Question Bank • QB_FM_GATE_005 NAT • 2 Marks

Section 2: Farm Machinery

Machine Design

Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints

Question Statement:

A solid circular steel shaft transmits $30\text{ kW}$ power at $750\text{ rpm}$. The allowable shear stress is $50\text{ MPa}$. The minimum required shaft diameter is ________ $\text{mm}$ (round off to 1 decimal place).

Answer Key:

33.9

Step-by-Step Solution & Explanation

Shaft Torsion Formula: $T = \frac{60 \times 30000}{2\pi \times 750} = 381.97\text{ N}\cdot\text{m} = 381972\text{ N}\cdot\text{mm}$ $d = \left( \frac{16 \times 381972}{\pi \times 50} \right)^{1/3} \approx 33.88\text{ mm}$ Rounding to 1 decimal place: **33.9** $\text{mm}$.
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