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} sin^{2} θ + σ_{n} sin^{2} θ – σ_{y} cos^{2} θ + σ_{n} cos^{2} θ = 0

σ_{n} = σ_{x} sin^{2} θ + σ_{y} cos^{2} θ

Putting, cos^{2 }θ = (1 + cos 2θ)/ 2

and sin^{2 }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 cos^{2} θ + σ_{y} cos θ sin θ + τ sin^{2} θ = 0

τ = (σ_{x} + σ_{y}) sin θ cos θ

or τ = ((σ_{x} + σ_{y})/2)sin 2θ