a book of machine design · 2020. 5. 30. · cotter joint has mainly three components – spigot,...

MECHANICALNOTESHUB

A Book of Machine Design Chapter-2

NOTES FROM SOUVIK BHATTACHARJEE

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A Book of Machine Design

1

COTTER JOINTS:-

Cotter joint is used to connect two rods subjected to axial tensile or compressive loads. It is not

suitable to connect rotating shafts which transmit torque. Axes of the rods to be joined should be

collinear. There is no relative angular movement between rods.

Cotter Joint has mainly three components – spigot, socket and cotter as shown in Figure 2.1.

Spigot is formed on one of the rods and socket is formed on the other. The socket and the spigot

are provided with a narrow rectangular slot. The cotter is tightly fitted in this slot. Spigot fits

inside the socket and the cotter is passed through both the socket and the spigot. A cotter is a

wedge shaped piece made of a steel plate. It has uniform thickness and the width dimension is

given a slight taper. Taper is usually 1 in 24 and provides mainly two benefits: i) cotter becomes

tight in the slot due to wedge action. This ensures tightness of the joint in operating conditions

and prevents loosening of the parts. ii) Due to its taper shape, it is easy to remove the cotter and

dismantle the joint.

The construction of cotter joint, used to connect two rods subjected to tensile force P is shown in

the figure. When the cotter is inserted into the slot, the central portion of cotter comes in contact

with spigot and the spigot gets pushed into the socket till the collar of the spigot comes in contact

with the collar of socket. As shown in the figure, finally the cotter is in contact with the spigot on

one side having some clearance with the socket slot and is in contact with the socket on the other

side having some clearance with the spigot slot. Clearance provided is generally 1.5 to 3 mm.

Cotter gets locked because of the frictional force of the contacting surfaces.

Fig.-2.1


2

DESIGN OF SPIGOT & SOCKET COTTER JOINT:-

Let,

σt = Permissible tensile stress for the rods material,

τ = Permissible shear stress for the cotter material, and

σc = Permissible crushing stress for the cotter material.

Assumption for stress analysis of Cotter Joint :

I. The rods are subjected to axial tensile force.

II. The effect of stress concentration due to holes is neglected

III. The force is uniformly distributed in different parts.

1. Design of shaft:-

Let, d= diameter of shaft in mm.

As the rods may fail due to tension so,

Area of failure A= 𝜋

4× 𝑑2

Hence, P =π

4× d2 × σt

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

2. Design of Spigot:-

Let,

d2 = diameter of spigot or inside diameter of socket in mm

t = thickness of cotter in mm

Since the weakest section of the spigot is that section which

has a slot in it for the cotter, as shown in Fig. 2.2, therefore Fig.-2.2

Failure area A = 𝜋

4× (𝑑2)2 − (𝑑2 × 𝑡)

Hence, P = {𝜋

4× (𝑑2)2 − (𝑑2 × 𝑡)} × 𝜎𝑡 ……(ii)


3

Now according to empirical relation t = d2 / 4 ……(iii)

By equating (ii) and (iii) we can calculate the diameter of spigot.

Checking the failure of spigot due to crushing:-

We know that the area that resists crushing of a rod or cotter =d2 × t

∴ Crushing strength = d2 × t × σc

Equating this to load (P), we have

P =d2 × t × σc

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

[Note:-1. If calculated σc is more than allowable then design is not safe. For safe design,

Again we write, P =d2 × t × σc ………… in this equation put t = d2 / 4 and calculate the value of

d2 and then calculate t.

2. If calculated σc is less than allowable then design is safe.]

3. Design of Socket:-

Let,

d1 = outside diameter of socket in mm

the resisting area of the socket across the slot, as shown in

Fig. 2.3

Fig.-2.3

A =π

4[(d1)2 − (d2)2] − (d1 − d2)t

Hence, P = {π

4[(d1)2 − (d2)2] − (d1 − d2)t} × σt ……. From this equation d1 is

calculated.


4

4. Design of Spigot collar:-

Let,

d3 = diameter of spigot-collar in mm

We know that area that resists crushing of the collar

A =π

4[(d3)2 − (d2)2]

Equating this to load (P) , we have Fig.-2.4

P =π

4[(d3)2 − (d2)2]σc

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

5. Design of Socket collar:-

Let,

d4 = Diameter of socket collar in mm.

We know that area that resists crushing of socket collar

=(d4 – d2) t

and crushing strength =(d4 – d2) t × σc

Equating this to load (P), we have Fig.-2.5

P =(d4 – d2) t × σc

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

6. Width of the Cotter:-

Let,

b = Width of cotter in mm.


5

Since the cotter is in double

shear, therefore shearing area of the cotter = 2 b × t

and shearing strength of the cotter =2 b × t × τ


P =2 b × t × τ ………… From this equation, width of cotter (b) is determined.

7. Shear faliure of Socket end/Spigot end:-

a = distance from end of slot to the end of spigot in mm

c = axial distance from slot to end of socket collar in mm

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

=2 × (d4 – d2) c

and shearing strength of socket collar

=2 × (d4 – d2) c × τ


P =2 × (d4 – d2) c × τ

From this equation, (c) or (a) may be obtained.

8. Bending failure of cottar:-

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


6

Fig.-2.6

the cotter in the rod end is uniformly distributed while in the socket end it varies from zero at the

outer diameter (d4) and maximum at the inner diameter (d2), as shown in Fig. 2.6.

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

Mmax =P

2(

1

3×

d4 − d2

2+

d2

2) −

P

2×

d2

4

=P

2(

d4 − d2

6+

d2

2−

d2

4) =

P

2(

d4 − d2

6+

d2

4)

We know that section modulus of the cotter,

Z = t × b2/6

Bending stress induced in the cotter,

σb =Mmax

Z=

P2 (

d4 − d2

6 +d2

4 )

t × b2/6=

P(d4 + 0.5d2)

2t × b2

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

cotter.

9. Shear failure of Spigot Collar:-

Considering the failure of the spigot collar in shearing as shown in

Fig. 2.7. We know that area that resists shearing of the collar

A = π d2 × t1

Hence, P = π d2 × t1 × τ

From this equation, the thickness of spigot collar (t1) may be

obtained.

Fig.-2.7

APPLICATION:-


7

1. Joint between the piston rod and and the cross head of the steam engine.

2. Joint between the slide spindle and the fork of the valve mechanism.

3. Joint between the piston rod and the tail.

4. Connecting two halves of flywheel.

KNUCKLE JOINTS:-

Knuckle joint is used to connect two rods subjected to axial tensile loads. It may also be used to

support the compressive load if the joint is guided. It is not suitable to connect rotating shafts

which transmit torque. Axes of the shafts to be joined should lie in the same plane and may

coincide or intersect. Its construction permits limited relative angular movement between rods,

about the axis of the pin. Knuckle joint is widely used to connect valve rod and eccentric rod, in

the link of a cycle chain, levers, tie rod joint for roof truss and many other links.

Knuckle Joint has mainly three components – eye, fork and pin as shown in Figure 2.8. Eye is

formed on one of the rods and fork is formed on the other. Eye fits inside the fork and the pin is

passed through both the fork and the eye. This pin is secured in its place by means of a split-pin.

The ends of the rods are made octagonal to some distance for better grip and are made square for

some portion before it is forged to make the eye and fork shapes.

Fig.-2.8

DESIGN PROCEDURE OF KNUCKLE JOINTS:-

Let,

D = diameter of each rod in mm

D1 = enlarged diameter of each rod in mm


8

d = diameter of knuckle pin in mm

d0 = outside diameter of eye or fork in mm

d1 = diameter of pin head in mm

a = thickness of each eye of fork in mm

b = thickness of eye end of rod B in mm

x = distance of the centre of fork radius R from the eye in mm

σt, τ and σc = Permissible stresses for the joint material in tension, shear and crushing

respectively in MPa .

Assumption for stress analysis of Knuckle Joint:-

I. The rods are subjected to axial tensile force.

II. The effect of stress concentration due to holes is neglected

III. The force is uniformly distributed in different parts.

1. Design of rod:-

As the rod fails due to tensile force,

Failure area A= π

4× D2

Hence, P =π

4× D2 × σt

From that equation D is calculated.

So, according to empirical relation

D1 = 1.1 D a = 0.75 D b = 1.25 D R = b/2

2. Design of knuckle pin:-

As the pin fails due to shearing and according to fig. pin has double shear

So shearing failure area A = 2 ×π

4× d2

Hence, P = 2 ×π

4× d2 × τ

From this equation d is calculated.


9

According to empirical relation d0 = 1.5 d and d1 = 2d

3. Failure Conditions for Eye:-

Tensile:- we know,

σt =P

b(d1 − d)

From this equation σt is

calculated. If (σt)cal is less than

allowable value then design is

safe.

Shear:- we know,

τ =P

b(d1 − d)

From this equation τ is

calculated. If (τ)cal is less than


safe.

Compressive:- We know,

σc =P

b × d

From this equation σc is

calculated. If (σc)cal is less

than allowable value then

design is safe.

4. Failure Conditions of Fork:-

Tensile:- we know,

σt =P

2a(d1 − d)

From this equation σt is

calculated. If (σt)cal is less than


safe.

Shear:- we know,

τ =P

2a(d1 − d)

From this equation τ is

calculated. If (τ)cal is less than


safe.


σc =P

2a × d

From this equation σc is

calculated. If (σc)cal is less

than allowable value then

design is safe.

TURN BUCKLE:-

A trunbuckle is a static device which is used for adjusting tension at any cable, tie-bars,tie-ropes.

Common applications are the ropes used for supporting power transmission cables belt tension

adjustment in tie rods. It consists of a tubular body called coupler which is mostly cast steel. The

central portion is hollow and at the ends there are two nuts in which the rods fit. Its exploded

view is shown below,


10

Fig.-2.9

According to fig.-2.9 there are left hand and right hand threads due to which when the coupler

is moved in one direction both rods move closer and vice versa.

DESIGN PROCEDURE OF TURN BUCKLE:-

Fig.-2.10

Consider a turnbuckle, subjected to an axial load P, as shown in Fig. 2.10.

The design is followed by through the chart below:-

Part Failure Equation To find

Design load Empirical Pd = 1.3 P Pd

Design of rod (dc,d)

Tensile

Empirical

Pd =π

4× (dc)2 × σt

d=1.19 × dc

dc

d

Design of Coupler nut

(D,l)

Tensile

Pd =π

4× (D2 − d2)

× σt

D


11

Shear

Pd = πdcl × τ

l

Design of Coupler

(D1,D2,L)

Empirical

Tensile

Empirical

D1=d+6

π

4× (D2

2 − D12) × σt

L=6d

D1=

D2=

L=

SOME EXAMPLES OF NUMERICALS:-

N-2.1 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 × 103 N ; σt = 50 MPa = 50 N / mm2 ; τ = 35 MPa = 35 N /

mm2 ; σc = 90 MPa = 90 N/mm2

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 × 103 =π

4× d2 × σ𝑡 =

π

4× d2 × 50 = 39.3d2

d2 = 30 × 103/39.3 = 763 or d = 27.6 say28 mm Ans.

2. Diameter of spigot and thickness of cotter

Let d2 = Diameter of spigot or inside diameter of socket, and


12

t = Thickness of cotter. It may be taken as d2/4.

Considering the failure of spigot in tension across the weakest section. We know that load (P) ,

30 × 103 = [π

4(d2)2 − d2 × t] σ𝑡 = [

π

4(d2)2 − d2 ×

d2

4] 50 = 26.8(d2)2

(d2)2 = 30 × 103/26.8 = 1119.4 or d2 = 33.4say34mm

and thickness of cotter, t =d2

4=

34

4= 8.5 mm

Let us now check the induced crushing stress. We know that load (P) ,

30 × 103 = d2 × t × σc = 34 × 8.5 × 0c = 289σc

σc = 30 × 103/289 = 103.8N/mm2

Since this value ofoc is more than the given value ofσc = 90N/mm2, therefore the dimensions

d2

= 34 mm and t = 8.5 mm are not safe. Now let us find the values ofd2 and t by substituting the

value of σc = 90N/mm2

30 × 103 = d2 ×d2

4× 90 = 22.5(d2)2

(d2)2 = 30 × 103/22.5 = 1333 or d2 = 36.5 say 40mm.

and t = d2/4 = 40/4 = 10 mm.

3. Outside diameter of socket

Let d1 = Outside diameter of socket.

Considering the failure of the socket in tension across the slot. We know that load (P) ,

30 × 103 = [π

4{(d1)2 − (d2)2} − (d1 − d2)t] σ𝑡

= [π

4{(d1)2 − (40)2} − (d1 − 40)10] 50

30 × 103/50 = 0.7854(d1)2 − 1256.6 − 10d1 + 400

or (d1)2 − 12.7d1 − 1854.6 = 0


13

d1 =12.7 ± √(127)2 + 4 × 1854.6

2=

12.7 ± 87.1

2

= 49.9 say 50 mm. ...(Taking +vesign)

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)

30 × 103 = 2b × t × τ = 2b × 10 × 35 = 700b

b = 30 × 103/700 = 43 mm.

5. Diameter of socket collar

Let d4 = Diameter ofsocket collar.

Considering the failure of the socket collar and cotter in crushing. We know that load (P) ,

30 × 103 = (d4 − d2)t × σc = (d4 − 40)10 × 90 = (d4 − 40)900

d4 − 40 = 30 × 103/900 = 33.3 or d4 = 33.3 + 40 = 73.3 say75 mm.

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 × 103 = 2(d4 − d2)c × τ = 2(75 − 40)c × 35 = 2450c

c = 30 × 103/2450 = 12 mm.

7. Distance from the end of the 𝐬𝐓𝐨𝐭 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 × 103 = 2a × d2 × τ = 2a × 40 × 35 = 2800cx


14

a = 30 × 103/2800 = 10.7 say 11 mm.

8. Diameter of spigot collar

Let d3 = Diameter ofspigot collar.

Considering the failure of spigot collar in crushing. We know that load (P) ,

30 × 103 =π

4[(d3)2 − (d2)2]σc =

π

4[(d3)2 − (40)2]90

or (d3)2 − (40)2 =30×103×4

90×π= 424

(d3)2 = 424 + (40)2 = 2024 or d3 = 45 mm.

9. Thickness of spigot collar

Let t1 = Thickness of spigot collar.

Considering the failure of spigot collar in shearing. We know that load (P),

30 × 103 = π d2 × t1 × τ = π × 40 × t1 × 35 = 4400 t1

∴ t1 = 30 × 103 / 4400 = 6.8 say 8 mm.

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

∴ l = 4 d = 4 × 28 = 112 mm.

11. The dimension e is taken as 1.2 d.

∴ e = 1.2 × 28 = 33.6 say 34 mm.

N-2.2 Design a knuckle joint to transmit 150 KN. The design stresses may be taken as 75 MPa in

tension, 60 MPa in shear and 150 MPa in compression.

Solution. Given : P = 150 KN = 150 × 103 N ; σt = 75 MPa = 75 N/mm2 ; τ = 60 MPa = 60

N/mm2 ; σc = 150 MPa = 150 N/mm2

Let,

D = diameter of each rod in mm

D1 = enlarged diameter of each rod in mm

d = diameter of knuckle pin in mm

d0 = outside diameter of eye or fork in mm

d1 = diameter of pin head in mm


15

a = thickness of each eye of fork in mm

b = thickness of eye end of rod B in mm

x = distance of the centre of fork radius R from the eye in mm

As the rod fails due to tensile force,

Failure area A= π

4× D2

Hence, P =π

4× D2 × σt or, 150 × 103 =

π

4× D2 × 75

Or, D = 50.4 mm = 52 mm

So, according to empirical relation

D1 = 1.1 D = 57.2 = 58 mm a = 0.75 D = 39 mm b = 1.25 D = 65 mm R = b/2 =26 mm

2. Design of knuckle pin:-

As the pin fails due to shearing and according to fig. pin has double shear

So shearing failure area A = 2 ×π

4× d2

Hence, P = 2 ×π

4× d2 × τ or, 150 × 103 = 2 ×

𝜋

4× d2 × 60

Or, d = 39.9 mm = 40 mm

According to empirical relation d0 = 1.5 d = 60 mm and d1 = 2d = 80 mm

3. Failure Conditions for Eye:-

Tensile:- we know,

σt =P

b(d1 − d)

σt =150 × 103

65(80 − 40)

= 57.7N/mm2

As (σt)cal is less than allowable

value then design is safe.

Shear:- we know,

τ =P

b(d1 − d)

τ =150 × 103

65(80 − 40)

= 57.7N/mm2

As (τ)cal is less than allowable



σc =P

b × d

σc =150 × 103

65 × 40= 57.7N/mm2

As (σc)cal is less than


safe.


16

4. Failure Conditions of Fork:-

Tensile:- we know,

σt =P

2a(d1 − d)

σt =150 × 103

2 × 39(80 − 40)

= 48.07N

/mm2

As (σt)cal is less than allowable


Shear:- we know,

τ =P

2a(d1 − d)

τ =150 × 103

2 × 39(80 − 40)

= 48.07N

/mm2

As (τ)cal is less than allowable



σc =P

2a × d

σc =150 × 103

2 × 39 × 40= 48.07 N

/mm2

As (σc)cal is less than

allowable value then design

is safe.

N-2.3 The pull in the tie rod of an iron roof truss is 50 KN. Design a suitable adjustable screwed

joint. The permissible stresses are 75 MPa in tension, 37.5 MPa in shear and 90 MPa in crushing.

Solution. Given : P = 50 KN = 50 × 103 N ; σt = 75 MPa = 75 N/mm2 ; τ = 37.5 MPa = 37.5

N/mm2

We know that the design load for the threaded section,

Pd = 1.3 P = 1.3 × 50 × 103 = 65 × 103 N

1. Diameter of the tie rod

Let d = Diameter of the tie rod, and

dc = Core diameter of threads on the tie rod.

Considering tearing of the threads on the tie rod at their roots.

We know that design load (Pd) ,

65 × 103 =π

4(dc)20f =

π

4(dc)275 = 59(dc)2

(dc)2 = 65 × 103/59 = 1100 or dc = 33.2mm

From Table 11.1 for coarse series, we find that the standard core diameter is 34.093 mm and the

corresponding nominal diameter of the threads or diameter of the rod,


17

d = 39 mm Ans.

2. Length of the coupler nut

Let /= Length of the coupler nut.

Considering the shearing of threads at their roots in the coupler nut. We know that design load

(Pd) ,

65 × 103 = (πdc. l)τ = π × 34.093 × l × 37.5 = 4107l

/= 65 × 103/4017 = 16.2mm

Since the length of the coupler nut is taken from d to 1.25 d, therefore we shall take

l = d = 39 mm

We shall now check the length of the coupler nut for crushing of threads.

From Table 11.1 for coarse series, we find that the pitch of the threads is 4 mm. Therefore the

number of threads per mm length,

n = 1/4 = 0.25

We know that design load (Pd) ,

65 × 103 =π

4[(d)2 − (dc)2]n × l × 0c

=π

4[(39)2 − (34.093)2]0.25 × 39 × 0c = 27500c

0c = 65 × 103/2750 = 23.6N/mm2 = 23.6MPa

Since the induced crushing stress in the threads of the coupler nut is less than the permissible

stress, therefore the design is satisfactory.

3. Outside diameter of the coupler nut

Let D = Outside diameter of the coupler nut

Considering tearing of the coupler nut. We know that axial load (P) ,

50 × 103 =π

4(D2 − d2)0f


18

=π

4[D2 − (39)2]75 = 59[D2 − (39)2]

or D2 − (39)2 = 50 × 103/59 = 848

Ij2 = 848 + (39)2 = 2369 or D = 48.7 say50 mm.

Since the minimum outside diameter of coupler nut is taken as 1.25 d(i.e. 1.25 × 39 = 48.75

mm),

therefore the above value ofD is satisfactory.

4. Outside diameter of the coupler

Let D2 = Outside diameter of the coupler, and

D1 = Inside diameter of the coupler = d + 6mm = 39 + 6 = 45 mm

Considering tearing of the coupler. We know that axial load (P) ,

50 × 103 =π

4[(D2)2 − (D1)2]0f =

π

4[(D2)2 − (45)2]75 = 59[(D2)2 − (45)2]

(D2)2 = 50 × 103/59 + (45)2 = 2873 or D2 = 53.6 mm

Since the minimum outside diameter of the coupler is taken as 1.5 d (i.e. 1.5 × 39 = 58.5 say

60 mm),

therefore we shall take

D2 = 60 mm.

5. Length of the coupler between nuts,

L = 6d = 6 × 39 = 234 mm.

6. Thiclςness of the coupler,

t1 = 0.75d = 0.75 × 39 = 29.25 say 30 mm.

and thickness of the coupler nut,

t = 0.5d = 0.5 × 39 = 19.5 say 20 mm.

EXERCISES

NUMERICALS:-


19

1. Two rod ends of a pump are joined by means of a cotter and spigot and socket at the ends.

Design the joint for an axial load of 100 KN which alternately changes from tensile to

compressive. The allowable stresses for the material used are 50 MPa in tension, 40 MPa in

shear and 100 MPa in crushing.

2. Design a cotter joint to connect a piston rod to the crosshead. The maximum steam pressure on

the piston rod is 35 KN. Assuming that all the parts are made of the same material having the

following permissible stresses : σt = 50 MPa ; τ = 60 MPa and σc = 90 MPa.

3. Design a knuckle joint to connect two mild steel bars under a tensile load of 25 KN. The

allowable stresses are 65 MPa in tension, 50 MPa in shear and 83 MPa in crushing.

4. The pull in the tie rod of a roof truss is 44 kN. Design a suitable adjustable screw joint. The

permissible tensile and shear stresses are 75 MPa and 37.5 MPa respectively. Draw full size two

suitable views of the joint.

THEORY:-

1. What is a cotter joint? Explain with the help of a neat sketch, how a cotter joint is made ?

2. What are the applications of a cottered joint ?

3. Discuss the design procedure of spigot and socket cotter joint.

4. Distinguish between cotter joint and knuckle joint.

5. Sketch two views of a knuckle joint and write the equations showing the strength of joint for

the most probable modes of failure.

6. Explain the purpose of a turn buckle. Describe its design procedure.

MCQ’S:-

01. For connecting the piston rod and cross-head in a steam engine, the joint used is:

a) Knuckle b).Cotter c) riveted d) welding

02. Cotter joint is used to join to rods under -------------types of forces.


20

a) tensile b) bending c) reciprocating d) fluctuating

03. Fork end is the part of :

a) coupling b) pulley c) cotter joint d) knuckle joint

04. In steam engine, Valve rod is connected to the eccentric by means of a

a) Knuckle joint b) universal joint c) flange coupling d) cotter joint

05. The failure of pin in knuckle joint is likely to be caused by

a) Shear b) Tear c) creep d) vibration

06. Calculations for the diameter of the rod in cotter joint are made by considering failure of the

rod in:

a) compression b) tension c) shear d) torsion

07. In a cotter joint, the width of the cotter at the centre is 50 mm and its thickness is 12 mm.

Shearing stress developed in the cotter if the load acting on it is 60 kN will be:

a) 120 N/mm2 b) 100 N/mm2 c) 75 N/mm2 d) 50 N/mm2

08. Identify the statement/statements valid for a knuckle joint:

i) used for joining rods subjected to tensile and compressive forces.

ii) the axis of the rods are not in alignment but meet at a point

iii) there exists a small angular movement of one rod relative to the other

iv) when easy connection/disconnection is desired

a) 1,2 and 4 b) 1,3 and 4 c) 2,3 and 4 d) 1,2,3 and4.

09. In a turn buckle, if one of the rods has left hand threads, then the other rod will have

(a) right hand threads (b) left hand threads (c) pointed threads (d) multiple threads


21

10. The taper on cotter varies from

(a) 1 in 15 to 1 in 10 (b) 1 in 24 to 1 in 20 (c) 1 in 32 to 1 in 24 (d) 1 in 48 to 1 in 24

MCQ’S ANSWER

1.b) 2.c) 3.d) 4.a) 5.a) 6.b)

7.d) 8.d) 9.a) 10.d)

a book of machine design · 2020. 5. 30. · cotter joint has mainly three components – spigot,...

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