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Q.
The upper end of a wire of radius 4 mm and length $100$ cm is clamped and its other end is twisted through an angle of $30^{\circ}$. The angle of shear is
Mechanical Properties of Solids
Solution:
Angle of twist at free end
$=30^{\circ}=\frac{30}{180} \times \pi $ rad $=\frac{\pi}{6}$ rad
Displacement of the free surface,
$\Delta L=\frac{2 \pi r}{2 \pi} \times \frac{\pi}{6}=\frac{\pi r}{6}=\frac{\pi \times 0.4}{6} cm$
Angle of shear or shearing strain $=\frac{\Delta L}{L}$
$=\frac{\pi \times 0.4}{6 \times 100}$ rad.
$=\frac{\pi \times 0.4}{6 \times 100} \times \frac{180}{\pi}$
degree $=0.12^{\circ}$