INTRODUCTION
The state of stress on any plane in a strained body is said to be
‘Compound Stress’, if, both Normal and Shear stresses are acting on
that plane. For, example, the state of stress on any vertical plane of a
beam subjected to transverse loads will, in general, be a Compound
Stress. In actual practice the state of Compound Stress is of more
common occurrence than Simple state of stress.
In a compound state of stress, the normal and shear stress may have
a greater magnitude on some planes which are inclined (or, Oblique)
to the given stress plane. Hence it is necessary, in a compound state of
stresses, to find (i) The normal and shear stress on a plane which is
inclined (or, Oblique) to the given stress plane; (ii) The inclination of
max. and min. normal stress planes and the values of the normal stress
(max. / min.) on them; (iii) The inclination of max. shear stress planes
and the values of the shear stress (max.) on them.
1
2
Consider an element ABCD subjected to a state
of compound stress as shown in the Fig. Let:
σx  Normal Stress in x- direction
σy  Normal Stress in y- direction
σx
τ Shear Stresses in x & y – directions
θ  Angle made by inclined plane AE wrt vertical
σn Normal Stress on inclined plane
τ
τn Shear Stress on inclined plane
σy
τ
C
E
B
θ
D
σx
A
σy
(i) Normal & Shear stress on plane inclined (Oblique) to given stress plane:
Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:
 x  y
   
2

   x  y
  
2
 
  x  y
   
2


 cos 2   sin 2    (1)


 sin 2   cos 2    (2)

2
(ii) The inclination of max. and min. normal stress planes and
the values of the normal stress (max. / min.) on them
Let, θP be the inclination of the plane of max. or min. normal stress
and σP be the value of the max. or min. normal stress on that plane,
then, from Eqn. (1):
 x  y
 P  
2

   x  y
  
2
 

 cos 2 P   sin 2 P - - - (1)

Thus, the condition for max. or
dσ
min. normal stress to occur on a
For σ P to be max.or min., P  0
dθ
plane is, shear stress on that plane
  x  y 
 2 sin 2 P    2 cos 2 P   0 should be zero. These planes on
 
 2 
which shear stress is zero and the
  x  y 
normal stress on them being either
sin 2 P    cos 2 P   0
 
the max. or the min. are called
 2 
   0
‘PRINCIPAL PLANES’.
P
2
From,  P  0, we have,
  x  y

 2


sin 2 P    cos 2 P   0  tan 2 P 
- - - (3)



 x
y 



 2 
 The above Eqn. (3), gives two values for θP, which differ
by 900. Thus, there are two mutually perpendicular
Principal planes, on which there are only normal stresses,
shear stress being zero on them.
 On one of them, the value of the normal stress is the max.; it is
called the ‘Major Principal plane’, the max. normal stress on it is
called the ‘Major Principal Stress’.
 On the other principal plane, the value of the normal stress is the
min.; it is called the ‘Minor Principal plane’, the min. normal stress
From t an 2 P 
2P
sin 2 P  
(σx-σy)/ 2
cos 2 P  
τ

  x  y

2




, we get ,

 ( 1   2 ) 
 2


2

 1   2  / 2
2
 ( 1   2 ) 
 2


2

2
Substituting for sin 2θP and cos 2θP in Eqn. (1), we get the equation
for principal stresses as:
P



x
 y 
2
  x  y 
 

2
2
   2

   (4)
The above equation (4) gives two values for principal stresses. The
numerically max. of the two values (+ ve or − ve) is the Major Principal
Stress, (σMajor or σMax) and the numerically min. (+ ve or − ve) is the
Minor Principal Stress (σMinor or σMin) .
(iii) Inclination of max. shear stress planes, Max. shear stress Equation.
Let, θS be the inclination of the plane of max. or min. shear stress
and τS be the value of the max. or min. shear stress on that plane,
then, from Eqn. (2):
  x  y
 S  
2


 sin 2 S   cos 2 S

d S
For S to be max.or min.,
0
dθS
  x  y 
2 cos 2 S     2 sin 2 S   0
 
 2 
  x  y 


2 
 tan 2 S   
- - - (5)

NOTE : We have tan2 P  tan 2 S  1
   (2)
Eqn. (5) gives two values for θS,
which differ by 900. Thus, there
are two mutually perpendicular
planes, on which shear stress are
max.; numerically equal but
opposite in sense.
The planes of Max. Shear
stresses are inclined at 450 to the
Principal planes.
  x  y

2
From t an 2 S   
cos 2 s  
(σx-σy)/ 2
2S
sin 2 S  
τ




 , we get ,
 ( 1   2 ) 
 2


2

 1   2  / 2
2
 ( 1   2 ) 
 2


2

2
Substituting for sin 2θS and cos 2θS in Eqn. (2), we get the equation
for Max. shear stresses as:
 max.
  x  y
  
2

2

   2

  max.    Major   Minor


    (6)


The above equation (6) gives two values for Max. shear stresses, which
are numerically equal but opposite in sense.
EQUATINS, NOTATIONS & SIGN CONVENTIONS :
2
Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:
  x  y    x  y 
  
 cos 2   sin 2    (1)
   
2  
2 

  x  y 
 sin 2   cos 2    (2)
   
2


Inclination of Principal planes θP . There are
two mutually perpendicular Principal planes
The equation for
Principal stresses are:
P


x
 y 
2
t an 2 P 
  x  y
 
2


  x  y

2




- - - - (3)
2

   2

   (4)
Inclination of Max. shear stress planes θS = θP + 450. Also, given by
Eqn (5). The two mutually perpendicular Max. shear stress planes have
equal & opposite shear stresses.
2
 max.
  x  y
  
2


   2

  max.    Major   Minor


    (6)


  x  y

2
t an 2 S   



 - - - - (5)
SIGN CONVENTIONS :
σx  Normal Stress in x- direction
σy  Normal Stress in y- direction
τ Shear Stresses in x & y – directions
σx
θ  Angle made by inclined plane wrt vertical
σθ Normal Stress on inclined plane AE
τθ Shear Stress on inclined plane AE
σy
τ
τθ E
C
B
θ
σθ
D
σx
A
τ
σy
All the parameters are shown in their +ve sense in the Fig.
Normal stresses, σ  Tensile stresses +ve.
Shear Stresses, τ, in x – direction & Inclined Plane  Clockwise +ve.
Shear Stresses, τ, in y – direction  Anti-Clockwise +ve.
Angle, θ  measured wrt vertical, Anti-Clockwise +ve.
1]. The state of stress at a point in a strained material is shown in Fig.
(i) Locate the principal planes and find the principal stresses,
(ii) Locate the Max. shear stress planes and find the max. shear stress.
32
Inclination of Principal planes θP wrt Vertical :

t an 2 P 

30
30
 80  (32) 


2


  x  y 


2


 2 P  28.180 or, P  14.090 & 104.090.
80
P
x
 y 
2


80  (32) 


2
A
30
  x  y 


80
D
Principal stresses, σP :



B
C
2
  

2
 80  ( 32) 
2

  30
2
2


  87.53MP a(T );- 39.53MP a (C)
2
32
Stresses are in MPa
32
Inclination of Max. shear stress planes θS = θP + 450.
θS = 14.090+ 450 = 59.090 &149.090. (wrt Vertical)
Max. shear stress τmax :
30
B
C
80
80
 max.    Major   Minor
D
A
30
  87.53  (39.53)  63.53 MP a.
32
2]. The state of stress at a point in an elastic material is shown in Fig.
Find the resultant stress on a plane AE inclined at 550 to horizontal.
40
Shear stress τ = 60 sin 20 = 20.52 Mpa
200
Normal stress σx = 60 cos 20 = 56.38 Mpa
Inclination of AE wrt Vertical θ = 90 – 55 = 350 60
C
E
B
550
A
D
40
Stresses are in MPa
60
200
Normal Stress, σθ, and Shear stress, τθ, on inclined plane AE:
 x  y
2

   x  y 
  
 cos 2   sin 2
2
 

 56.38  40   56.38  (40) 


 cos(2  35)  20.52sin(2  35)
2
2

 

  43.95 MP a (T )
   
  x  y
2


 sin 2   cos 2

 56.38  (40) 

 sin(2  35)  20.52 cos(2  35)
2


  38.26 MP a (Clockwise)
   
Resultant stress, fθ and its inclination α :
2
2
f   2   2  43.95  38.26
40
C
56.38
 f  58.27 MP a.
  
38.26 
  tan1 


43
.
95


 
   41.040
  tan1 
E
B
350
D
20.52
20.52
A
40
Stresses are in MPa
56.38
3]. The principal stresses at a point in a strained material are 80 MPa
and 40 MPa both tensile. Find the normal, tangential and resultant
stress on a plane inclined at 500 to the major principal plane.
Since σx and σy are principal stresses, the
shear stress, i.e, τ, on their planes is zero.
Normal Stress, σθ, and Shear stress, τθ,
on plane inclined at θ = 500 to plane of σx :
 x  y
2

   x  y 
  
 cos 2   sin 2
2
 

 80  40   80  40 


 cos(2  50)  0
 2   2 
  56.53 MP a (T )
   
40
B
C
500
80
D
A
40
Stresses are in MPa
  x  y 
 sin 2   cos 2
   
2


 80  40 

 sin(2  50)  0   19.68 MP a (Clockwise)
2


80
Resultant stress, fθ and its inclination α :
f 
 2   2 
40
56.532  19.682
 f  59.86 MP a.
B
C
500
80
  
 19.68 
  tan    tan1 

 56.53 
  
   19.190
D
1
80
A
40
Stresses are in MPa
4]. At a certain point in a shaft, the normal stresses across two planes
at right angles to each other are 60 MPa (C) and 40 MPa (T). The
major principal stress is known to be 150 MPa (C). Find the shear
40
stresses on the two said planes.
 major



x
 y 
2

  x  y

2

 60  40 
 150 
2
   130.77MP a.
2

   2

60
  60  40 
2

 
2


2
τ
C
B
D
A
τ
60
40
Stresses are
in MPa
5]. A 50mm×100mm tie member of a timber truss has a glued joint
(shown in Fig.) at an inclination of 400 to the longitudinal axis. If it is
subjected to an axial force of 200 kN, check whether there is a risk of
failure. The permissible normal and shear stress for the joint are 25
MPa and 16 MPa respectively.
P
P
Normal Stress σx = (200×103)/(50 ×100)
σx = 40 MPa (T)
40
Normal Stress, σθ, and Shear stress, τθ,

on plane AE inclined at θ = 500 to vert.
plane :
  x  y    x  y 
  
 cos 2   sin 2
   
2
2 

 
 40   40 
      cos(2  50)  0   16.5 MP a (T )
 2   2 
  x  y
2

   

 sin 2   cos 2

E
40
400
A
Stresses are in MPa
τθ > 16 MPa,
There is a risk
 40 

 sin(2  50)  0   19.68 MP a (Clockwise) of failure.
 2 
6]. Determine the max. safe uni-axial force that a strut of C/S area
60 ×103 mm2 can carry safely, if, permissible normal and shear stress are
25 MPa and 12.5 MPa respectively, on a critical plane (shown in Fig.)
P
P
inclined at 300 to the vertical.
Let P be the max. safe load and σx be the
stress under the load P.
Normal Stress, σθ, and Shear stress, τθ,
on plane AE inclined at θ = 300 to vert. :
 x  y
2

   x  y 
  
 cos 2   sin 2
2
 

  x    x 
 25  

 cos(2  30)  0
 2   2 
  x   33.33 MP a - - - - (1)
   
E
σx
300
σx
A
From (1) and (2),
max. safe σx = 28.87 MPa
P = σx × a => 28.87 × 60 ×103
P = 1732.2 ×103 N = 1732.2 kN
  x  y 
 sin 2   cos 2
   
2


  x 
12.5  
 sin(2  30)  0   x  28.87 MP a - -(2)
 2 
7]. At a transverse section of a beam of rectangular C/S, 100mm ×
240mm, the BM is 76.8 kNm (Hogging) and SF is 96 kN. Find at
that section, 40mm above the N.A. (i) Principal Stresses and (ii)
Max. shear stress.
M.I. → I = 100×2403
115.2×106
=
At 40mm above N.A.
(1) Bending stress σ and shear stress τ :
80
mm4
120
80
40
M
76.8 106
  y 
 40  26.67MPa(T )
I
115.2 106
F a y 96103  80100 80


 5.33MPa
6
Iby
115.2 10 100
120
 
100
(2) Principal stress σP and max. shear stress τmax :
P



x
 y 
2


26.67  0 


2
  x  y 


2
C/S
2
   2

 26.67  0 
2

  5.33
2


26.67
2
  27.69MP a (T );- 1.03MP a (C)
C
5.33
B 26.67
D
A
5.33
State of
stress
(2) Principal stress σP and max. shear stress τmax :
26.67
 max.    Major   Minor
  27.69  (1.03)  14.36MP a.
C
5.33
B 26.67
D
A
5.33
State of
stress
8]. A propeller shaft of 200mm external diameter and 100mm internal
diameter is subjected to a torque of 116.14 kNm, BM 10 kNm
(sagging) and a thrust of 100 kN. Find the Max. and Min. principal
stresses and inclinations of their planes.
(2) Stresses at outermost layer
(1) Geometrical Propertise of the C/S :
at the top :
C/S area  A 

PlaneM.I. I 
PolarM.I. J 
4


 2002  1002  23.56103 mm2

64

32




 2004  1004  73.63106 mm4
 2004  1004  147.26106 mm4
P
100103
Due to T hrust, d  
 4.24MPa (C)
A 23.56103
M
10106
Due to BM,  b   re 
100  13.58MPa (C)
I
73.63106
T
161.14106
Due to T orque,   re 
100  109.43MPa
J
147.26106
Net Normalstress, x   d   b  4.24  13.58  17.82MPa (C)
PROBLEM (8) CONTINUED :
(3) Principal stresses and Inclination of their planes :
Principal stresses, σP :
P



x
 y 
2

 17.82  0 

2
17.82
  x  y 


2
2
   2

  17.82  0 
2

  109.43
2


2
  118.7 MP a (C);  100.9 MP a (T )
Inclination of Principal planes θP wrt Vertical :
t an 2 P 


109.43
 17.82  0 


2


  x  y 


2


 2 P  85.340 or, P  42.670 & 132.670 .
C
109.43
B 17.82
D
A
109.43
State of
stress
MOHR’S CIRCLE :
DEFINITION : Mohr’s circle is a graphical method for solving
problems in compound stresses. It is defined as a circle drawn such a
way that, the X and Y – coordinates of a point on the circumference of
the circle representes, respectively, the normal and shear stresses, on
any plane at a point in a material subjected to a Two-Dimensional
stress system.
Sign Conventions: Shear stresses on all the planes, (Hz., Vert.,
Inclined), are taken as +ve, if, anti-clockwise.
All other sign conventions remain the same as for analytical method.
Consider an element ABCD subjected to a state of compound stress
as shown in the Fig. Let:
σ
σx  Normal Stress in x- direction
σy  Normal Stress in y- direction
C
τ Shear Stresses in x & y – directions
σx
θ  Angle made by inclined plane AE wrt vertical
D
τ
σθ Normal Stress on inclined plane
τθ Shear Stress on inclined plane
y
E
B
θ
A
σy
τ
σx
MOHR’S CIRCLE :
(I) CONSTRUCTION OF MOHR’S CIRCLE :
1]. Select a convenient point as the origine and draw the X and Y – axes.
Choose convenient scale to plot the stresses.
2]. Plot the stresses on the vertical plane, taking σx as x-coordinate and τ
as y-coordinate, to get the point ‘V’.
3]. Plot the stresses on the horizontal plane, taking σy as x-coordinate and
τ as y-coordinate, to get the point ‘H’.
4]. Join the points ‘V’ and ‘H’ by a straight lne and mark its point of
intersection with X-axis as ‘C’, to get the center of the Mohr’s Circle.
5]. With c as center and CV/CH as radius draw a circle to pass through
H and V, to get the Mohr’s Circle.
MOHR’S CIRCLE :
(II) TO FIND, PRINCIPAL STRESSES, THEIR PLANES AND MAX.
SHEAR STRESSES, THEIR PLANES :
1]. Mark the points A and B at which the Mohr’s Circle cuts the X-axis.
As y-coordinates of these points are zero, they represent the Principal
stresses.
2]. Measure the lengths OA and OB, the max. of the two represents, to
scale, the Major Principal stress and the min. represents the Minor
Principal stress.
3]. Half of the angle VCA and half of the angle VCB represent the angles
of the respective Principal planes wrt to the vertical.
4]. Mark the points D and E at the ends of the vertical diameter of the
Mohr’s Circle. As y-coordinates of these points are the max., they
represent the Max. shear stresses.
5]. Measure the lengths CD and CE, they represent, to scale, the Max.
shear stress.
6]. Half of the angle VCD and half of the angle VCE represent the angles
of the respective Max. shear stress planes wrt to the vertical.
MOHR’S CIRCLE :
(III) TO FIND NORMAL STRESS, SHEAR STRESS AND
RESULTANT STRESS ON ANY GIVEN INCLINED PLANE ‘AE’ :
1]. Draw the radial line CP in the Mohr’s Circle at an angle ‘2θ’ to CV.
2]. The x-coordinate of ‘P’ represents the Normal stress and
the y-coordinate of ‘P’ represents the Shear stress.
3]. The length ‘OP’ represents the resultant stress on given plane ‘AE’.
4]. The angle POA represents the angle of the resultant stress on the
given plane ‘AE’ wrt to the normal of the plane.