Principal Stress-Strain & Theories of Failure | Civil Engineering SSC JE (Technical) - Civil Engineering (CE) PDF Download

Analysis of Principal Stresses

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. 

Case-1 : If principal stresses acting on two mutually perpendicular planes are σ₁ and σ₂ then, normal and shear stresses on a plane n – n which is inclined at an angle θ with the plane of σ₁ are given by

Case-2 : 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 n-n inclined at an angle θ with the plane of σ_x are given by

Special case-1 : 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 σ₁ and σ₂ are given by

 σ₁ or σ₂ = (σ_x+σ_y)/2 ± √[(σ_x-σ_y/2)²+Τ²]

Special case-2 : 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 case-3 : In case of pure shear element, the principal stresses act at 45° to the plane of pure shear stress.

σ₁ = +  ζ_xy
σ₂ = – ζ_{xy ·}

• Properties of Mohr’s Circle for Stress :
•  Mohr’s circle is the locus on normal and shear stresses on an element with the changing angle of plane in 2 dimensional case.

The radius of Mohr’s circle is equal to maximum shear stress.
Radius,

• The centre of circle always lies on s axis and its co-ordinates are (σ_n, 0)

Note : Sum of normal stresses on two mutually perpendicular planes remain constant i.e.σ₁ + σ₂ = σ_x + σ_y = constant

• In case of pure shear element σ₁ = + ζ and σ₂ = – ζ therefore centre of the circle coincides with the origin,

• In case of an element inside the static fluid, ζ = 0 because principal stresses are equal and alike therefor Mohr’s circle reduces into a point

Combined Bending & Torsion

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.

•  The location of the principal planes through the point is given by tan

•  The principal stress es are given by

• The maximum shear stress is given by

• The position of principal planes is given by

EQUIVALENT BENDING MOMENT & EQUIVALENT TORQUE

• Let ‘M_e’ be the equivalent bending moment which acts alone producing the maximum tensi l e stress equal to σ₁, as produced by M and T.

 
Therefore

• Let ‘T_e’ be the equivalent torque, which acts alone producing the same maximum shearing stress tmax as produced by M and T.

Analysis of Principal Strains

•  Case-1 : If ε₁ and ε₂ are principal strains in two mutual perpendicular directions in plane stress problem then principal stresses are given by

• If ε₁ and ε₂ are principal strains in x and y directions respectively, then normal and shear strain in any other direction x' are given by

• If ε_x, ε_y and φ_xy are normal and shear strain in x – y plane the normal and shear strain in x' – y' plane are given by

Special case : If φ_x'y' = 0 then magnitude of principal strains and their plane are given by

• Properties of Mohr’s circle for strain

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.

Theories Of Elastic Failure

• Function of theories of elastic failure is to predict the behaviour of materials in simple tensile test when elastic failure will occur under any condition of applied stress.
• Maximum principal stress theory (Rankine) This assumes that max. principal stress in the complex system reaches the elastic limit stress in simple tension and failure occurs when σ₁ = σ_y for tension Failure can occur in compression when least principal stress (σ₃) reaches the elastic limit stress in compression i.e. σ₃ = σ_y  for compression
• It is well suited for brittle materials.

Failure envelope occurs when 
(a) σ₁ or σ₂ = σ_yt  or  σ_yc
(ii) σ₃ = 0

• Max shear stress theory (Guest - Tresca) This assumes that max shear stress in the  complex stress system becomes equal to that at the yield point in simple tensile test.

• This theory holds good for ductile materials. For like stresses in Ist and IIIrd quadrant σ₁ = σ_y or σ₂ = σ_y
   
• For unlike stresses in II^nd or IV^th quadrant

Note : Aluminium alloys & certain steels are not governed by the Guest theory.

• Max principal strain theory (Saint Venant) This assumes that failure occurs when max. strain in the complex stress system equals that at the yield point in the tensile test (σ₁ – µσ₂ – µσ₃) = σ_y
  Failure should occur at higher load because the Poisson's ratio reduces the effect in perpendicular directions

 RHOMBUS

•  Maximum strain energy theory (Haigue)or Total strain Energy Theory : -This assumes that failure occur when total strain energy in the complex system is equal to that at the yield point in tensile test.

It is fairly good for ductile materials.

   ELLIPSE

•  Maximum shear strain energy theory or distortion energy theory (Mises-Henky Theory).

The properties are similar in tension and compression

   ELLIPSE

•  Failure of most ductile materials is governed by the distortion energy criterion or Von mises theory.
  σ₁ = σ₂ = σ₃; cylinder
  σ₁ = σ₂ and σ₃ = 0; ellipse · 
• Factor of safety =  Tensile strength / Allowable working stress

Solved Numericals

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 cross-section when subjected to pure bending develops normal stress which is given by:

Circular cross-section 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 (σ₁) 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 σ₁ = +180 MPa, σ₂ = -100 MPa, σ₃ = 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