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STRESSES - STRENGTH OF MATERIAL BY IIT MADRAS GATE Notes | EduRev

GATE : STRESSES - STRENGTH OF MATERIAL BY IIT MADRAS GATE Notes | EduRev

``` Page 1

Strength of Materials Prof. M. S. Sivakumar

Stresses

Stress at a point
Stress Tensor
Equations of Equilibrium
Different states of stress
Transformation of plane stress
Principal stresses and maximum shear stress
Mohr's circle for plane stress

Page 2

Strength of Materials Prof. M. S. Sivakumar

Stresses

Stress at a point
Stress Tensor
Equations of Equilibrium
Different states of stress
Transformation of plane stress
Principal stresses and maximum shear stress
Mohr's circle for plane stress

Strength of Materials Prof. M. S. Sivakumar

Introduction
2.1 stress at a point

Figure 2.1

Consider a body in equilibrium under point and traction loads as shown in figure 2.1.
After cutting the body along section AA, take an infinitesimal area ?A lying on the surface
consisting a point C.
The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point
C can be defined,
A0
F
lim
A ??
?
s=
?
2.1
?F is resolved into ?F
n
and ?F
s
that are acting normal and tangent to ?A.
Normal stress,
n
n
A0
F
lim
A ??
?
s=
?

2.2
Shear Stress,
??
?
s=
?
s
s
A0
F
lim
A
2.3

Top

Page 3

Strength of Materials Prof. M. S. Sivakumar

Stresses

Stress at a point
Stress Tensor
Equations of Equilibrium
Different states of stress
Transformation of plane stress
Principal stresses and maximum shear stress
Mohr's circle for plane stress

Strength of Materials Prof. M. S. Sivakumar

Introduction
2.1 stress at a point

Figure 2.1

Consider a body in equilibrium under point and traction loads as shown in figure 2.1.
After cutting the body along section AA, take an infinitesimal area ?A lying on the surface
consisting a point C.
The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point
C can be defined,
A0
F
lim
A ??
?
s=
?
2.1
?F is resolved into ?F
n
and ?F
s
that are acting normal and tangent to ?A.
Normal stress,
n
n
A0
F
lim
A ??
?
s=
?

2.2
Shear Stress,
??
?
s=
?
s
s
A0
F
lim
A
2.3

Top

Strength of Materials Prof. M. S. Sivakumar

2.2 stress Tensor

Figure 2.2
Consider the free body diagram of an infinitesimally small cube inside the continuum as
shown in figure 2.2.
Stress on an arbitrary plane can be resolved into two shear stress components parallel to
the plane and one normal stress component perpendicular to the plane.
Thus, stresses acting on the cube can be represented as a second order tensor with nine
components.
xx xy xz
yx yy yz
zx zy zz
? ?
ss s
? ?
s= s s s
? ?
? ?
ss s
? ?
? ?

2.4

Is stress tensor symmetric?

Figure 2.3
Page 4

Strength of Materials Prof. M. S. Sivakumar

Stresses

Stress at a point
Stress Tensor
Equations of Equilibrium
Different states of stress
Transformation of plane stress
Principal stresses and maximum shear stress
Mohr's circle for plane stress

Strength of Materials Prof. M. S. Sivakumar

Introduction
2.1 stress at a point

Figure 2.1

Consider a body in equilibrium under point and traction loads as shown in figure 2.1.
After cutting the body along section AA, take an infinitesimal area ?A lying on the surface
consisting a point C.
The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point
C can be defined,
A0
F
lim
A ??
?
s=
?
2.1
?F is resolved into ?F
n
and ?F
s
that are acting normal and tangent to ?A.
Normal stress,
n
n
A0
F
lim
A ??
?
s=
?

2.2
Shear Stress,
??
?
s=
?
s
s
A0
F
lim
A
2.3

Top

Strength of Materials Prof. M. S. Sivakumar

2.2 stress Tensor

Figure 2.2
Consider the free body diagram of an infinitesimally small cube inside the continuum as
shown in figure 2.2.
Stress on an arbitrary plane can be resolved into two shear stress components parallel to
the plane and one normal stress component perpendicular to the plane.
Thus, stresses acting on the cube can be represented as a second order tensor with nine
components.
xx xy xz
yx yy yz
zx zy zz
? ?
ss s
? ?
s= s s s
? ?
? ?
ss s
? ?
? ?

2.4

Is stress tensor symmetric?

Figure 2.3
Strength of Materials Prof. M. S. Sivakumar

Consider a body under equilibrium with simple shear as shown in figure 2.3.
() ( )
zyxxz y xyyz x
Mddd ddd =t - t =0

xy yx
t =t

Similarly, .
xz zx yz zy
and t=t t =t
Hence, the stress tensor is symmetric and it can be represented with six
components, , instead of nine components.
xx yy zz xy xz yz
, , , , and ss s t t t
xx xy xz xx xy xz
yx yy yz xy yy yz
zx zy zz xz yz zz
???
ss s s t t
???
s= s s s = t s t
???
???
ss s t t s
???
???
?
?
?
?
?
?

2.5

Top

Page 5

Strength of Materials Prof. M. S. Sivakumar

Stresses

Stress at a point
Stress Tensor
Equations of Equilibrium
Different states of stress
Transformation of plane stress
Principal stresses and maximum shear stress
Mohr's circle for plane stress

Strength of Materials Prof. M. S. Sivakumar

Introduction
2.1 stress at a point

Figure 2.1

Consider a body in equilibrium under point and traction loads as shown in figure 2.1.
After cutting the body along section AA, take an infinitesimal area ?A lying on the surface
consisting a point C.
The interaction force between the cut sections 1 & 2, through ?A is ?F. Stress at the point
C can be defined,
A0
F
lim
A ??
?
s=
?
2.1
?F is resolved into ?F
n
and ?F
s
that are acting normal and tangent to ?A.
Normal stress,
n
n
A0
F
lim
A ??
?
s=
?

2.2
Shear Stress,
??
?
s=
?
s
s
A0
F
lim
A
2.3

Top

Strength of Materials Prof. M. S. Sivakumar

2.2 stress Tensor

Figure 2.2
Consider the free body diagram of an infinitesimally small cube inside the continuum as
shown in figure 2.2.
Stress on an arbitrary plane can be resolved into two shear stress components parallel to
the plane and one normal stress component perpendicular to the plane.
Thus, stresses acting on the cube can be represented as a second order tensor with nine
components.
xx xy xz
yx yy yz
zx zy zz
? ?
ss s
? ?
s= s s s
? ?
? ?
ss s
? ?
? ?

2.4

Is stress tensor symmetric?

Figure 2.3
Strength of Materials Prof. M. S. Sivakumar

Consider a body under equilibrium with simple shear as shown in figure 2.3.
() ( )
zyxxz y xyyz x
Mddd ddd =t - t =0

xy yx
t =t

Similarly, .
xz zx yz zy
and t=t t =t
Hence, the stress tensor is symmetric and it can be represented with six
components, , instead of nine components.
xx yy zz xy xz yz
, , , , and ss s t t t
xx xy xz xx xy xz
yx yy yz xy yy yz
zx zy zz xz yz zz
???
ss s s t t
???
s= s s s = t s t
???
???
ss s t t s
???
???
?
?
?
?
?
?

2.5

Top

Strength of Materials Prof. M. S. Sivakumar

2.3 Equations of Equilibrium

Figure 2.4

Consider an infinitesimal element of a body under equilibrium with sides dx as
shown in figure 2.4.
dy  1 × ×
B
x
, B
y
are the body forces like gravitational, inertia, magnetic, etc., acting on the element
through its centre of gravity.
x
F0 =
?
,
()
() () () ( )
yx
xx
xx y xx yx yx x
dx d 1 dy 1 dy dx 1 dx 1 B dx dy 1 0
xdy
?t ??
?s ??
s+ × -s × +t + × -t × + × × =
??
??
?
??
??

Similarly taking  and simplifying, equilibrium equations of the element in
differential form are obtained as,
y
F 0 =
?
yx
xx
x
xy yy
y
B0
xy
B0
xy
?t
?s
+ +=
??
?t ?s
+ +=
??

2.6

Extending this derivation to a three dimensional case, the differential equations of
equilibrium become,
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