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Cotter And Knuckle Joint (Part - 2) Mechanical Engineering Notes | EduRev

Mechanical Engineering : Cotter And Knuckle Joint (Part - 2) Mechanical Engineering Notes | EduRev

``` Page 1

Design of a cotter joint
If the allowable stresses in tension, compression and shear for the socket, rod
and cotter be
t
s ,
c
s and t respectively, assuming that they are all made of the
same material, we may write the following failure criteria:

1. Tension failure of rod at diameter d

2
t
dP
4
p
s=

4.2.2.1F- Tension failure of the rod (Ref.[6]).

2. Tension failure of rod across slot

2
11 t
ddt P
4
p??
-s=
??
??

4.2.2.2F- Tension failure of rod across slot (Ref.[6]).

t
Page 2

Design of a cotter joint
If the allowable stresses in tension, compression and shear for the socket, rod
and cotter be
t
s ,
c
s and t respectively, assuming that they are all made of the
same material, we may write the following failure criteria:

1. Tension failure of rod at diameter d

2
t
dP
4
p
s=

4.2.2.1F- Tension failure of the rod (Ref.[6]).

2. Tension failure of rod across slot

2
11 t
ddt P
4
p??
-s=
??
??

4.2.2.2F- Tension failure of rod across slot (Ref.[6]).

t
3. Tensile failure of socket across slot
22
21 2 1 t
(d d ) (d d )t P
4
p??
-- - s=
??
??

4.2.2.3F- Tensile failure of socket across slot.

4. Shear failure of cotter
2bt P t=

4.2.2.4F- Shear failure of cotter.

5. Shear failure of rod end
11
2d P t= l

4.2.2.5F- Shear failure of rod end

d
2
t
Page 3

Design of a cotter joint
If the allowable stresses in tension, compression and shear for the socket, rod
and cotter be
t
s ,
c
s and t respectively, assuming that they are all made of the
same material, we may write the following failure criteria:

1. Tension failure of rod at diameter d

2
t
dP
4
p
s=

4.2.2.1F- Tension failure of the rod (Ref.[6]).

2. Tension failure of rod across slot

2
11 t
ddt P
4
p??
-s=
??
??

4.2.2.2F- Tension failure of rod across slot (Ref.[6]).

t
3. Tensile failure of socket across slot
22
21 2 1 t
(d d ) (d d )t P
4
p??
-- - s=
??
??

4.2.2.3F- Tensile failure of socket across slot.

4. Shear failure of cotter
2bt P t=

4.2.2.4F- Shear failure of cotter.

5. Shear failure of rod end
11
2d P t= l

4.2.2.5F- Shear failure of rod end

d
2
t
6.  Shear failure of socket  end
()
31
2d d P -t= l

4.2.2.6F- Shear failure of socket end

7. Crushing failure of rod or cotter
1c
dt P s=

4.2.2.7F- Crushing failure of rod or cotter

8. Crushing failure of socket or rod
()
31 c
dd t P -s=

4.2.2.8F- Crushing failure of socket or rod
d
Page 4

Design of a cotter joint
If the allowable stresses in tension, compression and shear for the socket, rod
and cotter be
t
s ,
c
s and t respectively, assuming that they are all made of the
same material, we may write the following failure criteria:

1. Tension failure of rod at diameter d

2
t
dP
4
p
s=

4.2.2.1F- Tension failure of the rod (Ref.[6]).

2. Tension failure of rod across slot

2
11 t
ddt P
4
p??
-s=
??
??

4.2.2.2F- Tension failure of rod across slot (Ref.[6]).

t
3. Tensile failure of socket across slot
22
21 2 1 t
(d d ) (d d )t P
4
p??
-- - s=
??
??

4.2.2.3F- Tensile failure of socket across slot.

4. Shear failure of cotter
2bt P t=

4.2.2.4F- Shear failure of cotter.

5. Shear failure of rod end
11
2d P t= l

4.2.2.5F- Shear failure of rod end

d
2
t
6.  Shear failure of socket  end
()
31
2d d P -t= l

4.2.2.6F- Shear failure of socket end

7. Crushing failure of rod or cotter
1c
dt P s=

4.2.2.7F- Crushing failure of rod or cotter

8. Crushing failure of socket or rod
()
31 c
dd t P -s=

4.2.2.8F- Crushing failure of socket or rod
d

9. Crushing failure of collar
22
41 c
(d d ) P
4
p??
-s=
??
??

4.2.2.9F- Crushing failure of collar.

10. Shear failure of collar
11
dt P pt=

4.2.2.10F- Shear failure of collar.

Cotters may bend when driven into position. When this occurs, the bending
moment cannot be correctly estimated since the pressure distribution is not
known. However, if we assume a triangular pressure distribution over the rod, as
shown in figure-4.2.2.11 (a), we may approximate the loading as shown in figure-
4.2.2.11 (b)

Page 5

Design of a cotter joint
If the allowable stresses in tension, compression and shear for the socket, rod
and cotter be
t
s ,
c
s and t respectively, assuming that they are all made of the
same material, we may write the following failure criteria:

1. Tension failure of rod at diameter d

2
t
dP
4
p
s=

4.2.2.1F- Tension failure of the rod (Ref.[6]).

2. Tension failure of rod across slot

2
11 t
ddt P
4
p??
-s=
??
??

4.2.2.2F- Tension failure of rod across slot (Ref.[6]).

t
3. Tensile failure of socket across slot
22
21 2 1 t
(d d ) (d d )t P
4
p??
-- - s=
??
??

4.2.2.3F- Tensile failure of socket across slot.

4. Shear failure of cotter
2bt P t=

4.2.2.4F- Shear failure of cotter.

5. Shear failure of rod end
11
2d P t= l

4.2.2.5F- Shear failure of rod end

d
2
t
6.  Shear failure of socket  end
()
31
2d d P -t= l

4.2.2.6F- Shear failure of socket end

7. Crushing failure of rod or cotter
1c
dt P s=

4.2.2.7F- Crushing failure of rod or cotter

8. Crushing failure of socket or rod
()
31 c
dd t P -s=

4.2.2.8F- Crushing failure of socket or rod
d

9. Crushing failure of collar
22
41 c
(d d ) P
4
p??
-s=
??
??

4.2.2.9F- Crushing failure of collar.

10. Shear failure of collar
11
dt P pt=

4.2.2.10F- Shear failure of collar.

Cotters may bend when driven into position. When this occurs, the bending
moment cannot be correctly estimated since the pressure distribution is not
known. However, if we assume a triangular pressure distribution over the rod, as
shown in figure-4.2.2.11 (a), we may approximate the loading as shown in figure-
4.2.2.11 (b)

(a)        (b)
4.2.2.11F- Bending of the cotter

This gives maximum bending moment =
31 1
dd d P
26 4
-??
+
??
??
and

The bending stress,
31 31 11
b 3 2
dd dd dd Pb
3P
26 42 6 4
tb tb
12
-- ?? ? ?
++
?? ? ?
?? ? ?
s= =

Tightening of cotter introduces initial stresses which are again difficult to
estimate. Sometimes therefore it is necessary to use empirical proportions to
design the joint. Some typical proportions are given below:
1
d 1.21.d =
2
d 1.75.d =
d
3
= 2.4 d
4
d1.5.d =
t0.31d =
b 1.6d =
1
l l 0.75d ==
1
t0.45d =
s= clearance
d
3
b
d
1
P/2
P/2
P/2
P/2
-
+
31 1
dd d
64
31 1
dd d
64
-
+
1
d
4
1
4
d
```
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