Question

A piece of material is subjected to two perpendicular tensile stresses of 100 MPa and

60 MPa. Determine the plane on which the resultant stress has maximum obliquity

with the normal. Also, find the resultant stress on this plane.​

Answer 1

Answer:

50 MPa

The maximum shear stress = (70-10)/2 = 30. At maximum shear plane normal stress = sigma(Avg)= (70+10)/2= 40 . So the resultant stress is equal to √[30^2+40^2] = 50 Mpa. By drawing Mohr's circle you can clearly understand

Answer 2

resultant stress is 91.65 Mpa

Given:

two perpendicular tensile stresses

P1=100 Mpa

P2=60 Mpa

To find:

Determine the plane on which the resultant stress has maximum obliquity

and the resultant stress on this plane.​

Explanation:

Normal stress = σn/Pn

Pn/σn = [P1+P2/2] + [P1–P2/2] (cos 2θ)

          = (100+60)/2+[100–60/2] cos(2×30)

          =  90MPa

Tangential stress = σt/pt

pt/σt =  [P1–P2/2]sin2θ

         = (100–60/2)sin(2×30)

         = 17.32Mpa

Resultant stress = σR/PR

σR  = √(σn²+σt²)

      =  $\sqrt{90^{2} + 17.32^{2} }$

      =  91.65 Mpa

inclination ϕ=tan−1(σt/σn)

          = tan−1(17.32/90)

       ϕ=10.89° or 10°53’

hence, resultant stress is 91.65 Mpa