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## Q. 2.5

A plane element in a body is subjected to a normal stress of 25 MPa in +ve x-direction and counterclockwise shear of 100 MPa as shown in Fig. 2.15. Calculate the stresses at two planes, (i) 20° clockwise and (ii) 20° anticlockwise to x-axis.

## Verified Solution

(i) Plane 20° clockwise to x-axis. Inclination with vertical (Fig. 2.15), $\theta$ = 70° Normal stress,

$\sigma_{\acute{x}} =\frac{o_x+o_y}{2}+\frac{o_x-o_y}{2} \cos 2 \theta-\tau_{x y} \sin 2 \theta$ $=\frac{25}{2}+\frac{25}{2} \cos 140^{\circ}-(-100) \sin 140^{\circ}$

= 67.2 MPa (Tension)

Shear stress,

$\tau_{\acute{_x} \acute{_y}} =\frac{\sigma_x-\sigma_y}{2} \sin 2 \theta+\tau_{x y} \cos 2 \theta$ $=\frac{25}{2} \sin 140^{\circ}-100 \cos 140^{\circ}$

= 84.64 MPa (Clockwise)

(ii) Plane $20^{\circ}$ anticlockwise to x-axis.

Inclination with vertical (Fig. 2.15), $\theta=110^{\circ}$

Normal stress,

$\sigma_{\acute{x}} =\frac{25}{2}+\frac{25}{2} \cos 220^{\circ}-(-100) \sin 220^{\circ}$

= -61.35 MPa (Compression)

Shear Stress,

$\tau_{\acute{_x} \acute{_y}} =\frac{25}{2} \sin 220^{\circ}-100 \cos 220^{\circ}$

= 68.57 MPa (Clockwise)