**Socket and Spigot Cotter JointÂ :**

In a socket and spigot cotter joint, one end of the rods (say A) is provided with a socket type ofÂ end as shown in Fig. 1 and the other end of the other rod (say B) is inserted into a socket. The end

of the rod which goes into a socket is also called spigot. A rectangular hole is made in the socket andÂ spigot. A cotter is then driven tightly through a hole in order to make the temporary connectionÂ between the two rods. The load is usually acting axially, but it changes its direction and hence theÂ cotter joint must be designed to carry both the tensile and compressive loads. The compressive loadÂ is taken up by the collar on the spigot.

**Design of Socket and Spigot Cotter Joint :**

The socket and spigot cotter joint is shown in Fig. 1.

Let 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,

Ïƒt = Permissible tensile stress for the rods material,

Ï„ = Permissible shear stress for the cotter material, and

Ïƒc = Permissible crushing stress for the cotter material.

The dimensions for a socket andÂ spigot cotter joint may be obtained byÂ considering the various modes of failureÂ as discussed below :

**1. Failure of the rods in tension**

The rods may fail in tension due toÂ the tensile load P. We know that

Area resisting tearing

= Ï€Â Ă—Â d^2 / 4

âˆ´ Tearing strength of the rods,

= Ï€Â Ă—Â d^2 Ă— Ïƒt/ 4

Equating this to load (P), we have

P = Ï€Â Ă—Â d^2 Ă— Ïƒt/ 4

From this equation, diameter of theÂ rods ( d ) may be determined.

**2. Failure of spigot in tension across the weakest section (or slot)**

Since the weakest section of the spigot is that section whichÂ has a slot in it for the cotter, as shown in Fig. 2, therefore

Area resisting tearing of the spigot across the slot

=Â Ï€ (d2)^2 – d2Ă—t

and tearing strength of the spigot across the slot

= [Ï€ (d2)^2 – d2Ă—t]Â Ïƒt

Equating this to load (P), we have

P =Â [Ï€ (d2)^2 – d2Ă—t]Â Ïƒt

From this equation, the diameter of spigot or inside diameter of socket (d2) may be determined.

**Note** : In actual practice, the thickness of cotter is usually taken as d2 / 4.

**3. Failure of the rod or cotter in crushing**

We know that the area that resists crushing of a rod or cotter

=d2 Ă— t

âˆ´ Crushing strength = d2 Ă— t Ă— Ïƒc

Equating this to load (P), we have

P =d2 Ă— t Ă— Ïƒc

From this equation, the induced crushing stress may be checked.

**4. Failure of the socket in tension across the slot**

We know that the resisting area of the socket across theÂ slot, as shown in Fig. 3

=Â Ï€/4 [(d1)^2 – (d2)^2] – (d1 – d2)t

âˆ´ Tearing strength of the socket across the slot

= { Ï€/4 [(d1)^2 – (d2)^2] – (d1 – d2)t }Â Ïƒt

Equating this to load (P), we have

P = { Ï€/4 [(d1)^2 – (d2)^2] – (d1 – d2)t }Â Ïƒt

From this equation, outside diameter of socket (d1) may be determined.

**5. Failure of cotter in shear**

Considering the failure of cotter in shear as shown in Fig. 4. Since the cotter is in doubleÂ shear, therefore shearing area of the cotter

= 2 b Ă— t

and shearing strength of the cotter

=2 b Ă— t Ă— Ï„

Equating this to load (P), we have

P =2 b Ă— t Ă— Ï„

From this equation, width of cotter (b) is determined.

**6. Failure of the socket collar in crushing**

Considering the failure of socket collar in crushing as shown in

Fig. 5

We know that area that resists crushing of socket collar

=(d4 â€“ d2) t

and crushing strength =(d4 â€“ d2) t Ă— Ïƒc

Equating this to load (P), we have

P =(d4 â€“ d2) t Ă— Ïƒc

From this equation, the diameter of socket collar (d4) mayÂ be obtained.

**7. Failure of socket end in shearing**

Since the socket end is in double shear, therefore area thatÂ resists shearing of socket collar

=2 (d4 â€“ d2) c

and shearing strength of socket collar

=2 (d4 â€“ d2) c Ă— Ï„

Equating this to load (P), we have

P =2 (d4 â€“ d2) c Ă— Ï„

From this equation, the thickness of socket collar (c) may be obtained.

**8. Failure of rod end in shear**

Since the rod end is in double shear, therefore the area resisting shear of the rod end

= 2 a Ă— d2

and shear strength of the rod end

= 2 a Ă— d2 Ă— Ï„

Equating this to load (P), we have

P = 2 a Ă— d2 Ă— Ï„

From this equation, the distance from the end of the slot to the end of the rod (a) may beÂ obtained.

**9. Failure of spigot collar in crushing**

Considering the failure of the spigot collar in crushing asÂ shown in Fig. 6. We know that area that resists crushing of theÂ collar

=Â Ï€/4 [(d3)^2 – (d2)^2]

and crushing strength of the collar

=Â Ï€/4 [(d3)^2 – (d2)^2]Â Ïƒc

Equating this to load (P), we have

PÂ =Â Ï€/4 [(d3)^2 – (d2)^2]Â Ïƒc

From this equation, the diameter of the spigot collar (d3)Â may be obtained.

**10. Failure of the spigot collar in shearing**

Considering the failure of the spigot collar in shearing asÂ shown in Fig. 12.7. We know that area that resists shearing of theÂ collar

= Ï€ d2 Ă— t1

and shearing strength of the collar,

= Ï€ d2 Ă— t1 Ă— Ï„

Equating this to load (P) we have

P = Ï€ d2 Ă— t1 Ă— Ï„

From this equation, the thickness of spigotÂ collar (t1) may be obtained.

**11. Failure of cotter in bending**

In all the above relations, it is assumedÂ that the load is uniformly distributed over theÂ various cross-sections of the joint. But in actual

practice, this does not happen and the cotter isÂ subjected to bending. In order to find out theÂ bending stress induced, it is assumed that theÂ load on the cotter in the rod end is uniformlyÂ distributed while in the socket end it varies fromÂ zero at the outer diameter (d4) and maximum atÂ the inner diameter (d2), as shown in Fig. 8.

The maximum bending moment occurs at the centre of the cotter and is given by

Mmax = P/2 [1/3 Ă— (d4 – d2)/2 Ă— d2/2] – [P/2 Ă— d2/4]

=Â P/2 [(d4 – d2)/6 + (d2/2) – (d2/4)]

=Â P/2 [(d4 – d2)/6 + (d2/4)]

We know that section modulus of the cotter,

Z = t Ă— b^2 / 6

âˆ´ Bending stress induced in the cotter,

Ïƒb =Â Mmax / Z

=Â P/2 [(d4 – d2)/6 + (d2/4)] / [t Ă— b^2 / 6]

= P (d4 + 0.5d2) / 2t Ă— b^2

This bending stress induced in the cotter should be less than the allowable bending stress ofÂ the cotter.

12.The length of cotter (l) is taken as 4 d.

13. The taper in cotter should not exceed 1 in 24. In case the greater taper is required, then aÂ locking device must be provided.

14.The draw of cotter is generally taken as 2 to 3 mm.

**Notes:**

1. When all the parts of the joint are made of steel, the following proportions in terms of diameter of theÂ rod (d) are generally adopted :

d1 = 1.75 d , d2 = 1.21 d , d3 = 1.5 d , d4 = 2.4 d , a = c = 0.75 d , b = 1.3 d, l = 4 d , t = 0.31 d ,Â t1 = 0.45 d , e = 1.2 d.

Taper of cotter = 1 in 25, and draw of cotter = 2 to 3 mm.

2. If the rod and cotter are made of steel or wrought iron, then Ï„ = 0.8 Ïƒt and Ïƒc = 2 Ïƒt may be taken.

Reference A Textbook of Machine Design by R.S. Khurmi and J.K. Gupta

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