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Maximum power transmitted by belt drive

13/10/2021 Client: muhammad11 Deadline: 2 Day

POWER Transmission Questions- Mechanical Principles

Please could you answer all questions on page 3+4...

c, (i)(ii)

d, (i) (ii) (iii)

2, (i) (ii) (iii) (iv) (v) (vi) (vii)

Due December 10th 2018.

I have uploaded pages to help with the questions.

TOPIC TITLE : POWER TRANSMISSION

LESSON 1 : FLAT BELT DRIVES

MP - 3 - 1

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

retrieval system, or transmitted, in any form or by any means, electronic, mechanical,

photocopying, recording or otherwise without the prior permission

of the Copyright owner.

This book is sold subject to the condition that it shall not, by way of trade or

otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's

prior consent in any form of binding or cover other than that in which it is

published and without a similar condition including this

condition being imposed on the subsequent purchaser.

________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

Belt drive is a system using pulleys, which transmits power by the friction

between the belt and the wheel. To enable this, the belt must be tensioned to

grip the wheel. Belt drives are quiet, clean and relatively economical. The

belt is continuous and passes over pulleys which are keyed to their respective

shafts.

Belt drives are most often used to produce speed reduction between a motor

and the machine being driven, e.g. a motor driving an air compressor. Other

applications vary from small tape drives in video or audio recorders, normally

flat tape, to large multi-belt systems used in heavy industrial equipment. This

lesson only concentrates on flat belt drives.

The effectiveness of a belt drive depends upon the friction between the belt

and pulley and on the angle of contact between the belt and the pulley surface.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• list the factors that govern the power transmitted by a flat belt drive

• determine maximum power transmitted

• apply belt tension equations for flat belt drives.

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

BELT DRIVES ________________________________________________________________________________________

Before we study belt drives in detail, consider the following:

FIG. 1(a) FIG. 1(b)

FIGURE 1(a) shows a stationary pulley with a belt wrapped around it, each of

the two ends of the belt being anchored to a vertical surface. A horizontal

force F is applied to the pulley, putting the belt in tension, with equal forces

at the upper and lower lengths of belt. The force F = .

FIGURE 1(b) shows the same pulley and belt but the pulley is now being

gradually turned clockwise, still maintaining the force F. Due to friction

between the belt and pulley, the effect is to further tighten the lower length of

belt and slacken the upper length of belt to give values of F1 (tighter) and F2 (slacker). Hence F1 > F2, but for equilibrium F = F1 + F2. On rotating the

pulley further, a point is reached when the pulley slips on the belt and forces F1 and F2 will then have reached their respective maximum and minimum values.

The above simplified explanation forms the basis for belt drive calculations.

F F

2 2 +F

F 2

F F

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

POWER TRANSMITTED BY A BELT DRIVE ________________________________________________________________________________________

FIG. 2

FIGURE 2 shows a portion of a continuous belt in contact with a drive pulley

of effective radius r. The tension F1 is greater than the tension F2 due to

friction between the belt and the pulley.

F1 is referred to as the tension in the 'tight side' of the belt.

F2 is referred to as the tension in the 'slack side' of the belt.

The pulley drives the belt in a clockwise direction and, provided there is no slip,

the peripheral speed of the pulley will be equal to the linear speed of the belt.

Nett clockwise torque on pulley =

F r F r

T F

1 2

–

––

–

F r

P T

P F F r

1 2

( )

= ( )

Now power

Hence

Driver

Teesside University Open Learning (Engineering)

The difference in tensions (F1 – F2) may be called the effective tangential

force,

where ω = angular velocity of pulley (rad s–1) r = effective radius of pulley (m)

F1 = tension in tight side of belt (N)

F2 = tension in slack side of belt (N).

Also note that since

θ represents the angle of lap, i.e. angle of contact between

belt and pulley.

FIG. 3(a)

FIG. 3(b)

Driven

ω1

Driver

ω2

‘Slack’

‘Tight’

r θ

Pulley

Belt

where linear speed of belt (m s 1v = – ).

v r

P F F v

= ( )

ω then:

1 2–

Teesside University Open Learning (Engineering)

The angular contact between the belt and the pulley is called the angle of lap

(see FIGURE 3(a)) and the drive will be more effective if the angle of lap is as

large as possible. For this reason it is better to have the slack side of the belt

across the top of the pulleys wherever possible since the sag of the belt will

tend to increase the angle of lap. See FIGURE 3(b).

________________________________________________________________________________________

CROSSED-BELT DRIVE ________________________________________________________________________________________

FIG. 4

A crossed belt obviously increases the angle of lap and hence increases the

effectiveness of the drive but it also results in the pulleys rotating in opposite

directions. For these reasons such drives were often used for flat belts or ropes

but are seldom seen nowadays.

Example1

The tight and slack side tensions of a belt drive are 500 N and 180 N

respectively and the pulley diameter is 0.48 m. Calculate the power

transmitted when the speed is 360 rev min–1.

Driven

ω1

Driver

ω2

Slack

Tight

Teesside University Open Learning (Engineering)

Solution

Example 2

A motor transmits 4 kW to a belt-driven pulley, 140 mm effective diameter,

when the pulley speed is 1200 rev min–1. Determine the tensions in the tight

and slack side of the belt if they are in the ratio 3 to 1.

Solution

First of all we need to determine the nett torque:

P T

= × =

ω 1200 2 60

π π rad s

4000 W

1–

Angular speed

rad s

Power

ω = ×

360 2

π 60

π –

–P F FF r2

500 180 0 24 12

2895

( )

= ( ) × ×

– . π

Power 2.895 kkW

Teesside University Open Learning (Engineering)

It is now possible to determine the difference in tensions (F1 – F2), i.e. the

effective tangential force.

Now the ratio of tensions is 3 : 1, i.e.

F F

1 2

∴ =

Torque N m

= ( ) =

∴ =

F F r

1 2

31 83

– .

where radius mr

F F

= ( )

31 83 0 07

4541 2

. .

– .77 N ........................................... 1( )

Torque

N m

T P=

4000 40

31 83

Teesside University Open Learning (Engineering)

Hence, by substitution in (1):

Tensions in the belt are 682.1 N and 227.4 N.

3 454 7

2 454 7

227 4

2 2

F F

– .

∴ =

∴

F F1 2454 7

454 7 227 4

682 1

= +

. .

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

BELT TENSION EQUATIONS ________________________________________________________________________________________

If the belt velocity is constant, the power transmitted will depend upon the

difference in the belt tensions (F1 – F2). Hence if the ratio can be made

greater then more power will be transmitted. However, when the ratio

reaches a certain value, known as the limiting ratio, the belt will slip on

the pulley.

It thus seems likely that depends upon:

(a) the coefficient of friction (µ) between the belt and the pulley

(b) the angle of lap (θ) of the belt round the pulley,

and that an increase in either of the above values will lead to an increase in the

ratio .

It can be shown that the limiting ratio of tensions is given by:

where: F1 = tension in tight side of belt (N)

F2 = tension in slack side of belt (N)

e = the constant 2.718 which is the base of natural logarithms

µ = coefficient of friction between belt and pulley θ = angle of lap of belt round pulley (radians).

F e1

= µθ

F 1

⎛ ⎝⎜

⎞ ⎠⎟

F 1

⎛ ⎝⎜

⎞ ⎠⎟

F 1

⎛ ⎝⎜

⎞ ⎠⎟

F 1

⎛ ⎝⎜

⎞ ⎠⎟

F 1

⎛ ⎝⎜

⎞ ⎠⎟

Teesside University Open Learning (Engineering)

The proof of this formula is given in the Appendix at the end of this lesson and

should be studied. However, it will not be necessary for you to learn this proof

in order to complete any of the assignments.

Note that where an open belt connects two pulleys of different diameters the

angle of lap on the smaller pulley will be less than that on the larger pulley.

From FIGURE 5 it can be seen that θs is less than θl.

FIG. 5

Calculations, then, will be based on the smaller pulley, which normally is the

drive wheel.

The formula for = eµθ does not include the effect of centrifugal force which

has a tendency to push the belt away from the pulley. It is possible to allow for

this effect but it is not serious unless high speeds are involved, and in any case

it is beyond the scope of our studies at this stage.

F 1

θs θl

Teesside University Open Learning (Engineering)

Since the power transmitted P = (F1 – F2)v and from = e µθ we have

This is the maximum power that can be transmitted at a given speed with no

slip occurring.

Example 3

Calculate the maximum power that may be transmitted by a flat belt driving a

pulley 360 mm diameter which rotates at 180 rev min–1. The maximum belt

tension is 500 N and the angle of lap is 145°. Coefficient of friction between

belt and pulley is 0.35.

P F F v

F F

e v

vF e

= ( )

= ⎛ ⎝⎜

⎞ ⎠⎟

= ( )

1 2

1 1

–

– –

µθ

F 1

Teesside University Open Learning (Engineering)

Solution

Using

Example 4

A flat belt drive is required to transmit 5 kW at a speed of 240 rev min–1 with

the smaller pulley 300 mm effective diameter. The angle of lap between belt

and pulley is 150° and the coefficient of friction between belt and pulley is 0.3.

Determine the magnitude of the tensions in the tight and slack side of the belt.

Maximum possible power W= 997

P vF e

r F e

= ( )

= × × ×

0 18 2 180

–

µθ

µθω

π 60

00 1

997

0 35 2 5307× ( )

×– – . .e

Angle of lap 145

145 360

radians

2.5307 ra

= °

= ×

2π

ddians

Tight-side tension N= =F1 500

Teesside University Open Learning (Engineering)

Solution

= ×

∴ =

240 2 60

500

π rad s

Power

1–

P T

T P

00 8

198 9

1 2

Torque N m

= ( )

∴ =

–

T F F r

F F T

998 9 0 15

13261 2

. .

–F F = N ...........................................

Angle of lap

150 3

( )

=θ 660

2 618

. rad

Teesside University Open Learning (Engineering)

Substitute in (1):

This value of F1 may be calculated using the relationship F1 = 2.193F2

Tensions in the tight and slack sides are 2437 N and 1111 N respectively.

i.e.

F1 2 193 1111

2436 4

= ×

2 193 1326

1 193 1326

1326 1 193

111

2 2

. –

F F

= 11

1326

1326 1111

2437

1 2

∴ = +

= +

F F

F e

0 3 2 618

0 7854

2 193

∴

µθ

. .

11 22 193= . F

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

NOTES ________________________________________________________________________________________

....................................................................................................................................................

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. Determine the difference in tensions for a belt drive which transmits

800 watts when the belt speed is 240 m min–1.

2. A belt-driven pulley of diameter 420 mm and rotating at 240 rev min–1 is

to transmit 1.5 kW. If the tension in the tight side of the belt is 540 N

calculate the tension in the slack side.

3. The maximum permissible force in a belt material is 80 N per cm of belt

width. When transmitting 4 kW of power from a pulley 200 mm in

effective diameter the tension in the tight side of the belt is to be twice the

tension in the slack side. Determine the necessary width of belt required

to the nearest cm. The pulley speed is 900 rev min–1.

4. Calculate the maximum power which can be transmitted by a flat-belt

drive with an angle of lap of 135°, belt speed 18 m s–1 and coefficient of

friction between belt and pulley of 0.3 if maximum belt tension is 250 N.

5. An open flat-belt drive connects two pulleys each 360 mm effective

diameter with a coefficient of friction between belt and pulley surface of

0.4. Calculate the tight and slack side belt tensions if the drive transmits

5 kW when the pulley speed is 450 rev min–1.

6. A flat belt drive is to transmit 8 kW across two pulleys each 500 mm

diameter and rotating at 420 rev min–1. The coefficient of friction

between the belt and the pulley surface is 0.38. If the belt material is

manufactured in various widths in increments of 1 cm and a maximum

tension of 180 N cm–1 of belt width is permitted, calculate the width of a

suitable belt.

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

NOTES ________________________________________________________________________________________

....................................................................................................................................................

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

2. ω

= ×

∴ =

240 2 60

1500 8

rad s

Torq

1–

P T

T P

uue 59.68 N m=

Power

m m

P F F v

F F P

= ( )

∴ =

1 2

800

240

–

iin

m s

–

∴ =

240 60

800 4

200

1 2F F

ffference in tensions 200 N=

Teesside University Open Learning (Engineering)

3. Angular speed

rads

Power

ω = ×

900 2 60

30π –

P ==

∴ =

T P

ω Torque

N m

Now

4000

42 44

30π

T F F r

F F T

= ( )

∴ =

1 2

–

T F F r

F F T

= ( )

∴ =

1 2

59 68 0 21

284 2

–

. .

Now N

540

540 284 2

∴ =

– .

nnsion in slack side 255.8 N=

Teesside University Open Learning (Engineering)

Maximum tension = twice minimum tension:

Substituting into equation (1)

In a practical situation, the width of belt used would be the next available

size larger than 10.61 cm, i.e. 11 cm or 12 cm width.

∴ =

F F

1 1

0 5 424 4

– . .

. .

8848 8. N tension in tight side

belt width

∴ required 848.8 80

∴ = F F2 10 5.

∴ =

42 44 0 1

424 41 2

– . N .................F F .......................... 1( )

Teesside University Open Learning (Engineering)

4. Angle of lap 135

rad

= °

∴ = ×

θ 135 360

2 3562

. iians

Ratio of tensions F

F e

0 3 2 3562

µθ

. .

0 7069

2 0276

250

( ) =

∴

Now max N

P F F v

1 2

250 2 0276

123 3

250 123

= ( )

–

Power

..3 18

2281

( ) ×

watts

Maximum power 2.281 kW

Teesside University Open Learning (Engineering)

Now since pulleys are equal in diameter then angle of lap θ = 180° = π radians.

F e

0 4

1 2566

µθ

Pulley speed

rad s

Power

= ×

∴

450 2 60

π –

Torque

N m

T P

T F F r

= ( )

∴

5000 15

106 1

1 2

–

N .

F F T

F F

1 2

106 1 0 18

589 5

–

– .

∴ =

= .................................... 1( )

Teesside University Open Learning (Engineering)

substitute this in equation (1)

Tensions in tight and slack sides of belt at 824 N and 234.5 N respectively.

3 5136 589 5

2 5136 589 5

2 2

. – .

. .

F F

∴ =

∴

∴ = +

589 5 2 5136

234 5

589 5 234 5

824

. .

F F

1 2

3 5136

∴ =

Teesside University Open Learning (Engineering)

θ = ° =180 π radians (since the pulleys are of eequal diameter)

0.38µ =

F e1

= µθ

= ×

420 2 60

800

π rad s

Power

Torque

1–

P T

T P

00 14

181 9

1 2

T F F r

= ( )

∴

–

N m

Now torque

––

. .

– .

F T

F F

1 2

181 9 0 25

727 6

= N ................................... 1( )

Teesside University Open Learning (Engineering)

substitute in equation (1)

Belt width will be 6 cm.

F F

1 1

3 2996 727 6

0 3031 727 6

0 69

– .

– . .

∴ . 669 727 6

1044

104

Maximum belt tension 44 N

minimum belt width required 1044 180

∴ =

== 5 8. cm

∴ =

F e

F F

0 38

1 1938

2 1

3 2996

22996

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

You should now be familiar with the basic principles involved when designing

flat belt drives. The two factors which govern the difference in tensions are

the angle of lap and the coefficient of friction between the pulley surface and

belt. These factors are only variable to a certain extent since the angle of lap is

virtually decided as soon as the pulley diameters and distance between pulley

centres are decided, unless a crossed belt can be used, but crossed belts cause

alignment problems and so are rarely seen nowadays.

The coefficient of friction value is governed to a large extent by the fact that a

pulley surface has to be fairly smooth or the belt will wear too quickly.

Various semi-liquid compounds are available to help preserve belts and to

make the belt surface 'sticky' so that the coefficient of friction will remain at a

reasonable value.

In modern industry, belt drives are more commonly V-belts which can enlarge

the contact surface area to increase the friction. We shall study these in our

next lesson.

Teesside University Open Learning (Engineering)

________________________________________________________________________________________

APPENDIX ________________________________________________________________________________________

DERIVATION OF FORMULA FOR RATIO OF BELT TENSIONS

(a) Space Diagram (b) Force Vector Diagram

FIG. 6

Note that the difference in tensions (F1 – F2) takes place over the angle of

lap; hence at each small element there is a small element of difference in

tension from F to (F + δF).

Consider equilibrium of the piece of belt ab as shown in FIGURE 6(a) and

draw the force vector diagram FIGURE 6(b).

Let the tensions on either side be F and (F + δF) and let the radial reaction of the pulley on this piece of belt be δR. The difference between (F + δF) and F must be due to friction between the belt and pulley. Hence the maximum value

of δF occurs just as the belt is about to slip and is given by:

δ µδF R= ........................................... 1( )

F + δF

δR δθ

δθ F + δF F

δRω

Teesside University Open Learning (Engineering)

From the triangle of forces in FIGURE 6(b), which involves the three forces

acting on the portion of belt ab, we see that if angle δθ is very small then we can write:

(arc length = radius × angle in radians)

Note that (F + δF) and F and are considered nearly equal for very small angles. Substituting this value of δR into equation (1) we have:

Let ab be made smaller: then δθ approaches zero and the limiting expression becomes:

Integrating both sides between the limits of:

F = F1 (when θ = maximum angle θ) and F = F2 (when θ = 0) we have:

⌠

⌡

⎮ ⎮ ⎮⎮

⌠

⌡

⎮ ⎮ ⎮⎮

[ ] = [ ]

d d

µ θ

d d

F = µ θ

δ µδ

µ δθ

δ µδθ

F R

=or

δ δθR F= ........................................... 2( )

Teesside University Open Learning (Engineering)

θ being the angle of lap in radians µ being the coefficient of friction between belt and pulley.

∴ = ( )

∴ ⎛ ⎝⎜

⎞ ⎠⎟

ln ln

F F

1 2

0– –µ θ

µθ

F e1

= µθ

Teesside University Open Learning (Engineering)

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