Dr. S M Murigendrappa: Basic Equations of Elasticity;
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Basic Equations of Elasticity
INTRODUCTION
Elasticity is the property of a deformable body due to which the body recovers its original
shape upon the removal of the forces causing the deformation. Almost all engineering
materials possess to a certain extent the property of elasticity. In this chapter we will study
the basic equations of the theory of elasticity.
STRESS
Consider a body shown in Figure (3.1), is in equilibrium. Under the action of external forces,
P1 , P2 ,....., internal forces will be produced between the parts of the body. To study the
magnitude of these forces at any point ' O ', let us imagine the body is divided into two parts
1 and 2, by a cross-section ' mm ' through this point. It will be assumed that the internal
forces are continuously distributed over the cross-section ' mm '. The magnitude of such
forces is usually defined by their intensity, i.e., by the amount of force per unit area of the
surface
on
which
they
act.
This
intensity
is
called stress.
y
P3
P4
m
P2
O
n
∆P
2
1
P1
m
x
P7
P5
P6
z
Figure 3.1 Body with external forces
In general, if the stress is not uniformly distributed over cross-section ' mm ', taking small
area on the section and is ∆A ( = ∆z ∆y ) , the resultant force acting on it is ∆P (Figure 3.2).
∆P
gives the stress acting on the cross-section ' mm ' at the
∆A
point ' O '. The limiting direction of the resultant ∆P is the direction of the stress. In general
case, the direction of stress is inclined to the area ∆P on which it acts. We can resolve it
The limiting value of the ratio
into a normal or direct stress perpendicular to the area and other two, shear stresses acting in
the plane of the area ∆P .
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∆Py
y
∆Py
∆Pz
O
∆P
β
∆P
∆Px
γ
α
O
∆Y
∆Px
∆Pz
∆z
x
z
(Enlarged area)
Figure 3.2 Internal forces on the cross section ‘mm’
Notation of Stresses The letter σ used for normal stress and the letter τ for shearing stress.
To indicate the direction of the plane on which the stress is acting, subscripts to these letters
are used. The first subscript denotes the direction of the normal to the face and second
denotes the direction in which the stress component acts.
i.e.,
σ xx = σ x = lt
∆A→ 0
∆Px
∆A
---
(3.1)
The normal stress
(σ xx
or σ x ) component acting on the x − face will act in the
x − direction.
Similarly,
τ xy = lt
∆A→ 0
∆Py
∆A
∆Pz
∆A→0 ∆A
and τ xz = lt
---
(3.2)
τ xy : Shear stress component acting on the x − face in the y − direction
τ xz : Shear stress component acting on the x − face in the z − direction.
Components of Stress For each pair of parallel sides of a cubic element (Figure 3.3), one
symbol is needed to denote the normal component of stress and two more symbols to denote
the two components of shearing stress. To describe the stresses acting on the six sides of the
element, three symbols σ x , σ y and σ z are necessary for normal stresses and six symbols
τ xy , τ xz , τ yz , τ yx , τ zx , τ zy for shearing stresses.
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σy
y
τyx
τyz
σz
τxz
τzy
τxy
τzx
τzy
σx
dy
τxy
σx
τxz
τzx
σz
τyz
τyx
dz
σy
dx
x
z
Figure 3.3 Three-dimensional element
State of Stress in Two-dimensions or Plane Stress
A two-dimensional stated-stress exists when the stresses and body forces are independent of
one of the coordinates. Such a state is described by stresses σ x , σ y and τ xy , and the body
forces
Fx and Fy
( z − is taken as the
y
σy
τyx
independent coordinate axis). This combination
of stress components which are functions of
only x and y are called plane stress in the
xy − plane.
Example In a thin plate located in the plane of
the plate there will be no stress acting
perpendicular to the surface of the plate (Figure
3.4). The stress components acting on this
element are σ x , σ y and τ xy .
Differential Equations of Equilibrium
of Two-dimensional Element
τxy
σx
σx
dy
dx
τxy
τyx
σy
x
Figure 3.4 Two-dimensional plane with stresses
A body is said to be in equilibrium when under
the action of external forces, it is at rest or moving in a straight line with constant velocity.
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Consider an elementary block of unit thickness with cross-sectional area dx, dy as shown in
Figure 3.5. It is subjected to a general system of tensile stresses in x − and y − directions
and shear stresses as well as body forces.
σy +
y
∂σy
dy
∂y
τyx +
Fy
σx
dy
x
σx +
Fx
τxy
∂τyx
dy
∂y
∂τxy
τxy +
dx
∂x
∂σx
dx
∂x
dx
τyx
y
σy
O
x
Figure 3.5 Two-dimensional element with stresses
By two-dimensional stress system, i.e., σ x , σ y and τ xy are independent of z − axis and the
other stress components are zero. The body forces per unit volume, Fx and Fy to be
independent of z − axis and Fz = 0 . Such a state of stress is called plane stress.
For equilibrium of the system,
∑F
x
= 0 and
∑F
y
= 0 . Therefore, summing all forces in
the x − direction, we get
∂τ yx 

∂σ x 

 σ x + ∂x dx  dy − σ x dy +  τ yx + ∂y dy  dx − τ yx dx + Fx dxdy = 0




Simplifying
∂τ yx
∂σ x
dxdy +
dxdy + Fx dxdy = 0
∂x
∂y
Dividing by volume of the element, i.e., dx ⋅ dy ⋅1 , we get the equilibrium equation of the
system in x − direction as
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∂σ x ∂τ yx
+
+ Fx = 0
∂x
∂y
---
(3.3)
Similarly, summing all the forces in the y − direction, we get
∂σ y
∂τ xy 



⋅ dy  dx − σ y dx +  τ xy +
dx  dy − τ xy dy + Fy dx ⋅ dy = 0
σ y +
∂y
∂x




∂σ y
∂y
dxdy +
∂τ xy
∂x
dxdy + Fy dxdy = 0
Dividing by element volume we get another equilibrium equation of the system in
y − direction as
∂τ xy
∂x
+
∂σ y
∂y
+ Fy = 0
---
(3.4)
Taking moments of forces about ' O ' we obtain from the condition
(σ x dy )  y +

∑M
o
= 0 i.e.,
∂τ


dy  
∂σ
dy 
 
−  σ x + x dx  dy  y +  − (τ xy dy ) x +  τ xy + xy dx  dy ( x + dx )

2  
∂x
2 
∂x
 


∂σ y  
∂τ yx 

dx  
dx 

− (σ y dx )  x +  +  σ y +
dy  dx  x +  + (τ yx dx ) y − τ yx +
dy  dx ( y + dy )
2 
∂y
2
∂y

 


dy 
dx 


− ( Fx dxdy )  y +  + ( Fy dxdy )  x +  = 0
2 
2


Collecting terms and neglecting the higher-order terms, we get
(τ
xy
∂τ yx
 ∂σ

 ∂τ xy ∂σ y

− τ yx ) dxdy −  x +
+ Fx  ydxdy + 
+
+ Fy  xdxdy = 0
∂y
∂y
 ∂x

 ∂x

---
(3.5)
From equations (3.3) and (3.4), we see that the terms in the second and third parentheses of
(3.5) are zero.
∴τ xy − τ yx = 0
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τ xy = τ yx
or
---
(3.6)
Differential Equations of Equilibrium of Three-dimensional Element
Consider an infinitesimal element of size dx, dy, dz is subjected to tensile stresses in all
three directions as shown in Figure 3.6.
The differential equations of equilibrium are obtained, by resolving forces in x, y and
z − directions, we get
y
where
σy + dσy
τyx + dτyx
τyz + dτyz
σz
τzx
τxz
τzy + dτzy
σx
Fz
Fx
τzy
τzx + dτzx
τxy
dy
τxy + dτxy
Fy
σx + dσx
τxz + dτxz
σz + dσz
x
τyz
τyx
dz
σy
dx
z
Figure 3.6 Two-dimensional element with stresses
∂σ x
dx
∂x
∂σ y
dσ y =
dy
∂y
∂σ z
dσ z =
dz
∂z
∂τ
dτ xy = xy dx
∂x
∂τ yz
dτ yz =
dy
∂y
∂τ
dτ zx = zx dz
∂z
∂τ yx
dτ yx =
dy
∂y
∂τ
dτ zy = zy dz
∂z
∂τ xz
dτ xz =
dx
∂x
dσ x =
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∂σ x ∂τ yx ∂τ zx
+
+
+ Fx = 0
∂x
∂y
∂z
∂τ xy ∂σ y ∂τ zy
+
+
+ Fy = 0
∂x
∂y
∂z
∂τ xz ∂τ yz ∂σ z
+
+
+ Fz = 0
∂x
∂y
∂z









---(3.7a)
and using moments of forces, we get
τ xy = τ yx 

τ yz = τ zy 
τ zx = τ xz 
---
(3.7b)
STRAIN
The strain is defined as a measure of relative change in length or change in shape.
Consider a thin, continuous
body which lies entirely in the
xy − plane
and
which
undergoes
a
small
geometrically
compatible
deformation in the xy -plane
(Figure 3.7). This element to
be deformed in a state of
uniform strain.
C′
y
D′
C
D
D
dy
Deformed or
strained element
C
B′
B
For the deformed element
A′
Undeformed or
A′B′C ′D′ , let us express the
B
A
unstrained
element
deformation in the vicinity of
dx
point A′ quantitatively by
y
giving the changes in length of
x
the two lines AB and AD
x
and the relation of these lines
Figure 3.7 Two-dimensional element under stresses
relative to each. These two
aspect of the deformation can be define dimensionless quantities. The first gives measure of
the elongation or contraction of a line, will be called direct or normal strain component. The
second, which gives a measure of the relative rotation of two lines, will be called the shear
strain.
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The normal strain is defined as the fractional change in the original length of a line and is
designated by the symbol ∈ with subscript to indicate the original direction of the line for
which the strain is measured. The value of ∈x at point A in the x − direction is
∈x = lt
dx →0
A′B′ − AB
AB
---(3.8a)
Similarly, ∈y in the y-direction is
∈y = lt
dy → 0
A′D′ − AD
AD
---
(3.8b)
From definition, the normal strain is +ve, when the line elongative and − ve, when the line
contrast.
The shear strain is defined as the tangent of the change in angle between the two originally
perpendicular axes. This component is specified with respect to two axes which are
perpendicular on the undeformed body and is designated by the symbol, γ with two
subscripts to indicate these two axes.
γ xy = lt
dx →0
dy → 0
[ BAD −
B′A′D′]
π

= lt  − B′A′D′
dx →0 2

dy → 0 
---(3.8c)
Shear strain is +ve when the axes rotate so that the first and third quadrants become smaller
and it is -ve when first and third quadrants get larger (Figure 3.8).
II
III
I
II
IV
III
+γ
Figure 3.8 Quadrants
−γ
I
IV
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Plane Strain
A two-dimensional stated-strain exists when the strains are independent of one of the
coordinates. Such a state is described by strains ∈x , ∈y and γ xy , which are assumed to be
functions of only the x and y coordinates provided a cross-section of the body away from
the ends.
z
Example The assumptions of plane strain
is applicable for bodies that are long and
whose geometry and loading do not vary
significantly in the longitudinal direction
(Figure 3.9).
y
∈z = γzx = γyz = 0
(Longitudinal axis)
x
Relation between Strain and
Displacement in Plane Strain
Figure 3.9 Long beam
Consider a small rectangular element ABCD with sides dx and dy in the unstrained body.
After strain, the element is displaced to the position A′B′C ′D′ (Figure 3.10).
In this case, there are two basic change in deformations: one is a change in length and other
is change in angle.
In the rectangle element ABCD , before strain, the length of AD is dx , after strain, A is
displaced to A′ . Let us denote the xy components of the displacement of the particle at A
by u and v . As u and v vary from point to point in the body. Similarly, the displacements
from B to B′ can be written as u +
∂u
∂v
dx and v + dx
∂x
∂x
The x − projection of A′B′ is therefore, dx +
∂u
∂v
dx and the y − projection is
dx .
∂x
∂x
Therefore, length of line A′B′ can be expressed as
2
( A′B′ )
2
∂u   ∂v 

=  dx + dx  +  dx 
∂x   ∂x 

2
(3.9)
By definition, normal strain in x − direction, ∈x is given by
∈x =
A′B′ − AB
AB
---
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C′
∂u
dy
∂y
y
D′
∂v
dy
∂y
∂v
dy
∂y
v+
Strained
element
D
C
∂u
∂y
dy
∂v
∂x
B′
∂v
dx
∂x
A′
dx
v
x
u
A
B
∂u
dx
∂x
∂u
u+
dx
∂x
v+
∂v
dx
∂x
dx
y
x
O
Figure 3.10 Plane element with strains
AB ∈x = A′B′ − AB
A′B′ = (1+ ∈x ) AB
Squaring, we get
( A′B′ )
2
= (1+ ∈x )
2
( AB )
2
---
(3.10)
Equating (3.9) and (3.10), that is simplifying
2
2
∂u   ∂v 
2
2

dx  +  dx  = (1+ ∈x ) ( AB )
 dx +
∂x   ∂x 

2
or
But AB = dx
2
2
 ∂u   ∂v 
 1 +  +   = (1+ ∈x )
 ∂x   ∂x 
2
1+ 2
2
∂u  ∂u   ∂v 
+   +   = 1+ ∈2x +2 ∈x
∂x  ∂x   ∂x 
Ignoring higher order terms due to small quantities, we get
∂u
∂x
∂u
∴∈x =
∂x
2 ∈x = 2
---(3.11a)
Similarly, the longitudinal strain component in y − direction is obtained as
∈y =
(3.11b)
∂v
∂y
---
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In order to find shear strain, the line AB after strain or A′B′ , is inclined to its unstrained
direction by a small angle
∂v
and in the direction, A′D′ is inclined to AD by the small
∂x
∂u
. From this, by definition,
∂y
π
γ xy = − B′A′D′
2
π  π ∂u ∂v 
= − −
−
2  2 ∂y ∂x 
π π ∂u ∂v
= − + +
2 2 ∂y ∂x
∂u ∂v
∴ γ xy =
+
∂y ∂x
angle
---(3.11c)
Strains in Three-dimensional Element
The strain components in the three-dimensional case can be obtained in a similar manner as
the case plane-strain components, in this case we get:
Normal strains in the direction of x, y and z − axes are given by
∂u
∂x
∂v
∈y =
∂y
∂w
∈z =
∂z
∈x =









---(3.12a)
Shear-strains in the perpendicular axes, xy, yz and zx are given by
∂u ∂v
+
∂y ∂x
∂v ∂w
= +
∂z ∂y
∂u ∂w
=
+
∂z ∂x
γ xy =
γ yz
γ zx









---
(3.12b)
where u , v and w are longitudinal displacements of a point A on the element about axes
x, y and z respectively.
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STRESS-STRAIN RELATIONS
In order to subject the behaviour of engineering materials to a mathematical analysis,
irrespective of the independent nature of the equilibrium equations and strain-displacement
relations, in practice, it is essential to study the general behaviour of materials under applied
loads including these relations. This becomes necessary due to the application of a load,
stresses, deformations and hence strains will develop in a body. Therefore, in a threedimensional body, there will be 15 unknowns, namely, three displacements ( u , v, w ) , six
(
strains ∈x , ∈y , ∈z , γ xy , γ yz , γ zx
)
(
)
and six stresses σ x , σ y , σ z , τ xy , τ yz , τ zx . In order to
determine these fifteen unknowns, we have only nine equations, such as three equilibrium
equations and six strain-displacement equations. It is important to note that the six more
equations are required to find remaining six unknowns. Hence, the additional six equations
have to be based on the relationships between six stresses and six strains. The equations
obtained from six stresses and strains relate the relationship between stresses and strains.
Generalised Hooke's Law
Hooke's law of proportionality between forces and displacements can be related as
'Extension is proportional to force', and it refers to the average extension of a thin rod when
it is subjected to a tensile stress.
The Hooke's law can be written in the following form,
σ = D∈
---
(3.13)
where σ and ∈ are stress and strain respectively, D is a constant.
A natural generalised Hooke's law suggests that when more than one strain component exists
and when the elastic limit is not exceeded, then at every point of the medium each of the six
stress components may be expressed as a linear functions of the six components of strains
and
conversely.
This
statement
is
called
generalised
Hooke's law.
Expressed mathematically, we have the six stress-strain equations of the type
σ x = D11 ∈x + D12 ∈y + D13 ∈z + D14γ xy + D15γ yz + D16γ zx 
M
τ zx = D61 ∈x + D62 ∈y + D 63 ∈z + D64γ xy + D65γ yz + D66γ zx




---
(3.14)
or conversely, the six strain-stress equations of the type
∈x = D11′ σ x + D12′ σ y + D13′ σ z + D14′ τ xy + D15′ τ yz + D16′ τ zx 

M

′
′
′
′
′
′
γ zx = D61σ x + D62σ y + D 63σ z + D64τ xy + D65τ yz + D66τ zx 
(3.15)
---
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′ , D12′ ..... define the elastic properties of the material.
where D11 , D12 ..... and D11
Now, we will show generalised Hooke's law for the principal stresses (Planes on which no
shear strain components, i.e., only normal stresses, σ 1 , σ 2 and σ 3 exists) and principal
strains (planes on which no shear strain components, i.e., only normal strains, ∈1 , ∈2 and ∈3
exists) case for a three-dimensional body.
From the relation (3.14), the principal stresses in terms of principal strains are
of the form
σ 1 = D11 ∈1 + D12 ∈2 + D13 ∈3 

σ 2 = D21 ∈1 + D22 ∈2 + D23 ∈3 
σ 3 = D31 ∈1 + D32 ∈2 + D33 ∈3 
---
(3.16)
where σ 1 , σ 2 , σ 3 and ∈1 , ∈2 , ∈3 are principal stresses and principal strains respectively.
Dij represents the elastic property of the material relating the stress in the direction i to the
strain in the
j−
direction.
From the consideration of isotropy, i.e., elastic properties of the body are the same in all
directions about any point, the effect of an extension ∈1 on the stress σ 1 must be same as
the effect of ∈2 on σ 2 and the effect of ∈3 on σ 3 .
Thus, D11 = D22 = D33
Similarly, the effect on σ 1 of extensions ∈2 and ∈3 must be indistinguishable.
Thus, D12 = D13
Similarly, we can show, D21 = D23 and D31 = D32 .
Finally, we can denote the above terms with a and b constants as
D11 = D22 = D33 = a and D12 = D21 = D13 = D31 = D23 = D32 = b
Equation (3.16) is written as
σ 1 = a ∈1 +b (∈2 + ∈3 ) 

σ 2 = a ∈2 +b (∈1 + ∈3 ) 
---

σ 3 = a ∈3 +b (∈1 + ∈2 ) 
(3.17)
By letting a − b = 2 µ and b = λ and denoting e =∈1 + ∈2 + ∈3 , equation (3.17) can be
rewritten as
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σ 1 = λ e + 2 µ ∈1 

σ 2 = λ e + 2 µ ∈2 
σ 3 = λ e + 2 µ ∈3 
---
(3.18)
The constants µ and λ are called constants of Lame.
The stress-strain relations referring to an arbitrary system of Cartesian coordinate axes x, y
and z may be obtained by using the formulas for the transformation of stress and strain
components.
Let x, y and z be such arbitrary system coordinate axes and expressed through principal
axes 1, 2 and 3 through direction cosines (Figure 3.11).
2
x
y
β
γ
α
1
z
3
Figure 3.11 Principal and arbitrary coordinate systems
Since, 1, 2 and 3 are principal direction, the shearing stress and shearing strains referring to
these directions are zero. From the formulae for the transformation of stress components, on
the x − face, are
σ x = l12σ 1 + m12σ 2 + n12σ 3


and τ xy = l1l2σ 1 + m1m2σ 2 + n1n2σ 3 

---(3.19a)
Similarly, from the formulae for the transformation of straining


= l1l2 ∈1 + m1m2 ∈2 + n1n2 ∈3 
∈x = l12 ∈1 + m12 ∈2 + n12 ∈3
and γ xy
(3.19b)
---
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where l , m, n are direct cosines which define the x, y and z coordinates with respect to 1,
2 and 3 principal coordinates. The angles between them are α , β and γ .
Substitution of equation (3.18) into equation (3.19a), yields
σ x = ( λ e + 2µ ∈1 ) l12 + ( λ e + 2µ ∈2 ) m12 + ( λ e + 2µ ∈3 ) n12
(
)
(
)
= l12 + m12 + n12 λ e + l12 ∈1 + m12 ∈2 + n12 ∈3 2 µ
and
---(3.20a)
τ xy = l1l2 ( λ e + 2µ ∈1 ) + m1m2 ( λ e + 2µ ∈2 ) + n1n2 ( λ e + 2µ ∈3 )
= ( l1l2 + m1m2 + n1n2 ) λ e + ( l1l2 ∈1 + m1m2 ∈2 + n1n2 ∈3 ) 2 µ
---
(3.20b)
From the direction cosine relations
l12 + m12 + n12 = 1 and l1l2 + m1m2 + n1n2 = 0 and also
e =∈1 + ∈2 + ∈3 =∈x + ∈y + ∈z
Therefore, equation (3.20a) becomes,
σ x = λ e + 2 µ ∈x
---(3.21a)
and equation (3.20b) becomes
τ xy = µγ xy
---
(3.21b)
Similar relations can be obtained in an analogous manner for the other stress components
and we arrive at the following six relations which are the generalised Hooke's law for
isotropic materials referring to any arbitrary Cartesian coordinates x, y and z .
σ x = λ e + 2 µ ∈x 
σ y = λ e + 2µ ∈y 
σ z = λ e + 2 µ ∈z
τ xy = µγ xy
τ yz = µγ yz
τ zx = µγ zx






(3.22)
Solving the first three equations in (3.22) for ∈x , ∈y and ∈z . That is,
Let first equation of (3.22),
σ x = λ ( e ) + 2µ ∈x
---
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= λ (∈x + ∈y + ∈z ) + 2 µ ∈x
∴σ x = ( λ + 2 µ ) ∈x +λ ∈y + λ ∈z
Thus expressing all three stresses in the form,
σ x = ( λ + 2 µ ) ∈x + λ ∈y +λ ∈z 

σ y = ( λ + 2 µ ) ∈y + λ ∈x +λ ∈z 
---(3.23)

σ z = ( λ + 2 µ ) ∈z +λ ∈x + λ ∈y 
Equation (3.23) can be written in matrix form
λ
λ  ∈x  σ x 
λ + 2 µ
   
 λ
λ + 2µ
λ  ∈y  = σ y 

 λ
λ
λ + 2 µ  ∈z  σ z 
---
(3.24)
Using Gauss elimination method, ∈x , ∈y and ∈z can be obtained as
∈x =
∈y =
∈z =

λ+µ
λ
σx −
σ y + σ z )
(
µ ( 3λ + 2 µ )
2 µ ( 3λ + 2 µ )


λ+µ
λ
σy −
(σ x + σ z )
µ ( 3λ + 2 µ )
2 µ ( 3λ + 2 µ )

---(3.25a)

λ+µ
λ
σz −
σ x +σ y )
(
µ ( 3λ + 2 µ )
2 µ ( 3λ + 2 µ )

and from last three equations of (3.22), the shear strain components are
τ xy
µ
τ
γ yz = yz
µ
τ
γ xy = xy
µ
γ xy =









--(3.25b)
Generalised Hooke's Law in Terms of Engineering Elastic Constants
Let us consider an elementary block which is subjected to the action of normal stress σ x
uniformly distributed over x − face but no other stresses on the six faces. In such a case, the
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ratio of stress to strain is called the modulus of elasticity in tension and it is denoted by letter
E . The modulus of elasticity or Young's modulus symbolically is given by
E=
σx
∈x =
or
∈x
σx
E
---(3.26a)
From experimental results, it is observed that extension of the element in the
x − direction is accomplished by lateral contractions in the y − and z − directions and we
have,
∈y =∈z = −ν ∈x = −ν
σx
E
---(3.26b,c)
where v is a constant called Poisson's ratio.
From generalised Hooke's law, under this system of stresses, equation (3.25a) is reduced to
(σ
y
= σ z = 0)
∈x =
λ+µ
σ
µ ( 3λ + 2µ ) x
∈y =∈z = −
λ
σx
2 µ ( 3λ + 2 µ )
---(3.27a)
---(3.27b,c)
By comparing equations (3.26a) and (3.27a), we find that
E=
µ ( 3λ + 2 µ )
λ+µ
---(3.28a)
and by comparing equations (3.26a,c) and (3.27b,c), we find
ν=
λ
2 (λ + µ )
---
(3.28b)
The ratio between the shearing stress component and its corresponding shear strain
component is called the modulus of elasticity in shear or the modulus of rigidity and is
denoted by the letter G . That is,
G=
τ
γ
(3.29)
From equation (3.25b) for τ xy or τ yz or τ zx , we see that G = µ .
Solving for G from equations (3.28a) and (3.28b), in terms of E and ν , we have
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G=
E
2 (1 + ν )
---
(3.30)
Three-dimensional case
In terms of engineering elastic constants, the generalised Hooke's law may be written (i.e.,
equation (3.25)) as,
∈x =
λ+µ
λ
σx −
(σ y + σ z )
µ ( 3λ + 2 µ )
2µ ( 3λ + 2 µ )
Substituting value for E =
µ ( 3λ + 2µ )
2ν G
into above equation, we
, µ = G and λ =
1 − 2ν
λ+µ
get
∴
∈x =
1
ν
ν
σx − σy − σz
E
E
E
---(3.31a)
Similarly, we obtain other strains as
∈y =
1
ν
ν
σy − σx − σz
E
E
E
---
(3.31b)
1
ν
ν
σz − σx − σy
E
E
E
τ xy 2 (1 +ν )
τ xy
=
=
G
E
∈z =
---(3.31c)
γ xy
---
(3.31d)
γ yz =
γ zx =
τ yz
G
τ zx
G
=
=
2 (1 + ν )
E
2 (1 + ν )
E
τ yz
---(3.31e)
τ zx
---(3.31f)
Equation (3.31) can be expressed in terms of matrix form
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 1
 E

 −ν
∈
 x 
E
∈  
ν
 y  −
∈
E
 z 
γ  =  0
 xy  
γ yz  
 
0
γ zx  

 0


ν
ν
0
0
ν
0
0
ν
E
1
0
0
E
0
E
0
2 (1 + ν )
0
0
0
E
0
2 (1 + ν )
0
0
0
E
0
−
E
1
E
−
−
−
E



0 
 σ x 
0  σ y 
 
 σ z 
0  τ xy 
 τ yz 
 
0 
τ zx 


2 (1 + ν ) 
E 
0
---
(3.32)
Expressing in abbreviated matrix as
{∈}6×1 = [ D′]6×6 {σ }6×1
---
(3.33)
Equation (3.31) or (3.33) gives the strain components in terms of stress components, which
is said to be 'strain-stress relations'. Same equation can be modified to obtain stress
components in terms of strain, which is so called 'stress-strain relations'.
−1
{σ } = [ D′] {∈}
i.e.,
---
(3.34)
−1
For finding [ D′] matrix, the following procedure is used.
First three equations from (3.31) can be added to get stress components like
(σ
y
+ σ z ) , (σ z + σ x ) and (σ y + σ x ) .
i.e., from equation (3.31a), we have
∈x + ∈y + ∈z =
=
∴ (σ y + σ z ) =
1
2ν
σ x + σ y + σ z ) − (σ x + σ y + σ z )
(
E
E
1 − 2ν
(σ x + σ y + σ z )
E
E
(∈x + ∈y + ∈z ) − σ x
1 − 2ν
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Similarly, we obtain,
(σ x + σ z ) =
E
(∈x + ∈y + ∈z ) − σ y
1 − 2ν
(σ
E
(∈x + ∈y + ∈z ) − σ z
1 − 2ν
x
+σ y ) =
Equation (3.31) is rearranged such that above terms can be made to substitute and
simplifying further to obtain σ x , σ y and σ y . That is, from (3.31a), we have
∈x =
∈x =
σx ν
− (σ y + σ z )
E E
σx
∴
E
−
ν  E

∈x + ∈y + ∈z ) − σ x 
(

E 1 − 2ν

σx =
E
{(1 −ν ) ∈x +ν ∈y +ν ∈z }
(1 + ν )(1 − 2ν )
---(3.35a)
Similarly, we can obtain other two stresses as
σy =
E
{(1 −ν ) ∈y +ν ∈x +ν ∈z }
(1 +ν )(1 − 2ν )
---
σz =
E
{(1 −ν ) ∈z +ν ∈x +ν ∈y }
(1 +ν )(1 − 2ν )
---(3.35c)
(3.35b)
and including three shear-stress components from (3.31)
τ xy =
E
E
E
γ xy , τ yz =
γ yz , τ zx =
γ zx
2 (1 + ν )
2 (1 + ν )
2 (1 + ν )
(3.35d,e,f)
Equation (3.35) can be written in matrix form
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
ν
0
0
0
(1 −ν ) ν

0
0
0
 ν (1 −ν ) ν
σ x 

σ y 
ν
0
0
 ν
(1 −ν ) 0
E
σ z 

τ xy  =
1 − 2ν
0
0
0
0
τ  (1 + ν )(1 − 2ν )  0
yz
2

τ 
1 − 2ν
 zx 
 0
0
0
0
0

2
1 − 2ν
 0
0
0
0
0

2



 ∈
 ∈x 
 ∈ y 
 γ z 
 γ xy 
 γ yz 
  zx 



---
(3.36)
Expressing in abbriviated matrix form, we have
{σ }6×1 = [ D ]6×6 {∈}6×1
where [ D ] is the material property matrix.
Two-dimensional case
1. Plane stress
In the case of plane stress, we let σ z = τ zy = τ zx = 0 in equation (3.31) or in (3.32), we
obtain the strain-stress relations as
1
(σ x −νσ y )
E
1
∈y = (σ y −νσ x )
E
∈x =
---(3.37a)
---
(3.37b)
∈z = −
γ xy =
ν
E
τ xy
G
(σ
=
x
+σ y )
2 (1 + ν )
E
---(3.37c)
τ xy
---
(3.37d)
Expressing in the matrix form (excluding ∈z in the matrix)
∈x 
1
  1
∈y  =  −ν
γ  E  0

 xy 
(3.38)
−ν
1
0
 σ x 
 
0  σ y 
2 (1 + ν )  τ xy 
0
---
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or
{∈}3×1 = [ D′]3×3 {σ }3×1
---
(3.39)
Alternatively, equation (3.39) can be expressed in stress-strain relations as
−1
{σ } = [ D′] {∈}
---
(3.40)
−1
To find [ D′] , following procedure is used.
Adding only ∈x and ∈y terms from (3.37) and find σ x and σ y
1
ν
σ x + σ y ) − (σ x + σ y )
(
E
E
1 −ν
∈x + ∈y =
(σ x + σ y )
E
E
∴σ x =
(∈x + ∈y ) − σ y
1 −ν
E
Similarly, we have σ y =
(∈x + ∈y ) − σ x
1 −ν
∈x + ∈y =
i.e.,
Introducing value for σ y into equation (3.37a), we get
∈x =
Thus,
σx =
σx
E
−
ν  E

∈x + ∈y ) − σ x 
(

E 1 − ν

E
{∈x +ν ∈y }
1 −ν 2
(
)
---(3.41a)
Similarly, we obtain by introducing σ x into (3.37b),
σy =
E
{ν ∈x + ∈y }
1 −ν 2
(
)
---
(3.41b)
and the shear stress component is
τ xy =
E
γ xy
2 (1 + ν )
---(3.41c)
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Equation
(3.41)
σ x 
E
 
σ y  =
2
τ xy  1 −ν
(
can
)

 1

 ν


 0

be
ν
1
0
written

0 
 ∈ 
 x
0  ∈y 

 γ xy 
1 −ν 
2 
in
matrix
form
---
(3.42)
Expressing in abbriviated matrix from, we have
{σ }3×1 = [ D ]3×3 {∈}3×1
2.
Plane strain
In the case of plane strain, we let, ∈z = γ yz = γ zx = 0 in equation (3.35) or (3.36) and we
obtain stress-strain relations as
σx =
E
{(1 −ν ) ∈x +ν ∈y }
(1 +ν )(1 − 2ν )
---(3.43a)
σy =
E
{(1 −ν ) ∈y +ν ∈x }
(1 +ν )(1 − 2ν )
---
(3.43b)
τ xy = Gγ xy =
and
σz =
E
γ xy
2 (1 + ν )
νE
(1 +ν )(1 − 2ν )
(∈
x
---(3.43c)
+ ∈y )
---
(3.43d)
Taking only first three relations from (3.43), the matrix form of these relations is given by

(1 −ν ) ν
σ x 

E
 
1 −ν
σ y  =
 ν
τ  (1 + ν )(1 − 2ν ) 
 xy 
0
 0


 ∈x 
 
0  ∈y 
1 − 2ν  γ xy 

2 
0
---
(3.44)
or
{σ }3×1 = [ D ]3×3 {∈}3×1
(3.45)
The strain-stress relations are obtained from (3.45) as
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−1
{∈} = [ D ] {σ }
---
(3.46)
or
∈x 
  1 +ν
∈y  =
E
γ 
xy
 
(1 − ν )

 −ν
 0

0  σ x 
 
0  σ y 
2  τ xy 
−ν
(1 −ν )
0
---
(3.47)
If we compare equations (3.38) and (3.42), it is evident that if we assume state of plane
stress, we need not have a corresponding plane strain state and conversely.
TWO-DIMENSIONAL CYLINDRICAL (OR POLAR)
CO-ORDINATE SYSTEM
In elasticity problems, the proper choice of the co-ordinate system is very important, since
the choice establishes the complexity of the mathematical expressions employed to satisfy
the elasticity equations. In discussing problems with circular boundaries, it is more
convenient to use the cylindrical (or polar) co-ordinates r , θ and z . That is, by employing
polar co-ordinates reference frame, the equations of elasticity must be expressed in terms of
polar co-ordinates.
Differential Equations of Equilibrium of Plane Element
Consider the equilibrium of infinitesimal cylindrical element ABCD subjected to a general
system of positive two-dimensional stresses as well as body forces, is as shown in Figure
3.12 on θ , r plane and z − co-ordinate is perpendicular to the plane of paper.
In the case of plane stress, we have σ z = τ rz = τ θ z = 0 and other stress components are as
functions of r and θ only.
The equilibrium equations are obtained by resolving the forces in the radial direction as well
τθr +
y
σθ +
∂τθr
dθ
∂θ
∂σθ
dθ
∂θ
C
τrθ +
∂τrθ
dr
∂r
σr +
∂σr
dr
∂r
dr
Fθ
Fr
D
B
r
σr
τrθ A
τθr
σθ
dθ
θ
Figure 3.12 Two-dimensional element
x
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as tangential direction.
The radical components of the forces due to σ r and σ r +
∂σ r
dr are (assuming unit
∂r
thickness, i.e., dz = 1 )
∂σ r 

dr  ⋅ ( r + dr ) dθ ⋅1 − σ r r ⋅ dθ ⋅1
σ r +
∂r


---
(3.48)
The radial components of the forces due to τ θ r , τ θ r +
∂τθ r
∂σ
dθ , σ θ , σ θ + θ dθ can be
∂θ
∂θ
seen from the Figure 3.13 and as
−σ θ dr sin
∂σ
∂τ
dθ
dθ 
dθ 
dθ


− τ θ r dr cos
−  σ θ + θ dθ  dr sin
+  τ θ r + θ r dθ  dr cos
2
2 
∂θ
2 
∂θ
2


---
(3.49)
τ θr +
∂τθr
dθ dr
∂θ
dθ
2
dθ
dθ
2
2
+ve
−ve
τθr dr
∂ σθ
σθ +
dθ dr
∂θ
dθ
2
dθ
dθ
2
2
dθ
Figure 3.13 Radial components of the forces
Since, dθ is small, then
sin
dθ dθ
dθ
≈
and cos
≈1
2
2
2
Therefore, equation (3.49) becomes,
σθ dr
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−σ θ dr
∂σ
∂τ
dθ

 dθ 

− τ θ r dr −  σ θ + θ dθ  dr
+  τ θ r + θ r dθ  dr
2
∂θ
2 
∂θ



---
(3.50)
Let Fr and Fθ be the components of the body forces per unit volume in the radial and the
tangential directions, respectively.
The equilibrium equation in the radial direction is obtained by summing the radial forces,
i.e., equations (3.48) and (3.50) and Fr rdθ dr .
∂σ r
∂σ
drrdθ + r drdrdθ − σ r rdθ
∂r
∂r
∂τ
dθ
dθ ∂σ θ
dθ
−σ θ dr
− τθ r dr − σ θ dr
+
dθ
dr + τθ r dr + θ r dθ dr + Fr rdθ dr = 0
2
2
∂θ
2
∂θ
i.e., σ r rdθ + σ θ drdθ +
Neglecting higher-order terms and dividing this equation by volume = r ⋅ dθ ⋅ dr ⋅1 , we get
the equilibrium equation in the radial direction as
σ r ∂σ r σ θ 1 ∂τ θ r
∂σ r 1 ∂τ rθ σ r − σ θ
+
−
+
+ Fr = 0 or
+
+
+ Fr = 0
r
∂r
r r ∂θ
∂r r ∂θ
r
---(3.51)
The equation of equilibrium in the tangential direction can be obtained in the similar manner.
i.e., referring to Figure 3.14, resolving forces in tangential direction, we get
τrθ +
dr τθr +
∂τrθ
dr
∂r
(Sign)
∂τθr
dθ
∂θ
2
τθr rdθ
dθ
∂σθ
σθ +
dθ dr
∂θ
dθ
Fθ rdrdθ
σθ dr
2
τθr dr
dθ
dθ
2
2
Figure 3.14 Tangential force components
∂τ
∂σ
dθ 
dθ



−τ rθ rdθ +  τ rθ + rθ dr  ( r + dr ) dθ − σ θ dr cos
+  σ θ + θ dθ  dr ⋅ cos
+
∂r
2 
∂θ
2



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τθ r dr ⋅ sin
∂τ
dθ 
dθ

+  τ θ r + θ r dθ  dr ⋅ sin
+ Fθ r ⋅ dθ ⋅ dr = 0
2 
∂θ
2

Since, dθ << 1 , sin
dθ dθ
≈
2
2
and cos
dθ
≈ 1 , and ignoring higher orders terms,
2
we get
2τ rθ dr ⋅ dθ +
∂τ rθ
∂σ
dr ⋅ dθ ⋅ r + θ dθ dr + Fθ rdθ ⋅ dr = 0
∂r
∂θ
Dividing this equation by volume = r ⋅ dθ ⋅ dr , we obtain the equilibrium equation in
tangential direction as
1 ∂σ θ ∂τ rθ 2τ rθ
+
+
+ Fθ = 0
r ∂θ
∂r
r
---
(3.52)
If the body forces are neglected, the equilibrium equations (3.51) and (3.52) are modified to
∂σ r 1 ∂τ rθ σ r − σ θ

+
+
= 0

∂r r ∂θ
r

∂τ rθ 1 ∂σ θ 2τ rθ
+
+
=0 
∂r
r ∂θ
r

---
(3.53)
Strain-displacement Relationships
From the Figure 3.15, let A′B′C ′D′ be the position of the element ABCD after strain.
Let u , v be the displacements of the point A in the radial and tangential directions,


respectively. Then the displacements of B are  u +
∂u 
∂v 

dr  and  v + dr 
∂r 
∂r 

y
C′
∂u
rdθ
r∂θ
H
∂v
rdθ
r∂θ
r
∂v d
I′
∂r
v
D′
dr
B′
∂v
∂r
C
I
D
∂u
dr
∂r
A′
r
v
r
v
r
v
B
A
u
u+
∂u
dr
∂r
From diagram,
dθ
θ
x
O
Figure 3.15 Plane element
HA′D′ =
∂u
rdθ
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∂v
∂r
HA′I = 90°
A′I ′ parallel to AB
v
IOB = = IA′I ′
r
B′A′I ′ =
Then, square of length A′B′ is
2
∂u
∂v
2
( A′B′ ) =  dr + dr  +  dr 
∂r   ∂r 

2
---
(3.54)
By strain definition, the radial strain, ∈r is given by
∈r =
or
A′B′ − AB
AB
A′B′ = (1+ ∈r ) AB = (1+ ∈r ) dr
Squaring above term, we get
( A′B′ )
2
2
= (1+ ∈r ) dr 2
---
(3.55)
Equating ((3.54) and (3.55), we get
2
(1+ ∈r )
2
∂u   ∂v 

dr 2 =  dr + dr  +  dr 
∂r   ∂r 

2
or
∂u  ∂u   ∂v 
1 + 2 ∈r + ∈ = 1 + 2 +   +  
∂r  ∂r   ∂r 
2
2
2
r
Neglecting higher order terms, finally we get the radial strain as
∈r =
∂u
∂r
---
(3.56)
The longitudinal-strain component in the tangential direction depends on both
u and v .
Before strain,
AD = r ⋅ dθ
(3.57)
---
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After strain, owing to the displacement u , the length of AD becomes ( r + u ) dθ . At the
same time the point A′ has undergone a tangential-displacement v and the point D′ a


tangential-displacement  v +
∂v

rdθ 
r ∂θ

Hence, length of A′D′ is obtained from diagram,
2
( A′D′ )
2
∂v

  ∂u

rdθ  + 
rdθ 
= ( r + u ) dθ +
r
∂
θ
r
∂
θ

 

2
---
(3.58)
By definition of strain in the tangential direction, the tangential strain, ∈θ is given by
∈θ =
A′D′ − AD
AD
A′D′ = (1+ ∈θ ) AD = (1+ ∈θ ) rdθ
or
Squaring on both sides we get
( A′D′ )
2
2
= (1+ ∈θ ) ( rdθ )
2
---
(3.59)
Equating
(3.58)
2
(1+ ∈θ ) ( rdθ )
2
and
(3.59)
and
2
2
2
2
∂v

  ∂u

= ( r + u ) dθ +
rdθ  + 
rdθ 
r
r
∂
θ
∂
θ

 

∂v
2
2
  ∂u

rdθ  + 
rdθ 
(1+ ∈θ ) ( rdθ ) = rdθ + udθ +
r∂θ

  r∂θ

2
u ∂v
∂u
2
(1+ ∈θ ) = 1 + +  +  
 r r ∂θ   r ∂θ 
2
simplifying.
2
2
u ∂v  u 
u ∂v
∂v  ∂v   ∂u 
1 + 2 ∈θ + ∈θ = 1 + 2 +
+  +2
+
+
 +

r r ∂θ  r 
r r ∂θ r ∂θ  r∂θ   r ∂θ 
2
2
After neglecting higher order terms, we get
∴
∴
u
∂v
2 ∈θ = 2 + 2
r
r ∂θ
∈θ =
u 1 ∂v
+
r r ∂θ
---
(3.60)
Therefore, equations (3.56) and (3.60) are normal strains in the radial and tangential
directions, respectively.
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Let us now consider the shearing strain γ rθ . The angle between the directions AB and A′B′
∂v
∂u
, and the angle between AD and A′D′ or angle D′A′H is
. The
∂r
r ∂θ
change in the angle DAB or the shear strain γ rθ is therefore,
or angle B′A′I ′ is
γ rθ = D′A′H + B′A′I
= D′A′H + B′A′I ′ − IA′I ′
γ rθ =
∂u ∂v v
+ −
r ∂θ ∂r r
---
(3.61)
Stress-strain Relations in Two-dimensions
i)
For plane stress case: σ z = τ zθ = τ zr = 0
a)
Strains in terms of stresses;
1
(σ r −νσ θ ) 
E

1
∈θ = (σ θ −νσ r ) 

E

τ rθ

γ rθ =

G

ν
∈z = − (σ r + σ θ ) 
E

∈r =
---
(3.62)
b)
Stresses in terms of strains;
σr =
σθ =
E
(∈r +ν ∈θ )
1 −ν 2
(
)
E
(∈θ +ν ∈r )
1 −ν 2
(
)
τ rθ = Gγ rθ









(3.63)
ii)
For plane strain case: ∈z = γ zθ = γ zr = 0 .
a)
Strains in terms of stresses:
---
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1 +ν
E
1 +ν
∈θ =
E
∈r =
γ rθ =
(1 − ν ) σ r −νσ θ 
(1 −ν ) σ θ −νσ r 
τ rθ
G









---
(3.64)
b)
Stresses in terms of strains:
E

(1 −ν ) ∈r +ν ∈θ  
(1 +ν )(1 − 2ν )


E
(1 −ν ) ∈θ +ν ∈r  
σθ =
(1 +ν )(1 − 2ν )


τ rθ = γ rθ G

Eν

∈r + ∈θ ]
σz =
[

(1 +ν ) (1 − 2ν )

σr =
---
(3.65)
Strain-stress Relations in Three-dimensions
Consider an infinitesimal three-dimensional element defined by r , θ
and
z − co-ordinate systems. Let u , v and w are the displacement components in radial (r ) ,
circumferential ( θ ) and axial ( z ) directions.
The strain-displacement relations are given by
∂u
∂r
u 1 ∂v
∈θ = +
r r ∂θ
∂w
∈z =
∂z
∈r =
1 ∂u ∂v v
+ −
r ∂θ ∂r r
∂v 1 ∂w
+
γθ z =
∂z r ∂θ
∂u ∂w
+
γ zr =
∂z ∂r
γ rθ =
The stress-strain relations in matrix form









---(3.66)
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

ν
0
0
0 
 1 −ν ν


 ν 1 −ν ν

0
0
0
σ r 

 ∈r 
σ θ 

 ∈θ 
1
−
0
0
0
ν
ν
ν
σ 

 ∈z 
E
 z

 γ 
τ  =
1 − 2ν
 rθ  (1 + ν )(1 − 2ν )  0
0
0
0
0   rθ 
2

 γ θ z 
τ θ z 

 γ 
τ zr 
1 − 2ν
0
0
0
0   zr 
 0
2


1 − 2ν 
 0
0
0
0
0

2 
---
(3.67)
The strain-stress relations in matrix form

∈r 

∈θ 

∈  1 
 z
γ  = 
 rθ  E 
γ θ z 

γ zr 


1
−ν
−ν
0
0
0
−ν
1
−ν
0
0
0
−ν
−ν
1
0
0
0
0
0
0 
 σ r 
0
0
0  σ 
θ
0
0
0  σ z 
 
2 (1 + ν ) 0
0  τ rθ 
0 2 (1 + ν ) 0  τθ z 
 τ 
0
0 2 (1 + ν )  zr 
---(3.68)
Differential Equations of Equilibrium of Three Dimensional Element
The equilibrium equations for three – dimensional stress (Figure 3.16) may be
generalised from equations (3.51) and (3.52) as by resolving forces in r , θ and
z − directions, we have
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σr+dσr
τ
τ rz+d rz τzr+dτzr
τzθ+dτzθ
σz+dσz
τrθ+dτrθ
σθ +dσ τrθ+dτrθ
θ
τzθ+dτθz
Fθ
Fr
Fz
σz
τzr
τθz
τzθ
τrz
τrθ
τθr
σθ
σr
r
dθ
z
θ
Figure 3.16 Two-dimensional element
∂σ r 1 ∂τ rθ ∂τ rz σ r − σ θ

+
+
+
+ Fr = 0 
r
∂r r ∂θ
∂z

∂τ rθ 1 ∂σ θ ∂τ θ z 2τ rθ

+
+
+
+ Fθ = 0 
∂r
r ∂θ
∂z
r

∂τ rz 1 ∂τ zθ ∂σ z τ rz

+
+
+
+ Fz = 0 
r
∂r r ∂θ
∂z

---
(3.69)
If the body forces Fr , Fθ and Fz are ignored then the equilibrium equations without body
forces are given by
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∂σ r 1 ∂τ rθ ∂τ rz σ r − σ θ

+
+
+
= 0
∂r r ∂θ
∂z
r

∂τ rθ 1 ∂σ θ ∂τ θ z 2τ rθ

+
+
+
=0
∂r
r ∂θ
∂z
r

∂τ rz 1 ∂τ zθ ∂σ z τ rz

+
+
+
=0 
∂r r ∂θ
∂z
r

---
(3.70)
STRESS-STRAIN RELATIONS IN AXISYMMETRIC SOLIDS
A solid of revolution is generated by revolving a planar area about an axis in the same plane,
example includes, boiler shell, pipes, nozzle, etc. Convenient co-ordinates for axisymmetric
geometry are r , θ and z representing radial, circumferential and axial respectively. When
geometry, elastic properties, loads and boundary conditions are all axisymmetric, and
independent in the θ − direction. The material points have only the displacement
components u and w in the radial and axial directions, respectively. Thus, the problem is
mathematically two-dimensional.
Figure 3.17 depicts an example of axisymmetric solid about its axis subjected to loads in
radial and axial directions.
a
Infinitesim
τzr, γzr
σθ
l element
σr, ∈r
σz, ∈z
z
θ
Hollow cylinder
r
Figure 3.17 Axisymmetric solid
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To derive strain-displacement, stress-strain and strain-stress relations for axisymmetrical
solid and symmetric about its z − axis, the stress and strain components are independent with
respect to circumferential co-ordinate θ . Thereby, all derivatives with respect to θ vanish.
That is, the displacement v and strain components γ rθ and γ θ z are equal to zero.
The strain-displacement relations can be obtained for axisymmetric solid by putting,
v = γ rθ = γ θ z = 0 into three-dimensional strain-displacement equation (3.66), we get
∂u
u
,
∈θ =
∂r
r
∂w
∂u ∂w
∈z =
γ rz =
+
∂z
∂z ∂r
∈r =





---
(3.71)
The
stress-strain
relations
expressed
ν
ν
1 −ν
σ r 
 ν
ν
1 −ν
σ 

E
 θ
 ν
 =
ν 1 −ν
σ z  (1 + ν )(1 − 2ν ) 
 0
τ rz 
0
0

in
matrix

∈r 

0  
∈θ 
0  
 ∈z 
1 − 2ν   
γ rz 
2 
form
0
---
(3.72)
Thus, the strain-stress relations of the form
1
∈r 

∈ 
 θ  1  −ν
 = 
∈z  E  −ν
γ rz 
0
−ν
1
−ν
0
−ν
−ν
1
0
0  σ r 
0  σ θ 
 
0  σ z 

2 (1 + ν )  τ rz 
(3.73)
QUESTIONS
1.
Define stress and strain.
2.
What is plane stress? Give an example.
3.
What is plane strain? Give an example.
4.
Derive differential equations of equilibrium of two-dimensional element.
5.
State the generalised Hooke's law in terms of engineering elastic constants.
6.
Define an axisymmetric solid.
7.
Explain plane stress and plane strain methods with relevant equations.
---
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8.
Draw a typical three-dimensional element and indicate state of stress in their positive
senses.
9.
Derive the equations of equilibrium in case of a three-dimensional stress system.
10. Give strain-displacement relations in case of a three-dimensional elasticity problem.
11. Explain the term, ‘Isotropic’ as applied to material properties.