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° by applying a torque. If the maximum shear strain in the rod is p × 10^–3, then p is equal to___________ (round off to two decimal places)

The correct answer is 0.872.

The maximum shear strain ($\gamma$) in a cylindrical rod subjected to torsion occurs at the outer surface (at radius $R$) and is given by the relation:

$\gamma =\frac{R\theta }{L}$

Where:
- $R$ is the radius of the rod.
- $\theta$ is the angle of twist in radians.
- $L$ is the length of the rod.

Given data:
- Diameter of the rod, $d=10\text{mm}$, which gives a radius of $R=\frac{d}{2}=5\text{mm}=5\times {10}^{-3}\text{m}$.
- Length of the rod, $L=1.0\text{m}$.
- Angle of twist, $\theta ={10}^{\circ }$.

First, convert the angle of twist from degrees to radians:
$\theta ={10}^{\circ }\times \frac{\pi }{{180}^{\circ }}=\frac{\pi }{18}\text{rad}$

Now, substitute the values into the shear strain formula:
$\gamma =\frac{(5\times {10}^{-3})\times \left(\frac{\pi }{18}\right)}{1.0}$

Simplifying the expression:
$\gamma =5\times {10}^{-3}\times 0.05526$
$\gamma \approx 0.8727\times {10}^{-3}$

Comparing this with the given format for the maximum shear strain, $p\times {10}^{-3}$, we find:
$p\approx 0.872$ (or $0.87$ when rounded off to two decimal places).