The document describes the design of a cotter joint to withstand a maximum tensile load of 6KN. It provides definitions of the variables involved in the joint design. It then outlines 10 steps to size the different parts of the joint based on the material properties and load value, determining values for the diameters, thicknesses, and distances. The final section provides the results of the full design process.

1. 22564 EMD 2.1 Design of Joints, levers, offset links
Design a cotter joint for a maximum tensile load of 6KN. The
cotter joint parts are made of 30C6 steel. The permissible
stresses for it as follows t = 60 N/ mm2 , c = 100 N / mm2,  =
40 N / mm2.

2. 22564 EMD 2.1 Design of Joints, levers, offset links
Socket and Spigot Cotter Joint
P = Load carried by the rods,
d = Diameter of the rods,
d1 = Outside diameter of socket,
d2 = Diameter of spigot or inside diameter of
socket,
d3 = Outside diameter of spigot collar,
t1 = Thickness of spigot collar,
d4 = Diameter of socket collar,
c = Thickness of socket collar,
b = Mean width of cotter,
t = Thickness of cotter,
l = Length of cotter,
a = Distance from the end of the slot to the end
of rod,
σt = Permissible tensile stress for the rods
material,
τ = Permissible shear stress for the cotter
material, and
σc = Permissible crushing stress for the cotter
material.

3. 22564 EMD 2.1 Design of Joints, levers, offset links
1. Failure of the rods in tension
The rods may fail in tension due to the tensile load P. We know that
t = 60Mpa, P = 6KN = 6000N
Area resisting tearing =
Tearing strength of the rods, =
d2 = 127.23
d=11.28  12mm
From this equation, diameter of the rods ( d ) may be determined.

4. 22564 EMD 2.1 Design of Joints, levers, offset links
2. Failure of spigot in tension across the weakest section
Consider thickness of cotter as per
standard relations
t = 0.31 d = 3.6 = 4mm
find the value of d2 from above
equation
d2 = 38.31mm = 40mm
From this equation, the diameter of spigot or inside diameter of socket (d2)
determined.
t = 60Mpa, P = 6KN = 6000N find d2 and t

5. 3. Failure of the Spigot in crushing
d
2
t
d2 = 15mm is smaller than previous d2 = 40mm
Therefore d2 = 40mm is safe value
c = 100Mpa, P = 6KN = 6000N check d2

6. 22564 EMD 2.1 Design of Joints, levers, offset links
4. Failure of the socket in tension across the slot
we know d2=40mm , t=4mm, t = 60Mpa,
P = 6KN 6000N find d1
d1
2- 5.095 d1-1523.566 = 0
d1 = 41.66 mm  42 mm
From this outer diameter of socket is determined

7. 22564 EMD 2.1 Design of Joints, levers, offset links
5. Failure of cotter in shear
Area under failure
A = 2 x b x 4
We know t = 4mm,  = 40Mpa, P = 6KN 6000N find b
b = 18.75  20mm

8. 22564 EMD 2.1 Design of Joints, levers, offset links
6. Failure of the socket collar in crushing
We know d2=40mm, t=4mm c = 100Mpa, P = 6KN find d4
d4 – 40 =15
d4 = 55mm  56mm
Outer diameter of socket collar is 56 mm

9. 22564 EMD 2.1 Design of Joints, levers, offset links
7. Failure of socket end in shearing
We know d2=40mm, d4 =56mm, ,  = 40Mpa, P = 6KN
= 6000N find c
c = 4.68 mm  6mm
From this the distance ‘c’ of the socket is determined

10. 22564 EMD 2.1 Design of Joints, levers, offset links
8. Failure of Spigot end in shear
Resisting Area
We know that d2 = 40mm ,  = 40 Mpa ,  = 40Mpa, P =
6KN 6000N find a
a = 1.875 mm = 2 mm
The distance ‘a’ is determined here

11. 22564 EMD 2.1 Design of Joints, levers, offset links
9. Failure of spigot collar in crushing
We know d2 = 40 mm c =100Mpa , P = 6KN
=6000N find d3
d3
2 - 402=76.39
d3 =40.94  42mm
The outer diameter of collar can be determined here

12. 22564 EMD 2.1 Design of Joints, levers, offset links
10. Failure of the spigot collar in shearing
We know d2 = 40 mm,  = 40Mpa, P =
6KN 6000N find t2
Area
t1 = 1.19 mm  2mm
Thickness of collar can be determined

13. 22564 EMD 2.1 Design of Joints, levers, offset links
Final Design Results
d = 12 mm
d1 = 42mm
d2 = 40 mm
d3 = 42 mm
d4 = 56 mm
t = 4mm
t1 = 2 mm
a = 2mm
b = 20mm
c = 6mm

14. 22564 EMD 2.1 Design of Joints, levers, offset links