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Lecture 2 principal stress and strain
1.
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. 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
4. Stresses and strains
Compressive load Failure in shear
Stresses are acting normal to the surface yet the material failed in a different plane
5. 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
6. Stresses in oblique plane
In real life stresses does not act in normal direction but
rather in inclined planes.
Normal Plane Oblique Plane
7. Stresses in oblique plane
P
σ=
θ A
P =Axial Force
A=Cross-sectional area
perpendicular to force
€ 2
σn Unit depth
€ σn = σ cos θ
σ
€ σt = sin2θ
2
σt €
€ €
8. Stresses in oblique plane
σ1 σ1
Member subjected to direct stress in one plane
σ2
€ €
Member subjected to direct stress in two mutually σ1 σ1
perpendicular plane
€
σ2
τ
€ €
τ τ
Member subjected to simple shear stress.
€ € τ
σ2 € τ
€
τ
Member subjected to direct stress in two σ1 σ1
€
mutually perpendicular directions + simple shear τ
stress €
€
τ σ
€2
€ €
€
€ €
9. Stresses in oblique plane
Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
σ1 + σ2 σ1 − σ2
σn = + cos2θ + τ sin2θ
2 2
σ1 − σ2
σt = sin2θ − τ cos2θ
2
€
€
10. 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
σ1 − σ2
σt = sin2θ − τ cos2θ = 0
2
2τ
tan2θ =
σ1 − σ2
€
11. Stresses in oblique plane
Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
POSITION OF PRINCIPAL PLANES
2τ
tan2θ =
σ1 − σ 2
2τ 2τ
sin2θ = 2
(σ1 − σ 2 ) + 4τ 2 θ
€
cos2θ =
(σ1 − σ 2 ) σ1 − σ 2
2
(σ1 − σ 2 ) + 4τ 2 €
€
€
€ €
12. Stresses in oblique plane
Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
2
σ1 + σ2 ⎛ σ1 − σ2 ⎞
Major principal Stress = + ⎜ ⎟ +τ2
2 ⎝ 2 ⎠
2
σ1 + σ2 ⎛ σ1 − σ2 ⎞ 2
Minor principal Stress = − ⎜ ⎟ + τ
€ 2 ⎝ 2 ⎠
€
13. Stresses in oblique plane
Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
MAX SHEAR STRESS
d
dθ
(σ t ) = 0
d ⎡σ1 − σ 2 ⎤ σ1 − σ 2
⎢ 2 sin2θ − τ cos2θ ⎥ = 0 tan2θ =
dθ ⎣ ⎦ 2τ
€
€
€
14. Stresses in oblique plane
Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
MAX SHEAR STRESS
Evaluate the following equation at
σ1 − σ2
σt = sin2θ − τ cos2θ
2
σ1 − σ 2
tan2θ =
2τ
1 2
€ (σt )max =
2
(σ1 − σ2 ) + 4τ 2
€
15. 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
16. Stresses in oblique plane
Member subjected to direct stress in one plane
σ1 + σ2 σ1 − σ2
σn = + cos2θ + τ sin2θ
2 2
σ1 − σ2
σt = sin2θ − τ cos2θ
2
Stress in one direction and no shear stress σ2 = 0 τ = 0
€
σ1 σ1
σn = + cos2θ = σ1 cos2 θ
€ 2 2
€ σ €
1
σt = sin2θ
2
€
17. Stresses in oblique plane
Member subjected to direct stress in two mutually
perpendicular plane
σ1 + σ2 σ1 − σ2
σn = + cos2θ + τ sin2θ
2 2
σ1 − σ2
σt = sin2θ − τ cos2θ
2
Stress in two direction and no shear stress τ =0
€
σ1 + σ2 σ1 − σ2
σn = + cos2θ
€ 2 2
€σ − σ
σt = 1 2
sin2θ
2
€
18. Stresses in oblique plane
Member subjected to simple shear stress.
σ1 + σ2 σ1 − σ2
σn = + cos2θ + τ sin2θ
2 2
σ1 − σ2
σt = sin2θ − τ cos2θ
2
No stress in axial direction but only shear stress σ1 = σ 2 = 0
€
σn = τ sin2θ
€ σt = € τ cos2θ
−
€
€
19. 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.
20. 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.