If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.
A)40mm
B)50mm
C)60mm
D)70mm

Question

Quizplus · Accepted Answer

The crushing failure of the pin in the fork occurs when the compressive stress exceeds the allowable limit. The compressive stress ($\sigma$) is given by the formula $\sigma = \frac{P}{A}$, where $P$ is the load and $A$ is the area resisting the load. For a knuckle joint, the area resisting the load due to crushing in the fork is given by $A = d \times t$, where $d$ is the diameter of the pin and $t$ is the thickness of one eye of the fork. Since there are two eyes in the fork, the total area is $2dt$.Given:- $P = 50kN = 50000N$- Allowable compressive stress ($\sigma_{allow}$) = $25N/mm^2$- Thickness of each eye of fork ($t$) = $25mm$The allowable load $P$ can be equated to the product of the allowable stress and the area, so:$$P = \sigma_{allow} \times A$$$$50000 = 25 \times 2 \times d \times 25$$$$50000 = 1250d$$$$d = \frac{50000}{1250} = 40mm$$Therefore, the correct diameter of the knuckle pin for the given conditions is 40mm.

If Knuckle Joint Is to Fail by Crushing Failure of Pin