecting terms, we obtain 
j [(Eh yav) _ K y )8 Y + (EIzX<M - K X )8X] d* + 

+ EI 1 [(Y"8Y , -Y , "8Y)] + EI 2 [(X f '8X f -X , "SX)] = 0. 

The integral gives the equilibrium equations (20.4), since the variations 8X 
and 8 Y are arbitrary. The integrated terms give the boundary conditions on 
these equations ; for example, at a free end the variations 8X, 8 Y, 8X', 8 Y' 
are arbitrary, and the corresponding conditions (20.11) are obtained. Also, 
the coefficients of 8X and 8Y in these terms give the expressions (20.5) for 
the components of the shearing force, and those of 8X' and 8Y' give the 
expressions (20.3) for the components of the bending moment. 

Finally, the equations of equilibrium (20.14) in the presence of a tension 
force T can be obtained by the same method if we include in the energy a 
term TAL = \T\ (X' 2 + Y' 2 ) dsr, which is the work done by the force T over a 
distance AL equal to the extension of the rod. 

PROBLEMS 

Problem 1. Determine the shape of a rod (of length /) bent by its own weight, for various 
modes of support at the ends. 



94 The Equilibrium of Rods and Plates §20 

Solution. The required shape is given by a solution of the equation £(' v > = qlEI, where 
q is the weight per unit length, with the appropriate boundary conditions at its ends, as shown 
in the text. The following shapes and maximum displacements are obtained for various 
modes of support at the ends of the rod. The origin is at one end of the rod in each case. 

(a) Both ends clamped: 

£ = qz\z-lf\2\EI y £(£/) = qPfiMEI. 

(b) Both ends supported : 

£ = qz(z3-2lz* + P)/24EI, £(£Z) = 5ql*/384EI. 

(c) One end (z = I) clamped, the other supported: 

£ = qz(2z* - 3Z* 2 + /3)/48£7, £(0-42/) = 0-0054#/£7. 

(d) One end (z — 0) clamped, the other free : 

£ = qz\z 2 -\lz+W)\2\EI, £(Z) = ql*J8EI. 

Problem 2. Determine the shape of a rod bent by a force /applied to its mid-point. 

Solution. We have £( lv) = everywhere except at z = \l. The boundary conditions 
at the ends of the rod (z = and z — I) are determined by the mode of support ; at z = £/, 
£, £' and £" must be continuous, and the discontinuity in the shearing force F = —Ell,'" 
must be equal to/. 

The shape of the rod (for < z ^ £/) and the maximum displacement are given by the 
following formulae: 

(a) Both ends clamped : 

£ = > 2 (3Z- 4*)/48£7, £(£Z) = fP/192EI. 

(b) Both ends supported: 

£ = fz(3P - 4# 2 )/48£/, £(|Z) = //3/48£7. 

The rod is symmetrical about its mid-point, so that the functions £(#) in \l ^ z s$ / are 
obtained simply by replacing z by l—z. 

Problem 3. The same as Problem 2, but for a rod clamped at one end (z = 0) and free 
at the other end (z = I), to which a force/ is applied. 

Solution. At all points of the rod F = constant = /, so that £'" == —f/EI. Using the 
conditions £ = 0, £' = for z = 0, £" = for z = I, we obtain 

£ = fz*(3l-z)l6EI, £(/) = /73/3E7. 

Problem 4. Determine the shape of a rod with fixed ends, bent by a couple at its 
mid-point. 

Solution. At all points of the rod £ (lv) = 0, and at z = \l the moment M = Eli" has 
a discontinuity equal to the moment m of the applied couple. The results are: 

(a) Both ends clamped: 

£ = mz\l -2z)l8EIl for ^ z ^ \l, 
£ = -m(l-zf[l-2(l-z)]ISEIl for #<*</. 

(b) Both ends hinged: 

£ = mz(l 2 -4z 2 )j24EIl for < z ^ £/, 
£ = -m(/-#)[Z 2 -4(Z-#) 2 ]/24£/Z for J/ < z < Z. 



§20 Small deflections of rods 95 

The rod is bent in opposite directions on the two sides of z = \l. 

Problem 5. The same as Problem 4, but for the case where one end is clamped and the 
other end free, the couple being applied at the latter end. 

Solution. At all points of the rod M = Elt," = m, and at z = we have £ = 0, £' = 0. 
The shape is given by £ = mz 2 l2EI, 

Problem 6. Determine the shape of a circular rod with hinged ends stretched by a force 
T and bent by a force / applied at its mid-point. 

Solution. On the segment < z =£ J/ the shearing force is £/, so that (20.15) gives the 
equation 

£"-T£'IEI = -fj2EI. 

The boundary conditions are £ = C = for z = and I; £' = for z = \l (since £' is 
continuous). The shape of the rod (in the segment s$ z ^ \V) is given by 

c = n z _ sinh ** ), k = v(w. 

fe 2r\ kcoshiki!' vv 

For small ft this gives the result obtained in Problem 2 (b). For large k it becomes £ = fzjlT, 
i.e., in accordance with equations (20.17), a flexible wire under a force / takes the form of 
two straight pieces intersecting at z = \l. 

If the force T is due to the stretching of the rod by the transverse force, it must be deter- 
mined by formula (20.16). Substituting the above result, we obtain the equation 

1 r3 1 1 3 1 1 8E 2 / 3 
— l-+-tanh 2 -&/ tanh * 



H- 



*»L2 2 2 kl 2 J PS ' 

which determines T as an implicit function off. 

Problem 7. A circular rod of infinite length lies in an elastic substance, i.e. when it is 
bent a force K = — oc£ proportional to the deflection acts on it. Determine the shape of the 
rod when a concentrated force /acts on it. 

Solution. We take the origin at the point where the force / is applied. The equation 
E/£(iv) _ _ a £ holds everywhere except at z = 0. The solution must satisfy the condition 
J=0at2= ±oo, and at z = £' and £" must be continuous; the difference between the 
shearing forces F = —Eli"' for * -»> 0+ and z ->- 0— must be/. The required solution is 

Problem 8. Derive the equation of equilibrium for a slightly bent thin circular rod which, 
in its undeformed state, is an arc of a circle and is bent in its plane by radial forces. 

Solution. Taking the origin of polar co-ordinates r, <f> at the centre of the circle, we write 
the equation of the deformed rod as r = a + £(<£), where a is the radius of the arc and I a small 
radial displacement. Using the expression for the radius of curvature in polar co-ordinates, 
we find as far as the first order in £ 

1 r 2_ rr " + 2r'2 1 £+£" 



R ( r 2 + r '2)3/2 a a 2 

where the prime denotes differentiation with respect to <f>. According to (18.11), the elastic 
bending energy is 

r 1 1 1\ 2 EI r 



96 



The Equilibrium of Rods and Plates 



§20 



^o being the angle subtended by the arc at its centre. The equation of equilibrium is obtained 
from the variational principle 



00 



SF r od- jSCKradfi** 0, 



where K r is the external radial force per unit length, with the auxiliary condition 

00 



JCd^-0, 







which is, in this approximation, the statement of the fact that the total length of the rod is 
unchanged, i.e. it undergoes no general extension. Using Lagrange's method, we put 



4>o 



&Frod- jaKM d<f> + aoLJ8£ d<f> = 0, 



where a is a constant. Varying the integrand in F to< i and integrating the 8 £" term twice by 
parts, we obtain 

f{— (£ + 2^ + £ (iv) )-«^r + aaW d<f> + 

+^[«+D8n-^[«'+r)8a = o. 

a 3 a 3 

Hence we find the equation of equilibriumt 

EI(P*>+2C" + Qla*-Kr+* = 0, (1) 

the shearing force F= —EI(t,'+t,'")la*, and the bending moment M = £!/({ + £")/«*; 
cf. the end of §20. The constant a is determined from the condition that the rod as a whole 
is not stretched. 

Problem 9. Determine the deformation of a circular ring bent by two forces / applied 
along a diameter (Fig. 18). 




t In the absence of external forces, K r = and a = 0; the non-zero solutions of the resulting 
homogeneous equation correspond to a simple rotation or translation of the whole rod. 



§21 The stability of elastic systems 97 

Solution. Integrating equation (1), Problem 8, along the circumference of the ring, we 
have 27T<xa = J K r a d<f> = 2/. We have equation (1) with K, = everywhere except at 
<f> — and <f> — tt: 

£Uv) + 21" + 1 +fa?J7TEI = 0. 

The required deformation of the ring is symmetrical about the diameters AB and CD, and 
so we must have £' = at A, B, C and D. The difference in the shearing forces for <f> -> ± 
must be/. The solution of the equation of equilibrium which satisfies these conditions is 



< IT. 



£ = ( — h-<£cos<£ — 7rcos</> — sin<£), <: <f> 

In particular, the points .4 and B approach through a distance 

IROHJWI-y;-;)- 



§21. The stability of elastic systems 

The behaviour of a rod subject to longitudinal compressing forces is the 
simplest example of the important phenomenon of elastic instability, first 
discovered by L. Euler. 

In the absence of transverse bending forces K x , K y , the equations of 
equilibrium (20.14) for a compressed rod have the evident solution 
X = Y = 0, which corresponds to the rod's remaining straight under a 
longitudinal force |T|. This solution, however, gives a stable equilibrium 
of the rod only if the compressing force | T\ is less than a certain critical value 
Tor- For | T\ < T CT , the straight rod is stable with respect to any small pertur- 
bation. In other words, if the rod is slightly bent by some small force, it will 
tend to return to its original position when that force ceases to act. 

If, on the other hand, \T\ > T CT , the straight rod is in unstable equilibrium. 
An infinitesimal bending suffices to destroy the equilibrium, and a large 
bending of the rod results. It is clear that, if this is so, the compressed rod 
cannot actually remain straight. 

The behaviour of the rod after it ceases to be stable must satisfy the equa- 
tions for bending with large deflections. The value T cr of the critical load, 
however, can be obtained from the equations for small deflections. For 
\T\ = T cr , the straight rod is in neutral equilibrium. This means that, besides 
the solution X = Y = 0, there must also be states where the rod is slightly 
bent but still in equilibrium. Hence the critical value of T CI is the value of 
|T| for which the equations 

EI 2 X(W+ \T\X" = 0, EhY<™+ \T\Y" = (21.1) 

have a non-zero solution. This solution gives also the nature of the deforma- 
tion of the rod immediately after it ceases to be stable. 



98 The Equilibrium of Rods and Plates §21 

The following Problems give some typical cases of the loss of stability in 
various elastic systems. 

PROBLEMS 

Problem 1 . Determine the critical compression force for a rod with hinged ends. 

Solution. Since we are seeking the smallest value of \T\ for which equations (21.1) have 
a non-zero solution, it is sufficient to consider only the equation which contains the smaller 
of /j and I 2 . Let I % < I x . Then we seek a solution of the equation EI z X (lv ^-\-\T\X" = 
in the form X = A+Bz+C sin kz+D cos kz, where k = \/(\T\/EI 2 ). The non-zero 
solution which satisfies the conditions X = X" = for z = and z = / is X = C sin kz, 
with sin kl = 0. Hence we find the required critical force to be T cr = tt'EIJP. On ceasing 
to be stable, the rod takes the form shown in Fig. 19a. 



(a) 




Fig. 19 

Problem 2. The same as Problem 1, but for a rod with clamped ends (Fig. 19b). 
Solution. T cr = WEIJl 2 . 

Problem 3. The same as Problem 1, but for a rod with one end clamped and the other 
free (Fig. 19c). 

Solution. T cr = n*EJJW. 

Problem 4. Determine the critical compression force for a circular rod with hinged ends 
in an elastic medium (see §20, Problem 7). 

Solution. The equations (21.1) must now be replaced by EIXW + \T\X"+ctX = 0. 
A similar treatment gives the solution X = A sin nnzjl, 



tPEI/ od* \ 

Z2 \ nhflEl) 



where n is the integer for which T cr is least. When a is large, n > 1, i.e. the rod exhibits 
several undulations as soon as it ceases to be stable. 

Problem 5. A circular rod is subjected to torsion, its ends being clamped. Determine 
the critical torsion beyond which the straight rod becomes unstable. 

Solution. The critical value of the torsion angle is determined by the appearance of 
non-zero solutions of the equations for slight bending of a twisted rod. To derive these 
equations, we substitute the expression (19.7) M = EltXdt/dl+Crt, where t is the constant 
torsion angle, in equation (19.3). This gives 

d 2 t dt 



§21 The stability of elastic systems 99 

We differentiate ; since the bending is not large, t may be regarded as a constant vector t 
along the axis of the rod (the #-axis) in differentiating the first and third terms. Since also 
dF/d/ = (there being no external forces except at the ends of the rod), we obtain 

d3t d 2 t 

or, in components, 

Y^-kX'" = 0, 
X^+kY'" = 0, 

where k = CrjEI. Taking as the unknown function £ = X+iY,we obtain £< lv )— ii<£'" = 0. 
We seek a solution which satisfies the conditions £ = 0, £' = for z = and z = I, in the 
form £ = a(l+iKZ— e iKZ )+bz*, and obtain as the compatibility condition of the equations 
for a and b the relation e iKl = (2+iW)/(2— tW), whence \kI — tan \kI. The smallest root 
of this equation is \kI = 4-49, so that r cr = S-9SEIJCI. 

Problem 6. The same as Problem 5, but for a rod with hinged ends. 

Solution. In this case we have £ = a(l— e iKZ — i^z^+bz, where k is given by 

e iKl = 1, i.e. kI = 2tt. 

Hence the required critical torsion angle is t ci = 2-itEIJCI. 

Problem 7. Determine the limit of stability of a vertical rod under its own weight, the 
ower end being clamped. 

Solution. If the longitudinal stress F s = T varies along the rod, dF t /dl ^ in the 
first term of (20.1), and equations (20.14) are replaced by 

hEX^-{TX')'-K x = 0, 
hEYW-^TYJ-Ky = 0. 

In the case considered, there are no transverse bending forces anywhere in the rod, and 
T = — q(l— z), where q is the weight of the rod per unit length and z is measured from the 
lower end. Assuming that I t < I lt we consider the equation 



hEX'" = TX' = -q{l-z)X'\ 

illy. The general integral of 1 



for z = I, X'" — automatically. The general integral of this equation for the function 
« = X' is 



where 

The boundary conditions X' = for z — and X" = for z — I give for the function 
«(i?) the conditions u = for t) = 170 = $V(ql 3 JEh)> u'r] 113 = for 17 = 0. In order to satisfy 
these conditions we must put b = and J-i(-rjo) = 0- The smallest root of this equation 
is 170 = 1 -87, and so the critical length is l cr ~ 1 -98(£/ 2 /g) 1/3 . 

Problem 8. A rod has an elongated cross-section, so that I t ^> I x . One end is clamped 
and a force/ is applied to the other end, which is free, so as to bend it in the principal *«-plane 
(in which the flexural rigidity is EI 2 ). Determine the critical force / cr at which the rod bent 
in a plane becomes unstable and the rod is bent sideways (in the yz-plane), at the same time 
undergoing torsion. 



100 The Equilibrium of Rods and Plates §21 

Solution. Since the rigidity EI % is large compared with EI X (and with the torsional 
rigidity C),t the instability as regards sideways bending occurs while the deflection in the 
war-plane is still small. To determine the point where instability sets in, we must form the 
equations for slight sideways bending of the rod, retaining the terms proportional to the 
products of the force / in the xsr-plane and the small displacements. Since there is a concen- 
trated force only at the free end of the rod, we have F = f at all points, and at the free end 
(ar = I) the moment M = 0; from formula (19.6) we find the components of the moment 
relative to a fixed system of co-ordinates x, y, z: M x = 0, M v = {l—z)f, M z — (Y— Y )f, 
where Y — Y(l). Taking the components along co-ordinate axes £ , t], £ fixed at each point 
to the rod, we obtain as far as the first-order terms in the displacements Mg — <f>(l—z)f, 
M n = (/— z)f, M c = (l-z)fd Y/dz+f(Y— Y„), where <f> is the total angle of rotation of a 
cross-section of the rod under torsion; the torsion angle t = d<f>ldz is not constant along 
the rod. According to (18.6) and (18.9), however, we have for a small deflection 

M g = -EhY", M v = EI 2 X", M^ = Cf ; 

comparing, we obtain the equations of equilibrium 

EhX" = {l-z)f y EhY" = -<f>(l-z)f, 

Ccf>' = (l-z)fY' + (Y-Y )f. 

The first of these equations gives the main bending of the rod, in the xz-plane ; we require 
the value of/ for which non-zero solutions of the second and third equations appear. Eliminat- 
ing Y, we find 

<f>" + k\l- zf $ = 0, £2 = pfEhC. 

The general integral of this equation is 

* = «V(*- *!/*&*('- *)»]+ V('-*)/-i[P('-*) 2 ]. 

At the clamped end (z = 0) we must have <f> = 0, and at the free end the twisting moment 
C<f>' = 0. From the second condition we have a = 0, and then the first gives J-^kl 2 ) = 0. 
The smallest root of this equation is %kl 2 = 2-006, whence f CT = 4 -01 -\/^EIxC) 1 7*. 



t For example, for a narrow rectangular cross-section of sides b and h (i > h), we have 
EI X = bh 3 Ell2, ET t = VhE\\2, C = bh 3 fil3. 



CHAPTER III 

ELASTIC WAVES 

§22. Elastic waves in an isotropic medium 

If motion occurs in a deformed body, its temperature is not in general 
constant, but varies in both time and space. This considerably complicates 
the exact equations of motion in the general case of arbitrary motions. 

Usually, however, matters are simplified in that the transfer of heat from 
one part of the body to another (by simple thermal conduction) occurs very 
slowly. If the heat exchange during times of the order of the period of 
oscillatory motions in the body is negligible, we can regard any part of the 
body as thermally insulated, i.e. the motion is adiabatic. In adiabatic defor- 
mations, however, ow is given in terms of w$& by the usual formulae, the 
only difference being that the ordinary (isothermal) values of E and a must be 
replaced by their adiabatic values (see §6). We shall assume in what follows 
that this condition is fulfilled, and accordingly E and a in this chapter will be 
understood to have their adiabatic values. 

In order to obtain the equations of motion for an elastic medium, we must 
equate the internal stress force daucjdxjc to the product of the acceleration 
tit and the mass per unit volume of the body, i.e. its density p : 

pu t = da ik fdx k . (22.1) 

This is the general equation of motion. 

In particular, the equations of motion for an isotropic elastic medium can 
be written down at once by analogy with the equation of equilibrium (7.2). 
We have 

F E 

pu = A u + grad div u. (22.2) 

2(1 + a) 2(l + <r)(l-2a) S V 

Since all deformations are supposed small, the motions considered in the 
theory of elasticity are small elastic oscillations or elastic waves. We shall 
begin by discussing a plane elastic wave in an infinite isotropic medium, i.e. 
a wave in which the deformation u is a function only of one co-ordinate 
{x, say) and of the time. All derivatives with respect to y and z in equations 
(22.2) are then zero, and we obtain for the components of the vector u the 
equations 

dhi x 1 d 2 u x ^ d 2 u y 1 d 2 u y ,„„ „v 

— - = 0, — = (22.3) 

dx 2 ci 2 dt 2 dx 2 Ct 2 dt 2 

101 



102 Elastic Waves §22 

(the equation for u z is the same as that for u y ) ; heref 

Cl ~ Vp(l + a)(l-2a)' Ct = fe' (22 ' 4) 

Equations (22.3) are ordinary wave equations in one dimension, and the 
quantities ci and c t which appear in them are the velocities of propagation of 
the wave. We see that the velocity of propagation for the component u x is 
different from that for u y and u z . 

Thus an elastic wave is essentially two waves propagated independently. 
In one (u x ) the displacement is in the direction of propagation; this is called 
the longitudinal wave, and is propagated with velocity cj. In the other wave 
(%> «z) the displacement is in a plane perpendicular to the direction of propa- 
gation; this is called the transverse wave, and is propagated with velocity ct. 
It is seen from (22.4) that the velocity of longitudinal waves is always greater 
than that of transverse waves: we always have J 

c t > V(4/3)^. (22.5) 

The velocities ci and ct are often called the longitudinal and transverse veloci- 
ties of sound. 

We know that the volume change in a deformation is given by the sum of 
the diagonal terms in the strain tensor, i.e. by ua = div u. In the transverse 
wave there is no component u x , and, since the other components do not 
depend on y or z, div u = for such a wave. Thus transverse waves do not 
involve any change in volume of the parts of the body. For longitudinal 
waves, however, div u ^ 0, and these waves involve compressions and 
expansions in the body. 

The separation of the wave into two parts propagated independently with 
different velocities can also be effected in the general case of an arbitrary 
(not plane) elastic wave in an infinite medium. We rewrite equation (22.2) in 
terms of the velocities ci and ct : 

u = c t 2 A u + (q 2 - c t 2 ) grad div u. (22.6) 

We then represent the vector u as the sum of two parts : 

u = ui + ut, (22.7) 

of which one satisfies 

div u t = (22.8) 

and the other satisfies 

curl ui = 0. (22.9) 

We know from vector analysis that this representation (i.e. the expression of 



t We may give also expressions for Cj and ct in terms of the moduli of compression and rigidity 
and the Lam6 coefficients: c, = V{(3i^+V)/3p} = V{ (A+ 2fi)/p }, c t = V(m/p)- 

J Since a actually varies only between and J (see the second footnote to §5), we always have 



§22 Elastic waves in an isotropic medium 103 

a vector as the sum of the curl of a vector and the gradient of a scalar) is 
always possible. 

Substituting u = uj + u f in (22.6), we obtain 

Uj+ii* = c t 2 A(ui + u t ) + (ci 2 -ct 2 ) grad divuj. (22.10) 

We take the divergence of both sides. Since div u t = 0, the result is 
diviij = c t 2 A div ui + (c^-c t 2 ) A divuj, 

or div(ui-ci 2 /\ui) = 0. The curl of the expression in parentheses is also 
zero, by (22.9). If the curl and divergence of a vector both vanish in all 
space, that vector must be zero identically. Thus 

— - 'Muz = 0. (22.11) 
dp 

Similarly, taking the curl of equation (22.10) we have, since the curls of u t 
and of any gradient are zero, curl (u t -c t 2 A^t) = 0. Since the divergence 
of the expression in parentheses is also zero, we obtain an equation of the 
same form as (22.11): 

— - tfAut = 0. (22.12) 
dt 2 

Equations (22.11) and (22.12) are ordinary wave equations in three dimen- 
sions. Each of them represents the propagation of an elastic wave, with 
velocity c\ and c t respectively. One wave (u t ) does not involve a change in 
volume (since div u t = 0), while the other (u{) is accompanied by volume 
compressions and expansions. 

In a monochromatic elastic wave, the displacement vector is 

u = re{uo(r)*-^}, (22.13) 

where uo is a function of the co-ordinates which satisfies the equation 

c t 2 A uo + {ci 2 - c t 2 ) grad div uo + <u 2 uo = 0, (22. 14) 

obtained by substituting (22.13) in (22.6). The longitudinal and transverse 
parts of a monochromatic wave satisfy the equations 

Aui + J^ui = 0, Au, + £t 2 ut= 0, (22.15) 

where ki = co/cj, k t = cojc t are the wave numbers of the longitudinal and 
transverse waves. 

Finally, let us consider the reflection and refraction of a plane mono- 
chromatic elastic wave at the boundary between two different elastic media. 
It must be borne in mind that the nature of the wave is in general changed 
when it is reflected or refracted. If a purely transverse or purely longitudinal 
wave is incident on a surface of separation, the result is a mixed wave con- 
taining both transverse and longitudinal parts. The nature of the wave 



104 



Elastic Waves 



§22 



remains unchanged (as we see from symmetry) only when it is incident 
normally on the surface of separation, or when a transverse wave whose 
oscillations are parallel to the plane of separation is incident (at any angle). 
The relations giving the directions of the reflected and refracted waves can 
be obtained immediately from the constancy of the frequency and of the 
tangential components of the wave vector, f Let 6 and 6' be the angles of 
incidence and reflection (or refraction) and c, c' the velocities of the two waves. 
Then 

sin# c 

~^¥ = V (22 ' 16) 

sina c 
For example, let the incident wave be transverse. Then c = ct\ is the 
velocity of transverse waves in medium 1. For the transverse reflected wave 
we have c' = c t \ also, so that (22.16) gives 6 = 0', i.e. the angle of incidence 
is equal to the angle of reflection. For the longitudinal reflected wave, 
however, c' = en, and so 

sin 6 ca 

sin 6' en 
For the transverse part of the refracted wave c' = ct2> and for a transverse 
incident wave 

sin 6 en 

sin d' c t 2 
Similarly, for the longitudinal refracted wave 

sin 9 ct\ 

sin 6' C12 



PROBLEMS 
Problem 1. Determine the reflection coefficient for a longitudinal monochromatic wave 
incident at any angle on the surface of a body (with a vacuum outside).* 




Fig. 20 
Solution. When the wave is reflected, there are in general both longitudinal and trans- 
verse reflected waves. It is clear from symmetry that the displacement vector in the trans- 
verse reflected wave lies in the plane of incidence (Fig. 20, where n , nj and nj are unit 



f See Fluid Mechanics, §65. The arguments given there are applicable in their entirety. 

I The more general case of the reflection of sound waves from a solid-liquid interface, and the 
similar problem of the reflection of a wave incident from a liquid on to a solid, are discussed by L. M. 
Brekhovskikh, Waves in Layered Media, §4, Academic Press, New York 1960. 



§22 Elastic waves in an isotropic medium 105 

vectors in the direction of propagation of the incident, longitudinal reflected and transverse 
reflected waves, and u , u t , Ut the corresponding displacement vectors). The total displace- 
ment in the body is given by the sum (omitting the common factor e~ im for brevity) 

u = A noe ik *- r +Ainie ik rr+A t aXnte ik f r , 

where a is a unit vector perpendicular to the plane of incidence. The magnitudes of the wave 
vectors are k = k t = w/cj, kt = ojct, and the angles of incidence 9 and of reflection 9 U 
&t are related by 0j = 9 , sin dt = (ct/cj) sin O . For the components of the strain tensor at 
the boundary we obtain 

u xx = ik<s{AQ + Ai) cos 2 #o + iAtkt cos 0* sin 0*, uu = iko(Ao + At), 
u X y = iko(Ao — Ai) sin do cos 6o+ ^iAtk t (cos 2 6 t — sin 2 0*), 

again omitting the common exponential factor. The components of the stress tensor can be 
calculated from the general formula (5.11), which can here be conveniently written 

o ilc = 2pC t 2 Uik + p(cp -2c t 2 )uu8 ik . 

The boundary conditions at the free surface of the medium are a<k«* = 0, whence 

<*xx — a yx — 0, 
giving two equations which express Ai and At in terms of A . The result is 

c t 2 sin 20* sin 20 o - c t 2 cos 2 20* 



A x = A 



A t = -A G 



c t 2 sin 20* sin 20 o + Q 2 cos 2 2d t 
2ciCt sin 20o cos 20* 



c t 2 sin 20* sin 20 o + c t 2 cos 2 20* 



For 9 = we have A t = — A , At = 0, i.e. the wave is reflected as a purely longitudinal 
wave. The ratio of the energy flux density components normal to the surface in the reflected 
and incident longitudinal waves is Ri = |^4jM | a - The corresponding ratio for the reflected 
transverse wave is 



Rt = 



c t cos 0* 



A 



|2 



Ci COS 00 
The sum of R t and Rt is, of course, 1. 

Problem 2. The same as Problem 1, but for a transverse incident wave (with the oscilla- 
tions in the plane of incidence). -f 

Solution. The wave is reflected as a transverse and a longitudinal wave, with 0< = o , 
Ct sin 9 % = Cj sin 9 . The total displacement vector is 

u = aXno^oe* k »- r +n^ze <k '' r +aXn*^*e* k '-«\ 

The expressions for the amplitudes of the reflected waves are 

A t c t 2 sin 20; sin 20 o - ci 2 cos 2 20 o 

Aq ~~ c t 2 sin 20j sin 20 o + ci 2 cos 2 20 o ' 
Ai 2ciCf sin 20o cos 20o 

Aq ~ c t 2 sin 20; sin 20 o + ci 2 cos 2 20 o ' 



t If the oscillations are perpendicular to the plane of incidence, the wave is entirely reflected a* a 
wave of the same kind, and so Rt = 1 . 



106 Elastic Waves §23 

Problem 3. Determine the characteristic frequencies of radial vibrations of an elastic 
phere of radius R. 

Solution. We take spherical polar co-ordinates, with the origin at the centre of the sphere. 
For radial vibrations, u is along the radius, and is a function of r and t only. Hence curl u = 0. 
We define the displacement "potential" 4> by u r = u = 8<f>ldr. The equation of motion, 
expressed in terms of <f>, is just the wave equation ci s A<A = $, or, for oscillations periodic in 
time (~e~ i<Jit ), 

The solution which is finite at the origin is <j> = (A/r) sin kr (the time factor is omitted). The 
radial stress is 



a rr = pi (ci 2 — 2ct 2 )uu + 2c t 2 U rr 
= p\{c l 2 -2c t 2 )/\<l> + 2c t 2 <l>" 



or, using (1), 

Orrlp= -o> 2 <f>-W<f>'lr. (2) 

The boundary condition a rr (R) = leads to the equation 

tan kR 1 



kR l-(kRcil2c t ) 2 ' (3) 

whose roots determine the characteristic frequencies w = kci of the vibrations. 

Problem 4. Determine the frequency of radial vibrations of a spherical cavity in an infinite 
elastic medium for which ci !> c t (M. A. Isakovich 1949). 

Solution. In an infinite medium, radial oscillations of the cavity are accompanied by the 
emission of longitudinal sound waves, leading to loss of energy and hence to damping of the 
oscillations. When ci ^> c t (i.e. K ^> /x), this emission is weak, and we can speak of the charac- 
teristic frequencies of oscillations with a small coefficient of damping. 

We seek a solution of equation (1), Problem 3, in the form of an outgoing spherical wave 
ff> = Ae ikr lr, k = oi\ci and, using (2), obtain from the boundary condition o rr (R) = the 
result {kRcijctY = 4(1 —ikR). Hence, when c? ^> c t , 

lc t I c t \ 
co — — 1 1— i — . 
R\ ci) 

The real part of ou gives the characteristic frequency of oscillation ; the imaginary part gives 
the damping coefficient. In an incompressible medium (ci ~> oo) there would of course be no 
damping. These vibrations are specifically due to the shear resistance of the medium (fi #0). 
Il should be noticed that they have kR = 2ct\ci <^. 1 , i.e. the corresponding wavelength is 
large compared with R ; it is interesting to compare this with the result for vibrations of an 
elastic sphere, where with ci ^> ct the first characteristic frequency is given by (3) : kR = -it. 

§23. Elastic waves in crystals 

The propagation of elastic waves in anisotropic media, i.e. in crystals, is 
more complicated than for the case of isotropic media. To investigate such 
waves, we must return to the general equations of motion piii = dotje/dxk 
and use for o-^ the general expression (10.3) 0% = \MmUim- According to 



§23 Elastic waves in crystals 107 

what was said at the beginning of §22, A^zm always denotes the adiabatic 
moduli of elasticity. 

Substituting for o^ in the equations of motion, we obtain 

duim v d l dui du 



piii = XiMm— — = l^iklm 

OXjc OXfc 



l dui du m \ 



d 2 u t d 2 u m 

= iMklmrz — H f Mklmr 



dx]cdx m dxjcdxi 

Since the tensor Xikim is symmetrical with respect to the suffixes / and m 
we can interchange these in the first term, which then becomes identical 
with the second term. Thus the equations of motion are 

d 2 Um 
piii = Xikim— — - — • (23.1) 

OXkOXi 

Let us consider a monochromatic elastic wave in a crystal. We can seek 
a solution of the equations of motion in the form Ui = uoie i{k ' r ~ a)t \ where the 
uoi are constants, the relation between the wave vector k and the frequency (o 
being such that this function actually satisfies equation (23.1). Differentiation 
of m with respect to time results in multiplication by — ico, and differentia- 
tion with respect to xjc leads to multiplication by ikjc. Hence the above substi- 
tution converts equation (23.1) into pcohii = AijcimkkkiUm. Puttings = 8i m Um, 
we can write this as 

(p(o 2 him-\ucimkkki)U m = 0. (23.2) 

This is a set of three homogeneous equations of the first degree for the 
unknowns u%, u y , u z . Such equations have non-zero solutions only if the 
determinant of the coefficients is zero. Thus we must have 

\hkimkkh— poi 2 8 im \ — 0. (23.3) 

This is a cubic equation in co 2 . It has three roots, which are in general 
different. Each root gives the frequency as a function of the wave vector k.f 

Substituting each in turn in equation (23.2), we obtain equations giving 
the components of the corresponding displacement w* (since the equations 
are homogeneous, of course, only the ratios of the three components «* are 
obtained, and not their absolute values, so that all the m can be multiplied 
by an arbitrary constant). 

The velocity of propagation of the wave (the group velocity) is given by the 
derivative of the frequency with respect to the wave vector. In an isotropic 
body, the frequency is proportional to the magnitude of k, and so the direc- 
tion of the velocity U = dco/dk is the same as that of k. In crystals this 
relation does not hold, and the direction of propagation of the wave is there- 
fore not the same as that of its wave vector. 



t In an isotropic body, equation (23.3) gives the result previously obtained: one root u>* = c*k* 
and two coincident roots to 2 = C( 2 A a . 



108 Elastic Waves §23 

It is seen from equation (23.3) that o> is a homogeneous function, of degree 
one, of the components ki ; if the unknown quantity is taken as the ratio cufk, 
the coefficients in the equation do not depend on k. Hence the velocity of 
propagation 9cu/8k is a homogeneous function, of degree zero, of ki. Thus 
the velocity of a wave is a function of its direction, but not of its frequency. 

Since there are three possible relations between o» and k for any direction 
in the crystal, there are in general three different velocities of propagation 
of elastic waves. These velocities are the same only in a few exceptional 
directions. 

In an isotropic medium, purely longitudinal and purely transverse waves 
correspond to two different velocities of propagation. In a crystal, on the 
other hand, to each velocity of propagation there corresponds a wave in which 
the displacement vector has components both parallel and perpendicular 
to the direction of propagation. 

Finally, we may notice the following. For any given wave vector k in a 
crystal there can be three waves, with different frequencies and velocities of 
propagation. It is easy to see that the displacement vectors u in these three 
waves are mutually perpendicular. For, when k is given, equation (23.3) may 
be regarded as determining the principal values pco 2 of a tensor of rank two, 
Xiicimkkki, which is symmetrical with respect to the suffixes i, m.f Equations 

(23.2) then give the principal axes of this tensor, which we know are mutually 
perpendicular. 

PROBLEM 

Determine the frequency as a function of the wave vector for elastic waves propagated in a 
crystal of the hexagonal system. 

Solution. The non-zero components of the tensor A»&jm in the co-ordinates x, y, z are 
related to those in the co-ordinates £, ■q, z (see §10) by 

Axxxx — Ayyyy = a + b, Axxyy — a ~ b y Axyxy = b, 

^XXZZ = Ayyzz = C, &XZXZ = AyZyZ = U, &ZZZZ = /> 

where we have put 

The ar-axis is along the sixth-order axis of symmetry; the directions of the x and y axes are 
arbitrary. We take the xz-plane such that it contains the wave vector k. Then k x = k sin 6, 
k y — 0, kz = k cos 0, where 6 is the angle between k and the ar-axis. Forming the equation 

(23.3) and solving it, we obtain three different dependences of to on k: 

co! 2 = k 2 (bs'm2d + dcos 2 e)! P , 

k? 
C02 3 2 = — {(a + b) sinW+f cosW+d± ^([(a+ b - d) sinW + (d-f) cosWf + 
2 P 

+ 4(c+rf) 2 sm 2 0cos 2 0)}. 



t By the symmetry of the tensor \ncim, we have Xikimkkki = hktmikjcki = Xmihikicki- The latter 
expression differs from Xmkiikjcki only by the naming of the suffixes k and /, so that the tensor 
Xikimkkki has the symmetry stated. 



§24 Surface waves 109 

§24. Surface waves 

A particular kind of elastic waves are those propagated near the surface 
of a body without penetrating into it (Rayleigh waves). We write the equation 
of motion in the form (22.11) and (22.12): 

— - c*Au = 0, (24.1) 

dt 2 

where u is any component of the vectors uj, u t , and c is the corresponding 
velocity c\ or c t , and srek solutions corresponding to these surface waves. 
The surface of the elastic medium is supposed plane and of infinite extent. 
We take this plane as the ry-plane; let the medium be in z < 0. 

Let us consider a plane monochromatic surface wave propagated along the 
#-axis. Accordingly u = «<(**-**>/(*). Substituting this expression in (24.1), we 
obtain for the function f(z) the equation 



dy 

d*2 






If k 2 - ofijc 2 < 0, this equation gives a periodic function /, i.e. we obtain 
an ordinary plane wave which is not damped inside the body. We must 
therefore suppose that k 2 -w 2 ]c 2 > 0. Then the solutions for /are 

f(z) = constant x exp { ± J \k 2 -m. 

The solution with the minus sign would correspond to an unlimited increase 
in the deformation for z -» — oo. This solution is clearly impossible, and 
so the plus sign must be taken. 

Thus we have the following solution of the equations of motion: 

u = constant x «***-**>««, (24.2) 

where 

K = ^(k 2 -a> 2 !c 2 ). (24.3) 

It corresponds to a wave which is exponentially damped towards the interior 
of the medium, i.e. is propagated only near the surface. The quantity k 
determines the rapidity of the damping. 

The true displacement vector u in the wave is the sum of the vectors uj and 
Uf, the components of each of which satisfy the equation (24.1) with c = ci 
for ui and ct for u$. For volume waves in an infinite medium, the two parts 
are independently propagated waves. For surface waves, however, this 
division into two independent parts is not possible, on account of the boundary 
conditions. The displacement vector u must be a definite linear combination 
of the vectors u* and Uf . It should also be mentioned that these latter vectors 
have no longer the simple significance of the displacement components 
parallel and perpendicular to the direction of propagation. 



110 Elastic Waves §24 

To determine the linear combination of the vectors uj and ut which gives 
the true displacement u, we must use the conditions at the boundary of the 
body. These give a relation between the wave vector k and the frequency &>, 
and therefore the velocity of propagation of the wave. At the free surface 
we must have cr^n fc = 0. Since the normal vector n is parallel to the s-axis, 
it follows that a xz = o yz = a zz = 0, whence 

U X z = 0, Uyz = 0, v(u Z x + Uyy) + (l-v)Uzz = 0. (24.4) 

Since all quantities are independent of the co-ordinate y, the second of 
these conditions gives 

1 iduy du z \ 
Uyz = - \-^- + -J~) = % du vl dz = °* 

Using (24.2), we therefore have 

u y = 0. (24.5) 

Thus the displacement vector u in a surface wave is in a plane through the 
direction of propagation perpendicular to the surface. 

The transverse part u$ of the wave must satisfy the condition (22.8) 
div M t = 0, or 

du tx du tz 

.+-—. = 0. 



dx dz 

The dependence of utx and Utz on x and z is determined by the factor 
e ikx+K t z , where t<t is given by the expression (24.3) with c = ct, i.e. 

K = \/(k 2 —C0 2 jCt 2 ). 

Hence the above condition leads to the equation 

ikutx+KtUtz = 0, or ut x /ut z = —K t Jik. 
Thus we can write 

u tx = K t ae ikx + K t z - i(0 t y u tz = -ikae ikx+K t z - icot , (24.6) 

where a is some constant. 

The longitudinal part u* satisfies the condition (22.9) curl uj = 0, or 

dui x dui z ^ 
dz dx 

whence 

ikui z —KiUix = (ki = \Z[k 2 —co 2 Jci 2 ]). 
Thus we must have 

uix = kbe ikx+K i z - ia,t , ui z = -iKibe ikx+K i z - iu)t , (24.7) 

where b is a constant. 



§24 Surface waves 111 

We now use the first and third conditions (24.4). Expressing um in terms 
of the derivatives of U{, and using the velocities ci, ct, we can write these 
conditions as 

du x du z 
+ = 0, 

dZ dX (24.8) 

du z du x 

oz ox 

Here we must substitute u x = ui x + ut%, u z = ttiz + utz- The result is that 
the first condition (24.8) gives 

a(k? + Kt*) + 2bkKi = 0. (24.9) 

The second condition leads to the equation 

2ac?K& + b[c?(ic?-k*) + 2c?k*] = 0. 

Dividing this equation by Ct 2 and substituting 

we can write it as 

2aic£ + b(&+K?) = 0. (24.10) 

The condition for the two homogeneous equations (24.9) and (24.10) 
to be compatible is (k 2 + Kt 2 ) 2 = 4k 2 K t Ki or, squaring and substituting 
the values of K t 2 and ki 2 , 

I to 2 \ 4 / co 2 \ / ft) 2 \ 

From this equation we obtain the relation between o> and k. It is convenient 
to put 

co = cM\ (24.12) 

k 8 then cancels from both sides of the equation, and, expanding, we obtain 
for £ the equation 

$«-8£*+S€ 2 (3-2—\-l6(\ - —\ = 0. (24.13) 

Hence we see that £ depends only on the ratio ctjci, which is a constant 
characteristic of any given substance and in turn depends only on Poisson's 
ratio : 

CtjCi= V((l-2<r)/2(l-a)}. 



112 



Elastic Waves 



§24 



The quantity £ must, of course, be real and positive, and £ < 1 (so that 
Kt and ki are real). Equation (24.13) has only one root satisfying these con- 
ditions, and so a single value of f is obtained for any given value of ct/ci. 

Thus, for both surface waves and volume waves, the frequency is pro- 
portional to the wave number. The proportionality coefficient is the velocity 
of propagation of the wave, 

U = c&. (24.14) 

This gives the velocity of propagation of surface waves in terms of the 
velocities ct and ci of the transverse and longitudinal volume waves. The 
ratio of the amplitudes of the transverse and longitudinal parts of the wave 
is given in terms of £ by the formula 



2-|2 



2V(1-| 2 ) 



(24.15) 



The ratio ct\ci actually varies from \jy/2 to for various substances, 
corresponding to the variation of a from to \\ £ then varies from 0-874 to 
0-955. Fig. 21 shows a graph of f as a function of a. 



100 



0-95 



0-90 



085 




PROBLEM 

A plane- parallel slab of thickness h (medium 1) lies on an elastic half-space (medium 2). 
Determine the frequency as a function of the wave number for transverse waves in the slab 
whose direction of oscillation is parallel to its boundaries. 

Solution. We take the plane separating the slab from the half-space as the xy-plane, 
the half-space being in z < and the slab in s£ z <; h. In the slab we have 

Uzl = u zl = 0, u yl = f{z)eM*-<>>t\ 

and in medium 2 a damped wave : 

Ux2 = U z2 = 0, % 2 = ^eV« i(M) , K2 = V(k 2 -0 2 ICt2 2 ). 

For the function f(z) we have the equation 



/"+ K?f =0, Kl = V(«> 2 fCtl 2 -k*) 



§25 Vibration of rods and plates ' 113 

(we shall see below that k^ > 0), whence f(z) = B sin k x z+C cos k x z. At the free surface 
of the slab {z — h) we must have o ty — 0, i.e. duyjdz = 0. At the boundary between the 
two media (z = 0) the conditions are u yl = u vt , frdtij/Jdz = fi^du^dz, ^ and /i, being the 
moduli of rigidity for the two media. From these conditions we find three equations for 
A, B, C, and the compatibility condition is tan k-Ji = iH K ilfh. K i- This equation gives <o 
as an implicit function of k ; it has solutions only for real #c x and k 2 , and so Ct% > (o/k > en. 
Hence we see that such waves can be propagated only if c*2 > c«i. 

§25. Vibration of rods and plates 

Waves propagated in thin rods and plates are fundamentally different 
from those propagated in a medium infinite in all directions. Here we are 
speaking of waves of length large compared with the thickness of the rod or 
plate. If the wavelength is small compared with this thickness, the rod or 
plate is effectively infinite in all directions as regards the propagation of the 
wave, and we return to the results obtained for infinite media. 

Waves in which the oscillations are parallel to the axis of the rod or the 
plane of the plate must be distinguished from those in which they are per- 
pendicular to it. We shall begin by studying longitudinal waves in rods. 

A longitudinal deformation of the rod (uniform over any cross-section), with 
no external force on the sides of the rod, is a simple extension or compression. 
Thus longitudinal waves in a rod are simple extensions or compressions 
propagated along its length. In a simple extension, however, only the com- 
ponent a zz of the stress tensor (the ar-axis being along the rod) is different 
from zero; it is related to the strain tensor by a zz = Eu zz = Edu z jdz 
(see §5). Substituting this in the general equation of motion pu z = 9<t z a;/3^a;» 
we find 

^_Z_^ = . (25.1) 

This is the equation of longitudinal vibrations in rods. We see that it is an 
ordinary wave equation. The velocity of propagation of longitudinal waves 
in rods is 

V(E/p). (25.2) 

Comparing this with the expression (22.4) for q, we see that it is less than 
the velocity of propagation of longitudinal waves in an infinite medium. 

Let us now consider longitudinal waves in thin plates. The equations of 
motion for such vibrations can be written down at once by substituting 
— phdhiz/dt 2 and —phdhiyjdt 2 for P x and P y in the equilibrium equations 
(13.4): 



p d 2 u z 1 d 2 u x 1 d 2 u x 1 d 2 u y 

~E~W = l-a 2 1hP + 2(\ + a) Hy* + 2{\-o)lh&y' 

p d 2 u y 1 d 2 u y 1 B 2 Uy 1 d 2 u z 

•+t— : — _ „ +• 



(25.3) 



E dt 2 1 - a 2 dy 2 2(1 + a) Bx 2 2(1 - a) dxdy 



114 Elastic Waves §25 

We take the case of a plane wave propagated along the #-axis, i.e. a wave in 
which the deformation depends only on the co-ordinate x, and not on y. 
Then equations (25.3) are much simplified, becoming 

&U % E d*U X = d*Uy E d*Uy = 

d& pO—a 2 ) d*2 ' dfi 2p(l + a) dx* ' ^ ' ' 

We thus again obtain wave equations. The coefficients are different for u x 
and u y . The velocity of propagation of a wave with oscillations parallel to the 
direction of propagation (u x ) is 

ViElrfl-6*)]. (25.5) 

The velocity for a wave (%) with oscillations perpendicular to the direction 
of propagation (but still in the plane of the plate) is equal to the velocity ct of 
transverse waves in an infinite medium. 

Thus we see that longitudinal waves in rods and plates are of the same 
nature as in an infinite medium, only the velocity being different; as before, 
it is independent of the frequency. Entirely different results are obtained for 
bending waves in rods and plates, for which the oscillations are in a direction 
perpendicular to the axis of the rod or the plane of the plate, i.e. involve 
bending. 

The equations for free oscillations of a plate can be written down at once 
from the equilibrium equation (12.5). To do so, we must replace — P by the 
acceleration £ multiplied by the mass ph per unit area of the plate. This 
gives 

9 2 £ Eh* 

»w + W^) AH = °' (25 ' 6) 

where A is the two-dimensional Laplacian. 

Let us consider a monochromatic elastic wave, and accordingly seek a 
solution of equation (25.6) in the form 

£ = constant xe« k "- wi ), (25.7) 

where the wave vector k has, of course, only two components, k x and k y . 
Substituting in (25.6), we obtain the equation 

-pco 2 + Eh*k*/12(l-o2) = 0. 

Hence we have the following relation between the frequency and the wave 
number : 

a) = kW{Eh*ll2 P (l - <x2)}. (25.8) 

Thus the frequency is proportional to the square of the wave number, whereas 
in waves in an infinite medium it is proportional to the wave number itself. 



§25 Vibration of rods and plates 115 

Knowing the relation between the frequency and the wave number, we can 
determine the velocity of propagation of the wave from the formula 

U = dcojdk. 

The derivatives of k 2 with respect to the components k Xi k y are respectively 
2k x , 2k y . The velocity of propagation of the wave is therefore 

U = kV{Eh*/3 P (l - a 2 )}. (25.9) 

It is proportional to the wave vector, and not a constant as it is for waves in 
a medium infinite in three dimensions.^ 

Similar results are obtained for bending waves in thin rods. The bending 
deflections of the rod are supposed small. The equations of motion are 
obtained by replacing — K x and — K y in the equations of equilibrium for a 
slightly bent rod (20.4) by the product of the acceleration X or Y and the 
mass pS per unit length of the rod (S being its cross-sectional area). Thus 

P SX = EI y d*XJdz\ P SY = ElaPY/dz*. (25.10) 

We again seek solutions of these equations in the form 

X = constant x e«*z-<*« Y = constant x e i{kz - mt \ 

Substituting in (25.10), we obtain the following relations between the fre- 
quency and the wave number : 

co = kty(EIyl P S), co = kW(EIxlpS), (25.11) 

for vibrations in the x and y directions respectively. The corresponding 
velocities of propagation are 

U<*> = Ik^iEIyfpS), W> = 2kV(EI x l P S). (25.12) 

Finally, there is a particular case of vibration of rods called torsional 
vibration. The corresponding equations of motion are derived by equating 
Cdrjdz (see §18) to the time derivative of the angular momentum of the rod 
per unit length. This angular momentum is pld<f>fdt y where d<j>jdt is the 
angular velocity (<f> being the angle of rotation of the cross-section considered) 
and / = J (x 2 +y 2 ) d/ is the moment of inertia of the cross-section about its 
centre of mass; for pure torsional vibration each cross-section of the rod 
performs rotary vibrations about its centre of mass, which remains at rest. 
Putting t = d<f>/dz, we obtain the equation of motion in the form 

Cd 2 <f>/dz 2 = pidmet 2 . (25.13) 



f The wave number k = 2-ir/X, where A is the wavelength. Hence the velocity of propagation 
should increase without limit as A tends to zero. This physically impossible result is obtained because 
formula (25.9) is not valid for short waves. 



116 Elastic Waves §25 

Hence we see that the velocity of propagation of torsional oscillations along 
the rod is 

V(C/ P I). (25.14) 

PROBLEMS 

Problem 1. Determine the characteristic frequencies of longitudinal vibrations of a rod 
of length /, with one end fixed and the other free. 

Solution. At the fixed end (z = 0) we must have u t = 0, and at the free end (z = /) 
<*« = Eutg = 0, i.e. duzldz = 0. We seek a solution of equation (25.1) in the form 

# 2 = A cos(a>t + (x) sinks, 

where k = <o\Z(p/E). From the condition at z = I we have cos kl — 0, whence the charac- 
teristic frequencies are 

eo = V(Elp)(2n+l)7rl2l, 

n being any integer. 

Problem 2. The same as Problem 1, but for a rod with both ends free or both fixed. 
Solution. In either case <o = \/(E/p) tin/ 1. 

Problem 3. Determine the characteristic frequencies of vibration of a string of length /. 
Solution. The equation of motion of the string is 

d*X oS d*X 
L == 0; 

cf. the equilibrium equation (20.17). The boundary conditions are that X — f or z — 
and /. The characteristic frequencies are o> = ■y/(pS/T)nn/l. 

Problem 4. Determine the characteristic transverse vibrations of a rod (of length /) with 
clamped ends. 

Solution. Equation (25.10), on substituting X = X (z) cos(a>t +<x), becomes 

d^o/d* 4 = k*Xo, 

where k 1 = uPpS/EIy. The general integral of this equation is 

Xq = A cos kz + B sin kz+C cosh kz+D sinh kz. 

The constants A, B, C and D are determined from the boundary conditions that X = dX/dz 
= f or z = and /. The result is 

Xq = Aftsmfcl— sinh «r/)(cos kz— cosh kz) — 
— (cos kI— cosh «/)(sin kz — sinh kz)}, 

and the equation coskZ cosh#tZ = 1, the roots of which give the characteristic frequencies 
The smallest characteristic frequency is 

224 l Eh 

COmin = 



■ I Ely 

V pS 



/2 V P S 
Problem 5. The same as Problem 4, but for a rod with supported ends. 



§25 Vibration of rods and plates 117 

Solution. In the same way as in Problem 4, we obtain X = A sin kz, and the frequencies 
are given by sin*/ = 0, i.e. k = nn/l(n = 1, 2, ...). The smallest frequency is 



y-87 I Ely 



Problem 6. The same as Problem 4, but for a rod with one end clamped and the other 
free. 

Solution. We have for the displacement 

Xo = A{(cos kI+ cosh k/)(cos kz— cosh kz) 

+ (sin kI— sinh /c/)(sin kz — sinh kz)} 

(the clamped end being at z — and the free end atz = I), and for the characteristic fre- 
quencies the equation cos kI cosh kZ+1 = 0. The smallest frequency is 

3-52 J Ely 

COmin 



3-5Z I Ely 
= ~¥~J~p~S~' 



P S 

Problem 7. Determine the characteristic vibrations of a rectangular plate of sides a and b, 
with its edges supported. 

Solution. Equation (25.6), on substituting £ = l (x, y) cos(a>t+a), becomes 

where #c* = 12p(l — a^wPjEh*. We take the co-ordinate axes along the sides of the plate. 
The boundary conditions (12.11) become £ = 9 2 £/3* 8 = for x = and a, 

for y = and b. The solution which satisfies these conditions is 

Co = A sm(m7rx/a) s'mfoiry/b), 

where m and n are integers. The frequencies are given by 

E _ [m 2 w 2 l 

12o(l-o- 2 ) 7 



CO = h / 7T 2 — + — . 

Vl2p(l-a2) [ a 2 #jj 



Problem 8. Determine the characteristic frequencies for the vibration of a rectangular 
membrane of sides a and b. 

Solution. The equation for the vibration of a membrane is TA £ — phi', cf. the equili- 
brium equation (14.9). The edges of the membrane must be fixed, so that £ — 0. The 
corresponding solution for a rectangular membrane is 

£ = A sm(niTTxla) sin(»7ry/#) cos cot, 

where the characteristic frequencies are given by 

Ttt 2 /m 2 n 2 \ 

ft) 2 = — — +— I, 

ph W bV 
m and n being integers. 



118 Elastic Waves §26 

Problem 9. Determine the velocity of propagation of torsional vibrations in a rod whose 
cross-section is a circle, an ellipse, or an equilateral triangle, and in a rod in the form of a 
long thin rectangular plate. 

Solution. For a circular cross-section of radius R, the moment of inertia is / = \ttR*', 
C is given in §16, Problem 1, and we find the velocity to be V(plp), which is the same as the 
velocity ct. 

Similarly (using the results of §16, Problems 2 to 4), we find for a rod of elliptical cross- 
section the velocity [2a6/(a 2 +fc 2 )] vX^/p). for one with an equilateral triangular cross-section 
V(3l*l5p), and for one which is a long rectangular plate (2h/d)\/(fjL/p). All these are less than a. 

Problem 10. The surface of an incompressible fluid of infinite depth is covered by a thin 
elastic plate. Determine the relation between the wave number and the frequency for waves 
which are simultaneously propagated in the plate and near the surface of the fluid. 

Solution. We take the plane of the plate as 2 = 0, and the x-axis in the direction of 
propagation of the wave ; let the fluid be in z < 0. The equation of motion of the plate alone 
would be 

d% _ Em d*£ 

po Jfi ~ ~ 12(1-0*) a? 

where p is the volume density of the plate. When the fluid is present, the right-hand side 
of this equation must also include the force exerted by the fluid on unit area of the plate, 
i.e. the pressure p of the fluid. The pressure in the wave, however, can be expressed in 
terms of the velocity potential by p = —pd<f>ldt (we neglect gravity). Hence we obtain 

a 2 £ _ Em wz, r d<f>i 
poh ~dF = ~ 12(1 - CT 2) W " l Tt\z=o (1) 

Next, the normal component of the fluid velocity at the surface must be equal to that of the 
plate, whence 

dydt = [ty/a*],..o. (2) 

The potential <f> must satisfy everywhere in the fluid the equation 

~+- J ~ = 0. (3) 

dx 2 dz 2 

We seek £ in the form of a travelling wave £ = £ e ikx ~ ia)t ; accordingly, we take as the 
solution of equation (3) the surface wave <f> = <f> e i( - lcz ~ M '>e lcz , which is damped in the interior 
of the fluid. Substituting these expressions in (1) and (2), we obtain two equations for ^ 
and £o» and the compatibility condition is 

Eh^ & 



12(1 -a 2 ) p + hpok 

§26. Anharmonic vibrations 

The whole of the theory of elastic vibrations given above is approximate 
to the extent that any theory of elasticity is so which is based on Hooke's 
law. It should be recalled that the theory begins from an expansion of the 
elastic energy as a power series with respect to the strain tensor, which 
includes terms up to and including the second order. The components of 



§26 Anharmonic vibrations 119 

the stress tensor are then linear functions of those of the strain tensor, and 
the equations of motion are linear. 

The most characteristic property of elastic waves in this approximation is 
that any wave can be obtained by simple superposition (i.e. as a linear com- 
bination) of separate monochromatic waves. Each of these is propagated 
independently, and could exist by itself without involving any other motion. 
We may say that the various monochromatic waves which are simultaneously 
propagated in a single medium do not interact with one another. 

These properties, however, no longer hold in subsequent approximations. 
The effects which appear in these approximations, though small, may be of 
importance as regards certain phenomena. They are usually called anharmonic 
effects, since the corresponding equations of motion are non-linear and do 
not admit simple periodic (harmonic) solutions. 

We shall consider here anharmonic effects of the third order, arising from 
terms in the elastic energy which are cubic in the strains. It would be too 
cumbersome to write out the corresponding equations of motion in their 
general form. However, the nature of the resulting effects can be ascertained 
as follows. The cubic terms in the elastic energy give quadratic terms in the 
stress tensor, and therefore in the equations of motion. Let us suppose that 
all the linear terms in these equations are on the left-hand side, and all the 
quadratic terms on the right-hand side. Solving these equations by the 
method of successive approximations, we omit the quadratic terms in the 
first approximation. This leaves the ordinary linear equations, whose solution 
uo can be put in the form of a superposition of monochromatic travelling 
waves: constant xe i{k ' r ~ (0t) , with definite relations between co and k. On 
going to the second approximation, we must put u = uo + ui and retain only 
the terms in uo on the right-hand sides of the equations (the quadratic terms). 
Since uo, by definition, satisfies the homogeneous linear equations obtained 
by putting the right-hand sides equal to zero, the terms in uo on the left-hand 
sides will cancel. The result is a set of inhomogeneous linear equations for the 
components of the vector ui, where the right-hand sides contain only known 
functions of the co-ordinates and time. These functions, which are obtained 
by substituting uo for u in the right-hand sides of the original equations, are 
sums of terms each of which is proportional to 

^•[(ki-k,).!— (o^-w,)*] 

or 

where coi, o>2, ki, k2 are the frequencies and wave vectors of any two mono- 
chromatic waves in the first approximation. 

A particular integral of linear equations of this type is a sum of terms 
containing similar exponential factors to those in the free terms (the right- 
hand sides) of the equations, with suitably chosen coefficients. Each such 
term corresponds to a travelling wave with frequency coi ± a>2 and wave 



120 Elastic Waves §26 

vector ki + k2. Frequencies equal to the sum or difference of the frequencies 
of the original waves are called combination frequencies. 

Thus the anharmonic effects in the third order have the result that the set 
of fundamental monochromatic waves (with frequencies a>i, a>2, ... and wave 
vectors ki, k2, ...) has superposed on it other "waves" of small intensity, 
whose frequencies are the combination frequencies such as to± + to2, and 
whose wave vectors are such as ki±k2. We call these "waves" in quotation 
marks because they are a correction effect and cannot exist alone except in 
certain special cases (see below). The values to\ ± a>2 and ki + k2 do not in 
general satisfy the relations which hold between the frequencies and wave 
vectors for ordinary monochromatic waves. 

It is clear, however, that there may happen to be particular values of a>i, ki 
and a>2, k2 such that one of the relations for monochromatic waves in the 
medium considered also holds for a>i+o>2 and ki + k2 (for definiteness, we 
shall discuss sums and not differences). Putting toz = coi+102, ks = ki + k2, 
we can say that, mathematically, 103 and k3 then correspond to waves which 
satisfy the homogeneous linear equations of motion (with zero on the right- 
hand side) in the first approximation. If the right-hand sides in the second 
approximation contain terms proportional to £ < < k »' r-w » ( > > then a particular 
integral will be a wave with the same frequency and an amplitude which 
increases indefinitely with time. 

Thus the superposition of two monochromatic waves with values of to±, ki 
and co2, k2 whose sum C03, k3 satisfies the above condition leads, by the 
anharmonic effects, to resonance: a new monochromatic wave (with para- 
meters C03, h.3) is formed, whose amplitude increases with time and eventually 
is no longer small. It is evident that, if a wave with 003, k3 is formed on super- 
position of those with a>i, ki and C02, k2, then the superposition of waves with 
wi, ki and 003, k3 will also give a resonance with to2 = 103 — toi, k2 = k3 — ki, 
and similarly 102, k2 and C03, k3 lead to to\, ki. 

In particular, for an isotropic body to and k are related by to = ctk or 
to = cik, with ci > ct. It is easy to see in which cases either of these relations 
can hold for each of the three combinations 



wi, ki; 0)2, k2; a> 3 = CO1+C02, k3 = ki + k2. 

If ki and k2 are not in the same direction, A3 < ki + &2, and so it is clear that 
resonance can then occur only in the following two cases: (1) the waves with 
toi, ki and 002, k2 are transverse and that with C03, k3 longitudinal ; (2) one of 
the waves with toi, ki and a>2, k2 is transverse and the other longitudinal, and 
that with a>3, k3 is longitudinal. If the vectors ki and k2 are in the same 
direction, however, resonance is possible when all three waves are longi- 
tudinal or all three are transverse. 

The anharmonic effect involving resonance occurs not only when several 
monochromatic waves are superposed, but also when there is only one wave, 



§26 Anharmonic vibrations 121 

with parameters a>i, ki. In this case the right-hand sides of the equations of 
motion contain terms proportional to e 2 ^ k i- r-0, i () . if a)1 and ki satisfy the 
usual condition, however, then 2o>i and 2ki do so too, since this condition is 
homogeneous and of degree one. Thus the anharmonic effect results in the 
appearance, besides the monochromatic waves with o>i, ki previously ob- 
tained, of waves with 2oji, 2ki, i.e. with twice the frequency and twice the 
wave vector, and amplitude increasing with time. 

Finally, we may briefly discuss how we can set up the equations of motion, 
allowing for the anharmonic terms. The strain tensor must now be given 
by the complete expression (1.3): 

_\(dui eto* dui dui \ 

2\dxjc dxi dxi dxjeJ* 

in which the terms quadratic in ut can not be neglected. Next, the general 
expression for the energy densityf <?, in bodies having a given symmetry, 
must be written as a scalar formed from the components of the tensor um 
and some constant tensors characteristic of the substance involved; this 
scalar will contain terms up to a given power of tint. Substituting the ex- 
pression (26.1) for Uik and omitting terms in m of higher orders than that 
power, we find the energy $ as a function of the derivatives dui/dxjc to the 
required accuracy. 

In order to obtain the equations of motion, we notice the following result. 
The variation 8$ may be written 

S^ = 8—, 

d(duijdx]c) dxk 

or, putting 

d& 

<*ik — ; » (26.2) 

d(dui/dxk) 

dStii d dcrijc 

o& = a ilc — = — — (ar ik Sui) — 8ui— . 

dxjc 0X]c oxjc 

The coefficients of — 8ut are the components of the force per unit volume of 
the body. They formally appear the same as before, and so the equations of 
motion can again be written 

poiii = doujdxjc, (26.3) 



f We here use the internal energy S, and not the free energy F, since adiabatic vibrations are 
involved. 
5 



122 Elastic Waves §26 

where po is the density of the undeformed body, and the components of the 
tensor atk are now given by (26.2), with g correct to the required accuracy. 
The tensor ow is no longer symmetrical, f 

PROBLEM 

Write down the general expression for the elastic energy of an isotropic body in the third 
approximation. 

Solution. From the components of a symmetrical tensor of rank two we can form two 
quadratic scalars (urn* and «n 2 ) and three cubic scalars («u 8 , whw** 8 and uucUnujci). Hence the 
most general scalar containing terms quadratic and cubic in wie, with scalar coefficients 
(since the body is isotropic), is 

6 = pUiic 2 + {$K-$fi)uu 2 + $Aui k Uiiu k i+Bui k 2 uii + lCuuZ; 

the coefficients of utk* and tin* have been expressed in terms of the moduli of compression 
and rigidity, and A, B, C are three new constants. Substituting the expression (26.1) for 
in* and retaining terms up to and including the third order, we find the elastic energy to be 



dui dujA 2 v/^ M A 2 



\dxic Cxi I \dxil 



+ 

2 



dui dui dui dui /dui\'' 

OXjc OXi OXjc OXi \OXjcJ 

du t du k dui ,dui du k du t ,^/3«A 3 

+h A +iB +$c — . 

dx k dxi dxi dx k dxt dxi \dxiJ 



f It should be emphasised that <7<fc is no longer the momentum flux density (the stress tensor). 
In the ordinary theory this interpretation was derived by integrating the body force density 
doikjdxic over the volume of the body. This derivation depended on the fact that, in performing the 
integration, we made no distinction between the co-ordinates of points in the body before and after 
the deformation. In subsequent approximations, however, this distinction must be made, and the 
surface bounding the region of integration is not the same as the actual surface of the region considered 
after the deformation. 

It has been shown in §2 that the symmetry of the tensor am is due to the conservation of angular 
momentum. This result no longer holds, since the angular momentum density is not Xiiijc — **m< 
but (Xi+ujuk -(**+«*)«<• 



CHAPTER IV 

DISLOCATIONS! 

§27. Elastic deformations in the presence of a dislocation 

Elastic deformations in a crystal may arise not only by the action of external 
forces on it but also because of internal structural defects present in the crystal. 
The principal type of defect that influences the mechanical properties of cry- 
stals is called a dislocation. The study of the properties of dislocations on the 
atomic or microscopic scale is not, of course, within the scope of this book ; we 
shall here consider only purely macroscopic aspects of the phenomenon as it 
affects elasticity theory. For a better understanding of the physical significance 
of the relations obtained, however, we shall first give two simple examples to 
show what is the nature of dislocation defects as regards the structure of the 
crystal lattice. 

Let us imagine that an "extra" half-plane is put into a crystal lattice of 
which a cross-section is shown in Fig. 22 ; in this diagram, the added half-plane 



Fig. 22 

is the upper half of the _y#-plane. The edge of this half -plane (the #-axis, at 
right angles to the plane of the diagram) is then called an edge dislocation. In 
the immediate neighbourhood of the dislocation the crystal lattice is greatly 
distorted, but even at a distance of a few lattice periods the crystal planes fit 
together in an almost regular manner. The deformation nevertheless exists 
even far from the dislocation. It is clearly seen on going round a closed circuit 
of lattice points in the xy-plane, with the origin within the circuit: if the 



| This chapter was written jointly with A. M. Kosevich. 

123 



124 



Dislocations 



§27 



displacement of each point from its position intheideal lattice is denoted by the 
vector u, the total increment of this vector around the circuit will not be zero, 
but equals one lattice period in the x-direction. 

Another type of dislocation may be visualised as the result of "cutting" the 
lattice along a half-plane and then shifting the parts of the lattice on either 
side of the cut in opposite directions to a distance of one lattice period parallel 
to the edge of the cut (then called a screw dislocation). Such a dislocation 
converts the lattice planes into a helicoidal surface, like a spiral staircase 
without the steps. In a complete circuit round the dislocation line (the axis of 
the helicoidal surface) the lattice point displacement vector increment is one 
lattice period along that axis. Figure 23 shows a diagram of such a cut. 




Fig. 23 

Microscopically, a dislocation deformation of a crystal regarded as a 
continuous medium has the following general property; after a passage round 
any closed contour L which encloses the dislocation line D, the elastic 
displacement vector u receives a certain finite increment b which is equal to 
one of the lattice vectors in magnitude and direction; the constant vector b is 
called the Burgers vector of the dislocation concerned. This property may be 
•expressed as 

)6ui = q) dx/c = — bf, (27.1) 

J dxje 



where the direction in which the contour is traversed and the chosen direction 
of the tangent vector t to the dislocation line are assumed to be related by the 
corkscrew rulef (Fig. 24). The dislocation line itself is a line of singularities of 
the deformation field. 

It is evident that the Burgers vector b is necessarily constant along the dis- 
location line, and also that this line cannot simply terminate within the crystal: 
it must either reach the surface of the crystal at both ends or (as usually hap- 
pens in actual cases) form a closed loop. 



f The simple cases of edge and screw dislocations mentioned above correspond to straight lines D 
-with x J_ b and t || b. We may also note that in the representation given by Fig. 22 edge dislocations 
-with opposite directions of b differ in that the "extra" crystal half-plane lies above or below the xz- 
plane; such dislocations are said to have opposite signs. 



§27 Elastic deformations in the presence of a dislocation 125 

The condition (27.1) signifies, therefore, that in the presence of a dislocation 
the displacement vector is not a single-valued function of the co-ordinates, 
but receives a certain increment in a passage round the dislocation line. 




Physically, of course, there is no ambiguity: the increment b denotes an addi- 
tional displacement of the lattice points equal to a lattice vector, and this does 
not affect the lattice itself. 

In the subsequent discussion it is convenient to use the notation 

w>ik — dujcjdxi, (27.2) 

so that the condition (27.1) becomes 



I 



WW dxt = - bk. (27.3) 



The (unsymmetrical) tensor wm is called the distortion tensor. Its symmetrical 
part gives the ordinary strain tensor: 

uik = l(mk + mi)- (27.4) 

According to the foregoing discussion the tensors zvtk and utk, and therefore 
the stress tensor o-^, are single-valued functions of the co-ordinates, unlike 
the function u(r). 

The condition (27.3) may also be written in a differential form. To do so, 
we transform the integral round the contour L into one over a surface Sl 
spanning this contour :f 



r C 8w mk 

<bzv m jc ax m = earn— - — aft- 

v »/ C/XT 



dwmk 
■Urn 

~L St. 



dxi 
The constant vector bk is written as an integral over the same surface by 



t The transformation is made, according to Stokes' theorem, by replacing dx m by the operator 
d/ietim djdxi, where eam is the antisymmetric unit tensor. 



126 Dislocations §27 

means of the two-dimensional delta function : 

h= (nhSftdfi, (27.5) 

i 

where \ is the two-dimensional radius vector taken from the axis of the dis- 
location in the plane perpendicular to the vector t at the point considered. 
Since the contour L is arbitrary, the integrals can be equal only if the inte- 
grands are equal: 

eumdwmkldxi= -TihSfe). (27.6) 

This is the required differential form.f 

The displacement field u(r) around the dislocation can be expressed in a 
general form if we know the Green's tensor G^(r) of the equations of equilib- 
rium of the anisotropic medium considered, i.e. the function which determines 
the displacement component Ui produced in an infinite medium by a unit 
force applied at the origin along the a^-axis (see §8). This can easily be done 
by using the following formal device. 

Instead of seeking many-valued solutions of the equations of equilibrium, 
we shall regard u(r) as a single-valued function, which undergoes a fixed 
discontinuity b on some arbitrarily chosen surface Sd spanning the dislocation 
loop D. Then the strain tensor formally defined by (27.4) will have a delta- 
function singularity on the "surface of discontinuity": 

Wt™ = l{nib k + n k bi)8(Q, (27.7) 

where £ is a co-ordinate measured from the surf ace Sd along the normal n 
(which is in the direction relative to t shown in Fig. 24). 

Since there is no actual physical singularity in the space around the dis- 
location, the stress tensor ow must, as already mentioned, be a single-valued 
and everywhere continuous function. The strain tensor (27.7), however, is 
formally related to a stress tensor anc^ = Xncim uimS s \ which also has a 
singularity on the surface Sd- In order to eliminate this we must define ficti- 
tious body forces distributed over the surface Sd with a certain density f( s K The 
equations of equilibrium in the presence of body forces are doijc/dxjc +ft^ = 
(cf. (2.7)). Hence it is clear that we must put 

ft*) = JL_ = _ X mm -^—. (27.8) 

OXjc CXjc 

Thus the problem of finding the many-valued function u(r) is equivalent to 
that of finding a single-valued but discontinuous function in the presence of 



f To avoid misunderstanding it should be noted that on the dislocation line itself (( -*■ 0), which is 
a line of singularities, the representation of the ttHk as the derivatives (27.2) is no longer meaningful. 



§27 Elastic deformations in the presence of a dislocation 127 

body forces given by formulae (27.7) and (27.8). We can now use the formula 

^(r) = J^(r-r')/^)(r')dF. 

We substitute (27.8) and integrate by parts; the integration with the delta 
function is then trivial, giving 



C d 
u t (r) = - Xjumbm « r — G*y(r - r')d/'. (27.9) 

J OXlc 



This solves the problem. -f 

The deformation (27.9) has its simplest form far from the closed dislocation 
loop. If we imagine the loop to be situated near the origin, then at distances r 
large compared with the linear dimensions of the loop we have 

u t {t) = -Xjkimdi m dGij(r)jdx k , (27.10) 

where 

dm = Sib k , Si = n t df= le m i>x k dxi, (27.11) 

s D d 

and enci is the antisymmetric unit tensor. The axial vector S has components 
equal to the areas bounded by the projections of the loop D on planes perpen- 
dicular to the corresponding co-ordinate axes ; the tensor due may be called the 
dislocation moment tensor. The components of the tensor Gy are first-order 
homogeneous functions of the co-ordinates x, y, z (see §8, Problem). We 
therefore see from (27.10) that Ui~\/r 2 , and the corresponding stress field 

It is also easy to ascertain the way in which the elastic stresses vary with 
distance near a straight dislocation. In cylindrical polar co-ordinates z, r, <f> 
(with the s-axis along the dislocation line) the deformation will depend only 
on r and <f>. The integral (27.3) must, in particular, be unchanged by an 
arbitrary change in the size of any contour in the xy-plane which leaves the 
shape of the contour the same. It is clear that this can be true only if all the 
woc^X jr. The tensor w^, and therefore the stresses cr^, will be proportional 
to the same power, 1/r.J 



f The tensor Gn for an anisotropic medium has been derived in the paper by I. M. Lifshitz and 
L. N. RozentsveIg quoted in §8, Problem. This tensor is in general very complicated. For a straight 
dislocation, which corresponds to a two-dimensional problem of elasticity theory, it may be simpler to 
solve the equations of equilibrium directly. 

% Attention is drawn to a certain analogy between the elastic deformation field round a dislocation 
line and the magnetic field of constant line currents. The current is replaced by the Burgers vector, 
which must be constant along the dislocation line, like the current. Similar analogies will also be readily 
seen in the relations given below. However, quite apart from the entirely different nature of the two 
physical effects, these analogies are not far-reaching, because the tensor character of the corresponding 
quantities is different. 



128 Dislocations §27 

Although we have hitherto spoken only of dislocations, the formulae de- 
rived are applicable also to deformations caused by other kinds of defect in 
the crystal structure. Dislocations are linear defects; there exist also defects in 
which the regular structure is interrupted through a region near a given 
surface. | Such a defect can be macroscopically described as a surface of dis- 
continuity on which the displacement vector u is discontinuous but the stresses 
aik are continuous, by virtue of the equilibrium conditions. If the discontinuity 
b is the same everywhere on the surface, the resulting strain is just the same as 
that due to a dislocation along the edge of the surface. The only difference is 
that the vector b is not equal to a lattice vector. However, the position of the 
surface Sd discussed above is no longer arbitrary; it must coincide with the 
actual physical discontinuity. Such a surface of discontinuity involves a certain 
additional energy which may be described by means of an appropriate surface- 
tension coefficient. 

PROBLEMS 

Problem 1 . Derive the differential equations of equilibrium for a dislocation deformation 
in an isotropic medium, expressed in terms of the displacement vector.^ 

Solution. In terms of the stress tensor or strain tensor the equations of equilibrium have the 
usual form daikjdxu = or, substituting a i]c from (5.11), 

duM a du n 

dxjc l—2a dxi 

To convert to the vector u we must use the differential condition (27.6). Multiplying (27.6) 
by even and summing over * and k, we obtain§ 

dWnk dwjcjc 

= -(TXb)»8(|). (2) 



Writing (1) in the form 



dxjc dx n 



2 o ' 2~ ~ l"~ Z Z = " 



dx^ dxjc 1 — 2ct dx t 
and substituting (2), we find 

dzvjci 1 dzon 

dxjc l—2a dxi 

Now changing to u in accordance with (27.2), we find the required equation for the multi- 
valued function u(r) : 

1 

Au+— — - grad div u = xXbS(^). (3) 

1 — La 



t A well-known example of a defect of this type is a narrow twinned layer in a crystal. 

\ The physical meaning of this and other problems relating to an isotropic medium is purely 
conventional, since actual dislocations by their nature occur only in crystals, i.e. in anisotropic media. 
Such problems have illustrative value, however. 

§ Using also the formula eumetkn = Si k S mn — Sj„fi mfc . 



§27 Elastic deformations in the presence of a dislocation 129 

Problem 2. Determine the deformation near a straight screw dislocation in an isotropic 
medium. 

Solution. We take cylindrical polar co-ordinates z, r, ^, with the sr-axis along the disloca- 
tion line; the Burgers vector is b x = b y = 0, b z = b. It is evident from symmetry that the 
displacement u is parallel to the #-axis and is independent of the co-ordinate z. The equation 
of equilibrium (3), Problem 1, reduces to A«z = 0. The solution which satisfies the condition 
(27.1) isf u z = b<f>j2ir. The only non-zero components of the tensors utk and 0% are m z <j> = 
bjAirr, o z $ = fiil2irr, and the deformation is therefore a pure shear. 

The free energy of the dislocation (per unit length) is given by the integral 



fib 2 rdr 
4tt J r ' 



which diverges logarithmically at both limits. As the lower limit we must take the order of 
magnitude of the interatomic distances (~b), at which the deformation is large and the macro- 
scopic theory is inapplicable. The upper limit is determined by a dimension of the order of the 
length L of the dislocation. Then F = (\tb 2 \A-n) log (L/b). The energy of the deformation in the 
"core' ' of the dislocation near its axis (in a region of cross-sectional area ~ b 2 ) can be estimated 
as ~ fib 2 . When log (Ljb) ^> 1 this energy is small in comparison with that of the elastic 
deformation field. % 

Problem 3. Determine the internal stresses in an anisotropic medium near a screw disloca- 
tion which is perpendicular to a plane of symmetry of the crystal. 

Solution. We take co-ordinates x, y, z so that the #-axis is along the dislocation line, and 
again write b z = b. The vector u again has only the component u z = u(x,y). Since the xy- 
plane is a plane of symmetry, all the components of the tensor Xikim are zero which contain the 
suffix z an odd number of times. Thus only two components of the tensor aw are non-zero : 



o X z — A-xzxz~^ V^xzyz 



du du 

- — \-^xzyz~ , 

ox cy 



du du 

Vyz — Ayzxz-7- + Ayzyz~z~- 

ox cy 

We define a two-dimensional vector a and a two-dimensional tensor A a g: <r a = o az , A a $ = 
A<x«3 3 (a = 1, 2). Then <r a = \ a &8ul8x$, and the equation of equilibrium becomes div o = 0. 
The required solution of this equation must satisfy the condition (27.1) : $ grad u • dl = b. 

In this form, the problem is the same as that of finding the magnetic induction and magnetic 
field (represented by o and grad u) in an anisotropic medium of magnetic permeability A a p 
near a straight current of strength I = cb\\-n. Using the solution derived in electrodynamics, 
we obtain§ 



o\,, = — 



b KftepyzXy 



2n VH-*«V*«'*/ ' 

where |A| is the determinant of the tensor A a p. 



t In all the problems on straight dislocations we take the vector t in the negative ar-direction. 

t These estimates are general ones and are valid in order of magnitude for any dislocation (and not 
only for a screw dislocation). 

It should be noted that in practice the values of log (L/b) are usually not very large, and the energy 
of the "core" is therefore a considerable fraction of the total energy of the dislocation. 

§ See Electrodynamics of Continuous Media, §29, Problem 5. 



130 Dislocations §27 

Problem 4. Determine the deformation near a straight edge dislocation in an isotropic 
medium. 

Solution. Let the s-axis be along the dislocation line, and the Burgers vector be b x = b, 
by = b z = 0. It is evident from the symmetry of the problem that the displacement vector 
lies in the xy-plane and is independent of z, so that the problem is a two-dimensional one. In 
the rest of this solution all vectors and vector operations are two-dimensional in the xy-plane. 

We shall seek a solution of the equation 

AuH grad div u = -&jS(r) 

1 — 2ct 

(see Problem 1 ; j is a unit vector along the j>-axis) in the form u = u(°> + w, where u<°> is a 
vector with components u^ x = b<l>l2ir, w<% = (fc/2-n-) log r; these are the imaginary and real 
parts of (6/2tt) log (x+iy), r and <f> being polar co-ordinates in the xy-plane. This vector 
satisfies the condition (27.1). The problem therefore reduces to finding the single- valued 
function w. Since, as is easily verified, div u<°> = 0, Au<°> = £j8(r), it follows that w satisfies 
the equation 

Aw + grad div w = -2£j8(r). 

1 — 2cr 

This is the equation of equilibrium under forces concentrated along the sr-axis with volume 
density £6jS(r)/2(l + a) ; cf. §8, Problem, equation (1). By means of the Green's tensor found 
in that problem for an infinite medium, the calculation of w is reduced to that of the integral 



W = 



a) J [ r m 



r(l-a) 

u 

R = ^2 + ^2). 



The result is 



b j y 1 xy 

u x = — — {tan -1 --r- 



2tt 1 x 2(1 — a) x 2 +y 2 j 

b i 1-2<t 1 x 2 | 

The stress tensor calculated from this has Cartesian components 

y(3x 2 +y 2 ) 



vxx = - bD 

Oyy = bD 



o X y = bD 



(x 2 +y 2 ) 2 
y(x 2 — y 2 ) 
(x 2 +y 2 ) 2 
x(x 2 — y 2 ) 
(x 2 +y 2 ) 2 



and polar components 



where D = /*/27r(l — a). 



<*rr = o H = -(bD/r) sin <f>, 
a r4> = (bDjr) cos <f>, 



§28 The action of a stress field on a dislocation 131 

Problem 5. An infinity of identical parallel straight edge dislocations in an isotropic 
medium lie in one plane perpendicular to their Burgers vectors and at equal distances h apart. 
Find the shear stresses due to such a "dislocation wall" at distances large compared with h. 

Solution. Let the dislocations be in the ys-plane and parallel to the sr-axis. According to 
the results of Problem 4, the total stress due to all the dislocations at the point (x, y) is given by 
the sum 

a xy (x, y) = bDx> — — — -. 



This may be written in the form 
a xy 



a 

= -bD- 

h 



■ 3/(«,0) 

7(a, p) + a- 



da. 



where 






' + {^-nf 
n = — °° v 

According to Poisson's summation formula 



00 00 00 

Y/(«) = V j f( x )e 2 " ikx dx, 



k = — °° — °° 



find 



d£ ^ . re****idg 



f df ^ re* 



77 Z7T 



77 Z.77 -«r— v 

= -+ > e-2^^«cos 27rfyS. 

a a ^ 



ft = i 
When a = »/A^> 1 only the first term need be retained in the sum over k, and the result is 

bx 

a xy = ^ 7T 2]T > _ e -27rx/h C0S(277}>//*). 

h 2 

Thus the stresses decrease exponentially away from the wall. 



§28. The action of a stress field on a dislocation 

Let us consider a dislocation loop D in a field of elastic stresses a\^ 
created by given external loads, and calculate the force on the loop in such a 
field. 

According to the general rules, this must be done by finding the work 8R 
done by internal stresses in an infinitesimal displacement of the loop D. If 



1 32 Dislocations §28 

Sttjfc is the change in the strain tensor due to this displacement, we have from 
(3.1)f 



sr = - o ik mui k dv. 



~ 



dujc 
* -dV 



Since the distribution of the stresses o^ e) is assumed independent of the 
position of the dislocation, we can take the difference symbol S outside the 
integral. Using also the symmetry of the tensor ot k < e ) and the equation of 
equilibrium dai k ^\dx k = 0, we can write 

8R = -8L ik ^u ik dV 

J CXi 

= -h[- {a ik ^u k )dV. (28.1) 

J OXf 

As explained in §27, we shall regard the displacement u as a single-valued 
function having a discontinuity on some surface Sd spanning the line D. Then 
the volume integral in (28.1) can be transformed into an integral over a closed 
surface consisting of the upper and lower surfaces of the cut Sd, joined by a 
tubular lateral surface of infinitesimal width enclosing the line D. The values 
of the continuous quantities <Ji k ^ are the same on both surfaces, but the values 
of u differ by a given amount b. We therefore obtain^ 



jcr^dfi. (28.2) 



8R = -b k 8 

x 

Let each element of length dl of the dislocation be displaced by an amount 
Sr. This displacement causes a change in the area of the surface Sd, the 
elementary change being Sf = Sr X dl, i.e. 8fi = ei mn 8x m dl n = ei mn 8x m r n dl. 
The work (28.2) therefore becomes a line integral round the dislocation loop : 

8R = — Sb k ei mn cr k i^8x m rndl, 



where t is the tangent vector to D. 

The coefficient of 8x m in the integrand is minus the force f m on unit length 



f To avoid misunderstanding we must emphasise that Suae in this formula is to be taken (in accordance 
with the sense of this quantity in (3.1)) as the total (geometrical) change in the deformation following an 
infinitesimal movement of the dislocation. It comprises both elastic and plastic (see §29) parts. 

\ The integral over the tubular lateral surface of radius p vanishes as p — ► 0, since the uie become 
infinite more slowly than 1/p. 



§28 The action of a stress field on a dislocation 133 

of the dislocation line. Thus 

fi = e iklTk°lrn e) bm (28.3) 

(M. Peach and J. S. Koehler 1950). We may note that the force f is perpen- 
dicular to the vector t, i.e. to the dislocation line, and also to the vector 

oik (e) bk- 

The plane which is defined by the vectors t and b at each point of the dis- 
location is called the slip plane of the corresponding element of the dislocation ; 
for every element this plane of course touches the slip surface of the whole 
dislocation, which is a cylindrical surface with generators parallel to the Bur- 
gers vector b of the dislocation. The distinctive physical property of the slip 
plane is that it is the only one in which a comparatively easy mechanical dis- 
placement of the dislocation is possible.-f- For this reason it is of interest to 
determine the force (28.3) on this plane. 

Let k be a vector normal to the dislocation line in the slip plane. Then the 
required force component (f ±t say) is f ± = Kifa = eikiKiT k b m ai m < e \ or 

f x = vtomPbm, (28.4) 

where v = k Xt is a vector normal to the slip plane. Since the vectors b and v 
are perpendicular, we see that the force f x is determined by only one compo- 
nent oimS e ) if two of the co-ordinate axes are taken along these vectors. 
The total force acting on the whole dislocation loop is 

Fi = encib m ixri m {e) dx k . (28.5) 

This is zero except for a non-uniform stress field ; when a\ m ^ = constant, 
the integral is §dx k = 0. If the stress field varies only slightly over the loop, 
we can write 

ri = enab m — (pa^d^A;, 

P D 

the loop being regarded as situated near the origin .This force can be expressed 
in terms of the dislocation moment du defined by (27.11): 

F t = d k idad e) ldxi. (28.6) 

PROBLEMS 

Problem 1. Find the force of interaction between two parallel screw dislocations in an 
isotropic medium. 



f This fact follows from the microscopic form of a dislocation defect. For example, to move the 
edge dislocation shown in Fig. 22 in its slip plane (the xs-plane) comparatively slight movements of the 
atoms are sufficient, which make crystal planes farther and farther from the yar-plane (but still parallel to 
it) into "extra" planes. 

The movement of the dislocation in other directions can occur only by diffusion processes. For 
example, the dislocation shown in Fig. 22 can move in the yar-plane only when atoms leave the "extra" 
half-plane by diffusion. Such a process can be of practical importance only at fairly high temperatures. 



134 Dislocations §29 

Solution. The force per unit length acting on one dislocation in the stress field due to the 
other dislocation is determined from formula (28.4), using the results of §27, Problem 2. It is 
a radial force of magnitude/ = [ib\b 2 \2m : Dislocations of like sign (b x b 2 > 0) repel, while 
those of unlike sign (bib 2 < 0) attract. 

Problem 2. A straight screw dislocation lies parallel to the plane free surface of an iso- 
tropic medium. Find the force acting on the dislocation. 

Solution. Let the yz-plane be the surface of the body, and let the dislocation be parallel to 
the .sr-axis with co-ordinates * = xo, y — 0. 

The stress field which leaves the surface of the medium a free surface is described by the sum 
of the fields of the dislocation and its image in the y^-plane, considered to lie in an infinite 
medium : 



fxb 



y y 



[ib 

Gyz = — 



2ttL(x— xo) 2 +y 2 (x + xo) 2 +y 2 

X — Xq X + Xq 



2tt L (x — xo) 2 +y 2 (x+xo) 2 +y 2 _ 

Such a field exerts a force on the dislocation considered which is equal to the attraction exerted 
by its image, i.e. the dislocation is attracted to the surface of the medium by a force 
/ = nb'ftnxo. 

Problem 3. Find the force of interaction between two parallel edge dislocations in an 
isotropic medium which are in parallel slip planes. 

Solution. Let the slip planes be parallel to the xsr-plane and let the ar-axis be parallel to the 
dislocation lines ; as in §27, Problem 4, we put t s = — l,b x = b. Then the force on unit length 
of the dislocation in the field of elastic stresses ow has components f x = bo xy ,f y = — bo xx . 
In the case considered, oik is determined by the expressions derived in §27, Problem 4. If one 
dislocation is along the z-axis, it exerts on the other dislocation (passing through the point 
(x, y, 0)) a force whose polar components are f r = b\b 2 Djr, f<\> = {bib 2 Djr) sin 2<f>, D = 
nl2ir(l — a). The component of this force in the slip plane is f x = (bib 2 Dlr) cos <f> cos 2<j>, 
which is zero when <f> = \-n or \tt. The former position corresponds to stable equilibrium when 
6i&2 > 0» the latter when bib 2 < 0. 

§29. A continuous distribution of dislocations 

If a crystal contains several dislocations at the same time which are at 
relatively short distances apart (although far apart compared with the lattice 
constant, of course), it is useful to treat them by means of an averaging process : 
we consider "physically infinitesimal" volume elements in the crystal with a 
large number of dislocation lines through each. 

An equation which expressed a fundamental property of dislocation deforma- 
tions can be formulated by a natural generalisation of equation (27.6). We 
define a tensor pw (the dislocation density tensor) such that its integral over a 
surface spanning any contour L is equal to the sum b of the Burgers vectors 
of all the dislocation lines embraced by the contour : 



f. 



PiJcdft = b k . (29.1) 

The continuous functions pw describe the distribution of dislocations in the 



§29 A continuous distribution of dislocations 135 

crystal. This tensor now replaces the expression on the right of equation 
(27.6): 

eumdWmk/ %*% = - pik- (29.2) 

This equation shows that the tensor p ik must satisfy the condition 

8paldx t = Q; (29.3) 

for a single dislocation, this condition simply states that the Burgers vector is 
constant along the dislocation line. 

When the dislocations are treated in this way, the tensor w ik becomes a 
primary quantity describing the deformation and determining the strain ten- 
sor through (27.4). A displacement vector u related to w i1c by the definition 
(27.2) cannot exist; this is clear from the fact that with such a definition the 
left-hand side of equation (29.2) would be identically zero throughout the 
crystal. 

So far we have assumed the dislocations to be at rest. Let us now see how a 
set of equations may be formulated so as to allow in principle elastic deforma- 
tions and stresses in a medium where dislocations are moving in a given man- 
nerf (E. Kroner and G. Rieder 1956). 

Equation (29.2) is independent of whether the dislocations are at rest or in 
motion. The tensor w ilc still determines the elastic deformation; its symmetri- 
cal part is the elastic strain tensor, which is related to the stress tensor in the 
usual way, by Hooke's law. 

This equation, however, is now insufficient for a complete formulation of 
the problem. The full set of equations must also determine the velocity v of 
the points in the medium. 

It must be borne in mind that the movement of dislocations causes not only 
a change in the elastic deformation but also a change in the shape of the crystal 
which does not involve stresses, i.e. aplastic deformation. The motion of dis- 
locations is in fact a mechanism of plastic deformation. This is clearly illustra- 
ted by Fig. 25, where the passage of the edge dislocation from left to right 
causes the part of the crystal above the slip plane to be shifted to the right by 
one lattice period; since the lattice is then regular, the crystal remains un- 
stressed. Unlike an elastic deformation, which is uniquely defined by the 
thermodynamic state of the body, a plastic deformation depends on the process 
which occurs. In considering dislocations at rest we have no need to distin- 
guish elastic and plastic deformations, since we are concerned only with 
stresses which are independent of the previous history of the crystal. 

Let u be the geometrical displacement vector of points in the medium, 
measured, say, from their position before the deformation process begins; its 

f We shall not discuss here the problem of determining this motion itself from the forces applied to 
the body. The solution of such a problem requires a detailed study of the microscopic mechanism of the 
motion of dislocations and their retardation by various defects, which must take account of the 
conditions occurring in actual crystals. 



136 



Dislocations 



§29 



time derivative ii = v. If the "total distortion" tensor W ik = dut/dxt is 
formed from the vector u, its "plastic part" «^< pl ) is obtained by subtracting 




Fig. 25 

from Wik the "elastic distortion" tensor, which is the same as the tensor zcijc in 
(29.2). We use the notation 

-ja=fai*Wlto; (29.4) 

the symmetrical part ofjw gives the rate of variation of the plastic deformation 
tensor: the change in e^& (pl) in an infinitesimal time interval 8t is 

Si^CPD = -ft/tt+AOS*. (29.5) 

We may note, in particular, that, if a plastic deformation occurs without 
destroying the continuity of the body, the trace of the tensor jm is zero : a 
plastic deformation causes no extension or compression of the body (which 
would always involve the appearance of internal stresses), i.e. u^^ = 0, and 
therefore;^ = -du kk W/dt = 0. 



§29 A continuous distribution of dislocations 137 

Substituting in the definition (29.4) Wik^ = Wuc — ww, we can write it as 

= -r-+ja, (29. f^ 



dt dxt 

an equation which relates the rates of change of the elastic and plastic deforma- 
tions. Here the /^ must be regarded as given quantities which must satisfy 
conditions ensuring the compatibility of equations (29.6) and (29.2). These 
conditions are found by differentiating (29.2) with respect to time and sub- 
stituting (29.6), and are 

^+^ = 0. (29.7) 

Ot OX\ 

The complete set of equations is given by (29.2) and (29.6), together with the 
dynamical equations 

pvi = datjcldxjc, (29.8) 

where a ilc = XtjcimUim = At^m^zm-The tensors puc and fa which appear in 
these equations are given functions of the co-ordinates (and time) which 
describe the distribution and movement of the dislocations. These functions 
must satisfy the compatibility conditions of equations (29.2) with one another 
and with (29.6), which are given by (29.3) and (29.7). 

The condition (29.7) may be regarded as a differential expression of the "law 
of conservation of the Burgers vector" in the medium: integrating both sides 
of this equation over a surface spanning some closed line L, defining by (29.1) 
the total Burgers vector b of the dislocations embraced by L, and using 
Stokes' theorem, we obtain 



~df 



= -jjikdxi. (29.9) 



The form of this equation shows that the integral on the right gives the "flux" 
of the Burgers vector through the contour L per unit time, i.e. the Burgers 
vector carried across L by moving dislocations. We may therefore caliy^ the 
dislocation flux density tensor. 

In particular, it is clear that for an isolated dislocation loop the tensor jw has 
the form 

jik = eumpucVm 

= e am TiV m b k o{l), (29.10) 

puc being given by (27.6), and V being the velocity of the dislocation line at a 
particular point on it. The flux vector through the element dl of the contour 
L is jijcdli and is proportional to dl^x XV = V«dl Xt, i.e. the component of V 
in a direction perpendicular to both dl and t ; from geometrical considerations 
6 



138 Dislocations §29 

it is evident that this is correct, since only that velocity component causes the 
dislocation to intersect the element dl. 

We may note that the trace of the tensor (29.10) is proportional to the com- 
ponent of the velocity of the dislocation along the normal to its slip plane. It 
has been mentioned above that the absence of any inelastic change in density 
of the medium is ensured by the condition/^ = 0. We see that for an indivi- 
dual dislocation this condition signifies motion in the slip plane, in accordance 
with the previous discussion of the physical nature of the movement of dis- 
locations; see the last footnote to §28. 

Finally, let us consider the case where dislocation loops are distributed in 
the crystal in such a way that their total Burgers vector (denoted by B) is zero.f 




Fig. 26 

This condition signifies that integration over any cross-section of the body 
gives 



pikdfi = 0. 



(29.11) 



From this it follows that the dislocation density in this case can be written as 

pik = eumBPmkl dxi (29 . 1 2) 

(F. Kroupa 1962); then the integral (29.11) becomes an integral along a con- 
tour outside the body, and is zero. It may also be noted that the expression 
(29.12) necessarily satisfies the condition (29.3). 

It is easy to see that the tensor P^ thus defined represents the dislocation 
moment density in the deformed crystal, and may therefore be called the 
"dislocation polarisation": the total dislocation moment Afc of the crystal is, 
by definition, 



Dm — ^ Stbjc = leum£hi>xidx f 

D 

= J eamXlpmkdV, 



t The presence of a dislocation involves a certain bending of the crystal, as shown diagrammatically 
in Fig. 26 (greatly exaggerated). The condition B = means that there is no macroscopic bending of 
the crystal as a whole. 



§30 Distribution of interacting dislocations 139 

where the summation is over all dislocation loops and the integration is over 
the whole volume of the crystal. Substituting (29.12), we obtain 

dP a ic 



Dm = \ eu m e mpg xi— — dV 

J OXp 

-If. 



uXi OXm/ 



and, after integrating by parts in each term, 

D ik = jp ik dV. (29.13) 

The dislocation flux density is given in terms of the same tensor Pi k by 

jiJc=-dPi k l8L (29.14) 

This is easily seen, for example, by calculating the integral jj\ k dV over an 
arbitrary part of the volume of the body, using the expression (29.10), to give 
a sum over all dislocation loops within that volume. We may note that the 
expression (29.14) together with (29.12) automatically satisfies the condition 
(29.7). 

A comparison of (29.14) with (29.4) shows that Sto tk ^ = &P ik . If we 
agree to regard the plastic deformation as absent in the state with Pm = 0, 
then wifctoU = P^,f and 

Wik = Wit-to*®) = dujcldxt-Pijc, (29.15) 

where u k is again the vector of the total geometrical displacement from the 
position in the undeformed state. Equation (29.6) is then satisfied identically, 
and the dynamical equation (29.8) becomes 

pUi-Xm m d 2 u m ldx k dxi = -X ik i m dPi m jdx k . (29.16) 

Thus the determination of the elastic deformation due to moving dislocations 
with B = reduces to a problem of ordinary elasticity theory with body forces 
distributed in the crystal with density —hkimdPimfixk (A. M. Kosevich 
1963). 

§30. Distribution of interacting dislocations 

Let us consider a large number of similar straight dislocations lying 
parallel in the same slip plane, and derive an equation to determine their 
equilibrium distribution. Let the sr-axis be parallel to the dislocations, and 
the xs-plane be the slip plane. 



t It is assumed that the entire deformation process occurs with B = 0. This point must be empha- 
sised, since there is a fundamental difference between the tensors P ik and w Jfc (pl) : whereas P ik is a func- 
tion of the state of the body, the tensor «.- Mr (pl> is not, but depends on the process which has brought the 
body into that state. 



140 Dislocations §30 

We shall suppose for definiteness that the Burgers vectors of the disloca- 
tions are in the ^-direction. Then the force in the slip plane on unit length of 
a dislocation is ba xyy where a xy is the stress at the position of the dislocation. 

The stresses created by one straight dislocation (and acting on another 
dislocation) decrease inversely as the distance from it. The stress at a point x 
due to a dislocation at a point x' is therefore bD/(x — x'), where D is a constant 
of the order of the elastic moduli of the crystal. It may be shown that this 
constant D is positive, i.e. two like dislocations in the same slip plane repel 
each other.f 

Let p(x) be the line density of dislocations on a segment (a±, a^) of the 
x-axis ; p(x)dx is the sum of the Burgers vectors of dislocations passing through 
points in the interval dx. Then the total stress at a point x on the x-axis due to 
all the dislocations is given by the integral 



a> 



a xy {x)^-D[^p-^. (30.1) 

J £ — X 

For points in the segment («i, a?) this integral must be taken as a principal 
value in order to exclude the physically meaningless action of a dislocation on 
itself. 

If the crystal is also subjected to a two-dimensional stress field a xy ^ e \x, y) 
in the ry-plane, caused by given external loads, each dislocation will be sub- 
jected to a force b(a xy +p(x)), where for brevity p{x) denotes o xy W(x, 0). The 
condition of equilibrium is that this force should be zero: a xy +p — 0, i.e. 

D ? />(£)<*£ p(x) 

P — = — — = w(x), (30.2) 

J tj — x D 

ai 

where P denotes, as usual, the principal value. This is an integral equation to 
determine the equilibrium distribution p(x). It is a singular equation with a 
Cauchy kernel. 

The solution of such an equation is equivalent to a problem in the theory of 
functions of a complex variable which may be formulated as follows. 

Let Q,(z) denote a function defined throughout the complex #-plane (cut 
from a± to a?) as the integral 



Q(*) = J"! 



''^^ (30.3) 



«l 



I" 



Let Q. + (x) and Q~(x) denote the limiting values of Q(#) on the upper and lower 
edges of the cut. They are equal to similar integrals along the segment («i, #2) 



f For an isotropic medium this has been proved in §28, Problem 3. 



§30 Distribution of interacting dislocations 141 

with an indentation in the form of an infinitesimal semicircle below or above 
the point z = x respectively, i.e. 

a 2 
a ±(*) = P mLl. + i^x). (30.4) 

ax 

If p(£) satisfies equation (30.2), the principal value of the integral is w(x), and 
we therefore have 

Q+(x) + Qr(x) = 2(v(x), (30.5) 

Q+(x) - Q-(x) = 2iiT P {x). (30.6) 

Thus the problem of solving equation (30.2) is equivalent to that of finding an 
analytic function £l(z) with the property (30.5); p(x) is then given by (30.6). 
The physical conditions of the problem in question also require that Q( oo) = ; 
this follows because far from the dislocations (#-»+ oo) the stresses a X y must 
be zero (by the definition (30.3), a xy {x) = — DQ.(x) outside the segment 

(fli, «2)). 

Let us first consider the case where there are no external stresses (j>(x) = 0), 
and the dislocations are constrained by some obstacles (lattice defects) at the 
ends of the segment (a\> a%). When co(x) = we have from (30.5) CI +(x) = 
— Q.~(x), i.e. the function Q,(z) must change sign in a passage round each of 
the points «i, a%. This condition is satisfied by any function of the form 

P(z) 

Q(z) = — , (30.7) 

V[{<*2-z)(z-ai)] 

where P(z) is a polynomial. The condition Q,(co) = means that we must 
take P(z) = 1 (apart from a constant coefficient), so that 

°<*> = -7i7 Tr ^ (30 ' 8) 

y[{a 2 -z)(z-ai)] 

The required function p(x) will, according to (30.6), have the same form. The 
coefficient is determined from the condition 

a* 

j P (£)d£ = B, (30.9) 

ay 
where B is the sum of the Burgers vectors of all the dislocations, and so we have 

P(*)= /r , B v rr- (30.10) 

77 y [(«2 — x){x — a\) J 

We see that the dislocations pile up towards the obstacles at the ends of the 



142 Dislocations §30 

segment, with density inversely proportional to the square root of the distance 
from the obstacle. The stress outside the segment (ai, a 2 ) increases in the same 
manner as the ends of the segment are approached, e.g. for x>a 2 

BD 



\/[(x-a 2 )(a 2 -ai)] 

In other words, the concentration of dislocations at the boundary leads to a 
stress concentration beyond the boundary. 

Let us now suppose that under the same conditions (obstacles at the fixed 
ends of the segment) there is also an external stress field p(x). Let Q.o(z) 
denote a function of the form (30.7), and let us rewrite equation (30.5) 
divided by Qo + = — Qo~ as 

Q+(x) Q-(x) 2w(x) 



C1q + (x) Q,o~(x) Qq + (x) 
A comparison of this with (30.6) shows that 



a* 



Q(*) 1 f off) d| 



«i 



where P(z) is a polynomial. A solution which satisfies the condition D( oo) 
= is obtained by taking as Qo(#) the function (30.8) and putting P(z) = C> 
a constant. The required function p(x) is hence found by means of (30.6), and 
the result is 



u>2 

M— ,V[(^-l)(.- ai )] i 'J a ' tf)V[(fl2 - f)tf - ai) ^ + 



+ . (30.12) 

V[( a 2-x)(x-a{)] 

The constant C is determined by the condition (30.9). Here also p(x) increases 
as (^2 — x)~ 1/2 when x-^a 2 (and similarly when x->ai), and a similar concentra- 
tion of stresses occurs on the other side of the boundary. 

If there is an obstacle only on one side (at a 2 , say) the required solution 
must satisfy the condition of finite stress for all x<a 2 , including the point x= 
a\ ; the position of the latter point is not known beforehand and must be deter- 
mined by solving the problem. With respect to £i{z) this means that Q.(a{) 
must be finite. Such a function (satisfying also the condition Q(oo) = 0) is 
obtained from the same formula (30.11) by taking for Q,q(z) the function 



§30 Distribution of interacting dislocations 143 

\Z[(z-ai)l(a 2 -z)], which is also of the form (30.7), and putting P(z) = in 
(30.11). The result is 

^ )= _I/^p]/^4^. (30.13) 

«i 

When x^ai, p(x) tends to zero as <\/(x-ai). The total stress a xy (x)+p(x) 
tends to zero according to a similar law on the other side of the point a\. 

Finally, let there be no obstacle at either end of the segment, and let the 
dislocations be constrained only by external stresses />(#). The corresponding 
Q(*) is obtained by putting in (30.11) Cl (z) = V[(«2-*)(*-«i)], P( z ) = °- 
The condition ii(oo) = 0, however, here requires the fulfilment of a further 
condition: taking the limit as z~>co in (30.11), we find 



a 2 



i 



(30.14) 



y[(«s-£)(£-«i)] 

The function p(x) is given by 

*)- 4^-^- ai)]p J v^-f)«-^] p? (30 - 15) 

tti 

the co-ordinates «i and «2 of the ends of the segment being determined by the 
conditions (30.9) and (30.14). 

PROBLEM 

Find the distribution of dislocations in a uniform stress field p(x) = po over a segment 
with obstacles at one or both ends. 

Solution. When there is an obstacle at one end (a 2 ) the calculation of the integral (30.13) 
gives 

p0 IX- fli 



" w - sy 



«2 — X 



The condition (30.9) determines the length of the segment occupied by dislocations :a 2 — a\ = 
IBDjPo. Beyond the obstacle there is a concentration of stresses near it according to 

/ A2 - « 1 

<Txy=p0 • 

V x — a<i 

For a segment of length 2L bounded by two obstacles we take the origin of * at the midpoint 
and obtain from (30.12) 



pfx) = l—x + b) . 

HK } 77a/(L2-*2) \ D J 



144 Dislocations §31 

§31. Equilibrium of a crack in an elastic medium 

The problem of the equilibrium of a crack is somewhat distinctive among 
the problems of elasticity theory. From the point of view of that theory, a 
crack is a cavity in an elastic medium, which exists when internal stresses are 
present in the medium and closes up when the load is removed. The shape and 
size of the crack depend considerably on the stresses acting on it. The mathe- 
matical feature of the problem is therefore that the boundary conditions are 
given on a surface which is initially unknown and must itself be determined in 
solving the problem.f 

Let us consider a crack in an isotropic medium, of infinite length and uni- 
form in the ^-direction and in a plane stress field vik {e) (x, y) ; this is a two- 
dimensional problem of elasticity theory. We shall suppose that the stresses 
are symmetrical about the centre of the cross-section of the crack. Then the 
outline of the cross-section will also be symmetrical (Fig. 27). Let its length be 
2L and its variable width h(x) ; since the crack is symmetrical, h( — x) = h(x) 




We shall assume the crack to be thin (h<^L). Then the boundary conditions 
on its surface can be applied to the corresponding segment of the #-axis. Thus 
the crack is regarded as a line of discontinuity (in the xy-plane) on which the 
normal component of the displacement u y = + \h is discontinuous. 

Instead of h(x) we define a new unknown function p(x) by the formulae 

L 
h(x)= lp(x)dx, p(-x)=- P (x). (31.1) 

x 

The function p(x) may be conveniently, though purely formally, interpreted 
as a density of straight dislocations lying in the s'-direction and continuously 
distributed along the x-axis, with their Burgers vectors in the ^-direction.J 
It has been shown in §27 that a dislocation line may be regarded as the edge of 
a surface of discontinuity on which the displacement u has a discontinuity b. 
In the form (31.1) the discontinuity h of the normal displacement at the point 

f The quantitative theory of cracks discussed here is due to G. I. Barenblatt (1959). 
% It is for this reason that the theory of cracks is described here in the chapter on dislocations, 
although physically the phenomena are quite different. 



§31 Equilibrium of a crack in an elastic medium 145 

x is regarded as the sum of the Burgers vectors of all the dislocations lying to 
the right of that point; the equation p{-x) = -p{x) signifies that the dis- 
locations to the right and to the left of the point * = have opposite signs. 

By means of this representation we can write down immediately an expression 
for the normal stresses (a yy ) on the *-axis. These consist of the stresses 
oyy {e) (x> 0) resulting from the external loads (which for brevity we denote by 
p(x)) and the stresses o yy W{x) due to the deformation caused by the crack. 
Regarding the latter stresses as being due to dislocations distributed over the 
segment (-L, L), we obtain (similarly to (30.1)) 

ayrKx)= J[jm, (31.2) 

j L i- x 

for points in the segment ( - L, L) itself, the integral must be taken as a princi- 
pal value. For an isotropic medium, 

27r(l-cr) 4tt(1-<7 2 ) 

see §28, Problem 3. The stresses a xy due to such dislocations in an isotropic 
medium are zero on the #-axis. 

The boundary condition on the free surface of the crack, applied (as already 
mentioned) to the corresponding segment of the *-axis, requires that the 
normal stresses a yy = a 2/2/ < cr > +p(x) should be zero. This condition, however, 
needs to be made more precise, for the following reason. 

Let us make the assumption (which will be confirmed by the result) that 
the edges of the crack join smoothly near its ends, so that the surfaces approach 
very closely. Then it is necessary to take into account the forces of molecular 
attraction between the surfaces ; the action of these forces extends to a distance 
r large compared with interatomic distances. These forces will be of impor- 
tance in a narrow region near the end of the crack where h<ro; the length of 
this region will be denoted by d in order of magnitude, and will be estimated 
later. 

Let G be the force of molecular cohesion per unit area of the crack; it 
depends on the distance h between the surfaces.! When these forces are taken 
into account, the boundary condition becomes 

a yy (rt+p(x)-G = 0. (31.4) 

It is reasonable to suppose that the shape of the crack near its end is deter- 
mined by the nature of the cohesion forces and does not depend on the external 
loads applied to the body. Then, in finding the shape of the main part of the 
crack from the external forces p(x), the quantity G becomes a given function 
G(x) independent of p(x) (over the region d, outside which it is unimportant). 

f In the macroscopic theory, the function G(x) is to be regarded as increasing smoothly, asL-x 
decreases, up to a maximum value at the end of the crack. 

6* 



146 Dislocations §31 

Substituting a yy ^) from (31.2) in (31.4), we thus obtain the following 
integral equation for p(x) : 

L Cp(i)d$ 1 1 

P ^^ = -rftx)-^*) - «(*). (31-5) 

J t — x D D 

Since the ends of the crack are assumed not fixed, the stresses must remain 
finite there. This means that, in solving the integral equation (31.5), we now 
have the last of the cases discussed in §30, for which the solution is given by 
(30.15). With the origin at the midpoint of the segment ( — L,L) this formula 
becomes 

pw= __ v(i2 _, 2) pj__y___. (31 . 6) 

— Jj 

The condition (30.14) must be satisfied, which in this case gives 



l , , . l 



C P( x ) dx _ C G(x)dx = q 



(where the integrals from — L to L have been replaced by integrals from to 
L, using the symmetry of the problem). Since G(x) is zero except in the range 
L — x ~ d y in the second integral we can put L 2 — x 2, ^ 2L(L — x) ; the condition 
(31.7) then becomes 






p(x)dx M 

FK } (31.8) 



q V(L 2 -x*) V(2L) 



where M denotes the constant 






which depends on the medium concerned. This constant can be expressed in 
terms of the ordinary macroscopic properties of the body, its elastic moduli 
and surface tension a ; as will be shown later, the relation is 

M=vT™£/(l-a 2 )]. (31.10) 

The equation (31.8) determines the length 2L of the crack from the given 
stress distribution p{x). For example, for a crack widened by concentrated 



§31 Equilibrium of a crack in an elastic medium 147 

forces /applied to the midpoints of the sides (p(x) =fB(x)) we find 

2L=/ 2 /Af2 

= /2(l-cT 2 )/7ra£'. (31.11) 

It must be remembered, however, that stable equilibrium of a crack is not 
possible for every distribution p(x). For instance, with uniform widening 
stresses (p(x) = constant = po) (31.8) gives 

1L = 4M 2 /t7 V 

= 4a£/77(l-(72)/>o2. ( 31 - 12 ) 

This inverse relation (L decreasing when^o increases) shows that the state is 
unstable. The value of L determined by (31.12) corresponds to unstable 
equilibrium and gives the "critical" crack length: longer cracks grow spon- 
taneously, but shorter ones close up, a result first derived by A. A. Griffith 
(1920). 

Let us now return to the consideration of the shape of the crack. WhenL -x<d, 
the region L - 1 ~ d is the most important in the integral in (3 1 . 6) . The integral 
can then be replaced by its limiting value as x^L ; the result is p = constant x 
x \/(L — x), whencef 

h(x) = constant x(L-xf 12 (L-x~d). (31.13) 

We see that over the terminal region d the two sides of the crack in fact join 
smoothly. The value of the coefficient in (31.13) depends on the properties of 
the cohesion forces and can not be expressed in terms of the ordinary macro- 
scopic parameters. J 

For the part farther from the end, where d<L-x<^L, the region L-£~d 
is again the most important in the integral in (31.6), and o>(£)^ -G(0I D - * n 
addition to putting L 2 -x*z2L(L-x), L 2 -£ 2 z2L(L-£), we can here 
replace g-x by L-x, obtaining /> = MJt^D^/{L-x), where M is the same 
constant as in (31.9), (31.10). Hence 

h(x) = 2MV(L-x)J7rW (d4L-x<L). (31.14) 

Thus the end of the crack has a shape independent of the applied forces (and 
therefore of the length of the crack) throughout the range L-x^L: when 
L - x > d the shape is given by (3 1 . 14), and when L - x ~ d it has an infinitely 



t In order to proceed to the limit we must first divide the integral in (31.6) into two integrals with 
numerators a>(f) — m(L) and u(L); the second integral makes no contribution to the limiting value. 

X An estimate of the coefficient in (3 1 . 1 3) gives a value of the order of Va/d, where a is the dimension 
of an atom (using a ~ aE, M ~ EVa). An estimate of the length d is obtained from the condition 
h{d) ~ ro, whence d ~ ro 2 /a > ro. It should be mentioned, however, that in practice the required 
inequalities are satisfied only by a small margin, so that the resulting shape of the terminal projection 
of the crack is not to be taken as exact. 



148 Dislocations §31 

sharp projection (31.13) (Fig. 28). The shape of the remainder of the crack 
does depend on the applied forces. 



Y*-*;.- 



Fig. 28 

If we ignore details, of the order of the radius of the action of the cohesion 
forces, the crack therefore has a smooth outline with ends rounded according 
to the parabolas (31.14), and this shape is entirely determined by the applied 
forces and the ordinary macroscopic parameters. The small (~d) terminal 
projections which actually occur are of fundamental significance, however, 
since they ensure that the stresses remain finite at the ends of the crack. 

The stresses caused by the crack on the continuation of the #-axis are given 
by formula (31.2). At distances x—L such that d<^x—L<^L, we havej- 

Oyy^(Tyy^)^MlTT^{x-L). (31.15) 

The increase in the stresses as the edge of the crack is approached continues 
according to this law up to distances x—L~d, and a vy then drops to zero at 
the point x — L. 

It remains to derive the formula (31.10) already given above, which relates 
the constant M to the ordinary macroscopic quantities. To do this, we write 
down the condition for the total free energy to be a minimum by equating to 
zero its variation under a change in the length L. 

Firstly, when the length of the crack increases by SL the surface energy at 
its two free surfaces increases by &F S urf = 2aSL. Secondly, the "opening" of 
the crack end reduces the elastic energy F e \ by %$oyy(x)r)(x)dx, where rj(x) is 
the difference in width between the displaced and undisplaced crack shapes. 
Since the shape of the crack end is independent of its length, r)(x) = h(x — 8L) — 
— h{x). The stress a yy = for x<L, and h(x) = for x> L. Hence 

L+SL 



8F e i= — § \o yy (x)h(x — 8L)dx. 



f The integral is easily calculated directly, but it is not necessary to do this if we use the relation 
between the functions p(x) for x < L and a ay (cr) for x > L, which is evident from the results of §30. 



§31 Equilibrium of a crack in an elastic medium 149 

Substituting (31.14) and (31.15), we find 

L+dL 

M 2 r iL+SL-x, 



L 



8L 

M 2 r Vydy 



M *SL. 



Finally, the condition SF^t+BFei = gives the relation M 2 = 47r 2 oD, and 
hence we have (31.10).| 



t It may be noted that the theory described above, including the relation (31.10), is in fact applicable 
as it stands only to ideally brittle bodies, i.e. those which remain linearly elastic up to fracture, such as 
glass and fused quartz. In bodies which exhibit plasticity the formation of the crack may be accompanied 
by plastic deformation at its ends. 



CHAPTER V 

THERMAL CONDUCTION AND VISCOSITY IN SOLIDS 

§32. The equation of thermal conduction in solids 

Non-uniform heating of a solid does not cause convection as it generally 
does in fluids. Hence the transfer of heat is effected in solids by thermal 
conduction alone. The processes of thermal conduction in solids are there- 
fore described by somewhat simpler equations than those for fluids, where 
they are complicated by convection. 

The equation of thermal conduction in a solid can be derived immediately 
from the law of conservation of energy in the form of an "equation of con- 
tinuity for heat". The amount of heat absorbed per unit time in unit volume 
of the body is TdS/dt, where S is the entropy per unit volume. This must 
be put equal to — div q, where q is the heat flux density. This flux can 
almost always be written as — k grad T, i.e. it is proportional to the tempera- 
ture gradient (k being the thermal conductivity). Thus 

TdSJdt = div(/c grad T). (32.1) 

According to formula (6.4), the entropy can be written as 

S = S (T) + K<x.u iiy 

where a is the thermal expansion coefficient and So the entropy in the 
undeformed state. We shall suppose that, as usually happens, the tempera- 
ture differences in the body are so small that quantities such as k, a, etc. 
may be regarded as constants. Then equation (32.1), after substitution of 
the above expression for S, becomes 

dSn dtiu 

T— + olKT— = *AT. 
dt dt 

According to a well-known formula of thermodynamics, we have 

C p —C v = KofiT, 
whence 

olKT = (C p —C v )l<x. 

The time derivative of So can be written as (dSo/dT) • (dT/dt), where the 
derivative dSojdT is taken for uu = div u = 0, i.e. at constant volume, and 
therefore is equal to CJT. 

150 



§32 The equation of thermal conduction in solids 151 

The resulting equation of thermal conduction is 

C ?— + Cv - Cv - divu = kAT. (32.2) 

dt a dt 

In order to obtain a complete system of equations, it is necessary to add an 
equation describing the deformation of a non-uniformly heated body. This is 
the equilibrium equation (7.8) : 

2(1 _ a ) grad div u- (1 - 2a) curl curl u = |a(l + a) grad T. (32.3) 

From equation (32.3) we can in principle determine the deformation of the 
body for any given temperature distribution. Substituting the expression for 
divu thus obtained in equation (32.2), we derive an equation giving the 
temperature distribution, in which the only unknown function is T (x, y, z, t). 
For example, let us consider thermal conduction in an infinite solid in 
which the temperature distribution satisfies only one condition: at infinity, 
the temperature tends to a constant value T , and there is no deformation. 
In such a case equation (32.3) leads to the following relation between div u 
and T (see §7, Problem 8) : 

1 + cr 

divu = — -a(T-To). 

3(1 - a) 

Substituting this expression in (32.2), we obtain 

(l + a)C p + 2(l-2a)C v W_ =KATy (32 . 4) 

3(1 -a) dt 

which is the ordinary equation of thermal conduction. 

An equation of this type also describes the temperature distribution along 
a thin straight rod, if one (or both) of its ends is free. The temperature may 
be assumed constant over any transverse cross-section, so that T is a function 
only of the co-ordinate x along the rod and of the time. The thermal expan- 
sion of such a rod causes a change in its length, but no departure from straight- 
ness and no internal stresses. Hence it is clear that the derivative dSjdt in 
the general equation (32.1) must be taken at constant pressure and, since 
(dSldt) p = Cp/T, the temperature distribution will satisfy the one-dimen- 
sional thermal conduction equation C v dTjdt = Kd 2 T\dx 2 . 

It should be mentioned, however, that the temperature distribution m a 
solid can in practice always be determined, with sufficient accuracy, by a 
simple thermal conduction equation. The reason is that the second term on 
the left-hand side of equation (32.2) is a correction of order {C P ~C V )IC V 
relative to the first term. In solids, however, the difference between the two 
specific heats is usually very small, and if it is neglected the equation of 
thermal conduction in solids can always be written 

BTIft = x* T > ( 32 ' 5) 



152 Thermal Conduction and Viscosity in Solids §33 

where x is the thermometric conductivity, defined as the ratio of the ther- 
mal conductivity k to some mean specific heat per unit volume C. 

§33. Thermal conduction in crystals 

In an anisotropic body, the direction of the heat flux q is not in general 
that of the temperature gradient. Hence, instead of the formula 

q = _ K grad T 

relating q to the temperature gradient, we have in a crystal the more general 
relation 

qt = -K tk dT/dx k . (33.1) 

The tensor K ik) of rank two, is called the thermal conductivity tensor of the 
crystal. In accordance with (33.1), the equation of thermal conduction (32.5) 
has also a more general form, 

dT d*T 

c ir - * w < 33 - 2 > 

A general theorem can be stated: the thermal conductivity tensor is 
symmetrical, i.e. 

Kik = K ki - (33.3) 

This relation, which we shall now prove, is a consequence of the symmetry 
of the kinetic coefficients.f 

The rate of increase of the total entropy of the body by irreversible pro- 
cesses of thermal conduction is 

C div q C Q C 1 

St0t = " J T dV = " J div Y dF+ J q-grad-dF. 

The first integral, on being transformed into a surface integral, is seen to be 
zero. Thus 



= Jq.grad— dV = - j 



Stot = | q-grad^dF = - | q '**£ T dV, 



or 

1 dT 



ito,= -Jr^ dF - < 33 - + > 

In accordance with the general definition of the kinetic coefficients,:!: we 



t See Statistical Physics, §122. 

J We here use the definition given in Fluid Mechanics, §58. 



34 Viscosity of solids 153 

can deduce from (33.4) that in the case considered the coefficients T 2 K tk in 

1 dT 



/l dT\ 



are kinetic coefficients. Hence the result (33.3) follows immediately from the 
symmetry of the kinetic coefficients. 
The quadratic form 

dT dT dT 

dxi dxt dxjc 

must be positive, since the time derivative (33.4) of the entropy must be 
positive. The condition for a quadratic form to be positive is that the eigen- 
values of the matrix of its coefficients are positive. Hence all the principal 
values of the thermal conductivity tensor K ilc are always positive; this is 
evident also from simple considerations regarding the direction of the heat 
flux. 

The number of independent components of the tensor km depends on the 
symmetry of the crystal. Since the tensor K ik is symmetrical, this number is 
evidently the same as the number for the thermal expansion tensor (§10), 
which is also a symmetrical tensor of rank two. 

§34. Viscosity of solids 

In discussing motion in elastic bodies, we have so far assumed that the 
deformation is reversible. In reality, this process is thermodynamically 
reversible only if it occurs with infinitesimal speed, so that thermodynamic 
equilibrium is established in the body at every instant. An actual motion, 
however, has finite velocities ; the body is not in equilibrium at every instant, 
and therefore processes will take place in it which tend to return it to equili- 
brium. The existence of these processes has the result that the motion is 
irreversible, and, in particular, mechanical energyf is dissipated, ultimately 
into heat. 

The dissipation of energy occurs by two means. Firstly, when the tempera- 
ture at different points in the body is different, irreversible processes of thermal 
conduction take place in it. Secondly, if any internal motion occurs in the 
body, there are irreversible processes arising from the finite velocity of 
that motion. This means of energy dissipation may be referred to, as in 
fluids, as internal friction or viscosity. 

In most cases the velocity of macroscopic motions in the body is so small 
that the energy dissipation is not considerable. Such "almost irreversible" 
processes can be described by means of what is called the dissipative function.% 



f By mechanical energy we here mean the sum of the kinetic energy of the macroscopic motion in 
the elastic body and its (elastic) potential energy arising from the deformation. 
% See Statistical Physics, §123. 



154 Thermal Conduction and Viscosity in Solids §34 

If we have a mechanical system whose motion involves the dissipation of 
energy, this motion can be described by the ordinary equations of motion, 
with the forces acting on the system augmented by the dissipative forces or 
frictional forces, which are linear functions of the velocities. These forces 
can be written as the velocity derivatives of a certain quadratic function T 
of the velocities, called the dissipative function. The frictional force f a 
corresponding to a generalised co-ordinate q a of the system is then given by 
fa — - WlHa' The dissipative function T is a positive quadratic form in 
the velocities q a . The above relation is equivalent to 

SY = ~2/«%> (34.1) 

a 

where ST is the change in the dissipative function caused by an infinitesimal 
change in the velocities. It can also be shown that the dissipative function is 
half the decrease in the mechanical energy of the system per unit time. 

It is easy to generalise equation (34.1) to the case of motion with friction 
in a continuous medium. The state of the system is then defined by a con- 
tinuum of generalised co-ordinates. These are the displacement vector u at 
each point in the body. Accordingly, the relation (34.1) can be written in 
the integral form 

SJTdF= - jfiSutdV, (34.2) 

where ft are the components of the dissipative force vector f per unit volume 
of the body; we write the total dissipative function for the body as JT dV, 
where T is the dissipative function per unit volume. 

Let us now determine the general form of the dissipative function T for 
deformed bodies. The function T, which describes the internal friction, 
must be zero if there is no internal friction, and in particular if the body 
executes only a general translatory or rotary motion. In other words, the dis- 
sipative function must be zero if u = constant or u = SlXr. This means that it 
must depend not on the velocity itself but on its gradient, and can contain only 
such combinations of the derivatives as vanish when u = SI X r. These are 
the sums 



dxjc dxi 

i.e. the time derivatives u\u of the components of the strain tensor, f Thus 
the dissipative function must be a quadratic function of w^. The most 
general form of such a function is 

^ = IViklmUikUim- (34.3) 



t Cf. the entirely analogous arguments on the viscosity of fluids in Fluid Mechanics, §15. 



§35 The absorption of sound in solids 155 

The tensor rj ik i mt of rank four, may be called the viscosity tensor. It has the 
following evident symmetry properties: 

T)iklm = Vlmik = f]kilm = 7 ]ikmh ("•*) 

The expression (34.3) is exactly analogous to the expression (10.1) for the 
free energy of a crystal: the elastic modulus tensor is replaced by the tensor 
rjikim, and u ik by u ik . Hence the results obtained in §10 for the tensor X ik i m 
in crystals of various symmetries are wholly valid for the tensor r]m m also. 

In particular, the tensor rjmm in an isotropic body has only two independent 
components, and Y can be written in a form analogous to the expression 
(4.3) for the elastic energy of an isotropic body: 

Y = ^ite- ^imif+mi 2 , (34.5) 

where rj and £ are the two coefficients of viscosity. Since T is a positive 
function, the coefficients r\ and £ must be positive. 

The relation (34.2) is entirely analogous to that for the elastic free energy, 
8 $FdV = -IFihm dV, where F t = da ik \dx k is the force per unit volume. 
Hence the expression for the dissipative force ft in terms of the tensor u ik 
can be written down at once by analogy with the expression for F t in terms 
of ui k . We have 

fi = da'ikldxic, (34.6) 

where the dissipative stress tensor a'a is defined by 

a'ijc = BYjdUiJc = rjmmUim. (34.7) 

The viscosity can therefore be taken into account in the equations of motion 
by simply replacing the stress tensor <j ik in those equations by the sum 

oi ic + cr'i k . 

In an isotropic body, 

a'i k = 27](ui k -l8i k uu) + ^uiiB ik . (34.8) 

This expression is, as we should expect, formally identical with that for the 
viscosity stress tensor in a fluid. 

§35. The absorption of sound in solids 

The absorption of sound in solids can be calculated in a manner entirely 
analogous to that used for fluids.f Here we shall give the calculations for an 
isotropic body. The thermal-conduction part of the energy dissipation 
£nech is given by the integral -(k/T)/ (grad Tf dV. On account of viscosity, 
an amount of energy 2Y is dissipated per unit time and volume, so that the 
total viscosity part of E m ech is -2 J Y dV. Using the expression (34.5), we 



t See Fluid Mechanics, §77. 



156 Thermal Conduction and Viscosity in Solids §35 

therefore have 

imecn = - j /(grad Tf dV-2r)j(ui k -ls8 ik my dV-^ju^ dV. (35.1) 

To calculate the temperature gradient, we use the fact that sound oscilla- 
tions are adiabatic in the first approximation. Using the expression (6.4) for 
the entropy, we can write the adiabatic condition as So(T) + Koaiu = So(Tq), 
where To is the temperature in the undeformed state. Expanding the differ- 
ence So(T) — So(To) in powers of T—To, we have as far as the first-order 
terms S (T)-So(T ) = (T-T ) dSo/dT = C v {T-T )jT . The derivative 
of the entropy is taken for uu = 0, i.e. at constant volume. Thus 

T— T = — T(x.KuujC v . 

Using also the relations K = Kiso = C V K^\C V and K a a/p = cp-\c?\?>, 
we can rewrite this result as 

T*p(c?-4cflZ) 
T-T = — uu. (35.2) 

Let us first consider the absorption of transverse sound waves. The 
thermal conduction cannot result in the absorption of these waves (in the 
approximation considered). For, in a transverse wave, we have uu — 0, and 
therefore the temperature in it is constant, by (35.2). Let the wave be propa- 
gated along the #-axis; then 

u x = 0, u y = uoycos(kx—cot), u z = uo z cos(kx—cot), 

and the only non-zero components of the deformation tensor are 
u X y = — \kuoy sin(&K— exit), u X z = — \hioz sin(A#— cot). 

We shall consider the energy dissipation per unit volume of the body ; the 
(time) average value of this quantity is, from (35.1), 

^mech = -ii7 ft)4 ( M Oy 2 + Moz 2 )M 2 , 

where we have put k = co/ct. The total mean energy of the wave is twice 
the mean kinetic energy, i.e. 

E = pjtfdV; 

for unit volume we have 

£ = |/oco 2 (woy 2 + «oz 2 )- 

The sound absorption coefficient is defined as the ratio of the mean energy 
dissipation to twice the mean energy flux in the wave; this quantity gives 
the manner of variation of the wave amplitude with distance. The amplitude 



§35 The absorption of sound in solids 157 

decreases as e~ y *. Thus we find the following expression for the absorption 
coefficient for transverse waves: 

yt = illmecul/^fi = -n^Ppcf. (35.3) 

In a longitudinal sound wave u z - uocos(kx-cot), u y = u z = 0. A 
similar calculation, using formulae (35.1) and (35.2), gives 



CO 



[&♦«)« -sn- <■"> 



71 ~2 P c? 

These formulae relate, strictly speaking, only to a completely isotropic and 
amorphous body. They give, however, the correct order of magnitude for 
the absorption of sound in anisotropic single crystals also. 

The absorption of sound in polycrystalline bodies exhibits peculiar proper- 
ties. If the wavelength A of the sound is small in comparison with the 
dimensions a of the individual crystallites, then the sound is absorbed in 
each crystallite in the same way as in a large crystal, and the absorption 
coefficient is proportional to a> 2 . 

If A > «, however, the nature of the absorption is different. In such a 
wave we can assume that each crystallite is subject to a uniformly distributed 
pressure. However, since the crystallites are anisotropic, and so are the 
boundary conditions at their surfaces of contact, the resulting deformation is 
not uniform. It varies considerably (by an amount of the same order as 
itself) over the dimension of a crystallite, and not over one wavelength as in a 
homogeneous body. When sound is absorbed, the rates of change of the 
deformation (u ik ) and the temperature gradients are of importance. Of 
these, the former are still of the usual order of magnitude. The temperature 
gradients within each crystallite are anomalously large, however. Hence the 
absorption due to thermal conduction will be large compared with that due to 
viscosity, and only the former need be calculated. 

Let us consider two limiting cases. The time during which the temperature 
is equalised by thermal conduction over distances ~ a (the relaxation time 
for thermal conduction) is of the order of a 2 )x- Let us first assume that 
<o ^ xl<* 2 - Tnis means that the relaxation time is small compared with the 
period of the oscillations in the wave, and so thermal equilibrium is nearly 
established in each crystallite; in this case we have almost isothermal oscilla- 
tions. 

Let T be the temperature difference in a crystallite, and T ' the corres- 
ponding difference in an adiabatic process. The heat transferred by thermal 
conduction per unit volume is -divq = k[\T ~ kT/o 2 . The amount of 
heat evolved in the deformation is of the order of Tq'C ~ ojTo'C, where 
C is the specific heat. Equating the two, we obtain T ~ To'oja 2 jx- The 
temperature varies by an amount of the order of T over the dimension of 
the crystallite, and so its gradient is of magnitude ~ T'\a. Finally, TV is 



158 Thermal Conduction and Viscosity in Solids §35 

found from (35.2), with uu ~ ku ~ coufc (u being the amplitude of the 
displacement vector): 

T ' ~ Txpccou/C] (35.5) 

in obtaining orders of magnitude, we naturally neglect the difference between 
the various velocities of sound. Using these results, we can calculate the 
energy dissipated per unit volume : 

- K K (T'\ 2 

£ m ax~y(gradT)2~-^-j. 

Dividing this by the energy flux cE ~ cpa> 2 u 2 , we find the damping coefficient 
to be 

y ~ TaL 2 pca 2 a) 2 lxC for co < x/a 2 (35.6) 

(C. Zener 1938). Comparing this expression with the general expressions 
(35.3) and (35.4), we can say that, in the case considered, the absorption of 
sound by a polycrystalline body is the same as if it had a viscosity 

which is much larger than the actual viscosity of the component crystallites. 
Next, let us consider the opposite limiting case, where o> > xl a2 - I* 1 other 
words, the relaxation time is large compared with the period of oscillations 
in the wave, and no noticeable equalisation of the temperature differences 
due to the deformation can occur in one period. It would be incorrect, 
however, to suppose that the temperature gradients which determine the 
absorption of sound are of the order of To' /a. This assumption would take 
into account only thermal conduction in each crystallite, whereas heat ex- 
change between neighbouring crystallites must be of importance in the case in 
question (M. A. Isakovich 1948). If the crystallites were thermally insulated 
the temperature differences occurring at their boundaries would be of the 
same order To' as those within each individual crystallite. In reality, however, 
the boundary conditions require the continuity of the temperature across 
the surface separating two crystallites. We therefore have "temperature 
waves" propagated away from the boundary into the crystallite; these are 
damped at a distance^ 8 ~ vXx/ 60 )- I* 1 tne case under consideration 8 <^ a, 
i.e. the main temperature gradient is of the order of To'/8 and occurs over 
distances small compared with the total dimension of a crystallite. The cor- 
responding fraction of the volume of the crystallite is ~ a 2 8 ; taking the ratio 



t It may be recalled that, if a thermally conducting medium is bounded by the plane x — 0, at 
which the excess temperature varies periodically according to T" = T 'e r ~* tt) ', then the temperature 
distribution in the medium is given by the "temperature wave" T" = T^e-i^t e-(l+t)xv / (««'/2x); see 
Fluid Mechanics, §52. 



§35 The absorption of sound in solids 159 

of this to the total volume - a 3 , we find the mean energy dissipation 



Emech ~ ~f\T) a* ~ Ta8 ' 

Substituting for T ' the expression (35.5) and dividing by cE ~ cpuPu\ we 
obtain the required absorption coefficient: 

y ~ T<x. 2 pcV( X co)laC for co > x/a 2 . (35.7) 

It is proportional to the square root of the frequency-! 

Thus the sound absorption coefficient in a polycrystalline body varies as 
a? at very low frequencies (co <^ x /« 2 )*> for xl<* 2 < °> < c l a [t varies as V w » 
and for a> > c/a it again varies as a> 2 . 

Similar considerations hold for the damping of transverse waves in thin 
rods and plates (C. Zener 1938). If h is the thickness of the rod or plate, 
then for A > h the transverse temperature gradient is important, and the 
damping is mainly due to thermal conduction (see the Problems). If also 
o> ^ x jh 2 , the oscillations may be regarded as isothermal, and therefore, in 
determining (for example) the characteristic frequencies of vibrations of the 
rod or plate, the isothermal values of the moduli of elasticity must be used. 

PROBLEMS 
Problem 1. Determine the damping coefficient for longitudinal vibrations ofa rod. _ 
Solution. The damping coefficient for the vibrations is denned as £ = |£mech|/2-E; 
the amplitude of the vibrations diminishes with time as e'fiK 

In a longitudinal wave, any short section of the rod is subject to simple extension or com- 
pression; the components of the strain tensor are u zz = du z Idz, u xx = u yy = —o^duzjdz. 
We put u z = Mo cos kz cos wt, where k = «>/ V(#ad/p)- Calculations similar to those given 
above lead to the following expression for the damping coefficient: 

Here we have written £ a d and <* a d in terms of the velocities c u c t by means of formulae (22.4). 

Problem 2. The same as Problem 1, but for longitudinal oscillations of a plate. 

Solution. For waves whose direction of oscillation is parallel to that of their propagation 
(the w-axis, say) the non-zero components of the strain tensor are 

U X x = dUxfix, U ZZ = -[o-ad/tl-o-acOFM*/ 9 *; 

see (13.1). The velocity of propagation of these waves is \/[£'ad/p(l - ff ad 2 )]- A calculation 



gives 



_ afito 3cfi + 4c£-6c?c? W KT^p z (l + a & a) 2 \ 

P = Tp[Z cM{&-<*) + <#(<#- <$) + 9C v 2 '' 



t The same frequency dependence is found for the absorption of sound propagated in a fluid near 
a solid wall (in a pipe, for instance); see Fluid Mechanics, §77, Problems. 



160 Thermal Conduction and Viscosity in Solids §35 

For waves whose direction of oscillation is perpendicular to the direction of propagation, 
«n = 0, and the damping is caused only by the viscosity 17. In this case the damping coeffi- 
cient is fi = r)w*l2pct*. This applies also to the damping of torsional vibrations of rods. 

Problem 3. Determine the damping coefficient for transverse vibrations of a rod (with 
frequencies such that a> > x/h', where h is the thickness of the rod). 

Solution. The damping is due mainly to thermal conduction. According to §17, we have 
for each volume element in the rod Uzz = x/R, u xx = u yy = —a^xIR (for bending in the 
*ar-plane) ; for at > x/h', the vibrations are adiabatic. For small deflections the radius of 
curvature R = \\X", so that u ti = (1 — 2o & a)xX", the prime denoting differentiation with 
respect to z. The temperature varies most rapidly across the rod, and so (grad T) J 
«(3773*)*. Using (35.1) and (35.2), we obtain for the total mean energy dissipation in the 
rod — (KTa*E & &*SI9C v *) / X"* dz, where S is th e cros s-sectional area of the rod. The mean 
total energy is twice the potential energy E^l y \X" % dz. The damping coefficient is 

j8 = KT<x?SE aa /l8IyC p Z. 

Problem 4. The same as Problem 3, but for transverse vibrations of a plate. 
Solution. According to (11.4), we have for any volume element in the plate 

l-2a ad BH 
m = _ z 

1 — o"ad OX* 

for bending in the roar-plane. The energy dissipation is found from formulae ( 35 .1) and (5.2) 
and the mean total energy is twice the expression (11.6). The damping coefficient is 

2 K T^E &a 1 + gad _ 2kTv* p (3^-4^)2^2 
P ~~ ZC P W 1-crad " 3C,«A« ' W-cflcfi ' 

Problem 5. Determine the change in the characteristic frequencies of transverse vibrations 
of a rod due to the fact that the vibrations are not adiabatic. The rod is in the form of a 
long plate of thickness h. The surface of the rod is supposed thermally insulated. 

Solution. Let T & &(x, t) be the temperature distribution in the rod for adiabatic vibra- 
tions, and T(x, t) the actual temperature distribution; x is a co-ordinate across the thickness 
of the rod, and the temperature variation in the yz-ptene is neglected. Since, for T = T a d, 
there is no heat exchange between various parts of the body, it is clear that the thermal con- 
duction equation must be 

d 3 2 r 

u {T - T ** ) = x w- 

For periodic vibrations of frequency a>, the differences r a( j = T a d~" "^o» f — T—T from 
the equilibrium temperature T are proportional to e~ iM , and we have r" -\-imr\x = i<^T&Cilx> 
the prime denoting differentiation with respect to x. Since, by (35.2), r a( j is proportional 
to un, and the components «,# are proportional to x (see §17), it follows that T a d = Ax, where 
A is a constant which need not be calculated, since it does not appear in the final result. The 
solution of the equation T"+iair/x = iwAx/x, with the boundary condition t' = for 
x = ±ih( the surface of the rod being insulated), is 



(Sill nX \ 
*~i — in- ' k = (i+iWH2 x ). 
K COS ~hnfl / 



The moment M y of the internal stress forces in a rod bent in the xsr-plane is composed of 
the isothermal part M y< i s0 (i.e. the value for isothermal bending) and the part due to the 



§36 Highly viscous fluids 161 

non-uniform heating of the rod. If M„, a d is the moment in adiabatic bending, the second 
part of the moment is reduced from My.^a— M v ,iao in the ratio 

hh ift 

1 +f(o>) = j zr dz/j zr & a dz. 

-ift -ih 

Defining the Young's modulus # w for any frequency a» as the coefficient of proportionality 
between M„ and IJR (see (17.8)), and noticing that E^-E = E l Ta*l9C 9 (see (6.8); E is 
the isothermal Young's modulus), we can put 

E^E+ll+fMWTotpCj,. 

A calculation shows that /(o>) = (24/#V» 8 )(ita -tan \kh). For a. -> oo we obtain / = 1, 
which is correct, since Eoo = £ad» and for <a -*■ 0, / = and E = E. 

The frequencies of the characteristic vibrations are proportional to the square root of the 
Young's modulus (see §25, Problems 4-6). Hence 

"="o[l+/(*>o)— ], 

where a» are the characteristic frequencies for adiabatic vibrations. This value of to is 
complex. Separating the real and imaginary parts (<o = o'+tf), we find the characteristic 
frequencies 

r ETa? 1 sinh£-sin£l 

"' = °T " ~3C^'T*' cosh £+ cos d 

and the damping coefficient 



2ETa?x\ 1 sinh^+sinn 
^ = 3CpA2 [ ~ i'coshl+cosd' 



where £ = A\/( a> o/2x)« 

For large £ the frequency a> tends to w , as it should, and the damping coefficient to 
2£roc 2 x/3C,A 2 , in accordance with the result of Problem 3. 

Small values of £ correspond to almost isothermal conditions ; in this case 



co ^ coo\ 1 - 1S „ ) « cooV(EIE a d), 
and the damping coefficient /3 = i?T , a2A 2 w 2 /180C,X' 

§36. Highly viscous fluids 

For typical fluids, the Navier-Stokes equations are valid if the periods of 
the motion are large compared with times characterising the molecules. This, 
however, is not true for very viscous fluids. In such fluids, the usual equations 
of fluid mechanics become invalid for much larger periods of the motion. 
There are viscous fluids which, during short intervals of time (though these 
are long compared with molecular times), behave as solids (for instance, 
glycerine and resin). Amorphous solids (for instance, glass) may be regarded 
as a limiting case of such fluids having a very large viscosity. 

The properties of these fluids can be described by the following method, 
due to Maxwell. They are elastically deformed during short intervals of time. 



162 Thermal Conduction and Viscosity in Solids §36 

When the deformation ceases, shear stresses remain in them, although these 
are damped in the course of time, so that after a sufficiently long time almost 
no internal stress remains in the fluid. Let r be of the order of the time during 
which the stresses are damped (sometimes called the Maxwellian relaxation 
time). Let us suppose that the fluid is subjected to some variable external forces, 
which vary periodically in time with frequency w. If the period l/a» is large 
compared with the relaxation time t, i.e. wr <^ 1, the fluid under consideration 
will behave as an ordinary viscous fluid. If, however, the frequency co is suffi- 
ciently large (so that cot > 1), the fluid will behave as an amorphous solid. 

In accordance with these "intermediate" properties, the fluids in question 
can be characterised by both a viscosity coefficient r\ and a modulus of 
rigidity \l. It is easy to obtain a relation between the orders of magnitude of 
rj, fi and the relaxation time r. When periodic forces of sufficiently small 
frequency act, and so the fluid behaves like an ordinary fluid, the stress tensor 
is given by the usual expression for viscosity stresses in a fluid, i.e. 

o'ik = ^]Uik = —licorjuijc. 

In the opposite limit of large frequencies, the fluid behaves like a solid, and 
the internal stresses must be given by the formulae of the theory of elasticity, 
i.e. one = 2[m.ik\ we are speaking of pure shear deformations, i.e. we assume 
that ua — an = 0. For frequencies co ~ 1/r, the stresses given by these 
two expressions must be of the same order of magnitude. T1ius?7m/At ~ /x-m/A, 
whence 

7] ~ T[Jb. (36.1) 

This is the required relation. 

Finally, let us derive the equation of motion which qualitatively describes 

the behaviour of these fluids. To do so, we make a very simple assumption 

concerning the damping of the internal stresses (when motion ceases): 

namely, that they are damped exponentially, i.e. dcr^/d* = — ct^/t. In a 

solid, however, we have o% = 2/xm^, and so daacjdt = Ifidutjc/dt. It is easy 

to see that the equation 

da ilc a i1c dune 

= Zu. (3o.Z) 

dt T ^ dt ^ ' 

gives the correct result in both limiting cases of slow and rapid motions, and 
may therefore serve as an interpolatory equation for intermediate cases. 

For example, in periodic motion, where uiu and cruc depend on the time 
through a factor e~ iwt , we have from (36.2) —ioianc + oadr = —2iu)fj,Uijc, 
whence 

0ik = - — - — • (36.3) 

1 + l/COT 

For cot > 1, this formula gives aa = I^um, i.e. the usual expression for 
solid bodies, while for cut <^ 1 we have ow = —2ioi[irUik = 2/x,t«m, the 
usual expression for a fluid of viscosity /jlt. 



INDEX 



Absorption of sound 

in polycrystalline bodies 157-159 

in solids 155-161 
Adiabatic moduli 16-17 
Anharmonic effects 119-122 



Bending 

energy 58, 62—65 

moment 78 

of plates 44-53, 58-62 

principal planes of 78 

of rods 75-97 

of shells 62-68 

strip 64 

waves 114-115 
Biharmonic equation 18, 24 
Bulk modulus 1 1 
Burgers vector 124 



Coefficient 

of compression 12 

of extension 1 3 

of hydrostatic compression 12 

of thermal expansion 16 
in crystals 42 

of unilateral compression 1 5 
Combination frequencies 120 
Compression 

modulus of 11 

of rods 13-15 

uniform (hydrostatic) 6, 10 
coefficient of 12 
modulus of 11 

unilateral 1 5 
coefficient of 15 
Contact problems 30-37 
Crack in elastic medium 144-149 
Crystals 

biaxial 42 

elastic moduli of 37—43 

elastic properties of 37-43 

elastic waves in 106-108 

free energy of 37-41 

thermal conduction in 152-153 

thermal expansion of 42 

uniaxial 42 



Damping, see Absorption 
Deformation 1 



Deformation (continued) 
adiabatic 16-17, 101 
on contact 30-37 
due to dislocation 123-131 
elastic 8 

homogeneous 1 3-1 5 

of infinite medium 29 

isothermal 16-17 

plane 19-20, 24 

plastic 8, 135 

residual 8 

of shells 62-68 

with change of temperature 15-17 

thermodynamic relation for 9 

thermodynamics of 8-9 

torsional 69 
Dislocations (IV) 123-149 

continuous distribution of 134-139 

density tensor 134 

edge 123 

flux density tensor 137 

interacting 139-143 

moment tensor 127 

polarisation 138 

screw 124 

sign of 124n. 

in stress field 131-134 

wall 131 
Displacement vector 1 
Dissipative 

forces 1 54 

function 153 
Distortion tensor 125 



Elastic 

instability 97 

line 77 

modulus tensor 37 ; see also Modulus 

oscillations, see Elastic waves 

plane 53 

waves (III) 101-122 

absorption of 155-161 

anharmonic 1 1 9-1 22 

bending 114-115 

in crystals 106-108 

damping of 155-161 

in isotropic media 101-106 

longitudinal 102, 113-114 

Rayleigh 109-113 

reflection and refraction of 103-105 

in rods and plates 113-118 



163 



164 



Index 



Elastic waves (continued) 
surface 109-113 
torsional 115-116 
transverse 102, 114-115 
Energy 

bending 58 
for shell 62-65 

stretching 58 
for shell 62-65 

see also Free energy- 
Equilibrium 

equations of 7 

for infinite medium 29 
for isotropic bodies 18 
for membranes 61 
for plates 49, 54, 60, 61 
for rods 82-83, 85, 89 

of elastic half-space 25-29 

of isotropic bodies 17-25 

of rods and plates (II) 44-100 
Extension 

coefficient of 13 

modulus of 13 

of rods 13-15 



Free energy 

of deformed body 1 0-1 1 , 1 6 , 1 22 

of deformed crystal 37-41 

elastic lln. 

of bent plate 46, 59 

of bent rod 77, 80 

of unilaterally compressed rod 1 5 

of stretched rod 14 

of twisted rod 72, 80 
Frictional forces 154 



Group velocity of waves 107 



Hooke's law 12 

Hydrostatic compression 6, 10 

coefficient of 12 

modulus of 11 



Isothermal moduli 1 6-1 7 



Modulus (continued) 
of compression 1 1 
elastic 

for crystals 37-43 

for polycrystalline bodies 41-42 
of extension 1 3 
of hydrostatic compression 1 1 
isothermal 16-17 
of rigidity 1 1 
shear 1 1 
Young's 13 

for cubic crystals 43 



Neutral surface 
Notation viii 



44,75 



49 



Plates 

bending of 44-53, 58-62 
equation of equilibrium for 

clamped 49-50 

large deflections of 58-62 

longitudinal deformations of 53-57 

supported 51 

thin 44 

vibration of 113-118,159 
Poisson's ratio 13 



Rayleigh waves 109-113 

Reflection and refraction of elastic waves 

103-105 
Rigidity 

cylindrical 49n. 

flexural 49n., 90 

modulus of 11 

torsional 72 
Rods 

bending of 75-97 

clamped 83 

equations of equilibrium for 82-83, 85, 
89 

extension and compression of 1 3-1 5 

hinged 84 

small deflections of 89-97 

supported 84 

torsion of 68-75, 78-97 

vertical, deformation of 20 

vibration of 113-118, 159-161 



Lam6 coefficients 10 



Maxwellian relaxation time 162 

Membrane 61, 63 

Modulus 

adiabatic 16-17 

bulk 11 



Shear 

modulus 1 1 

pure 10 
Shearing force 89 
Shells 62 

deformation of 62-68 
Slip plane 133 
Stability of elastic systems 



65, 97-100 



Index 



165 



Strain tensor 2-4 

in cylindrical co-ordinates 4 

diagonalisation of 2 

principal axes of 2 

principal values of 2 

in spherical co-ordinates 3 

in terms of stress tensor 12, 14, 15 
Stress function 20, 54 
Stress, plane 54n. 
Stress tensor 5-7, 11 

mean value of 7 

in terms of strain tensor 11 , 14, 37 
Stresses 

concentration of 25, 57 

internal 4 

moment of 5-6 
Strings 92 
Summation rule In. 
Surface waves 109-113 



Tensor ellipsoid 42 
Theory of elasticity 1 

fundamental equations of (I) 1-43 
Thermal conduction 

in crystals 152-153 

in solids 150-153 
Thermal conductivity tensor 152 
Thermal expansion 15-17 

in crystals 42 



Thermal expansion coefficient 16 

in crystals 42 
Torsion 

angle 69, 79 

function 69 

of rods 68-75,78-97 
Torsional 

rigidity 72 

vibration 115-116 



Velocity of sound 

longitudinal 102 

transverse 102 
Vibration 

anharmonic 119-122 

of rods and plates 113-118 

torsional 115-116 
Viscosity 

high, in fluids 161-162 

of solids 153-155 

tensor 155 



Waves, see Elastic waves 



Young's modulus 13 
for cubic crystals 43 



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