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# Lecture 2 principal stress and strain

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principal planes and principal stresses

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• proof that principal stresses and strains occur in the same direction please?

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### Lecture 2 principal stress and strain

1. 1. Unit 1- Stress and StrainTopics Covered  Lecture -1 - Introduction, state of plane stress  Lecture -2 - Principle Stresses and Strains  Lecture -3 - Mohrs Stress Circle and Theory of Failure  Lecture -4- 3-D stress and strain, Equilibrium equations and impact loading  Lecture -5 - Generalized Hooks law and Castiglionos
2. 2. 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
3. 3. Stresses and strains Compressive load Failure in shearStresses are acting normal to the surface yet the material failed in a different plane
4. 4. 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
5. 5. Stresses in oblique plane  In real life stresses does not act in normal direction but rather in inclined planes. Normal Plane Oblique Plane
6. 6. 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 €€ €
7. 7. 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 € € € € €
8. 8. 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€ €
9. 9. 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€
10. 10. 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 €€ €€ €
11. 11. 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 ⎞ 2Minor principal Stress = − ⎜ ⎟ + τ € 2 ⎝ 2 ⎠ €
12. 12. 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 €€ €
13. 13. 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 €
14. 14. 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
15. 15. 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 €
16. 16. 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 €
17. 17. 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 − € €
18. 18. 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.
19. 19. 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. Find1)The direction and magnitude of each of the principal stress.2) Magnitude of the greatest shear stress.