Question

Obtain the expression for normal and tangential stresses on a plane BE inclined at an angle θ with BC subjected to compound 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 BCE.

Let σ_n and τ be the normal and shear stresses on the plane BE,

∑Fx = 0 

σ_x (BC) – τ_xy (EC) – τ (EB) cos (90 – θ) − σ_n (EB) cos θ = 0 

∑Fy = 0 

σ_x (EC) + τ_xy (BC) + τ (EB) sin (90 – θ) − σ_n (EB) sin θ = 0 

Dividing both equations by BE replacing

sin θ = EC/ BE and cos θ = BC/ BE

σ_x cos θ + τ_xy sin θ – τ sin θ − σ_n cos θ = 0 ...(i)

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

Multiplying equation (i) and cos θ, (ii) by sin θ and then adding, 

σ_x cos² θ + τ_xy sin θ cos θ – σ_n cos² θ + σ_y sin² θ + τ_xy cos θ sin θ − σ_n sin² θ = 0 

σ_x cos² θ + σ_y sin² θ + 2τ_xy sin θ cos θ = σ_n 

or σ_n = σ_x cos² θ + σ_y sin² θ + τ_xy sin 2θ

Putting 

cos² θ = (1 + cos 2θ)/2 and sin² θ = (1 - cos 2θ)/2

Multiplying equation (i) by sin θ, (ii) by cos θ and then subtracting (ii) from (i) 

(σ_x cos θ sin θ + τ_xy sin² θ – τ sin² θ) (σ_y sin θ cos θ + τ_xy cos² θ + τ cos² θ = 0 (σ_x – σ_y) cos θ sin θ + τ_xy (sin² θ − cos² θ) − τ = 0 

τ = (σ_x + σ_y)/ 2 sin 2θ – τ_xy cos θ (cos² θ – sin² θ = cos 2θ)