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Introduction to Continuum Mechanics
c Romesh C. Batra, 1998, 2000
i
CONTENTS
Chapter 1 Introduction
1. What is Mechanics? 1-1
2. Continuum Mechanics 1-2
3. An example of an ad-hoc approach 1-3
Chapter 2 Mathematical Preliminaries
1. Summation convention, Dummy Indices 2-1
2. Free Indices 2-3
3. Kronecker Delta 2-4
4. Index Notation 2-6
5. Permutation Symbol 2-7
6. Manipulation with the Indicial Notation 2-9
7. Translation and Rotation of Coordinate Axes 2-12
8. Tensors 2-19
Chapter 3 Kinematics
1. Description of Motion of a Continuum 3-1
2. Referential and Spatial Descriptions 3-3
3. Displacement Vector 3-5
4. Restrictions on Continuous Deformation of aDeformable Body 3-6
5. Material Derivative 3-9
6. Finding Acceleration of a Particle from a givenVelocity Field 3-11
7. Deformation Gradient 3-14
8. Strain Tensors 3-21
9. Principal Strains 3-24
10. Deformation of Areas and Volumes 3-33
11. Mass Density. Equation of Continuity 3-35
12. Rate of Deformation 3-38
13. Polar Decomposition 3-45
14. Infinitesimal Deformations 3-50
ii
Chapter 4 The Stress Tensor
1. Kinetics of a Continuous Media 4-1
2. Boundary Conditions for the Stress Tensor 4-10
3. Nominal Stress Tensor 4-13
4. Transformation of Stress Tensor under Rotation of Axes 4-15
5. Principal Stresses. Maximum Shear Stress 4-20
Chaper 5 The Linear Elastic Material
1. Introduction 5-1
2. Linear Elastic Solid. Hookean Material 5-1
3. Equations of the Infinitesimal Theory of Elasticity 5-7
4. Principle of Superposition 5-10
5. A Uniqueness Theorem 5-1ll
6. Compatibility Equations Expressed in terms of the StressComponents for an Isotropic, Homogeneous, Linear, Elastic Solid 51-4
7. Some Examples. 5-18
a) Vibration of an Infinite Plate
b) Torsion of a Circular Shaft 5-22
c) Torsion of Non-Circular Cylinders 5-26
Chaper 6 The Linear Elastic Material
1. Constitutive Relation 6-1
2. Formulation of an Initial-Boundary-Varlue Problem 6-3
3. Examples 6-4
iii
1 Introduction
The major objective of our study of Mechanics is the formulation and solution of initial-boundary-
value problems that model as realistically as possible a physical phenomenon. There are two
equally attractive approaches to Mechanics. One is the ad-hoc approach, which takes up specific
problems, and devises problem-dependent methods of solution, introducing simplifying assump-
tions as needed. (This approach is used in Strength of Materials where problems of bending,
torsion, pressure vessel are individually set up under varying assumptions and then solved.) The
other is the general approach, which explores the general features of a concept or a theory and
considers specific applications at a later stage. By and large, the latter is the quicker way to learn
about an entire field, but the former is more concrete and sometimes more easily understood. We
will study the general approach in this course.
1.1 What is Mechanics?
Mechanics is the study of the motion of matter and the forces required to cause its motion. Mechan-
ics is based on the concepts of time, space, force, energy, and matter. A knowledge of mechanics
is needed for the study of all branches of physics, chemistry, biology and engineering.
The consideration of all aspects of mechanics would be too large a task for us. Instead, in this
course, we shall study only the classical mechanics of non-polar continua. (A nonpolar continuum
is one whose material particles have only three translational degrees of freedom.) We shall concern
ourselves with the basic principles common to fluids and solids.
1.2 Continuum Mechanics
Matter is formed of molecules which in turn consist of atoms and subatomic particles. Thus matter
is not continuous. However, there are many aspects of everyday experience regarding the behavior
of materials, such as the amount of lengthening of a steel bar under the action of given forces,
the rate of discharge of water in a pipe under a given pressure difference or the drag force ex-
perienced by a body moving in air etc., which can be described and predicted with theories that
1
pay no attention to the molecular structure of materials. The theory which describes relationships
between gross phenomena, neglecting the structure of materials on a smaller scale, is known as
the continuum theory. The continuum theory regards matter as indefinitely divisible. Thus, within
the theory, one accepts the idea of an infinitesimal volume of material referred to as a particle in
the continuum, and in every neighborhood of a particle there are always infinitely many particles
present. Whether the continuum theory is justified or not depends upon the given situation. For
example, the molecular dimension of water is about 1 A� (10�8 cm); hence, if we are concerned
about the liquid water in a problem in which we never have to consider dimensions less than say
10�5 cm, we are safe to treat water as a continuum. The mean free path of the molecules of air on
the surface of the earth at room temperature is about 5 � 10�6 cm; hence, if we consider the flow
of air about an airplane, we may treat air as a continuum. The diameter of a red blood cell in our
body is about 8:5� 10�4 cm; hence, we can treat our blood as a continuum if we consider the flow
in arteries of diameter say 0.5 mm.
Thus the concept of a material continuum as a mathematical idealization of the real world
is applicable to problems in which the fine structure of the matter can be ignored. When the
consideration of fine structure is important, we should use principles of particle physics, statistical
mechanics, or a theory of micropolar continuum.
1.3 An example of an ad-hoc approach.
Consider the problem of the bending of a beam usually studied in the first course on Mechanics of
Deforms. This is generally based on the following assumptions:
i) The beam is initially straight.
ii) The cross-section is uniform.
iii) The beam is made of a homogeneous and isotropic material which obeys Hooke’s law.
iv) Plane sections remain plane.
v) The beam is subjected to a pure bending moment M applied at the ends.
2
Under these assumptions, one can derive the formula
� =My
I(1.3.1)
in which � is the longitudinal bending stress, y the distance from the neutral axis which passes
through the centroid of the cross-section and I the moment of inertia of the cross-section about the
neutral axis. The derivation of (1.3.1) makes no reference to other components of stress acting at a
point. Of course, if the beam were initially curved or were one interested in finding the transverse
shear stress at a point, one would start essentially from scratch.
1.4 Continuum Mechanics
In Continuum Mechanics, we first establish principles which are applicable to all media, both
fluids and solids, under all kinds of loading conditions. We then study constitutive equations
which define classes of idealized materials. Finally specific problems are analyzed, and results are
compared with experimental observations.
2 Mathematical Preliminaries
2.1 Summation Convention, Dummy Indices
Consider the sum
s = a1x1 + a2x2 + : : :+ anxn : (2.1.1)
We can write it in a compact form as
s =nXi=1
aixi =nXj=1
ajxj =nX
m=1
amxm : (2.1.2)
It is obvious that the index i; j or m in eqn. (2.1.2) is dummy in the sense that the sum is indepen-
dent of the letter used. This is analogous to the dummy variable in an integral of a function over a
finite interval.
I =
Z b
a
f(x)dx =
Z b
a
f(y)dy =
Z b
a
f(t)dt :
3
We can further simplify the writing of eqn. (2.1.2) by adopting the following convention, some-
times known as Einstein’s summation convention. Whenever an index is repeated once in the same
term, it implies summation over the specified range of the index.
Using the summation convention, eqn. (2.1.2) shortens to s = aixi = ajxj = amxm.
Note that expressions such as aibixi are not defined within this convention. That is, an index
should never be repeated more than once in the same term for the summation convention to be
used. Therefore, an expression of the form
nXi=1
aibixi
must retain its summation sign.
In the following, unless otherwise specified, we shall always take n to be 3 so that, for example,
aixi = amxm = a1x1 + a2x2 + a3x3 ;
aii = amm = a11 + a22 + a33 :
The summation convention can obviously be used to express a double sum, a triple sum, etc.
For example, we can write
3Xi=1
3Xj=1
aijxixj
simply as aijxixj . This expression equals the sum of nine terms:
aijxixj = ai1xix1 + ai2xix2 + ai3xix3 ;
= a11x1x1 + a21x2x1 + a31x3x1
+ a12x1x2 + a22x2x2 + a32x3x2
+ a13x1x3 + a23x2x3 + a33x3x3 :
Similarly, the triple sum3Xi=1
3Xj=1
3Xk=1
aibjckxixjxk will simply be written as aibjckxixjxk, and it
represents the sum of 27 terms.
We emphasize again that expressions such as aiixixjxj or aijxixjxixj are not defined in the
summation convention.
4
Exercise: Given
[aij] =
24 1 0 2
0 1 23 0 3
35 :
Evaluate (a) aii, (b) aijaij, and (c) ajkakj.
2.2 Free Indices
Consider the following system of three equations:
y1 = a11x1 + a12x2 + a13x3 = a1ixi;
y2 = a21x1 + a22x2 + a23x3 = a2ixi;
y3 = a31x1 + a32x2 + a33x3 = a3ixi:
(2.2.1)
These can be shortened to
yi = aijxj; i = 1; 2; 3: (2.2.2)
An index which appears only once in each term of an equation such as the index i in eqn. (2.2.2)
is called a “free index”. A free index takes on the integral number 1,2, or 3 one at a time. Thus
eqn. (2.2.2) is a short way of writing three equations each having the sum of three terms on its
right-hand side.
Note that the free index appearing in every term of an equation must be the same. Thus
ai = bj
is a meaningless equation. However, the following equations are meaningful.
ai + ki = ci;
ai + bicjdj = 0:
If there are two free indices appearing in an equation such as
Tij = AimAjm; i = 1; 2; 3; j = 1; 2; 3; (2.2.3)
then it is a short way of writing 9 equations. For example, eqn. (2.2.3) represents 9 equations; each
5
one has the sum of 3 terms on the right-hand side. In fact
T11 = A1mA1m = A11A11 + A12A12 + A13A13;T12 = A1mA2m = A11A21 + A12A22 + A13A23;T13 = A1mA3m = A11A31 + A12A32 + A13A33;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .T33 = A3mA3m = A31A31 + A32A32 + A33A33:
Again, equations such as
Tij = Tjk; Ti` = AimA``
are meaningless.
2.3 Kronecker Delta.
The Kronecker Delta, denoted by �ij, is defined as
�ij =
�1 if i = j;0 if i 6= j:
That is,
�11 = �22 = �33 = 1;
�12 = �13 = �21 = �23 = �31 = �32 = 0:
In other words, the matrix 24 �11 �12 �13�21 �22 �23�31 �32 �33
35
is the identity matrix 24 1 0 0
0 1 00 0 1
35 :
We note the following relations
(a) �ii = �11 + �22 + �33 = 1 + 1 + 1 = 3;
(b) �1mam = �11a1 + �12a2 + �13a3 = a1;
�2mam = �21a1 + �22a2 + �23a3 = a2;
�3mam = �31a1 + �32a2 + �33a3 = a3:
6
Or, in general
�imam = ai:
Similarly, one can show that
�imTmj = Tij:
In particular,
�im�mj = �ij; �im�mj�jn = �in:
2.4 Index Notation
Usually, rectangular Cartesian coordinates of a point are denoted by (x; y; z) and the unit vectors
along x; y and z-axes by i; j, and k respectively. In this coordinate system, the components of a
vector u along x; y, and z axes are denoted by ux; uy, and uz. The vector u has the representation
u = uxi+ uyj+ uzk:
This notation does not lend itself to any abbreviation. Therefore, instead of denoting the coordinate
axes by x; y; z we will denote them by x1; x2; x3. Also we will denote unit vectors
along x1; x2 and x3 axes by e1; e2, and e3 respectively. Naturally then components of a vector u
along x1; x2 and x3 axes will be indicated by u1; u2, and u3 respectively. Hence we can write
u = u1e1 + u2e2 + u3e3;
= ujej:(2.4.1)
Similarly,
v = v1e1 + v2e2 + v3e3;
= vjej:
7
The dot product u � v can simply be written as
u � v = u1v1 + u2v2 + u3v3 = uivi: (2.4.2)
Since e1; e2; e3 are mutually orthogonal unit vectors, therefore,
e1 � e1 = e2 � e2 = e3 � e3 = 1;
e1 � e2 = e2 � e3 = e1 � e3 = 0:
These equations can be summarized as
ei � ej = �ij: (2.4.3)
Exercise. Using the index notation, write expressions for
(1) the magnitude of a vector u,
(2) cos �; � being the angle between vectors u and v.
As another illustration of the use of the index notation, consider a line element with components
dx1; dx2; dx3. The square of the length, ds, of the line element is given by
ds2 = dx21 + dx22 + dx23;
= dxidxi
= �ijdxidxj:
Finally, we note that the differential of a function f(x1; x2; x3) can be written as
df =@f
@x1dx1 +
@f
@x2dx2 +
@f
@x3dx3;
=@f
@xidxi:
2.5 Permutation Symbol
The permutation symbol, denoted by �ijk, is defined by
�ijk =
8<:
1�10
9=; if i; j; k form
8<:
an evenan oddnot a
9=;
permutation of1; 2; 3:
(2.5.1)
8
That is,
�123 = �231 = �312 = 1;
�132 = �321 = �213 = �1;
�111 = �211 = �133 = : : : = 0:
We note that
�ijk = �jki = �kij = ��jik = ��ikj = ��kji:
If e1; e2; e3 form a right-handed triad, then
e1 � e2 = e3; e2 � e3 = e1; etc:
which can be written as
ei � ej = �ijkek: (2.5.2)
Now, if u = uiei; v = viei, then
u� v = uiei � vjej = uivjei � ej= uivj�ijkek = �ijkuivjek: (2.5.3)
Exercise. Using the index notation write an expression for j sin �j; � being the angle between
vectors u and v.
Exercise. Show that
(1) �ijk�jki = 6;
(2) �ijkAjAk = 0:
The following useful identity, which can be verified by long-hand calculations should be mem-
orized.
�ijm�klm = �ik�jl � �il�jk: (2.5.4)
9
Now by using this identity let us prove the vector identity
u� (v �w) = (u �w)v � (u � v)w:
Proof: Let v �w = a. Then a = �ijkviwjek, and
u� a = �ijkuiajek;
aj = a � ej = �ilmviwlem � ej;
= �ilmviwl�mj = �iljviwl:
u� (v �w) = �ijkui(�pljvpwl)ek;
= �ijk�pljuivpwlek;
= �kij�pljuivpwlek;
(use (2:5:4)) = (�kp�il � �kl�ip)uivpwlek;
= �kp�iluivpwlek � �kl�ipuivpwlek;
= ulwlvkek � upvpwkek;
(use (2:4:2)) = (u �w)v � (u � v)w:
Exercise. Show that
(a) If �ijkTjk = 0, then Tij = Tji,
(b) �ilm�jlm = 2�ij, and
(c) if Tij = �Tji, then Tijaiaj = 0.
We now write detAij in the index notation.
det [Aij] = det
24 A11 A12 A13
A21 A22 A23
A31 A32 A33
35 ;
= A11(A22A33 � A32A23)� A21(A12A33 � A32A13) + A31(A12A23 � A22A13);
= A11(�1jkAj2Ak3)� A21(��2jkAj2Ak3) + A31(�3jkAj2Ak3);
= Ai1�ijkAj2Ak3;
= �ijkAi1Aj2Ak3:
10
Example. Show that �ijkAilAjmAkn = (detA)�lmn.
2.6 Manipulations with the Indicial Notations
(a) Substitution
If
ai = vimbm (2.6.1)
and
bi = vimcm; (2.6.2)
then, in order to substitute for bi’s from (2.6.2) into (2.6.1) we first change the dummy index from
m to some other letter, say n and then the free index in (2.6.2) from i to m, so that
bm = vmncn: (2.6.3)
Now (2.6.1) and (2.6.3) give
ai = vimvmncn: (2.6.4)
Note that (2.6.4) represents three equations each having the sum of nine terms on its right-hand
side.
(b) Multiplication
If
p = ambm (2.6.5)
and
q = cmdm (2.6.6)
then
pq = ambmcndn: (2.6.7)
11
It is important to note that pq 6= ambmcmdm. In fact the right-hand side of this expression is not
even defined in the summation convention and further it is obvious that
pq 6=3X
m=1
ambmcmdm:
(c) Factoring
If
Tijnj � �ni = 0; (2.6.8)
then, using the Kronecker delta, we can write
ni = �ijnj; (2.6.9)
so that (2.6.8) becomes
Tijnj � ��ijnj = 0:
Thus
(Tij � ��ij)nj = 0:
(d) Contraction
The operation of identifying two indices and so summing on them is known as contraction. For
example, Tii is the contraction of Tij ,
Tii = T11 + T22 + T33;
and Tijj is a contraction of Tijk,
Tijj = Ti11 + Ti22 + Ti33:
If
Tij = �Ekk�ij + 2�Eij;
then
Tii = �Ekk�ii + 2�Eii;
= (3�+ 2�)Ekk:
12
Exercise. Given that Tij = �Ekk�ij + 2�Eij, W = 1
2TijEij; P = TijTij , show that
W = �EijEij +�
2(Ekk)
2;
P = 4�2EijEij + (Ekk)2(4��+ 3�2):
2.7 Translation and Rotation of Coordinate Axes
Consider two sets of rectangular Cartesian frames of reference O�x1x2 and O0�x01x02 in a plane.
If the frame of reference O0 � x01x02 is obtained from O � x1x2 by a shift of the origin without a
change in orientation, then, the transformation is a translation.
If a point P has coordinates (x1; x2) and (x01; x02) with respect to O � x1x2 and O0 � x01x
02 respec-
tively and (C1; C2) are the coordinates of 00 with respect to O � x1x2, then
x1 = x01 + C1;
x2 = x02 + C2;
or briefly
xi = x0i + Ci; i = 1; 2: (2.7.1)
If the origin remains fixed, and the new axes Ox01; Ox02 are obtained by rotating Ox1 and Ox2
through an angle � in the counter-clockwise direction, then
13
the transformation of axes is a rotation. Let the point P have coordinates (x1; x2) and (x01; x02)
relative to O � x1x2 and O � x01x02 respectively. Then,
x1 = OA = OB � CD;
= OD cos � � PD sin �;
= x01 cos � � x02 sin �;
x2 = AP = BD + CP;
= x01 sin � + x02 cos �:
We can write (x01; x02) in terms of (x1; x2) as
x01 = x1 cos � + x2 sin �;
x02 = �x1 sin � + x2 cos �:(2.7.2)
Using the index notation, the set of eqns. (2.7.2) can be written as
x0i = aijxj; i = 1; 2; j = 1; 2; (2.7.3)
where aij are elements of the matrix [aij];
�a11 a12a21 a22
�=
�cos � sin �� sin � cos �
�:
Before we generalize (2.7.1) and (2.7.2) to three dimensions we give below an alternate method
of arriving at (2.7.2). Let e01; e02 denote unit vectors along x01 and x02-axes and e1; e2 unit vectors
along x1 and x2-axes. Then
�!OP = x1e1 + x2e2;
= x01e01 + x02e
02:
Also
e01 = cos � e1 + sin � e2;
e02 = � sin � e1 + cos � e2:
14
Therefore,
x01 =�!OP � e01;
= (x1e1 + x2e2) � (cos �e1 + sin �e2)
= x1 cos � + x2 sin �;
x02 =�!OP � e02;
= � x1 sin � + x2 cos �:
This latter approach can more easily be adopted to the 3-dimensional case. If the primed axes O�x01x
02x
03 are obtained from the unprimed axes O�x1x2x3 just by a translation, then the coordinates
of a point with respect to the two sets of axes are related by (2.7.1) wherein the index i ranges from
1 to 3. Now let us assume that the primed axes are obtained from the unprimed axes by a rotation
only. Let us denote unit vectors along x1; x2; x3 by e1; e2, and e3 respectively and those along
x01; x02; x
03 by e01; e
02 and e03 respectively. If
a1j = cosine of the angle between e01 and ej;
then a11; a12 and a13 are the direction cosines of e01 with respect to the unprimed axes. We can
write
e01 = a11e1 + a12e2 + a13e3;
= a1iei:
Similarly,
e02 = a2iei; e03 = a3iei:
Or
e0i = aijej: (2.7.4)
Note that the matrix aij is 3� 3. Since
e0i � e0j = �ij; (2.7.5)
15
therefore,
�ij = aikek � ajpep = aikajpek � ep;
= aikajp�kp = aikajk: (2.7.6)
Equations (2.7.6) are equivalent to the following six equations.
a211 + a212 + a213 = 1;
a221 + a222 + a223 = 1;
a231 + a232 + a233 = 1;
a11a21 + a12a22 + a13a23 = 0;
a21a31 + a22a32 + a23a33 = 0;
a31a11 + a32a12 + a33a13 = 0:
(2.7.7)
The first three equations are equivalent to the statement that e01; e02 and e03 are unit vectors; the last
three equations are equivalent to the statement that e01; e02; e
03 are mutually orthogonal. Of course,
we can write ei’s in terms of e0i’s. Since
aj1 = cosine of the angle between e1 and e0j;
therefore,
e1 = aj1e0j or ei = ajie
0j: (2.7.8)
From the point of view of the solution of a set of simultaneous linear equations, the matrix aji in
(2.7.8) must be identified as the inverse of the matrix aij:
[aij]�1 = [aji] = [aij]
T : (2.7.9)
Here [aij]T is the transpose of the matrix [aij]. A matrix [aij], which satisfies eqn. (2.7.9) is called
an orthogonal matrix. That is, the transpose of an orthogonal matrix equals its inverse. A transfor-
mation is said to be orthogonal if the associated matrix is orthogonal. The matrix [aij] in (2.7.4)
defining a rotation of coordinate axes is orthogonal.
16
For an orthogonal matrix we have
[a][a]T = 1:
Therefore
det([a][a]T ) = 1;
or det[a] det[a]T = 1;
or det[a] det[a] = 1;
and thus
det[a] = �1:
An orthogonal matrix whose determinant equals +1 is called proper orthogonal and the one whose
determinant equals �1 is called improper orthogonal. A proper orthogonal matrix transforms a
right-handed triad of axes into a right-handed set of axes whereas an improper orthogonal matrix
transforms a right-handed set of axes into a left-handed set of axes or vice-versa.
Exercise: Consider a cube formed by the orthonormal vectors e01; e02 and e03. By setting the volume
of this cube equal to 1, show that det[aij] = 1.
Consider a vector OP emanating from the origin O and ending at a point P . With respect to
the primed and unprimed axes,
OP = x01e01 + x02e
02 + x03e
03;
= x0je0j;
= xjej:
Similarly,
x0i =OP � e0i;
= (xjej) � (aikek);
= xjaik�jk;
= aijxj:
17
Example: The components of a vector A with respect to unprimed axes are Ai = (0; 1; 1). Con-
sider a set of primed coordinate axes obtained by rotating the unprimed axes through an angle of
30� about the x3-axis (see Fig. ). What are the components, A0i, of this vector with respect to the
primed set of axes?
Solution:
e03 = e3; e01 = cos 30e1 + sin 30e2;
e02 = � sin 30e1 + cos 30e2:
Therefore
[aij] = [e0i � ej] =24 cos 30 sin 30 0� sin 30 cos 30 0
0 0 1
35
Now
A0i = aijAj
can be written as 24 A0
1
A02
A03
35 =
24 cos 30 sin 30 0� sin 30 cos 30 0
0 0 1
3524 0
11
35 =
24 sin 30
cos 301
35 :
Hence
A0i = (0:5; 0:866; 1):
18
Summarizing our discussion of the transformation of coordinate axes, we note that a general
transformation from unprimed to primed axes combines both a translation and a rotation of the
axes. This can be written as
x0i = aijxj + ci (2.7.10)
where aij is an orthogonal matrix and ci is a constant. Under this transformation, the components
of a vector A in the two sets of axes are related as
A0i = aijAj: (2.7.11)
2.8 Tensors
2.8.1 A Linear Transformation
Let T be a transformation from a vector space into the same vector space. That is, for any vector
u, Tu is also a vector of the same dimensions as u. Then T is linear if and only if
T(�u+ �w) = �Tu+ �Tw for every real � and � (2.8.1.1)
[NOTE: f is a linear function of x if and only if f(x) = �x where � is a real number. For example
f(x) = 2x + 3 is not a linear function of x even though it is often referred to as such.]
A linear transformation from a vector space into another vector space is also called a second-order
tensor.
2.8.2 Tensor Product Between Two Vectors
The tensor product (or the dyadic product) between two vectors a and b is defined as
(a b)c = (b � c)a for every vector c: (2.8.2.1)
Thus (a b) transforms a vector c into a vector parallel to a. Since it transforms a vector into a
vector and obeys (2.8.1.1), it is a linear transformation. Note that
a b 6= b a: (2.8.2.2)
19
2.8.3 Components of a Second-Order Tensor
Let e1; e2; e3 be a set of orthonormal basis vectors (i.e. e1; e2; e3 are mutually orthogonal unit
vectors). For any vector a,
a = aiei: (2.8.3.1)
Let b = Ta. Then
b = T(aj ej) = aj(Tej); (2.8.3.2)
or
biei = aj(Tej): (2.8.3.3)
Taking the inner product of both sides of this eqn. with ek, we obtain
bk = ek � aj(Tej) = ajek � (Tej) = ajTkj (2.8.3.4)
where
Tkj = ek � (Tej) (2.8.3.5)
is called the component of T with respect to the basis ei.
For computation purposes, eqn. (2.8.3.4) is written as8<:
b1b2b3
9=; =
24 T11 T12 T13T21 T22 T23T31 T32 T33
358<:
a1a2a3
9=; : (2.8.3.6)
Analogous to the representation (2.8.3.1) for vector a, we have the following representation for
second-order tensor T.
T = Tij ei ej: (2.8.3.7)
Because of (2.8.2.2), Tij need not equal Tji. In order to see that (2.8.3.7) is equivalent to (2.8.3.5),
20
we evaluate Ta.
b = Ta = (Tij ei ej)(akek);
= Tijak(ei ej)ek;
= Tijakei(ej � ek);
= Tijakei�jk = Tijaj ei; (2.8.3.8)
which is equivalent to bi = Tijaj or (2.8.3.4).
It is clear from (2.8.3.7) that the components Tij of T depend upon the choice of basis ei. Let
e0i = Qijej (2.8.3.9)
where Q is an orthogonal matrix (i.e. QQT = 1). Then
T = T 0ij e
0i e0j = T 0
ij(Qikek) (Qjlel) =
= T 0ijQikQjl(ek el) = Tklek el: (2.8.3.10)
Hence
Tkl = T 0ijQikQjl; (2.8.3.11)
and in matrix notation,
[T ] = [Q]T [T 0][Q]; (2.8.3.12)
and since Q is orthogonal,
[T 0] = [Q][T ][Q]T : (2.8.3.13)
The transpose TT of a second-order tenseor T is defined by
a � (TTb) = b � (Ta) for every vector a and b: (2.8.3.14)
The components of T and TT are related by
(T T )ij = Tji (2.8.3.15)
21
2.8.4 Tensors of Higher-Order
A third-order tensor is a linear transformation from the space of second-order tensors into vectors
or vectors into second-order tensors, and can be represented as
A = Aijkei ej ek: (2.8.4.1)
Under a change of basis (2.8.3.9), the transformation rule for its components can be derived as
follows.
A = A0ijke
0i e0j e0k
= A0ijkQilel Qjmem Qknekn
= A0ijkQilQjmQknel em en
= Almnel em en: (2.8.4.2)
Thus
Almn = QilQjmQknA0ijk (2.8.4.3)
or
A0ijk = QilQjmQknAlmn: (2.8.4.4)
A fourth-order tensor is a linear transformation from the space of second-order tensors to
second-order tensors, and has the representation
C = Cijklei ej ek el: (2.8.4.5)
Under the transformation (2.8.3.9) its components will transform as follows.
C 0ijkl = QipQjqQkrQlsCpqrs: (2.8.4.6)
3 Kinematics
3.1 Description of Motion of a Continuum.
Let us suppose that a body, at time t = t0, occupies a region of the physical space. The position
of a particle at this time can be described by its coordinates Xi with respect to a fixed rectangular
22
Cartesian coordinate system.
Let the body undergo a motion and point P move to P 0 whose coordinates with respect to the same
fixed axes are xi. Then an equation of the form
xi = xi(X1; X2; X3; t) (3.1.1)
describes the path of the particle which at t = t0 is located at Xi. In eqn. (3.1.1) the triplet
(X1; X2; X3) serves to identify different particles of the body and is known as reference coordi-
nates. The triplet (x1; x2; x3) gives the present position of the particle which at time t = t0 was
at the place Xi. Note that for a specific particle eqn. (3.1.1) defines the path line (or trajectory) of
the particle. Of course
Xi = xi(X1; X2; X3; t0);
which merely verifies the fact that the particle under consideration occupied the place Xi at t = t0.
Example: Consider the motion
x1 = X1 + 0 � 2tX2;
x2 = X2;
x3 = X3;
23
where (X1; X2; X3) gives the position of a particle at t = 0. Sketch the configuration at time
t = 2 for the body which at t = 0 has the shape of a cube of unit sides with one corner at the
origin.
Solution: For the particle which at t = 0 was at the origin,
xi = 0 for all t:
Thus this particle stays at the origin at all times.
Similarly, the particle which at time t = 0 was at the position (X1; 0; 0) will move to xi = X1�1i.
That is, particles on line OA do not move.
A particle (X1; 1; 0) on line CB will occupy, at time t = 2, the position
xi = (X1 + 0 � 2(2) (1))�1i + (1)�2i:
Thus every particle on line CB is displaced horizontally to the right through a distance (0 �2) (2) =0 � 4.
A particle (0; X2; 0) on line OC moves to
xi = (0 + 0 � 2(2)X2)�1i +X2�2i ;
so that every particle on the line OC moves horizontally to the right through a distance linearly
proportional to its height, that is, it remains a straight line. A similar situation prevails for the line
BA.
24
Thus at time t = 2, the side view of the cube changes from a square to a parallelogram as
shown.
The motion given in this example is known as simple shearing motion.
3.2 Referential and Spatial Descriptions
When a continuum is in motion, quantities (such as temperature �, velocity v) that are associated
with specific particles change with time. There are two ways to describe these changes.
(i) Following the particles, that is, we express �, vi as functions of the coordinates of a particle
in a fixed reference configuration and time t. Said differently, we express
� = �(X1; X2; X3; t);
vi = vi(X1; X2; X3; t):
Such a description is known as Lagrangian or referential or material description.
(ii) Observing the changes at fixed locations, that is, we express �; vi etc. as functions of xi and
t. Thus
� = �(x1; x2; x3; t);
vi = vi(x1; x2; x3; t):
Such a description is known as spatial or Eulerian. We note that in this description, what is
described (or measured) is the change of quantities at a fixed point in space (not a specific
material particle) as a function of time. The same spatial position is occupied by different
particles at different times. Therefore, the spatial description does not provide direct infor-
mation regarding the changes in the values of a quantity associated with a material particle as
it moves about in space.
The referential and spatial descriptions are, of course, related by the motion. That is, if the
motion is known then, one description can be obtained from the other as illustrated by the following
example.
25
Example: Given the motion of a body to be
xi = Xi + 0 � 2tX2�1i: (i)
For the temperature field given by
� = 2x1 + (x2)2; (ii)
(a) find the material description of temperature, and
(b) the rate of change of temperature of a particle which at time t = 0 was at the place (0; 1; 0).
Solution
(a) By substituting (i) into (ii), we obtain
� = 2x1 + (x2)2;
= 2(X1 + 0 � 2tX2) + (X2)2;
= 2X1 + (X2 + 0 � 4t)X2: (iii)
(b) The temperature of the desired particle at different times is given by
� = (1 + 0 � 4t):
)
d�
dt= 0 � 4:
Note that even though the temperature is independent of time in the spatial description, a particle
experiences a change in temperature as it moves from one spatial position to another; this becomes
clear from eqn. (iii).
Whereas spatial description is often used in fluid mechanics, referential description is employed
in solid mechanics and in formulating laws of mechanics.
3.3 Displacement Vector
By definition, the displacement vector ui of a particle is the difference between its position vectors
at time t and at time t = t0 (or 0).
ui = xi �Xi: (3.3.1)
26
In the Lagrangian description, the displacement ui is specified as a function of Xi and t. For
example, consider the motion
x1 = X1t2 + 2X2t+X1;
x2 = 2X1t2 +X2t+X2; (3.3.2)
x3 =1
2X3t+X3:
The corresponding components of the displacement are given by
u1 = X1t2 + 2X2t;
u2 = 2X1t2 +X2t;
u3 =1
2X3t:
In the Eulerian description, ui will be expressed as a function of xi and t. Solving (3.3.2) for Xi in
terms of xi and t and substituting that in (3.3.1) we obtain
u1 = x1 �X1 = x1 � x1(1 + t)� 2x2t
�3t3 + t2 + t+ 1;
u2 = x2 �X2 = x2 � x2(1 + t2)� 2x1t2
�3t3 + t2 + t + 1;
u3 = x3 �X3 = x3 � 2x3t+ 2
:
3.4 Restrictions on Continuous Deformation of a Deformable Body.
For t = 1, eqns. (3.3.2) give
x1 = 2(X1 +X2);
x2 = 2(X1 +X2);
x3 =3
2X3:
Thus points (a; �a; 0) in the reference configuration move to (0; 0; 0) at time t = 1. This implies
that distinct particles which occupy different places in the reference configuration are deformed
into the same place in the present configuration (at time t = 1). This amounts to the collision of
various material particles. Even though in particle mechanics collisions are allowed, in continuum
27
mechanics, such a possibility is assumed to be ruled out to start with. Thus in continuum mechanics
it is assumed that different material particles always occupy distinct places. This is equivalent to
the requirement that
xi = xi(X1; X2; X3; t)
be a one-to-one mapping of R onto the present configuration. Since in Continuum Mechanics we
will need to differentiate functions with respect to xi or Xi, we will henceforth assume that the
mapping
xi = xi(X1; X2; X3; t)
is continuously differentiable and has a continuously differentiable inverse given by
Xi = Xi(x1; x2; x3; t):
This is so if and only if the Jacobian J defined by
J = det
266666664
@x1@X1
@x1@X2
@x1@X3
@x2@X1
@x2@X2
@x2@X3
@x3@X1
@x3@X2
@x3@X3
377777775
(3.4.1)
is non-zero for all points in R and for every value of t. Since
J(X1; X2; X3; 0) = 1;
and J is a continuous function of t, therefore, J must be positive for every t. Using the index
notation, we can write J as
J = det
�@xi@Xj
�= det
��ij +
@ui@Xj
�: (3.4.2)
In summary, we can conclude from the preceding discussion that a necessary and sufficient con-
dition for a continuous deformation to be physically admissible is that the Jacobian J be greater
than zero.
28
A displacement field satisfying the condition J > 0 is said to be proper and admissible, or
simply admissible.
Thus, for an admissible deformation of a medium the displacement components (u1; u2; u3)
must satisfy J > 0. For example, a piece of rubber can not be subjected to displacement compo-
nents u1 = �2X1; u2 = 0; u3 = 0 since then J = �1. This displacement is called a reflection
about the (X2; X3) plane, since the point (x1; x2; x3) is the image of the point (X1; X2; X3) in
a mirror that lies in the plane x1 = 0.
Exercise: Determine whether or not
u1 = k(X2 �X1); u2 = k(X1 �X2); u3 = kX1X3;
where k is a constant, are possible displacement components for a continuous medium.
3.5 Material Derivative
The time rate of change of a quantity such as temperature or velocity of a material particle is
known as a material derivative, and is denoted by a superimposed dot or by D=Dt.
(i) When referential or Lagrangian description of a quantity is used, i.e.,
� = �(X1; X2; X3; t)
then
_� =D�
Dt=@�
@t
����Xi�fixed
: (3.5.1)
(ii) When spatial or Eulerian description of a quantity is used, i.e.,
� = �(x1; x2; x3; t);
where
xi = xi(X1; X2; X3; t);
then
_� =D�
Dt=@�
@t
����xi�fixed
+@�
@xj
@xj@t
����Xi�fixed
: (3.5.2)
29
In rectangular Cartesian coordinates
vj =@xj@t
����Xi�fixed
(3.5.3)
is the jth component of the velocity of a material particle. Therefore, in rectangular Cartesian
coordinates,
_� =@�
@t+
@�
@xjvj: (3.5.4)
Note that in (3.5.4) � is given in the spatial description.
Example: Given the motion
xi = Xi(1 + t); t � 0;
find the spatial description of the velocity field.
Solution
xi = Xi(1 + t)
vi = _xi = Xi =xi
1 + t:
Example: Given
� = 2(x21 + x22) where xi = Xi(1 + t):
Find, at t = 1, the rate of change of temperature of the material particle which in the reference
configuration was at (1; 1; 1).
Solution
(i) � = 2(x21 + x22)
= 2[X21 (1 + t)2 +X2
2 (1 + t)2]
_� = 2[2X21 (1 + t) + 2X2
2 (1 + t)]
) _� at t = 1 for the material particle (1; 1; 1) = 16:
(ii) _� =@�
@t+
@�
@xj
@xj@t
= 0 + 4x1x1
1 + t+ 4x2
x21 + t
:
30
At t = 1, for the material particle (1; 1; 1)
xi = 2(�i1 + �i2 + �i3)
)_� =
4(2)2
1 + 1+
4(2)2
1 + 1= 16:
Exercise: The motion of a continuous medium is defined by the equations
x1 =1
2(X1 +X2)e
t +1
2(X1 �X2)e
�t; x2 =1
2(X1 +X2)e
t � 1
2(X1 �X2)e
�t;
x3 =X3; 0 � t < constant:
(a) Express the velocity components in terms of referential coordinates and time.
(b) Express the velocity components in terms of spatial coordinates and time.
Exercise: Given the motion of a body to be xi = (X1 + ktX2)�i1 + X2�i2 + X3�i3, and the
temperature field by � = x1 + x2, find _� for the particle which currently is at the place (1; 1; 1).
3.6 Finding Acceleration of a Particle from a Given Velocity Field.
The acceleration of a material particle is the rate of change of its velocity. If the motion of a
continuum is given by
xi = xi(X1; X2; X3; t) with xi(X1; X2; X3; 0) = Xi
then the velocity v, at time t, of a particle Xj is given by
vi =@xi@t
����Xj�fixed
= _xi;
and the acceleration a, at time t, of a particle Xj is given by
ai =@vi@t
����Xj�fixed
= _vi:
Thus, if the material description of velocity v(X1; X2; X3; t) is known, then the accelera-
tion is computed simply by taking the partial derivative with respect to time of the function
v(X1; X2; X3; t). On the other hand, if only the spatial description of velocity (i.e. vi =
31
vi(x1; x2; x3; t)) is known, then the computation of acceleration involves the use of eqn. (3.5.2).
That is,
ai =@vi@t
����Xj�fixed
=@vi@t
����xj�fixed
+@vi@xk
@xk@t
����Xj�fixed
;
=@vi@t
����xj�fixed
+@vi@xk
vk:
The part of acceleration given by vk@vi@xk
is called convective acceleration. When the motion is
only along x1-axis, i.e. v2 = v3 = 0 and v1 = v1(x1; t), then
a1 =@v1@t
+ v1@v1@x1
:
Example: A fluid rotates as a rigid body with a constant angular velocity !e3.
(a) Write out explicitly the components of the velocity of a material particle in the Lagrangian
and Eulerian descriptions of motion.
(b) Compute the acceleration field in the Eulerian description.
Solution:
(a) Recall that
v = ! � r:
32
Thus
vi = "ijk!jxk = "ijk!�j3xk;
= !"i3kxk;
is the Eulerian description of velocity.
To convert these expressions into the Lagrangian description, let the fluid particle which
presently is at place P be at place Q at t = 0. Then, referring to Fig. 3.6.1,
� � � = !t;
� = � + !t = tan�1�X2
X1
�+ !t:
Since
x1 = OP cos � ;
= OQ cos � ;
=qX2
1 +X22 cos � ;
v1 = !"132x2 = �!qX2
1 +X22 sin(tan�1
�X2
X1
�+ !t) ;
v2 = !"231x1 = !qX2
1 +X22 cos(tan�1
�X2
X1
�+ !t) :
33
(b)
ai =@vi@t
+@vi@xj
vj ;
= 0 + !"i3k�kjvj ;
= !"i3j(!"j3mxm) ;
= !2"i3j"3mjxm ;
= !2(�i3�3m � �im�33)xm ;
ai = !2(�i3x3 � xi) :
) a1 = � !2x1 ;
a2 = � !2x2 ;
a3 = 0 :
Exercise: Given
v =x1e1 + x2e2x21 + x22
; � = 2(x21 + x22) ;
determine the acceleration and the rate of change of temperature of the material particle which
currently is at the place (1; 1).
Exercise: For the motion
xi = Xi + sin(�t) sin(�X1)�i2 ;
find the velocity and acceleration in referential and spatial descriptions.
3.7 Deformation Gradient
As pointed out earlier, in continuum mechanics, two different material particles always occupy two
distinct places. Consider two material particles P (Xi) and Q(Xi + dXi).
34
Here (Xi) stands for the triplet (X1; X2; X3) used to identify particles in the reference configura-
tion. Let points P and Q move to P 0 and Q0 respectively so that the vector PQ is deformed into
the vector P0Q0. The vector P0Q0 need not, and in general will not, equal PQ. This means that
the length and direction of P0Q0 may be different from that of PQ. Given PQ or P0Q0 and the
motion, our problem is to find the other vector.
Let the motion of the body be given by
xi = xi(X1; X2; X3; t): (3.7.1)
Then
P0Q0 = [xi(X1 + dX1; X2 + dX2; X3 + dX3; t)� xi(X1; X2; X3; t)]ei: (3.7.2)
Using Taylor’s theorem for series expansion, we get
P0Q0 =�@xi@X1
dX1 +@xi@X2
dX2 +@xi@X3
dX3
�ei +O(jdXj2); (3.7.3)
=@xi@Xj
dXjei +O(jdXj2): (3.7.4)
If points P and Q are close together in the reference configuration, then higher order terms can be
neglected as compared to terms linear in dXi. Using this approximation, we get
P0Q0 =�@xi@Xj
����P
dXj
�ei: (3.7.5)
Here the suffix��P
reminds us that@xi@Xj
is evaluated at the point P . In section 2.9, it was mentioned
that a comma followed by an index i will be used to indicate partial differentiation with respect to
xi. Now if we were to use that notation, then it will not be clear whether the partial differentiation
35
is with respect to xi or Xi. To clarify the situation we will, henceforth, use upper case latin indices
for X . That is, the triplet (X1; X2; X3) will be denoted by XA instead of Xi and eqn. (3.7.5) will
be written as
P0Q0 =@xi@XA
����P
dXAei = FiA��PdXAei: (3.7.6)
In component form,
(P 0Q0)j = P0Q0 � ej = FiAjP dXAei � ej ;
= FjA��PdXA ;
= FjA��P(PQ)A : (3.7.7)
Thus FjA relates the components of vector PQ in the reference configuration to the components
of the vector P0Q0 into which PQ is deformed. Since in Continuum Mechanics we assume that
J = det jFiAj > 0
therefore, FjA is an invertible matrix. This implies that once the motion (3.7.1) is known, we can
find P0Q0 from PQ and vice-a-versa.
Since the motion (3.7.1) gives how the body deforms and FiA is the gradient of the motion, FiA
is called deformation gradient. FiA relates a vectorPQ in the reference configuration to the vector
P0Q0 (in the present configuration) into which PQ is deformed. In terms of the displacement u,
FiA can be written as follows.
ui = xi �XA�iA ;
@ui@XA
=@xi@XA
� �iA ;
FiA = xi;A = �iA + ui;A :
(3.7.8)
The gradient ui;A of the displacement u is known as the displacement gradient.
Example: The deformation of a body is given by
u1 = (3X21 +X2); u2 = (2X2
2 +X3); u3 = (4X23 +X1):
Compute the vector into which the vector 10�2�1
3;1
3;1
3
�passing through the point (1; 1; 1) in
the reference configuration is deformed.
36
Solution
Here the point P is (1; 1; 1) and the vector PQ has components
�10�2
3;10�2
3;10�2
3
�. The
deformation gradient at any point is given by
[FiA] =
24 1 + 6X1 1 0
0 1 + 4X2 11 0 1 + 8X3
35 :
The deformation gradient evaluated at point P is
[FiA]��P=
24 7 1 0
0 5 11 0 9
35 :
Therefore, the components of the vector P0Q0 are given by
2664
(P 0Q0)1
(P 0Q0)2
(P 0Q0)3
3775 =
2664
7 1 0
0 5 1
1 0 9
3775
26666664
10�2
310�2
310�2
3
37777775=
10�2
3
2664
8
6
10
3775 :
Note that, in the preceding example, the vector P0Q0 is neither parallel to PQ nor is it equal to
PQ in magnitude. The ratio of the length of P0Q0 to that of PQ is called stretch in the direction
of PQ.
It is emphasized that in going from (3.7.4) to (3.7.5) we assumed that the length of the vector
PQ is infinitesimal. However, no assumption was made as to the magnitude1 of FiA. Thus (3.7.5)
is valid no matter how large or small the components of deformation gradient are. Said differently
(3.7.5) is valid both for small and large deformations so long as we study the deformation of
infinitesimal vectors passing through the point P . Thus FiA describes the deformation of material
particles in an infinitesimal neighborhood of P .
A special case in which (3.7.5) follows exactly from (3.7.4) is that when FiA is constant. That
is, each of the nine quantities @xi=@XA is a constant. A deformation for which FiA is a constant is
called a homogeneous deformation.
Example: Given the following displacement components
u1 = 0 � 1X22 ; u2 = u3 = 0:
1The magnitude of FiA is defined asp1=2FiAFiA.
37
a) Is this deformation possible in a continuously deformable body? Prove your answer.
b) Find the deformed vectors of the material vectors 0:01e, and 0:015e2 which pass through the
point P (1; 1; 0) in the reference configuration.
c) Determine the stretches at the point (1; 1; 0) in the X1 and X2 directions.
d) Determine the change in angle between lines through the point P (1; 1; 0) that were parallel
to X1 and X2 axes.
Solution:
a)
[FiA] =
24 1 0:2X2 0
0 1 00 0 1
35
det [FiA] = 1 > 0:
Therefore, the given deformation is possible in a continuously deformable body.
b) The material vector 0 � 01e1 through the material point (1; 1; 0) is deformed into the vector
given by 24 1 0:2(1) 0
0 1 00 0 1
3524 0:01
00
35 =
24 0:01
00
35 ;
and the material vector 0:015e2 through the material point (1; 1; 0) is deformed into24 1 0:2(1) 0
0 1 00 0 1
3524 0
0:0150
35 =
24 0:003
0:0150
35 :
c) Stretch at P (1; 1; 0) in X1-direction =0:01
0:01= 1.
Stretch at P (1; 1; 0) in X2-direction =
p0:0032 + :0152
0:015= 1:02.
d) Angle between the vectors into which vectors 0:01e1 and 0:015e2 through the point (1; 1; 0)
are deformed = cos�1(0:01)(0:003 + 0 + 0)
(0:01)(0:0032 + 0:0152)1=2= 78:7�.
Change in angle = 11:3�.
38
Example: Given the following displacement components
u1 = 2X21 +X1X2; u2 = X2
2 ; u3 = 0;
and that for points in the body, X1 � 0; X2 � 0.
a) Find the vector in the reference configuration that ends up parallel to x1-axis through the point
(1; 0; 0) in the current configuration.
b) Find stretch of a line element that ends up parallel to x1-axis through the point (1; 0; 0) in the
present configuration.
Solution: For the given displacement components,
x1 = X1 + 2X21 +X1X2; x2 = X2 +X2
2 ; x3 = X3:
Therefore the undeformed position of point (1; 0; 0) is obtained by solving
1 = X1 + 2X21 +X1X2;
0 = X2 +X22 ;
0 = X3:
The solution of these equations which satisfies X1 � 0 is (1=2; 0; 0). The material particle which
currently is at P 0(1; 0; 0) was at P (1=2; 0; 0) in the reference configuration.
[FiA] =
24 1 + 4X1 +X2 X1 0
0 1 + 2X2 00 0 1
35
�FiA
��P
�=
24 3 0:5 0
0 1 00 0 1
35
)
24 1
00
35 =
24 3 0:5 0
0 1 00 0 1
3524 dX1
dX2
dX3
35
where (dX1; dX2; dX3) is the vector in the reference configuration that ends up into a vector
(1; 0; 0) through the point (1; 0; 0) in the current configuration.24 dX1
dX2
dX3
35 = 1=3
24 1 �0:5 0
0 3 00 0 3
3524 1
00
35 =
24 1=3
00
35 :
39
Thus a vector parallel to x1-axis through (1=2; 0; 0) in the reference configuration ends up into a
vector parallel to x1-axis through (1; 0; 0) in the current configuration.
b) Stretch along the desired line =
p12 + 0 + 0p
(1=3)2 + 0 + 0= 3.
Exercise: Given the displacement field
u1 = 10�2(2X1 +X22 ); u2 = 10�2(X2
1 �X22 ); u3 = 0:
Find the stretches and the change of angle for the material lines (0:1; 0; 0) and (0; 0:1; 0) that
emanate from the material particle (1;�1; 0).Exercise: The displacement components for a body are
u1 = 2X1 +X2; u2 = X3; u3 = X3 �X2 :
a) Verify that this displacement vector is possible for a continuously deformed body.
b) Is this deformation homogeneous?
c) Determine the stretch in the direction (1=3; 1=3; 1=3) through the point (1; 1; 1) in the refer-
ence configuration.
d) Determine the direction cosines of the line element in the reference configuration that ends
up in the x3-direction at the place (1; 1; 0).
e) Determine the change in angle between the lines through the point (1; 1; 1) (in the reference
configuration) whose directions in the reference configuration are 1; 0; 0 and 1=3; 1=3; 1=3.
3.8 Strain Tensors
As seen in the previous section, during the motion of a deformable continuum, material lines
originating from a material point are rotated and stretched. Whenever the material lines emanating
from a material point are stretched and/or the angle between two different material lines passing
through a material point changes, the body is said to be strained or deformed. We have seen in the
previous section that the deformation gradient is a measure of the stretch and rotation of various
40
material vectors. In this section we will introduce two more measures of deformation which, of
course, will be related to the deformation gradient.
Consider two material vectors PQ and PR originating from the material point P (XA) in the
reference configuration. Let PQ be deformed into P0Q0 and PR into P0R0. Then
(P 0Q0)j = FjA
��P
(PQ)A;
(P 0R0)j = FjA
��P
(PR)A:
Henceforth we will drop the suffix jP to shorten the notation. Of course, FiA is to be evaluated at
the material point P . Therefore,
P0Q0 �P0R0 = FjAFjB(PQ)A(PR)B;
= (PQ)A CAB (PR)B; (3.8.1)
where
CAB = FjAFjB or [C] = [F]T [F]: (3.8.2)
Note that CAB = CBA, that is the matrix [C] is symmetric. To obtain a physical interpretation of
various components of [C], let
PQ = 10�2(1; 0; 0);
PR = 10�2(1; 0; 0):
41
Then (3.8.1) gives
P0Q0 �P0Q0 = 10�4C11;
so that
jP0Q0j = 10�2pC11 or jP0Q0j=jPQj =
pC11: (3.8.3)
Thus C11 is a measure of the stretch along X1-axis. Similarly C22 and C33 measure stretches along
X2 and X3 axes respectively. Now take
PQ = 10�2(1; 0; 0);
PR = 10�2(0; 1; 0):
Then (3.8.1) gives
P0Q0 �P0R0 = 10�4C12 :
Since jP0Q0j = 10�2pC11; jP0R0j = 10�2
pC22;
therefore;P0Q0 �P0R0
jP0Q0j jP0R0j =C12p
C11
pC22
: (3.8.4)
The left-hand side of (3.8.4) equals cosine of the angle betweenP0Q0 andP0R0. Thus C12 provides
a measure of the change in angle between two material lines passing through the point P that in
the reference configuration were parallel to X1 and X2-axes. Similarly C23 measures the change
in angle at the material point P between two material lines that in the reference configuration were
parallel to X2 and X3 axes.
In terms of displacement components CAB can be written as follows.
CAB = FiAFiB = (�iA + ui;A)(�iB + ui;B);
= �AB + uB;A + uA;B + ui;Aui;B: (3.8.5)
CAB is called the right Cauchy-Green tensor. The tensor
EAB = 1=2(CAB � �AB);
= 1=2(uA;B + uB;A + ui;Aui;B) ; (3.8.6)
42
is known as the Green-St. Venant strain tensor. We note from (3.8.1) that
P0Q0 �P0R0 �PQ �PR = (PQ)A(CAB � �AB)(PR)B;
= 2(PQ)AEAB(PR)B; (3.8.7)
so that EAB measures the change in lengths of various material line elements and the change in
angles between different material lines emanating in the reference configuration from the same
material point. It follows from eqn. (3.8.7) that
P0Q0 �P0R0 �PQ �PRjPQj jPRj = 2MAEABNB;
jP0Q0j2 � jPQj2jPQj2 = 2MAEABMB; (3.8.8)
where M and N are unit vectors parallel to PQ and PR respectively.
Since (PQ)A = (F�1)Ai(P 0Q0)i, therefore
PQ �PR = (F�1)Ai(F�1)Bj(P
0Q0)i(P0R0)j;
= (B�1)ij(P 0Q0)i(P 0R0)j; (3.8.9)
where
Bij = FiAFjA; [B] = [F][FT ]; (3.8.10)
is the left Cauchy-Green tensor. The tensor
�ij =1
2(�ij � (B�1)ij);
=1
2(ui;j + uj;i � uk;iuk;j); (3.8.11)
is called the Almansi-Hamel tensor. It follows from (3.8.9) and (3.8.11) that
P0Q0 �P0R0 �PQ �PRjP0Q0j jP0R0j = 2�ijminj; (3.8.12)
wherem and n are unit vectors parallel toP0Q0 andP0R0 in the present configuration. It is evident
from eqn. (3.8.12) that �ij also measures changes in lengths of line elements and changes in angles
between different line elements in the present configuration.
43
In the referential description of motion, EAB is used as a measure of strain. However, �ij is
used to measure strain in the spatial description of motion. Note that each one of these two tensors
vanishes when there is no deformation.
Exercise: By taking PQ = 10�2(1; 0; 0); PR = 10�2(1; 0; 0), and using (3.8.7), prove that
jP0Q0j =�p
1 + 2E11
�10�2:
Hence find the engineering strain (elongation/original length) along X1-axis.
Exercise: By settingPQ = 10�2(1; 0; 0) andPR = 10�2(0; 1; 0) into (3.8.7), and using the result
of the previous exercise, obtain an expression for the cosine of the angle between P0Q0 and P0R0.
3.9 Principal Strains
Having learned how to find the deformation of various material vectors emanating from a material
point P , we now investigate which of these lines is stretched the most. Let
PQ = 10�3(N1; N2; N3) (3.9.1)
where (N1; N2; N3) are components of a unit vector along PQ. That is, N = (N1; N2; N3) is a
unit vector along PQ. Our aim is to find N such that the stretch at P along N is maximum or
minimum. The vector P0Q0 into which PQ is deformed is given by
(P 0Q0)j = FjA(PQ)A = 10�3FjANA: (3.9.2)
Therefore,
jP0Q0j2jPQj2 = FjANAFjBNB = CABNANB: (3.9.3)
Thus the problem reduces to finding a unit vector N such that CABNANB is maximum. By using
the method of Lagrange multipliers we need to find N such that
CABNANB � �(NANA � 1) (3.9.4)
takes on extreme values for all N. In (3.9.4) � is a Lagrange multiplier. Such an N is given by
@
@NI[CABNANB � �(NANA � 1)] = 0
44
which is equivalent to
CABNB � �NA = 0;
or
(CAB � ��AB)NB = 0: (3.9.5)
These are three linear homogeneous equations in N1; N2; N3. A non-trivial solution of these equa-
tions exists if and only if
det[CAB � ��AB] = 0 ;
or
�3 � I�2 + II�� III = 0 ; (3.9.6)
where
I = CAA;
II = 1=2[�CABCBA + CAACBB] = (det[C])(C�1)AA;
III = det[C]:
I; II; III are called principal invariants of [C]. Equation (3.9.6) is cubic in � and will have three
roots. Since the matrix [C] is symmetric and positive definite, all three roots of (3.9.6) are positive.
Let us denote the three roots of (3.9.6) by �21; �22; �
23.
2 For each one of these roots of (3.9.6) we
find the correspondingN from (3.9.5). That is, once the three roots of (3.9.6) have been found, we
use (3.9.5) to find the directions N(1); N(2); N(3) along which stretches assume extreme values.
For example, N(1) is obtained by solving
(CAB � �21�AB)N(1)B = 0; N
(1)1 N
(1)1 +N
(1)2 N
(1)2 +N
(1)3 N
(1)3 = 1: (3.9.7)
2�1; �2�3 should not be confused with the components of a vector.
45
Having found N(1), we use (3.9.3) to find the stretch in this direction.
Stretch along N(1) =
qCABN
(1)A N (1)
B ;
=
q�21�ABN
(1)A N
(1)B =
q�21N
(1)A N
(1)A ;
= �1:
Thus the three roots of (3.9.6) are squares of extreme values of stretches at the point P . Let us now
assume that �21 6= �22 6= �23 and find the angle between directions N(1) and N(2). Rewriting (3.9.7)
as
CABN(1)B = �21�ABN
(1)B = �21N
(1)A ;
we obtain
N(2)A CABN
(1)B = �21N
(2)A N
(1)A : (3.9.8)
Similarly
N(1)A CABN
(2)B = �22N
(1)A N
(2)A : (3.9.9)
Since CAB = CBA, therefore, the left-hand sides of (3.9.8) and (3.9.9) are equal. Thus
(�21 � �22)N(1)A N (2)
A = 0:
Since �21 6= �22 by assumption,
N(1)A N
(2)A = 0: (3.9.10)
Thus N(1) and N(2) are perpendicular to each other. By using the same argument for N(2) and
N(3), we conclude that whenever �21 6= �22 6= �23, N(1); N(2) and N(3) are mutually perpendicular
to each other.
Let us now find the change in the angle between the linesN(1) andN(2) during the deformation.
46
If � is the angle between the deformed positions of N(1) and N(2), then by (3.8.1) and (3.8.4)
cos � =CFGN
(1)F N
(2)G
(CABN(1)A N
(1)B )1=2(CDEN
(2)D N
(2)E )1=2
;
=�21�ABN
(1)B N
(2)A
�1�2=�1N
(1)A N
(2)A
�2= 0:
) � = 90�: (3.9.11)
That is, the angle between directions N(1) and N(2) does not change during the deformation. The
same holds true for directions N(2) and N(3). Because of this property, the directions N(1); N(2)
and N(3) are called principal axes of stretch and �1; �2; �3 are called principal stretches.
Whenever any two or all roots of eqn. (3.9.6) are equal, equations (3.9.10) and (3.9.11) hold for
suitable choices of N(1); N(2) and N(3). If �21 = �22 6= �23, then N(3) is uniquely determined but
N(1) and N(2) can be taken as any two directions in a plane perpendicular to N(3). When �21 =
�22 = �23, then any three linearly independent directions which need not be mutually perpendicular
to each other can be taken as N(1); N(2), and N(3).
Recalling that
2EAB = CAB � �AB;
we obtain
2EABN(1)B = CABN
(1)B � �ABN
(1)B ;
= (�21 � 1)N(1)A : (3.9.12)
To arrive at this equation, we used (3.9.7). ThusN(1); N(2) andN(3) are the directions along which
EABNANB takes on extreme values
(�21 � 1)=2; (�22 � 1)=2; (�23 � 1)=2 (3.9.13)
respectively. If we define strain in the direction PQ as
(jP0Q0j2 � jPQj2)=jPQj2 (3.9.14)
47
instead of the engineering strain
(jP0Q0j � jPQj)=jPQj; (3.9.15)
we see from (3.8.7) that EABNANB gives strain in the direction N. Thus the three numbers given
by (3.9.13) are the extreme values of strains and since the angle between the mutually orthogonal
directions N(1);N(2);N(3) does not change, (3.9.13) are the principal strains and N(1);N(2);N(3)
are the axes of principal strain. Note that there is no shear strain between any two of the three
directions N(1); N(2) and N(3).
APPENDIX: SOLUTION OF A CUBIC EQUATION
The solution of a general cubic equation, which may have imaginary roots, in a closed form
involves the use of hyperbolic functions. However, the cubic equation obtained from
det[CAB � ��AB] = 0 (A1)
in which CAB = CBA has only real roots, and can be solved as follows.
The cubic equation
y3 + py2 + qy + s = 0; (A2)
can be reduced to the form
x3 + ax + b = 0 (A3)
by substituting
y = x+ p=3: (A4)
In (A3)
a = (3q � p2)=3 and b = (2p3 � 9pq + 27s)=27 :
The reduced cubic equation (A3) can be solved by transforming it to the trignometric identity
4 cos3 � � 3 cos � � cos 3� = 0:
48
(See pages 93-95, CRC Standard Mathematical Tables). The solution of the equation
�3 � I�2 + II�� III = 0
obtained by this method is
� = r cos � + I=3;
where
cos 3� = (2I3 � 9(I)(II) + 27III)=2(I2 � 3II)3=2 ;
r = 2(I2 � 3II)1=2=3:
Example: The deformation of a body is given by
u1 = 3X21 +X2; u2 = 2X2
2 +X3; u3 = 4X23 +X1:
a) Find the principal strains at the material point (1,1,1) in the reference configuration.
b) Find the axis of the maximum strain through the material point (1,1,1) in the reference con-
figuration. Also find the direction cosines of the line into which this axis is deformed.
Solution:
(a) At any point
[FiA] =
24 1 + 6X1 1 0
0 1 + 4X2 11 0 1 + 8X3
35 :
Thus
[FiAj(1;1;1)] =24 7 1 0
0 5 11 0 9
35 ;
[CAB] = [F]T [F] =
24 7 0 1
1 5 00 1 9
3524 7 1 0
0 5 11 0 9
35 ;
=
24 50 7 9
7 26 59 5 82
35 :
49
The principal invariants of [C] are given by
I = CAA = 50 + 26 + 82 = 158 ;
III = det[C] = 50[26(82)� 25]� 7[7(82)� 45] + 9[35� 234] = 99856 ;
II = det[C](C�1)AA =
���� 26 55 82
���� +���� 50 9
9 82
����+���� 50 7
7 26
���� = 7377 ;
r = 2(I2 � 3II)1=2=3 = 35:484 ;
cos 3� = (2I3 � 9I(II) + 27III)=2(I2 � 3II)3=2 = 0:31382 ;
3� = 71:7�; 360� 71:7�; 360 + 71:7� ;
� = 23:9�; 96:1�; 143:9� ;
�21 = 85:11; �22 = 48:9; �23 = 24 :
) Principal strains at the material point (1; 1; 1) are 42.06, 23.95, 11.5.
(b) To obtain N(1) we need to solve
(CAB � �21�AB)N(1)B = 0;
N(1)B N
(1)B = 1:
�35:11N (1)1 + 7N
(1)2 + 9N
(1)3 = 0 ; (i)
7N(1)1 � 59:11N
(1)2 + 5N
(1)3 = 0 ; (ii)
9N(1)1 + 5N
(1)2 � 3:11N
(1)3 = 0 ; (iii)
N(1)1 N
(1)1 +N
(1)2 N
(1)2 +N
(1)3 N
(1)3 = 1 : (iv)
7(i) + 35:11(ii) gives
� 2026:35N(1)2 + 238:55N
(1)3 = 0 ;
N(1)3 = 8:494N
(1)2 :
5(i)� 9(ii) gives
� 238:55N(1)1 + 567N
(1)2 = 0 ;
N(1)1 = 2:377N
(1)2 :
50
Substituting for N (1)3 and N (1)
1 into (iv), we get
(2:377)2N (1)2 N (1)
2 +N (1)2 N (1)
2 + (8:494)2N (1)2 N (1)
2 = 1:
N(1)2 = 0:1126
) N(1)3 = 0:957; N
(1)1 = 0:268 :
) The axis of the maximum strain at the material point P (1; 1; 1) is (0:268; 0:1126; 0:957).
To find the direction cosines of the line into which this is deformed, let
PQ = ds(0:268; 0:1126; 0:957):
Then (P 0Q0)j = FjA(PQ)A,
[P0Q0] = ds
24 7 1 0
0 5 11 0 9
3524 0:268
0:11260:957
35 = ds
24 1:989
1:528:881
35
) Direction cosines of P0Q0 are (0:216; 0:164; 0:962).
Exercise: Given the displacement components
u1 = 0:1X22 ; u2 = u3 = 0:
a) Find the principal strains at the material point (1; 1; 0) in the reference configuration.
b) Find the axis of the maximum strain at the point (1; 1; 0) in the reference configuration.
Exercise: Given the following displacement components
u1 = 2X21 +X1X2; u2 = X2
2 ; u3 = 0:
Find the principal strains and their axes at the material point (1=2; 0; 0) in the reference configura-
tion.
3.10 Deformation of Areas and Volumes
Consider two different infinitesimal line elementsPQ andPR emanating from a pointP in the ref-
erence configuation. During the deformation lines PQ and PR are deformed into
51
P0Q0 and P0R0 respectively. Hence the parallelogram whose adjacent sides are PQ and PR in
the reference configuration is deformed into the one with adjacent sides as P0Q0 and P0R0. Let us
denote the areas of these by dA and da respectively. Then
dA = PQ�PR
) dAB = "BCD(PQ)C(PR)D :
Also
da = P0Q0 �P0R0 ;
dai = "ijk(P0Q0)j(P 0R0)k :
Recalling that (P 0Q0)j = FjC(PQ)C , we obtain
dai = "ijkFjC(PQ)CFkD(PR)D ;
= "pjkFjCFkD(FpB(F�1Bi ))(PQ)C(PR)D ;
= J"BCD(F�1)Bi(PQ)C(PR)D ;
= J(F�1)BidAB:
Hence
da = J(F�1)TdA: (3.10.1)
Now consider the parallelepiped formed by three nonplanar infinitesimal line elements PQ,
PR andPS passing through a point P in the reference configuration. Because of the deformation,
52
the parallelepiped is deformed into the one whose three concurrent sides are P0Q0, P0R0 and
P0S0. If dV and dv denote the volumes of these in the reference and the current configurations
respectively, then
dV = PQ�PR �PS = "BCD(PQ)B(PR)C(PS)D:
Similarly
dv = P0Q0 �P0R0 �P0S0 = "ijk(P0Q0)i(P
0R0)j(P0S 0)k:
Substituting
(P 0Q0)i = FiB(PQ)B etc:; we get
dv = "ijkFiB(PQ)BFjC(PR)CFkD(PS)D ;
= J"BCD(PQ)B(PR)C(PS)D ;
= J dV :
Hence
dv = J dV : (3.10.2)
A deformation such that
dv = dV
at every material point in the body is called an isochoric (volume preserving) deformation. Thus
for an isochoric deformation, J = 1 at each material point of the body. Examples of isochoric
deformations are the simple shearing deformation given in the example problem on page 3-2 and
the one given in the exercise on page 3-31. Note that the latter is not a homogeneous deformation
even though it is isochoric.
Exercise: Given the following deformation
u1 = 2X21 +X1X2; u2 = X2
2 ; u3 = 0:
53
At the material point (1=2; 0; 0) in the reference configuration, consider an infinitesimal plane
formed by the vectors 10�2(1; 0; 0) and 10�2(1; 1; 0). Find the (vector) area of the element into
which this plane is deformed.
Exercise: The displacement components for a body are u1 = 2X1+X2; u2 = X3; u3 = X3�X2.
At the material point (1; 0; 0) on the surface of the body in the reference configuration, an element
of area has components 10�2(1; 1; 1). Find the components of the area into which this is deformed.
3.11 Mass density. Equation of continuity.
Consider a sphere of infinitesimal radius centered at a point P in the reference configuration. The
material contained within the sphere has mass �M . Let �V denote the volume of the sphere. The
mass density �0 at point P in the reference configuration is defined as
�0 = lim�V!0
�M
�V: (3.11.1)
One of the assumptions in continuum mechanics is that the limit on the right-hand side of (3.11.1)
exists at every point of the body. Since both �M and �V are positive, therefore, the mass density
�0(X1; X2; X3) is positive. Note that from the point of view of Atomic Physics, the assumption
that the right-hand side of (3.11.1) is well defined may not be justified. One can always envisage
infinitesimal volume elements surrounding a point which contain no atomic particles at some time
and hence will make the mass density at the point to be zero at that time. However, in contin-
uum mechanics, we are concerned with gross effects or macroeffects of deformation, and lengths
considered are much larger than the distance between adjacent atoms.
We now make the assumption that the mass of the material contained in every small volume el-
ement at P is conserved. That is, the mass of the matter enclosed in the infinitesimal parallelepiped
at P equals the mass of the matter contained in the infinitesimal parallelepiped at P 0 into which
the former is deformed. Denoting the mass density in the present configuration by �, we have
�dv = �0dV:
54
Substituting for dv from (3.10.2), we arrive at
�J = �0: (3.11.2)
or
�(X1; X2; X3; t)J(X1; X2; X3; t) = �0(X1; X2; X3):
Equation (3.11.2) which relates the mass density in the present configuration to the mass density
in the reference configuration is the equation of continuity or the conservation of mass in the
Lagrangian description.
To obtain the equation of continuity in the spatial description, we take the material derivative of
(3.11.2) and thereby obtain
_�J + � _J = 0: (3.11.3)
Since
J = "ABCF1AF2BF3C ;
therefore,
_J = "ABC( _F1AF2BF3C + F1A_F2BF3C + F1AF2B
_F3C): (3.11.4)
However,
"ABC _F1AF2BF3C = "ABC@ _x1@XA
F2BF3C ;
= "ABC
�@v1@x1
@x1@XA
+@v1@x2
@x2@XA
+@v1@x3
@x3@XA
�F2BF3C ;
= "ABC@v1@x1
F1AF2BF3C + 0 + 0 ;
= J@v1@x1
: (3.11.5)
Similarly, one can show that
"ABCF1A_F2BF3C = J
@v2@x2
;
"ABCF1AF2B_F3C = J
@v3@x3
:
(3.11.6)
55
Substituting from (3.11.5) and (3.11.6) into (3.11.4) and then the result into (3.11.3) we get
_�J + �J@vi@xi
= 0
and thus conclude that
_�+ �@vi@xi
= 0: (3.11.7)
This equation can equivalently be written as
@�
@t+
@
@xi(�vi) = 0; (3.11.8)
which is the continuity equation in spatial description.
For an isochoric deformation
J = 1
and hence
_J = 0; _� = 0: (3.11.9)
Equations (3.11.9) and (3.11.7) imply that for an isochoric deformation, the continuity equation
assumes the form
@vi@xi
= 0: (3.11.10)
Exercise: Show that
v1 = � 2x1x2x3(x21 + x22)
2; v2 =
(x21 � x22)x3(x21 + x22)
2; v3 =
x2x21 + x22
;
are the components of a velocity field in an isochoric deformation.
Example: For the velocity field given by
vi =xi
1 + t; t � 0;
find the density of a material particle as a function of time.
Solution: For the given velocity field,
@vi@xi
=3
1 + t:
56
Therefore, from the conservation of mass, we get
_� = � �@vi@xi
= � 3�
1 + t:
)
d�
�= � 3dt
1 + t:
Integration of this equation gives
`n� = �3`n(1 + t) + A (�)
where A is a constant of integration. If � = �0 at t = 0, then
`n�0 = �3`n1 + A or A = `n�0:
Thus eqn. (�) becomes
� = �0=(1 + t)3:
Exercise: Given the velocity field
vi = x1t�i1 + x2t�i2;
determine how the fluid density varies with time.
Exercise: In the spatial description the density of an incompressible fluid is given by � = kx2.
Find a permissible form for the velocity field, with v3 = 0, in order that the conservation of mass
equation be satisfied.
3.12 Rate of Deformation
Let us consider a material element P0Q0 emanating from a material point located at P 0(x1; x2; x3)
in the present configuration at time t. We wish to computeD(P0Q0)
Dt, the rate of change of
length and direction of P0Q0. Let P 0 be the deformed position of the material point located at
P (X1; X2; X3) in the reference configuration and Q0 that of Q(XA + dXA). Then
(P 0Q0)i = xi(XA + dXA; t)� xi(XA; t):
57
Thus,
D
Dt(P 0Q0)i = vi(XA + dXA; t)� vi(XA; t) ;
' @vi@XA
dXA;
= vi;AdXA:
(3.12.1)
Equation (3.12.1) expressesD(P0Q0)
Dtin a material description. To obtain
D(P0Q0)Dt
in a spatial
description, we note that since vi(XA; t) is the velocity of a material point P presently at the
position xi, therefore, if a spatial description of velocity is employed, this velocity is given by
vi = vi(xj; t). Note that vi(xj; t) and vi(XA; t) are, in general, different functions. Thus,
D(P0Q0)iDt
= vi(xj + dxj; t)� vi(xj; t);
' @vi@xj
dxj = vi;jdxj; (3.12.2)
= [1=2(vi;j + vj;i) + 1=2(vi;j � vj;i)]dxj;
= (Dij +Wij)dxj; (3.12.3)
where we have set
Dij = 1=2(vi;j + vj;i); (3.12.4)
Wij = 1=2(vi;j � vj;i):
Dij , the symmetric part of the velocity gradient vi;j , is known as the rate of deformation tensor or
the strain-rate tensor, and Wij , the antisymmetric part of the velocity gradient vi;j , is known as the
spin tensor.
In the following, we give a geometric interpretation of the elements of Dij . Let
P0Q0 = dsn
where n is a unit vector in the direction of P0Q0 = dx. Then
D
Dt(ds2) =
D
Dt(dxidxi) = 2dxi
D
Dt(dxi) ;
or dsD
Dt(ds) = dxi(Dij +Wij)dxj ;
58
where we have substituted forD
Dt(dxi) from (3.12.3). Since Wij = �Wji, therefore
dxiWijdxj = 0 ;
and we get
dsD
Dt(ds) = dxiDijdxj ;
or
1
ds
D(ds)
Dt= niDijnj : (3.12.6)
If n = (1; 0; 0), the right-hand side of (3.12.6) equals D11. Thus D11 gives the rate of change of
length per unit length known as stretching or rate of extension for a material line presently parallel
to x1-axis. Similarly D22 and D33 give, respectively, the stretching of a material line presently in
the x2- and x3-direction. To obtain a physical interpretation of the off-diagonal elements of Dij,
let
P0Q0 = ds1n and P0R0 = ds2m :
Then
1
ds1ds2
D
Dt(P0Q0 �P0R0) = mi(Dij +Wij)nj + ni(Dij +Wij)mj ;
1
ds1ds2
D(ds1ds2)
Dtcos �(m;n) � sin �(m;n)
_�(m;n) = 2miDijnj : (3.12.7)
Here �(m;n) is the angle between directionsm and n. Thus for m = (1; 0; 0) and n = (0; 1; 0), the
right-hand side of (3.12.7) equals 2D12 and the left-hand side equals � _�(m;n). Hence 2D12 equals
the rate of decrease of angle from�
2of two line elements presently parallel to x1 and x2-axes,
known as shearing or rate of shear. A similar interpretation holds for D23 and D31.
Since Dij is symmetric, we also have the result that there always exist three mutually perpen-
dicular directions (eigenvectors of Dij) along which the stretching (an eigenvalue of Dij) is either
maximum or minimum among stretchings for all differential elements extending from the material
particle which currently is at P 0.
59
Note that the strain-rate tensor does not, in general, equal the time rate of change of the strain
tensor. To prove this, we first conclude from eqns. (3.12.2) and (3.7.7) that
_FiAdXA = vi;jFjAdXA ;
which must hold for all choices of dXA. Thus
vi;j = _FiA(F�1)Aj; or L = _FF�1 ; (3.12.8)
where L is the spatial velocity gradient. Differentiation of both sides of (3.8.2) with respect to time
and the definition (3.8.6)1 of the strain tensor E gives
_E = 2 _C = _FTF + FT _F = FT (LT + L)F ;
= 2FTDF : (3.12.9)
Hence _E 6= D, and the strain-rate tensor D should not be confused with the time rate of change of
the strain tensor E.
We now attempt to provide a physical interpretation of the spin tensor. Let n be a unit eigen-
vector of D, i.e., Dn = �n where � is the eigenvalue corresponding to n. From (3.12.3), we
have
(dsni)� = _dsni + ds _ni = (Dij +Wij)dsnj ;
_ni = Dijnj +Wijnj � (nkDklnl)ni ;
=Wijnj ; (3.12.10)
where we have used (3.12.6). It states that the spin tensor operating on a unit eigenvector of D
gives the rate of change of that unit vector. Thus the spin equals the angular velocity of the principal
axes of stretching.
The axial vector
wi = ��ijkWjk or w = curlv ;
is the vorticity vector; its direction is the axis of spin, and its magnitude is the vorticity magnitude,
60
w.
w =pwiwi =
psWjkWjk :
We now show that the spin tensor defined by (3.12.4)2 does not equal the rate of change of the
rotation matrix R. Equations (3.12.9) and (3.13.1)1 give
L = ( _RU +R _U)(U�1RT ) ; (3.13.15)
or
D+W = _RRT +R _UU�1RT ;
= _RRT +1
2R( _UU�1 +U�1 _U)RT +
1
2R( _UU�1 �U�1 _U)RT ; (3.13.16)
where we have added and subtracted1
2RU�1 _URT to the right-hand side. Equating symmetric and
skew-symmetric tensors on both sides, we obtain
D =1
2R( _UU�1 +U�1 _U)RT ;
W = _RRT +1
2R( _UU�1 �U�1 _U)RT :
(3.13.17)
These equations clearly evince that D 6= _U and W 6= _R.
Taking the material derivative of both sides of eqn. (3.10.2) we obtain
D
Dt(dv) = _J dV; (3.12.11)
= J
�@vi@xi
�dV; (3.12.12)
=@vi@xi
dv;
and, therefore,
1
dv
�dv
=@vi@xi
= Dii = ID: (3.12.13)
In going from (3.12.11) to (3.12.12) we used
_J = J@vi@xi
= JID
61
which can be obtained from eqns. (3.11.4), (3.11.5) and (3.11.6). Thus the first principal invariant
ID of the rate of deformation tensor Dij gives the rate of change of volume per unit volume.
We now derive an expression for the rate of change of an area element. Rewriting eqn. (3.10.1)
as FTda = JdA and taking the time derivative of both sides, we obtain
_FTda+ FT _da = _JdA ;
FTLTda + FT _da = JIDdA ;
or
_da = (ID � LT )da; _dai =
�ID�ij � @vj
@xi
�daj : (3.12.14)
In an isochoric deformation, _dv = 0 but _da 6= 0 in general.
Example: Given the velocity field
vi = 2x2�1i;
find
(a) the rate of deformation and the spin tensors,
(b) the rate of extension per unit length of the line element P0Q0 = 10�2(1; 2; 0; ),
(c) the maximum and the minimum rates of extension.
Solution
(a) The matrix of the velocity gradient is
[vi;j] =
24 0 2 0
0 0 00 0 0
35 :
Therefore,
Dij = 1=2[vi;j + vj;i] =
24 0 1 0
1 0 00 0 0
35 ;
Wij = 1=2[vi;j � vj;i] =
24 0 1 0�1 0 00 0 0
35 :
62
(b) GivenP0Q0 = 10�2(1; 2; 0) = 10�2(p5)
�1p5;2p5; 0
�. Thus n = (1=
p5; 2=
p5; 0). From
eqn. (3.12.6)
1
ds
D(ds)
Dt= niDijnj = fngT [D]fng
= (1=p5 2=
p5 0)
24 0 1 0
1 0 00 0 0
3524 1=
p5
2=p5
0
35 = 4=5 :
(d) From the characteristic equation
det[Dij � ��ij] = 0;
we determine the eigenvalues of the tensor Dij as � = 0; �1. Therefore, 1 is the maximum
and �1 the minimum rate of extension. The eigenvector n(1) (for �1 = 1) determined from
Dijn(1)j � �1�ijn
(1)j = 0
and n(1)j n
(1)j = 1
is n(1) = (1=p2)(1; 1; 0). Similarly n(2) corresponding to �2 = �1 is (1=
p2)(1;�1; 0). The
third eigenvector is n(3) = (0; 0; 1).
Exercise: For the velocity field
vi = 2x22�i1;
find
a) the rate of extension per unit length of a material line element which in the present configu-
ration is given by 10�2(1; 1; 0) through the point (5; 3; 0),
b) the deformation corresponding to the given velocity field if xi = Xi at time t = 0,
c) the principal stretches and the deformed position of the axis of the maximum principal stretch
for the material particle which at t = 1 is at (0; 1=2; 0),
d) principal stretchings and their axes for the material particle which currently is at (0; 1=2; 0).
63
3.13 Polar Decomposition
The polar decomposition theorem of Cauchy3 states that a non-singular matrix equals an orthog-
onal matrix either pre or post multiplied by a positive definite symmetric matrix. If we apply this
theorem to the deformation gradient F, we get
F = RU = VR (3.13.1)
in which R is a proper orthogonal matrix and U and V are positive definite symmetric matrices.
Note that the decomposition (3.13.1) of F is unique in that R, U and V are uniquely determined
by F. From (3.13.1) it follows that
J = detF = detU = detV: (3.13.2)
Since C = FTF and B = FFT , therefore
C = FTF = (RU)TRU = U2; (3.13.3)
B = FFT = VR(VR)T = V2: (3.13.4)
We note that
V = RURT (3.13.5)
and
B = FFT = RU(RU)T = RCRT : (3.13.6)
Since U is symmetric, it has at least one orthogonal triad of eigenvectors. Let N(1) be an
eigenvector of U and �1 be the corresponding eigenvalue so that
UN(1) = �1N(1): (3.13.7)
Therefore
CN(1) = UUN(1) = �1UN(1) = (�1)
2N(1): (3.13.8)
3This theorem is proved in any book on linear algebra, e.g. on pg. 83 of P.R. Halmos, Finite Dimensional Vector Spaces, 2nd ed. Van Nostrand,Princeton, Toronto, and London, 1958.
64
Thus N(1) is an eigenvector of C and the corresponding eigenvalue is (�1)2. Since eqn. (3.13.8)
holds for every eigenvector of U, we conclude that eigenvectors of U and C are the same and the
eigenvalues ofC are equal to the squares of the eigenvalues ofU. In section 3.9, we proved that the
eigenvalues of C are the squares of the principal stretches and the eigenvectors of C are the axes
of principal stretches also usually called principal axes of stretch in the reference configuration.
Thus eigenvectors of U are the principal axes of stretch in the reference configuration and the
eigenvalues ofU are the princpal stretches. Let us now find the deformed position of an eigenvector
N(1) of U. Since
FN(1) = RUN(1) = �1RN(1); (3.13.9)
therefore
n(1) = RN(1) (3.13.10)
points in the direction of the vector into which N(1) is deformed. Note that
n(1) � n(1) = n(1)Tn(1) = N(1)TRTRN(1) = N(1)TN(1) = 1; (3.13.11)
therefore, n(1) is a unit vector in the direction of the deformed position of N(1). Now
FN(1) = VRN(1) = Vn(1): (3.13.12)
Equations (3.13.9), (3.13.10) and (3.13.12) when combined together give
Vn(1) = �1n(1): (3.13.13)
Thus �1 is an eigenvalue ofVwithn(1) as the eigenvector. This exercise proves that the eigenvalues
ofU andV are equal and that the eigenvectors ofV are the deformed positions of the eigenvectors
of U. Thus eigenvectors of V are the deformed position of the principal axes of stretch. Said
differently, eigenvectors of V are the principal axes of stretch in the deformed configuration or in
the present configuration. Of course,
Bn(1) = VVn(1) = (�1)2n(1) : (3.13.14)
65
From (3.13.14), (3.13.8), (3.13.9) and (3.13.10), we conclude that eigenvalues of B and C are
equal, and the eigenvectors of B are the principal axes of stretch in the deformed configuration.
Corresponding to the two decompositions of F given by (3.13.1) we can view the deformation
of the triadN(1); N(2); N(3), the eigenvectors ofU, as a stretch of these axes followed by a rotation
or a rotation of these axes followed by their stretch. This is schematically shown in the Fig. below.
Example: For simple shear
xi = Xi + kX1�2i;
a) find the principal stretches,
b) show that the angle � through which the principal axes of stretch in the reference configuration
are rotated so as to become the principal axes of stretch in the present configuration is given
by tan � = k=2.
66
Solution: For the given deformation
[F] =
24 1 0 0k 1 00 0 1
35
[C] =
24 1 + k2 k 0
k 1 00 0 1
35 ;
[B] =
24 1 k 0k 1 + k2 00 0 1
35 :
The squares of the principal stretches �1; �2; �3 are roots of the equation det([C]� �[1]) = 0
) [(1 + k2 � �)(1� �)� k2](1� �) = 0:
) (�1)2 = 1 + 1=2k2 + k
p1 + 1=4k2
(�2)2 = 1 + 1=2k2 � k
p1 + 1=4k2 = 1=(�1)
2
�3 = 1:
Note that the given deformation is a plane strain deformation. By looking at the matrices [C]
and [B] we see that X3-axis and x3-axis are eigenvectors of [C] and [B] corresponding to the
eigenvalue one. Also, x3-axis coincides with the X3-axis for the given deformation. Thus the
rotation of principal axes of stretch takes place about the X3-axis. Let this angle of rotation be � in
the clockwise direction. Then
[R] =
24 cos � � sin � 0
sin � cos � 00 0 1
35
Substituting for [B], [C] and [R] in eqn. (3.13.6) we arrive at24 1 k 0k 1 + k2 00 0 1
35 =
24 1 + k2 cos2 � � 2k sin � cos � ——— 0k(cos2 � � sin2 �) + k2 sin � cos � 1 + k2 sin2 � + 2k sin � cos � 0
0 0 1
35
Therefore
1 = 1 + k2 cos2 � � 2k sin � cos �
which gives tan � = k=2.
67
3.14 Infinitesimal Deformations
When the displacements and displacement gradients are small, we can neglect second order terms
in (3.8.6) and approximate the strain tensor by
eAB = 1=2(uA;B + uB;A): (3.14.1)
Since
@uA@XB
=@uA@xj
@xj@XB
;
=@uA@xj
��jB +
@uj@XB
�;
=@uA@xj
�jB +@uA@xj
@uj@XB
;
' @uA@xj
�jB; (3.14.2)
where we have neglected the second order term in the displacement gradients; for infinitesimal
deformations,
eAB =1
2
�@uA@XB
+@uB@XA
�=
1
2
�@ui@xj
+@uj@xi
��iA�jB: (3.14.3)
Thus one can differentiate displacements with respect to the current coordinates or the referential
coordinates to evaluate the infinitesimal strain tensor.
Since eAB is symmetric, therefore, it has at least one orthogonal triad of eigenvectors. The
eigenvectors of eAB are the principal axes of engineering strain and its eigenvalues are principal
infinitesimal strains. From equations (3.8.3) and (3.8.6) we obtain
jP0Q0j=jPQj =p
1 + 2E11 'p1 + 2e11 ' 1 + e11: (3.14.4)
Recalling that PQ = ds (1; 0; 0) we see that e11 equals the change in length per unit length of an
infinitesimal line element parallel to X1-axis. Similar interpretation holds for e22 and e33. From
equations (3.8.4) and (3.8.6) we conclude that
cos � =2E12p
1 + 2E11
p1 + 2E22
' 2e12(1 + e11)(1 + e22)
' 2e12 : (3.14.5)
68
Here � is the angle between the deformed positions of lines initially parallel to X1 and X2 axes. If
the change in the angle is small so that � = �2� , then eqn. (3.14.5) becomes
sin ' = 2e12 : (3.14.6)
Thus 2e12 equals the infinitesimal change in the angle between two lines originally parallel to X1
and X2 axes. This change in the angle is called the shearing strain.
From equations (3.9.3) and (3.8.6) we obtain
jP0Q0jjPQj = 1 + eABNANB : (3.14.7)
Thus the engineering strain in any directionN equals NAeABNB. One can similarly show that the
shearing strain between two orthogonal directionsM and N equals 2MAeABNB . Recalling that
FiA = �iA + ui;A = �iA +HiA ;
and that
CAB = FiAFiB ;
we obtain
[C] = [F]T [F] ' [1] + [H] + [H]T ; (3.14.8)
[U] = [C]1=2 ' [1] +1
2([H] + [H]T ) ;
= [1] + [e] ; (3.14.9)
[R] = [F][U]�1
= ([1] + [H])([1] + [e])�1
' [1] +1
2([H]� [H]T ) : (3.14.10)
The skew symmetric tensor
!AB =1
2(uA;B � uB;A) (3.14.11)
gives the infinitesimal rotation. For any infinitesimal vector PQ we have
fP0Q0g = [F]fPQg = ([1] + [e] + [!])fPQg
69
and therefore
fP0Q0g � fPQg = [e]fPQg+ [!]fPQg : (3.14.12)
Thus the deformation of a line element PQ equals the sum of the deformations caused by the
infinitesimal strain tensor and the infinitesimal rotation tensor. Similarly, if
u = u1 + u2 ; (3.14.13)
then
[e] = [e1] + [e2] ; (3.14.14)
[!] = [!1] + [!2] : (3.14.15)
Thus the infinitesimal strains and rotations caused by a given displacement equal the sum of the
infinitesimal strains and rotations caused by the components of the given displacement.
Exercise: Prove that for small deformations
J = 1 + eAA; (3.14.16)
� = �0(1� eAA); (3.14.17)
and for isochoric deformations
eAA = 0: (3.14.18)
Exercise: Consider the displacement field
uA = k[(2X21 +X1X2)�A1 +X2
2�A2]:
Using both the infinitesimal strain theory and the finite strain theory, find the change in length per
unit length for the material line elementPQ = ds (1; 1; 0) that emanates from the material particle
P (1; 1; 1) in the reference configuration for k = 10�4; 10�3; 10�2; 10�1; 1; 10. Plot these changes
as a function of k.
Exercise: With reference to a rectangular Cartesian coordinate system, the state of strain at a point
70
is given by the matrix.
[e] = 10�4
24 5 3 0
3 4 �10 �1 2
35 :
a) What is the engineering strain in the direction 2e1 + 2e2 + e3 at the point (1; 1; 1) in the
reference configuration?
b) What is the shearing strain between two perpendicular lines (in the reference configuration)
emanating from the point (1; 1; 1) in the directions of 2e1 + 2e2 + e3 and �3e1 + 6e3?
For an infinitesimal rigid body motion, the displacement vector u is given by
uA = cA + bABXB (3.14.19)
in which cA is a constant and bAB = �bBA is a skew-symmetric tensor. For the displacement given
by (3.14.19), eAB = 0. Naturally the following question arises: Is the rigid body motion the only
deformation for which the infinitesimal strain tensor vanishes identically? The answer, as proved
in the following example, is yes.
Example: Regarding eAB = 0 as six partial differential equations, solve for u.
Solution: eAB = 0 corresponds to
@u1@X1
=@u2@X2
=@u3@X3
= 0; (3.14.20)
and
@u1@X2
+@u2@X1
=@u2@X3
+@u3@X2
=@u3@X1
+@u1@X3
= 0: (3.14.21)
By differentiating (3.14.21)1 with respect to X2 and (3.14.21)3 with respect to X3 and making use
of (3.14.20)2;3 we obtain
@2u1@X2
2
=@2u1@X2
3
= 0 :
This when combined with (3.14.20)1 yields
u1 = c1 + b12X2 + b13X3 + a1X2X3; (3.14.22)
71
in which c1; b12; b13 and a1 are constants. By following a procedure similar to that used to obtain
(3.14.22) for u1, we obtain for u2 and u3 the following.
u2 = c2 + b21X1 + b23X3 + a2X1X3; (3.14.23)
u3 = c3 + b31X1 + b32X2 + a3X1X2: (3.14.24)
Substituting from (3.14.22), (3.14.23) and (3.14.24) into (3.14.21) we get
b12 + b21 + (a1 + a2)X3 = 0;
b23 + b32 + (a2 + a3)X1 = 0;
b31 + b13 + (a3 + a1)X2 = 0:
Since these equations hold for all values of X1; X2; X3 which correspond to various points in the
body, therefore,
b12 + b21 = b23 + b32 = b31 + b13 = 0;
a1 + a2 = a2 + a3 = a3 + a1 = 0:
The last set of equations implies that a1 = a2 = a3 = 0. From the other set of equations, it follows
that bAB = �bBA. Hence equations (3.14.22), (3.14.23) and (3.14.24) which are a solution of
eAB = 0 reduce to
uA = cA + bABXB:
Recalling equations (3.14.13) and (3.14.14) we see that the strain caused by displacements u1
and u1 + u2 is the same if u2 is a rigid body motion. To prevent the rigid motion of a body, one
needs to fix three noncolinear points of the body.
The strain field given by eqns. (3.14.20) and (3.14.21) is a very special one in that it is identically
zero throughout the body. What if we were given
e11 = f(X1; X2);
e22 = g(X1; X2);
e12 = h(X1; X2);
(3.14.25)
72
and asked to find the corresponding two-dimensional displacement field? Can we always find a
displacement field that will produce the strains specified by (3.14.25)? The answer is of course no.
Since
e11 =@u1@X1
; e22 =@u2@X2
; 2e12 =@u1@X2
+@u2@X1
;
therefore,
2@2e12
@X1@X2=
@3u1@X2
2@X1+
@3u2@X2
1@X2=@2e11@X2
2
+@2e22@X2
1
:
If we substitute for e11; e22 and e12 from (3.14.25) we obtain
2@2h
@X1@X2=
@2f
@X22
+@2g
@X21
: (3.14.26)
Thus unless the given functions f; g, and h satisfy (3.14.26) we will not be able to find a displace-
ment field that will produce the desired strain field.4 Another way of saying essentially the same
thing is that we have three equations (3.14.25)1;2;3 for two unknowns u1 and u2. Unless the given
expressions for e11, e22 and e12 are related somehow, we will not, in general, be able to find u1
and u2. That relation is the equation (3.14.26) which is known as a compatability condition. In
the three dimensional case we can derive compatability conditions like (3.14.26) in a similar way.
However, another, perhaps neater, approach to the problem is the following.
Given the displacement uP of a point P in the body and the strain field eAB in the neighborhood
of P , we would like to find the displacement uQ of a neighboring point Q. Now
uQA � uPA =
Z Q
P
duA =
Z Q
P
uA;BdXB;
=
Z Q
P
(eAB + !AB)dXB: (3.14.27)
Note that
Z Q
P
!ABdXB =
Z Q
P
!ABd(XB �XQB );
= !AB(XB �XQB )
����Q
P
�ZP
(XB �XQB )!AB;CdXC ; (3.14.28)
4This requirement is in addition to the condition that J > 0 for the given strain field.
73
where we have integrated by parts. Combining (3.14.27) and (3.14.28) we obtain
uQA = uPA � !PAB(X
PB �XQ
B ) +
Z Q
P
RACdXC ; (3.14.29)
where
RAC = eAC � (XB �XQB )!AB;C : (3.14.30)
Since the path of integration in (3.14.29) from point P to point Q is arbitrary, in order to obtain
a unique value of uQ, the integral in eqn. (3.14.29) must be path independent. The necessary and
sufficient condition for this is
RAC;D = RAD;C : (3.14.31)
Substituting from (3.14.30) into (3.14.31) and noting that
!AB;C =1
2(uA;BC � uB;AC) +
1
2(uC;AB � uC;AB) = eAC;B � eBC;A;
we arrive at
eAB;CD + eCD;AB � eAD;BC � eBC;AD = 0: (3.14.32)
Even though there are 81 equations given by (3.14.32) only the following six are non-trivial.
e11;22 + e22;11 � 2e12;12 = 0;
e22;33 + e33;22 � 2e23;23 = 0;
e33;11 + e11;33 � 2e13;13 = 0;
e11;23 + e23;11 � e13;12 � e12;13 = 0;
e22;31 + e31;22 � e21;23 � e23;21 = 0;
e33;12 + e12;33 � e32;31 � e31;32 = 0:
(3.14.33)
Out of these six equations only three are linearly independent. However, we will not prove that
here.
Example: For the two-dimensional small strain theory, the strains for a cantilever beam are given
by
e11 = AX1X2; e22 = ��AX1X2; 2e12 = A(1 + �)(a2 �X22 ); (3.14.31)
74
where A, a, � are positive constants and � � 12. Assume that the displacements u1, u2 relative to
axes X1; X2 are functions of X1; X2.
a) Show that continuous single-valued displacements (u1; u2) are possible.
b) Hence derive formulas for (u1; u2) as explicit functions of (X1; X2) with the conditions u1 =
u2 = u1;2 = 0 for X1 = L; X2 = 0.
Solution: a) From the given expressions for e11; e22 and e12 we obtain
e11;22 = e22;11 = e12;12 = 0
so that the only non-trivial compatability equation (3.14.33)1 is satisfied. J = 1 + e11 + e22 =
1 + A(1 � �)X1X2. For the given constraints on A; a; �; X1 and X2, J > 0. Thus continuous
single-valued displacements are possible.
b) Integrating
@u1@X1
= AX1X2;@u2@X2
= ��AX1X2;
we obtain
u1 =A
2X2
1X2 + f(X2); u2 = ��2AX1X
22 + g(X1):
Substituting for u1 and u2 into
@u1@X2
+@u2@X1
= A(1 + �)(a2 �X22 );
we obtain
df
dX2� �
2AX2
2 + A(1 + �)X22 = A(1 + �)a2 � dg
dX1� A
2X2
1 :
75
Since the left-hand side of this equation is a function of X2 and the right-hand side a function of
X1, for the two sides to be always equal, each must equal a constant say c. Thus
df
dX2� �
2AX2
2 + A(1 + �)X22 = c
A(1 + �)a2 � dg
dX1� A
2X2
1 = c:
Therefore
f =�
6AX3
2 �A
3(1 + �)X3
2 + cX2 + d;
g = A(1 + �)a2X1 � A
6X3
1 � cX1 + e;
where d and e are constants of integration. Thus
u1 =A
2X2
1X2 +�
6AX3
2 �A
3(1 + �)X3
2 + cX2 + d;
u2 = ��2AX1X
22 + A(1 + �)a2X1 � A
6X3
1 � cX1 + e:
(3.14.32)
In order that u1 = u2 = u1;2 = 0 for X1 = L; X2 = 0 we must have
d = 0; c = �A2L2; e = �A(1 + �)a2L+
A
6L3 � A
2L3:
Substituting for d; c and e into (3.14.32) we arrive at the following.
u1 =A
2X2
1X2 � A
3
�1 +
�
2
�X3
2 �A
2L2X2;
u2 = ��2AX1X
22 + A(1 + �)a2(X1 � L) +
A
6(L3 �X3
1 ) +PAL2
2(X1 � L):
(3.14.34)
For points on the plane X2 = 0,
u1 = 0; u2 = A(1 + �)a2(X1 � L) +A
6(L3 �X3
1 ) +AL2
2(X1 � L) : (3.14.33)
Thus longitudinal lines on the plane X2 = 0 which is the neutral surface are not stretched. How-
ever, points on the longitudinal line do move vertically. Equation (3.14.34) corresponds to the
deflection equation usually studied in the first Strength of Materials or Mechanics of Deforms
course. However, in that course the following boundary conditions are used. For
X1 = L; X2 = 0; u1 = u2 = u2;1 = 0:
76
With these, eqn. (3.14.32) gives
u1 =A
2X2
1X2 + A(1 + �)a2X2 � A
3
�1 +
�
2
�X3
2 �AL2
2X2 ;
u2 = ��2AX1X
22 �
A
6X3
1 +AL2
2X1 � AL3
3:
Thus
u2(0; 0) = �AL3
3
which agrees with the strength of materials solution.
Exercise: Check whether or not the following distribution of the state of infinitesimal strain satis-
fies the compatability conditions:
[e] =
24 X2
1 X22 +X2
3 X1X3
X22 +X2
3 0 X1
X1X3 X1 X22
35 :
Exercise: Given the strain field
e12 = e21 = X1X2;
and all other eAB = 0,
a) Does it satisfy the equations of compatability?
b) By attempting to integrate the strain field, show that it cannot correspond to a displacement
field.
4 The Stress Tensor
4.1 Kinetics of a Continuous Media
In this section we will study the laws of motion applicable to a continuous medium similar to
Newton’s laws of motion studied in particle mechanics. We first review Newton’s laws of motion
below.
1. Newton’s first law of motion: a free particle continues in its state of rest or of uniform
motion.
77
2. Newton’s second law of motion: In an inertial frame, the rate of change of linear momentum
of a particle equals the resultant force acting on the particle. That is
�F =d
dt(mv) = ma: (4.1.1)
3. Newton’s third law of motion: To every action there is an equal and opposite reaction.
Newton’s first law of motion defines an inertial frame. That is, an inertial frame is one in which
Newton’s first law of motion holds. Usually, it is taken as a frame attached to the Sun. However,
in most engineering problems, one can take the co–ordinate axes fixed to the earth as an inertial
frame without introducing any appreciable error. Hereafter, we will take an inertial frame as the
frame of reference.
To write the laws of motion for a continuum we note that the linear momentum of the material
enclosed in an infinitesimal volume dv is (� dv)v where � is the mass density and v is the velocity.
Hence the linear momentum of the shaded portion isR(�dv)v in which the integration is over the
shaded region. To find the resultant force acting on this region of interest, we observe that we have
two kinds of forces.
78
Body forces are forces that act on all particles in a body as a result of some external body or ef-
fect not in direct contact with the body under consideration. An example of this is the gravitational
force exerted on a body. This type of force is defined as a force intensity per unit mass or per unit
volume at a point in the continuum. Thus, if the body force per unit mass is gi, then the body force
on the material enclosed in the shaded region will be dm gi.
Surface forces are contact forces that act across a surface of the body, which may be internal
or external. In non-polar continuum mechanics we assume that the action of that part of the body
which is exterior to the shaded region on the body enclosed in the shaded region is equipollent
to a system of forces acting on the bounding surface of the shaded region. The assumption that
the contact force is of this kind is the cut principle of Cauchy: Within the shape of a body at any
given time, conceive a smooth, closed diaphragm; then the action of the part of the body outside
that diaphragm and adjacent to it on that inside is equipollent to that of a field of vectors defined
on the diaphragm. Note that no point moments are assumed to be exerted by one part of the body
on its adjacent part across the common surface. Thus in nonpolar continuum mechanics, moments
are caused by the forces. The contact force at a point P on a surface is usually given as a force f
acting on a unit area surrounding P and lying on the surface. Through a given point in the body,
there are infinitely many surfaces. The intensity of the contact force at the point P on each of these
surfaces will, in general, be different. How does f at the point P depend upon the surface through
P ? In the classical continuum mechanics, it is assumed that the intensity of the contact force on
all surfaces with a common tangent plane at P is the same. That is, f at P is assumed to depend
upon the surface through P only through the oriented normal n of the surface at P .
f = f(xi; nj): (4.1.2)
This is Cauchy’s Postulate. A unit normal to the surface which points out of the body is taken as
positive. Thus f(x;n) is the intensity of the contact force at P which the unshaded portion of the
body exerts on the shaded portion and f(x; �n) is the intesntiy of the contact force at P which
the shaded portion exerts on the unshaded one. The intensity of the contact force is also known
as surface traction or traction or stress vector. Denoting the magnitude of the element of area on
79
the surface by da, the total contact force on the shaded region isRf1da. Thus eqn. (4.1.1) for the
material contained in the shaded region takes the form
d
dt
Zdm vi =
Zfi da+
Zdm gi: (4.1.3)
Equation (4.1.3) is known as the conservation of linear momentum. A similar equation
d
dt
Z"ijkxj(dm vk) =
Z"ijkxjfkda+
Z"ijkxj(dm gk) (4.1.4)
for the moment of momentum is known as the conservation of moment of momentum. Equations
(4.1.3) and (4.1.4) are the BASIC LAWS OF MOTION of Continuum Mechanics, as far as this
course is concerned.
Since ddt(dm) = 0, therefore, equations (4.1.3) and (4.1.4) can also be written as
Zdm ai =
Zfi da+
Zdm gi; (4.1.5)Z
"ijkxj(dm ak) =
Z"ijkxjfkda+
Z"ijkxj(dm gk): (4.1.6)
We now study the dependence of f upon n in some detail. Consider a cylinder of radius R and
height " with the top and bottom faces perpendicular to n and the point P lying on one of the end
faces. We apply the balance of linear momentum (4.1.5) to the material
contained within this cylinder. Using the mean-value theorem of calculus, we obtain
(�R2"�)�ai =
Z
Top
Surface
fi(x;n)da+
Z
Bottom
Surface
fi(x;�n)da +Z
mantle
fi(x;n)da+ (�R2"�)�gi : (4.1.7)
80
Here �ai and �gi denote the values of ai and gi evaluated at some point in the cylinder. Let the height
" of the cylinder go to zero and assume that the fields ai and gi are bounded. Then in the limit, eqn.
(4.1.7) becomes
0 =
Z[fi(x;n) + fi(x;�n)]da:
Since this equation has to hold for all values of R, therefore, the integrand must be zero. That is
fi(x;�n) = �fi(x;n): (4.1.8)
This is known as Cauchy’s Fundamental Lemma and states that f is an odd function of n. We
now show that f is in fact linear in n. Consider a tetrahedron, three sides of which are mutually
orthogonal, the fourth having outward unit normal n.
Let the area of the inclined plane ABC be A. Then the areas of planes PBC, PAC and PAB
are An1, An2 and An3 respectively. On applying eqn. (4.1.5) to the material contained in the
tetrahedron, and using the mean-value theorem, we obtain
�V �ai = �fi(x;n)A+ �fi(x;�e1)An1 + �fi(x;�e2)An2 + �fi(x;�e3)An3 + �V �gi : (4.1.9)
In eqn. (4.1.9), the superimposed bars indicate quantities evaluated at some point in the tetrahedron
or at some point on a plane bounding the tetrahedron. In eqn. (4.1.9), dividing throughout by A,
taking the limit as the tetrahedron shrinks to the point P , and using (4.1.8) we arrive at
fi(x;n) = fi(x; e1)n1 + fi(x; e2)n2 + fi(x; e3)n3 ; (4.1.10)
81
where all stress vectors are evaluated at the point P . Setting n = �p+ �q where p and q are unit
vectors, we obtain
fi(x; �p+ �q) = �[fi(x; e1)p1 + fi(x; e2)p2 + fi(x; e3)p3]
+ �[fi(x; e1)q1 + fi(x; e2)q2 + fi(x; e3)q3] ;
= �fi(x;p) + �fi(x;q) : (4.1.11)
Thus f is a linear function of n and we can write
fi(x;n) = Tij(x)nj: (4.1.12)
By comparing the right-hand sides of (4.1.10) and (4.1.12) we get
Tij(x) = fi(x; ej): (4.1.13)
Thus Ti1; Ti2 and T13 denote, respectively, the surface tractions on planes whose outer normal
points in the positive x1; x2 and x3-directions. Tij is called the stress tensor. The plane whose
outer normal points in the positive x1-direction is simply known as the x1-plane. Thus Ti1; Ti2,
and Ti3 are, respectively, the surface tractions on the x1; x2 and x3-planes. Since
T11(x) = f1(x; e1); T21 = f2(x; e1); T31 = f3(x; e1);
T11; T21 and T31 are, respectively, the normal and shearing stresses on the x1-plane. Note that the
resultant shear stress on the x1-plane ispT 221 + T 2
31. Also the positive values of T11; T21 and T31
point in the positive direction of the axes on the positive x1-plane. Because of (4.1.8), positive
values of T11; T21 and T31 point in the negative direction of the axes on the negative x1-plane.
From equation (4.1.13) it is clear that stress vectors on three mutually perpendicular planes
at a point determine the stress tensor at that point. Because of eqn. (4.1.12) or (4.1.10), stress
vectors on three mutually perpendicular planes at a point also determine the stress vector on any
other plane. This proves Cauchy’s fundamental theorem: From the stress vectors acting on three
mutually perpendicular planes at a point, stress vectors on every plane through the point can be
determined; they are given by (4.1.12) as linear functions of the stress tensor Tij .
82
A glance at equations (4.1.5) and (4.1.6) reveals that one integration in each equation is over
the surface area whereas others are over the region under consideration. We now transform this
surface integral into the volume integral by using the divergence theorem. Note thatZ
fida =
ZTijnjda =
ZTij;jdv;
andZ
"ijkxjfkda =
Z"ijkxjTkpnpda =
Z("ijkxjTkp);p dv ;
=
Z"ijk(�jpTkp + xjTkp;p)dv:
Thus equations (4.1.5) and (4.1.6) can be written asZ
(�ai � Tij;j � �gi)dv = 0;
andZ
"ijk[xj(�ak � Tkp;p � �gk)� Tkj]dv = 0:
Since both these equations must hold for every region in the body, therefore, if the integrand is
continuous throughout the body, then it must vanish. This gives
�ai = Tij;j + �gi; (4.1.14)
"ijkTkj = 0 or Tij = Tji: (4.1.15)
These are Cauchy’s laws of motion. Equation (4.1.14) expresses the balance of linear momentum
and eqn. (4.1.15) the balance of moment of momentum. We remark that these hold in an inertial
frame. Thus on the assumption that the balance of linear momentum is satisfied, the balance of
moment of momentum reduces to the requirement that the stress tensor be symmetric. In classi-
cal continuum mechanics, Tij is always taken to be symmetric so that the balance of moment of
momentum is identically satisfied.
For static problems ai = 0 and eqn. (4.1.14) gives
Tij;j + �gi = 0 (4.1.16)
83
as the three equations of equilibrium. If a given stress field satisfies eqn. (4.1.16), we may or may
not be able to produce that stress field statically in a continuous body. However, if a given stress
field does not satisfy even only one of the three equations of equilibrium, then it certainly cannot
be produced statically in a continuous body.
Example: Show that the following stress field
T11 = x22 + ��x21 � x22
�; T12 = �2�x1x2;
T22 = x21 + ��x22 � x21
�; T23 = T13 = 0;
T33 = ��x21 + x22
�;
satisfies equations of equilibrium with zero body forces.
Solution
T11;1 = 2�x1; T12;2 = �2�x1; T13;3 = 0;
T21;1 = �2�x2; T22;2 = 2�x2; T23;3 = 0;
T31;1 = 0; T32;2 = 0; T33;3 = 0:
Thus
T11;1 + T12;2 + T13;3 = 2�x1 � 2�x1 + 0 = 0;
T21;1 + T22;2 + T23;3 = �2�x2 + 2�x2 + 0 = 0;
T31;1 + T32;2 + T33;3 = 0 + 0 + 0 = 0;
and the equations of equilibrium
Tij;j = 0
with zero body force are satisfied.
Exercise: Suppose that the body force is g = �ge3, where g is a constant. Consider the following
stress tensor
[T] = �
24 x2 �x3 0�x3 0 �x20 �x2 T33
35 :
84
Find an expression for T33 so that Tij satisfies the equations of equilibrium.
Exercise: Suppose that the stress distribution has the form (called plane stress)
[Tij] =
24 T11(x1; x2) T12(x1; x2) 0T12(x1; x2) T22(x1; x2) 0
0 0 0
35 :
(a) What are the equilibrium equations in this special case?
(b) If we introduce a function �(x1; x2) such that
T11 = �;22 ; T22 = �;11 ; T12 = ��;12 ;
will this stress distribution be in equilibrium with zero body force?
4.2 Boundary Conditions for the Stress Tensor
Applied distributed forces on the surface of a body are called surface tractions. We wish to find
the relation between the surface tractions and the stress field defined within the body. This relation
is eqn. (4.1.12) in which the left-hand side represents the surface tractions applied to the bounding
surface of the body. The equation
Tij(x)nj = fi(x)
for points x on the bounding surface of the body is called the boundary condition.
Example: A long prismatic dam is subjected to water pressure that increases linearly with the
depth. The dam has thickness 2b and height h. Write the traction boundary conditions for the
traction-type bounding surfaces of the dam.
85
Solution: At the top surface n = (�1; 0; 0). Since there is no applied force on this surface,
therefore
Tijnj = �Ti1 = 0 on the plane x1 = 0:
On the plane x2 = �b; n = (0;�1; 0). There is no force applied on this surface, therefore,
Tijnj = �Ti2 = 0 on the plane x2 = �b:
On the plane x2 = b; n = (0; 1; 0). The water pressure at any point on this plane exerts a normal
force equal to ��wgx1e2 in which �w is the mass density of water, g is the gravitational constant.
Therefore,
Tijnj = Ti2 = ��wgx1�i2 on the plane x2 = b:
The lower surface of the dam is in contact with the ground and the boundary conditions on it depend
upon whether the ground is taken as deformable or rigid; we will not discuss these boundary
conditions here.
Example: Write the traction boundary conditions at the inner and the outer surfaces of a cylindrical
pressure vessel subjected to an internal pressure p1 and external pressure p2.
Solution: At a point on the inner surface n =��x1
a; �x2
a; 0�. Therefore,
Tijnj = fi = �p1ni
86
simplifies to
T11x1 + T12x2 = �p1x1;
T21x1 + T22x2 = �p1x2;
T31x1 + T32x2 = 0:
At a point on the outer surface n =�x1b; x2b; 0�. Therefore, the traction boundary condition on the
outer surface are
T11x1 + T12x2 = �p2x1;
T21x1 + T22x2 = �p2x2;
T31x1 + T32x2 = 0:
Exercise: Given the following stress distribution
[T] =
24 x1 + x2 T12(x1; x2) 0T12(x1; x2) x1 � 2x2 0
0 0 x2
35� 103;
find T12 in order that the stress distribution is in equilibrium with zero body force, and that the
stress vector on the plane x1 = 1 is given by [(1 + x2)e1 + (5� x2)e2]103.
Exercise: Consider the following stress distribution for a certain circular cylindrical bar
[T ] =
24 0 ��x3 �x2��x3 0 0�x2 0 0
35 ;
where � is a constant.
(a) What is the distribution of the stress vector on the surfaces defined by x22 + x23 = 4; x1 = 0,
and x1 = `?
� Find the total resultant force and moment on the end face x1 = `.
4.3 Nominal Stress Tensor
Solid mechanics problems are easy to formulate in the referential description. However the balance
laws (4.1.14) and (4.1.15) derived earlier are in the spatial description. We note that the stress-
vector f in equations (4.1.5) and (4.1.6) is acting on a unit area in the present configuration of the
87
body. If we can express this f in terms of the area in the reference configuration, we would then be
able to write the balance laws in the referential description. Note that
fida = Tijnjda = Tijdaj = TijJ(F�1)BjdAB (4.3.1)
where we have substituted for dai in terms of dAB from eqn. (3.10.1). With the following definition
SiB = JTij(F�1)Bj; (4.3.2)
equation (4.3.1) becomes
fida = SiBdAB = SiBNBdA: (4.3.3)
Thus SiB provides a measure of the present force acting on a unit area in the reference configu-
ration. For this reason, SiB is called the nominal stress tensor, engineering stress tensor, or first
Piola-Kirchhoff stress tensor. Note that SiBNB equals the present
value of the force acting on the area in the present configuration at P 0 into which a unit area
perpendicular to N at P is deformed.
The stress tensors SiB and Tij are related by eqn. (4.3.2). However, to find one stress tensor
from the other, one must know the deformation gradient. In the Mechanics of Materials course, the
nominal stress tensor SiB and the engineering strain eAB are used to plot the stress strain curve. In
general, the graph of the true stress T11 versus the true strain �11 will look quite different from that
of S11 versus e11 in a simple tension test in which the load is applied in the X1-direction, and x1
and X1 axes point in the same direction.
Substituting for fida from (4.3.3) into (4.1.5) and writing dm = �0dV , we obtainZ�0aidV =
ZSiBNBdA+
Z�0gi dV: (4.3.4)
88
Now using the divergence theorem, we convert the surface integral on the right-hand side of (4.3.4)
into the volume integral and thereby obtainZ
(�0ai � SiB;B � �0gi)dV = 0: (4.3.5)
This equation is to hold for all infinitesimal volume elements in the body. If the integrand is
continuous which is assumed to be the case, then (4.3.5) can hold for every volume element in the
body if and only if
�0ai = SiB;B + �0gi; (4.3.6)
or
�0ai =@SiB@XB
+ �0gi:
This is Cauchy’s first law of motion in the referential description. From (4.1.15) and (4.3.2) we
obtain the following for the second law of motion.
[S][F]T = [F][S]T : (4.3.7)
Thus the nominal stress tensor SiB need not be and, in general, is not symmetric.
The traction type boundary conditions in terms of SiB are written as
SiBNB = fi (4.3.8)
in which fi is the force acting at points on the boundary in the deformed configuration but is
measured per unit area in the reference configuration.
The following stress tensor is introduced in the engineering literature:
~SAB = JF�1Ai Tij(F
�1)Bj
or
[~S] = J [F�1][T][F�1]T = [F�1][S]: (4.3.9)
The tensor ~S known as the second Piola-Kirchhoff stress tensor is symmetric; the symmetry of ~S
is equivalent to the Cauchy’s second law of motion. There is no easy physical interpretation of ~S.
89
4.4 Transformation of Stress Tensor under Rotation of Axes
Let us consider two sets of co-ordinate axes xi and x0i related by eqn. (2.7.4), that is,
e0i = aijej: (2.7.4)
Consider a plane the outer normal to which points in the positive x01 direction. Therefore, a unit
normal n to this plane can be written as
n = e01 = a1jej (4.4.1)
or
ni = a1i: (4.4.2)
The stress vector f 0 on this plane is given by
f 0i = Tijnj = Tija1j: (4.4.3)
Recalling the interpretations of various components T11; T12 etc. of the stress tensor given after
eqn. (4.1.3), we note that T 011; T
021; T
031 equal respectively, the components of f 0 in the positive
90
x01; x02, and x03 directions. Thus
T 011 = f 0 � e01 = a1iTija1j;
T 021 = f 0 � e02 = a2iTija1j; (4.4.4)
T 031 = f 0 � e03 = a3iTija1j:
These three equations can be written as
T 0k1 = akiTija1j: (4.4.5)
Similarly, now consider planes whose outer normals point in the positive x02; x03 directions and
proceed the way it was done to arrive at (4.4.5). The result is
T 0k2 = akiTija2j; (4.4.6)
T 0k3 = akiTija3j: (4.4.7)
Equations (4.4.5), (4.4.6) and (4.4.7) can be collectively written as
T 0kp = akiTijapj; [T
0] = [a][T][a]T : (4.4.8)
A comparison of this equation with eqn. (2.8.4) reveals that T is indeed a second order tensor.
An application of eqn. (4.4.8) is the problem of finding the normal and the shear stress on an
oblique plane when the stress-state on horizontal and vertical planes in the plane-stress problem is
known.
91
For the axes shown,
[a] =
24 cos � sin � 0� sin � cos � 0
0 0 1
35 :
Therefore24 T 0
11 T 012 T 0
13
T 021 T 0
22 T 023
T 031 T 0
32 T 033
35 =
24 cos � sin � 0� sin � cos � 0
0 0 1
3524 T11 T12 0T12 T22 00 0 0
3524 cos � � sin � 0
sin � cos � 00 0 1
35
=
24 T11 cos
2 � + T22 sin2 � + 2T12 sin � cos � —————— 0
T12(cos2 � � sin2 �) + (T22 � T11) sin � cos � T11 sin
2 � + T22 cos2 � � 2T12 sin � cos � 0
0 0 0
35
Thus one can find T 011; T
012, the normal and shear stresses on the oblique plane. Equations relating
T 011; T
012 to T11; T12 are called “stress transformation equations” in the Mechanics of Deformable
Bodies course.
Exercise: The stress matrix T referred to x1x2x3-axes is shown below at the left (in ksi). New
x01x02x
03-axes are chosen by rotating the unprimed axes as shown to the right.
[T] =
24 50 37:5 0
37:5 �25 �500 �50 25
35
x01 x02 x03
x1 0:6 0 0:8
x2 0 1 0
x3 �0:8 0 0:6
a) Determine the traction vectors on each of the new coordinate (x0i) planes in terms of com-
ponents referred to the old (xi) axes. For example, determine f on the x01 plane in the form
f = (�)e1 + (�)e2 + (�)e3.
(b) Now project each of the vectors obtained in (a) onto the three new coordinate axes, and verify
that the nine new components thus obtained for the stress matrix T0 are the same as those
given by the formulae for the transformation of a second-order tensor under the rotation of
axes.
Exercise: The work W done by external forces during the deformation of a continuous body is
given by
W =
Zf � uda+
Z�g � udv;
92
By substituting for f from eqn. (4.1.12), using the divergence theorem and assuming that the body
is deformed quasi-statically (that is, the inertia force �ai is negligible), show that
W =
ZTijui;jdv:
Hence show that for small deformations
W =
ZTijeijdv:
Exercise: The power P of external forces is defined as
P =
Zf � vda+
Z�g � vdv:
By substituting for f from eqn. (4.1.12), using the divergence theorem and Cauchy’s first law of
motion, show that
P = _K +
ZTijDijdv
where
K =1
2
Z�vividv
is the kinetic energy of the body.
4.5 Principal Stresses. Maximum Shear Stress
In stress analysis problems one is interested in finding the maximum and the minimum normal
stress and the maximum shearing stress at a point. In the following discussion, the point x in the
present configuration is kept fixed.
At a point x in the present configuration of the body, consider a plane whose outer unit normal
is n. The stress vector f on this plane given by equation (4.1.12) is Tijnj. The normal stress on
this plane is given by
Tnormal = n � f = niTijnj: (4.5.1)
Thus the problem of finding the maximum or the minimum normal stress at the point x reduces
to finding the unit vector n for which Tnormal is maximum or minimum. Using the method of
93
Lagrange multiplier, the problem becomes that of finding extreme values of the function
F (n) = niTijnj � �(nini � 1) (4.5.2)
in which � is an arbitrary scalar. The vector n which makes F assume a maximum or minimum
value is given by
@F
@ni= 0;
@F
@�= 0: (4.5.3)
That is
(Tij � ��ij)nj = 0; nini � 1 = 0: (4.5.4)
A nontrivial solution of eqn. (4.5.4)1 exists if and only if
det [Tij � ��ij] = 0 (4.5.5)
which gives the cubic equation
�3 � IT�2 + IIT�� IIIT = 0: (4.5.6)
Here IT ; IIT and IIIT are the principal invariants of Tij . Since Tij is symmetric, therefore, eqn.
(4.5.6) has three real eigenvalues �(1); �(2); �(3). Corresponding to each of these eigenvalues we
can find a unit vector n from eqn. (4.5.4). For example, n(1) corresponding to �(1) is given by
(Tij � �(1)�ij)n(1)j = 0; n
(1)j n
(1)j = 1: (4.5.7)
n(1); n(2); n(3) are normals to planes on which Tnormal assumes extreme values. Since
T(1)normal = n
(1)i Tijn
(1)j = n
(1)i �(1)�ijn
(1)j = �(1); (4.5.8)
therefore �(1); �(2) and �(3) are extreme values of normal stresses. Also the stress vector on the
plane with outer unit normal n(1) is given by
fi = Tijn(1)j = �(1)�ijn
(1)j = �(1)n
(1)i : (4.5.9)
Let e a unit vector in this plane. Then the shear stress in the direction of e on the n(1) plane is
given by
Tshear = f � e = �(1)ein(1)i = 0: (4.5.10)
94
The plane on which the shear stress is zero is called a principal plane and the normal stress on
a principal plane is called a principal stress. Thus �(1); �(2) and �(3) are principal stresses and
n(1); n(2) and n(3) are normals to principal planes. n(1); n(2) and n(3) are called principal axes of
the stress.
Whenever �(1) 6= �(2) 6= �(3); n(1); n(2) and n(3) are uniquely determined and are mutually
orthogonal. However, when any two or all three of the principal stresses are equal, then n(1); n(2)
and n(3) are not uniquely determined but can still be found so that they are mutually orthogonal.
Henceforth, in this section, we will assume that the principal axes of the stress tensor are mutually
orthogonal.
Taking x1; x2; x3 axes along the principal axes of the stress, we see that with respect to these
axes Tij has the form
[Tij] =
24 �(1) 0 0
0 �(2) 00 0 �(3)
35 : (4.5.11)
We now find the plane of the maximum shear stress. Let an outer unit normal to this plane be
n. Then, taking principal axes of stress as the coordinate axes,
fi = Tijnj = �(1)n1�i1 + �(2)n2�i2 + �(3)n3�i3 (4.5.12)
gives the traction on this plane. Therefore the shear stress on this plane is given by
(Tshear)2 = fifi � (fini)
2;
= �(1)2
n21 + �(2)2
n22 + �(3)2
n23 � (�(1)n21 + �(2)n22 + �(3)n23)2: (4.5.13)
Thus the problem of determining maximum Tshear reduces to that of finding extreme values of the
function
G(n) = T 2shear + �(nini � 1) (4.5.14)
in which � is a Lagrange multiplier. The necessary conditions are
@G
@ni= 0;
@G
@�= 0;
95
which are equivalent to
�(i)2
ni � (�(1)n21 + �(2)n22 + �(3)n23)(2�(i)i )ni + �ni = 0 (no summation on i) ; (4.5.15)
nini = 1 : (4.5.16)
Three equations given by (4.5.15) for i = 1; 2 and 3 and the eqn. (4.5.16) determine ni and �.
A solution of these equations for which two out of n1; n2 and n3 are zero is not interesting since
this corresponds to finding Tshear on a principal plane which is zero. The solution of (4.5.15) and
(4.5.16) for which n1 6= n2 6= n3 6= 0 exists only if �(1) = �(2) = �(3). In this case too, eqn.
(4.5.13) gives Tshear = 0.
The remaining possibility is that one out of n1; n2 and n3 be zero, say n1 = 0, n2 6= 0 6= n3.
For this case, eqn. (4.5.15) for i = 1 is satisfied identically. For i = 2 and 3 we obtain
�(2)2
n2 � (�(2)n22 + �(3)n23)2�(2)n2 + �n2 = 0;
�(3)2
n3 � (�(2)n22 + �(3)n23)2�(3)n3 + �n3 = 0;
and, of course,
n22 + n23 = 1:
A solution of these is
n2 = n3 = � 1p2: (4.5.17)
For the values of n2 and n3 given by (4.5.17) and n1 = 0, eqn. (4.5.13) gives
T 2shear =
1
4(�(2) � �(3))2;
and hence
Tshear = �1
2(�(2) � �(3)): (4.5.18)
Similarly for n2 = 0, a solution of (4.5.15) and (4.5.16) is n1 = n3 = � 1p2
and the shear stress on
this plane is
Tshear = �1
2(�(1) � �(3)): (4.5.19)
96
For n3 = 0, a solution of (4.5.15) and (4.5.16) is n1 = n2 = � 1p2
and the shear stress on this plane
is
Tshear = �1
2(�(1) � �(2)): (4.5.20)
Thus the extreme values of the shear stress are given by (4.5.18), (4.5.19) and (4.5.20). Note that
on planes of maximum shear stress, the normal stress need not be zero. In fact, the normal stress
on the plane for which
n =
�0;
1p2;
1p2
�is
1
2(�(2) + �(3)):
5 The Linear Elastic Material
5.1 Introduction
So far we have studied the kinematics of deformation, the description of the state of stress and three
basic conservation laws of continuum mechanics, viz., the conservation of mass (eqn. 3.11.8), the
conservation of linear momentum (eqn. 4.1.14), and the conservation of moment of momentum
(eqn. 4.1.15). All these relations are valid for every continuum, indeed no mention was made of
any material in their derivations.
These equations are, however, not sufficient to describe the response of a body to a given load-
ing. We know from experience that under the same loading conditions, identical specimens of
steel and rubber deform differently. Furthermore, for a given body, the deformations vary with the
loading conditions. For example, for moderate loadings, the deformation in steel caused by the
application of loads disappears with the removal of loads. This aspect of the material behavior
is known as elasticity. Beyond a certain loading, there will be permanent deformations, or even
fracture exhibiting behavior quite different from that of elasticity. In this chapter, we shall study
an idealized linear elastic material for which the stress-strain relationship is linear. Using this
stress-strain law, we will then study some dynamic and static problems.
Henceforth in this chapter we will assume that the deformations are small so that eAB is an
adequate measure of strain.
97
5.2 Linear Elastic Solid. Hookean Material
For a linear elastic solid or a Hookean material, it is assumed that the Cauchy stress is a linear
function of the infinitesimal strain. That is,
T11 = C1111e11 + C1112e12 + � � �C1133e33;
T12 = C1211e11 + C1212e12 + � � �C1233e33;
: : : : : : : : : : : : : : : : : : : : : : : : : : : :
T33 = C3311e11 + C3312e12 + � � �C3333e33:
(5.2.1)
We will assume that Tij is symmetric and since eAB is also symmetric, the above six equations can
be written as
Tij = Cijklekl (5.2.2)
in which
Cijkl = Cjikl = Cijlk: (5.2.3)
Since Tij and eij are components of second order tensors,Cijkl are components of a fourth-order
tensor. It is known as the elasticity tensor. It is this tensor which characterizes the mechanical
properties of a particular anisotropic Hookean elastic solid. The anisotropy of the material is
represented by the fact that the components of Cijkl are in general different for different choices
of coordinate axes. If the body is homogeneous, that is, the mechanical properties are the same for
every particle of the body, then Cijkl are constants (i.e. independent of position). We shall only
study homogeneous bodies.
Because of the symmetry relations (5.2.3), the fourth order tensor Cijkl has 36 independent
components. Thus, for a linear anisotropic elastic material, we need no more than 36 material
constants to specify its mechanical properties.
It follows from (5.2.1) that whenever eij = 0; Tij = 0. Thus, in the reference configuration,
there is no stress. This implies that there are no initial stresses present.
A material is said to be isotropic if its mechanical properties can be described without reference
to direction. That is, the components of the elasticity tensor Cijkl remain the same regardless of
98
how the rectangular Cartesian coordinate axes are rotated. In other words,
Cijkl = C 0ijkl (5.2.4)
under all orthogonal transformations of coordinate axes. A tensor having the same components
with respect to every orthonormal basis is known as an isotropic tensor. An example of an isotropic
tensor is �ij. It is obvious that the following three fourth-order tensors are isotropic:
Aijkl = �ij�kl; Bijkl = �ik�jl; Dijkl = �il�jk:
In fact, it can be shown that any fourth order isotropic tensor can be represented as a linear com-
bination of the above three tensors. Thus, for an isotropic, linear elastic material, the stress-strain
law (5.2.2) can be written as
Tij = (�Aijkl + �Bijkl + �Dijkl)ekl:
In order for the symmetry relation (5.2.3)2 to hold, � = �. Thus
Tij = (��ij�kl + ��ik�jl + ��il�jk)ekl;
= �ekk�ij + 2�eij: (5.2.5)
In the preceding equations � and � are constants. Equation (5.2.5) is the constitutive equation
for a linear elastic isotropic material. The two material constants � and � are known as Lame’s
constants. Since eij are dimensionless, � and � are of the same dimensions as the stress tensor,
force per unit area. For a given material, the values of Lame’s constants are to be determined from
suitable experiments.
We now write (5.2.5) in a form usually studied in the Mechanics of Deformable Bodies course.
Taking the trace of both sides of (5.2.5) we obtain
Tii = (3�+ 2�)ekk: (5.2.6)
Assuming that
(3�+ 2�) 6= 0; � 6= 0; (5.2.7)
99
we get
ekk =1
(3�+ 2�)Tkk;
and hence
eij =1
2�
�Tij � �
3�+ 2�Tkk�ij
�: (5.2.8)
To get a physical interpretation of Lame’s constants in terms of Young’s modulus E and Poisson’s
ratio �, we note that, in a simple tension test, with the axial load P applied along X1-axis,
Tij =P
A�i1�j1 ;
where A is the area of cross-section of the prismatic body. Also
e11 =T11E
; � = �e22e11
: (5.2.9)
Substituting for e11 and e22 from (5.2.8) into (5.2.9) and assuming that (�+ �) 6= 0 we arrive at
E =�(3�+ 2�)
�+ �; � =
�
2(�+ �): (5.2.10)
The elimination of � from these two equations gives
� =E
2(1 + �): (5.2.11)
In order for a tensile load to produce extension in the direction of loading, it is clear from (5.2.9)1
that E must be > 0.
Another stress state, called simple shear, is the one for which all stress components except one
pair of off-diagonal elements vanish. In particular, we choose T12 = T21 6= 0. Equation (5.2.8)
gives
e12 =T122�
: (5.2.12)
In order for a simple shear force on X2 plane to produce a sliding of the X2 plane in the direction
of the applied load, e12 and T12 must be of the same sign. For that to be true, � must be > 0.
100
For E and � to be positive, it follows from (5.2.11) that �1 < �. Now consider a stress state
called, hydrostatic stress, for which Tij = �p�ij . For this case, equation (5.2.8) gives
ekk = � 1��+ 2
3��p : (5.2.13)
Since ekk = dv�dVdV
, therefore, in order for the hydrostatic pressure to produce a decrease in volume,
we must have
�+ (2=3)� > 0 : (5.2.14)
Substituting in this equation from (5.2.10), we obtain
2�
3
�1 + �
1� 2�
�> 0 : (5.2.15)
On the assumption that � > 0, � > �1, the inequality (5.2.15) implies that � must be � 1=2.
Rewriting (5.2.13) as
ekk = � 3
2�
1� 2�
1 + �p; (5.2.16)
we see that for � = 1=2; ekk = 0. That is, there is no change in volume. Since for incompressible
materials ekk = 0, therefore, � = 1=2 for incompressible materials.
Equations (5.2.10), (5.2.11) and (5.2.5) yield
Tij =2��
1� 2�ekk�ij + 2�eij : (5.2.17)
For an incompressible material, the first term on the right-hand side of (5.2.17) is of the form %
and hence is indeterminate. It is usually denoted by �p�ij and the constitutive relation for an
incompressible linear elastic material becomes
Tij = �p�ij + 2�eij : (5.2.18)
Here p is called the hydrostatic pressure, and it can not be determined from the strain field. How-
ever, whenever surface tractions are prescribed on at least a part of the boundary, p can be uniquely
determined.
Example:
101
a) For an isotropic Hookean material, show that the principal axes of stress and strain coincide.
b) Find a relation between the principal values of stress and strain.
Solution: Note that eigenvectors of Tij and eij are the principal axes of stress and strain respec-
tively.
Let n(1) be an eigenvector of eij and e(1) be the corresponding eigenvalue. That is
eijn(1)j = e(1)n(1)i :
By Hooke’s law we have
Tijn(1)j = (�ekk�ij + 2�eij)n
(1)j ;
= �ekkn(1)i + 2�eijn
(1)j ;
= �ekkn(1)i + 2�e(1)n
(1)i ;
= (�ekk + 2�e(1))n(1)i : (�)
Therefore, n(1) is also an eigenvector of Tij and the corresponding eigenvalue is �ekk + 2�e(1).
b) It is clear from eqn. (�) that the eigenvalue T (1) of Tij corresponding to the eigenvector n(1) is
�ekk + 2�e(1). Since ekk = e(1) + e(2) + e(3), therefore,
T (1) = �(e(1) + e(2) + e(3)) + 2�e(1) :
Similarly,
T (2) = �(e(1) + e(2) + e(3)) + 2�e(2);
T (3) = �(e(1) + e(2) + e(3)) + 2�e(3) :
Exercise: (Recall the exercise given on page 2-11). For W = �eijeij +�2(ekk)
2, show that @W@eij
=
Tij .
Hint: It is advisable to do the problem long-hand. That is, first expand the given expression for W .
Substitute e21 = e12 = (e12 + e21)=2 etc. in it and then carry out the differentiation with respect to
e11; e22; e12 etc.
W is called the stored energy function or the strain energy density.
102
5.3 Equations of the Infinitesimal Theory of Elasticity
We shall consider only the case of small strains, and infinitesimal velocities and accelerations as
compared to some reference values. Thus every particle is always in a small neighborhood of the
reference configuration. The reference configuration in which Tij = 0 is also called a natural state.
Thus, if xi denotes the position in the natural state of a typical material particle, we assume that
xi ' Xi, and that the magnitude of the components of the displacement gradient @ui@Xj
is much
smaller than unity. Since
xi = Xi + ui;
we have
@Tij@X1
=@Tij@xk
@xk@X1
;
=@Tij@x1
�1 +
@u1@X1
�+@Tij@x2
@u2@X1
+@Tij@x3
@u3@X1
;
' @Tij@x1
:
Similarly,
@Tij@X2
� @Tij@x2
;@Tij@X3
� @Tij@x3
:
Therefore, for small deformations,
@Tij@Xk
� @Tij@xk
: (5.3.1)
Since
ai =@2xi@t2
����Xj��xed
=@2ui@t2
����Xj��xed
and
� = �0(1� ekk) ;
we have
�ai = �0(1� ekk)@2ui@t2
����Xj��xed
;
� �0@2ui@t2
:
(5.3.2)
103
Also
�gi ' �0gi :
Thus the balance laws or conservation laws for small deformations of a body take the following
form.
Balance of mass: � = �0(1� ekk) (5.3.3)
Balance of linear momentum: �0@2ui@t2
= �0gi +@Tij@Xj
: (5.3.4)
For a linear elastic homogeneous and isotropic body, substitution from (5.2.5) and (3.14.3) into
(5.3.4) gives
�0@2ui@t2
= �0gi + (�+ �)uk;ki + �ui;jj: (5.3.5)
These are three equations for the three unknowns u1; u2, and u3. After a solution of (5.3.5)
has been obtained, one can find the present mass density from eqn. (5.3.3). Note that eqn. (5.3.5)
is a system of three coupled partial differential equations. In order to find a solution of (5.3.5)
applicable to a given problem, side conditions such as initial conditions and boundary conditions
are needed.
In a dynamic problem, one needs the values of ui(Xj; 0) and _ui(Xj; 0). That is, the initial
displacement and the initial velocity field should be given as smooth functions throughout the
body. Note that these initial conditions are not needed in a static problem. However, in both static
and dynamic problems one needs boundary conditions which can be one of the following three
types. In the boundary condition of traction the stress vector is prescribed at the boundary points
of the body. That is, at the points on the boundary
Tijnj = �uk;kni + �(ui;j + uj;i)nj = fi(X; t); (5.3.6)
in which fi is a known function.
In a displacement type boundary condition, displacements are prescribed on the boundary
points. For example, a part of the boundary of a body could be glued to a rigid support. In this case,
104
displacements for these boundary points will be zero. The third type of boundary condition is the
one in which surface tractions are prescribed on one part and the displacements on the remainder
or at a boundary point, tangential components of the stress-vector and the normal component of the
displacement vector (or vice-versa) are prescribed. These are known as the mixed type boundary
conditions.
5.4 Principle of Superposition
Let u(1)i and u(2)i be two possible displacement fields corresponding to the body force fields g(1)i
and g(2)i and surface tractions f (1)i and f
(2)i respectively. Let T (1)
ij and T(2)ij be the stress fields
corresponding to displacement fields u(1)i and u(2)i . Then
�0@2u
(1)i
@t2= �0g
(1)i + (�+ �)u(1)k;ki + �u(1)i;jj; (5.4.1)
�0@2u
(2)i
@t2= �0g
(2)i + (�+ �)u
(2)k;ki + �u
(2)i;jj; (5.4.2)
either u(1)i = h(1)i (Xj; t); or T (1)
ij nj = f(1)i ; on the boundary; (5.4.3)
either u(2)i = h(2)i (Xj; t); or T (2)ij nj = f (2)i ; on the boundary: (5.4.4)
Adding (5.4.1) to (5.4.2), and (5.4.3) to (5.4.4) we obtain
�0@2(u
(1)i + u
(2)i )
@t2= �0(g
(1)i + g
(2)i ) + (�+ �)(u
(1)k + u
(2)k );ki + �(u
(1)i + u
(2)i );jj ;
either u(1)i + u(2)i = h
(1)i + h
(2)i ; or (T (1)
ij + T(2)ij )nj = f
(1)i + f
(2)i on the boundary:
Thus u(1)i + u(2)i is a possible motion for the same linear elastic body corresponding to the body
force g(1)i + g
(2)i and surface tractions f (1)i + f
(2)i . This is the principle of superposition and is
frequently used in the Mechanics of Materials course when solving problems for the combined
loads.
One application of this principle in linear elastic problems is that in a given problem, we shall
often assume that the body force is absent having in mind that its effect, if not negligible, can
always be obtained separately and then superposed onto the solution of the problem with vanishing
body force.
105
5.5 A Uniqueness Theorem
Equations (5.3.5) subject to given initial and boundary conditions have a unique solution.
Assume that there exist two solutions u(1)i and u(2)i of eqn. (5.3.5) subject to the same initial and
boundary conditions. Then
wi = u(1)i � u(2)i will be a solution of
�0@2wi
@t2= (�+ �)wk;ki + �wi;jj; (5.5.1)
and, on the boundary, either
T �ijnj = �wk;kni + �(wi;j + wj;i)nj = 0 (5.5.2)
or
wi = 0; (5.5.3)
and
wi(Xj; 0) = 0; _wi(Xj;0) = 0: (5.5.4)
Taking the inner product of (5.5.1) with _wi(Xj; t), integrating the resulting equation over the
region R occupied by the body in the reference configuration and by using the divergence theorem,
we arrive at
ZT �ijnj _widA�
ZT �ij _wj;idV =
d
dt
Z�0
_wi _wi
2dV: (5.5.5)
The first integral vanishes because of (5.5.2) and (5.5.3). The second integral can be rearranged to
read
d
dt
Z ��
2wk;kwi;i + �e�ije
�ij
�dV;
in which e�ij = (wi;j + wj;i)=2. Thus eqn. (5.5.5) becomes
d
dt
Z ��
2e�kke
�ii + �e�ije
�ij +
�02
_wi _wi
�dV = 0: (5.5.6)
106
Integrating this and making use of the initial conditions (5.5.4) for wi, we obtain
Z ��
2e�kke
�ii + �e�ije
�ij +
�02
_wi _wi
�dV = 0: (5.5.7)
Noting that the integrand can be written as
��
2+�
3
�e�kke
�ii + �e�dij e
�dij +
�02
_wi _wi (5.5.8)
in which e�dij = e�ij � e�kk�ij3
, we see that every term in (5.5.8) is positive provided that
3�+ 2� > 0; � > 0: (5.5.9)
We will henceforth assume that � and � satisfy (5.5.9). Thus for eqn. (5.5.7) to hold,
e�dij = 0; e�ii = 0; _wi = 0: (5.5.10)
Since wi = 0 initially, therefore, wi(Xj; t) = 0 which implies that u(1)i (Xj; t) = u(2)i (Xj; t).
Thus if somehow one can find a solution of eqn. (5.3.5) that satisfies the prescribed initial and
boundary conditions, then that is the only solution of eqn. (5.3.5). There are very few dynamic
problems that have been solved.
In a static problem or, more appropriately, in a quasi-static problem, the left-hand side of eqn.
(5.3.5) becomes zero and one needs only the prescribed boundary conditions. Thus the difference
solution wi = u(1)i � u
(2)i will satisfy
0 = (�+ �)wk;ki + �wi;jj; (5.5.11)
and either (5.5.2) or (5.5.3) on the boundary. Taking the inner product of (5.5.11) with wi, inte-
grating the resulting equation over R, and using the divergence theorem we arrive at
ZT �ijnjwidA�
ZT �ijwj;idV = 0: (5.5.12)
The first integral vanishes because of (5.5.2) and (5.5.3) and the second integral can be written as
Z ���
2+�
3
�e�kke
�ii + �e�dij e
�dij
�dV = 0: (5.5.13)
107
For this equation to hold,
e�dij = 0; e�ii = 0;
and hence
e�ij = 0: (5.5.14)
A solution of eqn. (5.5.14) given in Section 3.14 is
wi = ai + bijXj (5.5.15)
in which ai and bij = �bji are constants. If displacements are prescribed at three noncolinear
points on the boundary, then wi = 0 and the solution of the given boundary value problem is
unique. However, if surface tractions are prescribed on all of the boundary, then different solutions
of the same boundary value problem can differ at most by a rigid body motion. Even though the
two displacement fields differ by a rigid body motion, the stress fields and the strain fields obtained
from such displacement fields are identical.
5.6 Compatibility Equations Expressed in terms of the Stress Components for an Isotropic,Homogeneous, Linear Elastic Solid
In static problems where the surface tractions are prescribed on the entire boundary, it is convenient
to solve the equations of equilibrium
Tij;j + �0gi = 0 (5.6.1)
in terms of the stress components. Having obtained Tij which satisfy (5.6.1) and the assigned
boundary conditions, we solve for strains from eqn. (5.2.8). In order that eij give a unique dis-
placement field ui; eij must satisfy the compatibility conditions (3.14.32). Substitution for eij
from (5.2.8) into (3.14.32) yields
1
2�[Tij;kp + Tkj;ij � Tip;jk � Tjk;ip]
� �
2�(3�+ 2�)[�ijTmm;kp + Tmm;ij�kp � Tmm;ip�jk � Tmm;jk�ip] = 0: (5.6.2)
108
There are only 6 independent equations as there were only 6 independent equations expressed by
(3.14.32). In (5.6.2) set k = p and sum with respect to the common index to obtain
Tij;kk + Tkk;ij � Tik;jk � Tjk;ik =�
3�+ 2�[Tmm;kk�ij + Tmm;ij]: (5.6.3)
Out of these 9 equations only six are independent because Tij = Tji. Consequently, in combining
linearly some of the equations (5.6.2), the number of independent equations has not been reduced,
and hence the resulting set (5.6.3) of equations is equivalent to the original set (5.6.2).
In solid mechanics problems, it is usual to neglect the effect of gravity and write (5.6.1) as
Tij;j = 0: (5.6.4)
Whenever (5.6.4) holds, the third and fourth terms on the left hand side of (5.6.3) vanish and it
simplifies to
Tij;kk +2(�+ �)
3�+ 2�Tkk;ij � �
3�+ 2�Tmm;kk�ij = 0: (5.6.5)
Setting i = j and summing over the repeated index, we get Tmm;kk = 0 and hence (5.6.5) reduces
to
Tij;kk +2(�+ �)
3�+ 2�Tkk;ij = 0: (5.6.6)
These are the compatibility equations in the absence of body forces. Thus for a stress field to be
a solution of a static problem with zero body force for a homogeneous and isotropic linear elastic
body, it must satisfy
(a) equations of equilibrium (5.6.4),
(b) compatibility conditions (5.6.6), and
(c) the appropriate boundary conditions.
Note that if Tij satisfies (a) and (c) but not (b), then that Tij will not correspond to a stress field
in a linear elastic body since eij corresponding to such a Tij will not result in a unique ui.
109
Exercise: Can the following stress field represent a possible solution of a static problem with zero
body force for a homogeneous and isotropic linear elastic body?
T11 = c(X22 + �(X2
1 �X22 )); c = constant 6= 0;
T22 = c(X21 + �(X2
2 �X21 ));
T33 = c�(X21 +X2
2 );
T12 = � 2c�X1X2; T23 = T31 = 0:
Example: For a plane stress state, express the compatibility conditions in terms of a stress function.
Solution: Recall the second exercise given on page 4-9. For a plane stress state,
T11 = �;22 ; T22 = �;11 ; T12 = ��;12 ;
� = �(X1; X2) ; T13 = T23 = T33 = 0:(5.6.7)
As was proved in that exercise, the stress field thus obtained from � satisfies the equilibrium equa-
tions (5.6.4) for every choice of the function �. For this state of stress there is only one independent
compatibility condition. This we obtain from (5.6.6) by setting i = j and summing over the re-
peated index. The result is
Tii;kk = 0: (5.6.8)
Now substitution from (5.6.7) into (5.6.8) gives
(�;22 + �;11);11 + (�;22 + �;11);22 = 0
or
r2(r2�) = r4� = 0: (5.6.9)
Thus the problem of solving a static plane stress problem reduces to finding a solution of (5.6.9)
under the appropriate boundary conditions.
Exercise: Show that for plane strain problems (i.e., those for which u1 = u1(X1; X2); u2 =
u2(X1; X2); u3 = 0) with zero body force there is only one independent compatibility condition in
terms of the components of the stress and this can be written as
r2(T11 + T22) = 0: (5.6.10)
110
Exercise: (a) Simplify the equations of equilibrium with zero body force for plane strain problems.
(b) Introduce a function
�(X1; X2) such that T11 = �;22; T22 = �;11; T12 = ��;12: (5.6.11)
Will this stress distribution satisfy the equilibrium equations obtained in part (a)?
Note that for plane strain case T33 need not and, in general, will not vanish. If we substitute the
stress field given by (5.6.11) into (5.6.10) we arrive at (5.6.9). Thus the task of finding a solution
of a static plane strain problem also reduces to that of finding a solution of (5.6.9) which satisfies
the pertinent boundary conditions.
5.7 Some Examples
(a) Vibration of an Infinite Plate
Consider an infinite plate bounded by the planes X1 = 0 and X1 = `. The vibrations of the
plate are caused by a prescribed motion of these bounding planes or by prescribed surface tractions
on these planes. Since we study steady state vibration of the plate, the initial displacement and
velocity fields in the plate are not required. We will neglect the effect of gravity in these problems.
We begin by assuming that the displacement field in the plate is of the form
u1 = u1(X1; t); u2 = u3 = 0: (5.7.1)
For this displacement field, the equations of motion (5.3.5) reduce to the following equation
�0@2u1@t2
= (�+ 2�)@2u1@X2
1
; (5.7.2)
which can be rewritten as
@2u1@t2
= C2L
@2u1@X2
1
; C2L =
�+ 2�
�0: (5.7.3)
A steady state vibration solution to this equation is of the form
u1 = (A cos�X1 +B sin�X1)(C cosCL�t+D sinCL�t); (5.7.4)
where the constants A; B; C; D and � are determined by boundary conditions. This vibration
mode is sometimes termed a “thickness stretch” because the plate is being stretched through its
111
thickness. It is analogous to acoustic vibration of organ pipes and to the longitudinal vibration of
slender rods.
Another vibration mode can be obtained by assuming the displacement field
u2 = u2(X1; t); u1 = u3 = 0: (5.7.5)
In this case, the displacement field must satisfy the equation
C2T
@2u2@X2
1
=@2u2@t2
; C2T =
�
�; (5.7.6)
and the solution is of the same form as in the previous case.
This vibration is termed “thickness shear” and it is analogous to the motion of a vibrating
string.
Example 5.7.1: (a) Find the thickness-stretch vibration of a plate, where the left face (X1 = 0) is
subjected to a forced displacement ui = (� cos!t)�i1 and the right face (X1 = `) is fixed.
(b) Determine the values of ! that give resonance.
Solution: (a) The boundary condition on the left face X1 = 0 and eqn. (5.7.4) give
� cos!t = A(C cosCL�t +D sinCL�t):
Therefore,
AC = �; � =!
CL; D = 0:
The second boundary condition gives
0 = u1(`; t) =
�� cos
!`
CL+BC sin
!`
CL
�cos!t:
Therefore,
BC = �� cot!`
CL
and the vibration is given by
u1(X1; t) = �
2664cos !
CLX1 �
sin!
C1X1
tan!`
CL
3775 cos!t:
112
(b) Resonance is indicated by unbounded displacements. This occurs in part (a) for forcing fre-
quencies corresponding to tan!`
CL
= 0, that is, when
! =n�CL
`; n = 1; 2; 3; : : :
Example 5.7.2 (a) Find the thickness-shear vibration of an infinite plate which has an applied
surface traction fi = �� cos!t�i2 on the plane X1 = 0 and is fixed on the plane X1 = `.
(b) Determine the resonant frequencies.
Solution: On the plane X1 = 0, n = (�1; 0; 0). Therefore
fi = Tijnj = �Ti1
gives
T21��X1=0
= � cos!t:
This shearing stress forces a vibration of the form
u2 = (A cos�X1 +B sin�X1)(C cosCT�t+D sinCT�t):
Using Hooke’s law (5.2.5), we have
T12 = 2�e12 = �@u2@X1
:
Therefore
�@u2@X1
����X1=0
= � cos!t
�(B�)(C cosCT�t+D sinCT�t) = � cos!t:
Thus
D = 0; � =!
CT; BC =
�CT
�!:
The boundary condition at X1 = ` gives u2(`; t) = 0.
) (A cos�`+B sin�`)C cosCT�t = 0:
113
Hence
A = �B tan�`:
Thus
u2(X1; t) =�CT
!�
�sin
!
CTX1 � tan
!`
CTcos
!
CTX1
�cos!t:
(b) Resonance occurs for
tan!`
CT=1
or
! =n�CT
2`; n = 1; 3; 5; : : :
(B) Torsion of a Circular Shaft
Consider elastic deformations of a cylindrical bar of circular cross-section of radius a and length
L, that is being twisted by an end moment Mt at the right end and is fixed at the left end. We
choose the X3-axis to coincide with the axis of the cylinder and the left-hand and right-hand faces
to correspond to the planes X3 = 0 and X3 = L respectively.
This problem involves the solution of equilibrium equations
(�+ �)uj;ji + �ui;jj = 0 (5.7.7)
114
subject to the boundary conditions
Tijnj = 0 on the lateral surface,RTijnjdA = 0 on the plane X3 = L;R"ijkXjTkpnpdA = Mt�i3 on the plane X3 = L;
ui = 0 on the plane X3 = 0:
9>>>>>=>>>>>;
(5.7.8)
Note that if the problem is formulated in terms of displacements, then compatibility conditions
(5.6.6) are not needed. We use St. Venant’s semi-inverse method to solve the problem in an inverse
way. That is, we make a kinematic assumption about the displacement field and then ensure that
eqns. (5.7.7) and (5.7.8) are satisfied.
Because of the symmetry of the problem, it is reasonable to assume that the motion of each
cross-sectional plane induced by the end moments is a rigid body motion about the X3-axis. This
motion is similar to that of a stack of coins in which each coin is rotated by a slightly different
angle than the previous coin. We will see that this assumption results in a displacement field that
satisfies (5.7.7) and (5.7.8). To ensure that the deformations are small, we will assume that the
angle � of rotation of any section with respect to the left-end is very small as compared to one.
Under the preceding assumptions, the displacement of any point can be calculated as follows.
u1 = r cos(� + �)� r cos �
' ��X2;
u2 = r sin(� + �)� r sin�
' �X1;
' ��X2;
u3 = 0:
For this displacement field to satisfy the equations of equilibrium (5.7.7), we must have
d2�
dX23
= 0:
Thus
d�
dX3= � = constant: (5.7.9)
115
That is, the angle of twist per unit length is the same over the entire length of the shaft. From
(5.7.9) and recalling that � = 0 at X3 = 0, we get
� = �X3
and, therefore,
u1 = ��X2X3; u2 = �X1X3; u3 = 0: (5.7.10)
The strains and stresses associated with these displacements can be calculated from eqns. (3.14.3)
and (5.2.5). The non-zero components of strains are
e23 = e32 = �x1=2;
e31 = e13 = ��X2=2;(5.7.11)
and the non-zero components of stress are
T13 = T31 = ���X2;
T23 = T32 = ��X1:(5.7.12)
To see whether the boundary conditions are satisfied or not, we note that on the lateral surface
a unit outer normal n =1
a(X1; X2; 0). Therefore, on the lateral surface,
Tijnj = Ti1X1
a+ Ti2
X2
a;
and the calculated stress field does satisfy (5.7.8)1. On the end plane X3 = L, n = (0; 0; 1) and
in order that (5.7.8)2 be satisfied,RTi3dA = 0 which is true because
RX1dA = 0,
RX2dA = 0.
The boundary condition (5.7.8)3 requires that, on the end plane X3 = L,
Z"ijkXjTk3dA = Mt�i3 :
This gives
��
Z(X2
1 +X22 )dA = Mt ;
or
��Jp = Mt; (5.7.13)
116
in which Jp is the polar moment of inertia and equals �a4=2. The boundary condition (5.7.8)4 is
clearly satisfied by the displacement field (5.7.10).
Substituting for � from (5.7.13) into (5.7.10), (5.7.11) and (5.7.12) we obtain the displace-
ment components, non-zero strain components and non-zero stress components at any point in the
cylindrical bar. In terms of the twisting moment Mt, the stress tensor becomes
[Tij] =Mt
Jp
2664
0 0 �X2
0 0 X1
�X2 X1 0
3775 (5.7.14)
At any point (X1; X2; b) of a cross-section X3 = b; n = (0; 0; 1) and the stress vector f is given
by
fi = Tijnj = ��(�X2�1i +X1�2i) :
Note that f lies in the plane X3 = b implying thereby that there is no normal stress at any point on
the plane X3 = b. Also f is perpendicular to the radius vector joining the point (X1; X2; b) with
the center (0; 0; b) of the cross-section. The magnitude of f is
jf j = ��r =Mt
Jpr; r =
qX2
1 +X22 :
From this we see that the maximum stress is a tangential stress that acts on the boundary of the
cylinder and has the magnitudeMt
Jpa.
(c) Torsion of Non-Circular Cylinders
For cross-sections other than circular, the stress field (5.7.14) does not satisfy the boundary
conditions of zero tractions on the mantle of the cylinder. We will see that in order for this boundary
condition to be satisfied, the cross-sections will not remain plane.
We begin by assuming a displacement field that still rotates each cross-section by a small angle
�, but in addition there may be a displacement in the axial direction. This warping of a cross-
sectional plane will be defined by u3 = � (X1; X2). Thus we assume that each cross-section is
warped in the same way. For this to be true the left end of the cylinder can not be fixed to a rigid
flat support but is subjected to a torque equal and opposite to that applied to the right end. The
rigid motion is removed by applying suitable constraints.
117
The assumed displacement field has the form
u1 = �X2�(X3); u2 = X1�(X3); u3 = �(X1; X2): (5.7.15)
For this displacement field to satisfy the equations of equilibrium (5.7.7), we must have
d2�
dX23
= 0;
@2�
@X21
+@2�
@X22
= 0: (5.7.16)
Therefore, the angle � of twist per unit length is the same over the entire length of the cylinder.
Setting � = 0 at X3 = 0, we get � = �X3 and therefore the displacement field (5.7.15) can be
rewritten as
u1 = ��X2X3; u2 = �X1X3; u3 = ��(X1; X2); (5.7.17)
where � = ��. For equilibrium, � must satisfy
@2�
@X21
+@2�
@X22
= r2� = 0: (5.7.18)
Stresses corresponding to the displacements (5.7.17) are
T11 = T22 = T33 = T12 = 0;
T23 = ��
�@�
@X2
+X1
�; T13 = ��
�@�
@X1
�X2
�:
(5.7.19)
Since the bar is cylindrical, the unit normal to the lateral surface has the form n = n1e1 + n2e2
and the associated surface traction is given by
fi = ��
��@�
@X1
�X2
�n1 +
�@�
@X2
+X1
�n2
��i3:
For the lateral surface to be traction free, � must satisfy
d�
dn=
@�
@X1n1 +
@�
@X2n2 = X2n1 �X1n2; (5.7.20)
on the boundary.
In order that the boundary condition (5.7.8)2 be satisfied,Z
T13dA = 0;
ZT23dA = 0:
118
NowZ
T13dA = ��
Z �@�
@X1
�X2
�dA;
= ��
Z �@
@X1
�X1
�@�
@X1�X2
��+
@
@X2
�X1
�@�
@X2+X1
���dA;
= ��
IX1
��@�
@X1�X2
�n1 +
�@�
@X2+X1
�n2
�ds;
= 0
because of (5.7.20). Using a similar argument one can show thatRT23dA = 0. The boundary
condition (5.7.8)3 gives
Mt =
Z(X1T23 �X2T13)dA;
= ��
Z �X2
1 +X22 +X1
@�
@X2�X2
@�
@X1
�dA;
= D�; (5.7.21)
where
D = �
Z �X2
1 +X22 +X1
@�
@X2�X2
@�
@X1
�dA: (5.7.22)
The formula (5.7.21) shows that the twisting moment or torque Mt is proportional to the angle of
twist per unit length, so that the constant D provides a measure of the rigidity of a bar subjected to
torsion. D is called the torsional rigidity of the bar.
Hence the torsion problem is solved once � is determined by solving (5.7.18) and (5.7.20).
For an elliptic cylindrical bar with the cross-section given by
X21
a2+X2
2
b2= 1; (5.7.23)
we see that on the boundary,
n =1
F
�X1
a2e1 +
X2
b2e2
�; F 2 =
X21
a4+X2
2
b4; (5.7.24)
and, therefore, eqn. (5.7.20) becomes
@�
@X1X1b
2 +@�
@X2X2a
2 = X1X2(b2 � a2): (5.7.25)
119
This suggests that
� =b2 � a2
b2 + a2X1X2: (5.7.26)
This choice of � satisfies (5.7.25) and (5.7.18). Substituting for � in (5.7.22) and then the result in
(5.7.21) we arrive at
Mt =2�
b2 + a2[b2I22 + a2I11]�
in which I11 and I22 are, respectively, the second moments of area aboutX1 and X2 axes. Recalling
that for an ellipse I11 = �ab3=4; I22 = �a3b=4, we obtain
� = Mtb2 + a2
�a3b3�: (5.7.27)
Substitution for � into (5.7.17) and (5.7.19) yields
u1 = ��X2X3; u2 = �X1X3; u3 = �b2 � a2
b2 + a2X1X2; (5.7.28)
and
T11 = T22 = T33 = T12 = 0;
T23 = ��
�2b2
b2 + a2
�X1; T13 = ��
� �2a2b2 + a2
�X2: (5.7.29)
At any point (X1; X2; c) of a cross-section X3 = c; n = (0; 0; 1) and the stress vector f is given
by
fi = Tijnj =2��
b2 + a2(�a2X2�1i + b2X1�2i):
Since f lies in the plane X3 = c, therefore, there is no normal stress at the point on the plane
X3 = c. The magnitude of f gives the shear stress � at the point.
� =2��
a2 + b2(a4X2
2 + b4X21 )
1=2: (5.7.30)
A point in the interior of the cross-section where � takes on extremum values is given by
@�
@X1=
@�
@X2= 0:
120
This gives X1 = X2 = 0 and at this point � = 0. To find points on the boundary where � may
have extremum values we first write (5.7.30) as
� =2��
a2 + b2(a2(a2 � b2)X2
2 + b4a2)1=2: (5.7.31)
Since a2 > b2, it is obvious that � is minimum at X2 = 0 and has the value2��ab2
a2 + b2. � is maximum
at X2 = �b and has the value2��a2b
a2 + b2. Thus, the maximum shear stress occurs at the extremities
of the minor axis of the ellipse, contrary to the intuitive expectation that the maximum shear stress
would occur at points of maximum curvature.
It is clear from (5.7.28)3 that the axial displacement of points in the first and third quadrant
will be along the �X3 axis and that of points in the second and fourth quadrant will be along
the X3-axis. The points which have the same displacement in X3-direction lie on a rectangular
hyperbola.
6 The Linear Viscous Fluid
6.1 Constitutive Relation
For a viscous material, the Cauchy stress tensor depends not only on the deformation gradient but
also on its time derivative. That is
T = T(F; _F): (6.1.1)
Most fluids are isotropic and homogeneous; an exception being a liquid crystal which is anisotropic.
Here we will study isotropic fluids only. For an isotropic fluid, equation (6.1.1) reduces to
T = T(�;D); (6.1.2)
and for a linear viscous fluid, we have
T = �p(�)1 + �(trD)1+ 2�D: (6.1.3)
Here p is the hydrostatic pressure, � the bulk viscosity and � the shear viscosity. The word linear in
linear viscous fluid signifies that the viscous part of the stress (the last two terms on the right-hand
121
side of (6.1.3)) depends linearly upon the strain-rate tensor D. A perfect fluid or an ideal gas has
no viscosity, and therefore can be described by the constitutive relation
T = �p(�)1: (6.1.4)
Thus a perfect fluid is a nonlinear elastic material in the sense that the Cauchy stress for it depends
only on the present value of the deformation gradient. For an ideal gas,
p = RT� (6.1.5)
where R is the universal gas constant and T is the temperature in degrees Kelvin. Thus for defor-
mations of an ideal gas at a constant temperature
T = const. �1 (6.1.6)
and the value of the const. depends upon the gas and its temperature.
Equation (6.1.3) for D = 0 gives
T = �p(�)1: (6.1.7)
Hence in a fluid at rest, the state of stress is a hydrostatic pressure. Note that the state of stress
in a perfect fluid or an ideal gas is always that of hydrostatic pressure whether or not it is being
deformed. However, in a viscous fluid, no shear stress exists if and only if it is at rest. This is
sometimes taken as the definition of a viscous fluid; viz. a viscous fluid at rest can not support any
shear stresses. On the other hand, a solid body when subjected to shear or tangential tractions can
stay stationary.
For a homogeneous fluid the viscosities � and � are constants. Equation (6.1.3) implies that the
principal axes of the stretching tensor or the strain-rate tensor D coincide with the principal axes
of the stress tensor T.
Whereas the constitutive relation (5.2.1) or (5.2.2) for a linear elastic material describes well
its infinitesimal deformations, no such restriction is imposed on (6.1.3). Said differently, equation
(6.1.3) describes deformations of a viscous fluid for all values of the stretching tensor D. The
material characterized by equation (6.1.3) is called a Navier-Stokes fluid. Usually fluids and gases
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are assumed to be incompressible. The constitutive relations for an incompressible ideal gas and
an incompressible Navier-Stokes fluid are respectively
T = � p1; (6.1.8)
T = � p1+ 2�D; (6.1.9)
where the hydrostatic pressure p is not determined by the deformation of the fluid. It is, however,
determined by the boundary conditions. Even for a homogeneous fluid, the pressure p is a function
of the spatial coordinate x and time t. In (6.1.9) we have used the continuity condition, trD = 0,
for an incompressible fluid.
6.2 Formulation of an Initial-Boundary-Value Problem
One generally uses the spatial description of motion for a fluid. Thus the balance of mass, and the
balance of linear momentum for a compressible Navier-Stokes fluid are
@�
@t+@(�vi)
@xi= 0; (6.2.1)
�
�@vi@t
+@vi@xj
vj
�= �@p
@�
@�
@xi+ (�+ �)
@2vk@xi@xk
+ �@2vi
@xk@xk+ �gi: (6.2.2)
Equation (6.2.2) is obtained by substituting into (4.1.14) for the acceleration from (3.6.3) and for
the Cauchy stress from (6.1.3). Because of the presence of the convective part of the acceleration
on the left-hand side of (6.2.2), equations governing the motion of a fluid are nonlinear. In the
referential description of motion used to describe the deformations of a solid, the acceleration is
linear in displacements. Were we to use the referential description of motion for studying the
deformations of a fluid, the expression for the strain-rate tensor D in terms of the spatial gradients
of the velocity field will be more involved. The nonlinearity in (6.2.2) implies that we can no
longer use the principle of superposition. Whereas for solids, the deformations caused by gravity
are generally negligible and hence are ignored, gravity is the main driving force for a fluid. A
familiar example is the flow of a fluid in a river. Note that equations governing the deformations of
a solid were expressed in terms of displacements, those for a fluid are written in terms of velocities.
For an incompressible Navier-Stokes fluid, equations expressing the balance of mass and the
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balance of linear momentum are
@vi@xi
= 0; (6.2.3)
�
�@vi@t
+@vi@xj
vj
�= � @p
@xi+ �
@2vi@xk@xk
+ �gi: (6.2.4)
The mass density � is a constant, and the pressure field p is an arbitrary function of x and time t.
Equations (6.2.1) and (6.2.2) for a compressible Navier-Stokes fluid or (6.2.3) and (6.2.4) for
an incompressible Navier-Stokes fluid are supplemented by the following initial and boundary
conditions.
�(x; 0) = �0(x) in ;
vi(x; 0) = v0i (x) in ; (6.2.5)
[�p�ij + �vk;k�ij + �(vi;j + vj;i)]nj = fi(x; t) on @1� (0; T );
vi(x; t) = vpi (x; t) on @2� (0; T ): (6.2.6)
Here is the domain of study which may equal only a part of the region occupied by the fluid.
Equations (6.2.5) specifying the mass density and the velocity at time t = 0 are the initial con-
ditions, and (6.2.6) prescribing surface tractions on a part of the boundary and velocities on the
remainder of the boundary are boundary conditions. If the velocity field is prescribed on the entire
boundary, then for an incompressible fluid, the pressure field is indeterminate.
For a steady flow, the field variables �; p and v in the spatial description are independent of
time t. Thus@�
@t= 0 =
@v
@tin equations (6.2.1) through (6.2.4), and initial conditions (6.2.5) are
not required.
A flow is called irrotational if the spin tensor or the vorticity vanishes everywhere in the domain.
Governing equations for the steady irrotational flow of an incompressible Navier-Stokes fluid are
@vi@xi
= 0; (6.2.7)
@vi@xj
vj = � @p
@xi+ gi: (6.2.8)
In writing (6.2.8) we have absorbed the constant mass density in p. Thus the viscosity of a fluid
does not influence the steady irrotational flow of an incompressible Navier-Stokes fluid.
124
6.3 Examples
6.3.1 Steady Flow Between Two Parallel Plates
Consider the steady flow of a compressible Navier-Stokes fluid between two parallel horizontal
plates with the lower plate kept stationary and the upper one moved in the positive x1-direction at
a uniform speed of v0 m=s. Assume that the fluid sticks to the plates and that vi(x; t) = v(x2)�i1
where the x2-axis is perpendicular to the plate surfaces, and the fluid extends to infinity in the
x3-direction. Thus we have a two-dimensional problem in the x1 � x2 plane with the fluid flowing
only in the x1-direction. It is therefore reasonable to assume that the mass density and the pressure
field also depend only upon x1 and x2. For the presumed velocity field, vi;i = 0. Thus the motion
is isochoric and the mass density of a material particle does not change. Also, both the local and
the convective parts of the acceleration identically vanish.
Nontrivial equations governing the flow of the fluid are
@�
@x1v = 0; (6.3.1)
0 = � @p
@�
@�
@x1+ �
@2v
@x22; (6.3.2)
0 = � @p
@�
@�
@x2+ �g; (6.3.3)
v(x1; 0) = 0; v(x1; h) = v0; (6.3.4)
Tijnj
����x1=�L
= �p(x2)�i1 ; Tijnj
����x1=L
= �p0(x2)�i1 (6.3.5)
Whereas equation (6.3.1) expresses the balance of mass, equations (6.3.2) and (6.3.3) are the re-
duced forms of the balance of linear momentum in the x1 and x2 directions. Equation (6.3.4) states
the essential boundary conditions on the surfaces of the two plates, and equation (6.3.5) gives nat-
ural boundary conditions on the vertical surfaces x1 = �L. It follows from (6.3.1) that � = �(x2).
Equations (6.3.2) and (6.3.4) give
v =x2hv0 : (6.3.6)
Thus the velocity field is known. It follows from (6.3.3) that
d�
dx2=
�g
@p=@�: (6.3.7)
125
The determination of � as a function of x2 requires that the compressibility of the fluid be known.
Said differently, the constitutive equation p = p(�) must be given. As an example, we take
p = c� (6.3.8)
where c is a constant. Then, equations (6.3.7) and (6.3.8) yield
� = degcx2 (6.3.9)
where d is a constant. Equations (6.3.8) and (6.3.9) imply that p can depend only on x2. The stress
tensor in the fluid computed from equations (6.1.3), (6.3.6), (6.3.8) and (6.3.9) is given by
Tij = �cdegx2=c�ij + �v0h
(�2j�i1 + �2i�j1) ; (6.3.10)
and is only a function of x2. Thus �p(x2) must equal p0(x2) in equation (6.3.5); otherwise there
is no solution for the problem. That is, the pressure distribution must be same on every plane
x1 = const. From equations (6.3.5) and (6.3.10) we conclude that
�p(x2) = cdegx2=c : (6.3.11)
That is, for the fluid being studied, surface tractions on planes x1 = �L can not be arbitrarily
prescribed, but must be of the form (6.3.11). From the given value of �p(x2), we find the constant
of integration d. The fields of density and velocity are respectively given by (6.3.9) and (6.3.6).
In linear elasticity, the uniquenes theorem guarantees that the solution obtained by a semi-
inverse method is the only solution of the problem. However, in fluid mechanics, a uniqueness
theorem can not be proved because of the nonlinear governing equations. It implies that there may
be other solutions for the problem studied.
6.3.2 Steady Flow of an Incompressible Navier-Stokes Fluid down an Inclined Plane
In order to study this problem, it is more convenient to choose coordinate axes with x1-axis along
the inclined plane and x2-axis perpendicular to it. We assume that the pressure and velocity fields
are independent of x3, and
vi(x1; x2) = v(x2)�i1 : (6.3.12)
126
That is, the fluid is flowing parallel to the inclined plane. Thus, the top surface of the fluid will be
parallel to the inclined plane. Such a flow is called laminar.
The balance of mass or the continuity equation, vi;i = 0, is identically satisfied. Also, the
convective part of acceleration, vjvi;j, vanishes. Thus equations expressing the balance of linear
momentum are
0 = � @p
@x1+ �
d2v
dx22+ �g sin � ; (6.3.13)
0 = � @p
@x2� �g cos � : (6.3.14)
The associated boundary conditions are
vi(x1; 0) = 0; Tijnj
����x2=h
= �pani;
Tijnj
����x1=0
= ti(x2); Tijnj
����x1=L
= qi(x2):
(6.3.15)
That is, the fluid adheres to the stationary inclined plane, and the surface tractions are prescribed
on the top surface and two arbitrarily chosen surfaces x1 = 0 and x1 = L.
A solution of equation (6.3.14) is
p = ��g cos �x2 + f(x1); (6.3.16)
where f is an arbitrary smooth function of x1. Substitution of (6.3.16) into (6.3.13) gives
df
dx1= �
d2v
dx22+ �g sin �: (6.3.17)
127
Since the left-hand side of (6.3.17) is a function of x1 and the right-hand side a function of x2,
therefore each must equal a constant b. Thus
f = bx1 + c;
v =b� �g sin �
�
x222
+ e1x2 + e2;(6.3.18)
where c; e1 and e2 are constants of integration. Boundary condition (6.3.15)1 requires that e2 = 0.
Substitution from (6.3.12), (6.3.18) and (6.3.16) into the constitutive relation (6.1.9) gives
Tij = � (��gx2 cos � + bx1 + c)�ij
+ (�i1�2j + �j1�2i)
�b� �g sin �
�x2 + e1
�: (6.3.19)
Equations (6.3.15)2 and (6.3.19) yield
�(��gh cos � + bx1 + c)�i2 + �i1
�b� �g sin �
�h+ e1
�= �pa�i2 : (6.3.20)
Hence
� (��gh cos � + bx1 + c) = �pa ;b� �g sin �
�h+ e1 = 0
(6.3.21)
Because (6.3.21)1 must hold for all values of x1, therefore
b = 0; c = �gh cos � + pa; e1 =�g sin �
�h : (6.3.22)
With these values of b; c and e1, equation (6.3.19) gives
Tij = �(�g cos �(�x2 + h) + pa)�ij + (�i1�2j + �j1�2i)�g sin �
�(�x2 + h) : (6.3.23)
Noting that on the surface x1 = 0; nj = ��1j , equations (6.3.23) and (6.3.15) give
(�g cos �(�x2 + h) + pa)�i1 + �2i�g sin �
�(x2 � h) = ti(x2): (6.3.24)
One can similarly find that qi(x2) = �ti(x2). Unless normal and tangential tractions given by
(6.3.24) are supplied on the planes x1 = const., the assumed flow vi = v(x2)�i1 can not be
maintained on the inclined plane. The inclined plane is usually assumed to be infinitely long and
tractions necessary to maintain the flow are presumed to act on the planes x1 = const.
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