Question Bank • QB_FM_022 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 drive shaft in an agricultural machine is subjected to a maximum bending moment of $M = 1200\text{ N}\cdot\text{m}$ and a twisting moment of $T = 1600\text{ N}\cdot\text{m}$. The allowable shear stress for the shaft material is $\tau_{all} = 50\text{ MPa}$. According to the maximum shear stress theory (Guest's theory), the minimum required shaft diameter in $\text{mm}$ is ________ (round off to 1 decimal place). Take $\pi = 3.1416$.

Answer Key:

58.8

Step-by-Step Solution & Explanation

1. Equivalent twisting moment ($T_e$): $T_e = \sqrt{M^2 + T^2} = \sqrt{1200^2 + 1600^2} = \sqrt{1,440,000 + 2,560,000} = \sqrt{4,000,000} = 2000\text{ N}\cdot\text{m}$ 2. Maximum shear stress equation for solid shaft: $T_e = \frac{\pi}{16} \tau_{all} d^3$ $d^3 = \frac{16 T_e}{\pi \tau_{all}} = \frac{16 \times 2000}{\pi \times (50 \times 10^6\text{ N/m}^2)} = \frac{32,000}{1.5708 \times 10^8} \approx 2.0372 \times 10^{-4}\text{ m}^3$ $d = (2.0372 \times 10^{-4})^{1/3} = 0.05884\text{ m} \approx 58.8\text{ mm}$
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