GATE 2019 ME (ME1) – Question 30
A cylindrical rod 10 mm in diameter and 1.0 m long is fixed at one end. A torque twists the other by 10°. Its maximum shear strain is $p\times10^{-3}$. Find p, to two decimal places.
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Correct answer: .80 to .90
Explanation
Torsional shear strain is $\gamma=r\theta/L$. Convert angle to radians: $\theta=10\pi/180$. Then $\gamma=0.005(10\pi/180)/1=0.872665\times10^{-3}$. Thus **$p=0.87$**.