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37
Design of Temporary
Connections UNIT 2 DESIGN OF TEMPORARY
CONNECTIONS
Structure
2.1 Introduction
Objectives
2.2 Stresses
2.3 Calculation of Diameter of a Bar
2.4 Knuckle Joint
2.5 Cotter Joint
2.6 Summary
2.7 Key Words
2.8 Answers to SAQs
2.1 INTRODUCTION
We should understand that designing will be treated as finding dimensions or single
dimensions of a part. The input is normally available in terms of load or force to be
carried and material in which part is to be made. From the knowledge of material, mainly
the ultimate tensile strength and yield strength become known. It is understood that you
are familiar with stress and its types – tensile, compressive (or bearing) and shear. The
design will be based upon the relationship between load and stress and the stress value
which is allowed to occur in a part is only fraction of yield strength or ultimate tensile
strength the joining of two parts which carry force and may be either stationary or
permanent is a common engineering practice. Bars may be joined, plates may be
connected or a pulley may be connected to shafts. Pins passing through plates may
connect them and the plates can be pulled apart by equal and opposite forces applied on
two plates. It is example of a temporary joint. Other joints will be described and sizes
calculated.
Objectives
After studying this unit, you should be able to
correlate stress and load,
learn how rods can be connected, and
know types of joints, and causes of their failures.
2.2 STRESSES
The simple definition of stress is that is force divided by area. If the force is
perpendicular to the area and pulling away from it, the stress is tensile. If the force is
perpendicular to area and pushing towards it, the stress is compressive. Both tensile and
compressive stresses come under general category of direct stress. If the force is parallel
to area to cause sliding of one area over other the stress is shearing. If two bodies are in
contact and pressed against each other the stress is bearing. The magnitude of bearing
stress will be the compressing force divided by contact area between two bodies. The
bearing stress is compressive in nature and is also called crushing stress.
38
Machine Design
Three stresses described above are illustrated in Figure 2.1.
(a) (b)
(c)
Figure 2.1 : Three Types of Stresses
(a) Force P causing tensile stress on cross-section of a rectangular bar.
(b) Force P causing shearing stress on a rectangular section.
(c) Force P acting on a circular bar, presses it against a rectangular plate,
causing compressive or bearing stress on circular section of the bar.
The shearing stress and crushing stress are very common in circular cross-section pin
when it passes through two plates and plates are pulled apart. Figure 2.2(a) shows such
plates connected by a pin. The length of the pin equal to thickness t, of plate is subjected
to crushing force P on two cylindrical surfaces as shown in Figure 2.2(b). This surface is
pressed by cylindrical surface of hole in the plate as shown in Figure 2.2(c). Both the
surfaces of the hole in the plates and of the cylinder are subjected to the crushing stress.
The magnitude of this crushing stress is calculated as
Projected area of cylinder c
P . . . (2.1)
(a) (b) (c)
Figure 2.2
The projected area of cylinder in contact over a length t
= dt
where d = Diameter of cylinder (pin), and
t = Thickness of plate.
A
P
P
A
A
P= A
AS
P
P= A
P=c Ac
AC
Plate 1
Plate 2
P
t
P
B
B
P
P
P
AS
l
P
2
P
2
39
Design of Temporary
Connections Also note that the two forces P, acting opposite to each other on pin as shown in
Figure 2.2(b) are separated by the pin cross-section of area As. This is the area in the
plane of contact of two plates in the Figure 2.2(a). The area As is subjected to shearing
stress .
s
P
A . . . (2.2)
and 2
4
sA d
The plates has been cut along the line BB and shown in Figure 2.2(c). The area upon
which force P acts normally is seen as two rectangles. This is the cross-section of plate
subjected to tensile stress, .
P
A . . . (2.3)
( ) A l d t
l = Width of the plate.
2.3 CALCULATION OF DIAMETER OF A BAR
If a bar is subjected to a tensile force P along its axis and the maximum stress which can
be induced on the cross-section of the bar is limited to t, we can find the diameter of the
bar from definition of stress. Let A be the area of cross-section of bar, so that
2
4
sA d
If d is the diameter of the bar. The stress in the bar
2
4
t
P P
A d
2 4 t
Pd
For example, if a fan weighing 100 N is suspended at the end of a steel bar in which
stress is not allowed to exceed 10 N/mm2, find the diameter of the bar.
The diameter is calculated from
2 4 t
Pd
in which P = 100 N, t = 10 N/mm2
2 4 10012.73
10
d
or d = 3.6 mm
This finding of diameter is the design of the bar.
Example 2.1
Two plates 100 mm wide and 1 mm thick are connected with a pin passing
through them as shown in Figure 2.2(a). The plates are pulled apart by a force of
10000 N. If the tensile stress in plate is t, shearing stress in pin is s and crushing
stress in pin is c, the limiting values are given below.
t = 20 N/mm2, s = 10 N/mm
2 and c = 22 N/mm
2
Find thickness of plate and diameter of pin.
40
Machine Design
Solution
10,000
or 20( ) (100 )
t
P
l d t d t . . . (2.4)
2
2
4 10,000or 10
4
s
P
dd
. . . (2.5)
10,000
or 22 c
P
dt dt . . . (2.6)
Since in (2.4) and (2.6) there are two unknown, use (2.5) to find
2 4000d
d = 36.25 mm . . . (2.7)
Use this value of d in (2.4) so that
10,000
20(100 36.25) t
10,000
7.8 mm64 20
t
. . . (2.8)
So both d and t have been found. But we should check that c is not more than
22 N/mm2. From (2.5)
210,00035.6 N/mm
36 7.8c
Since this stress is greater than permissible stress of 22 N/mm2. The value of d and
t are not acceptable.
From Eq. (2.6), dt = 554 which can be obtained with by increasing d or t or both.
t = 13 mm and d = 43 mm gives dt = 559 mm2 which is safe.
Check 210,00013.5 N/mm (Safe)
(100 43) 13t
Check 2
2
10,0006.9 N/mm (Safe)
(43)4
s
Thus, d = 43 mm and t = 13 mm are finally selected dimension. The example
shows how design problems reed reiteration.
Example 2.2
Two steel rods are proposed to be connected by a pin in the same way as plates.
The force of tension to be carried by the joint is 8000 N. The joint is made by
removing metal from rod ends to create flat surfaces of depth of half the diameter
of the rod as shown in Figure 2.3. A hole is drilled in the centre of the flat in each
rod, the two parts are matched and a pin is passed through the hole. Find diameters
of rod and pin. The permissible stresses in tension, shear and crushing are :
t = 12 N/mm2, s = 7 N/mm
2 and c = 14 N/mm
2
41
Design of Temporary
Connections
Figure 2.3 : A Pin Joint between Two Bars
Solution
Recognise following ways in which joint can fail.
(a) Shearing of pin
(b) Crushing of pin against hole surface
(c) Tensile failure of bar along section through pin.
Shearing of Pin
2
4
sP d . . . (2.9)
28000 74
d
38.2 mmd . . . (2.10)
Crushing of Pin
The pin comes in contact with the hole surface over half the diameter of the
bar. It can be seen by taking section through centre line of hole for pin.
(See Figure 2.4)
Figure 2.4 : Section of Bars 1 and 2 Through Pin Hole
1
2
1
2
D
d
1
2
42
Machine Design
If bar 1 is pulled out of plane of paper, the bar 2 will be pulled in the
opposite direction. Only half circular section will carry the tensile force.
Similarly only half pin will compress against the half circle. Hence, area
resisting crushing is
1
2cA D d
D is he diameter of bar.
The area resisting tensile force
21
2 4
A D Dd
(Half hatched area).
The crushing strength of pin
1
2 cP Dd
22 2 80001142.9 mm
14
c
PDd
and tensile strength of rod
21
2 4
tP D Dd
or by using value of D d
2 2 80001142.9 1333.33
4 12
D
i.e. 2 4(1333.33 1142.9) 3153
D
D = 56.15 mm . . . (2.11)
It will be good idea to check what is the tensile stress in the rod when joint
has been made. The area of cross-section of rod is
2 2 2(56.15) 2476.4 mm4 4
D
Hence, when joint carries a force of 8000 N, the stress in the rod
2 280003.23 N/mm
2476.4
4
PD
Note that permissible stress in the rod is 12 N/mm2, hence it is very
uneconomical as it carries almost one fourth of stress of its full capacity.
Thus, the design of Example 2.2 is uneconomical though it is easy to make.
2.4 KNUCKLE JOINT
Two joints – one between plates and other between rods can be easily understood as
temporary joints as they can be dismantled by removing pin. Knuckle joint which is
practically used to join two bars being pulled apart is similarly a temporary joint. This
joint also consists of three parts – two rods and pin. But the ends of rods are made to
have specific shapes obtained by forging. These ends merge into circular rod. The Figure
2.5 shows a knuckle joint in which 1 is fork, 2 is eye and 3 is pin. Figure 2.6 shows three
parts separated.
43
Design of Temporary
Connections
Figure 2.5 : A Knuckle Joint
Figure 2.6 : Parts of the Knuckle Joint
To hold the pin in assembly it is made with a round head at one end and a collar is
placed at the other end. The pin is tightened on the end with the help of a taper pin.
Figure 2.7 : Assembly of a Knuckle Joint
In designing a knuckle joint we have to determine following dimensions :
Diameter of Pin
Note that pin is under shear. Also note that pin passes through two surfaces of
contact between eye and fork. Thus, at two cross-sections the pin is subjected to
Rod
1. Fork
2. Eye
Rod
3. Pin
4
5
2
3
1
Taper Pin
Sq. Sec.
Fork End
PIN
Collar
Eye End
1.2 d
4 d 4.5 d
0.6 d
1.5 d
2 d
1.1
d
3 d
0.7
5d
1.1
d
d
0.7
5d
d
P P d
Octagon
Split Pin
0.4
d
44
Machine Design
shearing stress. Ideally the pin subjected to shearing stress at two cross-sections
must have twice the permissible stress is single shear but practically permissible
stress in double shear is 1.75 of that in single shear. Eq. (2.9) may be used for
calculating, d.
Diameter of Eye
The hole in the eye has the same diameter as the pin. This is d. the outside
diameter of the eye is D and its thickness is t as shown in Figure 2.8. The force P
pulls the eye to the right and force equal P is exerted by pin on the inside surface
of the hole. Thus, the section BB is subjected to tensile stress. The area of
section BB (hatched area) is (D – d) t. D is the outside diameter of the eye. If
tensile stress produced in the section is t then
( ) tP D d t . . . (2.12)
In this equation D and t are unknown.
Figure 2.8 : The Eye of Knuckle Joint
But remember the pin is compressed against inside surface of hole and contact
length is t. Hence, the projected area for crushing is dt. If c is crushing stress,
then tP dt . . . (2.13)
From this equation t is determined. t is then placed in Eq. (2.12) and D is
determined.
Thickness of Fork
The outside diameter of fork is same as that of eye. The pin hole diameter is also
same. The fork appears same as eye in the elevation as can be seen in Figure 2.9.
But in side view the prongs of fork will appear. Each will have diameter D and
thickness t1. The force P will pull the fork to the left which will be opposed by
crushing force P developed between pin and inside hole surfaces of prongs of
fork. Thus, section BB of the fork will be under tensile stress. Section BB is shown
in Figure 2.10 as side view. The area of tensile stress in
1( ) 2 A D d t
1( ) 2 tP D d t . . . (2.14)
In this equation D and d are known, hence t1 can be calculated. Comparing
Eqs. (2.12) and (2.14) it can be concluded that
12
t
t
The dimensions of head of pin, collar and taper pin for collar are conveniently
chosen and calculated.
t
d D P
B
P
B
Section BB
45
Design of Temporary
Connections
Figure 2.9 : The Fork of Knuckle Joint
Bending of Pin
Pin in the fork and eye tends to act like a beam. Its has three regions of
loading – two in the fork and one in the eye. It is normally fitted tight in the
eye and slightly loose in the fork. This causes uniform pressure
P
t in the
eye but the pressure in the fork varies from zero to P
t over the length
12
P
t
per unit length. The force on pin is shown in Figure 2.10. For convenience
the pin is shown as loaded beam at (b) in the same figure. The force on
beam is due to eye and fork will provide reaction to the force on beam in
the eye.
(a)
(b)
Figure 2.10 : Pin Loaded as a Beam in the Fork and the Eye
The maximum bending moment will occur in the middle of the span which
is equal to 12
3
tt . The BM at the middle will be due to
2
P at a distance
of 1
2 3
tt and due to udl over a length of t.
2
1max
2 2 3 8
tP t P tM
t
For circular section beam, bending stress
13 3
32 132
8 6
tM tP
d d . . . (2.15)
This stress should be less than t for design to be safe. The equation for is
used as a check.
t t1
P
B t1
D P
B
Fork
t1 t1 t
Fork Eye
P/ t per unit length
P/ 2 t1 per unit length
t1 / 3 t1 / 3
P / 2 P / 2
t
P/ t per unit length
46
Machine Design
Example 2.3
Design a knuckle joint for a tie rod of a circular section to sustain a maximum pull
of 70 kN. The ultimate tensile strength of the rod material is 420 N/mm2. The
ultimate tensile and shearing strength of pin material are respectively 720 N/mm2
and 390 N/mm2. A factor of safety of 5 is to be used. The permissible stresses in
tension and compression are equal.
Solution
First find permissible stress Ultimate strength
=Factor of safety
Permissible stress in tension for rod, eye and fork
242084 N/mm
5 t
Permissible shearing stress in pin
239078 N/mm
5 s
Permissible compressive stress in pin
2720144 N/mm
5 c
Permissible compressive stress in eye and fork
284 N/mm c t
P = 70,000 N
Pin Diameter, d
The area of shearing 2
4
d
The pin is in double shear
21.754
sP d
or 270000 1.75 784
d
or 2 70000652.9
107.2 d
d = 25.6 mm . . . (2.16)
Thickness of the Eye, t
t is determined by considering crushing. Note that while pin is under
compression from cylindrical surface of hole in the eye, opposite of it is
also true, i.e. the cylindrical surface is under compression against pin. The
surface which has lesser compressive strength is likely to fail. We have
found that permissible compressive stress for pin is 144 N/mm2, the same
for eye is 84 N/mm2. Hence, crushing of eye is to be considered.
cP dt
or 70000 84 25.6 t
t = 32.6 mm . . . (2.17)
Diameter of Eye, D
See Figure 2.8, the section carries tensile stress.
( ) tP D d t
47
Design of Temporary
Connections or 70000 84 ( 25.6) 32.6 D
or 70000
25.684 32.6
D
D = 51.2 mm . . . (2.18)
Diameter of the fork will be 51.2 mm
Thickness of fork will be 32.6
16.3 mm2
. . . (2.19)
Diameter of the Rod, D1
21
4
tP D
2170000 84
4
D
D1 = 32.6 mm . . . (2.20)
Check for bending stress in the pin.
Eq. (2.15),
13
132
8 6
ttP
d
3
32.6 32.6 132 70000
8 12 (25.6)
713014
(4.075 2.72)16777.2
= 289.8 N/mm2
It can be seen that this stress is higher than permissible stress 284 N/mm t . Keeping 284 N/mm in Eq. (2.15), we may
calculate d.
i.e. 3
32.6 32.6 184 32 70000
8 12
d
3 1713014 (4.075 2.72) 57023.7
84 d
d = 38.5 mm . . . (2.21)
This diameter will be safe against shearing of pin. But the outer diameter of
eye and fork found at Eq. (2.19) from Eq. (2.15) will be affected. Hence, we
calculate that diameter again. Using d = 38.5 mm in Eq. (2.15).
70000 ( 38.5) 84 32.6 D
70000
38.584 32.6
D
or D = 64.1 mm . . . (2.22)
Thus, the dimensions of the joints are :
48
Machine Design
Diameter of pin, d = 38.5 mm. Outside diameter of eye and fork = 64.1 mm.
Thickness of eye, t = 32.6 mm. Thickness of form t1 = 16.3 mm. Diameter
of rod, D1 = 32.6 mm.
SAQ 1
(a) What is a temporary joint?
(b) If two rods are joined through a pin, show the section, that carries tensile
stress.
(c) Show the area of eye of a knuckle joint which is subjected to tensile stress.
(d) How would you calculate the width of fork in knuckle joint?
(e) Two mild steel rods are connected in a knuckle joint to carry a tensile load
of 150 kN Design the joint. Use permissible stresses in tension,
t = 77.5 N/mm2, shear, s = 38 N/mm
2, compression, c = 150 N/mm
2.
2.5 COTTER JOINT
You can imagine cotter to be a flat pin as shown in Figure 2.11(a). Imagine that in
Figure 2.11 it is a flat cotter that passes through a corresponding rectangular hole and
plates are being pulled. If it is difficult for you to imagine then look at Figure 2.11(b).
The force P applied on plates tends to pull two halves of cotter in opposite direction,
causing it to shear along plane of contact of two plates. The upper plate has been
removed in Figure 2.11(c) but the force P with which upper plate pushes the cotter is
shown. The cotter may shear along area in the contact surface as shown at. If the area of
shear is As, then As = b t where b is the width of cotter and t is its thickness. The
difference between the joint in Figure 2.11 and the pin joint in Figure 2.11 can be seen
that plates in Figure 2.11 can turn about pin but in Figure 2.11 plates can not turn. This
turning does not permit pin joint to carry compression but a cotter joint can carry
compression. A cotter can, likewise, replace the pin in Figure 2.12 for joining two rods.
But note that making a rectangular hole is more difficult and costlier than making a
circular hole.
(a) Cotter (b) Two Plates Connected by a Cotter
(c) Top Plate Removed and Force Exerted by it is P
Figure 2.11
P P
b t
P
P
Area of Shear of Cotter
49
Design of Temporary
Connections
Figure 2.12 : Three Parts of Cotter Joint
We do not make any calculations for joint of Figure 2.12 with a cotter in place of pin. If
you would do, you will find that the diameter of rod turns out large and joint becomes
uneconomical. A better proposition is to make a socket in one end of the rod and insert
the rod in the socket. Both the socket and inserted end of rod will have the slot in one
line through which cotter passes. The cotter is made with a slight taper so that it does not
just pass through the hole but is held in the hole.
A cotter joint for connecting two rods along which tension or compression act is made
with ends of the rod especially made. One rod end carries a spigot (the end to be
inserted) while the other rod end is finished in form of a socket. These ends are shown in
Figure 2.12. The ends are normally made by forging and rectangular hole is also created.
The rectangular hole (slot), the internal surface of socket and external surface of spigot
are finished by machining. The three parts that make a cotter joint are shown in
Figure 2.12. It is not difficult to see that to make the joint spigot is inserted into the
socket. The slots are coincided and cotter passed through the slots. You may note
features of spigot and socket. Collar of large diameter is provided at the mouth of the
socket. The internal diameter of socket (D) is same as external diameter of spigot. The
slot in the socket often passes through the collar and it is the straight edge of the cotter
that makes contact with the collar of the socket. If direction of force is reversed, the
inclined edge of cotter will not be able to make contact with the slot surface. Hence, a
collar is provided on the spigot which contacts the open surface of the mouth of the
socket. Thus, two collars bear against each other and crushing stress between them will
decide the diameter of collar on spigot (D1). The mean width of taper cotter is b which is
not much smaller than larger width as taper is 1 in 48. Parts shown in Figure 2.12 are
assembled in Figure 2.13.
The cotter joint was earlier used in steam engines to connect piston rod with cross-head.
It is still used as connector in a pump rod.
Figure 2.13 : Sectional Elevation of the Cotter Joint
Failure of Cotter Joint
Like any other machine part designer must become aware of various ways in
which this joint can fail. Remember that identifying a mode of failure will require
the critical area on which failure is likely to occur and the stress which is acting on
Collar (D1)
Collar (D2)
Rod (d)
Rod
Slot Cotter
Slot
Spigot (d1) Socket
a c e f
1 1
2 2
P d D1
P d d1 D D2
b
50
Machine Design
this area. The weakest area is the smallest area. The area of a section becomes
small if a hole or a slot passes through it. If no hole or slot passes, then whole
sectional area may be considered. For example, in spigot and socket the weakest
area is through the slot. For rod the area of section is the only area to be
considered. The weakness may arise where area of section changes. So the collars
on spigot and socket may be weak. The cotter may shear along the surfaces of
contact, hence, its area along contact surfaces may be weak. The stresses that may
cause failure may be tensile, crushing or shear. We will have to be careful to
examine if force P is pulling on one area to cause tensile stress or pushing on an
area to cause crushing or is parallel to the area to cause shearing. To make matters
easy we identify such modes of failure for cotter joint. Each will be used to
determine some dimension.
(a) The rod may fail in tension or compression. The area of section is a
circle of diameter, d and shown in Figure 2.14(a). The equation for
force is also written by the side.
2
4
tP d . . . (2.23)
(a)
Figure 2.14
(b) The spigot may fail due to tension or compression. Note, though the
spigot is a cylinder, it has a slot of width t. t is the thickness of the
cotter. So the weakest area of spigot will be through slot.
Figure 2.14(b) shows this area. Take help of Figures 2.12 and 2.13 to
understand how you draw Figure 2.14(b).
21 1
4
tP d d t . . . (2.24)
(b)
Figure 2.14
t
d1
51
Design of Temporary
Connections (c) Failure of socket under tension or compression. Note from
Figures 2.12 and 2.13 that socket is a hollow cylinder. Its inner
diameter is same as outer diameter, d1 of spigot. The weakest section
is where slot is made.
2 21 1( ) ( )
4
tP D d D d t . . . (2.25)
(c)
Figure 2.14
(d) The cotter will fail under shear at two sections which are along
surfaces of contact between the outer surface of spigot and inner
surface of socket. For your understanding these areas have been
marked as 1-1 and 2-2 in Figure 2.13.
(1.75 ) sP bt . . . (2.26)
1.75 s is the permissible shearing stress in double shear if s is same
in single shear.
(e) Failure of spigot under crushing against cotter or crushing of cotter
against spigot. The area over which crushing occurs is the slot area
shown at (b) of Figure 2.14 and the force on this area is P as shown at
Figure 2.14(d).
1 cP d t . . . (2.27)
(d)
Figure 2.14
(f) Failure of collar under compression. The collar on socket is larger
than the spigot collar and they come under crushing during
compression loading of the joint. The annular area on the spigot
D d1
t
52
Machine Design
collar that is crushed is 2 21 1( )
4
D d as can be judged from
Figure 2.13, and shown in Figure 2.14(e).
2 21 1( )
4
cP D d . . . (2.28)
This equation may be used to determine D1. For determining length of
the collar, a, we look at another possible way of failure. It is shearing
of collar on cylinder of diameter d1. The surface of this cylinder
(broken line) is shaded in Figure 2.14(e). The cylinder has diameter
d1 (the diameter of spigot) and length a. The area on which collar can
shear is 1 d a .
1 sP d a . . . (2.29)
(e)
Figure 2.14
(g) Shearing of socket collar against other. The cotter pushes the collar
over the area (D – d1) t as shown in Figure 2.14(c) and also in
Figure 2.14(f). The cotter may cut through the thickness, c of collar if
collar fails in shear. The cotter will face shearing resistance on its two
sides. One side is shown facing the reader in Figure 2.14(f). The other
side is in the back. One of two shaded areas is 2 1( )
2
D dc . There are
four such areas. Hence, shearing of socket collar against cotter will be
resisted along the area 2 (D2 – d1) c.
2 12( ) sP D d c . . . (2.30)
This equation will help determine c.
(f)
Figure 2.14
Similar situation exists at the tail of the spigot where the cotter may
shear tail over a length, e.
12 sP d e . . . (2.31)
P
Rod
Dia D1
Dia d1
Spigot Collar of length a. Force P uniformly
distributed annular area
P/2
P/2
P/2
C
Spigot dia. d1
Collar dia., D2
53
Design of Temporary
Connections In both Eqs. (2.30) and (2.31) 2 can be replaced by 1.75 as it is the
case of double shear.
(h) The thickness of the end of socket where rod begins is shown as f in
Figure 2.14.
There is a likelihood that if force is compressive the rod may pierce
into the end of the socket. The resistance will be offered by shearing
stress acting on cylindrical surface of diameter d and length f.
sP d f . . . (2.32)
The use of understanding and Eqs. (2.23) and (2.32) will become
clear through a solved example. Everytime you solve a design
problem you must draw figures to show the area on which stress
(tensile, compressive or shear) is acting. P can be calculated.
It may be advisable to check cotter in bending in the same way as pin in knuckle joint. It
will be shown in solved example.
Example 2.4
Write equations for tensile and crushing failure of the spigot of a cotter joint.
Equating strengths of spigot in tension and crushing against cotter show that
1
6
t
Pd
where d1 is the diameter of spigot, t is the permissible tensile stress in spigot,
P is the force acting on joint. The permissible compressive stress in spigot, c is
twice t.
If P = 40 kN, ultimate tensile strength = 650 MPa and f.s = 5 find diameter of the
rod and thickness of the cotter.
Solution
See Figure 2.14(b) and use Eq. (2.24) to write the strength of spigot in tension,
21 1
4
tP d d t
Also the strength of spigot against crushing from Eq. (2.27)
1 cP d t
You may note that left hand side of the above two equations are same, but
remember P is a symbol. If we really make both strengths equal, then we create a
co-relationship between any two dimensions. Such design is known as economical
design.
If we wish to solve first equation we cannot because it contains two unknowns
– d1 and t. so we use the second equations and find
12
c t
P Pd t
Since 2 c t
Then substitute for d1 t in first equation to obtain
21
4 2
t
t
PP d
or 21
3
4 2 2
t
P Pd P
54
Machine Design
1
6
t
Pd . . . (2.33)
Note that 2650130 MPa or N/mm
5 t
1
6 4000024.24 mm
130
d . . . (2.33)
Thickness of the cotter is found from 2nd
equation by putting d1 = 24.24 mm.
40000 24.24 2 130 t
40000
6.3 mm24.24 260
t . . . (2.34)
Example 2.5
Two rods are to be joined in a cotter joint to carry 90 kN of axial force which may
change from tension to compression and vice-versa. The ultimate strengths in
tension, compression and shear respectively are 255 N/mm2, 510 N/mm
2 and
130 N/mm2. Choose a factor of safety of 5.
Solution
The permissible stresses in tension, shear and compression are
225551 N/mm
5 t
213026 N/mm
5 s
2510102 N/mm
5 c
Diameter of Rod
Use Eq. (2.23) with P = 90000 N, t = 51 N/mm2
2 4 4 900002246.89
51
t
Pd
d = 47.4 mm say 47.5 mm . . . (2.35)
Spigot Diameter d1 and Thickness of Cotter t and Collar Dimensions
In Eq. (2.27) put c = 102 N/mm2
i.e. 1
90000882.35
102 c
Pd t . . . (2.36)
Use this value of d1 t in Eq. (2.24) to obtain
2190000 882.35 51
4
d
21
90000882.35 2647
4 51
d
1
4 264758.1 mm
d . . . (2.37)
Use the value of d1 in Eq. (2.36) to obtain
882.35
15.2 mm58.1
t . . . (2.38)
55
Design of Temporary
Connections Diameter of Collar of Spigot, D1
Use Eq, (2.28) with c = 102 N/mm2
2 21 190000 ( ) 102
4
D d
21
4 90000(58.1) 44.99
102D
or D1 = 67.1 mm . . . (2.39)
Length of Collar, a
a is determined from Eq. (2.29). Also see Figure (2.14(e)).
Use s = 26 N/mm2 and d1 = 58.1 mm.
1
90000
58.1 26 s
Pa
d
or a = 19 mm . . . (2.40)
Use Eq. (2.32) to obtain the length of tail of spigot
1
90000
2 2 58.1 26
s
Pe
d
or l = 29.8 mm . . . (2.41)
Thus, all dimensions, i.e. d1, D, t and a of spigot are determined.
Socket Rod and Socket Dimension
There is no need to calculate diameter of rod of socket. It is same as
diameter of spigot rod, d = 47.5 mm
The outside diameter of socket D is calculated from Eq. (2.25). Also see
Figure 2.14(c).
2 21 1( ) ( )
4
tP D d D d t
Use d1 = 58.1 mm, t = 15.2 mm.
2 290000 ( 58.1 ) ( 58.1) 15.2 514
D D
or 24 900003375.6 19.35 1124.42
51D D
or 2 19.85 4498 0 D D
1
9.925 394 17992.3 9.925 67.82
D
D = 77.72 mm . . . (2.42)
The cotter compresses against the collar of the socket. The area over which
cotter bears (compresses) on cotter is made of two rectangles similar to that
shown in Figure 2.14(c) with outer diameter being D2, the diameter of
collar.
2 1( ) cP D d t
2
9000058.1
102 15.2
D
or D2 = 116.2 mm . . . (2.43)
56
Machine Design
The length of the collar, c, is found by considering penetration of cotter into
collar of socket as shown in Figure 2.14(f) and expressed in Eq. (2.30)
2 12 ( ) sP D d c
or 90000
(116.2 58.1)2 26
c
c = 29.8 mm . . . (2.44)
The thickness f of the bottom of hollow of socket is found from Eq. (2.32)
s
Pf
d
90000
4.75 26
= 2.32 mm
Thus, all dimensions of socket, i.e. d1, D1, D2, c and f are determined.
Cotter Width, b
The thickness of cotter has been determined. the cotter fails in shear along
two sections, as shown in Figure 2.14(b) and the strength expressed in
Eq. (2.28).
90000 1.75 15.2 26 b
90000
130 mm1.75 15.2 26
b . . . (2.45)
The length of cotter may 1 mm more than the diameter of spigot collar on
either side. Hence, length of cotter = 136 mm. Hence, width of cotter on
top = 130 + 1.3 = 131.1 mm and at bottom 126 – 1.3 = 124.7 mm.
The bottom of the spigot should not touch the bottom of the hollow of the
socket. A clearance of 10 mm is desired. Thus, the length of the socket from
rod is c + e + f + b.
= 29.8 + 29.8 + 23.2 + 130 + 10
= 222.8 mm . . . (2.46)
The spigot part of the joint has length
= a + c + b + e
= 19 + 29.8 + 130 + 29.8
= 208.6 mm . . . (2.47)
It is difficult to imagine if cotter will bend like a beam, although it has
already been assumed to act as beam (Figure 2.14(d)). The force as shown
to act uniformly distributed between 1 and 2 are not exactly correct because
cotter is taper on right hand side. However, we take the distribution as
shown in Figure 2.14(d) and calculate bending stress with assumptions
similar to those made for pin in case of knuckle joint. So draw the beam in
Figure 2.15.
The uniformly distributed force is acting over a length of d1 (the diameter of
spigot). The cotter is supported against inside of collar of diameter D2.
Hence, force 2
P is distributed over length of 1 1
2
D d on two sides of the
57
Design of Temporary
Connections cotter. This distribution is assumed as triangular. Therefore, it will act at a
distance of 1 11
3 2
D d from the point where d1 begins. Hence, maximum
bending moment occurs at the central section of the cotter, where width = b
and thickness = t.
2
1 1 1 1max
12 6 2 8
D d d dP PM
d
Figure 2.15 : Cotter Idealised as Beam
Use D2 = 116.2 mm, d1 = 58.1 mm, P = 90000 N
max
90000 116.2 58.1 58.1 9000058.1
2 6 2 58.1 8
M
= 1743000 – 653625
= 1089375 N-mm
Bending stress,
3
3 61 (130) 15.2. , 2.783 10
2 12 12
M bI b t
I
6
2
6
1.1 10 13025.72 N/mm
22.78 10
This stress is less than permissible tensile stress, 51 N/mm2 hence, cotter is
safe in bending.
Dimensions Calculated (Figure 2.13)
Spigot : Diameter of rod, d = 47.4 mm
Diameter of spigot collar, D1 = 67.1 mm
Width of collar, a = 19 mm
Diameter of spigot, d1 = 58.1 mm
Length of collar, 208.6 mm
Socket : Diameter of socket, D2 = 116.2 mm
Width of collar, c = 29.8 mm
Outside diameter of socket, D = 77.72 mm
Length of socket = 22.8 mm
Inside diameter of socket, d1 = 58.1 mm
Cotter : Width = 130 mm
P/2
P/ d, per unit length
d1
D2
D2 − d1
2
D2 − d1
2
P/2
58
Machine Design
Thickness = 15.2 mm
Length = 136 mm
SAQ 2
(a) Describe three parts of cotter joint and sketch them separately.
(b) What materials will be used in making cotter joint?
(c) To what stresses spigot is subjected? Draw the areas of sections of spigot on
which tensile and compressive stresses act.
(d) If the permissible compressive stress is 1.75 of the permissible tensile
stress, then by equating tensile strength and crushing strength show
diameter d1 of spigot is given by
1
2
t
Pd
(e) A cotter joint is required to carry 30 kN axial force. Following dimension
have been found.
Diameter of spigot, d1 = 34 mm
Diameter of socket collar, D2 = 75 mm
Width of cotter at mid section = 43 mm
Thickness of cotter = 10 mm
Calculate the bending stress in cotter (maximum value).
2.6 SUMMARY
In this unit, you have learnt the calculations of diameter of a bar. Knuckle joint and
cotter joint have been described in this unit. Knuckle joint is practically used to join two
bars being pulled apart. This joint consists tow rods and pin. A cotter joint is used to
connect rigidly two co-axial rods or bars which are subjected to axial tensile or
compressive forces. It is a temporary fastening.
2.7 ANSWERS TO SAQs
SAQ 1
(e) t = 77.5 N/mm2, c = 150 N/mm
2, s = 38 N/mm
2, P = 150 10
3 N
Pin Diameter, d
3 2150 10 384
d
42 44 15 10
0.5 1038
d
or d = 71 mm . . . (2.48)
59
Design of Temporary
Connections Thickness of Eye, t
3150 10 150 71 c d t t
or 3150 10
14.1 mm150 71
t . . . (2.49)
Diameter of Eye, D
3150 10 ( ) 77.5 14.1 ( 71) t D d t D
or 3150 10
71 208.3 mm77.5 14.1
D . . . (2.50)
Diameter of Fork, D
Same . . . (2.51)
Thickness of the Fork
t1 = 7.1 mm . . . (2.52)
Check for Bending Stress
13
132
8 6
b
ttP
d
332 150 10 14.1 7.1
357911 8 6
4.3 (1.85 1.18)
213 N/mm b . . . (2.53)
This is less than t. Hence, design is safe.
SAQ 2
(d) Figure 2.14(b) and use Eqs. (2.26) and (2.29)
21 1
4
tP d d t . . . (2.54)
and 1 1 1.75 c tP d t d t . . . (2.55)
11.75
t
Pd t
Use this value of d1 t in Eq. (2.54)
21
4 1.75
t
PP d
21
1.57 4
t
Pd
1
2
t
Pd
(e) If cotter is assumed as beam as shown in Figure 2.15
22 1 1 1
max12 6 2 8
D d d PdPM
d
Use P = 30000 N, D2 = 75 mm, d1 = 34 mm
60
Machine Design
max
75 34 34 30000 3415000
6 2 8
M
5 53.57 10 1.275 10
52.3 10 N-mm
21
12I b t
Use b = 43 mm, t = 10 mm
21(4.3) 10
12I
1
2
Zb
2
31 (4.3) 10 23081.7 mm
12 43
Z
5
max
3
2.3 10
3.08 10
b
M
Z
274.7 N/mm b