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

Stresses and strains   In last lecture we looked at stresses were acting in a

plane that was at right angles/parallel to the action of force.

Tensile Stress Shear Stress

Stresses and strains Compressive load Failure in shear

Stresses are acting normal to the surface yet the material failed in a different plane

Principal stresses and strains

  What are principal stresses.

  Planes that have no shear stress are called as principal planes.

  Principal planes carry only normal stresses

Stresses in oblique plane   In real life stresses does not act in normal direction but

rather in inclined planes.

Normal Plane Oblique Plane

Stresses in oblique plane

€

σn

€

σt

Unit depth

€

σn =σcos2θ

€

σt =σ2sin2θ

€

σ =PA

P =Axial Force A=Cross-sectional area perpendicular to force

€

θ

Stresses in oblique plane   Member subjected to direct stress in one plane

  Member subjected to direct stress in two mutually perpendicular plane

  Member subjected to simple shear stress.

  Member subjected to direct stress in two mutually perpendicular directions + simple shear stress

€

σ1

€

σ1

€

σ1

€

σ1

€

σ2

€

σ2

€

τ

€

τ

€

τ

€

τ

€

σ1

€

σ2

€

σ2€

σ1

€

τ

€

τ€

τ

€

τ

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular directions + simple shear stress

€

σn =σ1 +σ22

+σ1 −σ22

cos2θ +τ sin2θ

€

σt =σ1 −σ22

sin2θ −τ cos2θ

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular directions + simple shear stress

  POSITION OF PRINCIPAL PLANES

  Shear stress should be zero

€

tan2θ =2τ

σ1 −σ2

€

σt =σ1 −σ22

sin2θ −τ cos2θ = 0

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular directions + simple shear stress

  POSITION OF PRINCIPAL PLANES

€

tan2θ =2τ

σ1 −σ 2

€

2τ

€

σ1 −σ 2

€

θ

€

sin2θ =2τ

σ1 −σ 2( )2 + 4τ 2

€

cos2θ =σ1 −σ 2( )

σ1 −σ 2( )2 + 4τ 2

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular directions + simple shear stress

€

=σ1 +σ22

+σ1 −σ22

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+τ 2Major principal Stress

€

=σ1 +σ22

−σ1 −σ22

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+τ 2Minor principal Stress

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular directions + simple shear stress

  MAX SHEAR STRESS

€

ddθ

σ1 −σ 2

2sin2θ −τ cos2θ

⎡

⎣ ⎢ ⎤

⎦ ⎥ = 0

€

ddθ

σ t( ) = 0

€

tan2θ =σ1 −σ 2

2τ

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular directions + simple shear stress

  MAX SHEAR STRESS

€

tan2θ =σ1 −σ 2

2τ

€

σt =σ1 −σ22

sin2θ −τ cos2θ

€

σt( )max =12

σ1 −σ2( )2 + 4τ 2

Evaluate the following equation at

Stresses in oblique plane   Member subjected to direct stress in one plane

  Member subjected to direct stress in two mutually perpendicular plane

  Member subjected to simple shear stress.

  Member subjected to direct stress in two mutually perpendicular directions + simple shear stress

Stresses in oblique plane   Member subjected to direct stress in one plane

€

σn =σ1 +σ22

+σ1 −σ22

cos2θ +τ sin2θ

€

σt =σ1 −σ22

sin2θ −τ cos2θ

Stress in one direction and no shear stress

€

σ2 = 0

€

τ = 0

€

σn =σ12

+σ12cos2θ =σ1 cos

2θ

€

σt =σ12sin2θ

Stresses in oblique plane   Member subjected to direct stress in two mutually

perpendicular plane

€

σn =σ1 +σ22

+σ1 −σ22

cos2θ +τ sin2θ

€

σt =σ1 −σ22

sin2θ −τ cos2θ

Stress in two direction and no shear stress

€

τ = 0

€

σn =σ1 +σ22

+σ1 −σ22

cos2θ

€

σt =σ1 −σ22

sin2θ

Stresses in oblique plane   Member subjected to simple shear stress.

€

σn =σ1 +σ22

+σ1 −σ22

cos2θ +τ sin2θ

€

σt =σ1 −σ22

sin2θ −τ cos2θ

No stress in axial direction but only shear stress

€

σ1 =σ 2 = 0

€

σn = τ sin2θ

€

σt = −τ cos2θ

Principal stresses and strains

  PROBLEM- The tensile stresses at a point across two mutually perpendicular planes are 120N/mm2 and 60 N/mm2. Determine the normal, tangential and resultant stresses on a plane inclined at 30deg to the minor stress.

Principal stresses and strains

  PROBLEM- A rectangular block of material is subjected to a tensile stress of 110 N/mm2 on one plane and a tensile stress of 47 N/mm2 on the plane at right angles to the former. Each of the above stresses is accompanied by a shear stress of 63 N/mm2 and that associated with the former tensile stress tends to rotate the block anticlockwise. Find

1)The direction and magnitude of each of the principal stress.

2) Magnitude of the greatest shear stress.