2. Unit 1- Stress and Strain
Topics Covered
 Lecture -1 - Introduction, state of plane stress
 Lecture -2 - Principle Stresses and Strains
 Lecture -3 - Mohr's Stress Circle and Theory of
Failure
 Lecture -4- 3-D stress and strain, Equilibrium
equations and impact loading
 Lecture -5 - Generalized Hook's law and Castigliono's

3. Mohr Stress Circle
We derived these two equations- These equations represent the equation of
a circle
σ1 + σ2 σ1 − σ2
σn = + cos2θ + τ sin2θ
2 2
σ1 − σ2
σt = sin2θ − τ cos2θ
2
€ ⎛ σ1 + σ2 ⎞ 2 ⎛ σ1 − σ2 ⎞ 2
⎜σn − ⎟ = ⎜ cos2θ + τ sin2θ ⎟
€
⎝ 2 ⎠ ⎝ 2 ⎠
2 ⎛ σ1 − σ2 ⎞ 2
(σt ) = ⎜
⎝ 2
sin2θ − τ cos2θ ⎟
⎠
€

4. Mohr Stress Circle
⎛ σ1 + σ2 ⎞ 2 ⎛ σ1 − σ2 ⎞ 2
⎜σn − ⎟ = ⎜ cos2θ + τ sin2θ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
2 ⎛ σ1 − σ2 ⎞ 2
(σt ) = ⎜
⎝ 2
sin2θ − τ cos2θ ⎟
⎠
Add above 2 equations. We will equation of circle.
€ 2 2
⎛ σ + σ ⎞
1 2
⎛ σ − σ ⎞
2 1 2 2
⎜σ −
n ⎟ + σ = ⎜ t ⎟ + (τ )
€ ⎝ 2 ⎠ ⎝ 2 ⎠
2 2
( x − a) y r 2
Equation of circle
€

5. Mohr Stress Circle
 Graphical method to determine stresses.
 Body subjected to two mutually perpendicular principal
stresses of unequal magnitude.
 Body subjected to two mutually perpendicular principal
stresses of unequal magnitude and unlike (one tensile
and other compressive).
 Body subjected to two mutually perpendicular principal
stresses + simple shear stress.

6. Mohr Stress Circle
 Body subjected to two mutually perpendicular
principal stresses of unequal magnitude
(σ1 - σ2 ) length AD = Normal stress on oblique plane
σt
= σn
E length ED = Tangential stress on Oblique plane
= σt
2θ
B length AE = Resultant stress on Oblique plane
θ
σn
O D σ1 2 2
A C = σ +σt n
€
σ2
σ1 €
€

7. Mohr Stress Circle
 Body subjected to two mutually perpendicular principal
stresses of unequal magnitude and unlike (one tensile
and other compressive).
(σ1+σ2 )
length AD = Normal stress on oblique plane
σt E = σn
length ED = Tangential stress on Oblique plane
= σt
2θ
length AE = Resultant stress on Oblique plane
C θ
B
σn 2 2
A O D = σt + σn
_
€
+ σ1
σ2
€
€

8. Mohr Stress Circle
 Body subjected to two mutually perpendicular principal
stresses + simple shear stress.
σt length AD = Normal stress on oblique plane
= σn
E length ED = Tangential stress on Oblique plane
2θ
= σt
length AE = Resultant stress on Oblique plane
B M σ
n
L C D σ1 2 2
A O = σ +σ
t n
€
σ2
σ1 €
€

9. Theories of failure
 Maximum principal stress (Rankine theory)
 Maximum principal strain (Saint Venant theory)
 Maximum shear stress (Guest theory)
 Maximum strain energy (Haigh theory)
 Maximum shear strain energy (Mises & Henky
theory)

10. 1. Maximum principal
stress theory
σ1,σ2 ,σ3 =principal stresses in 3 perpendicular
directions
*
max(σ1,σ2 ,σ3 ) ≤ σ
€
Maximum principal stress should be less than the max stress (yield stress) that material
can bear in tension or compression.
*
σ = max tensile or compressive strength of material
€
σ*
max principal stress=
safety _ factor

11. 2. Maximum principal
strain theory
σ1,σ2 ,σ3 =principal stresses in 3 perpendicular
directions
σ1 υσ2 υσ3 σ2 υσ1 υσ3 σ3 υσ1 υσ2
e1 = − − e2 = − − e3 = − −
E E E E E E E E E
* σ*
€ max(e1,e2 ,e3 ) ≤ e *
e =
E
*
σ = max tensile or compressive strength of material
€ Maximum principal strain should be less than the max strain (yield strain) that material
€ €
can bear in tension or compression. €
σ*
€ max principal stress=
safety _ factor

12. 3. Maximum shear stress
theory
max shear stress =half the difference of max and min principal stresses
1
= (σ1 − σ3 )
2 In simple tension the stress
To prevent failure max shear stress should be less that shear is existing in one direction
stress in simple tension at elastic limit
1 * *
= (σt − 0) σt
max shear stress at elastic limit = max tensile of material
€ 2
(σ1 − σ 3 ) ≤ σ *
t
Maximum shear stress should be less than the max shear stress in simple tension (at
elastic limit) that material can bear.
€ €
σ*t
allowable stress =
€ safety _ factor

13. 4. Maximum strain
energy theory
Strain energy per unit volume should be less than the strain energy per unit volume in
simple tension (at elastic limit) that material can bear.
2
[σ 2
1 + σ + σ − 2υ (σ1σ2 + σ1σ3 + σ2σ3 ) ≤ (σ
2
2
2
3 ] *
t )
σ*t
max allowable stress=
safety _ factor
€
€

14. 5. Maximum shear strain
energy theory
Shear strain energy per unit volume should be less than the shear strain energy per unit
volume in simple tension (at elastic limit).
2 2 2 2
(σ1 − σ2 ) + (σ1 − σ3 ) + (σ2 − σ3 ) ≤ 2 * (σ *
t )
σ*t
max allowable stress=
safety _ factor
€
€

15. Important points
 Brittle material -> Max principal stress
 Brittle material do not fail in shear
 Ductile material -> Max shear stress/max shear strain
energy
 Ductile material fail in shear because their yield
strength is high.

16. Failure Theory
 PROBLEM- The principal stresses at a point in an
elastic material are 200 N/mm2 (tensile), 100 N/
mm2 (tensile) and 50 N/mm2 (compressive). If the
stresses at the elastic limit in simple tension is 200
N/mm2, determine whether the failure of the
material will occur according to different failure
theory. (take Poisson's ratio =0.3)
 Max principal strain theory
 Max shear stress theory
 Max strain energy theory
 Max shear strain energy theory