Question Bank • QB_FM_071 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 steel shaft transmits $45\text{ kW}$ power at $300\text{ rpm}$. If the maximum allowable shear stress in the shaft material is $50\text{ MPa}$, calculate the minimum required shaft diameter in $\text{mm}$.

Answer Key:

52.68

Step-by-Step Solution & Explanation

Given: - Power $P = 45\text{ kW} = 45000\text{ W}$ - Speed $N = 300\text{ rpm}$ - Torque $T = \frac{P \times 60}{2 \pi N} = \frac{45000 \times 60}{2 \pi \times 300} = \frac{2700000}{1884.95} = 1432.39\text{ N}\cdot\text{m} = 1432.39 \times 10^3\text{ N}\cdot\text{mm}$ From torsion formula: $\tau = \frac{16 T}{\pi d^3} \le 50\text{ MPa}$ $d^3 = \frac{16 \times 1432.39 \times 10^3}{\pi \times 50} = \frac{22918240}{157.08} = 145901.7\text{ mm}^3$ $d = (145901.7)^{1/3} \approx 52.68\text{ mm}$
← Previous (QB_FM_025) Next (QB_FM_090) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test