# Cotter Joint: Use, Application, Construction, Design

A cotter is a flat wedge-shaped piece of rectangular cross-section and its width is tapered from one end to another for an easy adjustment. It is usually made of mild steel or wrought iron.

Use of Cotter Joint

A cotter joint is used to connect two co-axial rods, which are subjected to either axial tensile force or axial compressive force. It is not used for connecting shafts that rotate and transmit torque.

Application of Cotter Joints

• Joints between the piston rod and the cross-head of a steam engine.

• Joint between the piston rod and the tail or pump rod.

• Joint between the slide spindle and the fork of the valve mechanism.

• Foundation Bolt.

Principle of Cotter Joint

• The principle of wedge action is used in a cotter joint.

• The joint is tightened and adjusted by means of a wedge action of the cotter.

• Cotter has uniform thickness and the width dimension has a slight taper.

• The taper is usually 1 in 24. Due to the taper shape, it is easy to remove the cotter and dismantle the joint.

• The taper of the cotter as well as slots is on one side because machining a taper on two sides is more difficult.

• A clearance of 1.5 mm to 3 mm is provided between the slots and the cotter.

Draw of Cotter: The amount by which the two rods are drawn together is called the draw of the cotter {alertInfo}

• Assembly and dismantling the parts of the cotter joint is quick and simple.

• The wedge action develops a very high tightening force, which prevents loosening of parts in service.

• Joint is simple to design and manufacture.

Construction of Cotter Joint

The construction of a cotter joint, used to connect two rods A and B is shown above figure. Rod-A is provided with a socket and while Rod-B is provided with a spigot end. The socket end of Rod-A fits over the spigot end of Rod-B. The socket as well as the spigot is provided with a narrow rectangular slot. A cotter is tightly fitted in this slot passing through the socket and the spigot.

NOTE : Taper in the cotter is provided to facilitate its removal when it fails due to shear. {alertInfo}

Design procedure for Cotter Joint

Assumption for stress analysis:

• The rod's size is subjected to axial tensile force.

• Effect of stress concentration due to slot is neglected.

• Stress due to initial tightening of the cotter is neglected.

Here, P = Tensile force acting on rod

d = Diameter of each rod

d₁ = Outside diameter of socket

d₂ = Diameter of spigot or inside diameter of socket

d₃ = Diameter of spigot cotter

d₄ = Diameter of socket cotter

a = distance from end of slot to the end of spigot on Rod-B

b = mean width of cotter

c = axial distance from slot to end of socket cotter

t = thickness of cotter

t₁ = thickness of spigot-cotter

l = length of cotter

Tensile failure of rods

Permissible tensile stress  $$\sigma_{t}=\frac{P}{\frac{\pi}{4} d^{2}} \quad \Rightarrow \quad d=\sqrt{\frac{4 P}{\pi \sigma_{t}}}$$

Tensile failure of spigot

The weakest cross-section is at X-X of the spigot end, which is subjected to tensile stress.

Area of section at $$X-X=\frac{\pi}{4} d_{2}^{2}-d_{2} t$$

$$\therefore$$ Permissible tensile stress,$$\sigma_{t}=\frac{P}{\frac{\pi}{4} d_{2}^{2}-d_{2} t}$$

Thickness of cotter is usually determined by empirical relationships

t=0.31 d₂

So, we can find the inner diameter (d₂) of the socket.

Tensile failure of socket

From the above figure, weakest section is Y-Y

Area of section at Y-Y=$$\left[\frac{\pi}{4}\left(d_{1}^{2}-d_{2}^{2}\right)-\left(d_{1}-d_{2}\right) t\right]$$

∴ Permissible tensile stress, $$\sigma_{t}=\frac{P}{\frac{\pi}{4}\left(d_{1}^{2}-d_{2}^{2}\right)-\left(d_{1}-d_{2}\right) t}$$

Because ' d₂ ' and ' t ' are known, we can find the outside diameter of socket (d₁).

Shear failure of Cotter

The cotter is subjected to double shear

Area of shear plane =b t

∴ Permissible shear stress, $$\quad \tau=\frac{P}{2 b t}$$

∴ Mean width of cotter (b) can be calculated.

Shear failure of Spigot End

Spigot end is subjected to double shear. Area of plane that resist shear failure =a.d₂

∴ Permissible shear stress,$$\tau=\frac{P}{2\left(a d_{2}\right)}$$

Thus, we can calculate 'a' from the above equation.

Shear failure of socket end

Socket end is subject to double shear.

Area subjected to shear $$=\left(d_{4}-d_{2}\right) C$$

∴ Permissible shear stress,$$\tau=\frac{P}{2\left(d_{4}-d_{2}\right) C}$$

Thus we can find out C.

Crushing failure of Spigot end

Area, subjected to compressive stress $$=\tau \times d_{2}$$

∴ Permissible compressive stress, $$\sigma_{c}=\frac{P}{t \times d_{2}}$$

Crushing failure of socket end

Area subjected to compressive stress= $$\left (d {_4}- d_{2} \right )\times t$$

Permissible comprehensive stress = $$\sigma _{c}=\frac{P}{\left (d {_4}- d_{2} \right )\times t}$$

Calculation of a, c, d₃, d₄, t₁ by empirical relationship

d₃=1.5d

d₄=2.4d

a=c=0.75d

t₁=0.45

If material of Rod is not given

Take material - Plain carbon steel [3068]

$$\sigma_{yt}$$=400N/mm²