1. Principal Stresses and Strain and Theories of Failure
Strength of Materials
Prof. A. S. PATIL
Department of Mechanical Engineering
Sinhgad Academy of Engineering, Pune
Strength of Materials

2. 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.
Maximum Principal Stress Theory
Maximum Shear Stress Theory
Maximum Distortion Energy Theory
Maximum Strain Energy Theory
Strength of Materials

3. 1. Maximum principal stress theory
Maximum principal stress = Yield Stress
Rankine
brittle materials
based on a limiting normal stress
σ1 = σyp
Strength of Materials

4. 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
Strength of Materials

5. Maximum Shear Stress Theory (Tresca Criterion)
In a tension specimen:
tmax sy
The diameter of the Mohr circle = sy
For Plane Stress:
(s3=0)
s1, s2 have same signs.
s1, s2 have opposite signs.
Strength of Materials

6. 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
Strength of Materials

7. 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:
Max. Shear Stress Theory
Max. Normal Stress Theory
Take Syt = 380 N/mm2
Strength of Materials

8. σx = 120 Mpa , σy = - 40 Mpa , τxy = 80 N/mm2 Syt = 380 N/mm2
Given:-
σx = 120 Mpa , σy = - 40 Mpa , τxy = 80 N/mm2
Syt = 380 N/mm2
Solution :-
Principal Stress
Strength of Materials

9. σ1 = 153.13 N/mm2 , σ2 = -73.14 N/mm2 Maximum shear stress
Strength of Materials

10. Maximum Shear stress theory:
Factor of Safety by:-
Maximum Shear stress theory:
= 1.68
2) Maximum Normal stress theory:
= 2.48
Strength of Materials

11. 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
Maximum principal stress theory
Maximum shear stress theory
Take elastic limit in simple tension =270 N/mm2
Poisson’s Ratio = 0.3. Adopt a factor of safety = 3
Strength of Materials

12. Safe axial tensile stress = f = 270/3 = 90 N/mm2
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
Maximum principal stress theory
Strength of Materials

13. - Safe axial tensile stress
Strength of Materials

14. Maximum shear stress theory 𝜎1= 4000 𝐴 + 5000 𝐴 = 9000 𝐴 N/mm2
Strength of Materials

15. Safe shear stress = 90 2 = 45 N/mm2 5000 𝐴 = 45 ∴𝐴=111.11 mm2
d = mm
Strength of Materials