Mean stress and stress deviator: The stress in Eq. (1.36) can be
decomposed into
hydrostatic pressure and deviatoric stress. The former is related
to the change in
volume, while the latter is related to the change in shape. The
hydrostatic pressure,
often called the mean stress, can be defined using the trace of the
stress tensor as
p ¼ σm ¼ 1
3 tr σð Þ ¼ 1
3 σ11 þ σ22 þ σ33ð Þ: ð1:42Þ
Note that the hydrostatic pressure is invariant on coordinate
transformation in
Eq. (1.19), that is, for σ in xyz coordinates and σ0 for x0y0z0
coordinates, tr(σ)¼ tr(σ0). Therefore, the mean stress has the
property of frame indifference.
1.3 Stress and Strain 21
On the other hand, the stress deviator is defined by subtracting
the mean stress
from the original stress tensor as
s ¼ σ σm1 ¼ σ11 σm σ12 σ13
σ12 σ22 σm σ23 σ13 σ23 σ33 σm
2 4
3 5: ð1:43Þ
Note that tr(s)¼ 0. Therefore, the stress deviator is called
trace-free. The mean
stress and stress deviator are important in representing the
plastic behavior of a
material beyond the yield point.
For a formal definition, the stress deviator can be defined by
contracting the
original stress with the unit deviatoric tensor of rank-4:
s ¼ Idev : σ;
Idev ¼ I 1
3 1 1; ð1:44Þ
where I is a unit symmetric tensor of rank-4, which is defined
as
Iijkl¼ (δikδjl+ δilδjk)/2. Note that since Idev is trace-free, it
is easy to show that
Idev : 1¼ 0. In addition, the unit deviatoric tensor preserves a
deviatoric tensor,
that is, Idev : s¼ s for a deviatoric rank-2 tensor s.
Principal stresses: The normal and shear stresses acting on a
plane, which passes
through a given point in a solid, change as the orientation of the
plane is changed.
Then a natural question is: Is there a plane on which the normal
stress becomes the
maximum? Similarly, we would also like to find the plane on which
the shear stress
attains a maximum. These questions have significance in predicting
the failure of
the material at a point. In the following, we will provide some
answers to the above
questions, without furnishing the proofs. The interested reader is
referred to books
on continuum mechanics, e.g., Malvern [6] or Boresi [7] for a more
detailed
treatment of the subject.
It can be shown that, at every point in a solid, there are at least
three mutually
perpendicular planes on which the normal stress attains an extremum
(maximum or
minimum) value. On all of these planes, the shear stresses vanish.
Thus, the traction
vector, t(n), will be parallel to the normal vector, n, on these
planes, i.e., t(n)¼ σnn. Of these three planes, one plane
corresponds to the global maximum value of the
normal stress and another corresponds to the global minimum. The
third plane will
carry the intermediate normal stress. These special normal stresses
are called the
principal stresses at that point, the planes on which they act are
called the principal stress planes and the corresponding normal
vectors are called the principal stress
directions. The principal stresses are denoted by σ1, σ2, and σ3,
such that
σ1 σ2 σ3.
22 1 Preliminary Concepts
Based on the above observations, the principal stresses can be
calculated, as
follows. When the normal direction to a plane is the principal
direction, the surface
normal and the surface traction are in the same direction, i.e.,
(t(n) || n). Thus, the surface traction on a plane can be
represented by the product of the normal stress,
σn, and the normal vector, n, as
t nð Þ ¼ σnn: ð1:45Þ
By combining Eq. (1.45) with Eq. (1.38) for the surface traction,
we obtain
σ n ¼ σnn: ð1:46Þ
Equation (1.46) represents the eigenvalue problem, where σn is the
eigenvalue and n is the corresponding eigenvector. Equation (1.46)
can be rearranged as
σ σn1ð Þ n ¼ 0: ð1:47Þ
In the component form, the above equation can be written as
σ11 σn σ12 σ13 σ12 σ22 σn σ23 σ13 σ23 σ33 σn
2 4
9= ;: ð1:48Þ
Note that a solution, n¼ 0, is not only a trivial solution to the
above equation, but
also not physically possible as ||n|| must be equal to unity. The
above set of linear
simultaneous equations will have a nontrivial physically meaningful
solution if and
only if the determinant of the coefficient matrix is zero,
i.e.,
σ11 σn σ12 σ13 σ12 σ22 σn σ23 σ13 σ23 σ33 σn
¼ 0: ð1:49Þ
By expanding this determinant, we obtain the following cubic
equation in terms of
σn:
σ3n I1σ 2 n þ I2σn I3 ¼ 0; ð1:50Þ
where
I2 ¼ σ11 σ12 σ12 σ22
þ σ22 σ23
¼ σ11σ22 þ σ22σ33 þ σ33σ11 σ212 σ223 σ213, I3 ¼ σj j ¼ σ11σ22σ33 þ
2σ12σ23σ13 σ11σ223 σ22σ213 σ33σ212:
ð1:51Þ
1.3 Stress and Strain 23
In the above equation, I1, I2, and I3 are the three invariants of
the stress, which can
be shown to be independent of the coordinate system. The three
roots of the cubic
equation (1.50) correspond to the three principal stresses. We will
denote them by
σ1, σ2, and σ3 in the order of σ1 σ2 σ3. Once the principal
stresses have been computed, we can substitute them, one at
a time, into Eq. (1.48) to obtain n. We will get a principal
direction that will
be denoted as n1, n2, and n3, which each corresponds to a principal
value.
Note that n is a unit vector, and hence its components must satisfy
the following
relation:
2 ¼ 1, i ¼ 1, 2, 3: ð1:52Þ
It can be shown that the planes on which the principal stresses act
are mutually
perpendicular. Let us consider any two principal directions ni and
nj, with i 6¼ j. If σi and σj are the corresponding principal
stresses, then they satisfy the following
equations:
ð1:53Þ
By multiplying the first equation by nj and the second equation by
ni, we obtain
nj σ ni ¼ σinj ni, ni σ nj ¼ σjni nj: ð1:54Þ
Considering the symmetry of σ and the rule for inner product, one
can show that
nj σ ni¼ ni σ nj. Then subtracting the first equation from the
second in
Eq. (1.54), we obtain
ni nj ¼ 0: ð1:55Þ
This implies that if the principal stresses are distinct, i.e., σi
6¼ σj, then
ni nj ¼ 0; ð1:56Þ
which means that ni and nj are orthogonal. The three planes, on
which the principal
stresses act, are mutually perpendicular.
24 1 Preliminary Concepts
There are three different possibilities for principal stresses and
directions:
(a) σ1, σ2, and σ3 are distinct) principal stress directions are
three unique
mutually orthogonal unit vectors.
(b) σ1¼ σ2 6¼ σ3) n3 is a unique principal stress direction, and
any two orthog-
onal directions on the plane that is perpendicular to n3 are the
other
principal directions.
(c) σ1¼ σ2¼ σ3) any three orthogonal directions are principal
stress direc-
tions. This state of stress is called hydrostatic or isotropic
state of stress.
Example 1.8 (Principal stresses and principal directions) For the
Cartesian stress
components given below, determine the principal stresses and
principal directions.
σ ¼ 3 1 1
1 0 2
1 2 0
2 4
3 5:
Solution Setting the determinant of the coefficient matrix to zero
yields
3 σn 1 1
By expanding the determinant, we obtain the following
characteristic equation:
3 σnð Þ σ2n 4 σn 2ð Þ þ 2þ σnð Þ ¼ σn þ 2ð Þ σn 1ð Þ σn 4ð Þ ¼
0:
Three roots of the above equation are the principal stresses. They
are
σ1 ¼ 4, σ2 ¼ 1, σ3 ¼ 2:
For the case when σn¼ σ3¼2, we may obtain the following
simultaneous equa-
tions, by using the form of Eq. (1.48):
5nx þ ny þ nz ¼ 0,
nx þ 2ny þ 2nz ¼ 0,
nx þ 2ny þ 2nz ¼ 0:
We note that the three equations are not independent; in fact, the
second and third
equations are identical. From the first two equations, we can
obtain the following
ratios between components:
1.3 Stress and Strain 25
By using Eq. (1.52), a unique solution of the following form can be
obtained:
n 3ð Þ ¼ 1ffiffiffi 2
p 0
1
1
8< :
9= ;:
The same process can be repeated for σ1 and σ2 to obtain the
following two
principal directions:
p 2
3 p
1.3.2 Strain
When a solid is subjected to forces, it deforms. A measure of the
deformation is
provided by strains. Imagine an infinitesimal line segment in an
arbitrary direction
which passes through a point in a solid. As the solid deforms, the
length of the line
segment changes. The strain, specifically the normal strain, in the
original direction
of the line segment is defined as the change in length divided by
the original length.
However, the strain at the same point will be different in
different directions. In the
following, the concept of strain in a three-dimensional body is
developed.
Figure 1.8 shows a body before and after deformation. Let the
points, P, Q, and R, in the undeformed body move to P0, Q0, and R0,
respectively, after deformation.
For the convenience of notation, the three coordinate directions
are denoted by
P(x1,x2,x3) Q
Q'
R'
x1
x2
x3
Δx1
Δx2
26 1 Preliminary Concepts
x1x2x3 coordinates instead of using the xyz coordinates. The
displacement of P can
be represented by three displacement components, u1, u2, and u3 in
the x1-, x2-, and x3-directions. Thus, the coordinates of P
0 are (x1 + u1, x2 + u2, x3 + u3). The functions u1(x1,x2,x3),
u2(x1,x2,x3), and u3(x1,x2,x3) are components of a vector field
that is
referred to as the deformation field or the displacement field. The
displacements of
the point, Q, will be slightly different from that of P. They can
be written as
uQ 1 ¼ u1 þ ∂u1
∂x1 Δx1,
∂x1 Δx1,
∂x1 Δx1:
uR 1 ¼ u1 þ ∂u1
∂x2 Δx2,
∂x2 Δx2,
∂x2 Δx2:
ð1:58Þ
The coordinates of P, Q, and R before and after deformation are as
follows:
P : x1; x2; x3ð Þ, Q : x1 þ Δx1, x2, x3ð Þ, R : x1, x1 þ Δx2, x3ð
Þ, P
0 : x1 þ uP
3
¼ x1 þ u1, x2 þ u2, x3 þ u3ð Þ, Q
0 : x1 þ Δx1 þ uQ
1 , x2 þ uQ 2 , x3 þ uQ
3
∂x1 Δx1, x2 þ u2 þ ∂u2
∂x1 Δx1, x3 þ u3 þ ∂u3
∂x1 Δx1
1 , x2 þ Δx2 þ uR 2 , x3 þ uR
3
∂x2 Δx2, x2 þ Δx2 þ u2 þ ∂u2
∂x2 Δx2, x3 þ u3 þ ∂u3
∂x2 Δx2
:
The length of the line segment P0Q0 can be calculated as
P 0 Q
1.3 Stress and Strain 27
By substituting for the coordinates of P0 and Q0, we obtain
P 0 Q
0 ¼ Δx1
þ ∂u1 ∂x1
2 þ ∂u2
ð1:60Þ
It may be noted that we have used a two-term binomial expansion in
deriving an
approximate expression for the change in length. In this chapter,
we will consider
only small deformations such that all deformation gradients are
very small when
compared to unity, e.g., ∂u1/∂x1 1, ∂u2/∂x1 1. Then we can neglect
the
higher-order terms in Eq. (1.60) to obtain
P 0 Q
: ð1:61Þ
Now we invoke the definition of normal strain as the ratio of the
change in length to
the original length in order to derive the expression for strain
as
ε11 ¼ P 0 Q
0 PQ
∂x1 : ð1:62Þ
Thus, the normal strain, ε11, at a point can be defined as the
change in length per unit length of an infinitesimally long line
segment, originally parallel to the x1-axis. Similarly, we can
derive normal strains in the x2- and x3-directions as
ε22 ¼ ∂u2 ∂x2
, ε33 ¼ ∂u3 ∂x3
: ð1:63Þ
The shear strain, say γ12, is defined as the change in angle
between a pair of
infinitesimal line segments that were originally parallel to the
x1- and x2-axes. From Fig. 1.8, the angle between PQ and P0Q0 can
be derived as
θ1 ¼ xQ 0
θ2 ¼ xR 0
γ12 ¼ θ1 þ θ2 ¼ ∂u1 ∂x2
þ ∂u2 ∂x1
: ð1:66Þ
Similarly, we can derive shear strains in the x2x3- and x3x2-planes
as
γ23 ¼ ∂u2 ∂x3
ð1:67Þ
The shear strains, γ12, γ23, and γ13, are called engineering shear
strains. From the
definition in Eq. (1.66), it is clear that γ12¼ γ21. We define
tensorial shear strains as
ε12 ¼ 1
ð1:68Þ
It may be noted that the tensorial shear strains are one-half of
the corresponding
engineering shear strains. It can be shown that the normal strains
and the tensorial
shear strains transform from one coordinate system to another by
following the
tensor transformation rule in Eq. (1.19).
In the general three-dimensional case, the strain tensor can be
defined using the
dyadic product, as
where the components of the strain tensor are defined as
ε½ ¼ ε11 ε12 ε13 ε12 ε22 ε23 ε13 ε23 ε33
2 4
3 5: ð1:70Þ
As is clear from the definition in Eq. (1.68), the strain tensor is
symmetric. Similar
to the stress tensor, the symmetric strain tensor can be
represented as a pseudo
vector
8>>>>>>< >>>>>>:
9>>>>>>= >>>>>>;
¼
8>>>>>>< >>>>>>:
1.3 Stress and Strain 29
The six components of strain completely define the deformation at a
point. Since
strain is a tensor, it has properties similar to the stress tensor.
For example, the
normal strain in any arbitrary direction at that point and also the
shear strain in any
arbitrary plane passing through the point can be calculated using
the same process
as in the stress tensor. Similarly, the transformation of strain,
principal strains, and
corresponding principal strain directions can be determined using
the procedures
we described for stresses.
Decomposition of strain: The strain tensor can be decomposed into a
volumetric
and a distortional part. The former changes the volume of an
infinitesimal element,
while the latter changes the shape of the element. For volumetric
strain, consider a
unit cube in Fig. 1.9, which undergoes three normal strains (ε11,
ε22, and ε33). Since there is no shape change, all shear strains
are zero, for now. Then, the deformed
volume
V0
¼ 1þ ε11ð Þ 1þ ε22ð Þ 1þ ε33ð Þ 1 ε11 þ ε22 þ ε33:
Since the magnitudes of strain components are small, the
higher-order terms may be
ignored. Therefore, the volumetric strain can be defined as
εV ¼ ε11 þ ε22 þ ε33 ¼ εkk: ð1:72Þ
Or, in tensor notation, εV ¼ 1 : ε, with 1 being the rank-2
identity tensor. Note that
the volumetric strain is a scalar that is three times the value of
the average normal
strain.
The deviatoric part of strain can be defined by subtracting the
average normal
strain from the diagonal components of the original strain. The
deviatoric strain
tensor can be defined as
e ¼ ε 1
ð1:73Þ
For a formal definition, the deviatoric strain can be defined by
contracting the
original strain with the unit deviatoric tensor of rank-4.
Fig. 1.9 Volume change of
a unit cube
30 1 Preliminary Concepts
e ¼ Idev : ε;
where Idev is the unit deviatoric tensor of rank-4, defined in Eq.
(1.44).
1.3.3 Stress–Strain Relationship
Finding a relationship between the loads acting on a structure and
its deflection has
been of great interest to scientists since the seventeenth century
[8]. Robert Hooke,
Jacob Bernoulli, and Leonard Euler are some of the pioneers who
developed
various theories to explain the bending of beams and stretching of
bars. Forces
applied to a solid create stresses within the body in order to
satisfy equilibrium.
These stresses also cause deformation or strains. Accumulation of
strains over the
volume of a body manifests as deflections or a gross deformation of
the body.
Hence, it is clear that a fundamental knowledge of the relationship
between stresses
and strains is necessary in order to understand the global
behavior. Navier tried to
explain deformations considering the forces between neighboring
particles in a
body, as they tend to separate and come closer. Later this approach
was abandoned
in favor of Cauchy’s stresses and strains. Robert Hooke was the
first one to propose the linear uniaxial stress–strain relation,
which states that the stress is proportional
to strain. Later, the general relation between the six components
of strains and
stresses called the generalized Hooke’s law was developed. The
generalized
Hooke’s law states that each component of stress is a linear
combination of strains.
It should be mentioned that stress–strain relations are called
phenomenological
models or theories as they are based on commonly observed behavior
of materials
which may be verified through experimentation. Only recently, with
the advance-
ment of computers and computational techniques, have we started to
model the
behavior of materials based on the first principles or based on the
fundamentals of
atomistic behavior. This new field of study is called computational
materials, and it
involves techniques such as molecular dynamic simulations and
multiscale model-
ing. Stress–strain relations are also called constitutive relations
as they describe the
constitution of the material.
A cylindrical test specimen is loaded along its axis as shown in
Fig. 1.10. This
type of loading ensures that the specimen is subjected to a
uniaxial state of stress. If
the stress–strain relation of the uniaxial tension test in Fig.
1.10 is plotted, then a
typical ductile material may show a behavior as in Fig. 1.11. The
explanation of the
terms in the figure is summarized in Table 1.2.
After the material yields, the shape of the structure permanently
changes. Hence,
many engineering structures are designed such that the maximum
stress is smaller
than the yield stress of the material. Under this range of the
stress, the stress–strain
FF
Fig. 1.10 Uniaxial tension test
1.3 Stress and Strain 31
relation can be linearly approximated. The main interest of this
text is to study the
behavior of materials beyond the linear relation. However, in this
and the following
sections, we will focus on the linear relationship between stress
and strain.
Stress–strain relationship for isotropic material: The
one-dimensional stress–
strain relation can be extended to the three-dimensional state of
stress. The linear
elastic material means the relationship between the stress and
strain is linear. Since
both stress and strain tensors are rank-2, the relationship between
them requires a
rank-4 tensor. For a general linear elastic material, the
stress–strain relationship can
be written as
σ ¼ D : ε, σij ¼ Dijklεkl: ð1:74Þ
The rank-4 tensor, D, is called the elasticity tensor. A general
rank-4 tensor in three
dimensions has 81 components. Since the stress and strain tensors
are symmetric,
it can be shown that D must be symmetric1; hence, the number of
independent
coefficients or elastic constants for an anisotropic material is
only 21.
Proportional limit
Yield stress
Ultimate stress
Strain hardening
material in tension
Terms Explanation
Proportional
limit
The greatest stress for which the stress is still proportional to
the strain
Elastic limit The greatest stress without resulting in any
permanent strain on release of
stress
Slope of the linear portion of the stress–strain curve
Yield stress The stress required to produce 0.2 % plastic
strain
Strain hardening A region where more stress is required to deform
the material
Ultimate stress The maximum stress the material can resist
Necking Cross section of the specimen reduces during
deformation
1More specifically, D has major and minor symmetry.
32 1 Preliminary Concepts
Many composite materials that are naturally occurring, such as wood
or bone, and
man-made materials, such as fiber-reinforced composites, can be
modeled as an
orthotropic material with nine independent elastic constants. Some
composites
are transversely isotropic and require only five independent
elastic constants.
For isotropic materials, the 21 constants in the elasticity tensor
can be expressed in
terms of two independent constants called engineering elastic
constants. Therefore,
the elasticity tensor for an isotropic material can be written
as
D ¼ λ1 1þ 2μI; ð1:75Þ
where λ and μ are Lame’s constants. In fact, μ is also called the
shear modulus. The
Lame’s constants are related to the nominal engineering elastic
constants: Young’s modulus, E, and Poisson’s ratio, ν, as
ν¼ λ
,
2 1þ νð Þ: ð1:76Þ
Using index notation, the components of the rank-4 elasticity
tensor can be
written as Dijkl¼ λδijδkl + μ(δikδjl+ δilδjk). As the stress and
strain tensors are
decomposed into volumetric and deviatoric parts, the elasticity
tensor can also
be decomposed as
1 1þ 2μIdev; ð1:77Þ
where I