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}


Advantage of Cotter Joint

  • 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


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.

Design procedure for cotter joint

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.


Tensile failure of spigot

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

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 cotter

Shear failure of Spigot End


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