Answer to Question #137240 in Mechanical Engineering for Sudip

Question

Design a socket and spigot type cotter joint to withstand a reciprocating load of 50 kN.
Material: MS; Allowable tensile stress = 80 N/mm2
; Allowable compressive stress = 100
N/mm2
;Allowable shear stress = 50 N/mm2
.

Answer 1

Load=50 kN, shear stress= $\tau$ =50 $\frac{N}{mm^2}$ , Allowable tensile stress=$\sigma_t=$ 80 $\frac{N}{mm^2}$

Allowable compressive load =$\sigma_c= 100 \frac{N}{mm^2}$

(i) For Diameter of rod

$P=\frac{\pi}{4}\times d^2\times \sigma_t$

$50\times1000=\frac{\pi}{4}\times d^2\times 80$

$d=28.21 mm$ so we take dia as 30 mm

Failure of spigot in tension

$P=\frac{\pi}{4}\times d^2\times \sigma_t- dt \sigma_t, t= d/4$

$50000=\frac{\pi}{4}\times d^2\times 80- d^2\times 20$

d=34.179 mm so we take dia as 35 mm

(ii) Failure of spigot end in shear

$P=2\times a\times d\times \tau$

$50000=2\times a\times 35\times 50$

a=14.285 mm

Failure of spigot collar in shear

$P=\pi\times d\times t_1 \times \tau$

$50000=3.142\times 35 \times t_1 \times 50$

$t_1=$ 9.09 mm

(iii) Failure of socket in tension

P=$\frac{\pi}{4}(d_1^2-d_2^2)-(d_1-d_2)\times t\times \sigma_t$

$50000=\frac{3.14}{4}(d_1^2-35^2)-(d_1-35)\times 10\times 80$

$50000=0.785 d_1^2-961.625-800d_1+2800$

on solving this quadratic we get

d₁=39.44 so we take diameter as 40 mm