# Derive the expression for normal and shear stress on a plane AE inclined at an angle B with AB subjected to direct stresses

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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 σx and σy on two mutually perpendicular stresses as shown in fig. by (63.9k points)
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Consider the unit thickness of the element. Applying equations of equilibrium to the free body diagram of wedge ABE.

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

∑Fx = 0

σx (BE) – τ (AE) cos θ − σn (AE) cos (90 – θ) = 0

σx (BE) + τ (AE) cos θ + σn (AE) sin θ = 0 ...(i)

∑Fy = 0

σy (AB) + τ (AE) sin θ − σn (AE) sin (90 – θ) = 0

σy (AB) + τ (AE) sin θ − σn (AE) cos θ = 0 ...(ii)

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

sin θ = BE/ AE

and cos θ = AB/ AE

– σx sin θ + τ cos θ + σn sin θ = 0 ...(iii)

– σx cos θ + τ sin θ – σn cos θ = 0 ...(iv)

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

– σx sin2 θ + σn sin2 θ – σy cos2 θ + σn cos2 θ = 0

σn = σx sin2 θ + σy cos2 θ

Putting, cosθ = (1 + cos 2θ)/ 2

and sinθ = (1 - cos 2θ)/2 Also multiplying (iii) by cos θ and 4 by sin θ and then adding.

− σx sin θ cos θ + τ cos2 θ + σy cos θ sin θ + τ sin2 θ = 0

τ = (σx + σy) sin θ cos θ

or τ = ((σx + σy)/2)sin 2θ

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