Principal stresses and strains (Mos)

This seminar is about Principal stresses and strains.

1. AHEMEDABAD INSTITUTE OF TECHNOLOGY SEMINAR Principal stresses and strains

2. CIVIL ENGINEERING (3 rd sem )B.E SUBJECT: Mos Submitted by: Patel Bhavik Patel Parth Patel Garvish Jigar Milan

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 .

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. Stresses in oblique plane In real life stresses does not act in normal direction but rather in inclined planes.

6. = P = Axial forces A = cross sectional area = θ = sin2 θ  

7. 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.

8. Member subjected to direct stress in two mutually perpendicular directions + simple shear stress σn = + cos2θ+τsin2θ = sin2 θ−τ cos2 θ  

9. Member subjected to direct stress in two mutually perpendicular directions + simple shear stress POSITION OF PRINCIPAL PLANES Shear stress should be zero = sin2 θ−τ cos2 θ=0 tan2 θ = 2T/( - )  

10. Member subjected to direct stress in two mutually perpendicular directions + simple shear stress . Major principal Stress= + + T Minor principal Stress = + + T  

11. Member subjected to direct stress in two mutually perpendicular directions + simple shear stress MAX SHEAR STRESS ( ) = 0 [ sin2 θ−τ cos2 θ ] = 0 tan2 θ =  

12. Member subjected to direct stress in two mutually perpendicular directions + simple shear stress MAX SHEAR STRESS = sin2 θ−τ cos2 θ tan2 θ = = ( + 4  

13. 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

14. Member subjected to direct stress in one plane = + cos2 θ+τ sin2 θ = sin2 θ−τ cos2 θ Stress in one direction and no shear stress σ2 = 0,τ=0 = + cos2 θ = σ1 = sin2 θ  

15. Member subjected to direct stress in two mutually perpendicular plane = + cos2 θ+τ sin2 θ = sin2 θ−τ cos2 θ Stress in two direction and no shear stress τ=0 = + cos2 θ = sin2 θ  

16. Member subjected to simple shear stress. = + cos2 θ+τ sin2 θ = sin2 θ−τ cos2 θ No stress in axial direction but only shear stress σ1=σ2 =0 = τ sin2 θ = −τ cos2 θ  

17. Thank You