Strength of Materials
Prof. A. S. PATIL
Department of Mechanical Engineering
Sinhgad Academy of Engineering, Pune
Strength of Materials 1

•
Normal & shear stresses on any oblique plane. Concept of principal planes,
• derivation of expression for principal stresses & maximum shear stress,
•
Position of principal planes & planes of maximum shear.
•
Graphical solution using Mohr’s circle of stresses.
•
Principal stresses in shaft subjected to torsion, bending moment & axial thrust (solid as well as hollow), Concept of equivalent torsional and bending moments.
•
Theories of elastic failure: Maximum principal stress theory, maximum shear stress theory, maximum distortion energy theory, maximum strain theory -their applications & limitations.
Strength of Materials 2

Case 1 – Member subjected to axial load
Normal and Shear force on the plane at an angle Ɵ :-
F

P cos

V

P sin

Normal and Shear stress on the plane at an angle Ɵ






F
A

V
A



P cos

A
0
P cos sin


A
0 cos



P
A
0
P
A
0 cos
2 sin

 cos

Strength of Materials 3

Case 2 :- A body subjected to general two dimensional stress system
Stress element showing two-dimensional state of stress
METHODS FOR DETERMINATION OF THE STRESSES
ON AN OBLIQUE SECTION OF A BODY
1. Analytical method
2. Graphical method (Mohr’s circle)
Strength of Materials 4

σ y
σ x

Normal Stress in x- direction
σ y

Normal Stress in y- direction
τ 
Shear Stresses in x & y – directions
θ 
Angle made by inclined plane wrt vertical
σ
θ

Normal Stress on inclined plane AE
τ
θ

Shear Stress on inclined plane AE
σ x
C
τ
D
σ y
θ
P
 Inclination of Principal planes
σ
P
 Principal stresses
θ
S
 Inclination of Max. shear stress planes [θ
S
= θ
P
+ 45 0 ].
All the parameters are shown in their +ve sense in the Fig.
E
τ
θ
σ
B
θ
τ
θ
A
Strength of Materials
σ x
5

σ x
C
σ y
E
τ
θ
σ
B
θ
τ
θ
A
σ x
τ
D
σ y
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 w r t vertical, Anti-Clockwise +ve.
All the parameters are shown in their +ve sense in the Fig .
Strength of Materials 6

σ x
C
τ
D
σ y
E
τ
θ
σ
B
θ
τ
θ
A
σ x
Normal stress on plane AE =


  x cos
2
   y sin
2
   sin




 x
2

1
 cos 2
 


2 y

1
 cos 2
 cos


 
  cos
 sin
 sin 2
 cos




  x
  y
2



  x
  y
2

 cos 2
   xy sin 2

σ y

Shear stress on plane AE =

  x cos
 sin
   y sin
 cos
   sin
2
   cos
2






  x

2
 y


 s in 2
   xy c os 2

Strength of Materials 7

•
There are no shear stresses on principal planes
• the planes where the normal stress (

) is the maximum or minimum
• the orientations of the principal planes (
 p
τ = 0
) are given by equating
At
 p





 x

2
 y

 s in 2

P
  xy c os 2

P

0
.
. .
 p

1
2 tan

1
 x
2

 xy
 y
Which gives two values of
Ɵ differing by 90°.
Thus two principal planes are mutually perpendicular
Strength of Materials 8

•
Principal stresses are the normal stresses (

) acting on the principal planes (planes which are at an angle of
Ɵp and
Ɵp+90, where the shear stress is zero).
 max
 min
 
1
 
2

  x
  y
2

  x
  y
2



R



R where R

  x
  y
2


2
  xy
2
Strength of Materials 9

 max
•
To find maximum value for shear stress and its plane (
 s
), differentiate the equation of shear stress and equate to zero



  x
  y
2

 s in 2

P
  xy c os 2

P d d



 

  
 x y cos 2
 
2
 xy sin 2
 
0 tan 2
  
 x

2

 y
• orientations of the two planes (
 s
) are given by:
Strength of Materials 10

 max
 s

1
2 tan

1 
   x
2
 y gives two values (Ɵs1 and Ɵs2) differs by 90°
•
Thus maximum shear stress occurs on two mutually perpendicular planes
 max



 x

2
 y


2
  2
In terms of principal stresses
 max


1
 
2
2
•
Also, 𝜃 𝑠
= 𝜃 𝑝
+45°
Strength of Materials 11

σ y





 x

2
 y





 x

2
 y

 cos 2

σ x
C E
τ
θ
σ
B
θ
θ
σ x
D
A



  x
  y
2

 s in 2

σ y
•
Principal stresses are at
Ɵp
=0 and
Ɵp
=90
σ1 , σ2 = 𝜎 𝑥 , 𝜎 𝑦
•
Max. shear stress
 max


1
 
2
2

 x
  y
2
Strength of Materials 12

•
Case 4 – Member subjected to simple shear stress ( 𝜎 𝑥
, 𝜎 𝑦
=0)
C E
τ
θ
σ
B
θ
τ
θ
A


  sin 2

τ
D


  cos 2

•
For Principal stress,
• 𝜃 𝑝
=45,135


  cos 2
 
 
1

0
   
2
Strength of Materials 13

Maximum Shear
 s2
90

 s1 x
Strength of Materials 14

45  Principal plane x
Maximum shear plane
 p
=  s
45 
Strength of Materials 15