Allied & ICAR Exam Bank • ALLIED_0129 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 knuckle joint pin connects two rods subjected to an axial pull of $24\text{ kN}$. The pin is in double shear. If the allowable shear stress is $60\text{ MPa}$, the minimum pin diameter in mm is (rounded off to one decimal place):

Answer Key:

16.0

Step-by-Step Solution & Explanation

In double shear, total shear area $= 2 \times \frac{\pi}{4} d^2 = \frac{\pi}{2} d^2$. $P = \tau \times \frac{\pi}{2} d^2 \implies 24000 = 60 \times 10^6 \times \frac{\pi}{2} d^2 \implies d^2 = \frac{24000}{94247780} = 2.546 \times 10^{-4}\text{ m}^2 \implies d = 0.01595\text{ m} \approx 16.0\text{ mm}$.
← Previous (ALLIED_0128) Next (ALLIED_0130) →

Practice In Authentic CBT Exam Interface

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

Launch CBT Mock Test