Unit 1- Stress and Strain

  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

Topics Covered

Mohr Stress Circle

€

σn =σ1 +σ22

+σ1 −σ22

cos2θ +τ sin2θ

€

σt =σ1 −σ22

sin2θ −τ cos2θ

We derived these two equations- These equations represent the equation of a circle

€

σn −σ1 +σ22

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

=σ1 −σ22

cos2θ +τ sin2θ⎛

⎝ ⎜

⎞

⎠ ⎟ 2

€

σt( )2 =σ1 −σ22

sin2θ −τ cos2θ⎛

⎝ ⎜

⎞

⎠ ⎟ 2

Mohr Stress Circle

€

σn −σ1 +σ22

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

=σ1 −σ22

cos2θ +τ sin2θ⎛

⎝ ⎜

⎞

⎠ ⎟ 2

€

σt( )2 =σ1 −σ22

sin2θ −τ cos2θ⎛

⎝ ⎜

⎞

⎠ ⎟ 2

Add above 2 equations. We will equation of circle.

€

σn −σ1 +σ22

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+σt2 =

σ1 −σ22

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+ τ( )2

€

x − a( )2

€

y 2

€

r2 Equation of circle

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.

Mohr Stress Circle   Body subjected to two mutually perpendicular

principal stresses of unequal magnitude

O σ1

σ2

σ1

2θ σn

σt

θ A C

B D

E

(σ1 - σ2 ) length AD =

€

=σn

Normal stress on oblique plane

length ED = Tangential stress on Oblique plane

length AE = Resultant stress on Oblique plane

€

=σt

€

= σt2 +σn

2

Mohr Stress Circle   Body subjected to two mutually perpendicular principal

stresses of unequal magnitude and unlike (one tensile and other compressive).

σ2 σ1

2θ σn

σt

θ A

C B

D

E

(σ1+σ2 ) length AD =

€

=σn

Normal stress on oblique plane

length ED = Tangential stress on Oblique plane

length AE = Resultant stress on Oblique plane

€

=σt

€

= σt2 +σn

2O

+ _

Mohr Stress Circle   Body subjected to two mutually perpendicular principal

stresses + simple shear stress.

σ1

σ2

σ1

2θ σn

σt

A C

B D

E

length AD =

€

=σn

Normal stress on oblique plane

length ED = Tangential stress on Oblique plane

length AE = Resultant stress on Oblique plane

€

=σt

€

= σt2 +σn

2L

M

O

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)

1. Maximum principal stress theory

€

σ1,σ2,σ3 =principal stresses in 3 perpendicular directions

€

σ*

Maximum principal stress should be less than the max stress (yield stress) that material can bear in tension or compression.

€

max(σ1,σ2,σ3) ≤σ*

= max tensile or compressive strength of material

max principal stress=

€

σ*

safety _ factor

2. Maximum principal strain theory

€

σ1,σ2,σ3 =principal stresses in 3 perpendicular directions

€

σ*Maximum principal strain should be less than the max strain (yield strain) that material can bear in tension or compression.

€

e1 =σ1E−υσ2E

−υσ3E

= max tensile or compressive strength of material

max principal stress=

€

σ*

safety _ factor

€

e2 =σ2E−υσ1E

−υσ3E

€

e3 =σ3E−υσ1E

−υσ2E

€

max(e1,e2,e3) ≤ e*

€

e* =σ*

E

3. Maximum shear stress theory

€

=12σ1 −σ3( )

€

σt*

Maximum shear stress should be less than the max shear stress in simple tension (at elastic limit) that material can bear.

= max tensile of material

allowable stress =

€

σt*

safety _ factor€

=12(σt

* − 0)

max shear stress =half the difference of max and min principal stresses

To prevent failure max shear stress should be less that shear stress in simple tension at elastic limit

max shear stress at elastic limit

€

(σ1 −σ3) ≤σt*

In simple tension the stress is existing in one direction

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.

max allowable stress=

€

σt*

safety _ factor€

σ12 +σ2

2 +σ32 − 2υ σ1σ2 +σ1σ3 +σ2σ3( )[ ] ≤ σt

*( )2

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).

max allowable stress=

€

σt*

safety _ factor€

σ1 −σ2( )2 + σ1 −σ3( )2 + σ2 −σ3( )2 ≤ 2* σt*( )2

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.

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