Answer to Question #239560 in Mechanical Engineering for Pkl

Question

Design sleeve and cotter joint to withstand a load of 60kN. All parts of the
joint are made of same material and the permissible stresses are 60 N/mm2
in tension, 125 N/mm2
in compression and 70 N/mm2
in shear

Answer 1

Given :

$σc = 125 MPa = 125 N/mm^2$

1. Diameter of the rods

Let d = Diameter of the rods. Considering the failure of the rods in tension.

We know that load (P),

$60 × 103 = π x d2 x σt$

4

$d2 = 60 × 103/ 47.13 = 1273$ = 35.7 say 36 mm

Diameter of enlarged end of rod and thickness of cotter

Let     d2 = Diameter of enlarged end of rod, and

t = Thickness of cotter. It may be taken as d2 / 4.

Considering the failure of the rod in tension across the weakest section (i.e. slot). We know that

load (P),

$60 × 103 = [ π (d2)2 - d2 x t]σt4$

$= [ π (d2)2 - d2 x (d2/4)] 604$ = 32.13

$(d2)^2 = 60 × 103 / 32.13 = 1867$ d2     

$= 43.2 say 44 mm$

Thickness of cotter,

$= (d2/4) = 44/4$

=11 mm

Let us now check the induced crushing stress in the rod or cotter. We know that load (P), 60 × 10³

$= d2 × t ×σc$

=44 × 11 × σc

=484 σc

$σc = 60 × 103 / 484$

= 124 N/mm2

Since the induced crushing stress is less than the given value of 125 N/mm2, therefore the dimensions d2 and t are within safe limits.