Answer to Question #201649 in Mechanical Engineering for Maryam

Question

Design a typical cotter joint to transmit a load of 50 kN in tension or

compression. Consider that the rod, socket and cotter are all made of a

material with the following allowable stresses:

Allowable tensile stress sy = 150 MPa

Allowable crushing stress sc = 110 MPa

Allowable shear stress ty = 110 MPa.

Answer 1

Axial load 2

y P d

4

p = s . On substitution this gives d=20 mm. In general

standard shaft size in mm are

6 mm to 22 mm diameter 2 mm in increment

25 mm to 60 mm diameter 5 mm in increment

60 mm to 110 mm diameter 10 mm in increment

110 mm to 140 mm diameter 15 mm in increment

140 mm to 160 mm diameter 20 mm in increment

500 mm to 600 mm diameter 30 mm in increment

We, therefore, choose a suitable rod size to be 25 mm.

For tension failure across slot 2

1 y d dt P

4

? p ? ? - ?s =

? ?

. This gives 4

1 d t =1.58x10-

m2.From empirical relations we may take t=0.4d i.e. 10 mm and this gives

d1= 15.8 mm. Maintaining the proportion let d1= 1.2 d = 30 mm.

The tensile failure of socket across slot 2 2 ( )

2 1 2 1 y d d d d t P

4

?? p ? ? ?? - ? - - ?s =

?? ??

This gives d2 = 37 mm. Let d2 = 40 mm

For shear failure of cotter 2btt = P. On substitution, this gives b = 22.72

mm.

Let b = 25 mm.

For shear failure of rod end 2l1d1t = P and this gives l1 = 7.57 mm. Let l1 =

10 mm.

For shear failure of socket end 2l(d2-d1)t = P. This gives l= 22.72 mm. Let

l=25 mm

For crushing failure of socket or rod (d3-d1)tsc = P. This gives d3 = 75.5

mm. Let d3 = 77 mm.

For crushing failure of the collar ( 2 2 )

4 1 c d d P

4

p- s =. On substitution, this gives

d4= 38.4 mm. Let d4= 40 mm.

For shear failure of collar pd1t1t = P which gives t1= 4.8 mm. Let t1 = 5

mm.

Therefore the final chosen values of dimensions are

d= 25 mm; d1= 30 mm; d2 = 40 mm; d3 = 77 mm; d4 = 40 mm; t= 10 mm;

t1= 5 mm; l= 25 mm; l1= 10 mm; b= 27 mm.