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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3. Maximum Distortion Energy Theory
(von Mises Criterion)
• Failure occurs when the distortional energy associated
with the principal stresses equals or exceeds the
distortional energy corresponding to that for the yield
strength of the material in uniaxial tension.
• ductile material
• based on a limiting energy of distortion (shear strains)
• 0.5 ( (σ1 - σ2)2 + (σ2 - σ3)2 + (σ3 - σ1)2) < σyp2
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Maximum Distortion Energy Theory
(von Mises Criterion)
For the three principal stresses;
1
u   1 1   2 2   3 3  
2
1
2
2
2
 1   2   3  2  1 2   3 2   1 3 
2E


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After taking out the hydrostatic stress (ave=(1+2+3)/3)
Now substitute 1, 2 and 3 with
 1   ave ,  2   ave ,  3   ave 
1  2
For plane stress; ud 
 1   1 2   22
3E


1  2
ud 
 Y in an uniaxial tension test.
3E
   1 2     Y
2
1
2
2
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4
Graphical Representation of maximum
distortion theory
σ2
σ1
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4. Maximum Strain Theory
• By Saint Venant
• Equivalent direct stress acting alone which
produces strain equal to greater principal strain
shall not exceed the safe limit.
σ1
σ2
𝑒1 =
−
𝐸
mE
• Let f be the stress acting alone producing the
strain
𝑓 σ1
σ2
𝑒1 = =
−
𝐸
𝐸
Strength of Materials
mE
6
Maximum Strain Theory
f = σ1 −
σ2
mE
This stress f should not exceed the safe limit.
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NUMERICAL
1. Stresses induced at a critical point in a machine
component made of steel are as follows:
σ1 = 120 Mpa , σ2 = - 40 Mpa , τxy = 80 N/mm2
Calculate the factor of safety by:
1) Max. Shear Stress Theory
2) Max. Normal Stress Theory
3) Max. Distortion Energy 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
9
• 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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3) Maximum Distortion Energy Theory:-
 1   1   1
2
2


2  




yt 



2
fos
FOS = 1.9
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2. A bolt is under an axial pull of 24 kN together with a
transverse shear force of 5 kN. Calculate the diameter if
bolt using
i) Max. principle stress theory
ii) Max. shear stress theory
iii) Strain energy theory
Take, elastic limit of bolt material as 250 MPa and
µ = 0.3. Factor of safety is 2.5
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Given:d = diameter of the bolt in mm
P = 24 kN , Shear force S = 5 kN
σy = 250 N/mm2 FOS = 2.5 µ = 0.3
Solution:-
 1, 2 
x  y
2
1

2
x  y 
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 4xy
2
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12000 13000
𝜎𝑛 =
±
𝐴
𝐴
𝜎1 =
25000
𝐴
𝜎2 =
,
Safe Stress =
𝜎𝑦
𝐹𝑂𝑆
1000
𝐴
= 100 N/mm2
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1) Maximum principle stress theory
25000
𝐴
≤ 100
A ≤ 100
d = 17.84 mm
2) Maximum shear stress theory
𝜎1- 𝜎2 = 𝜎𝑦𝑝
𝐹𝑂𝑆
25000
𝐴
-
1000
−
𝐴
= 100
∴ 𝐝 = 𝟏𝟕. 𝟗𝟓 𝐦𝐦
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Examples to solve
1. A solid circular shaft subjected to bending moment of 60 kNm
and a torque of 15 kNm. Design the diameter of the shaft using.
i) Max. principle stress theory
ii) Max. shear stress theory
iii) Max. strain energy theory
Take µ = 0.28, yield strength of shaft is 225 MPa and factor of
safety = 2.5
2. A square pin is required to resist a pull of 40kN. Derive a
suitable section according to strain energy theory. Maximum
tensile stress is 350 MPA and Poisson’s ratio is 0.3. Take factor of
safety of 2.5
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Numerical
• A shaft is supported in a bearing 4 m apart and
transmits 60 kW at 160 rpm. At 1.2 m from one
bearing the shaft carries a pulley transmitting a load
of 50 kN on the shaft. Find the suitable diameter for
the shaft for each of the following cases.
i) Maximum direct stress shall not exceed 120 N/mm2
ii) Maximum shear stress shall not exceed 60 N/mm2
Ans:- i) D= 133.2 mm
ii) D= 134.5 mm
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