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