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GATE 2020 PI – Question 47

General Engineering · Theory of Machines and Design · 2 marks · Numerical answer

A solid shaft transmits50 kW at1910 rpm, with no bending. Allowable shear stress is80 MPa. Find minimum diameter in cm (one decimal).

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Correct answer: 2.0 to 3.0

Explanation

Angular speed is2π(1910)/60 rad/s and torque T=P/ω≈250 N m. For a solid shaft τmax=16T/(πd³). Thus d=[16T/(π×80×10⁶)]^(1/3)≈0.0252 m=2.52 cm, or2.5 cm.

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