Question #137240

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

.

Material: MS; Allowable tensile stress = 80 N/mm2

; Allowable compressive stress = 100

N/mm2

;Allowable shear stress = 50 N/mm2

.

Expert's answer

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

d_{1}=39.44 so we take diameter as 40 mm

Learn more about our help with Assignments: Mechanical Engineering

## Comments

## Leave a comment