Question

Prove the when stress are unequal and alike

(i) Normal stress  

(ii) Shear stress

Answer 1

By drawing the F.B.D. of wedge BCE, considering unit thickness of element.

Let σ_n and τ be normal and shear stresses on the plane BE, Applying Equilibrium conditions to the wedge BCE:

∑Fx = 0 

σ_x (BC) – σ_n (BE) cos θ − τ (BE) cos (90 – θ) = 0 

σ_x (BC) – σ_n (BE) cos θ − τ (BE) sin θ = 0 ...(i) 

∑Fy = 0 

σ_y (EC) – σ_n (BE) sin θ + τ (BE) sin (90 – θ) = 0 

σ_x EC – σ_n (BE) sin θ + τ (BE) cos θ = 0 ...(ii) 

Dividing equations (i) and (ii) by BE and replacing

BC/ BE = cos θ and EC/ BE = sin θ 

σ_x cos θ – σ_n cos θ – τ sin θ = 0 ...(iii) 

σ_y sin θ – σ_n sin θ + τ cos θ = 0 ...(iv) 

Multiplying equation (iii) by cos θ, (iv) by sin θ and adding 

σ_x cos² θ – σ_n cos² θ + σ_y sin² θ − σ_n sin² θ = 0

σ_n = σ_x cos² θ + σ_y sin² θ

putting

Multiplying equation (iii) by sin θ and (iv) by cos θ and then subtracting. 

σ_x sin θ cos θ – τ sin² θ − σ_y sin θ cos θ – τ cos² θ = 0 [∵ (sin² θ + cos² θ) = τ] 

or (σ_x – σ_y) sin θ cos θ – τ = 0 

or τ = (σ_x – σ_y) sin θ cos θ 

τ = 1/2 (σ_x − σ_y)sin²θ