Cotter and Knuckle Joints
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Fig. 12.6
Fig. 12.7
Fig. 12.8
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 × d
2
and shear strength of the rod end
=2 a × d
2
× ∀
Equating this to load (P), we have
P =2 a × d
2
× ∀
From this equation, the distance from the end of the slot to the end of the rod (a) may be
obtained.
9. Failure of spigot collar in crushing
Considering the failure of the spigot collar in crushing as
shown in Fig. 12.6. We know that area that resists crushing of the
collar
=
22
32
()–()
4
dd
#
&∋
∗+
and crushing strength of the collar
=
22
32
()–()
4
c
dd
#
&∋
!
∗+
Equating this to load (P), we have
P =
22
32
()–()
4
c
dd
#
&∋
!
∗+
From this equation, the diameter of the spigot collar (d
3
)
may be obtained.
10. Failure of the spigot collar in shearing
Considering the failure of the spigot collar in shearing as
shown in Fig. 12.7. We know that area that resists shearing of the
collar
= # d
2
× t
1
and shearing strength of the collar,
= # d
2
× t
1
× ∀
Equating this to load (P) we have
P = #/d
2
× t
1
× ∀
From this equation, the thickness of spigot
collar (t
1
) may be obtained.
11. Failure of cotter in bending
In all the above relations, it is assumed
that the load is uniformly distributed over the
various cross-sections of the joint. But in actual
practice, this does not happen and the cotter is
subjected to bending. In order to find out the
bending stress induced, it is assumed that the
load on the cotter in the rod end is uniformly
distributed while in the socket end it varies from
zero at the outer diameter (d
4
) and maximum at
the inner diameter (d
2
), as shown in Fig. 12.8.
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The maximum bending moment occurs at the centre of the cotter and is given by
M
max
=
42 2 2
–
1
–
23 2 2 2 4
ddd d
PP
01
∃2 ∃
34
56
=
4222 422
––
–
26 2426 4
dddd ddd
PP
0101
27 2
3434
5656
We know that section modulus of the cotter,
Z = t × b
2
/ 6
% Bending stress induced in the cotter,
!
b
=
42 2
42
22
–
(0.5)
26 4
/6 2
max
dddP
M
Pd d
Z
tb tb
01
2
34
2
56
77
∃∃
This bending stress induced in the cotter should be less than the allowable bending stress of
the cotter.
12.The length of cotter (l) in taken as 4 d.
13. The taper in cotter should not exceed 1 in 24. In case the greater taper is required, then a
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 the
rod (d) are generally adopted :
d
1
= 1.75 d , d
2
= 1.21 d , d
3
= 1.5 d , d
4
= 2.4 d , a = c = 0.75 d , b = 1.3 d, l = 4 d , t = 0.31 d ,
t
1
= 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 ∀ = 0.8 !
t
and !
c
= 2 !
t
may be taken.
Example 12.1. Design and draw a cotter joint to support a load varying from 30 kN in
compression to 30 kN in tension. The material used is carbon steel for which the following
allowable stresses may be used. The load is applied statically.
Tensile stress = compressive stress = 50 MPa ; shear stress = 35 MPa and crushing stress
= 90 MPa.
Solution. Given : P = 30 kN = 30 × 10
3
N ; !
t
= 50 MPa = 50 N / mm
2
; ∀ = 35 MPa = 35 N / mm
2
;
!
c
= 90 MPa = 90 N/mm
2
Accessories for hand operated sockets.
Cotter and Knuckle Joints
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437
The cotter joint is shown in Fig. 12.1. The joint is designed as discussed below :
1. Diameter of the rods
Let d = Diameter of the rods.
Considering the failure of the rod in tension. We know that load (P),
30 × 10
3
=
22
50
44
t
dd
##
∃∃!7∃∃
= 39.3 d
2
% d
2
= 30 × 10
3
/ 39.3 = 763 or d = 27.6 say 28 mm
Ans.
2. Diameter of spigot and thickness of cotter
Let d
2
= Diameter of spigot or inside diameter of socket, and
t = Thickness of cotter. It may be taken as d
2
/ 4.
Considering the failure of spigot in tension across the weakest section. We know that load (P),
30 × 10
3
=
22
2
22 22
()– ()– 50
444
t
d
ddt dd
##
&∋
&∋
∃!7 ∃
()
()
∗+∗ +
= 26.8 (d
2
)
2
% (d
2
)
2
= 30 × 10
3
/ 26.8 = 1119.4 or d
2
= 33.4 say 34 mm
and thickness of cotter, t =
2
34
44
7
d
= 8.5 mm
Let us now check the induced crushing stress. We know that load (P),
30 × 10
3
= d
2
× t × !
c
= 34 × 8.5 × !
c
= 289 !
c
%!
c
= 30 × 10
3
/ 289 = 103.8 N/mm
2
Since this value of !
c
is more than the given value of !
c
= 90 N/mm
2
, therefore the dimensions d
2
= 34 mm and t = 8.5 mm are not safe. Now let us find the values of d
2
and t by substituting the value of
!
c
= 90 N/mm
2
in the above expression, i.e.
30 × 10
3
=
2
2
90
4
d
d
∃∃
= 22.5 (d
2
)
2
% (d
2
)
2
= 30 × 10
3
/ 22.5 = 1333 or d
2
= 36.5 say 40 mm
Ans.
and t = d
2
/ 4 = 40 / 4 = 10 mm Ans.
3. Outside diameter of socket
Let d
1
= Outside diameter of socket.
Considering the failure of the socket in tension across the slot. We know that load (P),
30 × 10
3
=
−.
22
12 12
() () –( )
4
t
dd ddt
#
&∋
,,!
()
∗+
=
−.
22
11
() (40) –( 40)1050
4
dd
#
&∋
,,
()
∗+
30 × 10
3
/50 = 0.7854 (d
1
)
2
– 1256.6 – 10 d
1
+ 400
or (d
1
)
2
– 12.7 d
1
– 1854.6 = 0
% d
1
=
2
12.7 (12.7) 4 1854.6
12.7 87.1
22
82∃
8
7
= 49.9 say 50 mm
Ans. (Taking +ve sign)
4. Width of cotter
Let b = Width of cotter.
Considering the failure of the cotter in shear. Since the cotter is in double shear, therefore load (P),
438
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A Textbook of Machine Design
30 × 10
3
=2 b × t × ∀ = 2 b × 10 × 35 = 700 b
% b = 30 × 10
3
/ 700 = 43 mm
Ans.
5. Diameter of socket collar
Let d
4
= Diameter of socket collar.
Considering the failure of the socket collar and cotter in crushing. We know that load (P),
30 × 10
3
=(d
4
– d
2
) t × !
c
= (d
4
– 40)10 × 90 = (d
4
– 40) 900
% d
4
– 40 = 30 × 10
3
/ 900 = 33.3 or d
4
= 33.3 + 40 = 73.3 say 75 mm
Ans.
6. Thickness of socket collar
Let c = Thickness of socket collar.
Considering the failure of the socket end in shearing. Since the socket end is in double shear,
therefore load (P),
30 × 10
3
=2(d
4
– d
2
) c × ∀ = 2 (75 – 40 ) c × 35 = 2450 c
% c = 30 × 10
3
/ 2450 = 12 mm
Ans.
7. Distance from the end of the slot to the end of the rod
Let a = Distance from the end of slot to the end of the rod.
Considering the failure of the rod end in shear. Since the rod end is in double shear, therefore
load (P),
30 × 10
3
=2 a × d
2
× ∀ = 2a × 40 × 35 = 2800 a
% a = 30 × 10
3
/ 2800 = 10.7 say 11 mm
Ans.
8. Diameter of spigot collar
Let d
3
= Diameter of spigot collar.
Considering the failure of spigot collar in crushing. We know that load (P),
30 × 10
3
=
22 22
32 3
( ) ( ) ( ) (40) 90
44
c
dd d
##
&∋&∋
,!7 ,
∗+∗+
or (d
3
)
2
– (40)
2
=
3
30 10 4
90
∃∃
∃#
= 424
% (d
3
)
2
= 424 + (40)
2
= 2024 or d
3
= 45 mm
Ans.
A. T. Handle, B. Universal Joint
A.
B.
Cotter and Knuckle Joints
n
439
9. Thickness of spigot collar
Let t
1
= Thickness of spigot collar.
Considering the failure of spigot collar in shearing. We know that load (P),
30 × 10
3
= # d
2
× t
1
× ∀ = #/× 40 × t
1
× 35 = 4400 t
1
% t
1
= 30 × 10
3
/ 4400 = 6.8 say 8 mm
Ans.
10. The length of cotter ( l ) is taken as 4 d.
% l =4 d = 4 × 28 = 112 mm
Ans.
11.
The dimension e is taken as 1.2 d.
% e = 1.2 × 28 = 33.6 say 34 mm
Ans.
12.512.5
12.512.5
12.5
Sleeve and Cotter Joint Sleeve and Cotter Joint
Sleeve and Cotter Joint Sleeve and Cotter Joint
Sleeve and Cotter Joint
Sometimes, a sleeve and cotter joint as shown in Fig. 12.9, is used to connect two round rods or
bars. In this type of joint, a sleeve or muff is used over the two rods and then two cotters (one on each
rod end) are inserted in the holes provided for them in the sleeve and rods. The taper of cotter is
usually 1 in 24. It may be noted that the taper sides of the two cotters should face each other as shown
in Fig. 12.9. The clearance is so adjusted that when the cotters are driven in, the two rods come closer
to each other thus making the joint tight.
Fig. 12.9. Sleeve and cotter joint.
The various proportions for the sleeve and cotter joint in terms of the diameter of rod (d ) are as
follows :
Outside diameter of sleeve,
d
1
= 2.5 d
Diameter of enlarged end of rod,
d
2
= Inside diameter of sleeve = 1.25 d
Length of sleeve, L =8 d
Thickness of cotter, t = d
2
/4 or 0.31 d
Width of cotter, b = 1.25 d
Length of cotter, l =4 d
Distance of the rod end (a) from the beginning to the cotter hole (inside the sleeve end)
= Distance of the rod end (c) from its end to the cotter hole
= 1.25 d
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12.612.6
12.612.6
12.6
Design of Sleeve and Cotter Joint Design of Sleeve and Cotter Joint
Design of Sleeve and Cotter Joint Design of Sleeve and Cotter Joint
Design of Sleeve and Cotter Joint
The sleeve and cotter joint is shown in Fig. 12.9.
Let P = Load carried by the rods,
d = Diameter of the rods,
d
1
= Outside diameter of sleeve,
d
2
= Diameter of the enlarged end of rod,
t = Thickness of cotter,
l = Length of cotter,
b = Width of cotter,
a = Distance of the rod end from the beginning to the cotter hole
(inside the sleeve end),
c = Distance of the rod end from its end to the cotter hole,
!
t
, /∀ and !
c
= Permissible tensile, shear and crushing stresses respectively
for the material of the rods and cotter.
The dimensions for a sleeve and cotter joint may be obtained by considering the various modes
of failure as discussed below :
1. Failure of the rods in tension
The rods may fail in tension due to the tensile load P. We know that
Area resisting tearing =
2
4
d
#
∃
% Tearing strength of the rods
=
2
4
t
d
#
∃∃!
Equating this to load (P), we have
P =
2
4
t
d
#
∃∃!
From this equation, diameter of the rods (d) may be obtained.
2. Failure of the rod in tension across the weakest section (i.e. slot)
Since the weakest section is that section of the rod which has a slot in it for the cotter, therefore
area resisting tearing of the rod across the slot
=
2
22
()–
4
ddt
#
∃
and tearing strength of the rod across the slot
=
2
22
()–
4
t
ddt
#
&∋
∃!
()
∗+
Equating this to load (P), we have
P =
2
22
()–
4
t
ddt
#
&∋
∃!
()
∗+
From this equation, the diameter of enlarged end of the rod (d
2
) may be obtained.
Note: The thickness of cotter is usually taken as d
2
/ 4.
Cotter and Knuckle Joints
n
441
3. Failure of the rod or cotter in crushing
We know that the area that resists crushing of a rod or cotter
= d
2
× t
% Crushing strength = d
2
× t × !
c
Equating this to load (P), we have
P = d
2
× t × !
c
From this equation, the induced crushing stress may be checked.
4. Failure of sleeve in tension across the slot
We know that the resisting area of sleeve across the slot
=
22
12 12
()–() –( )
4
dd ddt
#
&∋
,
()
∗+
% Tearing strength of the sleeve across the slot
=
22
12 12
[( ) – ( ) ] – ( )
4
t
dd ddt
#
&∋
,!
()
∗+
Equating this to load (P), we have
P =
22
12 12
[( ) – ( ) ] – ( )
4
t
dd ddt
#
&∋
,!
()
∗+
From this equation, the outside diameter of sleeve (d
1
) may be obtained.
5. Failure of cotter in shear
Since the cotter is in double shear, therefore shearing area of the cotter
=2b × t
and shear strength of the cotter
=2b × t × ∀
Equating this to load (P), we have
P =2b × t × ∀
From this equation, width of cotter (b) may be determined.
6. Failure of rod end in shear
Since the rod end is in double shear, therefore area resisting shear of the rod end
=2 a × d
2
Offset handles.
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and shear strength of the rod end
=2 a × d
2
× ∀
Equating this to load (P), we have
P =2 a × d
2
× ∀
From this equation, distance (a) may be determined.
7. Failure of sleeve end in shear
Since the sleeve end is in double shear, therefore the area resisting shear of the sleeve end
=2 (d
1
– d
2
) c
and shear strength of the sleeve end
=2 (d
1
– d
2
) c × ∀
Equating this to load (P), we have
P =2 (d
1
– d
2
) c × ∀
From this equation, distance (c) may be determined.
Example 12.2. Design a sleeve and cotter joint to resist a tensile load of 60 kN. All parts of the
joint are made of the same material with the following allowable stresses :
!
t
= 60 MPa ; ∀ = 70 MPa ; and !
c
= 125 MPa.
Solution. Given : P = 60 kN = 60 × 10
3
N ; !
t
= 60 MPa = 60 N/mm
2
; ∀ = 70 MPa = 70 N/mm
2
;
!
c
= 125 MPa = 125 N/mm
2
1. Diameter of the rods
Let d = Diameter of the rods.
Considering the failure of the rods in tension. We know that load (P),
60 × 10
3
=
22
60
44
t
dd
##
∃∃!7∃∃
= 47.13 d
2
% d
2
= 60 × 10
3
/ 47.13 = 1273 or d = 35.7 say 36 mm
Ans.
2. Diameter of enlarged end of rod and thickness of cotter
Let d
2
= Diameter of enlarged end of rod, and
t = Thickness of cotter. It may be taken as d
2
/ 4.
Considering the failure of the rod in tension across the weakest section (i.e. slot). We know that
load (P),
60 × 10
3
=
22
2
22 22
()– ()– 60
444
t
d
ddt dd
##
&∋&∋
∃!7 ∃
()
()
∗+∗ +
= 32.13 (d
2
)
2
% (d
2
)
2
= 60 × 10
3
/ 32.13 = 1867 or d
2
= 43.2 say 44 mm
Ans.
and thickness of cotter,
t =
2
44
44
d
7
= 11 mm
Ans.
Let us now check the induced crushing stress in the rod or cotter. We know that load (P),
60 × 10
3
= d
2
× t × !
c
= 44 × 11 × !
c
= 484 !
c
%!
c
= 60 × 10
3
/ 484 = 124 N/mm
2
Since the induced crushing stress is less than the given value of 125 N/mm
2
, therefore the
dimensions d
2
and t are within safe limits.
3. Outside diameter of sleeve
Let d
1
= Outside diameter of sleeve.
Cotter and Knuckle Joints
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443
Considering the failure of sleeve in tension across the slot. We know that load (P)
60 × 10
3
=
22
12 12
[( ) ( ) ] ( )
4
t
dd ddt
#
&∋
,,,!
()
∗+
=
22
11
[( ) (44) ] ( 44) 11 60
4
dd
#
&∋
,,,
()
∗+
% 60 × 10
3
/ 60 = 0.7854 (d
1
)
2
– 1520.7 – 11 d
1
+ 484
or (d
1
)
2
– 14 d
1
– 2593 = 0
% d
1
=
2
14 (14) 4 2593
14 102.8
22
82∃
8
7
= 58.4 say 60 mm
Ans. (Taking +ve sign)
4. Width of cotter
Let b = Width of cotter.
Considering the failure of cotter in shear. Since the cotter is in double shear, therefore load (P),
60 × 10
3
=2 b × t × ∀ = 2 × b × 11 × 70 = 1540 b
% b = 60 × 10
3
/ 1540 = 38.96 say 40 mm
Ans.
5. Distance of the rod from the beginning to the cotter hole (inside the sleeve end)
Let a = Required distance.
Considering the failure of the rod end in shear. Since the rod end is in double shear, therefore
load (P),
60 × 10
3
=2 a × d
2
× ∀ = 2 a × 44 × 70 = 6160 a
% a = 60 × 10
3
/ 6160 = 9.74 say 10 mm
Ans.
6. Distance of the rod end from its end to the cotter hole
Let c = Required distance.
Considering the failure of the sleeve end in shear. Since the sleeve end is in double shear,
therefore load (P),
60 × 10
3
=2 (d
1
– d
2
) c × ∀ = 2 (60 – 44) c × 70 = 2240 c
% c = 60 × 10
3
/ 2240 = 26.78 say 28 mm
Ans.
12.712.7
12.712.7
12.7
Gib and Cotter Joint Gib and Cotter Joint
Gib and Cotter Joint Gib and Cotter Joint
Gib and Cotter Joint
Fig. 12.10. Gib and cotter joint for strap end of a connecting rod.
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A *gib and cotter joint is usually used in strap end (or big end) of a connecting rod as shown in
Fig. 12.10. In such cases, when the cotter alone (i.e. without gib) is driven, the friction between its
ends and the inside of the slots in the strap tends to cause the sides of the strap to spring open (or
spread) outwards as shown dotted in Fig. 12.11 (a). In order to prevent this, gibs as shown in
Fig. 12.11 (b) and (c), are used which hold together the ends of the strap. Moreover, gibs
provide a larger bearing surface for the cotter to slide on, due to the increased holding power. Thus,
the tendency of cotter to slacken back owing to friction is considerably decreased. The jib, also,
enables parallel holes to be used.
Fig. 12.11. Gib and cotter Joints.
Notes : 1. When one gib is used, the cotter with one side tapered is provided and the gib is always on the outside
as shown in Fig. 12.11 (b).
2. When two jibs are used, the cotter with both sides tapered is provided.
3. Sometimes to prevent loosening of cotter, a small set screw is used through the rod jamming against the
cotter.
12.812.8
12.812.8
12.8
Design of a Gib and Cotter Joint for Strap End of a Connecting Rod Design of a Gib and Cotter Joint for Strap End of a Connecting Rod
Design of a Gib and Cotter Joint for Strap End of a Connecting Rod Design of a Gib and Cotter Joint for Strap End of a Connecting Rod
Design of a Gib and Cotter Joint for Strap End of a Connecting Rod
Fig. 12.12. Gib and cotter joint for strap end of a connecting rod.
Consider a gib and cotter joint for strap end (or big end) of a connecting rod as shown in
Fig. 12.12. The connecting rod is subjected to tensile and compressive loads.
* A gib is a piece of mild steel having the same thickness and taper as the cotter.
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