Answer to Question #239560 in Mechanical Engineering for Pkl

Question #239560
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
1
Expert's answer
2021-09-20T06:06:47-0400


Given :

"\u03c3c = 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 \u00d7 103 = \u03c0 x d2 x \u03c3t"

4


"d2 = 60 \u00d7 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 \u00d7 103 = [ \u03c0 (d2)2 - d2 x t]\u03c3t4"


"= [ \u03c0 (d2)2 - d2 x (d2\/4)] 604" = 32.13


"(d2)^2 = 60 \u00d7 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 \u00d7 t \u00d7\u03c3c"

=44 × 11 × σc


=484 σc

"\u03c3c = 60 \u00d7 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.




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