Let the cross sectional area of the bar be A. Consider the equilibrium of the plane aa' . A force F must be acting on this plane making an angle (π/2)-θ with the normal ON. Resolving F into components, along the plane and normal to the plane

F_{p} = F cosθ

F_{N} = Sinθ

Let the area of the face aa' be A' , then

(A/A') = sinθ

A' = A/Sinθ

The tensile stress T = (Fsinθ/A') = (F/A)sin^{2}θ and the shearing stress Z = (F cosθ)/A' = (F/A)cosθ sinθ = (F sin2θ)/2A. Maximum tensile stress is when θ = π/2 and maximum shearing stress when 2θ = π/2 or θ = π/4.