Principal Stresses and Strain
and
Theories of Failure
Strength of Materials
Prof. A. S. PATIL
Department of Mechanical Engineering
Sinhgad Academy of Engineering, Pune
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6.4 THEORIES OF FAILURE
• To predict elastic failure under any conditions from the
behavior of materials in a simple test.
• Different theories for different materials and parameters that
are considered.
1. Maximum Principal Stress Theory
2. Maximum Shear Stress Theory
3. Maximum Distortion Energy Theory
4. Maximum Strain Energy Theory
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1. Maximum principal stress theory
•
•
•
•
•
Maximum principal stress = Yield Stress
Rankine
brittle materials
based on a limiting normal stress
σ1 = σyp
1 
x  y
2
 yp
1

2
x  y 
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 4xy
2
3
2. Maximum Shear stress theory
• Failure occurs when maximum shear stress in complex stress
system is equal to the value of maximum shear stress in simple
tension.
• Tresca, Guest, Coulomb
• ductile Material
• based on the concept of limiting shearing stress
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Maximum Shear Stress Theory
(Tresca Criterion)
In a tension specimen:
 max 
 max   min
2

The diameter of the Mohr circle =
For Plane Stress: 
1
(3=0)
 Y
 2  Y
max
y
2
s
y
y
1, 2 have same signs.
1   2   Y
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1, 2 have opposite signs.
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Graphical Representation
The boundaries of the
governing
equation is
represented
by
inclined
parallel lines .
Material will reach its elastic
limit when point (σ1, σ2) falls
outside the hexagonal region
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NUMERICAL
1. Stresses induced at a critical point in a machine
component made of steel are as follows:
σx = 120 Mpa , σy = - 40 Mpa , τxy = 80 N/mm2
Calculate the factor of safety by:
1) Max. Shear Stress Theory
2) Max. Normal Stress Theory
Take Syt = 380 N/mm2
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• Given:σx = 120 Mpa , σy = - 40 Mpa , τxy = 80 N/mm2
Syt = 380 N/mm2
Solution :Principal Stress
 1, 2 
x  y
2
1

2
x  y 
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 4xy 2
8
• σ1 = 153.13 N/mm2 ,
σ2 = -73.14 N/mm2
• Maximum shear stress
 max 
 max   min
2
= 113.135 N/mm2
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• Factor of Safety by:1) Maximum Shear stress theory:
yt
fos 
1- 2
= 1.68
2) Maximum Normal stress theory:
yt
fos 
1
= 2.48
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Numerical
2. A bolt is subjected to an axial pull of 8 kN and
a transverse shear force of 3 kN. Determine the
diameter of the bolt required based on
i) Maximum principal stress theory
ii) Maximum shear stress theory
Take elastic limit in simple tension =270 N/mm2
Poisson’s Ratio = 0.3. Adopt a factor of safety = 3
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• Solution:Safe axial tensile stress = f = 270/3 = 90 N/mm2
Direct stress on bolt section = 8000/A N/mm2
Shear stress on bolt section = 3000/ A N/mm2
Where ‘A’ is cross sectional area of bolt
i) Maximum principal stress theory
x  y 1
2
x  y   4xy 2
 1, 2 

2
2
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2
4000 1  8000 
3000
1 


 4
A
2  A 
A





2




9000

N / mm 2  90 N / mm 2
A
- Safe axial tensile stress
𝜋𝑑2
A=
4
∴ 𝑑 = 11.28 𝑚𝑚
= 100
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• Maximum shear stress theory
𝜎1 =
𝜎2 =
4000 5000 9000
+
=
N/mm2
𝐴
𝐴
𝐴
4000 5000
1000
=N/mm2
𝐴
𝐴
𝐴
Maximum shear stress
 1   2 5000
 max 

N / mm 2
2
A
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• Safe shear stress =
5000
𝐴
90
2
= 45 N/mm2
= 45 ∴ 𝐴 = 111.11 mm2
d = 11.89 mm
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