Question

Derive the expression for normal and shear stress on a plane AE inclined at an angle B with AB subjected to direct stresses of compressive nature (both) of σ_x and σ_y on two mutually perpendicular stresses as shown in fig.

Answer 1

Consider the unit thickness of the element. Applying equations of equilibrium to the free body diagram of wedge ABE. 

Let σ_n and τ be the normal and shear stresses on the plane AE.

∑Fx = 0 

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

σ_x (BE) + τ (AE) cos θ + σ_n (AE) sin θ = 0 ...(i) 

∑Fy = 0 

σ_y (AB) + τ (AE) sin θ − σ_n (AE) sin (90 – θ) = 0 

σ_y (AB) + τ (AE) sin θ − σ_n (AE) cos θ = 0 ...(ii) 

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

sin θ = BE/ AE 

and cos θ = AB/ AE 

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

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

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

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

σ_n = σ_x sin² θ + σ_y cos² θ 

Putting, cos² θ = (1 + cos 2θ)/ 2

and sin² θ = (1 - cos 2θ)/2 

Also multiplying (iii) by cos θ and 4 by sin θ and then adding. 

− σ_x sin θ cos θ + τ cos² θ + σ_y cos θ sin θ + τ sin² θ = 0 

τ = (σ_x + σ_y) sin θ cos θ 

or τ = ((σ_x + σ_y)/2)sin 2θ