Pergamon
Unified Engineering
Series
GENERAL E D I T O R S
Thomas F. Irvine, Jr.
State University of N e w York at Stony Brook
James P. Hartnett
University of Illinois at Chicago Circle
EDITORS
William F. Hughes
Carnegie-Mellon University
Arthur T. Murphy
Widener College
Daniel Rosenthal
University of California, Los Angeles
SECTIONS
Continuous Media Section
Engineering Design Section
Engineering Systems Section
Humanities and Social Sciences Section
Information Dynamics Section
Materials Engineering Section
Engineering Laboratory Section
Elasticity
Theory and Applications
A d d S. Saada, lng., E.C.P., Ph.D.
Professor of Civil Engineering
Case Western Reserve University
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Library of Congress Cataloging in Publication Data
Saada, Adel S
Elasticity: theory and applications.
(Pergamon unified engineering series, 16).
1. Elasticity. I. Title.
QA931.S2 1973
620.1'1232
72-86670
ISBN 0-08-017972-X
ISBN 0 - 0 8 - 0 1 7 0 5 3 - 6 (lib. bdg.)
Printed in the United States of America
Preface
This book is an outgrowth of notes used by the author during the past
few years in a course on solid mechanics. It is intended to give advanced
undergraduate a n d graduate students sound foundations on which to build
advanced courses such as mathematical elasticity, plasticity, plates a n d
shells, a n d those branches of mechanics which require the analysis of strain
and stress. T h e book is divided into three parts: Part I is concerned with
the kinematics of continuous media, Part II with the analysis of stress, a n d
Part III with the theory of elasticity a n d its applications to engineering
problems.
In Part / , the use of the notion of linear transformation of points makes
it possible to present the geometry of deformation in a language that is
easily understood by the majority of engineering students. It is agreed that
tensor calculus is the most elegant tool available to mechanicists, but
experience has shown that most engineering students are not ready to
accept it without a reasonable a m o u n t of preparation. T h e study of finite
a n d linear strains, using the notion of linear transformation, gradually
introduces the tensor concept and removes part of the abstraction c o m m o n ly associated with it. Orthogonal curvi-linear coordinates are examined in
detail a n d the results extensively used throughout the text.
In Part II, the study of stress proceeds along the same lines as that of
strain, a n d the similarities between the two are pointed out. All seven
chapters of Parts I a n d II are essential to the understanding of Part III a n d
serve as a c o m m o n base for all branches of mechanics.
In Part III, Chapter 8 covers the three-dimeinsional theory of linear
elasticity a n d the requirements for the solution of elasticity problems. The
method of potentials is presented in Chapter 9. Torsion is discussed in
Chapter 10 a n d topics related to cylinders, disks, a n d spheres are treated in
Chapter 11. Straight and curved beams are analysed in Chapters 12 a n d 13
respectively, a n d the answers of the elementary theories are compared to
the more rigorous results of the theory of elasticity. In Chapter 14, the semi-
infinite elastic medium a n d some of its related problems are studied using
the results of Chapter 9.
Energy principles and variational methods are presented in Chapter 15
and their application illustrated by a large n u m b e r of simple examples.
Columns and beam-columns are discussed in Chapter 16 a n d the bending
of thin flat plates in Chapter 17. Chapter 18 is more than an introduction
to the theory of thin shells. It includes a relatively detailed presentation of
the theory of surfaces which is necessary for the full understanding of the
analysis of thin shells. In this Chapter, as well as throughout this text,
geometry a n d the relations between strain and displacement are emphasized since it is my conviction that once geometry is mastered most of the
difficulties in studying the mechanics of solids will have disappeared.
The material in this text is suitable for two successive courses on solid
mechanics and elasticity. A first course would include Chapters 1 to 5, some
results from Chapter 6 a n d Chapter 7 to 13. A second course would include
Chapter 6 a n d Chapters 14 to 18. Chapters 10 to 18 can be read
independently from one another.
I wish to express my gratitude to Dr. T. P. Kicher who read the
manuscript and m a d e useful suggestions and to Dr. G. P. Sendeckyj with
w h o m many sections were discussed. Thanks are due to Professor W. F.
Hughes, technical editor of the Unified Engineering Series, for his patience
and support during the preparation of the final manuscript, a n d to the J o h n
T. Wiley Educational F u n d of Case Western Reserve University for
financial support. Mrs. W. Reeves very ably handled the typing.
Last but not least, I wish to acknowledge the encouragement and
understanding of my wife N a n c y during the various stages of writing this
book.
Adel S. Saada
About the Author
Adel S. S a a d a ( P h . D . , P r i n c e t o n University) is presently Professor of
Civil E n g i n e e r i n g at the C a s e Institute of T e c h n o l o g y of C a s e W e s t e r n
Reserve University, Cleveland, O h i o . D r . S a a d a received his Ingenieur
des Arts et Manufactures
degree from Ecole C e n t r a l e des A r t s et
M a n u f a c t u r e s d e Paris, F r a n c e a n d the equivalent of a M a s t e r of
Science degree from the University of G r e n o b l e , F r a n c e . Before c o m i n g
to P r i n c e t o n University the a u t h o r w a s a practicing structural engineer
in F r a n c e . D r . S a a d a ' s t e a c h i n g activities are in two major a r e a s : the
first is t h a t of the m e c h a n i c s of solids a n d in p a r t i c u l a r elasticity; the
s e c o n d is t h a t of m e c h a n i c s a p p l i e d to soils a n d f o u n d a t i o n s . His
research activities are primarily in the a r e a of stress-strain relations a n d
failure of transversely isotropic materials, in p a r t i c u l a r clay soils. M u c h
of his research w o r k h a s b e e n s u p p o r t e d b y p e r s o n a l grants from the
N a t i o n a l Science F o u n d a t i o n . D r . S a a d a is a m e m b e r of several
professional societies, a consulting engineer, a n d the a u t h o r of m a n y
p a p e r s o n soil m e c h a n i c s p u b l i s h e d in b o t h n a t i o n a l a n d i n t e r n a t i o n a l
journals.
Chapter 1
INTRODUCTION TO THE KINEMATICS
OF CONTINUOUS MEDIA
1.1
Formulation of the Problem
T h e theory of d e f o r m a t i o n of c o n t i n u o u s m e d i a is a purely m a t h e matical o n e . It is c o n c e r n e d with the study of the intrinsic properties of
the d e f o r m a t i o n s i n d e p e n d e n t of their physical causes. It is m o s t
conveniently expressed b y the n o t i o n of t r a n s f o r m a t i o n , w h i c h implies
d i s p l a c e m e n t a n d c h a n g e in s h a p e . T h e p r o b l e m is f o r m u l a t e d as
follows: G i v e n the positions of the points of a b o d y in its initial state
(i.e., before t r a n s f o r m a t i o n ) a n d in its final state (i.e., after t r a n s f o r m a tion), it is r e q u i r e d to d e t e r m i n e the c h a n g e in length a n d in direction
of a line element j o i n i n g two a r b i t r a r y points originally at a n infinitesim a l distance from o n e a n o t h e r .
In the following, we shall m a k e use primarily of o r t h o g o n a l sets of
cartesian c o o r d i n a t e s . Let xx, x 2, x3 b e the c o o r d i n a t e s of a p o i n t M of
: t r a n s f o r m a t i o n , this p o i n t
a b o d y B before t r a n s f o r m a t i o n . After
b e c o m e s M * with c o o r d i n a t e s £ l5 £2> £ 3
X
£1 =
\
+
U
\
£2 = x2 + u2
u
x
£3 =
3
+
(1.1.1)
3>
where w 1? u2, u3 are the projections of MM* o n the three axes OXx,
OX2, OX3 (Fig. 1.1). W e shall a s s u m e t h a t ux, u2, u3, as well as their
3
4
Kinematics of Continuous Media
*3
partial derivatives with respect to xl9 x2, x3, are c o n t i n u o u s functions of
x{, x2, x3. Eqs. (1.1.1) c a n therefore b e written a s :
|]
x+
= X
U (x^,X ,X )
]
23
£2 = x2 + w 2( x 1, x 2, x 3)
(1.1.2)
Let us consider two points, M(xx,x2,x3)
a n d N(xx + rfxj,x2 + rfx2,
x 3 -I- rfx3), infinitesimally n e a r o n e a n o t h e r . A s a result of the transform a t i o n , M is displaced to M * ( £ l £52, £ 3) a n d TV is displaced to
N*(£x + d£l9
£2 + ^ 2 ^ 3 + d& (Fig- 1.2). T h e c o o r d i n a t e s of N* are
5
Introduction
given b y :
£j + dix = xx + dxx + Wj 4- ^/wj
=
X
u
£2 + ^£2
2+ ^ 2+ 2 +
(1.1.3)
| 3 + d £ 3= x 3 + dx3 + w3+ <iw3.
Because of t h e a s s u m p t i o n s o n ux, u2, w 3, w e c a n write t h e d i s p l a c e m e n t
of TV u n d e r the form of a T a y l o r series in the n e i g h b o r h o o d of M :
«,+du,=(«,)*+(§-) ^,+i^tX ^X •
dX2+
••
dxi+
M
dxx
W3 + du3 = < « 3) „ + ( | f )M
+ ( | ^ ) ^ 2 +
( | | ) ^ .3
+ • • •
(1.1.4)
If we substitute E q s . (1.1.4) in E q s . (1.1.3), a n d s u b t r a c t E q s . (1.1.1)
from the resulting e q u a t i o n s , w e o b t a i n :
3
+
3
+ . . .
3 / W
(1.1.5)
3
+
• • •,
If, in E q s . (1.1.5), we neglect the h i g h e r - o r d e r t e r m s of T a y l o r ' s series,
the relations b e t w e e n
, d£2, d£3 a n d dxx, dx2, dx3 b e c o m e linear. T h e
J
6
Kinematics of Continuous Media
system of e q u a t i o n s c a n b e l o o k e d u p o n as a n o p e r a t i o n w h i c h
t r a n s f o r m s a vector dx {dxx,dx2,dx3)
of length ds to a vector d\
(d^,d^2,d^3)
of length ds*. This type of t r a n s f o r m a t i o n is called linear
transformation.
It is the linearization of E q s . (1.1.5) t h a t allows us to
a s s u m e t h a t the vector dx is t r a n s f o r m e d to a vector d\ a n d n o t to a
curve. T h e properties of linear t r a n s f o r m a t i o n s a r e discussed in C h a p t e r
3. If we o m i t the subscript M , Eqs. (1.1.5) are written as:
(1.1.6)
p r o v i d e d we k e e p in m i n d t h a t the partial derivatives of the functions
U\, u2, u3 are t a k e n at the p o i n t M.
In essentially static p r o b l e m s , while little c o n s i d e r a t i o n is given to
rigid b o d y displacements, p a r t i c u l a r a t t e n t i o n is given to the c h a n g e s in
length a n d in o r i e n t a t i o n of elements like ds. T h e s e c h a n g e s a r e
described b y the three c o m p o n e n t s of the relative d i s p l a c e m e n t vector
du , du , du (Fig. 1.3):
}
2
3
du
3
Introduction
d
du
x
= d£
-
x
dx
x
= g^-^i
3w
7>
+ ^
W
i
9w
3
+ ~—dx
2
7
7
<zw3 = a £ 3 — dx3 = ~—dxx
0X
x
2
9 2
2
vX
X
7
2
+ a^"*C3
W
^ 2
j
(\ 1 7 \
3w
+ ^ —3« x 3 .
^^3
T h e k i n e m a t i c s of c o n t i n u o u s m e d i a is c e n t e r e d o n the t w o sets of
Eqs. (1.1.6) a n d (1.1.7). W i t h i n the scope of this text, the necessary
m a t h e m a t i c a l tool r e q u i r e d to s t u d y these e q u a t i o n s is the n o t i o n of
linear t r a n s f o r m a t i o n . Since m a t r i x a l g e b r a w a s d e v e l o p e d primarily to
express linear t r a n s f o r m a t i o n s in a concise a n d lucid m a n n e r , it is
n a t u r a l t h a t it s h o u l d b e e m p l o y e d in the f o r m u l a t i o n a n d the solution
of k i n e m a t i c s p r o b l e m s . A brief review of m a t r i x a l g e b r a is given in
C h a p t e r 2.
1.2
Notation
T h e following system of n o t a t i o n will b e a d h e r e d to t h r o u g h o u t this
text:
9
In Eqs. (1.2.1), the e s r e m a i n u n c h a n g e d a n d the co's c h a n g e sign w h e n
the indices are i n t e r c h a n g e d . T h u s ,
dU
e
(
2
\dx
\2
e
\3
e
l
=
*31 =
(
\ \
e
23 = 32 =
\
^ 3
dx
(
\3x
x
+
+
+
2
3wj
dx
2
du
{
3
dx
du
2
3
dx
(1.2.2)
8
Kinematics of Continuous Media
and
(1.2.3)
W i t h these n o t a t i o n s , Eqs. (1.1.6) b e c o m e :
x = (1 4 eu)dxx
4 (e
)dx
x2 - co2X
2
d£
)1
^ + (1 +
xl 4 ( o 2dx
dii
= (e
h
=
d
0l3 ~
<*\l) \
+
4 (e
^22)^2 +
(*23 +
X3 4
(^23
^32)^2 +
0
u )dx
X3 3
~ w 3)2
rfx3
+
(1.2.4)
^33)^3-
Eqs. (1.1.7) b e c o m e :
du
e dx
du
(e
X2
(e
)dx
X3 - cox3
x
2
du
3
xx x
4 (e
)dx
X2 - u2X
2
4 u )dx
2X x
4 e dx
22 2
4 (e
4 (e
X3 4
u )dx
4 (e
-
u )dx
23
)dx
23 4 u32
2
X3 3
32 3
4
(1.2.5)
e dx .
33 3
In all the previous e q u a t i o n s , the c o o r d i n a t e s of the p o i n t s of the
b o d y in the t r a n s f o r m e d state are expressed in t e r m s of their c o o r d i nates in the initial state. This is k n o w n as the L a g r a n g i a n M e t h o d of
describing the t r a n s f o r m a t i o n of a c o n t i n u o u s m e d i u m . A n o t h e r m e t h od, the E u l e r i a n M e t h o d , expresses the c o o r d i n a t e s in the initial state in
t e r m s of t h e c o o r d i n a t e s in the final state. E a c h m e t h o d h a s its
a d v a n t a g e s . It is, however, m o r e c o n v e n i e n t in the study of the
m e c h a n i c s of solids to use the L a g r a n g i a n a p p r o a c h b e c a u s e the initial
state of the b o d y often possesses s y m m e t r i e s which m a k e it susceptible
to description in a simple system of c o o r d i n a t e s . T h e L a g r a n g i a n
M e t h o d is exclusively used in this text.
Chapter 2
REVIEW OF MATRIX ALGEBRA
2.1
Introduction
T h e use of matrices in m e c h a n i c s i n t r o d u c e s a n o t a t i o n t h a t enables
o n e to see the c o m p o n e n t s of the entities b e i n g studied in their totality,
while p r o v i d i n g great conciseness. I n this c h a p t e r , the basic definitions
a n d the o p e r a t i o n s of m a t r i x algebra w h i c h will b e n e e d e d in this text
are given.
2.2
Definition of a Matrix. Special Matrices
A matrix is a n a r r a y of elements a r r a n g e d in rows a n d c o l u m n s . F o r
instance, a m a t r i x of m rows a n d n c o l u m n s is w r i t t e n :
[a] =
a
0\2
a
•
•
\n
2\
22
•
•
ln
a
a
(2.2.1)
a
m\
ai
m
•
a n d is called a n (m X n) m a t r i x . T h e first subscript / of e a c h e l e m e n t atj
represents the n u m b e r of the row, a n d the s e c o n d subscript j represents
the n u m b e r of the c o l u m n . T h e a^s c a n b e p u r e n u m b e r s , functions,
instructions to a c o m p u t e r , or o t h e r m a t r i c e s . I n this text, the elements
are all real. A s q u a r e m a t r i x with n rows a n d c o l u m n s is said to be of
o r d e r n.
A symmetric matrix h a s elements which satisfy the c o n d i t i o n ay = a^.
This m e a n s t h a t elements symmetrically l o c a t e d with respect to the
9
10
Kinematics of Continuous Media
m a i n d i a g o n a l of the m a t r i x are e q u a l in m a g n i t u d e a n d sign.
An antisymmetric
or skew symmetric matrix has elements which satisfy
the c o n d i t i o n atj = — . T h i s m e a n s t h a t elements symmetrically located with respect to the m a i n d i a g o n a l are e q u a l in m a g n i t u d e a n d
opposite in sign, a n d t h a t the elements of the d i a g o n a l are equal to zero.
A diagonal matrix is a m a t r i x w h o s e elements atj vanish except for / = j .
T h e s e n o n - v a n i s h i n g elements constitute the m a i n d i a g o n a l of the
matrix.
A unit matrix is a d i a g o n a l m a t r i x w h o s e elements are e q u a l to unity. It
is written [1].
A null matrix h a s all its elements e q u a l to zero. It is written [0],
A column matrix h a s m rows a n d o n e c o l u m n , it is also called a c o l u m n
vector a n d is written:
0 l
2
{d}
=
A row matrix is a m a t r i x with o n e r o w a n d n c o l u m n s . It is also called
a row vector a n d is written:
[a] = [au .
..aln
].
The transpose of a matrix [a] is a m a t r i x [a] ', w h o s e rows are the s a m e
as the c o l u m n s of [a]. T h u s , a s y m m e t r i c matrix is its o w n t r a n s p o s e a n d
the t r a n s p o s e of a c o l u m n m a t r i x is a row matrix.
A scalar matrix is a d i a g o n a l m a t r i x w h o s e elements are identical.
2.3
Index Notation and Summation Convention
T h e i n t r o d u c t i o n of n u m e r i c a l subscripts in C h a p t e r 1 to d e n o t e the
reference axes m a k e s the use of indices in writing the c o m p o n e n t s of
vectors quite n a t u r a l . W h e n writing relations b e t w e e n vectors or o t h e r
directional quantities (such as tensors), a great deal of space is saved
w h e n a s h o r t h a n d n o t a t i o n is i n t r o d u c e d . In this text, the only indices
to b e used are subscripts a n d the following c o n v e n t i o n s will b e a d h e r e d
to:
Review of Matrix Algebra
11
The range convention: W h e n e v e r a subscript is r e p e a t e d in a term, it is
u n d e r s t o o d to represent a s u m m a t i o n over the r a n g e 1, 2, 3 unless
otherwise stated. Also, a n index never a p p e a r s m o r e t h a n twice in the
s a m e term. F o r e x a m p l e , the expression
(2-3.1)
Zi = aijXj
c o n t a i n s , in the r i g h t - h a n d term, the index j which is r e p e a t e d . T h e r e fore, taking the values of / = 1, 2, 3 in turn, we o b t a i n the three linear
equations:
xa
£l =
xa
+
\ \ \
\2 2 +
+^ 1x3 *a3
+ a 22 * 2
i 2 = alx
xx
23 3
£3 = ^31*1 + ^ 3 2 * 2 +
x
a
(2.3.2)
33 3-
/ is the identifying index a n d j is the summation index. W e notice t h a t the
s u m m a t i o n i n d e x c a n b e c h a n g e d at will a n d is therefore called a
dummy index. T h u s , E q s . (2.3.2) c a n also b e written:
=
£/
ax
ik k -
T h e index k is similar to the d u m m y variable of integration in a definite
integral a n d c a n b e c h a n g e d freely.
F o r c o n v e n i e n c e , it is s o m e t i m e s useful to i n t r o d u c e the two following
symbols:
The Kronecker delta, 8tj, which b y definition is such t h a t :
<5,lJ; = 1, w h e n / = / a n d
8tJ = 0, w h e n i =^y.
The alternating
symbol,
eijk
, which by definition is such t h a t :
e
=
ijk
E
0, w h e n a n y two of
ijk
~
(2.3.3)
k are e q u a l
1' w h e n / , 7 , k are different a n d in cyclic o r d e r ( 1 , 2 , 3 ,
1,2,3,...)
eijk = — 1, w h e n
(2.3.4)
k are different a n d n o t in cyclic o r d e r ( 1 ,
3,2,1,3,2,...).
12
Kinematics of Continuous Media
Examples
1). 8ik
xk
for / = 1 is equal t o :
X
8\\X
X
x
+ 8\2 2
~*~ ^13 3
x
~
\
~
x
i-
2). § , = 5 „ + 5 22 + 5 33 = 3.
3). A vector2 x w h o s e c o m p o n e n t s are X\,x2,x3,
has a m a g n i t u d e
|JC| = \ A i + x\ + x\ = \Jxixi
. Its direction cosines are given by
l = Xi /yJxjXj
t
.
4). T h e s u m of the d i a g o n a l elements of a m a t r i x [a] is called the trace
aa a
of [a] a n d is written au.
5). T h e d e t e r m i n a n t of the m a t r i x [a] is written
^ijk \i 2j 3k-
2.4
Equality of Matrices. Addition and Subtraction
Let us t u r n n o w to the rules governing the m a n i p u l a t i o n of the a r r a y s
of elements forming a matrix. T w o matrices [a] a n d [b] of the s a m e o r d e r
are said to be equal if, a n d only if, their c o r r e s p o n d i n g elements are
identical; that is, we h a v e :
(2.4.1)
provided that
a
=
u
bj; for all /' a n d /
(2.4.2)
If [a] a n d [b] are matrices of the s a m e order, then the s u m of [a] a n d [b]
is defined to be a m a t r i x [c], the typical element of which is ctj
= atj + by. In other w o r d s , b y definition:
[c] = [a] + [b],
(2.4.3)
provided
(2.4.4)
In a similar m a n n e r , we h a v e :
[d] = [a] - [b],
(2.4.5)
provided
dij
=
fly
-
/>//.
(2.4.6)
13
Review of Matrix Algebra
F r o m the a b o v e definitions,
o p e r a t i o n s are valid:
it c a n be s h o w n
that
the
following
[a] + [b] = [b] + [a]
(2.4.7)
([a] + [b]) + [c] = [a] + ([b] + [c]).
(2.4.8)
A n i m p o r t a n t p r o p e r t y of s q u a r e matrices, which follows from the
laws of a d d i t i o n a n d s u b t r a c t i o n , is t h a t a n y square m a t r i x m a y b e given
as the s u m of a s y m m e t r i c a n d of a n a n t i s y m m e t r i c matrix. I n d e e d , if
[a] is a s q u a r e matrix, then
[a] + [a]< +[a)-[a]'
[ ]a =
2.5
)
Multiplication of Matrices
T h e p r o d u c t of a matrix [a] by a m a t r i x [b] is defined by the e q u a t i o n
(2.5.1)
Mb] = [c],
where the elements of [c] are given b y :
c
ij
a
(2.5.2)
~ ik t>kj •
Thus
b
a
aa
X2 a 13
23
a22
b a bn
b2\ a bb22
3\
32
3\
33
a
b ;
bx
=
23
32
b33
a
b
xx bn
a
a 2Xbn
_ 3\
b
bn
+ \2 22 + a\3 32
b
+
#23*32
+
a
+
^32*22 + #33*32
2 22
2
+
^12*21 +
a b
+
^ 2 2a* 2 1 +
ab
+ 32
2\ +
a
b
X33X xx \2
233X a2XbbX2
ab
333x a3X \2
+ ^12*23 + ^13*33
\\b\3
# 2 1 * 13 a
+b # 2 2 * 2 3 +
3\
a
\3
+ #32*23 +
#23*33
#33*33
T w o matrices can be multiplied by each other only if they are
conformable,
which m e a n s t h a t the n u m b e r of the c o l u m n s of the first is
equal to the n u m b e r of the rows of the second. T h u s , if [a] is a n (m X p)
matrix a n d [b] is a (p X n) matrix, t h e n [c] is a n (m X n) matrix.
T w o n o n z e r o matrices c a n b e multiplied b y e a c h o t h e r a n d result in a
zero matrix. F o r e x a m p l e ,
14
Kinematics of Continuous Media
"1
1
0" "0
0
0"
0
0
0
0
0
0
0
0
0
1
0
0
=
0"
"0
0
0
0
0
0
0
0
A p e r m u t a t i o n of the matrices will lead to a different result:
"0
0
0 " "1
1
0"
0
0
0
0
0
0
1
0
0
0
0
0
=
"0
0
0"
0
0
0
1
1
0
T h e p r o d u c t [b] [a] is, in general, n o t equal to [a] [b]. Therefore, it is
necessary to differentiate b e t w e e n premultiplication,
as w h e n [b] is
premultiplied by [a] to yield the p r o d u c t [a] [b], a n d
postmultiplication,
as w h e n [b] is postmultiplied b y [a] to yield [b] [a]. If we h a v e two
matrices which are such t h a t
[a][b] = [b][a],
(2.5.3)
these matrices are said to commute or to be p e r m u t a b l e .
Of particular i m p o r t a n c e is the associative law of c o n t i n u e d p r o d u c t s ,
[d] = ([a][b])[c] = [a]([b][c]l
(2.5.4)
which allows o n e to dispense with p a r e n t h e s e s a n d to write [a] [b] [c]
w i t h o u t a m b i g u i t y since the d o u b l e s u m m a t i o n
dy =
hc
aik ki ij
(2.5.5)
c a n b e carried o u t in either of the orders indicated. It m u s t b e n o t i c e d
that the p r o d u c t of a c h a i n of matrices will h a v e m e a n i n g only if the
adjacent matrices are c o n f o r m a b l e .
T h e p r o d u c t of matrices is distributive, t h a t is
[a]([b] + [c]) = [a][b] + [a][c].
(2.5.6)
T h e multiplication of a matrix [a] by a scalar k is defined b y :
k[a] = [b],
where
(2.5.7)
15
Review of Matrix Algebra
U s i n g the definition of the t r a n s p o s e a n d the laws of a d d i t i o n a n d
multiplication of matrices, it c a n b e s h o w n t h a t :
([a] + [b])' = [a]' + [b]'
(2.5.8)
(*[*])' = k[a]'
(2.5.9)
(Mb])'
= [b]'[a]' (note t h e o r d e r ) .
(2.5.10)
F o r t h e case of t h e unit matrix, we h a v e :
W[l]
= [1]W = W
(2-5.11)
a n d , if k is a c o n s t a n t ,
[a]k[l] = k[a][l] = k[a] = k[\][a].
(2.5.12)
A n i m p o r t a n t result in t h e theory of matrices is that t h e d e t e r m i n a n t of
the p r o d u c t of t w o square matrices is equal to the p r o d u c t of their
determinants. Thus,
\[a][b]\ = (m|)(|[&]|)
= (|[6]|)(Ml).
(15
-
13)
A m o n g t h e special matrices defined in Sec. 2.2, the d i a g o n a l m a t r i x
plays a n i m p o r t a n t p a r t in o p e r a t i o n s involving matrices. T h e p r e m u l t i plication of a m a t r i x [a] b y a d i a g o n a l m a t r i x [d] p r o d u c e s a m a t r i x
whose rows a r e those of [a] multiplied b y t h e e l e m e n t in the c o r r e s p o n d ing r o w of [d\:
a
0
o"
0
d2
0
0
0
d3
a
a
a\3~
a23
\2
\ \
a
^32
_ 3\
33_
=
dxau
dd2aa2l
dxaX2
dd2aa22
3 3\
3 32
da
d\a \ 3
d2a23
.
(2.5.14)
3 33
T h e p o s t m u l t i p l i c a t i o n of [a] b y [d] p r o d u c e s a m a t r i x whose c o l u m n s
are those of [a] multiplied b y the element in t h e c o r r e s p o n d i n g c o l u m n
of [d\:
a
_ 31
0
o"
0
«12
«13
«22
«23
0
di
«33_
0
0
«32
d
K
ddxau
da
2 \2
= d\ <*21 add2aa22
\
3\
2 32
da
d3a\3
d3a23
3 33_
.
(2.5.15)
16
Kinematics of Continuous Media
T h e d i a g o n a l m a t r i x [d] is therefore a c o n v e n i e n t tool for writing g r o u p s
of e q u a t i o n s u n d e r t h e form of o n e single m a t r i x e q u a t i o n . F o r e x a m p l e ,
a whole g r o u p of systems of e q u a t i o n s , such as
[a]{x} = A,{8}, [a]{y} = \2{c},
[a]{z] =
\3{e},
c a n b e written:
[«][/] = [h][d],
w h e r e [/] h a s c o l u m n s f o r m e d b y {x}9{y}, (z); [h] h a s c o l u m n s fao r mn
ed by
{*}, {c}, [e}\ a n d [d] is a d i a g o n a l m a t r i x with elements A,, A2> ^ A 3. If
[a] a n d [*] a r e d i a g o n a l m a t r i c e s of the s a m e order, they a r e c o m m u t a tive with each o t h e r so that [a][b] = [b][a].
2.6
Matrix Division. The Inverse Matrix
1
If t h e d e t e r m i n a n t \[a]\ of a m a t r i x [a] does n o t vanish, [a] is said
to
b e n o n s i n g u l a r a n d possesses a reciprocal or inverse m a t r i x [ a ] " , such
that
[a][a]- 1
[l] = [*]-'[4
(2.6.1)
The cofactor matrix of a n y s q u a r e m a t r i x [a] is t h e matrix o b t a i n e d b y
replacing e a c h e l e m e n t of [a] b y its cofactor. It will b e r e m e m b e r e d t h a t
the cofactor Ay of a n y e l e m e n t ay of a d e t e r m i n a n t \[a]\9is t h e m i n o r of
that e l e m e n t with a sign a t t a c h e d to it d e t e r m i n e d b y the n u m b e r s / a n d
j w h i c h fix the position of ay in the d e t e r m i n a n t . T h e sign is given b y
the e q u a t i o n giving t h e cofactor Ay-:
+
Aij =
(-iy JMiJ
,
(2.6.2)
w h e r e My is the m i n o r of t h e e l e m e n t ay. F o r e x a m p l e , the m i n o r
of t h e d e t e r m i n a n t of the m a t r i x
#11
#12
#13
#21
#22
#23
#31
#32
#33
A23
(2.6.3)
17
Review of Matrix Algebra
is
A23 =
(-iy
au
#31
a n d the cofactor m a t r i x [C0A]
[C0A]
aX2
#32
= ana3X
axx
a32
\
-
(2.6.4)
of the s q u a r e m a t r i x [A] is:
=
A
A
An
2\
A 22 A 32
A 23 ^ 3 3
A3l
13
(2.6.5)
If the d e t e r m i n a n t \[a]\ of the m a t r i x [a] is n o t e q u a l to zero, in o t h e r
w o r d s if [a] is n o n s i n g u l a r , t h e n
1 J
^21
^31
M L
M L
A22
IWI
(2.6.6)
ll«JI
M L
^13
A32
'23
^33
IWI
IM
T h e t r a n s p o s e of the cofactor m a t r i x is also called the adjoint of [a].
Thus,
1
J
IWI
(2.6.7)
T h e previous e q u a t i o n c a n b e verified b y direct s u b s t i t u t i o n in E q .
(2.6.1). V a r i o u s m e t h o d s are available for the inversion of m a t r i c e s . T h e
b i b l i o g r a p h y at the e n d of this c h a p t e r gives detailed i n f o r m a t i o n o n the
subject.
I n m a t r i x algebra, multiplication b y the inverse of a m a t r i x plays the
s a m e role as division in o r d i n a r y algebra. T h a t is, if we h a v e :
[a][b] = [c][e],
(2.6.8)
1
where [a] is a n o n s i n g u l a r m a t r i x , t h e n o n p r e m u l t i p l y i n g b y [a] , t h e
inverse of [a], we o b t a i n :
(2.6.9)
18
Kinematics of Continuous Media
a n d , because of Eq. (2.6.1),
]
(2.6.10)
[b] = [a]- [c][e].
Using the definition of the t r a n s p o s e a n d that of the inverse, it c a n be
shown t h a t
61 1
{[«]-'}'= Mr .
1
(2- - )
which m e a n s that the t r a n s p o s e of the inverse of a matrix is equal to the
inverse of its t r a n s p o s e . Also,
(2.6.12)
{[a][b]}-' = [b]-*[a]-^
which m e a n s t h a t the inverse of the p r o d u c t of two matrices is equal to
the p r o d u c t of the inverse of the second by the inverse of the first. T h e
inverse of a diagonal m a t r i x is a diagonal m a t r i x whose elements are the
reciprocals of those of the matrix itself. T h e inverse of a s y m m e t r i c
matrix is a symmetric matrix.
PROBLEMS
1.
Given
"3
-2
5"
6
0
3
1
5
4
[«] =
2.
a n d [b] =
"2
3
- 1 "
4
1
0
5
2
-1
(a) c o m p u t e [a] + [b] a n d [a] - [b].
(b) Verify: [a] + ([b] - [c]) = ([a] + [b]) - [c],
(c) Split [a] into its s y m m e t r i c a n d its a n t i s y m m e t r i c p a r t s .
Given
1
[a]
-3
2
2
1
-3
4
-3
-1
1
,[b] =
4
1
0
2
1
1
1
1
-2
1
2
and
[c] =
2
1
3
-2
- 1 - 2
-1
2 - 5 - 1
- 1 , show that [a][b] = [a][c]
0
Review of Matrix Algebra
3.
4.
5.
19
in spite of the fact that [b] =^= [c].
If [b] = [a][a]', show t h a t [b] = [b]'.
If [a] is a c o l u m n matrix, show t h a t {5){a}' = [c], where [c] is a square
matrix with t h e p r o p e r t y that [c] = [c]'.
Given
"5
w
=
-2
" 3 "
0~
-2
3
-1
0
-1
1
and
[b]
-2
=
1
6.
c o m p u t e t h e p r o d u c t [a]{b}.
If [a] is a s q u a r e matrix of order 3, show that its d e t e r m i n a n t is given
aua2ja3k
(ij\k9
= 1,2,3).
by eijk
7.
W r i t e o u t in full t h e following expressions :
8.
a)
aiJ
xixJ
b)
8ijxixJ
c) ani = ojt(j
d)
o'y = 4 ljm
okm
e)
a /7 = 2/xe/y +
X8yekk
T h e subscripts
k, a n d m take the values 1, 2, a n d 3.
F i n d t h e inverse of t h e matrices
2
-2
2
3
2
-1
1
-1
2
3"
0
1
5
4
0
3
"1
4 "
and
REFERENCES
1. R. A. Frazier, W. J. Duncan, A. R. Collar, Elementary
Matrices,
MacMillan, N e w York,
N . Y., 1947.
2. S. Pedis, Theory of Matrices, Addison-Wesley, Reading, Mass., 1952.
3. F. B. Hildebrand, Methods of Applied Mathematics, Prentice-Hall, N e w York, N . Y., 1952.
4. L. Fox, An Introduction to Numerical Linear Algebra, Oxford University Press, N e w York,
N . Y., 1965.
5. J. B. Scarborough, Numerical Mathematical Analysis, The Johns Hopkins Press, Baltimore,
Md., 1966.
CHAPTER 3
LINEAR TRANSFORMATION OF POINTS
3.1
Introduction
T h e i m p o r t a n c e of linear t r a n s f o r m a t i o n s for the study of k i n e m a t i c s
was i n d i c a t e d in C h a p t e r 1. In the present chapter, this k i n d of
t r a n s f o r m a t i o n is e x a m i n e d in detail, a n d w h e n e v e r possible the results
are i n t e r p r e t e d geometrically. This i n t e r p r e t a t i o n is essential if o n e is to
visualize the d e f o r m a t i o n of c o n t i n u o u s m e d i a .
In a d d i t i o n to k i n e m a t i c s , topics such as stress, m o m e n t s of inertia of
surfaces a n d volumes, a n d c u r v a t u r e of surfaces, to m e n t i o n only a few,
involve linear t r a n s f o r m a t i o n s . This c h a p t e r serves, therefore, as a
f o u n d a t i o n c o m m o n to a w i d e variety of subjects in m e c h a n i c s .
3.2
Definitions and Elementary Operations
In a trirectangular system of c o o r d i n a t e s OXx, OX2, OX3 (Fig. 3.1),
consider the linear e q u a t i o n s giving the c o o r d i n a t e s of a p o i n t M * ( £ 1?
£ 2, £ 3 ) in t e r m s of those of M(xx,
x2,
x3):
£1 = axx
xx
+ aX2
x2
+
i 2 = a2X
xx
+ a22
x2
+ a23
x3
£3 =
+
*31*1
^32*2 +
aX3
x3
(3.2.1)
#33-*3>
w h e r e the
s are c o n s t a n t s . T h e s e e q u a t i o n s are said to t r a n s f o r m the
p o i n t M to the p o i n t M * . O n e m a y choose to consider t h a t E q s . (3.2.1)
t r a n s f o r m the vector OM to the vector OM*. I n such a case, however,
20
Linear Transformation of Points
21
these two vectors are tied to the p o i n t O a n d are n o t free vectors. T h e
t r a n s f o r m a t i o n expressed by E q s . (3.2.1) is called a p o i n t - t o - p o i n t linear
t r a n s f o r m a t i o n or, simply, a linear t r a n s f o r m a t i o n . It c a n b e written in
a n y of the following forms:
=
a
a2\
a
a22
23
3\
32
33
£3
{OM*}
=
[a]{OM},
x
a
a
x3
a
v j-
=
W e c a n look u p o n the m a t r i x [a] as a n o p e r a t o r acting o n the c o l u m n
vector {OM} to give the c o l u m n vector {OM*}. T h e inverse of this
t r a n s f o r m a t i o n gives {OM} in terms of {OM*}, p r o v i d e d the m a t r i x [a]
is n o n s i n g u l a r :
l
{OM} =
T h e vector MM*
given b y :
"2
_»3_
is called the d i s p l a c e m e n t of M . Its c o m p o n e n t s are
-
«l"
=
ii £3
(3.2.2)
[a]~ {OM*}.
-
xx
2
x3_
=
x on a
a2l
3\
1
an
a22 - 1
«32
a« 1 3
(3.2.3)
23
a33
x3
22
Kinematics of Continuous Media
If, from the origin O (Fig. 3.1), we d r a w a vector OK parallel to M M *
a n d whose c o m p o n e n t s are ux, u2, a n d w 3, Eqs. (3.2.3) can be looked
u p o n as t r a n s f o r m i n g the point M to the point K. Such a t r a n s f o r m a t i o n
from M to K is called the hodograph of the t r a n s f o r m a t i o n from M to
M*.
_
_
Eqs. (3.2.1) show that if Xx a n d X2 are two vectors tied to the origin
0 , then
[a]{Xx + *2)
=
+ [a]{X2].
(3.2.4)
Using Eq. (3.2.4) (n — 1) times, we get for a vector X:
[a]{nX] =
(3.2.5)
w h e r e n is an integer. Eq. (3.2.5) can be generalized for fractional values
of n.
Eq. (3.2.5) shows that a linear t r a n s f o r m a t i o n of points transforms a
straight line OAB t h r o u g h the origin to a n o t h e r straight line Oa/3
t h r o u g h the origin, a n d that (Fig. 3.2):
21 =
Oa
QB
OA '
Eq. (3.2.4) shows that a parallelogram OACB is transformed by Eqs.
(3.2.1) to a p a r a l l e l o g r a m Oayfi (Fig. 3.3). T h u s , a free vector AC is
transformed to a n o t h e r free vector ay, a parallelogram to a parallelogram, a p l a n e to a p l a n e , a n d a parallelepiped to a parallelepiped.
If a m a t r i x [a] t r a n s f o r m s a vector OP to OYlx, a n d a m a t r i x [6]
transforms OP to O n 2 — i.e., if
Fig. 3.2
Linear Transformation of Points
23
C
Fig. 3.3
[a]{OP] = (OTI,}
(3.2.6)
[b]{OP] = {On 2},
(3.2.7)
and
then O i l j + OU2 = OH is defined as the sum of the two
(Fig. 3.4). Therefore,
transformations
[a]{OP] + [b]{OP) = [c]{OP] = OH,
(3.2.8)
[c] = [a] + [b].
(3.2.9)
where
Fig. 3.4
24
Kinematics of Continuous Media
F i g . 3.5
If a m a t r i x [a] transforms a vector OP to OIix, a n d a n o t h e r m a t r i x [b]
t r a n s f o r m s OUx to OTI12
, the t r a n s f o r m a t i o n which brings OP to
OHn
is defined as the product of the two transformations
(Fig. 3.5). T h u s ,
[b][a]{OP]
=
[c]{OP]
= {On }.
i2
(3.2.10)
A s s h o w n in the previous chapter, in general,
[b][a]{OP) ^
[a][b}{OP}
= (On },
21
(3.2.11)
a n d the two points I I 21 a n d I I 12 d o n o t coincide (Fig. 3.5).
A small transformation
is o n e w h o s e m a t r i x is nearly e q u a l to the
identity or unit m a t r i x . F o r e x a m p l e , the t r a n s f o r m a t i o n (Fig. 3.6)
Linear Transformation of Points
[[1] + [a]]{OP] =
"1 a +au
fl12
«13
2\
1 + a22
«23
«31
«32
1 + a 33
x2
25
(3.2.12)
*3
is said to b e small w h e n all the
s are small with respect to unity. The
product of two small transformations
gives:
[[1] + [a]][[l] + [b]){OP} = [[1] + [a] + [b] + [a][b]]{OP).
(3.2.13)
If the t e r m s of the s e c o n d o r d e r are neglected, t h e n
[[1] + M i l ]
+ [b]]{OP} = [[1] + [a] + [b]]{OP}.
(3.2.14)
W e notice t h a t the o r d e r n o longer intervenes, which m e a n s t h a t the
o p e r a t i o n is c o m m u t a t i v e .
3.3
Conjugate and Principal Directions and Planes in a Linear
Transformation
C o n s i d e r a b o d y B which is t r a n s f o r m e d to /? b y a p o i n t - t o - p o i n t
linear t r a n s f o r m a t i o n (Fig. 3.7). A p l a n e P a n d a straight line D in B a r e
t r a n s f o r m e d to a p l a n e IT a n d a straight line A in /?. By definition, if A
is n o r m a l to I I , P a n d D are called conjugate. If P a n d Z), as well as A
a n d I I , are p e r p e n d i c u l a r to o n e a n o t h e r , t h e n P a n d D are called
principal plane a n d principal direction, respectively.
26
Kinematics of Continuous Media
Let us find the s h a p e of the surface w h o s e p o i n t s after t r a n s f o r m a t i o n
fall o n a sphere of radius R a n d which is c e n t e r e d at the origin. T h e
e q u a t i o n of the sphere is (Fig. 3.8):
2
(3.3.1)
# + % + % = R
a n d t h a t of the initial surface is:
2
( 0 1 1 * 1 + # 1 2 * 2 + #13 * 3 )
+
(#31*1 +
2
+
(#21*1 + #22*2 +
2
#32*2 + #33*3>
or
2
2
j
(ajiXi)
l
#23*3)
R
=
2
= R.
(3.3.2)
Eq. (3.3.2) is that of a n ellipsoid called the characteristic ellipsoid.
Recalling the definitions set forth at the b e g i n n i n g of this section, we
c o n c l u d e t h a t every radius vector a n d the t a n g e n t p l a n e at its p o i n t of
intersection with the ellipsoid are conjugate. Also the three principal
axes of the ellipsoid a n d the t a n g e n t planes at their extremities keep
their o r t h o g o n a l i t y after t r a n s f o r m a t i o n . Therefore, they are principal
directions a n d principal planes of the t r a n s f o r m a t i o n . T h e r e are three
possible cases for the characteristic ellipsoid:
1. If the three principal axes are n o t equal, there exist three principal
directions: This is the general case.
Linear Transformation of Points
27
2. If two of the three principal axes are equal, the characteristic ellipsoid
is a n ellipsoid of revolution. If, for instance, OXx is the axis of
revolution, all the axes of the ellipsoid n o r m a l to OXx are principal
directions a n d all the planes parallel to OXx are principal planes.
3. If the three principal axes of the ellipsoid are equal, it b e c o m e s a
sphere. All radii a n d all p l a n e s are principal axes a n d principal p l a n e s .
3.4
Orthogonal Transformations
Let us e x a m i n e w h a t c o n d i t i o n s are to be i m p o s e d on the m a t r i x [a]
in the p o i n t - t o - p o i n t t r a n s f o r m a t i o n (Fig. 3.9):
*3
M
(3.4.1)
so that the length of the vector OM r e m a i n s u n c h a n g e d . This obviously
would c o r r e s p o n d to a rotation or a r o t a t i o n followed by a reflection. If
the length of x is to be u n c h a n g e d , then
(3.4.2)
Substituting Eq. (3.4.1) in Eq. (3.4.2), we get:
XX
i i
x
=
(<*ijXj)(<*ik k)>
(3.4.3)
28
Kinematics of Continuous Media
w h e r e the d u m m y index in o n e of the b r a c k e t s h a s b e e n c h a n g e d from
j to k to c o n f o r m with the s u m m a t i o n c o n v e n t i o n . Eq. (3.4.3) c a n also
b e written:
aa x
= Sjk
xjxk9
ij ik*j k
since
x x
xx
(3.4.4)
xx
= i i •
fyk j k = k k
(3-4.5)
E q u a t i n g the coefficients of like p r o d u c t s in Eq. (3.4.4), we o b t a i n the
following six e q u a t i o n s :
a\x + a\x + a\x = 1
=
#?2 + #22 + #32
2
2
#?3 + #2 3 + #3 3 =
+
# 1 2 #13
#22 #23 +
1
1
0
#32 #33 =
0
#13 #11 + #23 #21 + #33 #31 =
0
+
#11 # 1 2
)
#21 # 2 2 + #31 # 3 2 =
or
8
<*ij*ik = jk-
(3.4.6a)
T h e s e e q u a t i o n s are the c o n s e q u e n c e s of the hypothesis that the length
of x r e m a i n s c o n s t a n t . E q s . (3.4.6) c a n also b e written in m a t r i x n o t a t i o n
as follows:
[a]'[a] = [1].
(3.4.6b)
Since
m - ' m = mTherefore,
1
[a]'
= [a]'.
F r o m the rules of multiplication of d e t e r m i n a n t s ,
(M'lXIN) = i,
(3.4.7)
Linear Transformation of Points
29
a n d since the value of a d e t e r m i n a n t does n o t c h a n g e w h e n the rows
a n d c o l u m n s are i n t e r c h a n g e d , t h e n
(N1XNI) = I N
2
= i.
(3-4-8)
T h u s , in a linear t r a n s f o r m a t i o n in which the length of the vector x
r e m a i n s c o n s t a n t , the inverse of the m a t r i x of the t r a n s f o r m a t i o n is
e q u a l to its t r a n s p o s e a n d the s q u a r e of its d e t e r m i n a n t is e q u a l to unity.
Xj in which a^aik = 8jk is called a n
A linear t r a n s f o r m a t i o n
= atJ
orthogonal transformation.
It is called a t r a n s f o r m a t i o n of r o t a t i o n w h e n
\[a\\ = + 1 , a n d the m a t r i x [a] is referred to as a proper orthogonal
matrix.
It is called a t r a n s f o r m a t i o n of reflection w h e n \[a]\ = — 1, a n d the
m a t r i x [a] is referred to as an improper orthogonal
matrix.
Since
[«]-'[*] ^ m m -
1
= [i],
then
[a][a] = [1]
(3.4.9)
a aj kii = 8jk.
(3.4.10)
and
X3
Fig. 3 . 1 0
30
Kinematics of Continuous Media
Eqs. (3.4.6) a n d (3.4.10) show that b o t h the c o l u m n s a n d the rows of the
o r t h o g o n a l m a t r i x [a] form a system of trirectangular unit vectors. T h e
t r a n s p o s e of [a], a n d c o n s e q u e n t l y its inverse, is also a n o r t h o g o n a l
matrix.
Example
In Fig. 3.10, consider the vector OP which is t r a n s f o r m e d by r o t a t i o n
a r o u n d the OX3 axis to Oil. T h e e q u a t i o n s of t r a n s f o r m a t i o n are
written:
=
cos 6
— sin 0
0
sin 6
cos 9
0
0
*3
l
x2
0 + 1
x3
T h e d e t e r m i n a n t of the matrix of the t r a n s f o r m a t i o n is equal to + 1 . T h e
vectors formed by its c o l u m n s are m u t u a l l y o r t h o g o n a l . This is also the
case for the vectors formed by its rows. If the t r a n s f o r m a t i o n from OP
to O i l is followed by a reflection with respect to the p l a n e Xx OX2, o n e
o b t a i n s the vector OH', which is of the s a m e length as OP. T h e matrix
[R] describing the reflection is written:
[R] =
1
0
0
0
1
0
0
0 - 1
a n d the c o o r d i n a t e s of I I ' are given by
&
—
"l
0
0 "
0
1
0
0
0
-1
" Si "
=
A
cos 9
— sin 9
sin e
cos i1
0
0
0 '
X\~
0
x*2
1
3
T h e d e t e r m i n a n t of the t r a n s f o r m a t i o n matrix is equal to — 1.
Linear Transformation of Points
31
Fig. 3.11
3.5
Changes of Axes in a Linear Transformation
In a trirectangular system of c o o r d i n a t e s OXx, OX2, OX3 (Fig. 3.11),
let the linear t r a n s f o r m a t i o n
{1} = [a]{x}
(3.5.1)
transform t h e vector x to | . This m e a n s t h a t the m a t r i x [a] o p e r a t i n g o n
the vector x gives the vector | in the base OXx, OX2, OX3. Let us n o w
consider a second system of axes OX\, OX'2, OX'3, o b t a i n e d b y m e a n s
of a n y r o t a t i o n of t h e first system a r o u n d O, a n d ask the q u e s t i o n : In
this s e c o n d system, w h a t is the form of the m a t r i x which transforms
x to | ? In o t h e r w o r d s , if in the second system we write:
{?'} = [fl'Kn
(3.5.2)
w h a t a r e the values of the a\f si gx, g2y g3, and x\, x2, x3 are the
c o m p o n e n t s of the two vectors £ a n d x in the s e c o n d system. T h e s e c o n d
system of c o o r d i n a t e s is often defined by m e a n s of the direction cosines
of its axes with respect to t h e first system. If t h e direction cosines of
OX\ a r e (1XX
JX2
JX3
),
those of OX2 are (l2X
J22
J23
),
a n d those of
OX3 a r e (f3]
J32
J33
);
then
x\ = lxx
xx
+ lx2
x2
+
(X3
x3
x'2 = !2X
x
=x
+ l22
x2
+ l23
x3
*3
+ ^32-^2 +
^33*3-
(3.5.3)
32
Kinematics of Continuous Media
In m a t r i x n o t a t i o n ,
{*'} =
(3.5.3a)
w h e r e [/] is the m a t r i x of the direction cosines. Also,
{£'} = [/]{?}•
(3.5.4)
Substituting Eq. (3.5.1) in Eq. (3.5.4), we o b t a i n :
{?'} = [ « * } •
(3.5.5)
F r o m E q . (3.5.3a), we o b t a i n :
{x} = [ / ] - ' {*'}.
(3.5.6)
If we substitute Eq. (3.5.6) in Eq. (3.5.5), we o b t a i n :
x
(3.5.7)
{!'} = [l\aV]~ {x'},
which m e a n s that the m a t r i x [a'] of Eq. (3.5.2) is given b y :
X
(3-5.8)
W] = [f][aVr .
This a n s w e r s the question asked at the b e g i n n i n g of this section.
It is c u s t o m a r y to set
[/]-'
[m],
=
so t h a t Eq. (3.5.8) b e c o m e s :
l
[a'] = [m]- [a][m],
(3.5.9)
T h e m a t r i x [/] is formed b y the direction cosines of the n e w axes with
respect to the old ones. Therefore, the c o m p o n e n t s of [/] satisfy the
o r t h o g o n a l i t y relation
T h e t r a n s p o s e of [/] is equal to its inverse, a n d b o t h are o r t h o g o n a l
matrices.
In index n o t a t i o n , Eq. (3.5.6) c a n n o w b e written:
x
*m
=
'nm 'n-
(3.5.6a)
Linear Transformation of Points
33
E q . (3.5.7) b e c o m e s :
(3.5.7a)
Eq. (3.5.8) b e c o m e s :
(3.5.8a)
Eq. (3.5.9) b e c o m e s :
(3.5.9a)
Eq. (3.5.9) is a p a r t i c u l a r case of a class of t r a n s f o r m a t i o n s called
similarity t r a n s f o r m a t i o n s . In general, if there exists a n o n s i n g u l a r
m a t r i x [s] such t h a t
l
[s]- [aM
= [b]
(3.5.11)
for a n y t w o s q u a r e matrices [a] a n d [b] of the s a m e order, t h e n [a] a n d
[b] are called similar. T h e t r a n s f o r m a t i o n of [a] to [b] is a similarity
t r a n s f o r m a t i o n . T a k i n g the d e t e r m i n a n t of b o t h sides of Eq. (3.5.11), we
obtain:
l
\[s]- [a][s}\
= (N-'IXMIXIMI) = Ml = \[b]\.
(3.5.12)
T h u s the d e t e r m i n a n t s of t w o similar m a t r i c e s a r e equal.
3.6
Characteristic Equations and Eigenvalues
I n Sec. 3.2, it w a s m e n t i o n e d t h a t in the t r a n s f o r m a t i o n
= a^Xj, the
m a t r i x [a] c o u l d b e l o o k e d u p o n as a n o p e r a t o r acting o n the vector x
to give the vector | . Let us ask the following q u e s t i o n : F o r a given
t r a n s f o r m a t i o n m a t r i x [a], does there exist a vector (or vectors) x which
w h e n t r a n s f o r m e d r e m a i n s parallel to itself (Fig. 3.12)? T h e a n s w e r c a n
b e f o u n d b y writing t h a t the vector | is parallel to x—in other w o r d s , is
a scalar multiple of x. T h u s ,
{|} = A{x} = [a]{*} = A[l]{x}
or
= Ax, =
OijXj,
a n d the scalar multiplier A m u s t b e d e t e r m i n e d .
(3.6.1)
34
Kinematics of Continuous Media
*3
Fig. 3 . 1 2
Eq. (3.6.1) is w r i t t e n :
M-X[l]]{x}
(3.6.2)
= 0
or
{atj - SyXjXj = 0.
This is a set of three linear h o m o g e n e o u s e q u a t i o n s with three u n k n o w n s , a nontrivial solution of w h i c h exists only w h e n t h e d e t e r m i n a n t
of the m a t r i x [[a] — A[l]] vanishes, or
\[[a]~
X[l]]\ = 0.
(3.6.3)
By e x p a n d i n g the d e t e r m i n a n t , w e o b t a i n a p o l y n o m i a l in A, which
w h e n e q u a t e d to zero gives t h e characteristic e q u a t i o n of the m a t r i x [a]:
3
2
A - (au + a22 + a33
)A
a aa
~ a Cl )\
l3 3l
~ (\
+ ^13^21^32 -
\ 22 33
+ (au a22 + a22
a33
aa
+
\2 23 3\
+ a33
au
a
a
^ 1 2 ^ 2 1 ^ 3 3 ~ # 1 1 ^ 2 3 ^ 3 2 ~ # 1 3 22 3\)
ax
- al2
a2{
n
-
a23
a32
(3.6.4)
= °-
It is thus a p p a r e n t that the scalar multiplier A m u s t b e a root of E q .
(3.6.4). T h e r e a r e three such roots, real or imaginary, with possible e q u a l
roots. T h e s e roots a r e called eigenvalues or characteristic r o o t s . E a c h
root Xh w h e n substituted in Eq. (3.6.2), gives a set of three linear
Linear Transformation of Points
35
e q u a t i o n s which are n o t all i n d e p e n d e n t . By a s s u m i n g o n e value for o n e
of the c o m p o n e n t s of x (or m o r e t h a n o n e value w h e n there are equal
roots) a n d d i s c a r d i n g o n e of the e q u a t i o n s , o n e c a n solve for the two
other c o m p o n e n t s of x. In o t h e r w o r d s , the eigenvector x (also called
characteristic vector) is d e t e r m i n e d within a n a r b i t r a r y multiplicative
c o n s t a n t . This is d u e to the h o m o g e n e i t y of the e q u a t i o n s : If x is a
solution of Eq. (3.6.1), then Kx is also a solution. Therefore, the a n s w e r
to the question is that three directions exist (of which at least o n e is real)
which r e m a i n parallel to themselves after t r a n s f o r m a t i o n .
Let us a s s u m e for the present that the characteristic e q u a t i o n (3.6.4)
h a s three distinct roots. In such a case, there are three distinct
eigenvectors {x}t that satisfy Eq. (3.6.1); t h a t is,
*3I
is the eigenvector c o r r e s p o n d i n g to the eigenvalue X{.
X
{ }2 =
x
x\2
x22
32
is the eigenvector c o r r e s p o n d i n g to the eigenvalue A 2.
x* 1 3
x23
33
is the eigenvector c o r r e s p o n d i n g to the eigenvalue A 3. T h e eigenvectors
are generally expressed in n o r m a l i z e d form; t h a t is, the elements of the
vector xx, for e x a m p l e , are chosen in such a way that
x
\ \ + 2\
x
+
x
3\
1.
(3.6.5)
This is possible b e c a u s e we h a v e a free choice of o n e of the c o m p o n e n t s
of each eigenvector.
Let a s q u a r e m a t r i x [M] be c o n s t r u c t e d from the eigenvector c o l u m n s
{x}i in the following m a n n e r :
36
Kinematics of Continuous Media
[M] =
x
x\ \
x2\
x
\2
\3
22
23
3\
32
33
x
x
x
x
x
(3.6.6)
t h a t is, the c o l u m n s of [M] a r e the eigenvectors of [a]. T h e m a t r i x [M] is
called the m o d a l m a t r i x of [a]. T h e d i a g o n a l m a t r i x formed b y t h e
eigenvalues of [a] is called the spectral m a t r i x :
A,
0
0
0
A2 0
0
0
(3.6.7)
A,
Recalling the r e m a r k s m a d e in Sec. 2.5 r e g a r d i n g the use of d i a g o n a l
matrices t o write g r o u p s of e q u a t i o n s u n d e r the form of o n e m a t r i x
e q u a t i o n , t h e three g r o u p s of three e q u a t i o n s r e p r e s e n t e d by E q . (3.6.1)
can be written:
(3.6.8)
[a][M] = [M][D].
- 1
Premultiplying b o t h sides b y [ A / ] , we o b t a i n :
1
(3.6.9)
[Z)] = [ M ] - [ a ] [ M ] .
This is a similarity t r a n s f o r m a t i o n of [a] to a d i a g o n a l m a t r i x [D]
t h r o u g h t h e use of t h e m o d a l m a t r i x of [a] a n d its inverse. T h e
t r a n s f o r m a t i o n expressed b y E q . (3.6.9) is referred to as the diagonalization of t h e m a t r i x [a].
A n i m p o r t a n t p r o p e r t y of linear t r a n s f o r m a t i o n s is that the eigenvalues obtained from a matrix similar to [a] are equal to those obtained
from [a] itself T o p r o v e this p r o p e r t y , let us replace [a] in E q . (3.6.3) b y
-1
a similar m a t r i x [j] [#][.?]:
l
\[s]- [a][s]
- A[l]| =
|XN -
A[l]|)(|[5]|) =
\[a] - A[l]|.
(3.6.10)
It follows t h a t t h e characteristic e q u a t i o n associated with M ^ F T F P ] is
the s a m e as the o n e associated with [a], a n d h e n c e their roots a r e
identical.
If we n o w consider the t r a n s p o s e of [a] instead of [a] itself in E q .
(3.6.3), we o b t a i n the s a m e characteristic E q . (3.6.4) a n d t h e s a m e
eigenvalues. Let t h e m o d a l m a t r i x of [a]' b e called [N]. Its spectral
m a t r i x is still [D], T h e n ,
Linear Transformation of Points
[a]'[N) = [N][D].
37
(3.6.11)
T a k i n g the t r a n s p o s e of Eq. (3.6.11), we o b t a i n :
[N]'[a] = [DIN]'.
(3.6.12)
If we n o w p r e m u l t i p l y b o t h sides of Eq. (3.6.8) b y [N]' a n d p o s t m u l t i p l y
b o t h sides of Eq. (3.6.12) b y [Af], we o b t a i n :
[N]'[a][M)
= [N]'[M][D]
(3.6.13)
[N]'[a][M]
= [D][N]'[M].
(3.6.14)
and
T h e c o m p a r i s o n of E q s . (3.6.13) a n d (3.6.14) shows t h a t the p r o d u c t of
the two matrices [A^ ' a n d [Af] m u s t result in a d i a g o n a l m a t r i x . Every
o n e of the c o l u m n vectors of these t w o m o d a l matrices c a n b e
multiplied b y a suitable c o n s t a n t a n d adjusted so t h a t the resulting
d i a g o n a l m a t r i x is the unit m a t r i x [1]. T h u s ,
[N]'[M]
= [1].
(3-6.15)
T h e vectors of [TV] a n d [Af] are said to b e n o r m a l i z e d . Eq. (3.6.15) c a n
be written in index n o t a t i o n a s :
NijMik = 8jk.
(3.6.15a)
This is the o r t h o g o n a l i t y relation of the eigenvectors of the m a t r i x [a]
a n d of its t r a n s p o s e [a]'. If the m a t r i x [a] is s y m m e t r i c so t h a t [a] — [a]\
t h e n [TV] = [Af], a n d Eq. (3.6.15a) b e c o m e s :
MyMik = 8jk
,
(3.6.16)
which is the o r t h o g o n a l i t y c o n d i t i o n for the m a t r i x [Af]. Therefore the
eigenvectors of a symmetric matrix are orthogonal.
A n o t h e r property of real symmetric matrices is that their eigenvalues are
always real. T o p r o v e this p r o p e r t y , let {x]x b e a n eigenvector of the
m a t r i x [a] associated with the eigenvalue Xx. T h e n ,
[a]{x}x =
Xx{x)v
If Xx is a c o m p l e x n u m b e r a n d the elements of [a] are real, t h e n the
elements of {x}x m u s t b e c o m p l e x n u m b e r s . Let the c o m p l e x quantities
of the previous e q u a t i o n b e r e p l a c e d b y their c o m p l e x conjugates. T h e n ,
38
Kinematics of Continuous Media
[a]{x}l
= A,{20
P
where the b a r u n d e r the c o m p l e x q u a n t i t y d e n o t e s its conjugate. If we
n o w premultiply the two previous e q u a t i o n s by {X}\ a n d {x}\, respectively, we o b t a i n :
{*}',[«, = ® i M * } i
(3.6.17)
and
={x}\\]{x}].
(3.6.18)
Let us take the t r a n s p o s e of b o t h m e m b e r s of Eq. (3.6.18):
S u b t r a c t i n g Eq. (3.6.17) from the previous equation, we o b t a i n :
(A, - A, ) © ' , { * } , = 0 .
Since
zero, then
is the sum of positive quantities a n d c a n n o t be equal to
A,=A,,
a n d therefore A! m u s t be real.
T h e eigenvalue p r o b l e m , i n t r o d u c e d here t h r o u g h the idea of a vector
r e m a i n i n g parallel to itself after t r a n s f o r m a t i o n , arises in m a n y b r a n c h es of m e c h a n i c s . T h e elements of the m a t r i x [a] can represent, a m o n g
others, stresses, strains, m o m e n t s of inertia, a n d couples. This p r o b l e m
will b e e n c o u n t e r e d several times in the c o m i n g sections. Various
m e t h o d s for o b t a i n i n g eigenvalues a n d eigenvectors h a v e been devised.
T h e y c a n b e found in the bibliography at the e n d of C h a p t e r 2.
3.7 Invariants of the Transformation Matrix in a Linear Transformation
T h e r e are some interesting relations a m o n g the characteristic values
a n d the coefficients of A in Eq. (3.6.4). F r o m the theory of e q u a t i o n s , we
k n o w that: In a n e q u a t i o n in A of degree n in which the coefficient of
X -is 1unity, the s u m of the roots equals the negative of the coefficient of
2 p r o d u c t s of the roots two at a time equals the
A " , the s u m of- the
coefficient of A " , the s u m of the p r o d u c t s of the roots three at a time
39
Linear Transformation of Points
- 3
equals the n e g a t i v e of the coefficient of A " , etc.; finally the p r o d u c t of
the roots equals the c o n s t a n t term or its negative, d e p e n d i n g o n w h e t h e r
n is even or o d d . T h u s , if the r o o t s of the characteristic e q u a t i o n a r e
Aj, A 2, A 3, we h a v e :
A! + A 2 + A 3 = axx + a22 + a33 = ati = Ix
AjA 2 + A 2A 3 + A 3A, = a ua 22 + a22
a33
- a23
a32
XX
X2X3
= axx
a22
a33
- aX3
a3X = I2
+ a 1 a 22 a 331 +
- axx
a23
a32
+ a33
axx
^ 1 3 ^ 2 1 ^32 ~
- aX3
a22
a3X
= eiJk
aXi
a2J
a3k
(3.7.1)
-
aX2
a2X
aa
a
n 2\ 33
=
I3.
In Sec. 3.5, it was s h o w n that, in a c h a n g e of c o o r d i n a t e s , the e l e m e n t s
of the t r a n s f o r m a t i o n m a t r i x c o u l d b e o b t a i n e d t h r o u g h t h e similarity
t r a n s f o r m a t i o n of Eq. (3.5.9); in Sec. 3.6, it was s h o w n t h a t c h a r a c t e r i s tic e q u a t i o n s associated with similar m a t r i c e s were the s a m e a n d led to
the s a m e eigenvalues: T h u s , w h a t e v e r b e the system of c o o r d i n a t e s a t
the start, the e l e m e n t s of the m a t r i x of a given linear t r a n s f o r m a t i o n
satisfy the three relations (3.7.1), (3.7.2), a n d (3.7.3). T h e s e three
c o m b i n a t i o n s of the elements of the m a t r i x are, respectively, called the
first, second, a n d third i n v a r i a n t of the t r a n s f o r m a t i o n . F o r c o n v e n ience, Eq. (3.6.4) is w r i t t e n :
3
2
A - 7,A + 7 2A - I3 = 0.
3.8
(3.7.4)
Invariant Directions of a Linear Transformation
A straight line is said to h a v e a n i n v a r i a n t direction if it keeps the
s a m e o r i e n t a t i o n u n d e r a linear t r a n s f o r m a t i o n . If the line initially
passes t h r o u g h the origin, b o t h the line a n d its t r a n s f o r m coincide (Fig.
3.13). If the line OM, for e x a m p l e , h a s a n i n v a r i a n t direction u n d e r a
linear t r a n s f o r m a t i o n , the c o o r d i n a t e s of M* m u s t b e given b y :
f = ayXj;
£, = Xx
(3.8.1)
t h a t is,
(ay
-
XS^Xj
= 0.
(3.8.2)
X,
Fig. 3 . 1 3
T h e solution of this set of three linear h o m o g e n e o u s e q u a t i o n s in
xx, x2, x3 h a s already b e e n e x a m i n e d in Sec. 3.6. It was used to
i n t r o d u c e the eigenvalue p r o b l e m a n d h a s yielded three possible values
for A, e a c h o n e c o r r e s p o n d i n g to a given eigenvector {*}.. If the roots of
the characteristic e q u a t i o n A, are all real, we o b t a i n three invariant
directions such that (Fig. 3.14):
OM*
OM,
= A,
2_
OM 7~
OM*3
'UM7
= A,
2
(3.8.3)
T h e three directions are n o t necessarily o r t h o g o n a l . T h e y define three
invariant p l a n e s — i n o t h e r w o r d s , three planes which d o n o t c h a n g e
Fig. 3 . 1 4
Linear Transformation of Points
41
Fig. 3.15
d u r i n g the t r a n s f o r m a t i o n . If o n e of the roots is real a n d the t w o o t h e r
are complex, we o b t a i n only o n e i n v a r i a n t direction.
It is instructive to e x a m i n e geometrically the previous analysis: I n a n
i n v a r i a n t p l a n e defined b y the two i n v a r i a n t directions c o r r e s p o n d i n g to
X{ a n d A2 (Fig. 3.15), the vector OP is t r a n s f o r m e d to OHj(i = 1,2). If
Xx = A 2, p o i n t P m o v e s to Hl. If Xx > A 2, OP rotates t o w a r d OMx to
O n 2. If A2 > A 1? OP rotates t o w a r d s OM2 to OU3. In other w o r d s , the
r o t a t i o n o c c u r s t o w a r d s the line c o r r e s p o n d i n g to the larger A.
T h e s a m e i n v a r i a n t directions are o b t a i n e d if the h o d o g r a p h of the
t r a n s f o r m a t i o n (3.8.1) is used in place of the t r a n s f o r m a t i o n itself. In
such a case we write:
u, =
- x f = (A -
\)XJ = (au - Sijfy;
(3.8.4)
t h a t is,
(ay -
XS^Xj
= 0,
(3.8.5)
which is the s a m e as Eq. (3.8.2).
3.9
Antisymmetric Linear Transformations
A linear t r a n s f o r m a t i o n is called a n t i s y m m e t r i c or a s y m m e t r i c w h e n
its h o d o g r a p h is expressed by a n a n t i s y m m e t r i c m a t r i x . T h u s (Fig. 3.16),
the t r a n s f o r m a t i o n of M to M* is a n t i s y m m e t r i c if
42
Kinematics of Continuous Media
x,
0
"1
"2
«3
=
o
12
~«13
- a n
0
23
a,
3
X,
-a23
a
0
(3.9.1)
_-*3_
It is of interest to find w h a t type of d i s p l a c e m e n t this t r a n s f o r m a t i o n
gives to a point M(xx,x2,x3).
Let OH be a vector whose c o m p o n e n t s
7 j , , ?y , r] (Fig. 3.17) a r e :
2 3
^1=^23'
^ 2 = «13»
Fig. 3 . 1 7
^3 = ^ 1 2 -
(3.9.2)
43
Linear Transformation of Points
T h e t r a n s f o r m a t i o n expressed by Eqs. (3.9.1) w h e n applied to H gives:
~ux
0
- a n
ai2 a 0
—
=
a i3
a
0
23
13
_ « 3_
"o"
023
«13
=
_«12_
(3.9.3)
0
0
which m e a n s that H does not c h a n g e u n d e r the a n t i s y m m e t r i c transform a t i o n . Therefore, OH has a n i n v a r i a n t direction a n d all p o i n t s on it
are fixed d u r i n g the t r a n s f o r m a t i o n . W h e n the t r a n s f o r m a t i o n (3.9.1) is
applied to a n y p o i n t M with c o o r d i n a t e s x{, x2, x3, we o b t a i n the
c o m p o n e n t s of MM*; n a m e l y :
a
a
0
==
"2
"13
*1
0 ) 3 * 3 ~~ _ 1 2 *x2
0
-«23
* 2
a0 1 x2 * 1
023
0
* 3
23 2
~ \2
« I2
_ « 3_
23 3
a
. (3.9.4)
~ 013*1
T h e m a g n i t u d e a n d direction of MM* with respect to OH a n d OM c a n
be f o u n d by c o m p u t i n g the vector p r o d u c t OH X OM a n d the scalar
p r o d u c t s OM • MM* a n d OH • MM*:
/
023
"13
a'3
\2
x
*i
X
3
'i
OHXOM
=
=
2
['1 ( 0 1 3 * 3
+
OA/ • MM*
OH
•
MM*
=
=
2
~
\2 l)
'3(023*2
-
013
# 1 2 * )2
+ x3(a23
x2
-
1
( 0
22
3 * 2
* 2 ( 0 1 21 *
+
0 1 2 *)2
-
^23*3)
*l)J
"23*3)
= 0
0 , 3 * , )
^ 2 3 ( 0 1 3 *-
O
' 2 ( 0 1 21 *
+
X | ( a | 3j c 3 —
+
(3.9.5)
x a
+
0|3*l)
0 1 3 ( 0 1 21 * -
= °-
0 2 3 *) 3
(3.9.6)
(3.9.7)
Eqs. (3.9.4) to (3.9.7) show that MM* is n o r m a l to the p l a n e (OH, OM),
a n d that it is given in m a g n i t u d e a n d direction by OH X OM.
In Fig. 3.17, let MQ b e the n o r m a l from M on OH a n d a b e the angle
b e t w e e n QM a n d QM*. T h e following relations d e d u c e d from the
geometry of Fig. 3.17 c a n be written:
44
Kinematics of Continuous Media
(
\MAf*\ = (OH)(OM)sin
<f> = (OH)(OM) ^^
= (OH)(QM)
2
tan
c so
a=
=0H
= V F E )
2
« =
2
+
2
(au)
2
+ (*,2)
(3.9.8).
399
(''>
2
/i
/
o
*, o
, =tt(3.9.10)
V L + (a23
)
+ ( a 1 )3 + ( a 1 )2
W e n o t e t h a t the angle a d e p e n d s only o n t h e values of the coefficients
of the t r a n s f o r m a t i o n matrix. It is therefore the s a m e for all p o i n t s like
M , a n d always varies b e t w e e n O a n d Wi in a direction following t h e
r i g h t - h a n d rule with the t h u m b in the direction of OH. T h u s , u n d e r t h e
t r a n s f o r m a t i o n , M rotates b y a n angle a a r o u n d OH in a p l a n e n o r m a l
to OH. H o w e v e r , this r o t a t i o n does n o t o c c u r alone, b u t is associated
with a radial d i s p l a c e m e n t n o r m a l to OH. T h e unit value of this r a d i a l
d i s p l a c e m e n t is given b y :
e
r
= QM* - QM
M
2
=
V L + (*23)
=i - cos a
cos Q
a
2
+ (*13)
)
2
+ (*12) " I -
In s u m m a r y , the t r a n s f o r m a t i o n expressed b y a n a n t i s y m m e t r i c
matrix is the p r o d u c t of a r o t a t i o n a a n d a cylindrical dilatation a r o u n d
a n axis OH w h o s e direction ratios a r e given b y the coefficients of the
matrix. T h e p r o d u c t of the r o t a t i o n a n d the dilatation is c o m m u t a t i v e .
OH is n o t only i n v a r i a n t b u t is also a principal direction. A n y other
2 a principal
direction in the p l a n e (QM,QM*)
n o r m a l to OH is also
direction. If the angle of r o t a t i o n a is small, so t h a t a is very small
c o m p a r e d to unity, er c a n b e neglected since it is of the s e c o n d o r d e r
with respect to a. This is s h o w n b y :
er =
3.10
cos a _ 1 cos a
1 + ^ T - 7 T
2!
4J + - - - ^ OR2
(3.9.12)
1 - — +
2!
Symmetric Transformations. Definitions and General Theorems
A linear t r a n s f o r m a t i o n is said to b e s y m m e t r i c if the m a t r i x of t h e
t r a n s f o r m a t i o n is s y m m e t r i c . T h u s ,
{1} = [<*Kx}9 with [a] = [a]\
(3.10.1)
Linear Transformation of Points
45
represents a linear s y m m e t r i c t r a n s f o r m a t i o n . I n index n o t a t i o n , we
write:
= ayXj,
with atj = ajt.
(3.10.2)
L i n e a r s y m m e t r i c t r a n s f o r m a t i o n s o c c u r c o m m o n l y in the s t u d y of
stresses a n d d e f o r m a t i o n s in c o n t i n u o u s m e d i a . I n t h e following, we
shall p r o v e s o m e t h e o r e m s related to this p a r t i c u l a r type of linear
transformations.
Theorem I: The symmetric transformations
are the only ones to possess
the property of reciprocity. By reciprocity w e m e a n t h a t if a vector OMx
is t r a n s f o r m e d into OM*x a n d a n o t h e r vector OM2 is t r a n s f o r m e d i n t o
OM*~2, t h e n
OMx • OM*2 = OM2 • O F 1 .
(3.10.3)
T o p r o v e this t h e o r e m , let us c o n s i d e r a general m a t r i x [a] o p e r a t i n g o n
b o t h OMx(xl9
x29
x3)
a n d C>M2(yx,y2,y3)
to give OM*x(£x^2,£3)
and
OM*2(r}x,ri2,7]3)y
respectively. I n t a b u l a r form, this is w r i t t e n :
OM2
OMx
y2
x
x,
x2
3
a«11
«12
013
Vi
u
12
12
2\
022
023
V3
*3
031
032
033
U s i n g t h e results of E q . (3.10.4), we get:
OMx • OM*2 = xx(axx
yx
+ aX2
y2
+
aX3
y3)
+ x2(a2X
y{
+ #22^2 + #23^3)
+ x3(a3x
yx
+ #32/2
(3.10.5)
+ a 3 ^33)
and
OM2 • OM*x =yl(axl
xl
+ aX2
x2
+
aX3
x3)
+ ^2(^21^1 + #22*2 + #23*3)
+ y3(a3X
xx
+ #32x2 +
(3.10.6)
a33
x3).
a=
W e see t h aat =E q . a(3.10.5)
caa nnn o t b e equal to E q . (3.10.6) unless
aX2 = a 2i » i 3 3\> 2 3 32'•> * ° t h e r w o r d s , unless [a] is a s y m m e t r i c
matrix. T h e p r e v i o u s analysis is b a s e d o n scalar p r o d u c t s a n d is
46
Kinematics of Continuous Media
Fig. 3 . 1 8
i n d e p e n d e n t of the chosen r e c t a n g u l a r system of c o o r d i n a t e s .
Theorem II: The symmetric
transformations
are non-rotational:
Any
invariant direction is a principal direction. Let Z b e a vector a l o n g an
invariant direction. A linear s y m m e t r i c t r a n s f o r m a t i o n transforms Z to
a vector f a l o n g the s a m e direction (Fig. 3.18). Let P be a p l a n e n o r m a l
to Z , a n d Y b e a n y vector in it. T h e s a m e linear s y m m e t r i c t r a n s f o r m a tion transforms Y to 77. Since Y • f = 0, then 77 • Z = 0 b e c a u s e of the
p r o p e r t y of reciprocity. Therefore, 77 lies in the p l a n e P. W h e n Y sweeps
the p l a n e P, 77 sweeps the s a m e p l a n e . Therefore, Z is a principal
direction since, w h e n c o u p l e d with a n y vector like 7, the angle b e t w e e n
t h e m r e m a i n s a right angle after t r a n s f o r m a t i o n .
Fig. 3 . 1 9
Linear Transformation of Points
47
J
0
a
A
Fig. 3 . 2 0
Let us n o w consider two other vectors X a n d Y a l o n g the two o t h e r
principal directions (Fig. 3.19). T h e y c a n n o t r o t a t e to | a n d 77 in the
t r a n s f o r m a t i o n since the p r o p e r t y of reciprocity would not be satisfied:
I n d e e d , X • 77 is negative a n d Y • | is positive. T h u s the two principal
directions X a n d Y are also invariant.
Theorem I I I : Non-rotational
transformations
are the only ones to have
the property of reciprocity. Let us consider a general linear t r a n s f o r m a t i o n — i n other words, a t r a n s f o r m a t i o n which is n o t necessarily s y m m e t ric. If OA a n d OB are two vectors (Fig. 3.20) along two invariant
directions, points A a n d B will m o v e to a a n d /? after t r a n s f o r m a t i o n in
a w a y such t h a t :
Oa =
W
=
XX
(OA)
X2(OB).
(3.10.7)
(3.10.8)
Since A, 7^ A 2 a n d cos $ 7^ 0, then
OA
Oft ^
OB - Oa,
which m e a n s that the p r o p e r t y of reciprocity is n o t satisfied. If
cos <f> = 0, the two invariant directions are also principal, which m e a n s
that the linear t r a n s f o r m a t i o n is a s y m m e t r i c o n e a n d consequently
possesses the p r o p e r t y of reciprocity.
In Fig. 3.21, let the three c o o r d i n a t e axes OXx, OX2, OX3 lie along the
principal directions of a linear s y m m e t r i c t r a n s f o r m a t i o n . T h e s e three
directions are invariant a n d d o n o t rotate. U n d e r the t r a n s f o r m a t i o n ,
a n y vector OP will b e c o m e OQ t h r o u g h a rotation a n d a c h a n g e in
48
Kinematics of Continuous Media
Fig. 3.21
length. A vector OP' s y m m e t r i c of OP with respect to a n y of the three
principal planes, will b e c o m e OQ' s y m m e t r i c of OQ with respect to the
s a m e p l a n e . T h u s , in a s y m m e t r i c t r a n s f o r m a t i o n all the directions
rotate except the three principal ones. T h e r o t a t i o n s , however, c o m p e n sate o n e a n o t h e r symmetrically with respect to the three o r t h o g o n a l
principal p l a n e s . T h e expression, s y m m e t r i c t r a n s f o r m a t i o n , therefore
implies the s y m m e t r y of the m a t r i x as well as the s y m m e t r y of r o t a t i o n
of the various directions. F r o m the previous t h e o r e m s , we c o n c l u d e t h a t
the terms n o n - r o t a t i o n a l , reciprocal, p u r e d e f o r m a t i o n , a n d s y m m e t r i c
t r a n s f o r m a t i o n are all equivalent.
3.11
Principal Directions and Principal Unit Displacements of a
Symmetric Transformation
In Sec. 3.3, it was s h o w n that in a linear t r a n s f o r m a t i o n three
o r t h o g o n a l directions exist t h a t r e m a i n o r t h o g o n a l after the t r a n s f o r m a tion. T h e s e directions are called the principal directions. In Sec. 3.10, it
was s h o w n t h a t for a s y m m e t r i c t r a n s f o r m a t i o n these directions were
also i n v a r i a n t directions of the t r a n s f o r m a t i o n . T h u s the search for
Linear Transformation of Points
49
Fig. 3 . 2 2
principal directions, in this case, is equivalent to the search for i n v a r i a n t
directions. This search was e x a m i n e d in b o t h Sees. 3.6 a n d 3.8.
F o r c o n v e n i e n c e , let us write the t r a n s f o r m a t i o n w h i c h brings P to I I
(Fig. 3.22), as follows:
1 +fl„
$1
=
£3
1 a
+
«13
a
a\3
«12
a
22
23
1 +
23
a 33
x2
(3.11.1)
x3
or
with ay = a^.. T h e h o d o g r a p h of the t r a n s f o r m a t i o n (in other w o r d s , the
c o m p o n e n t s of the d i s p l a c e m e n t ) is given b y :
(3.11.2)
T h e i n v a r i a n t directions are o b t a i n e d b y writing:
i: = (1 +
X)xj,
(3.11.3)
50
Kinematics of Continuous Media
Fig. 3 . 2 3
in which it is seen t h a t A represents the unit d i s p l a c e m e n t a l o n g the
i n v a r i a n t (also principal) directions. T h e characteristic e q u a t i o n h a s
three roots, A 1, A 2, a n d A 3, which are t h e n used to find the three
principal directions. T h e s a m e results are o b t a i n e d if we use Eq. (3.11.2)
in conjunction with
(3.11.4)
If A 3 < A 2 < X\, A 3 is called the m i n o r principal unit displacement, A 2
is called the i n t e r m e d i a t e principal unit d i s p l a c e m e n t , a n d A! is called
the m a j o r principal unit d i s p l a c e m e n t . T h e principal directions define a
trirectangular system of c o o r d i n a t e s , which is often very c o n v e n i e n t to
use. Let the axes of such a system b e called OXj, OX2, OX'3 (Fig. 3.23),
a n d let the c o m p o n e n t s of UP, PIT, a n d OlT in it b e
(x\,x2,x3),
(u\, u2, u3), a n d (Ij, | 2 > £3 )> respectively. T h u s ,
Si
1 + A
=
0
0
0
0
1 + A2
0
0
1 + A3
(3.11.5)
x'3
Linear Transformation of Points
51
and
0
0
u'2 =
0
0
_"3_
x\
(3.11.6)
0
A 3_
0
_*3_
In the principal system of axes, the t r a n s f o r m a t i o n is seen to b e m a d e
b y a d i a g o n a l m a t r i x . T h e d i s p l a c e m e n t of P is o b t a i n e d b y simple
extension o r c o n t r a c t i o n of the projections of OP o n the principal axes:
I n d e e d , those directions are also i n v a r i a n t in the t r a n s f o r m a t i o n . T h e
principal directions form three p l a n e s called the principal p l a n e s of the
t r a n s f o r m a t i o n a n d , regardless of w h i c h system of c o o r d i n a t e s we start
from, those directions will always b e the s a m e .
If the linear s y m m e t r i c t r a n s f o r m a t i o n (3.11.1) is such t h a t o n e of the
roots of the characteristic e q u a t i o n is r e p e a t e d , three different eigenvectors c a n still b e f o u n d . Consider, for e x a m p l e , the h o m o g e n e o u s system:
3 - A
0
1
X,
0
2 - A
0
x2
3 - A
x3
1
0
= 0,
(3.11-7)
w h o s e characteristic e q u a t i o n s is
3
2
A - 8A + 20A -
16 = 0.
(3.11.8)
T h e roots of this e q u a t i o n are X{ = A 2 = 2 a n d A 3 = 4. F o r A = 2, w e
o b t a i n o n e e q u a t i o n for d e t e r m i n g {x}{, n a m e l y
xu + JC31 = 0.
(3.11.9)
In Sec. 3.6, it w a s i n d i c a t e d t h a t in t h e eigenvalue p r o b l e m , o n e of the
e q u a t i o n s is d i s c a r d e d a n d r e p l a c e d b y a free choice of a c o m p o n e n t .
A s a choice, it is c u s t o m a r y to use the n o r m a l i z a t i o n c o n d i t i o n :
2
2
V * ! + x\x + * 3, = 1,
(3.11.10)
w h i c h r e d u c e s the c o m p o n e n t s of the eigenvector to direction cosines.
In o u r case, we are entitled to a s e c o n d free choice of a c o m p o n e n t since
o n e of the eigenvalues is r e p e a t e d . W e shall c h o o s e xn = l / \ / 2 . T h e s e
two choices, together with Eq. (3.11.9), give:
52
Kinematics of Continuous Media
(3.11.11)
0
V2
T h e o t h e r eigenvector c o r r e s p o n d i n g to A = 2 m u s t b e such t h a t
x 12 + JC32 = 0
(3.11.12)
a n d , since it is o r t h o g o n a l to {x},, t h e n
(xu)(xu)
+ (x2]
)(x22
)
+ (x3]
)(xn)
= 0
(3.11.13)
or
- J = x 12 + ( 0 ) ( x 2 ) 2 - ^ x 32 = 0.
(3.11.14)
E q s . (3.11.12), (3.11.14), a n d the n o r m a l i z a t i o n c o n d i t i o n show that
0
{x}2 =
1
(3.11.15)
0
F o r A = 4, we h a v e t h e system of e q u a t i o n s
— x )3 + x 33 = 0
— 2 x 23 = 0,
(3.11.16)
a d d e d to the n o r m a l i z a t i o n c o n d i t i o n (3.11.10). T h e solution of this
system is:
J_
V2
M =
3
0
_L
V2
(3.11.17)
Linear Transformation of Points
3.12
53
Quadratic Forms
If x a n d y are two sets of n variables (three in o u r case), a function
which is linear a n d h o m o g e n e o u s in the variables of each set separately
is called a bilinear form. T h u s ,
aX3
a ~xx a
aa n
an
2\
22
a
a32
3l
(3.12.1)
23
a33
_*3_
is a bilinear form. W h e n the sets of variables are identical so that
[y] = {x}\ the bilinear form b e c o m e s a q u a d r a t i c form. It was shown in
Sec. 2.4 t h a t a n y s q u a r e m a t r i x c a n b e given as the s u m of a s y m m e t r i c
a n d of a n a n t i s y m m e t r i c c o m p o n e n t . Setting [ y] = {x}' a n d d e c o m p o s ing the m a t r i x [a] into its two c o m p o n e n t s , we o b t a i n for A ( x , x) the
sum:
a
a
\2
[xx x2 x3 ]
+ 2\
aX2 + a2x
al3
+ a3l
3\
+
a
23 +
a22
a23 +
a a
\3
X
+[*\
2 3]
a32
x3
"33
a
\2
2\
\3
0
~ \3
32 ~
aa
23
a
3\
a
2
a2l -X al2
a
a
3\
*2
2
a
0
a
32
23 ~
32
(3.12.2)
a
a xx
2
x2
0
x3
T h e second t e r m is equal to zero, while the first o n e with its s y m m e t r i c
s q u a r e m a t r i x is f o u n d to b e equal to {x}'[tf]{X}. Therefore, in a q u a d r a t i c
form, the a n t i s y m m e t r i c c o m p o n e n t has n o effect, a n d a c o n v e n i e n t
expression for A(x,x)
becomes:
A(x,x)
=
{x}'[a]{x},
(3.12.3)
where [a] is a s y m m e t r i c matrix. In index n o t a t i o n , the scalar A(x,x)
written:
A(x,
x)
=
ajjXjXj.
is
(3.12.4)
54
Kinematics of Continuous Media
If A is a c o n s t a n t , Eq. (3.12.4) represents a q u a d r i c surface with its
center at the origin. T h e n a t u r e of the q u a d r i c d e p e n d s o n the value of
the e l e m e n t s atj. If the d e t e r m i n a n t of [a] does n o t vanish, the q u a d r i c
is either a n ellipsoid or a h y p e r b o l o i d . If the d e t e r m i n a n t of [a] vanishes,
the surface d e g e n e r a t e s into a cylinder of the elliptic or h y p e r b o l i c type
or else into two parallel p l a n e s symmetrically situated with respect to
the origin.
Let us a s s u m e t h a t the q u a d r i c surface is a n ellipsoid. T h i s ellipsoid
will, in general, h a v e three principal axes different in length so t h a t their
direction is uniquely d e t e r m i n e d (Fig. 3.24). A suitable r o t a t i o n of t h e
Fig. 3 . 2 4
system of reference axes c a n b e m a d e to b r i n g it in c o i n c i d e n c e with t h e
p r i n c i p a l axes of the ellipsoid. In this n e w system, the expression for A
becomes:
2
2
A(x',x')
= Xx(x\)
+ A 2( J ^ ) + A 3( ^ )
2
(3.12.5)
or
A(x\x')
= {xJ[D]{x'l
(3.12.6)
w h e r e [D] is a d i a g o n a l m a t r i x with elements Aj, A 2, a n d A 3. T h e c h a n g e
of axes from the system OX\, O Y 2, OX'3 t o OXx, OX2, OX3 c a n b e
b r o u g h t a b o u t b y (Sec. 3.5):
Linear Transformation of Points
55
(3.12.7)
{x} = [m}{x'l
w h e r e [m] is a m a t r i x whose c o l u m n s a r e the direction cosines of t h e
n e w system with respect to the o l d o n e . I n t r o d u c i n g E q . (3.12.7) into E q .
(3.12.3), we o b t a i n :
A(x9x) = {xJ[mY[a][m]ix'}.
(3.12.8)
E q u a t i n g E q . (3.12.6) to E q . (3.12.8), we get:
f
{x'}'[D){x- } =
{?}'WM44
Thus,
(3.12.9)
l
[D] = [m]'[a][m] = [m]- [a][m].
(3.12.10)
T h e o p e r a t i o n of E q . (3.12.10) diagonalizes t h e m a t r i x [a] so t h a t t h e
elements of [D] a r e t h e eigenvalues. T h e c o l u m n s of [m] a r e t h e
eigenvectors giving the directions of the n e w system of axes ( M o d a l
M a t r i x ) . T h e y a r e i m m e d i a t e l y o b t a i n e d o n c e t h e X's a r e d e t e r m i n e d
from t h e characteristic e q u a t i o n of [a]. If, for instance, w e set A = 1, t h e
lengths of the principal d i a m e t e r s of the ellipsoid a r e given b y 2 A / X 7 ,
2/vx^*, 2/yx^.
T h u s far w e h a v e a s s u m e d t h a t t h e principal axes of t h e ellipsoid a r e
of u n e q u a l length. W h e n t w o o u t of t h e three eigenvalues a r e equal, t h e
ellipsoid is of revolution a r o u n d o n e of its principal axes, a n d a n y o t h e r
axis n o r m a l t o it is a principal axis. W h e n t h e three eigenvalues a r e
equal, the ellipsoid degenerates into a s p h e r e : A n y three m u t u a l l y
p e r p e n d i c u l a r axes are principal axes.
M a n y p r o b l e m s associated with q u a d r a t i c forms a r e intimately related to p r o b l e m s associated with sets of linear e q u a t i o n s . W e m a y notice
that if we write:
(3.12.11)
we o b t a i n t h e e q u a t i o n s :
= anxx
+ a nx 2 +
ai3
x3
= anxx
+ a22
x2
+
a23
x3
a 1 x3, + a23
x2
+
a33
x3
£3'=
(3.12.12)
56
Kinematics of Continuous Media
or
= ctjjXj, with atj = a^. Eqs. (3.12.12) are those of the s y m m e t r i c
t r a n s f o r m a t i o n of {x} to {£}. W h e n A is formed by a s u m of s q u a r e s as
in Eq. (3.12.5), with n o c r o s s - p r o d u c t terms, we say t h a t A is r e d u c e d to
a c a n o n i c a l form. T o r e d u c e a q u a d r a t i c form to its c a n o n i c a l form, the
system of Eqs. (3.12.12) c a n first b e o b t a i n e d a n d w o u l d c o r r e s p o n d to
a s y m m e t r i c t r a n s f o r m a t i o n . aT hne <m a t r i x of the t r a n s f o r m a t i o n is then
^ ^ 3 - T h e e x a m p l e given in Sec. 3.11 c a n
diagonalized to give A 1? A 2,
b e used to d e m o n s t r a t e the r e d u c t i o n of a q u a d r a t i c form to its
c a n o n i c a l form. Starting from the q u a d r a t i c form:
A(x, x) = 3xf
+ 2x\
+ 3x
3 +
2x x ,
x3
Eq. (3.12.11) gives:
£1
£
2
£3
= 3xj + 0 x 2 +
3
+ 2x2 + 0x3
=x x
= 0x
\ + 0*2 +
3x
"3
0
1"
£2
0
2
0
£3
1
0
3
x
or
x
3
3
T h e eigenvalues a n d eigenvectors are o b t a i n e d as s h o w n in Sec. 3.11
a n d the c a n o n i c a l form is:
2
A(x',x')
3.13
= 2[(x\)
2
+ (x' ) ]
2
2
+
4(x ) .
3
Normal and Tangential Displacements in a Symmetric
Transformation. Mohr's Representation
located o n a unit sphere centered at the
C o n s i d e r a p o i n t P(xx,x2,x3)
origin (Fig. 3.25). U n d e r a linear s y m m e t r i c t r a n s f o r m a t i o n , P goes to
n . T h e c o m p o n e n t s of the d i s p l a c e m e n t PU are u u w 2, a n d u3. T h e
e q u a t i o n of the sphere is:
2
2
x 1 + x | + x 3 = 1.
(3.13.1)
In the p l a n e 0 / T I , the vector PN = n is the projection of PH o n the line
OPN a n d the vector TT = 1 is the projection of PTL o n the t a n g e n t p l a n e
to the sphere at P. n a n d 1 are respectively called the normal and the
tangential
components
of the displacement
of P. Since the m a g n i t u d e of
Linear Transformation of Points
57
OP is unity, the m a g n i t uud e of n is given b y the scalar p r o d u c t of
OP(xX9
x2,x3)
a n d PH(ux,w2, 2>)U s i n g the principal axes as axes of
reference, i.e., using E q s . (3.11.6), we o b t a i n :
n = PU cos a = \xx{
+ A 2JC| + \3x%.
(3.13.2)
T h e m a g n i t u d e of 1 is given b y the m a g n i t u d e (but n o t b y the direction)
of the vector p r o d u c t of ~OP a n d PH. T h e c o m p o n e n t s of this vector
product are:
(A 3 - A 2) x 2x 3,
Therefore,
2
t
(A, - \3)x3xx,
2 2
= (PU) sin a
(A 2 - \ l ) x l x 2.
2
(3.13.3)
2
= (A - \ ) xfx%
2 2 22 x
+ (Aj -
\3) x]x^
(3.13.4)
+ ( A 3- A 2) x 2 x 3
and
n2
+ ,2
=2X2 X
2 2
X A+ 2+
X2 X
)
T h u s , in a given linear s y m m e t r i c t r a n s f o r m a t i o n c h a r a c t e r i z e d by
specific values of A 1 A?2, a n d A 3, o n e c a n c o m p u t e the n o r m a l a n d
t a n g e n t i a l d i s p l a c e m e n t s of a n y p o i n t P(xx,x2,x3)
of the unit sphere.
T h e s a m e o p e r a t i o n c a n b e m a d e graphically by m e a n s of a c o n s t r u c tion d u e to O. M o h r . F o r that, o n e h a s to solve E q s . (3.13.1), (3.13.2),
a n d (3.13.5), a n d o b t a i n the expression of xx, x2, x3 in terms of n a n d /.
T h e s e are w r i t t e n :
2
t
+ (n - \2)(n - \ 3)
(A, - A 2)(A, - A 3)
(3.13.6a)
58
Kinematics of Continuous Media
2
2 _ t + (n - X3)(n - A,)
*2
( A 2- A 3) ( A 2- A , )
(3.13.6b)
2
/ + ( n - A , ) ( K - A 2)
( A 3- A , ) ( A 3- A 2)
•
(3.13.6c)
M o h r ' s c o n s t r u c t i o n establishes a c o r r e s p o n d e n c e b e t w e e n p o i n t s o n the
unit s p h e r e a n d p o i n t s in the n , t p l a n e (Fig. 3.26). O n l y the s q u a r e s of
^displacement
PTT
Fig. 3 . 2 6
x2, x3, a n d / a p p e a r in E q s . (3.13.6) so
a n d one-half of the n , / p l a n e a r e r e q u i r e d
Let us a s s u m e \ x > A 2 > A 3. In Figs.
c o r r e s p o n d i n g to the principal circle AB is
E q . (3.13.6c). Therefore,
t h a t one-eighth of the sphere
to s t u d y the c o r r e s p o n d e n c e .
3.26a a n d 3.26b, the curve
o b t a i n e d b y setting x3 = 0 in
2
t
+ (n -
- A 2) = 0.
(3.13.7)
This is the e q u a t i o n of a circle AjA 2, w h o s e r a d i u s is ( \ x — A 2) / 2 . T h e
center is yx, which is given by o'yx = (Xx + A 2) / 2 . In the s a m e way, the
circle A 2A 3 c o r r e s p o n d s to 2?C, a n d the circle Xx A 3 c o r r e s p o n d s to AC.
T h u s , all the p o i n t s o n the s p h e r e h a v e in the (n , /) p l a n e a n i m a g e
which falls in the h a t c h e d region limited b y the three circles.
If, o n the surface of the sphere, we consider a circle AXBX parallel to
the p l a n e OXx, OX2 a n d at a height x3 = h, the c o r r e s p o n d i n g curve in
the n , t p l a n e is o b t a i n e d b y setting x3 = h in Eq. (3.13.6c). T h i s gives:
Linear Transformation of Points
2
t
2
+ (n - \x)(n
- A2) = h (X3 - XX)(X3 -
A2).
59
(3.13.8)
This is the e q u a t i o n of a circle of center yx a n d of r a d i u s
(3.13.9)
Points o n a m e r i d i a n t h r o u g h C a n d G satisfy the e q u a t i o n :
(3.13.10)
x2 = Kxx
E q s . (3.13.6a), (3.13.6b), a n d (3.13.10) give, in the n , / p l a n e , a circle
c e n t e r e d o n the o'n axis a n d passing t h r o u g h g a n d A3 (Fig. 3.26b). W e
thus h a v e three families of parallels a n d three families of m e r i d i a n s
whose images are circles o n M o h r ' s d i a g r a m .
In Fig. 3.26a, let the angle AOG =
T h e c o o r d i n a t e s of p o i n t G a r e :
xXG = cos (3,
x2G = sin /?,
x3G = 0.
(3.13.11)
Eqs. (3.13.2) a n d (3.13.4) give for G:
nG =
'g|
=
1
Ai + A?
2 +
| ^ 2^
2
Ai
—1AT 2
2 cos 2/?
n s 2i
^l'
(3.13.12)
(3.13.13)
in which the sign of tG is n o t c o n s i d e r e d at the p r e s e n t time. Eqs.
(3.13.12) a n d (3.13.13) show that p o i n t G defined b y /? in Fig. 3.26a has
for a n image the p o i n t g defined b y 2/? in Fig. 3.26b. T h u s , w h e n G
describes the q u a r t e r of a circle AB , g describes half a circle o n M o h r ' s
d i a g r a m . T h i s p r o p e r t y is only true o n the three principal circles AB ,
BC , a n d CA .
Every p o i n t P on the sphere h a s a n i m a g e o n the (n, t) p l a n e . This
i m a g e falls in the area limited b y the three circles. It is o b t a i n e d b y
plotting the i m a g e of the parallel a n d t h a t of the m e r i d i a n passing
t h r o u g h P. This is a simple m a t t e r o n c e o n e h a s p l o t t e d the three
principal circles c o r r e s p o n d i n g to AB, BC, a n d CA . Since the sphere
h a s a unit r a d i u s , the c o o r d i n a t e s of P , xx, x2 -> x^ are also its direction
cosines lx, / 2, / 3. Eq. (3.13.9) gives the r a d i u s of the circle c o r r e s p o n d i n g
to Ax Bx, a n d /3 is o b t a i n e d from
cos [3 =
(3.13.14)
60
Kinematics of Continuous Media
T h e p o i n t p o n the n , t p l a n e c o r r e s p o n d i n g to P o n the sphere is the
intersection of the circle ax bx c e n t e r e d at yx a n d of the circle A 3g w h o s e
center falls o n the axis o'n. T h e abcissa a n d the o r d i n a t e of P give the
n o r m a l a n d tangential d i s p l a c e m e n t s of P .
M o h r ' s c o n s t r u c t i o n shows t h a t :
1. A m o n g all the p o i n t s located o n a unit sphere, the o n e which h a s
the largest n o r m a l d i s p l a c e m e n t u n d e r a linear s y m m e t r i c transform a t i o n is the p o i n t A located o n the major principal axis (Fig. 3.26).
2. In each principal p l a n e , the p o i n t located o n the bisector of the
principal axes h a s the largest tangential displacement. It is equal in
m a g n i t u d e to one-half of the difference b e t w e e n the two principal
n o r m a l d i s p l a c e m e n t s (Fig. 3.26b).
3. T h e m a x i m u m tangential d i s p l a c e m e n t occurs in the p l a n e of the
major a n d m i n o r principal axes.
4. If the s a m e q u a n t i t y h is a d d e d to the three principal unit displacem e n t s , M o h r ' s circles in the n , / p l a n e keep the s a m e d i a m e t e r a n d
are simply displaced by h o n the o'n axis. All the n o r m a l displacea o hnrtn' s( coeo n s t r u c t i o n
m e n t s n b e c o m e n + h. In particular, if h = 1, M
l
t h displacegives the t r a n s f o r m e d vector itself ( £ i , £ 2, £ 3)
m e n t vector ( w 1 w?2, w 3) (Fig. 3.26b).
In all the previous e q u a t i o n s , the c o o r d i n a t e s xx, x2, x3 of P o n the
unit sphere c a n b y replaced b y the direction cosines fx, /2, l3 of the line
OP.
3.14
Spherical Dilatation and Deviation in a Linear Symmetric
Transformation
Let us i n t r o d u c e the following n o t a t i o n :
m
(3.14.1)
\ x + A2 + A3 _
3
A 1 A=m + A;,
A 2 = A m+ A 2,
A 3 = A W+ A^.
(3.14.2)
W e notice t h a t
\ \ + A 2 + A 3 = 0.
(3.14.3)
U s i n g the principal directions as c o o r d i n a t e axes, the substitution of
E q s . (3.14.1) a n d (3.14.2) in Eqs. (3.11.6) gives:
Linear Transformation of Points
AM
"1
"2
_"3_
=
0
0
x
0
0
AM 0
0 A„,
\
x2
+
x3
A;
0
0
A^> 0
0 A^
0
61
0
x2
x3
or
PU = PW
+
(3.14.4)
PU".
Eq. (3.14.4) shows t h a t the d i s p l a c e m e n t of a p o i n t P u n d e r a linear
s y m m e t r i c t r a n s f o r m a t i o n is the s u m of the two vectors:
(1) A vector a l o n g OP equal to PU'
tt
(*)
«
Fig. 3 . 2 7
(2) A vector PU" c h a r a c t e r i z e d by a t r a n s f o r m a t i o n m a t r i x w h o s e
trace is equal to zero.
T h e first vector PU' is called the spherical d i l a t a t i o n (or dilation)
b e c a u s e it c o r r e s p o n d s to a n extension or a c o n t r a c t i o n a l o n g the
original vector OP. T h e s e c o n d vector is called the deviation. Figs. 3.27a
a n d 3.27b show e a c h o n e of the c o m p o n e n t s . T h e two c o m p o n e n t s are
a d d e d in a d i s p l a c e m e n t space in Fig. 3.27c.
T h e previous discussion c a n easily b e r e p r e s e n t e d o n a M o h r diag r a m . K n o w i n g A 1?A 2, a n d A 3, o n e c a n plot the three c o r r e s p o n d i n g
M o h r circles (Fig. 3.28). T h e p o i n t o" is t a k e n o n o'n such that
o'o" = \ m. T h e d i a g r a m with the origin at o' is used to find the n o r m a l
a n d tangential unit d i s p l a c e m e n t s for a n y vector OP with k n o w n
direction cosines. T h e s a m e d i a g r a m with origin at o" allows o n e to
o b t a i n t h e n o r m a l a n d t a n g e n t i a l u n i t d i s p l a c e m e n t s d u e to the deviation alone. O n M o h r ' s d i a g r a m , the c h a n g e of origin does n o t affect the
tangential c o m p o n e n t s ; only the n o r m a l ones are affected.
62
Kinematics of Continuous Media
t
Fig. 3 . 2 8
3.15
Geometrical Meaning of the a^s
Transformation
in a Linear Symmetric
Let us consider a unit c u b e w h o s e edges OA, OB, OC (Fig. 3.29)
coincide with the c o o r d i n a t e axes, a n d let us apply the t r a n s f o r m a t i o n
(3.11.1) to A , B , a n d C . T h e c o m p o n e n t s of their d i s p l a c e m e n t s a r e :
A
U\
B
"2
«n
a
a
"3
«13
«23
n
C
n «13
22 « 2 3
a
«33
(3.15.1)
Linear Transformation of Points
63
axx is seen to b e the n o r m a l c o m p o n e n t of the d i s p l a c e m e n t of A .
aX2 a n d aX3 c o n t r i b u t e only s e c o n d - o r d e r t e r m s to the c h a n g e in length
of OA . Therefore, for small t r a n s f o r m a t i o n s , axx gives to a g o o d
a p p r o x i m a t i o n the c h a n g e in length p e r unit length of a vector initially
parallel to OXx. Similarly, a22 a n d a33 give the c h a n g e in length per unit
length of vectors initially parallel to OX2 a n d OX3, respectively.
ax2 is seen to b e the projection o n OX2 of the tangential c o m p o n e n t
of the d i s p l a c e m e n t of A; it is also the projection o n OXx of the
tangential c o m p o an e nnt of the d i s p l a c e m e n t of B . Fig. 3.29 shows t h a t
axx
, aX3
, a22
,
d #23 c o n t r i b u t e only s e c o n d - o r d e r terms to the c h a n g e
in the right angle A OB . Therefore, for small t r a n s f o r m a t i o n s , 2 a 12 gives
the c h a n g e in the right angle b e t w e e n two vectors initially parallel to
OXx a n d OX2, respectively. Similarly, 2aX3 gives the c h a n g e in the right
angle b e t w e e n t w o vectors initially parallel to OXx a n d OX3, a n d 2 a 23
gives the c h a n g e in the right angle b e t w e e n two vectors initially parallel
to OX2 a n d OX3.
3.16
Linear Symmetric Transformation in Two Dimensions
Let I I b e a principal p l a n e of a s y m m e t r i c t r a n s f o r m a t i o n , a n d let
OXx a n d OX2 b e two reference axes in this p l a n e (Fig. 3.30); OX3 is the
principal direction n o r m a l to I I . Since I I is also invariant, the transform a t i o n of its p o i n t s is such t h a t
a
_"2_
_ \2
a
22_
Fig. 3 . 3 0
_*2_
(3.16.1)
64
Kinematics of Continuous Media
a n d u3 = 0. If9 the reference axes a r e r o t a t e d b y a n angle 9 a r o u n d
OX3, the atj s are t r a n s f o r m e d a c c o r d i n g to E q . (3.5.8a) with
( u = / 22 = cos 9
(n = - / 21 = sin 9
'13 = *23 = hi = hi = 0
i 33 = 1
. Thus,
=
tf'n
c2
2
+ < 2 2 s2i n # + 2 # 1 s2i n 9 cos 9
tfn °s 0
=
#22
2
(3.16.3)
2
sin 0 +
Un
a'n = -
(3.16.2)
tf22
cos 0
- 2 # 1 s2i n 0 cos 9
(3.16.4)
n
, °
sin 29 + a 1 c2o s 20.
(3-16.5)
T h e eigenvalue p r o b l e m in the p l a n e yields two principal directions,
0 1 and<92 given by (Fig. 3.30):
t
v
a ^ n ^a=2^ _ a ,2
u
2
(3.16.6)
a n d two p r i n c i p a l unit d i s p l a c e m e n t s , A) a n d A 2, given b y :
M
A,
A,
? ! L ^ ±^ l L l ^ y
+ .
(3.16.7)
fl?2
Linear Transformation of Points
65
T h e n o r m a l a n d tangential c o m p o n e n t s of the d i s p l a c e m e n t of a n y
point P (Fig. 3.31a) located o n a unit circle c e n t e r e d at O are o b t a i n e d
from Eqs. (3.13.2) a n d (3.13.4); in these e q u a t i o n s , the values of
xx, JC2, a n d x3 are set equal to cos /3, sin /?, a n d zero, respectively. T h u s ,
2
2
n = A! c o s / ? + A 2s i n / ?
|,| = m ^ s i n 2/?|.
W h e n referred to the system of axes OXx, OX2, n a n d t are given b y :
n
«l!+*22
=
11
«ll-fl22
2
/ = —
+
2
S ^C
Q
+
^
2 sin 20 + a 1 c2o s 29.
If Aj a n d A 2 are k n o w n , M o h r ' s circle for the p l a n e IT c a n be
c o n s t r u c t e d (Fig. 3.31b). Every p o i n t o n the unit circle of Fig. 3.31a h a s
a n image o n M o h r ' s circle. F o r e x a m p l e , the image of P is /?, w h e r e yxp
m a k e s a n angle 2/? with o'n; the image of a p o i n t o n the OXx axis is xx,
w h e r e yxxx m a k e s a n angle 2<f>x with o'n in a clockwise direction; the
i m a g e of a p o i n t o n the OX2 axis is diametrically o p p o s i t e to the
previous o n e . In Fig. 3.31b, o n e c a n verify t h a t o'b = a u, o'd = a22
,
a n d foci = tfi2. T h u s , k n o w i n g the elements of the t r a n s f o r m a t i o n
m a t r i x in a system of c o o r d i n a t e s OXx, OX2, o n e c a n d r a w M o h r ' s circle
a n d o b t a i n the principal directions a n d the principal unit d i s p l a c e m e n t s .
T h e c o n v e n t i o n is that w h e n axx a n d a22 are positive they are p l o t t e d to
the right of o' o n the o'n axis, a n d w h e n they are negative they are
p l o t t e d to its left; w h e n aX2 is positive it is p l o t t e d vertically u n d e r b
(Fig. 3.31b) a n d w h e n it is negative it is p l o t t e d vertically a b o v e b. In
Sec. 3.15, it was s h o w n t h a t 2 a 12 is a m e a s u r e of the c h a n g e in the right
angle b e t w e e n two vectors initially parallel to the c o o r d i n a t e axes; w h e n
aX2 is positive the angle b e c o m e s a c u t e a n d w h e n it is negative the angle
becomes obtuse.
Example
G i v e n a linear t r a n s f o r m a t i o n whose h o d o g r a p h is expressed b y :
M,"
_"2_
0.1
0.04"
0.04
0.04
x
x\
_ 2_
d r a w M o h r ' s circle a n d o b t a i n the principal directions a n d the principal
unit d i s p l a c e m e n t s . F i n d the n o r m a l a n d tangential d i s p l a c e m e n t s of a
66
Kinematics of Continuous Media
(b)
Fig. 3 . 3 2
point P whose c o o r d i n a t e s are ( l / \ / 2 ? l/V^)- W h a t are the coefficients
of the t r a n s f o r m a t i o n m a t r i x in a c o o r d i n a t e system formed by OP a n d
OQ (Fig. 3.32a)?
Fig. 3.32b shows M o h r ' s circle for the given d a t a . T h e principal
directions m a k e 26.6° a n d 116.6° with the OXx axis. T h e principal unit
d i s p l a c e m e n t s are 0.12 a n d 0.02. T h e n o r m a l a n d tangential displacem e n t s of P are 0.11 a n d 0.03, respectively. In the c o o r d i n a t e system
formed by OP a n d OQ , the t r a n s f o r m a t i o n matrix is:
"0.11
-0.03"
-0.03
0.03
'
N o t i c e t h a t the sign of axl in this n e w c o o r d i n a t e system is negative,
since p is a b o v e o'n. T h e right angle b e t w e e n two vectors parallel to
OXx a n d OX2 b e c o m e s acute after t r a n s f o r m a t i o n . O n the other h a n d ,
the right angle b e t w e e n two vectors parallel to OP a n d OQ b e c o m e s
obtuse.
PROBLEMS
1.
2.
F o r ajm = am
j, write in full the six e q u a t i o n s given in a c o n d e n s e d
form b y Eq. (3.5.8a).
F i n d the invariant directions of the linear t r a n s f o r m a t i o n whose
matrix is
" 2 - 2
1
1
1
3
3 "
1
.
-1
Linear Transformation of Points
3.
4.
W h a t is the angle b e t w e e n those directions?
Show t h a t the linear t r a n s f o r m a t i o n w h o s e matrix is
6.
-2
1
-1
1
-2
10
1
-2
-2
2
9.
1
2
+ 4 x + 2*3 — 4xx x2 + 4xx x3 + 4 x 2x 3 = A,
d e t e r m i n e the associated linear t r a n s f o r m a t i o n . If A = 1, find the
m a g n i t u d e s a n d directions of the principal axes of the q u a d r i c
surface.
R e d u c e the q u a d r a t i c form,
25xj + 34x1 + 4\xj
8.
whose
If [M] a n d [D] are its lm o d a l a n d spectral matrices, respectively,
verify t h a t [D] =
[M]- [a][M].
If the axis of r o t a t i o n OH in Sec. 3.9, coincides with the OX3 axis,
find the angle of r o t a t i o n a n d the radial extension d u e to a n t i s y m metric t r a n s f o r m a t i o n .
G i v e n the q u a d r a t i c form:
Ax
7.
1
does n o t possess a real i n v a r i a n t direction.
F i n d the principal directions of the linear t r a n s f o r m a t i o n
matrix
[a] =
5.
67
- 24x2x3
= A,
to its c a n o n i c a l form.
F i n d the eigenvalues a n d eigenvectors of the following system:
"3
0
2"
0
5
0
x2
2
0
3
_*3_
In a c o o r d i n a t e system OXx, OX2,
xxx
= A x2
_ i_
OX3, the system
68
Kinematics of Continuous Media
=
£3
"1
1
1
2
3
1
r
3
2
x
_ 3_
transforms {x} to {£}.
(a) F i n d the elements of the t r a n s f o r m a t i o n matrix in a n e w
system of c o o r d i n a t e s whose axes have direction cosines (0, 0,
1), ( 1 / V 2 , 1 / V 5 , 0 ) , ( l / \ / 2 , - l / \ / 2 , 0 ) .
(b)
F i n d the principal directions of the t r a n s f o r m a t i o n a n d the
principal unit displacements.
(c) F i n d the n o r m a l a n d tangential displacements of the point
whose c o o r d i n a t e s are l / \ / 3 , l / \ / 3 , l / \ / 3 . Show the position
of this point o n M o h r ' s d i a g r a m .
10. G i v e n a linear t r a n s f o r m a t i o n whose h o d o g r a p h is
"1
"2
0.13
-0.045"
-0.045
0.05
xX\~
_ 2_
find the principal directions a n d the principal unit d i s p l a c e m e n t s .
W h a t are the n o r m a l a n d0 tangential displacements of a point P such
that OP is inclined 30 on the OXx axis a n d its length is equal to 5?
CHAPTER 4
GENERAL ANALYSIS OF STRAIN IN
CARTESIAN COORDINATES
4.1
Introduction
In this c h a p t e r , the properties of linear t r a n s f o r m a t i o n s are used to
study the p r o b l e m f o r m u l a t e d in C h a p t e r 1. T h e two sets of Eqs. (1.2.4)
a n d (1.2.5) express the c h a n g e in length a n d direction of a n element MN
9 sets
at M (Fig. 1.3). Every p o i n t of the b o d y is associated with two such
9
of e q u a t i o n s a n d is the origin of a linear t r a n s f o r m a t i o n . T h e etj s a n d
uy s are n u m b e r s specific to the point, a n d in general they vary from
o n e p o i n t to a n o t h e r . In m a t r i x form, Eqs. (1.2.4) a n d (1.2.5) are written
as follows:
=
dux
du2
du3
=
1 + eu
e
+
X2
e
l3 -
21
(o
13
\3
1 +
e
23 +
el2 -
<M1
el2
eX2 -
co
+
co
-
Wj3
21
21 eX3 +
e
22
23 <o
32 1 +
co
e
(o i
e22
e
23
+
co
2
32
eX3 +
co
dxx
<o
dx2
13
32
e>
33
co
-
13
(4.1.1)
dx3
dxx
e
dx2
33
dx3
(4.1.2)
If the e l e m e n t MN is of unit length, dxx, dx2, a n d dx3 b e c o m e its
direction cosines.
In a b o d y t h a t is subjected to large t r a n s f o r m a t i o n s , a straight
element s e l d o m r e m a i n s straight: A c u r v e d e l e m e n t is m o r e likely to
69
70
Kinematics of Continuous Media
S3
0
Fig. 4.1
result (Fig. 4.1). T h e use of the linearized E q s . (4.1.1) a n d (4.1.2) to
express the t r a n s f o r m a t i o n of MN a m o u n t s to a substitution of the
t a n g e n t to the curve at Af* for the curve itself. This is w h y the
t r a n s f o r m a t i o n is sometimes called a linear tangent transformation.
It is
obvious t h a t the smaller t h e e l e m e n t ds, the better the a p p r o x i m a t i o n of
M*N* b y its t a n g e n t Af * 7 V * . A t every p o i n t :
1) T h e r e a r e three principal directions a n d three principal p l a n e s .
A l t h o u g h these directions a n d p l a n e s c a n b e o b t a i n e d from t h e e q u a t i o n
of the characteristic ellipsoid, this m e t h o d is operationally tedious a n d
a m o r e direct o n e will b e given in a later section.
2) T h e r e are three i n v a r i a n t directions a n d three invariant planes.
T h e s e directions a n d p l a n e s m a y or m a y n o t all b e real; they a r e
o b t a i n e d as s h o w n in Sec. 3.6.
3) T h e t r a n s f o r m a t i o n c a n b e split i n t o a s y m m e t r i c a n d a n a n t i s y m metric o n e . Eq. (4.1.2) b e c o m e s :
dux
du2
e
=
du3
+
e-11
e\2
'12
\3
23
22
e dx{
e<?23 e dx2
dx3
33
l3
0
-W
<o
0
21
13
-co
21
"32
<0
(4.1.3)
13
32
-co
dxx
dx2
0
dx3
T h e d i s p l a c e m e n t of p o i n t TV (with respect to A F ) is the s u m of t h e
d i s p l a c e m e n t s caused b y the t w o t r a n s f o r m a t i o n s ; each h a s b e e n
71
General Analysis of Strain in Cartesian Coordinates
studied in detail in C h a p t e r 3 a n d their s u m m a t i o n is t o b e m a d e as
i n d i c a t e d in Sec. 3.2.
I n t h e following sections, t h e c h a n g e s in length a n d direction of t h e
elements of a b o d y u n d e r t r a n s f o r m a t i o n will b e d e t e r m i n e d , t h e
c o n c e p t of strain a t a p o i n t will b e i n t r o d u c e d , a n d the various
definitions of strain will b e discussed. F o r clarity, cartesian c o o r d i n a t e s
are solely u s e d in this c h a p t e r ; analyses in curvilinear c o o r d i n a t e s a r e
p r e s e n t e d in C h a p t e r 6.
4.2
Changes in Length and Directions of Elements Initially Parallel to
the Coordinate A x e s
Let us a s s u m e t h a t t h e e l e m e n t MN of F i g . 1.2 is parallel to t h e OXx
axis. After a linear t r a n s f o r m a t i o n , t h e c o m p o n e n t s of M*N* a r e given
b y E q . (4.1.1) a s :
)dxx
dHx = (I + exx
(4.2.1)
d£2 = (eX2+ oo2X
)dxx
(4.2.2)
00X3
)dxx.
(--)
d^3 = (eX3
~
423
T h e e l o n g a t i o n of MN p e r u n i t length is given b y :
MN -
M*N*
— MN _
~
T h e direction of t h e t r a n s f o r m e d e l e m e n t M*N*
direction cosines:
)dxx
/ _ (1 + exx
(1 + EMN
)dxx
)dxx
h = (eX2+ u2X
(1 + EMN
)dxx
/3 =
(g
13~
Q3 i)dx
(\+EMN
)dxx
X x
(A 9 a\
nh
x\ •
V*-**)
is given b y its three
_ 1 + exx
1 + ExX
oj
eX2+
2X
=
\+ExX
e
X3=
co
13
\+Ex .X
)
(426)
4 2) ?
Similar e q u a t i o n s c a n b e written for e l e m e n t s MP a n d MQ initially
parallel to OX2 a n d OX3. U n l e s s OXx, OX2, OX3 a r e a l o n g p r i n c i p a l
directions of t h e t r a n s f o r m a t i o n , t h e e l e m e n t s will n o t r e m a i n o r t h o g o nal (Fig. 4.2). T h e i r direction cosines a r e listed in t h e following t a b l e :
72
Kinematics of Continuous Media
Initially
Initially
A l o n g OXx
A
e long
L
*\
e
1 + e u
1
+ExX
\2
12 + w 21
Exl
\3
~ "13
1
+Exl
Initially
A l o n g OX-
OX2
e
~ "21
1
+Ex2
1 + e22
1
+Ex2
e
e1 +
h
M*Q*
M*P*
M*N*
(4.2.8)
e1 + ^ 3
~ "32
2i
1 + Ex3
1 + e 33
1 + Ex3
23 + "32
1 +
13 + w , 3
Ex2
Since
2
M*N*
= V(^i)
2
42 9
(^3) ,
( - « >
2
+ ( ^ 2)
+
then
2
= V ( l + e u)
1 +
2
2
+ ( e 12 + <o 2)1 + ( e -i <3o 1 ) 3 . (4.2.10)
Similarly,
2
21
1 + Ex2 = V(«12 - "21 ) + (1 + ^22? + (^23 + "32)"
2
2
2
(4- -
0
4 21 2
x3 = V(^13 + " 1 3 > + (*23 - " 3 2 > + (• + ^ 3 3 ) • ( " - >
1 + E
73
General Analysis of Strain in Cartesian Coordinates
4.3
Components of the State of Strain at a Point
In Fig. 1.2, let us c o m p u t e the c h a n g e in length of the e l e m e n t s ds
after a linear t r a n s f o r m a t i o n :
2
(ds*)
2
2
2 (rf£/) " 2
2
- (ds) =
(d*i)
= SyMidtj
(4.3.1)
- dxtdxj).
I n t r o d u c i n g E q s . (1.1.6) in Eq. (4.3.1), we o b t a i n :
2
(ds*)
2
2
2
- (ds) = 2[exx
(dxx)
2
+ e22
(dx2)
+ e33
(dx3)
+ 2ex2
dxx dx2
(4.3.2)
+ 2623^X2^X3],
+ 2eX3
dxxdx3
w h e r e the following substitutions of Eqs. (4.3.3) h a v e b e e n m a d e :
11
dxx
2 LA 9*i /
2
= exx+ \[(exx
)
e
_
22
9«2.
i 17^
8x2
2 | _ \ a * 2/
^22 + |[(<?12 -
e
33
-
V .l
( ^ l
+ («22)
+
2
+
j / 8wj
2 \ 3x9
2\dx
2
du2
3xj
13)
+
3xj 3 x ?
3.X!
2
= 5 ^ 1 2 + ^ll(^l2 -
w
(^y]
w\ 2ox3/ J 2
+ ( 23 ~
dux3wj
+
+ (^23 + " 3 2 > ]
2 LA o x 3 / w 2 \ 0x3
e /
ax3
V l
\ 3 x 2/ J 2
"2 " \ 3 x 2/ 2
"21 )
(4.3.3)
2
+ (eX3
- c o 1 ) 3]
^ .r(^y (^)
^33 + 5 [(^13 +
€ 2l
\ dxx ) J
\ dxx )
2
+ (eX2
+ co2x
)
32)
+ 033 ) ]
3w 2 3w 2 ^ +
3w 3 3w 3 \
3.x,^ 3 x 9
3
2
33.x,
x 3.x? /
2 l ) + ^22(^12 + <*>2l)
+ ( e 13 - <0 1)(<?
3 23 + ( 0 3 ) 2]
e 32
1 / 3w 2
2 \ 3x3
2e
= \[ 23
3w 3 3t/j 3wj
3 x 2 3 x 23 x 3
e
e
+ ( \2 ~ <*2\)( \3
+ ^33(^23 + W32)]
3w 2 3w 2
3 x 23 x 3
8^3 3w 3 \
3 x 23 x 3/
w
+ ^ 3 ) + ^22<>23 - 3 2 )
74
Kinematics of Continuous Media
6 1 3 1 / 3w 3
dux
2x3^!
dx
e3e
l e
= \ \
3 \ + \\( \3
3w 2 3w 2 + 3 w 3 du3 \
dux dux
3x
dx dx
3 x
+ (0 ) + (>
1 3
23
3 3^j
-
dx
3 dxx /
C0 )(e
3 2 12 +
C0 )
2 1
+ ^33(^13 ~ <*>13>]-
In E q s . (4.3.3), if the subscripts of the e's are i n t e r c h a n g e d , the righth a n d sides r e m a i n the s a m e ; in o t h e r w o r d s ,
ey^eji.
(4.3.4)
T h u s , in i n d e x n o t a t i o n s , Eq. (4.3.2) is w r i t t e n :
2
(ds*)
2
- (ds)
(4.3.5)
= 2eiJ
dxidxJ,
a n d E q s . (4.3.3) are w r i t t e n :
u
2 L oxj
oxt
J
ox; oxj
T h e e l o n g a t i o n of the e l e m e n t ds p e r unit length is given b y :
therefore,
ds* = (EMN + \)ds,
(4.3.8)
a n d E q . (4.3.5) b e c o m e s :
2
(ds*)
2
- (ds)
2
= EMN
(EMN
+ 2)(ds)
2
D i v i d i n g b o t h sides of E q . (4.3.9) b y 2(ds) ,
= 2eijdxidxj.
(4.3.9)
we o b t a i n :
k2
[ ( f ) - ' ] - » ( - ¥ ) = ^
2
2
2
= e 1 / 1 + e 2 / 2 + e 33 / 3 +
(4.3.10)
2e1/1
2/ 2
+ 2 e 2/ 2
3/ 3 + 2 e 13 / 1/ 3.
T h e quantities ^x, /ds = /, a r e the direction cosines of the e l e m e n t ds in
the system OXx, OX2, OX3. E q . (4.3.10) shows that, in o r d e r to calculate
the e l o n g a t i o n per unit length EMN of a n e l e m e n t MN, it is sufficient to
General Analysis of Strain in Cartesian Coordinates
75
k n o w the six quantities etJ at M. T h e s e six quantities are called the
c o m p o n e n t s of the state of strain at the p o i n t M. T h e y describe the state
of d e f o r m a t i o n of the b o d y .
T h e r i g h t - h a n d side of Eq. (4.3.10) is a q u a d r a t i c form in the variables
/ 1? / 2, / 3, with the following matrix of coefficients (see Sec. 3.12):
e
el l
e\2
12
13 e
e
22
23 £
_ 13
23
33
e
e
e
(4.3.11)
This m a t r i x is called the m a t r i x of the state of strain.
4.4
Geometrical Meaning of the Strain Components eijmStrain of a
Line Element
Let us take the element ds of Fig. 1.2 parallel to the OXx axis (Fig.
4.2). After transformation, its elongation per unit length ExX c a n be
o b t a i n e d from Eq. (4.3.10). Setting /, = 1, / 2 = / 3 = 0, we o b t a i n :
(4.4.1)
Consequently,
(4.4.2)
In a similar m a n n e r ,
ExX = V I + 2*ii - 1 .
e
x2
=
v m ^ - i
(4.4.3)
(4.4.4)
Ex3 = V I + 2 e 33 - 1 ,
where Ex2 a n d Ex3 are the elongations per unit length of elements
initially parallel to the OX2 a n d OX3 axes, respectively. T h u s , the strain
c o m p o n e n t s e n, e 2 , 2e 33 characterize (or describe) the relative elongation of those line elements t h r o u g h M parallel to the three axes. T h e y
are called t h e n o r m a l c o m p o n e n t s of the state of strain.
T h e m e a n i n g of e 1 , 2e 1 , 3e 23 is o b t a i n e d b y writing the expressions
after t r a n s f o r m a t i o n for the angles b e t w e e n the elements dxx, dx2,
dx3
originally parallel to the c o o r d i n a t e axes. T h i s is a simple m a t t e r since
the direction cosines of the t r a n s f o r m e d elements are k n o w n . F r o m
table (4.2.8) a n d Fig. 4.2, we o b t a i n :
76
Kinematics of Continuous Media
c o s ( A / * * * , M * / > * ) = 1(
+^ (E\
)
+ )E - xc o si ( | - <f>12
) (
4
4
= sin <J>12
,
where <£>12 is the c h a n g e in the right angle between dx{ and dx2.
Similarly,
c o s ( M * / V . , A / * < r ) = 1(
)E+= xc o 3s ( | - * „ )
+f yE
)
(
4
) (
4
4
= sin <J>13
c o s ( M * / > * , M * e * ) = 1(
+ ^£
^
+= c o s ( f -
fe)
4
= sin <f> .
23
T h e angles <£ 1,2<^ 1,3a n d <#>23 are called the shear angles. T h u s e 1 , 2e 1 , 3
e 23 characterize (or describe) the c h a n g e in the right angle b e t w e e n
elements originally parallel to the axes. T h e y are referred to as the
shearing c o m p o n e n t s of the state of strain. If e 12 = £13 = e 23 = 0, the
elements dxx, dx2, dx3 in Fig. 4.2 r e m a i n o r t h o g o n a l after t r a n s f o r m a tion. This is precisely w h a t is required for their directions to be three
principal directions of the linear t r a n s f o r m a t i o n .
T h e previous analysis shows that a rigid b o d y m o t i o n is characterized
by etJ= 0.
A c o m p a r i s o n of E q s . (4.4.1) to (4.4.4) to Eq. (4.3.10) shows t h a t the
latter c a n b e written:
EmnV
L
+ - f
)
+ 2e
= eMN = znl?
+
*v.tl + % + ^nh
«33 (4.4.8)
2 / 23/ 3 + 2 « , 3 / , / ,
or
Thus,
?MN = ^\+2eMN
-1,
(4.4.9)
a n d eMNcharacterizes (or describes) the c h a n g e in length per unit length
of a n element A/TV. eMN is called the strain at A/ of the e l e m e n t MN.
General Analysis of Strain in Cartesian Coordinates
77
This strain is completely defined o n c e the elements of the m a t r i x
(4.3.11) are k n o w n at the p o i n t M.
In engineering practice, the strain of a n e l e m e n t MN is defined as the
c h a n g e in length per unit length of t h a t e l e m e n t : In o t h e r w o r d s , it is
the q u a n t i t y EMN a n d n o t eMN t h a t is called the strain of MN. It will b e
s h o w n in Sec. 4.8 t h a t for small strains the two quantities are nearly
equal.
4.5
Components of the State of Strain under a Change of Coordinate
System
E q s . (4.3.3) give the definitions of the c o m p o n e n t s of the state of
strain Ey referred to a system of c o o r d i n a t e s OXx, OX2, OX3. Let us
consider a n o t h e r trirectangular system OX\, OX2, OX3, o b t a i n e d b y a
r o t a t i o n a r o u n d O (see Sec. 3.5). T h e position of the n e w system is
defined b y the direction cosines of its axes ly with respect to the initial
ones (Fig. 4.3). F r o m Eqs. (3.5.3), we h a v e :
Fig. 4 . 3
dx\
(4.5.1)
Therefore,
(4.5.2)
Dividing b o t h sides of Eq. (4.5.2) b y ds, we o b t a i n :
78
Kinematics of Continuous Media
d
f&L = / i ^L.
ds
j
(4.5.3)
ds '
i.e.,
T h e left-hand sides of E q s . (4.3.9) a n d (4.3.10) a r e obviously i n d e p e n d ent of t h e system of c o o r d i n a t e axes. T h u s ,
2
(ds*)
2
-
f
(ds)
= ItijdXidXj
= 2e' dx dx'
rs r s
(4.5.5)
and
e
= E
MN
E
ij~ds~^
=
~ U^^J
E=
MN
e
(4.5.6)
'rsKh>
w h e r e l\ a r e the direction cosines of the element MN in t h e n e w system
of c o o r d i n a t e s . T h e substitution of E q . (4.5.4) in E q . (4.5.6) gives:
4 ^ = ^4/4^4-
(4.5.7)
Therefore,
(e' - 1 1 e )f f = 0
\ rs ri ^sj^ij J* r s ^'
c
l
1
Since fp 7^ 0, the rule of t r a n s f o r m a t i o n of £y is
For example,
2
e'u =
2
2 22 + Jf3e33 + 2 / u/ 1 £212
/, ,e„ + /, e
+ 2/, , 7 , 3 6 , 3 +
e'12
= Allien +
+ e, (/n
2
2/, /, e23
2 3
h i h i ^ i
+
A2 + '12'21) +
+ £ 2 (3/ l2 / 23 + / , 3 / 2 )2,
a n d so o n for the other c o m p o n e n t s .
(4.5.9)
h i h ^ ^
e (/„/23
1 3
(4.5.10)
+
79
General Analysis of Strain in Cartesian Coordinates
4.6
Principal A x e s of Strain
T h e three principal axes of the linear t r a n s f o r m a t i o n (4.1.1) are such
that eX2= eX3= e23= 0 (see Sec. 4.4). C o n s e q u e n t l y , they are also called
the principal axes of the strain m a t r i x (4.3.11). W h e n the c o o r d i n a t e
axes are a l o n g the principal directions, the unit elongations are d e n o t e d
b y Ex, E2, a n d E3; the n o r m a l c o m p o n e n t s of the state of strain are
d e n o t e d by e l5 e2, a n d e 3 a n d are referred to as the three principal
strains. T h u s , in this sytem of c o o r d i n a t e s :
Ex = a / 1 + 2ex - 1
(4.6.1)
E2 = V l + 2 e 2 - 1
(4.6.2)
E3 = V l + 2e3 - 1
(4.6.3)
T h r e e e l e m e n t s MNX, MN2, a n d MN3 initially a l o n g the principal
axes will experience unit e l o n g a t i o n s given b y E q s . (4.6.1) to (4.6.3);
they will r o t a t e in space, b u t will k e e p their o r t h o g o n a l i t y . A n e l e m e n t
MN with direction cosines (x, i 2, a n d i3 with respect to the principal
axes will experience a unit elongation,
4
Emn = V l + ^mn
"I,
( -6.5)
w h e r e eMN is given b y Eq. (4.6.4). F i n d i n g the principal directions
a m o u n t s to finding the system of o r t h o g o n a l axes in which the q u a d r a t i c
form in the r i g h t - h a n d side of Eq. (4.3.10) is r e d u c e d to its c a n o n i c a l
form—i.e., to the r i g h t - h a n d side of Eq. (4.6.4); in this system, t h e
m a t r i x (4.3.11), w h i c h is associated with the q u a d r a t i c form, b e c o m e s a
d i a g o n a l matrix. This p r o b l e m h a s b e e n solved in Sec. 3.12. T h e set of
linear h o m o g e n e o u s e q u a t i o n s :
( e n -e)Ix + £ 1 / £2 + e 1 / 3 = 0
*12 h + (*22 ~ ) ^ 2 + *23 h = 0
*13 A + £23^2 +
^ 3 3 - ^ 3
=
(4.6.6)
0,
a n d the c o n d i t i o n t h a t
/2 +
/2 +
/2 =
h
(4.6.7)
80
Kinematics of Continuous Media
allow o n e to o b t a i n the principal strains a n d their directions: A
nontrivial solution of Eqs. (4.6.6) exists if, a n d only if, the d e t e r m i n a n t
of the m a t r i x formed by the coefficients is equal to zero. All the
properties established in Sec. 3.6 apply here. T h e e x p a n s i o n of the
d e t e r m i n a n t yields a cubic e q u a t i o n in e.
3
2
- /
£
e
+ /
1
a n
whose roots are all real. I XJ 7 2,
the state of strain:
€
h
1 +
=
e
2 +
=e £ e e
1 2
h
=
+
e
e
n 22
2 3 +
+ e
e e
h
4.7
l
~
2 3
e3 = 0,
ea r
called the three invariants of
e
£
e
1 1 + 22
e
+
469
33
( - - )
3 l
e
(4.6.8)
- /
d h
£
£
=
3
2
2
2 33
2
+
e
e
£11
e1 2
•12
-13£
22
23e
13
23
33
3 n3
-
(e
)
1 2
2
-
(e
)
1 3
2
-
e
e
(e
)
2 3
(4.6.10)
(4.6.11)
Volumetric Strain
Let us consider a parallelepiped with edges dxx, dx2, a n d dx3. A c c o r d i n g to Eqs. (1.1.6), this r e c t a n g u l a r parallelepiped is t r a n s f o r m e d to
a n oblique o n e whose three edges h a v e projections on OXx, OX2,
OX3
given respectively b y :
« 2 - ^ l
di2-(l+%l)dx2
d
d&
= ^
x
\
di=2
d
^ 3
=
^
x
2
&
T h e v o l u m e of the t r a n s f o r m e d parallelepiped is given by the triple
scalar p r o d u c t :
General Analysis of Strain in Cartesian Coordinates
i +
3WI
du
dx
dx
1 +
y* =
3W3
du
x
3
du
2
dx
3
dx
3W2
dx
2
3
1 +
dx~>
dxi
= Ddxx dx2
Since V = dxx dx2 dx3,
by:
du
x
2
x
81
du
dxx dx2 dx3
(4.7.2)
3
dx*
dx3.
the c h a n g e of v o l u m e per unit v o l u m e is given
V* — V
(4.7.3)
D - 1.
= E=
c a n b e expressed in terms of the c o m p o n e n t s of the state of strain b y
s q u a r i n g the d e t e r m i n a n t of Eq. (4.7.2) a n d substituting the c o r r e s p o n d ing values from E q s . (4.3.3). T h u s ,
(1 + Evf
=
1 + 2exx
2eX2
2e
2eX2
1 + 2 e 22
2 e 23
2e ;
2 e 13
2
=
D.
(4.7.4)
1 + 2 e 3| 3
W h e n e x p a n d e d a n d the t e r m s g r o u p e d , Eq. (4.7.4) b e c o m e s :
2
(1 + Ev)
(4.7.5)
= 1 + 27, + 4 / 2 + 8 / 3,
w h e r e Ix, I2, a n d 7 3 are the three i n v a r i a n t s of the state of strain. I n
t e r m s of p r i n c i p a l strains, E q . (4.7.4) b e c o m e s :
1 + 2e,
£y=
0
0
0
2
1 + 2e 2
0
= Z> .
0
1 + 2e3
(4.7.6)
0
v
F o l l o w i n g the definition of strain used for elements of length, the
v o l u m e t r i c strain is defined a s :
(i +
2
ev = Ev(l
+
= ¥ °
-
D = A + 2 / 2 + 4 / 3,
(4.7.7)
a n d Ev is therefore given b y :
Ev = V ( l + 2 e , ) ( l + 2 e 2) ( l + 2 e 3) - 1
(4.7.8)
82
Kinematics of Continuous Media
or
Ev = V l + 2/j + 4 / 2 + 8 / 3 - 1 .
(4.7.9)
W h e n the principal directions are chosen as the reference r e c t a n g u l a r
axes at a p o i n t O of the u n d e f o r m e d b o d y , a unit c u b e is t r a n s f o r m e d
to a r e c t a n g u l a r parallelepiped with its three d i m e n s i o n s given by
1 + Ex, 1 + E2, 1 + £3, respectively. T h e v o l u m e of the t r a n s f o r m e d
unit c u b e is given b y :
V* = (1 + Ex)(l
+ E2)(\
+ £ 3)
(4.7.10)
and
Ev = (l + £ , ) ( ! + £ 2) ( 1 + £ 3 ) -
1-
(4.7.H)
T h e expression for Ev is i n d e p e n d e n t of the initial s h a p e of the c h o s e n
element at O.
4.8
Small Strain
In engineering practice, the m a g n i t u d e s of the unit elongations
ExX
, Ex2
, a n d Ex3
,
as well as the changes in right angles </>12
, <^ 2,3
and
are
generally
very
small.
F
o
r
the
usual
engineering
materials,
13
a n d in m o s t structures, the m a g n i 3
t u d e s of the unit elongations a n d of
the strains are on the o r d e r of 1 0 ~ : Such strains certainly deserve to be
called small.
T h e a s s u m p t i o n t h a t the strains are small i n t r o d u c e s substantial
simplifications in Eqs. (4.4.2) to (4.4.7): If
<f>
Exl «
1 < 2,
(4.8.1)
e x ^ xE x . X
(4.8.2)
Eq. (4.4.1) yields:
Similarly,
e
22 =
F o r a n y element
c
E
x2>
33 =
£*33-
4
( -8.3)
MN,
eMN =
(4.8.4)
General Analysis of Strain in Cartesian Coordinates
83
a n d the definition of strain given in Sec. 4.4 does n o t differ appreciably
from the engineering definition.
In Eq. (4.4.5), the sine of the angle c a n b e replaced by the angle itself;
also, the d e n o m i n a t o r in its r i g h t - h a n d side c a n be replaced b y unity.
Therefore,
<t>\2
M2
(4.8.5)
2
Similarly,
4>13
_ <t>23
(4.8.6)
T h e expression of the c h a n g e of v o l u m e per unit v o l u m e Ev as given
b y Eq. (4.7.9) b e c o m e s simplified b e c a u s e the invariants I2 a n d 7 3 are of
the o r d e r of Z ^ a n d Z ^ , respectively, a n d c a n be neglected with
respect to I{, which is of the o r d e r of Exl
. Also, if only the first-order
terms in the b i n o m i a l e x p a n s i o n of \/l + 2IX are retained, then
v
£
£
E
e
= h = n + 22 + 33-
4
( -8.7)
T h u s , u n d e r the a s s u m p t i o n of small strains, the first invariant is the
c h a n g e of v o l u m e p e r unit v o l u m e . Because of the o r d e r of m a g n i t u d e
of the strains, it c a n b e a s s u m e d t h a t
Ev «
1 < 2,
(4.8.8)
so t h a t Eq. (4.7.7) yields:
ev ~
Ev.
T h u s , the definition of volumetric strain given in Sec. 4.7 coincides with
the engineering definition.
4.9
Linear Strain
A further restriction t h a t is m a d e in the analysis of strain involves the
m a g n i t u d e s of the r o t a t i o n a n d of the cylindrical dilatation p r o d u c e d b y
the a n t i s y m m e t r i c p a r t of Eq. (4.1.3). T h e angle of r o t a t i o n h a s the s a m e
value for all the elements at M. If this angle is a s s u m e d to b e small so
that its s q u a r e is very small c o m p a r e d to unity, the cylindrical dilatation
c a n b e neglected since it is of the s e c o n d o r d e r with respect to the angle
(Sec. 3.9). I n such a case, the c h a n g e s in length of the elements at M , as
84
Kinematics of Continuous Media
well as the c h a n g e s in the right angles b e t w e e n m u t u a l l y p e r p e n d i c u l a r
elements, d e p e n d o n the s y m m e t r i c p a r t of Eq. (4.1.3) alone.
N o w let us consider Eqs. (4.3.3). If we a s s u m e t h a t the derivatives
diij/dxj
are small e n o u g h so t h a t their s q u a r e s a n d p r o d u c t s a r9e
9
negligible c o m p a r e d to the
derivatives themselves, the ey s
b e c o m e equal to the ey s. T h u s , the c h a n g e s in length of the
elements at M as in Eq. (4.3.9), a n d the c h a n g e s in the right angles
9
b e t w e e n m u t u a l l y p e r p e n d i c u l a r elements,
as in E q s . (4.4.5) to (4.4.7),
9
are completely d e t e r m i n e d o n c e the ey s are k n o w n . Therefore, the use
of the e{ s in p l a c e of the ty s does n o t only imply t h a t the derivatives
dui/dxj
are small, b u t also t h a t the r o t a t i o n d u e to the a n t i s y m m e t r i c
p a r t of the t r a n s f o r m a t i o n is small, a n d t h a t the d e f o r m a t i o n is
described solely b y the s y m m e t r i c p a r t :
dux
du2
9
du3
\\ n n
e
e
e
e
\2
e
e
22
23
A
h
e
(4.9.1)
h
X3 e23 e33
T h e e{ s are called linear strains. e n, e22
, a n d e33 are called the linear
n o r m a l strains, a n d e 1 , 2eX3
, a n d e23 are called the linear tangential or
shearing strains. Since linear strains are exclusively used in the classical
theories of9 elasticity a n d plasticity, the w o r d linear is usually o m i t t e d
a n d the ey s are referred to as n o r m a l a n d shearing strains.
All the properties of linear s y m m e t r i c t r a n s f o r m a t i o n s d e d u c e d in
C h a p t e r 3 a p p l y to linear strain. T h e e q u a t i o n s of Sees. 3.10 to 3.16 c a n
b e rewritten here with the following c h a n g e s in n o t a t i o n :
(a)
JCJ, x2, x3 are r e p l a c e d b y the direction cosines / j , / 2, / 3.
(b)
T h e ay s are r e p l a c e d b y ey
s.
(c)
A l5 A2> ^ 3 are r e p l a c e d by ex, e2, e3, a n d are called major,
i n t e r m e d i a t e , a n d m i n o r principal strains.
(d)
n a n d t are r e p l a c e d b y en a n d en a n d are called n o r m a l a n d
tangential (or shearing) strains.
In view of the restrictions p l a c e d o n the derivatives, the linear strains
are necessarily small strains a n d the t r a n s f o r m a t i o n expressed b y Eq.
(4.9.1)9 is a small s y m m e t r i c t r a n s f o r m9a t i o n . T h e geometrical m e a n i n g of
the ey s is the s a m e as t h a t of the ay s of Sec. 3.15: exx
, e22
, e33 are the
General Analysis of Strain in Cartesian Coordinates
85
unit elongations of elements initially parallel to the three axes, a n d
2 e 1 , 22 e 1 , 32e23 are the c h a n g e s in the right angles b e t w e e n those
elements. T h e following is essentially a list of e q u a t i o n s a n d results
o b t a i n e d b y m a k i n g the a b o v e substitutions in C h a p t e r 3 :
1. Characteristic equation and invariants
3
2
e - Ixe
Il=el
+ I2e -
+ e2 + e3 = exx + e22 +
I2 = exe2 + e2e3 +
= exx
e22
(4.9.3)
e33
(4.9.4)
e3ex
+ e22
e33 + e33
exx
e e
(4.9.2)
0
I3
e
- e\2 -
2
e
k = l 2 *3 = *11 22 33 +
e\3
e
^ 1 2 ^23 \3
eh
e e
~ 33 \2
495
~ ?22 ^13
(--)
I n a system of principal axes, Eq. (4.9.1) b e c o m e s :
du2
=
du3
2. Normal
and tangential
M
0
0
*2
0
0
e0
3
(4.9.6)
(Fig.
4.4)
2
= <?\ 1 A + *22 A + *33 % +
+
A
A
A
strains of an element
2
(e N)n
0
2*33/2/3
Fig. 4.4
A A + ^1)
A A
(4.9.7)
86
Kinematics of Continuous Media
2
(eMN
\
=
2
V ( ^ i )
2
(du2)
+
2
+ (du3)
-
(4.9.8)
{eMN
)n
3. Mean Strain. Spherical and deviatoric components
T h e m e a n strain em is defined b y :
of linear
strain
e
e„, =
+ 22 + ^33
(4.9.9)
T h e r i g h t - h a n d side of E q . (4.9.1) c a n b e d e c o m p o s e d into t w o p a r t s :
e
dux
m
=
du2
0
0
0
du3
A
A
A
0
0
e
0
+
e'11
e\2
'12
-13
22
23
\3
23
33
e
e
e
, (4-9.10)
e
with
e\x + e'22+ e'33= 0.
T h e first m a t r i x in the r i g h t - h a n d side of E q . (4.9.10) is called t h e matrix
of the spherical c o m p o n e n t of strain, a n d the second o n e is called t h e
matrix of t h e deviatoric c o m p o n e n t of strain.
4. Volumetric
strain
3e
e
e
f
e
ev = m = e\ + 2 + 3 = \\
^22 + ^33 = h •
(4.9.11)
5. Components of the linear strain in a change of coordinates
If the direction cosines of t h e n e w system with respect to the old o n e
are ( / n, / 1 , 2/ 1 ) 3, ( / 2, , / 2 , 2/ 2 ) 3, a n d ( / 3, , / 3 , 2/ 3 ) 3,
(4.9.12)
4 = 4 4 ^
Thus,
f?i
eu = exx
+ e22
f?2
+ e33
%
+ 2eX2
(4.9.13)
Ai A2 + 2*i3 Ai
+
2*23'12'13
l
2
+ ^22 A 2 + ^33 A 3 +
e'll = « 1 1 2\
+
2
2*23^2^23
2
e'n = eu / 3!
2
+ 2*23 A2 A3
\2*2l
l2 e
22 +
\ 3 Al
2
+ * 2/ 2 + « 3 A3 +
2 3
2e
3
A3
2 * 1 / 23i hi + 2 e , 3/ } i
A
3
(4.9.14)
A3 (4.9.15)
87
General Analysis of Strain in Cartesian Coordinates
* i 2 = (*ll'll + * 1 2 ' l 2 + *13'l3)'21
e
e
e
+ ( \lh\
+ 22 A2 + e 2 / 3
, 3) / 22
+ (<?I3 A 1
+ «23A2 +
23 =
(4.9.16)
e
33^\i)h3
+ ^12^2 + < ? 1 / 3
2 ) 3/ 31
+ («i2 Ai + e12hi + e23 / 2 ) 3/ 32
+ ( e l /321 + e 23 / 22 + 3e / 2
/ 33
3) 3
«13 = tell Al + <?12 A2
(4.9.17)
+ ^13A3)Al
+ tenAi + «22A2 + e 2A3)A
3
2
+ tenAi + e 2 / 3
+
e
/
)
/
]2
3 3
1 333
6. Octahedral
The normal
principal axes
o b t a i n e d from
(4-9.18)
normal and shearing strains
strain of a n element MN equally inclined to the three
is called the o c t a h e d r a l n o r m a l strain, eoct
,
a n d is
Eq. (4.9.7):
4919
= eoct = |tei + e2 + e3) =
)n
(eMN
(-- )
T h e tangential strain c o r r e s p o n d i n g to the s a m e element is o b t a i n e d
from Eq. (4.9.8), a n d is written:
(e„ ), =
N
7
-f
2
2
= | [ t e , - e2f
+ (e2 - e3)
+ (e3 - e , ) ] * .
492 0
( - -
)
y o t cis referred to as the o c t a h e d r a l shearing strain. W h e n the system of
reference is n o t a principal system,
492 1
*oc,-i(«..+e22
+ «33)-£
( - -
>
and
2
^
2
2
= | [ t e l . - ^22) + te22 - ^ 3 3 ) + te33 - eu)
2
2
2 ]/i
+ 6e 2 + 6e 3 +
7. Linear strain in two
dimensions
6e 23
].
)
(
88
Kinematics of Continuous Media
Sec. 3.16 c o n t a i n s all the e q u a t i o n s p e r t a i n i n g to this case of linear
strain. T h e s e e q u a t i o n s , as well as the r e p r e s e n t a t i o n o n M o h r ' s
d i a g r a m , c a n directly be used here after the a p p r o p r i a t e c h a n g e s in
n o t a t i o n are m a d e .
4.10
Compatibility Relations for Linear Strains
In E q s . (1.2.1), we defined the c o m p o n e n t s of the linear strain in
terms of the c o m p o n e n t s of the d i s p l a c e m e n t b y :
(4.10.1)
If the d i s p l a c e m e n t s u{, w 2, a n d w3 are prescribed c o n t i n u o u s functions
of xl,x2,andx3,
t h e n the strain c o m p o n e n t s etj c a n b e u n i q u e l y
d e t e r m i n e d . O n the o t h e r h a n d , if the strain c o m p o n e n t s are p r e s c r i b e d
functions of the c o o r d i n a t e s , it will n o t b e possible to find u n i q u e values
for the d i s p l a c e m e n t s b e c a u s e the strains represent p u r e d e f o r m a t i o n
whereas the d i s p l a c e m e n t s i n c l u d e b o t h d e f o r m a t i o n a n d rigid b o d y
m o t i o n . T h i s difficulty is o v e r c o m e b y specifying the rigid b o d y m o t i o n
of a p o i n t M of the b o d y , i.e., specifying its d i s p l a c e m e n t s ut a n d the
elements of its r o t a t i o n co^. T h e s t r a i n - d i s p l a c e m e n t relations (4.10.1)
a npartial differential e q u a t i o n s with only three
form a system of six
u n k n o w n s , « j , w 2, d u3\ it is o b v i o u s t h a t s o m e restrictions m u s t b e
p l a c e d o n the strains in o r d e r t h a t Eq. (4.10.1) h a v e a solution. T h e s e
restrictions are called the compatibility relations (or conditions).
A physical i n t e r p r e t a t i o n of the c o n d i t i o n s of compatibility c a n b e
o b t a i n e d b y e x a m i n i n g the d e f o r m e d b o d y . Let M(x{, x2, x3) b e a point
of a c o n t i n u o u s b o d y at w h i c h the d i s p l a c e m e n t s ut a n d the r o t a t i o n s o?^
are k n o w n . T h e d i s p l a c e m e n t s u\ of a n a r b i t r a r y p o i n t M'(x\
,x2,x3)
c a n b e o b t a i n e d in t e r m s of the k n o w n functions etj by m e a n s of a line
integral a l o n g a c o n t i n u o u s curve C j o i n i n g M a n d M ':
(4.10.2)
M
If the process of d e f o r m a t i o n does n o t create cracks or holes, in o t h e r
w o r d s , if the b o d y r e m a i n s c o n t i n u o u s , u\ should b e i n d e p e n d e n t of the
p a t h of integration; t h a t is, u\ should h a v e the s a m e value regardless of
General Analysis of Strain in Cartesian Coordinates
89
w h e t h e r the i n t e g r a t i o n is a l o n g curve a, b, or a n y o t h e r p a t h (Fig. 4.5).
F r o m E q s . (1.2.1), we h a v e :
dut = ^
dxj = (ey + o>gj)dxj.
(4.10.3)
(ey + 0)y)dXj .
(4.10.4)
Therefore,
U\ =
:+ f
M
I n t e g r a t i o n b y p a r t of the s e c o n d t e r m in E q . (4.10.4), yields:
M
«; = «, + WijXj
]M
+ f '
(eik - xj^)dxk,
(4-10.5)
w h e r e the d u m m y i n d e x j of ey h a s b e e n c h a n g e d to k. F r o m E q s .
(1.2.1), it c a n b e verified t h a t
H
dx
k
a n d h e n c e Eq. (4.10.5) b e c o m e s :
9 ? * =_ ^
dxj
(4.10.6)
dx
h
90
Kinematics of Continuous Media
< - + ^ f M
[.. - * { % - * * ) K
* f
<4-i^
F o r c o n v e n i e n c e , let us set:
".-.-*X^-£>
' -
<4 10
8)
T h e two first terms in the r i g h t - h a n d side of Eq. (4.10.7) are i n d e p e n d ent of the p a t h of integration. It is s h o w n in the theory of line integrals
that, for the third term to b e i n d e p e n d e n t of the p a t h , the i n t e g r a n d s
Uik
dxk
m u s t be exact differentials. Therefore, if the d i s p l a c e m e n t s u\
are to b e i n d e p e n d e n t of the p a t h of integration we m u s t h a v e :
dxj dx .
dE///
=
^Ujk
(4.10.9)
k
Now,
dxj-J^-ff^-aJ^L-^)
j \ dxj dx, dxi dx, ) \ dxj
W„ -Jlfii.-^L.)de ( d e J^L-^iL)
(4.10.10)
dXf
and
2
2
dx
k
dx
k
i(
j
dU _dU
dXj dx
k
k
\ 3 x /3 x
dxk
)
de
(4-10.11)
j
\ dxj
dx
t
)
7
jk
dxjdxk
/
de \
2
2
jl= J
1K
dxj
k
S
\ dxj dx
Therefore,
3 <?
it
2
+eu
_
d
dxjdxf
_
2,
ek
dxidxk
d
/
Jf
(4.10.12)
= 0.
Since the xf s are i n d e p e n d e n t , the necessary a n d sufficient c o n d i t i o n s
that u\ b e i n d e p e n d e n t of the p a t h of integration a r e :
de
2
dx^Xf
ilf^_i ^_i!^
2
jk
dxjdx
k
dxjdxf
dXjdx
k
= 0
(4.10.13)
T h e s e are the compatibility relations. A l t h o u g h Eq. (4.10.13) results in
81 e q u a t i o n s on a c c o u n t of its four different subscripts, only six deserve
3x,
General Analysis of Strain in Cartesian Coordinates
91
c o n s i d e r a t i o n ; the others a r e identically satisfied or a r e repetitions
resulting from the s y m m e t r y of
In detailed form, we h a v e :
2 g
d n
dx
dx
92
<?22
2 g
x
dx, dx
z 2
9 e,,
3xj 3 x 2
2
3 e 23
22
d e
3e
+ 12
dx
9x
2
9d 3e 9 e
)
3\
de
= - 9 _ ( _ ^ 33l!
dx
dx
\
9 33
dx
23 +
dx
{
dx
3
3 x 33 x |
de
_ 3 _=
dx
3
3 e 31 _
"8x33^
\2
+ <*fj2
+ +2« 23 3 \
+
dx
dxi
Ox,
2
3 8 9dx
2
7
te )
de
_3_
l2 + ^ 2 3 1 3 l
=
dx
dx
3^1
3x2
z3
Z
3
9 e2n
3 <?-,-,
2
)
+
32x
dx
2
3 e 22
2 _1_ 3 e233
3jt 3 1 3 x
2
9 %
2
3xj
+
3 e„
3JC|
Eqs. (4.10.14) a r e the necessary a n d sufficient c o n d i t i o n s for t h e strain
c o m p o n e n t s to give single valued d i s p l a c e m e n t s for a simply c o n n e c t e d
region.*
A region of space is said to be simply connected if an arbitrary closed curve lying in the
region can be shrunk to a point, by continuous deformation, without passing outside of the
boundaries.
PROBLEMS
1.
T h e d i s p l a c e m e n t c o m p o n e n t s a t t h e points of a b o d y a r e :
=
W]
:= =
C j X\ ,
14
2
C*2 X ,
2
U
3 — C3X3 .
(a)
2.
Find the components
of the strain matrix, a n d the value of
the three i n v a r i a n t s of t h e state of strain.
(b) W h a t is t h e value of t h e v o l u m e t r i c strain e^?
(c) If t h e c o n s t a n t s cl9 c2, a n d c3 a r e so small t h a t their squares
a n d p r o d u c t s a r e negligible, show t h a t t h e c o m p o n e n t s of t h e
strain m a t r i x Etj b e c o m e equal to the c o m p o n e n t s of the linear
strain m a t r i x etj.
Solve P r o b l e m 1 for d i s p l a c e m e n t c o m p o n e n t s given by
u
\
xc
~
1 2^
u2 — u3 — 0.
92
3.
Kinematics of Continuous Media
D r a w sketches showing a cubic element at a point, a n d with its
edges parallel to the reference axes, before a n d after t r a n s f o r m a t i o n .
Let
u
ux = C(2xx + x2)y
2
4.
5.
u2 = C(xx — 3*2 )> 3 = 0,
w h e r e C = 1 0 ~ , b e the expressions of the displacements of a
certain b o d y .
(a) Show the distorted shape of a t w o - d i m e n s i o n a l element of
a r e a w h o s e sides dxx a n d dx2 are initially parallel to the
c o o r d i n a t e axes; the two elements are at a p o i n t M whose
c o o r d i n a t e s are (2, 1, 0).
(b) D e t e r m i n e the c o o r d i n a t e s of M after t r a n s f o r m a t i o n .
(c) D e c o m p o s e the m a t r i x of the t r a n s f o r m a t i o n at M into its
symmetric and antisymmetric components.
(d) F i n d the angle of r o t a t i o n a n d the cylindrical dilatation of the
two elements dxx a n d dx2.
In P r o b l e m 3, c o m p u t e the strain eMN of a n element MN w h o s e
direction cosines are ( l / \ / 3 , l/\/3> l/V^)- W h a t are the principal
directions a n d the principal strains?
G i v e n the d i s p l a c e m e n t c o m p o n e n t s
2
ux = cxx(x2
6.
+ x3) ,
2
u2 = cx2(x3
u3 — cx3(xx
+
x2)
w h e r e c is a c o n s t a n t :
(a) F i n d the c o m p o n e n t s of the linear strain.
(b) F i n d the c o m p o n e n t s of the rotation.
(c) F i n d the principal elongations per unit length Ex, E2, a n d E3
at a p o i n t M whose c o o r d i n a t e s are ( 1 , 1, 1).
T h e c o m p o n e n t s of linear strain in a b o d y are given b y :
0
[ey] =
7.
2
+ xx) ,
00
00
0
:JC,
— CX
2 CX\
— cx2
cxx
0
w h e r e c is a c o n s t a n t . F i n d the principal strains a n d the principal
directions at the p o i n t ( 1 , 2 , 4).
D e t e r m i n e the volumetric strain ev for the following state of strain:
General Analysis of Strain in Cartesian Coordinates
[ . e. ] =
0.5
1
0
1
2
0.5
0
0.5
0
93
C o m p a r e the result to the unit c h a n g e of v o l u m e Ev, a n d to the first
invariant.
.
O005in.
Fig. 4 . 6
8.
9.
A plate w h o s e thickness is Vfe in. is stretched as s h o w n in Fig. 4.6.
F i n d the principal strains, ex, e2, a n d the m a x i m u m shearing strain
in the p l a t e .
I n a t w o - d i m e n s i o n a l state of strain,
6
6
6
e n = 800 X 1 0 " , e22 = 100 X 1 0 " , eX2 = - 8 0 0 X 1 0 ~ .
F i n d the m a g n i t u d e a n d direction of the principal strains, ex a n d e2,
b o t h analytically a n d t h r o u g h the use of M o h r ' s d i a g r a m . D r a w a
sketch s h o w i n g the d e f o r m a t i o n of a unit s q u a r e with edges initially
a l o n g OXx a n d OX2.
10. If
6
6
eu = - 8 0 0 X 1 0 - , e22 = - 2 0 0 X 1 0 ~ ,
= - 6 0 0 X 10" ,
6
s h o w in a suitable sketch the position of the axes with which the
m a x i m u m shearing strain is associated.
11. A r e the following states of strain possible?
94
Kinematics of Continuous Media
2
en = C(x, + x\)
e u = Cx3(xf
e22 = Cxf
e 22 =
e
e l2
= 2Cxj.x
2 e
33 = *13 = 23 =
0
e
i2
+ xf)
Cx\x3
= 2CXJ.X2.X3
0
*33 = *13 = ^23 =
C is a c o n s t a n t .
12. Show b y differentiation of the strain-displacement relations (4.10.1)
t h a t the compatibility relations (4.10.4) are necessary c o n d i t i o n s for
the existence of c o n t i n u o u s single-valued d i s p l a c e m e n t s .
13. Establish by differentiation a set of compatibility relations involving
b o t h the e~ s a n d the co^' s.
CHAPTER 5
CARTESIAN TENSORS
5.1
Introduction
A tensor is a q u a n t i t y which describes a physical state or a physical
p h e n o m e n o n a n d which is invariant, i.e., r e m a i n s u n c h a n g e d w h e n the
frame of reference within which the q u a n t i t y is defined is c h a n g e d . In
this chapter, we shall limit ourselves to cartesian frames of reference. If
the value of the q u a n t i t y at a p o i n t in space c a n b e described b y a single
n u m b e r , the q u a n t i t y is a scalar or a tensor of r a n k zero; if three
n u m b e r s are n e e d e d , the q u a n t i t y is a vector or a tensor of r a n k o n e ; if
nine n u m b enr s are n e e d e d , the q u a n t i t y is a tensor of r a n k two. I n
general, if 3 n u m b e r s are n e e d e d to describe the value of the q u a n t i t y
at a p o i n t in space, the q u a n t i t y is a tensor of r a n k n.
Before we p r o c e e d to o b t a i n s o m e of the i m p o r t a n t properties of
tensors, the reasons for which the tensor c o n c e p t is b e i n g i n t r o d u c e d at
this stage are w o r t h giving: (1) Since tensors are quantities describing
the s a m e p h e n o m e n o n regardless of the c o o r d i n a t e system used, they
p r o v i d e a n i m p o r t a n t guide in the formulation of the correct form of
physical laws. E q u a t i o n s describing physical laws m u s t b e tensorially
h o m o g e n e o u s , which m e a n s t h a t every term of the e q u a t i o n m u s t b e a
tensor of the s a m e r a n k . (2) T h e tensor c o n c e p t provides a c o n v e n i e n t
m e a n s of t r a n s f o r m i n g a n e q u a t i o n from o n e system of c o o r d i n a t e s to
a n o t h e r . (3) A decisive a d v a n t a g e of the use of cartesian tensors is t h a t
o n c e the properties of a tensor of a certain r a n k h a v e b e e n established,
they h o l d for all such tensors regardless of the physical p h e n o m e n a they
represent. I n the study of m e c h a n i c s , for e x a m p l e , o n e c a n n o t h e l p b u t
notice t h a t principal directions, invariants, a n d M o h r ' s r e p r e s e n t a t i o n
95
96
Kinematics of Continuous Media
a p p e a r in the analyses of strain, stress, inertia properties of rigid b o d i e s ,
a n d c u r v a t u r e of plates so t h a t there m u s t b e a b o n d c o m m o n to all:
T h e b o n d is t h a t they all are s y m m e t r i c tensors of r a n k two. I n C h a p t e r
3, it w a s m e n t i o n e d t h a t linear t r a n s f o r m a t i o n was the b o n d u n i t i n g the
quantities m e n t i o n e d a b o v e : I n d e e d , u n d e r l y i n g all the o p e r a t i o n s in
C h a p t e r 3 is the c o n c e p t of tensor. It is generally agreed t h a t this
c o n c e p t is the m o s t a d e q u a t e analytical tool for the study of d e f o r m a tion; if this is the case, t h e n w h y d i d we go t h r o u g h C h a p t e r s 3 a n d 4
w i t h o u t m e n t i o n i n g it at all? T h e first r e a s o n is t h a t the use of linear
t r a n s f o r m a t i o n m a k e s it possible to give a geometrical i n t e r p r e t a t i o n to
linear o p e r a t i o n s in a l a n g u a g e easily u n d e r s t o o d b y engineering
s t u d e n t s ; the process of d e f o r m a t i o n as expressed b y sets of linear
e q u a t i o n s is m o r e readily visualized a n d such a visualization is of
p r i m a r y i m p o r t a n c e in the study of m e c h a n i c s . T h e s e c o n d r e a s o n is
t h a t engineering s t u d e n t s d o n o t find a n y difficulty m a n i p u l a t i n g
m a t r i c e s ; such is n o t the case w h e n it c o m e s to a m a t h e m a t i c a l b e i n g
called tensor, written in a c o n d e n s e d n o t a t i o n , a n d defined t h r o u g h a
rule of c o o r d i n a t e t r a n s f o r m a t i o n . H o w e v e r , h a v i n g used it implicitly
a n d h a v i n g established s o m e of its i m p o r t a n t properties in p r e c e d i n g
c h a p t e r s , they h a v e n o r e l u c t a n c e a c c e p t i n g it as a necessary p a r t of the
study of m e c h a n i c s .
5.2
Scalars and Vectors
U n d e r a t r a n s f o r m a t i o n of c o o r d i n a t e axes, a scalar, such as the
density or the t e m p e r a t u r e , r e m a i n s u n c h a n g e d . This m e a n s t h a t a
scalar is a n i n v a r i a n t u n d e r a c o o r d i n a t e t r a n s f o r m a t i o n . Scalars are
called tensors of zero rank.
N o w consider a vector x w h o s e c o m p o n e n t s in a system of axes
OX{, OX2, OX3 are xx, x2, x3. I n a n e w system of r e c t a n g u l a r axes
OX\, OX2, OX3, the c o m p o n e n t s of x are given b y x\, x2, x3, with
x\
41*1 + 42*2 + 43*3
x'2
4i*i
x'3
41*1
(5.2.1)
x
+ hl 2
+ 43*3
or
x\ = jfyXj
(ij
=
1,2,3).
(5.2.2)
Cartesian Tensors
97
ly a r e t h e direction cosines of t h e axes of t h e n e w system with respect
to t h e o l d o n e . By definition, quantities which t r a n s f o r m a c c o r d i n g to
the relationship (5.2.2) a r e vectors o r tensors of the first rank. T e n s o r s of
the first r a n k n e e d only o n e subscript for their r e p r e s e n t a t i o n . Clearly,
the multiplication of a first-rank tensor b y a z e r o - r a n k tensor (i.e., t h e
multiplication of a vector b y a scalar) yields a first-rank tensor. T h u s ,
mx\
Also, if
= mlyXj
= iy(mxj).
(5.2.3)
a n d xt a r e t w o tensors of t h e first r a n k , t h e n
£
+ x' = lijtj
t
+ l xj
tj
= 1^% + ).
Xj
(5.2.4)
Therefore, t h e s u m of t w o tensors of t h e first r a n k is a tensor of t h e first
r a n k since a c c o r d i n g t o E q . (5.2.4) it t r a n s f o r m s as o n e .
5.3
Higher Rank Tensors
C o n s i d e r t w o tensors of t h e first r a n k , ut with c o m p o n e n t s ux, u2, w 3,
a n d Vj with c o m p o n e n t s vx, v2, v3. I n a n e w system of c o o r d i n a t e s
OX\, OX'2, OX3, t h e p r o d u c t
V
U
'j = (4 k)Vjm
u V
O
= 4 Ijm k m •
(5-3.1)
Eq. (5.3.1) c a n b e written a s :
w
{ {
! f = ik jm™km'
532
( - - )
T h e quantities Wy = u\ v'j a n d wkm= ukvm represent the general
product
of t h e first-rank tensors uk a n d vmin t h e (OX\9 OX'2, OX'3) system a n d
in t h e (OXl9
OX2,OX3)
system, respectively. H o w e v e r , p r o d u c t s of t w o
first-rank tensors a r e n o t t h e only quantities satisfying t h e rule of E q .
(5.3.2). I n general, a set of nine quantities wkmreferred t o a set of axes
a n d w h i c h t r a n s f o r m s t o a n o t h e r set a c c o r d i n g t o E q . (5.3.2) is defined
as a tensor of the second rank. F o r e x a m p l e , E q . (4.5.8), giving t h e
t r a n s f o r m a t i o n l a w of t h e c o m p o n e n t s of t h e state of strain, is of t h e
s a m e form as E q . (5.3.2). T h u s , these c o m p o n e n t s a r e t h e c o m p o n e n t s
of a tensor of t h e s e c o n d r a n k . T h e s a m e c a n b e said a b o u t t h e
c o m p o n e n t s of t h e linear t r a n s f o r m a t i o n m a t r i x a n d t h e linear strain
m a t r i x since E q s . (3.5.8a) a n d (4.9.12) a r e of t h e s a m e form as E q .
(5.3.2).
98
Kinematics of Continuous Media
In a similar fashion, a tensor of the third rank can be formed by
multiplying together three first-rank tensors or o n e first-rank tensor a n d
o n e s e c o n d - r a n k tensor. In general, however, a set of 27 quantities wrst
referred to a set of axes, a n d which transforms to a n o t h e r set a c c o r d i n g
to
is defined as a tensor of the third r a n k .
In the theory of elasticity, we shall use tensors of the fourth rank. T h e s e
tensors c a n be g e n e r a t e d by multiplying together a n u m b e r of lowerr a n k tensors which are such that the s u m of their r a n k is equal to four.
In general, however, a set of 81 quantities S m ,n which
p q transforms
a c c o r d i n g to
=
hm ^jn hp hq ^mnpq'
Sijkf
(5.3.4)
is defined as a tensor of the fourth r a n k .
5.4
O n Tensors and Matrices
T h e r e are great similarities b e t w e e n the rules governing the b e h a v i o r
of s q u a r e matrices a n d those governing the b e h a v i o r of tensors. Yet,
while a m a t r i x is n o t h i n g b u t a n a r r a y of elements a r r a n g e d in rows a n d
c o l u m n s , the c o m p o n e n t s of a tensor m u s t satisfy specific c o n d i t i o n s
w h e n passing from o n e c o o r d i n a t e system to a n o t h e r . F o r example, a
tensor of the second r a n k , wik
, c a n be symbolically represented by a
s q u a r e m a t r i x of the third o r d e r :
n
w
2\
w
w32
w3l
w 13
^23
w
(5.4.1)
33
b u t n o t every square m a t r i x is the matrix of an even-rank tensor. In Sec.
5.2, we h a v e seen that a d d i n g two tensors a m o u n t s to a d d i n g their
c o r r e s p o n d i n g c o m p o n e n t s to o b t a i n a n o t h e r tensor, a n d that the
multiplication of a tensor by a scalar a m o u n t s to multiplying each of its
c o m p o n e n t s b y the s a m e scalar; the s a m e is true for matrices, as we
h a v e seen in C h a p t e r 2. I n Sec. 2.4, it was s h o w n that a square m a t r i x
c a n b e split into a s y m m e t r i c a n d a n a n t i s y m m e t r i c c o m p o n e n t ; the
s a m e can b e said a b o u t a s e c o n d - r a n k tensor: If, in Eq. (5.3.2), we
i n t e r c h a n g e / a n d j, we get:
Cartesian Tensors
w
ji
w
=
99
(5.4.2)
ljk'im -km '
Since k a n d m are r e p e a t e d indices—in other w o r d s , d u m m i e s — t h e y
c a n be i n t e r c h a n g e d . T h u s ,
^ji
(5.4.3)
ym hk ^mk '
, a n d is
wmk is seen to t r a n s f o r m a c c o r d i n g to the s a m e rules as wkm
therefore a tensor of the s e c o n d r a n k . T h e tensor wmk is said to b e the
conjugate of wkm
. T h e two sets \{wkm + wmk
) a n d \{wkm - wmk
) are also
tensors of the s e c o n d r a n k . T h e set \(wkm + wmk
) is u n a l t e r e d if k a n d
m are i n t e r c h a n g e d , a n d is called a symmetric
tensor. T h e set
\(wkm
— wmk
)
has its c o m p o n e n t s reversed in sign w h e n k a n d m are
i n t e r c h a n g e d , a n d is called a n antisymmetric
tensor. T h e s u m of the two
sets is equal to wkm
. T h u s , we c a n consider a n y tensor of the s e c o n d
r a n k as the s u m of a s y m m e t r i c tensor a n d of a n a n t i s y m m e t r i c o n e .
Since the a n t i s y m m e t r i c tensor h a s only three c o m p o n e n t s , it c a n b e
associated with a vector. (This, in fact, was d o n e in the study of
a n t i s y m m e t r i c t r a n s f o r m a t i o n s in C h a p t e r 3.)
T h e general p r o d u c t of t w o first-rank tensors c a n also b e p r e s e n t e d in
m a t r i x form. F o r e x a m p l e , the general p r o d u c t ukvm = wkm of the two
vectors u(ux,u29
u3)
a n d v(vx,v29
v3)
c a n b e written a s :
u
u
x
u h
2
u
3
^2
"3]
=
v
v
uv
u
\ \
\ 2
\ 3
U v
u v
u
v
2 x 2 2 2 3
u v
3 x u3 v2 u3 v3
11
w 12
13
2X
w
22
w
32
^33
w
w 31
23
(5.4.4)
a n d yields a tensor of the s e c o n d r a n k .
Finally, it m u s t b e r e m e m b e r e d t h a t a tensor is defined b y a given
formula of t r a n s f o r m a t i o n a n d is a t t a c h e d to a specific p o i n t in a given
s p a c e ; its c o m p o n e n t s are all related to a given system of c o o r d i n a t e s in
this space a n d d o n o t s t r a d d l e o n two or m o r e systems. A set of n i n e
quantities like the
' s defining the position of o n e c o o r d i n a t e system
with respect to the other, are n o t the c o m p o n e n t s of a s e c o n d - r a n k
tensor.
5.5 The Kronecker Delta and the Alternating Symbol. Isotropic Tensors
In C h a p t e r 2 we i n t r o d u c e d two s y m b o l s : Sy9 which we called the
K r o n e c k e r D e l t a ; a n d eijk
, which we called the A l t e r n a t i n g S y m b o l .
T h e s e two symbols were used to simplify the writing of s o m e e q u a t i o n s .
100
Kinematics of Continuous Media
e
S:
In the following, we shall p r o v e t h a t 8y is a tensor of the s e c o n d r a n k
while
W e know that
ijk is of the third r a n k . Let us first consider
tJ
8;j = 1 for / =j a n d 8tj = 0 for / ^ j. If 8tj was a tensor of the s e c o n d
r a n k , it w o u l d t r a n s f o r m a c c o r d i n g to Eq. (5.3.2) in a c h a n g e of
cartesian c o o r d i n a t e s . Also, in the n e w system, 8y w o u l d still b e such
that
1 for /
0 for i
=j,
^j.
(5.5.1)
F r o m Eq. (5.3.2), we w o u l d h a v e :
r
= W
km
=
-
By = (
1 for /
=j,
.^
r f
oy
552
0
(--)
T h u s , the q u a n t i t y 8^ t r a n s f o r m s into itself a n d is a tensor of the s e c o n d
r a n k . 8tj is also called the substitution tensor.
T o prove that
is a tensor of the third r a n k , we m u s t p r o v e t h a t it
t r a n s f o r m s a c c o r d i n g t o the general e q u a t i o n :
I n other w o r d s , w e m u s t p r o v e t h a t
H e r e , too, w e find b y writing the detailed form of Eq. (5.5.3)
e
=
0' w h e n t w o of
'ijk
e'ij
e'jj
k
= 1, w h e n
k
= — 1, w h e n
k are e q u a l ;
k
k
are different a n d in cycle o r d e r ;
are different a n d n o t in cycle o r d e r .
T h u s , etjk transforms i n t o itself in a c h a n g e of cartesian c o o r d i n a t e s a n d
is a t h i r d - r a n k tensor.
T h e substitution a n d the a l t e r n a t i n g tensors a r e the exceptions t o the
rule t h a t tensors m u s t describe a physical p h e n o m e n o n . T h e i r c o m p o n e n t s r e m a i n u n c h a n g e d d u r i n g a t r a n s f o r m a t i o n of c o o r d i n a t e s . A n y
tensor w h o s e c o m p o n e n t s r e m a i n u n c h a n g e d d u r i n g a t r a n s f o r m a t i o n of
c o o r d i n a t e s is called a n isotropic tensor. Such a tensor possesses n o
directional properties. Therefore, a vector c a n never b e isotropic, b u t
tensors of a n y r a n k other t h a n o n e c a n be.
Cartesian Tensors
5.6
101
Function of a Tensor. Invariants
w
q = df/dw
q
q\j = df/dwy.
V(y»ij)dw
A general p r o p e r t y of tensors is the following: If /(w^-) is a function
of a tensor
ij9 a n d if
tj
ij9 t h e n
tj is also a tensor which
c h a n g e of c o o r d i n a t e s is given b y
This property can
p r o v e d as follows:
=
km
{56l)
Also, from E q . (5.3.2), we h a v e :
4 ^jm ^km '
Multiplying b o t h sides of this e q u a t i o n b y l i l rj ,s we o b t a i n :
W=
44 U
444 jm
{
5
W
km
W
= *rJt vm km
= ^
•
562
( - - )
Thus,
(5.6.3)
and
44*^
''
(5 6 4)
w h i c h p r o v e s the p r o p e r t y .
I n t h e s t u d y of linear t r a n s f o r m a t i o n s ( C h a p t e r 3), we established t h e
existence of three invariants. T h e s a m e c a n b e d o n e for s e c o n d - r a n k
tensors. W e shall p r o v e t h a t a n y tensor of t h e s e c o n d r a n k , 7^, h a s three
i n v a r i a n t s which d o n o t c h a n g e w h e n we pass from a system of
cartesian c o o r d i n a t e s OXl9 OX2, OX3 to a n o t h e r system of c a r t e s i a n
c o o r d i n a t e s OX\, OX2, OX3. T h e s e i n v a r i a n t s a r e :
Ttt = T'„
Tljlj^ry^,
(5.6.5)
(5.6.6)
TljTjtTu^T'yTjtT'u.
Eq. (5.6.5) is easily p r o v e d b y writing:
102
Kinematics of Continuous Media
a n d setting
Eq. (5.6.6) is p r o v e d b y writing:
^ij ^j*
=
4 Tmn) =
^ ^ki
kn ®$m kl
mn
fik tin tjf $j
m^kl ^mn
mn nm
*//' *ji •
Eq. (5.6.7) is p r o v e d by writing:
^0 ^jk ^ ki ~~ i^im ^jn ^mn)(^jr hs ^rs)^kt
J
=
°mp°nr°st
^ip ^tp)
J
mn rs tp
mn *ns sm
Tij Tjk T .
ki
In Sec. 3.7, c o m b i n a t i o n s of coefficients of the linear t r a n s f o r m a t i o n s
were f o u n d invariant in a c h a n g e of c o o r d i n a t e s . T h e s e c o m b i n a t i o n s
are equivalent to those of E q s . (5.6.5), (5.6.6), a n d (5.6.7). I n d e e d , the
coefficients ay in the linear t r a n s f o r m a t i o n were shown to be c o m p o n e n t s of a s e c o n d - r a n k tensor. T h e s a m e c a n b e said a b o u t the
invariants of the state of strain in E q s . (4.9.3), (4.9.4), a n d (4.9.5). T h e y
c a n b e expressed in a form similar to E q s . (5.6.5), (5.6.6), a n d (5.6.7), as
follows:
1
1e e
1 ^
2 ij jkeki
5.7
1 i
= 3/1 - hh
+ h-
Contraction
A n o p e r a t i o n which is often d o n e in tensor m a n i p u l a t i o n s is the
o p e r a t i o n of c o n t r a c t i o n . It simply consists of setting t w o free indices
equal to each other, thus d r o p p i n g the r a n k of the tensor by two. T h e
free indices b e c o m e d u m m i e s . F o r example, a s e c o n d - r a n k tensor, 7^,
b e c o m e s Tu u p o n c o n t r a c t i o n , w h e r e
Tu=Tn
+
T22
+Tn.
Cartesian Tensors
103
In other w o r d s , it is r e d u c e d to a scalar or a tensor of r a n k zero. N o w
consider two s e c o n d - r a n k tensors, A a n d B. T h e general p r o d u c t of A
by B gives a tensor of the fourth r a n k with 81 c o m p o n e n t s ,
ATJ
Bkm
{iJ,
k,m = 1,2,3). If this general p r o d u c t of two tensors is c o n t r a c t e d , a
s e c o n d - r a n k tensor will result. This s e c o n d - r a n k tensor m a y h a v e a n y of
the four f o r m s :
,
AijBkl
AyBik
,
Ai}Bkj
,
AtjBjk.
(5.7.1)
T h e c o n t r a c t i o n s AitBkm a n d AyBkk are p r o d u c t s of the scalars Au a n d
Bkk a n d the tensors BKMa n d Ay, respectively. T h e index n o t a t i o n m a k e s
quite clear which c o n t r a c t i o n is involved; however, the matrix n o t a t i o n
is s o m e t i m e s quite useful. T h e nine c o m p o n e n t s of the p r o d u c t
AtjBjk
= Cik can b e written in m a t r i x n o t a t i o n as [A] [B]. T h e four forms of
Eq. (5.7.1) c a n t h u s b e written:
[B][A], [A]'[B], [A][B]\ [A][B].
(5.7.2)
In the study of linear t r a n s f o r m a t i o n s , we were c o n t i n u o u s l y faced
with the p r o d u c t of the t r a n s f o r m a t i o n m a t r i x [a] by the vector {OAf}.
T h e t r a n s f o r m a t i o n m a t r i x [a] was shown to be the m a t r i x of a secondr a n k tensor (see Sec. 5.3). T h e general p r o d u c t of the s e c o n d - r a n k
tensor, atJ
, a n d the vector, xk, is a t h i r d - r a n k tensor, a(jXk.
Upon
c o n t r a c t i o n , w e o b t a i n a first-rank tensor or a vector. This c o n t r a c t i o n
w o u l d either give a^Xj or aijXi (aH
xk is the p r o d u c t of a vector by a
scalar). In m a t r i x n o t a t i o n , the c o m p o n e n t s of the p r o d u c t atjXj = | , c a n
b e written [a]{x} = {£}. T h u s , it a p p e a r s that all the properties of linear
t r a n s f o r m a t i o n s studied in C h a p t e r 3 are also those of the c o n t r a c t e d or
inner p r o d u c t of a s e c o n d - r a n k tensor by a first-rank tensor. All the
subjects discussed in C h a p t e r 3 apply to a s e c o n d - r a n k tensor a n d c a n
be generalized to include higher e v e n - r a n k tensors: Existence of principal directions, characteristic e q u a t i o n s a n d eigenvalues, invariants a n d
i n v a r i a n t directions, a n t i s y m m e t r i c a n d s y m m e t r i c t r a n s f o r m a t i o n s ,
M o h r ' s d i a g r a m , etc. . . . c a n b e discussed directly within the framew o r k of tensor analysis. C h a p t e r 3 c a n practically b e r e r e a d substituting
the w o r d tensor for the w o r d matrix. Finally, it will b e recalled t h a t in
Eqs. (3.14.4) a n d (4.9.10) the m a t r i x of the t r a n s f o r m a t i o n was d e c o m p o s e d i n t o t w o p a r t s referred to as spherical a n d deviatoric; the s a m e
c a n b e d o n e with a n y tensor Tj-:
(5.7.3)
104
Kinematics of Continuous Media
8yTaa is t h e s a m e in a n y system of c o o r d i n a t e s a n d , as such, is a n
isotropic tensor. Ty c a n b e symbolically r e p r e s e n t e d b y a m a t r i x with
zero trace, a n d is called a deviator.
5.8
The Quotient Rule of Tensors
S u p p o s e we k n o w n i n e quantities ay, a n d we wish to establish
w h e t h e r they are the c o m p o n e n t s of a tensor of r a n k t w o or n o t , w i t h o u t
going to the t r o u b l e of d e t e r m i n i n g t h e law of t r a n s f o r m a t i o n . I n m a n y
cases, the q u o t i e n t rule of tensors is a c o n v e n i e n t m e t h o d to use for this
purpose.
Let xi b e a n a r b i t r a r y tensor of r a n k o n e . If the p r o d u c t ayXj is
k n o w n to yield a tensor of r a n k o n e ,
t h e n t h e ay ' s a r e t h e
c o m p o n e n t s of a tensor of r a n k t w o . T h e proof of this s t a t e m e n t is
o b t a i n e d b y m a k i n g a r o t a t i o n of c o o r d i n a t e s a n d showing t h a t ay
t r a n s f o r m s a c c o r d i n g t o E q . (5.3.2):
a
X =
'y J
#
=
4&
x
= /
a x
=
4
km m
•
(-- 0
5
8
But
so t h a t
ax
'ij )=
x
a x
582
khm km n>
( - - )
and
( ^ ~ 4 C ^ m K
= 0.
Since xt is arbitrary, then
(5.8.3)
584
<*'in = lik*nm<*km>
( - - )
which shows t h a t ay is a tensor of r a n k t w o . (This is a n o t h e r proof t h a t
the m a t r i c e s of t h e linear t r a n s f o r m a t i o n s studied in C h a p t e r 3 a r e t h e
matrices of tensors of r a n k two.) I n t h e s a m e w a y , we c a n p r o v e t h a t if
the p r o d u c t ayX(Xj
is k n o w n t o yield a tensor of r a n k zero (a scalar), ay
is a tensor of r a n k t w o . T h e q u o t i e n t rule of tensors h o l d s for tensors of
any rank.
Cartesian Tensors
105
PROBLEMS
1.
F i n d the c o m p o n e n t s of the tensor of r a n k t w o Cy = by + dy, w h e n
2
8
-3
0
4
2.
3.
-7
-3
2
6 - 2
-7
2
8
0 - 2
9
1
F i n d the c o m p o n e n t s of the tensor wtJ resulting from the general
p r o d u c t of the two tensors of r a n k o n e ^ = ( 1 , - 2 , 3 ) a n d u(
= (-2,3,4).
Verify the e — 8 identity
£
8
^ijk imn
4.
6
jm kn
Show 8 t h a t
00
tijk jk £
=
8 8
jn
8
km '
0
=
(b) £tjh ju
25//
(c) eijk
xjxk
= 0.
5. F i n d the c o m p o n e n t s of the tensor tik resulting from the c o n t r a c t e d
p r o d u c t tik = bydjk w h e n by a n d dy are the s a m e as in P r o b l e m 1.
6. Show t h a t the scalar p r o d u c t of two vectors is n o t h i n g b u t the
c o n t r a c t e d p r o d u c t of these two vectors.
7. By writing d o w n the expression of the cosine of the angle b e t w e e n
two lines w h o s e direction cosines are /, a n d mh show t h a t 8y is a
tensor of r a n k two.
8. S h o w t h a t the vector p r o d u c t ct of t w o vectors at a n d bt c a n b e
written as ct = eijkaj bk.
9. Let the three vectors ai9 bt, a n d ct form the three edges of a
parallelepiped. By writing the expression of the v o l u m e , s h o w t h a t
Eyk is a tensor of r a n k three.
10. G i v e n the tensor of r a n k two
2
-2
3
1
1
1
1
3
-1
106
Kinematics of Continuous Media
(a)
(b)
F i n d its s y m m e t r i c a n d a n t i s y m m e t r i c c o m p o n e n t s .
F i n d the invariants a n d the principal directions of its s y m m e t ric c o m p o n e n t .
(c) D e c o m p o s e the s y m m e t r i c c o m p o n e n t into its isotropic a n d
deviatoric c o m p o n e n t s , W h a t are the principal directions of the
deviatoric c o m p o n e n t ?
CHAPTER 6
ORTHOGONAL CURVILINEAR
COORDINATES
6.1
Introduction
In m a n y p r o b l e m s of m e c h a n i c s , g e o m e t r y suggests the use of
nonrectilinear c o o r d i n a t e systems. F o r e x a m p l e , it seems n a t u r a l to
study the m e c h a n i c a l b e h a v i o r of a x i s y m m e t r i c a n d spherical objects
using cylindrical a n d spherical c o o r d i n a t e s , respectively: T h e formulation of the p r o b l e m s a n d their solutions are substantially simplified.
T h e s e t w o systems are special cases of the general curvilinear c o o r d i n a t e
systems.
T h e a i m of this c h a p t e r is to p r e s e n t in a simple w a y the basic
o p e r a t i o n s involved in the use of o r t h o g o n a l curvilinear c o o r d i n a t e s .
T h e expressions of such quantities as gradient, divergence, curl, a n d
L a p l a c i a n are o b t a i n e d , a n d the general expressions of the c o m p o n e n t s
of the strain tensor a n d the s t r a i n - d i s p l a c e m e n t relations a r e e s t a b lished. T h o s e expressions a n d relations are c o n t i n u o u s l y referred to in
future c h a p t e r s . Since we h a v e limited ourselves to o r t h o g o n a l systems,
the n o t i o n s of c o v a r i a n c e a n d c o n t r a v a r i a n c e , Christoffel's s y m b o l , a n d
R i e m a n n ' s tensor, c o m m o n l y used in the study of curvilinear coordinates, are n o t n e e d e d . T o avoid u n n e c e s s a r y difficulties, they will
neither b e i n t r o d u c e d n o r defined.
6.2
Curvilinear Coordinates
Let us refer a region of space to a set of o r t h o g o n a l cartesian axes
OXx, OX2, OX3. T h e c o o r d i n a t e s of a n y p o i n t P in the space are
107
108
Kinematics of Continuous Media
x
x
ew
m a
•*i> 2> 3 (Pig- 6-1)- ^
k e a t r a n s f o r m a t i o n of c o o r d i n a t e s from
this cartesian system to a n o t h e r system, functional relations b e t w e e n the
two m u s t b e given. Let these relations b e :
x
= y\( \
y\
>*2>*3)
yi = ^ 2 ( ^ 1 ^ 2 ^ 3 )
(6.2.1)
y* = ^ 3 ( ^ 1 ^ 2 ^ 3 ) -
W e shall a s s u m e that the f u n c t i o n s ^ ( x j ,x2,x3)
are single valued a n d
c o n t i n u o u s l y differentiable at all points of the region, a n d t h a t Eqs.
(6.2.1) c a n b e solved to yield the inverse t r a n s f o r m a t i o n :
x
\ =
^1(71^2^3)
*i
x
= x (y\^2^3)
3
=
2
622
x
( - - )
3(y\^2^31
in which t h e functions xi(yx,y2,y3)
are single valued a n d c o n t i n u o u s l y
differentiable with respect to the variables y t . T h e passage from E q s .
(6.2.1) to (6.2.2) a n d vice versa requires t h a t the j a c o b i a n 1 3 ^ / 9 ^ 1 ^ 0.
C o o r d i n a t e t r a n s f o r m a t i o n s with the a b o v e properties are called a d m i s sible t r a n s f o r m a t i o n s . If we set y x = cx in Eqs. (6.2.1), where cx is a
c o n s t a n t , the e q u a t i o n
= c,
(6.2.3)
y (x ,x ,x )
2 x23
= c
(6.2
y (x ,x ,x )
= c
y (x ,x ,x )
x x23
represents a surface Sx.
Similarly,
2
A)
and
3 x23
3
(6.2.5)
represent surfaces 52 a n d S3. T h e s e surfaces (Fig. 6.1) intersect at the
p o i n t w h o s e c o o r d i n a t e s a r e o b t a i n e d b y solving Eqs. (6.2.3), (6.2.4),
a n d (6.2.5). T h e surfaces St are called the c o o r d i n a t e surfaces a n d their
intersection pair b y pair are the c o o r d i n a t e lines Yx, Y2 a n d Y3. T h e Yx
c o o r d i n a t e line is the intersection of the two surfaces y 2 = c2 a n d
y 3 = c3. A l o n g this line, the only variable that changes is j ^ . Similarly,
Orthogonal Curvilinear Coordinates
109
Fig. 6.1
a l o n g the Y2 a n d the Y3 c o o r d i n a t e lines the only variables t h a t c h a n g e
are y2 a n d j> 3, respectively. By c h a n g i n g the values of the c o n s t a n t s
q , c 2, a n d c 3, o t h e r p o i n t s such as Q a n d R c a n b e l o c a t e d in the Yt
c o o r d i n a t e system.
A s a n e x a m p l e , consider the transformation
to a cylindrical
coordinates
system with variables
= r
y\
-> yi = 0> ^ 3 = -
z
(62.6)
en e9 a n d ez are the three-unit vectors in the radial, tangential, a n d axial
directions (Fig. 6.2).
F i g . 6.2
110
Kinematics of Continuous Media
Eqs. (6.2.2) are written:
xx = r cos 6,
x2 = r sin 6,
x3 = z.
(6.2.7)
T h e inverse of Eqs. (6.2.7) is:
a n d is single valued for 0 < 0 < 2H a n d r > 0. T h e surface r = cx is a
circular cylinder xf 4- x\ = cf w h o s e axis coincides with the OA^ axis
(Fig. 6.2). T h e surface 6 = c2 is the p l a n e x2 = xx tan c2 c o n t a i n i n g the
OX3 axis. T h e surface z = c 3 is the p l a n e x3 = c3 p e r p e n d i c u l a r to the
OX3 axis.
As a n o t h e r e x a m p l e , consider the transformation
to a spherical polar
coordinates system with variables (Fig. 6.3)
629
y\
= <t>, yi
= o>
a n d the three unit vectors e^, e9, a n d
Eqs. (6.2.2) are written:
xx = p sin
cos 0,
y3 = P>
ep.
x2 = p sin </> sin 0,
T h e inverse of Eqs. (6.2.10) is:
( - - )
x 3 = p cos <j>. (6.2.10)
Orthogonal Curvilinear Coordinates
111
a n d is single valued f o r p > 0 , 0 < < / > < n , 0 < # < 211. T h e surface
p = c, is a sphere. T h e surface <j> = c2 is a cone. T h e surface 9 = c3 is
a plane. T h e c o o r d i n a t e lines are the meridians, the parallels, a n d the
radial lines.
Consider a scalar function U defined in a cartesian system of
coordinates OX{, OX2, OX3, as well as in a curvilinear system of
c o o r d i n a t e s Yx, Y2, Y3. Let the functional relation b e t w e e n the t w o
systems [Eqs. (6.2.1), (6.2.2)] b e k n o w n . W e h a v e :
d U = W-dy
a
Since each of >>,, .y 2, ° d
+
ILL
dy
+ W dy
= W dy
.2.12)
is a function of xx, x2, x3, therefore,
9^ = 9JZ^.
dxj
dyj dxt
(6.2.13)
W e thus h a v e the following relation for the o p e r a t o r d/dx( :
dx;
(6.2.14)
dx; dyj '
T o find the s e c o n d derivatives, we write:
2
_3_
9
dxy \ dx )
t
dxm dxt
= -*-(
±\
dxm \ 3.x, dyj /
(6.2.15)
^
(6.2.16)
Therefore,
d2
2
3
3x
yj
wdx;
3
dxm dxt dyj
9>i
o*
dxm dx( dysdyj
F o r example, in the passage to cylindrical c o o r d i n a t e s we use t h e
functional relations expressed b y E q s . (6.2.6) to (6.2.8), a n d o b t a i n :
_3
a 3 _ _ sin
_S _ 9_ 3
C=0 _ O
_3
cos 9 ^3
= s • i na d^ +. —
s^i M \
(6.2.17)
9 i ox
(6.2.18)
c
(6.2.19)
112
Kinematics of Continuous Media
2
3 2
3 x
3
2
2_
2
2
___ _ 2C
„ 3
°
2
2 sin 9 cos 9 3 + , 2 sin 02 cos 0 3
'
drW
30
3r
2
2
2
3x 2 "
2
, 2 sin fl cos
3r
A3
'
3r '
/-
2 sin
3r30
2
2
2
2
r
, cos 3 , cos 0 3 2
2
90^
r
2
, s i n 0 3 . s i n20 3 2
' 23r
,30
„:.2/i 3
f6 9
02 cos
r
0 3
(6 2
21)
30
30
3
3x, 3 x 9
f 2k - (
3x
c
o
^ - ™ )
s
<6
''
2
23)
3
Similar relations c a n b e written for spherical c o o r d i n a t e s .
6.3
Metric Coefficients
In a c o o r d i n a t e system, the m o s t i m p o r t a n t thing to k n o w is h o w to
m e a s u r e lengths. T h i s i n f o r m a t i o n is given by the metric coefficients.
b e a n y p o i n t referred to a set of cartesian axes OXt.
Let Px(xx,x2,x3)
T h e position vector r of Px is w r i t t e n :
r = xlll+
x2l2
+ x 37 3,
(6.3.1)
w h e r e l u z 2, i3 are the unit b a s e vectors (Fig. 6.4). T h e s q u a r e of the
e l e m e n t of a r c ds a l o n g a curve c b e t w e e n Px(xx,x2,x3)
and
P2(xx
4- dxx,x2 + dx2,x3 + dx3) is given b y
2
(ds)
2
= (dxx)
2
4- (dx2)
2
4- (dx3)
= dx{dx{.
(6.3.2)
T h e vector dr b e t w e e n Px a n d P2 is given b y
df = dxxix
4- ^ / x 2/ 2 4- dx3i3.
(6.3.3)
Orthogonal Curvilinear Coordinates
Therefore,
113
2
(ds)
(6.3.4)
= df • dr.
This scalar p r o d u c t is, of course, a n i n v a r i a n t i n d e p e n d e n t of the
c o o r d i n a t e system used. In a n e w system Yj, Y2, Y3, defined b y
yiR = ^ /( J C 1, X 2, X 3) ,
the two p o i n t s Px a n d P2 h a v e c o o r d i n a t e s Px(yx,y2,y3)
+ dyx,y2 + dy2,y3 + rfy3) a n d
(6.3.5)
and
P2(yx
T h e s y m b o l 3 r / 3 / / d e n o t e s the derivative of r with respect to a
p a r t i c u l a r variable yt(i = 1,2,3) w h e n the r e m a i n i n g variables are held
c o n s t a n t . T h u s , if we fix the variables y2 a n d y3, r b e c o m e s a function
of yx a l o n e a n d the t e r m i n u s of r is c o n s t r a i n e d to m o v e along the Yx
c o o r d i n a t e line in the Yt c o o r d i n a t e system d e t e r m i n e d by Eq. (6.3.5).
C o n s e q u e n t l y , the vector dr/dyx is t a n g e n t to the c o o r d i n a t e line Yj.
Similarly, the vectors dr/dy2 a n d 3773y 3 are t a n g e n t to the Y2 a n d Y3
c o o r d i n a t e lines, respectively (Fig. 6.4). If we d e n o t e these vectors by ah
so t h a t
114
Kinematics of Continuous Media
*>~wr
(6 3 7)
--
df = atdyiK
(6.3.8)
t h e n from (6.3.6)
a n d Eq. (6.3.4) c a n b e written:
2
(ds)
• (ajdyj) - a,. • ajdy^yj.
= (atdyi
)
(6.3.9)
If we n o w define the scalar p r o d u c t a, • ~dj b y
gtj = a, • a, = a, • a, = g,,,
we c a n write E q . (6.3.9) as
2
(ds)
E x p a n d e d , this quadratic
2
(ds)
2
= gn(dy})
+
(6.3.10)
differential
2
+ g11
(dy1)
# i 324>i ^ 3 + 8i3
(6.3.11)
= gijdyidyj.
form r e a d s (see Sec. 3.12):
+ feC^) + 2g rfv rfv
2
12
2dy
2
dy .
3
1
2
(o.J.
)
12
T h e coefficients gy are called metric coefficients. A s c a n b e seen from
Eq. (6.3.12), they are the link b e t w e e n the length of a n e l e m e n t a n d the
differentials dyt. T h e s e coefficients are the c o m p o n e n t s of a s e c o n d - r a n k
Orthogonal Curvilinear Coordinates
115
tensor called the metric tensor (see Sec. 5.8). T h e vectors at, which were
f o u n d to b e t a n g e n t to the c o o r d i n a t e lines Yt at a given p o i n t P, are
called b a s e vectors in the curvilinear system Yt. A n y vector D (Fig. 6.5)
with its origin at P c a n b e resolved i n t o three c o m p o n e n t s a l o n g ax, a2,
a n d a3:
D = Dxax
+ D2a2
(6.3.13)
+ D3a3.
Dxax,
D2a2,
a n d D3a3 form t h e edges of a parellelepiped w h o s e
d i a g o n a l is D. T h u s , in the Y{ system the b a s e vectors ax, a2, a3 play the
s a m e role as the vectors ix, i2,13 play in the OXt system. However, while
the magnitude and directions of the cartesian base vectors are fixed, the
base vectors a(, in general, vary from point to point in space. F o r e x a m p l e ,
p o i n t B (Fig. 6.5) is defined b y the vector 1 a n d a set of c o n s t a n t s
cx, c2, c3 [see E q s . (6.2.3) to (6.2.5)] different from the ones used to
locate p o i n t P. T h e b a s e vectors at B c o u l d b e called bt a n d a r e given
b y dl/dyt; they a r e different from at in b o t h m a g n i t u d e a n d direction.
T h e c o m p o n e n t s of the metric tensor at B are therefore different from
the c o m p o n e n t s of the m e t r i c tensor at P.
F r o m the definition of E q . (6.3.10), we see o n setting / = j = 1 t h a t
the length of ax is
(6.3.14)
Similarly, for i =j
= 2, 3, we get:
(6.3.15)
(6.3.16)
1^3 I = Vfttf •
T h e s e vectors are o r t h o g o n a l if, a n d only if,
g\2 = 8i\ =
5
• a2 = 0
#13 = £ 3 1 = i * ^3 =
0
0
(6.3.17)
^23 = ^32 = ^2 ' ^3 =
A curvilinear system for w h i c h these relations h o l d is called o r t h o g o n a l .
In such a system:
2
(ds)
2
= gxx
(dyx)
2
+ g22
(dy2)
2
+ g33
(dy3) .
(6.3.18)
116
Kinematics of Continuous Media
a n <ew
T o o b t a i n the m e a n i n g of the coefficients gxx
, g22
,
^ £33
chose the
element ds a l o n g the Yx c o o r d i n a t e line. Therefore, dy2 a n d dy3 are
equal to zero since y 2 a n d y 3 d o n o t c h a n g e along the Yx line. In this
case, Eq. (6.3.12) gives:
2
2
(ds )
=
x
g (dy ) .
xx x
Therefore,
(6.3.19)
dsx = V^dyx.
T h u s , the length of the element of arc dsx along the Yx c o o r d i n a t e line
is o b t a i n e d b y multiplying the differential of y x b y \fg^\. Similarly, the
differentials of a r c a l o n g the Y2 a n d Y3 c o o r d i n a t e lines are (Fig. 6.4):
=
V#22 dy2
(6.3.20)
ds3 = V^33 dy3
9
(6.3.21)
d*2
9
Since b o t h the dst s a n d t h e dyi s are real, w e c o n c l u d e t h a t :
* n > 0,
g
22 > 0,
g 33 > 0.
If ex, e2, e3 are u n i t vectors a l o n g ax, a2, a3, t h e n Eq. (6.3.13) c a n b e
written:
D
D = *\Vgn
D
\ + eiVgn
D
2 + hVto
3-
(6.3.22)
I n case of a n o r t h o g o n a l system of c o o r d i n a t e s in which the n o r m a l
projections of D o n the vectors at are dx, d2, a n d d3, Eq. (6.3.22) is
written:
D = ex dx + e2 d2 + e 3J 3
(6.3.23)
and
Di =
An
element
of volume
(no s u m o n i) .
dv in general
curvilinear
(6.3.24)
coordinates
the triple scalar p r o d u c t :
dv = \a
x - a2Xa3
\dy
x
dy dy .
23
is given b y
117
Orthogonal Curvilinear Coordinates
F o r a n o r t h o g o n a l system, this e q u a t i o n r e d u c e s t o :
dv
= Vgii£22g33
dy{ dy2 dy3.
(6.3.25)
W h e n a curvilinear c o o r d i n a t e system Yt is d e t e r m i n e d b y a relation
such as E q . (6.3.5), the inverse is written xt = xi(y{9
y2,y3)
and
2
But, in cartesian c o o r d i n a t e s , (ds)
<*>
=
therefore,
= dx^dx^
v ^ r ^ -
=
(6
-3
26)
C o m p a r i n g E q s . (6.3.11) a n d (6.3.26), w e c o n c l u d e that
„.. = J ^ ^ L .
(6.3.27)
This f o r m u l a allows o n e to calculate the m e t r i c coefficients.
Example
1. Case of Cylindrical
Coordinates
F r o m E q s . (6.2.6), (6.2.7), a n d (6.3.27), we o b t a i n :
«" -
+
(If )
2 +
(If) - °
*» - ( I f )
2 +
(if)
2 +
(if)
2
c
s 2
- ^
2
*
+ s m H +
^
+
° °
+
1
<6
-"
--
8)
2
(6.3.29)
1
+ ,
* - ( ^ )
( ^ )
#12 = #13 = £23 =
°-
T h e expression for (ds)
2
(ds)
+
l
( f e )
+0
-
+
' - '
(6,,0,
2
is
2
2
= g^-rf^rfjj- = (dr)
2
2
+ r (^/<9) + (</z) .
(6.3.31)
T h e e l e m e n t of v o l u m e is given by
dv = rdrdOdz
(6.3.32)
118
Kinematics of Continuous Media
Example
2. Case of Spherical
Polar
Coordinates
F r o m E q s . (6.2.9), (6.2.10), a n d (6.3.27), we o b t a i n :
2
2 2
Su = P >
S22 = P sin <£,
£ 1 2 = £ 1 3 = £23 =
2
T h e expression for (ds)
2
(ds)
is
2
= (dp)
2
(6.3.33)
g 33 = 1
(6.3.34)
°-
2
2
2
2
+ p (d<j>) + p sin <K</0)
(6.3.35)
T h e e l e m e n t of v o l u m e is given b y
2
dv = p s i n <j)dpd<j>d0.
(6.3.36)
T o simplify the writing, it is s o m e t i m e s c o n v e n i e n t for o r t h o g o n a l
systems to write:
9
gu = hf ( n o sum),
(6.3.37)
w h e r e the ht s are called the scale factors.
Remark
Spaces in which it is possible to c o n s t r u c t a c o o r d i n a t e system such
t h a t the q u a d r a t i c differential form (6.3.11) reduces to a s u m of squares
of the c o o r d i n a t e differentials are called E u c l i d e a n spaces. I n o t h e r
w o r d s , in a E u c l i d e a n space the length of a line s e g m e n t c a n always b e
given by the formula of P y t h a g o r a s . If o n e specifies the rule for the
m e a s u r e m e n t b e t w e e n p o i n t s (i.e., if o n e specifies gy), the space is called
metric.
Spaces in w h i c h n o c o o r d i n a t e system c a n b e found such t h a t the
formula of P y t h a g o r a s c a n b e applied to the length of a line segment,
are called N o n - E u c l i d e a n spaces. A surface represents the only variety
of N o n - E u c l i d e a n space c a p a b l e of actual visualization. E l e m e n t s of the
t h e o r y of surfaces will b e p r e s e n t e d in C h a p t e r 18 within the scope of
the theory of thin shells.
6.4
Gradient, Divergence, Curl, and Laplacian in Orthogonal
Curvilinear Coordinates
A s s u m e a n o r t h o g o n a l curvilinear c o o r d i n a t e system defined by
119
Orthogonal Curvilinear Coordinates
w h e r e the variables xt a r e cartesian. Since yt a r e o r t h o g o n a l , t h e n
2
(ds)
2
= gxx
(dyx)
2
2
+ g22
(dy2)
+ g33(^3) -
(
6
A
)
W e d e n o t e the u n i t b a s e vectors a l o n g the t a n g e n t s to t h e Yt axes at P
by eX
y e2, a n d e3, a n d express a vector M at P b y
M = exmx
+ e2m2
+ e3
ra3.
(6.4.2)
m j , ra2. a n d m3 are the o r t h o g o n a l projections of M o n the u n i t b a s e
vectors.
Fig.6.6
T h e v o l u m e e l e m e n t dv, f o r m e d b y t h e c o o r d i n a t e surfaces yt =
c o n s t a n t a n d yt + dyt = c o n s t a n t (Fig. 6.6), h a s the s h a p e of a curvilinear p a r a l l e l e p i p e d with edges dsi = ^/g^ dyt ( n o s u m o n /). T h e a r e a s of
its faces a r e given b y
dAX2 = VS\\822
dyx dy2
(6.4.3)
dAX3 = V s n £ 3 3 dyx dy3
(6.4.4)
dA23 = Vg^g^
(6.4.5)
dy2dy3,
1
120
Kinematics of Continuous Media
a n d its v o l u m e is
dv
= VSn £ 2 2 ^ 3 3
d
d
<ty\ yi y?>
(6A.6)
•
To compute the divergence of a vector M at a point
definition of the divergence at a p o i n t :
divM = lim N-
J I M • ndA
P, we use the
,
(6.4.7)
w h e r e « is the o u t w a r d unit vector n o r m a l to dA. T h e c o n t r i b u t i o n to
the integral ffA M - ndA t h r o u g h the area PFGH in the direction of the
while t h a t t h r o u g h J COB is
o u t w a r d n o r m a l is — mxh2h3dy2dy3,
mxh2h3dy2dy3
+ d/dyx(mxh2h3)dy2dy3dyx.
F r o m these, a n d the corres p o n d i n g expressions for the other two pairs of surfaces, we h a v e :
M • ndA
A
=
a
oyx
w—(mlh2h3)dy]dy2dy3
+ ^-(m2hlh,)dyldy2dy3
(6.4.8)
hh
+
-^{^ \ i¥y\dy2dyz.
F r o m Eqs. (6.4.6), (6.4.7), a n d (6.4.8), we o b t a i n :
(6.4.9)
7 b compute the curl of a vector M at a point P, we use Stokes' t h e o r e m :
<f)M
-~ds = j
curXM • ndA,
(6.4.10)
w h e r e f is the integral t a k e n a l o n g a closed c o n t o u r . T h e first c o m p o n e n t of the curl of M is o b t a i n e d b y a p p l y i n g Stokes' t h e o r e m to the
surface PFGH (Fig. 6.6):
121
Orthogonal Curvilinear Coordinates
F
r
WM~ds=\
J
G
r
r
M ~ds +
JP
rn
I
M ~ds +
\
JF
rP
M ~ds +
JG
\
M -ds
JH
6
= m2 h2 dy2 + \m3 h3 dy3 + ^ - (m3 h3 )dy2 dy3 J
A
(
)
- [m2 h2 dy2 + ^ - (m2 h2 )dy2 dy3 J - m3 h3 dy3
= \j^{m3h3)
-
^(m2h2)\dy2dy3.
By Stokes' t h e o r e m , Eq. (6.4.11) is e q u a l to the first c o m p o n e n t of the
curl a l o n g ex multiplied b y the a r e a PFGH . H e n c e ,
(curl AOi - j^[^-2(m3h3)
_9_
dy3
-
^{m2h2)\.
(6.4.12)
T h e t w o o t h e r c o m p o n e n t s a l o n g e2 a n d a l o n g e3 c a n b e o b t a i n e d b y a
similar r e a s o n i n g or directly b y a cyclic p e r m u t a t i o n of the indices.
Thus,
curl M
1 Vd(h3m3)
h2h3
dy2
L
•p
-
1
h3
1
L
8 ( / ! 2m 2) " |
J
3y3
3(/j 3 m3)
3/3
dyx
Vd(h2m2)
d(hxmx)
h2 L
dy2
3>>i
T h i s expression c a n b e written a s :
curl M =
1
hxh2h3
hxex
h2e2
_3_
9yi
Aj^!
_9_
3^2
h2m2
h3e3
_9_
9
(6.4.13)
^ 3
h3m3
To compute the gradient of a scalar U(yx ,y2,y3),
we use the definition
of the g r a d i e n t . T h u s , the c o m p o n e n t in the direction of ex w h i c h is
t a n g e n t to the e l e m e n t PO is (Fig. 6.6):
1
1
122
Kinematics of Continuous Media
(grade/), -
lin,
~
- J ™ .
a n cn a
T h e t w o o t h e r c o m p o n e n t s a l o n g e2
reasoning. T h u s ,
g6 ^ t f - j ^
d ^3
(6.4,4)
b e o b t a i n e d b y a similar
+ jM^ + j ^ .
(6.4.15)
/*! 9yj
A 2 dy2
h3 dy3
I n vector analysis a certain vector differential o p e r a t o r V (read del or
n a b l a ) , defined in o r t h o g o n a l cartesian c o o r d i n a t e s b y
7
-^
+ i
>4
+ ?
>i5?
< 6 A 1 6 )
plays a p r o m i n e n t role. T h e g r a d i e n t of a scalar U is written:
i
™
-
+
^
^
!+
3 f ;
< " • » >
a n d the divergence of a vector M is written:
V •M = ^
+ 3-2- + 3-^-.
OXj
OX
(6.4.18)
OX3
2
T h e L a p l a c e e q u a t i o n is written:
2
V . Vtf = | ^ + f - ^ + f - ^ =
OXj
0-^2
V £/=0.
(6.4.19)
^-^3
C o m p a r i n g E q s . (6.4.15) a n d (6.4.17), we see t h a t in curvilinear o r t h o g o n a l c o o r d i n a t e s the differential operator V is written:
V = ? I 3 h- + 3 h
/z, 9yi
/ i 23 / 2
I n a s m u c h as
9
+ 3/ 3
"3
2
div(grad U) - V • V t / = V 1 / ,
2
/Ae expression for V U
E q . (6.4.9). T h u s ,
2
V72U r / _=
(6.4.20)
(6.4.21)
is o b t a i n e d b y substituting V t / instead of M in
1
/ii ^ 2^3 L9yi V hx 3 / , /
3/2 V h2 dy2/
+ J L (
3t/Y|
3^3 V ^ 3 3 / / J"
3
^6
4) 2
2
Orthogonal Curvilinear Coordinates
123
9
e
r
Fig. 6.7
Example
1 . Cylindrical
Coordinates
(Fig. 6.7)
I n this case, we h a v e (see Sec. 6.3):
g\\ = 1
hx = 1
1
g 33 = 1
gi2 = r
h2 = r
h3 = 1
(6.4.24)
Therefore, the divergence of a vector M(mr,m6,mz)
,.
(6.4.23)
1 a
1 3m fl
is given b y :
3m7
(6.4.25)
T h e c o m p o n e n t s of the curl of a vector M are given b y :
, _
i 3m7
dm*
(curl Af), = 7-9/ - -37->
(curlA7), =
,
, _
3mr
(^rl M)9= -gf
3w2
~3T
(6.4.26)
i [ ^ ( r m f ) l- ^ ] .
T h e c o m p o n e n t s of the g r a d i e n t of a scalar U(r,6,z)
are given b y :
w
(grad U\
= ^ ,
(grad {/), = i ^ ,
(grad U\
= ^
.
(6.4.27)
T h e L a p l a c i a n of a scalar £/(/-, 0, z) is given by
r
2
3r\
3r /
2
30
8 i t2/
r 3z
-
(6.4.28)
124
Kinematics of Continuous Media
Fig. 6 . 8
Example
2. Spherical
(Fig. 6.8)
Polar Coordinates
In this case, w e h a v e (see Sec. 6.3):
2
2
2
8\\ = P
= P sin 4>
g33= 1
h\ = P
h2 = p sin <J> h3 = 1
Therefore, t h e divergence of a vector M(m^,m9,mp)
2
2
v
J
(6.4.29)
(6.4.30)
is given b y :
p
div M = - L f ( p m p) + - 4 — - l - K s i n <#>) + — J —
p 3p
psin<f>94> *
p sin <J> o0
T h e c o m p o n e n t s of t h e curl of a vector M(m^,,mg,mp)
lT 1
(curl M ) p=
9
W
(6.4.31)
a r e given b y :
P 3(pmg)1
L^ "
(m
Sin
^-^ J "
T h e c o m p o n e n t s of t h e g r a d i e n t of a scalar U(<j>,0,p) a r e given b y :
d
(g-
"WW.
(grad £/)p=
r da
(s
^
= p-inUW«
( 64 3 3 )
Orthogonal Curvilinear Coordinates
125
T h e L a p l a c i a n of a scalar [/(<£, 0,p) is given b y :
2
6.5
(6.4.34)
2 12
8 C2/
p sin <£ 9 0 '
,
Rate of Change of the Vectors aj and of the Unit Vectors ej in an
Orthogonal Curvilinear Coordinate System
In Fig. 6.6, the unit vectors e x, e 2, a n d e3 (in the directions of
increasing y x , y 2 , and>> 3) b e i n g m u t u a l l y p e r p e n d i c u l a r , satisfy the s a m e
relations a m o n g themselves as the unit vectors ix, i2, a n d i3 of the
cartesian system, n a m e l y :
ex • ex = e2 • e2 = e3 - e3 = 1
(6.5.1)
*i ' e2 = e2 • e3 = e3 • ex = 0
(6.5.2)
e
x
X e
e X e
x
2
=
e ,
x
=
e
e
3
2X e 3
2X e 2
=
e
=
e,
e X e
x
3X e 3
3
=
0.
x
= e
2
(6.5.3)
(6.5.4)
Referring ourselves to Fig. 6.1, ex is n o r m a l to the surface Sx, e2 is
n o r m a l to the surface S2, a n d e3 is n o r m a l to the surface S3. F o r the
vectors ax, a2, a n d a3, we c a n also write:
ax - a2 = a2 - a3 = a3 - ax = 0.
(6.5.5)
Differentiating E q s . (6.5.5) with respect to j ^ , ^ , a n d j>2> respectively,
we get:
da
x
3
dy
da
2
dy
x
da
3
dy
2
da
• a
+
a
-a
3
+
a
- a
x+
2
x
2
a
3'
2 = 0
3
(6.5.6)
3 = 0
x
(6.5.7)
x = 0.
2
(6.5.8)
dy
da
dy
da
dy
S u b t r a c t i n g E q . (6.5.8) from (6.5.7), we get:
126
Kinematics of Continuous Media
(
da
da
x\
7
5 f - ^ J
But, from E q . (6.3.7):
as,
9a3 _
da*
+
~
s& ^
a
5 9)
^ ' ^ - 3 ^ " " -
a / a? \
' '
5 10)
--
a / a? \
3j»--M¥)'MT)'%
therefore,
^1
a n d E q . (6.5.9) b e c o m e s
= ^1
9yi
da*
(6.5.11)
9
^ 2 '
1
1
da*
„
,
c
c
a2 • Tr ~ IT ' "I = °(6.5.12)
9ji
9^2
C o m p a r i n g E q s . (6.5.12) a n d (6.5.6), a n d t a k i n g E q . (6.5.10) i n t o
account, we conclude that
1
a, •
da?
da*
da*
„
= a 2 • 3 - ^ = a* • 3 - ^ = 0.
9y3
dy
dy
x
2
5
~.
(6.5.13) t
N o w , b y differentiating,
a
x- a x
(6.5.14)
= hl
with respect t o j> 1?j>2> a n d j> 3, a n d using E q . (6.5.10), w e g e t :
da.
"\
1
3at
9j>2
da
x
dy*
=
h
3Aj
(6.5.15)
x9ji
= 3i
= 5!
Differentiating E q . (6.5.5) with respect ioyx,
(6.5.17), w e g e t :
f*L
9^2
= Al
(6.5.16)
9^3
-A,!*!-.
(6.5.17)
9yi
a n d using E q s . (6.5.16) a n d
(6.5.18)
Orthogonal Curvilinear Coordinates
1
da
dh
x
3
x
= -ax •
= -^i^a3 • ^
da
127
c I Q\
(6.5.19)
. T w o sets of e q u a t i o n s similar t o E q s . (6.5.18) a n d (6.5.19) c a n b e
derived b y differentiating a2 • a2 a n d a3 - a3. T h e s e relations c a n ,
however, b e directly written d o w n b y cyclical p e r m u t a t i o n of E q s .
(6.5.18) a n d (6.5.19):
a3
da<>
,
U
da>x
3
l
—
*2
' 9^2
' 9j>
9^: 3
2
da
da
Tr±2 = -a2 • ^ x = -
9j>
9j
2
2 .5.20)
(6
dh
.
dh
A s o n
2 /(6.5.21)
2
9^
2
and
^ . - ^ . ^ . - / A
9^3
^
9j>
a
2
3
J (6.5.22)
9^!
9^3
- ^ . ^ = - / ,
=
9j
3
3^ 3 .
9j>
J
(6.5.23)
2
H a v i n g established t h e previous relations, t h e derivatives of at c a n
now be obtained.
C o n s i d e r t h e vector dax /dyx;
its three c o m p o n e n t s a l o n g e x, e 2, a n d
e are
3
^L.g,
8y,
"
e
3y,
2
* '
! ? L . ?,
3y, >
e
(6.5.24)
or
.
9^i
9yi
^3
<^h_.
^1.^2
^i '
^ 2 '
9y
(6.5.25)
1 / * 3'
which b y virtue of E q s . (6.5.15), (6.5.18), a n d (6.5.19) a r e e q u a l to
9/*i
x '
dy
Jh^Jh
/z 29 / 2
A 3dy3
(6.5.26)
'
H e n c e , w e m a y write:
(6.5.27)
128
Kinematics of Continuous Media
R e p e a t i n g the previous steps, two e q u a t i o n s similar to Eq. (6.5.27) c a n
b e derived: T h e y can, however, b e directly written d o w n by cyclical
p e r m u t a t i o n of E q . (6.5.27):
dh
2
2
dy
"
3^2
2
a
dh
3
3
dy
"
3^3
3
a
da
2
h
dh
2 3_
dh
2
2
(6.5.28)
~ / | 3 ^
h
da
I n the s a m e way,
h
2
h
3
3
h
3
h
dh
dh
3
3
3
(6.5.29)
c a n b e resolved into the three c o m p o n e n t s :
da /dy
2 3
3/3
3
? « 2 . g ,2
3/3
* '
?I
3/3
(6.5.30)
* '
which, b e c a u s e of E q s . (6.5.13), (6.5.20), a n d (6.5.23), b e c o m e :
0,
*±
dy
3
Hence,
(6.5.31)
p-.
dy
2
d
3f*2
_h. ^1 + ^hh.
=
3^3 h
2 fy2
3^3
(6.5.32)
h'
3
By cyclical p e r m u t a t i o n of Eq. (6.5.32), we o b t a i n the two
relations:
=3 / ^ « 3
dyx h 3
d**3
dyx
9^1
3y 2
T h e derivatives of
Thus,
=
3j>2h \
+
(6.5.33)
+ ^ll^l
3YI V
(6.5.34)
<^h<h
3j>3h x
c a n easily b e d e d u c e d from the derivatives of a,.
1
3/,
3/,\/!,/
3?! _
dh
2 e2
other
1
L 9yi
de
x
_ _ dh
3
e
9yi J
3 / 2/ ! 2
9
/ 3h 3
3
(6.5.36)
9V, /i! '
3/2 " S ^ V
9VJ
3e2 _ 3 7 ^
377 ~ 3 / 2 ^ 2 '
3?2 = _ A f i _ 3 A £ l
3/2
3/3 h
3/i A, '
2
9
3^_ = 9/«3 f3.
3/3
3 / 2 ^2
(6.5.37)
Orthogonal Curvilinear Coordinates
9e 3
9Vi
6.6
=
\ h
9y 3 3 '
e 9 g
3
9j2
=9^2 £ 2
3y 3 / z 3'
129
9g / z
9fi
9j>3
_^3_fj_ _
=
3^
3 2
(6.5.38)
9^ 2 h2
The Strain Tensor in Orthogonal Curvilinear Coordinates
T h e p r o b l e m f o r m u l a t e d in Sec. 1.1, a n d s t u d i e d in C h a p t e r 4 in
c a r t e s i a n c o o r d i n a t e s , will n o w b e a n a l y z e d using curvilinear c o o r d i n a t e s . I n Fig. 6.9, c o n s i d e r t h e p o i n t - t o - p o i n t t r a n s f o r m a t i o n of t h e
b o d y B to B*. Points M a n d N, w h i c h are infintesimally n e a r o n e
a n o t h e r , a r e t r a n s f o r m e d to M * a n d N*, respectively. T h e v e c t o r dx of
length ds is t r a n s f o r m e d to the vector rff of length ds* . I n a d d i t i o n to
b e i n g l o c a t e d in the c a r t e s i a n system OXx, OX2, OX3 b y xx, x2 x3, p o i n t
M is also l o c a t e d in a curvilinear system Yx, Y2, Y3 b y yx, y2, y3. P o i n t
TV is l o c a t e d b y xx + dxx, x2 + dx2, x3 + dx3 in the c a r t e s i a n system,
a n d b y yx + dyx,y2 + dy2,y3 + dy3 in the curvilinear system. It is
i m p o r t a n t t o k e e p in m i n d that, w h e n w e say t h a t t h e c o o r d i n a t e s of M
a r e y x, y2, y3, this m e a n s t h a t three n u m e r i c a l values a r e a l l o c a t e d to M
a n d t h a t w h e n these values are i n t r o d u c e d in E q s . (6.2.3) to (6.2.5) we
o b t a i n three surfaces w h i c h c a n b e d r a w n in the c a r t e s i a n system a n d
130
Kinematics of Continuous Media
w h o s e intersection gives M . If these n u m e r i c a l values a r e cx, c 2, c 3, the
three surfaces a r e :
c
y\ = \ =y\(xx,x2,xs\
y3 = 3C=
y2 =
c2=y2(xX9
x3,x3)9
y3(xx,x2,x3).
(6.6.1)
Also, w h e n we say t h a t the c o o r d i n a t e s of TV a r e yx + dyx,y2 + rfy2,
^ 3 + dy3, this m e a n s t h a t three n u m e r i c a l values cx + dcx, c2 + d c 2,
c 3 + dc3, slightly larger t h a n c{9 c2, c3, a r e allocated to N, a n d t h a t
w h e n these values are i n t r o d u c e d in E q s . (6.2.3) to (6.2.5) w e o b t a i n
three surfaces w h o s e intersection gives TV at a close d i s t a n c e from M :
T h e s e surfaces a r e :
yx + dyx = cx + dcx =
yx(xX9
x2x3\
y2 + rfy2 = c 2 + dc2 = ^ 2( x l5 x 2, x 3)
(6.6.2)
y3 + rf^3 = c 3 + rfc3 = > ^ 3( x 1, x 2, x 3) .
P o i n t M * is l o c a t e d in the cartesian system b y ^ , | 2, | 3, a n d in a n o t h e r
curvilinear system Z 1? Z 2, Z 3 b y z 1? z 2, z 3. P o i n t N* is l o c a t e d in the
cartesian system b y £j +
, £ 2 + d£2, £ 3 4- rf£3, a n d in the Z y curvilinear system b y zx 9
+ rfzj, z 2 + d z 29, z 3 + dz3. H e r e , t o o , we insist o n the
fact t h a t t h e zi s a n d the dzt s are n u m e r i c a l values. R e l a t i o n s of the
form (6.2.1) a n d (6.2.2) exist b e t w e e n the cartesian system a n d the t w o
curvilinear systems. A l s o the t r a n s f o r m a t i o n from B to B* is expressed
b y a n admissible (see Sec. 6.2) relation b e t w e e n the c o o r d i n a t e s of the
p o i n t s in the two curvilinear systems, n a m e l y :
d
a zn
t = Zi(y\>yi>ys)-
yi = yi(z\>zi>zy)
(6.6.3)
A t M9 in the system Yi9 the b a s e vectors a n d the m e t r i c tensor a r e
ax,a2,a3,
a n d gy. A t Af*, in the system Z , , t h e b a s e vectors a n d the
m e t r i c tensor are bx, b2, Z?3, a n d Gy. T h e s q u a r e of the length of MN in
the original configuation B is
2
(ds)
= gydyi
dyp
(6.6.4)
w h e r e gy is e v a l u a t e d at M in the system Y(. T h e s q u a r e of the length of
M*7V* in the t r a n s f o r m e d configuration B* is
2
(ds*)
=
Gijdztdzj,
(6.6.5)
Orthogonal Curvilinear Coordinates
131
w h e r e Gy is e v a l u a t e d at M* in the system Z , . A s s t a t e d in Sec. 1.2, the
L a n g r a n g i a n M e t h o d will b e u s e d ; in o t h e r w o r d s , w e shall a t t e m p t to
express o u r variables in t e r m s of the c o n d i t i o n s prior to t r a n s f o r m a t i o n .
F r o m E q . (6.6.3), we h a v e :
so t h a t E q . (6.6.5) b e c o m e s :
After c h a n g i n g s o m e of the symbols for d u m m y indices, the c h a n g e in
length of the e l e m e n t MN c a n b e expressed b y
(ds*?
- (dsf
= ( G r
^ t - g^dyj.
(6.6.8)
If w e set t h e b r a c k e t e d q u a n t i t y in the r i g h t - h a n d side of E q . (6.6.8)
e q u a l to 2yy, t h e n
2
2
(ds*)
2
2
= 2ytj dyt dyj.
- (ds)
(6.6.9)
T h e q u a n t i t y [(ds*) — (ds) ] is a n i n v a r i a n t a n d yy = yJ( is a s y m m e t r i c
tensor of the s e c o n d r a n k called the strain tensor.
So far we h a v e used t w o systems of curvilinear c o o r d i n a t e s , o n e at M
a n d o n e at M* c o n n e c t e d b y E q s . (6.6.3). Let us n o w distort the Z,
frame of reference in the t r a n s f o r m e d configuration of the b o d y in a
w a y such t h a t the c o o r d i n a t e s z{, z 2, z3 of a p o i n t there, h a v e the s a m e
n u m e r i c a l values yx, y2, y3 as in the original configuration. F o r p o i n t
Af*, for e x a m p l e , w e w o u l d h a v e zx = yx = cx, z2 = y2 = c2, z3 = y3
= c3. Also, the differentials dzt a n d dyt w o u l d b e equal a n d the partial
derivatives dzr/dyt
w o u l d b e equal to 8 n, the K r o n e c k e r D e l t a . W h e n
such a n o p e r a t i o n is p e r f o r m e d , all the i n f o r m a t i o n r e g a r d i n g the
c h a n g e in distances b e t w e e n adjacent p o i n t s a n d the c h a n g e in angle
b e t w e e n t h e various elements is c o n t a i n e d in the c h a n g e of the m e t r i c
tensor from gy to Gy as the b o d y is b e i n g t r a n s f o r m e d . E q . (6.6.5) n o w
becomes:
2
(ds*)
=
Gydy.dyj
(6.6.10)
132
Kinematics of Continuous Media
and
2
2
(ds*)
- (ds) = (Gy - )dyi
dyj.
gij
(6.6.11)
T h e expression for t h e strain tensor b e c o m e s :
2yy = Gy-gy.
(6.6.12)
In t h e following, we shall consider that the Yt system in the original
configuration is c h o s e n t o b e o r t h o g o n a l .
Since Gy = bt • bj a n d gy = at •
we h a v e :
2yy = (\bMbj\)cos(9y)b
- ( | S f| ) ( | S y| ) c o s ( ^ ) fl ( n o s u m ) , (6.6.13)
w h e r e (9y)a a n d (0y)b a r e the angles b e t w e e n the base vectors in t h e
initial a n d t h e t r a n s f o r m e d states. T h e c h a n g e in length p e r u n i t length
of MN is given b y
h
F
M_ ds*
N -ds
~
ds
_ \d\\ ~ \dx\
~
\dx\
'
)
Thus,
\d\\ = ( 1 4 - EMN
)\dx\.
(6.6.15)
If the element MN is n o w a l o n g a base vector a{, for example, its
lengths before a n d after t r a n s f o r m a t i o n a r e given b y E q s . (6.6.4) a n d
(6.6.10) in which we set / =j = 1. T h u s , ds = V#i7 dyx a n d ds*
= V ^ i 7 dyx. C o n s e q u e n t l y , the c h a n g e in length p e r unit length of this
element is
E
(Vfr7
= ~VgTx)dyx
6
)1 6
Therefore,
VG,,
= ( 1 + £ a, ) v ^ u
(6-6.17)
or
= (1 +EaX
)\ax\.
(6.6.18)
Similarly,
\b2\
= (1 + Ea2
)\a2\,
| 5 3| = ( 1 + Ea3
)\a3\.
(6.6.19)
Orthogonal Curvilinear Coordinates
133
Eq. (6.6.13) c a n n o w be written as follows:
+ ^)(l^l)(l^l)cos(^.)/>
2ytj = (1 + Eai
)(l
-
(6.6.20)
!)(!«/l)cos((9y)a( n o sum).
Therefore,
){\
= (1 + Eai
+
E ^ c o ^ j ) ,
-
cosifiyX
( n o s u m ) . (6.6.21)
E q . (6.6.21) will n o w be used to o b t a i n a physical m e a n i n g for the
quantities y,-,. In this e q u a t i o n , setting / = j a n d taking into a c c o u n t the
o r t h o g o n a l i t y of the 5 , ' s , we get,
2
^
oii
= (1 + Ea,)
1 (no s u m )
(6.6.22)
-1 ( n o sum).
(6.6.23)
-
or
Eq. (6.6.23) shows t h a t y n, y 2 , 2733 characterize the c h a n g e in length
p e r unit length of those elements t h r o u g h M a l o n g the three b a s e vectors
a
i9 a2, a3. Let us n o w set (9iJ)b = IT/2 - 4>ij9 so t h a t <f>tj represents the
c h a n g e in the right angle b e t w e e n a t a n d aj after t r a n s f o r m a t i o n . Eq.
(6.6.21) gives:
sm
J =
r -
(1
+
E^y/g-i
r—
(1
+
E )y/fy
aj
=
( n o sum, /
VG^
VG~j
6
)6
U s i n g Eq. (6.6.22), Eq. (6.6.24) c a n be written as follows [ c o m p a r e with
Eqs. (4.4.5) to (4.4.7)]:
S
m
^
=
^,, +
2 X
Y
+^
'<^>- ^
Eq. (6.6.25) shows that y 1 , 2y 1 , 3y 23 characterize the c h a n g e in the right
angle b e t w e e n elements a l o n g the three b a s e vectors (see Sec. 4.4).
y n, y 2 , 2y 33 are called the n o r m a l c o m p o n e n t s of the strain tensor, a n d
y 1 , 2y 1 , 3y 23 are called the shearing c o m p o n e n t s of the strain tensor.
F r o m Sees. 4.3 a n d 4.4, the strain eMN in a n y direction was given by
2
4
134
Kinematics of Continuous Media
U s i n g the s a m e definition for curvilinear coordinates, we h a v e from E q .
(6.6.11)
d
M
N
=y..±L A
y
(6.6.27)
'J ds ds '
T h e derivatives dyx /ds,
direction cosines / }, / 2,
unit b a s e vectors ex, e2,
dy2 /ds, dy3 /ds c a n b e expressed in terms of the
( 3 of the element MN relative to the o r t h o g o n a l
e3. I n d e e d , since
/, = 6 , ^ 1
L —A,—
A - A , ^ -
(6.6.28)
E q . (6.6.27) c a n b e written a s :
.2
£
MN
//\2
/ / \2
= „ ( ^ y +. 2 ( l ) - 3 3 ( ^ y +^ ( ^ )
(6.6.29)
Let us set
6 2,
Eq.
e
MN
(6.6.29)
W
33
YL3
"A,
A3'
£
23
Y33
2
A3 '
(6.6.30)
Y23
h2h3
c a n n o w b e written in the s a m e form as Eq. (4.4.8), n a m e l y
=E
2
LL
y? +
6.7
6 31
YL2
"
8
722
Yll
2
€ 2 / 2 + e 3 / 3 + 2 6 1 / 21 / 2 + i e x t x3h + 2 € /22/ 3.
(6.6.31)
Strain-Displacement Relations in Orthogonal Curvilinear
Coordinates
I n Fig. 6.9, the d i s p l a c e m e n t of p o i n t M is given b y
u = l-
x.
(6.7.1)
Orthogonal Curvilinear Coordinates
135
R e c a l l i n g t h e distortion of t h e reference frame a t M * a n d its c o n s e q u e n c e s , w h i c h were discussed in t h e previous section, we h a v e :
_9w
= =
^i_te
*Vi _ '
3*
(6.7.2)
1 l
= = 5
T h e m e t r i c tensor Gy of t h e t r a n s f o r m e d configuration c a n b e o b t a i n e d
in t e r m s of t h e d i s p l a c e m e n t vector u from t h e scalar p r o d u c t of bt b y
by T h u s ,
«.-^-(* £)-0* f|)
+
, _
+
du .
du . du
(6.7.3)
du
and
~
^
_
3w , 3w , 3w 3w
2yy = G i - jg ij = a r
+
W a Jj- W
i - W/
+
tan
A\
j(6.7.4)
W
T h e v e c t o r u c a n b e written in t h e o r t h o g o n a l c o o r d i n a t e system ^ a t
M as follows:
(6.7.5)
u = uxex + u2e2 + u3e3 = uteh
w h e r e et is t h e u n i t vector a l o n g at. Substituting in E q . (6.7.4), w e
obtain:
2
y=
y
{
"^j
uk)
^
n
+
J '
+
"
(6-7 6)
Wi * '4
(Ur
r)
(M,5,)
or
+ a - 9^7("1^1 + u2e2 + u3e3)
8 W
+ 9^( l^l +
W
(6.7.7)
3 W
2 ^ 2 + "3*3> * 3 ^ ( 1 ^ 1 + "2^2 +
w
3^3)-
136
Kinematics of Continuous Media
U s i n g E q s . (6.5.35) to (6.5.38), we o b t a i n :
3«i ^1 hx u2 9/2]
9vi
A 2 dy2
Yn
~
j(p- rT
+
+l
y
+
2 \ 8y,
A 2 dy2)
2 \ 9/2
2\3y3
Y12
h+2ux
hx
2 \ dy2
dh2
2
dy{
(6.7.9)
/J
V + jY^l _ ^ y
9
"2
39_y3/
2 \ 9_y2
2
hx dyx /
h+3 u2 dh3
h2 dy2
2
(6.7.10)
£p. + + £p)
hx 3y,
h2 dy2 I
h dy J
x
« 2 3^2
hx dyx
V
A
3 ^ +
dyx
A
3^2
2
2
w 9/*i \
3^2 2
1 / 3 w 2 _ ux dhx \/du2
h2 dy2 A 3>>2
h3 dy3
h dy J
2\3j>3
x
2\dyx
2 \ 3^
/z 3 9 y 3 /
£p.
£p)
h3 6y3 + hx 6yx +/
1/f
+ , 3w 2 W
= 2 ^ ^
^ 3 ^ -
t 1/
6 7 8
2 \ 9/,
+h3 ux dh3
hx dyx
+ tfp.
2 \ 3y 3
<-->
h3 dy3 J
h3 dy3
+ j^h.
, 3 9"3
" 9.y3
rT )
h2u+3 dh2
dy2
dy3
L 2
fc2 9 j 2
+ 2l(p.
\ dy2
u
7 33
h3
2 \ 8/,
2_h 2 29^2
-
L
w3
^ ;
3Aj_ +M 3 9^1 \
h2 3y 2
h3 dy3 /
ux dh2
u3 dh2 \
«2
hx dyx
^3 3^3 /
h3 dy3 )
(6.7.11)
Orthogonal Curvilinear Coordinates
+if-
^1
2 \ 3y 3
+2if\
+ if
+
- flL^LV^1
A 3 3y 3 / \ 3>>3
_ ^1,3*1 y
^ 1
2\9y!
1 /,
!f3 3*3 V 3«1_ ^3 3*I_
hx dyx / \ 3y,
3^2 _
. 3w 2
y
y
\
+ ^1^1 + ^ <!M *
/i 3 3>>3
A 2 3y 2 /
hx dyx
U
h2 dy2 /
3 3*3
* 23 j > 2/ \ 3 . y 3
3w 3
"2
+ 3*1
137
(6J
\
h2dy2J
3/z2
\
^ =2 ^ 3 ^ + *2^--2^-"3a^J
^2 ^ 3 ^ 2 +^3*2 \
+ if ^ 1 _ ^ 3*3
2 \ 3>>3
1 / 3w 3
2 \ 3^2
+ if ^ 1
2\3y2
+
A 2 9y 2 A dy2
-
u2 dh2 \ / 3w 3
/z 3 3y 3 / \ 3y 3
^3*2
3«i _
h3 dy3
w 2 3/* 3
hx dyx /
wj 3/* 3\ (6.7.13)
hx 3y, J
h2 dy2
"3 3*3
hxdyx/\dy3
\
hxdyx/'
Eqs. (6.7.8) to (6.7.13) c a n b e sutstituted in Eq. (6.6.29) to give the value
. T h e expressions for etj are o b t a i n e d by dividing the ytj ' s by hthj
of eMN
as i n d i c a t e d b y E q s . (6.6.30).
Example
1. Cartesian
Coordinates
In this case, hx = h2 = h3 = 1 a n d
to E q s . (4.3.3).
Example
2. Cylindrical
= e,-,-. Eqs. (6.7.8) to (6.7.13) r e d u c e
(Fig. 6.2)
Coordinates
In this case, hx = 1, h2 = r, a n d h3 = 1. T h e subscripts /*, 0, z
substituted for 1, 2, 3. T h e d i s p l a c e m e n t vector u is written u =
+ u9ee + uzez. E q . (6.6.11) b e c o m e s :
2
\[(ds*)
2
- (ds) ]
2
= (^)
Yrr
2
+ yg9
(d0)
4- 2yrz
drdz
2
+ yZ2
(dz)
4- 2 y Bdz0 d z .
+
2lr9
drd6
are
urer
12)
138
Kinematics of Continuous Media
Substituting (6.6.30) into the previous e q u a t i o n , we o b t a i n :
2
\[(ds*)
eggirdff)
2
2
2
2
= Err
(dr)
- (ds) ]
+
+ 2er0
(rdr
dO) + 2erz
(drdz)
+ e2Z
(dz)
+
2e0z
(rd9dz).
E q u a t i o n (6.6.31) b e c o m e s :
E e 2
0 0 ^ 0 + zz4 + 2e gi (
lrlz
r r e + 2erz
+
(6.7.16)
+ 2e9z
t9fz.
U s i n g E q s . (6.6.30), the set of relations (6.7.8) to (6.7.13) b e c o m e s :
£ *[(&)'•(£)']
+
(1X0
<6
WW "
t zz
e
"
(1X1)
y
- 3z
"f^[(^)
9
2
= R
'
(r)(l)
2
+
+
3r
I
I
"/-
dr
L 3z
I
dr dz
dO
+
/-
(^) ]
2
30
r
r
dz \
30
V
/
(6 ? 2Q)
du du l
3ugdue
dr dz
z
2
dr dz J
'
r
3z
r
/
r
30
/
">
< 6 J
3t/rz i 3M
r 2"I
dr
30 J
«,\
/
99 M
2[_dz
£ cf
2
3«e / ir9i(j
3r V 30
M
l p
fe)
+
r
YRZ
7l7)
2_ (6.7.18)
* (lXr) " 2 L 3 r 3r
(1)(1)
-
3z
r
\\
(612
(6.7.22)
30 _ •
If, in E q s . (6.7.17) to (6.7.22), the s e c o n d - o r d e r terms are neglected, we
o b t a i n the expressions used in the linear theory:
e
= ^
M= I ^ e + ( ^
e
(6.7.23)
Orthogonal Curvilinear Coordinates
_ 1 / J_9W,
dUg
ez _ 1 / 9 ^ ,
°
~ 2\dz
Example
l ^ zr
\
d9
r
^
3. Spherical
Ug\
t
=
Polar Coordinates
]V^r
139
,
(Fig. 6.3)
I n this case, hx = p, h2 = p sin <J>, a n d /z 3 = 1. T h e subscripts <J>, 0, p a r e
s u b s t i t u t e d for 1, 2, 3. T h e d i s p l a c e m e n t vector u is written u = u^e^
+ ueee + upep. E q . (6.6.11) b e c o m e s :
2
2
2
- (ds) ]
\[(ds*)
+ y^(</0) +
= y„(dtf
2
ypp
(dp)
+ 2y<t>g
d<}>d0 + 2yp<j)
d<l>dp 4-
2y0p
d9dp.
Substituting E q . (6.6.30) in E q . (6.7.25), we o b t a i n :
2
2
i[(^*)
-
2
(<fe) ] = 6 ^
^ )
2
2
2
+ „ (£p sin cj> d9)
+ e p(</p)
p
4- 2 e ^ p s i n 4> d<f> d9 + 2e^pp
(6.7.26)
d<f> dp
4- 2e0 pp sin <>
| a?0 */p.
E q . (6.6.31) b e c o m e s :
MN <M>
e
=e
l\ + em 1} + epp(j + 2e^ / +
+ 2£ p
< >/
u
i iy1 ^ _ M . (i ^* + jl) s :
U s i n g E q s . (6.6.30), the set of relations (6.7.8) to (6.7.13) b e c o m e(6.7.28)
1a«..
2
«P
2-
9
+
1 « * .
\P
]
.
«P .
3<?>/
3wfl
p sin <J> 30
9 M
Ji
iv1 MP
p
P
<f.
M
X
2
\
, ^ iH
^
"P
2
140
Kinematics of Continuous Media
£
^ " 2lp~sTn^l^ "
+
£
<f>p
TC O +
* Pp~dj
s m ^ |AP 3 * * T A P ¥ ~
2 \ P 3<f>
^
s m
P
3P
L 3p \ P d<p
3p \ P 3<J>
P /
3p P 9<J> J /
j
( 6
P /
?
3
(6.7.32)
+
21 3P
psin*~30
P~
(6.7.33)
If, in E q s . (6.7.28) to (6.7.33), the s e c o n d - o r d e r terms are neglected,
we o b t a i n the expressions used in the linear theory:
=
«*
e
Pl-f
_ 9»p
pp ~
9p
+
T '
^
=e
+
p-iin^^
^
+c
T
o
^
t
)(
6
?
3
Orthogonal Curvilinear Coordinates
141
1 d"g\
&
**»
P
2\psin$80
2\P
P
d<t>
(6.7.35)
dp
dp /'
eg
p
6.8
2\
dp
p sin<|> 90
80
P /
Components of the Rotation in Orthogonal Curvilinear Coordinates
F r o m the definition given in E q . (1.2.1), we notice t h a t the three
c o m p o n e n t s of the r o t a t i o n co 3, 2uX3
, a n d w 21 are n o t h i n g b u t one-half
of the c o m p o n e n t s of the curl of the d i s p l a c e m e n t vector u. If (ux
,u2,u3)
are the c o m p o n e n t s of u in a n y o r t h o g o n a l curvilinear system of
c o o r d i n a t e s , then, from Eq. (6.4.13), we h a v e :
i Curl u =
F o r a cylindrical
system of
2hxh2h3
3
dyx
hxux
h2e2
h3e3
_3_
dy2
_3_
dy3
h2 u2
(6.8.1)
h3 u3\
coordinates:
(6.8.2)
F o r a spherical system of
coordinates:
(6.8.3)
142
Kinematics of Continuous Media
6.9
Equations of Compatibility for Linear Strains in Orthogonal
Curvilinear Coordinates
T h e classical m e t h o d of o b t a i n i n g t h e compatibility relations in
curvilinear c o o r d i n a t e s involves t h e use of t h e R e i m a n n tensor a b o u t
which n o t h i n g h a s b e e n said in this chapter. T h e i n t r o d u c t i o n of this
tensor a n d t h e discussion of its properties fall outside t h e scope of this
text. However, o n e c a n still o b t a i n the compatibility relations in a n y
system of o r t h o g o n a l curvilinear c o o r d i n a t e s first b y writing t h e six
expressions of t h e strains in terms of t h e displacements, t h e n b y
eliminating from these expressions the three c o m p o n e n t s of t h e disp l a c e m e n t s . F o r example, if w e eliminate un ug a n d uz from E q s .
(6.7.24), w e o b t a i n t h e compatibility
relations
in cylindrical
coordinates.
T h e s e relations a r e written a s follows:
2
d gflfl
, l ^ f r r
+
, 2^99_ _
1 9f,r
z,
J
=
± ^ £ ^
'9/*
r
g "V
r d \ r
d r d g#
2 r
2 , J 2_ ^ f z £2 . 1 9^z rJ l&fte
d eo9
++ r
=
2
dr
2
9z
r
90
3r
V
3z90
2
f l ^2 + fliL^^thL
9r
r
drde
1 3 e zz
2
•2 90
1 3 g, r
' 909z
2
3 ^
3r3z
_ 9 _ / Jr_ 3 ^
9 z \ 30
9 / 1r dezr
9 A 30
1 3frr
>* 9z
9z
1
>* 9z
de
/
(6 9 2)
9z /
(6.9.3)
3z9r
9ffc _ 3 f ^ \ _
d _r(
(
3r
9z /
9z\
/
9z
9r
(6 9 1)
de
r9
\
9z /
\£_(_\tezL
=r
9 0 \ ' 90
a
/ e
3 ^
9z
9z\
9A ' /
9
9r
9z /
PROBLEMS
Elliptical cylindrical c o o r d i n a t e s m a y b e defined b y
xx = a cosh
cos y 2,
r62 (9 ) 5
+ ^ £ 4- ^ ^
+ IJL
1.
6)
x2 = a sinh y ] sin >>2, x 3 = y 3,
(6.9.6)
9
Orthogonal Curvilinear Coordinates
2.
143
w h e r e yx > 0 a n d 0 < y2 < 211.
(a) Show t h a t this system of c o o r d i n a t e s is o r t h o g o n a l .
(b) S h o w that, in the OXx, OX2 p l a n e , a curve yx = c o n s t a n t is a n
ellipse with semi-axes (a cosh j ^ ) in the OXx direction, a n d
(a sinh yx) in the OX2 direction.
(c) S h o w t h a t a curve y2 = c o n s t a n t is half of o n e b r a n c h of a n
h y p e r b o l a with semi-axes (a cos y2) a n d (a sin y2).
(d) U s i n g the metric coefficients a p p r o p r i a t e to this system of
c o o r d i n a t e s , o b t a i n the expressions of the gradient, the divergence, the curl, a n d the L a p l a c i a n .
(e) W r i t e d o w n the s t r a i n - d i s p l a c e m e n t relations.
P a r a b o l i c cylindrical c o o r d i n a t e s m a y b e defined b y
*\ = \(y\
- yl\
x
x
i = y\yi>
3 = y*>
w h e r e — oo < yx < oo a n d y2 > 0.
(a) S h o w t h a t this system is o r t h o g o n a l .
(b) S h o w t h a t in the OXx, OX2 p l a n e a curve y2 = c o n s t a n t is a
p a r a b o l a s y m m e t r i c a l with respect to the OXx axis a n d o p e n ing to the right, while a curve yx = c o n s t a n t is one-half of a
similar p a r a b o l a o p e n i n g to the left.
(c) U s i n g the metric coefficients a p p r o p r i a t e to this system of
c o o r d i n a t e s , o b t a i n the expressions of the gradient, the divergence, the curl, a n d the L a p l a c i a n .
(d) W r i t e d o w n the s t r a i n - d i s p l a c e m e n t relations.
CHAPTER 7
ANALYSIS OF STRESS
7.1
Introduction
W h e n a b o d y is subjected to external forces, its b e h a v i o r d e p e n d s
u p o n the m a g n i t u d e of the forces, u p o n their direction, a n d u p o n the
i n h e r e n t strength of the m a t e r i a l of which it is m a d e . Structural a n d
m e c h a n i c a l c o n s t r u c t i o n units are usually subjected to various c o m b i n a t i o n s of forces, s o m e h a v i n g m o r e d e t r i m e n t a l effects t h a n others. It
is therefore necessary to consider h o w forces are t r a n s m i t t e d t h r o u g h
the m a t e r i a l c o n s t i t u t i n g these units.
In this c h a p t e r , the c o n c e p t s of stress vector o n a surface a n d state of
stress at a p o i n t will b e i n t r o d u c e d . It will b e s h o w n t h a t the c o m p o n e n t s of the stress vector c a n b e o b t a i n e d t h r o u g h a linear s y m m e t r i c
t r a n s f o r m a t i o n with a m a t r i x w h o s e elements are the c o m p o n e n t s of a
tensor of r a n k t w o called the stress tensor. All the properties of linear
s y m m e t r i c t r a n s f o r m a t i o n s will b e applied to stress the s a m e w a y they
h a v e b e e n applied to linear strain. H o w e v e r , while the c o m p o n e n t s of
the linear strain tensor h a v e to satisfy six compatibility relations of the
s e c o n d order, it will b e s h o w n t h a t the c o m p o n e n t s of the stress tensor
m u s t satisfy three partial differential e q u a t i o n s of the first order, called
the differential e q u a t i o n s of equilibrium. T h e s e e q u a t i o n s will b e
derived in b o t h cartesian a n d o r t h o g o n a l curvilinear c o o r d i n a t e systems.
7.2
Stress on a Plane at a Point. Notation and Sign Convention
Let us consider a b o d y in equilibrium u n d e r a system of external
forces Qx . . . Qn, a n d let us pass a fictitious p l a n e P t h r o u g h a p o i n t O
147
148
Theory of Stress
Fig. 7.1
in the interior of this b o d y (Fig. 7.1). P a r t A of the b o d y is in
equilibrium u n d e r ~Q\, Q2, Q3, a n d the effect of p a r t B. W e shall a s s u m e
this effect is c o n t i n u o u s l y d i s t r i b u t e d over the surface of intersection.
A r o u n d the p o i n t 0 , let us consider a small surface AA a n d a n o u t w a r d
unit n o r m a l vector n. T h e effect of B o n this small surface c a n b e
r e d u c e d to a force Q a n d a c o u p l e C. N o w let AA shrink in size t o w a r d
zero in a m a n n e r such t h a t p o i n t O always r e m a i n s inside a n d n r e m a i n s
the n o r m a l vector. It will b e a s s u m e d t h a t Q/AA t e n d s to a definite limit
a a n d t h a t C/AA t e n d s to z e r o as AA t e n d s to zero. T h u s ,
(7.2.1)
(7.2.2)
Fig. 7.2
Analysis of Stress
149
T h e vector o is called the stress vector on P at O. T h e projection of o o n
the n o r m a l n is called the n o r m a l stress on (Fig. 7.2). T h e projection of
o o n the p l a n e P, in the p l a n e of n a n d a, is called the tangential or
shearing stress ot. Therefore,
2
(\o\)
2
(7.2.3)
= o + oj.
ot can, in t u r n , b e projected o n two o r t h o g o n a l directions in the p l a n e
P. A stress in t h e direction of t h e o u t w a r d n o r m a l is c o n s i d e r e d positive
a n d is called a tensile stress. A stress in the opposite direction is
c o n s i d e r e d negative a n d is called a compressive stress.
F o r c e s like Qx, Q2,...
Qn acting over the surface of a b o d y are called
surface forces. L o a d s applied to a b o d y are never idealized p o i n t forces;
they are, in reality, forces p e r unit a r e a a p p l i e d over s o m e finite area.
T h e s e external forces p e r unit a r e a are called tractions. F o r c e s distributed across the v o l u m e of a b o d y such as gravitational forces a n d
m a g n e t i c forces a r e called body forces.
a
Fig. 7.3
If P a n d P' are a n y t w o parallel p l a n e s t h r o u g h a n y t w o p o i n t s O a n d
O' of a c o n t i n u o u s b o d y , a n d if t h e stress o n P at O is e q u a l t o the stress
o n P' at 0\ the state of stress in the b o d y is said to b e a homogeneous
state of stress (Fig. 7.3).
In a trirectangular system of c o o r d i n a t e s (Fig. 7.4a), the stress
c o m p o n e n t n o r m a l to a p l a n e t h a t is p e r p e n d i c u l a r to the OXx axis is
d e n o t e d b y o u. T h e first subscript indicates the direction of the axis
p e r p e n d i c u l a r to the p l a n e in question, a n d the s e c o n d indicates the
direction of t h e stress. O n the s a m e p l a n e , ol2 indicates the t a n g e n t i a l
stress in the direction of the OX2 axis, a n d a 13 the t a n g e n t i a l stress in
the direction of the OX3 axis. T h e s a m e c o n v e n t i o n applies to two
p l a n e s p e r p e n d i c u l a r to the OX2 a n d the OX3 axes, respectively (Fig.
150
Theory of Stress
7.4b): a22 is parallel to OX2, o2l is parallel to OX{, a n d a 23 is parallel to
O X 3; also a 33 is parallel to OX3, a 31 is parallel to OXx, a n d a 32 is parallel
to OX2.
A p l a n e w h o s e o u t w a r d n o r m a l p o i n t s in the direction of a positive
axis is a positive p l a n e . A n o r m a l stress in the direction of this o u t w a r d
n o r m a l is c o n s i d e r e d positive (tension). T h u s , in Fig. 7.4b, all the
n o r m a l stresses s h o w n — a n, a 2 , 2o33 — are positive. A stress tangential
to a positive p l a n e , a n d p o i n t i n g in the direction of a positive axis, is a
positive tangential stress. O n the other h a n d , a stress tangential to a
negative p l a n e , p o i n t i n g in the direction of a positive axis, is a negative
tangential stress. In Fig. 7.4b, all the tangential stresses s h o w n are
positive. T h e sign c o n v e n t i o n previously defined will b e a d h e r e d to
t h r o u g h o u t this text.
7.3
State of Stress at a Point. The Stress Tensor
In Sec. 7.2, we h a v e seen that, o n a p l a n e P (Fig. 7.1) passing t h r o u g h
0 , there acts a stress vector defined by Eq. (7.2.1). O n a n o t h e r p l a n e
t h r o u g h O a different stress vector will act. W e shall prove that the
stress vector on a n y p l a n e t h r o u g h O c a n b e o b t a i n e d o n c e the stress
vectors o n three p l a n e s n o r m a l to the c o o r d i n a t e axes a n d passing
t h r o u g h O are k n o w n (Fig. 7.5). A c c o r d i n g to the c o n v e n t i o n e s t a b lished in Sec. 7.2, the stress vector o n the p l a n e Px has three c o m p o -
Analysis of Stress
A
X
151
2
Fig.7.5
n ac stress
a n vector
ec s s to n r thee psl a n e P2 h a s three
n e n t s , a u, a 12 a n d a 1 ;a3 the
* *
vector o n the p l a n e P3 h a s
c o m p o n e n t s , o2\, cr 2, 2 * 2 3 >
three c o m p o n e n t s , a 3 , 1a 3 , 2a n d a 3 . 3T h e m a t r i x
an
o2l
o3l
°\2
°22
°32
°\3
°23
°33
(7.3.1)
w h o s e c o l u m n s are the c o m p o n e n t s of the three stress vectors, is called
the m a t r i x of the state of stress at O.
X3
Fig.7.6
T o find the stress vector o(onl
,on2
,on3
)
on an oblique plane whose
n o r m a l is n, let us isolate from the c o n t i n u o u s b o d y a small t e t r a h e d r o n
OABC (Fig. 7.6), a n d write t h a t the forces a c t i n g o n it are in equilibriu m . T h e p l a n e ABC is n o r m a l to n a n d at a small d i s t a n c e h from O.
Let the a r e a s ABC, OCB, OCA, a n d OAB b e d e n o t e d b y dS, dSx,
dS2,
152
Theory of Stress
a n d dS3, respectively. If we a s s u m e t h a t t h e stress vector varies in a
c o n t i n u o u s fashion, the c o m p o n e n t s of the force o n ABC a r e (ani
4- et)dS, w h e r e l i m ^ ^ e , - = 0; the c o m p o n e n t s of the forces acting o n
OCB, OCA, a n d OAB are ( - a , 7 4- e^dSj,
w h e r e l i m ^ e ^ = 0; the
e = w h e r e AK = \h(dS) is
c o m p o n e n t s of the b o d y force F are (Ft 4- €/)AK,
the v o l u m e of the t e t r a h e d r o n a n d l i m A_ > 0/
0- F o r equilibrium, we
must have:
dS
+ (F, + ej) f = 0.
(ani + e,) + ( - l0 +j ej,)-^
7 32
( - ' >
T h e relation b e t w e e n the a r e a s of t h e triangles ABC, OCB, OCA, a n d
OAB c a n b e o b t a i n e d as follows (Fig. 7.6): Let us write, vectorially,
OA=T , OB = f , OC = T ,
{
2
3
then
AB = r2 — rx,
AC = r3 — rx,
BC = r3 — r2,
a n d we h a v e :
AB XAC
= (r2 - rx) X (f3 = r2Xr3
+ r3Xrx
rx)
+
rxXr2.
(7.3.3)
N o w the vector p r o d u c t A X B of a n y t w o vectors A a n d 5 is a vector
p e r p e n d i c u l a r to A a n d 5 , w h o s e positive sense is d e t e r m i n e d b y the
r i g h t - h a n d rule, a n d w h o s e length is e q u a l to the a r e a of the parallelog r a m f o r m e d b y A a n d B as t w o sides. H e n c e , if w e d e n o t e b y vx, Z>2, v3
the unit vectors n o r m a l to the surfaces OCB, OCA, a n d OAB, respectively, Eq. (7.3.3) c a n b e written as:
n(ABQ
= vX
(OCB)
+ V2(OCA) 4- V3(OAB)
(7.3.4)
and
-N =- V
(OCB\
\ABC)
+
, - V(OCA\
\ABC)
+
, -v (OAB\
AABC)'
* 5\
}
(
If the direction cosines of n are / , , / 2> a n d / 3, t h e n
OCB
ABC
=l ' / QCA=L
ABC
l
'
QAB
_2
I
. 3 . 6)
3
Analysis of Stress
153
or
dSj
J
dS
*'
(7.3.7)
Substituting Eq. (7.3.7) into Eq. (7.3.2), a n d passing to the limit as h ->
0, we get:
o y - a ^ .
(7.3.8)
I n m a t r i x n o t a t i o n , E q . (7.3.8) is w r i t t e n :
=
_°n3_
a
al l
a1 2
°2\
31
22
32
13
23
33
a
a
h
a
a
a
(7.3.8a)
T h u s , k n o w i n g t h e m a t r i x of the state of stress a t O, we c a n find the
stress vector o n a n y p l a n e w h o s e n o r m a l n h a s direction cosines fl9 /2,
a n d / j . E q . (7.3.8) shows t h a t the m a t r i x of the state of stress is the
m a t r i x of a tensor of r a n k two, called the stress tensor (see Sec. 5.8).
This m a t r i x t r a n s f o r m s the vector n to the vector a. I n the next section,
it will b e s h o w n t h a t this linear t r a n s f o r m a t i o n is s y m m e t r i c ; in o t h e r
w o r d s , t h a t the stress tensor is a s y m m e t r i c tensor.
T h e m a g n i t u d e of t h e stress vector is:
2
2
2
1*1 = V f o , i ) + (<ta) + Oto) •
(7.3.9)
T h e n o r m a l c o m p o n e n t of t h e stress vector is given b y :
(7.3.10)
T h e t a n g e n t i a l c o m p o n e n t of t h e stress vector is given b y :
2
o? =
7.4
(\o\) -o2.
(7.3.11)
Equations of Equilibrium. Symmetry of the Stress Tensor. Boundary Conditions
So far we h a v e c o n s i d e r e d the state of stress at a p o i n t . If it is desired
to m o v e from o n e p o i n t to a n o t h e r , the stress c o m p o n e n t s will c h a n g e
in intensities a n d it is necessary to investigate the c o n d i t i o n s w h i c h
154
Theory of Stress
c o n t r o l the w a y in which they vary. T h e r e q u i r e m e n t t h a t the laws of
equilibrium m u s t b e o b e y e d gives us the m e a n s for d e t e r m i n i n g h o w the
stresses v a r y from p o i n t to p o i n t . C o n s i d e r a n a r b i t r a r y closed surface
S within a b o d y in equilibrium. T h e external forces o n the v o l u m e V
enclosed in S consist of surface forces a n d b o d y forces (Fig. 7.7). T h e
projection of the r e s u l t a n t b o d y force vector o n the OXx axis is:
Fig. 7.7
d Fv
/ /
v
I
i
7 A 1
-
(
>
T h e projection of the tractions o n the OXx axis is:
ffonX
dS
s
= f j
s
( A a n + / 2a 21 + l3o3X
)dS.
(7.4.2)
Since the r e s u l t a n t force o n a b o d y in equilibrium m u s t be equal to zero,
then
/ /
j
V
FxdV+
j j (txoxx
s
+ i2o2X + l3o3X
)dS
= 0.
(7.4.3)
A p p l y i n g the divergence t h e o r e m to the surface integral, we o b t a i n :
flf(T£ % + %} + «)"-*
+
Since this e q u a t i o n applies for a n y v o l u m e V of the b o d y , the i n t e g r a n d
m u s t vanish identically; t h a t is,
<
7A4)
Analysis of Stress
ox
ox
x
155
ox
2
3
In a similar m a n n e r , s u m m a t i o n s of forces in the OX2 a n d OX3
directions yield t w o m o r e e q u a t i o n s . T h e three e q u a t i o n s thus o b t a i n e d
are called t h e differential e q u a t i o n s of equilibrium of a d e f o r m a b l e
c o n t i n u o u s b o d y . T h e y a r e written:
9 a
9*11
-5
0X
X
^ l
ox
3a
x
13
OXj
,
H
2i
9 a
,
3i
H -3
0X
-5
p2 p2
ox
2
+ +
ox
2
3a
23
ox
2
ox
H -~
ox
R
,
1- t
x
3
3
33
+F
_
r
—U
0
n
1
z .4.
6)
3a
H
—U
or
^
+ ^. = 0.
(7.4.7)
In case of m o t i o n , Eq. (7.4.7) b e c o m e s
w h e r e Ax, A2, A3 a r e t h e c o m p o n e n t s of t h e acceleration vector, a n d p
is the m a s s p e r unit v o l u m e of the b o d y . E q s . (7.4.6) c a n also b e derived
b y s u m m a t i o n of t h e forces t h a t act o n t h e faces of a n e l e m e n t a r y
parallelepiped a n d c o n s i d e r a t i o n of the variation of the stresses t h r o u g h
the element. This will b e d o n e in a later section for the analysis in terms
of curvilinear c o o r d i n a t e s .
Let us n o w consider the e q u i l i b r i u m of m o m e n t s . T h e m o m e n t of t h e
b o d y forces with respect to the origin is given b y the integral over t h e
v o l u m e V of the vector p r o d u c t X X FdV (see Fig. 7.7); in index
n o t a t i o n , this vector p r o d u c t is written (see P r o b l e m 5.8)
etjkXjFkdV,
w h e r e x{9 x2, x3 a r e the c o o r d i n a t e s of p o i n t s inside the v o l u m e V. T h e
m o m e n t of t h e tractions with respect to the origin is given b y the
integral over the surface S of the vector p r o d u c t Y X odS (see Fig. 7.7);
a ervector p r o d u c t is written eijk
in index n o t a t i oan , nethis
yjOnk
dS,
where
°n\>°n2i
d °n3
^
c o m p o n e n t s of a, a n d yx,y2,y3
a r e the
156
Theory of Stress
c o o r d i n a t e s of p o i n t s o n t h e surface S. Since for equilibrium t h e
resultant m o m e n t d u e to b o d y a n d surface forces m u s t vanish, t h e n
/ /
WnkdS
+ / /
S
e XjF
/
dV = 0.
uk k
(7.4.9)
V
Substituting E q . (7.3.8) in t h e surface integral, a n d using t h e divergence
t h e o r e m , w e get:
e
/
/
ijkyj°nkdS
=
/
/
s
SijkyjOrktrdS
s
{
=
£ ~
111
kE aX ) v d
*
J
*
)
r
v
But from E q . (7.4.7),
q
d rfc
therefore,
/ /
= / /
tykyjOnkdS
/
e (-XjF
iJk k
+ ojk
)dV.
(7.4.11)
V
5
E q . (7.4.9) n o w b e c o m e s :
/ /
f
V
eyk
oJk
dV=0.
(7.4.12)
Since t h e stress tensor varies in a c o n t i n u o u s fashion a n d t h e v o l u m e V
is arbitrary, t h e n
*Vk°jk
= 0.
(7.4.13)
I n e x p a n d e d form, E q . (7.4.13) yields:
a
12
or
=
21»
a
= a= a
13
31»
= a a
23
32'
(7.4.14)
Analysis of Stress
157
which m e a n s t h a t the stress tensor is s y m m e t r i c . O n a c c o u n t of this
s y m m e t r y , the state of stress at every p o i n t — i n o t h e r w o r d s , the stress
field—is specified b y six i n s t e a d of n i n e functions of position.
In s u m m a r y , the six c o m p o n e n t s of the state of stress m u s t satisfy
three p a r t i a l differential e q u a t i o n s (7.4.6) within the b o d y , a n d the three
e q u a t i o n s (7.3.8a) o n the b o u n d i n g surface. E q s . (7.3.8a) are called the
b o u n d a r y c o n d i t i o n s . It is o b v i o u s t h a t the three e q u a t i o n s of equilibriu m d o n o t suffice for the d e t e r m i n a t i o n of the six functions t h a t specify
the stress field. T h i s m a y b e expressed b y the s t a t e m e n t t h a t the stress
field is statically i n d e t e r m i n a t e . T o d e t e r m i n e the stress field, the
e q u a t i o n s of e q u i l i b r i u m m u s t b e s u p p l e m e n t e d b y o t h e r relations t h a t
c a n n o t b e o b t a i n e d from statics c o n s i d e r a t i o n s .
E q s . (7.4.14) show t h a t the linear t r a n s f o r m a t i o n (7.3.8a) is s y m m e t ric. Therefore, all the p r o p e r t i e s of linear s y m m e t r i c t r a n s f o r m a t i o n s
studied in C h a p t e r 3 c a n b e a p p l i e d to the s t u d y of the state of stress at
a point, the s a m e w a y they h a v e b e e n applied to the study of linear
strain. Reciprocity, principal directions, invariants, M o h r ' s r e p r e s e n t a tion, etc. . . . are p r e s e n t e d in the next section within the f r a m e w o r k of
stress.
Remark
T h e c o m p o n e n t s of the state of stress exist only in the d e f o r m e d state
of the b o d y . Therefore, all the e q u a t i o n s of Sec. 7.4 are referred to the
d e f o r m e d b o d y . H o w e v e r , w h e n stresses are studied within the framew o r k of the classical t h e o r y of elasticity, n o distinction is m a d e b e t w e e n
the p r e d e f o r m a t i o n a n d p o s t d e f o r m a t i o n values of the m a g n i t u d e s a n d
directions of the a r e a s o n which they act. T h i s k i n d of a p p r o x i m a t i o n is
quite consistent with the theory of linear strain. I n d e e d , in this t h e o r y it
w a s a s s u m e d t h a t the edges of the d e f o r m e d e l e m e n t of v o l u m e u n d e r g o
negligible r o t a t i o n s a n d t h a t the lengths of the edges in the d e f o r m e d
state differ only b y a very small a m o u n t from their original lengths.
T h e s e factors m a k e the d e f o r m e d v o l u m e e l e m e n t indistinguishable
from the u n d e f o r m e d v o l u m e e l e m e n t as far as the analysis of stress is
concerned.
7.5
Application of the Properties of Linear Symmetric Transformations
to the Analysis of Stress
Let nx a n d n2 b e the o u t w a r d unit n o r m a l s to t w o planes t h r o u g h a
p o i n t O at w h i c h the state of stress is k n o w n (Fig. 7.8). If (o)x a n d ( a ) 2
are the stress vectors o n these two planes, then, b y the p r o p e r t y of
158
Theory of Stress
Fig.7.8
reciprocity, the projection of (o)x o n n2 is equal to the projection of (o)2
o n nx:
0 ) i ' n2 = (o)2 '
(7.5.1)
nx.
The principal directions of the linear transformation (7.3.8) are at the same
time invariant directions. While in the study of strains we were searching
for those directions which are associated solely with n o r m a l strains, we
are here searching for the directions (principal directions) a n d the
planes they define (principal planes) such t h a t those p l a n e s are subjected only to n o r m a l stresses (the tangential stresses vanish). Let the stress
o n a principal p l a n e b e called o. T h e direction cosines of the n o r m a l to
this p l a n e c a n b e o b t a i n e d b y substitution in Eq. (7.3.8), as follows:
aa
=
=
o/
_°n3_
3
xx a 12
a 13
a
a1 2
22
2 3a
13
23
33
Aa
ha
(7.5.2)
F o r a nontrivial solution, the d e t e r m i n a n t of the coefficients of the
h o m o g e n e o u s system (7.5.2) m u s t b e equal to z e r o :
oX
x ~ o
a
a
a 13
12
a-
22
23
Eq. (7.5.3) is a cubic e q u a t i o n in o:
a
a
13
a
23
33 ~
= 0.
0a
(7.5.3)
Analysis of Stress
T3 _
°
2
(°Un + °22 + ° 3 3 )
+ a(aua22
-
159
(o a
+ 033 a,,
+ o22
a33
a
u 2 2
33
0
+ 20
12
0,
23 3
-
a
2
°12
°23 - <>h)
023 -
°22°h
n
-
(7.5.4)
033^2)
= 0.
T h e three roots of this e q u a t i o n a r e t h e principal stresses ox, a 2, a n d a 3.
T h e r o o t s a r e i n d e p e n d e n t of the system of reference axes a n d so a r e
the coefficients of E q . (7.5.4). T h o s e coefficients a r e called the invariants
of the state of stress, a n d a r e w r i t t e n :
a =
=
J\
h
°\\ + °22 + 3 3
a
=
a
°\ + °2 +
aa
aa
l l 2 2 + 2 2 3 3 + 3 3 l l ~ °h
= ox a 2 + a 2a 3 +
(7.5.5)
3
oh
~ °23
(7.5.6)
o3ox
J3 = a na 2 a 233 + 2 a 1 a22 a313 - a na | 3 - a 2 a2f 3 - a 3 a3f 2 (7.5.7)
= a ! a 2a 3.
If o n e chooses t h e c o o r d i n a t e axes a l o n g t h e principal directions, t h e
m a t r i x of E q . (7.3.8) b e c o m e s a d i a g o n a l m a t r i x . I n this frame of
reference,
0
%\
°n2
=
0
0
_°«3_
0
0 " "A"
0
h
(7.5.8)
a 3
_
T h e major principal stress acts o n t h e major principal p l a n e , t h e
i n t e r m e d i a t e principal stress acts o n t h e i n t e r m e d i a t e principal p l a n e ,
a n d t h e m i n o r principal stress acts o n t h e m i n o r principal p l a n e . I n
t e r m s of principal stresses, E q s . (7.3.10) a n d (7.3.11) a r e written as
follows:
on = x0(\ + 20q + 1]03
2 2 2
of = (*i "
°2?H%
+ (o - a ) /
2
3
2 / 3 + ( a 3-
(7.5.9)
o Y%ll
x
(7.5.10)
with
t\ + % + H = !•
(7.5.11)
160
Theory of Stress
A t a p o i n t 0 , Mohr's diagram allows us to represent graphically the
a n stresses o n a n y p l a n e whose n o r m a l h a s direction
n o r m a l a n d tangential
cosines i 1? t2
^ h- T h e unit sphere (see Sec. 3.13) is c e n t e r e d at O
(Fig. 7.9). T h e unit element OH a l o n g the n o r m a l to the p l a n e u n d e r
n
Fig. 7.9
a n
c o n s i d e r a t i o n h a s direction cosines / i , / 2, d / 3. T h e c o o r d i n a t e s of H
are l\, t l9 a n d / 3. U n d e r the linear t r a n s f o r m a t i o n (7.3.8), OH is
t r a n s f o r m e d to a, w h o s e n o r m a l a n d tangential c o m p o n e n t s a l o n g OH
a n d the p l a n e are given b y E q s . (7.5.9) a n d (7.5.10), respectively. T h e
formulas of Sec. 3.13 are d u p l i c a t e d h e r e :
Solving E q s . (7.5.9), (7.5.10), a n d (7.5.11) simultaneously, o n e o b t a i n s :
A
2
2
A
f3
of +
(o
n
( " 1 - o2)(ox
(<J
2
-
a 3)
(°3 ~- °\)(P3
°l)(°3
(7.5.12)
3
°3)(°n q
o )(o
- i)
3 2
(o ~ °\)\°n
°tof ^+ y°n
n
=
- a )
a 2)
~- °2)
)
-~ a 2
°l)
(7.5.13)
(7.5.14)
T h e s e e q u a t i o n s a r e similar to E q s . (3.13.6). T h e entire discussion in Sec.
3.13 of the c o r r e s p o n d e n c e b e t w e e n points o n the sphere representing
different directions of p l a n e s t h r o u g h O a n d points o n M o h r ' s d i a g r a m ,
c a n b e r e p e a t e d h e r e a n d the s a m e conclusions d r a w n (Fig. 7.10):
1. A t a given point, a n d for a given state of stress characterized b y the
n i n e Oy ' s, the largest n o r m a l stress is the major principal stress a n d the
smallest n o r m a l stress is the m i n o r principal stress.
2. T h e p l a n e s subjected to the highest shearing stresses bisect the
angles between the principal p l a n e s . T h e r e are three such planes a n d the
Analysis of Stress
j
f
/
161
\
o"
o
0",
"*—cr
m
Fig. 7 . 1 0
m a x i m u m shearing stress acts o n the p l a n e bisecting the angle b e t w e e n
the major a n d m i n o r principal p l a n e s . T h e m a g n i t u d e s of those shearing
stresses, w h i c h are called the p r i n c i p a l shearing stresses, a r e :
a
u
\ i
_
lA^mlmax "
i
°\
[ K ) 1]3
max
~ °2
2
'
~
a
x !
_
_°2
~ °3
2
LV ^23Jmax
(7.5.15)
°3
2
3. If the s a m e q u a n t i t y a 0 is a d d e d to t h e t h r e e p r i n c i p a l stresses, t h e
M o h r circles d o n o t c h a n g e in size; they are j u s t shifted a l o n g the o'on
axis. T h e o t ' s d o n o t c h a n g e . If we define the m e a n stress b y :
m_
°
then
-
a
a
x
+ a
2
3
+ a
3
_
Q\\
+ o2
2
+
3
33
(7 5
°ji
=
16)
3 '
/
/
a 1 = a w + a 1,
<j2 = a w + a 2' ,
o3 = om + a 3
(7.5.17)
and
a / + a 2' + a 3' = 0.
Eq. (7.5.8) c a n n o w b e w r i t t e n :
(7.5.18)
162
Theory of Stress
0
0
0
°m
0
0
0
°m
°n\
=
°n2
°n3
0f
0"
0
o
0
0
0
'a\
+
2
(7.5.19)
0-3
T h e stress o n a p l a n e t h r o u g h a p o i n t O at which the state of stress is
k n o w n , is t h u s the s u m of t w o vectors (see Sec. 3.14):
1. A vector a l o n g the n o r m a l to the p l a n e : T h e m a g n i t u d e of this
vector is the s a m e w h a t e v e r b e the values of lx, / 2, a n d / 3. This is w h y
the first m a t r i x in the r i g h t - h a n d side of Eq. (7.5.19) is referred to as the
m a t r i x of the spherical component of the state of stress (also called
hydrostatic component).
2. A vector c h a r a c t e r i z e d b y a m a t r i x w h o s e trace is equal to z e r o :
This m a t r i x is referred to as the m a t r i x of the deviatoric component of the
state of stress or, simply, the deviator stress matrix.
I n a general system of cartesian c o o r d i n a t e s , the state of stress at a
p o i n t c a n b e expressed as:
a
12 a1 3
<J
A
a 12 a 22 a2 3
*11
_ 13
=
33 _
23
M 0
0
o
m
"A
<J
_0
0
A
+
0
o_
m
with
_ 13
a
^22 +
+
a
12 a1 3
aA 12 Aa 22 a2 3
0"
33
= 0.
23
(7.5.20)
33
=
(7.5.21)
In index n o t a t i o n , Eq. (7.5.20) is written:
/.
= l8
(7.5.22)
+
T h e deviatoric c o m p o n e n t of the state of stress at a p o i n t is
r e p r e s e n t e d o n M o h r ' s d i a g r a m by Fig. 7.10b, b u t with the origin shifted
to o" so t h a t o'o" = om
. B o t h the spherical a n d the deviatoric c o m p o n e n t s h a v e three invariants each. In terms of the invariants of the state
of stress defined in Eqs. (7.5.5), (7.5.6), a n d (7.5.7), the invariants of the
spherical components are written:
J
sX
The invariants
J
dx
= 0,
— 7j,
s2
of the deviatoric
=
J:
1
1
J —
3(^1) ^ 3 -
components
are
(A) , JM-JI-^W 1 /
2
27
(7.5.23)
•
written:
T
T \
2
+ YJW
I
3
(-- )
7
5
24
Analysis of Stress
7.6
163
Stress Quadric
In a trirectangular system of c o o r d i n a t e s , OXx, OX2, OX3,
the e q u a t i o n ,
oijxixj
consider
(7.6.1)
= ±K\
w h e r e AT is a c o n s t a n t . This e q u a t i o n represents a q u a d r i c surface with
its center at the origin O (Fig. 7.11). This q u a d r i c is called the stress
Fig. 7.11
q u a d r i c a n d is completely d e t e r m i n e d o n c e the state of stress Oy at a
p o i n t P is k n o w n . Let r b e the r a d i u s vector to a n y p o i n t B
(xx,x2,x3)
o n the q u a d r i c . T h e n , the length OB is a m e a s u r e of the n o r m a l stress
o n the p l a n e I I passing t h r o u g h P, a n d w h o s e n o r m a l is in the direction
of r: I n d e e d , since the direction cosines of r a r e :
l
L =
*
OB'
L = —
OB'
2
L =
9 3
OB
9
I = -^L
'
OB
(7.6.2)
the c o r r e s p o n d i n g n o r m a l stress is given by Eq. (7.3.10) a s :
=
°^
°yOBOB
=
-(dW'
±
A n o t h e r interesting p r o p e r t y of the stress q u a d r i c is t h a t the n o r m a l
to this surface at the e n d of the vector r (Fig. 7.11) is parallel to the
stress vector a acting o n the p l a n e I I . T o p r o v e this p r o p e r t y , let us write
164
Theory of Stress
E q . (7.6.1) in the f o r m :
2
= 0.
A = OijXiXjTK
(7.6.4)
T h e n t h e direction of t h e n o r m a l to t h e q u a d r i c is given b y t h e g r a d i e n t
of t h e scalar function A. T h e c o m p o n e n t s of the g r a d i e n t a r e :
=
OySfoXj
+ OyXiSjn
Xj.
= 2omj
(7.6.5)
S u b s t i t u t i n g E q . (7.6.2) in E q . (7.6.5), a n d recalling E q . (7.3.8), w e get:
= 2omjlj(OB)
= 2(OB)onm
.
(7.6.6)
E q . (7.6.6) shows t h a t t h e c o m p o n e n t s of t h e g r a d i e n t vector a r e e q u a l
to the c o m p o n e n t s of the vector o multiplied b y 2(02?), w h i c h m e a n s
t h a t b o t h vectors a r e parallel.
7.7
Further Graphical Representations of the State of Stress at a Point.
Stress Ellipsoid. Stress Director Surface
Besides M o h r ' s d i a g r a m , a n u m b e r of m e t h o d s h a v e b e e n devised to
h e l p visualize t h e state of stress a t a p o i n t a n d t o c o m p u t e t h e stresses
on oblique planes:
a. The Stress Ellipsoid ( L a m e ' s Ellipsoid)
Let t h e axes of reference OXx, OX2, OX3 b e t a k e n in t h e d i r e c t i o n of
the p r i n c i p a l stresses a t a p o i n t P. A l s o let t h e t h r e e c o m p o n e n t s of t h e
stress v e c t o r o(onl
, on2
, on3
) o n a p l a n e I I t h r o u g h this p o i n t b e m e a s u r e d
a l o n g these axes. E q . (7.5.8) c a n b e written a s :
Substituting E q . (7.7.1) i n t o E q . (7.5.11), w e o b t a i n :
4 + 4 + 4=1.
of
ai
oj
(7.7.2)
E q . (7.7.2) m e a n s t h a t for e a c h inclined p l a n e t h r o u g h 0 , the stress is
r e p r e s e n t e d b y a vector from 0 with c o m p o n e n t s a w, lon2
, a n d a „ 3, t h e
Analysis of Stress
165
e n d of w h i c h lies o n the surface of a n ellipsoid (Fig. 7.12). T h i s ellipsoid
is called t h e stress ellipsoid a n d its semi-axes a r e the p r i n c i p a l stresses.
F r o m this, it c a n b e c o n c l u d e d t h a t the m a x i m u m stress at a p o i n t is the
major principal stress. If two principal stresses are n u m e r i c a l l y equal,
the ellipsoid is of revolution. If all the principal stresses a r e n u m e r i c a l l y
equal, the ellipsoid b e c o m e s a sphere a n d a n y three p e r p e n d i c u l a r
directions c a n b e t a k e n as p r i n c i p a l axes. W h e n o n e of the principal
stresses is zero, the ellipsoid r e d u c e s to a n ellipse. W h e n two principal
stresses are e q u a l to zero, the ellipsoid r e d u c e s to a line,
b . The Stress Director
Surface
T h e radii of t h e stress ellipsoid r e p r e s e n t the stress o n o n e of the
p l a n e s t h r o u g h p o i n t P at w h i c h the state of stress is k n o w n . T o find t h e
p l a n e c o r r e s p o n d i n g to a given r a d i u s , w e use the stress director surface
defined b y the e q u a t i o n :
fL +
°\
+ d
°2
= 1
(7.7.3)
°3
It c a n b e s h o w n t h a t the stress r e p r e s e n t e d b y a r a d i u s of the stress
ellipsoid acts o n a p l a n e parallel to the p l a n e t a n g e n t to the stress
director surface at the p o i n t it is intersected b y this r a d i u s . This is
illustrated in Fig. 7.13 for the case in w h i c h o n e of the principal stresses,
n a m e l y a 3, is e q u a l to zero.
166
Theory of Stress
7.8
The Octahedral Normal and Octahedral Shearing Stresses
A n o c t a h e d r a l p l a n e is a p l a n e equally inclined o n the directions of
the three principal stresses. T h e directions cosines of the n o r m a l to this
p l a n e are l / \ / 3 , l / \ / 3 , l / \ / 3 (Fig. 7.14). T h e expression for the o c t a h e d r a l n o r m a l stress is o b t a i n e d b y substituting these direction cosines in
Eq. (7.5.9):
Analysis of Stress
a
_ o{ + q 2 + a 3
a n + a 22 + a 33 _
3
o c / ~"
3
°m
3•
167
(7.8.1)
T h e expression for the o c t a h e d r a l shearing stress is o b t a i n e d
substituting these direction cosines in E q . (7.5.10):
2
2
Toct =
~ °l)
by
+ <>2 -
^3)
+ <>3 -
°\Y
In a general system of cartesian c o o r d i n a t e s , the expression for roct is
written as follows:
2a
T
2
2
oct=
^VOll
7.9
The Haigh-Westergaard Stress Space
-
22)
+ <>22 ~
°3l)
+ 033 -
^ll)
+
6 a
12 +
6 a
6af 3.
(7.8.3)
23 +
A state of stress defined b y three principal stresses, ox, o 2, a n d a 3, is
r e p r e s e n t e d in a stress space 7b y a p o i n t h a v i n g cartesian c o o r d i n a t e s
°\> °2> °3
- 1 5 ) . T h e e q u a t i o n of the trisectrix A is ox = a 2 = a 3.
Fig. 7 . 1 5
C o n s i d e r a p o i n t C w h o s e c o o r d i n a t e s , Q , C 2, C 3, represent a state of
stress. T h e m e a n stress Cm a n d the three c o m p o n e n t s of the deviator a r e
given b y :
168
Theory of Stress
r
C,
=
+
Cj
+
c .
(7.9.1)
3
and
o
c
2
0
-
0
0
c„
0
^3 ~
Q
0
0
3
0
2 C
— C) —
2
C
0
3
3
0
(7.9.2)
2C,
0
-
C, -
C ,
3
I n the stress space, the three c o m p o n e n t s of the deviator h a v e a
r e s u l t a n t e q u a l to
2
{ i ^ C ,
-
C
2
-
C
)
2
+
3
( 2 C
2
-
C , -
( 2 C
3
C
2)
)
3
2
2
+
C
C,
]
2
+
C ? +
C ? -
3 C
(7.9.3)
}"
N o w c o n s i d e r a p l a n e I T t h r o u g h C a n d n o r m a l to the trisectrix A. S u c h
a p l a n e intersects A at M , w h e r e C M is n o r m a l to A. F r o m a n a l y t i c
geometry,
OM
and
=
v
^
(7.9.4)
Q
2
( C M )
2
=
C ? +
C\
+
C f
-
3 C
, .
(7.9.5)
Therefore, if from a p o i n t C representing a state of stress w e d r a w a
p e r p e n d i c u l a r C M to the trisectrix, the length of CM gives the m a g n i t u d e of the d e v i a t o r stress, a n d the length of OM divided b y y/3 gives
the m a g n i t u d e of the m e a n stress. All the p o i n t s o n a line A' t h r o u g h C
h a v e the s a m e d e v i a t o r a n d all the p o i n t s o n the p l a n e I I ' (which is a n
o c t a h e d r a l p l a n e ) h a v e the s a m e m e a n stress.
Analysis of Stress
169
T h e previous c o n s i d e r a t i o n s p o i n t to a g r a p h i c a l m e t h o d t h a t c a n b e
u s e d to d e c o m p o s e the state of stress r e p r e s e n t e d b y a p o i n t C ( Q , C 2,
C 3) i n t o its spherical a n d its deviatoric c o m p o n e n t s :
(a) D r a w a n o c t a h e d r a l p l a n e I I t h r o u g h 0 a n d a line A' t h r o u g h C
parallel to A. T h e p o i n t of intersection of A' a n d I I gives the p o i n t C ,
a n d the vector OC represents the d e v i a t o r stress.
(b) D r a w a p l a n e I I ' parallel to I I t h r o u g h C. T h e p o i n t of
intersection of I I ' a n d A gives the p o i n t Af, a n d 0Af represents the m e a n
stress a c c o r d i n g to E q . (7.9.4).
I n the s t u d y of the t h e o r y of plasticity, the yielding of various
materials is often expressed b y a relation a m o n g p r i n c i p a l stresses. Such
relations m a y b e graphically r e p r e s e n t e d b y m e a n s of surfaces in the
stress space. Since spherical states of stress are k n o w n to h a v e little or
n o influence o n the yielding of c o m m o n metals, a r e p r e s e n t a t i o n of the
deviator a l o n e is sought. S u c h a r e p r e s e n t a t i o n is o b t a i n e d b y projecting
the various possible states of stress o n a n o c t a h e d r a l p l a n e t h r o u g h the
origin. T h e directions of the three p r i n c i p a l stresses, 0 a 1? 0 a 2, 0 a 3,
w h e n projected o n the o c t a h e d r a l p l a n e I I m a k e a n angle of 120 ° with
respect to e a c h o t h e r (Fig. 7.16). Points o n the trisectrix a r e projected at
the origin 0 . P o i n t C is projected to C", a n d OC r e p r e s e n t s the deviator
stress. OC = p is called the intensity of the d e v i a t o r a n d <j> is called the
p h a s e a n g l e ; <f> c a n b e m e a s u r e d from a n y of the three axes 0 1 , 0 2 , a n d
Fig. 7 . 1 6
0 3 (Fig. 7.16). If the three c o m p o n e n t s of the d e v i a t o r are called Sx,
a n d S3,
oc = p = (s? + s% + s?y*,
a n d the 3 n o r m a l projections of OC (Fig. 7.16) a r e given b y :
S2>
(7.9.6)
170
Theory of Stress
(7.9.7)
F r o m the g e o m e t r y of the figure, we see t h a t :
OHx + OH2 + OH3 = 0 (in m a g n i t u d e a n d sign),
(7.9.8)
which is consistent with the fact t h a t Sx + S2 + S3 = 0. Projections o n
o c t a h e d r a l planes are also used to represent deviator strains a n d to
express geometrically relations b e t w e e n stresses a n d strains.
7.10
Components of the State of Stress at a Point in a Change of
Coordinates
T h e tensor c h a r a c t e r of stress was established in Sec. 7.3. Therefore,
in a r o t a t i o n of the reference system of c o o r d i n a t e axes, the c o m p o n e n t s
of the state of stress are given b y (Fig. 7.17):
f
•*2 (6/.&2,&j)
*2
%1
jXt if 11,^12,^,3)
Fig. 7 . 1 7
(7.10.1)
Thus,
2 o
Ai A2
+ 2<J / /
23 12 13
+
i 3 Ai
A3
(7.10.2)
Analysis of Stress
°22 =
a
l
*11 /?1 + ° l l l l
+
33^3
+
171
2 a
2
° \ l h \ hi
+
13
4l A>3
(7.10.3)
+ 2 a 23 / 22^23
<*33
22 ^2 +
a
= ^11^31 +
° 3 3 4?3 +
2a
12 h\
+ 2 a 23 / 32 / 33
<*12 = ( l l Al + 21 A2 + 31 A3V2I
+ {°\lh\
+ 22A2 +
°3lh3)hl
a
hi
+ 2 a 1 / 31
3 / 33
(7.10.4)
a
a
a
(7.10.5)
+ ( a 1 / „3 + a 2 / 12
3 + o3/1
3 ) 3/ 23
a
^23 =
fall
4l
+ °~21 A>2 +
31
*23)h\
+ ( a 1 / 21
2 + a 2 / 22
2 + aa32 / 2 ) 3/ 32
(7.10.6)
+ O13A1 + ^23^22 + 3 3 ^23^33
=a
<*13
( l l Al +
21
a
a
l
\l +
3 1 ^13
V31
+ (a1/n
2 + a 2 / 12
2 + a 3 / 12 ) 3/ 32
(7.10.7)
+ ( a 1 / „3 + a 2 / 12
3 + a 33 / 1 ) 37 3 . 3
7.11
Stress Analysis in Two Dimensions
T h i s section parallels Sec. 3.16. Let I I b e a p r i n c i p a l p l a n e t h r o u g h a
p o i n t O, a n d let OXx a n d OX2 b e t w o reference axes in this p l a n e ( F i g .
7.18); OX3 is t h e p r i n c i p a l direction n o r m a l t o I I . I n this system of axes,
a 13 a n d a 23 a r e e q u a l t o zero, a n d a n y p l a n e p a s s i n g t h r o u g h OX3 is
*2
Fig. 7.
172
Theory of Stress
subjected to a stress vector l o c a t e d in the p l a n e I I : This is o b v i o u s since
the stress vector results from the t r a n s f o r m a t i o n of a n o r m a l n l o c a t e d
in a n i n v a r i a n t p l a n e . T h e stress vector o n a n y p l a n e t h r o u g h OX3, a n d
w h o s e n o r m a l n h a s direction cosines lx = cos 0, / 2 = sin 0, a n d / 3 = 0,
is such t h a t
aa 12
a
%2_
_ 12
COS
0
(7.11.1)
sin 0
22_
a n d on3 = 0. T h e c o r r e s p o n d i n g n o r m a l a n d tangential stresses o n this
p l a n e are given b y :
=
2
2
O\\cos 0
°n
_ °\\
+ a 2 s 2i n 0 + 2 a 1 s2i n 0 cos 9
a
a
,
+ °22
ll ~
22
a
o
t
=
22
2
-
(7.11.2)
cos 29 + a 1 s2i n 29
(7.11.3)
sin 29 4- a 1 c2o s 29.
If the 9reference axes a r e r o t a t e d a n angle 9 a r o u n d OX3 (Fig. 7.19), the
atj s in E q . (7.11.1) b e c o m e :
=
°n
a
==
2
< 7 ns i n 0 + a 2 c2o s 0 - 2 a 1 s2i n 9 cos 9
a u + a 22
a n - a 22
cos 20 — a 1 s2i n 29
2
2
aa
22
°\2
2
c o s 0 4- a 2 s2i n 0 4- 2 a 1 s2i n 9 cos 9
a
a
a
+ 2 2 , n ~~ 2 2
cos 29 4- a 1 s2i n 29
2 2 ^
2
2
_
=
ll ~
-2
22
sin 29 4- a 1 c2o s 20.
(IMA)
(7.11.5)
(7.11.6)
T h e eigenvalue p r o b l e m in t h e p l a n e I I yields t w o p r i n c i p a l directions, 0 1 a n d 0 2 , given b y (Fig. 7.19):
v
t a n 2<b
=
2a 12
7=—_
ll
22 '
a
n
,
a
(7.11.7)
a n d t w o principal stresses given b y :
ai
?=n j p 2
^ _+ n ^ y
^
(7.11.8)
Analysis of Stress
ai
=
_
^(°nZ^J~2~
173
(7.11.9)
2
I n the r e p r e s e n t a t i o n o n M o h r ' s d i a g r a m , only o n e circle is of interest
(Fig. 7.20b). Let us a s s u m e t h a t we are given the c o m p o n e n t s of the
(b)
Fig. 7 . 2 0
a n c ta a
state of stress o n, a 2 , 2 * °\2
p o i n t O of a b o d y r e l a t e d t o a system
of c o o r d i n a t e s OXx, OX2. T h e c o n v e n t i o n for the n o r m a l stresses is t h a t
they are p l o t t e d positive to the right of the origin o n the o'on axis a n d
negative to its left. T h e c o n v e n t i o n for the t a n g e n t i a l or shearing stresses
is as follows: C o n s i d e r the t w o p l a n e s n o r m a l to the t w o axes OXx a n d
OX2, a n d a s s u m e t h a t o n e always goes from OXx to OX2 t h r o u g h a
174
Theory of Stress
counterclockwise m o t i o n . T h e p o i n t representing the stress oX2 o n the
p l a n e w h o s e n o r m a l is OXx is p l o t t e d at a distance a 12 b e l o w the o'on
axis if a 12 is positive, a n d a b o v e the o'on axis if a 12 is negative. Fig. 7.20
shows M o h r ' s circle a n d the n o r m a l a n d tangential stresses o n the pair
of p l a n e s w h o s e n o r m a l s OX\ a n d OX2 m a k e a n angle 9 (clockwise)
with OXx a n d OX2. In a system of axes OX\, OX2, the c o m p o n e n t s of
the state of stress a ' n, a 2 , 2a' 12 are directly r e a d in m a g n i t u d e a n d sign
o n the circle. T h e directions of the principal stresses m a k e <j>xa n d
<f>x + 90° with OXx. T h e m a x i m u m shearing stresses occur o n p l a n e s
w h o s e n o r m a l s m a k e 45 ° with the principal directions.
Example
Fig. 7.21
A sheet of metal is uniformly stressed in its o w n p l a n e , so t h a t the stress
c o m p o n e n t s at all its p o i n t s related to a set of axes OXx a n d OX2 a r e :
oxx = 10,000 psi,
a 22 = - 5 , 0 0 0 psi,
a 12 = 5,000 psi .
It is r e q u i r e d to find the stress c o m p o n e n t s associated with a set of axes
OX\ a n d OX2, inclined 45° clockwise to the OXx, OX2 set as s h o w n in
Fig. 7.21. It is also required to find the principal stresses a n d the
directions of the principal axes.
Fig. 7.21b shows M o h r ' s circle c o n s t r u c t e d from the given d a t a . T h e
m a g n i t u d e of the stresses o n the various planes c a n b e directly r e a d o n
the circle, a n d their a c t u a l direction as well as the direction of the
shearing a n d n o r m a l stresses c a n b e plotted (Fig. 7.21a) following the
previously established sign c o n v e n t i o n s . T h u s ,
Analysis of Stress
o\x = —2,500 psi,
o22 = 7,500 psi,
175
o\2 = + 7 , 5 0 0 psi
and
ox = 11,520 psi,
7.12
o2 = —6,520,
tan 2<j>x = ^ •
Equations of Equilibrium in Orthogonal Curvilinear Coordinates
T h e e q u a t i o n s of equilibrium will b e o b t a i n e d b y considering a n
e l e m e n t a r y curvilinear parallelepiped a n d writing the e q u i l i b r i u m of the
forces acting o n it (Fig. 7.22). Let the stress vectors acting o n the faces
0'y h h dy dy
l 23 2 a
a
y2
l h
d
h5d
yt '
h VyihihjdyzdLb
+
^yi 2^y dy dy .
f 23
-&y h,
3
h dy dy
2f 2
Fig. 7 . 2 2
OGFE, OADE, a n d OABG b e d e n o t e d by oyX
, oy2
, a n d oy3
, respectively.
T h e force acting o n OGFE is equal to — oyX
h2h3dy2dy3.
T h e force
acting o n A BCD is equal to
= oyX
h2h3dy2dy3
3a,
+ h2h3^dyxdy2dy3
dy
x
+
oyX
^—(h2h3)dyxdy2dy3
+
^
dy
x
^
1
dy
^
x
(
^
)
2
^
^
.
176
Theory of Stress
T h e last t e r m is a n infinitesimal of the fourth o r d e r a n d , as such, is
negligible. Therefore, the force o n face A BCD is e q u a l to
T h e s a m e steps c a n b e r e p e a t e d for t h e forces a c t i n g o n the
four o t h e r faces of the parallelepiped. T h e b o d y force vector is
a n d the inertia force v e c t o r in case of m o t i o n is
(Fhxh2h3dyxdy2dy3),
(Aphxh2h3dyxdy2dy3).
A is the acceleration vector a n d p the m a s s p e r
unit v o l u m e of the b o d y . T h e r e f o r e , the e q u a t i o n s of e q u i l i b r i u m in
v e c t o r form are w r i t t e n :
(ay i h2 h3 )dy, dy2 dy3 + ^ - (ay2 h, h3 )dy, dy2 dy3
h
d d d
(7.12.1)
+ 37-(Pyi \h) y\ yi y3
+ Fhxh2h3dy{
dy2dy3
— Aphx h2 h3 dyx dy2 dy3 = 0.
E a c h o n e of the stress vectors oyX
, oy2
, a n d oy3 c a n b e written in t e r m s
of its c o m p o n e n t s a l o n g Yx, Y2, a n d Y3 as follows:
o
yX
a
e
o
y2
e0
o
y3
exG3x + ^ 3 2 + ^3 3 3 •
x 2x + ^ a2 2 +
3°23
a
(7.12.2)
Substituting E q s . (7.12.2) i n t o E q . (7.12.1), we get:
(7.12.3)
Analysis of Stress
177
T h e e q u a t i o n s established in Sec. 6.5 a r e n o w u s e d in t a k i n g t h e p a r t i a l
derivatives of ex,e2,
a n d e3: T h u s ,
(7.12.4)
+
+
A
e 2 [ ^ ( o nh 2 h 3 ) - h 3 a u^ \
e\3\^ ( o X h 3
2 h 3 ) - h 2 o n^ \
?
a
g
aTT L 1 * 3 < 1 2 1 + 2 ° 2 2 +
dy
r
3°23)
3/i 2~|
9a
= 5 l L 3 ^ (
+
g
2 ^ ^ 3 ) - ^ 3 ^ 2
h
g
g 2[ ^ - ( 0 M )
22 3
+ 3°2l-fo
_ T 3
~ -
/ i 2( e , a 31 + g 2a 32 +
9^-(^1^2°3l) -
_ r
+
3
3^J
3/j 2
+
A 3A
2
(7.12.5)
l°23g^
2~\
e 3o 3 ) 3
]
M 3
3
(7.12.6)
2i
3
^h3
^h3 n
e 3[ g ^ ( M " 3 3 ) + M s i g ^ + * l « 3 2 g ^ _ | 2
By s u b s t i t u t i n g E q s . (7.12.4) to (7.12.6) i n t o E q . (7.12.3), a n d factoring
ex, e2, a n d e3, o n e o b t a i n s a vector e q u a t i o n of e q u i l i b r i u m in a f o r m
w h i c h expresses the projection of t h e forces in t h e ex direction, the e2
direction, a n d the e3 direction. T h i s allows us to write the three scalar
e q u a t i o n s of e q u i l i b r i u m a l o n g the t a n g e n t s t o t h e three curvilinear
c o o r d i n a t e s as follows:
178
Theory of Stress
3
+
9^(0.1*2*3)
,
+
' 3 « 2
3
+
9^(021*1*3)
a3/i,
a/z
9 ^ - - °2ih^
2
3
9^(031 * > *
3 3 * 2 g ^
+
3/?i
o x h 23^ (7.12.7)
a a/23
-
2)
+
hxh2h3(Fx
-
pAx)
3/Z0
^\
r\
9^(0.2*2*3)
+
0
9^(^22*1*3) +
3 ^ ( 0 3 2 * 1 * 2 ) +
^ 3 * 1 9 ^
(7.12.8)
0
CT
h
a
a
+
2 1 * 3 g ^
-
33"lg^-
~
n
3 ^
+
hxh2h3(F2
-
pA2)
0
g^-(o
)
*2*3)
+
3
9 ^ ( ^ 2 3 * 1 * 3 ) +
9^-(°33*l*2) +
31*2
3 ^
(7.12.9)
dh3
+
0 3 2 * 1 9^
adhx
~
H * 2 g ^
0.
dh2
-
<>22*1 g £ J +
* 1 * 2 * 3 ^ 3
~
P ^ )
x2
Fig. 7 . 2 3
Analysis of Stress
179
Example 1. Cylindrical Coordinates (Fig. 7.23)
I n this ease, hx = 1, h2 = r, a n d h3= \. T h e subscripts r, 0, z are
s u b s t i t u t e d for 1, 2, 3, a n d dr, dO, dz are substituted for dyx, dy2,
dy3,
respectively, in E q s . (7.12.7) to (7.12.9). T h e s e e q u a t i o n s b e c o m e :
dr
+
r^f
+ ^97 +
+
+
r~W
Hz
dr
Example
.
9
+
1 Ofc
r-ffi-
2. Spherical
~df
.
+
dazz
~
+
°oe) + Fr-
Ar
r°re + ^ - A8 P = 0
=P 0
(7.12.10)
i
+ o 7r2 + Fz - AZ P = 0.
Polar Coordinates
(Fig. 7.24)
Fig. 7 . 2 4
I n this case, hx — p, h2 = p sin <J>, a n d h3 = 1. T h e subscripts </>, 0, p are
substituted for 1, 2, 3, a n d d<f>, dO, dp are substituted for dyx, dy2,
dy3,
respectively, in E q s . (7.12.7) to (7.12.9). T h e m a s s per unit v o l u m e here
is called y. Eqs. (7.12.7) to (7.12.9) b e c o m e :
Fig . 7.25
180
Theory of Stress
9(
do
W
l W>
3p
P 3<f>
psincj) 3(9
- yA+ = 0
+
3a^
+
3 a*P0
dp
t
+1
P 8^
+
+
1
_
A
6>6>)
9
p+ + 02 cos <f>
p sin
P °
a
*
(7.12.11)
= 0
3 -PP
a ^ , i ^ 3Pa ^
3p
P 3<J)
H • ,
^
p sin 4> 30
-
3
cos <j>
P
*
'
p
sin <#>
ttO,A
d
i
3a« , coscfr
%
1
psiiKf) 3 ^
p sin </> P*
PROBLEMS
1.
A stress field is given b y :
3
a n = 20xj + x\
a
22 =
2
a 12 = x 3
+ 200
a
a 33 = 30.x + 3 0 x 3
2.
13
=
x\
3
a 23 = x .
W h a t a r e the c o m p o n e n t s of t h e b o d y force required t o insure
equilibrium?
T h e usual engineering e q u a t i o n s for the stresses d u e to the b e n d i n g
of a circular b e a m are (Fig. 7.25):
_ Mx2
°\\ a
a22
= 0
33 =
7
0
_
°\2
a
2
V(R
-
-
Yj
13 = 0
0
xj)
~
a
4
23 =
D o these e q u a t i o n s satisfy equilibrium? M is the b e n d i n g m o m e n t ,
V is the s h e a r i n g force, 7 is the m o m e n t of inertia a b o u t a d i a m e t e r
of t h e section, a n d R is the r a d i u s .
y
Analysis of Stress
3.
181
T h e stress field in a c o n t i n u o u s b o d y is given b y :
K-] =
1
0
0
1
2x2
4JCJ
3
io
2x2
4.x,
psi.
1
xF i n d the
+ stress
+
xvector a a t a p o i n t M ( 1 , 2, 3), acting o n a p l a n e
l
4.
*2
3
~
T h e state of stresses at a p o i n t is given b y :
Kl =
5.
2
io
10
5
-10
5
20
-15
-10
-15
-10
F i n d the m a g n i t u d e a n d direction of the stress vector acting o n a
p l a n e w h o s e n o r m a l h a s direction cosines ( 1 / 2 , 1 / 2 , l/y/2); w h a t are
the n o r m a l a n d tangential stresses a c t i n g o n this p l a n e ?
In a solid circular shaft subjected to p u r e torsion, the stress field is
given b y :
o
0
-Cx2
o
0
Cxx
Cx2
6.
psi.
Cxx
0
w h e r e C is a c o n s t a n t . A t the p o i n t w h o s e c o o r d i n a t e s are ( 1 , 2, 4),
find:
(a) the principal stresses
(b) the principal directions
(c) the m a x i m u m shearing stress a n d the p l a n e o n w h i c h it acts.
A t a p o i n t M of a c o n t i n u o u s b o d y , the c o m p o n e n t s of the stress
tensor a r e :
3
[Oy]
= 10
1
-3
-3
4
V2 -V2
(a)
(b)
V2
-V2
psi.
4
F i n d the p r i n c i p a l stresses a n d the principal directions.
D r a w M o h r ' s circles, a n d o b t a i n the n o r m a l a n d tangential
stresses o n a p l a n e w h o s e n o r m a l h a s direction cosines
182
Theory of Stress
(l/\/3, l/\/3,
with respect to the reference axes.
F i n d the o c t a h e d r a l n o r m a l a n d shearing stresses.
W h a t are the invariants of the spherical a n d the deviatoric
c o m p o n e n t s of this stress tensor?
(e) W h a t is the e q u a t i o n of the stress q u a d r i c ?
F i n d the c o m p o n e n t s of the stress tensor of P r o b l e m 4 in a system
of c o o r d i n a t e s w h o s e axes h a v e direction cosines (0, 0, 1), ( l / \ / 2 >
(c)
(d)
7.
1/V2,0), (1/V2,-1/V2,0).
8.
9.
A very thin plate is uniformly l o a d e d as s h o w n in Fig. 7.26. A m o n g
all the p l a n e s t h a t are n o r m a l to the p l a n e of the plate, which o n e s
are the principal p l a n e s a n d w h a t is the value of the stresses to
which they are subjected?
F o r the following states of stress at a point, use M o h r ' s circle to
o b t a i n the m a g n i t u d e a n d directions of the principal stresses:
(a)ou
= 4,000 psi
(b)ou = 14,000 psi
o22 = 0
a 22 = 5,000 psi
a 12 = 8,000 psi
°\3
(c)au = 12,000 psi
=
=
°23
33
a=
a 12 = - 6 , 0 0 0 psi
0
=
°\3
°23
=
°33
a 22 = 5,000 psi
=
a 12 = 10,000 psi
0
=
13
°23
=
=
°33
°-
10. O b t a i n the e q u a t i o n s of equilibrium in the two systems of c o o r d i nates defined in P r o b l e m s 1 a n d 2 of C h a p t e r 6.
P2=lOOOO
psi
Fig. 7.26
a
CHAPTER 8
ELASTIC STRESS-STRAIN RELATIONS
AND FORMULATION OF ELASTICITY
PROBLEMS
8.1
Introduction
In the p r e c e d i n g c h a p t e r s , the study of strain a n d the study of stress
were p u r s u e d i n d e p e n d e n t l y . A l t h o u g h certain engineering p r o b l e m s
c a n b e solved w i t h o u t relating the stresses to the strains, m a n y require
the s i m u l t a n e o u s c o n s i d e r a t i o n of stress a n d strain. Constitutive relations c o n n e c t i n g t h e stresses to the strains are therefore n e e d e d to solve
this class of p r o b l e m s . In this c h a p t e r , the elastic relations will b e
developed. T h e y will c o n t a i n experimentally d e t e r m i n e d c o n s t a n t s a n d
for simplicity will b e restricted in their applicability to linear strains.
T h i s restriction is n o t as drastic as it m a y a p p e a r since e x p e r i m e n t s h a v e
s h o w n that, in their w o r k i n g r a n g e , a large n u m b e r of structural
materials b e h a v e in a linearly elastic w a y with d e f o r m a t i o n s very
a d e q u a t e l y described b y the c o m p o n e n t s of linear strain.
8.2
Work, Energy, and the Existence of a Strain Energy Function
A b o d y subjected to external forces ( b o d y forces a n d surface forces)
will d e f o r m . T h e s e forces will d o s o m e w o r k a n d the b o d y will acquire
a n internal energy w h i c h will d e p e n d u p o n its s h a p e a n d its t e m p e r a t u r e
distribution. T h i s i n t e r n a l energy, also called strain energy, will b e
calculated with reference to a s t a n d a r d state of c h o s e n u n i f o r m t e m p e r a t u r e a n d zero strain.
185
186
The Theory of Elasticity
C o n s i d e r a b o d y o c c u p y i n g a v o l u m e V, enclosed in a surface S, a n d
in a d e f o r m e d state of equilibrium. D u r i n g the d e f o r m a t i o n , let dWB
a n d dWs d e n o t e the w o r k d o n e o n the v o l u m e V b y the b o d y forces a n d
by the surface forces, respectively. If the process of d e f o r m a t i o n is
a d i a b a t i c , the first law of t h e r m o d y n a m i c s yields:
dWB + dWs = d J f f
v
UdV=
J f f
(dU)dV,
(8.2.1)
v
w h e r e U is called the strain energy density. T h e w o r k d o n e b y the b o d y
forces is expressed b y the formula (see Fig. 7.7):
dWB = / / /
v
(8.2.2)
[Fxdux + F2du2 + F3du3]dV.
T h e w o r k d o n e b y the surface forces is expressed b y the f o r m u l a :
d
w
s = fj
s
(Pn\du{ + on2
du2
(8.2.3)
+ on3
du3)dS.
Substituting Eq. (7.3.8) into Eq. (8.2.3), we o b t a i n :
d
W
s =
j7
K V n + V21 +
h°3i)dux
+ a
(h °n + h°22 + h°3i)du2
( A i 3 + h°23 +
+
(8.2.4)
h°33)du3]dS.
By the divergence t h e o r e m , this integral c a n b e t r a n s f o r m e d i n t o the
following v o l u m e integral:
v
3i
3
9
3
I
Substituting Eqs. (8.2.2) to (8.2.5) into Eq. (8.2.1), a n d using E q s . (1.2.1)
a n d (7.4.6), we o b t a i n :
Elastic Stress-Strain Relations
/ / /
(dU)dV
= f f f
v
v
+ ol2
den
+
(oudeu
+ o22
de22
+
+ o2X
de2X + oX3
dex3
<*23^23 +
187
a33
de33
+
o3X
de31
(8.2.6)
o32
de32
)dV
where ^
represents the i n c r e m e n t s of the c o m p o n e n t s of the linear
strain. Therefore,
dexx
dU = axx
+ o22
de22
. . . o32
de32
=
aydey.
(8.2.7)
T h u s , the expression o n the r i g h t - h a n d side is a n exact differential a n d
a function U(etj) exists such that,
Since the stress tensor is s y m m e t r i c , t h e n
(8.2.9)
If, u n d e r the a p p l i c a t i o n of the external forces, the c h a n g e in state is
isothermal, it c a n b e s h o w n t h r o u g h the use of the s e c o n d law of
t h e r m o d y n a m i c s t h a t a function U with the properties expressed b y E q s .
(8.2.8) still exists. T h e function U is called the strain energy density
function. Therefore, the assumption
that a process is of a reversible
adiabatic or isothermal nature is implicit in the use of a strain energy
density function. L o a d s applied very slowly represent nearly i s o t h e r m a l
c o n d i t i o n s , a n d l o a d s applied very rapidly represent a d i a b a t i c c o n d i tions.
T h e use of a strain energy density function implies elasticity, in o t h e r
w o r d s , c o m p l e t e r e t u r n of a b o d y to its original s h a p e w h e n the stresses
a r e released. T h i s does n o t necessarily m e a n t h a t the relations b e t w e e n
stresses a n d strains are linear. Linearity is the a d d e d a s s u m p t i o n o n
which H o o k e ' s law is b a s e d . It will b e s h o w n in Sec. 8.7 t h a t this law
can b e arrived at b y neglecting all the t e r m s higher t h a n the q u a d r a t i c
in the expression of the strain energy density in t e r m s of the linear
strains.
188
The Theory of Elasticity
8.3
The Generalized Hooke's Law
F o r a large n u m b e r of h a r d solids, the m e a s u r e d strain is p r o p o r t i o n a l
to the load over a wide r a n g e of loads. This m e a n s t h a t w h e n the l o a d
increases, the m e a s u r e d strain increases in the s a m e ratio, a n d w h e n the
l o a d decreases, the m e a s u r e d strain decreases in the s a m e ratio. Also,
w h e n the l o a d is r e d u c e d to zero, the strain d i s a p p e a r s . T h e s e experim e n t a l facts lead b y inductive r e a s o n i n g to the generalized H o o k e ' s law
of the p r o p o r t i o n a l i t y of the stress a n d strain. T h e general form of the
law is expressed b y the s t a t e m e n t : E a c h of the c o m p o n e n t s of the state
of stress at a p o i n t is a linear function of the c o m p o n e n t s of the state of
strain at the p o i n t . M a t h e m a t i c a l l y , this is expressed b y :
=
°k(
e
(8.3.1)
Cktmn mn>
w h e r e the C k n
i amr e elasticity c o n s t a n t s . T h e r e are 81 such c o n s t a n t s
c o r r e s p o n d i n g to the indices k, / , m, n taking values equal to 1,2, a n d
3. F o r e x a m p l e ,
a
12
=
C
eC
\2\\ \\ +
1222^22 +
e
+
Q 233 ^ 3 3 +
e
Q213 13
+
C
eC
+
C\23\ 3\
+
\2\2 \2
^1223 ^23 +
1221
e C
e
2\
(5.3.2)
1232 32-
N o w , since the stress tensor is symmetric, i.e., since
=
°u
then
=e C
°U
Umn mn
Therefore,
(8.3.3)
°ik>
0 =e
Ik
Qkmn mn'
(S3
A)
=
Ckimn
(8.3.5)
Cfkmm
a n d the first pair of indices c a n b e freely i n t e r c h a n g e d . It is also possible
to p r o v e t h a t the s e c o n d pair of indices c a n b e freely i n t e r c h a n g e d . F o r
that, let us a s s u m e t h a t a b o d y is in a state of strain such t h a t the only
strain c o m p o n e n t different from zero is e n = ^21 • F o r this special
situation, E q . (8.3.1) is written:
e
°kt
-
Q/12^12
+
Q/21 21
(8.3.6)
Elastic Stress-Strain Relations
189
or
°U
=
( Q / 1 2 +
(8.3.7)
Q/2l)^12-
Let us i n t r o d u c e t h e n e w c o n s t a n t C k 2ndefined b y
Q / . 2
=
| ( Q / . 2
+
838
Q/2i)-
< - - )
W e see t h a t Ckn2 is s y m m e t r i c with respect to t h e t w o last indices. I n
terms of this n e w coefficient, E q . (8.3.7) is written as follows:
°ki
=
2
=
Q / 1 2 * 1 2
e
Q/12^12 +
Q / 2 1 2\
•
(8.3.9)
F r o m E q s . (8.3.6) a n d (8.3.9), we see that o n e c a n always consider t h a t
the c o n s t a n t s CUmn a r e also s y m m e t r i c with respect to t h e t w o last
indices: I n other w o r d s , these indices c a n b e freely i n t e r c h a n g e d . This
reduces t h e 81 elastic c o n s t a n t s to 36. F o r e x a m p l e , E q . (8.3.2) c a n also
b e written as follows:
°\2
=
e
C\2\\ \\
+
e
+
Q222 22 +
2(^1212 *12 +
Q213^13 +
^ 2 3 3 *33
C\22?> 2?)'
(o.3.1U)
T h e existence of a strain energy density function, weh e n t h e system is
a d i a b a t i c o r isothermal, allows us to g o o n e step further. I n d e e d , if such
a function exists, t h e n a c c o r d i n g t o E q s . (8.2.8),
-
W
dU
de22
Hence,
=
e22
ou - C u e uu + CU22
+ ...
=a
22
C22Ue n +
C
22 22* 2 2 + • • •
2
— c
3 f/
_ r
_
den de22— ^ 1 1 2 2
2211
+Cn32
e32
+
(8.3.11)
C 232*32-
2
r
a n d , in general,
^ve j veCumn - Cmnkf'
k
mn
Eq. (8.3.12) shows t h a t t h e elastic c o n s t a n t s Cktmn a r e s y m m e t r i c ; in
other w o r d s ,
(8.3.12)
190
T h e Theory of Elasticity
Q / m / z ~ Cmnk!'
(8.3.13)
Accordingly, t h e n u m b e r of i n d e p e n d e n t elastic coefficients for t h e
general a n i s o t r o p i c linearly elastic material is r e d u c e d t o 2 1 . I n a d d i tion, if certain symmetries exist in t h e material, this n u m b e r will b e
further r e d u c e d . T h e generalized H o o k e ' s law c a n n o w b e written in
matrix n o t a t i o n as follows:
Q u i
Q l 2 2
Ql33
Q l l 2
Q l l 3
Q123
^11 ~
°22
Q>211
Q222
^2233
Q212
Q 2 1 3
Q223
e^22
°33
Q311
Q322
Q333
Q312
C3313 C3323
33
°\2
Q 2 1 1
Q222
Q233
Q212
Q213
Q223
2el2
°\3
^1311
Q322
Q333
Q312
Q313
Q323
2e
2el3
°23
Q311
Q322
Q333
Q312
Q2313
Q323__
(8.3.14)
23
with C k ni = m
C
m . nT h eUm a t r i x of the elastic coefficients is a s y m m e t r i c
matrix. It is called t h e stiffness m a t r i x .
Since t h e c o m p o n e n t s of t h e stress a n d strain tensors a r e functions of
the o r i e n t a t i o n of t h e system of reference axes, t h e elastic coefficients in
Eq. (8.3.1) a r e also functions of this o r i e n t a t i o n . By t h e q u o t i e n t rule,
C
U n is
m a tensor of r a n k four called t h e stiffness tensor. Therefore, in a
n e w system of c o o r d i n a t e s OX\, OX2, OX'3 (Fig. 8.1):
=
Cprst
^pk h$ hm ^tn Cklmn
Fig. 8.1
'
(8.3.15)
Elastic Stress-Strain Relations
191
T h e stress-strain relations given b y E q . (8.3.1) c a n b e expressed in t h e
i n v e r t e d form,
(8.3.16)
kt ^klmn °mn >
e
C
where S k n
a r e c o n s t a n t s . It is evident t h a t S U n mh a s t h e s a m e
i m
s y m m e t r y p r o p e r t i eeas asr
kimn a n d it t r a n s f o r m s a c c o r d i n g to E q . (5.3.4).
c o m p o n e n t s of a tensor of t h e fourth o r d e r called t h e
c o m p l i a n c e tensor. I n m a t r i x n o t a t i o n , t h e generalized H o o k e ' s l a w c a n
b e w r i t t e n using t h e c o m p l i a n c e m a t r i x as follows:
Skimn
e
5 n l l
5*1122
^1133
5lH2
5lH3
5ll23
~°11
ee22
^2211
^2222
^2233
5*2212
52213
52223
°22
~ \\
33
5*3311
^3322
5*3333
53312
53313
53323
e\2
^1211
^1222
^1233
5l212
5i213
51223
e*\3
^1311
^1322
^1333
5l312
5l313
5l323
_ 23_
^2311
^2322
5*2333
52312
52313
52323
°33
2ol2
2ol3
2o23
(8.3.17)
with S U n m
= S,
>mnU'
8.4
Elastic S y m m e t r y
Cktmn ( SkMn)
A t y p e of s y m m e t r yor
is expressed b y t h e s t a t e m e n t t h a t t h e coefficients
remain invariant under the transformatio
n a t e s w h i c h describes this s y m m e t r y . W e shall c o n s i d e r t h e following
cases: (1) s y m m e t r y with respect to a p l a n e , (2) s y m m e t r y with respect
to t w o m u t u a l l y p e r p e n d i c u l a r p l a n e s , (3) s y m m e t r y of r o t a t i o n with
respect to o n e axis, a n d (4) s y m m e t r y of r o t a t i o n with respect to t w o
m u t u a l l y p e r p e n d i c u l a r a x e s — i n o t h e r w o r d s , isotropy.
(1) Symmetry
with Respect to One Plane: A m a t e r i a l w h i c h exhibits
s y m m e t r y of its elastic p r o p e r t i e s with respect to o n e p l a n e is called a
m o n o c l i n i c m a t e r i a l . L e t us t a k e this p l a n e to b e t h e OXx, OX2 p l a n e
(Fig. 8.2). T h i s s y m m e t r y is expressed b y t h e r e q u i r e m e n t t h a t t h e elastic
c o n s t a n t s d o n o t c h a n g e u n d e r a c h a n g e from t h e system OXx, OX2,
OX3 to t h e systems OX\, OX2, OX3. T h e direction cosines of t h e n e w
axes with respect t o t h e initial o n e s a r e ( 1 , 0, 0), (0, 1, 0), a n d (0, 0,-1).
F r o m E q . (8.3.15), w e m u s t , for e x a m p l e , h a v e :
hkhlhmhn^klmn
=
C'nil
=
192
The Theory of Elasticity
f*3
Fig. 8.2
w h i c h is true. T h e e x p a n s i o n in E q . (8.4.1) is simplified, since w e h a v e
only three n o n - z e r o direction cosines, n a m e l y :
Ai = 1.
hi = 1 ? ' 3 3 = - 1 -
(8-4.2)
I n a similar way, for this type of s y m m e t r y , we m u s t h a v e :
=
=
hkhihmhnCklmn
^1123
(8.4.3)
Ql23>
w h i c h is impossible since
hkhl
hm hn Ckimn = " Q 1 2 3 •
(8.4-4)
Therefore, CU2
3
m u s t b e e q u a l to zero. A similar r e a s o n i n g will s h o w
t h a t the n u m b e r of elements of the stiffness m a t r i x is r e d u c e d to
thirteen. T h e m a t r i x is written as follows:
Q u i
Ql22
Q133
Q112
0
0
0
Ql22
Q222
Q233
Q212
0
Ql33
Q233
Q333
Q312
0
0
Q l l 2
Q212
Q312
Q212
0
0
0
0
0
0
Q313
Q323
0
0
0
0
Q323
Q323
(8.4.5)
Elastic Stress-Strain Relations
193
It is to b e n o t i c e d t h a t a n y s u b s e q u e n t r o t a t i o n of axes will b r i n g in n o n zero t e r m s in the m a t r i x (8.4.5). T h e m a t r i x will, however, r e m a i n
s y m m e t r i c a n d the n u m b e r of i n d e p e n d e n t coefficients will r e m a i n 13.
A similar r e a s o n i n g in terms of c o m p l i a n c e leads to a m a t r i x of the s a m e
form as (8.4.5). In this case, the stress-strain relations are written as
follows:
e
22
e
e\ 2
el 3
^33
_ 23_
122 ^1133 S1112
Si 122 ^2222
S2212
Si
S3312
^1112
S3312 S1212
0
0
^2233
0
0
^3333
0
0
CT33
0
0
S1313
2ol2
Sl323
2on
Si 323
S2323_
Sllll
133
^2233
0
0
0
0
0
0
0
0
"
°22
(8.4.6)
_2<T 3_
2
Eq. (8.4.6) shows w h a t type of strains result w h e n each o n e of the
c o m p o n e n t s of the stress tensor is applied individually. F o r e x a m p l e , t h e
a p p l i c a t i o n of a stress o33 a l o n g the OX3 axis, results in three n o r m a l
strains e u, e22
, e33 a n d o n e shear strain el2
; b o t h el3 a n d e23 a r e e q u a l
to zero. T h e a p p l i c a t i o n of a stress a 13 will cause n o n o r m a l strain; j u s t
shear strains el3 a n d e23
.
/
/
^2>
^2.
Fig. 8.3
(2) Symmetry
with Respect to Two Orthogonal Planes: A m a t e r i a l
which exhibits s y m m e t r y of its elastic properties with respect to t w o
o r t h o g o n a l p l a n e s is called a n o r t h o t r o p i c m a t e r i a l . Let the t w o p l a n e s
b e the OXx, OX2 p l a n e a n d the OX2, OX3 p l a n e (Fig. 8.3). T h e direction
cosines of the n e w axes with respect to the initial ones are in this case
194
The Theory of Elasticity
(-1, 0, 0), (0, 1, 0), a n d (0, 0, -1). A s w a s d o n e in the case of m o n o c l i n i c
materials, the a p p l i c a t i o n of the t r a n s f o r m a t i o n law (8.3.15) with the
given direction cosines will lead to c o n t r a d i c t i o n s of the type s h o w n in
E q s . (8.4.3) a n d (8.4.4). T h e s e c o n t r a d i c t i o n s are a g a i n resolved b y
setting the elastic c o n s t a n t s e q u a l to zero. T h e n u m b e r of elastic
c o n s t a n t s is r e d u c e d to nine. T h e stiffness m a t r i x is written as follows:
Q l 2 2
Q 1 3 3
0
0
0
Ql22
Q 2 2 2
Q 2 3 3
0
0
0
Q l 3 3
Q 2 3 3
Q 3 3 3
0
0
0
0
0
0
Q 2 1 2
0
0
0
0
0
0
Q 3 1 3
0
0
0
0
0
0
Q 3 2 3
(8.4.7)
T h e c o m p l i a n c e m a t r i x h a s the s a m e form as the stiffness m a t r i x . U s i n g
the c o m p l i a n c e matrix, the stress-strain relations a r e written as follows:
e
~ n~
ee22
Si 122
122
^2222
33
^1133
S 233
0
*12
^llll
Si 133
0
0
0
^2233
0
0
0
22
^3333
0
0
0
033
0
0
S1212
0
0
2ai2
0
^1313
0
2(7,3
0
0
^l
2
e«13
0
0
0
23
0
0
0
S2323
~ <a» 1 1 "
2 a 23
This e q u a t i o n shows t h a t for o r t h o t r o p i c materials the a p p l i c a t i o n of
n o r m a l stresses results in n o r m a l strains alone, a n d the a p p l i c a t i o n of
shearing stresses results in shearing strains alone. This is only true,
however, in the system of axes with respect to w h i c h the symmetries are
defined.
(3) Symmetry of Rotation with Respect to One Axis: A m a t e r i a l w h i c h
possesses a n axis of s y m m e t r y , in the sense t h a t all rays at right angles
to this axis are equivalent, is called transversely isotropic or cross
anisotropic. T h e s y m m e t r y is expressed b y the r e q u i r e m e n t t h a t the
elastic c o n s t a n t s are u n a l t e r e d in a n y r o t a t i o n 0 a r o u n d the axis of
s y m m e t r y (Fig. 8.4). T a k i n g OX3 as the axis of s y m m e t r y , the direction
cosines of the n e w axes with respect to the initial ones are (cos 0, sin 0,
0), ( - s i n 0,cos 0,0), a n d (0, 0, 1).
I n s t e a d of starting from E q . (8.3.15), we shall start from the e q u a t i o n s
Elastic Stress-Strain Relations
195
XpXs
/
X;
i >
*X,
Fig. 8.4
of elasticity (8.3.1) in the initial a n d the t r a n s f o r m e d system a n d k e e p
the elasticity coefficients u n c h a n g e d . I n the OXx, OX2, OX3 system, the
elastic stress-strain relations are w r i t t e n :
°k( - Cklmn }
e
a n d in the OX\,
written:
OX3 system, the elastic stress-strain relations are
OX2,
N o t i c e t h a t in E q . (8.4.9)
a r o u n d the OX3 axis:
(8.4.9)
C
prst is u n p r i m e d . F o r a r o t a t i o n of axes
2
2
+ 2 cos 9 sin 9(eX2
) +
s\n 0(e22
)
e22 = s i n 0 ( e n) — 2 cos 0 sin 9(eX2
) +
c o s 9 ( e 2)2
e\i = cos 9(exx
)
2
e'33 =
e33
e'n = (e22 - exx
)cos
e
=s c o
\3
(8.4.8)
^13) + i
2
2
9 sin 9 + <? 1(cos
2 0 ns e
2
(8.4.10)
sin 0)
0( 23)
e'23 = - s i n 9(eX3
) + cos
9(e23
).
T h e c o m p o n e n t s of the stress tensor o u will t r a n s f o r m exactly in the
s a m e way. Let us, for i n s t a n c e , c o n s i d e r
196
T h e Theory of Elasticity
8
<^
3
= a
,
(
3 3
A
)
a relation which m a y b e written in the form:
£3311*11 + Q 3 2 2 * 2 2 + ^3333*33
+
=
2(C
3 e3i 21+ 2C 33 1 3 * 1 3
+
Q 3 1 1 * l l + Q322*22 +
+ 2(C
Q323*23)
(8.4.12)
Q333*33
3 e3121+ £23 3 1 3 * 1 3
f
+
£3323*23)-
Inserting in this e q u a t i o n the values of e y o b t a i n e d from Eqs. (8.4.10),
we find that
2
2
C 3 3[ c1o s1 0 ( e u) + 2 cos 9 sin 9(eX2
) + s i n 0 ( e 2 )2]
2
2
) + c o s 0 ( e 2 ) 2]
+ C 3 3[ s2i n2 9 ( e x ) x — 2 cos 6 sin 9(eX2
2
2
+ 2 C 3 3[ (1^ 22
2 — eu)cos
+ 2 C 3 3[ c1o s3 9(eu)]
2 Csn i
+
=
3323[~
0 sin 9 + el2
(cos 9
— sin 0)]
+ sin 9(e23
)}
cs
^(*13) +
£ 3 3 1 1 * 1 1 + £3322*22
°
(8-4.13)
%23)]
+ 2(C
3 e3121+ £23 3 1 3 * 1 3
+
£3323*23)-
E q u a t i n g t o zero the s u m of the coefficients of exxin this e q u a t i o n , w e
find t h a t for all values of #,
2
(£3311 -
£3322)sin 0
+ 2 sin 9 cos 0 C 3 23 =1 0,
(8.4.14)
from which it follows t h a t
=
£3311
£3322*
=
£3312
°-
(8.4.15)
If we e q u a t e the s u m of the coefficients of e22a n d eX2t o zero, we o b t a i n
the s a m e results. If we e q u a t e the s u m of the coefficients of eX3a n d e23
to zero, we find t h a t
£ 3 3 1 3 ~ £ 3 3 2 3 — 0.
(8.4.16)
1
1
Elastic Stress-Strain Relations
197
If we p e r f o r m similar c a l c u l a t i o n s for a 13 a n d a ' 1, 3we find, o n e q u a t i n g
the s u m of the coefficients of e n to zero, t h a t
=
Q311
=
Q322
=
Q312
=
Q311
0-
(8.4.17)
T h e s u m of the coefficients of e n, w h e n e q u a t e d to zero, leads to
=
Q311
C
1322'
=
Q312
C=
2312
°'
(8.4.18)
while the s u m of the coefficients of eX3 yields the relation:
C 2m = 0.
(8.4.19)
R e p e a t i n g these c a l c u l a t i o n s for au a n d a ' n, we find, o n e q u a t i n g to
zero the s u m of the coefficients of e u, e22
, a n d e 3 , 3t h a t
=
Q211
=
=
Q233
0'
Ql33
Q233>
=
Qlll
Q222-
(8.4.20)
T h e s u m of the coefficients of e l , 2w h e n e q u a t e d to zero, yields:
=
Q212
=
l ( Q i n " Q122X
Q222
0-
(8.4.21)
Finally, if w e c o n s i d e r the s u m of the coefficients of el3 in t h e e q u a t i o n
o b t a i n e d from a 23 a n d o23
, we get
=
Q323
Q313 •
(8.4.22)
S u b s t i t u t i n g E q s . (8.4.15) to (8.4.22) in the stiffness matrix, we get:
Qui
^1122
Ql33
C
0
0
0
0
0
0
0
0
0
Q313
0
0
Qm
Qui
1133
0
Ql33
Ql33
C3333
0
0
0
0
0
0
0
0
0
0
0
0
—
Q122)
2(^1111
(8.4.23)
Q313
T h e n u m b e r of i n d e p e n d e n t coefficients is r e d u c e d to five. A m a t r i x of
the s a m e f o r m c a n b e written for the c o m p l i a n c e s . U s i n g the c o m p l i a n c e
matrix, the stress-strain relations are w r i t t e n as follows:
198
The Theory of Elasticity
SUN
e
e21
122
Si
S I 133
122
SI
SIM
133
S I 133
0
0
0
S I 133
0
0
0
S3333
0
0
0
0
0
°33
2 o 12
0
2o,3
0
SI
0
e^3
0
0
0
0
Sl313
23
0
0
0
0
0
33
0
l(SIUI
_
S1122
°22
M313
_ 2 a 2 _3
(4) Isotropy: A n isotropic m a t e r i a l possesses elastic properties w h i c h
are i n d e p e n d e n t of the o r i e n t a t i o n of the axes. In o t h e r w o r d s , it is a
material which possesses a r o t a t i o n a l s y m m e t r y with respect to two
p e r p e n d i c u l a r axes. By r e p e a t i n g the a r g u m e n t s of tl\e previous subsection, we see that the elastic c o n s t a n t s of a n isotropic b o d y are given b y
a m a t r i x similar to Eq. (8.4.23) b u t with
_
Q313
2(Qin
Q333 ~ Q
(8.4.24)
Q133 ~" £1
1122'
so that, in fact, we only h a v e two i n d e p e n d e n t c o n s t a n t s . T h e stiffness
m a t r i x is written:
QUI
Q122
Q122
Q122
QUI
Q122
0
0
0
0
0
0
0
0
0 -
Q122
Q122
QUI
0
0
0
l(QIN
0
0
0
0 -
Q122)
0
0
i(QIN
0
0
0
0
0
Q122)
0
i(QIN
—
C1122)
(8.4.25)
A m a t r i x of the s a m e form c a n be written for the c o m p l i a n c e s . T h e
stress-strain relations in terms of the c o m p l i a n c e matrix are written as
follows:
Elastic Stress-Strain Relations
199
5
S
0
0
0
1122
^1122
l\22
S\122
0
Sun
S i 122
^1122
Sun
0
0
*13
0
0
0
0
*23
0
0
0
0
e
22
*33
^12
i(Si
0
0
0
0
0
*33
0
0
2a
11 ~~ S1122)
°22
12
13
_2a _
23
0
" S i 122
i(S ,i
n
0
2o
0
S1122)
(8.4.26)
T h e isotropic case is e x a m i n e d in detail in the following sections.
8.5
Elastic Stress-Strain Relations for Isotropic Media
T h e elastic c o n s t a n t s in t h e m a t r i x (8.4.25) a r e usually written in t h e
notation:
x
Q122 = K
Q212 = z ( Q i n - Q122) = ^
Qui =
2
851
+ ^
( - - )
T h e p a i r of c o n s t a n t s X a n d JU a r e called Lame's constants,
a n d /x is
referred to as t h e shear modulus
(also called G). T h e stress-strain
relations for a n isotropic m a t e r i a l a r e n o w written as follows:
°11
A
~A + 2ju
A
0
0
0~ " ^ 1 1
A
A + 2/x
A
0
0
0
33
A
A
A + 2/x
0
0
0
°12
0
0
0
0
0
0
0
M 0
0
0
0
0
0
a° 2 2
° n
_<*23_
0
0
e
22
2el2
.
(8.5.2)
2^13
_2e23
_
I n index n o t a t i o n , these e q u a t i o n s a r e written:
o^lfiey
+ XSye^.
(8.5.3)
200
The Theory of Elasticity
Eq. (8.5.3) c a n b e solved to yield the expressions of the strains in t e r m s
of the stresses:
A
8
a , 1
~ v
n
2JU <>•
^ ~ 2 K 3 A + 2jw) ™
Obviously, we m u s t require t h a t /x ^
form, we h a v e :
X +11
K3X + 2/i)
«11
(8.5.4)
0 a n d 3A + 2\x ^
-X
2/i(3X + 2/i)
-X
2/i(3X + 2ft)
-X
2/i(3X + 2/0
X+ n
K3X + 2/i)
-X
2/i(3X + 2/i)
-X
2/i(3X + 2/i)
-X
2/i(3X + 2/i)
X + /t
M(3X + 2/t)
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2an
0
20,3
±
4/i
2©23
4/i
e
23
0. I n m a t r i x
4/i
o
(8.5.5)
C o m p a r i n g E q s . (8.5.5) a n d (8.4.26), we see t h a t
, n , 1 X + /x
S u n~ —
/ x ( 3 X + 2|i)'
5
U22
_
iC'S'iin
_
S1122)
2ju(3X + 2/i)'
(8.5.6)
1
Eq. (8.5.5) shows that, for isotropic linearly elastic materials, a spherical
astate of= stress
=
=
aresultsa ewin a spherical state of strain. I n d e e d , if we set
n
°22
33
m>
obtain:
201
Elastic Stress-Strain Relations
Therefore, the c h a n g e of v o l u m e per unit v o l u m e d u e a spherical stress
om is given b y
^= ^ ,
e >> =^3A L
-h 2/x
K
9
(8.5.8)
where
3X
K =
+ 2
^
(8.5.9)
is called the Bulk Modulus. I n case w e h a v e a h y d r o s t a t i c c o m p r e s s i o n
of m a g n i t u d e P, E q . (8.5.8) is written:
Eq. (8.5.5) also shows t h a t for isotropic linearly elastic materials, a shear
stress results in n o t h i n g b u t a shear strain. I n general, a deviator stress
will c a u s e a deviator strain. I n d e e d , if, in E q . (8.5.5), w e set a 33
= — ( a n + a 2 ) 2, so t h a t the s u m of the three n o r m a l stresses is e q u a l to
zero (see the definition of a deviator stress), we o b t a i n the following
values for the n o r m a l strains:
€n
(A + / x ) a n
ti(3A + 2/x)
A q 22
2/x(3A + 2/x)
X(au + a 2 ) 2
2/x(3A + 2/x)
=an
~ °n
, (A + /x)a 22
2/x(3A + 2/x)
/x(3A + 2/x)
A ( q u + a 2 )2
2/x(3A + 2/x)
^22
=
X
6 22=
^
33 =
- X an
X a 22
2/x(3A
a -f 2/x) a 2/x(3A + 2/x)
= ( n + 22)
2/x
2/x
2/x
/ g 5 )1
5
1
) 12
_ (A + /x)(q n + a 2 ) 2
/x(3A + 2/x)
(8 5 13)
w h o s e s u m is equal to zero. T h e elastic stress-strain relations c a n thus
b e split i n t o t w o sets: First a relation b e t w e e n the spherical c o m p o n e n t s
of t h e stress a n d strain tensors, a n d s e c o n d a relation b e t w e e n the
deviatoric c o m p o n e n t s of the stress a n d strain tensors. T h e t w o following e q u a t i o n s c a n therefore replace E q . (8.5.3):
202
The Theory of Elasticity
(8.5.14)
onn = 3Kenn
(8.5.15)
N o t i c e t h a t E q . (8.5.14) is a scalar e q u a t i o n . I n t e r m s of //, a n d the b u l k
m o d u l u s K, Eq. (8.5.14) c a n also b e written as follows:
T h e two i n d e p e n d e n t c o n s t a n t s in engineering terminology are E,
Young's modulus a n d v, Poissorts ratio. U n d e r a uniaxial state of stress
e
= A + //, a 1 |
K3X + 2 M)
"
a
^
e.
=
On
X
=
a
5 1 )7
= 2A
t(3A + 2 J, ) "
- i "
3 3 -A
a 1i_
_
?a_
"2M3A+^)
£ " -
- '
Poisson's r a t i o p is thus the ratio b e t w e e n the lateral c o n t r a c t i o n a n d the
axial e l o n g a t i o n u n d e r a uniaxial stress c o n d i t i o n . Eqs. (8.5.17) define E
a n d v as
M3X + 2/0
„_
A + JLt
A ^
2(A + /x)
(8.5.18)
E q s . (8.5.18) c a n b e solved for A a n d /x to give:
A
=
^iJTfTy
(i + , ) 0 - 2 „ ) '
-
(8 5,19)
Eq. (8.5.3) c a n b e rewritten in terms of E a n d v as follows:
+
°v = T T ^
a
°
+,io-2,)V~
u
+e
(i
=
[ 1(
"
(1 + „)fl - 2 » )
22
=
~ ^
8 5 20
(8.5.21)
e +
V
(1 + )f\ v - 2V)\-
" 22 + f %>]
<-- )
n
^ ~ ")*22
+ " ^33]
(8.5.22)
Elastic Stress-Strain Relations
C
°33 " ( | + , j f l - „)1" '« +
+ (' " "> "1
2
< 8
203
' 5
M >
4)
•o-r^-..
° » ~ x h ' » -
"
E q . (8.5.4) c a n also be rewritten in t e r m s of E a n d v as follows:
etj = ^ [ ( 1 + v)aij - v
(8.5.25)
Syonn
]
or
e
a=
U
Jr( \\
va
v
~~
(8.5.26)
33)
°22 ~
(8.5.27)
e21 = - g ( - w n + a 22 - p a 3 ) 3
e 33 =
j*(-v
(8.5.28)
o n - v a 22 + a 3 ) 3
1 + v
"12
W e n e x t give in t a b u l a r f o r m the relations b e t w e e n t h e v a r i o u s elastic
constants:
Basic Pair
Constant
A, /i = G
A
A
jx = G
/i, G
K
E
V
3A + 2/i
3
/i(3A + 2/i)
A + /i
A
2(A + m)
E, v
v E
(1 + *>)(1 - 2v)
E
2(1 + v)
3K-2n
3
(8.5.30)
E
3(1 - 2v)
E
V
a:
3K + ii
3K-
2/i
6/^+2/1
F o r t h e shear m o d u l u s , the letters /x a n d G will b e u s e d i n t e r c h a n g e a bly in t h e c o m i n g sections. Since K is positive for all physical s u b s t a n c es, a n d E is positive, it follows t h a t v < 1/2. F o r a totally i n c o m p r e s s ible m a t e r i a l v = 1/2 a n d it = £ 7 3 . E q . (8.5.2) shows t h a t ft is a positive
c o n s t a n t , therefore 1 + j> > 0, a n d * > > - ! . T h u s , the v a l u e of Poisson's
(
8
204
The Theory of Elasticity
ratio seems to be limited at o n e e n d to 0.5 for incompressible materials
a n d at the other e n d t o - 1 . N e g a t i v e values of Poisson's ratio are
u n k n o w n in reality.
Remarks
All the e q u a t i o n s established in this section are also valid in a
cylindrical or a spherical system of c o o r d i n a t e s . In a cylindrical system,
the subscripts r, 0, z are substituted for 1, 2, 3, respectively, a n d in a
spherical system, the subscripts <J>, 0, p are substituted for 1, 2, 3,
respectively.
8.6
Thermoelastic Stress-Strain Relations for Isotropic Media
U n e q u a l h e a t i n g of the various p a r t s of a n elastic c o n t i n u o u s solid
p r o d u c e s stresses. T h e s e stresses w o u l d n o t b e p r e s e n t if each element
of the solid were allowed to e x p a n d freely; the c o n t i n u i t y of the m a t e r i a l
prevents free e x p a n s i o n a n d this in t u r n results in t h e r m a l stresses. Let
the t e r m p e r a t u r e T of a n elastic isotropic b o d y in a n a r b i t r a r y zero
configuration b e increased b y a small a m o u n t (A 7 ) . Since the b o d y is
isotropic, all infinitesimal line elements in the v o l u m e u n d e r g o e q u a l
e x p a n s i o n s . Also, all line elements m a i n t a i n their initial directions.
Therefore, if a is the coefficient of t h e r m a l expansion, the strain
c o m p o n e n t s d u e to the t e m p e r a t u r e c h a n g e (A 7 ) are
e
e
n
= 22 = *33 = <*(kT),
e n = e n = e23 = 0,
(8.6.1)
or
(8.6.2)
e& = Sya^T).
T h e thermally i n d u c e d strains c a n b e s u p e r i m p o s e d
i n d u c e d strains of E q s . (8.5.26) to (8.5.29) to give:
e
*12=
= Trfan ~ v o12 - v a 3 ) 3 +
22
= j?(-v
?33
=
to the
(8.6.3)
a(AT)
au + o22 - v a 3 ) 3 +
a(AT)
(8.6.4)
~ "o22 + a 3 ) 3 +
a(AT)
(8.6.5)
CT
1 +v
n
a
12>
a
e
\3
stress
-
1+ v
E
£
13>
23
e
1+ V
£
°23-
(8.6.6)
Elastic Stress-Strain Relations
205
In index n o t a t i o n , this set of e q u a t i o n s is written:
e,j =
+ "K
(8-6.7)
]
+ a 8&
(AT).
- v 6ijam
T h e inversion of E q s . (8.6.3) to (8.6.6) gives:
0 11
=
[ 1(
(1 + )(l
-
v
E
1 - 2v
22
2?)
v e
2 2 + " ^33]
=
e +0 )C22
+ ] e
" "
" *
[v
(1 + v)f\ - 2v)
a
£
"
n
(1 + ) (yl - 2 y ) ^ *
£
+ * *22 +
0
~
^33]
K
-a(Ar)
1 - 2v
e
°\2
=
TT~v ^
(8.6.8)
a(AT)
°
33 =
) 1e i +
~ "
a
e
°^
= TT^ ^
23 = iih;*23,
(8.6.9)
(8.6.10)
61 1
(8- - )
w h e r e the t e r m [(E/l — 2v)a(kT)] is the t h e r m a l stress i n d u c e d b y the
t e m p e r a t u r e c h a n g e . A n o t h e r w a y of writing the stresses in t e r m s of the
strains a n d the t e m p e r a t u r e c h a n g e is p r o v i d e d b y Eq. (8.5.3):
otj = 2fx etj + A ^enn - (3A + l^a{bT).
(8.6.12)
T h e stress-strain relation (8.6.12) is called the D u h a m e l - N e u m a n n law.
Remarks
(a) I n a cylindrical system of c o o r d i n a t e s , the subscripts r, 0, z are
s u b s t i t u t e d for 1, 2, 3, respectively; a n d in a spherical system of
c o o r d i n a t e s , the subscripts <£, 9, p are substituted for 1, 2, 3, respectively.
(b) I n t h e r m o e l a s t i c p r o b l e m s , the t e m p e r a t u r e d i s t r i b u t i o n m u s t first
b e d e t e r m i n e d from F o u r i e r ' s h e a t c o n d u c t i o n e q u a t i o n . T h e available
exact solutions of t h e h e a t c o n d u c t i o n p r o b l e m a r e limited.
206
The Theory of Elasticity
8.7
Strain Energy Density
Let us a s s u m e that the strain energy density U(etj) defined by E q s .
(8.2.8) can b e e x p a n d e d in a p o w e r series in terms of the etj ' s. T h e n
U=C
+C
+ - C
e
e
e
+
(8.7.1)
in which the C ' s are c o n s t a n t s . T h e c o n s t a n t C0 c a n b e disregarded
since the energy is m e a s u r e d from a n y arbitrary level. F r o m Eq. (8.2.9),
a n d neglecting the terms higher t h a n the q u a d r a t i c in Eq. (8.7.1), we
have:
oe
7Ty
8mi
8nj
ay — Cmn
I
=
C
J
+-C
s
+ ~ Cpqrs
[ers
8pj+
e
2v
s
r
+-C
2
cpq
8ri
8sj]
(8.7.2)
e
P^V
or
rj.. = r
+ -\C
4- C
k
(8.7.3)
F o r the stress to vanish in the a b s e n c e of strain, the c o n s t a n t Cy- m u s t
b e equal to zero. T h u s the expression for the strain energy density
reduces to
f/ = I c . rs ee
2
J
9l
r s
(8.7.4)
y
a n d t h a t of the stress to
o.. = I ( C .
+ C
)e
(8.7.5)
Eq. (8.7.5) c a n b e rewritten as follows:
e c
°ij =
ijrs rs>
(8.7.6)
w h e r e t h e coefficients Cijrsare symmetrized. Therefore, starting from
the a s s u m p t i o n t h a t a strain energy density function satisfying E q s .
(8.2.8) exists, a n d neglecting t h e t e r m s higher t h a n the q u a d r a t i c , w e
h a v e arrived at H o o k e ' s law a n d p r o v e d the s y m m e t r y of the stiffness
matrix. Substituting Eq. (8.7.6) into Eq. (8.7.4), we get:
U - \ o ve u.
(8-7.7)
Elastic Stress-Strain Relations
207
This f o r m u l a for the strain energy density is k n o w n as the Clapeyron
Formula. W h e n the stress-strain law is written, as in Eq. (8.3.16):
e
=
ij
(8.7.8)
Sijmn°mn>
the f o r m u l a of C l a p e y r o n gives:
u
=
1
so t h a t
S
W~
-
e
ijmn°mn
(8.7.10)
~ ij'
ij
This f o r m u l a is k n o w n as Castigliano's Formula. N o t i c e that, while E q s .
(8.2.8) d o n o t imply linearity, Eq. (8.7.10) does. T h e total strain energy
in a b o d y is
Ut = f J J
(8.7.11)
U dV.
v
In the following, we give several alternative forms of Eq. (8.7.7) for
isotropic bodies. F o r e x a m p l e , substituting Eq. (8.5.25) into Eq. (8.7.7),
we get:
2
U = ^ [ ( 1 + vfyOij
(8-7.12)
) ].
- v(onn
Explicitly, Eq. (8.7.12) has the form:
U
=f f
2lK
+
l'
a
33) ~
2
2 v+
2
a a
a
( ° l \ °22 +
22°33 +
2
33 ll)
(8.7.13)
+ 2(1 + * ' ) ( a 2 + a 23 + 0 1) ] .
Substituting Eq. (8.5.16) into Eq. (8.7.7), we get:
U
a
2
2
= ^\°ij ij
~ \^n) )
+ J^^nn) -
(8-7-14)
Substituting Eq. (8.5.3) into Eq. (8.7.7), we o b t a i n the expression of the
strain energy density in terms of the strains:
2
U = ju eyey + \\(enn
)
(8.7.15)
208
T h e Theory of Elasticity
Explicitly, E q . (8.7.15) h a s t h e form:
2
U = \\(en
+ e22 + e33
)
+ 2li(e\2 + e\3 +
+ /x(*ii + *22 + * 3 a )
(8.7.16)
e\3\
In case of t e m p e r a t u r e c h a n g e s , t h e term (3A + 2\£)a(bT\enn
)
s u b t r a c t e d from E q . (8.7.15), so t h a t
must be
2
U=n
e^j
- (3A + 2 M) « ( A 7 ) ( ^ ) .
)
+ \\(enn
W e sometimes n e e d to k n o w t h e strain energy associated with t h e
deviatoric c o m p o n e n t s a n d that associated with the spherical c o m p o n e n t s of t h e stress a n d strain tensors. It h a s b e e n s h o w n in Sec. 8.5 that,
for isotropic materials, spherical stresses result in spherical strains a n d
deviator stresses result in deviator strains. F o r this reason, t h e energy
associated with t h e spherical c o m p o n e n t s is called energy of dilatation
Us, a n d t h e energy associated with t h e deviatoric c o m p o n e n t s is called
energy of distortion Ud. I n terms of spherical a n d deviatoric c o m p o n e n t s ,
the expressions for the stress a n d strain tensors h a v e b e e n s h o w n to b e :
o, =
^
+
i v „ „
(8.7.17)
Substituting E q s . (8.7.17) a n d (8.7.18) into E q . (8.7.7), w e get:
f
j/ = I a
+-o
e e'-
(8.7.19)
W e notice that t h e first term of E q . (8.7.19) is t h e energy of dilatation
Us, a n d the second t e r m is t h e energy of distortion Ud. T h u s ,
U = Us + Ud9
(8.7.20)
a n d t h e total energy is t h e s u m of t h e energy d u e to t h e spherical
c o m p o n e n t s alone a n d t h a t d u e to t h e deviators alone. This superposition, however, does n o t h o l d for a n y general p a r t i t i o n i n g of t h e states of
stress o r of strain. T h e expressions of Us a n d Ud c a n b e written in t e r m s
either of t h e stresses or of t h e strains. U s i n g E q s . (8.5.14) a n d (8.5.15),
we get the following expressions in terms of the stresses:
Elastic Stress-Strain Relations
q
= Ia e .
11
Us
nn pp
Q
g
( »»)
( n= +
\ $
6 K
209
Q
22 +
(8.7.21)
33>
tf
K
u
2
* -
- ^<°W
2 () 2 8 J
= 4j [ * < / ^ - 5 ^ « ) ] -
-
Explicitly, E q . (8.7.22) is written:
a
2 A
-
Ud = T ^ K i i
22>
+
2
(°22
2
6
2 CT
~ o 3 )3 +
(033
-
n)
+
°12
+ 6 a 23 + 6 o 3]
or
Ud =
-
3j£,
AG
2G
(8.7.24)
=
In terms of the strains,
K(e ) K,
,
2
1
JT
m
.
0
2
x
(8.7.25)
Explicitly, E q . (8.7.26) is w r i t t e n :
2
U
d
=
-
^22>
+ 6el3 +
E
+
( 22 -
2
2
2E e
33>
+
( 33
~
«ll)
+
^ 1 2
(8.7.27)
6e 3]
or
2
( / , = - 6 G / , 2 = | G y (,
(8-7.28)
w h e r e Id2 is the s e c o n d i n v a r i a n t of the deviator strain tensor. Finally,
u p o n e x a m i n a t i o n of E q s . (8.7.4) a n d (8.7.16), o n e notices t h a t the strain
energy density is a q u a d r a t i c form in the strains (see Sec. 3.12). F o r
every set of values of the etj ' s, it takes only positive values. T h e s a m e
c a n b e said of
a n d Ud. T h u s , U is a positive definite q u a d r a t i c form
in the strains. This will b e used in establishing the u n i q u e n e s s of the
solution of elasticity p r o b l e m s .
210
The Theory of Elasticity
Remark
W h e n writing energy expressions in a cylindrical system of coordinates, the subscripts r, #, z are substituted for 1, 2, 3, respectively. I n a
spherical system of c o o r d i n a t e s , the subscripts (/>, 0, p are substituted for
1, 2, 3, respectively.
8.8
Formulation of Elasticity Problems. Boundary-Value Problems of
Elasticity
In general, a p r o b l e m of elasticity consists of finding the stresses a n d
the d i s p l a c e m e n t s in a n elastic b o d y subjected to surface forces, surface
displacements, a n d b o d y forces. T h e r e are six c o m p o n e n t s of the state
of stress at each point, a n d the three e q u a t i o n s of equilibrium are n o t
sufficient to o b t a i n the solution of the p r o b l e m . T h e six stress-strain
relations are therefore i n t r o d u c e d together with the six strain-displacem e n t relations (1.2.1). I n all, we h a v e 15 e q u a t i o n s a n d 15 u n k n o w n s
(oy^ey, w y). T o insure a u n i q u e value of the d i s p l a c e m e n t c o m p o n e n t s at
each point, the strains m u s t satisfy the compatibility relations.
T h e stresses a n d d i s p l a c e m e n t s are functions of the c o o r d i n a t e s .
W h e n the c o o r d i n a t e s of the points on the surface of the b o d y are
substituted into the expressions of the stresses a n d displacements, the
resulting values m u s t coincide with the externally i m p o s e d ones. W e are
thus faced with two types of b o u n d a r y value p r o b l e m s :
Problem 1 D e t e r m i n e the expressions of the stresses a n d displacem e n t s at all the p o i n t s in the interior of a n elastic b o d y in equilibrium
w h e n the b o d y forces are k n o w n a n d the surface forces are prescribed.
Problem 2 D e t e r m i n e the expressions of the stresses a n d displacem e n t s at all the p o i n t s in the interior of a n elastic b o d y in equilibrium
w h e n the b o d y forces are k n o w n a n d the surface d i s p l a c e m e n t s are
prescribed.
S o m e t i m e s the forces are prescribed o n o n e p o r t i o n of the b o u n d a r y
a n d the d i s p l a c e m e n t s o n the r e m a i n i n g o n e . This case is referred to as
the mixed b o u n d a r y value p r o b l e m . If stresses alone are i m p o s e d on the
b o u n d a r y of the elastic b o d y , it b e c o m e s desirable to express all the
e q u a t i o n s in terms of stresses. If displacements a l o n e are i m p o s e d to the
b o u n d a r y , then a formulation of the e q u a t i o n s in terms of displacem e n t s is generally m o r e useful. Both a p p r o a c h e s will b e e x a m i n e d in the
following sections.
Elastic Stress-Strain Relations
8.9
211
Elasticity Equations in Terms of Displacements
T h e original 15 e q u a t i o n s which a r e to b e solved in the analysis of a n
elasticity p r o b l e m c a n b e r e d u c e d to three e q u a t i o n s in t e r m s of the
c o m p o n e n t s of the d i s p l a c e m e n t s . T o o b t a i n these e q u a t i o n s , the set of
Eqs. (8.5.2) is substituted i n t o the e q u a t i o n s of equilibrium, t h e n the
strains are w r i t t e n in terms of the d i s p l a c e m e n t s . T h e o p e r a t i o n s are
m o s t c o n v e n i e n t l y m a d e in index n o t a t i o n . W e h a v e :
Stress-Strain:
2
°ij = V etj + A 80enn = 2/x etj + A 8tjev.
(8.9.1)
Equilibrium:
P-
+ F, = 0.
(8.9.2)
Strain-displacement:
S u b s t i t u t i n g E q s . (8.9.1) a n d (8.9.3) into E q . (8.9.2), we get:
^ V ^ + fA + ^
2
2
2
+ ^ O ,
(8.9.4)
2
w h e r e V = d /dxf
+ d /dxj
+ d /dxj
it form, E q . (8.9.4) is w r i t t e n :
is L a p l a c e ' s o p e r a t o r . I n explic-
2
(A +
x /) ^ + i t V W
l + F1 = 0
(8.9.5)
ox
x
2
(A + / x ) P - + it V u2 + F2 = 0
(8.9.6)
(A + ^
(8.9.7)
ox
2
+ ^ V ^
+ ^ O .
Eqs. (8.9.5), (8.9.6), a n d (8.9.7) are called N a v i e r ' s e q u a t i o n s of elasticity. T h e b o u n d a r y c o n d i t i o n s (7.3.8) c a n b e w r i t t e n in t e r m s of the
d i s p l a c e m e n t s a n d strains as follows:
212
T h e Theory of Elasticity
O n c e o n e h a s f o u n d a solution satisfying E q s . (8.9.5) to (8.9.7), t h e
strains a n d t h e stresses c a n easily b e o b t a i n e d . T h e b o u n d a r y c o n d i t i o n s
m u s t b e satisfied. T h e r e is n o n e e d t o check compatibility since t h e
strains a r e o b t a i n e d from t h e d i s p l a c e m e n t s a n d n o t vice versa. I n case
of t e m p e r a t u r e changes, E q s . (8.9.4) a n d (8.9.8) b e c o m e :
2
- (3A + 2ju)«g|-(Ar) + Ft = 0
juV «,. + (A +
oni + (3A +
2/xMAr)/,. =
jLt(|| + ^jtj
+ \ltev.
(8.9.9)
(8-9.10)
Eqs. (8.9.5) to (8.9.7) c a n b e written u n d e r t h e form of o n e vector
2 e is n o t2h i n g b u t t h e divergence of the d i s p l a c2e m e n t u,
e q u a t i o2n . Since
v
a n d V w 1? V w 2> a n d V w 3 a r e t h e c o m p o n e n t s of a vector V w, t h e n
2
(A + n) g r a d (div u) + /x V w + F - (3A + 2/x)a grad (AT) = 0.
(8.9.11)
T h i s vector e q u a t i o n c a n easily b e t r a n s l a t e d into c o m p o n e n t s in a n y
system of o r t h o g o n a l curvilinear c o o r d i n a t e s b y using t h e relations of
Sec. 6.4.
8.10
Elasticity Equations in Terms of Stresses
N o t every solution of t h e e q u a t i o n s of equilibrium c o r r e s p o n d s to a
possible state of strain b e c a u s e t h e c o m p o n e n t s of t h e strain m u s t
satisfy t h e e q u a t i o n of compatibility (4.10.14) to insure t h e existence of
single v a l u e d displacements. Let us consider, for example, t h e c o m p a t i bility e q u a t i o n :
a 2 g
,
23
9 2 g
22
'dx dx
dxj
23
=
9^33
+
(8.10.1)
dx%
H e r e w e shall u s e t h e n o t a t i o n :
0
= J
x
= o
u+
o
22 +
o
3. 3
(8.10.2)
Elastic Stress-Strain Relations
213
S u b s t i t u t i n g E q s . (8.5.26) to (8.5.29) i n t o Eq. (8.10.1), we get:
2
V
dx
dx
= 2(1 +
)
2
/
V 9.x
2
dx
39 2 ( 7 2
2 /
^ IQ ^
r
F r o m t h e s e c o n d a n d third e q u a t i o n s of equilibrium, w e h a v e :
3 a 23 _
3 a 32
dx
dx
3x2
dx
-
F2
(8.10.4)
— FX
.
(8-10.5)
x
3 a 13
3 a 33
3x]
3x3
2
2
3 a 12
3 a 22
3 a 32 __ 3 a 23 _
3x3
dx
3
Differentiating E q . (8.10.4) with respect to x2 a n d Eq. (8.10.5) with
respect to x3, a d d i n g t h e m together, a n d i n t r o d u c i n g the first e q u a t i o n
of equilibrium, we get:
d2
_9^23_
2
3 x 2dx3
°U
_
=
dx?
2 ( J
d2
_
°22
dxl
d
_ ^2
M\
+
33
9*i
dxj
(8.10.6)
_
9
9*2
*3 '
S u b s t i t u t i n g E q . (8.10.6) i n t o E q . (8.10.3), w e get:
<'->K' °..-g)-'(™-0)
!
fdFx _ 3 / ^ _
(8.10.7)
3 F 3\
U s i n g the two o t h e r relations of the type E q . (8.10.1) in Eq. (4.10.14),
two e q u a t i o n s similar to Eq. (8.10.7) c a n b e o b t a i n e d . T h e y c a n directly
b e d e d u c e d from E q . (8.10.7) b y cyclical p e r m u t a t i o n . A d d i n g these
three e q u a t i o n s together, we find t h a t
V2 0
=_ f W ^
^
x
1 — v \2 ax
2
ax
;>M + +
3
ox
(8.10.8)
/
Substituting this expression for V 0 i n t o E q . (8.10.7), we finally o b t a i n :
2
V
0
"
+1 9 0
i + *ax?
v_(^_
1 — v \ dxx
+
^
+
dx
2
^M_ M>.
2
dx
3
/
dx
x
(8.10.9)
214
The Theory of Elasticity
T w o similar e q u a t i o n s are o b t a i n e d b y circular p e r m u t a t i o n .
In a similar m a n n e r , the r e m a i n i n g three compatibility e q u a t i o n s c a n
be t r a n s f o r m e d into e q u a t i o n s of the form
23
^ °
( 3 ^ 2 ~*~ 3^3 ) '
~*~ 1 4- vdx2dx3
^
^
T w o similar e q u a t i o n s are again o b t a i n e d b y circular p e r m u t a t i o n .
G a t h e r i n g the results, the stresses are o b t a i n e d b y solving the system of
six e q u a t i o n s with 6 u n k n o w n s :
2
1 d 19
+ T~T—f~f = - r ^
1 + vdx
V %
div F - 2 ^ -
(8.10.11)
2
1 d9
_+ ! _ ^ = _
» ^ d i v F - 2 | ^
1 + Vdx\
1 —v
ax2
V2 0„
1 de
1 + vdx}
(8.10.12)
2
1 —
v
d i v F - 2 ^ dx-*
(8.10.13)
+ *5.)
(8.10.14)
2
2
d9
V o, „
n T+
J*5-
12
V 3 ^ 3x
1 +
\ 3-^2
2
a +
^
23
^
3i
9*1 '
+
]
^
a +
]
+
g^
(
)
^
^
^
^
+
+^ 3 x 33 x j
(3^!
3x )'
3
In index n o t a t i o n , these e q u a t i o n s are written:
1 + v dx, dxj
V
1
J L _ « fly y
+
(8.10.17)
In case of t e m p e r a t u r e c h a n g e s , Eq. (8.10.17) b e c o m e s :
2
V
a+ 1
v
l+v
_
d [9 +
Ea(AT)]
dXi
dxj
,
, d
1 - v '•>
(8.10.18)
8 l _ ( v| £ F
\ dxj
+p )
dxj /
_
5
(AT)]
^ ,Js
1- v
215
Elastic Stress-Strain Relations
E q s . (8.10.11) to (8.10.16) are k n o w n as the Beltrami-Michell
compatibility equations. T h u s , the state of stress in the interior of a n elastic b o d y
m u s t satisfy the three e q u a t i o n s of equilibrium, the Beltrami-Michell
compatibility e q u a t i o n s , a n d the b o u n d a r y c o n d i t i o n s (7.3.8).
8.11
The Principle of Superposition
Let the stresses in a n elastic b o d y subjected to surface forces Qi a n d
to b o d y forces Fi b e oj(. Let the stresses in the s a m e elastic b o d y w h e n
it is subjected to the surfaces forces Q] a n d to the b o d y forces F\ b e aj 7.
T h e n the stresses a y7 4- aj, will represent the stresses d u e to the surface
forces Qj + Q] a n d the b o d y forces Fi + F'r This h o l d s b e c a u s e all the
differential e q u a t i o n s a n d b o u n d a r y c o n d i t i o n s are linear. T h u s , a d d i n g
u p the two sets of e q u a t i o n s of equilibrium for the first a n d s e c o n d state
of stress,
da
£
; =+F0
(8.H.1)
and
P
+ F'^0,
(8-H-2)
we get:
^fy
+ ty + F. + F'.-O.
(8.11.3)
Also a d d i n g u p the two sets of b o u n d a r y c o n d i t i o n s for the first a n d
s e c o n d state of stress,
a„, = a,,4
(8.11.4)
^ + ^ = (^ + ^ H -
<- )
and
we get:
8
1L6
216
T h e Theory of Elasticity
T h e compatibility e q u a t i o n s c a n also b e c o m b i n e d in t h e s a m e m a n n e r .
T h e c o m p l e t e set of e q u a t i o n s shows t h a t (o^ + aj 7) satisfy all t h e
r e q u i r e m e n t s a n d c o n d i t i o n s d e t e r m i n i n g t h e stresses d u e t o t h e surface
forces Qt + Q\ a n d t o t h e b o d y forces Ft + F\. This is t h e principle of
superposition.
In o u r study, n o distinction w a s m a d e b e t w e e n t h e u n d e f o r m e d a n d
the d e f o r m e d shapes of t h e elements in equilibrium. C o n s e q u e n t l y , t h e
principle of superposition will only b e valid for t h e cases of small
d i s p l a c e m e n t s . A g a i n , all this is in line with the use of t h e linear t h e o r y
of strain.
8.12
Existence and Uniqueness of the Solution of an Elasticity Problem
T h e rigorous proof of t h e existence of solution is t o o lengthy a n d will
n o t b e e x a m i n e d h e r e ; also it is to b e r e m e m b e r e d t h a t w e a r e dealing
with linear strains a n d small d i s p l a c e m e n t s . I n o r d e r to establish t h e
u n i q u e n e s s of t h e solution of a n elasticity p r o b l e m , let us a s s u m e that it
is possible t o o b t a i n t w o solutions,
a ' n,
2 u'3
(8.12.1)
o j ^ o f c , . . . , u\, tiS,«S,
(8.12.2)
( J 2 2 , . • • , u\, u' ,
which satisfy t h e 15 elasticity e q u a t i o n s a n d t h e s a m e set of b o u n d a r y
c o n d i t i o n s . Therefore, for t h e first set of stresses, t h e e q u a t i o n s of
equilibrium,
^
+ ^ = 0,
(8-12.3)
are satisfied a s well as t h e following b o u n d a r y c o n d i t i o n s ,
a„, = o;.Z y,
(8.12.4)
if t h e surface forces a r e prescribed, o r
Uj = u'j
(8.12.5)
if t h e b o u n d a r y d i s p l a c e m e n t s a r e prescribed. F o r t h e s e c o n d state of
stress, w e h a v e :
Elastic Stress-Strain Relations
^L
1
+ F^O
217
(8-12.6)
dxj
a„, = a ; / y
(8.12.7)
w, = w';.
(8.12.8)
If we s u b t r a c t E q s . (8.12.6), (8.12.7), a n d (8.12.8) from E q s . (8.12.3),
(8.12.4), a n d (8.12.5), respectively, we find t h a t the stress distribution
defined b y a'y — a"j satisfied the e q u a t i o n s :
^ ( a ; 7- a ; ) = 0
(8.12.9)
= 0
(8.12.10)
(<£ - o^lj
u\ - u'] = 0.
(8.12.11)
This is a n e w " d i f f e r e n c e " stress distribution in which all the external
forces, the b o d y forces, a n d the b o u n d a r y d i s p l a c e m e n t s vanish. If there
are n o external forces or b o u n d a r y d i s p l a c e m e n t s , there is n o w o r k d o n e
a n d the strain energy stored in the b o d y m u s t b e equal to zero. It h a s
b e e n established in Sec. 8.7 t h a t the strain energy density is a positive
definite q u a d r a t i c form in the strain c o m p o n e n t s . It c a n n o t vanish
unless all the strains vanish. Therefore, if the strain energy stored in the
b o d y is zero, the strain c o m p o n e n t s a n d c o n s e q u e n t l y the stress
c o m p o n e n t s m u s t b e zero everywhere. C o n s e q u e n t l y , the difference
state of stress Oj; — o"^ m u s t b e z e r o a n d the t w o solutions m u s t b e
identical.
8.13
Saint-Venant's Principle
In 1885, B. d e S a i n t - V e n a n t in his m e m o i r o n torsion p r o p o s e d a
principle which c a n b e stated as follows: If a system of forces acting o n
a small p o r t i o n of the surface of a n elastic b o d y is r e p l a c e d b y a n o t h e r
statically equivalent system of forces acting o n the s a m e p o r t i o n of the
surface, the redistribution of l o a d i n g p r o d u c e s s u b s t a n t i a l c h a n g e s in
the stresses only in the i m m e d i a t e n e i g h b o r h o o d of the loading, a n d the
stresses are essentially the s a m e in the p a r t of the b o d y w h i c h are at
large distances in c o m p a r i s o n with the linear d i m e n s i o n of the surface
o n which the forces are c h a n g e d . By statically equivalent, we m e a n t h a t
the two distributions of forces h a v e the s a m e resultant force a n d
218
The Theory of Elasticity
m o m e n t . T h e principle of S a i n t - V e n a n t allows us to simplify the
solution of m a n y p r o b l e m s b y altering the b o u n d a r y c o n d i t i o n s while
keeping the systems of applied forces statically equivalent. A satisfactory a p p r o x i m a t e solution c a n thus b e o b t a i n e d .
8.14
One Dimensional Elasticity
Let us a s s u m e t h a t the b o d y forces are negligible. T h e r e are two states
to consider: a) a o n e d i m e n s i o n a l state of stress a n d b) a o n e
d i m e n s i o n a l state of d e f o r m a t i o n .
a) One dimensional state of stress. A o n e d i m e n s i o n a l state of stress
exists in a body, if, at all its points, the stress m a t r i x is of the form:
0
0
0
0
0
0
0
0
a 33
(8.14.1)
in which o33 is a function of x3 alone. T h e principal directions are the
OX3 direction a n d all directions in the OXx, OX2 p l a n e . T h e two first
e q u a t i o n s of equilibrium are identically satisfied a n d the third e q u a t i o n
yields:
(8.14.2)
a 33 = c o n s t a n t =
T h e stress-strain relations (8.5.25) b e c o m e :
e
e
e
\ \ = 22 = ~E°o>
e
\2
e
= \3
33
e
= 23 =
(8.14.3)
=
(8.14.4)
°'
T h e d i s p l a c e m e n t s are o b t a i n e d by integration of Eqs. (1.2.1):
du
x = e =
xx
x
(8.14.5)
dx
du
2 = e =
22
2
e
du
%
3
dx = 33 = E
3
v
dx
3w,
du
dx
3x,
2
2
(8.14.7)
du
du
du
3x3
9^1
dx
x
(8.14.6)
n
3
2
3
3«3
dx
2
(8.14.8)
Elastic Stress-Strain Relations
219
F o r n o rigid b o d y d i s p l a c e m e n t u{, u2, a n d u3 a r e given b y :
x
u
\ = -j,°o \,
u1 = -j,o0x1,
u3 = ^ x 3.
(8.14.9)
b) One dimensional state of deformation.
A o n e d i m e n s i o n a l state of
d e f o r m a t i o n exists in a b o d y , if, at all its points,
Wl = 0,
w 2 = 0,
is a function of x3 a l o n e .
u3 = u3{x3)
(8.14.10)
(8.14.11)
F r o m E q s . (1.2.1), we get:
= e22 = el2 = eX3 = e23 = 0
e 33 = e33
(x3)
is a function of x3 a l o n e .
(8.14.12)
(8.14.13)
T h e stress-strain relations (8.5.21) to (8.5.24) give:
E(\
-
o33 = ^
= a= +
°\2
13
°23
S1 au n c t l
v) e
_ 2^ 33
f
o n of x 3 a l o n e
=
0-
(8.14.15)
(8.14.16)
T h e t w o first e q u a t i o n s of e q u i l i b r i u m a r e identically satisfied a n d the
third o n e gives:
o33 = c o n s t a n t = o0.
(8.14.17)
<?33 = c o n s t a n t = eQ
.
(8.14.18)
Therefore,
F o r n o rigid b o d y d i s p l a c e m e n t ,
ux = u2 = 0
(8.14.19)
u3 = e0x3.
(8.14.20)
220
The Theory of Elasticity
8.15
Plane Elasticity
In a b o d y t h a t is being elastically deformed, let us consider, for
e x a m p l e , a n axis OX3: If all the p l a n e s initially n o r m a l to OX3 r e m a i n
n o r m a l to it after d e f o r m a t i o n , a n d if all the straight lines initially
parallel to OX3 r e m a i n parallel to it after d e f o r m a t i o n , a state of plane
deformation is said to exist in the b o d y . Analytically, these c o n d i t i o n s
are expressed b y :
ux = ux(xx,x2\
u2 = u2(xl,x2),
u3 =
u3(x3).
If u3 = u3(x3)
= 0, t h e b o d y is said to b e in a state of plane strain in the
OXx, OX2 p l a n e . T h u s , the state of p l a n e d e f o r m a t i o n results from the
superposition of a state of p l a n e strain a n d a state of o n e d i m e n s i o n a l
deformation.
If the state of stress in a b o d y is such that a 13 = o23 = o33 = 0, a n d
a
ll
a
= <*ll(*l>*2)>
2 2 = <*22(*1>*2)>
°\2
= * 1 2 ( * 1 > *2>>
the b o d y is said to b e in a state of plane stress.
If a c o n s t a n t strain eQ in the OX3 direction is s u p e r i m p o s e d to a state
of p l a n e strain w i t h o u t c h a n g i n g the stresses in the OXx, OX2 planes, the
b o d y is said to b e in a state of generalized plane strain.
In the case of a thin plate, we sometimes seek the average values
across the thickness of the c o m p o n e n t s of the d i s p l a c e m e n t vector a n d
of the stress tensor. T h e e q u a t i o n s in t e r m s of these averages are called
the e q u a t i o n s of generalized plane stress.
8.16
State of Plane Strain (Fig. 8.5)
Fig. 8.5
Elastic Stress-Strain Relations
221
A s i n d i c a t e d in Sec. 8.15, a state of p l a n e strain is c h a r a c t e r i z e d b y
u2 = u2(xx,x2),
u3 = 0. T h u s ,
ux = ux(xx,x2),
e
x)
xx = eu(xl9
2
=
e
\ ( 3WI
e
=
(x ,x )
22= e22
x2
3W
2\
jj^
8.16.1)
8.16.2)
8.16.3)
33 = e 13= e23= 0.
T h e stress-strain relations (8.5.3) b e c o m e :
o,, = 2n e u + X(en +
2
°22 = M ?22 + M^ii +
o 33 =
X(e
°\2
/* \2'
e )
{) +
= 2e
22
°\3
e )
8.16.4)
e )
8.16.5)
22
22
8.16.6)
=
23 °.
a
=
8.16.7)
T h e stress-strain relations (8.5.21) to (8.5.24) b e c o m e :
E
(1 + ")(1 - 2v)
8.16.8)
[y
E
*22 = (1( i ++ v){\
yx i
1 (
- 22v)) y
"
+
"^22]
33 = i ; fv -E2 , ) ( n + ^ 2 )
q
g
+
(
°13 = ^23 =
- 16
9)
(8-16.10)
) (
e
TT7 12'
«12 =
(8
0.
(8-16.11)
T h e stress-strain relations (8.5.26) t o (8.5.29) b e c o m e :
e\\
=
1 +
"[(1
-[(1-»*>.,-""22]
E
(8.16.12)
e
=
1 +
E "[(1
(8.16.13)
22
1 +
e\2 =
E
V
°\2
v)o22 - v oxx
]
e
e
e
33 = \3 = 23 = °-
(8.16.14)
Eq. (8.5.28) shows t h a t for t h e case of p l a n e strain, a 33 is p r o p o r t i o n a l
to ( a , , + a22) since
222
The Theory of Elasticity
a 33 = v(ou + a 2 ) 2
.
(8.16.15)
Eq. (8.16.6) shows t h a t a 33 is a function only of xx a n d
T h e e q u a t i o n s of equilibrium b e c o m e :
x2.
-p± + pL
d
(8.16.16)
+Fl=0
ax
d }
^l
ox2
p2
+ F
+
L
= 0
2
ox2
OXi
F3 = 0.
(8.16.17)
(8.16.18)
C o n s e q u e n t l y , in p l a n e strain, the b o d y forces are such that F3 = 0 a n d
Fj a n d F2 a r e i n d e p e n d e n t of x3.
Five out of the six compatibility relations are identically satisfied a n d
only o n e , n a m e l y ,
9 22 g
_ 9 *ii
2
9*19x2
9JCJ
i2
2
2
, 9 e 22
3X!
(8.16.19)
is to be considered.
T h e e q u a t i o n s of elasticity (8.9.5) to (8.9.7) b e c o m e :
(A +
9(^i 1 + ^ 2 2 )
)G
8
(A + G )
+ 2 W] G
( +e g) 2
"a
+VF
=0
(8.16.20)
22
+ GV u2 + F2= 0
(8.16.21)
F3 = 0,
(8.16.22)
where
U s i n g the stress-strain relations a n d the equilibrium e q u a t i o n s , Eq.
(8.16.19) can b e written in terms of stresses as follows:
1
2
\dx
dx^r
1 y
2 2
l-c\ax,
T h e b o u n d a r y c o n d i t i o n s (7.3.8) b e c o m e (Fig. 8.5):
8 x 2/
Elastic Stress-Strain Relations
°nl = V l 2 + 4 ° 2 2
(8.16.25)
a
°n3
F o r c o n s t a n t b o d y forces,
223
= h 33 •
2
V ( a n + a 2 ) 2 = 0,
(8.16.26)
which indicates t h a t ( o u 4- a 2 ) 2 is h a r m o n i c .
/AZ a system of cylindrical coordinates the strain-displacement relations
are given b y Eqs. (6.7.23) a n d (6.7.24):
K
.
^
(8.6.27)
\M
l(l^=
2 \ r 80
^_M
+
r
(8.16.28)
)
dr
ezz = erz = eBz = 0.
(8.16.29)
All the relations in Eqs. (8.16.4) to (8.16.15) r e m a i n the s a m e except that
r, 0, a n d z are substituted for 1, 2, a n d 3, respectively. T h e e q u a t i o n s of
equilibrium are given b y E q s . (7.12.10):
^
+
1
^
+ 2 ^ + ^ - 0
Fz = 0.
(8-16.31)
(8.16.32)
T h e e q u a t i o n of compatibility (8.16.19) b e c o m e s :
+
j _ o +^
2
d r
i
2
r
r W
9r
_ ] ^
r a
r
/
|
cz ^ j_cte^N
\ r ,-90
9 ^ 2 90
r /'
8
( 1) 6 3 3
8
1) 6 3 4
T h e e q u a t i o n s of elasticity in terms of d i s p l a c e m e n t s b e c o m e
(
X +
<
^ L ' f r
n) ,+
'
' l * J
224
The Theory of Elasticity
(A +
rK
)
rr
d$\_ dr
+
d9
.
(8.16.35)
* & ( ' £ ) * ( l S ? ) ] «-<>
+
+
Fz = 0.
(8.16.36)
U s i n g the relations established in Sec. 6.2, the compatibility relation
(8.16.24) b e c o m e s :
r
L
=
30
1 - v \ dr
r
(8.16.37)
J'
F o r c o n s t a n t b o d y forces, (prr + o00
) is h a r m o n i c .
8.17
State of Plane Stress
As m e n t i o n e d in Sec. 8.15, this state is defined b y
°n
= °ii(*i>*2)>
<>22 = <*22(*i>*2)>
a
i 2 = *i2(*i>*2)
a 33 = a 13 = a 23 = 0.
(8.17.1)
(8.17.2)
T h e stress-strain relations (8.5.3) b e c o m e :
ou
= 2/i e u +
= 2/x
e i i +
2
<*22 =
\(en
4-
2 M(
2im
M ^22 + M^ll
e22 4- e 3 ) 3
e i i +
^
+ ^22 + ^ 3 3 )
2 / 2x e+ 2/iX
2
2^n ^
(
i
+
,e 2 2_)
^33 = - 2 ^ 1 ( ^ 1 1 + ^22)
2
<*12 = ^ *12>
(8.17.3)
2 2 )
a a
13 = 23 = °-
8 1 74
( - - >
(8.17.5)
(8.17.6)
F o r m a l l y , Eqs. (8.17.3) a n d (8.17.4) b e c o m e identical to E q s . (8.16.4)
a n d (8.16.5) if o n e replaces the c o n s t a n t 2A/x/(A -h 2jut) = X b y A. T h e
stress-strain relations (8.5.21) to (8.5.24) b e c o m e :
Elastic Stress-Strain Relations
E
L
(8.17.7)
o u = - - 1 ( e n + ve22
)
1 — V
Eg
L
°22 = ,
1 —
^33 =
225
VE
( 22 +
-r^^ll
(8.17.8)
\\)
2
V
(8-17.9)
+^22)
e
«i2-TTT «'
° ' 3 = ° 2 3 = 0.
(8-17.10)
T h e stress-strain relations (8.5.26) to (8.5.29) b e c o m e :
(8-17.11)
e u = ^ o u- v c 2 ) 2
«22
|(-'"'ll+»22)
=
(8.17.12)
^33 = - f ( ° n + " 2 2 )
12 _ 1 + v
"
~E~
°12>
(8.17.13)
^ . 3 = ^23 = 0.
(8-17.14)
E q u a t i o n (8.17.13) shows t h a t e33 is only a function of xx a n d x 2. T h e
e q u a t i o n s of e q u i l i b r i u m b e c o m e :
^ii + ^ L
axx
ox2
^11
OXI
+ F0
+^ 2 2 _ +
ox2
(8-17.15)
=
F0
(8-17.16)
F3 = 0.
(8.17.17)
T h u s , in p l a n e stress, the b o d y forces a r e such t h a t F3 = 0, a n d Fx a n d
F2 a r e i n d e p e n d e n t of x3.
Of t h e six compatibility e q u a t i o n s (4.10.14), t w o a r e identically
satisfied a n d t h e four r e m a i n i n g ones b e c o m e :
2
2
d el2
d eu
3xj 3 x 2
2
3x^
2
d ^2
d =^
3JCJ
dx\
2
2
+
d e22
2
(8.17.18)
dxf
2
d=
e33
dx
x
dx
o =
(g.17.19)
2
Eqs. (8.17.19) d e m a n d that e33 b e given b y a n e q u a t i o n of the form
226
The Theory of Elasticity
e33 = C, + C 2x , + C3x2,
(8.17.20)
where Q , C 2 a n d C 3 are c o n s t a n t s . In general, however, only Eq.
(8.17.18) is taken into a c c o u n t in c o m m o n p r o b l e m s , a n d the requirem e n t of (8.17.20) is neglected. This results in a solution which, a l t h o u g h
a p p r o x i m a t e , is very satisfactory w h e n the d i m e n s i o n of the b o d y in the
OX3 direction is very small (thin plates).
U s i n g the stress-strain relations (8.17.11) to (8.17.14) a n d the equilibr i u m e q u a t i o n s (8.17.15) to (8.17.17), Eq. (8.17.18) can b e written in
terms of the stresses as follows:
T h e b o u n d a r y conditions (7.3.8) b e c o m e :
°nl
=
h°\2 h°22
(8.17.22)
+
n3 = 0.
o
F r o m E q s . (8.16.24) a n d (8.17.21) we see that, w h e n the b o d y forces are
constant, the distribution of stresses in the (OX{, OX2) p l a n e is the s a m e
for p l a n e strain a n d for p l a n e stress p r o b l e m s , a n d is governed by the
equilibrium e q u a t i o n s a n d Eq. (8.16.26). O n e m u s t n o t forget however
the a p p r o x i m a t e n a t u r e of the solution w h e n Eq. (8.17.20) is neglected.
In a system of cylindrical coordinates, the strain d i s p l a c e m e n t s relations are given by Eqs. (6.7.23) a n d (6.7.24):
d
+ ^
J k
e JlL
e
=e = e) ((8.17.23)
r 0
rB
r
z r
r
Bz
L + ^ l - ^ . )
= e „ =
(8.17.24)
e la 2( \i =^ d0
e 0
or
/
All the stress-strain relations in Eqs. (8.17.3) to (8.17.14) r e m a i n the
s a m e except that r, 0, a n d z are substituted for 1, 2, a n d 3, respectively.
T h e e q u a t i o n s of equilibrium are given by Eqs. (8.16.30) to (8.16.32).
T h e compatibility relation (8.17.21) b e c o m e s :
2
/ 3
2
1 8
1 3
o
\
(
F
F
1
r
°0
F
r \
+
{^2
+ 7^r
^)(<>»
+ °ee) = " 0
+ v) {^r
+
T)•
?W + (8.17.25)
Elastic Stress-Strain Relations
8.18
227
State of Generalized Plane Stress
Let us consider a thin p l a n e — i n other w o r d s , a b o d y with parallel
faces a n d a thickness 2h which is very small c o m p a r e d to the linear
d i m e n s i o n s of the faces. T h e m i d d l e surface of the plate is located
halfway b e t w e e n the faces a n d is t a k e n as the OXx, OX2 p l a n e (Fig. 8.6).
T h e g e n e r a t o r s forming the edges of the plate are n o r m a l to the two
parallel faces. All the l o a d i n g is applied to the edges, in planes parallel
*3
Fig. 8.6
to the OXx, OX2 p l a n e . This l o a d i n g is symmetrically distributed with
respect to the m i d d l e surface, a n d so are the b o d y forces Fx a n d F2; F3
is equal to zero. In view of the previous a s s u m p t i o n s , p o i n t s o n the
m i d d l e surface d o n o t suffer a n y d i s p l a c e m e n t u3 in the OX3 direction,
a n d for all other points u3 is very small; also the variations of the
c o m p o n e n t s ux a n d u2 of the d i s p l a c e m e n t t h r o u g h the thickness of the
plate will b e small. This suggests w o r k i n g with average values: In reality
these averages a r e often the only ones susceptible to e x p e r i m e n t a l
measurements. Thus,
(8.18.1)
(8.18.2)
228
T h e Theory of Elasticity
+h
1
f
= 2h J_h
u3(x]9
x2)
(8.18.3)
u3(xl9
xl9
x3)dx39
w h e r e b a r s over letters d e n o t e m e a n values. Because of s y m m e t r y ,
u3 = 0. Since the faces of the plate are a s s u m e d free of external loads,
(xl9
xl9
±h)
al3
= o23
(xl9
x29
±h)
= 0.
= o33
(xx,x2,±h)
(8.18.4)
Therefore,
dal3
(xl9
x29
±h)
dxx
do23
(xl9
x2,+h) =
3x2
(8.18.5)
=
Q
'
a n d , since F3 w a s a s s u m e d to b e equal to zero, the third e q u a t i o n of
equilibrium requires t h a t
(xl9
x29
±h)
do33
dx3
Q
(8.18.6)
=
Since o33 a n d its derivative with respect to x3 vanish o n the faces of t h e
plate, w e c o u l d consider with very little error t h a t o33 is zero everywhere. T h e a s s u m p t i ohn is, however, t o o stringent a n d it suffices to
consider t h a t \/2h $± h o33
(xl9
x29
x3)dx3
= 0. W e define the average
values of t h e stress c o m p o n e n t s a n, a 22 , a n d a 12 as follows:
i f
1
f
» 2h J-H +h ^ ° 2h J- ° ^
a
a i 1 DX
=
2LDX
22 =
(8.18.7)
H
=
1
f
°^
2hJ-h
T h e m e a n values of t h e b o d y forces a r e defined as
= 1
F
Fx
F hi X
=
1 Ff
^ 2 h ) - h
f
2hJ-h
^
is
dx~t
F
^
2
d
x
(8.18.8)
= 0.
3
I n t e g r a t i n g the e q u a t i o n s of equilibrium, taking into a c c o u n t t h e
definitions of the m e a n values, w e get:
3xi
+ J?! + p =o
3xo
x
(8.18.9)
Elastic Stress-Strain Relations
3(J|2
3xj
9^22
+ ^
3x-
i
+
= 0.
r
229
(8.18.10)
In terms of the m e a n values across the thickness, the stress-strain
relations (8.5.3) b e c o m e :
a n = 2[ien + X(eu + e22
)
(8.18.11)
o22 = 2\ie22 + X(eu + e22
)
(8.18.12)
a 12 = 2 / ^ 1 , 2
(8.18.13)
w h e r e A = 2Ajti/(A + 2\i). T h e s e three e q u a t i o n s , together with the two
e q u a t i o n s of equilibrium, serve to d e t e r m i n e the five u n k n o w n u{, u2,
a n, a 22 , a n d a 1 , 2 all a function of x{ a n d x2 only.
W h e n Eqs. (8.18.11) to (8.18.13) are substituted into E q s . (8.18.9) a n d
(8.18.10), we o b t a i n two e q u a t i o n s of elasticity in terms of average
strains a n d d i s p l a c e m e n t s :
(A +
/Oglj-O?!! +
2
^22)
+ f* V ^ + F, = 0
(8.18.14)
2
(A + M)
3^(^11
+
^22) +
^
" 2
+ 7-2 = 0.
(8.18.15)
T h e s e e q u a t i o n s serve to d e t e r m i n e the average c o m p o n e n t s of the
d i s p l a c e m e n t ux a n d u2 w h e n the average d i s p l a c e m e n t s are specified o n
the c o n t o u r .
T h e m e a n strains satisfy the compatibility e q u a t i o n (8.17.18). I n t e r m s
of stresses, this e q u a t i o n b e c o m e s :
8.19
S t a t e of Generalized P l a n e Strain
A s stated in Sec. 8.15, this state results from the superposition of a
c o n s t a n t strain eQ a l o n g OX3 to a state of p l a n e strain in the OXx,
OX2
plane:
e
\\ = ~ g ^ t O
~ v)an ~ v o22
] -
ve0
(8.19.1)
230
The Theory of Elasticity
e'22 =
[(1 -
«0*22 -
-
(8-19.2)
e'33= e0
(8.19.3)
*33 = e , 3 = «23 = 0,
(8.19.5)
w h e r e the e'y ' s are the resulting strains a n d the
' s a n d e,-, ' s are the
stresses a n d strains of the state of p l a n e strain. Solving for the stresses,
we get:
E
1 - v)e\
x + v{e'12 + e0
(1 + v)(\ -2v)
f
E
1 - v)e 22 + v(e\x + eQ
a22 =
(1 + v)(\ ~2v)
a
vE
e + e22
) + Ee0 = v(oxx + o22
) +
33 =
(1 + "XI - 2v) xx
n
E
a,2 =
1
+
v*
a
0 , 3 = 2 3 == 0.
0,,
=
(8.19.6)
(8.19.7)
Ee0 (8.19.8)
(8.19.9)
(8.19.10)
T h e a d d i t i o n of c o n s t a n t terms to the e q u a t i o n s of p l a n e strain does n o t
i n t r o d u c e a n y c h a n g e in the differential e q u a t i o n s of Sec. 8.16. T h e s e
e q u a t i o n s still h o l d in this case. T h e state of generalized p l a n e strain will
b e e n c o u n t e r e d in the study of r o t a t i n g long cylinders.
8.20
Solution of Elasticity P r o b l e m s
In S e c 8.8, it was stated t h a t the solution of a p r o b l e m of elasticity
requires the solution of a system of 15 e q u a t i o n s with 15 u n k n o w n s . A
systematic elimination of s o m e of the u n k n o w n s h a s allowed us to
r e d u c e the system to e q u a t i o n s , either in terms of the d i s p l a c e m e n t s in
Sec. 8.9, or in terms of the stresses in Sec. 8.10. This suggests various
m e t h o d s t h a t c a n be used in solving elasticity p r o b l e m s . S o m e of these
m e t h o d s d e p e n d primarily o n intuition, while others are b a s e d on a
systematic application of the techniques of applied m a t h e m a t i c s :
Elastic Stress-Strain Relations
231
1. T h e Inverse M e t h o d . In this m e t h o d , o n e guesses a solution
satisfying all the r e q u i r e m e n t s of the theory of elasticity a n d tries to find
to w h a t b o u n d a r y c o n d i t i o n s the solution c o r r e s p o n d s [1].
2. T h e Semi-Inverse M e t h o d . T h e r e , o n e guesses a p a r t of the solution
a s s u m i n g expressions for the stresses, the strains, or the d i s p l a c e m e n t s ;
e n o u g h freedom is left in the a s s u m p t i o n s so that the differential
e q u a t i o n s a n d the b o u n d a r y c o n d i t i o n s c a n b e satisfied [1].
3. T h e M e t h o d of Potentials. T o simplify the solution of the various
systems of e q u a t i o n s describing elasticity p r o b l e m s , a n u m b e r of
potentials h a v e b e e n i n t r o d u c e d . Potentials related to d i s p l a c e m e n t s
p r o v i d e solutions to N a v i e r ' s e q u a t i o n s while potentials related to
stresses g e n e r a t e systems of equilibrating stresses. S o m e of the m o s t
powerful potentials will b e studied in the next c h a p t e r [2,4,6].
4. Betti's M e t h o d [2,3].
5. T h e Integral T r a n s f o r m M e t h o d s [2].
6. T h e C o m p l e x Variables M e t h o d . This m e t h o d h a s b e e n devised
particularly for the solution of two d i m e n s i o n a l p r o b l e m s [4,5].
7. T h e V a r i a t i o n a l M e t h o d s . T h e s e m e t h o d s are b a s e d o n the fact t h a t
the g o v e r n i n g e q u a t i o n s of elasticity c a n b e o b t a i n e d from the m i n i m i zation of a n energy expression. T h u s , w e m a y seek a solution which will
m i n i m i z e t h e energy expression a n d avoid the difficulties involved in the
solving of the differential e q u a t i o n s . T h e use of these m e t h o d s will b e
discussed in C h a p t e r 15.
8. T h e N u m e r i c a l M e t h o d s [1,6].
I n the following c h a p t e r s , the inverse m e t h o d , the semi-inverse
m e t h o d , t h e m e t h o d of potentials, a n d the variational m e t h o d s will b e
used to solve a variety of p r o b l e m s of engineering i m p o r t a n c e . Betti's
m e t h o d , t h e integral t r a n s f o r m m e t h o d s , the c o m p l e x variables m e t h ods, a n d the n u m e r i c a l m e t h o d s are n o t e x a m i n e d in this text.
PROBLEMS
1.
Show t h a t the stress-strain relations for a p a n e l (Fig. 8.7) m a d e of
o r t h o t r o p i c m a t e r i a l u n d e r a c o n d i t i o n of p l a n e stress c a n b e written
in the following form which involves only four i n d e p e n d e n t constants:
<*11
a
22
0\2_
=
H
H
H\ 2
\2
22
0
0
0
0
e 22
2 G 1 _2 J\2_
232
The Theory of Elasticity
Fig. 8.7
Derive the expression of Hu, H22
, Hl2
, a n d GX2 in t e r m s of the
tensor Cijki
. In Fig. 8.7, the reference axes are parallel to the axes of
symmetry.
2.
S o m e t i m e s the c o m p o n e n t s of the c o m p l i a n c e matrix Sijki are
written in terms of the c o n s t a n t s E, v, a n d G in the following
manner:
^1111 —
^1212 —
1
4G 12
—
^2211
Z7
' •
w h e r e Ex is Y o u n g ' s m o d u l u s in the OXx direction, GX2 is the shear
m o d u l u s associated with the OXx, OX2 directions, a n d vX2 is Poisson's ratio for the strain in the OX2 direction c a u s e d by the stress in
the OXx direction. T h e inverse of the stress-strain relations in
P r o b l e m 1 is written as:
e22
*\2
3.
1
Ex
vn
a
vn
Ex
1
Ex
E2
0
0
ll
0
°22
1
2G 12
°\2
, a n d HX2 in terms of Ex, E2, a n d vX2
.
D e t e r m i n e Hu, H22
F i n d the coefficients of the m a t r i x of the elastic coefficients in
P r o b l e m 2, if the system of axes is r o t a t e d 30 degrees c o u n t e r c l o c k wise a r o u n d the OX3 axis.
Elastic Stress-Strain Relations
4.
A cubic material is a material in which the properties are the s a m e
a l o n g three o r t h o g o n a l directions. S h o w t h a t the m a t r i x of the
coefficients of elasticity c o n t a i n s three i n d e p e n d e n t c o n s t a n t s only:
C h o o s i n g the c o o r d i n a t e axes a l o n g these directions, the c o m p l i a n c e
m a t r i x c a n b e written as follows:
Sun
•^1122
6.
^1122
0
0
0
Sun
•^l 122
0
0
0
•^1122
0
0
0
^1122
0
0
Si 111
0
^1212
0
0
0
0
0
0
^1212
0
0
0
0
0
0
S\2\2
122
5.
Prove t h a t in a n isotropic, h o m o g e n e o u s , linearily elastic solid, the
principal axes of the stress tensor coincide with the principal axes of
the linear strain tensor.
C o u l d the following stress fields b e possible stress fields in a n elastic
solid, a n d , if so, u n d e r w h a t conditions?
2
a
Oil
7.
8.
233
= ax
x
+
bx
2
n = ax\x\
°22 = cxx +
dx
2
°22 =
°\2 = fxx +
gx
2
°\2
=
+
cx\
dx
x x2
bx
x
°\\ = a[x2 + b(xx — x 2)}
°22 = a[xx -h b(x2 — 4)]
°33 = ab(xx
+
x2
°\3 = a 23 = 0
°\3 = a 23 = 0
°\2 =
°33 = 0
°33 = 0
°\3 = °23 = °-
x)
2
2abx x
a, b, c, d, / , a n d g are c o n s t a n t s .
2 are 10 in. long is subjected to a uniform
A c u b e of iron w h o s e edges
pressure of 10 tons / i n o n two opposite faces; the other faces are
p r e v e n t e d from m o v i n g m o r e t h a n 0.002 in. by lateral pressure.
6 faces a n d the m a x i m u m shearing
D e t e r m i n e the pressures o n these
stress in the c u b e . E = 30 X 1 0 psi a n d v = 0.3.
A c u b e of D u r a l u m i n , w h o s e edges are 5 in. long, is subjected to a
uniform pressure of 15,000 psi o n the four faces n o r m a l to the OXx
a n d OX2 axes. T h e two faces n o r m a l to the OX3 axis are restricted
to a total d e f o r m a t i o n of 0.0006 in. D e t e r m i n e the stress a733 a n d the
c h a n g e in length of the d i a g o n a l of the c u b e . E = 10 psi a n d v
= 0.3.
234
9.
The Theory of Elasticity
I n P r o b l e m 8 of C h a p6t e r 7, find the c h a n g e in length of
d i a g o n a l s . E = 30 X 10 a n d v = 0.3.
the
10. A steel pulley is to b e fitted tightly a r o u n d a shaft. T h e i n t e r n a l
d i a m e t e r of the hole in the pulley is 0.998 in., while the o u t s i d e
d i a m e t e r of the shaft is 1.000 in. T h e pulley will b e a s s e m b l e d o n
the shaft b y h e a t i n g the pulley, t h e n allowing the assembly to r e a c h
a u n i f o r m t e m p e r a t u r e . W h a t is the t e m p e r a t u r e c h a n g e r e q u i r e d to
6 for easy assembly? F o r steel,
p r o d u c e a c l e6a r a n c e of 0.001 in.
a = 6.0 X 1 0 " / ° F , E = 30 X 1 0 psi, v = 0.3.
11. A weight of 20,000 lbs. is s u p p o r t e d o n two short lengths of
2
c o n c e n t r i c c o p p e r a n d steel t u b e s (Fig. 8.8). T h e thickness of these
tubes is such that b o t h tubes h a v e a cross-sectional a r e a of 2 i n .
D e t e r m i n e the a m o u n t of load carried by e a c h t u b e at r o o m
6
t e m p e r a t u r e a n d w h e n the t e m p e r a t u r e is raised
100 ° F a b o v e
r o o m 6t e m p e r a t u r e . F o r steel, E = 30 X 160 psi, v = 0.3, a = 6
_ 6
x 1 0 ~ / ° F , a n d for copper, E = 17 X 1 0 psi, v = 0.35, a = 9.2
X 1 0 / ° F . T h e t u b e s h a v e the s a m e length at r o o m t e m p e r a t u r e
when unloaded.
12. A p r i s m a t i c b a r of length / h a n g s u n d e r its o w n weight a n d is
s u p p o r t e d at its t o p by the u n i f o r m stress pgf w h e r e pg is the weight
per unit v o l u m e (Fig. 8.9). Show t h a t the solution,
20O00
lb
1 1!
l
steel
11
I
Fig. 8.8
=
°33
P£*>
a
ll
= °22 = °\2 = °\3
=
==
°23
°>
satisfies equilibrium, compatibility, a n d the p r e s c r i b e d b o u n d a r y
c o n d i t i o n s . If a n e l e m e n t at A a l o n g the OX3 axis is fixed, find the
Elastic Stress-Strain Relations
235
1*3
T
J
A!
0
X
2
Fig. 8.9
expressions of the d i s p l a c e m e n t s w,, u2, a n d w 3.
13. T h e stress distribution in a thin disk of r a d i u s b r o t a t i n g a t a n
a n g u l a r velocity w r a d . / s e c . is given b y :
^ - 3 4 ^ , ^ ( 1
- £ )
N e g l e c t i n g gravity forces, show t h a t this solution satisfies equilibriu m , compatibility, a n d the prescribed b o u n d a r y c o n d i t i o n s .
14. T h e solution of the p r o b l e m of the circular shaft fixed at o n e e n d
a n d subjected to a twisting m o m e n t at the other is given b y (see Fig.
10.5):
WJ = — ax2x3,
u2 = axxx3,
W 3= 0.
W h a t are the c o n d i t i o n s t h a t this solution imposes o n the applied
twisting m o m e n t s ? Is the shaft in a state of p l a n e strain or of p l a n e
stress?
REFERENCES
[1] S. Timoshenko and J. N . Goodier, Theory of Elasticity,
McGraw-Hill, N e w York, N . Y.,
1970.
[2] I. N . Sneddon and D . S. Berry, "The Classical Theory of Elasticity," Encyclopedia of
Physics, Vol. 6, Springer-Verlag, 1958.
[3] A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th ed., Dover, N e w
York, N . Y., 1927.
[4] Y. C. Fung, Foundation of Solid Mechanics, Prentice-Hall, Englewood Cliffs, N . J., 1965.
[5] N . J. Mushkelishvili, Some Basic Problems of the Mathematical
Theory of Elasticity,
Noordhoff, Groningen, 1953.
[6] I. S. Sokolnikoff, Mathematical Theory of Elasticity, McGraw-Hill, N e w York, N . Y., 1956.
CHAPTER 9
SOLUTION OF ELASTICITY PROBLEMS
BY POTENTIALS
9.1
Introduction
In this c h a p t e r we shall describe t w o types of p o t e n t i a l s :
(a) potentials related to d i s p l a c e m e n t s , n a m e l y , the scalar a n d
vector potentials, the G a l e r k i n Vectors, a n d the N e u b e r P a p k o v i c h functions;
(b) potentials t h a t g e n e r a t e systems of equilibrating stresses,
n a m e l y , the M a x w e l l stress function, the M o r e r a stress function, a n d the Airy stress function.
O n the subject of potentials, references [1] a n d [3] of C h a p t e r 8 give
extensive details as well as a large b i b l i o g r a p h y .
W e shall first s u m m a r i z e s o m e results of field theory. F o r a d d e d
clarity, a vector n o t a t i o n is used in m o s t of the o p e r a t i o n s involving the
displacements.
9.2
S o m e Results of Field T h e o r y
S u p p o s e that, associated with each p o i n t in a region R, there is a
scalar p o i n t function
T=
x2,x3)
T(xl9
= T(xt).
(9.2.1)
T h e resulting field is said to b e a scalar field. A n e x a m p l e of a scalar
field is the t e m p e r a t u r e distribution in a r o o m . In the study of the
236
Solution of Elasticity Problems by Potentials
237
d i s p l a c e m e n t s of the p o i n t s of a b o d y subjected to linear t r a n s f o r m a tion, t h e r e is, associated with every point, a vector
u = u(xx, x2, x3).
(9.2.2)
T h e resulting field is said to b e a vector field. T h e scalar field associated
with a given p o i n t is d e s i g n a t e d b y o n e n u m b e r while the vector field is
d e s i g n a t e d b y three n u m b e r s .
Let
b e a scalar p o i n t function, ^ b e a vector p o i n t function, a n d X
b e the position vector. T h e following identities are established in vector
analysis:
div curl $ = V • (V X
= 0
(9.2.3)
curl g r a d <j> = V X (V<|>) = 0
2
curl curl $ = c u r l $ = V X (V X
= V(V - $ ) - V $
2
(9.2.4)
2
(9.2.5)
div g r a d <j> = V • (V<f>) = V <£
2
(9.2.6)
2
div Lapl if = V • ( V ^ ) = V ( V •
2
2
= Lapl div $
(9.2.7)
L a p l g r a d <t> = V (V<J>) = V(V <£) = g r a d L a p l ^>
2
Lapl $
-X)
= V $ • X) = 2V • $ + J • V ^
div ( # ) = <|>V • $ + (V0) • $
2
L a p l fojf) = 2V</> + XV <#>.
If, t h r o u g h o u t a region
conditions
(9.2.8)
2
(9.2.9)
(9.2.10)
(9.2.11)
a vector p o i n t function \p satisfies
div ^ = V - $ = 0,
the
(9.2.12)
the field is said to b e solenoidal in t h a t region. If, at every p o i n t in R,
curl $ = V X $ = 0,
(9.2.13)
the field is said to b e i r r o t a t i o n a l or lamellar. F o r a lamellar field, there
exists a scalar p o i n t function <f> say, such t h a t
^ = g r a d <#> = V<#>.
(9.2.14)
238
The Theory of Elasticity
is d e t e r m i n e d within a n additive a r b i t r a r y c o n s t a n t since the gradient
of the latter is zero. Setting the c o n s t a n t c o m p o n e n t of <TJ equal to s o m e
chosen value (for e x a m p l e , zero) uniquely d e t e r m i n e s
<f> is called a
scalar potential. F o r a solenoidal field, it is always possible to i n t r o d u c e
a vector potential A, say, such that
^ = eurlZ = V x l
(9.2.15)
A is d e t e r m i n e d within a n additive a r b i t r a r y vector function representing a potential field, since the curl of the latter is zero [Eq. (9.2.4)]. If
the vector function A is i n t r o d u c e d b y Eq. (9.2.15) merely for c o n v e n ience, t h a t is, as a n auxiliary function in the course of a n analysis, this
difficulty m a y b e o v e r c o m e b y the further d e m a n d that
d i v Z = 0.
(9.2.16)
This is a n a l o g o u s to setting the c o n s t a n t c o m p o n e n t of
equal to zero
in Eq. (9.2.14). A n y other disposition of the value of div A a p p r o p r i a t e
to the c i r c u m s t a n c e s p e r t a i n i n g to a specific p r o b l e m m a y likewise be
m a d e . T a k i n g the curl of b o t h sides of Eq. (9.2.15), we o b t a i n :
2
curl xp = curl curl A = g r a d div A —
V A.
If div A is c h o s e n e q u a l to zero, the previous e q u a t i o n b e c o m e s :
2
curl$ = - V Z .
(9.2.17)
This is Poisson's e q u a t i o n , which c a n b e solved to give A.
Finally, we shall give the proof of a t h e o r e m which is of i m p o r t a n c e
in c o n n e c t i o n with the applications of scalar a n d vector potential
functions. This t h e o r e m is k n o w n as Helmholtz's
theorem, a n d c a n b e
stated as follows:
A vector field E with k n o w n divergence a n d curl, n o n e of which equal
to zero, a n d which is finite, uniform, a n d vanishes at infinity, m a y b e
expressed as the s u m of a lamellar vector U a n d a solenoidal vector V,
E = V + V
(9.2.18)
with
curl U = 0,
div V = 0.
(9.2.19)
Solution of Elasticity Problems by Potentials
239
T o p r o v e this s t a t e m e n t , we shall show t h a t b o t h U a n d V c a n b e f o u n d
w h e n E is given everywhere. Since curl U = 0, a scalar potential <f> exists
such t h a t
V = g r a d cf>.
(9.2.20)
Also, since div V = 0, a vector potential \p c a n b e i n t r o d u c e d such t h a t
F = curl$,
d i v ^ = 0.
R e t u r n i n g to Eq. (9.2.18), we h a v e :
(9.2.21)
2
div E = div 77 + div V = div g r a d
+ 0 = V <£.
Also,
(9.2.22)
2
curl E = curl 77 + curl K = 0 + curl curl $ = V ^
(9.2.23)
Eqs. (9.2.22) a n d (9.2.23) are t w o Poisson e q u a t i o n s w h o s e solutions give
<j> a n d \p. K n o w i n g
a n d $, U a n d V c a n b e o b t a i n e d . T h i s p r o v e s the
theorem.
9.3
The Homogenous Equations of Elasticity and the Search for
Particular Solutions
W h e n the b o d y forces are equal to zero, Eq. (8.9.4) is r e d u c e d to the
h o m o g e n e o u s form:
2
MV W|
(A
+ +
)M^ = 0.
(9.3.1)
W h e n e v e r a particular solution of Eq. (8.9.4) is found, the solution of
a n elasticity p r o b l e m c a n be r e d u c e d to the solution of a set of three
h o m o g e n e o u s e q u a t i o n s : By p a r t i c u l a r solution, we m e a n a solution
satisfying Eq. (8.9.4) b u t n o t the b o u n d a r y c o n d i t i o n s of the given
p r o b l e m . T o show t h a t this is the case, let
u\ = u\(xl9
x2,x3)9
u'2 = u2(x]9
xl9
x3)9
u3 = u3(x]9
x29
x3)
(9.3.2)
b e a p a r t i c u l a r solution of Eq. (8.9.4) in a p r o b l e m w h e r e the displacem e n t s a r e prescribed o n the b o u n d a r y ; the values t h a t u\9 u'l9 a n d u3
take o n the b o u n d a r y differ from the prescribed ones a n d c a n readily
240
The Theory of Elasticity
b e o b t a i n e d by i n t r o d u c i n g the c o o r d i n a t e s
b o u n d a r y , in E q s . ( 9 . 3 . 2 ) . Let us set
u\ = u { - u \ ,
u2 = u 2- u 2,
u3 = u 3- u 3,
of the p o i n t s o n
e"v = e v- e ' v,
the
(9.3.3)
a n d consider u'{, u2, a n d u3 as o u r n e w u n k n o w n s . If we substitute E q s .
( 9 . 3 . 2 ) in E q s . (8.9.4), w e get:
2
lxV u'i
+ (\ +
+ Ft = 0.
(9.3.4)
S u b t r a c t i n g Eq. (9.3.4) from E q . (8.9.4), we o b t a i n the e q u a t i o n :
)M^ - 0 .
MV V , + (X +
(9-3.5)
T h e p r o b l e m is n o w r e d u c e d to finding the solution of E q . ( 9 . 3 . 5 ) with
n e w b o u n d a r y c o n d i t i o n s in t e r m s of w"; these b o u n d a r y c o n d i t i o n s a r e
o b t a i n e d from E q s . ( 9 . 3 . 3 ) . O n c e the solution of this h o m o g e n e o u s
system is o b t a i n e d , the p a r t i c u l a r solution is a d d e d to it to give t h e
general solution.
It is often simple to find a p a r t i c u l a r solution to E q . (8.9.4) a n d the
m a i n difficulty resides in finding the solution to the h o m o g e n e o u s
system of e q u a t i o n s . Let us, for e x a m p l e , consider E q . (8.9.4) with the
b o d y forces deriving from a scalar p o t e n t i a l <f>, a n d let us find a
p a r t i c u l a r solution of the form:
u = g r a d M.
I n this case, Eq. (8.9.4), w h e n written in vector form, b e c o m e s :
2
g r a d [(A + 2 J U ) V M + <t>] = 0.
(9.3.6)
A p a r t i c u l a r solution of this e q u a t i o n is
w h i c h is a Poisson e q u a t i o n , for which a p a r t i c u l a r solution c a n b e
f o u n d . Let us a s s u m e the b o d y forces to b e gravity forces. T h e n
Fx = F2 = 0
F3 =
-pg
(9.3.8)
(9.3.9)
Solution of Elasticity Problems by Potentials
241
E q . (9.3.7) in this case b e c o m e s :
^ - .
V2 M = A
+ 2/x
(9.3.10)
A p a r t i c u l a r solution of E q . (9.3.10) is:
M =
(9.3.11)
PS 4
6(A + 2/x)'
T h u s , the p a r t i c u l a r solution of E q . (8.9.4) is:
u
_
0
W
2
0
_
" ° '
W 3
u
_ _ P g 4 _
u
(9.3.12)
" 2 ( X + 2/x)-
W h e n the b o d y forces derive from a scalar p o t e n t i a l <j>, E q . (8.9.4) c a n
be written in vector form a s :
2
(A + /x) curl curl u + (A + 2ii)V w + g r a d $ = 0.
(9.3.13)
T a k i n g t h e divergence of Eq. (9.3.13), we get:
2
V [(A + 2/x) div u + <$>] = 0.
T h e r e f o r e , [(A + 2/x) div w +
is a h a r m o n i c
cal cases, <f> is h a r m o n i c , therefore div u =
E q . (8.5.14), onn is also a h a r m o n i c function.
E q . (9.3.13), we get
2
2
(9.3.14)
function. I n m o s t p r a c t i ev is h a r m o n i c . Because of
T a k i n g the L a p l a c i a n of
4
V ( V w ) = V w = 0.
(9.3.15)
E q . (9.3.15) shows t h a t the c o m p o n e n t s of the d i s p l a c e m e n t vector u a r e
b i h a r m o n i c . T h e c o m p o n e n t s of the states of stress a n d strain, b e i n g
linear c o m b i n a t i o n s of the first derivatives of w 1? u2, a n d w 3, a r e also
b i h a r m o n i c . I n s u m m a r y , w h e n the b o d y forces Ft derive from a
p o t e n t i a l , a n d as a p a r t i c u l a r case w h e n they are e q u a l to zero, w e h a v e :
2
2
V e , = 0,
V a„ = 0
V W
/ = 0,
V a 0. = 0,
4
(9.3.16)
4
V % = 0;
(9.3.17)
in o t h e r w o r d s , all the basic functions of the t h e o r y of elasticity are
biharmonic.
242
T h e Theory of Elasticity
In the following sections, we shall c o n c e n t r a t e o n finding a solution
to t h e h o m o g e n o u s system. T h i s solution m u s t , of course, satisfy t h e
b o u n d a r y c o n d i t i o n s of the p r o b l e m . O n c e such a solution is f o u n d , the
u n i q u e n e s s t h e o r e m will e n s u r e t h a t it is the only solution to the
p r o b l e m at h a n d . T h e r e are r a r e cases in which N a v i e r ' s e q u a t i o n s c a n
b e directly i n t e g r a t e d to give the d i s p l a c e m e n t s : O n e such case is
e n c o u n t e r e d in t h e s t u d y of disks.
9.4
Scalar and Vector Potentials. Lame's Strain Potential
In vector form,
follows:
the h o m o g e n e o u s
equation
(9.3.1) is written
as
2
(X 4- G) g r a d (div u) + G L a p l u = (X + G)V(V • u) + GV u
= 0.
(9.4.1)
A c c o r d i n g to H e l m h o l t z ' s t h e o r e m , the vector field u c a n b e written in
t e r m s of its scalar a n d vector p o t e n t i a l s <j>(xl,x2,x3)
a n d \p(x{,x2,x3)
as
follows:
u = g r a d <t> + curl $ = Vcj> + V X
(9.4.2)
with
div $ = 0.
(9.4.3)
If we t a k e the divergence of b o t h sides of E q . (9.4.2), we get:
2
V . u = V <£ = ev.
(9.4.4)
If we t a k e the curl of b o t h sides of E q . (9.4.2), we get:
V x u = V X (V X $).
But the curl of u is n o t h i n g b u t twice the r o t a t i o n vector
c o m p o n e n t s are c o 3, 2< o 1, 3a n d c o 2. 1T h u s ,
(9.4.5)
whose
2
2o5 = V X (V X
= -V xp.
(9.4.6)
S u b s t i t u t i n g Eq. (9.4.2) into Eq. (9.4.1), we get:
2
2
(X + 2G)V(V 4>) 4- GV X V $ = 0.
(9.4.7)
Solution of Elasticity Problems by Potentials
243
A n y set of functions <j> a n d \p w h i c h satisfies Eq. (9.4.7) will p r o d u c e ,
w h e n s u b s t i t u t e d i n t o E q . (9.4.2), a d i s p l a c e m e n t field u which satisfies
N a v i e r ' s e q u a t i o n s . Conversely, for every u t h a t satisfies N a v i e r ' s
e q u a t i o n at least o n e set of <f> a n d \p exists w h i c h satisfies E q s . (9.4.2),
(9.4.3), a n d (9.4.7). Obviously, since u is related to <f> a n d \p b y m e a n s of
first derivatives, for a given w, <f> a n d $ are n o t u n i q u e .
S o m e p a r t i c u l a r solutions of Eq. (9.4.7) are functions w h i c h satisfy the
two e q u a t i o n s :
2
2
V 4> = c o n s t a n t ,
If we chose
V v// = c o n s t a n t .
(9.4.8)
2
V <|> = c o n s t a n t ,
$ = 0,
(9.4.9)
t h e n the function <j> is called the Lame strain potential. E q . (9.4.9) shows
t h a t a n y h a r m o n i c function m a y b e used as <f> a n d t h e resulting
d i s p l a c e m e n t field,
u=
V</>,
(9.4.10)
will satisfy N a v i e r ' s e q u a t i o n . F o r c o n v e n i e n c e , u is often written as
9
5 = 2
^
.
<
A
)
This form is the o n e we shall use in s u b s e q u e n t sections. M a n y solutions
satisfying p r a c t i c a l b o u n d a r y c o n d i t i o n s in cylindrical a n d spherical
c o o r d i n a t e s c a n b e g e n e r a t e d from this p o t e n t i a l . P l a n e strain axisymm e t r i c p r o b l e m s in p a r t i c u l a r h a v e b e e n studied b y L a m e using E q s .
(9.4.9). S o m e of these p r o b l e m s will b e e x a m i n e d in C h a p t e r 11.
T h e following h a r m o n i c functions are helpful in the solution of
practical p r o b l e m s :
^ = A(x\
- JC|) + 2Bxx x2
2
cj> = O*"cos(A20),
2
4>=C/fl£,
4> = C0,
2
r = xl + x 2
0 = tan-'g-
p2 =
$ = Cln(p
r = x\ + x\
2X+ 2 +
X 2 J C
+ x 3) ,
2
p = x\ + x\ + x\.
(9.4.12)
(9.4.13)
(9.4.14)
(9A15)
(9.4.16)
(9.4.17)
1
1
244
The Theory of Elasticity
O n e m a y also m e n t i o n the two functions of the Poisson type:
2
4> = Cr
2
<?> = C p .
(9.4.18)
(9.4.19)
If we take the divergence of Eq. (9.4.7), we o b t a i n :
4
V<f> = 0
(9.4.20)
Eqs. (9.4.4) a n d (9.4.20) s h o w that ev is a h a r m o n i c function (already
p r o v e n ) . By taking the curl of Eq. (9.4.7), we o b t a i n :
4
V ^ = 0.
(9.4.21)
Eqs. (9.4.6) a n d (9.4.21) s h o w t h a t co is a h a r m o n i c function. A solution
of Eq. (9.4.7) m u s t also b e a solution of Eqs. (9.4.20) a n d (9.4.21), b u t
the reverse is n o t necessarily true.
9.5
The Galerkin Vector. Love's Strain Function. Kelvin's and
Cerruti's Problems
In the previous section, the d i s p l a c e m e n t vector u was represented b y
the s u m of first derivatives of t w o f u n c t i o n s — n a m e l y , a scalar function
<f> a n d a vector function \p. T h e differential o p e r a t o r of o r d e r o n e , V, w a s
used for t h a t p u r p o s e in Eq. (9.4.2). In search for solutions of general
applicability, it is r e a s o n a b l e to try differential o p e r a t o r s of o r d e r t w o
2 second
which w o u l d express the d i s p l a c e m e n t vector u in terms of
derivatives. T w o such o p e r a t o r s are the L a p l a c e o p e r a t o r , V , a n d the
o p e r a t o r g r a d (div) = V(V •); b o t h c a n b e expressed in a n y system of
c o o r d i n a t e s a n d c a n b e applied to a vector function. T h u s , let us
consider a vector function V which is related to the d i s p l a c e m e n t vector
by
2
2Gu = 2(1 - v)V V
T h e vector
solution of
a n y vector
(9.5.1) [1].
obtain:
- V(V • V).
(9.5.1)
V is called a G a l e r k i n vector. T h i s vector supplies a general
N a v i e r ' s e q u a t i o n . This c a n b e p r o v e d b y showing that, for
function u, it is possible to find a vector V satisfying Eq.
Substituting Eq. (9.5.1) into N a v i e r ' s e q u a t i o n (9.4.1), we
2 2
V ( V ~ F ) = 0.
(9.5.2)
Solution of Elasticity Problems by Potentials
245
Therefore, a n y b i h a r m o n i c vector function m a y b e used as the G a l e r k i n
vector a n d the resulting d i s p l a c e m e n t vector given b y Eq. (9.5.1) will
always satisfy N a v i e r ' s e q u a t i o n s . T h u s E q s . (9.5.1) a n d (9.5.2) are
equivalent to N a v i e r ' s e q u a t i o n . C o m p a r i n g Eqs. (9.5.1) a n d (9.4.2), we
notice t h a t the G a l e r k i n vector is related to the scalar p o t e n t i a l <j> a n d
the vector p o t e n t i a l \p b y
<t> = - ^ ( V
V
x
f
^
V
^
• V)
.
(9.5.3)
(9-5.4)
If V is c h o s e n to b e n o t only b i h a r m o n i c b u t also h a r m o n i c , t h e n
V X $ = 0,
a n d <j> satisfies the e q u a t i o n
(9.5.5)
2
V <f>
= 0.
(9.5.6)
This <£> is L a m e ' s strain p o t e n t i a l of the previous section.
Let us consider the p a r t i c u l a r case in w h i c h V h a s only o n e
c o m p o n e n t V3. In this case, we h a v e w h a t is called Love's strain function:
F = / 3F 3,
Vx = V2 = 0.
(9.5.7)
T h e g o v e r n i n g e q u a t i o n (9.5.2) is n o w r e d u c e d t o :
2
2
V ( V K 3) = 0,
a n d Eq. (9.5.1) b e c o m e s :
2
2Gu = 2(1 - p)l3 V K3 -
v(|^).
(9.5.8)
(9-5.9)
In a system of cartesian c o o r d i n a t e s Eq. (9.5.9) is written in e x p a n d e d
form as follows:
(9.5.10)
246
The Theory of Elasticity
In cylindrical c o o r d i n a t e s , Eq. (9.5.9) is written in e x p a n d e d form as
follows:
2
W
G_
1Gu l
z
"' ~379l'
9
^
«-~-rmz>
)
2 d
2Guz =
2(l-v)V V2- -^.
U s i n g the s t r a i n - d i s p l a c e m e n t relations a n d the stress-strain relations in
cylindrical c o o r d i n a t e s , we o b t a i n the expressions for the stresses
c o r r e s p o n d i n g to E q s . (9.5.11). T h e y a r e :
31'
(„ ^-0)
V
(9-5.12)
]
dz
d_
[ 2-,)V^-0]
(
0 Br
r
drdOdz \
o, =
)
}|[(l-)V^-0]
*-h[«->w-%f\-
(9-5.14)
(9.5.15)
(^-16)
--
(9 5 ,7)
Love i n t r o d u c e d the strain function Vz{r,z) in studying solids of
revolution u n d e r a x i s y m m e t r i c loading. O n e a p p l i c a t i o n of Love's strain
function is the p r o b l e m of the single c o n c e n t r a t e d force acting in the
interior of a n infinite b o d y . This p r o b l e m is k n o w n as Kelvin's
problem
(Fig. 9.1):
Let 2P b e applied at the origin in the direction of OX3. T h e b o u n d a r y
c o n d i t i o n s a r e : (a) All the stresses vanish at infinity, a n d (b) the stress
singularity is equivalent to a c o n c e n t r a t e d force of m a g n i t u d e 2P. T h e
c o n c e n t r a t e d force m a y b e r e g a r d e d as the limit of a system of l o a d s
applied o n the surface of a small cavity at the origin. U s i n g cylindrical
c o o r d i n a t e s , Love's strain function will b e of the form:
^ 3 = Vz=
Vz(r,z),
(9.5.18)
Solution of Elasticity Problems by Potentials
247
Fig. 9.1
a n d m u s t b e b i h a r m o n i c . Its third partial derivative s h o u l d define
stresses t h a t v a n i s h at infinity a n d h a v e a singularity at the origin. A
function satisfying these c o n d i t i o n s is:
21 / z
2
Vz = Bp = B(z
+ r )
(9.5.19)
.
A p p l y i n g E q s . (9.5.11) to (9.5.17), we get:
2Gur =
Brz
2 1(
2 G ^ = 0,
2Guz = 2 ? [
) 2 V
p + 1 +
(9.5.20)
^]
(9.5.21)
(1 -
2v)Bz
(1 - 2v)z
„3
o.
7
=
(9.5.22)
3
„5
]
3 z
(9.5.23)
(9.5.24)
(9.5.25)
T h e stresses are singular at the origin a n d vanish at infinity. T o
d e t e r m i n e the c o n s t a n t B, we c o m p u t e the total vertical stress o n two
248
T h e Theory of Elasticity
planes z = c o n s t a n t falling o n b o t h sides of t h e origin w h e r e IP is
acting. Let us isolate a b a n d z = ±a a n d write the equilibrium of t h e
forces in the OX3 direction. Fig. 9.2 shows the stresses in their positive
direction a c c o r d i n g to the c o n v e n t i o n s of C h a p t e r 7. T h e e q u i l i b r i u m
e q u a t i o n is:
TTT TTT
2P
Fig. 9.2
X
2P =
mrdrif,a\__a
- fQ
2Urdr(o22
)2,+a
.
(9-5.26)
Substituting E q . (9.5.23) into E q . (9.5.26), a n d n o t i n g that rdr — pdp for
a given value of z, we get:
2P = 4115 [ ( 1 - 2v)a
j
pdp
3
(9.5.27)
+ 3a
(9.5.28)
= 8IXB(1 - v).
Therefore,
B =
P
411(1 -
(9.5.29)
v)'
This value of B is n o w substituted in E q s . (9.5.19) to (9.5.25) to yield t h e
expressions of the strain function, t h e displacements, a n d t h e stresses.
P r o b l e m s c a n also b e solved b y c o m b i n i n g several G a l e r k i n vectors
or b y c o m b i n i n g L a m e ' s strain potentials a n d G a l e r k i n vectors. F o r
e x a m p l e , CerrutPs problem of a tangential force acting o n the b o u n d a r y
of a semi-infinite solid (Fig. 9.3) c a n b e solved b y c o m b i n i n g t h e
G a l e r k i n vector V w h o s e c o m p o n e n t s a r e
Vx = Ap,
V2 = 09
V3 = Bxx ln{f> +
x3),
(9.5.30)
Solution of Elasticity Problems by Potentials
249
Fig. 9.3
a n d the L a m e strain p o t e n t i a l :
(9.5.31)
A9 B9 a n d C a r e c o n s t a n t s to b e d e t e r m i n e d from the b o u n d a r y
c o n d i t i o n s . T h e d i s p l a c e m e n t v e c t o r is o b t a i n e d from the superposition
of E q s . (9.4.11) a n d (9.5.1). H e n c e ,
2
2Gu = V<f> + 2(1 - v)V V
- V(V • V).
(9.5.32)
T h e strains are o b t a i n e d from the s t r a i n - d i s p l a c e m e n t relations (1.2.1)
a n d the stresses from the stress-strain relations (8.5.21) t o (8.5.24). T h e
c o n s t a n t s A, B, a n d C c a n b e o b t a i n e d from the following c o n d i t i o n s :
1. A t x3 = 0
a 33 = a23 = o.
2.
(9.5.33)
O n a n y h o r i z o n t a l p l a n e , at a d e p t h x3 from the surface, the s u m of
all the forces a l o n g OXx m u s t b a l a n c e P; i.e.,
+ 00
+00
/
/
— oo
— 00
9534
ol3
dXl
dx2
+ P = 0.
(-- )
T h e i n t e g r a t i o n in Eq. (9.5.34) is m o s t easily a c c o m p l i s h e d in cylindrical
c o o r d i n a t e s . T h e c o m p u t a t i o n s are lengthy b u t d o n o t p r e s e n t a n y
difficulty. T h e y yield:
250
The Theory of Elasticity
P
\ —
P
(
-
~
)l 2 v
BC -
P
)
° ~
l v
(9 5 35)
T h e values of the d i s p l a c e m e n t s a n d of the stresses c a n n o w
o b t a i n e d . T h e y are given b y :
w, =
"2
be
P
4UGp
P
=
+
"3
2
4UGpl
J
pp + x3
(9.5.38)
a,, =
-33 = -
a
ll
W+
q
22 +
(9.5.41)
q
33
_
1 + V3
nP
3
Fx,
(9.5.42)
7 21
" 2 T V
3Pxi
9.6
3Px]
x*) x^
Xi
The Neuber-Papkovich Representation. Boussinesq's Problem
T h i s r e p r e s e n t a t i o n uses a c o m b i n a t i o n of h a r m o n i c functions
represent the d i s p l a c e m e n t vector u. W e i n t r o d u c e the expression:
2Gu =
A
-
V[B + jfc^Q,
to
(9-6.1)
where
is a vector field, B a scalar field, a n d X is the position vector.
S u b s t i t u t i n g Eq. (9.6.1) into Eq. (9.4.1), we get:
Solution of Elasticity Problems by Potentials
GV A
2
2
2
- (X + 2 G ) V ( V S )
251
\V(X • V A)
-
= 0.
(9-6.2)
T h i s e q u a t i o n is satisfied if:
2
2
I A = 0,
VB
(9.6.3)
= 0.
Therefore, a n y four h a r m o n i c functions Al,A2,A3,
and B can be
substituted in E q . (9.6.1) a n d the resulting u satisfies N a v i e r ' s e q u a t i o n .
T h e s e four functions, however, are n o t completely i n d e p e n d e n t . It c a n
b e p r o v e d ([2], [3]) t h a t for a n a r b i t r a r y three d i m e n s i o n a l c o n v e x
d o m a i n , the n u m b e r of i n d e p e n d e n t functions is r e d u c e d to three. T h e
functions A a n d B c a n b e d e d u c e d from G a l e r k i n ' s vector V if we set
2
A = 2(\ -
v)V
V, B = V • V - A • X
4(1 -
y
(9.6.4)
v
A special form for A a n d B for p r o b l e m s with axial s y m m e t r y is:
Ar = Ag = 0,
B = B(r, z).
Az =
Az(r,z)
(9.6.5)
(9.6.6)
252
The Theory of Elasticity
F o r e x a m p l e , Boussinesq's problem (Fig. 9.4) of a force P acting in the
OX3 direction, at the origin of c o o r d i n a t e s , on a semi-infinite elastic
solid, h a s t h e solution:
Ar = A9 = 0,
(9-6-7)
Az = A(\-v)§
B = C Inffi + z).
(9.6.8)
T o s h o w this, let us substitute Eqs. (9.6.7) a n d (9.6.8) in Eq. (9.6.1):
_
4(1 - v)
fez-^[cHf>
u =
2G
+ z) + f ] ,
(9-6.9)
where ez is the unit vector in t h e OX3 direction. I n e x p a n d e d form, in
cylindrical c o o r d i n a t e s , this expression is written as follows:
L
ur = -^r-r 1—,
+
2Gp(p + z)
2Gp
(9-6.10)
(9.6.11)
u9 = 0
u7 =
3
(3 - 4v)K - C ,
2
Kz 3
2Gp
2Gp
(9.6.12)
T h e b o u n d a r y c o n d i t i o n s of the p r o b l e m a r e : O n the surface of t h e
semi-infinite solid (1) orz = 0 everywhere a n d (2) ozz is equal to zero
everywhere except at the origin. U s i n g Eqs. (6.7.24) a n d E q . (8.5.24), we
get:
arz = J l [ C - K(\
- 2v) -
(9.6.13)
T h e first b o u n d a r y c o n d i t i o n , w h e n substituted in Eq. (9.6.13), leads to
the following relation b e t w e e n C a n d K:
C = K(l
- 2v).
(9.6.14)
T h e expression for ozz is given b y E q . (8.5.23), a n d is f o u n d to b e :
ozz = - ^ f .
P
(9.6.15)
Solution of Elasticity Problems by Potentials
253
This q u a n t i t y is i n d e t e r m i n a t e at the origin O a n d Boussinesq did n o t
a t t e m p t to describe the p r o b l e m there. T o d e t e r m i n e K, let us consider
a h o r i z o n t a l p l a n e at a d i s t a n c e z from the origin a n d write t h a t the
resultant of all the vertical forces o n this p l a n e is equal to P (Fig. 9.5).
Fig. 9.5
3
3Kz J
p
=o
(9.6.16)
2Urdr.
T h i s i n t e g r a t i o n is easily p e r f o r m e d b y substitution: It yields:
K =
P
(9.6.17)
211'
E q u a t i o n (9.6.14) yields:
C =
P{\ 2TT
2v)
(9.6.18)
'
S u m m i n g u p the results of this i m p o r t a n t p r o b l e m we h a v e :
2
_ m
Z _ \n 2 _ 0 ~ ">"1Z
u
P+
4nGpLp
u9 = 0
u
*
4YLGp
(9.6.19)
[2(1 - v) + 4]
J
<--)
(9.6.20)
9 6 21
254
T h e Theory of Elasticity
2IV
_(\-2P)P[z
*
p
1
(9.6.23)
=
(9.6.25)
a m = \{orr +
+ oz2
) = - ^ 3 ^ ^ -
(9-6-26)
I n c a r t e s i a n c o o r d i n a t e s , we h a v e :
_
P
[" (1 - 2r)xi
411(7
2
"
L P(x3 + p)
f
4UG!
'L
~
p
(1^2^X2
P(X
3
"1
3
J
X3X2I
+
p
+ p)
3
P
6 28)
J
2
a,, =
(9.6.27)
2
f 3 x 33
x
fx3
(
W^"P ~"
^
+p 3
P
* ( 2 p + x 23) ] \
P(P + * 3 )
*
(9.6.30)
a 22 -
WlT"
+
(
^
P + x
3
p
0 21
< - -
2nP
=
p
Ji
(9.6.31)
963 2
3 F x5
|
-33 =
2
px+3)(
f " 3 x , x 2x 3
~iW
3Pxx x\
(1 - 2y)(2p + x3)x{x2~\
P < P ~ W
3Px2 x]
)
(Qfi-iVi
-I
_ fi
Solution of Elasticity Problems by Potentials
255
T h e extension of this p r o b l e m to include loads distributed over finite
areas will b e given in C h a p t e r 14.
9.7
Summary of Displacement Functions
T h e following c h a r t s u m m a r i z e s the d i s p l a c e m e n t functions p r e s e n t e d
in the previous sections, a n d the p r o b l e m s they h e l p e d solve:
Navier's Equation
1
Scalar and Vector Potentials Galerkin Vectors Papkovich-Neuber
I
1
Lame's Strain Potential Love's Strain Function Boussinesq's Problem
nz
Kelvin's Problem
To be used in Chapter 14
for the Semi-Infinite medium
Cerruti's Problem
To be used in Chapter 11
for Cylinders and Spheres
Reference [4] gives a m o r e c o m p l e t e c h a r t of d i s p l a c e m e n t functions
a n d their interrelation.
9.8
Stress Functions
In Sec. 8.10, we h a v e seen t h a t the stresses at different points of a n
elastic b o d y are g o v e r n e d b y the equilibrium e q u a t i o n s , the BeltramiMichell compatibility relations, a n d the b o u n d a r y c o n d i t i o n s . In a
m a n n e r similar to the study of d i s p l a c e m e n t functions, functions
g e n e r a t i n g systems of equilibrating stresses h a v e b e e n e x a m i n e d . H o w ever, while in the case of d i s p l a c e m e n t functions the q u a n t i t y s o u g h t
was a tensor of r a n k 1, n a m e l y , w, in this case the q u a n t i t y is a tensor
of r a n k two, o^. This tensor is s y m m e t r i c a n d the stress functions m u s t
reflect this p r o p e r t y . In the following, we shall neglect the b o d y forces,
a n d the e q u a t i o n s of equilibrium to b e satisfied a r e :
da
11
dx
d°\2
dx
9^21
^31
=
Q
(9.8.1)
, ^22 , ^32
=
0
(9.8.2)
dx*>
dx
+
dx-t
256
T h e Theory of Elasticity
a < J
i3
dx
^+2 3 _ + ^ 3 3 _ = q
dx
x
(9.8.3)
dx
2
3
Let us c h o o s e a set of a r b i t r a r y functions <f>iJ(xl,x2,x3),
and assume that:
with ^
= <f>ji9
2
Oil
dx
3
2
r H
0X\
dx
2
2
3JCJ 9 X
2
2
2
9 <J) 12
31
3 x 29 x 3
2
a
_
12
9 <#>23
dx
3
dxi
(9.8.5)
8 2 2< I > 1
2 _ 2
9 </>22
' 3.x
3xj 3 x 2
9 <#>31
a
3
2
_ 9 <J>n
°23
2 3 4»3i
3x33xt
_
dx
2
33
(9.8.4)
3x23x3
2
) 9 2 <2
_ 3 <f>33 _j
o
2
dx
9 <#>23
9 </>12
' 9xj
2
3.x
3
9 <J>23
dx
2
2
dx
{
9 <f>31
' 9x39x2
(9.8.6)
2
3 4>23
3x23x3
2
2
9 <f>22
3 x 33 ^ !
2
9 <#>31
3 «l>33
3xj 3 x 2
2
9<?M2
2
(9.8.7)
(9.8.8)
3x
2
dx
(9.8.9)
•
W e n o t i c e t h a t the s e c o n d a n d third relations c a n b e o b t a i n e d from t h e
first b y cyclical p e r m u t a t i o n , a n d t h a t the fifth a n d sixth relations c a n
b e o b t a i n e d from the f o u r t h b y cyclical p e r m u t a t i o n . It is easy to verify
t h a t t h e e q u i l i b r i u m E q s . (9.8.1), (9.8.2), a n d (9.8.3) a r e formally
satisfied b y t h e p r e v i o u s a s s u m p t i o n s o n the values of the stresses. T h e
six scalar functions
a r e n o t i n d e p e n d e n t however. T w o m e t h o d s of
g e n e r a t i n g c o m p l e t e solutions from <f>y are the m e t h o d s of M a x w e l l ' s
a n d M o r e r a ' s stress functions. O n setting <f>{2 = <j>23= <£31 = 0, w e
o= b t a i n =
the solution p r o p o s e d b y M a x w e l l ; a n d o n taking <j>u = (j>22
033
0 w e get the solution p r o p o s e d b y M o r e r a . E a c h of these t w o
solutions is c o m p l e t e ; i.e., for every stress d i s t r i b u t i o n t h a t satisfies t h e
e q u i l i b r i u m e q u a t i o n s , it is possible t o c o n s t r u c t a set of M a x w e l l
functions a n d a set of M o r e r a functions. T o p r o v e this s t a t e m e n t for t h e
case of M a x w e l l ' s function, for e x a m p l e , consider E q s . (9.8.4) to (9.8.9)
in w h i c h </>12
, <f>23
, a n d <j>3l a r e set e q u a l to z e r o :
Solution of Elasticity Problems by Potentials
257
2
a <j>22
dx\
+
(9.8.10)
dxj
2
2
9<f>33 1 3 <l>n
°22 =
1 dx]
dxf
(9.8.11)
2
2
9 <h.
°33 =
3^
+
2
9 022
(9.8.12)
2
3 <J> U
_
9 <j>22
^33
r9 8 13^
If we integrate E q s . (9.8.13) a n d substitute the values of </>n, $ 22
*
<J>3 t h u s o b t a i n e d in E q s . (9.8.10) to (9.8.12), the result s h o u l d be a n
identity. E q s . (9.8.13) w h e n i n t e g r a t e d give:
4>n = -
/ /
$33 = ~ j j
°23 dx dx ,
$22 = ~ f f
2 3
°3\ dx dx ,
3 x
a n c
^
o dx dx ,
l2 l 2
in w h i c h the c o n s t a n t s of i n t e g r a t i o n h a v e b e e n o m i t t e d since <j>y is
a r b i t r a r y . S u b s t i t u t i n g E q . (9.8.14) i n t o Eq. (9.8.10), we get:
Differentiating E q .
(9.8.15) with
respect to xx, we get:
dx
2
dx
3
(9.8.16)
w h i c h is a n identity. In the s a m e way, we c a n show t h a t the t w o o t h e r
E q s . (9.8.11) a n d (9.8.12) b e c o m e identities. T h e s a m e r e a s o n i n g c a n b e
r e p e a t e d to p r o v e t h a t M o r e r a ' s stress functions p r o v i d e a c o m p l e t e
solution for stress d i s t r i b u t i o n s satisfying equilibrium.
If we substitute E q s . (9.8.10) to (9.8.13) i n t o the Beltrami-Michell
compatibility relations w i t h o u t b o d y forces, we get:
258
The Theory of Elasticity
2
+
2
1
r
L +
'K^
+
(9.8.18)
3x,
dxj
<
(9.8.17)
dx
+
^
+
)
§ < ^ - « )
2
3 ^ - [ ( l
V g + /l] = 0
2
dxi
(9.8.20)
2
+ " ) V 0 n - V Q + /?] = 0
2
(9.8,9,
2
+ ^)V ^3 -
2
3
3x3
= 0
(9.8.21)
2
[(1 + v)V <t>22 ~V Q
+ R] = 0,
(9.8.22)
where
(2 = 011 + 0 2 2 + 033
(9.8.23)
and
2
2
= 9 0n
dx
2
2
9+022
dx
9 2
2
+0 3 3
(9.8.24)
dx
Solution of Eqs. (9.8.17) to (9.8.22) satisfy both equilibrium and
compatibility and are therefore possible stress states in an elastic body.
Both Maxwell's and Morera's stress functions can be particularized
to generate Prandtl's stress function for torsion problems and Airy's
stress function for plane problems [4]. Airy's stress function is examined
in the next two sections of this chapter and Prandtl's stress function will
be examined in Chapter 10.
9.9
Airy's Stress Function for Plane Strain Problems
It has been shown in Sec. 8.16 that, for plane strain problems, the
equations of equilibrium are reduced to two equations, namely Eq.
(8.16.16) and Eq. (8.16.17). In practice, body forces can usually be
expressed by a potential function A such that
Solution of Elasticity Problems by Potentials
259
The equilibrium equations then become:
+
^ t e
- 0) - 0.
If we n o w c h o o s e a stress function <K-*i>*2)
a
<».M>
s u c
-i 0 l+ ^ ,
^ that
a
1
- - g ^ - ,
2
(9.9.4)
we see t h a t t h e e q u i l i b r i u m e q u a t i o n s a r e identically satisfied. Substituting E q s . (9.9.4) in the c o m p a t i b i l i t y Eq. (8.16.24), we get:
4
4
4
d <t>
d <t>
2
2
3 <j>
i - 2 » / V1a
3x|
3x, 3x 2
3x1
+ " ^ 9*?
2
, 3 a\ _ n
.995^
'
or
2
V4 < ) | i+r= > v f l
^
1+
W h e n t h e r e are n o b o d y forces,
= 0.
(9-9.6)
V
4
V<?> = 0,
(9.9.7)
which shows t h a t <f> is a b i h a r m o n i c function. By the a b o v e analysis, the
p r o b l e m of elasticity in p l a n e strain h a s b e e n r e d u c e d to seeking a
solution to Eq. (9.9.5) such t h a t the stress c o m p o n e n t s satisfy the
boundary conditions.
In cylindrical coordinates, we shall a s s u m e t h a t the b o d y forces are
radial a n d derive from a p o t e n t i a l
w h i c h d e p e n d s only o n r. T h u s ,
r=
(9.9.8)
or
T h e e q u a t i o n s of e q u i l i b r i u m (8.16.30) a n d (8.16.31) b e c o m e :
(9.9.9)
260
The Theory of Elasticity
^
+ 1^
+ ^ - 0 .
(9-9.10)
T h e s e e q u a t i o n s are identically satisfied by a stress function <p(r, 9)
defined as:
2
l3<?>
2
l3 <|>
„_9 4>
3fi
_
d(\fy\
(9 9 11)
Substituting Eqs. (9.9.8) a n d (9.9.11) into Eq. (8.16.37), we get:
Var
2
r
3r
2
r 30 A 3r
2
2
' 3r
1
1 ~ v \ dr
2
r
2
30 /
r
(9.9.12)
dr /
or
2
l
z
i
2
^
v
V < + f 1 +) j> f l = 0.
W h e n there are n o b o d y forces,
4
V<f> = 0.
4
9.10
(9.9.13)
(9.9.14)
Airy's Stress Function for Plane Stress Problems
If the b o d y forces are derived from a potential function fi such that
9fi
L ,
F{ = - ™9xj
'
2
p/ r _ ___9fi
2 = | 9M
=x. 9'
2
(9.10.1)
the e q u a t i o n s of equilibrium (8.17.15) a n d (8.17.16) b e c o m e :
^ l
| +( 5„ 2 _ 2f
,i 0 . )
As was d o n e for the case of p l a n e strain, we choose a function
such that
(9.10.3,
<t>(xl9
x2)
(9.10.4)
Solution of Elasticity Problems by Potentials
261
W i t h this choice, the e q u a t i o n s of equilibrium are identically satisfied.
It w a s s h o w n in Sec. 8.17, t h a t two o u t of the six compatibility
c o n d i t i o n s are identically satisfied in p l a n e stress. This leaves E q s .
(8.17.18) a n d (8.17.19) for c o n s i d e r a t i o n . Eq. (8.17.18) gives:
^
d4
+ 2
2^ 2 + ^
dxj
dx dx 2
or
+ ( l - , ) {( ^ + ^ 2) = 0
\ dxj
dx )
4
(9.10.5)
2
V <J> + (1 - v)V U
= 0.
(9.10.6)
U s i n g E q s . (8.17.13) a n d (9.10.4), we see t h a t Eqs. (8.17.19) give:
2
2
20
+ | y(V ^>) = O
20
+ ^-(V <|,) = O
(9.10.7)
2
(9.10.8)
2
2 ^ # -
+ ^
—
(V <f>) = 0,
(9.10.9)
oxx dx2
oxx OX2
to2b e satisfied b y <f>(xx, x2). F o r zero or c o n s t a n t b o d y forces, we see t h a t
V <f> m u s t satisfy the c o n d i t i o n :
2
V </> = Ax xx + A2 x2 + A39
(9.10.10)
a n d A3 are a r b i t r a r y c o n s t a n t s . If o n e neglects the
w h e r e AX,A2,
c o n d i t i o n s t h a t E q s . (9.10.7) to (9.10.9) i m p o s e o n </>, the solution is only
a n a p p r o x i m a t e o n e . W h e n there are n o b o d y forces Eq. (9.10.6)
becomes
4
V<f> = 0,
(9.10.11)
which is identical to Eq. (9.9.7). Both p l a n e strain a n d p l a n e stress
p r o b l e m s h a v e the s a m e solution in this case.
In Cylindrical Coordinates, if the b o d y forces are radial a n d derive
from a p o t e n t i a l £2 = fi(r), then
Fr = ^
dr
(9.10.12)
262
T h e Theory of Elasticity
and
^
%
U
^
-
o
,
f
)
r2 6
9^9 ,1^01
3r
d0
=
+ 0
(9-10.13)
r r
(9.10.14)
0
T h e s e e q u a t i o n s are identically satisfied b y a stress function <f>(r,9)
defined a s :
l3<|>
2
2
l3 <J>
3 6
3fi
_
3/l9<A
(91015)
Substituting E q s . (9.10.12) a n d (9.10.15) into E q . (8.17.25), w e get:
2
\3r
2r
+
3A-
1
(
or
2
2
r 3 0 A 3/-
+
- " ) ^
4
2
' 3r
2
r 30 /
7 ^ j - o
2
V<#> + (1 - v)V tt
W h e n there are n o b o d y forces,
= 0.
(9.10.17)
4
V <t> = 0.
9.11
. 1 . 10 6 )
(9.10.18)
F o r m s of Airy's Stress Function
In looking for a suitable stress function p r o v i d i n g the solution of
p l a n e elastic p r o b l e m s , o n e c a n often m a k e a g o o d guess for various
types of b o u n d a r y c o n d i t i o n s . A c o m m o n m e t h o d is t o use a p o l y n o m i a l
a n d find t h e c o m b i n a t i o n of t e r m s which fit a p a r t i c u l a r set of b o u n d a r y
c o n d i t i o n s . F o r e x a m p l e , in t h e p o l y n o m i a l
<f> = a xx + b xx x2 + c x\ + d xx + e xxx2
+ /xxx2
+ g x\ +
h xf\ + j xx x2 4- k xx x\ 4- / xx x\ 4- m x\ 4- ft xf . . .
^
( 9
Solution of Elasticity Problems by Potentials
263
a n y t e r m c o n t a i n i n g xx4 or x2 u p to the third p o w e r will satisfy t h e
b i h a r m o n i c e q u a t i o n V <£ = 0. T e r m s c o n t a i n i n g x?\ or x\ a n d higher
p o w e r s m u s t h a v e , a m o n g their coefficients, relations satisfying the
b i h a r m o n i c e q u a t i o n . F o r d i s c o n t i n u o u s loads o n the b o u n d a r y , however, t h e p o l y n o m i a l a p p r o a c h h a s severe theoretical limitations since
discontinuous boundary conditions are not representable by polynomials.
In cylindrical c o o r d i n a t e s , the stress function is, in general, of t h e
form:
<j> = / ( r ) c o s n0,
$ = / ( r ) s i n n0,
(9.11.2)
w h e r e / ( r ) is a function of r a l o n e a n d n is a n integer. T h r e e o t h e r forms
deserve m e n t i o n i n g , n a m e l y :
2
<J> = Cr0 cos 0,
$ = Cr0 sin 0,
<j> = Cr 0,
(9.11.13)
w h e r e C is a c o n s t a n t . C o m b i n a t i o n s of the previous forms c a n of
course b e used.
For axially symmetric stress distribution ( a b o4u t the Z axis), $ does n o t
d e p e n d o n 0, a n d t h e b i h a r m o n i c e q u a t i o n V <£ = 0 b e c o m e s :
4
4
d <t>
2d^±3 _
dr
dr
2
r\_(P±2
3
r dr
Q
r dr
(9.11.4)
T h i s is Euler's differential e q u a t i o n . T h e solution is:
2
2
$ = Cxr inr
+ C 3/ w + C 4,
+ C2r
(9.11.5)
w h e r e Q , C 2, C 3, a n d C 4 a r e c o n s t a n t s of integration. C 4 plays n o p a r t
in the solution since the stresses a r e expressed in t e r m s of t h e derivatives
of <t>; C 1? C 2, a n d C 3 a r e to b e d e t e r m i n e d from the b o u n d a r y
c o n d i t i o n s . F o r a state of p l a n e stress, for e x a m p l e , the stresses a r e given
by E q s . (9.10.15). T h u s ,
XC
°rr =
+
^
+
o99 = Cx(3 + llnr)
or9 = 0.
2 / +m 2
*>
2
+ 2 C 2 - %L
r
C
(9.11.6)
(9.11.7)
(9.11.8)
=
264
The Theory of Elasticity
T h e strains a n d the d i s p l a c e m e n t s c a n b e o b t a i n e d b y m a k i n g use of
Eqs. (8.17.11), (8.17.12), (8.17.14) a n d Eqs. (8.17.23), (8.17.24). E q s .
(8.17.11) a n d (8.17.23) yield:
L
r
(9.11.9)
+ 2(1 -
p)C2\.
Integrating Eq. (9.11.9), we get:
=
u
C
i L-
r
^
3 + 2(1 - v)Ci rlnr - (1 + j»)C, r
+ 2(1
w h e r e fx{6)
-*-)C 2rJ
(9.11.10)
+ /,(*),
is a function of 9 only. E q s . (8.17.2) a n d (8.17.23) yield:
^
(9.1U1)
= r e m- u r ^ - f x { 0 ) .
I n t e g r a t i n g Eq. (9.11.11), we get:
w h e r e f2(r)
is a function of r only. E q s . (8.17.14) a n d (8.17.24) yield:
r
*
2G
dr
, 1 ^ _ ^r
30
'
) ={
'
(9.11.13)
since
= 0. Substitution of Eqs. (9.11.10) a n d (9.11.12) i n t o
(9.11.13) gives:
nir
+
?W
+
^/
'
= °-
Eq.
( 9 1 U 4 )
This e q u a t i o n can b e s e p a r a t e d into two e q u a t i o n s in the variables r a n d
9. T h u s ,
dm
dr
-f2{r)
= m,
and
^
+ /
Jx{9)d9
= -m.
CMU5)
Solution of Elasticity Problems by Potentials
265
T h e e q u a t i o n in 6 is only satisfied if the c o n s t a n t m = 0 so t h a t
f2(r) = Ar a n d fx(6) = B sin 0 + F cos 9, w h e r e A, B, a n d F are constants. Therefore,
ii r = | [ " 4^
C 3 + 2(1 - iOC, r/iir - (1 + v)Cx r
7
+ 2(1 - v)C2r^
(9.11.16)
+ 5 sin 0 + i cos 0
and
„ , « ^
4 r i++ * c o s * - F s i n * .
(9.11.17)
T h e c o n s t a n t s in Eqs. (9.11.16) a n d (9.11.17) are to b e d e t e r m i n e d for
each p a r t i c u l a r case.
PROBLEMS
1.
2.
2
$ = A(xx — x2)
n
<j> =
3.
4.
5.
2
G i v e n the scalar a n d vector potentials $ = x + 2x2 a n d \p = p / 3,
does the d i s p l a c e m e n t field g e n e r a t e d by <j> a n d \p satisfy N a v i e r ' s
e q u a t i o n s , a n d , if so, w h a t is it?
F i n d the d i s p l a c e m e n t s a n d the stresses defined by the following
L a m e strain p o t e n t i a l s :
+ 2Bxx x2
Cr cos(n9).
D e t e r m i n e the d i s p l a c e m e n t s a n d the stresses defined b y the
2
G a l e r k i n vectors:
2
/3
(a) F = C P 2
(b) V= ~Cp x2ix
+
Cp xxl2.
F i n d the stresses c o r r e s p o n d i n g to the L a m e strain p o t e n t i a l <j>
= Ctn(p + x 3) . W h a t is the p r o b l e m to which this potential furnishes a solution [1]?
Show t h a t the solution of Boussinesq's p r o b l e m c a n b e o b t a i n e d
t h r o u g h a c o m b i n a t i o n of a G a l e r k i n vector V = Bpi3 a n d a L a m e
strain p o t e n t i a l <j> = C(n(p + x3). Show t h a t C = —(1 — 2v)B a n d
5 =
P/2U.
266
6.
The Theory of Elasticity
W h a t are the stresses c o r r e s p o n d i n g to the following Airy stress
functions:
<j>
= ^xx
+ bxx x2 +
Fig. 9.6
7.
8.
A thin s q u a r e p l a t e w h o s e sides are parallel to the OXx a n d OX2
axes (Fig. 9.6) h a s in it stresses described b y oxx = cx2, a22 = cxx,
a n d possibly s o m e shearing stresses oX2
. c is a c o n s t a n t .
(a) F i n d the stress function b y integration, a n d the m o s t general
shearing stresses which c a n b e associated with the given oxx
a n d o22
.
(b) O b t a i n the strains a n d , by integration, d e d u c e the expressions
of the d i s p l a c e m e n t s ux a n d u2.
(c) F i n d the extension of the d i a g o n a l OC.
Show t h a t the stress function
1
<J> = C [ ( x ? + x ^ t a n - ^ -
xxx2^
provides the solution to the p r o b l e m of the semi-infinite elastic
m e d i u m acted u p o n b y a u n i f o r m pressure q o n o n e side of the
origin (Fig. 9.7).
9. Investigate w h a t p r o b l e m of p l a n e strain is solved b y the stress
function <J> = CrO sin 0.
3
10. Investigate the expression
= c o s 0 / r as a possible stress function.
Solution of Elasticity Problems by Potentials
267
REFERENCES
[1] H. M. Westergaard, Theory of Elasticity and Plasticity, Dover, N e w York, N . Y., 1964.
[2] E. Sternberg, "On Some Recent Developments in the Linear Theory of Elasticity,"
Structural Mechanics, Proceedings of the First Symposium on Naval Structural Mechanics, Pergamon Press, N e w York, N . Y., 1960.
[3.] R. A. Eubanks and E. Sternberg, "On the Completeness of Boussinesq-Papkovich Stress
Functions," J. Rat. Mech. Analy. Vol. 5, p. 735, 1956.
[4] P. C. Chou and N . J. Pagano, Elasticity, Van Nostrand, Princeton, N . J., 1967.
CHAPTER 10
THE TORSION PROBLEM
10.1
Introduction
In this chapter, the semi-inverse m e t h o d p r o p o s e d b y S a i n t - V e n a n t is
used to solve the p r o b l e m of non-circular p r i s m a t i c b a r s subjected to
t o r q u e . A stress f u n c t i o n — n a m e l y , P r a n d t l ' s stress function—is introd u c e d , a n d it suggests a n a n a l o g y which is utilized to o b t a i n solutions
for c o m p l i c a t e d shapes. T h e case of a circular p r i s m a t i c b a r is first
treated since it represents the first step of the intuitive solution
suggested b y S a i n t - V e n a n t .
10.2
Torsion of Circular Prismatic Bars
Let us consider a circular p r i s m a t i c b a r of length L a n d of r a d i u s a,
with o n e e n d fixed a n d the o t h e r e n d a c t e d u p o n b y a c o u p l e w h o s e
m o m e n t , M 33 = Mz, is a l o n g the OX3 axis. T h e b a r deforms a n d its
268
The Torsion Problem
269
g e n e r a t o r s are t r a n s f o r m e d from straight lines to helical curves (Fig. 10.
l a ) . O n a c c o u n t of s y m m e t r y [1], it is r e a s o n a b l e to a s s u m e t h a t cross
sections of the b a r n o r m a l to the OX3 axis r e m a i n p l a n e after d e f o r m a tion, a n d t h a t the couple rotates every section by a n angle 9 p r o p o r t i o n al to its d i s t a n c e from the fixed e n d x3 = 0 (proved b y e x p e r i m e n t ) .
Thus,
(10.2.1)
0 = ax3 = az,
w h e r e a is the twist per unit length, i.e., the relative a n g u l a r displacem e n t of t w o cross sections a unit distance a p a r t . F r o m the a b o v e
a s s u m p t i o n s , it c a n b e c o n c l u d e d t h a t the d i s p l a c e m e n t vector PP* of
a n y p o i n t P, in a cross section at a d i s t a n c e x3 = z from 0 (Fig. 10.1b),
h a s the following c o m p o n e n t s :
ur = r(\ — cos 9),
ue = r sin 9,
F o r small values of 9, sin 9 « 9 a n d cos 9 «
becomes:
ur = 0,
u9 = r9 = raz,
uz = 0.
(10.2.2)
1 so t h a t Eq. (10.2.2)
uz = 0.
(10.2.3)
K n o w i n g the c o m p o n e n t s of the d i s p l a c e m e n t , the strains c a n
o b t a i n e d from E q s . (6.7.23) a n d (6.7.24):
e
e
e
e
e
be
0
(10.2.4)
rr = 99 = zz = r0 = rz =
* fe = f .
00.2.5)
Substituting Eqs. (10.2.4) a n d (10.2.5) in the stress strain relations, we
get:
0
°rr = °M = °zz = °r9 = °rz =
(10.2.6)
o9z = Gar.
(10.2.7)
This state of stress satisfies the general e q u a t i o n s of equilibrium. It is
illustrated in Fig. 10.2, w h e r e o9z is the only stress c o m p o n e n t acting o n
the element referred to cylindrical c o o r d i n a t e s . Every cross section
including the e n d o n e is subjected to the shearing stress distribution
270
The Theory of Elasticity
Fig. 10.2
s h o w n in Fig. 10.2c. T h e radial direction is a principal direction with the
principal stress o2 = orr = 0. This state of stress is therefore a state of
p l a n e stress. T h e two o t h e r principal stresses, ox a n d a 3, lie in the
tangential p l a n e (eg,ez).
T h e r e p r e s e n t a t i o n o n M o h r ' s d i a g r a m is
s h o w n in Fig. 10.3. T h e s a m e state of stress exists at all p o i n t s of
f
\
Fig . 10.3
c o n c e n t r i c cylindrical surfaces which d o n o t interact with each other.
It r e m a i n s n o w to check if the p r o p o s e d solution satisfies the
b o u n d a r y c o n d i t i o n s at the lateral surface a n d at b o t h e n d s of the b a r .
The Torsion Problem
271
T h e direction cosines of the n o r m a l to the lateral surface with respect
to a system of cylindrical c o o r d i n a t e s are (1,0,0). This surface is free of
stress. T h e stress distribution of E q s . (10.2.6) a n d (10.2.7) satisfies the
b o u n d a r y c o n d i t i o n s (7.3.8). T h e direction cosines of the n o r m a l to the
p l a n e e n d of the b a r are (0,0,1). T h i s e n d surface is free from n o r m a l
forces a n d c o n s e q u e n t l y from n o r m a l stresses. H e r e , too, the stress
distribution of E q s . (10.2.6) a n d (10.2.7) satisfies this b o u n d a r y c o n d i tion. T h e r e is, however, a twisting m o m e n t Mz acting o n the e n d of the
b e a m , w h i c h m u s t b e in equilibrium with the resultant of the stress
distribution. T h u s (Fig. 10.2c),
Mz=
j\ur o dr
2
= aG
9z
j\uP
dr = a c ( ^ )
(10-2-8)
or
Mz = aGIz,
(10.2.9)
w h e r e Iz is the p o l a r m o m e n t of inertia of the cross sectional a r e a a b o u t
the axis of the b a r . Therefore, for a b a r of length L, w h i c h is l o a d e d at
the e n d b y a twisting m o m e n t M z, E q s . (10.2.1), (10.2.7), a n d (10.2.9)
can be written:
0G
=ML
og
=L
(10.2.10)
T h e factor b y which we divide the t o r q u e to o b t a i n a is called the
torsional rigidity.
S u m m a r i z i n g , we h a v e a s s u m e d a m o d e of d e f o r m a t i o n a n d d e d u c e d
the strains a n d t h e n the stresses w h i c h were f o u n d to satisfy equilibrium
a n d b o u n d a r y c o n d i t i o n s . Because of u n i q u e n e s s , this a s s u m e d solution
is the only solution to the p r o b l e m . T h e only s h o r t c o m i n g of this
solution o c c u r s at the e n d of the b a r w h e r e the externally applied
twisting m o m e n t m u s t b e d i s t r i b u t e d a c c o r d i n g to the p a t t e r n s h o w n in
Fig. 10.2c. I n practice, the stress distribution, a l t h o u g h statically equivalent to t h a t of Fig. 10.2c, is quite different from it. T h e influence of this
difference decreases quite rapidly as we m o v e a w a y from the e n d a n d
the solution p r e s e n t s a n excellent a p p r o x i m a t i o n starting from a dist a n c e of o n e or two d i a m e t e r s from the p l a n e of a p p l i c a t i o n of Mz
( S a i n t - V e n a n t ' s principle).
In the case of hollow circular bars, the value of Iz is given b y
272
The Theory of Elasticity
Fig. 10.4
/z = ^ ( l
- | ) ,
(10-2.11)
w h e r e a0 a n d a, are the outer a n d inner radii of the bar. All the
e q u a t i o n s previously given apply to this case with Iz given b y E q .
(10.2.11) (Fig. 10.4).
Fig. 10.5
The Torsion Problem
273
It is of interest to write s o m e of the previous e q u a t i o n s in a cartesian
system of c o o r d i n a t e s . T h e s e e q u a t i o n s will b e n e e d e d in the next
section; they are in fact the starting p o i n t in the search for a solution to
the p r o b l e m of n o n - c i r c u l a r p r i s m a t i c b a r s . T h e c o m p o n e n t s of the
d i s p l a c e m e n t vector are (Fig. 10.5):
ux = r cos(0 + /?) — r cos ji «
— x26
u2 = r sin(# + /?) — r sin /? ^
= —ax2x3
xx 6 = axx x3
(10.2.12)
(10.2.13)
u3 = 0.
(10.2.14)
T h e c o m p o n e n t s of the state of strain a r e :
e
e
0
e
(10.2.15)
*n = 22 = 33 = \2 =
e
\3
1a
aX
1
= -2 2>
23 =
2
T h e c o m p o n e n t s of the state of stress a r e :
a
X
e
l
0
n = °22 = °33 = °\2 =
a 13 = — Gax2,
(10.2.16)
a 23 = Gaxx.
(10.2.17)
(10.2.18)
T h e twisting m o m e n t M 33 m u s t b e in e q u i l i b r i u m with the stresses at the
e n d of the b a r . Therefore,
M
33
x
= f f
(G<* \ + Gax2)dxx
dx2 = Gal3.
(10.2.19)
T h e m a g n i t u d e of the stress vector is given b y
a, = Ga^/x[T~xJ
103
= Gar.
(10.2.20)
Torsion of Non-Circular Prismatic Bars
F o r n o n - c i r c u l a r p r i s m a t i c b a r s , N a v i e r tried to use the a s s u m p t i o n
m a d e for the case of circular b a r s — n a m e l y , t h a t the p l a n e cross sections
r e m a i n p l a n e ; this, led h i m to e r r o n e o u s conclusions. In fact, it c a n b e
p r o v e d t h a t the solution
oX3 = —Gax2,
o23 =
Gaxx,
(10.3.1)
274
The Theory of Elasticity
o b t a i n e d in the previous section, is only valid for circular prismatic b a r s .
I n d e e d , the lateral surface being free from stresses, the third b o u n d a r y
c o n d i t i o n of E q s . (7.3.8) gives (Fig. 10.6):
0 = — Gax2
dx2
—
_
dx
Gaxx x
ds
(10.3.2)
or
x2 dx2 + xx dx{ = 0.
This is the e q u a t i o n of a set of c o n c e n t r i c circles
xj + x\ = c o n s t a n t .
Fig. 10.7
(10.3.3)
The Torsion Problem
275
If the values of E q s . (10.3.1) were valid for n o n - c i r c u l a r bars, the stress
vectors w o u l d h a v e to b e t a n g e n t to c o n c e n t r i c circles, a n d at the
b o u n d a r y they w o u l d h a v e a c o m p o n e n t t a n g e n t to the c o n t o u r a n d o n e
n o r m a l to it (Fig. 10.7). This last o n e w o u l d be associated with a n o t h e r
shear stress o n the free outside surface of the bar, which does n o t exist.
Fig. 10.8
F o r e x a m p l e , the shear stress at the c o r n e r of a r e c t a n g u l a r b a r m u s t b e
zero b e c a u s e n o n e of its two p e r p e n d i c u l a r c o m p o n e n t s c a n exist. W e
are thus led to the a s s u m p t i o n t h a t for b a r s with n o n - c i r c u l a r sections,
p l a n e cross sections d o n o t r e m a i n p l a n e b u t are w a r p e d , a n d that all
cross sections are w a r p e d the s a m e w a y (Fig. 10.8).
Fig. 10.9
276
The Theory of Elasticity
Let us n o w consider a p r i s m a t i c b a r fixed at o n e e n d in the
(OXx,OX2)
p l a n e , while the o t h e r e n d is subjected to a couple w h o s e
m o m e n t M33 is a l o n g the OX3 axis (Fig. 10.9). OX3 passes t h r o u g h the
center of twist of e a c h section, i.e., t h r o u g h the p o i n t a b o u t which e a c h
cross section will r o t a t e . U s i n g the semi-inverse m e t h o d , S a i n t - V e n a n t
m a d e the following a s s u m p t i o n s :
1. E a c h cross section rotates b y a n angle 6 p r o p o r t i o n a l to its d i s t a n c e
from the fixed e n d with n o i n p l a n e distortion, i.e., with eX2 = 0; this
a s s u m p t i o n is similar to t h a t of circular b a r s .
2. All sections will w a r p the s a m e way, i.e., the w a r p i n g is i n d e p e n d ent of x3.
F o r a n y cross section at a distance x3 from the origin, these
a s s u m p t i o n s are analytically expressed b y :
ux = — ax2x3,
u2 = axxx3,
(10.3.4)
u3 = a\p(xx,x2)9
w h e r e \p defines the w a r p i n g a n d is a function of xx a n d x2 a l o n e . E q s .
(10.3.4) leave us e n o u g h freedom to try to satisfy the equilibrium
e q u a t i o n s a n d the b o u n d a r y c o n d i t i o n s . In d o i n g so, a n u m b e r of
restrictions to b e i m p o s e d o n \p will result. W i t h the a b o v e values of the
d i s p l a c e m e n t s , the strains are given b y :
e
e
0
e
* n = 22 = 33 = \2 =
(10.3.5)
6> < l 0 3
- 4
'13
+
* - i ( &
+
£ ) - i ( &
T h e stress-strain relations (8.5.2) give:
°n
' -
0
c
° » -
3 7)< l 0
4
(10.3.8)
= °22 = °33 = °\2 =
- 4
-
+
" ( H
, a M
4
<
>
A substitution of these values in the e q u a t i o n s of equilibrium with n o
b o d y forces shows t h a t the two first are identically satisfied while the
third o n e requires t h a t \p b e such t h a t
+
31) 0
'
t h r o u g h o u t each section. This is L a p l a c e ' s e q u a t i o n .
'
T h e Torsion Problem
277
Let us n o w investigate the b o u n d a r y c o n d i t i o n s , first o n t h e lateral
sides, t h e n a t the e n d s of the b a r . T h e direction cosines of t h e n o r m a l
t o t h e c o n t o u r C (Fig. 10.9) are (dx2 /ds, -dx} /ds, 0). T h e first t w o E q s .
(7.3.8) a r e identically satisfied a n d the third o n e gives:
"
(,0J )
O n t h e o t h e r t w o b o u n d a r y surfaces, i.e., t h e e n d s of the b a r defined b y
x 3 = 0 a n d x3 = L, the d i s t r i b u t i o n of stresses given b y E q s . (10.3.8) a n d
(10.3.9) m u s t h a v e n o r e s u l t a n t force a n d m u s t b e e q u i v a l e n t to a
torsional c o u p l e . Let us first p r o v e t h a t the r e s u l t a n t force is e q u a l t o
zero. T h e r e s u l t a n t force in the OXY d i r e c t i o n is given b y :
2
f j
dxx
ol3
dx2 = Ga j
j
R
~ * )
(10.3.12)
1
R
Eq. (10.3.12) c a n b e written a s :
ff
dxx
oXi
dx2 = Ga j
Rj
{giL
-
[ '(^
X
dx-,
x 2) ]
+ Xl
(10.3.13)
)]} ^ (/Xl
2
N o w , using the G r e e n - R e i m a n n f o r m u l a (see a p p e n d i x to this c h a p t e r )
Eq. (10.3.13) can b e written:
jj
R
°i3
d
d*\
*3
=
Ga(f)
[x, ( ~
- x 2)
dx2
(10.3.14)
- '(^
X
+
X
')^'] '
=0
where Eq. (10.3.11) h a s b e e n used. I n a similar way, w e c a n p r o v e t h a t
/ /
R
o23
dxxdx2
= 0,
(10.3.15)
so t h a t the r e s u l t a n t force a c t i n g o n the e n d s of the b a r vanishes.
T h e r e s u l t a n t torsional m o m e n t o n the e n d of the b a r d u e t o t h e
a s s u m e d stress distribution m u s t b e e q u a l to M 3 . 3T h u s ,
278
T h e Theory of Elasticity
^
M
=
/ R/ ^ ' ° ~ °^
2i
= Ga
X2
*
dXi
II (*?
+
d
2
"
+
(10.3.16)
*-
dXldx
1
R
T h e integral in Eq. (10.3.16) d e p e n d s o n ip, a n d h e n c e o n the s h a p e of
the cross section. Setting
2
11
R
2
{
*
~
*
1
X {
l x ~ )
l dd
+X
=X 1J 30
1
"
(
*
-
7
)
we h a v e
M 33 = GJa.
(10.3.18)
S u m m a r i z i n g , we h a v e a s s u m e d a m o d e of d e f o r m a t i o n a n d d e d u c e d
the strains a n d t h e n the stresses. T h o s e stresses were found to satisfy
equilibrium a n d b o u n d a r y c o n d i t i o n s p r o v i d e d the w a r p i n g function \p
satisfies E q s . (10.3.10) a n d (10.3.11).
It is worthwhile noticing t h a t Eq. (10.3.11) m a y b e written in a form
such t h a t the torsion p r o b l e m c a n b e classified as a special case of the
second b o u n d a r y value p r o b l e m of potential theory. This is d o n e as
follows: By definition, the g r a d i e n t (also called n o r m a l derivative) of the
function \p(xl, x2) at a p o i n t P is a vector directed along the n o r m a l n to
the level curve of the function t h r o u g h P, p r o v i d e d this curve possesses
a t a n g e n t at P (Fig. 10.10); the m a g n i t u d e of the g r a d i e n t vector is
k
r level
s
X2
I
^
curve
const
2
ds
Fig. 1 0 . 1 0
T h e Torsion Problem
279
d\p/dn a n d its c o m p o n e n t s a l o n g the OXx a n d OX2 axes a r e d\p/dxx a n d
d\p/dx2, respectively. T h e scalar p r o d u c t of the unit vector a l o n g n a n d
of the g r a d i e n t vector gives:
dxp _ dxp
dn
dxx
cos y
dx*
(10.3.19)
sin y,
which shows t h a t
dxx
dn
= y C dx~>
=O
^ '
Sdx2
=
~dn~ = - s i n y =
(10.3.20)
ds
Substituting E q s . (10.3.19) a n d (10.3.20) into Eq. (10.3.11), we get o n the
c o n t o u r C (Fig. 10.11):
2
"ds
'dn
Fig. 10.11
dn =
l^ L +X Jl+C ^ L X= I ^ {Xl
X l
O t ? + hx
X
ds
ds
2ds
*
'
F o r circular cross sections, xx+x2=
c o n s t a n t . In general, however,
c o n s t a n t a n d d\p/dn
g =, /,-v
2
)
(10.3.21)
= 0; i.e., \p =
( -- >
10
2
3
22
o n the b o u n d a r y . T h e r i g h t - h a n d of Eq. (10.3.22) is a f u n c t i o n / ( s ) of
the b o u n d i n g curve C. W e c a n n o w state t h a t the solution of t h e
p r o b l e m of torsion of p r i s m a t i c b a r s a m o u n t s to finding a function
*p(xx,x2)
which satisfies the e q u a t i o n s
280
T h e Theory of Elasticity
2
V ^ = 0
in R
^ « * 2/ i - * , W ( ' )
(10.3.23)
onC.
dO.3.24)
E q s . (10.3.23) a n d (10.3.24) define the N e u m a n n b o u n d a r y value
p r o b l e m w h i c h h a s b e e n extensively studied in p o t e n t i a l theory. T h e
s t a t e m e n t of N e u m a n n ' s p r o b l e m is as follows: T o d e t e r m i n e a function
\p(xx,x2)
which is h a r m o n i c a n d regular in a region R a n d o n its
b o u n d a r y C a n d such t h a t its n o r m a l derivative dxp/dn takes o n
p r e a s s i g n e d v a l u e s / ( s ) o n C*
s
x
A function \p(x\, x ) is said to be regular in a region R and o n its boundary C if in this region
2
*K*i > i) i uniform and possesses second derivatives which are continuous in R and finite o n
C. It is obvious that under these conditions both i>(x\,X2) and its first derivatives are
continuous in R.
T h e c o n d i t i o n for t h e existence of a solution to N e u m a n n ' s p r o b l e m is
t h a t t h e integral of t h e n o r m a l derivative d\p/dn calculated over t h e
entire b o u n d a r y C vanish, i.e., t h a t
^ d s
c an
= 0.
(10.3.25)
In the case of the torsion p r o b l e m , this c o n d i t i o n is satisfied since
(j)^ds
= <fi
[x2lx
- xx /2]ds
= <fi x2 dx2 + xx dxx = 0. (10.3.26)
T h e u n i q u e n e s s of the solution of N e u m a n n ' s p r o b l e m is easily e s t a b lished a n d c a n b e f o u n d in texts o n p o t e n t i a l theory.
10.4
Torsion of a n Elliptic B a r
Knowing that
m u s t satisfy E q s .
by trying various
they c o r r e s p o n d .
(10.3.10) is:
the w a r p i n g function \p(xx,x2)
for a n y prismatic b a r
(10.3.10) a n d (10.3.11), we c a n use the inverse m e t h o d
expressions for \p a n d finding t h e b o u n d a r i e s t o w h i c h
F o r e x a m p l e , the simplest solution of L a p l a c e ' s E q .
\p = C = c o n s t a n t ,
(10.4.1)
which w a s f o u n d in t h e previous section to solve the p r o b l e m of t h e
prism with circular cross section.
N o w consider the function:
The Torsion Problem
if, = Kxxx2,
281
(10.4.2)
w h e r e K is a c o n s t a n t . T h i s function, w h e n s u b s t i t u t e d in Eq. (10.3.11),
gives o n the b o u n d a r y :
< &2 -
- (K,,
f^f*)
or
+
- 0.+ (.0.4.3)
U p o n integration, E q . (10.4.3) gives:
x? +
(y=r§
)*2
(10.4.4)
= constant,
in w h i c h xx a n d x 2 are the c o o r d i n a t e s of a n y p o i n t o n the b o u n d a r y .
T h e e q u a t i o n of a n ellipse with center at the origin a n d semi-axes a a n d
b is
2
x
2
2
+ ^x 2
= a.
00.4.5)
E q s . (10.4.4) a n d (10.4.5) b e c o m e identical if:
2
o2. = 1 ~ K
b
1 + K-
(10.4.6)
By solving for K we get:
K^
b 22_ ^ _
b
2
(10.4.7)
+ a
Therefore, the w a r p i n g function for a n elliptical cylinder u n d e r torsion
is
<
1 >0
T h e c o n s t a n t J of E q . (10.3.17) is given b y :
/= jj
R
= (K+
2
4- x\
(JCJ
I)
ff
R
4-
Kx\ - Kx\)dxx
dx2
jc? dxx dx2 + (\ - K)
jj
R
(10.4.9)
xl
dxx dx2
or
J = (K + l ) / 2 + (1 - * ) / , =
z
a
(10-4.10)
4-
A
8
282
The Theory of Elasticity
w h e r e IX a n d I2 are the m o m e n t s of inertia with respect to the OXX a n d
OX2 axes, respectively. T h e torsional m o m e n t at the e n d of the b a r is
given b y :
33
2 b
Ua
M 33 = G
a +
2i
(10.4.11)
b
K n o w i n g the w a r p i n g function \p, the d i s p l a c e m e n t vector u c a n b e
c o m p u t e d from Eq. (10.3.4). T h e three c o m p o n e n t s of the d i s p l a c e m e n t
are:
ux = — ax2x3,
u2 =
ax3xx
( 2b ^ _ \ 2
u3 — axx x2
+
a)
\b
(10.4.12)
(10.4.13)
Eq. (10.4.13) shows t h a t the c o n t o u r lines defined b y u3 = c o n s t a n t are
h y p e r b o l a s (Fig. 10.12). If the b a r is twisted b y the t o r q u e M33 in a
Fig. 1 0 . 1 2
counterclockwise direction, the p a r t s w h e r e u3 is positive ( u p w a r d ) are
i n d i c a t e d b y solid lines a n d the p a r t s w h e r e u3 is negative ( d o w n w a r d )
are i n d i c a t e d b y d o t t e d lines. If o n e e n d of the b a r is restrained a n d
p r e v e n t e d from warping, n o r m a l stresses will b e i n d u c e d , positive in the
two q u a d r a n t s which otherwise w o u l d h a v e b e c o m e c o n c a v e , a n d
negative in the two others.
T h e c o m p o n e n t s of the state of strain are given by Eqs. (10.3.5) to
(10.3.7):
e
\\ =
e
2i
= e33 = eX2 = 0
(10.4.14)
(10.4.15)
The Torsion Problem
283
a X ,
- ( K + l ) = 4^V
+
(10-4-16)
T h e c o m p o n e n t s of the state of stress are given b y E q s . (10.3.8) a n d
(10.3.9):
a,, = a 22= a 33 = a 12 = 0
(10.4.17)
2
2Gaa
2 x2 2 _
a3
'
"
b
2+
_ 2Gab xx
a
~
nab*
_ 2M33
xx
T h e r e s u l t a n t shearing stress at a n y p o i n t P(xx,x2)
a -
J(a
2
V + (a
)
-
( 1 0 4 18)
2M33
x2
2 M
-i/^*?
33
o
4) 1 9
is given b y :
a 2 j C
•
2
(10.4.20)
a n d its direction is given b y :
tan</> = - ^ .
(10.4.21)
N o w , a l o n g every d i a m e t e r of the ellipse, the ratio xx /x2 is c o n s t a n t a n d
the direction of the t a n g e n t at a n y p o i n t of the ellipse's c o n t o u r is given
by (Fig. 10.13):
( ^ l ) =
\ dxx )
2
x2
(10.4.22)
a
Therefore, the r e s u l t a n t shearing stress at p o i n t s o n a given d i a m e t e r of
the ellipse increases linearly as we m o v e a w a y from its origin; it is
parallel to the t a n g e n t at the p o i n t of intersection of the d i a m e t e r a n d
the ellipse (Fig. 10.13).
T h e m a x i m u m value of ot occurs at the extremity B of the m i n o r axis.
O n e w a y to p r o v e this s t a t e m e n t is to express the stress at a n y p o i n t C
o n the b o u n d a r y in t e r m s of the c o o r d i n a t e s of D o n the conjugate
d i a m e t e r . F r o m analytic geometry, it is k n o w n t h a t the c o o r d i n a t e s of
D are:
(10.4.23)
284
The Theory of Elasticity
Fig. 10.13
A t p o i n t C, t h e r e s u l t a n t stress a, c a n b e r e w r i t t e n as follows:
2
(
\
n
_
3 3
M
-// /^2
Y
2
,
jy/\2
_
3 3
,/
M
(10.4.24)
T h e expression for (ot)c is m a x i m u m w h e n r" is m a x i m u m a n d this
o c c u r s w h e n C coincides with B or / / . Therefore, the m a x i m u m s h e a r i n g
stress o c c u r s at B a n d H.
10.5
Prandtl's Stress Function
I n Sec. 10.3, w e h a v e seen t h a t the solution of the torsion p r o b l e m
a m o u n t e d to finding a function \p(xx,x2)
w h i c h satisfied L a p l a c e ' s
equation
2
V xp
= 0
(10.5.1)
in a region R, a n d t h e c o n d i t i o n
(10.5.2)
o n the c o n t o u r C (Fig. 10.11). A n alternative p r o c e d u r e w h i c h leads to
simpler b o u n d a r y c o n d i t i o n s involves t h e i n t r o d u c t i o n of a stress
function <t>(xl9
x2) called P r a n d t l ' s stress function. T h i s function is
defined in t e r m s of xp b y the t w o following e q u a t i o n s :
T h e Torsion Problem
3<j>
''
<10 5 4)
•"•(eH)—»•
3xj
285
W i t h this definition, E q . (10.5.1) is identically satisfied a n d E q . (10.5.2)
becomes:
d^_dx2
d^dx^
dx2 ds
dxy ds
(10.5.5)
d$ = = 0
ds
T h u s t h e function <f> m u s t b e c o n s t a n t o n t h e c o n t o u r C (Fig. 10.11).
T h i s c o n s t a n t is a r b i t r a r y for solid b a r s a n d we shall t a k e it e q u a l t o
zero. E l i m i n a t i n g \p from E q s . (10.5.3) a n d (10.5.4), w e get:
T h i s is a Poisson e q u a t i o n , for w h i c h a solution c a n always b e f o u n d .
T h e solution of t h e torsion p r o b l e m is t h u s r e d u c e d t o finding a
function <f>(x 1,x 2) such t h a t ( F i g . 10.11):
2
V <t> = -2Ga
</> = 0
(10.5.7)
inR
o n C.
(10.5.8)
T h e expression of t h e twisting m o m e n t in t e r m s of <j> is o b t a i n e d b y
s u b s t i t u t i n g E q s . (10.5.3) a n d (10.5.4) in E q . (10.3.16):
*~-
f
M
f[ ^ ^] *
x
+X2
dx
dX2
1
R
=
-
/
+ 2
/
R
[ g f - ( * l * ) + 3 F ( * 2 * ) 1]
jj
2
<t> dx
x dx2.
N o w , b y t h e G r e e n - R i e m a n n formula, t h e integral (Fig. 10.11)
,L
, 0
5d 9x
(
-
)
286
The Theory of Elasticity
-
n\A A ^
M)+
R
{x2
dxxdx2
.
(10.5.10)
= — Q) <p[xxcos y + x2 sin y] ds = 0,
c
since $ = 0 o n C. Therefore,
M 33 = 2 / f
R
<f>dxxdx2,
(10.5.11)
a n d one-half of the t o r q u e is d u e to a 13 while the o t h e r half is d u e to a 2 . 3
In s o m e cases, it is q u i t e a d v a n t a g e o u s to express the torsion p r o b l e m
in terms of P r a n d t l ' s stress functions. F o r e x a m p l e , w h e n the e q u a t i o n
of the b o u n d a r y of the cross section is a simple function of xx a n d x2,
the stress function $ is c h o s e n such t h a t it c o n t a i n s the e q u a t i o n of the
b o u n d a r y a n d c o n s e q u e n t l y its value is always e q u a l to zero o n it;
a r b i t r a r y c o n s t a n t factors are i n c l u d e d in <j> so t h a t Eq. (10.5.7) c a n also
b e satisfied. T h i s a p p r o a c h to the p r o b l e m , a l t h o u g h n o t generally
applicable, h a s allowed us to find solutions for a n u m b e r of simple cases
two of w h i c h are e x a m i n e d in the next section. A n o t h e r interesting fact
n o t i c e d b y P r a n d t l is t h a t E q . (10.5.7) is the s a m e as the differential
e q u a t i o n for the s h a p e of a stretched m e m b r a n e originally flat w h i c h is
t h e n b l o w n u p b y air pressure from the b o t t o m . This allowed h i m to
d r a w analogies b e t w e e n the geometrical p a r a m e t e r s of the m e m b r a n e
a n d the stress-strain c o n d i t i o n s of the cross section of a b a r subjected
to twist. T h e m e m b r a n e a n a l o g y will b e e x a m i n e d in detail in
s u b s e q u e n t sections.
10.6
Two Simple Solutions Using Prandtl's Stress Function
The problem of a bar with an elliptical cross section, solved in Sec. 10.4,
c a n b e s t u d i e d b y starting with a stress function of the form:
6 )1
• - " • ( ^ - O -
-
This e q u a t i o n satisfies E q . (10.5.8), a n d w h e n substituted in Eq. (10.5.7)
gives:
m
2
2
_Gaa^_
=
a + b
. 6. 2 )
T h e Torsion Problem
287
Therefore, the stress function p r o v i d i n g the solution is:
22
2 b (A
2
Gaa
_
*
. l 2 ,\
x
2
a + b \a
b
(10.6.3)
r
The problem of a bar with a cross section in the form of an equilateral
triangle is solved b y starting with a stress function of the form (Fig.
10.14):
-2*
Fig. 1 0 . 1 4
0 = m(xx - V3 x2 + 2a)(x{ + ^
x2 + 2a){xx - a).
(10.6.4)
T h i s e q u a t i o n satisfies E q . (10.5.8), a n d w h e n substituted in E q . (10.5.7)
gives:
m = —
Got
6a
(10.6.5)
Therefore, the stress function p r o v i d i n g the solution is:
<f> = —
~
*2 + 2a)(xx + \ / 3 x2 + 2a)(xx — a).
(10.6.6)
M a k i n g use of E q s . (10.5.3) a n d (10.5.4), we get:
Got /x
x
°\3 = -a~( \
"23 = Ga,
a
~
)i
+ 2ax{ - x\).
(10.6.7)
(10.6.8)
F r o m these e q u a t i o n s , we see t h a t the shearing stress c o m p o n e n t a 13
vanishes a l o n g the OXx axis, while a 23 b e c o m e s :
X X
°23
= ^
\ (
\
2
+
) '
a
(10.6.9)
288
The Theory of Elasticity
*2
.A
0
Fig. 1 0 . 1 5
T h e distribution of a 23 a l o n g the OXx axis is s h o w n in Fig. 10.15. T h e
shearing stress is a m a x i m u m at the m i d p o i n t s of the sides of the
triangle, a n d is given b y :
a
( ;)max =
G
2
a
a
(10.6.10)
'
T h e shearing stresses vanish at the corners a n d at the origin O. T h e
twisting m o m e n t is given b y E q . (10.5.11) as:
M 33 = 2 j
j
-g(*!
- V3 x2 + 2a\xx
27
(xx — a) dxx dx2 = —-7= Gacft
5y3
+ ^ 3 x2 +
la)
(10.6.11)
or
M 33 =
^Gal3,
(10.6.12)
w h e r e I3 = 3^/3 dfr is the p o l a r m o m e n t of inertia of the triangle. T h e
w a r p i n g function is given b y :
(10.6.13)
w h e r e C is a c o n s t a n t e q u a l t o zero for n o rigid b o d y d i s p l a c e m e n t . T h e
three c o m p o n e n t s of the d i s p l a c e m e n t a r e :
The Torsion Problem
5 M 33
U{
"
3GI3
+5wM 33
XX
2 3'
289
2 —
3 ^
*1*3>
(10.6.14)
5Mr
Lines of e q u a l vertical d i s p l a c e m e n t are s h o w n in Fig. 10.16.
Fig. 10.16
10.7
Torsion of Rectangular Bars
C o n s i d e r a b a r with a r e c t a n g u l a r cross section, with its c e n t e r at t h e
origin, a n d with sides 2a a n d 2b (Fig. 10.17). A s stated in the previous
sections, the solution to the p r o b l e m a m o u n t s to finding a function
yp{xx,x2)
such that
Fig. 1 0 . 1 7
290
The Theory of Elasticity
2
= 0
V \p(xl9
x2)
in the rectangle
(10.7.1)
and
O n the t w o sides AB a n d C D , the direction cosines of the n o r m a l s are
(1,0,0) a n d (-1,0,0), respectively. O n the two sides BC a n d DA, the
direction cosines of the n o r m a l s are (0,1,0) a n d (0,-1,0), respectively.
T h e b o u n d a r y c o n d i t i o n (10.7.2) c a n thus b e written a s :
| ^ = *2
on
| £ = -*i
xx=±a
on
If we n o w i n t r o d u c e the function \px(xXy
x2)
h
x2 = ±b.
(10.7.3)
(10.7.4)
such t h a t [2]
(10.7.5)
=xxx2~^
Eq. (10.7.1) b e c o m e s
2
= 0
V ipx(xx,x2)
over the rectangle,
(10.7.6)
a n d Eq. (10.7.2) b e c o m e s
|^- = 0
dxx
on
x,=±a
(10.7.7)
and
j^i
= 2xx
on
x2 = ±b.
(10.7.8)
Let us a s s u m e t h a t the solution of E q . (10.7.6) c a n b e expressed in the
form of a n infinite series,
00
« M * i . * 2) = 2
n=0
En(Xi
)G„(x2),
(10.7.9)
1
T h e Torsion Problem
291
w h e r e e a c h t e r m of the series satisfies the differential Eq. (10.7.6),
En(xx)
a r e functions of xx a l o n e , a n d Gn(x2) a r e functions of x2 a l o n e .
S u b s t i t u t i n g E q . (10.7.9) i n t o Eq. (10.7.6), we get:
E"n(xx)Gn(x2)
= 0,
+ En{xx)G"n{x2)
(10.7.10)
for e a c h v a l u e of n. E q . (10.7.10) c a n also b e written as follows:
x
x
x \)
E"n(
En( \)
Gn(Xi)
n( l)
_
G
(10.7.11)
'
2
T h i s equality c a n n o t b e fulfilled unless b o t h sides of E q . (10.7.11)
are
e q u a l t o a c o n s t a n t . T h i s c o n s t a n t will b e t a k e n e q u a l to — k . T h i s leads
us to a p a i r of o r d i n a r y differential e q u a t i o n s :
d
2
- ^
dx}
+ kE
2
d G,2
dx
=0
(10.7.12)
= 0.
0°- - )
713
2
Y-k G„
T h e solution of these e q u a t i o n s is:
En = cxsin(knxx)
+ c2cos(knxx)
Gn = c 3sinh(/c: / X2)
I + c 4c o s h ( / c l Xl 2 ) .
(10.7.14)
(10.7.15)
T h e c o n s t a n t s cx, c2, c 3, a n d c 4 will b e d e t e r m i n e d from t h e b o u n d a r y
c o n d i t i o n s (10.7.7) a n d (10.7.8). Let us first c o n s i d e r the b o u n d a r y
c o n d i t i o n (10.7.8). W e see t h a t
1a ^ x- 2
n=0
2
En(xx)G'n(x2)
= 2xx
dO.7.16)
m u s t h a v e t h e s a m e value for x2 = +b a n d x2 = — b. T h e r e f o r e ,
m u s t b e a s y m m e t r i c function in x2. A l s o for x2 = ±b:
G'n(x2)
00
2
En(xx)G'n(b)
= 2xx.
(10.7.17)
Therefore, En{xx)
m u s t b e a n a n t i s y m m e t r i c function in xx. F r o m these
considerations, we find t h a t c2 = c4 = 0 in E q s . (10.7.14) a n d (10.7.15).
292
T h e Theory of Elasticity
T h e c o n d i t i o n (10.7.7) is satisfied if:
= 0
E' (±a)
n
(10.7.18)
or
cxkncos(kna)
= 0.
(10.7.19)
Thus,
k
n
(2n=+ l ) n
2a
(10.7.20)
cx a n d c 3 b e i n g a r b i t r a r y , E q . (10.7.9) c a n n o w b e written as follows:
00
^
= 2
(10.7.21)
Ansm(knxx)smh(knx2l
n=0
w h e r e kn is given b y E q . (10.7.20) a n d An is still t o b e d e t e r m i n e d from
b o u n d a r y c o n d i t i o n (10.7.8).
( ST")
\OX /x =±b
22
= 2
=o
Ankncosh(knb)sin(knxx)
00
= 2
n
= 2xx
(10.7.22)
Bnsin(knxxl
w h e r e w e h a v e set:
Bn = Ankncosh(knb).
T o d e t e r m i n e An, let u s multiply b o t h sides of E q . (10.7.22) b y
a n d i n t e g r a t e with respect t o xx. W e get:
+a
f
2xxsin(km
xx)dxx
oo
= §
2
f+a
C"
Bnsm(km
xx)s'm(knxx)dxx.
(10.7.23)
sin(km
xl)
(10.7.24)
H o w e v e r , since
r+a
/
rO if m
n
m
1
s i n ^ x O s i n ^ x O ^ i = 1 -f
>
"
La if m = n
(10.7.25)
then
2
£7
2 x 1s i n ( / c mx 1) ^ i = |_7
^sin ^*,)***! =
(10.7.26)
The Torsion Problem
293
I n t e g r a t i n g E q . (10.7.26), w e get:
2( - i r i 6 « 2
m
n ( 2 m + l) '
Therefore,
2
j
"
U\2n
so t h a t t h e w a r p i n g function $ is:
2
32a
3
^
3
^-x,x2
(
l ) " 3 32 a
+ l) cosh
=
(10.7.28)
knb'
n
29)
(~l) sm(knxx)sinh(knx2)
2 „ + i )n3 Cs(O
h(*„&)
"
T h e c o n s t a n t 7 is given by E q . (10.3.17) a s :
or
' -fx2 r r f r o *
J, 8 ^ 1 " !
"
3
L
, 96 «
1
n „tb (2/i +
4
4
l)
+
<
,
384a
5 «
%
Ub
o
-
j
3
o
)
t a n h ( £ „ f e5) l
{In + l )
n
)
J*
(
l
Since
1
oo
4 _ =i |
„?o(2n+ l)
4
(10.7.32)
96'
then
7
L3
- ^
5
n 6
5
3
2 £^§1
= Ao *.
„=o (2/i + l ) J
(10-7.33)
T h e series
~
tanh(/c„6)5
(2n + l )
c a n b e written a s :
«> tanhOfc^)
„?o
(2^Tr7
tan
H2j)
/ nA
+
5
3
tanh(£„Z>)
(2^Tl7-
™
( 1 0 J
-
3 4 )
n
0
294
The Theory of Elasticity
T h e first t e r m in the r i g h t - h a n d side of the previous e q u a t i o n gives the
value of the series to within 1/2 percent, a n d for all practical p u r p o s e s :
(10.7.35)
T h e shearing stresses are given b y :
_ M 33 / ^
_
•($-4
\6M2
a'--" «»
33
_
yn
°23
(2«
£o
_ A/ 33 / 9 ^
(10.7.36)
(( --1' )) "" 2 sinh(/c„x
sinh(/c„x22))
+
l)
cosh(k b)
n
\
(10.7.37)
=
^ 3 3 f~
16a «
^rL " W
2Xl
(-l)"co2
s h ( ^ x 2)
"I
^^>J-
sm
„?o (2» + l ) c o s h ( / t „ 6 )
T h e m a x i m u m shearing stress occurs at the m i d - p o i n t s of the long sides
JC, = ±a of the rectangle. Its value is given b y :
(
f W _f9 3 2
'
gA^ar 1
"
•/
= Ky
L
g>
8
n „tb ( 2 * +
2
21
]
.7.38)
l) cosh(/c„Z>)J
(10
Ma
33
Substituting the value of J from Eq. (10.7.33) in Eq. (10.7.38), we get:
M33
T h e following table gives the values of K, AT,, a n d K2 for different ratios
b/a:
|
K
K\
K
|
K
K
K
4.208
1.970
0.468
1.994
0.443
2
1.0
2.250
1.350
0.600
3.0
1.2
2.656
1.518
0.517
4.0
4.496
x
2
1.5
3.136
1.696
0.541
5.0
4.656
1.998
0.430
2.0
3.664
1.860
0.508
10.0
4.992
2.000
0.401
2.5
3.984
1.936
0.484
00
5.328
2.000
0.375
The Torsion Problem
295
m
Fig. 1 0 . 1 8
T h e expression for u3 is given b y w3 = a\p(xl,x2),
a n d the c o n t o u r lines
of the surface w3 = c o n s t a n t c a n easily b e d r a w n . F o r a s q u a r e bar, these
c o n t o u r lines are s h o w n in Fig. 10.18.
10.8
Prandtl's Membrane Analogy
In 1903, P r a n d t l o b s e r v e d t h a t the differential e q u a t i o n (10.5.6) of the
stress function is the s a m e as the differential e q u a t i o n of the s h a p e of a
stretched m e m b r a n e initially flat which is t h e n b l o w n u p b y air pressure
from the b o t t o m . This o b s e r v a t i o n will p r o v i d e us with a w a y of
visualizing the s h a p e of the stress function <£> a n d the stress distribution.
C o n s i d e r a thin weightless m e m b r a n e , initially with a large tension T9
h a v i n g the s a m e value in all directions. It is b l o w n u p from a flat s h a p e
i n t o a c u r v e d surface, b e i n g held at the edges b y a frame h a v i n g the
s a m e outline as the cross section of the b a r u n d e r torsion. W e shall
\*2
X3 I » [
I
ol—^TTTfN
P
Fig. 1 0 . 1 9
CI
296
T h e Theory of Elasticity
a s s u m e t h a t t h e air p r e s s u r e p is so small a n d t h e initial tension T so
large t h a t T does n o t c h a n g e d u r i n g the b l o w i n g u p process. Fig. 10.19
shows a m e m b r a n e w h o s e p e r i p h e r y is held d o w n in t h e p l a n e OXx,
OX2. Since p is small, u3 is also small. T h e e q u a t i o n of t h e m e m b r a n e ' s
surface is given b y u3 = u3(xx,x2),
a n d t h e slopes in t h e OXx a n d OX2
directions a r e respectively given b y du3/dxx
a n d du3/dx2.
T h e s e slopes
will also b e small. C o n s i d e r n o w a small element dxx, dx2 of t h e
m e m b r a n e (Fig. 10.19). A c t i n g o n it, a r e the forces Tdxx a n d Tdx2 as
well as pdxxdx2
in t h e OX3 direction. If w e resolve t h e forces in t h e
OXx, OX2, a n d OX3 directions, w e find, b e c a u s e of t h e smallness of t h e
slope, t h a t t h e equilibrium e q u a t i o n s in the OXx a n d OX2 directions a r e
a u t o m a t i c a l l y satisfied. T h e equilibrium in the OX3 direction gives (Fig.
10.20):
*3
X3
Tdx2
Tdxf
Tdx2
Fig. 1 0 . 2 0
Tdx / du3
+
Adx7
(10.8.1)
+ pdxx dx2 = 0
or
2
2
d u3
d u:
(10.8.2)
Therefore, t h e s u m of the c u r v a t u r e s in t w o p e r p e n d i c u l a r directions is
a c o n s t a n t for all p o i n t s of t h e m e m b r a n e . If w e n o w adjust p a n d T
such t h a t p/T is n u m e r i c a l l y e q u a l to 2Ga, E q . (10.8.2) b e c o m e s
The Torsion Problem
297
identical to E q . (10.5.7). If we a r r a n g e the m e m b r a n e so t h a t its height
u3 r e m a i n s zero at the b o u n d a r y c o n t o u r of the section, t h e n the heights
w3 of the m e m b r a n e are n u m e r i c a l l y equal to the stress function <f>. If we
h a v e the deflection surface of the m e m b r a n e r e p r e s e n t e d b y c o n t o u r
lines, several i m p o r t a n t conclusions r e g a r d i n g stress distribution in
torsion c a n b e o b t a i n e d :
1. T h e c o n t o u r lines w3 = c o n s t a n t are lines following the shearing
stress, i.e., t a n g e n t to the shearing stress vector (Fig. 10.21): A l o n g a
I
0
ds
c o n t o u r line, du3 /ds = 0, since u3 = c o n s t a n t . Therefore d<j>/ds = 0 a n d
(10.8.3)
Therefore,
CT13COS y — a 3 s i n y =
2
0.
(10.8.4)
I n o t h e r w o r d s , the projection of the stress vector on the n o r m a l n is
e q u a l to zero. Therefore, the stress vector is t a n g e n t to the c o n t o u r line
t h r o u g h a given p o i n t of the twisted bar.
298
T h e Theory of Elasticity
2. T h e shearing stress vector is given in m a g n i t u d e by t h e slope of t h e
m e m b r a n e n o r m a l to t h e c o n t o u r line (i.e., the m a x i m u m slope): Since
the shearing stress vector is directed a l o n g the c o n t o u r lines, its
m a g n i t u d e is given b y Fig. (10.21).
( dx2 dxdn2
9<j>
du3
dn
dxx \ _ d<f>
dx{ dn )
dn ( j Q 8 5)
d<$>
3. T h e m a x i m u m shearing stress acts a t the points w h e r e the c o n t o u r
lines a r e closest t o each other.
4. T h e m a x i m u m shearing stress acts a t t h e b o u n d a r y since t h e slope
is m a x i m u m there.
5. T h e twisting m o m e n t is e q u a l in m a g n i t u d e to twice t h e v o l u m e
u n d e r t h e m e m b r a n e since
M 33 = 2 / /
<$>dxxdx2= 2 f f
u3dxxdx2.
In s u m m a r y , t h e m e m b r a n e a n a l o g y allows u s :
a. t o m e a s u r e experimentally t h e shearing stress ot
b . to visualize intuitively t h e torsion p r o b l e m
c. t o solve t h e c o m p l e t e p r o b l e m w h e n it is easy to find w 3.
Fig. 1 0 . 2 2
(10.8.6)
T h e Torsion Problem
10.9
299
Application of the Membrane Analogy to Solid Sections
In this section, the m e m b r a n e a n a l o g y is a p p l i e d to b a r s with circular
a n d thin r e c t a n g u l a r cross sections, t h e n e x t e n d e d to thin o p e n sections.
1. Circular Cross Section
T h e c o m p u t a t i o n s are m o s t easily m a d e in cylindrical c o o r d i n a t e s . O n
a c c o u n t of s y m m e t r y , the height uz of the m e m b r a n e d e p e n d s only o n
r. C u t t i n g a c o n c e n t r i c circle o u t of the m e m b r a n e (Fig. 10.22) a n d
writing the equilibrium in the vertical direction, we get:
00.9.i)
-mrT^-Utp
p=
r.
If
duz
"dr
(10.9.2)
This m e a n s t h a t the slope of the m e m b r a n e is p r o p o r t i o n a l to the
d i s t a n c e r a n d h e n c e the shearing stresses follow this law. Substituting
2Ga to p/T we get:
(10.9.3)
ot = o9z = Gar,
w h i c h w a s o b t a i n e d in Sec. 10.2. T h e s h a p e of the m e m b r a n e is o b t a i n e d
by integrating Eq. (10.9.2).
P
uz = -
j
+ constant .
^ f rd = ~ ~
O n the p e r i p h e r y , uz = 0, so t h a t
uz = ^ ( / ?
2
(10-9.4)
2
- r ).
T h e v o l u m e u n d e r the m e m b r a n e is given b y
twisting m o m e n t b y twice this v a l u e :
(10.9.5)
I
2TLruzdr,
0
a n d the
R
Mz2 = 2
f
Jo
2Uruz dr = § ^ 4
4 T
=z I G
a
(10.9.6)
Therefore,
^T=GIZ,
which is the result o b t a i n e d in Sec. 10.2.
(10-9.7)
z
300
T h e Theory of Elasticity
2. Thin Rectangular
Section
C o n s i d e r a n a r r o w r e c t a n g u l a r cross section with sides b a n d /, in which
b is m u c h larger t h a n t (Fig. 10.23). Since b is m u c h larger t h a n /, w e
c a n d e d u c e that t h e s h a p e of t h e m e m b r a n e will b e t h e s a m e b e t w e e n
CC a n d DD. T h e m e m b r a n e flattens u p at b o t h e n d s . T h e c o n t o u r lines
in t h e central p o r t i o n a r e straight a n d parallel t o t h e OX2 axis. C u t t i n g
o u t a central piece of m e m b r a n e of d i m e n s i o n s 2xx a n d / , t h e third
e q u a t i o n of equilibrium gives:
\ \
x
A
—
^
V2.
77^
^
4£
B
J
A, -X,
Cfx
.
x
2
D
3
X2
b
sect. B B
Fig. 1 0 . 2 3
- 2 T l p -
dxx
=
2xxlp
XxP=
-lx~
7 -
1
(10.9.8)
9 9)
' '
T h i s m e a n s that t h e slope of t h e m e m b r a n e is p r o p o r t i o n a l to xx, a n d
h e n c e t h e shear stress follows this law. T h e m a x i m u m s h e a r stress
occurs a t xx = ± / / 2 , a n d is equal to
kux = f4 =
I n t e g r a t i n g E q . (10.9.9), w e get:
M3 »
O n t h e periphery, x 3 = 0, so t h a t :
+ constant .
(-->
10 9 10
00.9.11)
The Torsion Problem
_
P_(t±
2T\ 4
"3 ~
301
(10.9.12)
3
w h i c h is a p a r a b o l a w h o s e a r e a is pt /\2T.
I n calculating the v o l u m e
3 flattening n e a r the edges x2 = ±\ is neglected
u n d e r the m e m b r a n e , the
so t h a t the v o l u m e is pt b/\2T.
T h e twisting m o m e n t is thus given b y :
3
PGbt
2
33
(
\2T)
Therefore,
(10.9.13)
-a.
3
Gbt
(10.9.14)
C/.7,
where
(10.9.15)
3
T h e expression of the m a x i m u m shearing stress c a n b e written in t e r m s
of the twisting m o m e n t as follows:
a
( /)max
~
3M,
133
(10.9.16)
bt
3. Thin Open Sections
S u p p o s e the n a r r o w rectangle of Fig. 10.23 is b e n t b y 90 degrees so t h a t
the section b e c o m e s a thin-walled angle. T h e s h a p e of the m e m b r a n e
will n o t c h a n g e except for local effects at the c o r n e r s a n d the v o l u m e
u n d e r the m e m b r a n e will r e m a i n essentially the s a m e . Eq. (10.9.13)
Tl
T
T
J
Fig. 1 0 . 2 4
r e m a i n s valid for a n a n g u l a r section if b is t a k e n as the length of b o t h
legs c o m b i n e d (Fig. 10.24). T h e s a m e is true for T shapes, / shapes, or
a n y section built u p with rectangles; b o x sections will b e e x a m i n e d in
302
T h e Theory of Elasticity
the next section. If the w e b a n d the flanges d o n o t h a v e the s a m e
section, E q . (10.9.14) is applied to each p a r t alone a n d the results a d d e d
u p . T h e angle a is the s a m e for all p a r t s of the s a m e section a n d each
p a r t is subjected to a m o m e n t p r o p o r t i o n a l to its torsional rigidity:
For angle sections we h a v e , to a g o o d a p p r o x i m a t i o n :
3
J
+ b2t\
3
Vi
~
M
3
33
a
Mi
( a , ) mx a«
(10.9.17)
+ htl u
3
(10.9.18)
(10.9.19)
Gatit
where tt is the larger of ?, a n d t2.
For a channel or an I section, we h a v e :
3
7
M
+ 2y2
3
~
^33
a
3
3
V i
+ 2b
r 2t\
3
~ °
( a , ) mx a«
(10.9.20)
Ga/,-,
(10.9.21)
(10.9.22)
where
is the larger t.
For a trapezoidal section, the value of a is o b t a i n e d by a s s u m i n g that
the surface of the deflected m e m b r a n e is conical. F r o m Eq. (10.9.13), we
h a v e (Fig. 10.25):
0923
=
M33
V
f*
±Gat>dx2,
t>
H I —
Fig. 10.25
1
(» - - )
The Torsion Problem
303
with
(10.9.24)
Therefore,
^33 _
a
Kh r
+ h\t\
+ t\)
(10.9.25)
n
where
J =
(10.9.26)
12
T h e previous e q u a t i o n s are directly applicable to a n / section
sloping flanges (Fig. 10.26):
h—4a
H ft
with
t
m
Fig. 1 0 . 2 6
(10.9.27)
33
a
=
(^)max =
(10.9.28)
JG
^a/ ,
3
(10.9.29)
o c c u r r i n g a t p o i n t m.
It s h o u l d b e n o t e d that in all the previous cases a c o n s i d e r a b l e stress
c o n c e n t r a t i o n takes place at the r e - e n t r a n t corners, the m a g n i t u d e of
which d e p e n d s o n the r a d i u s of the fillets. F o r small radii of fillets,
Trefftz found the following a p p r o x i m a t e solution (Fig. 10.27):
304
T h e Theory of Elasticity
Fig. 10.27
3
(a,)
fillet =
in which / is the larger of tx a n d
1.74(0,),'max
(10.9.30)
t2.
10.10 Application of the Membrane Analogy to Thin Tubular Members
I n Sec. 10.5, it w a s s h o w n that, o n the b o u n d a r y of a b a r subjected to
torsion, <J> is c o n s t a n t . T h i s c o n s t a n t w a s c h o s e n e q u a l to zero. F o r the
case of a hollow section, the function <f> is c o n s t a n t o n the o u t e r a n d
i n n e r b o u n d a r i e s . H o w e v e r , while it c a n still b e c h o s e n e q u a l t o zero o n
the o u t e r o n e , a different c o n s t a n t m u s t b e assigned to its value o n t h e
inner o n e . I n a d d i t i o n to the t w o e q u a t i o n s (10.5.7) a n d (10.5.8), o n e
a d d i t i o n a l e q u a t i o n is n e e d e d to solve the p r o b l e m . T h i s a d d i t i o n a l
e q u a t i o n is o b t a i n e d b y writing t h a t the d i s p l a c e m e n t s m u s t b e single
v a l u e d . Fig. 10.28 shows w h a t is m e a n t b y this last s t a t e m e n t : Starting
Fig. 1 0 . 2 8
The Torsion Problem
305
from p o i n t M o n the c o n t o u r s2 a n d p r o c e e d i n g a r o u n d the hole, w e
m u s t e n d u p with the s a m e value of w 3 we started from. W e see t h a t this
is the case in Fig. 10.28a, while the slit in the t u b u l a r m e m b e r of Fig.
10.28b results in a discontinuity MM'. M a t h e m a t i c a l l y the c o n t i n u i t y or
compatibility r e q u i r e m e n t is written a s :
3w3
3x,
dxx +
3u3
dx~>
dx
,)-a
(10.10.1)
Let us n o w c o m p u t e t h e integral j> a,ds a r o u n d the i n n e r b o u n d a r y
U s i n g E q s . (10.3.4), (10.5.3), (10.5.4), a n d (10.8.5), we get:
i
+
=G
l
dx2 +Ga
(XId%1
'-
X1dxx) (10 10 3)
2
= Ga
Js
2
)a
-
$ (If" £ ) ^ ~
dxx+
s2.
(10.10.4)
(x{ dx2 — x2 dx\).
T h e integral in the r i g h t - h a n d side of Eq. (10.10.4) is e q u a l to twice the
a r e a A2 inside the c o n t o u r s2: I n d e e d , the first integral j>S2xxdx2
is
e q u a l to (Fig. 10.29):
.**
D
Fig. 1 0 . 2 9
(|) x, dx2 = - a r e a D'DEBB'
a n d the s e c o n d integral fs
CD -x2dxx
JSI
—x2dxx
= - ( a r e a EE'C'CB
A2,
(10.10.5)
CD) = A2,
(10.10.6)
+ a r e a B'BCDD'
=
is e q u a l t o :
- a r e a EEC
2
306
The Theory of Elasticity
so t h a t
(f) otds
= 2GaA2.
(10.10.7)
If we t r a n s l a t e E q . (10.10.7) in t e r m s of m e m b r a n e b y m e a n s of E q .
(10.8.5), w e get:
(10.10.8)
or
d
-T(^) ^ds=pA2.
(10.10.9)
E q . (10.10.9) r e p r e s e n t s the e q u a t i o n of e q u i l i b r i u m of a weightless flat
p l a t e covering the a r e a A2 a n d subjected to a n u p w a r d p r e s s u r e p a n d
a d o w n w a r d pull a r o u n d its c o n t o u r b y a m e m b r a n e with tension T a n d
slope — du3/dn
(Fig. 10.30). Therefore, in the case of hollow sections,
Fig. 1 0 . 3 0
the m e m b r a n e m a y b e c o n s i d e r e d as stretched b e t w e e n the o u t e r
c o n t o u r sx a n d a weightless flat p l a t e of the s a m e s h a p e as the i n n e r
c o n t o u r s2. If the hollow t u b e h a s t h i n walls, the m e m b r a n e b e t w e e n sx
a n d 5*2 will b e a p p r o x i m a t e l y straight. This m e a n s t h a t the s h e a r i n g
stress is uniformly d i s t r i b u t e d across the thickness of the wall. If the
The Torsion Problem
307
height of t h e p l a t e is h ( F i g . 10.30), t h e shearing stress at a n y p o i n t
w h e r e t h e thickness is / is given b y :
at
(10.10.10)
h
=
T h e applied twisting m o m e n t M 33 m u s t b e e q u a l t o t h e m o m e n t s of t h e
shearing stresses a r o u n d t h e center of r o t a t i o n of each section of t h e
hollow b a r . T h u s (Fig. 10.30),
M 33 = (j)
ottndS
= (j)
hndS
= 2/L4,
(10.10.11)
w h e r e S is t h e m e a n line b e t w e e n sx a n d s2 a n d A is t h e a r e a inside S.
Therefore, t h e twisting m o m e n t is e q u a l t o twice t h e v o l u m e u n d e r t h e
p l a t e a n d t h e m e m b r a n e . T h e shearing stress a t a n y p o i n t w h e r e t h e
thickness of t h e wall is / is given b y :
a -
°<
(10.10.12)
2At'
Eq. (10.10.7) gives:
^ftds
(10.10.13)
= 2G«A2.
Therefore, t h e torsional rigidity is given b y :
4
M33 _
2
*r A 2
4A
If / is c o n s t a n t , E q . (10.10.14) b e c o m e s :
^31
= 4 G^
=/ G
(10.10.15)
w h e r e L is t h e length of t h e m e a n line of t h e b o u n d a r i e s a n d
j = *Ah
(10.10.16)
Li
C o m b i n i n g E q s . (10.10.15) a n d (10.10.12), w e get t h e value of t h e angle
of twist p e r unit length in terms of ot:
a—
(io.io.i7)
2AG'
308
The Theory of Elasticity
T w o i m p o r t a n t conclusions c a n b e d r a w n from the previous e q u a t i o n s :
a) If a hollow t u b e is flattened (i.e., A « 0), the torsional rigidity t e n d s
to zero a n d the shearing stresses b e c o m e very high.
b) F o r a given length of c o n t o u r L, a circular s h a p e w o u l d give the
m a x i m u m possible rigidity since it c o r r e s p o n d s to the highest A.
10.11 Application of the Membrane Analogy to Multicellular Thin
Sections
If the cross section of a t u b u l a r m e m b e r h a s m o r e t h a n t w o
b o u n d a r i e s , the m e m b r a n e a n a l o g y involves several stiff weightless
plates w h i c h will h a v e to b e b l o w n u p with the air pressure p. T h e
,/*„). A s s u m i n g t h a t the
heights of the plates are u n k n o w n (hx,h2>...
thickness of the walls is small, the slope of the outside walls will b e h/t
while the slope of the inside walls will b e Ah/t, w h e r e
is the
difference in height of the t w o adjoining plates. W e c a n write o n e
e q u a t i o n of equilibrium for e a c h plate,
(10.11.1)
PAn=T(fifdS,
th
w h e r e An is the a r e a of the n plate, the integral extends a r o u n d t h a t
plate, a n d M is the height of the p l a t e in q u e s t i o n less the height of the
n e i g h b o r i n g plate. T h e r e a r e n such e q u a t i o n s a n d they c a n b e solved
for the heights h{9 h 2, . . . , hn. O n c e the heights are k n o w n , we translate
the m e m b r a n e p r o b l e m to the torsion p r o b l e m b y setting M/t = ot a n d
p/T = 2Ga. T h e twisting m o m e n t is given b y twice the v o l u m e u n d e r
the plates a n d m e m b r a n e :
^ 3 3 - 2 2 ^*1.-
(10.11.2)
T h e following e x a m p l e illustrates the use of the m e m b r a n e a n a l o g y for
a bicellular section.
Example
I n the section s h o w n in Fig. 10.31, let
BCD = SY,
DEB
= S2,
BD = S3.
(10.11.3)
The Torsion Problem
t
D
JHtlit
C
309
I
~ "JL
i_"2
"1—I H
t H
M *
P
Fig. 10.31
The slopes of the membrane are given by:
(10.11.4)
on =
hi - h2 _ txaa
-
t2a,2
(10.11.5)
The twisting m o m e n t is given by twice the volume under the plates and
membrane:
M 33 = 2 0 4 , * , * , , + A2t2at2
).
(10.11.6)
The vertical equilibrium of the two plates gives:
~A{
= anS{
= 2GaAx
(10.11.7)
j,A2
= al2 S2 - 0 , 3S3 = 2GaA2.
(10.11.8)
+ al3
S3
The solution of the four simultaneous Eqs. (10.11.5) to (10.11.8) gives
0 , (, 0,2, ol3
, and a. In the special case where the central wall is a plane
of symmetry of the cross section, hx = h2 and o, 3 = 0.
10.12 Torsion of Circular Shafts of Varying Cross Section
In the study of bodies of revolution, the computations are most
conveniently performed in cylindrical coordinates. Fig. 10.32 shows
such a b o d y subjected to terminal couples. The axis of the shaft will b e
taken as the OX3 or the Z axis. The general equations of equilibrium are
given by Eqs. (7.4.6) in which w e set the b o d y forces equal to zero. T h e
310
The Theory of Elasticity
Fig. 10.32
strain-displacement relations are given by E q s . (6.7.23) a n d (6.7.24). Let
us a s s u m e t h a t
ur = 0,
uz — 0,
(10.12.1)
a n d t h e n p r o v e t h a t the solution b a s e d o n this a s s u m p t i o n satisfies
equilibrium a n d b o u n d a r y c o n d i t i o n s . O n a c c o u n t of s y m m e t r y , ue
c a n n o t b e a function of 9 a n d will only d e p e n d o n r a n d z. Eqs. (6.7.23)
a n d (6.7.24) give:
err = eon = e77 = er=
(10.12.2)
0
ee
z€ r
'
" 2 \ dr
P °
2 dz '
(10.12.3)
F r o m H o o k e ' s law, we h a v e :
(10.12.4)
r0
(= dUg
9z
Ug\
\ ~dr ~ ~r~/' °
°
dUg_=
dz
(10.12.5)
W i t h these values of the stresses, two e q u a t i o n s of equilibrium are
identically satisfied while the third o n e gives:
r
dr
or
2
dz
2
| ( r a r, ) + j-z(r °9z
)
= 0.
U (10.12.6)
(10.12.7)
The Torsion Problem
311
E q . (10.12.7) is identically satisfied b y i n t r o d u c i n g a stress function H
(A*, Z), such t h a t
f
=^ ^ , f = - ^ -
00.12.8)
2
Since
a n d e9z are the only n o n - v a n i s h i n g strains, a n d since b o t h are
expressed in t e r m s of ug, compatibility is o b t a i n e d b y eliminating u9
from the t w o e q u a t i o n s of (10.12.3). This gives the following c o m p a t i bility r e l a t i o n :
e
^ r9
dz
e
^ 9z
= _
dr
r
Jz^
e
(10.12.9)
U s i n g E q s . (10.12.5) a n d (10.12.8), E q . (10.12.9) is written in t e r m s of
the stress function H as follows:
^ 2- 3 M
dr
dr
^+ 2r+
dz
(10.12.10)
0 =
Since the lateral surface of the shaft is free from external loads, it
follows t h a t the s h e a r i n g stress there m u s t b e directed a l o n g the t a n g e n t
to the b o u n d a r y of the axial section a n d its projection o n the n o r m a l to
the b o u n d a r y m u s t b e zero. Therefore,
c^cos a — a^sin a = 0
(10.12.11)
or
2
2
.±™dz_\Wdr
=\_dH
r dz ds
r dr ds
2
r
, 2)
ds
w h i c h gives
H = c o n s t a n t o n the b o u n d a r y .
(10.12.13)
T h e solution of the torsion p r o b l e m of a circular shaft with v a r i a b l e
d i a m e t e r t h u s r e d u c e s to finding a function H satisfying E q s . (10.12.10)
a n d (10.12.13).
T h e m a g n i t u d e of the twisting m o m e n t is given b y :
Mz = jl
oezr(2nr)dr
= 2U[H(a,z)
-
= 211 £
H(0,z)].
f
*
)
(
1
312
T h e Theory of Elasticity
r
Fig. 10. 33
F o r t h e case of a conical shaft (Fig. 10.33), we h a v e o n the b o u n d a r y
z
,
22 x i / > = cos /3 = c o n s t a n t .
,
(10.12.15)
T h u s , a n y function of this ratio will satisfy t h e b o u n d a r y c o n d i t i o n . O n e
c a n verify t h a t t h e function
z
H = A
2
-
2
±T
2
-
2
Tl
(10.12.16)
l(r + z )^
3L(A- + Z )^J
y
w h e r e C i s a c o n s t a n t , satisfies E q . (10.12.10). Substituting E q . (10.12.16)
i n t o E q . (10.12.14), t h e c o n s t a n t C is f o u n d to b e given b y :
C =
3M
3
^
=—.
2 n ( 2 - 3 c o s / ? + cos /?)
(10.12.17)
T h e shearing stresses a9z a n d or9 a r e given b y :
^ " V + V
w h e r e C is given b y E q . (10.12.17).
(1
°- 12
18)
The Torsion Problem
313
10.13 Torsion of Thin-Walled Members of Open Section in which some
Cross Section is Prevented from Warping
I n o u r s t u d y of thin o p e n sections in Sec. 10.9, it was a s s u m e d t h a t
e a c h section w a s free to w a r p . A s u b s t a n t i a l a m o u n t of torsional rigidity
c a n b e a d d e d to such b a r s if o n e or m o r e cross sections are p r e v e n t e d
from w a r p i n g . F o r b a r s of solid cross sections such as ellipses or
rectangles, the p r e v e n t i o n of w a r p i n g h a s a negligible effect o n the angle
of twist p e r unit length a [3]. O n the o t h e r h a n d , in thin o p e n sections
like / b e a m s a n d c h a n n e l s , the p r e v e n t i o n of w a r p i n g d u r i n g twist is
a c c o m p a n i e d b y a b e n d i n g of the flanges w h i c h m a y h a v e s u b s t a n t i a l
effect o n the angle of twist. This p r o b l e m h a s b e e n a n a l y z e d b y
T i m o s h e n k o [3, 4].
Let us consider, for e x a m p l e , a n / b e a m in w h i c h a section is
p r e v e n t e d from w a r p i n g . This c a n b e a c c o m p l i s h e d b y subjecting the
b e a m to a t o r q u e M 33 at e a c h e n d a n d a t o r q u e 2 M 33 at the center as
s h o w n in Fig. 10.34. T h e center of the b e a m r e m a i n s p l a n e b e c a u s e of
Fig. 1 0 . 3 4
s y m m e t r y a n d it m a y b e c o n s i d e r e d as built in with the t o r q u e M 33
a p p l i e d at the o t h e r e n d (Fig. 10.35). Let 0 b e the angle of r o t a t i o n of
a n y section of t h e b e a m at a d i s t a n c e x3 from t h e origin. T h e twisting
m o m e n t M 33 is t r a n s m i t t e d from o n e section to the o t h e r of the b e a m
in t w o w a y s :
(a) By the usual torsional stresses d i s t r i b u t e d over the / section.
(b) By m e a n s of shearing forces in the flanges d u e to b e n d i n g as
s h o w n in Fig. 10.35.
Therefore,
M 33 = JGa +Vxh
= JG^-
+ Vxh,
(10.13.1)
314
The Theory of Elasticity
Fig. 10.35
w h e r e JG is the torsional rigidity of the / section a n d VXis the s h e a r i n g
force in the lower flange in the system of axes of Fig. 10.35. C o n s i d e r i n g
e a c h flange as a b e a m with o n e e n d built in a n d the o t h e r free, we h a v e
(from the e l e m e n t a r y t h e o r y of b e a m s ) :
(10.13.2)
in which If is the m o m e n t of inertia of the flange a b o u t the OX2 axis.
T h e d i s p l a c e m e n t ux of the lower flange is given b y :
«, = * § .
(10.13.3)
Substituting E q s . (10.13.2) a n d (10.13.3) into E q . (10.13.1), we get:
-
"33 "
Bl£%
- JGa -
E4$J
(10.13.4)
or
(10.13.5)
T h i s is a linear differential e q u a t i o n in a, w h o s e solution is:
a =
^3
+
^-0*3+^3
(10.13.6)
T h e Torsion Problem
315
where
2 JG
2
b
h EIf
(10.13.7)
'
T h e c o n s t a n t s of i n t e g r a t i o n Cx a n d C 2 a r e o b t a i n e d from the b o u n d a r y
c o n d i t i o n s . T h e first b o u n d a r y c o n d i t i o n is t h a t at x3 = oo, d0/dx3 = a
m u s t r e m a i n finite so t h a t Cx = 0. E q . (10.13.6) b e c o m e s :
a - C ^ - f a a + ^ M .
(10.13.8)
T h e s e c o n d b o u n d a r y c o n d i t i o n is that, at x3 = 0, the slope
= 0, a n d h e n c e d0/dx3 = a = 0, so t h a t
c
°
2 - Jhk
"
dux/dx3
(10.13.9)
GJ '
T h e value of a is n o w given b y :
a d =i _
dx3
^ 3
=(
_
GJ
el- b
)X
i
E q . (10.13.10) c a n b e i n t e g r a t e d to give 9 as a function of
j
e+
3
* - ^ (
f
k
a
)
(10.13.10)
x3:
, +c
n)
'
-
T h e c o n s t a n t C3 c a n b e o b t a i n e d b y setting 0 = 0 for x3 = 0. H e n c e ,
0 = ^
*
3
+ l . (e - * * 3 _ l ) ] .
(10.13.12)
D u e to the p r e s e n c e of a s e c o n d t e r m in E q . (10.13.10), the angle of twist
p e r u n i t length varies a l o n g the length of the b e a m a l t h o u g h
M33
r e m a i n s c o n s t a n t : I n o t h e r w o r d s , the torsion is n o t u n i f o r m . O n c e a is
d e t e r m i n e d , b o t h p a r t s in the r i g h t - h a n d side of E q . (10.13.4) c a n b e
o b t a i n e d a n d t h a t p o r t i o n d u e to the b e n d i n g of the flanges d e t e r m i n e d .
A t the p o i n t x3 = 0, a = 0 so t h a t the entire t o r q u e is b a l a n c e d b y the
m o m e n t of t h e s h e a r i n g forces in the flanges a n d Vx = M33 /h. F o r very
large values of x3, E q . (10.13.12) gives:
<
i
a> ,
3
i
3
316
The Theory of Elasticity
Eq. (10.13.13) states t h a t the torsional stiffness of a long cantilever /
b e a m is equal to the torsional stiffness of a free b e a m of a length shorter
t h a n the cantilever b e a m b y a n a m o u n t 1 / b.
Example
Let us consider a steel / b e a m with the d i m e n s i o n s s h o w n in Fig. 10.36.
F o r such a n / b e a m , we h a v e :
V
1
H
1
-J
—I
Fig. 10.36
22
E I f= ^ -
GJ =
u — t1 ^[24G° ~ a
E
V
(10.13.14)
= Et a
^GL
(10.13.15)
= {i Ga
t1 , / 2 4 x 1 2 _
1
~ a
30
3.9a'
V
(10.13.16)
T h e expression of the b e n d i n g m o m e n t in the lower flange is given b y :
>- ~ h% -
M
- -#«--• < >
E!
,om7
with the m a x i m u m m o m e n t o c c u r r i n g at x 3 = 0. H e n c e ,
(^)
ma X = ^
= 3 . 9 M 3.3
00.13.18)
T h e shearing force in the lower flange Vx is o b t a i n e d from E q . (10.13.2).
Its m a x i m u m value is given b y :
v
^ =1
(10.13.19)
T h e Torsion Problem
317
T h e m a x i m u m n o r m a l stress d u e to b e n d i n g o c c u r s at JC3= 0, a n d is
e q u a l to
(
)o
l°33W
3
n
-
2
^=
/ 6t
=2 8 0+. 8+M 33
a~
3
•
a
0 3 10 2
T h e m a x i m u m s h e a r i n g stress d u e to b e n d i n g o c c u r s at x2 — 0, a n d is
equal to:
(
u„
)
o
, »
v 13 /max .B
33 3
a
1
8
M
(10.13.21)
'
T h e m a x i m u m s h e a r i n g stress d u e to twist o c c u r s at x3 = L, a n d is given
by:
22)
F r o m the p r e v i o u s calculations, it is clear t h a t the shearing stresses d u e
to b e n d i n g a r e negligible c o m p a r e d to those d u e to twist for long b e a m s .
Also, the m a x i m u m n o r m a l b e n d i n g stresses d o n o t o c c u r at the s a m e
section w h e r e the m a x i m u m shearing stresses o c c u r : F o r design p u r p o s es, the s h e a r i n g stresses are potentially m o r e d a n g e r o u s . Therefore, in
this e x a m p l e , the p r e v e n t i o n of w a r p i n g will affect the design i n a s m u c h
as it c a u s e s a s h o r t e n i n g of t h e b e a m b y 3.9a for p u r p o s e s of t o r s i o n a l
stiffness c o m p u t a t i o n s .
Remark
A d d i t i o n a l e x a m p l e s of torsion of thin o p e n sections c a n b e f o u n d in
[3].
A P P E N D I X T O C H A P T E R 10
A-10.1 The Green-Riemann Formula
T h e G r e e n - R i e m a n n f o r m u l a t r a n s f o r m s a line integral t a k e n a r o u n d
a closed c o n t o u r C (Fig. 10.37) i n t o a surface integral over the a r e a A
enclosed b y C. It is written as follows:
(
318
The Theory of Elasticity
Fig. 1 0 . 3 7
(j)c P(xl9
x2)dxx
= j
+ Q(xl9
x2)dx2
J
(-^-
+ ||)
dxx
dx2.
(A-10.1.1)
T o p r o v e this formula, let us c o m p u t e the first d o u b l e integral of the
r i g h t - h a n d side of (A-10.1.1):
I I — ~— dxx dx2 — — \
J
OX
J a
2
J
dxx
I
Jx =f (x )
2l l
~— dx2
2
OX
or
A
=
j
GDB
P(xx,x2)dxx—
j
P(xx,x2)dxx
P(xx,x2)dxx
=
GEB
I n a similar way, o n e c a n p r o v e t h a t :
dxx dx2 = ( p
Q(xx,
x2)dx2.
1
T h e Torsion Problem
319
PROBLEMS
1.
A circular shaft is m a d e of a n i n n e r circular solid cylinder w h o s e
m a t e r i a l h a s a shear m o d u l u s Gx a n d a n o u t e r circular a n n u l u s
w h o s e m a t e r i a l h a s a shear m o d u l u s G2 (Fig. 10.38). T h e m a t e r i a l s
are perfectly b o n d e d at the interface r{ a n d the shaft is subjected to
a twisting m o m e n t Mz:
(a)
(b)
F i n d the expression of the angle of r o t a t i o n p e r unit length a.
F i n d the d i s t r i b u t i o n of the shearing stresses o9z in t h e
cylinder a n d the a n n u l u s .
(c) H o w m u c h of the total twisting m o m e n t Mz d o e s the a n n u l u s
carry?
X
2
Fig. 1 0 . 3 9
320
2.
T h e Theory of Elasticity
Show t h a t the P r a n d t l stress function
2
(t>= m ( r
2
- b
) ( ^ f ^ -
l)
furnishes the solution to the p r o b l e m of t h e circular shaft with a
circular groove (Fig. 10.39). F i n d t h e value of t h e c o n s t a n t m a n d
the expressions of t h e stresses a 13 a n d a 23 o n t h e b o u n d a r i e s Cx a n d
3.
4.
5.
6.
c 2.
T h r e e b a r s — o n e with a s q u a r e cross section, o n e with a n equilateral
triangle cross section, a n d o n e with a circular s e c t i o n — h a v e e q u a l
cross sectional a r e a s a n d a r e subjected to e q u a l twisting m o m e n t s .
C o m p a r e the m a x i m u m shearing stresses a n d the torsional rigidities
of t h e b a r s .
A steel b a r h a v i n g a r e c t a n g u l a r cross section 1 in. wide a n d 2 in.
long is subjected to a twisting m o m e n t of 1,000 lb-in. C a l c u l a t e t h e
m a x i m u m shearing stress 6a n d t h e shearing stress at t h e center of the
short side. (G = 12 X 1 0 psi)
A steel b a r h a v i n g a slender r e c t a n g u l a r cross section \ in. w i d e a n d
6 in. long is subjected t o a twisting couple of 1,500 lb-in. F i n d t h e
m a x i m u m shearing stress a n d the angle of twist p e r unit length a
using t h e exact solution of Sec. 10.7 a n d t h e a p p r o x i m a t e solution
b a s e d o n the m e m b r a n e 6a n a l o g y . W h a t is the m a g n i t u d e of the error
involved? (G = 12 X 1 0 psi .)
A b r a s s b a r h a v i n g t h e cross section s h o w n in F i g . 10.40 is subjected
to a twisting c o u p l e of 600 lb-in. This b a r is 6 ft. long. W h a t is t h e
6 p e r unit length a n d the m a x i m u m shearing stress?
angle of twist
(G = 4 X 1 0 psi .)
K4 -
U
3
6
1
4
Fig. 1 0 . 4 0
The Torsion Problem
—
8
—
321
20
Fig. 1 0 . 4 1
7.
8.
A h o l l o w cylinder h a v i n g outside a n d inside d i a m e t e r s of 10 in. a n d
8 in., respectively, is subjected to a twisting m o m e n t Mz. C o m p a r e
the angle of twist p e r unit length to the o n e the cylinder w o u l d h a v e
if it were split longitudinally a l o n g a g e n e r a t o r .
A t h r e e - c o m p a r t m e n t thin-walled b o x section h a s a c o n s t a n t wall
thickness (Fig. 10.41). F i n d the values of the m a x i m u m shearing
stresses in the v a r i o u s elements if the b o x is subjected to a twisting
m o m e n t of 1,000,000 lb-in. H o w m u c h larger w o u l d the stresses in
6 were m a d e in
the walls of the central c o m p a r t m e n t get if t w o cuts
the walls of the side c o m p a r t m e n t s ? (G — 12 X 1 0 psi .)
9.
T h e section s h o w n in Fig. 10.42 is subjected to a t o r q u e of 1,000,000
lb-in. C a l c u l a 6t e the stresses in the different p a r t s of the section.
(G = 12 X 1 0 psi .)
10. T h e wide-flange / b e a m s h o w n in Fig. 10.43 is 20 ft. long. It is fixed
6 m o m e n t of 30,000 lb-in.
6
at o n e e n d a n d free at the other. A twisting
is a p p l i e d at the free e n d . If E = 30 X 1 0 psi a n d G = 12 X 1 0 psi,
,
If
r
4
4
4
4
1
-—20
/0
—*J^ "*f
F i g . 10.42
1
18
322
The Theory of Elasticity
h
10.5
.8+5'
10
Fig. 1 0 . 4 3
c o m p u t e the m a x i m u m n o r m a l a n d shearing stresses d u e to b e n d i n g
a n d the m a x i m u m s h e a r i n g stress d u e to twist. W h a t is the angle of
twist of the b e a m at the free e n d ?
REFERENCES
[1] S. H. Crandall and N . C. Dahl, An Introduction to the Mechanics of Solids, McGraw-Hill,
N e w York, N . Y., 1959.
[2] C. T. Wang, Applied Elasticity, McGraw-Hill, N e w York, N . Y., 1953.
[3] S. Timoshenko, Strength of Materials, Vol. 2, Van Nostrand, Princeton, N . J., 1955.
[4] S. Timoshenko and J. N . Goodier, Theory of Elasticity, McGraw-Hill, N e w York, N . Y.,
1970.
CHAPTER 11
THICK CYLINDERS, DISKS, AND
SPHERES
11.1
Introduction
I n the study of d i s p l a c e m e n t functions (Sec. 9.4), it was m e n t i o n e d
t h a t p a r t i c u l a r solutions of N a v i e r ' s e q u a t i o n s are o b t a i n e d b y m e a n s of
L a m e ' s strain p o t e n t i a l <f>; <j> c o u l d b e either a h a r m o n i c function or a
function satisfying Poisson's e q u a t i o n . Solutions c o r r e s p o n d i n g to p r a c tical b o u n d a r y c o n d i t i o n s h a v e b e e n g e n e r a t e d from this potential, a n d
it will b e u s e d in this c h a p t e r to investigate p r o b l e m s of cylinders, disks,
provides the solution to a
a n d spheres. W h e t h e r a c h o s e n 4>(xi,x2,x3)
given elasticity p r o b l e m or n o t d e p e n d s o n the b o u n d a r y c o n d i t i o n s .
A n o t h e r m e t h o d of solution we shall a p p l y is t h a t of Airy's stress
function. T h e forms suggested in Sec. 9.11 are well-suited to the
p r o b l e m s w h i c h are e x a m i n e d in the following sections.
11.2
Hollow Cylinder with Internal and External Pressures and Free
Ends
C o n s i d e r t h e strain function of E q . (9.4.14).
w h e r e C a n d K are c o n s t a n t s . This is a h a r m o n i c function a n d , as such,
S = i
^
(11-2.1)
323
324
The Theory of Elasticity
is a solution of N a v i e r ' s e q u a t i o n . T h e c o m p o n e n t s of the d e f o r m a t i o n
are given b y Eqs. (6.4.27):
1 C
u
r = 2QT->
=0> Uw2 = 0.
O
(11.2.2)
T h e c o m p o n e n t s of the state of stress are given b y :
C
o
C
(11.2.3)
(11.2.4)
Let a state of stress, in w h i c h orr = oee = D a n d all o t h e r c o m p o n e n t s
are e q u a l to zero, b e s u p e r i m p o s e d to the state of stress of E q s . (11.2.3)
a n d (11.2.4), so t h a t :
C
C
+ A
°rr =
°r9
= 0
ez = 0
O
Ogg
O
=
+ D,
zz = 0
O
rz = 0.
(11.2.5)
(11.2.6)
This superposition of stress will a d d a strain
a n d ezz
quantities
ur a n d w z,
err = eg9 = (1 - p)D/2 (1 + v)G
= -vD/G{\
+ ^) to the original o n e . C o n s e q u e n t l y , the
(1 - v)Dr/{\ + v)2G a n d -*>Z)z/G(l 4- v) will b e a d d e d to
respectively. T h e e q u a t i o n s for the d i s p l a c e m e n t s b e c o m e :
m
vDz
G(l + v)
Fig. 11.1
(11.2.7)
Thick Cylinders, Disks, and Spheres
325
N o w c o n s i d e r a hollow cylinder (Fig. 11.1) with i n n e r r a d i u s r = a, o u t e r
radius r = b, a n d free e n d s . Let a pressure Pt be applied o n the inner
surface, a n d a pressure P0 b e applied o n the outer surface. E q s . (11.2.5)
to (11.2.7) a p p l y to this p r o b l e m since they are such t h a t orr is c o n s t a n t
for r = a a n d r = b, a n d ozz is e q u a l to z e r o ; they also define values of
or0
, oa2, orz
, ug equal to zero, a n d a value of ezz e q u a l to a c o n s t a n t as
r e q u i r e d b y s y m m e t r y in the cylinder p r o b l e m . T h e c o n s t a n t s C a n d D
c a n b e o b t a i n e d b y setting orr = -Pt for r = a, a n d orr = -PQ for r = b
in E q . (11.2.5):
r„ _
t\
/>. = - ^C1 +t Z),
i
-p r>= - ^Cl + D
b
a
Therefore,
.8)
2
_
2 - p20b
n\p0~Pi
2)
6 -«
'
2
a -*
a
)
•
T h e s e values are n o w substituted into E q s . (11.2.5) to (11.2.7). T h e only
c o m p o n e n t s of the state of stress orr a n d ogg b e c o m e :
2
2
22
P.a 2 - P20b
b -a
^
i2 a b (P02 - 2P , )
r
b -a
,.(*)'-'
2
/-(f)
_ P 1 a 2 - ^ 2b
6 - a
!2 a
_
r
= p.-V-f^—p
'(*)-
< l u
-
, 0 )
2
2
°"~
1
^ 2 - 2P,)
b -a
v
y
n
-(f)
T h e s e are L a m e ' s formulas for the stresses. orr + o99 is a c o n s t a n t
t h r o u g h o u t the cylinder. T h e c o m p o n e n t s of the d i s p l a c e m e n t b e c o m e :
2
2
_ j _ a P,,p 2 62 P
E
- a
6
2 2
+, a 6 ( P21/ - J2P 0) i
£
(6 -« )
'
n
, 2)
2(
326
T h e Theory of Elasticity
r [„W
\+v
+
2
Uo
p
l +v
+
1
VJ
(
n
2
1
3
)
2
= 0 2vPi<*
=u -Pob
)
Eqs. (11.2.3) to (11.2.6) s h o w t h a t this is a case of p l a n e stress, even
t h o u g h the cylinder is n o t thin.
If the outer radius b is much larger than the inner radius a, t h e n
2
om = Pi(j) -P0[l
(11.2.15)
2
(f) ]
+
*-&M0 -4^(0 ]}
a
2
(1L216)
A t the i n n e r surface r = a, a n d E q s . (11.2.15) a n d (11.2.16) give:
arr =-/>,,
B = Pt-2Pa0, 9
" =^(^-TT7^)r
<-- )
n
2
17
If r is very large, E q s . (11.2.15) a n d (11.2.16) give:
(1 - v) PQ
r
n i 2 1R^
Eqs. (11.2.18) show t h a t if values of r are k e p t large at the o u t e r surface
c o m p a r e d to the i n n e r r a d i u s a, the outer surface m a y b e r e p l a c e d by
a n y p r i s m a t i c surface parallel to the OX3 or Z axis a n d subjected to a
c o n s t a n t pressure P0. F o r e x a m p l e , the cross section of the o u t e r surface
m a y b e a rectangle with edges parallel to the OXx a n d OX2 axes with
pressures o u = o12 = — P0. If the inside pressure Pi = 0, E q . (11.2.17)
shows t h a t o9B = — 2PQ
, twice the pressure t h a t w o u l d exist w i t h o u t the
hole.
IfP0 = 0, E q s . (11.2.10) a n d (11.2.11) give:
rr = - 2T — ^2 ( l " ^ rz )
a
b - a
\
r /
(always compressive)
(11.2. 19)
Thick Cylinders, Disks, a n d Spheres
=
°9B
+
p—^O
327
S ( ta e l n ws ai ly e )
'
~t)
(H-2.20)
Therefore,
+
(*»)ma* =
(°9eUa
=
^ 2
(0
T
2^
^
— a )
and
( e»)min
^
<
'>
2
_a
a
n t h al2LT
!
21 L2
~
( 9o)r=b
2a
2
P,2
(11.2.22)
b -a '
If Z? - a is very small, a good a p p r o x i m a t i o n (Fig. 11.2) is given b y :
F i g . 11.2
T h e m a x i m u m shearing stress occurs at the inner surface a n d is given
by:
96
(„\-
=- ( °
a
~
b2 P
"\
-
'
(11.2.23)
T h e radial d i s p l a c e m e n t at r = a is o b t a i n e d from Eq. (11.2.12)
2 2 )4
< " ^ - t ( ^ £
+ ')
' '
///>,. = 0, E q s . (11.2.10) a n d (11.2.11) give:
2
2Pb
b
(
^\ \ —
2 \
J
(always compressive)
(11.2.25)
328
The Theory of Elasticity
2
Ppb
_
b -a V^r J
2
2
2
(11.2.26)
(always compressive a n d larger in m a g n i t u d e t h a n
orr
).
Therefore,
(°ee)max
-
(°0o)r=a ~
yi °_
2
Q
>
(11.2.27)
a n d w h e n b/a increases, the m a x i m u m compressive stress tends to twice
the external pressure. T h e radial d i s p l a c e m e n t at r = b is o b t a i n e d from
Eq. (11.2.12):
O n e of the m a n y a p p l i c a t i o n s of the previous e q u a t i o n s is in the a r e a
of shrink fit. This o p e r a t i o n is used to p r o d u c e a c o n t a c t b e t w e e n a h u b
a n d a shaft: T h e initial r a d i u s of a n o u t e r cylinder is m a d e smaller t h a n
the outer r a d i u s of a n i n n e r cylinder b y a q u a n t i t y 8. T h e outer cylinder
is then h e a t e d a n d , u p o n cooling, a c o n t a c t pressure is applied b e t w e e n
the two p a r t s . T h e increase in the inner r a d i u s of the o u t e r cylinder
a d d e d to the decrease in the outer r a d i u s of the i n n e r cylinder m u s t b e
equal to S. E q s . (11.2.24) a n d (11.2.28) give (Fig. 11.3):
Fig. 11.3
w h e r e p is the pressure b e t w e e n the two p a r t s . Therefore,
Thick Cylinders, Disks, a n d Spheres
2
2 2
P _ E8(b ~ 2 <>
2 2)(c ~
b
329
2
b)
2b (c -a )
(11.2.30)
'
K n o w i n g p, o n e c a n get the stresses at e a c h p o i n t of the o u t e r a n d i n n e r
ring. T h e highest stress o c c u r s at the inner surface of the outer cylinder:
2
p(b
( 0e)r=b
a
2
2+
c^-b
=
c)
(11.2.31)
'
2 _ t.2 ' ' °rr ~ ~P-
T h e m a x i m u m shearing stress at this surface is:
2
a
( /)max
=
- 2
2pc*
c - b
2 2
E8c (b2
2b\c
2
- 2a )
- a)
(11.2.32)
'
If the t w o cylinders are different in length, stress c o n c e n t r a t i o n s will
o c c u r at the edges of the shorter cylinder (Fig. 11.4).
y
Fig. 11.4
Remark
T h e solution of the hollow cylinder p r o b l e m in t e r m s of Airy's stress
function c a n b e f o u n d in references [1] a n d [2].
11.3
Hollow Cylinder with Internal and External Pressures and Fixed
Ends
T h e p r o b l e m in this case is a p r o b l e m of p l a n e strain (Fig. 11.5). T h e
two c o m p o n e n t s of the state of stress orr a n d oeB k e e p the s a m e value as
2
in the previous case,
while ozz b e c o m e s 2e q u a l to 2vD. A s a result, err is
decreased b y 2v D/E
a n d the ur b y 2v Dr/E.
uz a n d ezz are equal to
zero. Therefore,
330
The Theory of Elasticity
r
i
m
i
|
H
3
\ n i
n
i
I I | j I 1t
•
X
I i|
m
M f l l l H
F i g . 11.5
2
2
22
r _Pia -P0b2
°'
6
2
i2 a b (P02 - 2P . )
6 - a
(11.3.1)
i2 a fe (Jg
2 - 2P.)
r
b - a
(11.3.2)
2
- a
r
2
e e_ P , a 2 - P20b
° ~
b -a
2
P,« -
2
2 2
P0b
(11.3.3)
(11.3.4)
and
2
r
2
^ r ^ ( 2^ - f2 , ) l 2
2GL
a - b
r
^
Fig. 11.6
2
2- ^ 2
( 1 _
b -a
K
'\
(, ,.3.5)
Thick Cylinders, Disks, a n d Spheres
u9 = 0,
331
(11.3.6)
uz = 0.
Eqs. (11.3.1) to (11.3.6) c o u l d h a v e b e e n o b t a i n e d b y starting directly
from the strain p o t e n t i a l
2
<t> = Cln-j^+
Hr ,
(11.3.7)
w h e r e C, K, a n d H are c o n s t a n t s . Eq. (11.3.7) is a c o m b i n a t i o n of E q s .
(9.4.14) a n d (9.4.18).
11.4
Hollow Sphere Subjected to Internal and External Pressures
Let a hollow sphere of inner radius a a n d o u t e r radius b b e subjected
to a n internal pressure Pt a n d a n external pressure PQ (Fig. 11.6).
Because of the spherical s y m m e t r y , it is a d v a n t a g e o u s to use spherical
p o l a r c o o r d i n a t e s (<|>, 0, p). I n such a system of c o o r d i n a t e s , all the shear
stresses a n d shear strains vanish. Of the three c o m p o n e n t s of the
d i s p l a c e m e n t vector
u9 w p, o n l y up is different from zero.
T h e solution to this p r o b l e m is furnished b y the c o m b i n a t i o n of the
t w o functions given in Eqs. (9.4.16) a n d (9.4.19), n a m e l y b y :
2
4> = ^ + D p .
(11.4.1)
Indeed,
satisfies Poisson's e q u a t i o n a n d gives stresses a n d d e f o r m a tions fulfilling all the s y m m e t r y c o n d i t i o n s . T h e d i s p l a c e m e n t s a r e given
by:
2Gup
(11.4.2)
p
(11.4.3)
E q s . (6.7.34) a n d (6.7.35) give:
(11.4.4)
(11.4.5)
e
%
=
0
P
e
= 4>p =
°-
(11.4.6)
332
The Theory of Elasticity
T h e stresses are given b y :
(11.4.7)
(11.4.8)
T h e c o n s t a n t s C a n d D are o b t a i n e d b y setting app=
°PP = -P0forr
= b in Eq. (11.4.8):
rp
- Pt for /- = a a n d
3 , ^ \ -\- v
— 2C
<
a
r
(11.4.9)
1 - 2?
3
°
(11.4.10)
1 - 2?
b
Therefore,
3 3
q f r (3P 0 - 3 />,•)
2C =
6 - a
(11.4.11)
3 3
3 a Ps-b
3 P0
! _ 2„
D =
2(1 +v)
b -a
(11.4.12)
'
T h e s e values are n o w substituted i n t o Eqs. (11.4.7) a n d (11.4.8):
33
3
3
a b (P0 3 - 3/>,.) j3 ^ a Pi 3 - 3b P0
2(6 -a ) p
b -a
+ 2
+2
P:\\^)
3
33
a b (P30
A - a
3
j 3 ^ a />, -
- 3 P^
(11.4.13)
6 P0
p
(11.4.14)
1
1
T h e s e are L a m e ' s formulas for the stresses. T h e c o m p o n e n t s of the
displacement become:
Thick Cylinders, Disks, a n d Spheres
i
U
f
p r
aH\P0 3- 3Pj) i 3
2GL
2(Z> -« ) p
, 3
i(b\
_p_r 2VP;
2
C
L
'
1
i
+
/ / ft is large compared
1 - 2 ^
2f
I ,
+
(s) -
i
(1 -2v)a Pi-b P0l[3
3
(1+") b - a
I ,
+
•
1
5 1
J
l / a \
w ]
2
- ( f )
( 1 1 4 1 6 )
J
to a, w e get:
V =^ =^ ( ^ ) - f [ ( ^ )
3
3
%
333
= - ^ )
3
+
2
]
(H.4.17)
2
- ^ [ l - ( ^ )
]
( „ . 4 . 1 )8
A t t h e i n n e r surface p = a, a n d E q s . (11.4.17) a n d (11.4.19) give:
p. =
=
(o )
= (°<M>)p=a
(a^)
90
)p=a
= ~2
-=
/
vVp=*
2GL2
3Po
2y~ '
,
_
=
22
((°pp)p=a
°Pp)p
(11.4.20)
^ 2 ( 1 +*>)_]*
If p is very large, E q s . (11.4.17) t o (11.4.19) give:
21
n
~ n
-
n
-
- P
u
-
~
°
p
p
P
(11.4.21)
Eqs. ( 1 1 . 4 . 2 1 ) show t h a t if values of p a r e k e p t large a t t h e o u t e r surface
c o m p a r e d t o t h e i n n e r r a d i u s a, t h e o u t e r surface m a y b e r e p l a c e d b y
a n y surface p r o v i d e d t h e u n i f o r m external p r e s s u r e PQis m a i n t a i n e d .
F o r e x a m p l e , t h e o u t e r surface m a y b e a c u b e with edges parallel t o t h e
OXx, OX2, a n d OX3 axes a n d with pressures oxx = a 22 = o33= — P0- If
the inside p r e s s u r e Pt = 0 , E q s . ( 1 1 . 4 . 2 0 ) s h o w t h a t o99=
= —3^/2,
\ times t h e p r e s s u r e t h a t w o u l d exist if there w a s n o spherical cavity.
IfP0 = 0 , E q s . (11.4.13) a n d (11.4.14) give:
*-*.-^('+&)><>
<
nA22
>
334
The Theory of Elasticity
3
_
Of
Pta
(11.4.23)
< 0.
Therefore,
(°Oe)'max
-
(°90)p=a
~
3
3
2 b - a
'
(11.4.24)
If b - a is small, a g o o d a p p r o x i m a t i o n for og9 is given b y :
a
a
P
oaa =
^
2(b -a)'
(
(11.4.25)
Similar e q u a t i o n s c a n b e written for Pi = 0.
11.5
Rotating Disks of Uniform Thickness
T h e p r o b l e m of the stresses a n d d e f o r m a t i o n s in disks r o t a t i n g at high
speed is of p r i m a r y i m p o r t a n c e in the design of a wide variety of
m a c h i n e s . Let us consider a thin disk of uniform thickness r o t a t i n g with
a c o n s t a n t a n g u l a r velocity <o r a d / s e c . T h e b o d y force is the centrifugal
force (Fig. 11.7):
Fig. 11.7
2
p<o r,
(11.5.1)
w h e r e p is the m a s s p e r u n i t v o l u m e of the m a t e r i a l of the disk.
A l t h o u g h the m e t h o d of L a m e ' s strain p o t e n t i a l c a n b e slightly modified
a n d applied to this case [3], we shall use a stress function in seeking a
solution to the p r o b l e m . F o r a thin disk, the e q u a t i o n s related to a state
of p l a n e stress in Sec. 9.10 apply. T h e force Fr derives from a p o t e n t i a l ,
Thick Cylinders, Disks, a n d Spheres
a =^ .
335
(11
-
52)
Because of circular s y m m e t r y , the equilibrium e q u a t i o n (9.10.13) b e comes:
+ p ^ r - 0
^ + » ( ^ - a w)
(".53)
or
while Eq. (9.10.14) is identically satisfied. A n y stress function </> will b e
a function of r alone. It is easy to verify that the stress function defined
by
22
r a „ = <J>,
^ =a ^
+ pco r ,
01.5.5)
satisfies the e q u a t i o n of equilibrium (11.5.4). Eqs. (6.7.23) give:
rr
e ~-^Ldr '
0
9
n = ^L'
r
e n
(11.5.6)
a n d a simplified compatibility relation is o b t a i n e d b y eliminating ur
from E q s . (11.5.6):
£(reffe
)-err
(11.5.7)
= 0.
U s i n g the stress-strain relations (8.17.11) a n d (8.17.12), a n d the stress
function defined by Eqs. (11.5.5), Eq. (11.5.7) b e c o m e s :
{3
{r )
i[7§ * ]
+ v2 =)r p
-
"
-
5 )8
-
By direct integration, we get:
^ - ^ - f W ^
+ C.f + Q l ,
(H.5.9)
w h e r e Q a n d C2 are c o n s t a n t s of integration. T h e c o r r e s p o n d i n g stress
components are:
22
an = £ = -^+ZpUr
+ ^
+ %
(11-5-10)
2
+ P^r
= - ^ P » V
+ § • - % .
(H-5-11)
336
The Theory of Elasticity
T h e c o n s t a n t s Cx a n d C 2 are o b t a i n e d from the b o u n d a r y c o n d i t i o n s :
a. Solid disks
2 2 at r = 0 so t h a t C = 0. A t r = b o = 0,
T h e stress c a n n o t b e infinite
2
y rr
so t h a t Cx = [(3 + v)/4]po b .
Therefore (Fig. 11.8),
T
r
Fig. 11.8
(11.5.12)
(11.5.13)
T h e m a x i m u m stress is at the center, w h e r e
V ...2£2
3 +
-pw
(11.5.14)
T h e radial d i s p l a c e m e n t ur is given b y :
2
2
2
(11.5.15)
_ pco r(f — 1)
[r (j- + 1) - 6 (*> + 3)].
8£
b . Disk with a circular hole of radius a
A t r = a, a n d at r = 6, o„. = 0. Therefore,
2
C, = ^ p < o ( 6
2
2
+ a ),
2 2 2
C2 = - 4 ^ p o
a
6
(H.5.16)
Thick Cylinders, Disks, a n d Spheres
337
and
(11.5.17)
(11.5.18)
T h e m a x i m u m stress occurs at the i n n e r b o u n d a r y , w h e r e
(11.5.19)
If the circular h o l e is very small, E q . (11.5.19) b e c o m e s :
o
99 =
(11.5.20)
^ P ^ b \
which is twice the value of the stress at the center of a solid disk [Eq.
(11.5.14)]. Therefore, b y m a k i n g a small circular hole at the center of a
r o t a t i n g disk, we shall d o u b l e the m a x i m u m stress. T h e r a d i a l displacem e n t ur is given b y :
1 + vr2
3 + v
(11.5.21)
Remarks
a) T h e solution of the p r o b l e m c o u l d h a v e b e e n as easily o b t a i n e d b y
solving N a v i e r ' s e q u a t i o n s a n d getting ur.
b) T h e a p p r o x i m a t e n a t u r e of the p l a n e stress solution m u s t b e k e p t
in m i n d . I n d e e d , of the six c o m p a t i b i l i t y relations (6.9.1) to (6.9.6) three
are identically satisfied—namely, E q s . (6.9.4) to (6.9.6); E q s . (6.9.1) to
(6.9.3) for the special case of this p r o b l e m b e c o m e :
(11.5.22)
r
dr
= 0.
zz
= 0.
(11.5.23)
While E q . (11.5.22) has b e e n satisfied, Eqs. (11.5.23) h a v e n o t b e e n
c o n s i d e r e d in the solution.
338
The Theory of Elasticity
11.6
Rotating Long Circular Cylinder
W e shall e x a m i n e two cases:
Case a. The rotating cylinder is not free to deform
longitudinally
T h e p r o b l e m is a p l a n e strain p r o b l e m . T h e stress strain relations
(8.16.12) to (8.16.14) are the ones to b e applied. U s i n g the stress
function defined in Eqs. (11.5.5), the compatibility relation (11.5.7)
becomes:
This e q u a t i o n is seen to b e different from its c o u n t e r p a r t of Sec. 11.5.
Direct integration gives:
9
_ ( 3 - 2v) P
8(1 - v)
Q
2Q
3
2
)
}
r '
'
T h e stress c o m p o n e n t s a r e :
y
8(1 - v) ™
om
-~±
'
'
2
'
r
y
dr
+ P<oV = - ^ r ^
' ^ '
8(1 - v)
+
' 2
'
r
(11.6.4)
T h e c o n s t a n t s of integration are n o w o b t a i n e d in the s a m e m a n n e r as
in the case of the disk. T h u s :
F o r a solid cylinder of r a d i u s b, we h a v e :
1
s o ^ ^ K
6 5)
- ^ )
- -
1
o2Z= v(on + o0e
) = 4 ^ ^ ) [ ( 3 - 2v)b
2
- 2r ],
(11.6.7)
a n d the m a x i m u m stress occurs at the center, a n d is
°rr = °*> =
%f^)f"> l> 2
2
68
(ll- - )
F o r a hollow cylinder of inner radius a a n d outer r a d i u s b, we h a v e :
6
2
339
Thick Cylinders, Disks, and Spheres
8(1 -v)
2
°09 -
3 -- 2v
q7I _-v) ^P^ (tf
8(1
3 -- 2v
vpj(b
4(1 -v)
2
_ \ ± ^ r
+ 2 a+
2
)
(11.6.10)
2
+ a - 33^),
(11-6.11)
a n d the m a x i m u m stress occurs at the inner surface, a n d is
H e r e , too, we see that the m a x i m u m stress is d o u b l e d w h e n a solid
cylinder h a s a small hole drilled t h r o u g h its center. It is to b e n o t i c e d
that, for the previous solution to b e valid, ozz m u s t act at the e n d s of the
cylinder a c c o r d i n g to Eqs. (11.6.7) a n d (11.6.11).
Case b. The rotating cylinder is free to deform
longitudinally
In the a b o v e discussion, the values of the axial stress ozz were so
adjusted t h a t there were n o longitudinal strains ezz a n d , c o n s e q u e n t l y ,
n o longitudinal d e f o r m a t i o n s uz. If the cylinder is n o w allowed to
d e f o r m freely longitudinally, t h e n b e c a u s e of s y m m e t r y the strain ezz
m u s t b e such that every cross section r e m a i n s p l a n e . T h e cylinder is in
a state of generalized p l a n e strain, i.e., the radial a n d transverse stresses
will n o t c h a n g e while the cylinder is c h a n g i n g length uniformly:
F o r a solid cylinder:
^Ki-g)
v
3
8(1 - v)^
"
3v l 22
a„„ =
~
^
2
" V
wh(
b
1-
8( 1 — v)
(1L6
+
1
2v
13)
(11.6.14)
d.\
\
-
3-2n '
2
° » - 4 T T ^ ) [ 3( ~ ^ 2 - 2 r 2 ]
Ee+
.
z2
01-6.15)
T h e value of ezz c a n b e o b t a i n e d from the c o n d i t i o n t h a t there is n o
resultant longitudinal force o n the e n d s H e n c e
.
,
b
fQ 2Urazz dr = 0.
(11.6.16)
Substituting Eq. (11.6.15) into Eq. (11.6.16) a n d integrating, we get:
2
ezz = e 0 = - ^
2
b
(11.6.17)
340
The Theory of Elasticity
and
^ 4 7 T ^ - ' > 2
2
2
( 1 L 6
-
1 8 )
F o r a hollow cylinder:
^-s ^ ^ *
°09
o.
7
3 - 2
^
8(1
4(i -
2 + 02
- ^ "'O
+
2
(b
y
2
(n.,20)
2
+ a - 2r ),
(11.6.21)
v
and
2
eZ2 = e0 = -^Ep<,Hb
11.7
(1L619)
2
+ a ).
(11.6.22)
Disks of Variable Thickness
T u r b i n e disks are usually m a d e thicker n e a r their h u b a n d t a p e r d o w n
to a smaller thickness t o w a r d s the periphery. T h e r e a s o n for this is the
high stress c o n c e n t r a t i o n n e a r the center of flat disks. T h e m e t h o d used
in the analysis of flat disks will b e applied h e r e with the difference t h a t
the thickness t will h a v e to b e i n c l u d e d in the calculations. T h e
thickness is a function of r a l o n e (Fig. 11.9). orr a n d o99 are the m e a n
341
Thick Cylinders, Disks, a n d Spheres
radial a n d transverse stresses at a n y distance r from the center line. T h e
e q u a t i o n of equilibrium (11.5.4) is valid for a flat disk of unit thickness.
T o apply t o this case, all its m e m b e r s m u s t b e multiplied b y /, so t h a t
we n o w h a v e :
22
£r(trorr)-to9e
+ pto> r
01.7.1)
= 0
as o u r e q u a t i o n of equilibrium. T h e stress function <f> is defined b y
d§
dr
a n d t h e governing e q u a t i o n (11.5.8) b e c o m e s :
^+(> - ?! ) 4
+
7
(fi
-
' > -
"
'
3 )
'
from which <J> c a n b e found, p r o v i d e d t = t (r) is given. F o r a h y p e r b o l i c
shape, E q . (11.7.3) c a n easily b e integrated. If t h e thickness varies
a c c o r d i n g t o t h e law
(11.7.4)
/ = Cr-P,
w h e r e C is a c o n s t a n t a n d p is a n y positive n u m b e r , E q . (11.7.3) reduces
to:
r2^f
dr
1 +
1 (+
y ^ _
p
dr
1 +(
p) p _) =((3 f
+ 2 pC 3)-rpp. o ) (11.7.5)
T h e solution of this e q u a t i o n is easily o b t a i n e d b y substitution. T h i s
solution is:
C
</> = Qr*
+ C r*
2
-
z_\^
2
P<»
v
)
rP 3
~ >
p
1 L6 7
(
- )
w h e r e qx a n d q2 a r e t h e roots of the e q u a t i o n
2
q +pq-{\
(11.7.7)
+ pp) = 09
i.e.,
l}--5 V(f)
±
+
(
l
+
v p )
-
( 1 1 J
-
8 )
T h e c o m p o n e n t s of the stress a r e therefore,
(11.7.9)
+
342
The Theory of Elasticity
o.-g
+
p
S
'
'
(
]
|)
Eq. (11.7.8) shows t h a t q2 + p will always b e a negative q u a n t i t y so that,
for a solid disk, C 2 m u s t b e equal to zero since orr a n d o99 c a n n o t b e
infinite for r = 0. F o r n o forces acting on the b o u n d a r y , orr = 0 for r =
b. Therefore,
r
C
8 — (3 +
v)p
Therefore, for a solid disk,
3 + v
8 - (3 + v)p
"^[(sP
7l2
' - ( i f ]
<"- >
F o r a disk of uniform thickness p = 0, a n d E q s . (11.7.12) a n d (11.7.13)
r e d u c e to Eqs. (11.5.12) a n d (11.5.13). Fig. 11.10 shows h y p e r b o l i c
profiles for various values of p. Similar e q u a t i o n s c a n b e established for
hollow disks. Cx /C a n d C 2 / C are o b t a i n e d b y writing t h a t orr = 0 for r
= a a n d r = b, respectively.
Fig. 1 1 . 1 0
7
Thick Cylinders, Disks, a n d Spheres
11.8
343
Thermal Stresses in Thin Disks
Let us c o n s i d e r a thin circular disk with a n u n e v e n t e m p e r a t u r e
d i s t r i b u t i o n AT = AT(r). T h e stress-strain relations in this case a r e those
of a p l a n e stress p r o b l e m . B e c a u s e of s y m m e t r y , the only relations of
interest are p r o v i d e d b y E q s . (8.6.3) a n d (8.6.4), w h i c h b e c o m e :
+ a(AT)
(11.8.1)
= ±(oee - vorr
) + a(AT).
(11.8.2)
err = ±(orr
-vo9e
)
T h e equilibrium equation,
d
°n
dr
=Q
°rr +
~ °00
r
'
is identically satisfied if we i n t r o d u c e the stress function
(11.8.3)
such t h a t
T h e c o m p a t i b i l i t y Eq. (11.5.7) b e c o m e s :
T h i s e q u a t i o n is easily i n t e g r a t e d to give t h e value of <f> at a n y p o i n t of
a circle of r a d i u s r. H e n c e ,
^ = -«E
£
AT)rdr (
+ ^
+ ^ ,
(11-8.6)
w h e r e Cx a n d C 2 a r e c o n s t a n t s of i n t e g r a t i o n a n d the lower limit a is
e q u a l to z e r o for a solid disk, a n d is e q u a l to the inner r a d i u s for a
hollow disk. T h e stresses c a n n o w b e o b t a i n e d b y substituting E q .
(11.8.6) i n t o E q . (11.8.4). T h i s gives:
a
(AT)rdr+
oeg = a £ [ - ( A r ) + i
±
ja
+
(11.8.7)
n
(AT)rdr]
+ ^
+ ^ .
(11.8.8)
344
The Theory of Elasticity
F o r a solid disk, the stresses m u s t b e finite at the origin so t h a t C 2 = 0.
If t h e r e a r e n o e x t e r n a l forces a p p l i e d o n t h e b o u n d a r y , orr = 0 for r =
b. T h e n ,
1
C , - ^
b
b
( (AT)rdr
Jo
( " ^ )
and
[t^
=aE
°09
L ( >*"? / < >*]
A7
r
(iL8io)
0
= a£[-(Ar) + ^
F o r a hollow
Therefore,
A7
disk
Jo
(Ar)rrfr + i
Jq
(A7>rfr].
(11.8.11)
with i n n e r r a d i u s a, orr = 0 at r = a a n d r = b.
and
1
0 „ - a£[-(Ar> + \
L
Y
f
J a
(HT)rdr
f
+ ^
u
— Q,
(4T>*
(11.8.14)
J a
O n c e A r = A7(A*) is k n o w n , t h e v a l u e of t h e stresses c a n b e c o m p u t e d .
11.9
Thermal Stresses in Long Circular Cylinders
Case a. The cylinder is not free to deform
longitudinally
T h e p r o b l e m is a p l a n e strain p r o b l e m . T h e stress-strain relations a r e
E q s . (8.6.3) t o (8.6.5) w h i c h b e c o m e :
Thick Cylinders, Disks, a n d Spheres
+
^K
r
eee
j 1L
1
er =
Pr/i
=
V
, a _
~ )°rr
~ ™99 +
"
-
™
,
. . e v a ™
CLE(AT)]
rr +
*E{bT)\
«zz = Mr + <%) ~ <*£(Ar).
345
(11.9.1)
01-9.2)
(11.9.3)
U s i n g t h e stress function defined in E q . (11.8.4), t h e c o m p a t i b i l i t y E q .
(11.5.7) b e c o m e s :
d_
dr
E x c e p t for t h e coefficient of d(AT)/dr,
(11.8.5). Therefore, its solution is:
this e q u a t i o n is t h e s a m e as E q .
+
* = - T ^ t r + ^
%
( i i
-
9 5 )
F o r a solid cylinder, C 2 = 0 so t h a t t h e stresses will b e finite a t t h e
origin, a n d if there a r e n o stresses o n t h e outer surface r = b,
f\
1
1
C = i 2aE_ 1
1 - vb
Jo
(11.9.6)
AT)rdr
Therefore, t h e stresses a r e :
* -A
li f
( A 7
A
<% - T
~ [~(
°» f=h[%
=
( A r
>* - ? r
r
A
) + £
/ 0 (
I!
H
l9 7>
7
> ^ + yi £
- ]-
{LT)rdr
°-
(Ar)
W>dr]
(11.9.8)
--
(ii 9 9)
ozz is t h e n o r m a l stress distribution w h i c h m u s t b e a p p l i e d t o k e e p
ezz = 0 t h r o u g h o u t .
F o r a hollow cyilinder with i n n e r r a d i u s a, orr= 0 a t r = a a n d a t r =
b. T h e r e f o r e ,
1
C, =
£2
"
1-
.
1
2
,
- a
C (AT)rdr
ya
("A"*)
(11.9.11)
346
T h e Theory of Elasticity
and
(11.9.12)
°90
(11.9.13)
a,, =
a 22 is the n o r m a l stress distribution which m u s t b e applied to k e e p
ezz = 0 t h r o u g h o u t .
Case b. The cylinder is free to deform
longitudinally
T h e r e a s o n i n g m a d e in Sec. 11.6 c a n b e r e p e a t e d here. T h e cylinder is
in a c o n d i t i o n of generalized p l a n e strain:
F o r a solid cylinder:
+ £<?0| dr = 0.
Therefore,
°-¥
Ee
L
(AT)rdr
and
^[jXV^-Ar].
"zz
J
a rr a n d
a r e given b y E q s . (11.9.7) a n d (11.9.8), respectively.
F o r a hollow cylinder:
01-9.16)
Thick Cylinders, Disks, and Spheres
347
and
•ft
aE
< i - » ) i
orr a n d O9Q are given b y E q s . (11.9.12) a n d (11.9.13), respectively.
11.10 Thermal Stresses in Spheres
C o n s i d e r a sphere in which the t e m p e r a t u r e distribution is s y m m e t r i cal with respect to the center a n d is therefore a function of p, the radial
distance, a l o n e . I n a spherical system of c o o r d i n a t e s , the n o n z e r o
stresses are a pp a n d oBe = a ^ , a n d the n o n z e r o d i s p l a c e m e n t is up. O n l y
the third e q u a t i o n of e q u i l i b r i u m is of significance, the t w o others b e i n g
identically satisfied. This e q u a t i o n b e c o m e s :
^
+ £ < a „ - < * ) - 0.
dl.10.1)
T h e s t r a i n - d i s p l a c e m e n t relations are given b y Eq. (6.7.34). In o u r case
they a r e :
(
1 2) L
I
0
'
T h e e q u i l i b r i u m Eq. (11.10.1) c a n b e written in terms of the displacem e n t w p. U s i n g the stress-strain relations (8.6.9) a n d (8.6.10), as well as
E q . (11.10.2), w e get:
T h e solution of Eq. (11.10.3) is:
m
» -
r
>
p2( A+ r c +
*
> o
%
)
w h e r e Q a n d C 2 are c o n s t a n t s of i n t e g r a t i o n a n d a is a c o n s t a n t t a k e n
e q u a l to zero for a solid sphere a n d e q u a l to the i n n e r r a d i u s for a
hollow sphere. E q s . (11.10.2) give the strains. T h e stresses are directly
o b t a i n e d from the stress-strain relation. T h e y a r e :
(
u
348
T h e Theory of Elasticity
W
VP
3
(1 ~
V)p
P
'
Ja
3
1-2?
(1 + ? ) p
1-
V
(11.10.6)
F o r a ^o/W sphere t h e stresses m u s t b e finite a t t h e origin so t h a t
C 2 = 0. If there a r e n o external forces o n t h e b o u n d a r y of t h e sphere,
°PP = 0 for p = Z>. T h e n ,
C
,
=
^
f
^
/
>
>
^
,
("..0.7,
<w - u r > * - ? r *]
and
(A7 2
(Ar)p2
2
(iiia8)
2
° » = - p ^ ; [ p - J0 ( A * > 4> + ^
/ 0" ( A 7 > dp - ( A r ) ] . (l 1.10.9)
/ b r a hollow sphere with i n n e r r a d i u s a, a pp = 0 a t p = a a n d a t p = b.
Therefore,
2
2 ±
_
1 +
1—2?
J>tf
E
3 Q
C
(11.10.10)
=
and
+
- f ^ ?
T
t
^
s
-
"
«•
" >
Solving for Cx a n d C 2, a n d substituting t h e results in E q s . (11.10.5) a n d
(11.10.6), w e get:
* - " [ f ^
r « ^ *-? / w * ]
<
n
,
a
,
2
)
Thick Cylinders, Disks, and Spheres
2aE
1 -
3 3
_k 3
L2(fe
vL2(b - a ) p A
v
(AT)]
3
2p
349
A
(11.10.13)
PROBLEMS
1.
F i n d the r a t i o of thickness to i n t e r n a l d i a m e t e r for a t u b e subjected
to i n t e r n a l pressure w h e n the pressure is e q u a l in m a g n i t u d e to | of
the m a x i m u m circumferential stress. If the i n t e r n a l d i a m e t e r of t h e
t u b e is 4 in., d e t e r m i n e the increase in the external d i a m e t e r w h e n
the i n t e r n a l p r e s s u r e is 12,000 6psi a n d the t u b e is p r e v e n t e d from
c h a n g i n g length. (E = 30 X 1 0 p s i , ? = 0.3.)
2. A solid b a r of u n i f o r m circular section is subjected t o u n i f o r m
r a d i a l p r e s s u r e . S h o w t h a t the* stress at a n y p o i n t in a p l a n e section
parallel to the axis of the b a r is compressive a n d e q u a l in m a g n i t u d e
to the radial stress.
3 . A steel b a r of 2 in. d i a m e t e r is pressed i n t o a steel sleeve so that,
w h e n a s s e m b l e d , the m a g n i t u d e of the r a d i a l stress b e t w e e n the t w o
is 2,000 psi, a n d t h a t of the circumferential stress at the inside of the
sleeve is 3,200 psi. A s s u m i n g a close fit a n d neglecting friction,
d e t e r m i n e the c h a n g e of radial stress w h e n the b a r is subjected t o
a n axial compressive l o a d of 15,000 lb. (v = 0.3.)
4. A short steel r o d of 2 in. d i a m e t e r is subjected to a n axial
compressive l o a d of 60,000 lb. It is s u r r o u n d e d b y a sleeve \ in.
thick, slightly shorter t h a n the r o d so t h a t the l o a d is carried only
b y the r o d . A s s u m i n g a close fit before t h e l o a d is a p p l i e d a n d
neglecting friction, find the p r e s s u r e b e t w e e n the sleeve a n d the r o d ,
a n d the m a x i m u m tensile stress in the sleeve, (v = 0.3.)
5. T h e external d i a m e t e r of a steel h u b is 10 in. a n d the i n t e r n a l
d i a m e t e r increases 0.005 in. w h e n s h r u n k o n to a solid steel shaft of
5 in. d i a m e t e r . F i n d the r e d u c t i o n in d i a m e t e r of the shaft, the r a d i a l
6 circumferential
pressure b e t w e e n the h u b a n d the shaft, a n d the
stress at the i n n e r surface of the h u b . (E 30 X 1 0 psi, v = 0.3.)
6. A steel cylinder of 8 in. external d i a m e t e r a n d 6 in. i n t e r n a l
d i a m e t e r h a s a n o t h e r steel cylinder of 10 in. external d i a m e t e r
s h r u n k o n t o it. If the m a x i m u m tensile stress i n d u c e d in the o u t e r
cylinder is 10,000 psi, find the r a d i a l compressive stress b e t w e e n the
350
7.
8.
9.
The Theory of Elasticity
cylinders. D e t e r m i n e the circumferential stresses at inner a n d o u t e r
d i a m e t e r of b o t h cylinders, a n d show o n a d i a g r a m h o w these
6
stresses vary with the r a d i u s . C a l c u l a t e the necessary
shrinkage
allowance at the c o m m o n radius. (E = 30 X 1 0 p s i , ? = 0.3.)
A steel hollow sphere, w h o s e inside d i a m e t e r is 5 in., is subjected to
a n internal pressure of 5,000 psi. D e t e r m i n e the thickness of the
m a t e r i a l if the m a g n i t u d e of the m a x i m u m stress is n o t to exceed
10,000 psi. C o m p a r e this thickness to t h a t o b t a i n e d from E q .
(11.4.25).
F i n d the expressions of the stresses a n d d i s p l a c e m e n t s for a hollow
sphere subjected to a n external pressure P0 a n d filled with a n
incompressible fluid such t h a t its inner d i a m e t e r does n o t c h a n g e .
D e t e r m i n e the greatest value of the radial a n d circumferential
stresses for a thin disk r o t a t i n g at a n a n g u l a r velocity of 150 r a d i a n s
p e r s e c ; the inner a n d o u t e r radii of the disk are 6 in. a n d 12 in.
3 a n d the m a s s6 per unit v o l u m e p of the m a t e r i a l is
respectively,
0.28 l b / i n . . (E = 30 X 1 0 p s i , ? = 0.3.)
10. A solid steel shaft of 8 in. d i a m e t e r h a s a steel cylinder of 16 in.
6
d i a m e t e r s h r u n k o n t o it. T h e inside d i a m e t e r of the cylinder
prior
to the shrink fit o p e r a t i o n w a s 7.992 in. (E = 30 X 1 0 p s i , ? = 0.3.)
(a) D e t e r m i n e the external pressure P0 o n the outside of the
cylinder which is r e q u i r e d to r e d u c e to zero the circumferential
stress at the i n n e r surface of the cylinder.
(b) D e t e r m i n e the radial pressure o n the surface of c o n t a c t d u e
3
to shrink fit.
(c) F i n d the speed of r o t a t i o n to loosen the fit. (p = 0.28 l b / i n . ) .
11. A solid steel shaft 36 in. in d i a m e t e r is r o t a t i n g at 200 r p m . If the
shaft c a n n o t d e f o r m longitudinally, calculate the total longitudinal
6
thrust over 3 a cross section
d u e to r o t a t i o n a l stresses, (p
= 0.28 l b / i n . , E = 30 X 1 0 psi, ? = 0.3.)
12. Show t h a t the radial d i s p l a c e m e n t in a r o t a t i n g solid cylinder,
w h o s e e n d s are free to deform, is given b y :
2
ur
2
p<o (l + ?)(! - 2?) I" (3 5v)b r
8£(1
L(l + v){\ - 2?)
13. A b r a s s r o d is
d i a m e t e r s are
a temperature
a temperature
- ?)
fitted firmly inside a steel t u b e w h o s e inner a n d o u t e r
1 in. a n d 2 in., respectively, w h e n the materials are at
of 60 ° F . If the r o d a n d the t u b e are b o t h h e a t e d to
of 300° F , d e t e r m i n e the m a x i m u m stress in the brass
Thick Cylinders, Disks, and Spheres
351
6 steel. T h e coefficients
6
a n d in the
of e x p a n s i o n for steel a n d b r a s s are
6
6 X 10~
a n d 10 X 1 0 "
p6e r degree F a h r e n h e i t , respectively.
Y o u n g ' s m o d u l u s is 30 X 1 0 psi for steel a n d 12.5 X 1 0 psi for
brass. Poisson's ratio is 0.3 for steel a n d 0.34 for brass.
, oge
, a n d ozz in a long hollow cylinder with
14. F i n d the expression of orr
fixed e n d s , which c o n d u c t s h e a t in steady state a c c o r d i n g t o
tn
a
ATa is a c o n s t a n t increase in the t e m p e r a t u r e of the i n n e r surface of
the cylinder a n d A Tb, smaller t h a n A Ta, is a c o n s t a n t increase in the
t e m p e r a t u r e of the o u t e r surface of the cylinder.
15. F i n d the expression of the axial stress in the cylinder of P r o b l e m 14
w h e n the e n d s are free.
REFERENCES
[1] S. Timoshenko and J. N . Goodier, Theory of Elasticity,
1970.
McGraw-Hill, N e w York N
[2] C. T. Wang, Applied Elasticity, McGraw-Hill, N e w York, N . Y., 1953.
[3] H. M. Westergaard, Theory of Elasticity and Plasticity, Dover, N e w York, N . Y., 1964.
Y
CHAPTER 12
STRAIGHT SIMPLE BEAMS
12.1
Introduction
I n this c h a p t e r , the a s s u m p t i o n s o n w h i c h the e l e m e n t a r y t h e o r y of
b e a m s is b a s e d a r e e n u m e r a t e d a n d the i m p o r t a n t e q u a t i o n s listed. T h e
inverse m e t h o d , Airy's stress function, a n d the semi-inverse m e t h o d a r e
u s e d to s t u d y the p u r e b e n d i n g of a p r i s m a t i c b a r , two cases of n a r r o w
b e a m s with r e c t a n g u l a r cross section, a n d S a i n t - V e n a n t ' s p r o b l e m of
the cantilever subjected to a n e n d l o a d : T h e results are c o m p a r e d with
those of the e l e m e n t a r y theory.
Recalling t h a t a positive face (i.e., cross section) is o n e w h o s e o u t w a r d
n o r m a l is in the positive direction, the following c o n v e n t i o n s for axial
forces, shearing forces, a n d b e n d i n g m o m e n t s will b e a d h e r e d t o :
1) An axial force N is t a k e n positive w h e n it causes n o r m a l stresses in
a positive direction o n a positive face, or n o r m a l stresses in a negative
direction o n a negative face; TV is t a k e n negative w h e n it causes n o r m a l
stresses in a negative direction o n a positive face, or n o r m a l stresses in
a positive direction o n a negative face (Fig. 12.1a).
2) A shearing force Vis t a k e n positive w h e n it causes shearing stresses
in a positive direction o n a positive face, or shearing stresses in a
negative direction o n a negative face; K i s t a k e n negative w h e n it causes
shearing stresses in a negative direction o n a positive face, or s h e a r i n g
stresses in a positive direction o n a negative face (Fig. 12.1a).
3) A bending moment M, acting o n a positive face, is t a k e n positive
w h e n it causes n o r m a l stresses in the positive direction o n the positive
352
Straight Simple Beams
353
0
Fig. 12.1
or lower half of the face (Fig. 12.1b) a n d n o r m a l stresses in the negative
direction o n the negative or u p p e r half of the face. T h e a s s u m p t i o n h e r e
is t h a t those halves fall a b o v e or b e l o w the n e u t r a l surface. O n a
negative face, M is t a k e n positive w h e n it causes n o r m a l stresses in the
negative direction o n the positive or lower half of the face, a n d n o r m a l
stresses in the positive direction o n the u p p e r half of the face (Fig.
12.1b).
T h e p r e v i o u s c o n v e n t i o n s o n forces a n d m o m e n t s are b a s e d o n the
signs of the stresses they g e n e r a t e . T h o s e stresses, in turn, follow the
rules established in Sec. 7.2, w h i c h are universally used in the theory of
elasticity. It is i m m a t e r i a l in w h i c h direction the axes of the r i g h t - h a n d
system p o i n t , these sign c o n v e n t i o n s always hold. T h e d i s p l a c e m e n t s are
positive w h e n they are in the positive directions of the axes. In the
t h e o r y of b e a m s , the deflection w h i c h is the vertical d i s p l a c e m e n t of the
center line follows the s a m e rule.
W h e n relations a r e established a m o n g forces, m o m e n t s , a n d deflections, the c u r v a t u r e of the b e a m is i n t r o d u c e d . T h e center of c u r v a t u r e
is always o n the c o n c a v e side of the b e a m . F o r small deflections, the
c u r v a t u r e is given to a g o o d a p p r o x i m a t i o n b y the rate of c h a n g e of the
slope. Therefore, a n u p w a r d concavity (in the system of axes of Fig.
12.1c), w h i c h is associated with positive b e n d i n g m o m e n t s , c o r r e s p o n d s
to a negative c u r v a t u r e a n d vice versa; h e n c e , o n e c a n a t t a c h a sign to
the r a d i u s of c u r v a t u r e d e p e n d i n g o n the side o n w h i c h the c o n c a v i t y is.
I n s u m m a r y , a n d with the previous sign c o n v e n t i o n s , a positive b e n d i n g
m o m e n t gives a negative c u r v a t u r e .
354
The Theory of Elasticity
12.2
T h e E l e m e n t a r y T h e o r y of B e a m s
T a k i n g the OXx axis as the line j o i n i n g the c e n t r o i d s of the cross
sections, the f u n d a m e n t a l a s s u m p t i o n s in the e l e m e n t a r y t h e o r y of
b e a m s are (Fig. 12.2):
F i g . 12.2
a) T h e b e a m is p r i s m a t i c a l a n d straight b u t n o restrictions are p l a c e d
o n the s h a p e of the cross section.
b) T h e l o a d i n g a n d , c o n s e q u e n t l y , the b e n d i n g m o m e n t Mx3 are
applied in a p l a n e c o n t a i n i n g o n e of the principal m o m e n t s of inertia of
the cross section. If the a p p l i e d b e n d i n g m o m e n t d o e s n o t lie in such a
p l a n e , it c a n always b e split into two c o m p o n e n t s e a c h of w h i c h satisfies
this a s s u m p t i o n . H e r e it is applied a b o u t the OX3 axis in the OXx,
OX2
plane.
c) P l a n e cross sections in t h e . u n s t r e s s e d b e a m r e m a i n p l a n e d u r i n g
bending deformations.
d) T h e deflection of e a c h b e a m e l e m e n t is a s s u m e d to b e in the form
of a circular a r c of r a d i u s R a n d deflections a n d slopes a r e small e n o u g h
so t h a t the c u r v a t u r e C is given b y
2
d u2
r - 1 =
R
*
MOT
dx
~ ^
(12.2.1)
e) T h e shearing stresses are a s s u m e d to b e uniformly d i s t r i b u t e d
across the w i d t h of the b e a m .
f) Lines parallel to the OX3 axis before d e f o r m a t i o n r e m a i n parallel
to this axis after d e f o r m a t i o n .
Straight Simple Beams
355
F r o m t h e a b o v e a s s u m p t i o n s a n d with t h e sign c o n v e n t i o n s a n d
s y m b o l s s h o w n in Figs. 12.1 a n d 12.2, the following w e l l - k n o w n
e q u a t i o n s h a v e b e e n d e d u c e d [1]:
2
d u2 _
dx}
d
M 13
'
EI33
Slope = 9(xx)
d M n ^
+y
dxx
= g
f
= -
'
2 (12.2.2)
Y k =- p
dx
^-dxl
Q
+
(12.2.3)
A
= + Y^
o u = ou(xl,x2)
a=
°22 = 3 3
02.2.4)
33
a
a
1 3 = 2 3 = °>
(12.2.6)
w h e r e M 13 is the b e n d i n g m o m e n t a b o u t the OX3 axis o n the face
n o r m a l t o OXu V2 is the s h e a r i n g force parallel to OX2, / 33 is t h e
m o m e n t of inertia of the section a b o u t the OX3 axis ( 7 33 = I3 is a
p r i n c i p a l m o m e n t of inertia), Q3 is t h e static m o m e n t of the a r e a A
a b o u t t h e OX3 axis, Cx a n d C2 a r e c o n s t a n t s of i n t e g r a t i o n to b e
d e t e r m i n e d from the b o u n d a r y c o n d i t i o n s , a is the d e p t h of t h e b e a m ,
a n d b is the w i d t h of the b e a m at a d i s t a n c e x2 from OX3 (Fig. 12.2).
T h e a b o v e e q u a t i o n s are extremely useful in design b e c a u s e of their
simplicity. I n m a n y cases, their a c c u r a c y is quite sufficient. U n d e r
certain c i r c u m s t a n c e s , h o w e v e r , it is necessary to use the m o r e a c c u r a t e
e q u a t i o n s p r o v i d e d b y the t h e o r y of elasticity, since, in the d e r i v a t i o n of
these e q u a t i o n s , c o m p a t i b i l i t y is i g n o r e d a n d often certain b o u n d a r y
c o n d i t i o n s are violated.
12.3
Pure Bending of Prismatical Bars
Let us c o n s i d e r a p r i s m a t i c b a r b e n t a b o u t the OX3 axis b y t w o
o p p o s i t e couples e q u a l to M (Fig. 12.3). Let us a s s u m e t h a t OX2 a n d
OX3 a r e p r i n c i p a l axes of inertia at the c e n t r o i d of the cross section, a n d
t h a t t h e couples act in the p r i n c i p a l p l a n e OXx, OX2. W e shall a s s u m e
that
a
ll
—
xxx
\\( \> 2i ?>)
a
(12.3.1)
356
The Theory of Elasticity
M
t
L -
H
x2
*
2
F i g . 12.3
=
=
°22
a=
13
°33
a=
23
=
a
12
0,
(12.3.2)
a n d s h o w t h a t this stress field satisfies the r e q u i r e m e n t s of elasticity.
F r o m H o o k e ' s law, w e h a v e :
e n = e23 = ei3 = 0.
(12.3.4)
W i t h these relations, t h e general e q u a t i o n s of e q u i l i b r i u m a r e r e d u c e d
to the e q u a t i o n :
^ 1 1 = 0,
(12.3.5)
dx
x
w h i c h m e a n s t h a t au is i n d e p e n d e n t of x { a n d only a function of x 2 a n d
x 3. T h e compatibility relations (4.10.14) r e d u c e t o :
9^11
= 0
(12.3.6)
dx]
a 2 ( T
n = o
9
a
n
dx
2
I n t e g r a t i n g E q . (12.3.6), we get:
(12.3.7)
3^3
o
=
(12.3.8)
Straight Simple Beams
357
1 329
Oll=*3/l(*2)+/ (*2)-
(
2
- - )
Substituting E q . (12.3.9) into Eq. (12.3.8), we get:
|
= 0
or
/ , ( * 2) = C „
w h e r e Cx is a c o n s t a n t . Substituting Eq. (12.3.9) i n t o Eq. (12.3.7), we
get:
0
= 0
/ 2( * 2) = C 2* 2 + C 3,
or
(12.3.10)
w h e r e C 2 a n d C 3 are c o n s t a n t s . Therefore,
oxx = Cxx3 + C 2x 2 + C 3.
(12.3.11)
T h e c o n s t a n t s Cx, C 2, a n d C 3 c a n b e d e t e r m i n e d from the c o n d i t i o n s
t h a t o n a n y p l a n e parallel to the OX2, OX3 p l a n e the r e s u l t a n t force in
the OXx direction m u s t b e e q u a l to zero, t h a t the r e s u l t a n t m o m e n t
a b o u t the OX3 axis m u s t b e equal to M , a n d t h a t the resultant m o m e n t
a b o u t the OX2 axis m u s t b e equal to z e r o :
J j
°\\dA
= 0,
j
J
= M,
oxx
x2dA
j
j
oxx
x3dA
=0.
(12.3.12)
Substituting Eq. (12.3.11) i n t o Eq. (12.3.12), the first c o n d i t i o n requires
that
C 3 = 0,
(12.3.13)
a n d the last c o n d i t i o n requires t h a t
hiCx
+ / 2 C32 = 0,
(12.3.14)
w h e r e 7 22 is the m o m e n t of inertia of the cross section of the b a r a b o u t
the OX2 axis a n d 7 23 is the p r o d u c t of inertia. Since OX2 a n d OX3 a r e
principal axes of inertia, 7 23 = 0 a n d , c o n s e q u e n t l y , Cx — 0. F r o m the
s e c o n d c o n d i t i o n we get:
C 2/ 33 = M
or
7
C2 = ^ ,
33
(12.3.15)
358
T h e Theory of Elasticity
w h e r e I33 is the m o m e n t of inertia a b o u t the OX3 axis. H e n c e ,
Mx
2
(12.3.16)
^33
T h i s result agrees with t h a t of t h e simple theory of b e a m s .
Let us n o w consider the d i s p l a c e m e n t s d u e t o simple b e n d i n g . T h e
strain d i s p l a c e m e n t relations (1.2.1), E q s . (12.3.3), (12.3.4), a n d (12.3.16)
give:
^
d
M
= # - * 2
u
2
dx2
d u
3
dx3
ox2
dxi
ox3
(12.3.17)
^33
=VM x2
EI33
(12.3.18)
= VM x2
EI33
(12.3.19)
ox2
ox3
axx
T h e integration of E q s . (12.3.17) to (12.3.20) to give ul9 w 2, a n d u3 does
n o t present a n y difficulty. It yields:
M2
+ d2x3 + d3x2 + d4
"i = *}*
*33
u
2
~
M
2
=
m
(Xi
0
y x
u3 = — ^j-x
23
^ 33
(12.3.21)
—
^ 3 + *>* ) — ^ 3 ^ 1
2
+
^i-X
— d2xx — dxx2 + d5,
3+ ^
(12.3.22)
(12.3.23)
w h e r e dx, d2, d3, d4, d5, a n d d6 a r e c o n s t a n t s of integration. T o
d e t e r m i n e these c o n s t a n t s , w e shall a s s u m e that the centroid of the b a r
together with a n element of the OXx axis a n d a n element of the OXx,
OX2 p l a n e a r e fixed at the origin. Therefore at xx = x2 = x3 = 0:
from which w e find:
dx = d2 = rf3 = d4 = rf5 = db = 0.
(12.3.25)
Straight Simple Beams
359
T h e d i s p l a c e m e n t s are, therefore,
M xx
\
(12.3.26)
i
u2 = - ^ ( 4 - vx\ + vxl)
M3
=-|g^2*3-
(12.3.27)
(12-3.28)
In the p l a n e x 2= 0, b o t h ux a n d u2 a r e e q u a l to z e r o . T h e deflection of
the axis of the b a r is o b t a i n e d from E q . ( 1 2 . 3 . 2 7 ) , in w h i c h x 2 a n d x 3
are set e q u a l to zero. T h e e q u a t i o n of the deflection curve is
1 23 2 9
u2 = - 2 l j £ * ? ,
( - - >
which is the e q u a t i o n of a p a r a b o l a . T h e c u r v a t u r e C is:
2
d u2
r =
« —M_
MOT *' '
(12.3.30)
53
if ML/EI33 is a small q u a n t i t y . This is the result t h a t is o b t a i n e d from
the e l e m e n t a r y theory.
N o w c o n s i d e r a p l a n e x { = K . After d e f o r m a t i o n , p o i n t s o n this cross
section will h a v e the following c o o r d i n a t e s :
X1 = K +
JC* =
WI = K ( I + ^ X 2 )
2
= * - ^ J f - ^ - wtf + wc§).
x + U2
(12.3.31)
(12.3.32)
C o m b i n i n g these t w o relations, a n d neglecting the t e r m s c o n t a i n i n g
degrees of M/EI33 higher t h a n the first, we get:
*0 wA
+
x*
°-
2 3M)
360
The Theory of Elasticity
Therefore, in p u r e b e n d i n g , a p l a n e cross section r e m a i n s p l a n e as
a s s u m e d in the e l e m e n t a r y theory. Let us n o w a s s u m e t h a t the cross
section is rectangular, a n d consider the sides x3 = ±b/2 (Fig. 12.3).
After b e n d i n g , we h a v e :
x
3
-
- 9
Therefore, in p u r e b e n d i n g the two sides r e m a i n straight b u t b e c o m e
inclined to their original position as s h o w n in Fig. 12.3. T h e t w o sides
x2 = ±h/2 b e c o m e :
1 - 4 + *-4-m;[*+<$-«*)]
(I2335)
T h e s e sides are, therefore, b e n t i n t o p a r a b o l i c curves as s h o w n in Fig.
12.3. T h i s effect is called anticlastic c u r v a t u r e .
12.4
Bending of a Narrow Rectangular Cantilever by an End Load
C o n s i d e r a cantilever h a v i n g a n a r r o w r e c t a n g u l a r cross section a n d
b e n t b y a n e n d l o a d P (Fig. 12.4). T h e p r o b l e m c a n b e t r e a t e d as o n e
of p l a n e stress with
1
h
;
0
A7
1
1
P
X
2
^1
Xi
Fig. 12.4
Straight Simple Beams
a 33 = a 13 = a 23 = 0.
361
(12.4.1)
T h e origin O is t a k e n at t h e c e n t r o i d of t h e cross section at t h e e n d
w h e r e the l o a d is applied. T h i s l o a d will b e l o o k e d u p o n as a shearing
force w h i c h is d i s t r i b u t e d in the OX2 direction in the s a m e w a y as the
shear stress a 12 is d i s t r i b u t e d in the b e a m . W e shall invoke SaintV e n a n t ' s p r i n c i p l e a n d c o n s i d e r t h a t a n y d i s t u r b a n c e of t h e stress
p a t t e r n w h i c h otherwise satisfies the p r o b l e m , dies o u t very rapidly
a w a y from the l o a d i n g s p o i n t (at a d i s t a n c e of h to 2h). N e g l e c t i n g the
b o d y forces, the solution to this p r o b l e m is given b y the Airy stress
function:
+
- A ( X
X$
- \ X I
X
# ) ,
(12-4.2)
I
. w h e r e A is a c o n t a n t a n d h is the d e p t h of t h e b e a m . F r o m E q . (9.10.4)
we get:
2
* „
=
Q
=
6 ^ ; f
2
,
o 22 = 0,
a 12 = - ^ ( 3 ^ - | ^
) .
(12.4.3)
T h e expression for a 12 leaves the l o n g i t u d i n a l sides free from stress as
r e q u i r e d b y the p r o b l e m . A t the e n d xx = 0, we m u s t h a v e :
2
f
^ bol2
dx2
= -
f
^ bA^xj
- | / z ) dx2 = -P,
(12.4.4)
from w h i c h
"~sp~4r
<12A5)
7 33 is the m o m e n t of inertia o f the cross section a b o u t OX3. Eqs. (12.4.2)
a n d (12.4.2) n o w b e c o m e :
i--^-(x }-lx #)
lX
J ^yi * ^
33
lX2
2
=O 02
(12.4.6)
. 4 . 7)
362
T h e Theory of Elasticity
T h e s e e q u a t i o n s coincide with t h e e q u a t i o n s of t h e e l e m e n t a r y theory.
It s h o u l d b e r e m e m b e r e d , however, t h a t P m u s t b e distributed a c c o r d ing t o E q . (12.4.8) a t t h e l o a d e d e n d .
Let u s n o w o b t a i n t h e d i s p l a c e m e n t c o m p o n e n t s c o r r e s p o n d i n g t o t h e
state of stress d e d u c e d a b o v e . T h e stress-strain relations (8.17.11),
(8.17.12), a n d (8.17.14) give:
P x x
\
1
XX
_ VP
o\2
2
_
P o (h
\
(\2AQ\
y 1
T h e c o m p o n e n t s of t h e d i s p l a c e m e n t ux a n d u2 a r e o b t a i n e d b y
integrating t h e previous e q u a t i o n s . I n d o i n g so, w e shall a s s u m e t h a t ux
a n d u2 a r e only functions of xx a n d x2. T h u s ,
w h e r e t h e functions fx a n d f2 a r e u n k n o w n . Substituting E q s . (12.4.10)
i n t o t h e third of E q s . (12.4.9), w e g e t :
+
S i - - i 4 ( t - 4
(I2A,,)
I n this e q u a t i o n , s o m e t e r m s a r e functions of xx alone, s o m e a r e
functions of x2 alone, a n d o n e is a c o n s t a n t . D e n o t i n g these terms b y
F(xx)y G(x2), a n d K, w e h a v e :
<'2A12)
F M- - £ l + n < * 0
^33
2
u2
Ph
K
^
(12.4.14)
8G/33
a n d E q . (12.4.11) is written:
F(xx)
+ G(x2) = K.
(12.4.15)
Such a n e q u a t i o n m e a n s t h a t F(xx) m u s t b e s o m e c o n s t a n t d a n d G(x2)
s o m e c o n s t a n t e. T h u s ,
d
e +
=-K * £ =-
(12.4.16)
Straight Simple Beams
363
and
P x
*
f -
Px
+d
f
a nx
The functions^(^l)
=
"%
ea
d/i( 2)
PX
+
^
(12.4.17)
+ e
r
then:
p3
x
(12A18)
h
f^ = - M M; > v
+
+ ex
+8
(12A19)
E q s . (12.4.10) n o w b e c o m e :
+
» > - W -
+
£ k
d + x h
<
)
-
T h e c o n s t a n t s e, g, d, a n d A can b e d e t e r m i n e d using Eq. (12.4.16) a n d
the three c o n d i t i o n s w h i c h are necessary to p r e v e n t the b e a m from
m o v i n g in the OXx, OX2 p l a n e . If we a s s u m e t h a t p o i n t A at the e n d
section is fixed, t h e n
U\ = u
2 = 0 at xx = L a n d x2 = 0,
a n d E q s . (12.4.20) a n d (12.4.21) give:
"--afe"*"
( , 2 A 2 2 )
T h e c o n s t a n t d c a n b e d e t e r m i n e d from the third c o n s t r a i n t at A. H e r e ,
we h a v e t w o possibilities:
a) A n e l e m e n t of the axis of the b e a m is fixed at the e n d A. I n this
case (Fig. 12.5):
a n d E q . (12.4.21) gives:
(&)r=r°
'--w
(,2A23>
<i2A24)
364
The Theory of Elasticity
F r o m Eq. (12.4.16),
2
e
2
PL
=
Ph
2F/33
(12.4.25)
8G/33
S u b s t i t u t i n g the c o n s t a n t s in the expressions of u{ a n d u2, we get:
2
2
_ vPxl
Px2x
2£Y 33
6 £ Y 33
2
f P X | x\
2 £ 7 33
Px\
^ „ ( PL
6G/33
Px\
6 £ 7 33
Ptixx
2 £ 7 33
i
3 £ 7 33 '
2
Ph
P L
(12 4 27)
T h e e q u a t i o n of the deflection c u r v e is:
fa)
l " 2 ^ 2= o
Px
P i L 2 xPI
+
*n + ?
/ 33 6 ££/ 33
32 £ / 3 ' 3
=
(12.4.28)
A t the l o a d e d e n d we h a v e :
( " 2) x 2= o
(12.4.29)
=m~-
T h i s coincides with the result b a s e d o n the e l e m e n t a r y t h e o r y of b e a m s .
T o illustrate the d i s t o r t i o n of t h e cross section p r o d u c e d b y t h e shearing
stresses, c o n s i d e r t h e d i s p l a c e m e n t a n d the slope at the fixed e n d
xx = L. E q . (12.4.26) gives:
2
("i)*,=z. -
vPx\
~
y 33
6 £+
< e,
Vi2)XL
-L
2\dx J
(en)
-
»-L
2
Px\
_Px2h
~ JQJ^-
=ir_i^i
2l
X!=L
2EI33
(12.4.30)
+
-^L_-Z^
_l
2
+
2 / 33G
8 G / 3 J3
(12.4.31)
(12-4.32)
T h e s h a p e of the cross section after distortion is as s h o w n in Fig. 12.5.
b) A vertical e l e m e n t of the cross section is fixed at the e n d A. I n this
case (Fig. 12.6):
(*)
o
r
( , 2, . 3 3 >
Straight Simple Beams
365
Fig. 12.5
l e a d i n g to
2
PL
e = 2EI-.
33
(12.4.34)
and
2
2
PL
2 £ 7 33
d =
Ph
8 G 7 33
(12.4.35)
S u b s t i t u t i n g the c o n s t a n t s in t h e expressions of ux a n d u2, we get:
U _y
~
Px2xf
~2ELi\~
PPX
"2
=
X X\
2EI-33
2
_ vPx\
Px\
+
6EI^
6GI^
PX\
_
6EI33
l
[
+Pl}x2
(12.4.36)
lElT
33
X
PI}
| _ 2 £ / 33
2
"I
8 G / 33 J
+
i
33
3EI,
P
h
P
I
1
2
(12.4.37)
Ph L
SGI 33
Therefore,
(2
" ^=°
+
~ 6EI^
*' LlEI^
SGI^j
PL'
P £ +
L
3 £ Y 33
SGI33
)
(
366
The Theory of Elasticity
and
2
(u )
PI}
2
Ph L
8G/33
(12.4.39)
2 we see that the vertical
C o m p a r i n g Eq. (12.4.28) to Eq. (12.4.38),
deflections of the axis are increased b y [Ph /SGI33
][L
— xx]. T h e s h a p e
of the cross section after distortion is s h o w n in Fig. 12.6.
Fig. 12.6
In practice, the c o n d i t i o n s are different from those s h o w n in Figs.
12.5 a n d 12.6. T h e cross section is usually n o t free to w a r p . H o w e v e r ,
b y invoking S a i n t - V e n a n t ' s principle, the distribution r e p r e s e n t e d b y
case (a) or (b) h o l d s for all sections at a small d i s t a n c e from the s u p p o r t
( a b o u t h).
A l t h o u g h the previous solutions represent a substantial i m p r o v e m e n t
over the e l e m e n t a r y theory, they suffer from the w e a k n e s s t h a t s o m e of
the compatibility relations h a v e n o t b e e n considered. I n a d d i t i o n , e33
a n d u3 h a v e n o t b e e n e x a m i n e d a n d b o t h ux a n d u2 were a s s u m e d to b e
i n d e p e n d e n t of w 3, t h u s neglecting the anticlastic c u r v a t u r e .
12.5
Bending of a Narrow Rectangular Beam by a Uniform Load
C o n s i d e r a n a r r o w b e a m of height h a n d width b, which is subjected
to a u n i f o r m l o a d i n g of intensity q per unit length (Fig. 12.7). T h e
p r o b l e m c a n b e t r e a t e d as o n e of p l a n e stress with
(12.5.1)
Straight Simple Beams
*3
//
0
1-
L
t
_
2
Is*
367
h 1H
1°
x2
x2
F i g . 12.7
T h e following c o n d i t i o n s m u s t a p p l y :
a t x2 =
at
X) =
°22
,L , I
±2
(12.5.2)
an= 0
o12=
a t x 2 = ~,
~
~h>
a 1 62a f x 2
PS
r
ah
ubdx2
23
aubx2dx2
- \5x x2h
(12.5.4)
= 0.
2 2
2
+
' 2'
2
= 0,J_h
T h e fifth p o w e r stress function,
$ = A{-Ax\ + 20x xl
.qL
=
+
/•-hi
J
(12.5.3)
= 0
\2
a
2
2
- 5x\L
+
2x\h
(12.5.5)
5x h ),
p r o v i d e s the solution to this p r o b l e m . It satisfies the b i h a r m o n i c Eq.
(9.10.11), a n d w h e n substituted into Eq. (9.10.4) it gives:
2
o „ = y 4 ^ - 8 0 ^ + 120JC 2(X, - ^ )
2
3
a22 = A[40xl - 30x2h
a 12 = A[-I20xxxj
+ 10/i ]
2
+ 30.x,/j ].
2
+
1 2 x 2/ i ]
(12.5.6)
(12.5.7)
(12.5.8)
C o n d i t i o n (12.5.3) gives:
-q 3
A = 20£ Z>'
(12.5.9)
368
T h e Theory of Elasticity
a n d t h e s e c o n d a n d third c o n d i t i o n s of E q s . (12.5.4) a r e satisfied. If t h e
m o m e n t of inertia of the cross section a b o u t the OX3 axis is 7 3 , 3E q s .
(12.5.5) to (12.5.8) b e c o m e :
v
2 4 0 / 33
(12.5.10)
23
-
5x h )
2 ?( Z
on = 8 f c *
o22 =
4)
"
-2^j(A
2
3
2
x+ 2 (
6 o t
- ^ x 2 + 4x1)
22 0
"
x3 ) x
2 1 h 25 U
2
<
- - )
(12.5.12)
2
ol2 = g ^ ( 4 x 2 - h )xx
(12.5.13)
2
T h e first t e r m in the expression of oxx [Le.9(q/&I33
)x2(l2
— 4x )] is t h e
usual e l e m e n t a r y t h e o r y t e r m . T h e s e c o n d o n e is a c o r r e c t i o n t e r m
w h i c h c o m e s from the c o n s i d e r a t i o n of a 22 acting o n the surface
x2 = —h/2, a n d from t h e fact t h a t the c o m p a t i b i l i t y e q u a t i o n s h a v e
b e e n p a r t l y satisfied (Sec. 9.10). But this t e r m is n o t i m p o r t a n t for l o n g
b e a m s in w h i c h the s p a n is large in c o m p a r i s o n with the d e p t h , it d o e s
n o t c o n t a i n xx a n d t h u s is c o n s t a n t a l o n g t h e b e a m . It is p l o t t e d in Fig.
12.8. a 22 d o e s n o t d e p e n d o n xx, a n d decreases from its m a x i m u m value
at the t o p edge t o zero at the b o t t o m edge as s h o w n in Fig. 12.8.
Fig. 12.8
Strictly speaking, the e n d s xx = ±L/2 s h o u l d b e w i t h o u t stress a n d
they a r e n o t . H o w e v e r , the r e s u l t a n t force a n d the resultant m o m e n t
over the entire section a r e e q u a l to zero. By S a i n t - V e n a n t ' s principle
(Sec. 8.13), the effect of the stress dies d o w n at a short d i s t a n c e from the
e n d so t h a t the stress d i s t r i b u t i o n of E q s . (12.5.11) to (12.5.13) d o e s
r e p r e s e n t t h e a c t u a l stresses in the large center p o r t i o n of the b e a m .
369
Straight Simple Beams
T h e expressions for the strains a n d the d i s p l a c e m e n t s a r e o b t a i n e d b y
following steps similar to t h o s e of Sec. 12.3. T h e b o u n d a r y c o n d i t i o n s
m a y b e given a s :
a t x 1= x 2 = 0,
ux = 0,
w2 = + S ,
a t x , = ± ^ a n d x 2 = 0,
u2 = 0.
^
= 0
oxx
2 the d i3s p l a c e m e n t a r e t h e n 2given b y :
T h e c o m p o n e n t s of
\
(
h
\
L X]
i
x
x
3 2
Mi
=
2EU
K
2
+ VX\
A
2
=
-
\
i
(12.5.16)
n)]
h x2
x8
2ELi L 1 2
(12.5.15)
3
h x\
'
"2
+X X2i
\3
~ ~10~J
~T~ ~ T r
(12.5.14)
12
+
-i l
r
x
3 3U 2
Eq. (12.5.16) shows t h a t the c e n t e r line h a s a tensile strain //=e q u a l to
vq/lEb.
T h i s is d u e to the c o m p r e s s i v e stress (o22)x2=o
~Q 2b.
The
value of 8 is o b t a i n e d b y s u b s t i t u t i n g E q s . (12.5.15) i n t o E q . (12.5.17).
T h i s gives:
-&&b+na d]+
'
<i2 5i8>
T h e first t e r m of E q . (12.5.18) c o r r e s p o n d s to t h e m a x i m u m deflection
as c a l c u l a t e d from the e l e m e n t a r y t h e o r y . T h e s e c o n d t e r m arises from
o u r t a k i n g i n t o a c c o u n t t h e d i s t r i b u t i o n of o 22 in t h e OX2 direction. T h i s
t e r m h a s a n a p p r e c i a b l e value only for b e a m s t h a t are very s h o r t a n d
d e e p w h e r e the value of h a p p r o a c h e s the length of the b e a m .
T h e previous solution suffers from the s a m e d r a w b a c k s m e n t i o n e d a t
the e n d of Sec. 12.4.
12.6
Cantilever Prismatic Bar of Irregular Cross Section Subjected to a
Transverse End Force
T h e b e n d i n g of a n a r r o w r e c t a n g u l a r cantilever b y a n e n d l o a d w a s
370
T h e Theory of Elasticity
studied in a previous section, a n d it was s h o w n that, in a d d i t i o n to
n o r m a l stresses p r o p o r t i o n a l to the b e n d i n g m o m e n t , shearing stresses
p r o p o r t i o n a l to the s h e a r i n g force act o n e a c h section. Let us n o w
c o n s i d e r the general case of the b e n d i n g of a cantilever b y a force P
a p p l i e d at the e n d a n d parallel to o n e of the principal axes of t h e cross
section. T h e b e a m is p r i s m a t i c in form a n d n o restrictions a r e a p p l i e d
as to its s h a p e . T h e OXx axis is t a k e n such t h a t it c o n n e c t s the c e n t r o i d s
of the cross sections at all p o i n t s ( c e n t r o i d a l axis), a n d OX2 a n d OX3 a r e
p r i n c i p a l axes of inertia. T h e solution of this p r o b l e m is d u e to SaintV e n a n t : T o g e t h e r with the torsion p r o b l e m it illustrates the use of t h e
semi-inverse m e t h o d in t h e t h e o r y of elasticity.
T h e b e n d i n g m o m e n t o n e a c h cross section is (Fig. 12.9):
F i g . 12.9
M 13 = -P(L
(12.6.1)
- xx)
F o l l o w i n g t h e e l e m e n t a r y theory, let us a s s u m e t h a t the n o r m a l stresses
over a cross section at a d i s t a n c e xx from the fixed e n d a r e given b y :
P
o
n
= -
{X L l
2
\
~
33
)
X
o
= o
22
33 =
0,
02.6.2)
and that
a 23 = 0.
(12.6.3)
W i t h these a s s u m p t i o n s , it will b e s h o w n t h a t a solution c a n b e r e a c h e d
w h i c h satisfies all the e q u a t i o n s of elasticity. N e g l e c t i n g t h e b o d y forces,
the e q u i l i b r i u m e q u a t i o n s b e c o m e :
Straight Simple Beams 371
foil
9jC|
9 a
12
3xt
^
dx
x
++ 3^13 = ^ 2
^33
dx
2 9 ( 37x 3
9^12
22
+
3x2
9*12
+
3^2
x
+9^13 =
d3
<^23_ = ^ 1 1 = 0
+
3^3
3^i
+ ^ 2 1+ ^ 3 3 = ^ 1 1
dx
3^3
3^i
0
(12.6.4)
(12.6.5)
= 0.
(12.6.6)
2
F r o m E q s . (12.6.5) a n d (12.6.6), w e c o n c l u d e t h a t t h e shearing stresses
a 12 a n d a 13 a r e i n d e p e n d e n t of xx, w h i c h m e a n s t h a t t h e total shearing
force o n t h e cross section of t h e b e a m is c o n s t a n t a s r e q u i r e d from t h e
l o a d i n g c o n d i t i o n s . L e t u s n o w consider t h e b o u n d a r y c o n d i t i o n s (7.3.8)
a n d a p p l y t h e m t o t h e lateral surface of t h e b a r w h i c h is free from
external forces (Fig. 12.9). T h e three c o n d i t i o n s r e d u c e t o o n e , n a m e l y :
T u r n i n g n o w t o t h e Beltrami-Michell compatibility e q u a t i o n s
(8.10.11) t o (8.10.16), w e see t h a t t h e three first, as well as t h e fifth, of
these e q u a t i o n s a r e identically satisfied, while t h e fourth a n d sixth give:
2
V72a 17 = —
2
1
' '
P
33
y
(12.6.8)
V a 13 = 0,
Therefore, t h e solution of t h e p r o b l e m of b e n d i n g of a p r i s m a t i c
cantilever b a r of a n y section is r e d u c e d t o finding a 12 a n d a 13 functions
of x2 a n d j t 3, satisfying t h e e q u i l i b r i u m E q . (12.6.4), t h e b o u n d a r y
c o n d i t i o n (12.6.7), a n d t h e compatibility relations (12.6.8). T h e solution
c a n b e o b t a i n e d using a stress function <$>(x2,x3) defined b y t h e t w o
equations:
d<t>
a 13 = - dx '
2
(12.6.10)
w h e r e f(x3) is only a function of x 3 t o b e d e t e r m i n e d from t h e
b o u n d a r y c o n d i t i o n s . T h e e q u a t i o n of e q u i l i b r i u m (12.6.4) is identically
satisfied, a n d t h e t w o compatibility relations b e c o m e :
372
T h e Theory of Elasticity
x
V J
(^)-4
3
2( ^ _V
< V 2
*
2
v
P _
d2
f( 3)
(12.6.11)
) = A
<12AI2)
E q . (12.6.12) shows t h a t V <£ m u s t b e i n d e p e n d e n t of x2.
integrating E q . (12.6.11) with respect to x3, we get:
Therefore,
w h e r e C is a c o n s t a n t of integration. T h e m e a n i n g of this c o n s t a n t is
o b t a i n e d as follows: T h e r o t a t i o n of a n e l e m e n t of a r e a in t h e p l a n e of
the cross section (from OX2 t o OX3) is given b y :
T h e r a t e of c h a n g e of this r o t a t i o n in the direction of the OXx axis is
3<o 32
8xi
2
L
\ dx2
2
L 3x2 \
dx3
3xi
/J
(12.6.15)
3x3/
3x3
V 3^!
dx2
/ J'
Therefore,
3co 32 _ 3 e 13
3^
dx
2
3 e 12
dx
23
i f
2GY
2G
L 3*2
JLKJ L
2
9 <fr
9 «J>2
9X|
8X
0*2
dx3
J
(12.6.16)
ff(*3)"|
<& 3 J '
a n d , from E q . (12.6.13), we get:
1 ^ ox
G
x
- v l p L =- C .
1 + v I33
(12.6.17)
T h u s the r a t e of r o t a t i o n a b o u t the OXx axis consists of t w o p a r t s : a) A
c o n s t a n t r a t e r e p r e s e n t e d b y C, w h i c h c o r r e s p o n d s to the u n i f o r m twist
of a cylindrical r o d u n d e r p u r e torsion, a n d b) a r a t e w h i c h is a function
of x3 a n d w h i c h c o r r e s p o n d s to a d i s t o r t i o n of t h e cross section in t h e
OX2, OX3 p l a n e . T h i s s e c o n d r a t e is similar t o the anticlastic c u r v a t u r e
Straight Simple Beams
373
of Sec. 12.3. I n w h a t follows, we shall position t h e force P in t h e OX2,
OX3 p l a n e , parallel to the p r i n c i p a l axis OX2 a n d such t h a t t h e c o u p l e
it exerts o n the e n d cross section cancels o u t the twist r e p r e s e n t e d b y C.
Therefore,
C = 0,
a n d E q . (12.6.13) b e c o m e s :
2
V <j> =
Px3
I + v I33
(12.6.18)
dx-fix*).
3
S u b s t i t u t i n g E q s . (12.6.9) a n d (12.6.10) i n t o the b o u n d a r y
(12.6.7), we get:
d(j> dx3
dx3 ds
+
3<J> dx2
dxo dx
dx3
d<f>
ds
33
~ds~
condition
(12.6.19)
T h e form of <f> o n the b o u n d a r y c a n b e c o m p u t e d from E q . (12.6.19) if
f(x3)
is suitably c h o s e n . W e shall c h o o s e the function f(x3)
such t h a t
the r i g h t - h a n d side of E q . (12.6.19) is equal t o zero for e a c h p a r t i c u l a r
case. <f> is t h e n c o n s t a n t o n t h e b o u n d a r y . If this c o n s t a n t is c h o s e n e q u a l
t o zero, t h e solution of t h e b e n d i n g p r o b l e m r e d u c e s t o solving E q .
(12.6.18) w i t h <f> = 0 o n t h e b o u n d a r y . T h e p r o b l e m is similar to t h a t m e t
in the s t u d y of torsion.
I n t h e following, we shall e x a m i n e the cases of t h e circular a n d t h e
elliptic cross sections. O t h e r cross sections c a n b e f o u n d in reference [2].
1) Circular Cross Section
T h e b o u n d a r y of the cross section (Fig. 12.10) is given b y t h e
equation:
0
X2
Fig. 1 2 . 1 0
374
T h e Theory of Elasticity
2
(12.6.20)
xj + xj = R .
T h e r i g h t - h a n d side of Eq. (12.6.19) vanishes if w e take
Z i
33
Substituting E q . (12.6.21) i n t o E q . (12.6.18), t h e stress function is
determined by:
= Vz^^i,
(12.6.22)
a n d t h e c o n d i t i o n t h a t (J> = 0 o n t h e b o u n d a r y . L e t u s t a k e
2
<t> = mx3(xj
+ x\ - R l
(12.6.23)
w h e r e m is a c o n s t a n t . <j> is t h e n e q u a l t o zero o n t h e b o u n d a r y a n d
satisfies E q . (12.6.22), p r o v i d e d
m =
1 + 2v P
(12.6.24)
1 + V 8/33
'
E q . (12.6.23) n o w b e c o m e s :
*
=
2
2
2
2
TTT87T*3(*2 + *3 " * ) .
33
(12.6.25)
F r o m E q s . (12.6.9) a n d (12.6.10) w e h a v e
J
•i3—4(V^K*>33
<12
"
7)
T h e o t h e r c o m p o n e n t s of t h e state of stress a r e :
P(L —
a
a
xx)x2
(12.6.28)
^33
a
22 = 33 = 23 = °-
(12.6.29)
T h e distribution of stresses a 12 a n d a 13 o n a n y cross section gives a
r e s u l t a n t a l o n g t h e vertical d i a m e t e r OX2:
Straight Simple Beams
375
jl2
dA=P
(12.6.30)
ol3
dA=0.
(12-6.31)
A l o n g t h e h o r i z o n t a l d i a m e t e r of the cross section (x2 = 0), w e find:
a
_ (3 + 2v)P [ R
' 2 - 8(1 + v)I33 l
X
2i - 2v 1 2
T+Tv ^\
o , 3 = 0.
(12632)
'
(12.6.33)
T h e m a x i m u m value of the s h e a r i n g stress o c c u r s a l o n g the OXx axis
a n d is e q u a l to
= (3 + 2v)P
8(1 + v)I33
^12 W
2
)
T h e s h e a r i n g stress a t the e n d of a h o r i z o n t a l d i a m e t e r (x3 = ±R) is:
l 2
(-
)
(12.6.35)
( + ")P&
=
D
O n e notices t h a t the m a g n i t u d e of t h e s h e a r i n g stress d e p e n d s o n
Poisson's ratio. F o r v = 0.3 E q s . (12.6.34) a n d (12.6.35) b e c o m e :
,
x
1.38P
,
x
1.23P
H 2 6 36^
where
is t h e cross sectional a r e a of t h e b e a m . T h e e l e m e n t a r y t h e o r y
b a s e d o n t h e a s s u m p t i o n t h a t t h e s h e a r i n g stress a 12 is uniformly
distributed a l o n g a h o r i z o n t a l d i a m e t e r of t h e cross section gives:
a 12 = | f
(12-6.37)
T h u s , t h e e l e m e n t a r y b e a m theory, in spite of t h e fact t h a t it violates
b o t h compatibility a n d b o u n d a r y c o n d i t i o n s , is in error only b y 3 t o 4
percent.
2) Elliptic Cross Section
T h e b o u n d a r y of the cross section (Fig. 12.11) is given b y t h e
equation:
^12 4. £ i2 ^ 1
a
b
(12.6.38)
376
The Theory of Elasticity
J x,
V
VJ
b
Fig. 12.11
T h e r i g h t - h a n d side of E q . (12.6.19) vanishes if we t a k e
Eq (12.6.13)
.
becomes
now
:
E q . (12.6.40) a n d t h e c o n d i t i o n t h a t <f> = 0 o n t h e b o u n d a r y d e t e r m i n e
the stress function <j>. Let us t a k e
2
<p = mx3(xi
+^xl-a ),
(12.6.41)
w h e r e m is a c o n s t a n t . <J> is t h e n equal to zero o n the b o u n d a r y a n d
satisfies E q . (12.6.40), p r o v i d e d
2
2
= p_ (1 + v)a +2 vb 2
/ 33 2(1 + *>)(3a + 6 ) *
2
62 4
E q . (12.6.41) n o w b e c o m e s :
<f> = —
^
—
2= — r ^ - Ji x\
2=
2(1 + v)(3a
+ b)
33 \
+2 2
-
b
1
3a
).
(12.6.43)
/
T h e stress c o m p o n e n t s are o b t a i n e d from Eqs. (12.6.9) a n d (12.6.10):
2
"(Tr^TF)2^L
a,2 _
2(1 + v)a
+ b>
f Pa
,2
*
(l-2,)q*
x 1
2 d + v)<? + b> >}
u
377
Straight Simple Beams
2
2
+ v)a 2 + ,b l2
13 = _(l
(1 + v)(3a + b )
Px2x3
h3
T h e o t h e r c o m p o n e n t s of the state of stress a r e :
P(L
-A
xx)x2
(12.6.46)
33
°22 = °33 = °23 = 0-
(12.6.47)
A l o n g the h o r i z o n t a l d i a m e t e r of the cross section (x2 = 0), we find:
2
2
2
2(1 + v)a
T
+ b
1
(1 - 2v)a
P2
7
a 13 = 0.
(12.6.49)
T h e m a x i m u m value of the shearing stress o c c u r s a l o n g the OXx axis,
a n d is e q u a l t o :
2
1
(„
)
- Pa
^i2W- /
2
3 3
a
f,
L
1
l
+ 2
2 (1
+
3a +
b
v b
(12.6.50)
-•
22
If b is small c o m p a r e d to a, the t e r m s c o n t a i n i n g b /a
neglected, a n d
(<.„>„»-fg-If
can
be
(12*51)
w h i c h coincides with the e l e m e n t a r y theory. If b is very large in
c o m p a r i s o n with a, t h e n
T h e stress at the e n d s of the h o r i z o n t a l d i a m e t e r , w h e n b is very large
c o m p a r e d to a, is
a
P
( i 2 ) * 2= o
X3
=±b
- (i % V
\A
^
'
(12.6.53)
'
T h e d i s t r i b u t i o n a l o n g a h o r i z o n t a l d i a m e t e r is c o n s i d e r a b l y far from
being u n i f o r m in this case, a n d for a Poisson r a t i o v = 0.3 we get
(oi2)max = 1-54^ a n d ( o 1 ) 2 X0 2 == 0.92 J .
X3 = ± / >
(12.6.54)
378
The Theory of Elasticity
12.7
Shear Center
If a b e a m ' s cross section h a s t w o axes of s y m m e t r y , a n d if the l o a d is
applied in such a m a n n e r t h a t the l o a d line passes t h r o u g h the c e n t r o i d
of the cross section, the b e a m will deflect w i t h o u t twisting (Fig. 12.12a).
P
P
T
C
X3
*2
Fig. 1 2 . 1 2
If the b e a m ' s cross section h a s only a single axis of s y m m e t r y a n d t h e
p l a n e of the l o a d i n g is such t h a t it d o e s n o t c o n t a i n the axis of
s y m m e t r y even t h o u g h the l o a d line passes t h r o u g h the c e n t r o i d of the
cross section, the b e a m will b e subjected to twisting (Fig. 12.12.b), b u t
it is possible to locate a p o i n t C o n the axis of s y m m e t r y , t h r o u g h w h i c h
the l o a d m u s t pass in o r d e r to eliminate the t e n d e n c y for twist. T h i s
p o i n t is called the shear center. If the b e a m ' s cross section h a s n o axis
of s y m m e t r y (Fig. 12.12c) it is also possible to locate the shear center.
T h e shear center m a y b e generally defined as the p o i n t o n the cross
sectional p l a n e of a b e a m t h r o u g h w h i c h the resultant of the transverse
l o a d (shear) m u s t b e applied in o r d e r t h a t the stresses in the b e a m m a y
b e d e t e r m i n e d only from the theories of p u r e b e n d i n g a n d transverse
shear. I n discussing the cantilever p r o b l e m of Sec. 12.6, it w a s a s s u m e d
t h a t the force P w a s parallel to the OX2 axis a n d at such a d i s t a n c e from
the c e n t r o i d t h a t twisting of the b a r did n o t occur. This d i s t a n c e
d e t e r m i n e s t h e l o c a t i o n of t h e center of shear a n d c a n b e o b t a i n e d o n c e
the shear stresses a 12 a n d a 13 ( c o m p u t e d w i t h o u t allowing for a n y
torsional twist) a r e k n o w n . F o r this p u r p o s e , we evaluate the m o m e n t
a b o u t t h e centroid p r o d u c e d b y a 12 a n d a 13 a n d this m o m e n t m u s t b e
e q u a l to t h a t of P. Therefore, (Fig. 12.13),
(— al2x3 +
ai3x2)dx2dxi
(12.7.1)
Straight Simple Beams
379
PLL
Fig. 1 2 . 1 3
and
e3 = ^ L .
(12.7.2)
F o r Mu positive, e3 m u s t b e t a k e n in the negative direction of OX3.
F o r sections t h a t are l o a d e d in p l a n e s n o t parallel to a p r i n c i p a l p l a n e
OX3,
of inertia (Fig. 12.14), we first locate the principal axes OX'2 a n d
t h e n d e c o m p o s e the shearing force P a l o n g these t w o axes. T h e p r o b l e m
is t h e n solved i n d e p e n d e n t l y for e a c h c o m p o n e n t leading to the values
of e2 a n d e3 w h i c h locate the shear center.
Fig. 1 2 . 1 4
380
The Theory of Elasticity
PROBLEMS
1.
2.
Solve the p r o b l e m of S e c 12.4 a s s u m i n g t h a t the r e c t a n g u l a r b e a m
is n o t n a r r o w a n d t h a t suitable restraints p r e v e n t d i s p l a c e m e n t s in
the OX3 direction w i t h o u t causing a n y friction o n the lateral faces.
F o r a Poisson ratio v = 0.3, find the ratio of length to height in o r d e r
t h a t the use of the e l e m e n t a r y theory of b e a m s does n o t p r o d u c e a n
error in excess of 2.5 p e r c e n t in the p r e d i c t i o n of the m a x i m u m
deflection of the b e a m of Sec. 12.5.
REFERENCES
[1] S. Timoshenko, Strength of Materials, Vol. 1, Van Nostrand, Princeton, N . J., 1955.
[2] S. Timoshenko and J. N . Goddier, Theory of Elasticity, McGraw Hill, N e w York, N . Y.,
1970.
CHAPTER 13
CURVED BEAMS
13.1
Introduction
I n the previous chapter, only straight b e a m s were considered, which
m e a n s t h a t all the axial fibers w e r e of the s a m e length before b e n d i n g .
In this c h a p t e r , we shall s t u d y cases in w h i c h the line j o i n i n g the
centroids of the cross sections, i.e., the center line, h a s a n initial
c u r v a t u r e . T h e center line is a p l a n e curve a n d the cross sections h a v e
a n axis of s y m m e t r y in this p l a n e . T h e b e a m s h a v e a c o n s t a n t cross
section. W e shall first s u m m a r i z e the results of the simplified t h e o r y d u e
to W i n k l e r , t h e n e x a m i n e m o r e a c c u r a t e solutions satisfying the c o n d i tions of the t h e o r y of elasticity. T h e n o t a t i o n a n d sign c o n v e n t i o n s
defined in Sec. 12.1 will b e followed in this c h a p t e r .
13.2
The Simplified Theory of Curved Beams
T a k i n g t h e OXx axis t a n g e n t to the line j o i n i n g t h e c e n t r o i d s of the
cross sections, the following a s s u m p t i o n s a r e m a d e (Fig. 13.1):
a - T h e cross section h a s a n axis of s y m m e t r y a n d the transverse
loadings, (consequently, the b e n d i n g m o m e n t s ) are applied in a p l a n e
c o n t a i n i n g the axis of the s y m m e t r y .
b - T r a n s v e r s e cross sections w h i c h were originally p l a n e a n d n o r m a l to
the center line r e m a i n p l a n e after b e n d i n g a n d axial d e f o r m a t i o n ,
c - T h e r e is n o lateral pressure b e t w e e n the longitudinal fibers,
d - T h e d i s t r i b u t i o n of shearing stresses over the cross section is the
s a m e as for straight b e a m s .
381
382
T h e Theory of Elasticity
_i
k
ctnrer
line-
Ac/0
(c)
(b)
Fig. 13.1
W i t h the a b o v e a s s u m p t i o n s a n d t h e s a m e sign c o n v e n t i o n s defined
in Fig. 12.1 ( a n d 13.1), t h e following w e l l - k n o w n e q u a t i o n s h a v e b e e n
d e d u c e d [1]:
Ml3
(x2
- e)
Ae(rc - x2)
M ds
Ad<j> = u
ercAE
c
e = rm
r
mA
-fi
_
_
TV
A
Ufa
(13.2.2)
AErc
m
+ 1
or
(13.2.1)
m =
x2 dA
x
(13.2.3)
(13.2.4)
2
V2Q3
where
V2 = the shearing force in the direction of OX2
Q3 = the static m o m e n t a b o u t t h e OX3 axis (see Sec. 12.2)
N = the l o n g i t u d i n a l force o n the cross section
M 13 = the b e n d i n g m o m e n t a b o u t the OX3 axis
e = the d i s t a n c e b e t w e e n t h e center line a n d the n e u t r a l axis
A = the cross sectional a r e a
rr — the r a d i u s of c u r v a t u r e of the center line
(13.2.5)
Curved Beams
383
mA = the modified a r e a
E = Young's modulus
kd<f> = t h e r o t a t i o n of the cross section d u e to b e n d i n g a n d to N (see
Fig. 13.2)
T h e d i s t r i b u t i o n of on d u e to the effect of b e n d i n g is h y p e r b o l i c as
s h o w n in Fig. 13.1c, a n d the n e u t r a l axis is displaced t o w a r d s the center
of c u r v a t u r e b y the a m o u n t e. T h e q u a n t i t y m c a n easily b e c o m p u t e d
for a n y cross section [1]. F o r r e c t a n g u l a r a n d t r a p e z o i d a l cross sections,
we respectively h a v e (Fig. 13.3a,b):
13.2.6)
F o r tee sections a n d circular sections, we h a v e (Fig. 13.3c,d):
384
T h e Theory of Elasticity
m
=
(13.2.7)
m
13.3
= 211/:
Pure Bending of Circular Arc Beams
Let us c o n s i d e r a circular a r c b e a m with a n a r r o w r e c t a n g u l a r cross
section b e n t in the p l a n e of c u r v a t u r e b y e n d couples M (Fig. 13.4). T h e
F i g . 13.4
b e a m h a s a c o n s t a n t cross section a n d the b e n d i n g m o m e n t is c o n s t a n t
along the b e a m . C o n s e q u e n t l y , it is n a t u r a l to expect t h a t the stress
distribution is the s a m e in all radial cross sections; in o t h e r w o r d s , is
i n d e p e n d e n t of 9. A solution in a system of cylindrical c o o r d i n a t e s is
readily available in t e r m s of a n A i r y stress function <f> given b y E q .
(9.11.5):
2
<j> = Cxr (nr
2
+ C2r
+ C3tnr
+ Q
(13.3.1)
w h e r e C 1? C 2, C 3, a n d C 4 are c o n s t a n t s of integration to b e d e t e r m i n e d
from the b o u n d a r y c o n d i t i o n s . T h e b o d y forces a r e neglected a n d since
the a s s u m p t i o n h a s b e e n m a d e t h a t the cross section is n a r r o w , p l a n e
stress c o n d i t i o n s will apply. F o r simplicity, the w i d t h of t h e cross
section will b e a s s u m e d e q u a l to unity. T h e b o u n d a r y c o n d i t i o n s a r e :
1) orrr = 0 for r = rt a n d r = r0
2) f ° oge dr = 0 over a n y cross section
Curved Beams
385
r
3) $ r° ra99 dr = —M over a n y cross section
4) org = 0 at the b o u n d a r y .
T h e stresses a r e given b y E q . (9.11.6) to (9.11.8). T h e first b o u n d a r y
c o n d i t i o n gives:
r
%
i
%
+ C , ( l + 2/w}) + 2 C 2 = 0
(13.3.2)
+ C , ( l + 21nr0) + 2 C 2 = 0.
(13.3.3)
T h e s e c o n d b o u n d a r y c o n d i t i o n gives:
fc[^T+C,(l
+ 2//ir 0) + 2 C 2]
J
" ' [^'
+
(13.3.4)
C
1 | +(
2 +
C =z
^
l
°'
which is a c o m b i n a t i o n of t h e two p r e v i o u s e q u a t i o n s . T h i s shows t h a t
the s e c o n d b o u n d a r y c o n d i t i o n will b e satisfied if the first o n e is. T h e
third b o u n d a r y c o n d i t i o n gives:
I ' "»* - I ° I?'*' 'l' (^)
rd
(13 3 5)
But, b e c a u s e of E q s . (13.3.2) a n d (13.3.3):
Therefore,
2
|Y|"
= C 3/ # i | + C ^ Z / i r , - A / / ^ . ) 4- C 2f e
- rf) = M.
(13.3.6)
T h e fourth b o u n d a r y c o n d i t i o n is identically satisfied. E q s . (13.3.2),
(13.3.3), a n d (13.3.6) c o m p l e t e l y d e t e r m i n e the three c o n s t a n t s Q , C 2,
a n d C 3:
386
T h e Theory of Elasticity
2
\r
C, =
^'i 'o
-
rf)
-M
to)
(13.3.7)
2\2
rf)
2
2
2{rf tnr, - r lnr0) - (r -
2
rf)
2 2
4rfr (lnf)
M
(13.3.8)
~{r -rf)
4r? o ""n
2-M.
4rft(ln\)
(13.3.9)
-tf-rf)
Substituting these values i n t o E q . (13.3.1), yields t h e expression of <j>.
T h e stresses a r e n o w o b t a i n e d from E q s . (9.11.6) to (9.11.8). Setting
D =
2
4#rV(/«|) - (r
2
2
(13.3.10)
- rf) ,
we get:
4M(
D \
°9t
rfrf
2
r
- rftnjr
™ { - ^ t n \ = 0.
-
rfln
(13.3.11)
r
r}in Ti - r}ln%
+ r} - rf)
(13.3.12)
(13.3.13)
Since this stress field satisfies equilibrium, compatibility, a n d b o u n d a r y
conditions, it gives the solution of t h e p r o b l e m p r o v i d e d the m o m e n t M
is applied t o the e n d of the b e a m b y external forces c o r r e s p o n d i n g to
the o99 distribution of E q . (13.3.12). This distribution is s h o w n in Fig.
13.5 together with t h a t of orr
. If, however, t h e m o m e n t is applied so as
to give a distribution different from t h a t s h o w n in Fig. 13.5, w e k n o w
from S a i n t - V e n a n t ' s principle t h a t at a distance from t h e e n d s of the
b e a m c o r r e s p o n d i n g to the b e a m ' s d e p t h r0 — ri9 E q s . (13.3.11) to
(13.3.13) will h o l d with a high degree of a c c u r a c y .
T h e strains are o b t a i n e d from the stress-strain relations (8.17.11),
(8.17.12), a n d (8.17.14). T h e displacements ur a n d u9 a r e o b t a i n e d from
Curved Beams
387
Fig. 13.5
E q s . (9.11.16) a n d (9.11.17), in w h i c h Q , C 2, a n d C 3 are given b y E q s .
(13.3.7) to (13.3.9). T o d e t e r m i n e the c o n s t a n t s A, B, a n d F, let us
consider the c e n t r o i d of the cross section from which 0 is m e a s u r e d , a n d
c h o o s e the e l e m e n t of r a d i u s at this p o i n t to b e fixed. T h u s ,
at r =
a n d 0 = 0, u = 0 a n d
r
= 0.
or
2
F r o m these b o u n d a r y c o n d i t i o n s , the following e q u a t i o n s result:
A = B = 0
• = 1 |~1 + v C - r2(1 - v)C rJnr
3
x c
E\_
c
(13.3.14)
+ (1 +
v)Cxrc
(13.3.15)
-2(1
Thus,
y )
Ur =
*
C 3 + 2(1 - v)Cxrlnr
- (1 + ^ ) Q r + 2(1 - *>)C 2r]
(13.3.16)
cos 0
[ ^ C 3
- 2(1 -
- 2(1 - *-)C,r c/«r c + (1 +
„ ) C 2r c]
v)Cxrc
388
The Theory of Elasticity
4
Q rO
*
sin 9 \ 1 + Lv ^ L
*
n
^ l
+ (1 + i O C , r f- 2 ( l
a
(13.3.17)
- ^ ) C 2r cJ .
Eq. (13.3.17) shows that the d i s p l a c e m e n t in the transverse direction of
a n y cross section consists of a translation — F sin 0, which is the s a m e
for all points of the cross section, a n d a r o t a t i o n 4 Q 6/E a b o u t the
center of c u r v a t u r e 0 (Fig. 13.1). Therefore, p l a n e cross sections r e m a i n
p l a n e as is a s s u m e d in the simplified theory.
Finally, it m u s t be r e m e m b e r e d t h a t some compatibility e q u a t i o n s
h a v e b e e n ignored a n d all the quantities a s s u m e d i n d e p e n d e n t of x3.
13.4
Circular Arc Cantilever Beam Bent by a Force at the End
Let the b e a m h a v e a r e c t a n g u l a r n a r r o w cross section, w h i c h for
simplicity will be a s s u m e d equal to unity. T h e b e n d i n g m o m e n t at a n y
cross section mn (Fig. 13.6) is p r o p o r t i o n a l to sin 9 a n d a c c o r d i n g to the
Fig. 13.6
simplified theory the n o r m a l stress o09 is p r o p o r t i o n a l to the b e n d i n g
m o m e n t . A n Airy stress function of the form
<J> = / ( r ) s i n 9
(13.4.1)
will therefore be tried. It will b e s h o w n t h a t this stress function d o e s
i n d e e d p r o v i d e us with the solution to the p r o b l e m . Eq. (13.4.1), w h e n
Curved Beams
389
4
s u b s t i t u t e d i n t o the b i h a r m o n i c V </> = 0 a n d the sin 0 divided out,
yields the o r d i n a r y differential e q u a t i o n :
T h e general solution is:
3
= Qr
f(f)
+ ^ +
C 3r + Q r / r c r ,
(13.4.3)
in w h i c h Q , C 2, C 3, a n d C 4 are c o n s t a n t s to b e d e t e r m i n e d from t h e
b o u n d a r y c o n d i t i o n s . T h e stress function <j> is n o w given b y :
3
$ = ^
r + ^
+ C3r
+ C 4r y / i r ) s i n 0.
(13.4.4)
U s i n g E q s . (9.10.15), we find the following expressions for the stress
components:
arr = (lCxr
- ^
= (6C, r + ^
a* = - ( l C x r -
+ ^)sin 0
(13.4.5)
)sin 0
(13.4.6)
+ ^ ) c o s 9.
(13.4.7)
+ ^
^
The boundary conditions are:
\)orr = or9 = 0
2)
f
A
ar^ dr = P
for r = r{ a n d r = r0
(13.4.8)
for 0 = 0.
(13.4.9)
T h e first c o n d i t i o n s give:
2
2 C , / i-
2
^ + ^
T h e s e c o n d c o n d i t i o n gives:
= 0,
2 C , r o- ^ + T
i
= 0.
(13.4.10)
390
The Theory of Elasticity
-
C
. 2 _
l
/Q
_
J °
nQ
= _p C =2i _ ,f. 2 ) e +
/
C ^ l ^ l
(13.4.11)
F r o m E q s . (13.4.10) a n d (13.4.11), w e get:
2 2
Pr
r
•* 'i 'o
2N
'
(13.4.12)
C4 =
where
N = $ - £ +
+
tfVn'f-.
(13.4.13)
T h e expressions for the stresses n o w b e c o m e :
(13.4.14)
(13.4.15)
(13.4.16)
F o r the u p p e r e n d of the b a r 9 = 0, then,
a
rr =
= 0
rr( r
„
°r9 ~ -Jf\
r r
? +
?
+ ~p
+
?
rr
Q \
J-—)•
F o r the lower e n d of t h e b a r 8 = n / 2 , then,
+
°rr-NV
„
-P(*r
°rff = 0.
—3
—
)
l±l\
(13.4.17)
391
Curved Beams
E q s . (13.4.14) to (13.4.16) constitute a n exact solution of the p r o b l e m
only w h e n the forces at the e n d of the c u r v e d b a r a r e distributed
a c c o r d i n g t o E q s . (13.4.17). I n a n y o t h e r distribution of forces, t h e
solution will b e valid at s o m e d i s t a n c e from the e n d s a c c o r d i n g to SaintV e n a n t ' s principle. H e r e , too, n u m e r i c a l c o m p u t a t i o n s s h o w t h a t t h e
results of t h e simplified t h e o r y are very close to t h o s e given b y the exact
theory.
Let us n o w c o n s i d e r t h e d i s p l a c e m e n t s p r o d u c e d b y t h e force P. T h e
elastic stress-strain relations for p l a n e stress a n d the strain displacem e n t s relations give:
^
- ^ [ 2 C , K 1 - 3.) - ^ ( 1
- ,)]
(13.4.18)
(13.4.19)
^
= rem - ur
30
due
S 9r_ dur
~ rd9
+ v) + %
r
Ug
(13.4.20)
dr
I n t e g r a t i o n of E q . (13.4.18) yields:
^
^
(
1
- 3?) +
2f-(l
+ v) + C ( l -
P)/™]
4
(13.4.21)
w h e r e fx(9) is a function of 9 only. Substituting E q s . (13.4.21) a n d
(6.7.23) i n t o E q . (13.4.19) a n d integrating, we get:
2
«9 = - ^ [h c , / - ( 5L + v) + % 1
+v)-
Q(l
-
v)(nr
r
(13.4.22)
C 4( l - " ) ] " /
M0)d9+f2(r),
in w h i c h f2(f) is a function of r only. Substituting E q s . (13.4.21) a n d
(13.4.22) i n t o E q . (13.4.20), w e o b t a i n the e q u a t i o n :
/
/,(9)d9
+ f\(9)
+ rf'2(r)
- f2(r)
=
4 C 4c o s 9
-=^.
S e p a r a t i n g the variables, we get the two e q u a t i o n s :
A
t\>> a T>\
(13.4.23)
392
T h e Theory of Elasticity
m 0 +d f \ « D - - t £E ^
rfi(.r)-f2(r)
(13-4-24)
(13.4.25)
= 0,
which are satisfied b y the following functions:
m= J^Ocos
fl
f2(r)
6 +
K 0 s +i L 0 n c
o(13.4.26)
s
(13.4.27)
= Hr.
AT, L, a n d H are c o n s t a n t s to b e d e t e r m i n e d from the b o u n d a r y
c o n d i t i o n s o n the d i s p l a c e m e n t s . E q s . (13.4.21) a n d (13.4.22) c a n n o w
b e written as follows:
Ur =
si|i ^
r 2 ( l
2Cd9cos9
_
3 p )
Q
+
( 1
+
v )
+
Q
(
1
_
y ) y / l r
J
+ AT sin 0 + L cos 0
(13.4.28)
2
sin 0 - ^
= ^ 9
[c,(5
+ »)r +
- Q(l -
„)//ir]
lC
+ ^ J~
^ cos 9 + K cos 0 — L sin 0 + / / r .
(13.4.29)
T a k i n g the c e n t r o i d of the cross section for 9 = TL/2, a n d also a n
e l e m e n t of the r a d i u s at this p o i n t as rigidly fixed, the b o u n d a r y
c o n d i t i o n s o n the d i s p l a c e m e n t s a r e :
Ur = gU= o, ^
= 0
for 9 = S a n d r = rc =
A p p l y i n g these to expressions (13.4.28) a n d (13.4.29), we get:
H-0,
L = ^
(13-4-30)
,
2
K = -i
[c, r c ( l - 3*) + % 1
T h e deflection at the u p p e r e n d is
+
+ C 4(l
-
(13.4.31)
Curved Beams
393
(13.4.32)
If (r0 ~ /}) is small c o m p a r e d to ri9 Eq. (13.4.32) b e c o m e s :
2>IirfP
Or)
w h i c h coincides with the expression o b t a i n e d from t h e
t h e o r y [1].
(13.4.33)
elementary
T h e p r e v i o u s solution gives the stresses a n d d i s p l a c e m e n t s for t h e
p r a c t i c a l case of the c r a n e h o o k lifting a l o a d P (Fig. 13.7). I n all the
P
F i g . 13.7
e q u a t i o n s , however, the sign of P is to b e reversed. A d d i t i o n a l solutions
c a n b e o b t a i n e d using superposition.
PROBLEMS
1.
T o e v a l u a t e the a p p r o x i m a t i o n involved in the use of the simplified
t h e o r y of c u r v e d b e a m s , c o n s i d e r the case studied in Sec. 13.3.
C o m p u t e a n d t a b u l a t e the m a x i m u m a n d m i n i m u m values of oee
a n d o n from E q s . (13.3.12) a n d (13.2.1), respectively, for the
394
2.
3.
4.
The Theory of Elasticity
following ratios of r0/rt\
1.2, 1.5, 2.0, 2.5. H o w a d e q u a t e is the
simplified t h e o r y for the design p u r p o s e s , a n d h o w m u c h is there to
b e g a i n e d b y using the results of Sec. 13.3?
K n o w i n g the results of Sees. 13.3 a n d 13.4, s h o w the s u p e r p o s i t i o n
s c h e m e to b e used w h e n the cantilever b e a m of Fig. 13.6 is
subjected at its u p p e r e n d to a couple a n d a vertical force. P r o v e
t h a t t h e solution of this p r o b l e m c a n b e o b t a i n e d using t h e A i r y
stress function, <j> = / ( r ) c o s 9.
A c u r v e d b a r of s q u a r e cross section 3 in. b y 3 in., a n d of m e a n
r a d i u s of c u r v a t u r e rc 4.5 in., is initially unstressed. If a b e n d i n g
m o m e n t of 60,000 lb-in. is a p p l i e d to t h e b a r in o r d e r to straighten
it, find the m a x i m u m a n d m i n i m u m circumferential stresses using
the e q u a t i o n s of Sec. 13.3 a n d those of the e l e m e n t a r y theory.
A s s u m i n g t h a t the angle s u b t e n d e d at the origin is 6 0 ° , a n d
k n o w i n g t h a t the solution of Sec. 13.3 is only valid at a c e r t a i n
d i s t a n c e from the e n d s of the b e a m , is the use of the e l e m e n t a r y
t h e o r y a p p r o p r i a t e for design p u r p o s e s ?
A c r a n e h o o k of r e c t a n g u l a r cross section a n d of unit thickness h a s
a n i n t e r n a l r a d i u s rxr = 3 in. a n d a n external r a d i u s r0 = 4 in.
D e t e r m i n e the m a x i m u m n o r m a l a n d shearing stresses a n d c o m p a r e
t h e m to those o b t a i n e d from the e l e m e n t a r y t h e o r y for a l o a d P of
5000 lb. (Fig. 13.7).
REFERENCES
[1] S. Timoshenko, Strength of Materials,
Vol. 1, Van Nostrand, Princeton, N . J., 1955.
CHAPTER 14
THE SEMI-INFINITE ELASTIC MEDIUM
AND RELATED PROBLEMS
14.1
Introduction
M a n y p r o b l e m s in stress a n d strain analysis, w h i c h are of p r a c t i c a l
i m p o r t a n c e , are c o n c e r n e d with the effect in semi-infinite m e d i a of
stresses a c t i n g o n their straight b o u n d a r i e s . In theory, the solution of
such p r o b l e m s c a n b e o b t a i n e d b y integration from the results of
Boussinesq a n d C e r r u t i w h i c h were p r e s e n t e d in Sees. 9.5 a n d 9.6. I n
t w o d i m e n s i o n a l cases, s o m e solutions c a n readily b e o b t a i n e d t h r o u g h
the use of a n Airy stress function. Since the i n t e g r a t i o n of t h e results of
Boussinesq a n d C e r r u t i c a n s o m e t i m e s b e c o m e extremely tedious,
solutions involving stress functions b e c o m e quite v a l u a b l e , at least from
a p r a c t i c a l p o i n t of view. T h e y fail, however, to s h o w t h e close
relationship b e t w e e n the s p a c e p r o b l e m a n d the t w o d i m e n s i o n a l
problem.
T h e p r o b l e m s e x a m i n e d in t h e following sections h a v e results t h a t a r e
extensively used b y engineers u n d e r the forms of tables a n d c h a r t s [1].
M a n y h a v e b e e n photoelastically c o n f i r m e d a n d s h o w n to b e a p p l i c a b l e
to finite b o d i e s as long as o n e r e m a i n s far e n o u g h from the b o u n d a r i e s .
T h e a i m of this c h a p t e r is to discuss a n d s h o w h o w s o m e of these results
have been obtained.
395
396
The Theory of Elasticity
14.2
Uniform Pressure Distributed over a Circular Area on the Surface
of a Semi-Infinite Solid
F i g . 14.1
W e shall first derive the expressions of the d i s p l a c e m e n t s at p o i n t s
u n d e r the c e n t e r of the l o a d e d a r e a (Fig. 14.1a) a n d at the surface of t h e
semi-infinite solid (Fig. 14.1bc). q is 2
the uniformly distributed pressure,
a n d Q is the total l o a d e q u a l to
Ila q.
At a point M under the center of the loaded area (Fig. 14.1a), w e h a v e ,
b e c a u s e of s y m m e t r y :
ur = u9 = 0.
(14.2.1)
U s i n g Eq. (9.6.21), the vertical d i s p l a c e m e n t of M is given b y :
(14.2.2)
2
c o s (S
I n t e g r a t i o n of E q . (14.2.2), yields
^gfy^z.j]^,.^
Therefore,
.
]. 04,,,
The Semi-Infinite Elastic M e d i u m
397
At the surface of the semi-infinite solid, two cases m u s t b e c o n s i d e r e d :
a) T h e p o i n t M is outside the l o a d e d a r e a (Fig. 14.b). Setting z = 0 a n d
s = p in E q . (9.6.21), we get:
x
- W
((u" A - o -
- **>
-U
s
(14.2.5)
E
N o w c o n s i d e r the deflection in the OX3 direction at a p o i n t M o n the
surface of t h e solid, a n d at a d i s t a n c e / from the center of the l o a d e d
area. T a k e a small element of l o a d e d a r e a b o u n d e d b y the two radii
enclosing t h e angle dxp a n d t w o arcs of circle with radii s a n d s + ds
c e n t e r e d at M. T h e l o a d o n this element is qs(ds)d\p a n d , using E q .
(14.2.5), t h e deflection of the p o i n t M is:
^ u - ^ E * -
If
(14-2.6)
2 s varies
2 2 b e t w e e n MD a n d A/C, a n d the length of the c h o r d
T h e distance
CD is 2\/a
— / s i n i / / . Therefore, E q . (14.2.6) b e c o m e s :
in w h i c h \px is t h e m a x i m u m value of \p. T h e calculation of the integral
in E q . (14.2.7) is simplified b y i n t r o d u c i n g the variable 0, w h e r e (Fig.
14.1b)
a sin 9 = / sin xp.
(14.2.8)
9 varies from 0 to I I / 2 w h e n \p varies from 0 to ^ .
substitution, Eq. (14.2.7) b e c o m e s :
- d
1
<£.) C
? u
d9
With
1 < -->
14
2
j
sin 0
this
2
9
7^jT" "
T h e s e are elliptic integrals, a n d their values for various a/1 c a n be f o u n d
in tables [2]. A t the b o u n d a r y / = a, a n d
398
T h e Theory of Elasticity
b) T h e p o i n t M is within t h e l o a d e d area (Fig. 14.1c). W e consider t h e
deflection in t h e OX3 direction of p o i n t M d u e t o t h e l o a d qs(ds)(dxp)
acting o n t h e s h a d e d area. T h i s deflection is given b y :
{
{u2)M =
±^q
j
j
(14-2.11)
d*ds.
T h e d i s t a n c e s varies b e t w e e n C a n d £>, a n d t h e length CD is e q u a l t o
2a cos 9. T h e angle xp varies b e t w e e n 0 a n d I I , a n d
a sin 9 = I sin \p.
Therefore,
T h e s y m m e t r y in F i g . 14.1c m a k e s it possible to c h a n g e the limits of t h e
previous integral to 0 a n d 11/2, so t h a t
< " , > „ - j ? V
1
- ( * y » "
2
* ^
2 )i
3<
H
'
T h e deflection c a n b e c o m p u t e d for a n y ratio I/a b y using tables of
elliptic integrals. F o r / = 0, E q . (14.2.4) is o b t a i n e d . T h e average
deflection of the l o a d e d a r e a is given b y :
w
( z)aver.
//
uz2Urdr
^ 2
_ \j q
\a - { 2 }y_ o > 5 1 4_ 62 y}(
£
. 1 4 )( 1 4 > 2
aE
Let us now compute the stresses under the center of the loaded area {Fig.
14.2).
F r o m E q . (9.6.24), w e h a v e :
T h e integration is easily a c c o m p l i s h e d b y setting r = z t a n /?. H e n c e ,
399
T h e Semi-Infinite Elastic M e d i u m
Fig. 14.2
3
=
f
3q t a n
3
cos fidfi
= L cos /|
L
4l
or
B e c a u s e of s y m m e t r y ,
(14.2.17)
and
(14.2.18)
T o c o m p u t e orn w e p r o c e e d as follows [8]: A c c o r d i n g to E q s . (9.6.22)
a n d (9.6.23), the t w o e l e m e n t s 1 a n d 2 in Fig. 14.2 give a t M :
DA
(1 -
( 9$h,2
=
2v)grd9dr
2
lip
2 2)o
b
- p t t ] -
- -
Also t h e t w o e l e m e n t s 3 a n d 4 give a t M:
(1 -
2v)qrd6d>
2
Up
|_P
p +
zj
(14.2.21)
400
The Theory of Elasticity
<**>*-
0"[-9 ^]-
QJ
+
-'
(L4 2 22)
w h e r e arr a n d a69 are s h o w n in Fig. 14.2. By s u m m a t i o n of Eqs. (14.2.19)
a n d (14.2.21) or of E q s . (14.2.20) a n d (14.2.22), w e get:
- *. -
- 3*].
04.2.23)
p r o d u c e d b y the entire load, we integrate
T o o b t a i n the stress orr =
Eq. (14.2.23) with respect to 8 from 0 to I I / 2 a n d with respect to r from
0 to a. H e n c e ,
04.2.24,
+
A t p o i n t 0 , we h a v e :
+
l v
„
- _ a
-
a-
^
)
(14.2.25)
T h e m a x i m u m shearing stress at a n y point o n the OX3 axis is given
by:
v
\{°ee
~ ozz
) = f [ ^ - ^
+ (1 +
3(
2V^
z
) ^ / + 22
.3-
(14.2.26)
VI
This expression b e c o m e s a m a x i m u m for
^)
2v
(14.2.27)
and
[ j ( * i f - ^ ) ] m x a= |
+ | d
+ *h/3(nn)].
(14.2.28)
If we set v = 0.3 in Eqs. (14.2.27) a n d (14.2.28), we get:
z = 0.638a,
U ( o w- o „ ) |
L ^
J max
=0.33?.
(14.2.29)
T h e Semi-Infinite Elastic M e d i u m
401
T h u s , t h e m a x i m u m s h e a r i n g stress o c c u r s a t a d e p t h a p p r o x i m a t e l y
e q u a l to two-thirds of t h e r a d i u s of t h e l o a d e d circle a, a n d t h e
m a g n i t u d e of this m a x i m u m is a b o u t o n e - t h i r d of t h e a p p l i e d u n i f o r m
pressure q.
Remark
T h e e q u a t i o n s given in this section h a v e b e e n u s e d b y N e w m a r k [3,
4] to d e v e l o p c h a r t s t h a t c a n b e u s e d to c o m p u t e stresses a n d displacem e n t s a t a n y p o i n t of a semi-infinite elastic m e d i u m d u e t o uniformly
d i s t r i b u t e d stresses acting o n a r e a s of a n y s h a p e .
143
Uniform Pressure Distributed over a Rectangular Area
T h e g e n e r a l expressions for t h e stresses a n d d i s p l a c e m e n t s p r o d u c e d
b y l o a d s d i s t r i b u t e d over r e c t a n g u l a r a r e a s c a n b e f o u n d in [5]. I n this
section, w e shall only give t h e expression of t h e vertical stress u n d e r t h e
c o r n e r of a r e c t a n g u l a r a r e a w h o s e d i m e n s i o n s a r e a a n d b ( F i g . 14.3),
a n d t h a t of t h e average vertical d i s p l a c e m e n t of t h e surface. The vertical
7
X ,Z
3
Fig. 14.3
stress a t M (0,0,d) d u e to a uniformly d i s t r i b u t e d p r e s s u r e q is given b y
Eq. (14.3.1):
2
q I"
(°zz)m
where
~
ATI
4n
2
V
2BV
2
Lv + B
2
V 2
+ 1
22BV 2
+ t a n -i
2 V
V - B
2
a + b2 + d
=
d
2
(14.3.2)
402
T h e Theory of Elasticity
O n e m a y find the stress ozz at a n y p o i n t u n d e r a r e c t a n g u l a r a r e a b y
dividing the area i n t o smaller rectangles, such t h a t the p o i n t b e n e a t h
which we a r e seeking the stresses is a c o r n e r c o m m o n to all smaller
rectangles (Fig. 14.3). E q . (14.3.1) c a n then b e applied to each of these
smaller rectangles a n d the s u m of the individual results yields the total
stress.
The average vertical displacement
of the surface of the uniformly
l o a d e d m e d i u m is given b y :
,
1
m
(2(1 - v )
,
^JJ
("3)aver. =
14J
3)
>
>
R
( -
w h e r e Q is the total load, A is the m a g n i t u d e of the l o a d e d area, a n d m
is a n u m e r i c a l factor d e p e n d i n g o n the ratio a / b. T h e following table
gives the value of m for various a / b. F o r c o m p a r i s o n p u r p o s e s , m for
a circular a r e a is also i n c l u d e d .
Rectangles with various a / b
Circle
m =
0.96
1
1.5
2
3
5
10
100
.95
.94
.92
.88
.82
.71
0.37
T h e a b o v e table shows t h a t for a given Q a n d A the deflection increases
w h e n the ratio of the p e r i m e t e r of the l o a d e d area to the area decreases.
14.4
Rigid Die in the Form of a Circular Cylinder
I n this case, the d i s p l a c e m e n t s are given a n d it is necessary to find the
c o r r e s p o n d i n g distribution of t apressures o n the b o u n d a r y p l a n e . T h e
vertical d i s p l a c e m e n t (w z)z=o
the surface is a c o n s t a n t u n d e r the die,
b u t the distribution of pressure (Fig. 14.4) is n o t ; its intensity is given
b y [6]:
(14.4.1)
T h e Semi-Infinite Elastic M e d i u m
403
Fig. 14.4
T h e pressure h a s its smallest value at the center, a n d is infinite o n the
edges. I n a c t u a l cases, we shall h a v e a yielding of the m a t e r i a l a l o n g the
b o u n d a r y . T h e d i s p l a c e m e n t s of the die ( w 2) z 0=c o r r e s p o n d i n g to the
distribution of E q . (14.4.1) is given b y :
( ) =
Mz
jj
2=0
qd*ds,
(14A2)
a n d is the s a m e for a n y p o i n t M (Fig. 14.4): I n Eq. (14.4.2), \p varies
from 0 to I I , a n d s varies b e t w e e n C a n d D. N o w
(14.4.3)
s = / cos \p + e
and
2
r
2
b
=
2
(14.4.4)
e,
4-
so t h a t Eq. (14.4.2) c a n b e written as follows:
("Jz=0-
2
aE
2
U
E q s . (14.4.5) is easily i n t e g r a t e d to give:
y2
,
^
- 6(! -
("z)z=o -
2aE
>
•
T h e value of ( w z) z 0=is n o t very different from t h a t of Eq. (14.2.14).
404
T h e Theory of Elasticity
14.5
Vertical Line Load on a Semi-Infinite Elastic Medium
T h e stresses in a n elastic semi-infinite m e d i u m subjected t o a line
l o a d c a n b e derived from E q s . (9.6.30) to (9.6.34) b y s u m m i n g t h e
stresses p r o d u c e d b y the e l e m e n t a l loads of a n infinite system of p o i n t
loads as s h o w n in Fig. 14.5. T h e p o i n t at w h i c h t h e stresses a r e to b e
F i g . 14.5
c o m p u t e d is p l a c e d in t h e p l a n e OX{, OX3 for c o n v e n i e n c e . Since the
line l o a d e x t e n d s to infinity o n b o t h sides of the origin, this choice
imposes n o restrictions o n the solution. T h e only c o m p o n e n t s of the
state of stress at a p o i n t M are a n, a22
, a 3 , 3 a n d a 1 ; 3 a 12 a n d o23 a r e
e q u a l to z e r o b e c a u s e of s y m m e t r y . T h e p r o b l e m is a p l a n e strain
p r o b l e m with n o d i s p l a c e m e n t s a l o n g the OX2 direction. T h e b o d y
forces are neglected. F r o m E q s . (9.6.30) to (9.6.34), we get:
2
°33
-
' 2n
2q
dx2
4
cos <f)
3qx%
II
f ' tsec
5 5xp dip _ _ 2qx\
4
Jo
t sec \p
lit
(14.5.1)
T h e Semi-Infinite Elastic M e d i u m
2
an =
a
22
[an]M
dx2
00
-I.
[<>22\M 2
=
K<>11 +
0 )
3 3=
2
-
2vqx
23
lit
2vq
COS 0
~T\x3
a
— 23
=
0-
z
2q
.
^
sin <J>0 cos <|>
ILc3
dx
+ 00
~
°\2
/
z
2qx 4 x3
Ut
+00
405
(14.5.3)
(14.5.4)
(14.5.5)
2
I n these e q u a t i o n s , t = x3 + x\; in o t h e r w o r d s , / is t h e r a d i a l d i s t a n c e
from O t o t h e p o i n t M in t h e OXx, OX3 p l a n e . T h e stresses a 12 a n d ol3
v a n i s h b e c a u s e of s y m m e t r y . T h e n o t a t i o n [ a n] Min E q . (14.5.3) d e n o t e s
the stress from Eq. (9.6.30). It is n o t evident from i n s p e c t i o n w h y a n in
the p l a n e strain case is i n d e p e n d e n t of Poisson's ratio, however, t h e
a c t u a l i n t e g r a t i o n shows t h a t the t e r m s w h i c h a r e multiplied b y (1 - 2v)
a the plane
in E q . (9.6.30) d o n o t m a k e a n y c o n t r i b u t i o n to t h e result in
strain case. Similarly, t h e i n t e g r a t i o n for a 22 yields a 22 = K n + ^ 3 3 ) I n cylindrical c o o r d i n a t e s , with r r e p l a c i n g t a n d 6 r e p l a c i n g <f> (Fig.
14.6), t h e c o m p o n e n t s of t h e state of stress a r e w r i t t e n as follows:
q
[y
Xi
<a
/ J
/
/
rr
Fig. 14.6
2 ? cos 9
n
°90
= 0
r
(14.5.6)
(14.5.7)
406
T h e Theory of Elasticity
zz _
~~
r
2qpC
Qs9
(14.5.8)
n
ore = °rz = o9z= 0.
(14.5.9)
T h e p r e v i o u s e q u a t i o n s s h o w t h a t t h e r a d i a l direction is a p r i n c i p a l
direction, a n d t h a t all p o i n t s o n a n y circle of d i a m e t e r D c e n t e r e d o n
the vertical axis a n d p a s s i n g t h r o u g h the p o i n t of l o a d i n g h a v e the s a m e
p r i n c i p a l stress= ox = — 2q/UD a n d t h e s a m e m a x i m u m s h e a r i n g stress
\(P\ ~ °i)
-q/HD>
E q s . (14.5.6) to (14.5.9) c a n also b e o b t a i n e d
t h r o u g h t h e use of the Airy stress function:
<?> = C T 0 s i n 0
(14.5.10)
with
1 4
5 1 1
c = - § .
(
-
>
H a v i n g t h e stresses, t h e strains a n d d i s p l a c e m e n t s c a n b e c o m p u t e d
at a n y p o i n t . It is m o r e c o n v e n i e n t to use cylindrical c o o r d i n a t e s for
s u c h c o m p u t a t i o n s . U s i n g E q s . (6.7.23), (6.7.24), a n d (8.16.12) t o
(8.16.14), w e get:
2
e
1 - 2 vrr _
E
°
_duL_
"~
dr
e
2(1 - p )gcos
HE
+
\du
e _< 2v{\ + v)gCos
r 0d ~
HE
_ur
ee ~ r
9
ezz = 0
Q
/ 14
5i2)r
r (14 5 13)
(14.5.14)
l/ljfor
9^ _ ^ )
erz = e9z = 0.
=0
(14.5.15)
(14.5.16)
I n t e g r a t i o n of E q . (14.5.12) yields:
Ur - V p
2
=
jn
cos 9lnr+M9).
451
O - - ?)
S u b s t i t u t i n g Eq. (14.5.17) i n t o Eq. (14.5.13) a n d integrating, w e get:
The Semi-Infinite Elastic M e d i u m
407
2
2v(l E
+ v)q
u»9 =
n r
.
.
n2(1 - E v )q Inr tsin
sin 9 + U
9
r
n
n
U
(14.5.18)
M9)d9+f2(r),
w h e r e fx{9) a n d / 2( r ) a r e functions of 9 a n d r, respectively. E q s . (14.5.17)
a n d (14.5.18) are n o w substituted in E q . (14.5.15), to give:
2
2
2(1 ~ r )q
.
, 1 d , , m , 2(1 - v )gSm
aim
9 . d , , ,
2
+ i j
m
mQ
2(1 - v )q
UE
2v{\ + v)qsjn 0
UE
r
M9)d9-±f2(r)
(14.5.19)
s
r
= 0.
S e p a r a t i n g the variables a n d solving, Eq. (14.5.19) yields:
fx(9)
= Asin6
Mr)
= Cr,
+ Bcos0-
^-(l
- 2v)(\
+ v)0 sin 9
(14.5.20)
(14.5.21)
w h e r e A, B, a n d C are c o n s t a n t s of integration w h i c h are to b e
d e t e r m i n e d from the b o u n d a r y c o n d i t i o n s . Substituting the values of
fx(9) a n d / 20 * ) in E q s . (14.5.17) a n d (14.5.18), we get the e q u a t i o n s of the
displacements:
2
ur =
4-
M
0
S
2(1 - v )q
YlE
0r Cl Oqn
~~ TlE^
A sin 0 + B cos 9
2v(\ + v)q
=
jj£
m
.
2X v
"
^
+ vm S 6
(14.5.22)
^
2
S - + v )q
n2(1
UE
m.S „
S+r * C„ Om. *
~
(14.5.23)
+
uz = 0.
- 2*0(1 + ^)[sin 0 - 9 cos 0] 4- Cr
(14.5.24)
Let us consider two sets of b o u n d a r y c o n d i t i o n s :
a) All points o n the OX3 axis d o n o t h a v e a n y lateral d i s p l a c e m e n t ,
a n d o n e p o i n t at a distance d o n this axis does n o t m o v e vertically.
S
408
The Theory of Elasticity
Therefore, ue = 0 for 0 = 0, a n d ur = 0 for 0 = 0 a n d r = d. T h e first
c o n d i t i o n leads to
A = C = 0,
(14.5.25)
a n d the s e c o n d c o n d i t i o n leads to
B-^p-qlnd.
(14-5.26)
T h e e q u a t i o n s of the d i s p l a c e m e n t s in this case b e c o m e :
ur =
COS
T7/ ^
2(1
2)
u = 9(1 + v) m
.
°
HE
9
,
n
7 ~ 1^(1
_
2f,
Xl +
s i n 0
2
s 2(1 - f ) ^ . d • n
n
~~
.
/n
/if £
sin 0
( -14
5
27)
(14.5.28)
- j ^ ( l - 2*-)(l + *06>cos 0
u, = 0.
(14.5.29)
O n the surface of the straight b o u n d a r y , we h a v e :
(ur)0=±1} = - ^ ( 1 - 2*0(1 + v)
(14.5.30)
b ) All p o i n t s o n the OX3 axis d o n o t h a v e a n y lateral d i s p l a c e m e n t ,
a n d a p o i n t at a d i s t a n c e d a l o n g the OXx axis d o e s n o t m o v e vertically.
Therefore, u9 = 0 for 0 = 0, a n d ^ = 0 for 0 = I I / 2 a n d r = d. T h e
first c o n d i t i o n leads to
A = C = 0,
(14.5.32)
a n d the s e c o n d c o n d i t i o n leads to
2
9(1 + v)
2(1 - r )
T h e e q u a t i o n s of t h e d i s p l a c e m e n t s therefore b e c o m e :
(U.5.33)
T h e Semi-Infinite Elastic M e d i u m
409
2
ur
q
2(1 ~ v )
q cos 0 In j
TIE
- ^ ( 1
- 2*<)(1 + v)0 sin 0
(14.5.34)
, «?(! + v)
cos 0
UE
<?(! + v)
sin 0
UE
(14.5.35)
u2 = 0.
(14.5.36)
O n the surface of the straight b o u n d a r y , w e h a v e :
(14.5.37)
(14.5.38)
F o r b o t h sets of b o u n d a r y c o n d i t i o n s , t h e q u a n t i t y d is i n d e t e r m i n a t e
a n d t h e r e is n o t h i n g in the analysis b y w h i c h it c a n b e f o u n d . It is
usually t a k e n to b e very large. B o t h E q s . (14.5.30) a n d (14.5.37) i n d i c a t e
a d i s p l a c e m e n t of the m a t e r i a l o n the surface t o w a r d s the origin. W e
m a y r e g a r d such a d i s p l a c e m e n t as a physical possibility if w e r e m o v e
a cylindrical surface of small r a d i u s a r o u n d the line of a p p l i c a t i o n of q
(Fig. 14.7a) a n d substitute to q a n equivalent system of stresses.
Actually, in this p o r t i o n , the m a t e r i a l is plastically d e f o r m e d a n d
p e r m i t s the d i s p l a c e m e n t s of E q s . (14.5.30) a n d (14.5.37). T h e solutions
p r e s e n t e d in this section are subject to the s a m e restriction i m p o s e d o n
the Boussinesq solution of the p o i n t l o a d — n a m e l y , t h a t their validity
starts at a small d i s t a n c e from the p o i n t of a p p l i c a t i o n of the load.
Finally, it is of interest to e x a m i n e the s h a p e of the lines (they a r e
actually surfaces e x t e n d i n g in the OX2 direction from -oo to + o o ) in the
m e d i u m , w h i c h at e a c h of their p o i n t s a r e t a n g e n t to the principal
stresses. Such lines are called principal stress trajectories a n d from E q s .
(14.5.6) to (14.5.9) are seen to b e straight lines c o n v e r g i n g at the p o i n t
of a p p l i c a t i o n of the l o a d a n d c o n c e n t r i c circles with centers at this
p o i n t (Fig. 14.7a). A s e c o n d set of lines, w h i c h is of interest in the study
of the t h e o r y of plasticity, is f o r m e d b y lines which, at e a c h p o i n t , a r e
t a n g e n t to the directions of the m a x i m u m shearing stresses. Such lines
are called m a x i m u m s h e a r i n g stress trajectories a n d are inclined 45° to
410
The Theory of Elasticity
Fig. 14.7
the principal stress trajectories. T h e y are therefore logarithmic spirals
(Fig. 14.7b) w h o s e e q u a t i o n is:
9
r =
Ce .
A third set of lines called isochromatics consists of the loci of equal
m a x i m u m shearing stresses. T h o s e loci h a v e a l r e a d y b e e n s h o w n to b e
circles passing t h r o u g h 0 a n d c e n t e r e d o n OX3 (Fig. 14.7c).
14.6
Vertical Line Load on a Semi-Infinite Elastic Plate
T h e p l a t e is a s s u m e d to b e of unit w i d t h so t h a t the l o a d is e q u a l to
q (Fig. 14.8). T h e p r o b l e m is a p l a n e stress p r o b l e m . Since the b o d y
forces are neglected, the stresses are the s a m e as those o b t a i n e d in the
previous case except t h a t a 22 = 0. T h u s ,
2
a 33 = - ^ g ^ ,
2
an = -^sin 0cos 0,
a 22 = 0 (14.6.1)
3
a 13 = ~
sin 9 c o s 0 ,
a 23 = a 12 = 0.
(14.6.2)
A n d in cylindrical c o o r d i n a t e s :
°rr
= _ggcos 9
Yl
(14.6.3)
r
°00 = °zz = °r0 = °rz = °9z = °«
(14.6.4)
The Semi-Infinite Elastic Medium
411
Fig. 14.8
T h e strains are given b y :
dur
e
»e
r
1q cos 0
"97
= uL
r
+
\dug_
2qv
r $q
j j £
=
r
c s oQ
r
*
2 \
(14.6.5)
HE
r
W
dr
)
(14.6.6)
(14.6.7)
(14.6.8)
= « fe = o.
T h e i n t e g r a t i o n of E q s . (14.6.5) a n d (14.6.8) p r o c e e d s a l o n g the s a m e
lines followed in the previous section. If we a s s u m e t h a t a p o i n t a l o n g
the OX3 axis a n d at a d e p t h d is fixed, the e q u a t i o n s of the displacements are:
l ( )
r
«r = TTE- cos Olni
HE
Ug =
-
^^
HE
a 2q
i
q(\ + v)
sin 6
UE~
" " " I H
q
r
0 sin 0
n S 9i d l
r -
n
(1 - v)q
UE
0 cos 0.
(14.6.9)
(14.6.10)
O n the straight b o u n d a r y of the plate, we h a v e :
("r)e=±9
(1 - v)q
2E
(14.6.11)
(14.6.12)
412
T h e Theory of Elasticity
T h e validity of the previous e q u a t i o n s starts at a small d i s t a n c e from the
p o i n t of a p p l i c a t i o n of the load. Principal stress trajectories, m a x i m u m
shearing stress trajectories, a n d i s o c h r o m a t i c s a r e as s h o w n in Fig. 14.7.
14.7
Tangential Line Load at the Surface of a Semi-Infinite Elastic
Medium
T h e stress resulting from a tangential line l o a d at the surface
semi-infinite m e d i u m c a n b e derived from Eqs. (9.5.39) to (9.5.44)
s u m m a t i o n of the stresses for the elemental h o r i z o n t a l p o i n t l o a d s
14.9). T h e steps in such a s u m m a t i o n a r e similar to those of Sec.
except t h a t the angle </> is m e a s u r e d from the OXx axis.
(j
33
3qx
-
a 13 -
n
q X3
x
4
Ut
X
\
+ oo
[o
- 00
Jo
12 =
0
23
2
— 2vqx,
= 0.
2qxxx\4
-2q
ILc,
II/
z
3
— 2vq
= JixJ
3
sin <|> cos <j>
2,
= - — i - s i n 6 cos <J>
Ux3
3
3
— 2qx,
—2q
,
u]M
dx2
= - ^ 4 -=
cos </> sin <j>
°22 = K ° n + 0 3 3 )
<J
dx2 _
s
c ns o i
*
*
of a
by a
(Fig.
14.5,
(14.7.1)
(14.7.2)
(14.7.3)
(14.7.4)
(14.7.5)
T h e Semi-Infinite Elastic M e d i u m
413
1
I n these e q u a t i o n s , t = x\ + x\.
I n cylindrical c o o r d i n a t e s , with r r e p l a c i n g / a n d 0 replacing <j>, the
c o m p o n e n t s of t h e state of stress a r e written as follows (Fig. 14.10):
<7
0
Fig. 14.10
2
ff cos 0
Ivq
cs g no
°r9 = °rz = °0z = °-
(1416)
(14.7.7)
T h e s e e q u a t i o n s are the s a m e as E q s . (14.5.6) to (14.5.9) with the
difference t h a t 0 is m e a s u r e d from the h o r i z o n t a l , i.e., from the direction
of the load. O n c e the b o u n d a r y c o n d i t i o n s are chosen, the displacem e n t s c a n b e c o m p u t e d in a m a n n e r similar to t h a t of Sec. 14.5.
T h e p r i n c i p a l stress trajectories a r e r a d i a l lines c o n v e r g i n g at the
p o i n t of a p p l i c a t i o n of the l o a d a n d c o n c e n t r i c circular arcs with centers
at the p o i n t of a p p l i c a t i o n of the load. T h e m a x i m u m shearing stress
trajectories are l o g a r i t h m i c spirals similar to those s h o w n in Fig. 14.7b.
T h e i s o c h r o m a t i c s are semi-circles, the centers of w h i c h lie o n OXx (Fig.
14.10).
Remark
T h e stresses a n d d i s p l a c e m e n t s in a semi-infinite elastic m e d i u m
subjected to inclined loads c a n b e o b t a i n e d b y superposition of the
vertical a n d h o r i z o n t a l cases. If the c o m p o n e n t s of the line l o a d are
q cos a a n d q sin a (Fig. 14.11), the stresses at a p o i n t M in cylindrical
c o o r d i n a t e s a r e given b y :
rr
°
=2qcos(6>
~Wr
~
°r9 = °rz = °0z = 0-
a )m
'
°
= zz
°'
°
= 2vq c ( o s ) a
~ T F
~
478
O - - )
(14.7.9)
414
The Theory of Elasticity
• Xa
Fig. 1 4 . 1 1
T h e difference b e t w e e n the three g r o u p s of E q s . (14.5.6) to (14.5.9),
(14.7.6) to (14.7.7), a n d (14.7.8) to (14.7.9) is in the d a t u m line from
which the angle is m e a s u r e d . This d a t u m line in each case is given b y
the direction of the applied line load.
14.8
Tangential Line Load on a Semi-Infinite Elastic Plate
T h e p l a t e is a s s u m e d to b e of unit w i d t h so t h a t the load is equal to
q (Fig. 14.12). T h e p r o b l e m is a p l a n e stress p r o b l e m . T h e stresses a r e
*3
Fig. 1 4 . 1 2
given b y Eqs. (14.7.1) to (14.7.5) in cartesian c o o r d i n a t e s , a n d b y E q s .
(14.7.6) a n d (14.7.7) in cylindrical c o o r d i n a t e s b u t with a 22 = ozz = 0.
H e r e , too, o n c e the b o u n d a r y c o n d i t i o n s are chosen, the d i s p l a c e m e n t s
c a n b e c o m p u t e d in a m a n n e r similar to t h a t of Sec. 14.5. T h e principal
stress trajectories, the m a x i m u m shearing stress trajectories, a n d the
isochromatics are the s a m e as those of Sec. 14.7.
T h e Semi-Infinite Elastic Medium
415
T h e effect of inclined l o a d s c a n b e o b t a i n e d b y superposition of the
h o r i z o n t a l a n d vertical cases. T h e expression of the stresses in cylindrical c o o r d i n a t e is the s a m e in all cases p r o v i d e d the angle is m e a s u r e d
from the direction of the load.
14.9
Uniformly Distributed Vertical Pressure on Part of the Boundary
of a Semi-Infinite Elastic Medium
Fig. 14.13
a
M
I n Fig. 14.13, the l o a d e d strip e x t e n d s to infinity o n b o t h sides of the
origin a l o n g the OX2 axis. T h e p r o b l e m is a p l a n e strain p r o b l e m . T h e
stresses at a n y p o i n t M defined b y 0X a n d 92 are o b t a i n e d b y i n t e g r a t i o n
of E q s . (14.5.6) to (14.5.9). F r o m Fig. 14.13:
1
dxx
cos 0
(14.9.1)
U s i n g the e q u a t i o n s of t r a n s f o r m a t i o n (7.11.4) to (7.11.6), together with
E q s . (14.5.6) to (14.5.9), we get:
(14.9.2)
yff [2(0 2 - 0 , ) + (sin 2 0 2 - sin 20,)]
(14.9.3)
2Yy[2(0 2 - 0 , ) - (sin 2 0 2 - sin 20,)]
416
The Theory of Elasticity
r0
2
o 13 =
Jg
sin 20d0
= - j f y l c o s 20x - cos 202]
(14.9.4)
o22 = v(an + a 3 ) 3 = - ^ ( 0 2 - ^ ) =
(14.9.5)
a 12 = a 23 = 0,
(14.9.6)
w h e r e a is e q u a l to 02 — 0X. T h e principal stresses at the p o i n t M are
given b y :
ox = - | ^ ( a + sin a ) ,
a 3 = - ^ ( a - sin a ) .
(14.9.7)
T h e angle a is c o n s t a n t for a n y circle c e n t e r e d o n the OX3 axis a n d
passing t h r o u g h Ox a n d 02 (Fig. 14.14). Therefore, the principal stresses
a n d their difference are the s a m e for all p o i n t s falling o n this circle. T h e
angle m a d e b y the principal stresses a n d the OXx axis is given b y E q .
(7.11.7) as
t a n 2<jf> =
a 2a 13 a
( n ~~
33)
cosn 20x —
s cos 2 0 2
= - t a n ( 0 , + 02).
i 2#i — sin 2 0 2
(14.9.8)
T h e Semi-Infinite Elastic M e d i u m
417
Therefore.
1 9
49
=
4>\
y
~~ \ ( (0
®X
\ + 02)
and
<t> = n
2
-
±(0, + 0 )-
2
^
- - )
E q s . (14.9.9) show t h a t at all p o i n t s o n the circle, the direction of
principal stresses is given b y the t w o lines passing t h r o u g h C a n d Z); in
o t h e r w o r d s , the directions of the principal stresses bisect the angle
b e t w e e n the two radii rx a n d r2. Therefore, the principal stress trajectories are families of confocal h y p e r b o l a s a n d confocal ellipses (Fig.
14.15a) with focii at Ox a n d 0 2. T h e i s o c h r o m a t i c s are circles c e n t e r e d
(b)
Fig. 14.15
o n OX3 a n d passing t h r o u g h Ox a n d 0 2 (Fig. 14.15b). O n each of these
circles, the m a x i m u m shearing stress is given b y :
(14.9.10)
T h e m a x i m u m value in Eq. (14.9.10) is for a = U / 2 c o r r e s p o n d i n g to
a circle c e n t e r e d at O.
T h e deflection of points at the surface of the semi-infinite solid c a n
b e o b t a i n e d b y integration of Eq. (14.5.31) or Eq. (14.5.38), d e p e n d i n g
o n the b o u n d a r y c o n d i t i o n s chosen. Let us consider Eq. (14.5.31), for
e x a m p l e . If the p o i n t w h o s e vertical deflection is sought is outside the
l o a d e d area, we h a v e (Fig. 14.16):
f
J a
J a
dr
(i4-9.il)
418
The Theory of Elasticity
T 3fr
I *0
X3
Fig. 1 4 . 1 6
or
(14.9.12)
^ ( 1
+ ,)(1 -
2v).
In the s a m e m a n n e r , for a p o i n t u n d e r the l o a d (Fig. 14.17):
J_
Fig. 1 4 . 1 7
= - n ^ ' h
-
" ^
+ „)(1 -
2
F
^
"
.
1)3
2v),
d is the distance from O of a p l a n e parallel to the surface a n d w h o s e
vertical d i s p l a c e m e n t is equal to zero.
A n u m b e r of p r o b l e m s related to the semi-infinite elastic m e d i u m in
p l a n e strain h a v e b e e n solved by Holl [7] w h o c o n s i d e r e d b o t h
The Semi-Infinite Elastic M e d i u m
419
h o r i z o n t a l a n d inclined surface p l a n e s . T i m o s h e n k o [8] h a s given the
stress functions p r o v i d i n g the solution for various l o a d i n g p a t t e r n s o n
the h o r i z o n t a l surface of a semi-infinite m e d i u m or plate.
14.10 Uniformly Distributed Vertical Pressure on Part of the Boundary
of a Semi-Infinite Elastic Plate
t*3
Fig. 1 4 . 1 8
T h e plate is a s s u m e d to b e of unit w i d t h (Fig. 14.18). T h e p r o b l e m is
a p l a n e stress p r o b l e m . T h e stresses are the s a m e as those given in Sec.
14.9, except for a 22 which is e q u a l to zero. T h e d i s p l a c e m e n t s of points
o n the surface of the plate c a n b e o b t a i n e d b y i n t e g r a t i o n of E q .
(14.6.12). If the p o i n t w h o s e vertical deflection is sought is o u t s i d e the
l o a d e d a r e a (Fig. 14.16), we h a v e :
^
ra+2b
ra+2b
'~mL "i*
d
MM-W"L
or
= -mV
(,410,)
wva~ 'A ~ ^ W ^ -
lb+a)ln
aln
( 1 4 I A 2 )
In the s a m e m a n n e r , for a p o i n t u n d e r the load (Fig. 14.17):
(u ) ^
e9
= -?±[(2b
- a)ln^-a
+ ain{\
-
^£±qb;
(14-10.3)
d is the d i s t a n c e from O to a p o i n t o n the OX3 axis w h o s e d i s p l a c e m e n t
is equal to zero.
420
The Theory of Elasticity
F r o c h t [9] h a s p r o v i d e d a p h o t o e l a s t i c c o n f i r m a t i o n for m a n y of t h e
solutions related to semi-infinite plates.
14.11 Rigid Strip at the Surface of a Semi-Infinite Elastic Medium
It c a n b e s h o w n that, w h e n a l o a d Q is applied t h r o u g h a rigid flat
die [10] so t h a t the deflection u n d e r it is c o n s t a n t (Fig. 14.19), t h e
distribution of pressure o n the die is given b y :
Fig. 1 4 . 1 9
2
9 = U^/b
(14.11.1)
- x[
T h i s expression shows t h a t q = Q/Ub
w h e n xx = 0, a n d b e c o m e s
infinite w h e n xx = b. I n a c t u a l cases, we shall h a v e a yielding of the
m a t e r i a l a l o n g t h e b o u n d a r y . T h e d i s p l a c e m e n t (u0)g=u/2
of the die
c o r r e s p o n d i n g to the distribution (14.11.1) is o b t a i n e d b y i n t e g r a t i n g
E q . (14.5.31). F o r e x a m p l e , u n d e r the center of the die we h a v e :
L >
(14.11.2)
+| F
) (14.11.3)
QDR
where q is given b y Eq. (14.11.1). T h u s ,
•2
( ^ = - ^ [ 2 ( 1 - ^ ")
M
/ _n( 1
T h e s a m e result is o b t a i n e d for a n y p o i n t u n d e r t h e die.
T h e Semi-Infinite Elastic M e d i u m
421
14.12 Rigid D i e at the Surface of a Semi-Infinite Elastic Plate
T h e distribution of pressure o n the die is given b y Eq. (14.11.1),
n a m e l y (Fig. 14.20):
*3
Fig. 1 4 . 2 0
«-;vfcr
<,412l)
T h e d i s p l a c e m e n t of the surface c o r r e s p o n d i n g to the
(14.12.1) is o b t a i n e d b y integrating E q . (14.6.12). T h u s ,
distribution
14.13 Radial Stresses in Wedges
T h e stresses in a w e d g e of infinite length subjected to a vertical l o a d
at the a p e x (Fig. 14.21) c a n b e o b t a i n e d from the Airy stress function:
<t> = Cr0 sin 0
(14.13.1)
T h e c o m p o n e n t s of the state of stress a r e :
2 C cos 0
O
n
=
, (700 = 0,?
°r0 = °0z = °rz = °-
2Cv cos 0 n
G
zz =
r i 4 13 2^
?
(14.13.3)
422
The Theory of Elasticity
Fig. 14.21
T h e s e stresses satisfy the b o u n d a r y c o n d i t i o n s in the w e d g e : T h e y
vanish at infinity a n d leave the straight edges free of n o r m a l a n d shear
stresses. In o r d e r to d e t e r m i n e the c o n s t a n t C in terms of the l o a d q, we
set o u t a sector of the w e d g e as a free b o d y a n d write the equilibrium
of the forces in the OX3 direction. Since the stress distribution for b o t h
p l a n e strain a n d p l a n e stress is the s a m e as far as orr
, OQQ, a n d or0 are
c o n c e r n e d , we shall consider a w e d g e of unit length in the OX2
direction. T h u s ,
2
q +
j
orr
cos
9{rd0)
= q + 2C
I
cos 9d9
= 0
(14.13.4)
and
C =
-
1
(14.13.5)
2a + sin 2a'
Therefore, the c o m p o n e n t s of the state of stress in the wedge a r e :
2q cos 9
9
ogg = 0,
r(2a 4- sin 2a)
°r0
=
°9z
=
°rz =
°-
ozz =
-
2qv cos 9
r(2a + sin 2a)
(14.13.6)
(14.13.7)
F o r a equal to I I / 2, Eqs. (14.5.6) to (14.5.9) are o b t a i n e d .
If the load is n o r m a l to the axis of the wedge (Fig. 14.22), the s a m e
stress function c a n b e used p r o v i d e d 9 is m e a s u r e d from the direction
The Semi-Infinite Elastic M e d i u m
423
Fig. 1 4 . 2 2
of the force. T h e c o n s t a n t C is f o u n d from the e q u a t i o n of e q u i l i b r i u m :
q +
/•?+«
I
a r cro s 0(rdO) = 0.
(14.13.8)
Thus,
q
c = - 2 a — sin 2 a
(14.13.9)
and
2q cos 0
a rr — —r(2a
— sin 2a)'
r0
a
=
=
°0z
°rz
=
a 7T =
2qv cos 0
- A*(2a — sin 2 a ) (14.13.10)
(14.13.11)
0-
T h e case of inclined loads at the a p e x of a w e d g e c a n easily b e studied
b y the superposition of the t w o previous cases.
If a w e d g e is cut from a thin plate, the sets of E q s . (14.13.6), (14.13.7),
(14.13.10), a n d (14.13.11) are valid except t h a t ozz = 0. It is of interest to
c o m p a r e the e q u a t i o n s in this case to those of the e l e m e n t a r y t h e o r y of
b e a m s : I n cartesian c o o r d i n a t e s , we h a v e [see E q s . (7.11.4) to (7.11.6)]
4
an =
2qxx x 3s i n 0
xx(2a — sin 2 a )
(14.13.12)
424
The Theory of Elasticity
4
2qx\ s i n 0
a 13
(14.13.13)
— sin 2a)
xx(2a
2
2
2qxx x 3s i n 0 c o s 0
xx(2a — sin 2 a )
a 33
(14.13.14)
E x p a n d i n g sin 2 a into a p o w e r series, we h a v e :
sm
3
so t h a t 2 a - sin 2 a = ( 2 a ) / 6 for small values of a.
If I22 is the m o m e n t of inertia of a n y cross section MN,
a n d (14.13.13) c a n b e written a s :
Eqs. (14.13.12)
(14.13.15)
(14.13.16)
4
F o r small3 values
of a, 0 is nearly e q u a l to I I / 2 a n d the factor
( t a n a / a ) s i n 0 is nearly e q u a l t o unity. T h e expression for o n b e c o m e s
equal to t h a t of the e l e m e n t a r y theory. T h e m a x i m u m shearing stress
o c c u r s at M a n d N a n d is twice as large as t h e s h e a r i n g stress t h e
e l e m e n t a r y theory gives for the center of a triangular b e a m with a
r e c t a n g u l a r cross section. F r o c h t [9] gives a series of g r a p h s illustrating
this situation.
14.14 M . Levy's Problems of the Triangular and Rectangular Retaining
Walls
Let us first consider a triangular retaining wall or d a m subjected to a
pressure linearly increasing with d e p t h (Fig. 14.23). T h e b o u n d a r y
conditions are:
1) O n O B,
Xx = 0
a 13 = 0.
ou =
3
(14.14.1)
ot = 0,
(14.14.2)
-yx
2) O n O A,
xx = x 3t a n /?,
on = 0,
The Semi-Infinite Elastic M e d i u m
425
Fig. 14.23
w h e r e y c a n b e l o o k e d u p o n as the unit weight of the liquid. Since the
n u m b e r of b o u n d a r y c o n d i t i o n s is four, a p o l y n o m i a l of the third degree
w o u l d p r o v i d e a suitable stress function (see Sec. 9.11). C o n s i d e r the
Airy stress function:
=dx\
^
-6~
+ ex\x3
~ 2 ~
+jxxx\
~X-
+ kx\
~T-
(14.14.3)
F r o m E q s . (9.9.4), we h a v e :
oxx = fxx + kx3 - pgx3
(14.14.4)
o33 = dxx + ex3 — pgx3
(14.14.5)
oX3 = ~exx — fx3,
(14.14.6)
w h e r e pg is the weight p e r unit v o l u m e of the m a t e r i a l of the wall. T h e
first set of b o u n d a r y c o n d i t i o n s (14.14.1) gives:
/ = 0
k = -y
(14.14.7)
+ pg.
(14.14.8)
Thus,
o„ = -y^3
(14.14.9)
a 33 = dx\ + ex3 — pgx3
(14.14.10)
al3 = -exl.
(14.14.11)
426
The Theory of Elasticity
T h e direction cosines of the n o r m a l n (Fig. 14.23) with OXx a n d OX3 a r e
(cos
— sin /?). A p p l y i n g the t r a n s f o r m a t i o n of axes of Sec. 7.11, we
get:
2
2
— 7 * 3 c o s / ? + 2exx sin /? cos ft (14.14.12)
on = (dxx + ex3 — pgx3)sin fi
^ _ ^-yx3-dxl-ex3
^_^ ^
xy
Pg
3
+
n
{HMU)
Setting X[ = x 3t a n /? a n d e q u a t i n g to zero, we o b t a i n the two e q u a t i o n s :
d t a n /3 + 3e = p g +
Z
2
tan /5
/ 22t atan
n / 3/?
\
d t a n /? + e\ ~
+ 1 ) = + p g - y.
(14.14.14)
(14.14.15)
Solving, we get:
e =
Y2
tan /?'
d =
Pg
tan/?
2y3
tan /?'
(14.14.16)
T h e stresses n o w a s s u m e the final form:
°u =
a33
-(
° 13 =
(14.14.17)
-7*3
pg
t a n ji
2y3
tan /?
pg)x3
2
(14.14.18)
(14.14.19)
tan /?
(14.14.20)
a 22 = v{au + a 2 ) 3
Fig. 1 4 . 2 4
T h e Semi-Infinite Elastic Medium
o n = *23 = 0.
427
(14.14.21)
T h e d i a g r a m s of stresses a 33 a n d a 13 over any h o r i z o n t a l section are
shown in Fig. 14.24. T h e e l e m e n t a r y theory gives the s a m e a n s w e r for
a 3 , 3 b u t a n essentially different o n e for a 1 . 3
F o r a r e c t a n g u l a r retaining wall (Fig. 14.25), the b o u n d a r y c o n d i t i o n s
are:
Fig. 14.25
1) o n O B,
x
0
\ =
a
i 3 = °>
°\\ = - 7 * 3
(14.14.22)
2) o n C D ,
xx=a
a 13 = 0,
a n= 0
(14.14.23)
x3 = 0
a 33 = 0,
a 13 = 0.
(14.14.24)
3) o n O C,
By taking a stress function in the form of a p o l y n o m i a l of the sixth
degree, M . Levy [11] d e d u c e d the following expressions for the stresses:
•,.--1*3(1 ~ i)
x
(1 + ^ r )
)
y4(^
/4x?
A
dxf
12.,
1 4) 2 6
A
-
(
428
The Theory of Elasticity
( l - ^ - ) ]
a21 = {au
v
(14.14.27)
(14.14.28)
+ 03) 3
(14.14.29)
= 0.
Eq. (14.14.27) does n o t satisfy the last b o u n d a r y c o n d i t i o n , w h i c h
specifies t h a t at the t o p of the wall a 13 = 0. I n d e e d , if we set x 3 = 0 in
Eq. (14.14.27), we get:
T h e s e stresses, however, r e d u c e to a b a l a n c e d system of forces since
their resultant is e q u a l to z e r o :
(14.14.31)
T h e y h a v e , therefore, only local significance a c c o r d i n g to Saint-Ven a n t ' s principle, a n d their i m p o r t a n c e is limited in practice since they
act at the t o p of the wall w h e r e a stress analysis is n o t usually required.
T h e expression of a 33 o b t a i n e d from the e l e m e n t a r y theory of b e a m s is:
(14.14.32)
which c o n t a i n s only the two first t e r m s of E q . (14.14.26).
REFERENCES
[1] R. F. Scott, Principles of Soil Mechanics, Addison-Wesley, Reading, Mass., 1963.
[2] E. Jahnke and F. Emde, Tables of Functions, Dover, N e w York, N . Y., 1945.
[3] N . M. Newmark, "Influence Charts for Computation of Stresses in Elastic Foundations,"
Engineering Experiment Station Bulletin, Series N o . 338, University of Illinois, Urbana,
Illinois, 1942.
[4] N . M. Newmark, Influence Charts for Computation of Vertical Displacements in Elastic
Foundations, Engineering Experiment Station Bulletin, Series N o . 367, University of
Illinois, Urbana, Illinois, 1947.
[5] A. E. H. Love, "The Stress Produced in a Semi-Infinite Solid by Pressure on Part of the
Boundary," Transactions of the Royal Society, London, Series A, Vol. 228, 1929.
[6] J. Boussinesq, Application des Potentiels, Gauthier-Villars, Paris, 1885.
[7] D . L. Holl, "Plane Distribution of Stress in Elastic Media," Engineering Experiment
Station
Bulletin, 148, Iowa State College, Ames, Iowa, 1941.
[8] S. Timoshenko and J. N . Goodier, Theory of Elasticity,
McGraw-Hill, N e w York, N . Y.,
1970.
[9] M. M. Frocht, Photoelasticity, Vol. 2, John Wiley & Son, N e w York, N . Y., 1948.
[10] M. Sadowsky, Zietschriftangew. Math. Mech., Vol. 8, p. 107, 1928.
[11]M. Levy, Comptes Rendus, Vol. 126, p. 1235, 1898.
CHAPTER 15
ENERGY PRINCIPLES AND
INTRODUCTION TO
VARIATIONAL METHODS
15.1
Introduction
In Sec. 8.20, it was stated t h a t the solution of a n elasticity p r o b l e m
a m o u n t e d to solving a system of 15 e q u a t i o n s with 15 u n k n o w n s .
V a r i o u s m e t h o d s of solution w e r e listed, a n d a m o n g t h e m the v a r i a t i o n al m e t h o d s w h i c h are b a s e d o n the fact t h a t the g o v e r n i n g o p e r a t i o n s of
elasticity c a n b e o b t a i n e d as a direct c o n s e q u e n c e of the m i n i m i z a t i o n
of a n energy expression. E n e r g y is a n invariant, i.e., a q u a n t i t y
i n d e p e n d e n t of the c o o r d i n a t e system of reference. It is a scalar a n d as
such is easy to m a n i p u l a t e . T h e use of m e t h o d s b a s e d o n energy avoids
the task of h a v i n g to solve, in a direct way, the fifteen partial differential
e q u a t i o n s of elasticity. T h e basis of the variational f o r m u l a t i o n is the
principle of virtual w o r k e n u n c i a t e d b y J o h n Bernoulli in 1717. T h i s
principle states t h a t if a particle is in e q u i l i b r i u m u n d e r n forces Qx,
Q2,
<2„, the total virtual w o r k d o n e d u r i n g a n y a r b i t r a r y virtual
d i s p l a c e m e n t of the particle is zero. A solid b o d y at rest m a y b e
c o n s i d e r e d as consisting of a system of particles in equilibrium u n d e r
the a c t i o n of surface a n d b o d y forces. T h e difference b e t w e e n a particle
a n d a solid b o d y is t h a t d u r i n g the virtual d i s p l a c e m e n t s the c o n t i n u i t y
of the m a t e r i a l as well as the b o u n d a r y c o n s t r a i n t s m u s t b e o b s e r v e d .
By expressing the virtual d i s p l a c e m e n t s in t e r m s of c o n t i n u o u s functions, the c o n d i t i o n of c o n t i n u i t y of the m a t e r i a l is satisfied. T h e
429
430
The Theory of Elasticity
b o u n d a r y constraints generally d e p e n d o n the type of structure. T h e
ends of a simple b e a m o n two supports, for example, c a n n o t m o v e in
the transverse direction so that the virtual displacement there m u s t be
taken as zero.
In this chapter, various energy t h e o r e m s for a solid c o n t i n u o u s b o d y
will b e derived, a n d a n u m b e r of simple examples will b e p r e s e n t e d to
illustrate their use. A brief i n t r o d u c t i o n to the calculus of variations is
also included.
15.2
Work, Strain and Complementary Energies. Clapeyron's Law.
Fig. 15.1
Let us consider a d e f o r m a b l e b o d y fixed at points A a n d B (Fig. 15.1).
T h e b o d y is acted u p o n b y a system of generalized forces Qx, Q2,
...,
Qn. A s a result of the generalized forces, generalized d i s p l a c e m e n t s qx,
q2, . . . , qn are p r o d u c e d . T h e t e r m "generalized d i s p l a c e m e n t s " is used
to m e a n b o t h linear displacements along the forces a n d a n g u l a r
rotations a b o u t a line p e r p e n d i c u l a r to the p l a n e of the couples. T h e
p r o d u c t of a generalized force b y a generalized d i s p l a c e m e n t represents
w o r k : W h e n b o t h generalized force a n d generalized d i s p l a c e m e n t h a v e
the s a m e direction, the w o r k is positive; w h e n they h a v e o p p o s i t e
directions, the w o r k is negative.
Let us n o w consider a b o d y acted u p o n b y o n e force Q (Fig. 15.2) T h e
d a s h e d line represents the d e f o r m e d s h a p e of the b o d y after Q h a s
r e a c h e d its final m a g n i t u d e . T h e relationship b e t w e e n Q a n d q is s h o w n
Energy Principles
431
Fig. 15.2
in Fig. 15.3. F o r a n infinitesimal i n c r e m e n t dq, the i n c r e m e n t in w o r k
d o n e b y Q is
dW=
Qdq.
(15.2.1)
A s s u m i n g negligible t e m p e r a t u r e c h a n g e s , this w o r k is stored u n d e r the
form of i n t e r n a l strain energy, so t h a t
dUt = dW=
Qdq,
(15.2.2)
w h e r e Ut is the strain energy. T h e total strain energy,
U.-tfQdq,
05.2.3)
432
The Theory of Elasticity
is given b y the area u n d e r the curve in Fig. 15.3. T h e a r e a over the curve
is given b y
U* = I/'
qdQ,
(15-2.4)
w h e r e 17* is called the c o m p l e m e n t a r y energy. If the relation b e t w e e n
Q a n d q is linear, the strain energy a n d the c o m p l e m e n t a r y energy a r e
numerically equal (Fig. 15.4). I n such a case, we c a n write:
Q = Kg,
(15.2.5)
where A' is a c o n s t a n t called the stiffness coefficient. Alternatively, we
c a n write:
q = CQ,
(15.2.6)
w h e r e C is a c o n s t a n t called the flexibility or c o m p l i a n c e coefficient.
Substituting Eq. (15.2.5) i n t o Eq. (15.2.3), the expression of the strain
energy for a linearly elastic b o d y b e c o m e s :
Ut = \ K q ^ \ Q . q
(15.2.7)
Eq. (15.2.7) is k n o w n as C l a p e y r o n ' s Law. [ C o m p a r e to Eq. (8.7.7)]. If
we substitute Eq. (15.2.6) into Eq. (15.2.4), the expression of the
c o m p l e m e n t a r y energy for a linearly elastic b o d y b e c o m e s :
Energy Principles
433
2
(15.2.8)
U*=±CQ .
Eqs. (15.2.6), (15.2.7), a n d (15.2.8) s h o w t h a t
(15.2.9)
K=C-K
T h e previous e q u a t i o n s c a n b e e x t e n d e d to a linearly elastic b o d y
subjected t o a system of ./V generalized forces. T h e i n c r e m e n t in strain
energy dU, is given b y :
dUt=
2
m=l
(15.2.10)
Qm
dqm
.
E a c h o n e of t h e generalized forces causes a generalized d i s p l a c e m e n t at
its p o i n t of a p p l i c a t i o n as well as a t all o t h e r points of t h e b o d y .
Therefore, w e c a n write:
qn = <lniQ\>Qi
(15.2.11)
Qn)
and
- J5T*.
+
—
+
(,52A2)
Since t h e b o d y is linearly elastic, t h e partial derivatives a r e c o n s t a n t s .
W e shall call t h e m t h e flexibility or c o m p l i a n c e influence coefficients:
C nm = J^L
(15.2.13)
dQ '
m
W e see t h a t Cnm is t h e d i s p l a c e m e n t at t h e p o i n t n c a u s e d b y a u n i t
force at t h e p o i n t m. E q . (15.2.12) c a n b e i n t e g r a t e d to give:
Qx + Cn2
Q2
qn = CnX
2
CnN
QN
=
m=\
Cnm
Qm
.
(15.2.14)
Alternatively, we c a n write:
Qn = Qn(qx,Qi
qN
)
(15.2.15)
and
G„=
S
m=l
Knm
gm
,
(15.2.16)
434
The Theory of Elasticity
where
K
— 30,
(15.2.17)
Mm
T h e Knm ' s are called the stiffness influence coefficients. Knm r e p r e s e n t s
the force acting at n w h i c h causes a unit d i s p l a c e m e n t at m. W e shall
p r o v e in a later section t h a t Cnm = Cmn a n d t h a t Knm = Kmn
.
Eqs.
(15.2.14) a n d (15.2.16) c a n , respectively, b e written in m a t r i x f o r m as
follows:
C,2
9\
42
Q2
=
JN_
-
•N2
-
Q/v
Qx
^2N
Qi
QNN
QN
(15.2.18)
or
(15.2.19)
{<?} = [ C ] { 0 ;
~Qx~
Qi
_QN_
^11 12
A^2
K
=
2\
AT^j
A,
A
2V
i
K,NN
K,N2
1\
<l2
(15.2.20)
IN
or
{Q} =
[K]{q).
(15.2.21)
F r o m E q s . (15.2.19) a n d (15.2.21), we d e d u c e t h a t
1
[ C ] - = [K].
(15.2.22)
[C] a n d [K] are called the flexibility m a t r i x a n d the stiffness m a t r i x ,
respectively.
T h e total strain energy c a n b e e v a l u a t e d b y integrating Eq. (15.2.10),
k e e p i n g in m i n d t h a t the b o d y u n d e r l o a d b e h a v e s in a linear a n d
conservative way. This strain energy d e p e n d s only o n the final state of
force a n d d i s p l a c e m e n t a n d n o t o n the w a y this state h a s b e e n r e a c h e d :
In o t h e r words, it is i n d e p e n d e n t of the p a t h of integration. T h e
i n t e g r a t i o n is simplified b y setting:
(15.2.23)
Energy Principles
435
w h e r e Bn is a factor of p r o p o r t i o n a l i t y a n d q is a function varying from
0 to 1. T h e linearity of the b e h a v i o r justifies the use of Eq. (15.2.23),
since the generalized d i s p l a c e m e n t s qn m a i n t a i n the s a m e p r o p o r t i o n at
all times. Substituting Eqs. (15.2.16) a n d (15.2.23) into Eq. (15.2.10) a n d
integrating, we get:
2
l/, = J !
2
Bm
Bnq
Kmn
= \
^ m = l n=\
i
§
^ m=\
Kmn
qm
qn.
(15.2.24)
n=\
Therefore,
Z
U, = \l
,2=1
Q„qn.
(15-2.25)
This f o r m u l a is C l a p e y r o n ' s L a w w h e n the linearly elastic b o d y is
subjected to a system of generalized forces a n d d i s p l a c e m e n t s .
F o l l o w i n g a similar reasoning, the c o m p l e m e n t a r y energy is given b y :
U*t =Z\ 2
m=\
2 Cmn
QmQn.
(15.2.26)
n=\
C l a p e y r o n ' s L a w c a n b e stated as follows: T h e w o r k d o n e b y a system
of forces acting o n a linearly elastic b o d y is i n d e p e n d e n t of the r a t e at
w h i c h the forces increase a n d the o r d e r in which they are applied. T h e
w o r k d o n e , which is stored as strain energy, is e q u a l to one-half the
p r o d u c t of the final m a g n i t u d e s of the generalized forces a n d their
c o r r e s p o n d i n g generalized d i s p l a c e m e n t . T h e b o d y m u s t b e initially
stress free a n d n o t subjected to t e m p e r a t u r e c h a n g e .
15.3
Principle of Virtual W o r k
C o n s i d e r a b o d y w h i c h is in a state of static equilibrium u n d e r the
action of surface forces a n d b o d y forces (Fig. 15.5). T h e s e forces result
in stresses a n d d e f o r m a t i o n s which satisfy the differential e q u a t i o n s of
equilibrium, the s t r a i n - d i s p l a c e m e n t relations, a n d the stress-strain
relations w h i c h are n o t necessarily elastic. N o w s u p p o s e t h a t the b o d y
is subjected to a small c h a n g e in s h a p e c a u s e d b y s o m e source o t h e r
t h a n the applied forces. S u c h a c h a n g e in s h a p e is called a virtual
distortion, a n d the w o r d virtual is used here to indicate t h a t the
distortion is i n d e p e n d e n t of the a c t u a l system of forces acting o n the
b o d y . O w i n g to this c h a n g e in shape, a n e l e m e n t of v o l u m e inside the
b o d y is d e f o r m e d , translated, a n d r o t a t e d . T h e stresses acting o n this
436
The Theory of Elasticity
yy^**i
^
n
j
3
Fig. 15.5
e l e m e n t m o v e a n d , therefore, d o virtual work. T h e d i s p l a c e m e n t of e a c h
e l e m e n t m u s t b e consistent with the geometrical c o n s t r a i n t s of the b o d y .
Let w l? u2, a n d u3 b e the c o m p o n e n t s of the actual d i s p l a c e m e n t of a n
e l e m e n t d u e to the surface a n d b o d y forces, a n d let Su{, 6w 2, a n d 8u3 b e
the c o m p o n e n t s of its virtual d i s p l a c e m e n t . T h e s e c o m p o n e n t s of the
virtual d i s p l a c e m e n t are a s s u m e d to b e infinitesimally small quantities
satisfying the c o n d i t i o n s of c o n t i n u i t y of the d e f o r m a t i o n , i.e., they a r e
c o n t i n u o u s functions of xx, x2, a n d x3. A d d i t i o n a l l y , the virtual
d i s p l a c e m e n t c a n n o t affect the equilibrium of the external forces a n d
their internally i n d u c e d stresses. Since the e l e m e n t is in equilibrium, the
r e s u l t a n t of the forces a c t i n g o n it is e q u a l to zero. I n a virtual
d i s p l a c e m e n t Sw, these forces d o work, b u t the n e t result is zero.
Therefore,
(15.3.1)
Eq. (15.3.1) c a n b e rewritten as follows:
Energy Principles
437
///[
3
+
Su
g^(21 l + <*222 +
a
w
5
8w
°23 3)
+ af^(31l + 32 2 + 33 3)] ^
a
8w
a
+ / / / ( ^ f i w i
5w
+ ^ ^ i
+ a
(
5w
+ ^ f i ^ ) ^
(15.3.2)
/ / / K°" 4 ° 4 4 )
4 4 4) "
+
=
a
+a22
21
+031
+a32
8
8ui
2
> 2
U s i n g the divergence t h e o r e m a n d recalling E q s . (7.3.8), the left-hand
side of E q . (15.3.2) c a n b e w r i t t e n :
8u
/ /
S
8u
(°n\ \
+ °n2 2
8
+
°n3
udS
3)
8
+ f f f
v
+ F28u2 +
(F\ U\
F38u3)dV.
R e c a l l i n g t h e definition of t h e linear strains given b y E q s . (1.2.1), w e
can write:
(15.3.3)
+
a n d t h e r i g h t - h a n d side of E q . (15.3.2) b e c o m e s :
/ / /fallen
8e
+
V
°2\ 2\
+ o32
8e32
+31^31 +
a
5
a
8e
°\2 \2
+
22^22
+ oX3
8eX3 + a 2 S3e 23 + o 3 &3? 3 ] 3
</K
Eq. (15.3.2) n o w a s s u m e s the form (in i n d e x n o t a t i o n ) :
/ /
{anM)dS
+ fff
{FM)dV
= / / /
(oySe^dV.
(15.3.4)
438
The Theory of Elasticity
T h e left-hand side of E q . (15.3.4) is physically interpreted as the virtual
work d o n e by the external forces, a n d the r i g h t - h a n d side as the virtual
w o r k d o n e b y the internal stresses. A l t h o u g h the b o d y forces act within
the v o l u m e of the b o d y , they are c o n s i d e r e d as external forces in the
sense that they cause stresses a n d strains. Eq. (15.3.4) is the m a t h e m a t ical s t a t e m e n t of the principle of virtual work. I n w o r d s , this principle
m a y b e stated as follows: If a b o d y is in equilibrium a n d r e m a i n s in
equilibrium while it is subjected to a virtual distortion c o m p a t i b l e with
the geometrical constraints, the virtual w o r k d o n e b y the external forces
is equal to the virtual work d o n e b y the internal stresses. It is to b e
noticed t h a t this principle holds i n d e p e n d e n t l y of the stress-strain
relations of the m a t e r i a l of the b o d y .
15.4
Variational Problems and Euler's Equations
Let us consider the p r o b l e m of d e t e r m i n i n g a function of y (x) which
m a k e s the integral
(15.4.1)
a n e x t r e m u m , i.e., stationary, a n d which satisfies the prescribed e n d
conditions,
(15.4.2)
y(x)x=b
=y(b)
y
Fig. 15.6
=yb.
(15.4.3)
Energy Principles
439
T h e value of t h e integral in E q . (15.4.1) d e p e n d s o n t h e choice of
y = y(x), h e n c e t h e n o t a t i o n / (y). T h e i n t e g r a n d F(x,y,/)
is a function
of x,y a n d t h e first derivative dy/dx = y'. L e t u s a s s u m e t h a t >> = y(x) is
the a c t u a l m i n i m i z i n g function ( F i g . 15.6), a n d find o u t w h a t h a p p e n s
to t h e integral / w h e n n e i g h b o r i n g functions a r e u s e d in F. T h e s e
functions a r e c o n s t r u c t e d b y a d d i n g t o y (x) a function er](x), w h e r e e is
a c o n s t a n t w h i c h c a n take different q u a n t i t a t i v e values a n d TJ (x) is a n
a r b i t r a r y function of x w h i c h vanishes at x = a a n d at x = b; t h a t is:
i){a) = 0
and
r)(b) = 0.
(15.4.4)
Therefore, y(x) + ey(x) will satisfy t h e e n d c o n d i t i o n s . If w e n o w
replace y(x) b y y(x) + e-q(x), the integral / b e c o m e s a function of e o n c e
T] (x) is c h o s e n , since y is the a c t u a l m i n i m i z i n g function. Therefore, w e
c a n write:
I(e)
= £
f(x,y
+ eq,/
+ eq^dx
(15.4.5)
This integral takes its m i n i m u m value w h e n e is zero, b u t this is possible
only if
^
de
= 0,
w h e n e = 0.
(15-4.6)
T o simplify t h e c o m p u t a t i o n s , w e set:
Fe = F(x,y
+ z<r),y' + cr?')
(15.4.7)
and
Y(x)
= y(x) + er](x).
(15.4.8)
Differentiating Y with respect to x, a n d Fe with respect to e, we get:
+
Y' =/(x)
EJ]\X)
(15.4.9)
and
§-I*> §^>+
Therefore, from E q s . (15.4.5) a n d (15.4.6), w e h a v e :
<
I5A,O
>
440
The Theory of Elasticity
^-/.'[$*>+M*-*
'"
(,5 4
)
T h e s e c o n d t e r m of Eq. (15.4.11) c a n b e t r a n s f o r m e d , t h r o u g h integration b y p a r t s , as follows:
f
- [$*<>]! - r
*<{w)
Since 17(0) = r}(b) = 0,
*•
< , 5 A i 2 )
< 1 5) A 1 3
a n d E q . (15.4.11) b e c o m e s :
A c c o r d i n g to E q . (15.4.6), this derivative m u s t vanish w h e n e a p p r o a c h es 0. But as e a p p r o a c h e s 0, Fe a p p r o a c h e s F, Y a p p r o a c h e s y, a n d Y
a p p r o a c h e s y '. Therefore, E q . (15.4.14) b e c o m e s :
r [if°-
It is possible to p r o v e rigorously that, if Eq. (15.4.15) is true for a n y
function TJ (x) which is twice differentiable in the interval (a,b) a n d zero
at the e n d s of t h a t interval, the coefficient of TJ (x) in the i n t e g r a n d m u s t
b e zero everywhere in (a b) [1] T h u s
.
,
9 Z _ A( <¥L.\ =
(15.4.16)
Q
dy
dx \ dy' /
Eq. (15.4.16) is called Euler's e q u a t i o n a n d is a necessary c o n d i t i o n for
y (x) to m a k e / a m i n i m u m or a m a x i m u m . In c o m p u t i n g d/dx it m u s t
b e r e m e m b e r e d t h a t y a n d y' are functions of x. T h u s ,
9
Energy Principles
441
Euler's e q u a t i o n b e c o m e s :
2
2
2
2
dF _ d F _ d F dy _ d^Fd^y
3/
dxdy'
3>>3/ dx
d/
dx
=
Q
(15.4.17)
T h i s is a s e c o n d - o r d e r differential e q u a t i o n . Its solution c o n t a i n s t w o
a r b i t r a r y c o n s t a n t s . T h e y a r e to b e d e t e r m i n e d b y t h e r e q u i r e m e n t t h a t
the curve passes t h r o u g h t h e e n d - p o i n t s A a n d B.
If t h e i n t e g r a n d in E q . (15.4.1) c o n t a i n s the s e c o n d derivative y",
Euler's e q u a t i o n b e c o m e s [1]:
2
+ AL|£
_ AM
dy
dx dy'
dx
= o
(15.4.18)
9/'
Eq. (15.4.18) c a n b e generalized to include higher derivatives of y.
Let u s n o w i n t r o d u c e t h e n o t a t i o n of v a r i a t i o n a n d establish t h e
a n a l o g y b e t w e e n t h e differential calculus a n d the calculus of variations.
W h i l e in t h e first w e deal with t h e differential of a function a l o n g a
p a r t i c u l a r curve, in t h e s e c o n d we deal with a v a r i a t i o n of a functional
from curve t o curve [1]. T o define a functional, let us consider a set S
of functions satisfying certain c o n d i t i o n s . A n y q u a n t i t y which takes o n
a specific n u m e r i c a l value c o r r e s p o n d i n g t o each function in S is said to
b e a functional o n t h e set S. T h u s ,
(15.4.19)
" F(x,y,/)dx
is a functional since, c o r r e s p o n d i n g to a n y function y (x), I takes a
definite n u m e r i c a l value. W i t h i n t h e s a m e context, it is justifiable to call
such quantities as
f[Ax)]>
g[x,y{x\y\x\
...
,/»>(*)]
f u n c t i o n a l in those cases w h e n t h e variable x is c o n s i d e r e d as fixed in
a given discussion a n d the f u n c t i o n ^ (x) is varied. T h u s , in E q . (15.4.1),
we h a v e c o n s i d e r e d a n i n t e g r a n d of t h e f o r m :
F=F(x9y,/)9
(15.4.20)
which for a fixed value of x d e p e n d s o n t h e function y (x) a n d its
derivative. It is, therefore, a functional.
442
The Theory of Elasticity
In Eq. (15.4.8), the c h a n g e ej](x) in y (x) is called t h e variation of y,
a n d is conventionally d e n o t e d b y Sy,
8y = eq(x).
(15.4.21)
C o r r e s p o n d i n g to the c h a n g e 8y in the function y, F c h a n g e s b y a n
a m o u n t AF, w h e r e
f
AF = F(x,y
+
+ eq') - F(x,y,y ).
(15.4.22)
If the r i g h t - h a n d m e m b e r is e x p a n d e d in p o w e r s of e, there follows:
AF
er
= l^ ? + f^7 7'
67
+ t e s r wm
(
^ h higher p o w e r s of e).
(15.4.23)
By a n a l o g y with the definition of the differential, the first t w o terms
in the r i g h t - h a n d m e m b e r of E q . (15.4.23) a r e defined to b e the variation
of F,
S ^ = | ^
+ GW.
(15.4.24)
W h e n F = y9 this definition is consistent with E q . (15.4.21) a n d w h e n F
= y , it yields:
8 / = eq',
(15.4.25)
so t h a t E q . (15.4.24) c a n b e written in the form:
SF=^Sy
(15.4.26)
+ ^S/.
F o r a c o m p l e t e a n a l o g y with the definition of the differential, o n e
w o u l d have a n t i c i p a t e d the definition:
8F
=^ py w 8x +
+
sy
(15A27)
But, in the m a n i p u l a t i o n s of F, x is n o t varied, so that:
Sx = 0
(15.4.28)
Energy Principles
443
a n d the a n a l o g y b e t w e e n differential a n d v a r i a t i o n is c o m p l e t e . F r o m
the definition, it follows t h a t the laws of variation of sums, p r o d u c t s ,
ratios, a n d so forth, are completely a n a l o g o u s to those of differentiation. T h u s , for e x a m p l e ,
8(FXF2)
(15.4.29)
= FX8F2 + F28FX
F28FX-FX8F2
8FX=
5
)4
F{
^2
w h e r e Fx a n d F2 are two different functions of x, y, a n d y'. F r o m E q s .
(15.4.21) a n d (15.4.25), we d e d u c e t h a t
T h a t is, the o p e r a t o r s 8 a n d d/dx are c o m m u t a t i v e . Let us n o w t u r n to
Eq. (15.4.1), a n d show t h a t the necessary c o n d i t i o n for / to b e s t a t i o n a r y
is t h a t its first v a r i a t i o n vanishes; t h a t is,
rb
81=
I
8F(x,y,/)dx
(15.4.32)
= 0.
Ja
Indeed,
/>*-/.*
<i5A33)
I n t e g r a t i n g the s e c o n d t e r m b y p a r t s , we h a v e :
(15.4.34)
Thus,
•b
b
r
3
0
444
The Theory of Elasticity
a n d this is precisely w h a t was o b t a i n e d in Eq. (15.4.15) [see E q .
(15.4.13)]. T h u s , a stationary function for a n integral function is o n e for
which the variation of t h a t integral is zero, j u s t as a stationary p o i n t of
a function is o n e at which the differential of the function is zero.
In the m o r e general case, w h e n the function to b e m i n i m i z e d or
m a x i m i z e d is of the form
F(x,y,
w,v, ux,u
vx,v
)dxdy,
(15.4.36)
w h e r e ux, uy, vx, vy indicate differentations of u a n d v with respect to x
a n d y, the c o n d i t i o n for a n e x t r e m u m is again
SI = 0.
(15.4.37)
H e r e x a n d y are i n d e p e n d e n t variables; the region R c o u l d b e a b o d y
in a state of p l a n e stress so t h a t u a n d v could b e the d i s p l a c e m e n t s . T h e
c o n d i t i o n (15.4.37) t h e n b e c o m e s :
w
-//[(f^" i^"- f>)
+
+
y
x
R
H e r e the v a r i a t i o n s 8u a n d 8v a r e to b e c o n t i n u o u s l y differ en tiable over
the region R a n d are to vanish over its b o u n d a r y w h e n u a n d v are
prescribed o n S, b u t otherwise a r e completely a r b i t r a r y . T o t r a n s f o r m
the terms involving the variations of the derivatives, we m a k e use of
G r e e n - R i e m a n n ' s t h e o r e m (see A p p e n d i x A-10.1). T h e general p r o c e d u r e m a y b e illustrated b y c o n s i d e r i n g the t r e a t m e n t of a typical t e r m .
If /j a n d jf2 are the direction cosines of the n o r m a l n to the b o u n d a r y S
(Fig. 15.7):
R
- II TM ") " "y - II s(sf )«"** <
s
R
x
R
15A3
Energy Principles
Following
becomes:
the s a m e p r o c e d u r e
for
the other
terms, Eq.
«-£ [(f'. ^>» (i? ^'>]
+
+
/ +
l
5
445
(15.4.38)
rfs
<5
i A4o>
+
If the v a r i a t i o n s 8u a n d 8v are i n d e p e n d e n t of e a c h o t h e r — t h a t is, if u
a n d v c a n b e varied i n d e p e n d e n t l y — t h e coefficients of 8u a n d 8v m u s t
e a c h vanish identically in R, giving the t w o Euler e q u a t i o n s :
+f ( M ) - f
\ 3w x /
f
dy \ dw^ /
+ f ( f
dx \ 3 ^ /
) - f
3j> \ di^ /
=0
(15.4.41)
=0.
(15.4.42)
du
df
W h e n w is n o t p r e s c r i b e d o n S,
+
0 o n S,
(15.4.43)
= 0z o n 5 .
(15.4.44)
a n d w h e n f is n o t p r e s c r i b e d o n 5 ,
1
+ 1^4
3^
3i^
446
The Theory of Elasticity
P
a
2
P
Fig. 15.8
15.5
The Reciprocal Laws of Betti and Maxwell
C o n s i d e r a linearly elastic b o d y which will b e l o a d e d in t u r n b y t w o
different systems of generalized forces: the Q system a n d the P system
(Fig. 15.8). T h e n u m b e r of the Q forces is L a n d t h a t of the P forces is
K. T h e d i s p l a c e m e n t s of the p o i n t s of application of the Q forces c a u s e d
b y the P forces is d e n o t e d b y A ^ . Similarly, the deflections of the
points of a p p l i c a t i o n of the P forces c a u s e d b y the Q forces is d e n o t e d
b y A j p . Let us apply the principle of virtual w o r k to o n e system of
forces at a time, while a s s u m i n g t h a t the virtual surface d i s p l a c e m e n t s
a n d the virtual internal strains are c a u s e d by the o t h e r system. T h u s , the
virtual w o r k d o n e b y the Q system is:
(15.5.1)
w h e r e 8e$> are the internal virtual strains c a u s e d b y the P forces a n d
of$ are the internal stresses c a u s e d by the Q forces. Similarly,
(15.5.2)
w h e r e Se^
are the internal virtual strains c a u s e d b y the Q forces a n d
oj£} are the internal stresses c a u s e d b y the P forces. F r o m Eq. (8.3.16),
we h a v e :
= S
(15.5.3)
Energy Principles
447
and
5^(0 = s
a<2)
(15.5.4)
T h e t w o E q s . (15.5.1) a n d (15.5.2) b e c o m e :
1 Qi*P
2
-
/
^S^c^dV
/ /
- / / /
o&S^atPdV.
(15-5.5)
(15-5.6)
T h e t w o integrals in the two last e q u a t i o n s are the s a m e b e c a u s e of the
symmetry property:
^mnrs
(15.5.7)
^rsmn'
Therefore,
2 G/AP= |
/=1
P tfp.
k
(15.5.8)
/c=l
Eq. (15.5.8) is the m a t h e m a t i c a l s t a t e m e n t of Betti's law: In a n y b o d y
which is linearly elastic, the w o r k d o n e b y a system of Q forces u n d e r a
d i s t o r t i o n c a u s e d b y a system of P forces is e q u a l to the w o r k d o n e b y
the system of P forces u n d e r a distortion c a u s e d b y the system of Q
forces. If we h a v e o n e Q force a n d o n e P force w h i c h are n u m e r i c a l l y
equal, we get (Fig. 15.9):
F i g . 15.9
448
The Theory of Elasticity
QA\V = QAP.
(15.5.9)
T h u s , if we d e n o t e the deflection at a p o i n t m c a u s e d b y a load at p o i n t
n b y A m„ , a n d the deflection at p o i n t n c a u s e d b y the s a m e m a g n i t u d e
of l o a d at p o i n t m b y A „ m, t h e n Eq. (15.5.9) b e c o m e s , in general:
A m„ = A „ m.
(15.5.10)
T h i s is the general f o r m u l a for M a x w e l l ' s law of reciprocal deflections:
In a n y b o d y w h i c h is linearly elastic, the generalized deflection at p o i n t
m c a u s e d b y a generalized l o a d Q at p o i n t n is n u m e r i c a l l y e q u a l to the
generalized deflection at p o i n t n c a u s e d b y the s a m e m a g n i t u d e of
generalized force Q at p o i n t m.
If we n o w t u r n to Eq. (15.2.18) we see, for e x a m p l e , t h a t C 12 is the
d i s p l a c e m e n t at p o i n t 1 d u e to a unit l o a d at p o i n t 2, a n d that C 21 is the
d i s p l a c e m e n t at p o i n t 2 d u e to a unit l o a d at p o i n t 1. A c c o r d i n g to
Maxwell's law, these t w o d i s p l a c e m e n t s m u s t b e e q u a l ; so that,
C 12 = C 2 . 1
(15.5.11)
In the s a m e way,
C 13 = C 3 , 1 C 14 = C 4 , 1 etc.;
(15.5.12)
in o t h e r w o r d s , the m a t r i x [C] is s y m m e t r i c ; consequently, the m a t r i x
[K] is also s y m m e t r i c .
15.6
Principle of Minimum Potential Energy
I n this section, we i n t r o d u c e a functional called the potential energy
of d e f o r m a t i o n a n d p r o v e t h a t it attains a n a b s o l u t e m i n i m u m w h e n the
d i s p l a c e m e n t s of the b o d y are those of the state of equilibrium. This
principle, which is called the principle of m i n i m u m p o t e n t i a l energy,
m a y b e l o o k e d u p o n as a special case of the principle of virtual w o r k :
It lies at the basis of several direct variational m e t h o d s of solution of
elastostatic p r o b l e m s . W e shall use the s y m b o l 5 to m e a n the v a r i a t i o n
of a function a c c o r d i n g to the calculus of variations (Sec. 15.4). L e t a
b o d y b e in equilibrium u n d e r the action of specified b o d y a n d surface
forces. A t a given p o i n t ( x 1? x 2, x 3) , Oy a n d ey are the states of stress
Energy Principles
449
e
x x respectively. Se^ represents a n y small c h a n g e of the function
a n d x strain
ij ( \> 2> 3)
satisfying the compatibility a n d b o u n d a r y c o n d i t i o n s . If a
strain energy density function U{etj) exists, so t h a t (see Sec. 8.2),
ij
t h e n the r i g h t - h a n d t e r m of Eq. (15.3.4):
v
1
iin
)= >-
=8
udv
V
su
(15.6.2)
In Eq. (15.3.4) oni a n d F( are c o n s t a n t s , so t h a t the principle of virtual
w o r k c a n n o w b e written a s :
> - UIJ
SU
S
- /// '»< - // ""A
UdV
F
dV
(,5.6.3)
= 0.
It is c u s t o m a r y to set:
W=
j
j
j
v
j
FiUl
dV+
j
(15.6.4)
oni
utdS,
s
so t h a t E q . (15.6.3) b e c o m e s :
8Up = 8(Ut T h e scalar
the b o d y :
W) = 0.
(15.6.5)
defined b y E q . (15.6.3) is called the p o t e n t i a l energy of
Up=Ut-W=
j
jv J
UdV-
j
j
J
FiUi
dV
(15.6.6)
~ j
j
oniUi
dS.
s
Eq. (15.6.5) states t h a t : A m o n g all the geometrically possible states of
d i s p l a c e m e n t satisfying the given b o u n d a r y c o n d i t i o n s , those w h i c h
satisfy the e q u a t i o n s of e q u i l i b r i u m results in a s t a t i o n a r y value of the
450
The Theory of Elasticity
p o t e n t i a l energy. T o show t h a t
m u s t b e a m i n i m u m for a state of
stable equilibrium, consider a n e i g h b o r i n g state TVp c h a r a c t e r i z e d b y
strains eu + 8eu a n d the c o r r e s p o n d i n g d i s p l a c e m e n t s ut + 8ut. H e n c e ,
n ; - up = j
J j
[U{etj + 8ey)
-
u{eij
)]dv
V
~ 111
8 F dVu
i
i
~ Jj
(15.6.7)
oni
8utdS.
E x p a n d i n g U{etj + 8etj) i n t o a p o w e r series, a n d neglecting t e r m s with
a n o r d e r higher t h a n the second, w e get:
U(ev +
8ev)
= U(ev)
$%8e„
+
+
•
(15-6.8)
A s u b s t i t u t i o n i n t o E q . (15.6.7) yields:
(15.6.9)
T h e s u m of the three first t e r m s in the r i g h t - h a n d side of Eq. (15.6.9)
vanishes o n a c c o u n t of E q . (15.3.4). T h e last t e r m is positive for
sufficiently small values of 8etj. T h i s c a n b e seen as follows: Let us set
etj = 0 in E q . (15.6.8). T h e c o n s t a n t t e r m c a n b e d i s r e g a r d e d since we
are only interested in derivatives of U. dU/de^j =
must vanish when
etj = 0. T h e r e f o r e , u p to the s e c o n d o r d e r :
^H^ ^* '2
5
1
--
(15 6 10)
a n d if U{8eij) is positive definite, t h e n 1/2 ( 3 U / d e y d e ^ 8etj8ekl is
positive definite. T h e r e f o r e 11^ — Iip is positive a n d 11^ is a m i n i m u m .
T h e principle of m i n i m u m p o t e n t i a l energy c a n n o w b e stated as
follows: A m o n g all the geometrically possible states of d i s p l a c e m e n t
satisfying t h e given b o u n d a r y c o n d i t i o n s , t h o s e w h i c h satisfy the
e q u a t i o n s of e q u i l i b r i u m results in a m i n i m u m value of the p o t e n t i a l
energy.
Energy Principles 4 5 1
It is of interest to s h o w t h a t the variational principle (15.6.3) gives the
e q u a t i o n s of elasticity. I n d e e d ,
s
n
»-
fJ'h ^
F
i77(f *)"•8c
V
- J j
dv
J
V
(15.6.11)
(O n,.S M
;)JS=0.
But
(15.6.12)
J
K
V
U s i n g t h e divergence t h e o r e m , we get:
(15.6.13)
(JO:,
T/
•/
Hence,
=
+
J/7
/
/ ^
- °^
) 8 dS
u+
>
(15.6.14)
= 0,
which c a n b e satisfied for a r b i t r a r y 8ut if
^ + F
OXj
a n d either
=0
in1
K,
(15.6.15)
452
The Theory of Elasticity
8ut = 0
o n S (where the surface d i s p l a c e m e n t s are prescribed)
(15.6.16)
or
JJ
o n S (where the surface stresses are prescribed) .
(15.6.17)
T h e b o u n d a r y c o n d i t i o n s which are prescribed in the p r o b l e m at h a n d
are called the forced b o u n d a r y c o n d i t i o n s : F o r e x a m p l e at the fixed e n d
of a cantilever b e a m , the forced b o u n d a r y c o n d i t i o n s are m a t h e m a t i c a l ly expressed b y writing 8u2 = 0 a n d 8(du2 /dxx) = 0. O n the other h a n d ,
some b o u n d a r y c o n d i t i o n s m a y be d e d u c e d as necessary c o n d i t i o n s for
m i n i m u m p o t e n t i a l energy. Such c o n d i t i o n s are called n a t u r a l b o u n d a r y
c o n d i t i o n s (See P r o b l e m 8).
T h u s we h a v e d e m o n s t r a t e d the o n e - t o - o n e c o r r e s p o n d e n c e b e t w e e n
the differential e q u a t i o n s of equilibrium a n d the variational e q u a t i o n :
W e h a v e derived Eq. (15.6.3) from the e q u a t i o n s of equilibrium [starting
from Eq. (15.3.1)] a n d then h a v e s h o w n that, conversely, Eqs. (15.6.15)
to (15.6.17) necessarily follow Eq. (15.6.3). It is to b e noticed t h a t at n o
time was linearity of the stress-strain relations of the b o d y i n v o k e d : All
t h a t was r e q u i r e d w a s the existence of the strain energy density
function. T h u s , the principle is valid for a n y elastic b o d y , linear or
nonlinear.
If the b o d y is linearily elastic, t h e n :
ay = 2 ^
+ \8tj enn
(15.6.18)
and
2
Ut=
J j
j
[i*^
+ |(0 ]
dV.
(15.6.19)
v
T o satisfy the c o n d i t i o n of c o n t i n u i t y of d i s p l a c e m e n t s over the entire
b o d y , the function ey m u s t b e expressed in terms of the d i s p l a c e m e n t
functions ui b y
a n d the variation of the strain energy Ut m u s t b e carried o u t with
respect to the d i s p l a c e m e n t functions.
Energy Principles
15.7
453
Castigliano's First Theorem
C o n s i d e r a b o d y in equilibrium u n d e r a set of generalized external
. If the strain energy Ut is expressed as a function of the
forces Qm
c o r r e s p o n d i n g d i s p l a c e m e n t s qm
, the principle of virtual w o r k c a n be
used to show t h a t
|£=G
05.7.1)
m=\,2,...,N.
m
T h e proof consists of allowing a virtual d i s p l a c e m e n t Sq to take place in
the b o d y in such a m a n n e r t h a t Sq is c o n t i n u o u s everywhere, b u t
. D u e to Sq, the strain
vanishes at all p o i n t s of l o a d i n g except u n d e r Qm
energy c h a n g e s b y a n a m o u n t 8Ut while the w o r k d o n e b y the external
forces is the p r o d u c t of Qm times 8qm
, i.e., Qm
Sqm
.
A c c o r d i n g to the
principle of virtual work,
(15.7.2)
8Ut = Qm
8qm
.
W h e n rewritten in differential form, Eq. (15.7.2) b e c o m e s Eq. (15.7.1).
Therefore, Castigliano's first t h e o r e m states t h a t : W h e n the strain
energy Ut c a n b e expressed as a function of a system of generalized
d i s p l a c e m e n t s qm
, t h e n the generalized force Qm is given b y
dUt/dqm
.
It is i m p o r t a n t to notice t h a t this t h e o r e m h a s b e e n p r o v e d using the
a s s u m p t i o n of the existence of Ut{qm
) a n d using the principle of virtual
w o r k as applied to a state of equilibrium. Linearity h a s n o t b e e n
invoked, a n d the t h e o r e m is applicable to elastic b o d i e s t h a t follow a
n o n l i n e a r l o a d - d i s p l a c e m e n t relationship.
15.8
Principle of Virtual Complementary Work
In c o n t r a s t to Sec. 15.3 w h e r e we h a d a s s u m e d a v a r i a t i o n of the
d e f o r m a t i o n s (virtual d e f o r m a t i o n s ) , let us a s s u m e here a variation of
the stresses (virtual c h a n g e in the stresses) in a b o d y held in equilibrium
u n d e r the b o d y forces per unit v o l u m e Ft a n d the surface forces per unit
a r e a oni
. Let otj b e the actual stress field w h i c h satisfies the e q u a t i o n s of
e q u i l i b r i u m a n d the b o u n d a r y c o n d i t i o n s .
^
+ ^. = 0
=
°ni
(15.8.1)
mV
n
o
the surface S.
(15.8.2)
454
The Theory of Elasticity
Let us consider a system of stress variations which also satisfy the
e q u a t i o n s of equilibrium a n d the stress b o u n d a r y conditions,
dXj
+ 8E = 0
(15.8.3)
(Sa y)/,.
7 = S a m-
(15.8.4)
T h e c o m p l e m e n t a r y virtual w o r k is given b y :
J j
j
j
(Ui
8F})dV+
v
j
s
uMidS
(15.8.5)
j
V
where
j
S
a r e the c o m p o n e n t s of the real d i s p l a c e m e n t s . T h u s ,
j
j
V
+ j
(uidFi)dV
j
s
uMmdS
- Ill ( °>£) ~ III
8
dV
i
r
^
W J*
(15.8.6)
J
V
V
:dS
J
s
a n d , using the divergence t h e o r e m , we get:
J j
j
j
{ui8Fi)dV+
V
j
s
u.SomdS
-IIJ{^w)
dv
< ->
l58 7
J
V
Therefore,
u 8
/ / / (
i
Fd
i )
v
+
j
J
ut8oni
dS=
j
f
f
eySoydV.
(15.8.8)
Energy Principles
455
E q . (15.8.8) m a y b e called the principle of virtual c o m p l e m e n t a r y work.
T h i s principle m a y b e s t a t e d as follows: T h e virtual c o m p l e m e n t a r y
w o r k d o n e u n d e r the a c t u a l state of strain a n d virtual stress c h a n g e Soy,
w h i c h satisfies the differential e q u a t i o n s of equilibrium, is e q u a l to the
c o m p l e m e n t a r y w o r k d o n e b y the virtual external forces. Eq. (15.8.8) is
applicable w i t h o u t restrictions regardless of the n a t u r e of the stressstrain relations g o v e r n i n g the b e h a v i o r of the b o d y .
This principle is c o m p l e m e n t a r y to the principle of virtual work. E a c h
represents a n alternative a p p r o a c h to the solution of m e c h a n i c s p r o b lems. W h i l e in the principle of virtual w o r k the quantities to b e varied
are the d i s p l a c e m e n t s , in the principle of virtual c o m p l e m e n t a r y w o r k
the quantities to b e varied are the i n t e r n a l stresses a n d the external
forces.
15.9
Principle of Minimum Complementary Energy
This principle m a y b e l o o k e d u p o n as a special case of the principle
of virtual c o m p l e m e n t a r y work. Let us a s s u m e the existence of a
function of the stresses called the c o m p l e m e n t a r y energy function
U*(o{j) with the p r o p e r t y t h a t [ c o m p a r e to Eq. (15.2.4)]:
d
u
* = .. e
day
v
(15 9 1)
U^-U
T h e principle of virtual c o m p l e m e n t a r y w o r k m a y b e written a s :
j
j Vj
+ j
(ui8Fi)dV
j
8oni
dS
Ui
ni^ = ffj * -
=
dv
d
u
S
(15.9.2)
dv
v
y
v
Since the v o l u m e is fixed a n d ut are n o t varied, E q . (15.9.2) c a n b e
written a s :
SIT* =
8[fff
U*dV
-
v
= 8(U* -
fff
v
W) = 0.
utFtdV
-
ff
s
oni
dS]
Ui
(15.9.3)
456
T h e Theory of Elasticity
T h e q u a n t i t y Tt* is a function of the stresses ay, of the surface forces
a n d of t h e b o d y forces: It is called the c o m p l e m e n t a r y energy a n d is
defined b y :
v
v
U* -
=
(15.9.4)
s
W.
W e , therefore, h a v e the following t h e o r e m : Of all the states of stress
w h i c h satisfy the e q u a t i o n s of e q u i l i b r i u m a n d b o u n d a r y c o n d i t i o n s
w h e r e stresses a r e prescribed, the actual o n e is t h a t w h i c h m a k e s the
c o m p l e m e n t a r y energy s t a t i o n a r y . T h e proof t h a t I I * is a m i n i m u m is
a n a l o g o u s to t h a t in Sec. 15.6:
Il*(Oi,-
+ SOy) ~ U*(Oy)
= / / /
v
-
[U*(oy
+ SOy) -
[o (o +
Say) -
U*((Jy)]
dV
1595
/ /
s
nl iJ
oMj^idS
< --)
-ffJ«,8F,dV.
V
E x p a n d i n g U*(<jy + Soy) in p o w e r series, a n d neglecting the terms of a n
o r d e r higher t h a n the second, we get:
n < ( . s + Sof) - n * ( « s)
-jfj
+ff f
y
y
-jf
s
[Soy^dS
-
j jJ
kl
J
(15.9.6)
*
oFtdV.
Ui
The quantity
a n d the o t h e r terms in the r i g h t - h a n d side of E q . (15.9.7), are e q u a l to
zero b e c a u s e of E q . (15.9.3). H e n c e ,
II*(oy
+ Say) - U*(ay)
= U*(8o&
).
(15.9.8)
Energy Principles
457
If U*(8oij) is positive definite, w e h a v e t h e following t h e o r e m : Of all
states w h i c h satisfy t h e e q u a t i o n s of e q u i l i b r i u m a n d b o u n d a r y c o n d i tions w h e r e stresses a r e p r e s c r i b e d , t h e a c t u a l o n e is t h a t w h i c h m a k e s
the c o m p l e m e n t a r y energy m i n i m u m . It is to b e n o t i c e d t h a t a t n o time
was linearity of the stress-strain relations of the b o d y i n v o k e d . All t h a t
w a s r e q u i r e d w a s the existence of t h e c o m p l e m e n t a r y energy function.
T h u s , t h e p r i n c i p l e is valid for a n y elastic b o d y , linear or n o n l i n e a r .
If the b o d y is linearly elastic, t h e n
^
(15-9.9)
-v8vam
]
= l [ ( l +v)aIJ
and
2
U*t=Ut=
J j
j
^ [ ( 1 + vfrjoy
- v(ann
) ]dV.
V
(15.9.10)
U s i n g t h e M a x w e l l a n d M o r e r a stress functions (see Sec. 9.8), it c a n
b e s h o w n [2] t h a t if 8H* = 0 for all v a r i a t i o n s of stresses Soy w h i c h
satisfy the e q u a t i o n s of e q u i l i b r i u m in the b o d y a n d o n the b o u n d a r y
w h e r e stresses a r e p r e s c r i b e d , t h e n Oy also satisfies the e q u a t i o n s of
compatibility.
15.10 Castigliano's S e c o n d T h e o r e m
Let us c o n s i d e r a linearly elastic b o d y subjected to TV generalized
forces Qm
, a n d a s s u m e t h a t the i n t e r n a l stresses h a v e b e e n expressed in
t e r m s of the generalized forces. T h e c o m p l e m e n t a r y energy IT* c a n b e
written as:
n* =
U,(Q1,Q2
QN
)
- 2
m=\
Qm
qm
,
(1510.1)
w h e r e qm is the generalized d i s p l a c e m e n t c o r r e s p o n d i n g to Qm
. According to the principle of m i n i m u m c o m p l e m e n t a r y energy,
SII* = 0 =
2
m=l
^ S Q m. 0(
(15.10.2)
^m
Since e a c h v a r i a t i o n 8Qm is i n d e p e n d e n t
(15.10.2) yields yV e q u a t i o n s :
of the others, t h e n
Eq.
458
T h e Theory of Elasticity
an*
3ft
= 0
m = 1, 2,
N.
(15.10.3)
N.
(15.10.4)
A p p l y i n g E q . (15.10.3) to E q . (15.10.1), we get:
9m
=
3a
m = 1, 2,
Eq. (15.10.4) is referred to as Castigliano's s e c o n d t h e o r e m , which
states: I n a linearly elastic structure, t h e partial derivative of t h e strain
energy with respect to a n externally applied generalized force is e q u a l
to t h e d i s p l a c e m e n t c o r r e s p o n d i n g to that force. This t h e o r e m is often
used in t h e c o m p u t a t i o n of t h e deflection of structures.
15.11 Theorem of Least Work
T h e strain energy Ut for a statically i n d e t e r m i n a t e linearly elastic
structure c a n n o t b e written in terms of t h e applied external forces Qm
a l o n e . It c a n , however, b e expressed in t e r m s of t h e external forces a n d
a n u m b e r of r e d u n d a n t internal forces o r reactions. If t h e s t r u c t u r e
X2
XN
), i n t e r n a l or
c o n t a i n s /V i n d e p e n d e n t r e d u n d a n t forces (X]9
at the b o u n d a r y , t h e expression for t h e c o m p l e m e n t a r y energy is:
n * = Ut(Ql9
Q2
Qm9
Xl9
X2
- —XN
) - Qm
qm
.
(15.11.1)
T h e principle of m i n i m u m c o m p l e m e n t a r y energy yields t w o e q u a t i o n s :
(15.11.2)
and
(15.11.3)
Energy Principles
459
Eq. (15.11.2) is Castigliano's s e c o n d t h e o r e m , while E q . (15.11.3) is
referred to as t h e t h e o r e m of least work. This t h e o r e m states: F o r a
statically i n d e t e r m i n a t e linearly elastic structure, t h e derivative of t h e
strain energy with respect to a n y r e d u n d a n t r e a c t i o n m u s t b e zero.
15.12 S u m m a r y of Energy T h e o r e m s
Before illustrating t h e use of t h e t h e o r e m s derived in t h e previous
sections, it is w o r t h w h i l e to p r e s e n t t h e m in a w a y t h a t brings o u t t h e
duality t h a t exists b e t w e e n strain a n d c o m p l e m e n t a r y energies:
Principle of virtual work
[Principle of complementary virtual work
Any Elastic Stress-Strain Relations
Strain Energy function U exists:
Complementary energy function U* exists:
i
Principle of minimum complementary energy
Principle of minimum potential energy
1
1
Castigliano's first theorem
Linear Elastic Stress-Strain Relations
Reciprocal law of Betti and Maxwell
Castigliano's second theorem
Theorem of least work
T h e t h e o r e m s b a s e d o n t h e principle of virtual w o r k a n d those b a s e d o n
the principle of c o m p l e m e n t a r y virtual w o r k offer t w o different a p p r o a c h e s t o t h e solution of elasticity p r o b l e m s . T h e a p p l i c a t i o n of these
t h e o r e m s to t h e s a m e p r o b l e m s will yield t h e s a m e answers if t h e
p r o b l e m c a n b e solved exactly, b u t if a p p r o x i m a t e answers a r e desired
the different a p p r o a c h e s will, in m a n y cases, yield slightly different
answers. T h e r e a d e r is referred t o W a s h i z u ' s treatise [3] for a detailed
e x a m i n a t i o n of t h e interrelations a m o n g t h e various t h e o r e m s .
460
The Theory of Elasticity
15.13 Working Form of the Strain Energy for Linearly Elastic Slender
Members
T h e n a m e , slender m e m b e r , is given to a solid g e n e r a t e d b y a p l a n e
section of c o n t o u r C a n d area A w h o s e c e n t r o i d G describes a curve S
in s p a c e : T h e r a d i u s of c u r v a t u r e of this curve is large c o m p a r e d to the
d i m e n s i o n s of A, a n d the solid is g e n e r a t e d in a way such t h a t the
principal axes of inertia of the cross section t h r o u g h its c e n t r o i d
m a i n t a i n a c o n s t a n t angle with the principal n o r m a l a n d the b i n o r m a l
of S. I n this m a n n e r , a n y p o i n t of the cross section will describe a fiber
of the solid parallel to S. In Fig. 15.10, the OXx axis is t a k e n t a n g e n t to
5 , a n d OX2 a n d OX3 are the principal axes of inertia t h r o u g h the
c e n t r o i d G. S is called the m e a n line of the m e m b e r .
In this section, we shall c o m p u t e the expression of the elastic strain
energy in t e r m s of the n o r m a l a n d shearing forces, as well as in t e r m s
of the b e n d i n g a n d twisting m o m e n t s . T h e only stresses to b e c o n s i d e r e d
are those acting o n the surface A: n a m e l y , a n, a 1 , 2 a n d a 1 ; 3 a 2 , 2 ^ 3 3 ,
a n d a 23 are a s s u m e d e q u a l to zero. W e shall first e x a m i n e the p a r t of
the strain energy c o n n e c t e d with the n o r m a l stresses, t h e n e x a m i n e the
p a r t c o n n e c t e d with the shearing stresses.
a. Parts of the strain energy due to ou
F o r a m e m b e r of length L, a n d cross section A, Eq. (8.7.13) gives:
« -0/ [ / / ^ * ] *
<i5i3i)
A
If the m e m b e r is subject to a n o r m a l force N, a n d to b e n d i n g m o m e n t s
Af 12 a n d M 13 at a distance S from the origin O (see Fig. 15.10), the
Energy Principles
461
n o r m a l stress a n is given b y :
TV
X3
+ Ml2
^ 1 3 * 2+
(15.13.2)
w h e r e I2 a n d I3 are the m o m e n t s of inertia a b o u t the two principal axes
GX2 a n d GX3. Substituting Eq. (15.13.2) i n t o E q . (15.13.1), a n d n o t i n g
that
A
and that
A
we get:
< i 5
7) i 3
-
o
b. Part of the strain energy due to a 12 and a 13
F r o m E q . (8.7.13), we h a v e :
L
<=fU J jG(°u
U
+ oh)M]dS.
05.13.8)
0
a 12 a n d a 13 are p r o d u c e d b y a p p l i e d couples a n d b y shearing forces
which m a y or m a y n o t b e a c t i n g at the shear center. Recalling E q s .
(10.5.3), (10.5.4), a n d (12.2.5), we h a v e (Fig. 15.10):
462
The Theory of Elasticity
<
i 9) 5
'
• » - ^ - &
<,5
' '
i3
io)
w h e r e <j> is P r a n d t l ' s stress function, a n d Q2 a n d Q3 a r e static m o m e n t s
a b o u t the OX2 a n d OX3 axes, respectively. A substitution of E q s .
(15.13.9) a n d (15.13.10) into E q . (15.13.8) yields:
U>=
JG
I iff
\ I3b3
/
[(&|) (&|)
+
dx3 I3b3
+
dx2 I2b2
(W)
J
( 1 5 1 3 1 1 )
J
Let us n o w consider the various t e r m s in the r i g h t - h a n d side of E q .
(15.13.11):
A
where
A
is a p r o p e r t y of the cross section.
A
where
i-//(^)
is a p r o p e r t y of the cross section.
!rfs!
-
<i5 ,3i5)
i
3
Energy Principles
2
463
r ((1*^-1*^)^
J
AJ
\dx
3
/ D
3 3
dx
2
Ib
22
/
(15.13.16)
a c c o r d i n g to the G r e e n - R i e m a n n formula (see A p p e n d i x A-10.1). But <j>
= 0 o n the c o n t o u r C, so t h a t the s u m in the r i g h t - h a n d side of E q .
(15.13.16) vanishes.
Eq. (15.13.11) c a n n o w b e written as follows:
2G
(15.13.17)
Recalling C h a p t e r 10, we c a n write:
0
w h e r e GJ is the torsional rigidity a n d Mx xis the twisting m o m e n t a b o u t
the OXx axis. A d d i n g E q s . (15.13.7) a n d (15.13.18), we get the expression
of the strain energy in a slender m e m b e r :
u
*
2
J
+
lAE
GC2
+
GC3
+
GJ
EI2
EI3 J
^
T e m p2e r a t u r e effects c a n b e i n c l u d e d in E q . (15.13.19) b y substituting
for N /AE
the q u a n t i t y N[N/AE
+ a(AT')], w h e r e a is the coefficient of
t h e r m a l e x p a n s i o n a n d A T is the c h a n g e in t e m p e r a t u r e .
F o r a r e c t a n g u l a r cross section, the c o n s t a n t s C2 a n d C 3 are e q u a l to
5A / 6. F o r a circular a n d a n elliptic cross section, they are e q u a l to
0.9,4.
464
The Theory of Elasticity
15.14 Strain Energy of a Linearly Elastic Slender Member in Terms of
the Unit Displacements of the Centroid G and of the Unit Rotations
U n d e r the effects of a n o r m a l force N a c t i n g at a d i s t a n c e S from t h e
origin (see Fig. 15.10), the m e a n line suffers a c h a n g e in length per u n i t
length eXG
, where
Therefore, in t e r m s of
•«,-&.
05.14..)
\ T E - \ ^ A B .
(15.1«)
,
eXG
F r o m the e l e m e n t a r y t h e o r y of b e a m s , w e k n o w t h a t the angle of
r o t a t i o n per unit length d u e t o b e n d i n g , ft, is given b y :
=
(15-14.3)
Therefore,
i t = i(ft>^,
\£-W*h.
<'.4.4>
5
I n C h a p t e r 10, it w a s s h o w n t h a t the angle of twist p e r unit length is
given b y :
„a ~
M
"
GJ '
(15.14.5)
Therefore,
IM
2 GJ
I ( aW
) 2=G;
'
2
U J
(15.14.6)
D u e t o a s h e a r i n g force Vt (i = 2 or 3), the c e n t r o i d G of A m o v e s a
d i s t a n c e eiG p e r unit length of S a l o n g a direction n o r m a l to S (Fig.
15.11). U s i n g E q s . (15.13.13) a n d (15.13.14), the w o r k p e r u n i t length
d o n e b y V2 a n d V3 is
Energy Principles
465
hi
F i g . 15.11
Therefore,
e G2
~
V
(15.14.8)
2
GCi
and
let A ^f
GC
(15.14.9)
Substituting the previous e q u a t i o n s i n t o E q . (15.13.19), we get:
L
Ut = \j
2
2
2
[EA(elG
)
+ GC2(e2G
)
+ GC3(e3G
)
+ £/ ( 8 ) + ^/ ( 83) ]^.
2
2
y
2
2
+ GJ(a)
^ J )Q
2
3
i
2
)
the
T e m p e r a t u r e effects c a n b e i n c l u d e d by substituting for EA(elG
q u a n t i t y EA elG
[elG + a ( A T ) ] , w h e r e a is the coefficient of t h e r m a l
e x p a n s i o n a n d A T is the c h a n g e in t e m p e r a t u r e .
15.15 A Working Form of the Principles of Virtual Work and of Virtual
Complementary Work for a Linearly Elastic Slender Member
I m a g i n e the b o d y s h o w n in Fig. 15.12 to b e subjected to a generalized
force F t h a t is i m a g i n a r y a n d n o t a p a r t of a n y force system o n the
b o d y : £ i s a virtual force. N o w s u p p o s e t h a t the b o d y u n d e r g o e s a small
c h a n g e in s h a p e as the result of s o m e a c t i o n o t h e r t h a n F. This c h a n g e
in s h a p e m a y b e d u e to a system of real a p p l i e d loads, c h a n g e s in
466
The Theory of Elasticity
Fig. 1 5 . 1 2
t e m p e r a t u r e , or o t h e r causes. Consistent with the c h a n g e in s h a p e ,
internal distortions will result such as elongations, r o t a t i o n s , a n d
shearing distortions. Let y b e the real distortion of a differential e l e m e n t
a n d let A b e the a c t u a l d i s p l a c e m e n t of a p o i n t A in the direction of the
virtual force F . Also l e t / b e the n o r m a l stress, m o m e n t , t o r q u e , or shear
stress a c t i n g o n the differential e l e m e n t as a result of the application of
F at A. A c c o r d i n g to the principle of virtual c o m p l e m e n t a r y work,
(15.15.1)
FA is the w o r k d o n e b y the i m a g i n a r y force F m o v i n g a distance A a l o n g
its line of action. If F = 1,
(15.15.2)
Therefore, to d e t e r m i n e a d i s p l a c e m e n t of a p o i n t A in a n y direction,
apply a l o a d F at the p o i n t in the desired direction. T o d e t e r m i n e the
a b s o l u t e direction of m o v e m e n t as well as the m a g n i t u d e , d e t e r m i n e
separately the m o v e m e n t in t w o o r t h o g o n a l directions. If it is desired to
d e t e r m i n e the r o t a t i o n at A, apply a virtual couple. Let us n o w apply
Eq. (15.15.1) to slender m e m b e r s :
F o r axial forces in b e a m s or m e m b e r s of a truss:
L
£A =
o
KeiG
dS
(15.15.3)
0
Energy Principles
467
w h e r e N is the n o r m a l force d u e to £ a n d eXG is the elongation p e r u n i t
length of the real system at a d i s t a n c e S from the origin (see Fig. 15.11).
F o r b e n d i n g m o m e n t s in b e a m s , frames, a n d a r c h e s :
L
£
= Aj
^£,
0
M
05-15.4)
w h e r e M is the b e n d i n g m o m e n t d u e to the a p p l i c a t i o n of £ , a n d
is the angle of r o t a t i o n p e r unit length of the real system.
F o r a t o r q u e Mx : x
M/EI
0
w h e r e Mxx is the twisting m o m e n t resulting from the a p p l i c a t i o n of F,
a n d Mxx /GJ is the angle of twist per unit length of the real system.
F o r shear:
L
FA=
J
0
V^§,
(15-15.6)
w h e r e V is the s h e a r i n g force d u e to the a p p l i c a t i o n of F, a n d V/GC is
the d i s p l a c e m e n t p e r unit length of the gentroid G along V.
A d d i n g Eqs. (15.15.3) to (15.15.6), we get the general w o r k i n g
relation:
It is to b e n o t i c e d t h a t the d i s p l a c e m e n t s c a u s e d b y the virtual force F
are a s s u m e d to b e very small, so t h a t the w o r k d o n e b y the real system
d u e to the d i s p l a c e m e n t s i n d u c e d b y F is negligible.
A l t h o u g h we h a v e applied a n i m a g i n a r y force F to o u r elastic system
a n d used the principle of virtual c o m p l e m e n t a r y work, we c a n consider
F as being the only force of the real system a n d consider the real
468
The Theory of Elasticity
d i s p l a c e m e n t s of the b o d y as i m a g i n a r y d i s p l a c e m e n t s : I n this case, the
principle of virtual w o r k yields Eq. (15.15.7). In the following sections,
the use of Eq. (15.15.7) will b e illustrated b y various examples.
15.16 Examples of Application of the Theorems of Virtual Work and
Virtual Complementary Work
A
f
\\\
\\oc
\ v—
v\\\ /
V
C
«/'
'
I
VL /
Q
'
L
1
1
-1
.
3_
<74
Fig. 1 5 . 1 3
1. Statically indeterminate
truss. Let us consider the p r o b l e m of a
p l a n e truss with three m e m b e r s a c t e d o n b y a l o a d Q as s h o w n in Fig.
15.13. T h e tensions in the three m e m b e r s c a n b e c o m p u t e d using the
principle of virtual w o r k a n d the c o n d i t i o n of geometrical compatibility
as follows:
„AIL
7 ~iBD = q~L
L
nr
AD
rp L
— CD
(15.16.1)
2
^AE
cos a
— 0.
l
'
(15.16.2)
w h e r e A is the cross section of the b a r s a n d E is Y o u n g ' s m o d u l u s . Let
us n o w i m p o s e a virtual d i s p l a c e m e n t 8q at D. A c c o r d i n g to the
principle of virtual work,
_ AE * ,
n» =
Q8q
q~j^8q + 2q
A£cos^a8q
j
•
(15.16.3)
Therefore,
q =
AE{\
LQ
3
4- 2 c o s a )
(15.16.4)
Energy Principles
and
469
2
TAD — T D
C
—
Q c o s a3
(15.16.5)
1 + 2 cos a
It is to b e n o t i c e d t h a t the equilibrium e q u a t i o n s were n o t i n t r o d u c e d in
the solution since they are implied in the principle of virtual work.
2. Deflection of a flexible elastic string. Let us consider a string AB
stretched b y forces S b e t w e e n fixed p o i n t s A a n d B (Fig. 15.14) a n d
l o a d e d b y a distributed vertical l o a d of intensity / , w h e r e / is a function
of xx. W e shall a s s u m e t h a t the initial tension of the string is so large
*X 2
Fig. 1 5 . 1 4
t h a t the increase in tensile force d u e to a d d i t i o n a l stretching d u r i n g t h e
deflection c a n b e neglected. T h e deflection itself is also a s s u m e d to b e
small. T h e length of the deflected string is given b y :
L
0
L
0
(15.16.6)
T h e stretching of the string is:
dx]
0
o
a n d the increase in the strain energy is given b y :
(15.16.7)
1
470
The Theory of Elasticity
1
,2
0
56 1
^
1
0
T h e principle of virtual w o r k gives the following e q u a t i o n :
0
0
C a l c u l a t i n g the v a r i a t i o n in the r i g h t - h a n d side of Eq. (15.16.9), we find:
L
L
o
o
(15.16.10)
0
I n t e g r a t i n g b y p a r t s , a n d r e m e m b e r i n g t h a t a t the e n d s of the string
8x2 = 0, we get:
L
2 0
(15.16.11)
2
d x2
-~ fl£t
* i0
2
8x
dx
Substituting Eq. (15.16.11) i n t o Eq. (15.16.9), we o b t a i n :
L
/(^M^ 0
u
1
<15
0
-
,6,2)
This e q u a t i o n will b e satisfied for a n y virtual d i s p l a c e m e n t Sx2 only if
+/(*,)-0.
(15.16.13)
This is the differential e q u a t i o n of the vertically l o a d e d string.
Energy Principles
471
3. Indeterminate
reaction in a slender member. C o n s i d e r the uniformly
l o a d e d b e a m of Fig. 15.15 with e n d A fixed a n d e n d B simply
s u p p o r t e d . It is desired to d e t e r m i n e the reaction at B. T w o e q u a t i o n s
of equilibrium c a n b e written for such a b e a m , n a m e l y
1
P
<£r-r=
-l-I
I T T
B&X,
L
Fig. 15.15
R
R
A + B=
L
(15.16.14)
P
P
(15.16.15)
- £ - R BL = MA.
T h e s e t w o e q u a t i o n s are n o t sufficient to d e t e r m i n e the three u n k n o w n s
RA, RB, a n d MA. T o a p p l y E q . (15.15.17), let us consider a n y section at
a d i s t a n c e x{ from the origin O. A t such a section, we h a v e :
N_
AE
'
up ( L - x ) ~ R
J ^ =
l
B
GC2
GC2
'
JS_
GC3
(15.16.16)
'
Rl
^LL = n
GJ
'
MXi U
_1 B(
EI3
^12=0
EI2
'
~ *x) ~
EI3
J (15.16.17)
2
T h e s e c a n b e c o n s i d e r e d as the i m a g i n a r y d i s p l a c e m e n t s of the auxiliary
system s h o w n in Fig. 15.16. T h i s system consists of a fixed e n d b e a m
subjected to a force F at B. Therefore,
N = 0,
M , , = 0,
Z3= 0
(15.16.18)
M 13 = + £ ( L - x , ) ,
(15.16.19)
V2 = -F,
M i 2 = 0,
v
472
The Theory of Elasticity
4
Bi
i*2
Fig. 15.16
with the b o u n d a r y c o n d i t i o n t h a t B d o e s n o t m o v e (A = 0 at B).
Substituting E q s . (15.16.16) to (15.16.19) i n t o E q s . (15.15.7), we get:
EA
f f-E
p ( L - x x ) - R B\
1
= ) { - ^ c 2
F(L —
Xj)
+
RB(L
(15.16.20)
p(L-xxf
— X])
]}
dxx = 0.
EL
I n t e g r a t i n g E q . (15.16.20), we o b t a i n :
J
L8EL 3 ^
2GC2S
L
,
J}_
GC2
3£Y 3
(15.16.21)
If we neglect the shear t e r m s in E q . (15.15.7), t h e n :
3pL
(15.16.22)
4. Displacements
in a slender member. T h e d i s p l a c e m e n t of the e n d
p o i n t of a cantilever b e a m u n d e r the effect of a c o n c e n t r a t e d l o a d P
(Fig. 15.17), c a n b e o b t a i n e d b y substituting in E q . (15.15.7):
V2
GO 2
P
GC2'
M 13 _
EI3
-P(L-Xl
)
EL
(15.16.23)
Energy Principles
473
Fig. 1 5 . 1 7
Fig. 1 5 . 1 8
a n d (Fig. 15.18):
V2 = F,
M 13 = —F(L — xx),
(15.16.24)
with all the o t h e r values equal to zero. T h u s ,
* - / [ €
+
T K
0
Therefore, the d i s p l a c e m e n t of A is given b y : .
(15J6
-
25)
474
The Theory of Elasticity
T h e first t e r m of Eq. (15.16.26) is the deflection d u e to the shearing
forces, a n d the s e c o n d is d u e to b e n d i n g . T o c o m p a r e the o r d e r of
m a g n i t u d e of the two c o m p o n e n t s , let us apply Eq. (15.16.26) to a steel
b e a m with a r e c t a n g u l a r cross section (Fig. 15.19). F o r such a b e a m :
I
X3
Fig. 15.19
C7 = -pbyb-, =
G
E
A 7
2
5
and
2
2
F o r b/L = 1/10, for e x a m p l e , 3 6 / 4 L = 0.0075, w h i c h is negligible
c o m p a r e d to unity. Therefore, the a p p r o x i m a t i o n which is involved in
neglecting the deflections d u e to shear in slender m e m b e r s is quite
justified.
15.17 Examples of Application of Castigliano's First and Second
Theorems
1. Deflection of a wire. Castigliano's first t h e o r e m c a n b e used to
d e t e r m i n e the relation b e t w e e n the force Q acting a t the c e n t e r of a n
Energy Principles
475
Fig. 1 5 . 2 0
elastic wire a n d the d i s p l a c e m e n t q (Fig. 15.20). S u c h a relation is elastic
b u t non-linear. T h e strain energy stored in the stretched wire is given b y
[see E q . (15.14.2)]:
u
=
( 1 5 1 7 1 )
( 2 r )'
2 l
'
M A2
w h e r e A is the cross section of the wire, a n d A is the e l o n g a t i o n of half
the initial length 2 L . W e h a v e :
A =
y/TFT?-L
(15.17.2)
a n d , in case of small deflections,
A-4, i(I)V]-L;
+
("•'«>
or, a p p r o x i m a t e l y ,
A
i =i
(15.17.4)
Therefore, the strain energy c a n b e written a s :
£ V
(15.17.5)
a n d from the Castigliano's first t h e o r e m , w e h a v e :
( _ )W L
dq
=
1}
'
(15.17.6)
476
The Theory of Elasticity
T h e elastic n o n l i n e a r i t y of this p r o b l e m results from g e o m e t r y a n d n o t
from m a t e r i a l properties.
2. Indeterminate
reaction in a slender member. I n Fig. 15.15 of Sec.
15.16, the r e a c t i o n R Bc a n b e o b t a i n e d t h r o u g h the use of Castigliano's
s e c o n d t h e o r e m . T h e system is linearly elastic a n d E q s . (15.16.16) a n d
(15.16.17) u s e d in conjunction with E q . (15.13.19), give:
2
[P(L - xx) GC2
R]
B
(15.17.7)
Eh' 3
J
dxx.
)
2
2]
^
A c c o r d i n g to Castigliano's s e c o n d t h e o r e m ,
dU, _
—
_ n_
- q b- 0 - J
f f-[p(L-xQ-
^
^
o
+
(L
-
RB
LP
X )[R (L
XB
- xx) ET3
(l
*
J
(15.17.8)
dx}
w h i c h is the s a m e as E q . (15.16.20).
3. Deflection of curved bars. Let us a s s u m e t h a t the b a r s h o w n in Fig.
15.21 h a s a circular profile, a n d t h a t its r a d i u s is large e n o u g h so t h a t
the t e r m s d u e to shear in the energy e q u a t i o n c a n b e neglected. It is
Fig. 15.21
Energy Principles
477
desired to find the vertical a n d h o r i z o n t a l d i s p l a c e m e n t s of B u n d e r the
effect of the force P. T h e strain energy is given b y :
(
i 9)5
'
o
a n d the b e n d i n g m o m e n t M 13 at a n y section mn is:
M 13 = -PR
cos </>.
(15.17.10)
Therefore, the vertical d i s p l a c e m e n t A F of B is:
23
2
%PR
cos <p
2£/3
^
3
(15.17.11)
_ UPR
3£A
T o find the h o r i z o n t a l d i s p l a c e m e n t of 5 , we c a n a s s u m e the existence
of a h o r i z o n t a l force Q a c t i n g a t B in a d d i t i o n to P, a n d o b t a i n the value
of
(dUt/dQ)Q=0
:
M 13 = -[PR
cos $ + QR(\
- sin <#>)]
(15.17.12)
and
A =
=
" ( H ) o = o
fe/o
W
#
L = o
8^=
° 517
13)
Remark
F o r c u r v e d b a r s w h o s e cross section is large c o m p a r e d to their r a d i u s ,
the strain energy d u e to n o r m a l a n d shearing forces c a n n o t b e neglected. Recalling the n o t a t i o n used in Sec. 13.2, it c a n b e s h o w n [4] t h a t the
strain energy in such cases is given b y :
In the case of the circular a r c previously e x a m i n e d (Fig. 15.21):
M 13 = -PR
N = -P
cos <f»
(15.17.15)
cos <f>
(15.17.16)
i
7
478
T h e Theory of Elasticity
(15.17.17)
V2 = Psin<t>
and
iy-Wifrik-tt
<I5I7I8)
w h e r e R = rc is t h e r a d i u s of t h e center line, a n d e is the distance
b e t w e e n the center line a n d the n e u t r a l axis.
15.18 Examples of Application of the Principles of Minimum Potential
Energy and Minimum Complementary Energy
x3
p
Qy
l l
M
1ZEZEZE
L
x
2
.
Fig. 1 5 . 2 2
1. Deflection of the mean line of a beam. T h e energy stored in the
uniformly l o a d e d b e a m s h o w n in Fig. 15.22, is given b y [see E q .
(15.14.4)]:
L
(15.18.1)
w h e r e the p a r t d u e to shear h a s b e e n neglected. Recalling E q s . (12.2.2),
we h a v e :
Energy Principles
479
N e g l e c t i n g the b o d y forces, the w o r k d o n e b y the external forces to
b r i n g the b e a m to its final position is [see (Eq. 15.6.4)]:
L
W=
j
(15.18.3)
pu2dxx.
o
T h e p o t e n t i a l energy of the b e a m is:
L
,2
(15-18.4) 1
J^(^)
n , - U , - W -
0
T a k i n g the v a r i a t i o n of EL, we get:
L
I n t e g r a t i n g the first integral b y p a r t s :
L
2 2
2
L
_ Vd u2d(8u2)l
1
L dx\
&i
^
C d u2d (8u2)
J dx[
dx\
Jo
dW\
L
L
_ - * g d p ) _
_ dxj
3
C d(8u2)d u2
J
dxx dx}
o
dxx
dx\
+
Jo
r
1
d£
^
(15.18.6)
1
1
(15 18J)
.
d£
J
T h e c o n d i t i o n 511^ = 0 b e c o m e s :
E dI
f
o
(
p 8
u i2
d
i -df- )
X
1
(15.18.8)
L
2
2
r
2
d u2d(8u2)
_ E d I' u 2
s
1 _0
Since MX3 = —EI3d u2/dx
= 0 a n d u2 = 0 at xx = 0 a n d xx = L,
8u2 = 0 at
= 0 a n d xx = L a n d the terms inside the b r a c k e t vanish o n
the b o u n d a r y xx = 0 a n d xx = L. I n b e t w e e n the b o u n d a r i e s , 8u2 is
a r b i t r a r y a n d the c o n d i t i o n
j
(El/-^
- p)su2dxl
= 0
(15-18.9)
480
The Theory of Elasticity
is possible only w h e n
E I ^
= P,
(15.18.10)
which is the governing differential e q u a t i o n in this case (see Sec. 12.2).
2. Torsion of prismatic bars. I n C h a p t e r 10, it was s h o w n t h a t there
exists a stress function <£> such that (Fig. 15.23):
Fig. 1 5 . 2 3
the other stress c o m p o n e n t s b e i n g equal to zero. T h e variation of the
stresses is therefore equivalent to the variation of the stress function.
T h e strain energy stored in a twisted b a r is equal t o :
2
Ur-jgf
f j
V
K ° i 3) +
2
(°23) ]dV
Energy Principles
481
w h e r e R d e n o t e s t h e cross section of t h e b a r . T h e lateral faces of the b a r
are free of forces. If at the e n d x3 = 0, the d i s p l a c e m e n t s are e q u a l to
zero, t h e w o r k d o n e b y the external forces to b r i n g t h e b a r to its final
position is:
W = Iff
(oX3
ux
(15.18.13)
+ o23
u2)dxxdx^
F o l l o w i n g S a i n t - V e n a n t ' s a s s u m p t i o n s , the d i s p l a c e m e n t s ux a n d u2 a t
x3 = L a r e
u2 = axxL.
ux = —ax2L,
(15.18.14)
Hence,
f
W=aL
f
(15.18.15) 2 1
( - ^ 2 ^ - - ^ ) d x x d x 2.
R
E q . (15.18.15) c a n b e t r a n s f o r m e d b y the G r e e n - R i e m a n n formula, to
give (see Sec. 10.5):
W = 2aL
f
f
<t>dxxdx2 - aL f)
R
(15.18.16)
(<t>xx dx2 - <t>x2dxx).
C
T h e total c o m p l e m e n t a r y energy of t h e system is, therefore,
n
- « - " ' - K / / [ ( ^ )
1
+
( ^ )
J
- ^ L] ^
R
(15.18.17)
(f) <j)(xxdx2 — x2
c
+aL
dxx)
and
*TT.
m
L
2 G )
f
f
J
R
H,2 3«#> d(84>)
a^^x7
I
+ d4>
2 d(8<j>) 4
9x7^T
"
G
a1
W ]
^
^
(15.18.18)
+ a L f)
c
8<j>(xx dx2 — x2
dxx).
482
The Theory of Elasticity
Since
dx
2
dx
dx
2
x
dx
dx
x
+
™* =
\
x
-B//(§
dx
/
x
dx
2
\
S «)^^
dx
2
/
^
15.18.19)
+2G
1
z
R
15.18.20)
+
O n the b o u n d a r y C, the surface forces are prescribed w h i c h m e a n s t h a t
<j> is p r e s c r i b e d so t h a t :
8<f>
= 0
onC.
(15.18.21)
T h e line integral of Eq. (15.18.20) t h u s vanishes. Since 8<j> is a r b i t r a r y in
R, the only w a y to m a k e S I P = 0, is to h a v e :
^
dxf
^ = _+ 2 G «
ox
in*.
(15.18.22)
2
As s h o w n in C h a p t e r 10, the torsion p r o b l e m is solved o n c e a function
<t> satisfying Eq. (15.18.22) in R, a n d equal to a c o n s t a n t o n the
b o u n d a r y , is found. This c o n s t a n t c a n b e chosen e q u a l to zero, a n d in
such a case the expression for the total c o m p l e m e n t a r y energy is:
R
Eq. (15.18.23) provides us with a n o t h e r a v e n u e to a p p r o a c h the torsion
p r o b l e m . This p r o b l e m c a n b e c o n s i d e r e d as the o n e seeking the
function <f> which will m i n i m i z e the total c o m p l e m e n t a r y energy I I * a n d
which satisfies the c o n d i t i o n <f> = 0 o n the b o u n d a r y .
15.19 Example of Application of the Theorem of Least Work
C o n s i d e r the b e a m s h o w n in Fig. 15.24. T h e r e are four u n k n o w n
reactions which c a n n o t b e directly d e t e r m i n e d , since we only h a v e t w o
Energy Principles
C i I * i * * l'i
i \
483
t~i~f
- f — < - — - i
1-
Fig. 1 5 . 2 4
e q u a t i o n s of static equilibrium. W e c a n consider RB a n d
r e d u n d a n t forces. T h e e q u a t i o n s of static equilibrium give:
R A + RB + R C + MA =
(RB + 2RC)L
as the
(15.19.1)
2pL
2
(15.19.2)
2pL .
T h e s e two e q u a t i o n s c a n b e solved to give MA a n d RA in terms of the
two r e d u n d a n t forces RB a n d Rc:
RA
=
(15.19.3)
2pL — RB — Rc
2
MA = 2pL
- L(RB +
(15.19.4)
2RC).
If we t a k e into a c c o u n t the b e n d i n g terms alone, the total strain energy
of the b e a m is given b y :
<
' " J
+
J
2EI3
o
P
rlL
JL
[ i ? cx , - -^-
+ RB{xx
2EL
-L)]
dx.
(15.19.5)
484
T h e Theory of Elasticity
A c c o r d i n g to the t h e o r e m of least work,
dU,
dRB
dU,
dR
= 0 = SRB + 20RC -
MpL
(15.19.6)
= 0 = 16RC + 5RB -
\2pL.
(15.19.7)
Solving E q s . (15.19.6) a n d (15.19.7) simultaneously, we get:
RB = §pL,
Rc = 2jpL;
(15.19.8)
a n d from E q s . (15.19.3) a n d (15.19.4):
15.20 The Rayieigh-Ritz Method
T h e R a y l e i g h - R i t z m e t h o d is a general p r o c e d u r e for o b t a i n i n g
a p p r o x i m a t e solutions of p r o b l e m s expressed in variational form. T h e
p r o c e d u r e consists of a s s u m i n g t h a t the desired s t a t i o n a r y function / (x)
of a given p r o b l e m (see Sec. 15.4) is a p p r o x i m a t e d b y a c o m b i n a t i o n of
suitably c h o s e n functions satisfying the b o u n d a r y c o n d i t i o n s b u t with
u n d e t e r m i n e d p a r a m e t e r s q. T h e relevant q u a n t i t y / is t h e n expressed
as a function of the c , ' s , w h i c h are so d e t e r m i n e d t h a t the resultant
expression is stationary. Therefore, i n s t e a d of using the calculus of
variations in a t t e m p t i n g to d e t e r m i n e t h a t function w h i c h r e n d e r s /
s t a t i o n a r y with reference to all admissible slightly varied functions, w e
consider only the family of all functions of the type a s s u m e d ; we t h e n
use o r d i n a r y differential calculus to seek the m e m b e r of t h a t family for
which / is s t a t i o n a r y with reference to slightly modified functions
b e l o n g i n g to the s a m e family. T h e efficiency of this p r o c e d u r e t h u s
d e p e n d s o n the choice of the functions c o m b i n e d to p r o v i d e the
s t a t i o n a r y function y (x).
Application to the deflection of the mean line of a beam. I n Sec. 15.18,
we h a v e seen t h a t the solution of the p r o b l e m of the uniformly l o a d e d
simply s u p p o r t e d b e a m c o u l d b e o b t a i n e d b y minimizing the p o t e n t i a l
energy 1 1 ^ ; this resulted in E q . (15.18.10), which c a n b e solved to give
the deflection u2 — u2{xx).
U s i n g the Rayleigh-Ritz m e t h o d , o n e c a n
Energy Principles
485
directly o b t a i n a n a p p r o x i m a t e solution t o t h e p r o b l e m . F o r that, let us
a s s u m e a deflection curve in the form of a t r i g o n o m e t r i c series:
Ux,
u2 = a,sin —j— + a2sm
2Ux,
nllx.
—j— + . . . +a„sin
^
+
.
nlix.
(15.20.1)
= 2 «„sin—jin which ax, a2
a„ a r e u n d e t e r m i n e d p a r a m e t e r s . T h u s , t h e deflection curve is o b t a i n e d b y superposition of sinusoidal curves ( F i g . 15.25)
e a c h of w h i c h satisfies the b o u n d a r y c o n d i t i o n s of the p r o b l e m . F r o m
Eq. (15.18.4), we h a v e :
*3
Fig. 1 5 . 2 5
,2
n„
2 02>
- J
'
}>«*>•
'
E q . (15.20.1) gives:
2
d u2
dx\
2
2
n
nx,
-a,-^-sin
2
-
„,
2
n
3 a3
• 3nx,
sin L
L
n
2
•
njc,
(15.20.3)
486
The Theory of Elasticity
W h e n squared, the r i g h t - h a n d side of Eq. (15.20.3) involves t e r m s of t w o
kinds, n a m e l y :
a2>W
sin
sd a2i annn a m ^ ^
2 *®*L
sin
(15-20.4)
H o w e v e r , since,
/o
z nllx
T f
. ~
x
s i n —j— dxx =
L
J
sm
a n dL
1
(15.20.5)
nUxx .
rallxi
^
sm —^-— axj = 0 if n ^
m,
o
Eq. (15.20.2) will r e d u c e t o :
4
EI3U
V
.
«=1
For
V
22pL
X
a,
A
(15.20.6)
/i-1,3,5—
to b e a m i n i m u m , we m u s t h a v e :
dUp
da
= o
(15.20.7)
n
so t h a t
E
^ ^ a n
- ^ 1
= 0 for i, o d d
(15-20.8)
and
^ 2 3^ a „
4L
= 0 for n e v e n .
05-20.9)
Hence,
4 p L
a =
*
for K o d d
(15.20.10)
Energy Principles
487
and
an = 0 for n even .
(15.20.11)
Therefore, the deflection curve c a n b e written a s :
"2 = ^
5
2
- V s i n ^ .
05.20.12)
T h e series is rapidly c o n v e r g e n t a n d only the first few t e r m s are
necessary to give a satisfactory a p p r o x i m a t i o n . F o r xx = L / 2 , w e h a v e :
4/7L /=
4
( \u
_ j_
j_ _
\
(15.20.13)
If we t a k e the first t e r m of this series, we o b t a i n :
PL
(u.)
=
(15.20.14)
*
T h e exact a n s w e r is:
p L
\
=
*
v( " 2uW
76.8F/,'
(15.20.15)
so t h a t the error involved is only 0.26 p e r c e n t .
O t h e r e x a m p l e s c a n b e f o u n d in [5,6,7].
PROBLEMS
1.
2.
D e t e r m i n e the curve b e t w e e n t w o p o i n t s A a n d B, w h i c h b y
revolution a b o u t the OXx axis generates the surface of least a r e a .
W h a t is the Euler e q u a t i o n of the p r o b l e m
8 j
j
F(x,y,u,ux,uy)dxdy
= 0,
R
3.
in w h i c h x a n d y are i n d e p e n d e n t variables a n d u (x,y) is p r e s c r i b e d
a l o n g the closed b o u n d a r y S of the region Rl
A u n i f o r m elastic b e a m is fixed at b o t h e n d s a n d carries a linearly
increasing d i s t r i b u t e d l o a d t h a t varies from zero at o n e e n d to qQ at
the other. O b t a i n the e q u a t i o n of the deflection of the b e a m using
the energy t e r m s d u e to b e n d i n g a l o n e a n d Euler's e q u a t i o n
488
4.
5.
The Theory of Elasticity
(15.4.18). E l i m i n a t e the c o n s t a n t s of integration using the b o u n d a r y
conditions.
W h a t is the expression of the strain energy p e r unit length of a
linearly elastic thick cylinder with i n n e r a n d o u t e r radii a a n d b,
with free ends, a n d which is subjected to a n internal pressure Ptl
(See Sec. 11.2.)
A short thick cylindrical h u b is shrunk-fit o n a short shaft, so t h a t
the r a d i a l pressure at the interface is p. If the i n n e r a n d o u t e r radii
of the h u b are a a n d b, a n d if the shear m o d u l u s is G, find t h e strain
energy density at a n y r a d i u s r of the h u b .
9
A
B
{.V
L
Fig. 1 5 . 2 6
6.
7.
8.
U s e the m e t h o d of virtual w o r k or of virtual c o m p l e m e n t a r y w o r k
to find the r e a c t i o n s a n d fixed-end m o m e n t s for a b e a m AB, of
length L, fixed at b o t h e n d s , subjected to a uniformly d i s t r i b u t e d
l o a d q, a n d w h o s e e n d B is given a small vertical d i s p l a c e m e n t uB
(Fig. 15.26). Discuss the influence of the shearing forces o n the
reactions a n d the fixed-end m o m e n t s .
A steel tube, 2 in. i n t e r n a l d i a m e t e r a n d \ in. thick, s t a n d s vertically
from a rigid b a s e . A t 3 ft. from the base, the t u b e is b e n t i n t o a
q u a d r a n t of a circle of 2 ft. r a d i u s , a n d at the e n d is s u s p e n d e d a
l o a d of 500 lbs. (Fig. 15.27). C o n s i d e r i n g the strain energy d u e to
b e n d i n g alone, use Castigliano's s e c o n d t h6e o r e m to calculate the
vertical deflection of the l o a d (E = 30 X 1 0 psi).
U s i n g the energy d u e to b e n d i n g a l o n e a n d the t h e o r e m of
m i n i m u m p o t e n t i a l energy, find the expression of the deflection of
a cantilever b e a m subjected to a n e n d l o a d P (Fig. 15.17). W r i t e
d o w n the forced a n d the n a t u r a l b o u n d a r y c o n d i t i o n s .
Energy Principles
500
489
lk>
Fig. 15.27
9.
A s s u m i n g t h a t the deflection in P r o b l e m 8 is given b y u2 = C ( l
— cos Uxx / 2 L ) , w h e r e C is a c o n s t a n t a n d xx is the d i s t a n c e from
t h e fixed e n d , o b t a i n t h e deflection at t h e free e n d u s i n g the
principle of m i n i m u m p o t e n t i a l energy. C o m p a r e this deflection t o
t h a t o b t a i n e d in P r o b l e m 8.
10. U s e the m e t h o d of least w o r k to solve P r o b l e m 6.
11. F o l l o w i n g steps similar to those of Sec. 15.20, use the R a y l e i g h - R i t z
m e t h o d to o b t a i n the solution of the simply s u p p o r t e d b e a m
subjected to a c o n c e n t r a t e d l o a d P a t m i d s p a n .
REFERENCES
[1] H. L. Langhaar, Energy Methods
in Applied
Mechanics,
John Wiley & Son, N e w York,
N . Y., 1962.
[2] Y. C. Fung, Foundations of Solid Mechanics,
[3] K. Washizu, Variational
Methods in Elasticity
Prentice-Hall, Englewood Cliffs, N . J., 1965.
and Plasticity,
Pergamon Press, N e w York,
N . Y., 1968.
[4] S. Timoshenko, Strength of Materials, Vol. 1, Van Nostrand, Princeton, N . J., 1955.
[5] S. Timoshenko, Strength of Materials, Vol. 2, Van Nostrand, Princeton, N . J., 1955.
[6] S. Timoshenko and J. N . Goodier, Theory of Elasticity, McGraw-Hill, N e w York, N . Y.,
1970.
[7] C. T. Wang, Applied Elasticity, McGraw-Hill, N e w York, N . Y., 1953.
CHAPTER 16
ELASTIC STABILITY: COLUMNS AND
BEAM COLUMNS
16.1
Introduction
In the previous c h a p t e r s , we discussed p r o b l e m s in which the stressd e f o r m a t i o n relationships were generally linear with d e f o r m a t i o n s quite
small c o m p a r e d to the smallest d i m e n s i o n of the b o d y . T h e forces were
in equilibrium a n d , for linearily elastic bodies, stress a n d d e f o r m a t i o n
p a t t e r n s c o u l d b e s u p e r i m p o s e d to p r o d u c e c o m p l e x configurations. I n
this chapter, we shall e x a m i n e w h a t h a p p e n s w h e n a b o d y in equilibriu m is slightly d i s t u r b e d from its configuration: D o e s it t e n d to r e t u r n to
its equilibrium position or does it t e n d to d e p a r t from it? F o r e x a m p l e ,
a slender r o d b e h a v e s n o r m a l l y w h e n l o a d e d in tension a n d c a n also
carry a small a m o u n t of c o m p r e s s i o n ; however, as the c o m p r e s s i o n l o a d
is increased, the r o d b e c o m e s u n s t a b l e a n d u n d e r g o e s large deflections.
T h e q u e s t i o n of stability of a c o m p r e s s e d b a r c a n b e investigated b y
using m e t h o d s a n a l a g o u s to those used in investigating the stability of
equilibrium configurations of rigid b o d i e s . Consider, for e x a m p l e , the
small weight o n the frictionless surface of Fig. 16.1. I n Fig. 16.1.a, the
l o a d is slightly displaced from its equilibrium position. T h e weight W
a n d the reaction yV are n o longer in b a l a n c e , b u t the resultant is a
restoring force which accelerates the particle to its equilibrium position.
Such a n equilibrium is called stable. In Fig. 16.1c, the resultant
u n b a l a n c e is a n upsetting force which accelerates the particle a w a y from
the equilibrium position. Such a n equilibrium is called u n s t a b l e . In Fig.
16.1b, there is n o t e n d e n c y to r e t u r n to the original position or to go
further. Such a n equilibrium is called neutral.
490
Elastic Stability
491
F i g . 16.1
Generalizing, a l o a d carrying structure is said to b e in a state of stable
equilibrium if, for all admissible small d i s p l a c e m e n t s from the equilibr i u m position, restoring forces arise w h i c h t e n d to accelerate the
structure b a c k t o w a r d s its e q u i l i b r i u m position. A classical e x a m p l e
used in illustrating the p r o b l e m of stability is t h a t r e p r e s e n t e d b y Fig.
16.2. T h e l o a d P acts o n the infinitely rigid b a r AB, w h i c h is h i n g e d at
P
A
Fig. 16.2
A. F o r small values of P, the vertical position of the b a r is stable.If a
d i s t u r b i n g force p r o d u c e s a lateral d i s p l a c e m e n t at B, the spring BC will
r e t u r n the b a r to its vertical position. I n this case, the m o m e n t a b o u t A
of the spring force is higher t h a n t h a t of P, i.e.,
2
KaL
> PaL
(16.1.1)
or
KL >
P.
(16.1.2)
492
The Theory of Elasticity
T h e force P , however, m a y increase to such a p o i n t t h a t the spring force
is n o t sufficient to restore the b a r to its original position after the
d i s t u r b a n c e h a s t a k e n place. I n such a case,
2
(16.1.3)
< PaL
Kal
or
KL<
(16.1.4)
P,
a n d B is accelerated further a w a y from its equilibrium position. T h e
value of P for n e u t r a l e q u i l i b r i u m is, therefore,
(16.1.5)
P = KL.
O n e c a n use a n energy m e t h o d to arrive at Eqs.(16.1.2), (16.1.4), a n d
(16.1.5). T h e system of Fig. 16.2 is stable if, d u e to the d i s t u r b a n c e , the
c h a n g e in p o t e n t i a l energy is positive; a n d u n s t a b l e if, d u e to the
d i s t u r b a n c e , the c h a n g e in p o t e n t i a l energy is negative. I n the first case,
4
n , _ ^ - P m -
c
„
s
„ ) ^ ^ - ^ > o
(M.14)
or
(16.1.7)
KL > P;
a n d in the s e c o n d case,
MI, - ^
- PKl
- cos « ) « ^
- l±£
< 0
(16-1.8)
or
KL < P.
(16.1.9)
T h e value of P for n e u t r a l equilibrium is o b t a i n e d b y writing:
ATlp = 0.
(16.1.10)
I n the following sections, we shall e x a m i n e the p r o b l e m s of p r i s m a t i c
b a r s subjected to axial c o m p r e s s i o n a n d to a c o m b i n a t i o n of axial
c o m p r e s s i o n a n d b e n d i n g . I n the first case, the b a r s will b e called
c o l u m n s a n d in the s e c o n d case they will b e called b e a m c o l u m n s .
Elastic Stability
16.2
493
Differential Equations of Columns and Beam-Columns
'X3
SI**)
P
A
Mf3*dMis
-||-
S»*
May
^
dx,
du,
(«0
Vt+dVi
Fig. 16.3
Fig. 16.3a shows a b e a m subjected to a longitudinal l o a d P as well as
to a transverse l o a d q{xx), a n d Fig. 16.3b shows a p o r t i o n of such a b e a m
of length dxx b e t w e e n two cross sections n o r m a l to the undeflected axis.
T h e relations a m o n g load, s h e a r i n g force, a n d b e n d i n g m o m e n t are
o b t a i n e d b y c o n s i d e r i n g the equilibrium of the e l e m e n t in Fig. 16.3b.
S u m m i n g the forces in the OX2 direction gives :
q =
-
dV2
(16.2.1)
dxx
T a k i n g m o m e n t s a b o u t m, a n d a s s u m i n g t h a t the deflection of the b e a m
is small, we get:
dM, 13
dx.
(16.2.2)
dx.
in which the terms of the s e c o n d o r d e r h a v e b e e n neglected. If we
neglect the effects of shear o n the d e f o r m a t i o n , the expression for the
c u r v a t u r e of the axis of the b e a m is:
2
d 2u
2
dx
A^3
(16.2.3)
EL
£ 7 3 is the flexural rigidity in the p l a n e of b e n d i n g , which is a s s u m e d to
b e a p l a n e of s y m m e t r y . C o m b i n i n g E q s . (16.2.1) to (16.2.3), we get :
J
U u2
Eh
~dxj
du
+ p dx.2 =
-K
(16.2.4)
494
The Theory of Elasticity
and
2
3
^4
+ P
d u2
(16.2.5)
Eqs. (16.2.1) to (16.2.5) are the basic differential e q u a t i o n s , for b e n d i n g
of b e a m - c o l u m n s . If the axial l o a d P is e q u a l to zero, these e q u a t i o n s
r e d u c e to the usual e q u a t i o n s for b e n d i n g b y lateral loads. If the
transverse l o a d q is e q u a l to zero, these e q u a t i o n s a p p l y to a n axially
loaded column.
16.3
Simple Columns
Let us consider the fixed e n d c o l u m n of Fig. 16.4, a n d a s s u m e t h a t it
h a s a u n i f o r m b e n d i n g stiffness EI3 a n d t h a t b u c k l i n g occurs in t h e
p l a n e OXx, OX2. If the c o l u m n is accidentally displaced from its straight
position a l o n g the OXx axis, the force P p r o d u c e s a m o m e n t a b o u t O
which t e n d s to b e n d the c o l u m n even further; o n the other h a n d , the
elastic forces in the c o l u m n t e n d to restore it to its original position. F o r
small values of P, the straight position is stable a n d the c o l u m n is
subjected to uniform c o m p r e s s i o n . F o r large values of P, the straight
position is u n s t a b l e a n d the c o l u m n buckles. If P is gradually increased,
a c o n d i t i o n is r e a c h e d w h e r e a n accidentally p r o d u c e d d i s p l a c e m e n t
does n o t d i s a p p e a r u p o n r e m o v a l of the disturbing agent. This value of
P is the critical l o a d w h i c h c a n b e defined [1] as the axial force which
is sufficient to k e e p the c o l u m n in a slightly b e n t s h a p e . T h e e q u a t i o n
g o v e r n i n g the deflection of the c o l u m n in Fig. 16.4 is:
Fig. 16.4
Elastic Stability 495
T h e b o u n d a r y c o n d i t i o n s t o b e satisfied a r e :
At
* , = 0 , K 2= 0 a n d ^ = 0.
(16.3.2)
At
x , = L, u2 = 5.
(16.3.3)
A t a n y section mn,
M 13 = -P(8 - u2),
(16.3.4)
so t h a t E q . (16.3.1) b e c o m e s :
2
2
+ K u2
^
= K 8,
(16.3.5)
where
2
*
1 36 6
= ^
-
<
- - >
T h e g e n e r a l s o l u t i o n of E q . ( 1 6 . 3 . 5 ) is :
u
2=
A cos Kx
x
+ B sin AJCJ + 8 ,
(16.3.7)
in w h i c h A a n d B a r e c o n s t a n t s of integration. T h e b o u n d a r y c o n d i t i o n s
( 1 6 . 3 . 2 ) give:
A = - 8 and 5 = 0.
(16.3.8)
w = 8(1 -
(16.3.9)
Therefore,
2
c o s Jfccj).
T h e b o u n d a r y c o n d i t i o n ( 1 6 . 3 . 3 ) yields:
S = 8 -
8 cos KL,
(16.3.10)
which requires t h a t either 8 = 0 o r c o s KL = 0 . If 8 = 0 , there is n o
deflection a n d therefore n o buckling. If c o s KL = 0 , w e m u s t h a v e :
*L = (2n-l)2,
(16.3.11)
496
T h e Theory of Elasticity
where n = 1,2,
E q . (16.3.11) d e t e r m i n e s the values of K a t which
a b u c k l e d s h a p e c a n exist. 8 r e m a i n s i n d e t e r m i n a t e a n d c a n t a k e a n y
value within t h e scope of small deflection theory [ r e m e m b e r t h a t E q .
(16.3.1) is b a s e d o n such a theory]. O n e h a s t o notice t h a t in this
p r o b l e m , as P increases, t h e m o m e n t increases, thus increasing t h e
deflection which in t u r n increases t h e m o m e n t : T h e p r o b l e m is n o
longer linear a n d t h e results b a s e d o n linear theory , such as E q . (16.3.7),
d o n o t c o n t a i n e n o u g h b o u n d a r y c o n d i t i o n s to p e r m i t a solution for t h e
exact value of 8. F o r values of P slightly higher t h a n t h e critical value,
the deflections b e c o m e s so high t h a t the linear theories d o n o t apply.
T h e smallest value of KL satisfying Eq.(16.3.11) c o r r e s p o n d s t o n = 1.
Thus
KL =
or
n
(16.3.12)
2
P = P =
U EI3
(16.3.13)
T h i s is the smallest axial force w h i c h c a n m a i n t a i n a slightly b e n t s h a p e :
T h e c o l u m n is in a n e u t r a l equilibrium position. O t h e r values of n
c o r r e s p o n d t o the deflection p a t t e r n s s h o w n in Fig. 16.5. T h o s e p a t t e r n s
P
//////
n-.3
25 1
Per* "'/ *
F i g . 16.5
are called b u c k l i n g m o d e s a n d c o r r e s p o n d t o values of P higher t h a n Pcr
.
A value of P c o r r e s p o n d i n g t o n = 2 c a n only b e a t t a i n e d if t h e m o d e
of b u c k l i n g c o r r e s p o n d i n g t o n = 1 is p r e v e n t e d . T h e high b u c k l i n g
m o d e s a r e , however, m a t h e m a t i c a l l y i m p o r t a n t . T h e fact t h a t n = 1
Elastic Stability
497
c o r r e s p o n d s to the smallest axial force t h a t c a n m a i n t a i n a slightly b e n t
shape, will b e exploited to o b t a i n a n energy solution to the b u c k l i n g
problem.
T h e previous results c o u l d h a v e b e e n o b t a i n e d b y starting from
Eq.( 16.2.5) in w h i c h q is set e q u a l to zero. T h e solution of this e q u a t i o n
is:
=
u
x
Q + Ci i
2
+ C 3s i n Kxx + Q c o s Kx{,
(16.3.14)
w h e r e Q , C 2, C 3 a n d C 4 are c o n s t a n t s of i n t e g r a t i o n to b e o b t a i n e d
from the following b o u n d a r y c o n d i t i o n s :
At
xx = 0, 2W= 0 a n d ^ - = 0
(16.3.15)
At
xx = L, M 13 = —EI3^^-
(16.3.16)
= 0 and
Substituting these four b o u n d a r y c o n d i t i o n s i n t o Eq. (16.3.14), we
o b t a i n the following s i m u l t a n e o u s e q u a t i o n s for the c o n s t a n t s of
integration:
C 1+ 0 + 0 + C 4 = 0
0 + C 2 + C3K+
0 + 0 - C 3AT sin AX - C4KcosKL
0 + C2P
0 = 0
= 0
(16.3.17)
+ 0 + 0 = 0.
This is a set of four h o m o g e n e o u s e q u a t i o n s with four u n k n o w n s , a n o n trivial solution of w h i c h exists only w h e n the d e t e r m i n a n t f o r m e d b y the
coefficients is e q u a l to zero. This type of p r o b l e m , the eigenvalue
p r o b l e m , h a s b e e n e x a m i n e d at length in C h a p t e r 3. Setting the
d e t e r m i n a n t e q u a l to zero yields:
cos KL = 0
(16.3.18)
or
KL = (2» -
1)2
(16.3.19)
498
The Theory of Elasticity
which is the s a m e as Eq. (16.3.11) . T o every value of n, there
c o r r e s p o n d s a value of K a n d c o n s e q u e n t l y a value of P, which p r o v i d e s
1
us with a possible solution. T h o s e values of P are the eigenvalues
of the
p r o b l e m , each of which c o r r e s p o n d s to a form of w2> i* o t h e r w o r d s , to
a b u c k l i n g m o d e : I n d e e d , substituting Eq. (16.3.2) into Eq. (16.3.14), we
get:
u2 =
Cx(l
cos
(16.3.20)
Kx{).
If we set u2 = 8 at xx = L, then
u2 = 5(1
cos
(16.3.21)
Kxx),
which is the s a m e as E q . (16.3.9). T h e r e are as m a n y expressions of u2
as there are values of K a n d , consequently, of P (Fig. 16.5).
T h e i n d e t e r m i n a c y in the value of 8 c a n b e lifted if the l o a d P is
slightly eccentric, thus i n d u c i n g a couple M 0 at the free e n d (Fig. 16.6).
In such a case, Eq. (16.3.5) b e c o m e s :
0
Fig. 16.6
2
+ K u2
2
= K8
(16.3.22)
+
T h e general solution of this> eeqquuaat it oi onn isis : :
u2 = A cos Kxx + B sin Kxx
4-
8 +
2M 0
K EI3 '
(16.3.23)
T h e b o u n d a r y c o n d i t i o n s (16.3.2), give:
5 = 0.
(16.3.24)
Elastic Stability
499
Therefore,
=
u2
5
(
+ (
2
1
- X C l O)
^ )
*
K EI
-
S
(16-3-25)
3
T h e b o u n d a r y c o n d i t i o n (16.3.3) yields :
=
MQ(
P
1 -— cos
cos KLKL\
cos KL
/"
V
(16.3.26)
T h e expression for u2 n o w b e c o m e s :
"2
= M0 ( 1 - cos Kx\
~P
cos KL
\
\
A
(16.3.27)
T h e b u c k l i n g l o a d of a c o l u m n is q u i t e sensitive to the n a t u r e of the
s u p p o r t s of the e n d s of the c o l u m n s :
a) For a bar with hinged ends (Fig. 16.7), the b o u n d a r y c o n d i t i o n s a r e :
IX,
P
1\
1
0
Fig. 16.7
u2 — 0
at Xj = 0
a n d x{ = L
2
d u2
0
at xx = 0
a n d xx = L.
T h e value of the b u c k l i n g l o a d is f o u n d to b e :
2
U EI2
3
(16.3.28)
L
This case is called the f u n d a m e n t a l case of b u c k l i n g of a p r i s m a t i c b a r .
b) For a bar with both ends built in (Fig. 16.8), the b o u n d a r y c o n d i t i o n s
are:
500
The Theory of Elasticity
Fig. 16.8
u2 = 0 at xx = 0 a n d xx = L
= 0 at xx = 0 a n d xx = L
T h e value of the b u c k l i n g l o a d is f o u n d to b e :
2
P
=
4U EL2
(16.3.29)
L
c) For a bar built in at one end and hinged at the other (Fig. 16.9), the
boundary conditions are:
Fig. 16.9
Elastic Stability
501
u2 = 0 at xx = 0 a n d xx = L
= 0 at xx = L.
In this case, a reactive force R is d e v e l o p e d at the p i n n e d e n d . T h e
critical l o a d is f o u n d to b e :
2
Per
16.4
U EI3 2
(16.3.30)
(0.7L)
Energy Solution of the Buckling Problem
Let us consider the e q u i l i b r i u m of a fixed e n d c o l u m n w h e n the
c o m p r e s s i o n l o a d is equal to the b u c k l i n g load. U n d e r such a load, ,the
c o l u m n c o u l d b e straight or, if d i s t u r b e d , it could a s s u m e a b e n t form
a n d k e e p it (Fig. 16.10). D u r i n g this c h a n g e of form, the c o l u m n is still
Fig. 1 6 . 1 0
in equilibrium, b u t its strain energy is i n c r e a s e d since the energy of
b e n d i n g of the c o l u m n will b e a d d e d to the energy of c o m p r e s s i o n .
T h u s , the total strain energy in the c o l u m n is given b y :
1
o
o
502
The Theory of Elasticity
w h e r e A is the cross section of the c o l u m n . T h e d i s p l a c e m e n t ux is given
by:
L
L
« . - 0/ ( * - * . ) - / h y i
+ ( ^ )
- * . ]
0
(16.4.2)
0
T h e w o r k d o n e by the external force P d u r i n g axial c o m p r e s s i o n a n d
b e n d i n g is:
o
o
T h e potential energy of the system prior to its a s s u m i n g a b e n t form is:
L
L
o
o
a n d its potential energy after it has a s s u m e d a b e n t form:
0
0
(16.4.5)
L
L
0
^
0
T h e c h a n g e in potential energy is, therefore,
2
2
A H , = AU, - AW=
i f
0
E h( ^ )
1
d x x
- f /
( ^ )
* „
0
(16.4.6)
a n d the value of P for n e u t r a l equilibrium is o b t a i n e d from writing
AIl^ = 0, in other w o r d s from:
Elastic Stability
503
dx]
(16.4.7)
P =
N o w , in Sec. 16.3, it was s h o w n t h a t there exists a n infinite n u m b e r of
values of P w h i c h are possible solutions of the eigenvalue p r o b l e m .
E a c h value c o r r e s p o n d s to a b u c k l i n g m o d e ; in o t h e r w o r d s , to a certain
u2 = u2(xl).
T h e o n e value of P we are interested in is Pcn w h i c h is the
smallest o n e a n d which c o r r e s p o n d s to the first m o d e . Therefore, the
b u c k l i n g l o a d will b e given b y the m i n i m u m value of Eq. (16.4.7);i.e.,by
(16.4.8)
T h e n u m e r a t o r in Eq. (16.4.7) is the increase in the strain energy of the
c o l u m n w h e n , u n d e r P, it a s s u m e s the b e n t form; the d e n o m i n a t o r is the
d i s p l a c e m e n t of P a l o n g its line of a c t i o n d u e to this b e n d i n g . Eq.
(16.4.8) is called Rayleigh's F o r m u l a .
T h e energy m e t h o d is used to o b t a i n a n a p p r o x i m a t e solution to
p r o b l e m s which c a n n o t b e solved exactly. F o r that, a n expression for
satisfying the b o u n d a r y c o n d i t i o n s is a s s u m e d , a n d P is d e d u c e d
u2(xx)
from Eq. (16.4.7). This value will always b e larger t h a n Pcr
. W h e n using
the energy m e t h o d , T i m o s h e n k o [2] n o t i c e d t h a t if the expression of the
strain energy d u e to b e n d i n g w a s written in terms of the m o m e n t , a
b e t t e r a p p r o x i m a t i o n w o u l d result. F r o m E q . (16.3.1) a n d for the case
of the c o l u m n in Fig. 16.10, we h a v e :
(16.4.9)
Substituting Eq. (16.4.9) into Eq. (16.4.6) a n d setting AIl^ = 0, we
obtain Timoshenko's formula:
P =
L
0
(16.4.10)
504
The Theory of Elasticity
and
(16.4.11)
Eq. (16.4.10) gives a better a p p r o x i m a t i o n t h a n E q . (16.4.7). W a n g [1]
gives a n analysis of the errors in the b u c k l i n g loads calculated b y the
energy m e t h o d .
W h e n a p o l y n o m i a l or a t r i g o n o m e t r i c series is i n t r o d u c e d to express
the value of u2(xl), the Rayleigh- Ritz m e t h o d is often used to m i n i m i z e
P a n d get as close as possible to Pcr
.
16.5
Examples of Calculation of Buckling Loads by the Energy Method
In this section, we shall illustrate the use of the energy m e t h o d by
m e a n s of t w o e x a m p l e s w h o s e correct answer is k n o w n . This will give
us a n idea a b o u t the a p p r o x i m a t i o n s involved in this m e t h o d .
1. Column with one end built in and the other end free. Let us a s s u m e t h a t
the deflection curve of the b u c k l e d b a r (Fig. 16.10) is given b y a n
e q u a t i o n of the form:
«2 = s ( l - c o s ^ - ) .
06-5.1)
A p p l y i n g E q . (16.4.7), we get: 2 4
n
a n
P - _64ZL
"
«ir£
=
2
32
4L
£
/
(16.5.2)
•
16L
This is the correct a n s w e r P = Rr, since we started b y a s s u m i n g a
b u c k l e d s h a p e which h a p p e n e d to b e the correct o n e .
Let us n o w a s s u m e t h a t u2 is given by the following e q u a t i o n :
u 2 ^ Q L - X ) .x
06.5.3)
T h i s is the e q u a t i o n of the deflected c o l u m n u n d e r the effect of a
h o r i z o n t a l l o a d acting at B. A p p l y i n g Eq.( 16.4.7), we get:
Elastic Stability 505
2
3EI38 3
2L
P 3 P
5 L
-
5 2E
h
=2
(16.5.4)
L
W h e n c o m p a r e d t o Eq. (16.5.2), it is f o u n d t h a t this result is in error b y
1.3 p e r c e n t . If we n o w apply T i m o s h e n k o ' s E q . (16.4.10), w e get:
P . J S .
(16.5.5)
2
17 S^L
L
70 £ / 3
w h i c h is i n error b y only 0.13 p e r c e n t .
Let u s n o w a s s u m e a very p o o r r e p r e s e n t a t i o n of u2(xx).
p a r a b o l a w h o s e e q u a t i o n is
u2 = ^
F o r that, a
(16-5.6)
is c h o s e n . T h i s r e p r e s e n t a t i o n d o e s n o t satisfy the e n d c o n d i t i o n s since
it results in a c o n s t a n t c u r v a t u r e a l o n g t h e c o l u m n . Y e t t h e resulting
a p p r o x i m a t e solution for P is quite satisfactory. I n d e e d , substituting
Eq.( 16.5.6) i n t o Eq.(16.4.10), we get:
5
'
r
2
E
(16.5.7)
h
=
T h e error is only 1.3 p e r c e n t .
Finally, w e c a n a s s u m e t h a t
u2
-
<.(£)'•
<,6
'5
8)
w h e r e c0 a n d cx a r e u n d e t e r m i n e d p a r a m e t e r s . T h i s form satisfies t h e
b o u n d a r y c o n d i t i o n s u2 = 0 a n d du2/dxx = 0 a t xx = 0. Substituting
Eq.(16.5.8) into Eq.(16.4.7), w e get :
4*7 l + Sr
+
^= k
£°
3
P
3
o
3
u
l
\" / , ,
c
(16.5.9)
506
T h e Theory of Elasticity
T h e m i n i m u m value of P is o b t a i n e d by differentiating with respect to
c{ /c0 a n d setting t h e result e q u a l to zero (Rayleigh -Ritz m e t h o d ) . T h u s
d p
-GO
=
0
=
1
8
( ^ )
2
+
2 2
^
+
5
'
( - °)
16
51
which gives cx /c0 = —0.3 a n d —0.92. Substituting these values i n t o
Eq.(16.5.9), we find t h a t cx /c0 = —0.3 gives the smaller P, so t h a t
2
P
A
9
= ^ .
(16.5.11)
T h e e r r o r is 0.92 p e r c e n t . Better a c c u r a c y c a n b e o b t a i n e d b y t a k i n g
more than two undetermined parameters.
2. Prismatic column with hinged ends. T h e b o u n d a r y c o n d i t i o n s in this
case a r e (Fig. 16.7):
2
u2 = 0 a n d d u2/dx}
= 0 at xx = 0 a n d xx = L.
T h e s e c o n d i t i o n s are satisfied b y a s s u m i n g t h a t the s h a p e of
deflected c o l u m n is r e p r e s e n t e d b y a trigonometric series:
the
1 56 1 2
u2 = 2
c
n
s i n ^ L .
<
' -
>
We have:
L
oo
and
L
0
Substituting E q s . (16.5.13) a n d (16.5.14) into Eq.(16.4.7),we get:
,
At/,J,_IPEI,
_ ii ^ 3
1 +
I 6
(^T)
2
/
+ 8
„
'(^T)
\2
+
"
(16.5.15)
Elastic Stability
507
T h e m i n i m u m P is o b t a i n e d by adjusting the u n d e t e r m i n e d p a r a m e t e r s
c 2/ q , c3/cl9 etc. U s i n g the Rayleigh- Ritz m e t h o d , we m u s t write
dP
_
dP
. . .
==
0
C a r r y i n g o u t the calculation, we find that these c o n d i t i o n s require that
Hence
1p
=
mm
TP ELI
j2
»
which is the value given b y the exact solution. This is the case b e c a u s e
the a s s u m e d form of the deflection curve h a p p e n e d to include the exact
solution.
Remark
F o r a w i d e variety of examples using the energy m e t h o d , the r e a d e r
is referred to T i m o s h e n k o a n d G e r e ' s classic treatise [2].
16.6 Combined Compression and Bending
1. Beam-column
with a concentrated
lateral load. Let us consider the
p r o b l e m of a strut AB with h i n g e d e n d s l o a d e d by a n axial force P a n d
a force Q at a distance a from B (Fig. 16.11). T h e differential e q u a t i o n s
of the two p o r t i o n s of the strut a r e :
508
The Theory of Elasticity
Left of Q : ElJ^-
= - P u 2- ^ - x x.
d2u
p
irT 2
u+
R i g h t of Q : EI3-^2~
=
~Pu2
g(L -
(16.6.1)
a)(L
j-
-
x)
{.
(16.6.2)
Setting
tf-^,
(16-6.3)
the solutions o f these e q u a t i o n s a r e :
Left of Q : u2 = C,COS(AJC,) + C 2sin(Ajc,) -
^pj^-
R i g h t of g : « 2= C j c o s ^ x , ) + C 4sin(AJc,)
(16.6.4)
1665
_ Q(L-a)(L-xQ
< --)
PL
w h e r e Cx, C 2, C 3, a n d C 4 a r e c o n s t a n t s of integration to b e d e t e r m i n e d
from t h e b o u n d a r y c o n d i t i o n s . A t xx = 0 a n d xx = L, u2 = 0. T h e r e f o r e ,
C, = 0 a n d C 3 = - C 4t a n AX.
(16.6.6)
A t the p o i n t of a p p l i c a t i o n of t h e l o a d Q, xx = L — a, the two p o r t i o n s
of the deflection c u r v e a s given b y E q s . (16.6.4) a n d (16.6.5) m u s t give
the s a m e deflection a n d slope. T h e n ,
C2sin[K(L
- a)] = C 4{sin[A"(L - a)]
-
C2Kcos[K(L
~
(16.6.7)
a)]}
a)\-p\
= CAK{co%\K(L
+,
tan(AX)cos[A"(L -
- a)] + tan(AX)sin[A"(£ - a)]} (16.6.8)
Q(L-a)
PL
•
Solving, we o b t a i n :
2
°
e s i n ( t a )
PKsm(KL)'
Q &w[K(L
4 - a)]
PA:tan(AX)
'
)
S u b s t i t u t i n g E q s . (16.6.6) a n d (16.6.9) i n t o E q s . (16.6.1) a n d (16.6.2), w e
o b t a i n for t h e p o r t i o n of the strut to the left of Q:
n
Elastic Stability
dh±
dx\
x
_QKsm(Ka)
Psin(KL)
s m
509
6
^ it>
)1 2
'
and for the portion of the strut to the right of Q;
2
"
L
Qsm[K(L-a)}
=
-)]S
PKsm(KL)
mQ(L[ - Ka)(L( - xQ
*'
du2
Q sm[K{L - a)]
PsinW)
dYx = -
^
, )3
PL
rL C t „Ov
-
S
- a)5+
lQ(L
^ P Z -
~
1 61 4
< - - >
In t h e p a r t i c u l a r case of a l o a d Q applied a t t h e center of t h e b e a m , t h e
deflection curve is symmetrical, a n d for xx = a = L/2
_ oe K f ) i l l
'
3
06 616>
48£/3
or
Q
=
'
L
Y
(16.6.17)
N u m e r i c a l values of x for v a r i o u s
c a n b e f o u n d in [2]. W h e n
a p p r o a c h e s Wz ,the deflection b e c o m e s infinite a n d
n
P
=
P =
3
2
j
g
/
(16.6.18)
which is t h e b u c k l i n g load. T o find t h e slope of t h e deflection curve a t
the e n d of t h e b e a m c o l u m n w h e n Q is a t t h e center, set a = ^ a n d
xx = 0 in E q . (16.6.11). This gives:
510
The Theory of Elasticity
f<M
_ g z ^ 2[i-cos(f)]_
1 66 1
Hem
^
-
>
w h e r e A is a function of
. T h e s a m e w a y the m a x i m u m b e n d i n g
m o m e n t is o b t a i n e d from Eq.(16.6.12) as
(f)
t na
<2L
(16.6.20)
AX
2
N u m e r i c a l values of A a n d tan ^ j n / ^ n c a n b e found in [2].
2. iteam—column with several concentrated
loads. Eqs.(16.6.10) a n d
(16.6.13) show that, for a given P, the deflections are linear functions of
Q. O n the o t h e r h a n d , P intervenes in these e q u a t i o n s in a r a t h e r
c o m p l i c a t e d way. F o r m o r e t h a n o n e lateral load, the principle of
superposition can b e applied b u t in a slightly modified form to take into
a c c o u n t the effect of P. T o p r o v e this statement, let us consider a b e a m c o l u m n subjected to two lateral l o a d s Qx a n d Q2 at distances ax a n d a2
from B (Fig. 16 .12). T o the left of Qx, the differential e q u a t i o n of the
deflection curve is:
Fig. 16.12
9
Elastic Stability
dh*
_Qja=L
axf
_
_ L^
511
L
)
C o n s i d e r n o w Qx a n d Q2 acting separately o n the axially c o m p r e s s e d
strut a n d d e n o t e b y (u2)x the deflection c a u s e d b y Qx , a n d b y (u2)2 the
deflection c a u s e d b y Q2. F o r the p o r t i o n of the b e a m left of C a n d for
Qx a c t i n g a l o n e , we h a v e :
d
E h
^ = - ^ - P ( u l
.)
(16.6.22)
{
F o r Q2 acting alone, we h a v e :
E l / ^
(16.6.23)
= - ^ X - Pl ( u 2 ) 2 .
By a d d i n g these t w o e q u a t i o n s , w e find:
2
rj 3 d [{u2)x
+ (u2)2]
dx\
_
Qxax
"
L
- P[(u2)x
X Q2a2 l
l
L
+
X
(16.6.24)
(u2)2].
Eq.(16.6.24), for the s u m of the deflections (u2)x a n d (u2)2, is the s a m e
as E q . (16.6.21). T h e s a m e c o n c l u s i o n h o l d s for p o i n t s b e t w e e n Qx a n d
Q2 a n d p o i n t s to the right of Q2. Therefore, w h e n there are several l o a d s
a c t i n g o n a b e a m c o l u m n , t h e r e s u l t a n t deflection c a n b e o b t a i n e d b y
s u p e r p o s i t i o n of the deflections p r o d u c e d separately b y e a c h lateral
l o a d a c t i n g in c o m b i n a t i o n with the compressive force P.
Fig. 1 6 . 1 3
512
T h e Theory of Elasticity
3.Beam column with a uniformly distributed load. C o n s i d e r t h e b e a m
s h o w n in F i g . 16.13, subjected t o a uniformly d i s t r i b u t e d transverse l o a d
q a n d t h e axial force P. T h e differential e q u a t i o n of the deflection curve
is:
q
u
2
q
O - -
Xl+
2
Setting K
662 5
El/-^=- -± 4-P»2.
)
= -gj-, t h e solution of this e q u a t i o n is:
3
2
= Q s i n C A x , ) + C COS(A"JC,) + jp(x
(16.6.26)
- Lx
x - -^).
2
T h e b o u n d a r y c o n d i t i o n s t h a t u2 = 0 a t xx = 0 a n d xx = L d e t e r m i n e C,
a n d C2 t o b e :
q
q2 \ \ - c o s ( A X ) ~ |
1
sin(AX)
PK l
C
J
-
Therefore,
2 1
{ [
L)
sin(°^
-
i, } n +( cK o x s
^
2
- LxxK
S
(
^]
+ K2 X
?
(16.6.27)
2^.
T h e deflection a t t h e m i d d l e of t h e b e a m c o l u m n is:
~ ~( f )
5 q L 4 (f)
2
2SeC
4
3
4
5(^j
(16.6.28)
TJ
_ 5
qL
~ 384 EI3 '
N u m e r i c a l values of TJ for various
c a n b e found in [2]. A s
a p p r o a c h e s W2, P a p p r o a c h e s its critical value given by Eq.(16.3.28).
T h e slope a t t h e e n d of t h e b e a m - c o l u m n is o b t a i n e d b y differentiating Eq.( 16.6.27). T h u s ,
513
Elastic Stability
9
i4,=o
24E%
16629
7T7V
24ET3X-
< -- )
T h e m a x i m u m b e n d i n g m o m e n t is o b t a i n e d b y differentiating Eq.
(16. 6.27) twice. T h u s ,
M
( 13)max
=
(16.6.30)
Fig. 1 6 . 1 4
4. Beam-column
with end couple. T h i s p r o b l e m c a n b e solved [2] b y
a s s u m i n g in Fig. 16.11 t h a t t h e d i s t a n c e a t e n d s to zero while Q
increases such t h a t t h e p r o d u c t ( Q a ) r e m a i n s finite a n d e q u a l t o MB. I n
Eq.(16.6.10), ( K a ) is s u b s t i t u t e d t o s i n ( K a ) a n d MB t o Q a . T h u s
(Fig.16.14),
^1 - ^ r ^ T & r - r l
and
P L sin(AX)
L J
6 )J
-
1
514
The Theory of Elasticity
d x
a -( ^\
=
=
M L
B
3
f
1
1
1
M BL
(16.6.32)
6 i ^
A - - f e ^
L
*
=
- U,
3^/3 ^ )
3
r
M
i
L^)
N u m e r i c a l values for <j> a n d \p for various
i
^
( f)
i
2 J (16-6.33)
are given in [2].
Remark
T h e four previous cases c a n b e s u p e r i m p o s e d to analyze b e a m c o l u m n s with various e n d c o n d i t i o n s . H o w e v e r , it is to b e r e m e m b e r e d
t h a t the axial force P m u s t b e a p a r t of every individual solution u s e d
in the superposition.
16.7
Lateral Buckling of Thin Rectangular Beams
h^MCosrCosfi
MCosr
li
M MCosY
F i g . 16.15
Sinfi
Elastic Stability
515
If a b e a m is very stiff against b e n d i n g in o n e p l a n e a n d very flexible
in a p e r p e n d i c u l a r p l a n e (like a ruler), a n d if the b e a m is l o a d e d in the
stiff p l a n e , it m a y b e c o m e u n s t a b l e at s o m e critical value of the l o a d
a n d b u c k l e sidewise as s h o w n in Fig. 16.15. This b e n d i n g in the less
flexible direction is always associated with a twist. In this section, we
shall consider t w o simple cases of lateral buckling.
1.Simply supported beam bent by couples. C o n s i d e r a simple b e a m with a
n a r r o w r e c t a n g u l a r cross section w h i c h is subjected to p u r e b e n d i n g in
the OXx, OX2 p l a n e (Fig. 16.15). Let us a s s u m e t h a t the b e a m is
d i s t u r b e d so as to h a v e a small lateral deflection u3. A t a n y p o i n t xx, the
b e n d i n g m o m e n t vector M 13 is directed a l o n g the — OX3 axis (using the
r i g h t - h a n d rule): Its m a g n i t u d e is M a n d it c a n b e d e c o m p o s e d into
three c o m p o n e n t s : n a m e l y (see Fig. 16.15), M sin y, which causes a
twisting of the b e a m ; a n d M cosy cos/? a n d M cosy sin/?, which causes
a b e n d i n g of the b e a m in t w o o r t h o g o n a l directions. Recalling the
e q u a t i o n s of the deflection a n d of the twisting of simple b e a m s , we
have:
E
— M cos y cos fi =
(16.7.1)
h ^ T
2
d u
-M
EI
cos y sin p = 2~^r
(16.7.2)
M sin y = GJa = GJ^-9
axj
(16.7.3)
w h e r e I2 a n d I3 are the m o m e n t s of inertia with respect to the OX2 a n d
OX3 axes, GJ is the torsional rigidity, a n d a = dfi/dxx is the angle of
r o t a t i o n p e r unit length of the b e a m following the sense of the twisting
moment.
F o r r e c t a n g u l a r cross sections, we h a v e :
/2 = ^ , / 3 = ^ , /
=
3^
(16.7.4) to
w h e r e K is given b y the table in Sec. 10.7. Since y a n d /? as well as the
d i s p l a c e m e n t s are small, t h e n
sin y «
cos y «
1, sin /? « /?, cos /? «
1.
(16.7.5)
516
The Theory of Elasticity
Eq.(16.7.1) gives the vertical d i s p l a c e m e n t of the b e a m a n d is of n o
2 case. Differentiating Eq. (16.7.3) with respect to x , a n d
interest in this
x
eliminating d u3/dxx
by using Eq.( 16.7.2), we get:
1 67 6
GJ^4
+ ¥rP
= 0.
( - - )
dxf
Ll2
If the b e a m h a s a c o n s t a n t cross section, the solution of Eq. (16.7.6) is:
w h e r e Q a n d C 2 are c o n s t a n t s to b e d e t e r m i n e d from the b o u n d a r y
c o n d i t i o n s : /? = 0 at xx = 0 a n d xx = L. Therefore,
C 2 = 0,
C l
n ^ /1 ^ L
S
= 0.
06.7.8)
If Q = 0, we h a v e the trivial solution c o r r e s p o n d i n g to the u n b u c k l e d
form. T h e r e f o r e the b u c k l e d form is only possible w h e n
sm
V^- 0
'-
<i6 7 9)
Therefore, the critical b e n d i n g m o m e n t necessary to k e e p the b e a m in
its laterally d e f o r m e d position is given by :
Fig. 16.16
Elastic Stability
( )„
M
= H^5
517
(16.7.10)
2. Cantilever beam bent by an end load. Let us consider the case of a
cantilever b e a m b e n t b y a n e n d l o a d P which is applied at the c e n t r o i d
of the cross section (Fig. 16.16) a n d in the OXx, OX2 p l a n e . Let us
a s s u m e t h a t the b e a m is d i s t u r b e d so as to h a v e a small lateral
d i s p l a c e m e n t 5. A t a n y d i s t a n c e xx from the origin, we h a v e a b e n d i n g
m o m e n t a n d a torsional m o m e n t respectively e q u a l in m a g n i t u d e to
P(L — xx) a n d P(8 — w 3). F o l l o w i n g the s a m e steps as in the previous
derivation, we get:
d
E
^I
dxf
= P- x ){
E
^ I=
dxf
P - X(
i )L
L
(16.7.11)
f
(16.7.12)
E l i m i n a t i n g w3 from E q s . (16.7.12) a n d (16.7.13), we get:
\
d \ x 2 GJEI
2 24
L /
Let i = (1 - xx /L) a n d t = P L /GJEI2.
Eq.(16.7.14) b e c o m e s :
2
= 0.
2 + H$
(16.7.15)
T h e solution of this e q u a t i o n is:
^VmlQJ^
+ C ^ ^ ) ] ,
06.7.16)
where
and
are Bessel functions of the first k i n d of o r d e r | a n d
- ^ r e s p e c t i v e l y : Cx a n d C 2 are c o n s t a n t s to b e d e t e r m i n e d from the
conditions:
/? = 0 at xx = 0 a n d
axx
= 0 at xx = L,
518
The Theory of Elasticity
since the torsional m o m e n t is zero there. I n t e r m s of £, these b o u n d a r y
conditions become:
/3
= o for £ = 1 a n d ~
= 0 for | = 0.
This last c o n d i t i o n requires t h a t Q = 0. T o o b t a i n a non-trival solution,
the s e c o n d c o n d i t i o n requires t h a t
/ _ , ( ! ) = 0.
(HS.7.17)
F r o m a table of zeros of Bessel functions of the first k i n d a n d of o r d e r
- | , we find t h a t the smallest t is given b y :
t = 4.013.
Therefore, the critical P necessary to keep the b e a m in its laterally
d e f o r m e d position is:
_ 4.013VG7^
far-
2
Remark
Reference [2] gives solutions for various types of b e a m s
different loading c o n d i t i o n s .
)
L
under
PROBLEMS
1.
Solve the p r o b l e m of the c o l u m n with o n e e n d built in a n d the other
e n d free ( S e c 16.5) b y a s s u m i n g that the deflection curve of the
b u c k l e d b a r , u2 = ^ ( ^ l ) ' is the s a m e as t h a t of a uniformly l o a d e d
b e a m built in at o n e e n d a n d free at the other. U s e b o t h Eqs.
(16.4.7) a n d (16.4.10), a n d c o m p a r e the axial load o b t a i n e d in each
case to the critical o n e .
2.
3.
Solve P r o b l e m 1 for the case of the c o l u m n h i n g e d at b o t h e n d s .
A coupling r o d is 10 ft. long, 2 in. wide, a n d 4 in. d e e p . It carries a n
axial compressive l o a d of 10 t o n s a n d a uniformly d i s t r i b u t e d l o a d
of 100 lb./ft.6 r u n . C a l c u l a t e the m a x i m u m stress in the r o d
(E = 30 X 1 0 p s i ) .
Elastic Stability
4.
5.
6.
7.
519
A long slender steel strut, originally straight a n d built in at o n e e n d
a n d free at the o t h e r e n d , is l o a d e d at the free e n d with a n eccentric
load w h o s e line of action is parallel to the original axis of the strut.
D e t e r m i n e the deviation of the free e n d from its original position
a n d the greatest compressive stress, if the length of the strut is 10 ft.,
its cross section is circular with 2 in. external d i a m e t e r a n d 1 in.
6 l o a d is 800 lb., a n d the original eccentricity is
internal diameter, the
3 in. (E = 30 X 1 0 p s i ) .
O b t a i n the expressions for the b e n d i n g m o m e n t s at the e n d s a n d
center of a u n i f o r m b e a m of length L, built in at b o t h e n d s , a n d
subjected to a u n i f o r m lateral l o a d of intensity q, a n d to e n d thrusts
of intensity P. Show, w i t h o u t e l a b o r a t e analysis, from the expressions derived, w h i c h of the two b e n d i n g m o m e n t s is n u m e r i c a l l y
greater t h a n the other.
A simply s u p p o r t e d b e a m of length L ( a n d originally straight) is
subjected to e n d thrusts P, together with a lateral l o a d w h i c h
increases uniformly in intensity from zero at o n e e n d to q p e r unit
length at the o t h e r e n d . D e d u c e a n expression from w h i c h the
b e n d i n g m o m e n t at a n y section of the b e a m m a y b e calculated.
A straight vertical c o l u m n is built in at the b a s e a n d free at the t o p .
It carries a vertical l o a d P at the t o p a n d a h o r i z o n t a l side l o a d i n g
which varies uniformly in intensity from zero at the t o p to q per u n i t
length at the b o t t o m . D e r i v e expressions for the m a x i m u m b e n d i n g
m o m e n t a n d m a x i m u m deflection.
REFERENCES
[1] C.T.Wang, Applied ElasticityMcGrdw-HiM,
N e w York, N.Y., 1953.
[2] S.Timoshenko and J.Gere. Theory of Elastic Stability, McGraw-Hill, N e w York, N.Y.,
1961.
CHAPTER 17
BENDING OF THIN FLAT PLATES
17.1
Introduction and Basic Assumptions. Strains and Stresses
A flat plate is a b o d y b o u n d e d b y two flat parallel surfaces, the
distance b e t w e e n these surfaces (called the thickness) being very small
in c o m p a r i s o n with the d i m e n s i o n s of the surfaces. T h e p l a n e parallel
to the two faces of the plate, a n d bisecting the thickness h, is called the
m i d d l e p l a n e . T h e c o o r d i n a t e axes are such t h a t the OXx a n d OX2 axes
are in the m i d d l e p l a n e a n d the OX3 axis is p e r p e n d i c u l a r to it (Fig.
17.1). In this chapter, the small deflection theory of thin plates is
presented. T h e a s s u m p t i o n s o n which this theory is b a s e d a n d their
implications are discussed, with the g e o m e t r y of d e f o r m a t i o n b e i n g
given special attention. T h e inverse m e t h o d is used to solve a few simple
p r o b l e m s gradually leading to N a v i e r ' s solution of the simply s u p p o r t e d
r e c t a n g u l a r plate. This solution is also o b t a i n e d using the principle of
m i n i m u m potential energy.
Fig. 17.1
520
Bending of Thin Flat Plates
521
If, w h e n subjected to a l o a d q = q(xx,x2\
the deflection of the thin
plate is small c o m p a r e d to its thickness, the following a s s u m p t i o n s
a t t r i b u t e d t o Kirchhoff m a y b e m a d e :
1. T h e m i d d l e p l a n e r e m a i n s u n s t r a i n e d . T h i s a s s u m p t i o n will m a k e
it u n n e c e s s a r y to consider the equilibrium of t h e forces a c t i n g o n a n
e l e m e n t of the p l a t e in the OXx a n d OX2 directions.
2. T h e n o r m a l strain e33 in the OX3 direction is small e n o u g h to be
neglected, a n d the n o r m a l stresses o33 is small c o m p a r e d to oxx a n d o22
so t h a t it c a n b e neglected in the stress-strain relations. Therefore,
e
du
3
33 = s T T3 = 0
9 m
^22 =
2
3 ^
=
V
I
*11 = ^ T T =
?
1
,
-
°22h
1
(17.1.1)
x
^(^22-^11).
3. T h e n o r m a l s to the m i d d l e p l a n e before b e n d i n g r e m a i n n o r m a l to
this p l a n e after b e n d i n g . T h i s m e a n s t h a t the out-of-plane shear strains
are small e n o u g h to b e neglected. Therefore,
T h e only shearing strain left is:
T h e p r e v i o u s e q u a t i o n s r e p r e s e n t a generalization of t h e e q u a t i o n s of
the simplified t h e o r y of b e a m s . T h e y a r e such t h a t all the strains a n d ,
c o n s e q u e n t l y , all the stresses c a n b e written in t e r m s of t h e deflection
u3 of the m i d d l e p l a n e . F o r e x a m p l e , Eq. (17.1.1) expresses the fact t h a t
u3 is o n l y a function of xx a n d x2, i.e.,
u3 = u3(xx,x2).
I n t e g r a t i n g E q s . (17.1.2), we get:
(17.1.4)
522
The Theory of Elasticity
and
1 L
7
"2 =
- * 3 ^ + / 2 ( * l > * 2 ) -
(
*
6
)
N o w , fx a n d f2 are t w o functions which represent d i s p l a c e m e n t s in the
m i d d l e p l a n e a n d , a c c o r d i n g to o u r first a s s u m p t i o n , these displacem e n t s are negligible. T h u s (Fig. 17.2):
Substituting E q s . (17.1.7) i n t o E q s . (17.1.1) a n d (17.1.3), we get:
2
i
d u->
e
=
u
- ^ J xJ
e 22 =X3
_
~ ~dxf
= ^ u - v o 2 ) 2
(17.1.8)
2
3 M
* I _^ 2i 2 - w u )
3
Solving for the stresses, the following e q u a t i o n s are o b t a i n e d :
(17.1.9)
(17.1.10)
Bending of Thin Flat Plates
523
)
1
• " - 1^ - ( V• H
°22 =
+ «
n
v
z
1 —v
>
(*22 +
~
T
^
e
\\) =
(
l
?
*
—
^
l
i
)
- ' ' "
L
- T ^ ~ 2
1 —V
I M
H I M I I I
0_
'
1
•—r
c r 0^
< y x f-
Fig. 17.3
2
M2
= 2 G e l2 =
£ ( i - 1y)
1 —v
a w3
(17.1.13)
3-^1 3^2
T h u s , at a given p o i n t , a n, a 2 , 2a n d a 12 vary linearly with x3 (Fig. 17.3).
w e neo u rg h n o t to affect the stressT h e fact t h a t a 33 w a s c o n s i d e r e d small
strain relations, a n d t h a t el3 a n d e 23
e neglected m a k e s it impossible
to d e t e r m i n e a 3 , 3a 1 , 3a n d a 23 from H o o k e ' s law. T h o s e three q u a n t i t i e s
c a n , however, b e d e t e r m i n e d from the differential e q u a t i o n s of equilibr i u m (7.4.6). N e g l e c t i n g the b o d y forces in E q s . (7.4.6), we get:
3a
n + pL
3x
9 a
12
,
+ p i = 0
3x
9 a 3
22
,
3x2 '
3 a 13
3x
3 a 23
3x9
3x3
3 a 33
3^o
(17.1.14)
= 0
(17.1.15)
= 0.
(17.1.16)
T h e b o u n d a r y c o n d i t i o n s for a 1 , 3a 2 , 3a n d a 33 are (Fig. 17.3):
(
1
524
The Theory of Elasticity
h
at x3
= +
a t x3
=
at x3
=
n
, h
h
= a 23 = 0
(17.1.17)
= 0
(17.1.18)
0 33
~2'
n
= -<?•
(17.1.19)
Substituting Eqs. (17.1.11) a n d (17.1.13) i n t o E q . (17.1.14), a n d integrating using the b o u n d a r y c o n d i t i o n (17.1.17), w e get:
--2^)(t-*0[£< 4
(17.1.20)
v2
Substituting Eqs. (17.1.12) a n d (17.1.13) i n t o E q . (17.1.15), a n d integrating using the b o u n d a r y c o n d i t i o n (17.1.17), we get:
=-^(£-*0fe< 4
v2
(17.1.21)
F r o m E q s . (17.1.20) a n d (17.1.21), the v a r i a t i o n of a 13 a n d a 23 with x 3
is seen to b e p a r a b o l i c . Substituting E q s . (17.1.20) a n d (17.1.21) i n t o E q .
(17.1.16), a n d integrating using the b o u n d a r y c o n d i t i o n (17.1.18), we
get:
(17.1.22)
2(1
This is the e q u a t i o n of a c u b i c p a r a b o l a (Fig. 17.4).
A t the u p p e r surface of the plate, the b o u n d a r y c o n d i t i o n (17.1.19)
gives:
V
^ = 1 2 ( f ^ )
4
" -
1 71 2 3
< - - )
Bending of Thin Flat Plates
h <7
525
-I
F i g . 17.4
4
Eq. (17.1.23) is a c o n d i t i o n of equilibrium
of the plate. Its c o u n t e r p a r t
in the t h e o r y of b e a m s is q = EI3 d u3 /dy\. T h e q u a n t i t y
L
12(1 — v )
is defined as the flexural rigidity of the plate. Eq. (17.1.23) is called
Lagrange's equation', in e x p a n d e d form, it is written:
(17.1.25)
L a g r a n g e ' s e q u a t i o n relates the vertical d e f o r m a t i o n u3 to the applied
l o a d q. It will b e derived in a different w a y in Sec. 17.4.
L o o k i n g b a c k at the previous a s s u m p t i o n s a n d derivations, we see
t h a t the stresses c a n b e g r o u p e d into three classes: the stresses parallel
to t h e m i d d l e p l a n e of the p l a t e o u, a 2 , 2a n d a 1 ; 2 t h e transverse n o r m a l
stress a 3 ; 3 a n d the transverse shear stresses a 13 a n d a 2 . 3 a 33 is of the
o r d e r of m a g n i t u d e of q, w h i c h rarely reaches values higher t h a n 50 psi.
It usually varies b e t w e e n 1 a n d 10 psi. This is negligible c o m p a r e d to
the h u n d r e d s of psi r e a c h e d b y a u a n d a 2 . 2 T h e total transverse l o a d
(Fig. 17.1) o n the p l a t e is of the o r d e r of qi}. F o r equilibrium, this l o a d
m u s t b e b a l a n c e d b y transverse shearing forces of the o r d e r of oX3
Lh or
Lh.
Therefore, a 13 a n d a 23 are of the o r d e r of q(L/h). If we consider
o23
the b e n d i n g of a strip of the p l a t e of u n i t width, t h e b e n d i n g m o m e n2t is
of the2 o r d e r of qi? a n d the resisting m o m e n t is of the o r d e r of o n h or
2
a 2/ 2
* (see Sec. 17.3). Therefore,
a n, a 22 (and, it is a s s u m e d , a 12 also),
are of the o r d e r of q(L/h) .
T h u s , since L / h is relatively large for thin
plates, then a n, a 2 , 2 .and a 12 are greater t h a n a 13 a n d a 2 , 3 a n d m u c h
526
The Theory of Elasticity
greater t h a n a 3 . 3 Since a 1 , 3 a 2 , 3 a n d a 33 are relatively small, o u r
neglecting of their effects o n the d i s p l a c e m e n t u3 was quite justified.
This, however, will lead to s o m e inconsistencies in the d e v e l o p m e n t of
the theory. O n e such inconsistency will a p p e a r w h e n writing the
b o u n d a r y c o n d i t i o n s at the free edge of a plate in Sec. 17.5.
17.2
Geometry of Surfaces with Small Curvatures
D u e to b e n d i n g , the m i d d l e p l a n e of a flat plate b e c o m e s slightly
curved. T h e resulting surface is described by the e q u a t i o n (Fig. 17.5):
u3 =
(17.2.1)
u3(xx,x2).
Let us first consider the slopes: A t a p o i n t P of the surface, a n d w h e n
when
p r o c e e d i n g in the OXx direction, the slope is given b y du3 /dxx;
p r o c e e d i n g in the OX2 direction, the slope is given b y 3W 3/3JC 2.
Recalling the definition of the g r a d i e n t of a furfction, we see t h a t the
two quantities 3w 3/dxx a n d 3w 3/dx2 are the c o m p o n e n t s of the vector
Q
dx,
Xf
)G
o/x
2
Q
refer en c e s y s tern
t
Fig. 17.5
Bending of Thin Flat Plates
—
,
_ du
_ du
V M3 = g r a d ^ - , , _ 3 + / , _3
527
(17.2.2)
w h o s e m a g n i t u d e is
^ 3
(17.2.3)
^ 3 )
6?«
3-x2 /
a n d w h o s e direction is a l o n g the n o r m a l n to the c o n t o u r lines (Fig.
17.5). T h e angle m a d e b y n with the OXx axis is given b y :
tan y =
du<i
ax
2
du~>
(17.2.4)
w^-/^ox
x
x2)
is given b y the
T h e m a x i m u m slope of the surface a t a p o i n t P(xXy
m a g n i t u d e of the g r a d i e n t vector at t h a t p o i n t ; the direction a l o n g
which this m a x i m u m slope occurs is t h a t of n a n d is given by Eq.
(17.2.4). T h e m i n i m u m slope of the surface at a p o i n t P(xx,x2)
is e q u a l
to zero, a n d it occurs in a direction 1 n o r m a l to n since it h a s to b e
t a n g e n t to the c o n t o u r line. T h e g r a d i e n t of u3 b e i n g a vector q u a n t i t y ,
it c a n b e r e p r e s e n t e d in a n y system of axes OX\, OX2 b y m e a n s of the
e q u a t i o n (see C h a p t e r 3):
3w
3
dx':
„l 3"3
* J dXj
ij=
1,2,
(17.2.5)
w h e r e itj are the direction cosines of the n e w system with respect to the
old o n e . I n t e r m s of the angle of r o t a t i o n 0 (Fig. 17.5), E q . (17.2.5)
becomes:
3w
3
dx\
d"3
3^2
cos 9
— sin 0
sin 6
cos 9
3w
3
x
3w
3
dx
2
dx
(17.2.6)
Eq. (17.2.6) gives the slopes w h e n p r o c e e d i n g in the directions of OX\
a n d OX2, respectively.
Let us now consider the curvatures'. I n the system of axes of Fig. 17.1,
the c u r v a t u r e of the surface at a p o i n t P a n d in a p l a n e parallel to the
OX{, OX3 p l a n e is a p p r o x i m a t e l y given by [see Eq. (12.2.1)]:
528
The Theory of Elasticity
(17.2.7)
I n a p l a n e parallel to the OX2, OX3 p l a n e , it is given b y :
(17.2.8)
C u a n d C 22 represent the r a t e at which the slope of u3 =
u3(x{,x2)
c h a n g e s w h e n p r o c e e d i n g in the OXx a n d OX2 directions, respectively.
In a d d i t i o n to the t w o s e c o n d derivatives of E q s . (17.2.7) a n d (17.2.8),
the following two m i x e d s e c o n d derivatives will b e n e e d e d :
(17.2.9)
and
(17.2.10)
C 12 represents the rate at w h i c h the slope 3 w 3/ 3 x 2 c h a n g e s w h e n
m o v i n g in the OXx direction, a n d C 21 represents the r a t e at which the
slope 3w 3/dxx c h a n g e s w h e n m o v i n g in the OX2 direction. T h e geometrical i n t e r p r e t a t i o n of C 12 a n d C 21 c a n b e o b t a i n e d as follows: C o n s i d e r
the s q u a r e element abed w h o s e sides are c h o s e n equal to unity. U n d e r
the effect of local twisting couples, it will take the s h a p e a'b'c'd\ s h o w n
in Fig. 17.6. O n this figure, we notice t h a t
u
3
™ " " «
=3w 3
+ a / 3 ^ \ ' - 3^"
dx2\dxj
W _du
3 3
"
(17.2.11)
3 ^
M U a _i 3w=3
~
dx~
2
8
"3 |
2
9
"3
(17.2.12)
(17.2.13)
(17.2.14)
Bending of Thin Flat Plates
529
Fig. 17.6
S u b t r a c t i n g E q . (17.2.11) from E q . (17.2.12), a n d E q . (17.2.13) from E q .
(17.2.14), w e o b t a i n the t w o following e q u a t i o n s :
2
(u3c + u3a
) - (u3b + u3d
) =
u
u
("3c + 3a) ~ ("3b + 3d) = 3
3 u
A^
2
= C 21
(17.2.15)
d u3
3
= C\2 •
(17.2.16)
2
Eqs. (17.2.15) a n d (12.7.16) s h o w t h a t C 12 = C 2 . 13 w 3/ 3 x 1 dx2 is called
the twist. Therefore, at a p o i n t P , a n d with respect to a system of axes
OXx, OX2, OX3, w e h a v e defined three q u a n t i t i e s — n a m e l y , t w o c u r v a tures a n d o n e twist. W e c a n p r o v e t h a t C N, C 2 , 2C 1 , 2a n d C 21 are t h e
c o m p o n e n t s of a tensor of the s e c o n d r a n k called the c u r v a t u r e tensor.
I n d e e d , let us c o n s i d e r the vector o p e r a t o r VQX d/dxxJ2d/dx2)
a n d the
vector g r a d i e n t Vu3(ix du3/dxx,l2du3/dx2).
I n a r o t a t i o n of c o o r d i n a t e s
a r o u n d the OX3 axis, the vector g r a d i e n t t r a n s f o r m s a c c o r d i n g to E q .
(17.2.5); n a m e l y , a c c o r d i n g to
530
T h e Theory of Elasticity
I n the s a m e w a y , t h e vector o p e r a t o r t r a n s f o r m s a c c o r d i n g to
im
/ m = 1 2
'
'
— = / —
dx\
dxm
I n this n e w system of c o o r d i n a t e s t h e c u r v a t u r e is written a s :
8
d
c
^
=
dx'fdx'j
k
J _ ( ^l) = / =
h^ 3
dx'iKdx'jJ
J dxm
dxk
9
im M
= /c l
(17.2.17)
^ jk^mk'
E q . (17.2.17) shows t h a t t h e CtJ
- s transform according to the law
g o v e r n i n g tensors of t h e s e c o n d r a n k . I n m a t r i x n o t a t i o n t h e c u r v a t u r e
tensor is written a s :
2
2
3 w3
d u3
2
2
3 w93
3 w3
3JC^
3*2 * i
(17.2.18)
Qi
Q2
T h e c u r v a t u r e tensor is s y m m e t r i c a n d , like t h e strain a n d stress tensors,
it c a n b e d i a g o n a l i z e d ; in o t h e r w o r d s , a system of reference axes c a n
b e f o u n d in w h i c h t h e off-diagonal terms C 12 a n d C 21 d i s a p p e a r . It is
also susceptible to a r e p r e s e n t a t i o n b y m e a n s of M o h r ' s circle a n d
enjoys t h e o t h e r p r o p e r t i e s p r e s e n t e d in C h a p t e r 3 o n linear s y m m e t r i c
transformations.
F o r a r o t a t i o n 9 a r o u n d t h e OX3 axis ( F i g . 17.5), E q . (17.2.17)
becomes:
2
2
C ' n = C nc o s 0 + C 2 s 2i n 0 + 2 C 1 s2i n 9 cos 9
2
2
(17.2.19)
C22 = C ns i n 0 + C 2 c2o s 0 - 2 C 1 s2i n 9 cos 9
(17.2.20)
C\2 = c 21 = - ( C u - C 2 ) 2s i n 9 cos 9 + C 1 c2o s 29.
(17.2.21)
T h e previous e q u a t i o n s a r e similar to those written for t w o d i m e n s i o n a l
states of stress a n d strain. T h e principal c u r v a t u r e s a r e given b y :
C
C, =
" + °
22
C
+ ^ (
C
" ~2
2
2
y + Ch
Q = ^ i ± ^ i _ ^ ( ^ r ^ y
+ cl,
(17.2.22)
(17.2.23)
X
Bending of Thin Flat Plates
531
a n d c o r r e s p o n d to the m a x i m u m a n d m i n i m u m c u r v a t u r e s . T h e y fall in
the principal planes of c u r v a t u r e whose directions are given by a n angle
8 = <(>, such t h a t
tan
2C 12
2(f)
c,
(17.2.24)
c 22
T h e invariants of the c u r v a t u r e tensor a r e :
=
Ql + Q2
C
C
Q l 22
2
~ ( 12)
C'n + Q 2
=
ll
CC
(17.2.25)
2
(17.2.26)
2 2 ~~ ( C i 2 ) -
T h e r e p r e s e n t a t i o n o n a M o h r d i a g r a m follows the s a m e c o n v e n t i o n s
established for the stresses in Sec. 7.11. Positive values of C u a n d C 22
f
are p l o t t e d o n the positive side of the o'Cn axis (Fig. 17.7), a n d negative
values o n the negative side. If C 12 is positive, it is plotted below the o Cn
axis for the c u r v a t u r e c o r r e s p o n d i n g to the m o r e clockwise of the two
p l a n e s in which Cu(i = 1,2, n o s u m ) is c o m p u t e d ; if C 12 is negative, it
is plotted a b o v e the o'Cn axis. F o r the less clockwise of the two planes,
the location of the p o i n t representing C 12 is reversed. Fig. 17.7 shows
the r e p r e s e n t a t i o n for C n,
C 2 , 2 a n d C 12 positive. A p l a n e
(OX\,
\f
d) °?
k
\\
F i e . 17.7
c,
532
The Theory of Elasticity
OX3)
making an angle 6 = 6X with (OXx, OX3) will have in it a positive
curvature equal to o'd a n d a negative twist equal to x\ d. T h e plane making
(j> with (0XX, OX3) is the major principal plane of curvature. In this plane,
the twist is equal to zero.
A surface which is convex d o w n w a r d will h a v e values of Cu a n d C 22
which are negative, while a surface which is convex u p w a r d will h a v e
values of C H a n d C 22 which are positive. Fig. 17.6 shows the direction
of a positive twist.
If the two principal c u r v a t u r e s Cx a n d C 2 are the same, M o h r ' s circle
shrinks to a point, the c u r v a t u r e is the s a m e in all directions, a n d there
is n o twist in a n y direction: T h e surface is purely spherical at this point.
If the two principal c u r v a t u r e s are equal in m a g n i t u d e a n d opposite in
sign, the result is a s a d d l e p o i n t : T h e r e are n o c u r v a t u r e s in p l a n e s
m a k i n g 45° with the principal planes, j u s t twists (Fig. 17.8).
Fig. 17.8
Bending of Thin Flat Plates
533
Finally, in terms of the c u r v a t u r e s a n d twist, the stress-strain relations
(17.1.11) to (17.1.13) c a n b e written as follows:
C
= 1,2).
T T - ^ u ~ SijY
^x,Cnn
(iJ,n
(17.2.27)
In m a t r i x n o t a t i o n , this e q u a t i o n c a n b e written in either of the two
forms:
aOil
_ 21
a 12
a 2 _2
Ex3
1 +
Qi
V
22
a
a 12
1
Ex-, 2
v
1 - v
0
(C„ +
v
Q
(17.2.28)
1
0
2) 0
1
0
1
0
Cl2_
_Ql
2
l-v
Q2
C 22
0
1
v
C,2
2
Oil
°22
a
\l
17.3
Ex3 1
1 - V
1
f
0
J>
1
0
'
9 "2 3 "
dx
2
(17.2.29)
d u3
dX\
2
0
0
1 -
V
3 «3
dx{ dx2
Stress Resultants and Stress Couples
Let us consider a n e l e m e n t of a plate u n d e r the a c t i o n of a n o r m a l
d i s t r i b u t e d l o a d q = q(xi,x2).
In a d d i t i o n to b e n d i n g a n d twisting
m o m e n t s acting o n the sides of the element, there will b e shearing forces
d u e to the shearing stresses a 13 a n d a 2 . 3 B o t h m o m e n t s a n d shearing
forces are expressed per unit length of p l a t e in the OXx a n d the OX2
directions.
T h e directions for positive a n d negative stresses established in Sec. 7.2
always h o l d a n d a p p l y to n o r m a l a n d shear forces o n the plate's cross
534
The Theory of Elasticity
sections. The convention for moments is that a positive bending
moment
produces a positive normal stress in the positive half of the plate and a
positive twisting moment causes a positive shear stress on the positive half
of the plate. Both m o m e n t s a n d shear forces carry the subscripts
c o r r e s p o n d i n g to the stresses they cause. If o n e wishes to use the righth a n d rule for the positive m o m e n t s , the t h u m b h a s to p o i n t in the
direction of the d o u b l e a r r o w s in Fig. 17.9.
M
2Z
Fig. 17.9
T h e b e n d i n g m o m e n t per unit length a l o n g the OXx direction is given
by:
Mi
=J
h
a x dx
u
3
3
=
- J_
1 7
k YZr^(Cu
+ "C22
)dx3
( -3-l)
or
M „ = -D(CU
+
(17.3.2)
vC22
).
Similarly, the b e n d i n g m o m e n t per unit length a l o n g the OX2 direction
is given by:
1733
r+i
22
= J
c22
x3dx3
= -D(C22
+ vCu).
< --)
T h e twisting m o m e n t per unit length a l o n g the OXx direction is given
by:
r+1
Since ox2 =
o2\,
(17.3.4)
Bending of Thin Flat Plates
21 =
I
= - Z ) ( l - v)C2l = M 1 . 2
h°2\*3dx3
535
(17.3.5)
N o t i c e t h a t the effects of o 13 a n d a 23 were neglected in the expressions
for M , 2 a n d M 2 , 1respectively. This will result in a n inconsistency in the
writing of the b o u n d a r y c o n d i t i o n s for a free edge in Sec. 17.5. I n index
n o t a t i o n , Eqs. (17.3.2) to (17.3.5) are written as follows:
M
M =
v
3
3
c _ o
£/?
Evh 2
r
12(1 + „ ) 0
*12(l-„ )
( 1 J , K = 1,2).
(17.3.6)
I n m a t r i x n o t a t i o n , we h a v e :
Mn
M 12
3
M1' 2
M 22
C, 2
£/i
12(1 + v)
12(1 -
C, 2 c 22
1
1
o
v)
o
1
(17.3.7)
or
M 22
_ M 1 _2
1
v
0
= — Z> y
1
0
0
0
1 - *<
C22
C, 2
2
M 22
A/,2
1
V
0
9 «3 "
dxf
2
= -£> v
1
0
3 " 23
3;c
2
0
0
1 -
V
(17.3.8)
a «3
3x, 3 x 2
T h e previous e q u a t i o n s show that the four quantities Mtj{iJ = 1 , 2 ) are
the c o m p o n e n t s of a s y m m e t r i c tensor of the s e c o n d r a n k w h o s e
properties are similar to those of the c u r v a t u r e tensor. T h e tensor is
called the m o m e n t tensor M . If Eq. (17.3.8) is inverted, we get:
536
The Theory of Elasticity
Cy
C 22
-12
1
1
Z)(l -
2 -v
v)
0
-v
0
My
1
0
M~22
0
1 4- v
Mi12
(17.3.9)
E q s . (17.3.8) a n d (17.3.9) represent the stress-strain relations for thin
plates in t e r m s of m o m e n t s a n d c u r v a t u r e s .
All the e q u a t i o n s written for the c u r v a t u r e tensor a p p l y to the
m o m e n t tensor. T h u s E q s . (17.2.17) to (17.2.26) c a n be rewritten here
with M replacing C. T h e r e p r e s e n t a t i o n o n M o h r ' s d i a g r a m follows the
c o n v e n t i o n s established for stresses. Positive values of Mxx a n d M22 (as
s h o w n in Fig. 17.9) are p l o t t e d o n the positive side of the o'Mn axis, a n d
negative values o n the negative side. If Ml2 is positive, it is p l o t t e d
b e l o w the o'Mn axis for the p l a n e w h o s e o u t w a r d n o r m a l is parallel to
the m o r e clockwise of the t w o axes (here OX2); if MX2 is negative, it is
p l o t t e d a b o v e the o'Mn axis. Fig. 17.10a shows the r e p r e s e n t a t i o n o n
M o h r ' s d i a g r a m for the positive values of M n, M22
, a n d MX2 in Fig.
17.9. A p l a n e w h o s e o u t w a r d n o r m a l m a k e s a n angle 0 = 6X with the
OXx axis (Fig. 17.10b), will h a v e a positive b e n d i n g m o m e n t M\x acting
o n it e q u a l to o'd a n d a negative twisting m o m e n t M\2 equal to x\ d. T h e
p l a n e w h o s e n o r m a l m a k e s <f> with OXx is subjected to the major
principal b e n d i n g m o m e n t a n d does n o t suffer a n y twisting m o m e n t .
T h e quantities in Fig. 17.10 are related to the quantities in Fig. 17.7
Fig. 1 7 . 1 0
b y m e a n s of the coefficients in Eq. (17.3.8). T h e negative sign in front
of the m a t r i x of coefficients in Eq. (17.3.8) m u s t b e kept in m i n d w h e n
trying to visualize m o m e n t s a n d c u r v a t u r e s .
T h e shearing forces p e r unit length o n planes n o r m a l to the OXx a n d
OX2 directions are given b y :
Bending of Thin Flat Plates
v
\3
= J ,
ol3
dx39
V23 = J
o23
dx
537
(17.3.10)
Substituting E q . (17.1.20) i n t o E q . (17.3.10) a n d integrating, we o b t a i n
:
the following expression for VX3
'13
Vi3
(17.3.11)
Similarly,
V2i = -D^-(V^)
=
3 M 12
dx
3^22
dx
x
(17.3.12)
2
2 t h a t V a n d V are the two c o m p o Eqs. (17.3.11) a n d (17.3.12) s h o w
X3
23
n e n t s of the vector [-D g r a d ( V w 3) ] . T h u s , in m a t r i x n o t a t i o n :
^3
=
-D
(17.3.13)
1
V3
2
E q s . (17.3.11) a n d (17.3.12) will b e derived in a different w a y in Sec.
17.4. K n o w i n g VX3 a n d V23 at a point, the vertical shearing forces per
unit length o n a pair of o r t h o g o n a l p l a n e s w h o s e n o r m a l s h a v e direction
cosines ttj(ij = 1,2) with OXx a n d OX2 c a n b e o b t a i n e d from the
t r a n s f o r m a t i o n f o r m u l a (see C h a p t e r 3):
(17.3.14)
Therefore, for the r o t a t i o n s h o w n in Fig. 17.10b, we h a v e :
>'l3~
cos 0X
sin 0X
— sin 0X cos 6X
(17.3.15)
V3
2
538
The Theory of Elasticity
O n c e the m o m e n t s a n d the shearing forces are k n o w n , the n o r m a l
a n d the shearing stresses c a n b e c o m p u t e d : Substituting Eq. (17.3.9)
into Eq. (17.2.29), we o b t a i n a n, a 2 , 2 a n d a 12 in terms of the b e n d i n g
a n d twisting m o m e n t s :
I2x,
3
h
My
(17.3.16)
M,22
12
°\2
F r o m E q s . (17.1.20), (17.1.21), (17.3.11), a n d (17.3.12), we o b t a i n a 13
a n d a 23 in terms of the shearing forces:
2
*h )[y2K
'
2hV
_a J
23
(17.3.17)
T h e similarities b e t w e e n the expressions for the stresses in plates a n d in
b e a m s are clearly seen in the two previous e q u a t i o n s .
17.4
Equations of Equilibrium of Laterally Loaded Thin Plates
T o study the equilibrium of a small element of plate, o n e m u s t
consider the variations in the m o m e n t s a n d shears a l o n g the OXx a n d
OX2 directions. Such variations are s h o w n in Fig. 17.11. S u m m i n g the
vertical forces leads to
dx
x
ox
2
(17.4.1)
539
Bending of Thin Flat Plates
T a k i n g m o m e n t s a b o u t t h e OX2 axis a n d neglecting higher
differentials leads t o
order
^LL +
(17.4.2)
n -V 0,
ox{
ox2
which is t h e s a m e as E q . (17.3.11). T a k i n g m o m e n t s a b o u t t h e OXx axis
a n d neglecting higher o r d e r differentials leads to
^
^
_+ %
, =0
07.4.3)
w h i c h is t h e s a m e a s E q . (17.3.12). If w e n o w substitute t h e values of Vu
a n d V23from E q s . (17.4.2) a n d (17.4.3) into E q . (17.4.1), w e get:
2
2
d M2
n
dx
2
3 +z
M 12
d M22 ^_
dxj
dxxdx2
( 1 7q4 4 )
-
In t e r m s of t h e c u r v a t u r e s , this e q u a t i o n b e c o m e s :
2
2
9 Q2 i
dx
2
+92^ 1 2
dx^x
+3 C a _ •?
dxl
d-
)
In t e r m s of t h e d i s p l a c e m e n t M 3, w e o b t a i n t h e following e q u a t i o n :
4
9 2" 3 2
d
_S.
34
3x 3x2
3S
3 ^
4 =
V 1 =
/)•
3
(17.4.6)
2
This is L a g r a n g e ' s e q u a t i o n , w h i c h w a s derived in a different w a y in
Sec. 17.1. It is t h e basic p l a t e e q u a t i o n a n d a n y u3(xx ,x2) satisfying it is
a solution of a plate p r o b l e m . O n e m u s t r e m e m b e r , however, t h a t in E q .
(17.4.6) t h e effects of Vl3 a n d V23o n u3 h a v e b e e n neglected since t h e
relations b e t w e e n m o m e n t s a n d c u r v a t u r e s d i d n o t a c c o u n t for a 13 a n d
o23
:
It is recalled t h a t this is t h e result of t h e third a s s u m p t i o n in Sec.
17.1.
17.5
Boundary Conditions
T h e edges of a plate m a y b e (1) built in, (2) simply s u p p o r t e d , t h a t is,
free t o r o t a t e a r o u n d t h e edge b u t n o t free to deflect there, or (3) free,
t h a t is u n s u p p o r t e d . Let u s consider e a c h case separately.
1) Built in or clamped edge (Fig. 17.12). A t t h e built in edge, w e h a v e :
540
T h e Theory of Elasticity
Fig. 17.12
5 1)
' '
(fel,=.-°-
w h e r e OXx is the n o r m a l to the c l a m p e d edge.
2) Simply supported edge (Fig. 17.13). A t a simply s u p p o r t e d edge, we
have:
Fig. 1 7 . 1 3
(u3)xi=0 = 0,
(Mu)xi=0
= -D(CU
+ vC21
) = 0,
(17.5.2)
2 s u p p o r t e d edge. H o w e v e r , since
w h e r e OXx is the n o r m a l to the simply
at xx = 0, u3 = 0, a n d du3/dx2 = d u3/dx2
= 0, the b o u n d a r y c o n d i tions for this case b e c o m e :
53
- ' >
» ( S f L o - ° -
3. Free erfge (F/g. 17.14). A t a free edge, there m u s t b e n o b e n d i n g or
twisting m o m e n t s as well as n o shearing forces. O n e , therefore, c o u l d
write:
(Mn)xi=a
= 0,
(Vl3
)x^a
= 0,
(Mn)X]=a
= 0.
(17.5.4)
Bending of Thin Flat Plates
541
Fig. 17.14
It w a s s h o w n , however, b y Kirchhoff t h a t two c o n d i t i o n s are sufficient
for the c o m p l e t e d e t e r m i n a t i o n of u3 satisfying Eq. (17.4.6). This
inconsistency is d u e to the a s s u m p t i o n t h a t eX3 = e23 = 0. This a s s u m p tion, w h i c h resulted in neglecting the effects of oX3 a n d a 23 o n the
deflection u3, was m a d e so t h a t all the strains, a n d c o n s e q u e n t l y the
stresses, c o u l d b e expressed in t e r m s of o n e d e p e n d e n t variable u3. T h e
twisting m o m e n t MX2 at xx = a c a n n o t b e specified i n d e p e n d e n t l y of
VX3
. T h a t o n e c a n n o t specify three c o n d i t i o n s a l o n g the b o u n d a r y c a n
b e seen b y e x a m i n i n g the solution of t h e b i h a r m o n i c e q u a t i o n [1]:
4
V u3 = 0.
(17.5.5)
This solution involves the use of c o m p l e x variable t h e o r y a n d we are
only interested h e r e in the final a n s w e r : If
z = xx + ix2 a n d z = xx — ix2,
w h e r e / = \ / - - T , the solution of Eq. (17.5.5) is:
u3 = R e a l p a r t of [zFx{z)
(17.5.6)
(17.5.7)
+ F2(z)].
T h e functions Fx a n d F2 are analytic functions a n d are i n d e p e n d e n t of
e a c h other. By m e a n s of Fx a n d F 2, w e c a n satisfy t w o a n d only t w o
i n d e p e n d e n t c o n d i t i o n s at the b o u n d a r y of a plate. U s i n g a variational
m e t h o d , Kirchhoff [2] s h o w e d t h a t t h e b o u n d a r y c o n d i t i o n s for the free
edge a r e :
-0,
^
= (^3 + ^
)
^
= 0,
(17.5.8)
w h e r e RX3 is the vertical r e a c t i o n at the edge. In effect, the s e c o n d
e q u a t i o n (17.5.8) states t h a t a distribution of twisting m o m e n t s Mx2
along a n edge is equivalent to a distribution of vertical shearing forces.
This equivalence is illustrated in Fig. (17.15): T h e twisting m o m e n t
542
The Theory of Elasticity
c o r r e s p o n d i n g to a distribution of shearing stresses a 12 in (a) is r e p l a c e d
b y a statically equivalent system in (b). This statically equivalent system
results in a distributed shearing force equal to 9 M 12/ 3 x 2 per unit length
(c a n d d) a n d t w o e n d forces e q u a l to M 1 . 2In t e r m s of u3, the b o u n d a r y
c o n d i t i o n s (17.5.8) are written a s :
Fig. 17.15
17.5.9)
Remark
In the case of plates with curvilinear b o u n d a r i e s , the expressions of
the slopes, curvatures, m o m e n t s , a n d shearing forces m u s t first b e
o b t a i n e d in a system of axes formed b y the n o r m a l a n d the t a n g e n t to
the b o u n d a r y . T h e formulas for such t r a n s f o r m a t i o n s of axes were given
in the previous sections.
17.6
S o m e Simple Solutions of Lagrange's Equation
U s i n g the inverse m e t h o d , a n u m b e r of simple p r o b l e m s related to
thin plates h a v e b e e n solved. A n expression for u3 = u3(xx,x2),
satisfy-
Bending of Thin Flat Plates
543
ing L a g r a n g e ' s e q u a t i o n , is e x a m i n e d a n d the type of b o u n d a r y c o n d i tions to which it applies is o b t a i n e d . L a g r a n g e ' s e q u a t i o n b e i n g linear,
superposition c a n b e used to g e n e r a t e solutions to n e w p r o b l e m s . Let us
consider the following cases:
1. q = 0, w 3 = — Bx\, w h e r e B is 4a c o n s t a n t . This expression for w3
satisfies the b i h a r m o n i c e q u a t i o n V w 3 = 0. It is, therefore, a possible
solution to the plate p r o b l e m . T h e s h a p e of the p l a t e (Fig. 17.16) is a
flat p a r a b o l i c trough. T h e c u r v a t u r e tensor is:
Fig. 1 7 . 1 6
C =
-IB
0
0
0
(17.6.1)
a n d is the s a m e everywhere. T h e m o m e n t tensor is:
M =
2BD
0
0
2BDv
(17.6.2)
T h e shearing forces Vl3 a n d V23 are b o t h e q u a l to zero. Therefore, if we
cut from a n infinite plate a rectangle of sides e a n d / (Fig. 17.6), the
outside l o a d i n g w o u l d consist of a uniformly distributed b e n d i n g
m o m e n t 2BD a l o n g the edge / a n d of a uniformly d i s t r i b u t e d b e n d i n g
m o m e n t 2BDv a l o n g the edge e. I n spite of the latter, the e d g e / r e m a i n s
straight: I n d e e d 2BDv is the m o m e n t necessary t o c o m p e n s a t e for t h e
Poisson effect, which w o u l d h a v e c a u s e d a n anticlastic surface if Mn
w a s the only m o m e n t acting o n the finite plate.
2 w3 = — Bx\.
Similar results are o b t a i n e2d if we set
2. q = 0, u3 = —B(x\ + x ) = —Br , w h e r e B is a c o n s t a n t . T h e
s h a p e of the d e f o r m e d plate is t h a t of a p a r a b o l o i d of revolution. u3
544
The Theory of Elasticity
4
satisfies the b i h a r m o n i c e q u a t i o n V w = 0. T h e c u r v a t u r e tensor is:
0
-2B
C =
0
(17.6.3)
-IB
T h e m o m e n t tensor is:
M =
+ v)
2BD{\
0
0
+ v)
2BD{\
(17.6.4)
T h e shearing forces VX3a n d V23 are b o t h e q u a l to zero. W e h a v e h e r e a
case of spherical b e n d i n g . W h i l e we started with a p a r a b o l o i d of
revolution, we o b t a i n c u r v a t u r e s w h i c h are c o n s t a n t at each p o i n t ; this
c o r r e s p o n d s to a spherical s h a p e . T h e inconsistency c o m e s from the
a p p r o x i m a t e expression of the c u r v a t u r e a d o p t e d in the t h e o r y of t h i n
plates.
3. q = 0, u3 = B(xx — x2), w h e r e B is a c o n s t a n t . T h e s h a p e of the
d e f o r m e d plate is t h a t of a saddle. Lines originally parallel to the OXx
axis curve d o w n w a r d s , a n d lines originally parallel to the OX2 axis
curve u p w a r d s (Fig. 17.8). T h e c u r v a t u r e tensor is the s a m e everywhere
a n d its c o m p o n e n t s a r e :
C =
2B
0
0
-2B
I—
(17.6.5)
_i
T h e r e p r e s e n t a t i o n of C o n M o h r ' s d i a g r a m is s h o w n in Fig. 17.8. T h e
m o m e n t tensor is:
M =
-2BD{\
0
-v)
0
2BD{\
-
v)
(17.6.6)
Lines m a k i n g 45° with OXx a n d OX2 r e m a i n straight after b e n d i n g ;
they are subjected only to twists. Therefore, if as s h o w n in Fig. 17.8, a
p l a t e abed is cut from the original plate, its sides will b e subjected to
twisting m o m e n t s only.
4. q = 0, u3 = Bxx x2, w h e r e B is a c o n s t a n t . W i t h i n a r o t a t i o n of
c o o r d i n a t e axes, this case is equivalent to the previous o n e . I n d e e d , t h e
c u r v a t u r e tensor is the s a m e everywhere a n d its c o m p o n e n t s a r e :
C =
T h e m o m e n t tensor is given b y :
0
B
B
0
(17.6.7)
Bending of Thin Flat Plates
M =
0
-BD
BD
0
545
(17.6.8)
After d e f o r m a t i o n , the plate takes the s h a p e s h o w n in Fig. 17.17a. T h e
X
2
W
1
"
^-f^
Fig. 1 7 . 1 7
(c)
r e p r e s e n t a t i o n o n M o h r ' s d i a g r a m for c u r v a t u r e s a n d m o m e n t s is s h o w n
in Fig. 17.17b a n d 17.17c, respectively. Because of the sign c o n v e n t i o n s
a d o p t e d in the study of flat plates, we see t h a t c o r r e s p o n d i n g p o i n t s o n
the t w o M o h r d i a g r a m s fall o n o p p o s i t e sides of the h o r i z o n t a l . F r o m
these t w o d i a g r a m s , we c o n c l u d e t h a t lines m a k i n g 45° with OXx a n d
OX2 are subjected to b e n d i n g m o m e n t s alone. Therefore, if a plate
w h o s e sides m a k e 45° with the axes is cut from the original o n e , it will
b e subjected to p u r e b e n d i n g o n its b o u n d a r i e s .
5. q — ?, u3 = B sm(Hxx /a)sm(Hx2 /b\ w h e r e B, a, a n d b are constants. T h e plate is infinite in extent a n d is b e n t in a d o u b l e sinusoidal
s h a p e (Fig. 17.18).
O
f
Fig. 1 7 . 1 8
J
546
T h e Theory of Elasticity
Substituting t h e expression of t/ 3 into L a g r a n g e ' s e q u a t i o n , w e get:
vS-n«(«-i±£)Vf
T h e d o u b l y sinusoidal shape will, therefore, b e o b t a i n e d if:
,- n«(«-i±£) *s,„^ i„^.
2
D
<»*'<»
S
Setting:
A
=
*(<?-±Jp)
,
BDU
(17-6.11)
the l o a d distribution is
2
^ = ^ s i n ^ - s i n ^ - = | M. 3
07.6.12)
F r o m E q s . (17.3.8), t h e m o m e n t s a r e given b y :
2
M „ = n £ > ( ^ + ^ ) M3
(17.6.13)
2
M 22 = n Z > ( ^ + ^ ) M3
M I2 = -^BD(\
-
(17.6.14)
cos
v)cos
(17.6.15)
T h e shearing forces a r e given b y :
1
b h ^ *
« 5 . " ^
£
( ?
+
? >
t
a
^
6 1) 6
* ? L
" '
!
^ -
(
'
7 A I 7 )
Eqs. (17.6.12) t o (17.6.17) show that t h e ratio of t h e loading to t h e
deflection is a c o n s t a n t which is t h e s a m e for all p o i n t s ; also, t h e ratio
of t h e b e n d i n g m o m e n t s is a c o n s t a n t .
A l t h o u g h t h e previous solution is, in itself, of little practical i m p o r tance, it d o e s p r o v i d e u s with t h e m e a n s of o b t a i n i n g a solution for a n y
a l t e r n a t e l o a d i n g using a F o u r i e r series of sine c o m p o n e n t s ; also, if w e
cut o u t of t h e infinite plate a rectangle of sides a a n d b, we h a v e :
< l
Bending of Thin Flat Plates
1/3 = 0 a n d Mxx = 0
for x x = 0 a n d x x = a
(17.6.18)
w 3 = 0 a n d M 22 = 0
for x 2 = 0 a n d x 2 = b.
(17.6.19)
547
T h e s e are the b o u n d a r y c o n d i t i o n s for a simply s u p p o r t e d r e c t a n g u l a r
plate of sides a a n d b l o a d e d a c c o r d i n g to Eq. (17.6.12). T h e previous
solution is therefore valid for a simply s u p p o r t e d r e c t a n g u l a r plate, if we
c a n p r o v i d e a l o n g the sides the twisting m o m e n t s a n d shearing forces
given b y E q s . (17.6.15) to (17.6.17) (Fig. 17.19). N o w , the twisting
m o m e n t s acting o n the sides of the rectangle parallel to the OXx axis are
statically equivalent to a c o n t i n u o u s l y distributed shear l o a d i n g e q u a l to
3 M 21 /dxx a n d to c o n c e n t r a t e d forces acting at the corners, e a c h equal
in m a g n i t u d e to M2X at these p o i n t s (see Sec. 17.5). I n the s a m e way, the
twisting m o m e n t s acting o n the sides parallel to the OX2 axis are
equivalent to a c o n t i n u o u s l y d i s t r i b u t e d shear l o a d i n g equal to
Fig. 1 7 . 2 0
548
The Theory of Elasticity
3 M 12 / 3 x 2 a n d to c o n c e n t r a t e d forces acting at the corners, each equal
2
in m a g n i t u d e to M 12 at these p o i n t s (Fig. 17.20). A t the four corners
of
the plate, the m a g n i t u d e of the c o n c e n t r a t e d force is
2(H /ab)BD(\
— v). T h e r e a c t i o n a l o n g side b is given b y :
T h e r e a c t i o n along side a is given b y :
T h e negative sign in the two previous e q u a t i o n s obviously indicates t h a t
the sinusoidally distributed r e a c t i o n acts u p w a r d s .
In s u m m a r y , we see t h a t a r e c t a n g u l a r plate l o a d e d a c c o r d i n g to E q .
(17.6.12) will h a v e m o m e n t s a n d shearing forces distribution given b y
Eqs. (17.6.13) to (17.6.17). T h e reactions a l o n g the sides are given b y
Eqs. (17.6.20) a n d (17.6.21) a n d , in a d d i t i o n , there will b e four c o n c e n t r a t e d forces acting at the four corners. T h e necessity of these c o n c e n t r a t e d forces is easy to visualize: W h e n l o a d e d , the plate takes a dishlike s h a p e a n d the corners t e n d to rise; they h a v e to b e pressed d o w n .
It is interesting to n o t e t h a t the s u m of the c o n c e n t r a t e d forces is e q u a l
to the s u m of the distributed shear l o a d i n g d u e to the twisting m o m e n t s .
Indeed:
* / (^L* - / (^L - ^ - '>- -'
<7, 622)
0
0
This solution c a n n o w b e used to calculate the deflection of a simply
s u p p o r t e d r e c t a n g u l a r plate u n d e r a n y loading.
17.7
Simply Supported Rectangular Plate. Navier's Solution
Let us consider a simply s u p p o r t e d r e c t a n g u l a r plate of sides a a n d b
subjected to arbitrary l o a d i n g (Fig. 17.21). T h e deflection w3 will b e
a s s u m e d to h a v e the form:
" 3 = 2 2
m=\
n=\
^ s i n
-s-i
2
sin - r - - .
(17.7.1)
Bending of Thin Flat Plates
549
Fig. 17. 21
This a s s u m p t i o n satisfies the b o u n d a r y c o n d i t i o n s (17.6.18) a n d
(17.6.19), a n d is d i c t a t e d b y the results o b t a i n e d in the previous section.
Substituting Eq. (17.7.1) i n t o L a g r a n g e ' s e q u a t i o n , we get:
w= i „=i
\
<r
b* /
0
Setting:
^-z>n<(£+
,2
07.7.3)
the l o a d d i s t r i b u t i o n is:
<7= 2
2
^ s i n - ^ s i n - ^ .
(17.7.4)
m = l n=\
By m e a n s of E q . (17.7.4), o n e c a n a p p r o x i m a t e a n y l o a d d i s t r i b u t i o n
q = q(x{,x2)
a n d easily o b t a i n the c o r r e s p o n d i n g d i s p l a c e m e n t s u3
= u3(xl,x2)
from Eqs. (17.7.3) a n d (17.7.1).
T h e first step is to find the value of the coefficients Amn of E q . (17.7.4)
as a function of q. F o r that, we m a k e use of the t w o following identities:
a
sm — ^
/
sm
nllxx
a
sm —^— dxx = 0 w h e n n ^
k
(17.7.5)
. klixx
,
^ ,
sm
rfxj g = ^ w h e n n = k.
(17.7.6)
550
T h e Theory of Elasticity
M u l t i p l y i n g b o t h sides of E q . (17.7.4) b y sin kUxx/a
from 0 t o a, we get:
a n d integrating
a
1 77
dxx = § 2 =14 « s m - 7 - ^ .
q(xx,x2)sm
0
M u l t i p l y i n g b o t h sides of E q . (17.7.7) b y sin jUx2/b
from 0 t o b, we get:
ft
ab
( - - )
"
a
55
0
s ni
4(*i > * 2 )
a n d integrating
s
^ T " ^1 ^ 2 =
i An
kj
•
1
(
78 J
- )
0
Therefore, given t h e function q(xx,x2),
we c a n find a n y coefficient
Akj
.
Substituting E q . (17.7.8) into E q . (17.7.3) a n d using the subscripts m a n d
n instead of k a n d 7 , w e get:
(17.7.9)
0
0
T h e deflection w 3, c o r r e s p o n d i n g to the l o a d i n g q(xx,x2),
can now be
o b t a i n e d using E q s . (17.7.1) a n d (17.7.9). Let us n o w consider t w o cases:
1. q(xx,x2)
= q0 is uniformly distributed over the area of the plate. In
this case, t h e d o u b l e integral in E q . (17.7.9) c a n b e split i n t o t w o single
integrals w h i c h are quite e l e m e n t a r y to evaluate. B MNis f o u n d to b e
given b y
(17.7.10)
w h e r e m a n d n t a k e only o d d values 1,3,5—00. T h e expression of w 3 is,
therefore,
Bending of Thin Flat Plates
mllxl
11
(
U m= 1 n= 1
.
nHx2
2
2
m
551
.
/7
\
with m = 1,3,5—oo a n d n = 1,3,5—oo. This series converges quite
rapidly. T h e m a x i m u m deflection occurs at the center of the plate, a n d
is:
(» )
3 mx a-
2, 2
/
2„ 2 y
(17.7.12)
U s i n g the expression of the deflection given b y E q . (17.7.11), we c a n
find b y differentiation the m o m e n t s a n d t h e shearing forces at a n y p o i n t
of the plate.
is a concentrated force P acting at any point xx = e, x2 = /
2. q(xx,x2)
(Fig. 17.22). I n thise case, we c a n replace the force P b y a c o n t i n u o u s
l o a d q0 d i s t r i b u t e d over a n infinitisimal a r e a dxx, dx2 such t h a t :
Fig. 1 7 . 2 2
1 71 3
*o = ^
-
< - - >
T h e function q(xx,x2)
in E q . (17.7.9) is zero everywhere except at the
p o i n t xx = e, x2 = / w h e r e it is e q u a l t o q0. T h e d o u b l e integral in E q .
(17.7.9) b e c o m e s P s i n ( m n e / a ) s i n ( m I I / / 6 ) a n d
4P sin
* ~ -
/
sin
V2
^
•
7 1) 4
2
- -
552
The Theory of Elasticity
T h e deflection of the m i d d l e p l a n e is, therefore,
•
q
=
C
V
V
(
mile
•
^¥
2
2
m2 , m 2\
a
1
sin
sinn— r ^ - .
(17.7.15)
b)
T h e m o m e n t s a n d the shearing forces c a n n o w b e o b t a i n e d at a n y p o i n t
of the plate b y differentiation.
Remark
A large n u m b e r of p r o b l e m s related to simply s u p p o r t e d r e c t a n g u l a r
plates c a n b e f o u n d in [2]. C a r e should b e exercised w h e n using the
e q u a t i o n s for the m o m e n t s a n d the shears, since sign c o n v e n t i o n s differ
from text to text.
17.8
Elliptic Plate with Clamped Edges under Uniform Load (Fig. 17.23)
Fig. 1 7 . 2 3
Let us a s s u m e t h a t the deflection u3 is given b y :
,2
, / -*1
. -*2
w3
w h e r e the e q u a t i o n of the b o u n d a r y of the ellipse is
42 + 4 =21
(17-8.2)
a
b
a n d A is a c o n s t a n t . « 3 is e q u a l to zero o n the b o u n d a r y a n d the two
c o m p o n e n t s of the gradient vector,
Bending of Thin Flat Plates
553
(17.8.3)
and
(17.8.4)
also vanish o n the b o u n d a r y . Therefore, the a s s u m e d deflection satisfies
the b o u n d a r y c o n d i t i o n s of a c l a m p e d plate. Substituting Eq. (17.8.1)
into L a g r a n g e ' s e q u a t i o n , we o b t a i n :
(17.8.5)
which shows t h a t q(xx,x2)
(17.8.5), we d e d u c e t h a t
is a c o n s t a n t q0. F r o m E q s . (17.8.1) a n d
(17.8.6)
is the solution to the p r o b l e m of the elliptic plate with c l a m p e d edges
subjected to a uniformly d i s t r i b u t e d l o a d q0.
T h e m o m e n t s at a n y p o i n t are o b t a i n e d from Eq. (17.3.8). F r o m these
values, the b e n d i n g a n d twisting m o m e n t s a r o u n d a n y two directions
c a n b e o b t a i n e d b y m e a n s of a t r a n s f o r m a t i o n of c o o r d i n a t e s (or
M o h r ' s circle).
T h e shearing forces at a n y p o i n t are o b t a i n e d from Eq. (17.3.13). T h e
t r a n s f o r m a t i o n expressed b y E q s . (17.3.15) allows o n e to find the
shearing forces o n a n y pair of o r t h o g o n a l p l a n e s w h o s e n o r m a l s h a v e
k n o w n direction cosines. F o r the case of the ellipse, t r a n s f o r m a t i o n s of
c o o r d i n a t e s are necessary to o b t a i n m o m e n t s , shearing forces, a n d
reactions at the b o u n d a r i e s .
17.9
Bending of Circular P l a t e s
In the discussion of b e n d i n g of circular plates, it is c o n v e n i e n t to use
cylindrical c o o r d i n a t e s . T h e c o o r d i n a t e s r a n d 9 will b e t a k e n as s h o w n
in Fig. 17.24. E q s . (6.2.17) to (6.2.24), c o u p l e d with the results of Sec.
6.4, allow us to write quite easily all the e q u a t i o n s of the b e n d i n g of thin
plates in cylindrical c o o r d i n a t e s . F r o m E q . (6.4.28), the expression of
the L a p l a c i a n of w3 = uz is:
554
T h e Theory of Elasticity
(17.9.1)
so t h a t L a g r a n g e ' s e q u a t i o n in cylindrical c o o r d i n a t e s is:
u
, +r21 & 2
z \
u 4 U _z (j_ 2 , I I . l 2ri i 2V ^ £2 .
~
\ dr
g_
=
D'
r 30 A
%r
3r
3r
r
30 /
(17.9.2)
Positive directions for b e n d i n g m o m e n t s , twisting m o m e n t s , a n d shearing forces a r e s h o w n in F i g . 17,24.
T o o b t a i n the expressions for Mrn M99
, M ^ , Vrz
, a n d V9z in terms of
u3 = uz, let us consider t h e element s h o w n in F i g . 17.24 a n d a s s u m e t h a t
the OXx axis coincides with t h e radial direction er. Therefore, Mrr
,
M99
,
, Vrz
, a n d V9zh a v e t h e s a m e values as M n, M 2 , 2A / 1 , 2VX3
, a n d V23at
Mr9
the s a m e point. Setting 9 = 0 in E q s . (6.2.20), (6.2.21), a n d (6.2.22), w e
get:
M
^ -
D
2
2
(d uz
d uz\
\ 2M
+
V
\d uz
W ^
2
(17.9.3)
1 9 "zY|
(idu
2
2
(d uz
d u2\
(17.9.4)
l
+2
r
dr
r
2
W
2
dr
J
555
Bending of Thin Flat Plates
M
rV
R0 =
- D ( \
y
- v)(
* )
\ d x , a3 x 2/ 9 - o
(17 9 5)
77*e boundary conditions at the edge of a circular plate of r a d i u s a are as
follows:
F o r a simply s u p p o r t e d e d g e :
u = 0 a n d M rr = 0.
(17.9.8)
For a clamped edge:
= 0 and ^
= 0.
(17-9-9)
or
F o r a free e d g e :
M„ = 0 a n d ^
=
+
07-9.10)
If the l o a d # is symmetrically d i s t r i b u t e d a b o u t the Z axis, uz is
i n d e p e n d e n t of 9 a n d L a g r a n g e ' s e q u a t i o n b e c o m e s :
lii'iM-W-i-
>
(,7A,,
This e q u a t i o n c a n easily b e i n t e g r a t e d w h e n q = q (r) is given:
1. #(r) = q0 = constant. E q . (17.9.11) yields:
=
- »> 4 ^
+C
+ C
NR + C
- '
(17 912)
w h e r e Q , C 2, C 3, a n d C 4 a r e c o n s t a n t s of i n t e g r a t i o n to b e d e t e r m i n e d
from the b o u n d a r y c o n d i t i o n s . F r o m E q . (17.9.6):
rz
V = -D<L(£±
dr\
1
dr
+ ^
r
) = - D ( ^+ ^ )
dr )
\2D
/'
r
(17-9.13)
F o r a circular p l a t e w i t h o u t a c e n t r a l hole, we see t h a t Vrz b e c o m e s
infinite for r = 0. Since this is impossible, C{ m u s t b e e q u a l to zero.
Since uz is finite for r = 0, C 3 m u s t also b e e q u a l t o zero. T h u s , for a
556
The Theory of Elasticity
uniformly l o a d e d circular plate w i t h o u t a center hole:
U -z ^ +
+2
+
u ~ 64D r ° ^ 4 + r * '
C
(17.9.14)
If we a s s u m e t h a t the edge is c l a m p e d , then a c c o r d i n g to Eq. (17.9.9),
we h a v e :
a4
C
,
%
a
2
a3
\
r
_
0
ac
,
%
_
i
0
(17.9.15)
Solving, we get:
a2
r
-
q
°
r
4%
(17.9.16)
-
Therefore, the deflection wz of a uniformly
c l a m p e d at the edge is given b y :
2
u2 = ^ { a
a
l o a d e d circular
plate
2
- r
(17.9.17)
? .
Substituting Eq. (17.9.17) i n t o Eqs. (17.9.3) a n d (17.9.4), we get:
2
Mrr = ^[a (l
2
+ p) - r (3
+ v)]
(17.9.18)
+ 3v)].
(17.9.19)
2
Mu = f g [ a ( l + v)-
r\l
Because of the s y m m e t r y , Mr9 = 0. Substituting r = a in the t w o
previous expressions, we find for the b e n d i n g m o m e n t s at the b o u n d a r y :
A t the center:
9 2 )1
(M„) r=o - ( * W _ o
T h e stresses a rr a n d
are given b y :
-
+ »•
- -
are equal at the center a n d , from Eq. (17.3.16),
orr = oee = ^ ( X
v).
+
(17-9.22)
A t the lower face z = h/2, a n d Eq. (17.9.22) b e c o m e s :
<V = om = ^ ( 1
+ v).
(17-9.23)
Bending of Thin Flat Plates
557
2. The plate is loaded by a single concentrated force P at the center. Eq.
(17.9.12), in w h i c h q0 is set e q u a l to zero, applies to this case except at
the center w h e r e q0 is infinite. W e k n o w from experience t h a t a solution
exists with a finite deflection uz, a n d t h a t the slope duz /dr at the center
is e q u a l to zero. F r o m Eq. (17.9.12), w e get:
^
= 0 + ^-{rlnr
- §) +
+
(17-9.24)
F o r duz /dr to b e e q u a l to zero for r = 0, the c o n s t a n t C 3 m u s t b e e q u a l
to zero. Let us n o w isolate a small vertical cylinder of r a d i u s r a r o u n d
P. E q u i l i b r i u m requires t h a t Vrz= -P/lRr.
F r o m E q . (17.9.13) Vrz
= -DCX /r so t h a t C, = P/2UD. E q . (17.9.12) b e c o m e s :
«; = £^Vnr-l)
C/j
(17-9.25)
C4. +
+
If we a s s u m e t h a t the edge is c l a m p e d , t h e n from E q . (17.9.9), we h a v e :
C ,7 =
and
K
?—(2ina
4UD
-
1)
1 P
4
C =
q
h (17.9.26)
16UD'
M o m e n t s a n d stresses c a n b e o b t a i n e d using E q s . (17.9.3) a n d (17.9.4),
together with E q . (17.3.16).
A w i d e variety of p r o b l e m s related to circular plates c a n b e f o u n d in
[2].
17.10 Strain Energy and Potential Energy of a Thin Plate in Bending
F r o m E q s . (8.7.7) a n d (8.7.11), the expression for t h e strain energy
stored in a n elastic b o d y is given b y :
Ut
e dV
=
\\I
j
° u u
(17.10.1)
v
I n the case of a thin plate in which a 3 , 3e 1 , 3a n d e 23 are neglected, this
expression is r e d u c e d t o :
v
d xd x d x
+ -^p^n]
\
2
3•
(17.10.2)
558
The Theory of Elasticity
Substituting Eq. (17.2.29) in the a b o v e e q u a t i o n , we o b t a i n :
A
dxf
dx2 /
(17.10.3)
dx] dx
w h e r e A is the a r e a of the plate. This is the expression of the strain
energy of a thin plate in b e n d i n g . It c a n b e p u t in a simpler form for
plates with c l a m p e d edges a n d for r e c t a n g u l a r plates with w3 = 0 a l o n g
the edges. I n d e e d , integrating twice b y p a r t s the last t e r m of E q .
(17.10.3), we o b t a i n :
(17.10.4)
w h e r e <>
/ is the integral t a k e n a r o u n d the c o n t o u r C of the plate. F o r
plates with c l a m p e d edges, the t w o c o m p o n e n t s of the g r a d i e n t vector
du3/dx{
a n d 3 w 3/ 3 x 2 a l o n g the edges vanish. F o r a r e c t a n g u l a r plate
with u3 = 0 o n the b o u n d a r y2, 3w 3/dxx = 0 a l o n g the edge parallel to the
OXx axis, a n d 3 w 3/ 3 x 2 = d u3/dx2
= 0 a l o n g the edge parallel to the
OX2 axis. Therefore, the two first integrals in the r i g h t - h a n d side of Eq.
(17.10.4) vanish in these two cases. W i t h this result, the expression of the
strain energy Ut b e c o m e s :
(17.10.5)
If a p l a t e h a s c l a m p e d edges or is simply s u p p o r t e d a n d is subjected to
a l o a d q = q{xx,x2)> the w o r k d o n e b y the external forces is:
Bending of Thin Flat Plates
W=
j
j
559
(17.10.6)
qu3dxxdx2,
a n d t h e p o t e n t i a l energy of t h e p l a t e is:
T h e m i n i m i z a t i o n of t h e p o t e n t i a l energy gives, in theory, the solution
of the p l a t e p r o b l e m . Such a m i n i m i z a t i o n is n o t easy to achieve a n d a n
a p p r o x i m a t e solution c a n b e o b t a i n e d using the Rayleigh-Ritz m e t h o d
(Sec. 15.20).
17.11 Application of the Principle of Minimum Potential Energy to
Simply Supported Rectangular Plates
In Sec. 17.7 it w a s seen t h a t the deflection of a simply s u p p o r t e d
r e c t a n g u l a r plate c a n b e r e p r e s e n t e d in the form of a d o u b l e trigonometric series:
00
00
m=\
n=\
Substituting Eq. (17.11.1) into Eq. (17.10.5), we get:
< "17
2)
, which d e t e r m i n e the s h a p e of the plate, c a n b e
T h e coefficients Bmn
o b t a i n e d b y m e a n s of the principle of m i n i m u m p o t e n t i a l energy:
1. q(xl,x2)
= q$ is uniformly distributed over the area of the plate. T h e
w o r k d o n e b y the applied loads is:
SM
w
= JL0
Io
%J
ZJ =1
™
00
=
4 ^ y
tt2
11
=\
=1 ^ «
"
00
y
mn = \m n
N S I
—a—
A
d xd x
\
i
(17.11.3)
560
T h e Theory of Elasticity
T h e potential energy 11^ is to b e m i n i m i z e d with respect to t h e
a m p l i t u d e t e r m Bmn
, so t h a t
4
sit
2
[u DabD
4
2 2
( m
. n
m
n
6
?o^~L»
\
_ 0
(17.11.4)
Thus,
B
~
16%
/™2
DIl mn\
2
„ 2 \
A
'
(17-11.5)
which is t h e s a m e as E q . (17.7.10).
2. q{xx, x2) is a concentrated force P acting at any point xx = e, x2 = /.
T h e w o r k d o n e b y the applied load is:
00
oo
m=l
n=\
^=^22
5 m„ s i n ^ s i n ^
07.11.6)
t>
and
4
an _ | n £ » a f c ft
2
/ m
2
, nY
. ,«IL?p •
«n/T
m
" (17.11.7)
= 0.
Thus,
4/ > sin ^
A
sin
^
^y,
07.11.8)
which is t h e s a m e as E q . (17.7.14).
PROBLEMS
1.
A thin r e c t a n g u l a r plate is subjected to the following u n i f o r m edge
m o m e n t s : Mxx = 15 in.-lb/in., M22 = — 10 in.-lb/in., a n d MX2 = 5
i n . - l b / i n . If t h e p l a t e is 1 inch thick:
(a) F i n d t h e c o m p o n e n t s of t h e c u r v a t u r e tensor.
(b) F i n d the directions of t h e p l a n e s c o r r e s p o n d i n g to t h e
principal curvatures.
Bending of Thin Flat Plates
2.
3.
4.
5.
561
(c) F i n d the m a g n i t u d e a n d directions of the p r i n c i p a l stresses.
(d) D r a w the M o h r circles r e p r e s e n t i n g the m o m e n t , the curva7 stress tensors. E a n d v for the m a t e r i a l of the
ture, a n d the
p l a t e are 10 psi a n d 0.3, respectively.
A thin r e c t a n g u l a r plate 20 in. long, 10 in, wide, a n d 1 in. thick lies
in the OX{, OX2 p l a n e with its long side parallel to the OXx axis.
F i n d the m a g n i t u d e s of the radii of c u r v a t u r e if the edges parallel
6 are subjected to a m o m e n t Mn = 2 0 i n . - l b / i n .
to the OX2 axis
( £ = 3 0 x 1 0 p s i , ? = 0.3).
A simply s u p p o r t e d r e c t a n g u l a r plate 20 in. long a n d 15 in. wide is
subjected at its center to a c o n c e n t r a t e d l o a d of 80 lbs. If the p l a t e
is j in. thick, find the c o m p o n e n t s of the m o m e n t tensor at the
7 a n d the c o r r e s p o n d i n g c o m p o n e n t s of the stress
center of the plate
tensor (E = 10 psi, v = 0.3).
A simply s u p p o r t e d r e c t a n g u l a r plate of sides a a n d b (see Fig.
17.21) a n d thickness h is subjected to a h y d r o s t a t i c p r e s s u r e
q(xx,x2)
= q0xx/a.
F i n d the expressions of the deflection, the
m o m e n t s , a n d the stresses in the plate.
Show t h a t
X
2
Fig. 17.25
i
u3 = 64aD
f — 3x2 — a(xx + x2)
+ YjO
^
562
6.
7.
8.
9.
The Theory of Elasticity
is the solution of the p r o b l e m of the simply s u p p o r t e d plate h a v i n g
the s h a p e of a n equilateral triangle, a n d being subjected to a
uniformly distributed l o a d q0 (Fig. 17.25). F i n d the m a x i m u m
values of Mx xa n d M22
.
F i n d the expressions of the deflection a n d of the stresses orn o09
,
a n d org for a circular simply s u p p o r t e d plate subjected to a
uniformly distributed l o a d q(r, 0) = q0.
Solve P r o b l e m 6 for a c o n c e n t r a t e d load P applied at the center of
the plate.
A circular plate with radius a has a c o n c e n t r i c circular hole with
radius b. T h e plate is subjected to a uniformly distributed l o a d q0
a n d h a s its i n n e r edge built in a n d its o u t e r edge free. F i n d the
m a x i m u m deflection, the m a x i m u m m o m e n t , a n d the m a x i m u m
b e n d i n g stresses, if b / a = 6.
A r e c t a n g u l a r plate of sides a a n d b is subjected to a uniformly
distributed load q0, a n d is simply s u p p o r t e d a l o n g its edges. A s s u m e
that the deflection is given b y
w h e r e / is the deflection at the center of the plate. U s i n g the
principle of virtual work, find the value of / a n d c o m p a r e it to t h a t
o b t a i n e d by N a v i e r ' s solution w h e n a = b.
10. A r e c t a n g u l a r plate subjected to a uniformly distributed l o a d q0 is
simply s u p p o r t e d a l o n g two sides a n d built in along the two others.
The equation
oo
00
satisfies the b o u n d a r y c o n d i t i o n s . D e t e r m i n e Bmn using the R a y leigh-Ritz m e t h o d .
REFERENCES
[1] W. Kaplan, Advanced Calculus, Addison-Wesley, Reading, Mass., 1952.
[2] S. Timoshenko and S. Woinowski-Kreiger, Theory of Plates and Shells, McGraw-Hill, N e w
York, N . Y., 1959.
CHAPTER 18
INTRODUCTION TO THE THEORY OF
THIN SHELLS
18.1
Introduction
A shell s t r u c t u r e m a y b e defined as a b o d y enclosed b e t w e e n t w o
closely s p a c e d a n d c u r v e d surfaces. If the thickness is small c o m p a r e d
with the overall d i m e n s i o n s of the b o u n d i n g surfaces, the shell is called
a thin shell. Before s t u d y i n g the t h e o r y of thin shells, it is i m p o r t a n t t h a t
o n e a c q u i r e s a clear u n d e r s t a n d i n g of the t h e o r y of surfaces a n d of the
curves t h a t a r e d r a w n o n t h e m . T h e s e topics b e l o n g to the subject of
differential g e o m e t r y [1].
In this c h a p t e r , we shall p r e s e n t a n d discuss those relations of
differential g e o m e t r y which are n e e d e d in the d e v e l o p m e n t of t h i n shell
theory. T h e relations a m o n g forces, m o m e n t s , a n d stresses will b e given,
a n d the e q u a t i o n s of equilibrium will b e derived. B o t h will t h e n b e
applied to a few p a r t i c u l a r cases of thin shells. It is r e c o m m e n d e d that,
prior to r e a d i n g this chapter, a t h o r o u g h u n d e r s t a n d i n g of C h a p t e r 6 b e
acquired.
18.2
S p a c e Curves
In vector analysis, a curve is defined as the locus of a p o i n t w h o s e
position vector r, relative to s o m e fixed origin O, is a function of o n e
p a r a m e t e r £ t. If the curve is e m b e d d e d in a t r i d i m e n s i o n a l cartesian
space, the c o o r d i n a t e s of a n y p o i n t o n the curve are given b y (Fig. 18.1):
563
564
T h e Theory of Elasticity
t
Fig. 18.1
x
\
= *l(£l)>
2
=
X
x
X2(£l)>
3
x
= 3^\
)•
(18.2.1)
E q s . (18.2.1) a r e t h e p a r a m e t r i c e q u a t i o n s of a space curve. xx, x2, a n d
x3 a r e a s s u m e d t o b e single v a l u e d functions of £j to insure t h a t each
value of £ t gives o n e single p o i n t o n t h e space curve.
Let r b e a vector giving t h e position of a p o i n t P o n a space curve
with respect to a n a r b i t r a r y origin O (Fig. 18.1), a n d let s b e a p a r a m e t e r
of the curve representing t h e d i s t a n c e of P m e a s u r e d a l o n g t h e curve
from a reference p o i n t PQ
. T h e vector r c a n b e c o n s i d e r e d as d e p e n d e n t
o n the p a r a m e t e r s. If 1 represents the unit vector at P in the direction
of increasing s, t h e n
t
dr
ds'
(18.2.2)
Let us define a unit vector,
n , - A & ,
08.2.3)
w h e r e di d e n o t e s the i n c r e m e n t in 1 in passing from t h e p o i n t P(s) to
F(s + ds), a n d A is a positive factor of p r o p o r t i o n a l i t y . Since 1 is also a
unit vector, di will b e p e r p e n d i c u l a r to 7. T h e t w o vectors 1 a n d nx
d e t e r m i n e a p l a n e called t h e osculating p l a n e of the curve a t P. T h e
osculating p l a n e is also defined as t h a t p l a n e c o n t a i n i n g three c o n s e c u tive p o i n t s o r t w o consecutive t a n g e n t s o n a space curve.
Introduction to the Theory of Thin Shells
565
Finally, let n2 b e a unit vector at P p e r p e n d i c u l a r to the osculating
p l a n e a n d in a direction such t h a t (Fig. 18.2):
Fig. 18.2
(18.2.4)
n2 = ~tXnx.
T h e three u n i t vectors 7, nx, a n d n2 c o n s t i t u t e a u n i t r i g h t - h a n d e d system
associated with P. 7 is called the (vector) t a n g e n t , nx the (vector)
p r i n c i p a l n o r m a l , a n d n2 the (vector) b i n o r m a l . Since
7 • nx = nx • n2 = n2 • 7 = 0,
(18.2.5)
then, b y differentiation, we get:
7 • dnx = —nx • diy
nx • dh2 = —n2 • dnx,
n2-
di = — 7 • d ^ 2-
(18.2.6)
must
But from E q . (18.2.3), 7?j a n d d7 are parallel; therefore, 7 z 2 a n d
b e p e r p e n d i c u l a r to e a c h other. A n d from Eq. (18.2.6), 7 a n d dn2 m u s t
b e p e r p e n d i c u l a r to e a c h other. N o w , since n2 is a unit vector, it m u s t
also b e p e r p e n d i c u l a r to dn2, so t h a t dn2 a n d nx m u s t b e collinear (Fig.
18.2).
Let us n o w define t w o characteristic quantities associated with the
curve, at P, b y the e q u a t i o n s :
2 1
c ; = m1 - - = m
ds
(18.2.7)
ds
CnX is seen to b e e q u a l to the reciprocal of the positive factor of
p r o p o r t i o n a l i t y A in E q . (18.2.3), a n d is called the first c u r v a t u r e or
566
The Theory of Elasticity
flexure (or j u s t the c u r v a t u r e ) of the curve at P; Cn2 is called the s e c o n d
c u r v a t u r e or torsion of the curve at P. CnX is always positive a n d , since
nx is a unit vector in the s a m e direction as di, t h e n CnX is e q u a l in
m a g n i t u d e to \di/ds\; also since 7 is a unit vector, CnX is e q u a l to the
angle t u r n e d t h r o u g h b y the t a n g e n t to the curve per unit d i s t a n c e
traveled a l o n g it (Fig. 1 8 . 1 ) . Cn2 c a n b e either positive or negative
d e p e n d i n g o n w h e t h e r dn2 is in the opposite or in the s a m e direction as
nx. M o r e o v e r , since nx is a unit vector, then Cn2 = +\dn2/ds\\
also, since
n2 is a unit vector, then Cn2 is numerically equal to the angle t u r n e d
t h r o u g h b y the osculating p l a n e per unit distance traversed along the
curve. T h e torsion Cn2 is r e g a r d e d positive w h e n the r o t a t i o n of the
osculating p l a n e (i.e., of the b i n o r m a l ) , as s increases, follows the righth a n d rule with the t h u m b in the direction of 7. T h e torsion of a p l a n e
curve is obviously e q u a l to z e r o .
T h e inverses of CnX a n d Cn2 are called the radii of c u r v a t u r e of the
curve at P b y :
R
ni
=
1
(
(18.2.9)
is called the r a d i u s of flexure, a n d
R
n2
= - r -
(18-2.10)
is called the r a d i u s of torsion. T h e p o i n t at a d i s t a n c e RnX from P in the
direction of nx is called the first center of c u r v a t u r e or the center of
flexure. T h e p o i n t at a d i s t a n c e Rn2 from P in the direction of dn2 (i.e.,
in the direction of ±nx) is called the s e c o n d center of c u r v a t u r e or the
center of torsion.
18.3
E l e m e n t s of the Theory of Surfaces
In this section, we shall give a short p r e s e n t a t i o n of the elements of
the theory of surfaces which will e n a b l e us to d e d u c e the results n e e d e d
in the study of the theory of thin shells.
1 ) Gaussian surface coordinates. First fundamental
form
I n vector analysis, a surface is defined as the locus of a p o i n t w h o s e
position vector r, relative to s o m e fixed origin O, is a function of t w o
i n d e p e n d e n t p a r a m e t e r s £ x a n d £ 2. If the surface is e m b e d d e d in a
t r i d i m e n s i o n a l cartesian space, the c o o r d i n a t e s of a n y p o i n t of the
surface are given b y :
Introduction to the Theory of Thin Shells
567
x
*\ = *l(£l>£2)>
2 = ^2(fl»fe)»
* 3 = X&l'tl)'
(18.3.1)
E q s . (18.3.1) a r e t h e p a r a m e t r i c e q u a t i o n s of a surface. If we eliminate
£i a n d £ 2 from E q s . (18.3.1), we o b t a i n the e q u a t i o n of the surface in
cartesian c o o r d i n a t e s , n a m e l y :
F(xl9xl9x3)
= 0.
(18.3.2)
It is u n d e r s t o o d t h a t the surface is regular in the sense t h a t the functions
t e their derivatives with
xl9 x2, a n d x3 of E q s . (18.3.1), togethero with
t h o r d e r r e q u i r e d in t h e
respect to t h e p a r a m e t e r s £j a n d | 2
following discussions, a r e c o n t i n u o u s . A n y relation b e t w e e n £j a n d | 2,
such as
* « i , * 2 ) = 0,
(18.3.3)
represents, together with E q s . (18.3.1) a curve o n the surface. If either
of the p a r a m e t e r s (say, £j) is held c o n s t a n t , t h e n E q s . (18.3.1) together
with t h e a d d i t i o n a l e q u a t i o n ^ = c o n s t a n t will r e p r e s e n t a line o n the
surface (Fig. 18.3). Such lines a r e called c o o r d i n a t e lines or p a r a m e t r i c
Fig. 18.3
curves. T h r o u g h e a c h p o i n t o n the surface there will pass t w o c o o r d i = o n e c o r r e s p o n d i n g to £j = c o n s t a n tusa n d o n e c o r r e s p o n d i n g
n a t e lines,
to | 2 c o n s t a n t . T h e p a r a m e t e r s £j a n d | 2 t h
c o n s t i t u t e a system of
curvilinear c o o r d i n a t e s for p o i n t s o n t h e surface (see Sec. 6.2). T h e
568
The Theory of Elasticity
curve a l o n g which £ 2 is c o n s t a n t a n d ^ varies is called a ^ c u r v e ; the
other, a l o n g w h i c h | j is c o n s t a n t a n d £ 2 varies, is called a £ 2 curve.
£i a n d £ 2 m a y b e c o n s i d e r e d as surface c o o r d i n a t e s of the p o i n t P: T h e y
are called Gaussian coordinates. As a n e x a m p l e , consider the vertical
circular cylinder of Fig. 18.4. If R is the r a d i u s of the cylinder, the
cartesian c o o r d i n a t e s of a p o i n t o n t h e cylinder a r e :
xx = R cos 0,
x2 = R sin 0,
x3 = z.
(18.3.4)
z a n d 0 are the p a r a m e t e r s w h i c h define points o n the surface, a n d they
m a y b e identified with | t a n d £ 2.
In Fig. 18.3, let r d e n o t e the position vector of a p o i n t P with respect
to a n y a r b i t r a r y origin 0> Since P is o n the surface, r c a n b e c o n s i d e r e d
as a function of £ t a n d | 2, so t h a t
dr = ^ d i x
+ | ^ 2,
(18.3.5)
where
d e n o t e s the differential i n c r e m e n t in r w h i c h occurs in passing
from P^xAi)
^ P'{£\ + d ! i , £ 2 + d£2)- T h e vector 3 ^ / 3 ^ d e n o t e s the
derivative of r with respect to i x w h e n £ 2 r e m a i n s c o n s t a n t ( c o m p a r e to
Sec. 6.3). Therefore, dr/d^x is t a n g e n t to the £ t curve. Similarly, 3 r / 3 £ 2
is t a n g e n t to the £ 2 curve. T h e s e t w o vectors will b e d e n o t e d b y :
Introduction to the Theory of Thin Shells
569
a, = # -
(18.3.6)
,2 = ^ ,
(18.3.7)
9*i
a n d m a y b e c o n s i d e r e d as c o n s t i t u t i n g a n ax, a2 b a s e system at p o i n t P.
A n y vector A associated with p o i n t P, w h i c h c a n b e r e p r e s e n t e d b y a
line t a n g e n t to the surface at P , is called a surface vector a n d c a n b e
r e p r e s e n t e d b y [ c o m p a r e to Eq. (6.3.13)]:
A = Axax
+ A2a2.
(18.3.8)
T h u s , for the differential i n c r e m e n t in r, we h a v e :
dr = axd£x
(18.3.9)
+ a2di2.
T h e s q u a r e of the m a g n i t u d e of dr is given b y :
2
(ds)
2
= dr - dr = ax1 • ax (d^x ) + 2ax • a2 d£x di2
2
+ a2 •
(18.3.10)
a2{d£2)
Setting
ax • ax = axx = E
(18.3.11)
ax • a2 = tf12 = F
(18.3.12)
a2 - a2 — a22 = G,
(18.3.13)
Eq. (18.3.10) c a n b e written a s :
2
(ds)
2
2
= aydtidZj
= E(dZx)
+ 2Fdixdi2
+ G ( ^ 2) .
(18.3.14)
T h e differential q u a d r a t i c form (18.3.14) is called the first
fundamental
form of the surface a n d E, F, G are called the first
fundamental
magnitudes. T h e s e m a g n i t u d e s are the metric coefficients for the surface
a n d , as c a n b e seen from Eq. (18.3.14), they are the link b e t w e e n the
length of a n e l e m e n t a n d the differentials d^. If dr is t a k e n a l o n g the £j
curve, Eq. (18.3.14) gives:
dsx =
d£x = V^7
d$x.
~
(18.3.15)
570
The Theory of Elasticity
If dr is t a k e n a l o n g the £ 2 curve, Eq. (18.3.14) gives:
d
ds2 = \/G di2 = V^22
(18.3.16)
ii •
T h e cosine of the angle b e t w e e n ax a n d a2 is given b y :
cos 6 =
a x • a2
aX2
\ax\\a2\
f
\Jaxx
a22
Since cos 6 < 1, then
EG'
(18.3.17)
2
(18.3.18)
EG > F ,
a n d the q u a n t i t y
2
H
2
(18.3.19)
= EG — F
is always positive. T h e p a r a m e t r i c curves £ x a n d £ 2 will form a n
o r t h o g o n a l system of curvilinear c o o r d i n a t e s if, all over the surface,
aX2 = F = 0. In this case, the first f u n d a m e n t a l form b e c o m e s :
2
(ds)
2
= E{dZx)
2
+
(18.3.20)
G{di2)
F r o m a given p o i n t (£j, £ 2) a n y direction o n the surface is d e t e r m i n e d
b y the i n c r e m e n t s d£x a n d di2. Let ds a n d ds' b e elements of a r c lengths
of two surface curves C a n d C intersecting at P (Fig. 18.5). T h e n ,
5/
Fig. 18.5
Curves
Introduction to the Theory of Thin Shells
571
(18.3.21)
(18.3.22)
T h e angle 0 b e t w e e n C a n d C is given b y :
C O S * - *
ds
ds
ds ds
\ ds ds
ds
ds /
(18.3.23)
F o r C a n d C to b e o r t h o g o n a l , we m u s t h a v e :
(18.3.24)
f
It is desirable to eliminate s a n d s from Eq. (18.3.24). F o r that, w e write:
(18.3.25)
(18.3.26)
e
a
r
a n d (d£{ /d£2)c
to b e c o m p u t e d a l o n g the curves
w h e r e (d£{ /d£2)c
C a n d C , respectively. Substituting E q s . (18.3.25) a n d (18.3.26) i n t o E q .
(18.3.24), we get:
+ G = 0,
(18.3.27)
w h i c h is t h e c o n d i t i o n for C a n d C to b e o r t h o g o n a l .
2) Second fundamental
form
A n o r m a l section of a surface a t a p o i n t P is t h e section defined b y a
p l a n e c o n t a i n i n g t h e n o r m a l t o t h e surface at t h e p o i n t (Fig. 18.6). T h i s
section is a p l a n e curve w h o s e p r i n c i p a l n o r m a l is collinear with t h e
n o r m a l t o t h e surface at t h a t p o i n t . T h e c u r v a t u r e of a n o r m a l section,
such as the c u r v e AB (Fig. 18.6), is called the normal curvature of the
surface at P in t h e direction of AB. T h e u n i t vector a l o n g the n o r m a l to
the surface at P is given b y :
+
572
The Theory of Elasticity
*(M2) = f ^ f T ,
(18-3.28)
w h e r e the q u a n t i t y in the d e n o m i n a t o r is the m a g n i t u d e of the vector
But
p r o d u c t ax Xa2.
E
|S, X a2\ = | Sl | | S 2| s i n 9 = ^EG
^
G
^
= **.
18329
< -- >
Therefore,
_
=o LX a 2
(18.3.30)
H
O n Fig. 18.5, in passing eta l osnug a curve C from
, £ 2) to the n e a r p o i n t
P\i\ + d l i , I2 +
l
a s s u m e t h a t n a n d r increase b y dn a n d dr.
Then,
^ = g-^, + | ^ ,
2
dr =
-g-d^+^d^ .
F o r m i n g the scalar p r o d u c t of dn a n d dr, we get:
2
= L(d^)
+ 2Mdixdi2
9r
^
9|i " 9|i '
_=
I f 9ft
2L9?i
-dn-dr
(18.3.31)
2
2
+ # ( d £ 2) ,
(18.3.32)
where
^
= 9A?
9|2
dr^dn
9£ 2
dr j
3|2 a^J
( 1 38b )3 3
Introduction to the Theory of Thin Shells
573
L, Af, a n d TV are called the fundamental
magnitudes of the second order.
It will b e s h o w n t h a t they are intimately c o n n e c t e d with the c u r v a t u r e
properties of the surface. T h e expression in the r i g h t - h a n d side of Eq.
(18.3.32) is called the second fundamental
form for the surface. T h e
coefficients L, Af, a n d N of this form c a n b e e v a l u a t e d as follows: Since
n is n o r m a l to the surface, a n d since dr/d^
a n d 3 r / 3 £ 2 are t a n g e n t to
the surface, therefore,
oil
0I2
a n d u p o n p a r t i a l differentiation
relations a r e f o u n d :
37?
of these expressions, the following
3?
3li ' 3 £ i
dn
3 | 2 " 9£ 2
dr
dn
dr
'3|,
dn
dr
31, ' 3 &
2
—n '
dr
(18.3.34)
2
—n
dr2
(18.3.35)
3|2
2
—n
—n
dr
31, 3 | 2
2
dr
3*, 3 | 2
(18.3.36)
(18.3.37)
H e n c e , from E q s . (18.3.33), we h a v e :
L = n
Af = n
3*
»Hr
3£i 3 | 2
# = «-fS.
-
(18 139)
(18.3.40)
T h u s , L, Af, a n d TV are the projections of the s e c o n d derivatives of r o n
the n o r m a l n to the surface.
Since n is a unit vector, dn is n o r m a l to n a n d , therefore, parallel to
the surface. C o n s e q u e n t l y , dn/d^
c a n b e expressed in t e r m s of its
c o m p o n e n t s a l o n g ax a n d a2:
574
The Theory of Elasticity
w h e r e p a n d q are u n k n o w n s t o b e d e t e r m i n e d . F o r m i n g the scalar
p r o d u c t of e a c h side of E q . (18.3.41) b y ax a n d a2, respectively, a n d
recalling E q s . (18.3.11) to (18.3.13), we get:
-L=pE
-M
=pF
+ qF
(18.3.42)
+ qG.
(18.3.43)
Solving for p a n d q, we o b t a i n :
P =
FM -2 LG
H
(18.3.44)
FL -EM
(18.3.45)
Therefore,
dn _ FM -2 LG dr
3£,
H
9li
FL
+ —2EM
H
dr
3|2
3 4^
Similarly,
dn
W
FN — MG dr
=
2
, FM - EN dr
— ^ — a l 7 —fl —^+
2
nR^dl\
--
(18 3 47)
F r o m Eqs. (18.3.30), (18.3.46), a n d (18.3.47), the values of the following
scalar triple products can be obtained:
£M
- FL
H
FM
- GL
EN
- FM
H^
(18.3.50)
^ . g L x i | = FN ~ - GM
H
(18.3.51)
to ^ ™-^
n
x =
- ^
g . g x | . a ^ H
|i « X # = ^
7 . 3&
a$i
(18.3.48)
(18.3.49)
Vf i
3) Curvature of a normal section. Meunier's
theorem
It w a s previously p o i n t e d o u t t h a t the f u n d a m e n t a l m a g n i t u d e s of the
s e c o n d o r d e r L, M , a n d TV are c o n n e c t e d with the c u r v a t u r e properties
Introduction to the Theory of Thin Shells
575
of the surface. Let us consider a n o r m a l section of a surface at p o i n t P
(Fig. 18.6); t h a t is to say, the section b y a p l a n e c o n t a i n i n g the n o r m a l
to the surface at P. Such a section AB is a p l a n e curve w h o s e p r i n c i p a l
n o r m a l nx (see S e c 18.2) is collinear with the n o r m a l to the surface. We
shall adopt the convention that the surface coordinates are chosen in a way
such that the normal n to the surface points in the same direction as the
principal normal nx of the section AB. W i t h this c o n v e n t i o n , the first
center of c u r v a t u r e of the section AB [see E q . (18.2.9)] falls o n the
positive side of n. I n w h a t follows, the c u r v a t u r e of a n o r m a l section will
b e called C„, a n d its r a d i u s of c u r v a t u r e will b e called Rn = \/Cn.
From
E q . (18.2.7), we h a v e :
C
"
=n-^±
2
(18.3.52)
ds '
w h e r e s is the d i s t a n c e m e a s u r e d a l o n g the n o r m a l section AB
18.6). U s i n g E q . (18.3.21), we h a v e :
2
2
d r2 _
ds
dr dHi2
3*i
^
3fe
ds
2
dr d i2
2
2
dh(d^)
+
^+ 2
d2r
3 | 2 V ds )
ds
(Fig.
"di di
ds
x2
ds
8
(3) 1 5
93
3{|" \ds J
Recalling E q s . (18.3.38) to (18.3.40), the expression for Cn b e c o m e s :
S u b s t i t u t i n g E q . (18.3.14) i n t o E q . (18.3.54), we o b t a i n the expression of
C„, the n o r m a l c u r v a t u r e , in terms of the first a n d s e c o n d f u n d a m e n t a l
forms:
c
_ L(d£ ? + 2M(d^m )
" E(d^) + 2F(d^m )
x
+ N{dj )
+ G(di f
2
2
2
2
2
( 1 8 3
55)
2
or
n
Cn = - ^ - J
dr • dr
.
(18.3.56)
2
Dividing t h e n u m e r a t o r a n d d e n o m i n a t o r of Eq. (18.3.55) b y (d{-2) ,
get:
we
3
576
The Theory of Elasticity
(18.3.57)
In Eq. (18.3.57), the quantities L, M , N9 E, F, G a r e functions of the
c o o r d i n a t e s £ 1? a n d £ 2, a n d h a v e a given c o n s t a n t value for e a c h p o i n t
P o n the surface. Therefore, at every point, the n o r m a l c u r v a t u r e
d e p e n d s only o n the ratio (d£x /d£2).
T h u s , it c a n b e stated that all
surface curves t h r o u g h a p o i n t P which are t a n g e n t to the s a m e
direction h a v e the s a m e n o r m a l c u r v a t u r e .
Let us a s s u m e n o w t h a t the section CD of a surface (Fig. 18.7) b y a
p l a n e I I , at a p o i n t P is n o t a n o r m a l section. T h e n nx, the principal
n o r m a l of the curve, is n o t parallel to n, the n o r m a l to the surface. F r o m
Eqs. (18.2.3) a n d (18.2.7), we d e d u c e that the principal n o r m a l c a n b e
written as
n
i
~ C ^ d ?
9
(18.3.58)
w h e r e CnX is the first c u r v a t u r e of the section. Let y b e the inclination
of the p l a n e of the section to the n o r m a l p l a n e I I w h i c h t o u c h e s the
curve at P. T h e n y is the angle b e t w e e n n a n d nx. H e n c e ,
Introduction to the Theory of Thin Shells
cos y = n - nx
n
Qi
dh2
ds '
577
(18.3.59)
Therefore,
(18.3.60)
cos y
or
C
= Cn cos y.
x
(18.3.61)
T h i s is M e u n i e r ' s t h e o r e m c o n n e c t i n g the n o r m a l c u r v a t u r e in a n y
direction with the c u r v a t u r e of a n y o t h e r section t h r o u g h t h e s a m e
t a n g e n t line.
4) Principal directions and lines of curvature
T h e n o r m a l s at consecutive p o i n t s of a surface d o n o t intersect in
general. H o w e v e r , at a n y p o i n t P of a surface, there are two directions
at right angle to e a c h other such t h a t the n o r m a l at a consecutive p o i n t
in either of these directions m e e t s the n o r m a l at P. T h e t w o directions
are called p r i n c i p a l directions at P. T o s h o w this, let r b e the position
vector of P, a n d n b e the unit n o r m a l to the surface there (Fig. 18.8).
F i g . 18.8
Let f + dr b e a n adjacent p o i n t in a direction defined b y dix a n d di2,
a n d n + dn b e the unit n o r m a l at this point. T h e n o r m a l s will intersect
if 7z, n 4- dn, a n d dr are c o p l a n a r ; t h a t is to say, if n, dn, a n d dr are
578
T h e Theory of Elasticity
c o p l a n a r (Fig. 18.8). T h e c o n d i t i o n t h a t three vectors b e c o p l a n a r is t h a t
their scalar triple p r o d u c t vanishes, so t h a t
ii-dnXdr
(18.3.62)
= 0.
T h i s c o n d i t i o n c a n b e e x p a n d e d i n terms of d£x a n d d£2 using E q s .
(18.3.31) a n d E q s . (18.3.48) t o (18.3.51), t o give:
2
(EM - FL)(d^)
+ (EN 2
+ (FN - GM)(d£2)
GL)dZxd£2
=0
or
(EM -
FL)(J^
+ {EN - GL)(jj^
+ (FN - GM) = 0. (18.3.63)
T h i s e q u a t i o n gives t w o values of the ratio (d£x/d^2) a n d , therefore, t w o
directions o n t h e surface for w h i c h t h e r e q u i r e d p r o p e r t y h o l d s . L e t
these t w o directions b e a l o n g t w o curves C a n d C intersecting a t P.
F r o m t h e t h e o r y of e q u a t i o n s , w e k n o w t h a t the s u m of the two roots of
Eq. (18.3.63) is given b y :
+
( i ! ) c
© c ~ f & ^ -
™
a n d t h a t t h e p r o d u c t of these t w o roots is given b y :
(*k)
(*k\
\dZ2)c\dZ2)c
= FN — GM
EM-FL*
)
Substituting E q s . (18.3.64) a n d (18.3.65) i n t o t h e o r t h o g o n a l i t y c o n d i tion (18.3.27), w e find t h a t this c o n d i t i o n is identically satisfied, w h i c h
m e a n s t h a t the p r i n c i p a l directions a r e o r t h o g o n a l . Since dr is p e r p e n dicular t o n, a n d dn is p e r p e n d i c u l a r t o n, t h e n dn is parallel t o dr.
Therefore, for a principal direction, dn/ds is parallel t o dr/ds.
A curve d r a w n o n a surface, a n d which possesses t h e p r o p e r t y t h a t
the n o r m a l s t o t h e surface a t consecutive p o i n t s intersect, is called a line
of c u r v a t u r e . Therefore, the direction of a line of c u r v a t u r e a t a n y p o i n t
is a p r i n c i p a l direction a t t h a t p o i n t . T h r o u g h e a c h p o i n t o n the surface
two lines of c u r v a t u r e pass, cutting each other a t right angle. O n t h e
surface, there a r e t w o systems of lines of c u r v a t u r e w h o s e differential
e q u a t i o n is (18.3.63). I t is interesting t o r e m a r k t h a t Eq. (18.3.63) gives
}
Introduction to the Theory of Thin Shells
579
the directions of the m a x i m u m a n d m i n i m u m n o r m a l c u r v a t u r e s at a
p o i n t P . I n d e e d , if we differentiate Cn in E q . (18.3.57) with respect to
(d^x /d^2) a n d e q u a t e the result to zero, we o b t a i n E q . (18.3.63). T h u s ,
the p r i n c i p a l directions at a p o i n t are the directions of greatest a n d least
normal curvatures.
It is c o n v e n i e n t in the d e v e l o p m e n t of the theory of thin shells to refer
to its lines of c u r v a t u r e as p a r a m e t r i c curves. If this is d o n e , the
differential e q u a t i o n (18.3.63) for the lines of c u r v a t u r e b e c o m e s
identical with the differential e q u a t i o n of the p a r a m e t r i c curves; t h a t is,
dZxd£2 = 0.
(18.3.66)
EM — FL = 0
(18.3.67)
FN — GM = 0
(18.3.68)
0.
(18.3.69)
H e n c e , we m u s t h a v e :
and
EN - GL^
Multiplying Eq. (18.3.67) b y N, a n d Eq. (18.3.68) b y L, a n d a d d i n g , we
get:
(EN - GL)M
= 0.
(18.3.70)
M u l t i p l y i n g Eq. (18.3.67) b y G, a n d Eq. (18.3.68) b y E, a n d a d d i n g , we
get:
(EN - GL)F = 0.
(18.3.71)
In view of E q . (18.3.69), the c o n d i t i o n s t h a t the p a r a m e t r i c curves also
b e lines of c u r v a t u r e a r e :
F=
M = 0.
(18.3.72)
5) Principal curvatures, first and second
curvatures
T h e p o i n t of intersection of consecutive n o r m a l s along a line of
c u r v a t u r e at P is called a center of c u r v a t u r e of the surface: Its d i s t a n c e
from P, m e a s u r e d in the direction of the unit n o r m a l n, is called a
principal r a d i u s of c u r v a t u r e of the surface at P. T h e reciprocal of a
principal r a d i u s of c u r v a t u r e is called a principal c u r v a t u r e . T h u s , at
each p o i n t of a surface two principal c u r v a t u r e s exist a n d these are the
580
The Theory of Elasticity
n o r m a l c u r v a t u r e s of t h e surface in the direction of the lines of
c u r v a t u r e . T h e y m u s t n o t b e confused with the (first) c u r v a t u r e s of the
line of c u r v a t u r e b e c a u s e the principal n o r m a l of a line of c u r v a t u r e is
not, in general, the n o r m a l to the surface. In o t h e r w o r d s , the osculating
p l a n e of a line of c u r v a t u r e does n o t as a rule give a n o r m a l section of
the surface; however, the c u r v a t u r e of a line of c u r v a t u r e is c o n n e c t e d
with the c o r r e s p o n d i n g principal c u r v a t u r e b y M e u n i e r ' s t h e o r e m
(18.3.61).
T h o s e p o r t i o n s of the surface o n which the t w o principal c u r v a t u r e s
h a v e the s a m e sign are called synclastic: for e x a m p l e , the surface of a
sphere is synclastic at all p o i n t s . If the principal c u r v a t u r e s h a v e
o p p o s i t e signs o n a n y p a r t of the surface, this part is said to b e
anticlastic: for e x a m p l e , t h e surface of a h y p e r b o l i c p a r a b o l o i d is
anticlastic at all points.
A t a n y p o i n t of a surface, there are two centers of c u r v a t u r e — o n e for
each principal direction. B o t h lie o n the n o r m a l to the surface. Let t h e
principal c u r v a t u r e s b e d e n o t e d b y C ( Q or C 2) , a n d the principal radii
of c u r v a t u r e b y R(R{ or R2). T o d e t e r m i n e the principal c u r v a t u r e s at
a n y point, we p r o c e e d as follows: Let r b e the position vector of the
p o i n t P o n the surface, n b e the unit n o r m a l there, a n d R(R{ or R2) b e
a principal r a d i u s of c u r v a t u r e (Fig. 18.9a). T h e n , the c o r r e s p o n d i n g
c e n t e r of c u r v a t u r e is given b y p, w h e r e
(R*dR)(f,+dn)
(b)
Fig. 18.9
Introduction to the Theory of Thin Shells
581
(18.3.73)
p = r + Rn.
If F is a p o i n t adjacent to P a l o n g a line of c u r v a t u r e of the surface
(Fig. 18.9b), then
= dr + Rdn
dp = dr + d(Rn)
+ ndR.
(18.3.74)
N o w , the vector (dr + R dn) is tangential to the surface since b o t h dr
a n d dn a r e . T h e vector dp h a s the direction of n (see Fig. 18.9b);
c o n s e q u e n t l y , we m u s t h a v e :
dr +
Rdn =
0.
If C (i.e., Cx or C 2) is the c o r r e s p o n d i n g principal c u r v a t u r e ,
(18.3.75)
Cdr + dn = 0.
Eq. (18.3.75) is called R o d r i g u e ' s formula. Inserting E q s . (18.3.31) i n t o
Eq. (18.3.75) a n d r e a r r a n g i n g terms, we get:
F o r m i n g t h e scalar p r o d u c t of this e q u a t i o n with dr/d^x
successively, we o b t a i n :
and 3r/3£2
(CE - L)d£x + (CF - M)d£2 = 0
(18.3.77)
+ (CG - N)d£2 = 0.
(18.3.78)
(CF - M)d£x
T h e s e t w o e q u a t i o n s d e t e r m i n e t h e p r i n c i p a l c u r v a t u r e s a n d t h e directions of the lines of c u r v a t u r e . E l i m i n a t i n g d£i /d£2, we get:
22
2
- (EN - 2FM 4- GL)C + (LN - M )
HC
= 0.
(18.3.79)
This is a q u a d r a t i c in C, w h o s e t w o roots are the p r i n c i p a l c u r v a t u r e s
Cx a n d C 2. W h e n the principal c u r v a t u r e s h a v e b e e n d e t e r m i n e d from
Eq. (18.3.79), the direction of the lines of c u r v a t u r e is given b y either
Eq. (18.3.77) or (18.3.78). T h u s , c o r r e s p o n d i n g to Q , the principal
direction is given b y :
D
(
A )
582
The Theory of Elasticity
a n d , c o r r e s p o n d i n g to C 2, the principal direction is given b y :
C2F-M
C2G-N
(18.3.81)
C2E-L
C2F~M'
\ d i 2h
T h e directions of the lines of c u r v a t u r e m a y also b e f o u n d b y eliminating C from E q s . ( 1 8 . 3 . 7 7 ) a n d ( 1 8 . 3 . 7 8 ) . This elimination leads t o :
2
(EM - FL)(d£x)
2
+ (EN - GL)dixdi2
= 0,
+ (FN - GM)(d£2)
(18.3.82)
which w a s previously o b t a i n e d in a different way. Eq. ( 1 8 . 3 . 8 2 ) fails to
d e t e r m i n e the principal directions if the coefficients vanish identically;
that is to say, w h e n
E : F : G = L : M : N.
In this case, the n o r m a l c u r v a t u r e as given b y Eq. ( 1 8 . 3 . 5 5 ) is i n d e p e n d ent of the ratio (d£x /d£2) a n d , consequently, h a s the s a m e value for all
directions t h r o u g h the point. Such a p o i n t is called a n umbilic o n the
surface.
T h e first c u r v a t u r e of the surface at a n y p o i n t is defined as the s u m
of the principal c u r v a t u r e s . It is d e n o t e d b y / . T h u s ,
EN
J =
Q
+
C
2
= ^
(
M2
F G
+
~
)'
L
(18.3.83)
T h e s e c o n d c u r v a t u r e of the surface, also called the G a u s s i a n or the
specific c u r v a t u r e , is defined as the p r o d u c t of the principal
c u r v a t u r e s . It is d e n o t e d b y K. T h u s ,
L
K=Ci
C2
=
M
N
- 2
2
(18-3.84)
.
6) Euler's theorem
T h e c o n d i t i o n t h a t the p a r a m e t r i c curves also b e lines of c u r v a t u r e
were f o u n d to b e ( 1 8 . 3 . 7 2 ) , F = M = 0 . F = 0 is the c o n d i t i o n of
o r t h o g o n a l i t y of the p a r a m e t r i c curves ( 1 8 . 3 . 1 7 ) . Euler's t h e o r e m expresses the n o r m a l c u r v a t u r e at a p o i n t in a n y direction in t e r m s of the
p r i n c i p a l c u r v a t u r e s . If, in the expression of the n o r m a l c u r v a t u r e
(18.3.55),
L
r
^
2
^ x ? 2 + 2MdZxdij2 + N(dij2)
E(dZx)
+ 2FdZxd£2 + Gd£
9( 1 8 3 8 5 )
Introduction to the Theory of Thin Shells
583
we set M = F = 0 a n d d £ 2 = 0 (since a l o n g the ^ curves £ 2 is c o n s t a n t ) ,
we o b t a i n :
r
=L
^-1
E77
= Ar>
R,
(18.3.86)
Similarly, with d£x = 0, we get:
2
~ G
(18.3.87)
R2
Fig. 1 8 . 1 0
Let us n o w c o n s i d e r a n o r m a l section of the surface in a direction
m a k i n g a n angle 9 with the £ x curve at P (Fig. 18.10). U s i n g E q s .
(18.3.15) a n d (18.3.16), we h a v e :
cos0 =
VE^-
(18.3.88)
sin0 =
VG^|,
(18.3.89)
so t h a t the n o r m a l c u r v a t u r e of this section is b y E q . (18.3.85):
so t h a t
2
2
Cn = k cos 9
"
E
+ % sin 0
G
2
(18.3.90)
2
(18.3.91)
C„ = Q c o s ^ + C 2s i n ^ .
T h i s is k n o w n as Euler's t h e o r e m o n n o r m a l c u r v a t u r e . F r o m this
t h e o r e m , it follows t h a t the s u m of n o r m a l c u r v a t u r e s in two directions
at right angle is c o n s t a n t , a n d e q u a l to the s u m of the principal
curvatures.
7) Rate of change of the vectors a, and the corresponding unit vectors
along the parametric
lines
584
T h e Theory of Elasticity
Recalling that t h e f u n d a m e n t a l m a g n i t u d e s L, M , a n d N a r e t h e
projections o n n of t h e s e c o n d derivatives of r, w e c a n write:
^ r - = Ln + bax + e a 2
1
Mn
=
3?
(18.3.92)
= ~ +
+ Pi
(18.3.93)
^ r - = Nn + d&j + y a 2,
where t h e coefficients b, c, d, e,f,j
(18.3.94)
c a n b e f o u n d as follows:
=
a i
*al7"29^ ^-^
(
2a^
)
(18
' '
3
95)
and
_
"2
9ai
a
_ v
• 917 3?7 ^
=
(
'*
1
2 >
1 a
_ ,
" 29|^ ' •
(a
9F
1 9£
9|7 ~ 2917•
=
n K( I, Q9 f6t)
°-
By forming t h e scalar p r o d u c t of each side of Eq. (18.3.92) with ax a n d
a2, respectively, w e g e t :
|£
- ±|£
2 d£
Solving for Z? a n d e, w e get:
= bE + eF
(18.3.97)
= bF + eG.
(18.3.98)
2
G
* - 2 ^ (
e
2
f
2
-
f<
f
£ £
f
f I f ) '
" 2 ^ (
1 3 89
'
'
) 9
8 3 1) 0 O
' '
By forming t h e scalar p r o d u c t of e a c h side of Eq. (18.3.93) with ax a n d
a2, respectively, t h e t w o following expressions for c a n d / a r e o b t a i n e d :
-wA Wr f}
f
E
F
-'
<,8 3 02)
By forming t h e scalar p r o d u c t of each side of Eq. (18.3.94) with ax a n d
a2, respectively, t h e t w o following expressions for d a n d y a r e o b t a i n e d :
(
Introduction to the Theory of Thin Shells
c
-M H j-M wr w
d
2G
E
f
l f -
I )
f
+
ff)-
2F
2
585
i s3 i o 3
< »
1 83 , 0 4
< - - >
Eqs. (18.3.92), (18.3.93), a n d (18.3.94), with the coefficients given b y
Eqs. (18.3.99) to (18.3.104), give the r e q u i r e d derivatives of a, a n d a2.
1
W h e n the p a r a m e t r i c curves are o r t h o g o n a l , the values
of the
coefficients b, c, d, e, f, j are simplified, since F = 0 a n d H = EG. T h e
derivatives b e c o m e :
dE
dE
dax
w
= Ln +
ai
ai
lE
"
dE
da2 =
9?T W
3fe
=
Nn-
dG
ai+
Mn =+
2£°'
dG
da2
(18.3.105)
2G
2G
(18.3.106)
BG
Ul
+
2E
2 G C
2.
(18.3.107)
If e, a n d e2 are the u n i t vectors parallel to ax a n d a2, then, from E q s .
(18.3.11) a n d (18.3.13), we h a v e :
= - k
ex=-J==
e2 = -%= = -%•
la~n
\[G
(18.3.108)
(18.3.109)
ex, e2, a n d n form a r i g h t - h a n d e d system of u n i t vectors m u t u a l l y
p e r p e n d i c u l a r . F r o m E q s . (18.3.105) t o (18.3.109), we easily d e d u c e t h a t
dex
dE
9 £e2_
2H >
L _
" v i " -
dG
M 3&
.
+
-Te"
e
2H *
917
dE
M — . 9£ 2e_
2F '
de2
9(7
9£i_
de2
N
-
(18.3.110)
(18.3.111)
(18.3.112)
(18.3.113)
586
T h e Theory of Elasticity
2
F r o m E q s . (18.3.46) a n d (18.3.47), in w h i c h w e set F = 0 a n d H
we get:
(
= EG,
i 3 8n ) 5
-
-
T h e p r e v i o u s derivatives c a n further b e simplified if t h e p a r a m e t r i c
curves a r e lines of c u r v a t u r e . I n such a case, M is set e q u a l t o zero a n d
the derivatives of t h e u n i t vectors e}, e2, a n d n b e c o m e :
M
3*1
L -
9f7 TjE ~
=
3?i
=
2
317
^
f
3*ei
2_
2H
=
=
=
in'
be
2/7 >
JL
n
1
VE_
"RT" ~ VG~W'
2
1
« « , n l f
(1
116)
°-
n 8
(
2
2
3*2 e
=
3V^_
VE~W
3*.
W
de2
n
.
1 1 T k
}
1 3V^3
- VGl*7
3G
_ ^llg _
1
}/£
(18.3.118)
n
_ _ L ^ g ,
~VE^ "^T*'
=
(18 3 119)
( 1 8 J
-
1 2 0 )
E q s . (18.3.116) t o (18.3.121) c o u l d h a v e b e e n directly o b t a i n e d from
E q s . (6.5.35) t o (6.5.38). T h i s is easily seen b y m a k i n g t h e following
substitutions in t h e latter g r o u p of e q u a t i o n s :
y uy 2, a n d y 3 a r e r e p l a c e d b y ^ , £2> a n d £ 3
e3 is r e p l a c e d b y n
hx is r e p l a c e d b y y/E ^ 1 — - ^ - ^
h2 is r e p l a c e d b y y/G ^ 1 — j ^ - ^
/z 3 is set e q u a l t o unity.
587
Introduction to the Theory of Thin Shells
T h e three last substitutions will b e justified in Sec. 18.4.
8) The Gauss-Codazzi
conditions
T h e six f u n d a m e n t a l m a g n i t u d e s E, F, G, L, M9 N a r e n o t functionally i n d e p e n d e n t b u t a r e c o n n e c t e d b y three differential relations. T h e s e
relations a r e to b e satisfied if the six m a g n i t u d e s a r e t o d e t e r m i n e a
surface uniquely, except for its p o s i t i o n a n d o r i e n t a t i o n in space
( c o m p a r e to c o m p a t i b i l i t y e q u a t i o n s of strain). W e shall restrict ourselves to t h e cases in w h i c h the p a r a m e t r i c curves a r e lines of c u r v a t u r e .
T h e f u n d a m e n t a l m a g n i t u d e s are t h u s r e d u c e d to four, since in this case
F = M = 0. T h e t h r e e relations a r e d e r i v e d b y writing t h e equality of
the m i x e d s e c o n d derivatives of the u n i t vectors. Let us first c o n s i d e r t h e
unit vector n. F r o m E q s . (18.3.120) a n d (18.3.121), we h a v e :
<
3
1
)1
8
'
If we carry the differentiation, we get:
§1
L4(^t)
"
r\^W)\
+
'
2
t i ^ T "
4(©]
= a
(18.3.123)
This e q u a t i o n is satisfied only if t h e coefficients of g, a n d e2 v a n i s h ;
hence,
R
2
3£ 2
I 3\/G _
RI 3 | ,
(18.3.124)
\Rj
3£ 2
( V G \
A
3|,
(18.3.125)
VR )
2
Eqs. (18.3.124) a n d (18.3.125) a r e k n o w n as t h e C o d a z z i c o n d i t i o n s .
If w e r e p e a t t h e p r e v i o u s steps with E q s . (18.3.116) a n d (18.3.117), w e
o b t a i n two c o n d i t i o n s of w h i c h only o n e is n e w — n a m e l y ,
J_(
_L_hLl) + J-( J-hH)
3|i \y/E
9*, /
3 £ 2\ y / G
3fe /
= _
^1^2
(18.3.126)
T h i s e q u a t i o n is k n o w n as the G a u s s c o n d i t i o n . It is useless to c o n s i d e r
the two e q u a t i o n s (18.3.118) a n d (18.3.119), since they d o n o t lead to
a n y n e w relations. T h e three c o n d i t i o n s (18.3.124), (18.3.125), a n d
(18.3.126) a r e k n o w n as the G a u s s - C o d a z z i c o n d i t i o n s . W e n o w state a
2
2
588
The Theory of Elasticity
f u n d a m e n t a l t h e o r e m of the t h e o r y of surfaces: If E, G, L, a n d N a r ee a
given functions of the real curvilinear c o o r d i n a t e s ^ a n d £2>
differentiable, a n d satisfy the G a u s s - C o d a z z i c o n d i t i o n s while E > 0
2
a n d G > 0, t h e n a real surface exists w h i c h is uniquely
det2
ermined
2
2
except for its position in s p a c e a n d which h a s [E(d^x)
+ G(d£2) ]
and
[L(di;x)
+ N(di;2) ]
as first a n d s e c o n d f u n d a m e n t a l forms. T h e G a u s s C o d a z z i c o n d i t i o n s are referred to as the compatibility c o n d i t i o n s of the
t h e o r y of surfaces.
9) Application to surfaces of revolution
A surface of revolution is o b t a i n e d b y r o t a t i o n of a p l a n e curve a b o u t
a n axis lying in the p l a n e of the curve. T h e curve is called the m e r i d i a n ,
a n d its p l a n e is the m e r i d i a n p l a n e . Let the axis of r o t a t i o n b e the OZ
axis, a n d let R0 b e the p e r p e n d i c u l a r from a n y p o i n t P o n the surface
to the OZ axis (Fig. 18.11). T h e e q u a t i o n of a m e r i d i a n is:
Fig. 18.11
R0 = R0(z),
(18.3.127)
a n d its p o s i t i o n is defined b y the angle 0 with the OXx, OZ p l a n e . T h e
intersections of the surface with p l a n e s p e r p e n d i c u l a r to the OZ axis a r e
circles called parallels. T h e position of a parallel is defined b y the
equation:
r
Introduction to the Theory of Thin Shells
z = constant.
589
(18.3.128)
T h e cartesian c o o r d i n a t e s of P a r e :
xx = R0cos 0,
x2 = i ^ s i n 0,
x3 = z.
(18.3.129)
T h e position vector of P is:
0 + l2R0sin
r = \R0qos
0 + l3z.
(18.3.130)
T h e p a r a m e t r i c e q u a t i o n s of the surface a r e :
xx = xx(0,z)
= R0(z)cos
0
(18.3.131)
x2 = x2(0,z)
= R0(z)sin
0
(18.3.132)
x3 = x3(0,z)
= z.
(18.3.133)
F o r each c o n s t a n t value of 0, there c o r r e s p o n d s a m e r i d i a n of t h e
surface; a n d for each c o n s t a n t of z, there c o r r e s p o n d s a parallel o n t h e
surface. L e t us take t h e m e r i d i a n s a n d t h e parallels as o u r p a r a m e t r i c
curves a n d identify z with £ x, a n d 0 with £ 2. T h e n
l
^
=
W
a 2 = |p
1 c s o 9= + s n i
=
'
^
0+
' 2 ^
1
= | | == - 7 j R0sin 0 + 7 2# 0c o s 0,
3
(18.3.134)
(18.3.135)
where
(18.3.136)
= ^2_m
dz
T h e first f u n d a m e n t a l m a g n i t u d e s a r e :
^ = ^ i ^ 2
F
=
=f'f=l
^ ' ^
=
(18.3.138)
2
= ^
(18.3.139)
H = ^EG - F = R ^/^TJR^
2
I
0
T h e first f u n d a m e n t a l form is:
(18.3.137)
= 0
£ ' §
G = a 2- a 2 = g.g
2
+ (^)
.
(18.3.140)
590
T h e Theory of Elasticity
(ds)
2
= [1 +
2
(R'0) ](dz)
2
2
2
+ R (d0) .
(18.3.141)
T h e n o r m a l is given b y :
_
=5
^
_ \ _ =(
h
R
S
g9 +C l Q
n2 _R0 - hgR oS l iK
(18.3.142)
a n d the second fundamental magnitudes are:
L =
_ R
= o K
n. ?r
m
M = n - ^ r
oi\
N =
. n¥ l
(18.3.143)
H
(18.3.144)
= 0
°?2
R
(18.3.145)
= i
where
Since b o t h F a n d M a r e e q u a l to zero, the p a r a m e t r i c curves a r e also
lines of c u r v a t u r e . T h e p r i n c i p a l radii of c u r v a t u r e are calculated from
Eqs. (18.3.86) a n d (18.3.87), with the result t h a t
2
R
E_
* ~ I "
[ l + ( * ; )
] i
(18.3.146)
Ro
W e see (Fig. 18.11) t h a t Rx is t h e r a d i u s of c u r v a t u r e of t h e g e n e r a t i n g
curve R0 = R0(z). R2 is t h e length of t h e n o r m a l i n t e r c e p t e d b e t w e e n P
a n d t h e OZ axis. B o t h Rx a n d R2 are positive quantities, w h i c h m e a n s
t h a t they a r e m e a s u r e d in t h e positive direction of n. Finally, it is easy
to check t h a t E, G, R{, a n d R2 define a valid surface b y substituting
their values in t h e G a u s s - C o d a z z i e q u a t i o n s w h i c h a r e satisfied.
A n a l t e r n a t e a n d useful description of a surface of revolution is b a s e d
o n t h e i n d e p e n d e n t variables <f> a n d 0, w h e r e </> is the angle b e t w e e n t h e
axis of revolution of the surface a n d the n o r m a l t o t h e surface at t h a t
p o i n t . I n this case, the first f u n d a m e n t a l form is:
2
2
(ds) = RKd$)
2
+ Rl(d0) ,
(18.3.148)
Introduction to the Theory of Thin Shells
591
w h e r e the first t e r m in the r i g h t - h a n d side represents the s q u a r e of the
differential length of a r c a l o n g a m e r i d i a n , a n d the s e c o n d t e r m
r e p r e s e n t s t h e s q u a r e of the differential length of a r c a l o n g a parallel. I n
this case, ^ is identified with <J>, a n d £22 with 0. T h e first f u n d a m e n t a l
m a g n i t u d e s in this case a r e Rx a n d R .
10) Important
remarks
a) T h e a n a l o g y b e t w e e n the m e t r i c p r o p e r t i e s of a surface [as
expressed t h r o u g h the m e t r i c coefficients ay of its first f u n d a m e n t a l
form] a n d the c o r r e s p o n d i n g m e t r i c p r o p e r t i e s of a three d i m e n s i o n a l
E u c l i d e a n space [as expressed t h r o u g h the m e t r i c coefficients gy of t h e
q u a d r a t i c differential f o r m (6.3.12) associated with a n y general system
of c o o r d i n a t e s ] , is evident. T h e expressions for gy a n d ay are t h e s a m e ,
except t h a t the indices of a t a k e the values 1 a n d 2, while the indices of
g t a k e the values of 1, 2, a n d 3. T h e quantities hx, h2 of C h a p t e r 6 c a n
b e identified with ^/E a n d yJ~G, respectively. A difference w h i c h is of
i m p o r t a n c e exists, however, b e t w e e n the t w o cases:
I n the case of a t h r e e d i m e n s i o n a l E u c l i d e a n space, it is always
possible, b y a c h a n g e of c o o r d i n a t e s , to r e d u c e the q u a d r a t i c differential
form to a s u m of squares of differential c o o r d i n a t e s ; i n d e e d , this is a
characteristic of E u c l i d e a n spaces. I n the case of a surface, unless t h e
G a u s s i a n c u r v a t u r e vanishes, it is n o t possible to r e d u c e its first
f u n d a m e n t a l f o r m to a s u m of squares of the differentials of its
G a u s s i a n c o o r d i n a t e s . Therefore, a surface for w h i c h the G a u s s i a n
c u r v a t u r e K does n o t v a n i s h is called a t w o d i m e n s i o n a l n o n - E u c l i d e a n
space. T h e G a u s s i a n c u r v a t u r e vanishes for a flat surface as well as for
a d e v e l o p a b l e surface, such as t h a t of a cylinder or of a c o n e .
b ) I n a system of o r t h o g o n a l curvilinear c o o r d i n a t e s , the curves of
intersection of the t h r e e c o o r d i n a t e surfaces Sx, S2, a n d . S 3 (Fig. 6.1) are
lines of c u r v a t u r e of these surfaces ( D u p i n ' s T h e o r e m [1]).
c) I n s o m e texts, the n o r m a l to the surface is c h o s e n in a w a y s u c h
t h a t it p o i n t s a w a y from the center of c u r v a t u r e . T h i s results in s o m e
c h a n g e s in sign in those expressions c o n t a i n i n g n. C a u t i o n s h o u l d b e
exercised w h e n c o m p a r i n g the e q u a t i o n s d e d u c e d in this text with t h o s e
in w h i c h such a c o n v e n t i o n is a d o p t e d .
18.4
Basic Assumptions and Reference System of Coordinates
I n developing the t h e o r y of thin elastic shells, the following a s s u m p tions (which are called Love's a s s u m p t i o n s ) are m a d e :
592
The Theory of Elasticity
1. T h e shell is thin. This m e a n s t h a t the thickness of t h e shell h is
small c o m p a r e d with the radii of c u r v a t u r e R{ a n d R2 of the m i d d l e
surface, so t h a t their r a t i o is small c o m p a r e d to unity.
2. T h e deflections of the shell a r e small, a n d the strains in t h e
d i r e c t i o n of t h e n o r m a l a r e small e n o u g h t o b e neglected. T h i s a s s u m p t i o n allows us to refer t h e analyses to the initial c o n f i g u r a t i o n of t h e
shell.
3. T h e n o r m a l stresses a c t i n g o n p l a n e s parallel to the m i d d l e surface
a r e negligible c o m p a r e d with o t h e r stress c o m p o n e n t s a n d m a y b e
neglected in t h e stress-strain relations. T h i s a s s u m p t i o n will generally b e
valid except in t h e vicinity of highly c o n c e n t r a t e d loads.
4. T h e c o m p o n e n t s of the d i s p l a c e m e n t s (ux a n d u2) a r e linearly
d i s t r i b u t e d across the thickness.
5. T h e shear strains w h i c h c a u s e the distortions of the n o r m a l s to the
)
c a n b e neglected. This, a d d e d to t h e
m i d d l e surface (el3 a n d e23
p r e v i o u s a s s u m p t i o n , allows us t o c o n c l u d e t h a t t h e n o r m a l s t o t h e
u n d e f o r m e d m i d d l e surface r e m a i n n o r m a l to it after d e f o r m a t i o n .
F r o m the s t u d y of the t h e o r y of surfaces, we k n o w t h a t a n y p o i n t o n
a shell c a n b e l o c a t e d b y m e a n s of three p a r a m e t e r s , two of w h i c h v a r y
o n t h e m i d d l e surface while t h e third o n e varies a l o n g t h e n o r m a l to t h e
m i d d l e surface. T h e m i d d l e surface will b e t h e reference surface a n d its
lines of c u r v a t u r e will b e c h o s e n as p a r a m e t r i c curves. T h o s e lines,
together with the n o r m a l , form o u r o r t h o g o n a l system of reference. A n
a r b i t r a r y p o i n t in the s p a c e o c c u p i e d b y the shell: c a n therefore b e
l o c a t e d b y m e a n s of t h e position v e c t o r
|(li,l2>^3)
(18.4.1)
w h e r e r is t h e p o s i t i o n v e c t o r of a c o r r e s p o n d i n g p o i n t o n the m i d d l e
surface, n is the u n i t n o r m a l vector, £ 3 is the d i s t a n c e of the a r b i t r a r y
p o i n t from the m i d d l e surface m e a s u r e d a l o n g n (Fig. 18.12). T h e
m a g n i t u d e of a n e l e m e n t of length is given b y :
1
(ds)
= dl-
dl = (dr + £3dn
+ nd£3)
• (dr + i3dn
+ nd£3).
(18.4.2)
If this scalar p r o d u c t is c a r r i e d out, k e e p i n g in m i n d the o r t h o g o n a l i t y
of the c o o r d i n a t e s , we get:
2
(ds)
2
= E\( I - | ) ( ^ )
2
G+ ( I - | )
(( ^
^ 2)
2
2
2
+ ( ^ 3) .
( 1 8 A 3
>
Introduction to the Theory of Thin Shells
593
This expression c o n t a i n s all t h e i n f o r m a t i o n n e e d e d t o m e a s u r e lengths,
areas, a n d v o l u m e s in a shell. I n o t h e r w o r d s , it c o n t a i n s all t h e m e t r i c
p r o p e r t i e s of t h e shell. T h e first t w o t e r m s in t h e r i g h t - h a n d side
r e p r e s e n t t h e first f u n d a m e n t a l form of a surface at a d i s t a n c e | 3 from
the m i d d l e surface (Fig. 18.12). T h e lengths of t h e edges of this e l e m e n t
of surface a r e :
1°
Fig. 18.12
( * 1) ^ = V ^ ( l - | - ) ^ i
(18.4.4)
(18.4.5)
a n d the differential a r e a s of t h e edge faces a r e :
594
T h e Theory of Elasticity
1 84 6
(dAx)h
= VE ( l - j | ) ^ i ^ 3
( - - )
(dA2)b
= VG ( l - ^ ) ^ 2 ^ 3 •
(18-4.7)
Therefore, the m e t r i c coefficients of a surface at a d i s t a n c e £ 3 from
the m i d d l e surface a r e :
( A i ) | = ( £ ) fe = £ ( l - | r )
G
(*2)f3 = ( < % =
=
( l - ] | )
1
(18.4.8)
2
-
(18-4.9)
(18.4.10)
W e a r e n o w in a position to derive t h e s t r a i n - d i s p l a c e m e n t relations a n d
the e q u i l i b r i u m e q u a t i o n s .
18.5
Strain-Displacement Relations
T h e s t r a i n - d i s p l a c e m e n t relations in o r t h o g o n a l curvilinear c o o r d i n a t e s were derived in Sees. 6.6 a n d 6.7. T h e expression for the strain ey
is [see E q s . (6.6.30)]
£ j i= *L ( n o s u m ) .
(18.5.1)
T h e values of yy, for / a n d j v a r y i n g from 1 to 3, a r e given b y E q s . (6.7.8)
to (6.7.13). If, in these e q u a t i o n s , we neglect all t h e s e c o n d o r d e r terms,
we a r e left with t h e expressions for the linear strains. T h e linearized
e q u a t i o n s a r e the only o n e s of interest t o us in this c h a p t e r . F o l l o w i n g
o u r previous c o n v e n t i o n s , t h e linear strains will b e called ey. F o r t h e
a p p l i c a t i o n to thin shells, we m a k e t h e following s u b s t i t u t i o n s :
y x is r e p l a c e d b y | ,
y 2 is r e p l a c e d b y £ 2
a
n u ea
r
(18.5.2)
y 3 is replaced b y £ 3.
W j , w2> d 3
the c o m p o n e n t s of the d i s p l a c e m e n t of a p o i n t a l o n g
the £ i , £ 2, a n d £ 3 curves. H e n c e , from E q s . (18.5.1) a n d (6.7.8) to
(6.7.13), we h a v e :
Introduction to the Theory of Thin Shells
h\ 3£i
e22 =
e 33 =
\(
/2
2
d£
£
1 3wi
'13
A3
3^3
(18.5.4)
u2 "
3A
"l
"1 3/z
" 33 +, "2
"3
3£i
M 2 3^2
(18.5.5)
u
3
1 3«2
1 3"3 .
A,
« 22
_|_ "M1 j _ o3A
1
A
A,
3li
3^3
2
2
1 9^3 9 ,
^3
+
3^2
, J f"3
3 _ ^dh
22
_
J_3«2
~ 2 V A 3 3£ 3
\(
M 2
595
A 2 3£ 2
8
1 "i
w
dhz2
» z2 —
"l
h\h2 3£,
w3 3/i 3
hA2h23A 3 3£ 2
"i
—1 9
M l
£i
"2
3A,
—. .
3*2
3A 2
A
A22/Al 33 3 { 3 /
"3
3A 3 1
—j
A 3A, 3«,
, 8 5 6^ r
K
' >
n x s K\
In E q s . (18.5.3) to (18.5.8), h\, h2, a n d h3 are the metric coefficients of
a surface at a distance £ 3 from the m i d d l e surface. F r o m E q s . (18.4.8)
t o ( 1 8 . 4 . 1 0 ) , hx> h2, a n d h3 are given by:
hx = Ve(\
- | - )
(18.5.9)
*2 = V ^ ( l
- J ^ )
(18.5.10)
A 3 = 1.
(18.5.11)
T h e a b o v e s t r a i n - d i s p l a c e m e n t relations are general a n d d o n o t reflect
the a s s u m p t i o n s of Sec. 18.4:
a) T h e s e c o n d a s s u m p t i o n , in c o n j u n c t i o n with Eqs. (18.5.5) a n d
(18.5.11), leads us t o :
~ = ^ 1 = 0
e 33
^
u,a
t h a t is to say:
(18.5.12)
v
)
w
£2)
u3 = u3(£l9
= 30>
(18.5.13)
w h e r e w 30 is the d i s p l a c e m e n t of the m i d d l e surface in the direction of
the n o r m a l .
l
6
596
The Theory of Elasticity
b) T h e fourth a s s u m p t i o n allows us to r e p r e s e n t the c o m p o n e n t s of
the d i s p l a c e m e n t at e a c h p o i n t as follows:
».-«.o + « 3 ( ^ ) f
e)
"2 = «20 + * 3 ( | ^
-
3
(18-5-14)
H
=
(18-5.15)
0
w 10 a n d w 20 r e p r e s e n t t h e c o m p o n e n t s of the d i s p l a c e m e n t s of p o i n t s
the m i d d l e surface £ 3 = 0. (du{ / 3 ^ 3) | 3 0= a :n d ( 9 w 2/ 9 £ 3) £ 3 0=r e p r e s e n t
angles of r o t a t i o n ( m o r e precisely their t a n g e n t ) of t h e n o r m a l to
p a r a m e t r i c curves £j a n d £ 2, w h i c h lie o n the m i d d l e surface. I n
following, w e shall refer to these t w o angles as /3X a n d / ? 2.
on
the
the
the
c) T h e fifth a s s u m p t i o n results in e12>
= e n = 0. T a k i n g i n t o a c c o u n t
E q s . (18.5.9) to (18.5.11), we h a v e :
3
8 2
e
-
if "
1
"
1
I
2
8
1-0
"
(18.5.16)
a n d setting £ 3 = 0, w e o b t a i n :
In the s a m e way, e I3 = 0 leads t o
Hence,
10 +=
«1 = «10 - ^ ( ^ ^
+ 5f)
"2 = "20 - l 3 ( ^ | |
+ ^ )
"
= "20 +
^
(18.5.19)
(18-5.20)
S u b s t i t u t i n g E q s . (18.5.9), (18.5.10), (18.5.11), (18.5.19), a n d (18.5.20)
i n t o t h e s t r a i n - d i s p l a c e m e n t relations (18.5.3) to (18.5.8), a n d neglecting
in the final result £ 3/Rx a n d £ 3/R2 c o m p a r e d t o u n i t y (first a s s u m p tion), we get:
Introduction to the Theory of Thin Shells
597
en = * 2 3 = * 3 3 = 0
(18.5.21)
*ii = * i i o - f e * i i
08.5.22)
^22 =
(18.5.23)
e
220
"
^3^22
(18.5.24)
^2 = ^120-^3^12,
w h e r e t h e subscript 0 refers t o t h e m i d d l e surface a n d
9w
i o +,
o= 1
w 9
20 V ^
" 7£W v^^-t
e
9 w
20
1
.
(18 5 25)
w
w
10
3
V ^ U
V^i/
11
--
3
Vc3fe/3fe
(18.5.28)
1
^
22
A
9
J
-08,)—TUW,)-^
1=
("20
\ .
1
V c ^ \ « 2
/ «io
1
d "
3
\
9Vg
V^U,
(18.5.29)
1
-(ft)--TUGS,)- ^
a
1 g9
8
f
20
"
8
, i
3
V ,AE a
m
/
" 10
,
I^YI
(18.5.30)
+
2LV
Ed^
\y/G)
V
G S & V V I / J -
598
The Theory of Elasticity
m
a
In the a b o v e expressions, e n , oe220
, ^120
Y b e i n t e r p r e t e d physically
as strains in the m i d d l e surface of the shell. Kxx a n d K22 represent the
c h a n g e s in c u r v a t u r e of the m i d d l e surface d u r i n g d e f o r m a t i o n . KX2
represents the c h a n g e in twist of the m i d d l e surface d u r i n g d e f o r m a t i o n .
E q s . (18.5.25) to (18.5.30) are the strain-displacement relations for thin
shells.
18.6
S t r e s s Resultants and S t r e s s Couples
I n the previous sections, the strains (therefore, the stresses) h a v e b e e n
s h o w n to b e linearly distributed across the thickness of the shell. It is
convenient, as was d o n e in the study of thin plates, to integrate the
stress distribution t h r o u g h the thickness, a n d to replace the stresses b y
equivalent stress resultants a n d stress couples. T h e variations with
respect to | 3 are t h u s completely eliminated. Let usn consider a n e l e m e n t
of a shell subjected to a lateral l o a d q = q(£x, £ 2) . I a d d i t i o n to b e n d i n g
a n d twisting m o m e n t s , there will b e n o r m a l a n d shearing forces acting
o n the sides of the element. B o t h m o m e n t s a n d forces are expressed p e r
u n i t length of shell a l o n g the £ x a n d £ 2 directions. T h e c o n v e n t i o n for
Fig. 18.13
m o m e n t s is t h a t positive m o m e n t s give positive stresses o n the positive
half of the shell. Stresses a n d forces follow the c o n v e n t i o n s established
in Sec. 7.2. Fig. 18.13 shows the directions for positive m o m e n t s a n d
forces. F o r m o m e n t s , the r i g h t - h a n d rule applies with the t h u m b
Introduction to the Theory of Thin Shells 599
p o i n t i n g i n t h e direction of t h e d o u b l e a r r o w . F o r c e s a n d m o m e n t s
c a r r y t h e subscript of t h e stresses they cause.
Recalling t h a t t h e lengths of t h e edges of t h e e l e m e n t in Fig. 18.13
(for a n y d i s t a n c e £ 3from t h e m i d d l e surface) a r e given b y E q s . (18.4.4)
a n d (18.4.5), w e h a v e :
r +—
/
\
or
Similarly,
N22
-I^O-IH
r +-
Nil-
''
<,8 6 3)
2
/
\
J_l °n(l - jtj d£ ,
(18.6.4)
3
^ . - / ? a 2 . ( l - | ) ^
^=X?^(l-|)^3,
(18.6.5)
^23=X?^23(l-|-)^3
(18.6.6)
M22
=
j7o {\ ~ | - ) « 3 ^ 3
22
600
T h e Theory of Elasticity
(18.6.7)
/ _ ? a 2 ( l1 - 0 3 ^ 3 -
M 21 =
F r o m the a b o v e e q u a t i o n s , we see that a l t h o u g h ol2 = o2\, Nl2 is n o t
e q u a l to N2\, a n d M 12 is n o t equal to M 21 b e c a u s e Rx is n o t necessarily
equal to R2. H o w e v e r , t a k i n g i n t o a c c o u n t the first a s s u m p t i o n of Sec.
18.4, the q u a n t i t i e s i;3/R\ a n d £-j/R2 are negligible c o m p a r e d to unity,
so t h a t Nl2 = N2l a n d M 12 = M 2 ; 1E q s . (18.6.2) to (18.6.7) t h e n b e c o m e
similar to those u s e d in t h e t h e o r y of thin flat plates.
T h e expressions for forces a n d m o m e n t s w h i c h involve a u, a22
, and
a 12 c a n b e written in terms of the strains using the stress-strain
relations:
a,, =
a 22 -
E
(e + j - e 2 ) 2
1 - V- u
E 2[e
+ ve22Q- £3 (Kx, +
1 - v u0
L
j{e
22 +
— v
II
1
2
V
-
fe>20 +
ven)
(18.6.8)
vK22
)]
(18.6.9)
_
£?(^22 +
^110
^ll)]
_
ey
€n
E
(18.6.10)
*3^12)+ v
- 1 + :(v \20
Because of the a s s u m p t i o n s o n e 3 , 3e i , 3 a n d e23
, H o o k e ' s law gives us
only three stress-strain relations. Substituting E q s . (18.6.8) to (18.6.10)
i n t o the expressions of Ny a n d My, a n d neglecting i3/Rx
a n d £-$/R2
c o m p a r e d to unity, we o b t a i n :
°12
=
Eh
( e m+
1 - v2
Eh „
Nl2 = N2l =
ve220
),
^22
=
Eh
1
(^220 + ^ 1 1 0 )
(18.6.11)
(18.6.12)
(18.6.13)
r+7
M „ = -D(KU
(18.6.14)
120
+ vK22
),
Ml2 = M 21 = -Z>(1 where
2
D M= = -D(K
v )' +
2212(1 - 22
v)Kl2
,
3
Eh
vKn)
(18.6.15)
Introduction to the Theory of Thin Shells
601
E q s . (18.6.11) to (18.6.14) are the force-strain relations for thin shells.
VX3 a n d V23 c a n n o t b e written in t e r m s of the strains. T h e y c a n ,
however, b e o b t a i n e d from the general e q u a t i o n s of equilibrium.
F o r the sake of clarity a n d easy c o m p a r i s o n with the e q u a t i o n s of thin
flat plates, Eqs. (18.6.8) to (18.6.14) will b e rewritten in m a t r i x form:
1
v
0
°22
v
1
0
°12
0
0
M22
ES3 2 v
Ky
1
0
K-22
0
0
1 - v
v
12
v
0
1
0
220
0
1—v
^120
1
v
0
-D v
1
0
0
0
1 - v
Mx2
(18.6.16)
0
1
Eh 2 v
1 - v
0
=
^120
v
1 -
Mu
220
1-?
1
_NX2
_
e
e
^.i
K22
(18.6.17)
(18.6.18)
12
F r o m E q . (18.6.17), o n e c a n o b t a i n the expression of the strains in t e r m s
of the forces:
e
e220
\20
1
J_
Eh
V
0
V
0
1
0
N22
0
1 + v
_NX2
_
(18.6.19)
A t this stage, it is a p p r o p r i a t e to m a k e a r e m a r k r e g a r d i n g L o v e ' s
a s s u m p t i o n s : W h e n investigating the d e f o r m a t i o n s of the surfaces
parallel to the m i d d l e surface, e n a n d e23 were e q u a t e d to z e r o ; in o t h e r
w o r d s , the effects of a 13 a n d o23 o n the d e f o r m a t i o n of these surfaces
were neglected. T h i s d o e s n o t m e a n t h a t VX3a n d V23 c a n b e neglected,
since these s h e a r i n g forces are essential to equilibrium.
18.7
Equations of Equilibrium of Loaded Thin Shells
T o derive the equilibrium e q u a t i o n s , let us c o n s i d e r a small e l e m e n t
s e p a r a t e d from a shell b y four sections p e r p e n d i c u l a r to the m i d d l e
surface. T h e external forces a c t i n g o n the e l e m e n t s are the b o d y forces,
602
The Theory of Elasticity
which will b e neglected, a n d the surface forces. T h e internal forces are
the stresses acting o n the sides of the elements. Both external a n d
internal forces are r e d u c e d to statically equivalent systems acting o n the
m i d d l e surface. A s w a s d o n e in Sec. 7.12 for the derivation of the
equilibrium e q u a t i o n s in o r t h o g o n a l curvilinear c o o r d i n a t e s , w e shall
express the equilibrium of forces a n d m o m e n t s in vector form, t h e n
write d o w n the c o m p o n e n t e q u a t i o n s . U s i n g the sign c o n v e n t i o n s of
Fig. 18.13, the sides of the e l e m e n t of the m i d d l e surface (Fig. 18.14a)
are a c t e d u p o n b y the following forces a n d m o m e n t s (Figs. 18.14b,
18.14c):
O n side O C , the force per unit length is
=
+ ^ 1 2 ^ 2 + v nn \
(18.7.1)
a n d the m o m e n t p e r unit length is
-Mix
= -(Mne2-
Mnex).
O n side OA, the force p e r unit length is
Fig. 1 8 . 1 4
(18.7.2)
Introduction to the Theory of Thin Shells
- Np
= -(N2X
ex
+ N22
e2
+ V23
n\
603
(18.7.3)
a n d t h e m o m e n t p e r unit length is
-Mi2
= -(M2X
e2
(18.7.4)
- M22
ex).
O n t h e side OB, the force p e r unit length is
fy + ^f-*..
< 18J
5)
a n d t h e m o m e n t p e r unit length is
6)
+
-
O n side BC, t h e force p e r unit length is
a n d t h e m o m e n t p e r unit length is
7 8)
*
»
- + ^ f * 2 .
I n a d d i t i o n , t h e e l e m e n t is subjected to t h e external surface l o a d i n g q
p e r unit a r e a :
q = qxex + q2e2 + q3n.
(18.7.9)
T h e a r e a of t h e e l e m e n t is given b y \[EG dix di2. T h e c o n d i t i o n t h a t t h e
resultant vector of all t h e forces acting o n the e l e m e n t of t h e m i d d l e
surface is to vanish, c a n b e written as follows:
(]V {1 +
)(y/G di2 +
+ (% + l ^ ^ ) ( v ^
+ q^/EGdt>x
di2
= 0.
d£2) - Nv VG d£2
dix + ^d^dh)
-
NaVE
d^
(18.7.10)
N e g l e c t i n g the infinitesimals of t h e third order, t h e previous e q u a t i o n is
reduced to:
604
The Theory of Elasticity
+ ^(VE
N&
)
+ VEG
N#)
(18.7.11)
q = 0.
R e p l a c i n g N$x, 7V| 2, a n d q b y their expressions in t e r m s of
c o m p o n e n t s we h a v e :
their
^[VG(N e +N e +V n)]
xx
x
X2
2
X3
1 7 18
+ ^[VE(N e
2l
+
{
#
2 2
e
+
22
K 7 2 )]
3
(
-
-
>
Eqs. (18.3.116) to (18.3.121) are n o w used in taking the partial derivatives of the unit vectors in Eq. (18.7.12), which b e c o m e s :
NU) + , | - ( V £ N2l
) + nJ-^-
[^(VG
/EG
N22
^
+ q \[EG e
x
x
N ) + ^-(VE N )
J^iVG
-
l2
22
N ^--N ^
+
2l
u
—-"l
VEG
~ V \-+
(18.7.13)
q VEGJe
23
2
[ i ^ ^ ) 4 ( ^ K
+
V
}
2
2
3
)
V ^ ( f +
+
+ # 3V £ G ] « = o.
In o r d e r t h a t this vector e q u a t i o n b e satisfied, the coefficients of ex, e2,
a n d n m u s t identically vanish. H e n c e , the following three differential
e q u a t i o n s of equilibrium of forces are o b t a i n e d :
£-(y/G Nu)
^
3
+
^
af-(V£
2)
J V
+
nJ-^
1
+ ^iV^ = 0
-
N J^
2
2
Introduction to the Theory of Thin Shells
605
(18.7.15)
(18.7.16)
T o find t h e e q u i l i b r i u m e q u a t i o n s for the m o m e n t s , we m u s t vectorially a d d the internal m o m e n t s s h o w n in Fig. 18.14c to those m o m e n t s
d u e to the forces s h o w n in Fig. 18.14b. T h e s u m of the internal m o m e n t s
s h o w n in Fig. 18.14c is:
+
l « ~w ^)
M
2+
d
[y
1
+
M
-4-^1
- &VE
dt
x
(18.7.17)
= [^(VG
M ) + ^(V£
ix
M )\di di ,
i2
x
2
in w h i c h the infinitesimals of the third o r d e r h a v e b e e n neglected. In
c o n s i d e r i n g the m o m e n t s a b o u t O of the forces in Fig. 18.14b, we shall
neglect the infinitesimals of the third order. In t h a t respect, we notice
that:
a) T h e m o m e n t d u e to the external force q is c o m p u t e d by multiplying q first b y the a r e a of the e l e m e n t (a s e c o n d o r d e r q u a n t i t y ) , t h e n by
the m o m e n t a r m (a first o r d e r q u a n t i t y ) . T h e result is t h a t q will b e
multiplied b y a n infinitesimal of the third o r d e r a n d , c o n s e q u e n t l y , its
m o m e n t c a n b e neglected.
b ) T h e c o m p9o n e n t s of t h e forces in the p l a n e of t h e m i d d l e surface
(i.e., the Ntj s) give m o m e n t s of significance only a b o u t the n o r m a l n.
O n the o t h e r h a n d , the c o m p o n e n t s of the forces n o r m a l to the m i d d l e
surface (i.e., t h e Vy ' s) give m o m e n t s of significance only a b o u t the | j
a n d £ 2 lines. W h e n multiplying a force b y its a p p r o p r i a t e m o m e n t a r m ,
we c a n neglect the differential forces since they lead to third o r d e r
t e r m s . T a k i n g t h e p r e v i o u s r e m a r k s i n t o a c c o u n t (in o t h e r w o r d s ,
606
T h e Theory of Elasticity
retaining only t h e s e c o n d o r d e r terms), w e o b t a i n t h e following expression for t h e m o m e n t of all t h e forces a b o u t 0:
[V23
ex
- VX3
e2 + (NX2- N2X
)n]^/EG
(18.7.18)
dlxdl2.
F o r e q u i l i b r i u m of m o m e n t s , t h e s u m of E q s . (18.7.17) a n d (18.7.18)
m u s t b e e q u a l to zero. T h u s ,
v
/
€v 1
3£i
9&
+ VEG [V23
ex
*
(18.7.19)
- VX3
e2 + (NX2- N2X
)n]
= 0.
R e p l a c i n g M^x a n d M^2 b y their expressions in t e r m s of their c o m p o nents, we have:
^-[y/G
(Mne2
- M 1 e2, ) ] + ^[\/E
+ VEG [V23
ex
(M2x
e2
- VX3
e2 + (NX2- N2X
)n]
- M22
ex)\
= 0.
T a k i n g t h e derivatives, E q . (18.7.20) b e c o m e s :
[ ~ ^ ( V G
M I )2- g | ( V ^ M 2 )2 - M 2
^ 4I M , , ^
+ y/EG F 2 3]g,
^" ^ir
M2i)+Mi2
+
M n ) +
w
(VE
2
*2
+ yfEG
[ - - ^
+
+ Nl2 - N2X
jn
M22
(18.7.21)
= 0.
In o r d e r t h a t this vector e q u a t i o n b e satisfied, t h e coefficients of ex, e2,
a n d n m u s t identically v a n i s h . H e n c e , t h e following three differential
e q u a t i o n s of equilibrium of m o m e n t s a r e o b t a i n e d :
Introduction to the Theory of Thin Shells
^ ( V G
M 1 ) 2+ ± ( V E
-
F 23 = 0
^(VG
M U) + ^
- ^/EG
Vn = 0
(
^
N n ~ N * + ^ - ^
M 2 ) 2+ M 2
^ 1- M n ^
M 2 )1
^ 2- M 2
+M 1
^ 2
= 0.
607
)
g
7 ) 2 (3 l
(18.7.24)
T h e six E q s . (18.7.14) to (18.7.16) a n d (18.7.22) to (18.7.24) are the
c o n d i t i o n s for the e q u i l i b r i u m of a small e l e m e n t of the shell. U p o n
e x a m i n a t i o n of these e q u a t i o n s , we notice t h a t E q . (18.7.24) is a n
identity. I n d e e d , if we i n t r o d u c e in it the definitions given in Eqs.
(18.6.4) a n d (18.6.7), we get:
+^
-/.I'O-fcX'-fcK-•*>'«>-*
(18.7.25)
since oX2 = o2X
. W e also notice t h a t VX3a n d V23 c a n b e e l i m i n a t e d from
the set of six e q u a t i o n s b y solving E q s . (18.7.22) a n d (18.7.23) for these
q u a n t i t i e s a n d substituting the resulting expressions i n t o the r e m a i n i n g
e q u a t i o n s . T h u s the definition of VX3 a n d V23 h a s n o b e a r i n g o n the
analysis.
T h e d e r i v a t i o n of the e q u a t i o n s of e q u i l i b r i u m d i d n o t involve a n y
equality b e t w e e n Nl2 a n d N2l or MX2 a n d M2X
. H o w e v e r , since these
e q u a t i o n s will b e u s e d in c o n j u n c t i o n with the force-strain relations a n d
the s t r a i n - d i s p l a c e m e n t relations in w h i c h those equalities w e r e ass u m e d , the e q u i l i b r i u m e q u a t i o n s m a y b e rewritten setting NX2 = N2X
a n d MX2 = M2X
.
In s u m m a r y , w e h a v e five e q u a t i o n s of equilibrium, six force-strain
relations [Eqs. (18.6.11) to (18.6.14)], a n d six s t r a i n - d i s p l a c e m e n t relations [Eqs. (18.5.25) to (18.5.30)] for a total of 17 e q u a t i o n s in t e r m s of
e17 v a r i aeb l e s : N , 7V , N = N , V , V , M , M , M = M , e ,
n
2 2 X2
2X X3 23 n
2 2 X2
2X x x o
2io> no>
K22
, KX2
, w 1 , 0w 2 , 0u30
. In theory, the p r o b l e m c a n b e
solved o n c e the b o u n d a r y c o n d i t i o n s are specified.
(
1
608
The Theory of Elasticity
18.8
Boundary Conditions
W e shall only e x a m i n e the case w h e r e the b o u n d a r i e s coincide with
the lines of c u r v a t u r e of the m i d d l e surface, a n d a s s u m e for the p u r p o s e
of this analysis t h a t the b o u n d a r y coincides with the £1 line (Fig. 18.15).
, N22
, and
A c t i n g o n this b o u n d a r y are the five quantities M 2 , 1M 2 , 2V23
N2X
, a n d o n e w o u l d think t h a t the n u m b e r of c o n d i t i o n s necessary to
completely d e t e r m i n e the solution m u s t b e five. I n the following, w e
shall p r o v e t h a t the p r o b l e m is completely defined b y four, a n d n o t five,
b o u n d a r y c o n d i t i o n s . T h e r e a s o n i n g closely follows t h a t m a d e in the
case of flat plates (see Sec. 17.5).
Let us consider a segment of the b o u n d a r y n e a r a p o i n t mx, a n d
a p p r o x i m a t e this segment b y two e q u a l c h o r d s mmx a n d mxm2.
The
value of the twisting m o m e n t per unit length at the m i d d l e of mmx is
M 2 , Ia n d t h a t at the m i d d l e of mx m2 is (M2X + 3 M 21 / 3 £ , d£x). T h e total
twisting m o m e n t s acting o n mmx a n d mxm2
a r e M2X
\[E
d^x a n d
( M 21 + 3Af 21 /d£x d£x)(\/E
d^x\
respectively. E a c h of these m o m e n t s
c a n b e r e p l a c e d b y t w o parallel forces e q u a l in m a g n i t u d e a n d o p p o s i t e
in direction at the e n d s of mmx a n d mxm2,
as s h o w n in Fig. 18.15. T h e
force at m is parallel to the n o r m a l to the c h o r d mmx, a n d t h e force at
m2 is parallel to the n o r m a l to the c h o r d mx m2. Projecting the forces at
mx a l o n g the n o r m a l , w e get:
Introduction to the Theory of Thin Shells
+ M 21 - M 2 ) 1c o s * «
609
(18.8.1)
a n d a l o n g the t a n g e n t , we get:
( M 21 + M 21 + - ^ 1 1 ^ , j s i n 4> «
[2M2i + - ^ ^ i
J^I^T
(18.8.2)
T h u s , a l o n g the edge of the shell, the twisting m o m e n t c a n b e r e p l a c e d
b y d i s t r i b u t e d shearing forces in the direction of the n o r m a l e q u a l t o :
1
a n d b y d i s t r i b u t e d shearing forces in the direction of the t a n g e n t e q u a l
to:
1
( 2iVE d£x
M
\ _
M 2I
0884)
Therefore, w h e n the b o u n d a r y is a ^ line, the four quantities,
s ofe the shell. T h e
completely d e t e r m i n e the state of stress at the needge
s a m e r e a s o n i n g c a n b e r e p e a t e d w h e n the £ 2 l i * t h b o u n d a r y line.
Therefore, the n u m b e r of b o u n d a r y c o n d i t i o n s at each edge m u s t b e
equal to four. It is o b v i o u s t h a t the b o u n d a r y c o n d i t i o n s are n o t always
expressed in t e r m s of forces a n d m o m e n t s , since o n e often prescribes
the d i s p l a c e m e n t s a n d the angles of r o t a t i o n . T h e total n u m b e r of
c o n d i t i o n s , however, c a n n o t exceed four:
1) Built-in or clamped edge
A t a built-in edge, we h a v e :
«, = 0,
u2 = 0,
« 3 = 0,
f}2 =
2) Simply supported edge
A t a simply s u p p o r t e d edge, we h a v e :
- ^
= 0
(18.8.6)
610
T h e Theory of Elasticity
ux = 0,
u2 = 0,
u3 = 0,
M22 = 0.
(18.8.7)
3) Free edge
A t a free e d g e :
N22 = 0,
N 2 - ^l
= 0,
3 + jV Jj ^ - 0 ,
M 22 = 0.
(18.8.8)
T h e c o n c e p t of b o u n d a r y c o n d i t i o n s loses its m e a n i n g w h e n the shell is
closed. T h e c o o r d i n a t e lines £j a n d £ 2 o n the m i d d l e surface a r e closed
curves a n d o n e periodically r e t u r n s to the s a m e p o i n t a l o n g a curve £ 2 =
c o n s t a n t or £ 2 = c o n s t a n t . I n such cases, o n e m u s t i m p o s e t h a t the
solution b e p e r i o d i c functions of £i a n d £ 2, a n d the b o u n d a r y c o n d i t i o n s
are r e p l a c e d b y t h e periodicity c o n d i t i o n s .
18.9
Membrane Theory of Shells
I n m a n y p r o b l e m s of thin shells, the loadings a r e such t h a t the
b e n d i n g a n d twisting m o m e n t s are zero, or so small t h a t they c a n be
neglected. T h e stresses in the shell are m a i n l y d u e t o the forces
7 V n, 7V 2, 2NX2
, a n d N2l
. T h e t h e o r y of thin shells, b a s e d o n the a s s u m p tion of zero stress couples, is called m e m b r a n e theory. T h e c o n d i t i o n s
of e q u i l i b r i u m for this case c a n b e o b t a i n e d b y setting:
Mxx = M22 = MX2 = M2X = 0
(18.9.1)
first in the differential e q u a t i o n s of equilibrium of m o m e n t s , a n d s e c o n d
in the differential e q u a t i o n s of equilibrium of forces. This gives:
(18.9.2)
V23 =VX3 = 0
NX2 = 7V 21
(18.9.3)
+ qx^EG
= 0
+
+
4- q2\fEG
(18.9.5)
= 0
Introduction to the Theory of Thin Shells
611
(18.9.6)
E q s . (18.9.2) to (18.9.6) s h o w t h a t the applied l o a d s a r e s u p p o r t e d b y
internal forces in the p l a n e of the shell. T h e y a r e the e q u a t i o n s of
e q u i l i b r i u m for a shell in a m e m b r a n e state of stress. T h e force-strain
relations are given b y E q . (18.6.17). T h e s t r a i n - d i s p l a c e m e n t s relations
are given b y E q . (18.5.25) to (18.5.27).
T h e three e q u a t i o n s of equilibrium (18.9.4) to (18.9.6) c o n t a i n three
, a n d Nl2 = N2l
. T h e p r o b l e m is, therefore, statiu n k n o w n s — N n, N22
cally d e t e r m i n a t e . O n c e those values are found, the strains c a n b e
o b t a i n e d from E q . (18.6.19). K n o w i n g the strains, the d i s p l a c e m e n t s in
the m e m b r a n e shell a r e o b t a i n e d b y i n t e g r a t i o n of E q s . (18.5.25) to
(18.5.27).
Finally, t h e b o u n d a r y c o n d i t i o n s a l o n g a line of c u r v a t u r e m u s t b e
limited to t w o . I n d e e d , if the a s s u m p t i o n s of the m e m b r a n e t h e o r y a r e
i n t r o d u c e d in E q . (18.8.5), the t w o r e m a i n i n g q u a n t i t i e s a r e N22 a n d N1X
,
to b e specified a l o n g a £j line. W h e n the b o u n d a r y c o n d i t i o n s a r e
specified in t e r m s of d i s p l a c e m e n t s , the four q u a n t i t i e s involved are
w 1 , 0w 2 , 0w 3, a n d / ? 2. H o w e v e r , it is n o t possible to i m p o s e c o n d i t i o n s o n
u3 a n d / 3 2, since this w o u l d affect the values of V23 a n d M22
. If we
specify, for e x a m p l e , t h a t u3 = P2 = 0 o n t h e b o u n d a r y , t h e c o n d i t i o n
t h a t V23 = M22 = 0 in m e m b r a n e shells c a n n o t b e satisfied. It follows
that, o n t h e edge of the m e m b r a n e o n e c a n only specify the c o m p o n e n t s
of
w the d i s p l a c e m e n t tangential to the m i d d l e s u r f a c e — n a m e l y , ul0 a n d
20-
18.10 Membrane Shells of Revolution
T h i n shells of revolution a r e extensively u s e d in v a r i o u s types of
structures, such as c o n t a i n e r s , t a n k s , na sned d o m e s . Let us t a k e t h e £j lines
a l o n g the m e r i d i a n s , a n d the £ 2 l i
a l o n g the circles in the p l a n e s
p e r p e n d i c u l a r to the axis of revolution (Fig. 18.16).
It w a s s h o w n in Sec. 18.3 t h a t the radii of c u r v a t u r e a t A of t h e ^ a n d
£ 2 lines lie o n the n o r m a l to the surface b u t h a v e different lengths. T h e
first f u n d a m e n t a l form, as given b y E q . (18.3.148), is:
2
(ds)
2
2
= R (d<}>) +
2
R (d9)
2
(18.10.1)
612
The Theory of Elasticity
Fig. 1 8 . 1 6
w h e r e w e h a v e r e p l a c e d the subscript 1 by <£. R e m e m b e r i n g t h a t R^ a n d
R0, (the first f u n d a m e n t a l m a g n i t u d e s ) are i n d e p e n d e n t of 9, a n d t h a t
dR0/d<$> = R^cos 4> (Fig. 18.6), the e q u a t i o n s of equilibrium (18.9.4) to
(18.9.6) b e c o m e :
^ ( * 0A W ) + K * " ^ - NnR+cos
os
+ R j $ -
^
+ ^
+ , 3 = 0.
+ AW V
<p + q^R+R0
= 0
RR
* + «e * o
(18.10.2)
0
=
(18.10.3)
(18.10.4)
T h e force-strain relations are given b y E q s . (18.6.17), in w h i c h the
subscripts 1 a n d 2 are c h a n g e d to 0 a n d 0.
T h e strain-displacement relations (18.5.25) to (18.5.27) b e c o m e :
Introduction to the Theory of Thin Shells
^
= ^ ( ^ - " 3 )
(18-10.5)
c so
e
m
+ "*o
= R~\'Q0~
613
sm
4» - " 3
<i>)
(18.10.6)
T o t h e previous e q u a t i o n s , o n e m u s t a d d t h e b o u n d a r y c o n d i t i o n s of
the type p r e s e n t e d in Sec. 18.9.
Let us n o w consider t h e special case in w h i c h t h e shell is loaded
symmetrically.
All t h e quantities in t h e previous e q u a t i o n s d o n o t
d e p e n d o n 0, a n d b o t h
a n d qg m u s t b e e q u a l t o zero. E q . (18.10.3)
is thus identically satisfied. By solving t h e t w o E q s . (18.10.2) a n d
(18.10.4), t h e values of
a n d N9B c a n b e calculated. N o t i c i n g t h a t
R0 = i ^ s i n <£, t h e value of Neg o b t a i n e d from E q . (18.10.4) is:
N
N„ = -
+ *
R
°
(18.10.8)
Substituting E q . (18.10.8) into E q . (18.10.2) a n d multiplying t h e result
b y sin <|>, w e find t h a t :
j^(R0N^m
4) + R0R+(q+sm
I n t e g r a t i n g with respect t o
2n^sin *
<t> + q3cos
<f>) = 0.
(18.10.9)
9w e o b t a i n :
H
2
<f>
* » W -
0 +
(18.10.10)
Since ( ^ s i n 0 + # 3c o s <f>) h a s a resultant qv w h i c h acts o n t h e a n n u l a r
a r e a (Fig. 18.16) 2IiR0R^d^>,
t h e integral in t h e r i g h t - h a n d side of E q .
(18.10.10) c a n b e replaced b y t h e resultant of t h e total l o a d acting o n
t h a t p a r t of t h e shell c o r r e s p o n d i n g to t h e angle <f>. If w e set:
Jo
2 n * 0/ ^ s i n </> + <? 3cos <j>)d<j> = F,
(18.10.11)
E q . (18.10.10) gives:
(18.10.12)
614
The Theory of Elasticity
I n s t e a d of solving E q s . (18.10.2) a n d (18.10.4), it is m o r e c o n v e n i e n t t o
obtain
from Eq. (18.10.12), t h e n o b t a i n N90 from E q . (18.10.8). T h e
strains c a n n o w b e o b t a i n e d from the force-strain relations a n d the
d i s p l a c e m e n t s b y integrating the s t r a i n - d i s p l a c e m e n t relations.
I n s t e a d of the e q u i l i b r i u m of a n element, t h e e q u i l i b r i u m of t h e
p o r t i o n of the shell a b o v e the parallel circle defined b y t h e angle
may
b e c o n s i d e r e d . If the r e s u l t a n t of the total l o a d o n t h a t p o r t i o n of the
shell is d e n o t e d b y F (Fig. 18.17), e q u i l i b r i u m requires t h a t
Fig. 1 8 . 1 7
2UR0N<H>
sm<t>
+
F=0,
(18.10.13)
w h i c h is the s a m e as (18.10.12)
Example
1. Spherical Dome of Constant
Thickness
under Its Own
Weight
Let y b e the unit weight of the m a t e r i a l from w h i c h the shell is m a d e .
T h e gravitational force p e r unit a r e a of the shell is yh. T h i s force h a s the
components
q+ = yh sin <£, qe = 0,
q3 = yh cos <j>.
(18.10.14)
If the r a d i u s of the m i d d l e surface of the d o m e is a (Fig. 18.18),
R
a,
(18.10.15)
E q . (18.10.12) gives:
yha
1 + cos </>'
(18.10.16)
Introduction to the Theory of Thin Shells
615
Fig. 1 8 . 1 8
w h e r e it w a s a s s u m e d t h a t n o external l o a d s a r e acting o n the d o m e .
Substituting this value of
into (18.10.8), we get:
^
=
- 4
o
s
^ T T U )
(18
-
IAI7)
is always compressive. NB9 is compressive for small values of <f> a n d
b e c o m e s tensile for </>> 51° 50'.
Example 2. Shell in the Form of an Ellipsoid of Revolution
Such shells are u s e d in the c o n s t r u c t i o n of the e n d s of cylindrical boilers
a n d air t a n k s (Fig. 18.19). T h e e q u a t i o n of the ellipse is:
*1
a-
2
(18.10.18)
b
T h e m a g n i t u d e s of the principal radii of c u r v a t u r e c a n b e c o m p u t e d
,2
from E q s . (18.3.146) a n d (18.3.147):
616
The Theory of Elasticity
2
_ (cfiz
42
2
+ b4R o)
*
a*b
42
(a*z 9
'
»
2+ b R o)
b
If p is the u n i f o r m p r e s s u r e in the air t a n k or t h e boiler, t h e n
qe =
= 0,
q<j>
(18.10.20)
q3 = -p.
F r o m Eq. (18.10.12), we h a v e :
42
_
**
URlp
_ R0 P_pR9
2riR 0sin<J>
2sin<|>
2
_p{a z
42
+ b2R )»
2b
io21) n 8
and
N„ = ~i?-NH + RgP
2<t>
42
ptfz
+2
2b
b R 0y*
(18.10.22)
42
42
( -—<&—)
2
\
a!
+ bR
r
A t the t o p of the shell, R0 = 0 a n d z = b : T h e r e f o r e ,
N
,2
« v M= £ L
(18.10.23)
At the e q u a t o r , R0 = a a n d z = 0: Therefore,
f,
N M= p a ( \ - ^ ) .
08.10.24)
18.11 Membrane Theory of Cylindrical Shells
In cylindrical shells, the g e n e r a t o r curve is a straight line parallel t o
the axis of revolution. Therefore, the angle <#> b e t w e e n the n o r m a l to the
g e n e r a t o r a n d the axis is I I / 2. I n the OXx, OX2 p l a n e the g e n e r a t o r
follows a p a t h (Fig. 18.20):
(18.11.1)
R0 = R0(9).
Since the g e n e r a t o r is a straight line,
= oo a n d
lim(/^rf<f>) = dz.
(18.11.2)
617
Introduction to the Theory of Thin Shells
Fig. 18.20
W e shall, therefore, associate ^ with z a n d | 2 with 0.
T h e first f u n d a m e n t a l f o r m is:
2
(ds)
2
2
2
+ R (d9) .
= (dz)
(18.11.3)
T h e t h r e e e q u a t i o n s of equilibrium (18.9.4) to (18.9.6) b e c o m e :
°^
-w
R
+ d
R
^ o ^
^
+ q3 = 0,
+
°«> = °
R
+
eR +o= 0q
-
(18 1L4)
(18.11.5)
(18.11.6)
w h e r e we h a v e r e p l a c e d the subscripts 1 a n d 2 b y z a n d 9, E b y 1, a n d
GbyRl
T h e s t r a i n - d i s p l a c e m e n t relations (18.5.25) to (18.5.27) b e c o m e :
e
"°
m
*
e
(18.11.7)
dz
-
18
"*°
RB d9
9
1 / %>
*">-2\~dT
"3
R0
+1
(18.11.8)
duz0
).
d9
T h e force-strain relations (18.6.17) b e c o m e :
(18.11.9)
618
The Theory of Elasticity
Nzz = - j - ^ r f e o + vem
)
(18.11.10)
0» =
(18.11.1 1)
e
N
^
< «0 + ^zzo)
= J ^ e z
.e
(18.11.12)
o
T h e three e q u a t i o n s of equilibrium a r e solved to give:
(18.11.13)
Nm = -R0q3
^
=
-
/
+
(18.11.14)
i
Nzz = -
j
(qz +
R- ^ )
dz + f2(9),
(18.11.15)
w h e r e f{(0) a n d / 2( # ) a r e t w o functions of 9.
T h e force-strain relations (18.6.19) b e c o m e :
ezz
o = ^h-(Nzz
-vNm
)
(18.11.16)
)
em = ^(N„-vNzz
(18.11.17)
ezeo =
(18.11.18)
L
^ N a.
T h e s t r a i n - d i s p l a c e m e n t relations result in:
U =
*
j^zz-vN99
)dz+f3(9)
(18.11.19)
=
"3
1 F
- f e ^ w - " ^ ) '
(18.11.21)
w h e r e f3(9) a n d / 4( 0 ) are two a d d i t i o n a l functions of 9. E q s . (18.11.13)
to (18.11.21) represent the c o m p l e t e solution of a cylindrical m e m b r a n e
shell. T h e four functions / i , / 2 , / 3 , a n d f4 require four b o u n d a r y
c o n d i t i o n s . T h e y are all functions of 9 a n d c a n only b e applied to edges
a l o n g which 9 varies; in other w o r d s , along which z is c o n s t a n t .
Therefore, o n e c a n n o t satisfy the c o n d i t i o n s at edges a l o n g which 9 is
c o n s t a n t . This d r a w b a c k c a n only b e r e m e d i e d b y including the b e n d i n g
Introduction to the Theory of Thin Shells
619
resistance of t h e shell. H o w e v e r , w h e n the shell is closed these difficulties d o n o t arise.
Example 1. Circular Tube Filled with Liquid and Supported at the Ends
(Fig. 18.21)
1
L
L
_ ji
_ j
2
1
0
2
iI
Fig. 18.21
T h e p r e s s u r e at a n y p o i n t of t h e shell is in a direction o p p o s i t e to t h a t
of t h e positive n o r m a l a n d e q u a l to t h e weight of t h e u n i t c o l u m n of t h e
liquid a t t h a t point. If y is t h e unit weight of t h e liquid,
q2 = qe = 0
q3 = -yR0(\
~ COS 0).
(18.11.22)
S u b s t i t u t i n g these values in E q s . (18.11.13) to (18.11.15), w e get:
2
NB6= yR 0(\
Nz9 =-YR0z
2
- cos 9)
sin 9 + M9)
yz cos 0
(18.11.23)
(18.11.24)
(18.11.25)
T h e s t r a i n - d i s p l a c e m e n t relations r e d u c e t o :
(18.11.26)
"90
I
(18.11.27)
620
The Theory of Elasticity
(18.11.28)
Let us a s s u m e t h a t the s u p p o r t s at e a c h e n d are such t h a t u90 = 0 a n d
t h a t Nzz — 0. T a k i n g the origin at the m i d d l e of the cylinder, the
boundary conditions are:
atz = ± |
A;2 = 0
atz = ± f
0 = 0.
T h e first two c o n d i t i o n s in c o n j u n c t i o n
(18.11.25) lead t o :
/
0l , =
(18.11.29)
U $
(18.11.30)
with E q s . (18.11.24)
— 1 m* % ± ± .
and
(18-11-31)
N o t i c e t h a t those t w o c o n d i t i o n s result in fx b e i n g a c o n s t a n t , b u t since
n o t o r q u e is applied to the cylinder this c o n s t a n t m u s t b e e q u a l to zero.
T h e third a n d fourth c o n d i t i o n s lead t o :
/3 = 0
*M ^ 7 ^
=
(18.11.32)
-&\ -
2 ( 2 +v)5
+
(18 1L33)
Substituting f\,f2,f$,
a n d / 4 in the values of the forces a n d
d i s p l a c e m e n t s , we get:
Neg = yRfrl
- cos 9)
(18.11.34)
sin 9
Nz0 = -yR0z
(18.11.35)
^
_ y c= o s J ( 2L_
"zO
^[^f^ - l
4) z
nz
=
"90 '
the
os9
2
)
2
- vR z{\
- cos 9)]
(18.11.37)
3=
"
SEh
Y C O S
V-4z )(2,
2
(18.11.39)
0
+^ ) - ^ ( l - c o s , )
Introduction to the Theory of Thin Shells
621
If the cylindrical t u b e is l o a d e d only b y a n i n t e r n a l p r e s s u r e q3 =
—p0,
N99 = P0R0,
Nz9 = Nzz = 0
Rz
v
o P o
j ) _=_ n
%
0,
^ — ,
uzQ =
(18.11.40)
R
l
n a n an
ju3 _= f Po
— ( 1f 8 . 1 1 . 4 1 )
E q s . (18.11.34) to (18.11.41) a r e only valid if n o force Nzz is a p p l i e d a t
the e n d s of the t u b e .
Fig. 1 8 . 2 2
Let u s n o w c o n s i d e r t h e case m w m u i i t h e e n d s of t h e t u b e a r e built
in a n d a s s u m e t h a t a t these e n d s e990= 0. T a k i n g t h e origin a t o n e e n d
of t h e cylinder (Fig. 18.22), the b o u n d a r y c o n d i t i o n s necessary to
d e t e r m i n e fx(9) a n d f2(0) a r e :
at z = 0 a n d z = L, N99 = vNzz
.
(18.11.42)
Thus,
/ 2( 0 ) = ^ ( l - c o s 0 ) ,
0 ) =/ ^l (s i n 0
C, + 0 8 . 1 1 - 4 3 )
w h e r e C is a c o n s t a n t of i n t e g r a t i o n . If w e substitute fx (9) i n t o E q .
(18.11.24), we see t h a t C r e p r e s e n t s a s h e a r i n g force uniformly distribu t e d a r o u n d the t u b e . If t h e r e is n o t o r q u e applied, such a force m u s t
b e e q u a l to zero. T h u s , with t h e b o u n d a r y c o n d i t i o n s (18.11.42), the
forces in t h e t u b e a r e :
2
N0o = yR o(l
-cos
Nz9 = yR0sm9^
Nzz
=-
YZC QS
2
6)
(18.11.44)
- z)
(18.11.45)
V
- z) + ^ - ( 1 - cos 9).
(18.11.46)
622
The Theory of Elasticity
If the s u p p o r t s are rigid a n d c a n n o t m o v e , there will b e n o c h a n g e in
the length of the generators, which m e a n s t h a t uz0 m u s t be equal to zero.
H o w e v e r , it is a p p a r e n t from Eq. (18.11.26) t h a t this is n o t the case.
Such a result indicates t h a t b e n d i n g will o c c u r in the shell a n d t h a t the
m e m b r a n e theory is n o t sufficient to describe the d e f o r m a t i o n .
If the cylinder is l o a d e d only by a n internal pressure q3 = —p0,
P
Nee=P0R0,
Example
N26 = 0,
2. Vertical Cylindrical
08.11.47)
N Z =Z - ^ .
Tanks Filled with
Liquid
Z
Fig. 18.23
C o n s i d e r a cylindrical t a n k of r a d i u s RQ a n d height L (Fig. 18.23),
which is filled with liquid of unit weight y a n d rigidly built in at the
b a s e . T h e pressure at a n y p o i n t of the shell is in a direction opposite to
t h a t of the positive n o r m a l , a n d h a s the c o m p o n e n t s :
qz = q9 = 0
(18.11.48)
q3 = -y(L-z).
Substituting these values i n t o E q s . (18.11.13) to (18.11.15), we get:
N99 = R0y(L-z),
Nz9
=M0),
Nzz = -f-^
+ F2{9).
T h e b o u n d a r y c o n d i t i o n s necessary to d e t e r m i n e fx(0)
(18.11.49)
a n d f2(9)
are:
at z = 0,
Nz9 = 0
(18.11.50)
at z = L,
Nzz = 0.
(18.11.51)
Introduction to the Theory of Thin Shells
623
Thus,
/ i ( 0 ) = / 2( 0 ) = O,
(18.11.52)
so t h a t
Nff9= R0y(L-z),
Nz9 = 0,
N2Z= 0.
(18.11.53)
T h e d i s p l a c e m e n t s a r e given b y E q s . (18.11.19) t o (18.11.21). T h u s ,
"
V
1
(pR0yLz
Eh
-
-^j^-)
+ / 3( 0 )
«*o = - ^ f | + / 4 ( 0 )
(18.11.55)
+
"3 = ~R~dT
T o d e t e r m i n e f3(9)
1Q-
(18.11.54)
~Eh^
z)
~ -
(18-11.56)
a n d / 4( 0 ) , w e h a v e t h e t w o b o u n d a r y c o n d i t i o n s :
a t z = 0,
uz0 = um = 0
(18.11.57)
Thus,
/ 3( 0 ) = / 4 ( 0 ) = O,
(18.11.58)
so t h a t
M0 z= - ^ ( 2 L - z ) ,
0 = 0,U u93 = - § l ( L - z ) .
08.11.59)
O n e notices t h a t a t z = 0, w 3 is n o t e q u a l t o zero, w h i c h c a n n o t b e t r u e
if t h e e n d is fixed. T h i s result indicates t h a t b e n d i n g will o c c u r in t h e
shell a n d t h a t t h e m e m b r a n e theory does n o t completely describe t h e
deformation.
Example 3. Cantilever Circular Cylindrical Shell under Its Own Weight
C y l i n d r i c a l shells a r e c o m m o n l y used as roofing structures a n d , as such,
they m a y either b e s u p p o r t e d a t t h e e n d s o r cantilevered as s h o w n in
Fig. 18.24. If t h e g r a v i t a t i o n a l force p e r unit a r e a is yh, this force h a s
the c o m p o n e n t s :
qz = 0,
q9 = yh sin 6,
q3 = yh cos 9.
(18.11.60)
S u b s t i t u t i n g these values i n t o E q s . (18.11.13) t o (18.11.15), w e get:
624
The Theory of Elasticity
Fig. 1 8 . 2 4
Nz9 = - 2 y h z sin 9 + /,(9)
N99 = -R0yhcos9,
(18.11.61)
(18.11.62)
K
T h e b o u n d a r y c o n d i t i o n s for this p r o b l e m are t h a t
at z = L,
(18.11.63)
Nzz = Nz9 = 0.
Thus,
fx(9)
= 2yhL sin 9,
f2(9)
=
7 h i } cos 9
(18.11.64)
so t h a t the forces are n o w
N99 = -R0yh
cos 0,
= 2yA(L - z)sin 9,
(18.11.65)
A s m e n t i o n e d at the b e g i n n i n g of this section, o n e c a n n o t satisfy the
c o n d i t i o n s o n the b o u n d a r i e s a l o n g which 9 is c o n s t a n t for this o p e n
shell. T h u s , at the edges 9 = ±9X,
Nge = -R0yhcos
NzB = ±2yh(L
0 , ^ 0
- z)sin 9X
(18.11.66)
0,
(18.11.67)
Introduction to the Theory of Thin Shells
625
a n d the c o n d i t i o n s o n these b o u n d a r i e s a r e violated. T h e m e m b r a n e
theory is unsatisfactory for this case a n d o n e h a s to use the general
theory w h i c h i n c o r p o r a t e s b e n d i n g .
18.12 General Theory of Circular Cylindrical Shells
T h e general e q u a t i o n s of the theory of circular cylindrical shells a r e
o b t a i n e d b y associating £j with z a n d £ 2 with 0, a n d b y setting \[E = 1
a n d \[G = RQ in the e q u a t i o n s of equilibrium (Sec. 18.7), the straind i s p l a c e m e n t relations (Sec. 18.5), a n d the stress resultants a n d stress
couples relations (Sec. 18.6). T h e subscripts 1 a n d 2 are r e p l a c e d b y z
a n d 9, respectively: Rx is infinite a n d R2 = R0. T h e first f u n d a m e n t a l
form is
2
(ds)
2
= (dz)
2
2
+ R (d9) .
(18.12.1)
T h e differential e q u a t i o n s of e q u i l i b r i u m of forces (18.7.14) to (18.7.16)
b e c o m e (Fig. 18.25):
Fig. 18.25
626
T h e Theory of Elasticity
o ^F
R
-W
d
+ d
^-^f+^W
°lt~
R
o^-°
R
08.12.2)
K +
~
^
+
+
«
s0
^ * *
08.12.3)
++
*
=
°>
N
R q
°-
08.12.4)
W e shall a s s u m e t h a t t h e r a t i o of t h e thickness t o R0 is small c o m p a r e d
to unity, so t h a t Mz6 = M9z a n d Nz0 = N9z
. T h e differential e q u a t i o n s of
e q u i l i b r i u m of m o m e n t s (18.7.22) a n d (18.7.23), b e c o m e :
dM
dMfi
R o - ^z9 + ^ - R o V n - O
dz
W
M
o^f zz
R
+
,
dM
^T- °6z*
R
(18.12.5)
-
V
= 0
08.12.6)
F r o m these t w o e q u a t i o n s , w e find t h a t
v =
1 dMnn
9Mrfl
+
<»
^ r
V _ dM
zz
^-~dz~
08.12.7)
i ^ r
j dM0z+
T 0^ ~ -
(18.12.8)
Substituting the values of VB3a n d Vz3i n t o E q s . (18.12.2) t o (18.12.4), w e
o b t a i n t h e three e q u i l i b r i u m e q u a t i o n s :
R
o ^ r
+
^"aP"
if
+0
+
" if
~d~0dY
+R = 0
o ^
+
^w
- i l f
R~ ~do
~
r
99 +
0
° >-°-=
R
q
08.12.9)
)
08.12.10
08.i2.il)
T h e s t r a i n - d i s p l a c e m e n t relations a r e o b t a i n e d from E q s . (18.5.22) to
(18.5.30):
e = e z -z ^ K0 Z2
e2Z
e
K
(18.12.12)
9e =
(18.12.13)
m - t i
9 8
ez0 = e z -e i 3oK z , e
(18.12.14)
Introduction to the Theory of Thin Shells
627
where
e
zz0
e
(18.12.15)
1 9«0=O
m
3
M
R0 dO
e
z0O
"z0
9
=
=
1
(18.12.16)
R0
( 9%)
1
3"z0
(18.12.17)
30 .
R0
V
22
dz
3 w3
(18.12.18)
dz2
=
K
m
3 u 3\
(18.12.19)
30 /
0
9
l (1 %)
Kzg = / ? \ 2
0
,
3H
2
3
(18.12.20)
303z
3z
Substituting the previous e q u a t i o n s into Eqs. (18.6.11) to (18.6.15), we
obtain:
Nzz =
Eh 2
1 - *>
Eh
1
M
*z9
d
u
v _ _ E h _ (
9 0 ,
= Afe = 2(1 + p)\^z~
1
zo\
9w
R~ ~WJ
n R i ? m
0
M.
T h e three e q u a t i o n s of equilibrium (18.12.9) to (18.12.11) c a n n o w b e
written in t e r m s of the d i s p l a c e m e n t s w z , 0
a n d u3 of the m i d d l e
surface. Substituting Eqs. (18.12.21) to (18.12.26) into Eqs. (18.12.9) to
(18.12.11), we get:
628
The Theory of Elasticity
2
2
d » z2
O
, (1 - 2 v)^u
3z
z02
2R
dO
2
*2
12R
2
h
9
";0
" 3z
9
2R0
y
12*2
V
12 9 t/gQ
2 _
9
R
R
2
„9 M
2
=
Eh
n
(18.12.27)
12 a « 3
30
R
d9
3
l2 ?V)
2
(\-
R0 dz
3z30
2
1 - d ueo
v2
2
dz
\303z
2
. ?+
z(l ~ v)
9 Mgo _ J L ^ .
2
(1 + v)d uz0
2 i ? 0 3z30
+
1 +
9z
2
30
fl f
/
(18
2
2
2
28)
2
1 3 %)^
i?2
..
, +
(1 -
)
W
q U
v)
Eh
» '
, 1 %) _
"*"/?„ 30
/?„
3 S4
3Z
2
h (
1
2 _A
_+ 32 %
R„dz d0
9 3
"2 » 0
3
1 3d %A3
,
1 2 U o3 z 3 0
i?
30
j _ 3 % \
Rl dO* )
(18 12 29)
2
|
1 ~
/"*"
f
»
,
3
t
£A
0
?
=
f
The problem of the circular cylindrical shell reduces in each particular
case to the solution of this system of three partial differential equations.
18.13 Circular Cylindrical Shell Loaded Symmetrically with Respect to
its Axis
Eqs. (18.12.27) to (18.12.29) are much simplified in the case of
circular cylindrical shells loaded symmetrically with respect to their
axes. Because of this symmetry, qe = 0, u90 = 0, and uz0 and w 3 are
functions of z only. Eq. (18.12.28) is identically satisfied, while Eqs.
(18.12.27) and (18.12.29) b e c o m e :
^
2
dz 2
r ^
4RQ dz
duz0
"3
h Rj , d u3
" ^ T " R~0 ~ n
° ^ F
+ ^
L
2 Eh
(1 +- vR)q
—Eh— ° *
^
Q
(18.13.1)
_
n
~ °-
„
-
1 .32) ( 1 8
n
-
If the cylinder is subjected to radial pressure only, qz = 0, then Eq.
(18.13.1) can be integrated to give:
Introduction to the Theory of Thin Shells
^ £ 0 _ ! ^ i = C
dz
R„
'
629
(18.13.3)
w h e r e C is a c o n s t a n t of integration. But, from E q . (18.12.21),
(18.13.4)
Eh_(du^_vu1\
i \ dz
v
R0 /
T h e r e f o r e , t h e forces iV„ a r e c o n s t a n t s . E l i m i n a t i n g du^/dz
from E q .
(18.13.2), t h e following linear differential e q u a t i o n of the fourth o r d e r is
obtained:
+
4 4 f ftl
^ 1
=+
(18.13.5)
where
P
3
02 2 ^
JT/,3
'
R ah
(18.13.6)
12(1-^)-
^
'
T h e solution of this differential e q u a t i2o n is
"3 = ^ " ^ ( Q c o s Pz + C 2s i n /iz) + ^ ( C 3c o s /?z + C 4s i n fiz)
(18.13.7)
+ /(*),
w h e r e C \ , C 2, C 3, a n d C 4 a r e c o n s t a n t s of integration, a n d f (z) is t h e
p a r t i c u l a r integral. O n c e w 3 h a s b e e n f o u n d , the forces a n d m o m e n t s p e r
u n i t length c a n b e o b t a i n e d from E q s . (18.12.21) to (18.12.26). Nzz h a s
a l r e a d y b e e n s h o w n to b e a c o n s t a n t a n d , if this c o n s t a n t is e q u a l to 0,
^zO
dz
=^R '
0
(18.13.8)
S u b s t i t u t i n g E q . (18.13.8) i n t o E q . (18.12.22), we get:
Also,
d
Nze = 0,
Mzz = - D
^ ,
M e = e- D / - ^ .
(18.13.10)
F r o m E q s . (18.12.7) a n d (18.12.8):
V0i = O,
K ^ - D ^ .
(18-13.11)
630
The Theory of Elasticity
Example 1: Vertical Cylincrical Tank Filled with Liquid
This p r o b l e m was analyzed in the previous section b y m e a n s of the
m e m b r a n e theory, a n d the conclusion was r e a c h e d that such a theory
does n o t completely describe the d e f o r m a t i o n . It is interesting to
c o m p a r e the results o b t a i n e d using the m e m b r a n e theory to those
o b t a i n e d using the general theory. T h e c o m p o n e n t s of the pressure at
a n y p o i n t of the shell are (Fig. 18.23):
<7z = % = 0,
(18.13.12)
q3 = -y(L-z).
Nzz is a c o n s t a n t t h r o u g h o u t the shell a n d , a s s u m i n g t h a t n o load is
applied to it in the vertical direction, Nzz is equal to zero. Eq. (18.13.5)
becomes:
^
B*»34
(
+=
1
8
-
1
3
.
1
3
)
a n d its particular integral is:
/ ( z ) = - Y( L - z ) § .
(18-13-14)
Eq. (18.13.7) has four c o n s t a n t s of integration to b e d e t e r m i n e d from
the b o u n d a r y c o n d i t i o n s . T h e c o n s t a n t /? is inversely p r o p o r t i o n a l to
^/R0h\
if the length L is large c o m p a r e d to \/R0hy
the t a n k m a y b e
considered as infinitely long. In this case, the two c o n s t a n t s C 3 a n d C 4
m u s t be equal to zero if o n e is to h a v e meaningful d i s p l a c e m e n t s at the
top. A t the b o t t o m of the tank, we h a v e the two c o n d i t i o n s :
a zt = 0
0,
3= M
^
= 0.
(18-13.15)
Substituting Eq. (18.13.15) into Eq. (18.13.7), we find:
C
'
- *|L
Y
Eh '
2
°
Z ^ | L/ L A =
Eh \
fi)'
(18.13.16)
T h e deflection u 3 then b e c o m e s :
"3
= -^{l
- z - e-*[L
cos fiz + (l
- ±)sin
fiz]}
(18.13.17)
and
Nzz = Nz6 = 0
(18.13.18)
Introduction to the Theory of Thin Shells
N„ =
jR {l
=
0
- z -
[ L cos j8z + ( L - ± ) s i n /3z]
631
j
(18.13.19)
(18.13.20)
_z
~dz^~
Vi2(i-^)
? ^
- s i n Rz + ( l - 4^)cos
L
^ ' ^ M ' " ^
Rz
(18.13.21)
^3 = °
3
(18.13.22)
3
1
V* = - Z J ^ X " = * "
" Jz
Vl2(l -
.>^[(2fiL
v )'
- l)cos £ z - sin /fe].
(18.13.23)
K n o w i n g t h e values of the forces a n d m o m e n t s , t h e stresses c a n easily
be calculated. T h e m a x i m u m m o m e n t a n d shearing force o c c u r at t h e
b o t t o m of the tank, a n d are given b y :
_ /
i \
yR0Lh
-«si - ') ,^* -
1 "^TWJW
1
<>&W -
V
>J)
( 1 8 1 3 2 4 )
O n e notices that the solution b a s e d o n m e m b r a n e theory is c o n t a i n e d
in the previous expressions for Ne9 a n d w 3. Close to t h e base, t h e results
differ quite substantially; the differences decrease as t h e value of z
increases. A m o r e severe criticism of t h e m e m b r a n e theory is t h a t it does
n o t give a n y i n d i c a t i o n of the n a t u r e or values of the b e n d i n g m o m e n t s
M z , zwhich give t h e m o s t critical stress c o n d i t i o n at the b a s e of the t a n k .
T a b l e s a n d charts which c o n s i d e r a b l y facilitate t h e n u m e r i c a l c o m p u t a tions involved in this p r o b l e m c a n b e f o u n d in [2].
Example 2: Cylindrical Pressure Tank with Rigid End Plates
632
T h e Theory of Elasticity
n
I
L
L
2
L
^
HV
; \ \
Fig. 18.26
Let us c o n s i d e r a cylindrical p r e s s u r e t a n k of r a d i u s R0 a n d of length L
(Fig. 18.26) subjected to a n i n t e r n a l air p r e s s u r e p, a n d let us c h o o s e t h e
origin at the c e n t e r of the t a n k . T h e c o m p o n e n t s of the p r e s s u r e at a n y
point are:
(18.13.25)
E q . (18.13.5) b e c o m e s :
4
(18.13.26)
rfz
T h e solution of E q . (18.13.26) is:
p
"3 = e " ^ ( C , c o s Bz + C 2s i n £ z ) + e ^ ( C 3c o s Bz + C 4s i n /?z) - - f - .
Eh
Eh
(18.13.27)
It is c o n v e n i e n t to r e p l a c e the e x p o n e n t i a l functions b y
functions a c c o r d i n g to t h e well k n o w n f o r m u l a s :
hyperbolic
2
eH = cosh Bz + sinh Bz,
e"^
= cosh Bz - sinh Bz.
(18.13.28)
Thus,
« 3 = ^ ( S i n Bz sinh Bz + A2sin
Bz cosh Bz
+ ^ 3c o s /?z sinh fiz + ^4 4cos fiz cosh /?z
p 2R
—
(18.13.29)
Introduction to the Theory of Thin Shells
633
w h e r e AX,A2,A3,
a n d A4 a r e c o n s t a n t s to b e o b t a i n e d from t h e
b o u n d a r y c o n d i t i o n s . Because of s y m m e t r y , t h e d i s p l a c e m e n t u3 m u s t
b e t h e s a m e a t e q u a l distances o n b o t h sides of t h e origin. T h e t w o terms
in E q . (18.13.29) t h a t a r e s y m m e t r i c a l with respect to t h e origin a r e
^ 4 | S i n fiz sinh fiz a n d ^ 4 c o s fiz cosh fiz, so t h a t A
2 a n d A3 m u s t b e
4
e q u a l to zero a n d
u3 = Axsin fiz sinh fiz + ^ 4c o s fiz cosh
-(18.13.30)
fiz
Since t h e e n d s of t h e shell a r e rigid, t h e b o u n d a r y c o n d i t i o n s a r e :
atz±§
(18.13.31) into
Substituting E q .
"3 = ^
Eq.
08-13.31)
= 0.
(18.13.30),
a n d setting:
« =^ ,
we get:
R
^P l
Eh
f sin a cosh a - cos a sinh
I
sin 2a + sinh 2a
-~Eh~l
f c o s a sinh a + sin a cosh
sin 2a + sinh 2a
R
Aj _ 2p l
*
(18.13.32)
ai
J
ai
J *
n c n
„.
U*-13.33)
/ i q i o * > i x
O- - )
8
13
34
T h e expressions of t h e stress resultants a n d stress couples c a n n o w
easily b e o b t a i n e d b y substitution of u3 i n t o E q s . (18.13.9) t o (18.13.11):
Eh
Neg = pR0 — - ^ - [ ^ s i n fiz sinh fiz + A^cos fiz cosh fiz]
NZ0= O
2
Mzz = — 2DB [AX cos Bz cosh Bz — A4sin
Bz sinh Bz]
Mgg = vMzz
^3 = 0
3
Vz3= -2DB [-(AX
+ A4)sin
Bz cosh Bz
+ (Ax — y 4 4) c o s Bz sinh Bz],
w h e r e Ax a n d A4 a r e given b y Eqs. (18.13.33) a n d
e x a m p l e s c a n b e f o u n d i n [2] a n d [3].
(18.13.35)
(18.13.36)
(18.13.37)
(18.13.38)
(18.13.39)
(18.13.40)
(18.13.34). A d d i t i o n a l
634
The Theory of Elasticity
PROBLEMS
1.
F i n d the principal directions a n d the principal c u r v a t u r e s o n t h e
surface
x b
x
x
=
a(Zx + £ 2)>
2
x
= (%\ ~ &)>
3
= £1 £2 •
W h a t are the expressions for the first a n d s e c o n d curvatures?
Fig. 1 8 . 2 7
2.
Derive the first f u n d a m e n t a l form a n d the s e c o n d f u n d a m e n t a l
m a g n i t u d e s of the following surfaces of revolution: 1) a flat circular
sheet 2) a c o n e 3) a sphere 4) a p a r a b o l i o d 5) a n ellipsoid.
—
_
V ,\ J /
Fig. 1 8 . 2 8
/
Introduction to the Theory of Thin Shells
3.
4.
5.
6.
635
A spherical t a n k , s u p p o r t e d a l o n g a parallel circle AA (Fig. 18.27),
is filled with a liquid of u n i t weight y. U s i n g the m e m b r a n e t h e o r y
of shells of revolution, find
a n d N99 in t e r m s of <j>, a, a n d y for
<j> < <f>0 and<|> > <t>0.
A conical shell, filled with a liquid of unit weight y is s u p p o r t e d b y
forces in the direction of the generatrices, as s h o w n in Fig. 18.28.
U s i n g t h e m e m b r a n e t h e o r y of shells, find
a n d N99 in t e r m s of
a, y, d a n d y. d is the d e p t h of the liquid a n d y is a n a r b i t r a r y
d i s t a n c e from the a p e x of the c o n e .
A h o r i z o n t a l thin circular cylinder h a s its e n d s built in a n d is u n d e r
its o w n weight. F i n d the m e m b r a n e forces.
D r a w the distribution of the m o m e n t s a n d forces with respect to
d e p t h for a t a n k similar to the o n e s h o w n in Fig. 18.23, full of water,
6 d i m e n s i o n s : R = 330 ft., L = 30 ft., h = 1
a n d h a v i n g the following
n
ft. (v = 0.3, £ = 3 X 1 0 p s i , y = 62.5 lb / f t )
REFERENCES
[1] D . J. Struik, Differential Geometry, Addison-Wesley, Reading, Pa., 1950.
[2] S. Timoshenko and S. Woinowsky-Kreiger, Theory of Plates and Shells, McGraw-Hill,
N e w York, N . Y., 1959.
[3] J. E. Gibson, Linear Elastic Theory of Thin Shells, Pergamon Press, N e w York, N . Y., 1965.
Index
Adiabatic deformation, 186
Angle of twist per unit length, 269
Anisotropic material, 190
Airy stress function, 258-265
Forms of, 262
in plane strain, 258
in plane stress, 260
for beams, 361, 384, 388
for the semi-infinite medium, 395,
406, 421
Alternating symbol, 11, 99
Anticlastic curvature, 360, 372
Associative law, 14
Beams, straight, 352-380
elementary theory of, 354
of irregular cross section and end
force, 369
curvature of, 354
(see also bending of beams)
Beam-columns, differential
equation of, 494
with concentrated latteral load,
507
with end couples, 513
with uniformly distributed load
512
Bending of beams, by couples, 355
of narrow rectangular section by
an end load, 360
636
of narrow rectangular section by
a uniform load, 366
of circular cross section, 373
of elliptical cross section, 375
Bernouilli, J., 429
Beltrami-Michell compatibility
equations, 215, 257, 371
Berri, D . S., 235
Betti's law, 447
Betti's method, 231
Biharmonic function, 241
Bilinear form, 53
Binormal, 565
Body forces, 149
Boundary conditions, 153, 157
natural, 452, 488
forced, 452, 488
for plates, 539
for shells, 608
Boundary value problems of
elasticity, 210
Boussinesq, J., 428
problem, 250, 252
Buckling, energy solution, 501, 504
latteral, 514
modes, 496
Bulk modulus, 201
Cantilever, bending of narrow
rectangular beams, 360
Index 637
Cantilever (Continued)
prismatic bar of irregular cross
section, 369
of circular cross section, 373
of elliptic cross section, 375
Cartesian tensors, 95-106
Castiglianos's first theorem, 453
application of, 474
formula, 257
Castigliano's second theorem, 457,
458
application of, 474
Cerruti's problem, 244, 248
Characteristic ellipsoid, 26
Chou, P. C , 267
Circular arc beam, cantilever, 388
pure bending of, 384
Circular bar, torsion of, 268
of varying cross section in
torsion, 309
Circular cylindrical shells, general
theory, 625
loaded symmetrically with respect
to its axis, 628
Clapeyron, formula, 207
law, 432, 435
Codazzi's conditions, 587
Coefficient of thermal expansion,
204
Cofactor matrix, 16
Collar, A. R., 19
Columns, differential equation of,
493
simple, 494
Compatibility relations, for linear
strains, 88
in orthogonal curvilinear
coordinates, 142
in two-dimensional problems, 222
Complementary energy, 432, 456
Complex variable method, 231
Confocal ellipses, 417
hyperbolas, 417
Conjugate diameter, 283
Continuous media, deformation of,
1
kinematics of, 1-142
Coordinate surfaces, 108
Coordinates, Curvilinear, 107-142
cylindrical, 109
spherical polar, 110
Crandall, S. H., 322
Cubic material, 233
Curl in orthogonal curvilinear
coordinates, 120
Curvature, of beam, 354
of a normal section, 574
of plates (principal), 530
tensor, 530
invariants, 531
Curved beams, 381-393
simplified theory of, 381
Cylinder, hollow with internal a n d
external pressure, 323, 329
rotating, 338
thermal stresses in, 344
Cylinders, disks, and spheres,
323-344
Dahl, N . C , 322
Deflection of straight beams, 359,
366, 369
Deformation of continuous media,
3, 84, 88
Deviation, 60, 61
Deviator (tensor), 104
Deviatoric component of, linear
strain, 86
stress, 152
Diagonalization, 36
Differential equations of
equilibrium, 153-155
in orthogonal curvilinear
coordinates, 175
in terms of displacements, 211
of laterally loaded plates, 538
of thin shells, 604, 607
Differential geometry, 328
Differential operator (Del), 122
Dilatation, cylindrical, 44
energy of, 208
spherical, 60, 61
Disks, rotating, 334
thermal stresses in, 343
638
Index
Disks (Continued)
of variable thickness, 340
Displacement, function, 225
vector, 21
relative, 6
Distortion, energy of, 208
Distributive law, 14
Divergence, in orthogonal
curvilinear coordinates, 120
theorem, 154
D u h a m e l - N e u m a n n law, 205
Duncan, W. J., 19
Dupin's theorem, 591
Eigenvalue problem, 33,64, 158, 497
Elastic constants, 188
Elastic, stress-strain relations, 191,
199
symmetry, 191
Elasticity, boundary value
problems of, 210
equations in terms of
displacements, 211
equations in terms of stresses, 212
isotropic, 198
one dimensional, state of
deformation, 219
state of stress, 218
plane, 220
strain, 220
generalized, 229
stress, 224
generalized, 227
solution of problems, 230
theory of, 185-231
uniqueness, 216
vector equations of, 212
Elliptic, cylindrical coordinates, 142
integrals, 397
Emde, F., 428
Energy, complementary, 432, 456
minimum, 455, 478
of dilatation, 208
of distortion, 208
potential, 448, 449, 502, 557
minimum, 448, 450, 478, 559
principles and variational
methods, 429-^87
strain, 430, 431, 460,463,464. 557
solution of buckling problems,
501, 504
summary of theorems, 459
Equilibrium equations, in cartesian
coordinates, 153
in cylindrical coordinates, 179
in orthogonal curvilinear
coordinates, 175
in spherical polar coordinates, 179
of plates, 538
of thin shells, 601
Eubanks, R. A., 267
Euclidean spaces, 118
Euler's equations, 438, 440
theorem on normal curvature, 583
Eulerian method, 8
Fields, lamellar, 237
scalar, 236
solenoidal, 237
theory of, 236
vector, 237
First curvature, of a curve, 565
of a surface, 579
First fundamental, form, 566, 569
magnitudes, 569
Flexibility matrix, 434
Flexure of a curve, 566
Flexural rigidity of a plate, 525
Fourier, 205
Fox, L., 19
Frazier, R. A., 19
Frocht, M. M., 420, 428
Functional defined, 441
F u n d a m e n t a l magnitudes of the
second order, 573
Fung, Y. C , 235, 489
Galerkin vector, 244
Gauss-Codazzi conditions, 581
Gaussian, coordinates, 566, 568
curvature, 582
General product, 97, 99
Gere, J., 519
Gibson, J. E., 635
Goodier, J. N., 235, 322, 380, 428,
489
Index
Gradient, in the torsion problem,
278
in orthogonal curvilinear
coordinates, 121
Green-Riemann formula, 285, 317,
444
Haigh-Westergaard stress space, 167
Harmonic functions, 241, 243
Helmholtz's theorem, 238
Hildebrand, F . B., 19
Hodograph, 22
Holl, D . L., 428
Hollow circular bar, torsion of, 271
Hollow cyliner with internal and
external pressures, 323, 329
Hollow sphere subjected to internal
and external pressures, 331
Homogeneous state of stress, 149
Hooke's law, generalized, 188
connection with strain energy
density, 206
Indices, dummy, 11
identifying, 11
summation, 11
Integral transform methods, 231
Invariant, directions of a linear
transformation, 39
planes of a linear transformation,
40
Invariants of, linear
transformation, 38
strain, finite, 80
small, 85
stress, 159
deviatoric component of, 162
spherical component of, 162
Inverse method, 231, 520, 542
Irrotational field, 237
Isochromatics, 410, 412, 413, 414,
417
Isothermal deformation, 187
Isotropic tensors, 99, 100
Isotropy defined, 198
Jacobian, 108
639
Jahnke, E., 428
Kaplan, W., 562
Kelvin's problem, 244, 246
Kinematics of continuous media,
1-142
Kirchhoff's assumptions, 521, 541
Kronecker delta, 11, 99
Lagrange's equation, 525, 539, 554
Lagrangian method, 8, 131
Lame, constants, 199
formulas, 325, 332
strain potential, 242, 323
Lamellar field, 237
Langhaar, H. L., 489
Laplacian in orthogonal curvilinear
coordinates, 122
Least work, Theorem of, 458
application of theorem of, 482
Levy, M., 424, 427, 428
Line of curvature. 577, 578
Linear tangent transformation, 70
Linear transformations 6, 20-66
antisymmetric, 41
change of axes, 31
characteristic equation, 33
characteristic roots, 34
cylindrical dilatation, 44
deviation, 60
eigenvalues, 33
eigenvectors, 35
hodograph, 49
invariants, 38
invariant directions, 39
invariant planes, 40
Mohr's construction, 58
normal displacement, 56
orthogonal, 27
principal directions, 25
principal displacements, 48, 50
principal planes, 25, 51
product of two, 24
reciprocity, 45
rotation, 44
small, 24
spherical dilatation of, 60
640
Index
Linear transformations (Continued)
sum of, 23
tangential displacement, 56
in two dimensions, 63
Love, A. E. H., 235, 428
assumptions for shells, 591
strain function, 244
Matrix, addition, 12
algebra, 9-18
antisymmetric, 10
column, 10
conformable, 13
diagonal, 13
division, 16
inverse, 16
modal, 36
multiplication, 13
null, 10
row, 10
scalar, 10
subtraction, 12
symmetric, 9
transpose, 10
unit, 10
Maximum shearing stress
trajectory, 409, 412, 413, 414
Maxwell's law, 448
stress function, 256, 457
M e m b r a n e analogy, application to,
circular cross sections, 299
multicellular thin sections, 308
thin open sections, 301
thin rectangular sections, 300
thin tubular members, 304
M e m b r a n e , shells of revolution, 611
theory of cylindrical shells, 616
theory of shells, 610
Metric coefficients, 112, 114
of a surface, 594
Metric tensor, 115
Meunier's theorem, 514
Middle plane, 520
Middle surface, 592
Minimum complementary energy,
principle of, 455
application of, 478
M i n i m u m potential energy,
principle of, 448, 450
application of, 478, 559
Modulus, Bulk, 201
shear, 199
Young's 202
Mohr's representation for linear
symmetric transformations, 56
for strain, 88
for stress, 160
for the curvature of plates, 531
for the moments on plates, 536
M o m e n t tensor, 535
Monoclinic material, 191
Morrera's stress function, 256, 457
Mushkelishvili, N . J., 235
Navier's equations, 211
Neuber-Papkovich representation,
250
N e u m a n n ' s boundary value
problem, 280
Newmark, N . M., 428
N o r m a l curvature of a surface, 571
N o r m a l section of a surface, 571
Normalized form, 35
Numerical methods, 231
Octahedral plane, 166
Octahedral strain, normal, 87
shearing, 87
Octahedral stress, normal, 166
shearing, 167
Orthogonal curvilinear coordinates,
107-142
Orthogonal transformation, 27
Orthotropic material, 193, 231
Osculating plane, 564
Pagano, N . J., 267
Papkovich, 250
Parabolic cylindrical coordinates,
143
Parametric curves, 567
Perlis, S., 19
Plane strain, 220
compatibility equations for, 222
Index 641
Plane strain (Continued)
generalized, 229
stress-strain relations in case of,
221
Plane stress, 224
generalized, 227
stress-strain relations in case of,
224
Plates, bending of thin, 520-560
boundary conditions of, 539
circular, 553
curvature of, 527
elliptic with clamped edges, 552
equations of equilibrium of, 538
flexural rigidity of, 525
Kirchhoff s assumptions for, 521
Lagrange's equation for, 525
rectangular simply supported, 548
strain energy of, 557
Poisson's effect, 543
ratio, 202
Potential energy (see Energy)
Potentials, method of, 231
scalar, 236, 238, 242
solution by, 236-265
strain, 242
theory of, 278
vector, 236, 238, 242
Prandtl's m e m b r a n e analogy (see
m e m b r a n e analogy), 295
stress function, 258, 268, 284
Principal curvatures, of plates, 530
of shells, 579
Principal directions, of a linear
transformation, 25
of strain, 79
of stress, 158
Principal normal to a curve, 565
Principal radii of curvature, 580
Principal strains, 79, 84
Principal stresses, 159
trajectories, 409, 412, 413, 414, 417
Principle, of Saint-Venant
of superposition, 217
Pure bending of prismatical bars,
355
of circular arc beams, 384
Quadratic differential form, 114,
569
Quadratic form, 53
Quadric surface, 163
Quotient rule, 104
Radius of flexure, 566
Radius of torsion, 566
Range convention, 11
Rayleigh's formula, 503
Rayleigh-Ritz method, 484
Reciprocal law of Maxwell and
Betti, 446
Retaining wall, rectangular, 427
triangular, 424
Rigidity, modulus of plates,
flexural, 525
torsional, 271
Rodrigue's formula, 581
Rotating, cylinders, 338
disks, 334
Rotation, 42, 83, 372
in orthogonal curvilinear
coordinates, 141
Saddle point, 532
Sadowski, M., 428
Saint-Venant's principle, 217, 368,
386, 391
Saint-Venant's solution of the
torsion problem 268, 273
Saint-Venant's solution of the
cantilever beam problem, 369
Scalar, 95
field, 236
Scarborough, J. B., 19
Scott, R. F , 428
Second curvature of a curve, 566
Second curvature of a surface, 582
Second fundamental form, 571, 573
Semi-infinite medium, 395-424
rigid circular die on a, 402
rigid strip at the surface of a, 420
uniform pressure over a circular
area on a, 396
uniform pressure over a
rectangular area on a, 401
642
Index
Semi-infinite medium (Continued)
vertical line load on a, 404
Semi-infinite plate, rigid die at the
surface of, 421
vertical line load on a, 410
vertical pressure on part of the
boundary of a, 419
tangential line load on a, 414
Semi-inverse method, 231, 268, 276
Shear center, 378
Shear modulus, 199
Shell theory, 563-634
Shell in the form of an ellipsoid of
revolution, 615
Shrink fit, 328
Similarity, 33
Simply connected region, 91
Sneddon, I. N . , 235
Sokolnikoff, I. S., 235
Space curves, 563
Specific curvature, 582
Spectral matrix, 36
Spheres, hollow with internal and
external pressures, 331
thermal stresses in, 347
Spherical dome, 614
Stability, elastic, 490-518
Sternberg, E . , 267
Solenoidal field, 231
Stiffness matrix, 190, 434
Stiffness tensor, 190
Strain, change of coordinates, 77, 86
characteristic equation of, 85
compatibility of, 88
components of, 73
deviatoric, 86
displacement relations, in
cartesian coordinates, 7, 73
in curvilinear coordinates, 134
of a surface, 594
energy, 431
energy density function, 187, 206
energy for linearly elastic slender
members, 463
general analysis of, 69-91
geometrical meaning of, 75
invariants, 80, 85
of line element, 75
linear, 83
normal, 84
tangential, 84
mean, 86
octahedral, normal, 87
shearing, 87
principal axes of, 79
shear angles, 76
shearing, 84
shearing components of, 76
spherical component of, 86
tensor, normal component of, 133
in orthogonal curvilinear
coordinates, 129
shearing components of, 133
in two dimensions, 87
volumetric, 80, 86
Stress, analysis of, 147-180
change of coordinates, 170
couples in plates, 533
couples for a shell, 598
deviatoric component of, 162
director surface, 165
ellipsoid, 164
function, 255(see Airy stress
function)
homogeneous state of, 149
invariants, 159, 162
Mohr's diagram for, 160
normal to a plane, 149
octahedral normal, 166
octahedral shearing, 166
principal, 159
quadric, 163
resultants for plates, 533
resultants for shells, 598
space, 167
spherical component of, 162
tangential to a plane, 149
tensor, 150, 153
in two dimensions, 171
vector, 149, 153
in wedges, 421
Struik, D. J . , 635
Superposition, principle of, 215
Surface loads, 149
Index
643
Surfaces, theory of, 566-591
Surfaces with small curvature, 526
Synclastic, 580
Uniqueness of solution, 216
Unit displacement, 50
Unit elongations, 74, 75
Taylor series, 5
Tensor, cartesian, 95-105
antisymmetric, 99
contraction, 102
curvature, 530
function of a, 101
invariants, 101
isotropic, 99
moment, 535
strain (see strain tensor)
stress (see stress tensor)
substitution, 100
symmetric, 99
Thermoelastic stress-strain
relations, 204
Thermal stresses, in thin disks, 343
in long circular cylinders, 344
in spheres, 347
Thin plates, bending of, 520-560
Thin shells, introduction to the
theory of, 563, 634
Timoshenko, S., 235, 322, 351. 380,
394, 428, 489, 503, 519, 562, 635,
Torsion, 268-319
of circular prismatic bar, 268
of circular shaft of varying cross
section, 304
of conical shaft, 312
of elliptic bar, 280, 286
of hollow circular bar, 271
of noncircular prismatic bar, 273
of rectangular bar, 289
of triangular bar, 287
Torsional rigidity, 271
Tractions, 149
Transformation, 1 (see linear
transformation)
admissible, 108
Transversely isotropic material, 194
Trefftz, E., 303
Twist, angle of, 269
of plates, 529
Variation, notation of, 441
Variational methods, 231, 429-487
Vector, 95, 96
Displacement, 21
Galerkin, 244
potential, 238, 242
stress, 149
Virtual complementary work,
principle of, 453, 455
application of theorem of, 468
Virtual work, principle of, 429, 435
application of theorem of, 468
Volumetric strain (see strain,
volumetric)
Umbilic, 582
Wang, C. T., 276, 322, 351, 489, 519
Warping function, 276
Washizu, K., 459, 484
Wedges, radial stresses in, 421
Westergaard, M. M., 267, 351
Winkler, 146
Woinowski-Krieger, S., 562, 635
Work, virtual, principle of, (see
virtual work)
complementary (see virtual
complementary work)
least, theorem of, 458
application of, 482
Young's modulus, 202
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UNIFIED
SERIES
1. W . H . D A V E N P O R T / D . R O S E N T H A L — E n g i n e e r i n g : Its Role and
Function in Human
Society
2 . M . T R I B U S — R a t i o n a l Descriptions, Decisions and Designs
3 . W . H . D A V E N P O R T — The One Culture
4 . W . A . W O O D — T h e Study of Metal Structures and Their
Mechan ical Properties
5 . M . S M Y T H — L i n e a r Engineering Systems: Tools and
Techniques
6 . L . M . M A X W E L L / M . B . R E E D — T h e Theory of Graphs: A Basis for
Network
Theory
7 . W . R . S P I L L E R S — A u t o m a t e d Structural Analysis:
An
Introduction
8 . J . J . A Z A R — M a t r i x Structural
Analysis
9 . S . S E E L Y — A n Introduction to Engineering
Systems
10. D.T.THOMAS—Engineering
Electromagnetics
1 1 . D . R O S E N T H A L — R e s i s t a n c e and Deformation of Solid Media
1 2 . S . J . B R I T V E C — The Stability of Elastic
Systems
1 3 . M . N O T O N — M o d e r n Control
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1 4 . B . M O R R I L L — A n Introduction to Equilibrium
Thermodynamics
1 5 . R . P A R K M A N — T h e Cybernetic
Society
1 6 . A . SAADA—Elasticity: Theory and
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1 7 . W . M . L A I / D . R U B I N / E . KREMPL—Introduction to
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1 8 . D . K . A N A N D — I n t r o d u c t i o n to Control
Systems