Table of contents  
Analysis of Principal Stresses  
Combined Bending & Torsion  
Analysis of Principal Strains  
Theories Of Elastic Failure  
Solved Numericals 
Principal stresses are direct normal stresses acting on mutually perpendicular planes on which shear stresses are zero. The planes which carry zero shear stresses are known as principal planes.
Case1 : If principal stresses acting on two mutually perpendicular planes are σ_{1} and σ_{2} then, normal and shear stresses on a plane n – n which is inclined at an angle θ with the plane of σ_{1} are given by
Case2 : If σ_{x} and σ_{y} are normal stresses and t_{xy} is shear stress acting on the mutually perpendicular planes then the normal and shear stresses on any plane nn inclined at an angle θ with the plane of σ_{x} are given by
Special case1 : If θ becomes such that ζ_{x'y'} on this plane becomes zero then this plane will be known as principal plane and the angle of principal plane is given by
The magnitude of principal stresses σ_{1} and σ_{2} are given by
σ_{1 }or σ_{2} = (σ_{x}+σ_{y})/2 ± √[(σ_{x}σ_{y}/2)^{2}+Τ^{2}]
Special case2 : The plane of maximum shear stress lies at 45° to the plane of principal stress and magnitude of ζ_{max }is given by
Note that planes of ζ_{max }carry equal and alike normal stresses. The normal stress on plane of ζ_{max} is given by
Therefore resultant stress on the plane of Т_{max} is
The angle of obliquity of σ_{r} with the direction of σ_{n} is given by
Special case3 : In case of pure shear element, the principal stresses act at 45° to the plane of pure shear stress.
σ_{1} = + ζ_{xy}
σ_{2 }= – ζ_{xy ·}
The radius of Mohr’s circle is equal to maximum shear stress.
Radius,
Note : Sum of normal stresses on two mutually perpendicular planes remain constant i.e.σ_{1} + σ_{2} = σ_{x }+ σ_{y} = constant
Let a shaft of diameter ‘d’ be subjected to bending moment ‘M’ and a twisting moment ‘T’ at a section. At any point in the section at radius ‘r’ and at a distance y from the neutral axis, the bending stress is given by
and shear stress is given by
Where I = Moment of inertia about its NA and I_{p} = Polar moment of Inertia.
EQUIVALENT BENDING MOMENT & EQUIVALENT TORQUE
Therefore
Special case : If φ_{x'y'} = 0 then magnitude of principal strains and their plane are given by
The radius of Mohr’s circle is half of maximum shear strain i.e.
Therefore Diameter of Mohr’s circle,
Static Loading & Dynamic Loading
When load is increased gradually from zero to P, it is called static loading. Under static loading the normal stress ’σ’ developed due to load P is given by
σ = (P/A)
When load is applied suddenly, then the normal stress ‘σ’ due to load P is given by
σ = (2P/A)
Hence, maximum stress intensity due to suddenly applied load is twice the stress intensity produced by the load of the same magnitude applied gradually.
Failure envelope occurs when
(a) σ_{1} or σ_{2} = σ_{yt } or σ_{yc}
(ii) σ_{3} = 0
Note : Aluminium alloys & certain steels are not governed by the Guest theory.
RHOMBUS
It is fairly good for ductile materials.
ELLIPSE
The properties are similar in tension and compression
ELLIPSE
Example 1: According to the maximum normal stress theory, the diameter of circular shaft subjected to bending moment M and torque T is
(where σy is the yield stress in the uniaxial tensile test and N is the factor of safety)
(a)
(b)
(c)
(d)
Ans: (d)
Solution:
As per this theory, for no failure maximum principal stress should be less than yield stress under uniaxial loading.
For design,
Where N is the factor of safety
Circular crosssection when subjected to pure bending develops normal stress which is given by:
Circular crosssection when subjected to pure twisting moment develops shear stress which is given by:
The combined effect of bending and torsion produces principal stress which is given by:
Example 2: The yield strength of bolt material is 300 MPa and factor of safety is 2.5. What is the maximum principal stress using maximum principal stress theory ?
(a) 750 MPa
(b) 120 MPa
(c) 27.38 MPa
(d) 10.95 MPa
Ans: (b)
Solution:
Maximum principal stress theory (Rankine’s theory)
According to this theory, failure happens under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.
For the design criterion, the maximum principal stress (σ_{1}) must not exceed the working stress ‘σ_{y}’ for the material.
Calculation:
Given:
σyt = 300 MPa, FOS = 2.5
Example 3: In metal forming operation when the material has just started yielding, the principle stresses are σ_{1} = +180 MPa, σ_{2} = 100 MPa, σ_{3 }= 0. Following the Von Mises criterion, the yield stress is ________ MPa.
Ans: 245  246
Solution:
According to von Mises theory, for yielding
Calculation:
Given:
σ1 = +180 MPa, σ2 = 100 MPa, σ3 = 0
Yield stress is:
σ_{yt} = 245.76 MPa
2 videos122 docs55 tests

1. What is the significance of analyzing principal stresses in combined bending and torsion? 
2. How do engineers analyze principal strains in structural components? 
3. What are the theories of elastic failure commonly used in civil engineering? 
4. Can you provide an example of a numerical problem involving principal stressstrain analysis? 
5. What are some common challenges faced by civil engineers when analyzing principal stresses and strains in structural components? 
2 videos122 docs55 tests


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