Question

A load carrying member is subjected to the following stress condition; 

Tensile stress σ_x = 400MPa; 

Tensile stress σ_y = – 300MPa; 

Shear stress τ_xy = 200MPa (Clock wise); 

Obtain 

(1) Principal stresses and their plane 

(2) Maximum shearing stress and its plane

Answer 1

Since Principal stresses are given as:

Major Principle stress = 

Given that; 

σ_x = 400MPa; 

σ_y = – 300MPa; 

τ_xy = 200MPa (Clock wise); 

σ_1,2 = ½ [(400 – 300) ± {(400 + 300)² + 4 x (200)² }^1/2] 

σ₁ = 453.11MPa  

σ₂ = – 353.11MPa

The direction of principal planes is: 

tan 2θ = 2τ_xy/ (σ_x − σ_y)

tan 2θ = (2 x 200)/(400 – (– 300))

2θ = 29.04º or 209.04º 

θ = 14.52º or 104.52º 

Since Maximum Shear stress is at θ = 45º 

τ = 1/ 2 (σ₁ – σ₂ ) sin 2θ = ½ (453.11 + 353.11) sin 90º 

= 403.11MPa

Now plane of maximum shear 

θ_s = θ_p + 45º 

θ_s = 14.52º + 45º or 104.52º + 45º 

θ_s = 59.52º or 149.52º