A body is subjected to a direct tensile stress (σ) in one plane. The shear stress is max at a section inclined at ____ to the normal of the section.

Concept:

Shear stress formula for complex loading is given by;

\({τ _θ } = \frac{1}{2}\;\left( {{σ _x} - {σ _y}} \right)\sin 2θ - {τ _{xy}}\cos 2θ \)

where σ_x = stress in x-direction, σ_y = stress in y-direction, τ_xy = shear stress in xy plane, τ_θ  = shear stress in a plane at angle θ 

Calculation:

Given:

σ_x = σ,  σ_y = 0, τ_xy = 0

Now the relation becomes 

\({τ _θ } = \frac{1}{2}\;\left( {{σ _x}} \right)\sin 2θ\)

Now the value of shear stress will be maximum for the maximum value of sin 2θ and for θ = 45° and θ = 135° value of sin 2θ will be maximum that is 1 hence shear stress will be maximum on these planes.