Date post: | 17-Jan-2015 |
Category: |
Health & Medicine |
Upload: | deepak223 |
View: | 54,293 times |
Download: | 12 times |
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.