A prismatic bar having cross-sectional areas A = 1200 mm^{2 }is compressed by an axial load P = 90 kN. The normal and shear stresses acting on inclined plane p – q cut through the bar at an angle θ = 25° are respectively

This question was previously asked in

HPPSC AE Civil 2016 (Deputy Manager) Official Paper

Option 1 : -61.6 MPa, 28.7 MPa

__Concept:__

ii) If σx and σy are normal stress on vertical and horizontal plane respectively and this plane is accompanied by shear stress τxy then normal stress and shear stress on plane a-a, which is inclined at an angle θ from plane of σx.

\(σ _{1\left( {a - a} \right)}' = \frac{{{σ _x} + {σ _y}}}{2} + \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right) \cdot \cos 2\theta + {τ _{xy}}\sin 2\theta \)

\(σ _{2\left( {a - a} \right)}' = \left( {\frac{{{σ _x} + {σ _y}}}{2}} \right) - \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right) \cdot \cos 2\theta - {τ _{xy}}\sin 2\theta \)

\({τ _{\left( {a - a} \right)}} = - \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)\sin 2\theta + {τ _{xy}}\cos 2\theta \)

**Calculation:**

Given,

P = 90 kN, A = 1200 mm^{2}

stress in x-direction, σ_{x} = -P/A (-ve because compressive force is given in the question)

⇒ \({σ _x} = - \frac{{90 \times 1000}}{{1200}} = - 75\ MPa\)

Stress in y-direction = 0

Normal stress on plane p-q is given by, \({σ _n} = \frac{{{σ _x}}}{2} + \left( {\frac{{{σ _x}}}{2}} \right)\cos 2\theta \)

⇒ \({σ _n} = \frac{{ - 75}}{2} + \left( {\frac{{ - 75}}{2}} \right)\cos \left( {2 \times 25} \right)\)

**σ _{n} = -61.60 MPa**

Shear stress on plane p-q is given by, \(τ = - \left( {\frac{{{\sigma _x}}}{2}} \right)\sin 2\theta\)

⇒ \(τ = - \left( {\frac{{ - 75}}{2}} \right)\sin \left( {2 \times 25} \right)\)

**τ = 28.7 MPa**