Question

A cylindrical rod of diameter 10 mm and length 1.0 m is fixed at one end. The other end is twisted by an angle of ${10}^0$ by applying a torque. If the maximum shear strain in the rod is p x ${10}^3$, then p is equal to (round off to two decimal places)

Answer 1

$d=10mm;\theta ={10}^o$
$L=1m;\varphi =P\times {10}^{-3}$
From torsion eq.
$\frac{T}{J}=\frac{{\tau }_{max}}{R}=\frac{G\theta }{L}\frac{{\tau }_{max}}{G}=\frac{R\theta }{L}\Rightarrow {\varphi }_{max}=\frac{R\theta }{L}\rho \times {10}^{-3}=\frac{5\times {10}^0}{1000}\times \frac{\pi }{180}=0.8726\times {10}^{-3}\rho =0.8726$