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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).

1. 0.87
2. 1.75
3. 8.73
4. 17.45

Correct answer: 0.87

Solution

gamma = r*theta/L with r=0.005 m, theta=10 deg=0.17453 rad, L=1 m gives gamma=0.005*0.17453=8.727e-4=0.8727e-3, so p=0.87. The stored 1.75 is wrong.

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