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 connecting two tie-rods in a tractor steering linkage carries an axial tensile load of $P = 40\text{ kN}$. The joint pin has a diameter of $d = 25\text{ mm}$ and is subjected to double shear. The average shear stress induced in the pin in $\text{MPa}$ is ________ (round off to 1 decimal place). Take $\pi = 3.1416$.
Answer Key:
40.7
Step-by-Step Solution & Explanation
In a knuckle joint, the pin fails in double shear across two circular cross-sections:
$A_s = 2 \times \left(\frac{\pi}{4} d^2\right) = \frac{\pi}{2} (25\text{ mm})^2 = 1.5708 \times 625 = 981.75\text{ mm}^2$
$\tau = \frac{P}{A_s} = \frac{40,000\text{ N}}{981.75\text{ mm}^2} \approx 40.74\text{ MPa} \approx 40.7\text{ MPa}$