A bar is subjected to a uniform tensile stress of 100 N/mm². Find the intensity of normal stress on a plane the normal to which is inclined 30° to the axis of the bar:

Question

Question

This question was previously asked in

JKSSB JE CE 2021 Official Paper Shift 2 (Held on 29 Oct 2021)

Answer 1

Concept:

F1 M.J Madhu 25.03.20 D10

Normal stress at inclined plane is given by

\({σ _n} = \frac{1}{2}\left[ {{σ _x} + {σ _y}} \right] + \frac{1}{2}\left[ {{σ _x} - {σ _y}} \right]\cos 2θ + {τ _{xy}}\sin 2θ \)

Shear stress at inclined plane is given by

\({τ _s} = - \frac{1}{2}\left[ {{σ _x} - {σ _y}} \right]\sin 2θ + {τ _{xy}}\cos 2θ \)

Calculation:

Given:

σ_x = 100 N/mm2, θ = 30°.

Normal stress at inclined plane is given by

\({σ _n} = \frac{1}{2}\left[ {{σ _x} + {σ _y}} \right] + \frac{1}{2}\left[ {{σ _x} - {σ _y}} \right]\cos 2θ + {τ _{xy}}\sin 2θ \)

\({σ _n} = \frac{σ _x}{2}+\; \frac{σ _x}{2}\cos 2θ \) [∵ σ_y = 0, τ_xy = 0]

\({σ _{30}} = \frac{100}{2}+\; \frac{100}{2}\cos 60\)

\({σ _{30}} = 50+\;(50\times 0.5) = 75\;N/mm^2\)