THEORIES OF FAILURE CLASS NOTES FOR MECHANICAL ENGINEERING
THEORIES OF FAILURE
CLASS NOTES FOR MECHANICAL
ENGINEERING
Failure means breaking or deforming.
Failure of ductile material is entirely different
from the failure of a brittle material under
same type of load. There are five well
known theories of elastic failures. These
theories explain material behavior under
simple or complex loaded
condition. It is elastic under five different
criterions. It does not meet a failure in four
of the five. Failure occurs in one of the
remaining criterions. The material is safe
and elastic in four remaining ways. For
example, cast iron fails under first theory
only. Mild steel fails under second theory
only.

Maximum Principal Theory (Rankine Theory)
Failure due to maximum principal stress

Maximum Shear Stress Theory (Guest Theory)
Failure due to maximum shear stress

Maximum principal strain theory (Saint Venant’s Theory)
Failure due to maximum principal strain

Maximum Principal Strain Energy Theory (Haigh Theory)
Failure due to maximum principal strain energy

Distortion Energy Theory (Maximum Shear Strain Energy Theory OR Von MisesHencky Theory)
Failure due to maximum shear strain energy
Theory
Assumed in all the theories
Firstly σ_{1 } is tensile
Secondly σ_{2} is tensile
Thirdly σ_{3} is tensile
Fourthly σ_{1} > σ_{2} > σ_{3}
FOR DUCTILE MATERIALS
σ_{allow} =σ_{yp}/FOS
FOR BRITTLE MATERIALS
σ_{allow} =σ_{utm}/FOS
ALLOWABLE VALUES for a 2dimensional stress system is σ_{allow}, σ_{allow}, and 0. It is because in 3dimensional stress system, one of the stresses becomes zero . It leads to 2dimensional stress system.
Maximum allowable shear stress
= (σ_{allow}— σ_{allow})/2,
(σ_{allow}—o)/2, (0–σ_{allow})/2
We know Stresses actually present in a 2dimensional stress system is σ_{1}, σ_{2} and σ_{3}=0
Actually present maximum shear stress
(σ_{1}– σ_{2})/2, (σ_{2}σ_{3})/2 and (σ_{3}– σ_{1})/2
Fig. MAXIMUM PRINCIPAL STRESS THEORY (RANKINE THEORY)
MAXIMUM PRINCIPAL STRESS THEORY (RANKINE THEORY)
ALLOWABLE stress VALUES σ_{allow}, σ_{allow}, and 0
Stresses actually present σ_{1}, σ_{2} and σ_{3}=0
Finally Maximum Principal Stress Actually present≤ σ_{ult}/FOS
MAXIMUM SHEAR STRESS THEORY (Guest Theory or Tresca Theory)
Fig. Maximum Shear Stress Theory (Guest Theory)
We know Maximum shear stress from allowable stress
(i) (σ_{allow}— σ_{allow})/2,
(ii) (σ_{allow}—o)/2,
(iii) (0–σ_{allow})/2
Principal Stresses actually present σ_{1}, σ_{2} and σ_{3}=0
Actually present maximum shear stress
(i) (σ_{1}– σ_{2})/2
(ii) (σ_{2}σ_{3})/2
(iii) (σ_{3}– σ_{1})/2
Actually present maximum shear stress≤ allowable shear stress
MAXIMUM PRINCIPAL STRAIN THEORY
(SAINT VENANT’S THEORY)
Fig. Maximum Principal Strain Theory (st. Venant’s Theory)
ALLOWABLE stress VALUES σ_{allow}, σ_{allow}, 0
Allowable strains (1/E) (σ_{allow}µ σ_{allow}),
(1/E)(σ_{allow}–0) and (1/E)(0–σ_{allow})
Principal Stresses actually present σ_{1}, σ_{2} and σ_{3}=0
Actual strains present
(i) (1/E)(σ_{1}– µσ_{2})
(ii) (1/E)(σ_{2}–µ σ_{3})
(iii) (1/E)(σ_{3}— µσ_{1})
If Actually present≤ Allowable strain, it is safe
MAXIMUM PRINCIPAL STRAIN ENERGY THEORYHAIGH THEORY
Fig. Maximum Principal Strain Energy Theory (Haigh Theory)
ALLOWABLE stress VALUES σ_{allow}, σ_{allow}, 0
Allowable strain energy
u=(1/2E)( σ^{2}_{allow}+ σ^{2}_{allow}–2µ σ_{allow} σ_{allow})
Stresses actually present σ_{1}, σ_{2} and σ_{3}=0
Strain energy ACTUALLY present
u=(1/2E)( σ_{1}^{2}+ σ_{2}^{2}–2µ σ_{1} σ_{2})
FINAL IS
( σ_{1}^{2}+ σ_{2}^{2}–2µ σ_{1} σ_{2})= σ^{2}_{allow}
MAXIMUM SHEAR STRAIN ENERGY THEORY (VON MISES HENCKY THEORY)
ALLOWABLE stress VALUES σ_{allow}, σ_{allow}, 0
Allowable shear strain energy
u_{SSE} =(1/12G)[( σ_{allow}—σ_{allow})^{2} +(σ_{allow}–0)^{2} + (0 σ_{allow})^{2}
u_{SSE} =(1/6G) σ^{2}_{allow}
Stresses actually present σ_{1}, σ_{2} and σ_{3}=0
Fig. Maximum Shear Strain Energy Theory (Von Mises Hencky Theory )
ACTUAL SHEAR STRAIN ENERGY PRESENT
u_{SS}_{E} =(1/12G)[( σ_{1}— σ_{2})^{2} +(σ_{2}–σ_{3})^{2} + (σ_{3}– σ_{1})^{2}]
Actually present≤ Allowable
FINAL SHEAR STRAIN ENERGY THEORY
[( σ_{1}—σ_{2})^{2} +(σ_{2}–0)^{2} + (0 σ_{1})^{2} = 2 σ^{2}_{allow}
GRAPHICAL REPRESENTATION OF VARIOUS THEORIES OF FAILURE
Fig. Theories of Failure (Combined)
ABCD Max Principal Stress Theory Boundary
EBFGDHE is Max. Shear Stress Boundary
IJKLI is Max. Principal Strain Boundary
Dark Ellipse Max. Shear Strain Energy Theory Boundary
Dotted Ellipse Max. Principal Strain Energy Theory Boundary
ULTIMATE FINAL RESULTS

Maximum principal theory
Largest of σ_{1}, σ_{2} and σ_{3} =σ_{yp}/FOS

Maximum shear stress theory
Largest of (σ_{1}– σ_{2})/2, (σ_{2}–σ_{3})/2 and (σ_{3}– σ_{1})/2 = σ_{yp}/2 FOS

Maximum principal strain theory
σ_{1}– µσ_{2} = σ_{Yp}/FOS

Maximum Principal Strain energy theory
( σ_{1}^{2}+ σ_{2}^{2}–2µ σ_{1} σ_{2})= (σ_{yp}/FOS)^{2}

MAXIMUM SHEAR STRAIN ENERGY THEORY
OR
MAXIMUM DISTORSION ENERGY THEORY
[( σ_{1}—σ_{2})^{2} +(σ_{2}–0)^{2} + (0 σ_{1})^{2} = 2 (σ_{yp }/FOS)^{2}_{ }
Conclusions
(i) In any theory, if the word ‘Principal’ exists, that will be applicable to brittle materials only.
(ii) In any theory, if the word ‘Shear’ exists, that will be applicable to ductile materials only.
(iii) A particular theory is applicable to few specific brittle material(s) or few specific ductile material(s).
(iv) Different theories won’t be applicable to the same particular brittle or same ductile material.
PROBLEM
A bolt is under an axial thrust of 9.6 k N together with a transverse force of 4.8 k N. Calculate the diameter according to:

Maximum principal stress theory

Maximum shear stress theory
Given factor of safety =3,
yield strength of the material=270 N/mm^{2}
Poisson’s ratio=0.3
SOLUTION
Let A be the area of the bolt.
σ =9600/A
τ = 4800/A
We know σ_{1,2} =σ/2 + (1/2)[(σ/2)^{2}+4τ^{2}]^{0.5}
Firstly σ_{1,2}=9600/2A/2)±[(9600/2A)^{2}+4x(4800/A)^{2}]^{0.5}
Secondly σ_{1}=11588.2/A
Thirdly σ_{2} =–1988.2/A
Allowable stress =σ_{yp}/FOS=270/3=90

As per maximum principal stress theory
11588.2/A=90
A=128.76
(π/4)d^{2}=128.76
d=12.8 mm ANS

As per maximum shear stress theory
[11588.2/A—(–1988.2/A)] =90
A = 150.8
(π/4)d^{2}=150.8
= 13.8 mm ANS
Final answer d=13.8 mm
https://www.mesubjects.net/wpadmin/post.php?post=4215&action=edit MCQ FOS
https://www.mesubjects.net/wpadmin/post.php?post=4378&action=edit MCQ theories of elastic failure