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.
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Maximum Principal Theory (Rankine Theory)
Failure due to maximum principal stress
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Maximum Shear Stress Theory (Guest Theory)
Failure due to maximum shear stress
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Maximum principal strain theory (Saint Venant’s Theory)
Failure due to maximum principal strain
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Maximum Principal Strain Energy Theory (Haigh Theory)
Failure due to maximum principal strain energy
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Distortion Energy Theory (Maximum Shear Strain Energy Theory OR Von Mises-Hencky Theory)
Failure due to maximum shear strain energy
Theory
Assumed in all the theories
Firstly σ1 is tensile
Secondly σ2 is tensile
Thirdly σ3 is tensile
Fourthly σ1 > σ2 > σ3
FOR DUCTILE MATERIALS
σallow =σyp/FOS
FOR BRITTLE MATERIALS
σallow =σutm/FOS
ALLOWABLE VALUES for a 2-dimensional stress system is σallow, σallow, and 0. It is because in 3-dimensional stress system, one of the stresses becomes zero . It leads to 2-dimensional stress system.
Maximum allowable shear stress
= (σallow— σallow)/2,
(σallow—o)/2, (0–σallow)/2
We know Stresses actually present in a 2-dimensional stress system is σ1, σ2 and σ3=0
Actually present maximum shear stress
(σ1– σ2)/2, (σ2-σ3)/2 and (σ3– σ1)/2
Fig. MAXIMUM PRINCIPAL STRESS THEORY (RANKINE THEORY)
MAXIMUM PRINCIPAL STRESS THEORY (RANKINE THEORY)
ALLOWABLE stress VALUES σallow, σallow, and 0
Stresses actually present σ1, σ2 and σ3=0
Finally Maximum Principal Stress Actually present≤ σult/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) (σallow— σallow)/2,
(ii) (σallow—o)/2,
(iii) (0–σallow)/2
Principal Stresses actually present σ1, σ2 and σ3=0
Actually present maximum shear stress
(i) (σ1– σ2)/2
(ii) (σ2-σ3)/2
(iii) (σ3– σ1)/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 σallow, σallow, 0
Allowable strains (1/E) (σallow-µ σallow),
(1/E)(σallow–0) and (1/E)(0–σallow)
Principal Stresses actually present σ1, σ2 and σ3=0
Actual strains present
(i) (1/E)(σ1– µσ2)
(ii) (1/E)(σ2–µ σ3)
(iii) (1/E)(σ3— µσ1)
If Actually present≤ Allowable strain, it is safe
MAXIMUM PRINCIPAL STRAIN ENERGY THEORY-HAIGH THEORY
Fig. Maximum Principal Strain Energy Theory (Haigh Theory)
ALLOWABLE stress VALUES σallow, σallow, 0
Allowable strain energy
u=(1/2E)( σ2allow+ σ2allow–2µ σallow σallow)
Stresses actually present σ1, σ2 and σ3=0
Strain energy ACTUALLY present
u=(1/2E)( σ12+ σ22–2µ σ1 σ2)
FINAL IS
( σ12+ σ22–2µ σ1 σ2)= σ2allow
MAXIMUM SHEAR STRAIN ENERGY THEORY (VON MISES HENCKY THEORY)
ALLOWABLE stress VALUES σallow, σallow, 0
Allowable shear strain energy
uSSE =(1/12G)[( σallow—σallow)2 +(σallow–0)2 + (0- σallow)2
uSSE =(1/6G) σ2allow
Stresses actually present σ1, σ2 and σ3=0
Fig. Maximum Shear Strain Energy Theory (Von Mises Hencky Theory )
ACTUAL SHEAR STRAIN ENERGY PRESENT
uSSE =(1/12G)[( σ1— σ2)2 +(σ2–σ3)2 + (σ3– σ1)2]
Actually present≤ Allowable
FINAL SHEAR STRAIN ENERGY THEORY
[( σ1—σ2)2 +(σ2–0)2 + (0- σ1)2 = 2 σ2allow
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
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Maximum principal theory
Largest of σ1, σ2 and σ3 =σyp/FOS
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Maximum shear stress theory
Largest of (σ1– σ2)/2, (σ2–σ3)/2 and (σ3– σ1)/2 = σyp/2 FOS
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Maximum principal strain theory
σ1– µσ2 = σYp/FOS
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Maximum Principal Strain energy theory
( σ12+ σ22–2µ σ1 σ2)= (σyp/FOS)2
-
MAXIMUM SHEAR STRAIN ENERGY THEORY
OR
MAXIMUM DISTORSION ENERGY THEORY
[( σ1—σ2)2 +(σ2–0)2 + (0- σ1)2 = 2 (σyp /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:
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Maximum principal stress theory
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Maximum shear stress theory
Given factor of safety =3,
yield strength of the material=270 N/mm2
Poisson’s ratio=0.3
SOLUTION
Let A be the area of the bolt.
σ =9600/A
τ = 4800/A
We know σ1,2 =σ/2 + (1/2)[(σ/2)2+4τ2]0.5
Firstly σ1,2=9600/2A/2)±[(9600/2A)2+4x(4800/A)2]0.5
Secondly σ1=11588.2/A
Thirdly σ2 =–1988.2/A
Allowable stress =σyp/FOS=270/3=90
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As per maximum principal stress theory
11588.2/A=90
A=128.76
(π/4)d2=128.76
d=12.8 mm ANS
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As per maximum shear stress theory
[11588.2/A—(–1988.2/A)] =90
A = 150.8
(π/4)d2=150.8
= 13.8 mm ANS
Final answer d=13.8 mm
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