clutches and brakes

7/29/2019 Clutches and Brakes

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1

CLUTCHES AND BRAKES

Clutches:

A Clutch is a mechanical device which is used to connect or disconnect the source of power

from the remaining parts so the power transmission system at the will of the operator. The

flow of mechanical power is controlled by the clutch.

Types of Clutches

(i) Positive Clutches (ii) Friction clutches

Positive Clutches: In this type of clutch, the engaging clutch surfaces interlock to produce

rigid joint they are suitable for situations requiring simple and rapid disconnection, although

they must be connected while shafts are stationery and unloaded, the engaging surfaces areusually of jaw type. The jaws may be square jaw type or spiral jaw type. They are designed

empirically by considering compressive strength of the material used.

The merits of the positive clutches are

. (i) Simple (ii) No slip (iii) No heat generated compact and low cost.

Friction Clutches: Friction Clutches work on the basis of the frictional forces developed

between the two or more surfaces in contact. Friction clutches are usually over the jaw

clutches due to their better performance. There is a slip in friction clutch. The merits are

(i) They friction surfaces can slip during engagement which enables the driver to

pickup and accelerate the load with minimum shock.

(ii) They can be used at high engagement speeds since they do not have jaw or teeth

(iii) Smooth engagement due to the gradual increase in normal force.

The major types of friction clutches are

(i) Plate clutch (Single plate) (multiple plate)

(ii) Cone clutch

(iii) Centrifugal clutch

(iv) Dry

(v)Magnetic current clutches

(vi)Eddy current clutches

Square-jaw clutch Spiral-jaw clutch


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2

We will be studying about single plate multi-plate and cone clutches.

Single plate clutch:

A single plate friction clutch consisting of two flanges shown in fig 2. One flange is rigidlykeyed in to the driving shaft, while the other is free to move along the driven shaft due to

spliced connection. The actuating force is provided by a spring, which forces the driven

flange to move towards the driving flange. The face of the drive flange is linked with frictionmaterial such as cork, leather or ferodo

Torque transmitted by plate or disc clutch

A friction disk of a single plate clutch is shown in above fig

The following notations are used in the derivationDo = Outer diameter of friction disc (mm)

Di = Inna diameter of friction disc (mm)

P = pressure of intensity N/mm2

F = Total operating force (N) (Axial force)

T = torque transmitted by friction (N-mm)

Consider an elemental ring of radius r and radial thickness dr

Area of elemental length = 2r. dr

Axial force length = 2r r. P


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3

( or f) friction force = 2r. drp

Friction torque = 2r dr P * r

Total axial force Fa = =/2D

/2Da

0

i

drpr2F ------------ (1)

Torque Transmitted by friction T = 2/

2/1

20

2

D

D

pdrr ------------ (2)

There are two criteria to obtain the torque capacity uniform pressure and uniform wear

1. Uniform pressure Theory:

In case of new clutches, un playing assumed to be uniformly distributed over the entire

surface area of the friction disc. With this assumption, P is regarded as constant.

Equation 1 becomes

=/2D

/2Da

0

i

drpr2F

==/2D

/2Da

0

i

drr2F p

2/

2/

2 0

12

2

D

D

rp

=

222

2F

2

1

2

0 DDpa

( )212

04

1F DDpa =

or[ ]2120

F4DD

aP

=

From Equation -2

T = 2/

2/1

20

2

D

D

drpr

T = 2/

2/1

20

2

D

D

drrp


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4

T =

2/

2/

3 0

1

32

D

D

rp

T =

223

2 33 DiDop

T = ( )3133

1DDop

Substituting the value of p from equation 3

T =( )

12

3

1

3 DD

( )31

3

4

DD

Fa

o

T =( )( )220

33

0Fa3

1

i

i

DD

DD

T =2

Fa Dm

Where DmDD

DD

i

i =

22

0

33

0

3

2mean diameter

Torque transmitted by n- friction surfaces

T =2

1 FaDmn

Axial force = ( ) ( )2202

1

2

24

iDDPRRP =

Uniform Wear Theory:

According to this theory, it is assumed that the wear is uniformly distributed over the entiresurface --- of the friction disc. This assumption is used for workout clutches. The axial wear

of the friction disc is import ional to the frictional work. The work done by the frictional force

( P) and subbing velocity (2rN) where N is speed in rpm. Assuming speed N and

coefficient of friction is constant for given configuration

Wear PrPr = constant C

When clutch plate is new and rigid. The wear at the outer radius will be more, which will

release the pressure at the outer edge due to the rigid pressure plate this will change the

pressure distribution. During running condition, the pressure distribution is adjusted in such a

manner that the product pressure is constant, C.


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5

From equation - (1)

Axial force Fa = 2/

2/

0

1

Pr2

D

D

dr

Fa = 2/

2/

0

1

2

D

D

drc

Fa = [ ] 2/2/

0

12

D

Drc

Fa =

222 10

DDc

C =

2

][2 10 DD

Fa

- (7)

From equation - (2)

T = =2/

2/

2/

2/

20

1

0

1

2Pr2

D

D

D

D

drrcdr

=

2/

2/

2 0

1

22

D

D

rc

=

8

2/

8

2/2 10

DDc

T = [ ]21208

DDc

Substitute the value of C from equation (7)

T = [ ][ ]10

2

1

2

04 DD

FaDD

T =[ ]

22

Fa 10 DD

T = 2

DmFa

- (8)


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6

Where Dm =2

10 DD + - (9)

Torque transmitted by n friction plates

T =2

1 DmFan

Axial force Fa =

Average pressure occurs at mean radius )2/( mm Dr =Q

Fa =2

)( 10 DDDmb - (10)

Maximum pressure occurs at inner radius

Fa = )(2

0 ii DD

Dp

- (11)

Note: The major portion of the life of friction lining comes under the uniform wear friction

lining comes under the uniform wear criterion in design of clutches uniform wear

theory is justified.

Problems:

1. A single plate friction clutch of both sides effective has 300 mm outer diameter and160 mm inner diameter. The coefficient of friction o.2 and it runs at 1000 rpm. Find

the power transmitted for uniform wear and uniform pressure distributions cases if

allowable maximum pressure is 0.08 Mpa.

Given:

N1 = I = 2, D0 = 300 mm D1 = 160mm = 0.2

N = 1000 rpm p = 0.08 Mpa = 0.08 N /mm2

Solution:

i. Uniform wear theory

Mean Diameter mmDD

Dm 2302

160300

2

10 =+

=

=

Axial Force

Fa = )(2

1101 DDDb

From DDH 13.32 or Equation - (11)


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7

Fa = )160300(16008.02

1

= 2814.87 N

Torque transmitted

T = DmFan1

2

1

T = 23087.281422.02

1

T = 129484 N mm

Power transmitted

P =61060

2

TM

P = =

61060

12948410002

P = 13.56 kW

ii. Uniform wear theory

Mean Diameter

== 2

1

2

0

3

1

3

0

3

2

DD

DDDm

== 22

33

160300

160300

3

2mD

Dm = 237.1 mm

Axial Force Fa =4

)(2

1

2

0 DDp

Fa =4

)160300(08.0 22

Fa = 4046.4 N

Torque transmitted

T = DmFan 2

11


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8

T = 1.2374.40462.02

12

T = 191880.3 N mm

Power transmitted

P =61060

2

Tn

P = =

61060

3.19188010002

P = 20.1 kW

2. A car engine develops maximum power of 15 kW at 1000 rpm. The clutch used is

single plate clutch both side effective having external diameter 1.25 times internal

diameter = 0.3. Mean axial pressure is not to exceed 0.085 N/ mm2. Determine the

dimension of the friction surface and the force necessary to engage the plates. Assume

uniform pressure condition.

Given p = 15 kW, n 1000rpm, I =2 both sides are effective D0 = 1.25 D1, = 0.3,

p = 0.085 N/mm2

Torque transmitted

P =61060

2

Tn

T =n

P

2

1060 6

T =10002

106015 6

mmN= 239.143


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9

Mean Diameter Dm

==

== 22

33

22

0

33

0

)25.1(

)25.1(

3

2

3

2

ii

ii

i

im

DD

DD

DD

DDD

= 1.13 Di

Axial Force Fa = )(4

22

0 iDDp

Fa = )25.1(085.04

22

ii DD

Fa = 0037552 2iD

Torque transmitted T = DmFai 2

1

143239 = ii DD 13.1037552.03.02

12 2

Di = 224 mm

D0 = 280 mm

Dm = 253 mm

Fa = 1884.21 N

Thickness of disc h = 2 mm

3. Design a single plate clutch consist of two pairs of contacting surfaces for a torque

capacity of 200 N-m. Due to space limitation the outside diameter of the clutch is to

be 250mm

Given:

Single plate clutch, Torque = 2 x 105

N-mm, D0 = 250mm I = 2 (since two pairs of

contacting surfaces)

Solution:

Assume suitable friction material leather = 0.3 to 0.5 P = varies from 0.07 to

0.29 Mpa select = 0.4, P = 0.135 Mpa N /mm2

1. Torque transmitted= 2 x 105

N-mm

2. Mean diameter

Assuming uniform wear theory


10/77

10

2

250

2

0 +=+

= iimDDD

D

3. Axial force :

For uniform, wear condition

Fa = = )(2

1iii DDDp

= )250(135.02

1ii DD

= 0.212 Di (250 Di)

Torque transmitted

T = iDFa m2

1

i.e 2)250(2

1)250(212.04.0

2

1102

5+== iii DDD

= 62500 D1- 0132.47169813

1 =D

By trial and error method

Inner dia Di = 85.46 mm is 86 mm

Outer dia D0 = 250 mm given, mean m = dia of friction surface

Dm = 168 mm

Fa = 0.212 x 86 (250 86) = 2990 N

Multiple plate clutch

Fig. shows a multiple plate clutch. The driving

discs are splined to the driving shaft so that they

are free to slip along the shaft but must rotate

with it. The driven discs drive the housing bymeans of bolts along which they are free to slide.

The housing is keyed to the driven shaft by a

sunk key. In the clutch shown there are five pairsof friction surfaces. The driving discs may be

pressed against the driven discs by a suitable

mechanism so that the torque may be transmittedby friction between the discs.


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11

Multi disc Clutch:

Equations derived for torque transmitting velocity of single plate are modified to account for

the number of pairs of contacting surfaces in the following way.

For uniform pressure T = =DmFai 121

)(4

22

2 iDDp

Uniform wear T = )( 220 iDD

Where I = number of pairs of contacting surfaces.

For uniform pressure theory

T = DmFai 12

1

Dm =)(

)(

3

222

0

33

0

i

i

DD

DD

Fa = )(4

22

0 iDDp

Where I = number of friction surfaces

Uniform wear theory

T = DmFai 12

1

Dm =2

)( 2 iDD +

Fa = )(2

10 im DDDb

Maximum pressure occurs at inner radius

Fa = )(2

1011 iDDDp

Problem: A multi plate clutch having effective diameter 250mm and 150mm has to transmit60 kW at 1200 rpm. The end thrust is 4.5 kN and coefficient of friction is 0.08 calculate the

number of plates easuming (i) Uniform wear and (ii) uniform pressure distribution on the

plates

Given: D0 = 250 mm Di = 150mm, P = 60 kW, N = 1200 rpm, Fa = 4.5 kN = 4500N, = 0.08


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12

P =61060

2

NT

Torque T = mmNP

=

47750012002

1060 6

(i) Uniform wear theory

Mean diameter Dm = mmDD

i 2002

150250

2

0 =+

=+

T = DmFai 12

1

477500 = 200450008.02

1

i

Number of friction plates, I = 13.26 1.4 (even numbers)

Total number of plates i + 1 = 14 + 1 = 15

Uniform pressure

Dm =)(

)(

3

222

0

33

0

i

i

DD

DD

= 204.17 mm

T = DmFai 2

1

477500 = 17.20408.045002

1i

i = 12.99 14 (even number)

Total number of plates = 14 + 1 = 15

Problem 4:

A multi plate clutch of alternate bronze and steel plates is to transmit 6 kW power at 800 rpm.

The inner radius is 38 mm and outer radius is 70 mm. The coefficient of friction is 0.1 andmaximum allowable pressure is 350 kN /m2 determine

(i) Axial force required

(ii) Total number of discs(iii) Average pressure and

(iv) Actual maximum pressure


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13

Given: P = 60 kW, N = 800 rpm, R1 = 38mm, Di= 76 mm R0 = 70, D0 = 140mm, = 0.1,

P = 350 kN / m2

= 0.35 N / mm2

1. Axial force

Fa = )(21 1011 DDDp - (13.32 DDH)

Fa = )76140(7635.02

1

= 2674.12 NTorque to be transmitted

P =61060

2

NT

T = mmNN

P == 71625800210606

21060

66

Assuming uniform wear theory

Mean Diameter Dm = mmDD

1082

76140

2

)( 12 =+

=+

Torque transmitted T = DmFai 12

1

10812.26741.02

171625 = n

n = 4.96 6 (even number)

Number of driving (steel) discs = 32

6

21 ===

nn

Number of driven (bronze) discs n2

= n1 + 1 = 3 + 1 = 4

3. Average pressure occurs at mean diameter

Axial force Fa = )(2

101 im DDDp

2674.12 = )76140(1082

11 p -

Average pressure p = 0.246 N/ mm

2


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14

4. For 6 friction surface, torque transmitted

T = DmFai 12

1

71625 = 1081.0216 1 Fa -

Fa = 2210.6 N

Maximum pressure occur at inner radius

Axial force = )(2

1011 iDDDp

2210.6 = )76140(7621 1 p -

Actual maximum pressure P = 0.289 N/ mm2

Problem 5:

In a maultilate clutch radial width of the friction material is to be 0.2 of maximum radius.

The coefficient of friction is 0.25. The clutch is 60KW at 3000 rpm. Its maximum diameteris 250mm and the axial force is limited is to 600N. Determine (i) Number of driving anddriven plates (ii) mean unit pressure on each contact surface assume uniform wear

Given: Radial width = 0.2 Ro, = 0.25, P = 60KW, N = 3000rpm, D0 = 250mm,Ro = 125mm, Fa = 600N uniform wear condition.

Solution b = Ro- Ri

0.2 Ro =RiRi= 0.8 Ro = 0.8x 125 = 100mm

Inner diameter 2x100 = 200mm

i) Number of disc

Torque transmitted T =N

P2

1060

6

mmN.19100030002

106060 6=

For uniform wear condition

Mean diameter ( ) ( ) mmDDD iom 2252002502

1)

2

1=+=+=

Torque Transmitted

FaDmnT 21=


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15

225600,25.02

119100 = n

i.e n = 11.32

Number of active surfaces n =12 (n must be even number)

Number of disc on the driver shaft

62

12

21 ===

nn

Number disc on the driven shaft

712

121

21 =+=+=

nn

Total number of plates : n1 + n2 = 617 = 13

ii) Mean unit pressure

Fa ( )iom DDDp = 2

1

( )2002502252

1600 = p

P = 0.34 N/ mm2

iii) for actual mean unit pressureActual axial force

FamDn

T

2= n

mDfaT 21=

/926.5652251225.0

1910002N=

= \

( )2002502252

1926.565 = mPD

Actual mean unit pressure P = 2/032.0 mmN

A Multiple plate clutch has steel on bronze is to transmit 8 KW at 1440 rpm. The inner

diameter of the contact is 80mm and outer diameter of contact is140 mm. The clutch operatesin oil with coefficient of friction of 0.1. The overage allowable pressure is 0.35Mpa. Assume

uniform wear theory and determine the following.

a) Number of steel and bronze platesb) Axial force required

c) Actual maximum pressure

Given P = 8KW, N = 1440 rpm, D1 = 80mm, Do = 140mm, = 0.1, P = 0.35 N / mm2

Uniform Wear Theory.


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16

Solution:

1) Number of steel and bronze platesFor uniform wear theory

Axial force Fa = ( )io DDp 2

1

( )801408035.02

1 =

Fa = N9.2638

Mean diameter ( ) ( ) mmDDDm i 110801402

1

2

12 =+=+=

Torque transmitted T =N

P

2

1060 6

T = mmN=

556.53055

14402

10608 6

Also T = FaDmn2

1

n= 11094.26382

1556.53055

655.3= n

No. of Active surface is 4Number discs on the driver shaft (Bronze)

22

4

21 ===

n

n

Number disc on the driven shaft (Steel)

312

41

22 =+=+=

nn

Total No. of disc 53221 =+=nn

b) Axial force required

62.2411110041.0

552.5305522=

==

mDn

TFa

c) Actual maximum pressure since maximum presser occur at inner diameter

Fa = ( )12

1ioi DDDP

( )8014080.2

162.2411 = P

P = 2/32.0 mmN


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17

Cone clutch

A simple form of a cone clutch is shown in fig. itconsists of ;i driver or cup and the follower or cone.The cup is keyed to the driving shaft by a sunk key andhas an inside conical surface or face which exactly fitsthe outside conical surface of the cone. The slope of thecone face is made small enough to give a high normalforce. The cone is fitted to the driven shaft by a featherkey. The follower may be shifted along the shaft by aforked shifting lever in order to engage the clutch bybringing the two conical surfaces in contact.

Advantages and disadvantages of cone clutch:

Advantages:

1. This clutch is simple in design.

2. Less axial force is required to engage the clutch.

Disadvantages :

1. There is a tendency to grab.2. There is some reluctance in disengagement.

Strict requirements made to the co-axiality of the shafts being connected

Torque transmitted by the cone clutch:-

Let Di= Inner diameter of cone, Ri = inner radius of cone,Do = Outer diameter of cone, Ro = Outer radius of cone,

Rm = mean radius of cone, Dm = mean diameter of cone,

= Semi cone angle or pitch cone angle or face angle,P = Intensity of normal pressure at contact surface,

= Coefficient of friction,Fa = Axial force

Fn=Normal force =sin

Fa


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18

Consider an elemental ring of radius r and thickness dr as shown in the figure

The sloping lengthsin

dr=

Area of elementary ring

sin

2 drr=

Normal force on the ring

sin

2dr

rP=

Axial component of the above force

Sin

Sin

drrp

.2

drpr2=

Total axial force Fa = drprRo

Ri2 ------------ (1)

Frictional force outer ring

Sin

drrp 2=

Moment of friction force about the axial

rSin

drrP =

2

Sin

drrp 22=

Total torque T =

Sin

drRo

Ri

2Pr2 ---------- (2)

Uniform pressure theory: p constant

Equation (1) becomes

Fa = rdpdrprRo

Ri

Ro

Ri

= 22

Fa = ( )22 io RRP

p =( )22 io RR

Fa

---------- (3)

Equation 2 becomes

T = ( ) =

Ro

Riio

Ro

Ri drrSin

P

Sin

dr

rRrp

2222 2

2


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19

T =3

233

ia RR

Sin

p =

Substitutes the value of P from equation ---------- (3)

T =( ) 3

233

2

1

2

ia

o

RR

RR

Fa

Sin

P

=

T =( ) ( )22

33

2

1

23

2

io

ia

o RR

RR

RR

Fa

Sin

Fa

=

T = Dm

Sin

Fa

DD

DD

Sin

Fa

ioo

io.

23

2

2

222

33

=

---------- (4)

Where

=

22

33

3

2

ioo

io

DD

DDmD

Axial force ( )22 io DDPFa = ---------- (5)

Uniform wear: For uniform wear condition Pr = C Constant

Equation (1) become Fa = rdPrdrPRo

Ri

Ro

Ri = 22

Fa = )io RRc 2 or

( )io

RR

FaC

=

2

Equation (2) become

T = =Ro

Ri

Ro

Ri

rdrSin

CSin

drrP

22 2

T =( )

2

222

ia RR

Sin

c =

Substitute for C

T =( )

( )ioia

RR

Fa

Sin

RR

2

2 22


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20

( )

Sin

mDDD

Sin

io

2

Fa.

22

Fa +=

Where Dm = Mean diameter2

io DDDm

+=

If the clutch is engaged when one member is stationary and other rotating, then the cone faces

will tend to slide on each other in the direction of an element of the cone. This will resist theengagement and then force

Axial load ( ) cosFaFa += Sini

Force widthSin

DDb io

2

=

Outer diameter SinbDmDo +=

Inner diameter SinbDmDi =

Problem:A cone clutch is to transmit 7.5 KW at 600 rpm. The face width is 50mm, mean diameter is

300mm and the face angle 15. Assuming co efficient of friction as 0.2, determine the axial

force necessary to hold the clutch parts together and the normal pressure on the cone surface.

Given P = 7.5 KW, N = 600 rpm, b = 50mm,

Dm = 300mm, = 15 = 0.2

Solution:

mmNN

P=

=

= 119375

6002

10605.7

2

1060T

66

Torque transmitted

Sin

Dm

2

FaT =

152300Fa2.0119375

Sin=

NFa 88.1029=

Also SinPbDm=Fa ---------- Equation 13.37 DDH

1550P30088.1029 Sin=

2/0844.0P mmN=


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21

A friction cone clutch has to transmit a torque of 200 N-m at 1440 rpm. The longer diameter

of the cone is 350mm. The cone pitch angle is 6.25the force width is 65mm. the coefficientof friction is 0.2. Determine i) the axial force required to transmit the torque. ii) The averagenormal pressure on the contact surface when maximum torque is transmitted.

Data T = 200 N-m, 2 10

5

N-mm N = 1440 rpmDo = 350, = 6.25 b = 65mm, = 0.2

SolutionI) Axial force

Outer diameter SinbDD mo +=

25.665350 SinDm +=

mmDm 92.342=

Torque transmitted

Sin

DmFa

2

1T =

25.6

Fa2.0

2

1102 5

Sin

Dm=

Axial force required N934.634Fa =

ii) Average normal pressure

SinPbDm=Fa

P25.66592.342.934.634 = Sin

Average Normal pressure

2/0833.0P mmN=

An engine developing 30 KW at 1250 rpm is filted with a cone clutch. The cone face

angle of 12.5. The mean diameter is 400 rpm = 0.3 and the normal pressure is not to

exceed 0.08 N / mm2. Design the clutch

Date: P = 30KW, N = 1250 rpm, = 12.5, Dm = 400mm, = 0.3, P = 0.08 N/ mm2

Solution

i) Torque transmitted

12502

106030

2

1060T

66

=

=

N

p


22/77

22

mmN = 229200T

ii) Axial force Fa

5.12

4003.0

2

1,229200

2

1T

Sin

Fa

Sin

DFa m =

Fa= 826.8 N

Dimensions

SinPbDm=Fa

5.1208.0400.8.826 Sinb =

mmb 38=

Inner diameter mmSinSinbDD mm 3925.1238400 ===

Outer diameter mmSinSinbDD mm 4085.1238400 =+=+=


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23

BRAKES

A brake is defined as a machine element used to control the motion by absorbing kinetic energyof a moving body or by absorbing potential energy of the objects being lowered by hoists,

elevators, etc. The obsorbed energy appears as heat energy which should be transferred to cooling

fluid such as water or surrounding air. The difference between a clutch and a brake is that whereas

in the former both the members to be engaged are in motion, the brake connects a moving member toa stationary member.

Block or shoe brake

A single-block brake is shown in fig. It consists of a short shoe which may be rigidly mounted

or pivoted to a lever. The block is pressed against the rotating wheel by an effort Fat one end of the

lever. The other end of the lever is pivoted on a fixed fulcrum O. The friclional force produced bythe block on the wheel will retard the rotation of the wheel. This type of brake is commonly used in

railway trains. When the brake is applied, the lever with the block can be considered as a free body

in equilibrium under the action of the following forces.

1. Applied force Fat the end of the lever.

2. Normal reaction Fn between the shoe and the wheel.

3. Frictional or tangential braking force F between the shoe and the wheel.4. Pin reaction.

Let F= Operating force

M1 = Torque on the wheelr = Radius of the wheel

2 = Angle of contact surface of the block

= Coefficient of friction

F = Tangential braking force =r

Mt

Fn = Normal force =

F

a= Distance between the fulcrum pin and the center of the shoe

b = Distance between the center of the shoe to the end of lever where theeffort is applied

c = Distance between the fulcrum pin and the line of action of F g

Consider the following three cases;

(i) Line of action of tangential force F0 passes through fulcrum


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24

Taking moments about O,

aF

aFbaF n

==+ )(

F

FnQ

Actuating force)( ba

aFF

+=

In this case the actuating force is the same whether the direction of tangential force is towards or

away from the fulcrum.

(ii) Line of action of tangential force F is in between the center of the drum and the fulcrum

(a)Direction ofF is towards the fulcurm :


aFcFbaF nn ==+ )(

cFaF

cFaFbaF n

==+ )(

FFnQ

Actuating force

+=

a

c

ba

aFF

1

)(

(b)Direction ofF is away from the fulcrum :


25/77

25


cFaF

cFaFbaF n

+==+ )(

FFnQ

Actuating force

+= a

c

ba

aFF

1

)(

(iii) Line of action of tangential force F is above the center of the drum and the fulcrum:

(a) Direction ofF is towards the fulcrum :


cFaF

cFaFbaF n

+==+ )(

F

FnQ

Actuating force

+

=ac

baaFF

1

)(

(b)Direction ofF is away from the fulcrum :


aaF

aFcFbaF

==++ )(

FFnQ

cFaF

baF

=+ )(

Actuating force

+= a

c

ba

aF

F

1

)(


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26

Note: If the direction of F0 is towards the fulcrum, use the clockwise rotation formula and if

the direction of F& is away from the fulcrum, use counter clockwise formula from the

data handbook.

When the angle of contact between the block and the wheel is less than 60, we assume that the

normal pressure is uniform between them. But when the angle of contact 2 is more than 60,we assume that the unit pressure normal to the surface of contact is less at the ends than at thecenter and the wear in the direction of applied force is uniform. In such case we employ the

equivalent coefficient of friction ', which is given by.

Equivalent coefficient of friction

2sin2

sin41

+=

Where = Actual coefficient of friction

= Semi block angle

For the given value of. The value of

2sin2

sin4

+can be found by using the chart given in

The brake is self -energizing when the friction force helps to apply the brake. If this effect is

great enough to apply the brake with zero external force, the brake is called self-locking i.e.,the brake is self locking when the applied force Fis zero or negative.

Normal pressure on the shoe psin2Pr wr

F

shoeofareaojected

loadNormalP n==

Where w = Width of the shoe

Example: The block type hand brake shown in fig. 3.1 la has a face width of 45 mm. The friction

material permits a maximum pressure of 0.6 MPa and a coefficient of friction of 0.24. Determine;1. Effort F, 2. Maximum torque, 3. Heat generated if the speed of the drum is 100 rpm and the

brake is applied for 5 sec. at full capacity to bring the shaft to stop.

Data: w = 45 mm, p = 0.6 MPa = 0.6 N/mm2, = 0.24, 2 = 90, = 45e, d = 300 mm,


27/77

27

r = 150 mm, n = 100 rpm

Solution:

Since 2 > 60, equivalent coefficient of friction

2sin2

sin41

+

=

264.0

90sin180

90

45sin424.0 =

+

=

Allowable pressuresin2 wr

Fp n=

i.e., 45sin1504526.0 =

nF

Normal force Fn = 5727.56 N

Tangential force F= ' Fn = 0.264 x 5727.56 = 1512.1 NThe various forces acting on the shoe are shown in fig. 3.11b.From the figure, a = 200 mm, b = 300 mm, c = 0

The tangential force F, passes through the fulcurm.

Effort)(' ba

aFF

+=

N1.2291)300200(264.0

2001.1512=

+

=

Torque on the drumM1 = F r = 1512.1 x 150 = 226815 N-mm = 226.815 N-m

Power absorbed sec/375.2375.29550

100815.226

9550

1 kJkWnM

N ==

==

Heat generated during 5 sec = 5 x 2.375 = 11.875 kJ

Example:

A 400 mm radius brake drum contacts a single shoe as shown in fjg.3.12a, and sustains 200 N-mtorque at 500 rpm. For a coefficient of friction 0.25, determine:

1. Normal force on the shoe.

2. Required force Fto apply the brake for clockwise rotation.

3. Required force Fto apply the brake for counter clockwise rotation.

4.The dimension c required to make the brake self-locking, assuming the other dimensionsremains the same.

5.

Heat generated.


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28

Data; r = 400 mm, M1- 200 N-m = 200 x 103

N-mm, n = 500 rpm, =0.25,a = 350 mm, i = 1000 mm, b = l - a = 1000 - 350 = 650 mm, c = 40 mm

Solution:

Tangential friction force Nr

MF 500

400

10200 31 =

==

Normal force on the NF

F 200025.0

500===

From the figure, the tangential force F lies between the fulcrum and the center of drum.. When the

force Facts towards the fulcrum (clockwise rotation),

Actuating force

+= a

c

ba

aF

F

10

N680350

40

25.0

1

650350

350500=

=

+

=

For anticlockwise rotation of the drum, the tangential force Fs acts away from the fulcrum as

shown in figure.

Actuating force

+

+=

a

c

ba

aFF

10

N720350

40

25.0

1

650350

350500=

+=

+

=

When the drum rotates in clockwise direction, self locking will occur. For self locking effort

F 0.


29/77

29

01

.,.0

+=

a

c

ba

aFei

or

ac

a

c ,

1

mmc 120025.0

350

Heat generated vFvApH ncg ==

W98.1071100060

500800200025.0 =

=

skJkW /472.10472.10 ==

Example: The layout of a brake to be rated at 250 N-m at 600 rpm is shown in figure. The drumdiameter is 200 mm and the angle of contact of each shoe is 120. The coefficient of friction may be

assumed as 0.3 Determine.

1. Spring force Frequired to set the brake.

2. Width of the shoe if the value of pv is 2 N-m/mm2-sec

Data: Mt = 250 N-m = 250 x 103

N-mm, n = 600 rpm, d = 200 mm, r = 100 mm,

2 =120, 0=60, = 0.3, pv = 2 N-m/mm2-sec


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30

Solution:

Since 26 > 60, equivalent coefficient of friction

22

4'

Sin

Sin

+=

351.0

120sin180

120

603.0 =

++

=

Sin

The various forces acting on the shoes are shown in figure.

Left hand shoe:

For left hand shoe, a = 150 mm, b = 150 mm, c = mm502

100100 =

The tangential force F1 lies above the center of the drum and fulcrum, and acting away from

the fulcrum (counter clockwise).

Spring force

+=

a

c

ba

aFF

101

+

=

150

50

351.0

1

150150

15001FF

Tangential force Fgi= 0.795 F

Right hand shoe:

The tangential force F2 lies below the center of the drum and the fulcrum, and acting towards

the fulcrum (clockwise).

Spring force

+=

a

c

ba

aFF

102

+

= 150

50

351.0

1

150150

15002F

F

Tangential force F02 = 0.6285 F

Torque rFFM )( 01011 +=

i.e., 100)6285.0795.0(10250 3 += FF

Spring force F = 1756.2 N

The maximum load occurs on left hand shoe.


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31

Tangential force F01 = 0.795 x 1756.2 = 1396.2 N

Normal force on left hand shoe NF

F 8.3977351.0

2.13960101 ===

Surface velocity of drum sec/283.6100060

600200

100060m

dnv =

=

=

By data pv = 2 N-m/mm2

sec

Normal pressure2/3183.0

283.6

22mmN

vp ===

Also pressuresin2

1

wrFp n=

1.e.,60sin1002

8.39773183.0

=

wr

Width of the shoe w = 72.15mm


32/77

32

Band brakes

A band brake consists of a band, generally made of metal, and embracing a part of thecircumference of the drum. The braking action is obtained by tightenting the band. The difference in

the tensions at each end of the band determines the torque capacity.

Simple band brakes: When one end of the band is connected to the fixed fulcrum, then the bandbrake is called simple band brake as shown in figure

Let T, = Tight side tension in NT2 = Slack side tension in N

= Angle of lap in radians

= Coefficient of friction

D = Diameter of brake drum in mmM1 = Torque on the drum in N-mm

Ratio of tensions

eT

T

=2

1

Braking forceD

MTTF t

221 ==

Clockwise rotation of the drum:

Taking moments about the fulcrum O,

bTaF = 2

Force at the end of levera

bTF 2=

Braking force )1(22221 ===

eTTeTTTF

= e

T

T

2

1Q

:. Slack side tension 12 =

e

F

T


33/77

33

Substitute the value of T2 in equation (1), we get

=

1

12ea

bTF

Counter clockwise rotation of the drum:

For counter clockwise rotation of the drum, the tensions T1 and T2 will exchange their places and

for the same braking torque, a larger force Fis required to operate the brake.


bTaF = 1

Force at the end of levera

bTF 1=

Braking force )1

1(11

121 e

Te

TTTTF ===

= e

T

T

2

1Q

Tight side tension1

2

=

e

eFT

Substitute the value of T1, in equation (i), we get

=

1

e

e

a

bFF

This type of brake does not have any self-energizing and self-locking properties.

Thickness of the band h = 0.005 D

Width of banddh

Tw

1=

where d is the allowable tensile stress in the band.


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34

Differential band brakes:

In differential band brake the two ends of the band are attached to pins on the lever at a distance ofbl and b2 from the pivot pin as shown in figure. It is to be noted that when b2 > b1, the force Fmust

act upwards in order to apply the brake. When b2 < b1, the force Fmust act downwards to apply the

brake.

Clockwise rotation of the drum:


aFbTbT += 1221

FabTbeT = 1222 )( 21

eTT =Q

Braking force )1(22221 ==

eTTeTTTF )( 21eTT =Q

Slack side tension1

2

=

e

FT

Substitute the value ofT2in equation (1), we get

=

1

12

e

beb

a

FF

For the brake to be self locking, the force F at the end of the lever must be equal to zero ornegative.

01

12

=

e

beb

a

F

The condition for self-locking is 12 beb


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35

Counter clockwise rotation of the drum:


aFbTbT += 1122

FaeTbT = 222 )( 21eTT =Q

Ora

ebbTF

)( 122

=

Braking force )1(22221 ===

eTTeTTTF )( 21eTT =Q

Slack side tension1

2

=

e

FT

Substitute the value ofT2in equation (1), we get

=

1

12

e

ebb

a

FF

For self-brake locking, the force Fmust be zero or negative.

i.e., 01

12

e

ebb

a

F

The condition for self-locking isebb 12

If b1 = b2 = b, the brake is called two way brake and is shown in figure. This type of brake can be

used in either direction of rotation with same effort.

Force at the end of lever

+=

1

1

e

e

a

bFF


36/77

36

Example:

A simple band brake operates on a drum 0.6 m in diameter rotating at 200 rpm. The coefficient of

friction is 0.25 and the angle of contact of the band is 270. One end of the band is fastened to afixed pin and the other end to 125 mm from the fixed pin. The brake arm is 750 mm long.

(i)

What is the minimum pull necessary at the end of the brake arm to .stop the wheel if 35kW is being absorbed? What is the direction of rotation for minimum pull?

(ii)Find the width of 2.4 mm thick steel band if the maximum tensile stress is not to exceed 55

N/mm2.

Data: D = 0.6 m = 600 mm, n = 200 rpm, p = 0.25, 0= 270, a = 750 mm,

b = 125 mm, N = 35kW, h = 2.4 mm, d = 55 N/mm2

Solution:

Frictional torquen

NM

95501 =

mmNmN ==

=3

1025.167125.1671200

359550

Braking for NDM

RMF 83.5570

6001025.167122

3

11 ====

Ratio of tensions 248.318027025.0

2

1 === ee

T

T

Clockwise rotation:

Force

=

11

ea

bFF


37/77

37

N02.4131248.3

1

750

12583.5570=

=

Counter clockwise rotation:

Force

=

1

e

e

a

bFF

N5.13411248.3

3248

750

12583.5570=

=

Therefore the minimum pull F= 413.02 N

The direction of rotation is clockwise.

Braking force )1

1(11

121 e

Te

TTTTF ===

i.e.,

=

248.3

1183.5570 1T

Tight side tension T1 = 8048.96N

Width of banddh

Tw

1=

mmmm 6298.60554.2

96.8048==

=

Problem:

In a simple band brake, the length of lever is 440 mm. The tight end of the band is attached to the

fulcrum of the lever and the slack end to a pin 50 mm from the fulcrum. The diameter of thebrake drum is 1 m and the arc of contact is 300. The coefficient of friction between the band and

the drum is 0.35. The brake drum is attached to a hoisting drum of diameter 0.65 m that sustains

a load of 20 kN. Determine;

1. Force required at the end of lever to just support the load.

2. Required force when the direction of rotation is reversed.

3. Width of stee! band if the tensile stress is limited to 50 N/mnr.


38/77

38

Data: a = 440 mm, b = 50 mm, D= 1 m = 1000 mm, Db = 0.65 m = 650 mm,

= 300, = 0.35, W = 20kN = 20x 10

3

N, d = 50N/mm

2

Solution:

Torque on hoisting drum2

1

b

b

WDRWM ==

mmN=

= 63

105.62

6501020

The hoisting drum and the brake drum are mounted on same shaft.

Torque on brake drum M1 = 6.5 x 106N-mm

Braking forD

M

R

MF 11

2==

N130001000

105.62 6=

=


2

1 === eeT

T

Clockwise rotation:


=

1

1

ea

bFF

N38.281125.6

1

440

5013000=

=

Counter clockwise rotation:


39/77

39

Force

=

1

e

e

a

bFF

N66.1758125.6

25.6

440

5013000=

=

Thickness of band h = 0.005 D

= 0.005 x 1000 = 5 mm

Braking force )1

1(11

121 eT

e

TTTTF ===

i.e.,

=

25.6

1113000 1T


Width of banddh

Tw

1=

mmmm 629.61505

2.15476==

=

Problem:

A band brake shown in figure uses a V-belt. The pitch diameter of the V-grooved pulley is 400

mm. The groove angle is 45 and the coefficient of friction is 0.3. Determine the power rating.

Data: D - 400 mm, 2a = 45, =22.5, -0.3, F= 100N, b = 400 mm,

a = 750 mm, = 180 = rad, n = 400 rpm


40/77

40

Solution:

Ratio of tensions for V-belt, SineT

T /

2

1 =

738.115.22/3.0 == Sine

For clockwise rotation,

Force

=

1

1

ea

bFF

i.e.,

=

1738.11

11

750

400100

F

Braking force F = 2013.4N

Torque on the drum2

1

DFRFM ==

mNmmN ==

= 68.4024026802

4004.2013

Power rating9550

1 nMN =

kW87.169550

40068.402 ==

Problem

Figure shows a two way band brake. It is so designed that it can operate equally well in both

clockwise and counter clockwise rotation of the brake drum. The diameter of the drum is 400 mmand the coefficient of friction between the band and the drum is 0.3. The angle of contact of band

brake is 270 and the torque absorbed in the band brake is 400 N-m. Calculate;

1.Force F required at the end of the lever.

2.Width of the band if the allowable stress in the band is 70 MPa.


41/77

41

Data: D = 400 mm, = 0.3, = 270, Mt = 400 N-m = 400 x 103

N-mm,

d = 70 MPa = 70 N/mm2, b = bl = b2 = 50mm, a = 1000mm

Solution:

Braking force NDMTTF t 2000

4001040022

3

21 ====


2

1 === eeT

T


+=

1

1

e

e

a

bFF

N28.1641112.411112.4

1000502000 =

+=

Band thickness h = 0.005 D

= 0.005 x 400 = 2 mm

Braking force )1

1(121 e

TTTF ==

= e

T

T

12

Q

i.e., 20001112.4

1

11 =

T


Width of banddh

Tw

1=

mmmm 2088.18702

84.2642==

=


42/77

42

Design of Belts, Ropes and Chains -- U8

INTRODUCTION

Power is transmitted from the prime mover to a machine by means of intermediate

mechanism called drives. This intermediate mechanism known as drives may be belt or chainor gears. Belt is used to transmit motion from one shaft to another shaft with the help of

pulleys, preferably if the centre distance is long. It is not positive drive since there is slip inbelt drive.

Three types of belt drives are commonly used. They are: Flat belt drive

V-belt drive

Rope or circular belt drive

FLAT BELT DRIVE

When the distance between two pulleys is around 10 meters and moderate power is required

then flat belt drive is preferred. This may be arranged in two ways Open belt drive Cross belt drive

When the direction of rotation of both the pulleys are required in the same direction , then we

can use open belt drive; if direction of rotation of pulleys are required in opposite directionthen cross belt is used. The pulleys which drives the belt is known as driver and the pulley

which follows driver is known as driven or follower.

MERITS AND DEMERITS OF FEAT BELT DRIVE

Merits:

Simplicity, low cost, smoothness of operation, ability to absorb shocks, flexibility andefficiency at high speeds.

Protect the driven mechanism against breakage in case of sudden overloads owing tobelt slipping.

Simplicity of care, low maintenance and service.

Possibility to transmit power over a moderately long distance.

Open belt drive


43/77

43

Cross belt drive

Demerits: It is not a positive drive.

Comparatively large size.

Stretching of belt calling for resewing when the centre distance is constant. Not suitable for short centre distance.

Belt joints reduce the life of the belt.

High bearing loads and belt stresses.

Less efficiency due to slip and creep.

Creep in Belts

Consider an open belt drive rotating in clockwise direction as shown in figure. The portion ofthe belt leaving the driven and entering the driver is known as tight side and portion of belt

leaving the driver and entering the driven is known as slack side.

During rotation there is an expansion of belt on tight side and contraction of belt on the slack

side. Due to this uneven expansion and contraction of the belt over the pulleys, there will be a

relative movement of the belt over the pulleys, this phenomenon is known as creep in belts.


44/77

44

Flat belt drive system

Serpentine belt system

Flat belt drive system

3D arrangement of belt drive

Velocity RatioThe ratio of angular velocity of the driver pulley to the angular velocity of the driven pulley

is known as velocity ratio or speed ratio or transmission ratio.

Let d1 = Speed of driver pulley

d2 = Speed of driver pulleyn1 = Speed of driver pulley

n2 = Speed of driver pulley

Neglecting slip and thickness of belt,

Linear speed of belt on driver = Linear speed of belt on driven


45/77

45

Velocity Ratio

The ratio of angular velocity of the driver pulley to the angular velocity of the driven

pulley is known as velocity ratio or speed ratio or transmission ratio.

Let d1 = Speed of driver pulleyd2 = Speed of driver pulley

n1 = Speed of driver pulleyn2 = Speed of driver pulley

Neglecting slip and thickness of belt,

Linear speed of belt on driver = Linear speed of belt on driven

i.e., d1 n1 = d2 n2

Considering the thickness of belt

Slip in BeltsConsider an open belt drive rotating in clockwise direction, this rotation of belt over the

pulleys is assumed to be due to firm frictional grip between the belt and pulleys.

When this frictional grip becomes in sufficient, there is a possibility of forward motion of

driver without carrying belt with it and there is also possibility of belt rotating withoutcarrying the driver pulley with it, this is known as slip in belt.

Therefore slip may be defined as the relative motion between the pulley and the belt in it. Thisreduces velocity ratio and usually expressed as a percentage.

Effect of Slip on Velocity Ratio

Let s1 = Percentage of slip between driver pulley rim and the belt.

s2 = Percentage of slip between the belt and the driven pulley rim.

Linear speed of driver = d1 n1


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46

Material Used for Belt

Belts used for power transmission must be strong, flexible, and durable and must have a

coefficient of friction. The most common belt materials are leather, fabric, rubber, balata,Camels hair and woven cotton.

Length of Open BeltConsider an open belt drive as shown in Figure.

Let, D = diameter of larger pulley

d = diameter of smaller pulleyC = distance between centers of pulley

L = length of belt


47/77

47

Where L and s are in radians.For equal diameter pulleys L = s = radians.

For unequal diameters pulleys, since slip will occur first on the smaller diameter pulley, it isnecessary to consider s while designing the belt.

Length of Cross Belt

Consider a cross-belt drive as shown in FigureLet, D = diameter of larger pulley

d = diameter of smaller pulley

L = Length of belt


48/77

48

Ratio of Belt TensionsConsider a driven pulley rotating in clockwise direction as shown in Figure.

Let, T1 = Tension on tight sideT2 = Tension on slack side

= Angle of lap

RN = Normal ReactionF = Frictional force = RN

Now consider a small elemental portion of the belt PQ subtending an angle at the centre.

The portion of the belt PQ is in equilibrium under the action of the following forces, (i)Tension T at P (ii) Tension T + T at Q (iii) Normal reaction RN (iv) Frictional force F =

RN


49/77

49

Centrifugal Tension

Consider a driver pulley rotating in clockwise direction, because of rotation of pulley therewill be centrifugal force which acts away from the pulley. The tensions created because of

this centrifugal force both on tight and slack side are known as centrifugal tension.

Let, m = Mass of belt per meter lengthv = Velocity in m/sec

TC = Centrifugal tension in N

r = Radius of pulleyFC = Centrifugal force

Consider a small elemental portion of the belt PQ subtending an angle shown in Figure.Now the mass of belt PQ = M = Mass per unit length x Arc length PQ = mrd


50/77

50

Effect of Centrifugal Tension on Ratio ofTensions

Ratio of belt tension considering the effect of centrifugal tension is

1

2

C

C

T T

T T


51/77

51

Initial Tension

The motion of the belt with the pulleys is assumed to be due to firm frictional grip betweenthe belt and pulleys surface. To increase this grip the belt is mounted on the pulleys with

some tension when the pulleys are stationary.

The tension provided in the belt while mounting on the pulley is Initial tension and is

represented by T0. Since in actual practice the belt is not perfectly elastic, C.G.Barth hasgiven the relation as

Design Procedure for Flat Belt Drive


52/77

52

3. Centrifugal stress

4. Capacity

Calculate e

LL and e

ss and take the smaller value as the capacity. If the coefficient of

friction is same for both pulleys then find only es since it is smaller than e

L


53/77

53

Example

A belt is required to transmit 18.5 kW from a pulley of 1.2 m diameter running at 250

rpm to another pulley which runs at 500 rpm.The distance between the centers of pulleys is

2.7 m. The following data refer to an open belt drive, = 0.25. Safe working stress forleather is 1.75 N/mm2. Thickness of belt = 10mm. Determine the width and length of belt

taking centrifugal tension into account. Also find the initial tension in the belt and absolute

power that can be transmitted by this belt and the speed at which this can be transmitted.


54/77

54


55/77

55

Area of cross section of belt

Also A = b x t1503.18

= b x 10

Therefore, b = 150.318 mm

Standard width b = 152 mm

2

2

Total given power 18.51503.18 mm

Power transmitted per mm area 0.01231= = =


56/77

56

V- BELT DRIVE

Introduction

When the distance between the shafts is less, then V-belts are preferred. These are endless and

of trapezoidal cross section as shown in Figure. It consists of central layer of fabric andmoulded in rubber or rubber like compound. This assembly is enclosed in an elastic wearing

cover. The belt will have contact at the two sides of the groove in the pulley. The wedging

action between the belt and groove will increase the coefficient of friction making the drive apositive one.


57/77

57

Types of V - belts

a) Endless V-belt

b) Assembling of V-beltc) Narrow V-belt

d) Wide V-belt with cogs

e) Narrow V-belt with cogsf) Double angle V-belt

g) Great angle V-belt

h) Vee-bandi) Pol-rib belt

Advantagesof V-belt over Flat belt

Advantages:

Compact and give high velocity ratio. Provides shock absorption between driver and driven shafts.

Positive and reliable drive.

Because of wedging action in the grooves, loss of power due to slip is less. There is no joints problem as the drive is of endless type.

Disadvantages: Initial cost is more as the fabrication of pulleys with V-grooves are complicated.

Cannot be used when the center distance is large.

Improper belt tensioning and mismatching of belt results in reduction in service life.


58/77

58

Ratio of belt tensions for V-belt or rope drive

V-Belt drive or rope drive runs in a V-grooved pulley as discussed earlier. The cross-sectionof V-belt is shown in Figure.

Let, 2 = angle of groove

RN = normal reaction between each side of grooveand the corresponding side of the belt strip PQ

From Figure Resolving Forces Vertically,

R = RN sin + RN sin = 2 RN sin

Total frictional force = RN + RN = 2 RN

In case of V-belt or rope, there are two normal reactions as

shown in Figure, so that the radial reaction R is 2 RN sin

and the total frictional force = 2( RN) = 2 RN

Consider a short length PQ of belt subtending angle at

the center of the pulley as shown in Figure.


59/77

59

Let, R = radial reaction between the belt lengthPQ and the pulley rim = 2RN sin

RN= Normal reaction between the belt length PQ and

the pulley rim.T = Tension on slackside of the shot strip PQT+T = Tension on tight side of short strip PQ

T= Difference in tension due to friction between the

length PQ and the surface pulley rim

= Coefficient of friction between the belt and pulleysurface

= effective coefficient of friction = /sin

The strip PQ will be in equilibrium (figure) under the action of four forces T, T+T, 2RNand R where 2 RN is the frictional force which is opposing the motion


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DESIGN PROCEDURE FOR V-BELT

1. Selection of belt c/s

Equivalent pitch diameter of smaller pulley

de = dp.Fb

Where , dp = d1Fb = smaller diameter factor

Based on de select the c/s of belt

If d is not given them based on power select the c/s of beltand diameter d1

2. Velocity

3. Power capacity

Based on the cross-section selected, calculated the power capacity N* from the formulas.

4. Number of V belts

N= Total power transmitted in kW

N*= power capacity

Fa= Service factorIf the condition is not given then assume medium duty and 10-16 hours duty per day.

or

12 cos2

D d

C

=

1180 - 2 sin2

D d

C

=


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Correction factor for angle of contact = Fd

(If the type is not given select V - V - Belts)

Therefore, Number of belts = i

Select a V-belt drive to transmit 10 kW of power froma pulley of 200 mm diameter mountedon an electric motor running at 720 rpm to another pulley mounted on compressor running at

200 rpm. The service is heavy duty varying from 10 hours to 14 hours per day and centre

distance between centre of pulleys is 600 mm.


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v. Number of Belts

The nearest standard value of nominal pitch length for the selected C-cross section belt L =2723 mm

Nominal inside length = 2667 mm

For nominal inside length = 2667 mm, and C-cross section belt, correction factor for length

Fe = 0.94


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ROPE DRIVESIntroduction

When power is to be transmitted over long distances then belts cannot be used due to the

heavy losses in power. In such cases ropes can be used. Ropes are used in elevators, mine

hoists, cranes, oil well drilling, aerial conveyors, tramways, haulage devices, lifts andsuspension bridges etc. two types of ropes are commonly used. They are fiber ropes and

metallic ropes. Fiber ropes are made of Manila, hemp, cotton, jute, nylon, coir etc., and are

normally used for transmitting power. Metallic ropes are made of steel, aluminium. alloys,copper, bronze or stainless steel and are mainly used in elevator, mine hoists, cranes, oil well

drilling, aerial conveyors, haulage devices and suspension bridges.

Hoisting tackle (Block and Tackle Mechanism)It consists of two pulley blocks one above the other. Each block has a series of sheavesmounted side by side on the same axle. The ropes used in hoisting tackle are

Cotton ropes Hemp ropes and

Manila ropes

The pulleys are manufactured in two designs i.e. fixed pulley and movable pulley.

Pulley system

A pulley system is a combination of several movable and fixed pulleys or sheaves.

The system can be used for a gain in force or for a gain in speed. Hoisting devices employpulleys for a gain in force predominantly. Pulley systems for a gain in forces are designed

with the rope running off a fixed pulley and with the rope running off a movable pulley.

Consider a hoisting tackle (block and tackle mechanism) as shown in fig.


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STEEL WIRE ROPESA wire rope is made up of stands and a strand is made up of one or more layers of wires as

shown in fig. the number of strands in a rope denotes the number of groups of wires that are

laid over the central core. For example a 6 19 construction means that the rope has 6 strands

and each strand is composed of 19(12/6/1) wires. The central part of the wire rope is calledthe core and may be of fiber, wire, plastic, paper or asbestos. The fiber core is very flexible

and very suitable for all conditions.

The points to be considered while selecting a wire rope are

Strength Abrasion resistance

Flexibility

Resistance of crushing

Fatigue strength Corrosion resistance.

Ropes having wire core are stronger than those having fiber core. Flexibility in rope ismore desirable when the number of bends in the rope is too many.

DESIGN PROCEDURE FOR WIRE ROPELet d=Diameter of rope

D=Diameter of sheave

H= Depth of mine or height of buildingW= total load

WR= Weight of rope

dw= Diameter of wireA= Area of c/s of rope

Pb= Bending load in the rope

Fa= allowable pull in the ropeFu= Ultimate of breaking load of rope

n= Factor of safety

Ws= Starting load


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problemSelect a wire rope to lift a load of 10kN through a height of 600m from a mine. The weight of

bucket is 2.5kN. the load should attain a maximum speed of 50m/min in 2 seconds.

Solution:

From table select the most commonly used type of rope i.e. 619

From table for 619 rope Fu= 500.8 d2 N where d in mmWeight per meter length = 36.310-3 d2 N/m where d in mm

Wire diameter dw = 0.063 d, mm

Area of c/s A = 0.38 d2, mm2Sheave diameter D= 45 d, mm

From table for 600 m depthF.O.S = n = 7

1. Total load

W= Load to be lifted + weight of skip = 10000+ 2500= 12500N


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CHAIN DRIVE

IntroductionChain is used to transmit motion from one shaft to another shaft with the help of

sprockets. Chain drives maintain a positive speed ratio between driving and driven

components, so tension on the slack side is considered is as zero. They are generally used forthe transmission of power in cycles, motor vehicles, agricultural machinery, road rollers etc.

Merits and demerits of chain drives Merits1. Chain drives are positive drives and can have high efficiency when operating under ideal

conditions.

2. It can de used for both relatively long centre distances.

3. Less load on shafts and compact in size as compared to belt drive.

Demerits

1. Relatively high production cost and noisy operation.2. Chain drives require more amounts of servicing and maintenance as compared to belt

drives.


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Velocity ratio in chain drive

Let n1= speed of driver sprocket in rpmn2 = speed of driven sprocket in rpm

z1= number of teeth on drivers sprocket

z2 = number of teeth on driven sprocket

Therefore Velocity ratio n1/n2= z1/z2

Chains for power transmissionThe different types of chain used for power transmission are:

i. Block chain ii. Roller chain iii. Inverted-tooth chain or silent chain.

ROLLER CHAIN

It consists of two rows of outer and inner plates. The outer row of plates in known as pin link

or coupling link whereas the inner row of plates is called roller link. A Pin passes through the

bush which is secured in the holes of the inner pair of links and is riveted to the outer pair oflinks as shown in Fig. Each bush is surrounded by a roller. The rollers run freely on the

bushes and the bushes turn freely on the pins.

A roller chain is extremely strong and simple in construction. It gives good service under

severe conditions. To avoid longer sprocket diameter, multi-row-roller chains or chains withmultiple strand width are used. Theoretically, the power capacity multistrand chain is equal to

the capacity of the single chain multiplied by the number of strand, but actually it is reduced

by 10 percent.

Inverted tooth chain or silent chain

It is as shown Fig. these chains are not exactly silent but these are much smoother and quieter

in action than a roller chain. These chains are made up of flat steel stamping, which make it

easy to built up any width desired. The links are so shaped that they engage directly withsprocket teeth. In design, the silent chains are more complex than brush roller types, more

expensive and require more careful maintenance.


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Chordal action

When a chain passes over a sprocket, it moves as a series of chords instead of a continuous

arc as in the case of a belt drive. Thus the center line of a chain is not a uniform radius. Whenthe driving sprocket moves at a constant speed, the driven sprocket rotates at a varying speed

due to the continually varying radius of chain line. This variation in sped ranges from

Where

n1= Speed of the driving sprocket in rpmd1 Pitch circle diameter of the driving sprocket in mmz1 = number of teeth on driving sprockets.

It is clear from above that for the same pitch, the variation in speed and articulationangle decreases, if the number of teeth in sprocket is increased. The average speed of thesprocket as

given by

Where p = pitch of the chain in mm and z = number of teeth in sprocket. This chordal action

of the chain is shown in Fig.


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DESIGN PROCEDURE FOR ROLLER CHAIN

Let p = Pitch

d1 = diameter of smaller sprocketd2 = diameter of larger sprocketn1 = speed of smaller sprocket

n2 = speed of larger sprocket

z1 = number of teeth on smaller sprocketz2 = Number of teeth on larger sprocket

L = Length of chain in pitches

C = Center diameter

Cp = Center distance in pitches

1. Pitch of chain

Where p in mm, and n1= speed of smaller sprocket

Select standard nearest value of pitch from Table

Chain numberBreaking load Fu

Measuring load w

2. Number of teeth on the sprockets

From Table for the given ratio select the number of teeth on the smaller sprocket (z1)Since

Number of teeth on larger sprocket = z2

3. Pitch diameters


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4. Velocity

5. Required pull

6. Allowable pull

7. Number of strands in a chain

8. Check for actual factor of safety

= pitch diameter of smaller sprocket

= pitch of larger sprocket


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6. Allowable pull

10. length of chain

11. Correct centre distance

Select a roller chain drive to transmit power of 10 kw from a shaft rotating at 750 rpm to

another shaft to run at 450 rpm. The distance between the shaft centers could be taken as 35pitches.

Data:

N= 10 kw; n1 = 750 rpm; n2 = 450 rpm; C = 35 pitches


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clutches and brakes

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