POWER Transmission Questions- mechanical Principles

MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : POWER TRANSMISSION

LESSON 2 : V-BELT DRIVES

MP - 3 - 2

© Teesside University 2011

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published and without a similar condition including this

condition being imposed on the subsequent purchaser.

A more efficient transmission of power is achieved by the use of

vee-belts and grooved pulleys. The above photograph, reproduced

by kind permission of Fenner's, shows a multi V-belt drive from an

electric motor to a pump.

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

We studied flat belt drives in the preceding lesson and it should be clear that

more power is transmitted if the 'effective tension', i.e. the difference in tension

between the tight and slack sides of the belt (F1 – F2) can be increased. This

can be achieved by using V-section belts and matching grooved pulleys.

V-belts grip on the side and not the bottom. The wedging effect increases the

reaction force between the pulley and the belt is due to increasing the friction

surface area. Therefore, more power can be transmitted compared with that

when using flat belts.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

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

• apply belt tension equations for V-belt drives.

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________________________________________________________________________________________

V-BELT DRIVES ________________________________________________________________________________________

If a V-belt in conjunction with a grooved pulley is used instead of a flat belt

then the 'normal' reaction between the belt and the pulley is increased by the

'wedging action' of the V-belt in the groove.

FIG. 1

The belt tension sets up a radial force (R) acting towards the centre of the

pulley. FIGURE 1(a) shows a section through the belt and the vector diagram

is shown in FIGURE 1(b).

From FIGURE 1(b) sin

sin

sin

N

N

N

α

α

α

=

∴ =

∴ =

∴

R

R

R R

R R

2

2

2

2RR R

N sin =

α

RN R RN

RN

RN

R

αα

2α

(a) (b)

2

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Now for flat belts, friction force

and for V-belts, friction force

Hence, the belt tension formula for flat belts:

is modified for V-belts and becomes:

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

F2 = tension in slack side of belt (N)

e = the constant 2.718

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

2α = included angle of pulley groove.

For practical reasons the groove angle is usually 40° and hence α is usually 20° when using this formula. It should also be noted that it is of the utmost

importance that the belt is gripped by the sides of the groove and that there is

no contact between the bottom of the belt and the bottom of the groove.

Problems are approached in a similar way to those involving flat belts which

were dealt with in the previous lesson and some examples are given in the

following text.

F

F e1

2

= µθ

αsin

F

F e1

2

= µθ

=

=

µ

µ

R

R

(see previous lesson)

2 (two friction N ssurfaces)

sin since 2

sinN = =

⎛ ⎝⎜

⎞ ⎠⎟

µ α α R

R R

3

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Example 1

Calculate the maximum power transmitted by a V-belt drive with a pulley

150 mm effective diameter, angle of lap 165°, speed of rotation 360 rev min–1,

maximum permissible belt tension 450 N, coefficient of friction 0.3 and pulley

groove angle 40°.

Solution

Let us start by summarising the given information.

Effective pulley diameter 150 mm 0.15 m

angle

= =

oof lap 165 165 360

radians

speed 36

θ = ° = × =

=

2 2 88π .

00 rev min

rad s

max tension

1

1

–

–∴ = × =ω 360 2 60

12 π π

F11 450

40

=

= °

∴ = °

N

pulley groove angle 2

20

R

α

α

aatio of tension sin F

F e

F e

1

2

2

0 3 2 88 450

=

= ×

µθ α

. .

ssin 20

2 526

12 51

°

=

=

e .

.

4

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Example 2

The mean diameter of the driving pulley for a V-belt drive with two belts is

110 mm. The groove angle is 40° and the drive transmits 4.4 kW at

1500 rev min–1. The coefficient of friction between belt and pulley is 0.32 and

the angle of lap is 160°. Determine:

(a) the driving torque

(b) the maximum stress in the belt material if the cross-sectional area of each

belt is 120 mm2.

Solution

(a) Using

where W

and

P T

P

=

=

ω

4400

rad s 1ω = × =1500 2 60

50 π π –

∴ =

=

= ( ) =

450

N

Torque

F

F F r

2

1 2

12 51

35 98

450

.

.

–

– 335 98 0 075

31 05

31 05 12

11

. .

.

.

( ) × =

=

= ×

=

N m

Power Tω

π

771 watts

Max power 1.171 kW=

5

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(b) Since there are 2 belts the torque transmitted by each belt will be 14 N m

(assuming the belts share the torque equally).

We must now determine the ratio between the tensions in order to solve

this equation:

F

F e1

2

= µθ

αsin

For each belt, torque

where

T F F r

r

= ( )1 2–

==

=

=

effective radius of pulley

Now mm

0.0

r 55

555 m

N ...........

F F T

r

F F

1 2

1 2

14 0 055

254 5

–

.

– .

=

=

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

∴ =

=

=

Torque

Torque N m

T P

ω

4400 50

28

π

6

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Substituting this in equation (1)

F F

F F

F

1 1

1 1

1

13 64 254 5

0 0733 254 5

0 9267 2

– .

.

– . .

.

=

=

= 554 5

254 5 0 9267

274 6

1

1

.

. .

.

F

F

=

=

∴

N

tension in tight side of each belt N= 274 6.

Coefficient of friction

angle of lap

µ

θ

=

=

0 32.

1160

160 2 360

2 793

°

= ×

=

π

. radians

Half groove anglle

sin 20

α = °

∴ =

=

∴

× °

20

1

2

0 32 2 793

2 613

F

F e

e

. .

.

Hence

F

F

F F

1

2

2 1

13 64

13 64

=

=

.

.

7

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Example 3

A multiple V-belt drive is required to transmit 30 kW from a pulley 160 mm

diameter rotating at 450 rev min–1. The angle of lap is 170° and the groove

angle is 40°. The coefficient of friction between belt and pulley is 0.35 and the

maximum permissible stress in the belt material is 2.6 MN m–2. If the cross-

sectional area of each belt is 600 mm2 calculate the minimum number of belts

required.

Solution

To solve this problem it will be necessary to determine the total tension in the tight

side of the drive and then divide by the maximum permissible tension per belt.

We will first find the torque and then the difference in tensions (F1 – F2).

P T

T P

P

=

∴ =

=

= ×

=

ω

ω

ω

torque

watts

450 2 60

30 000

1

π

55 π rad s 1–

Now stress force area

N

=

=

=

σ 274 6 120

2 288

.

. mmm

Maximum stress in belt material N

2–

.= 2 288 mm 2–

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We must now determine the ratio between the tensions in order to solve this

expression for F1.

F

F e1

2

0 35

170

170 360

2

=

=

= °

= ×

µθ α

µ

θ

sin

.where

π

== 2 967. radians

T

T

T F F r

=

=

= ( )

∴

30 000 15

636 6

1 2

π

torque N m

Now

.

–

where effective pulley ra

F F T

r

r

1 2– =

= ddius

80 mm

0.08 m

=

=

∴ =

=

F F

F F

1 2

1 2

636 6 0 08

7

– .

.

– 9958 N ......................................... 1( )

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Substitute this in equation (1):

F F

F F

F

F

1 1

1 1

1

20 83 7958

0 048 7958

0 952 7958

– .

– .

.

=

=

=

11

1

7958 0 952

8359

=

=

.

F N

Total tension in tight sside of drive 8359 N

Max permissible tension

=

per belt

where max permissible bel

=

=

σ

σ

A

tt stress

cross-sectional area of belt

Max

A =

ppermissible tension per belt = × × ×2 6 10 600 16. 00

1560

6–

= N

∴ =

=

∴

× °

F

F e

e

F

F

1

2

0 35 2 967

20

3 0363

1

. .

sin

.

22

2 1

20 83

20 83

=

∴ =

.

. F

F

10

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Hence 6 belts will be necessary.

Example 4

A V-belt drive consists of 5 belts with an included angle of 40°. The angle of

lap on the smaller pulley is 160° and the maximum tension per belt is 480 N.

If the coefficient of friction between the belts and pulley is 0.3 determine the

maximum power transmitted for a belt speed of 600 m min–1.

Solution

We will consider a single belt and first calculate the slack side tension (F2).

Angle of lap

rad

θ

θ

= °

∴ = ×

=

160

160 360

2

2 793

π

. iians

Now

sin F

F e

F e

1

2

2

0 32 2 480

=

∴ = ×

µθ α

. .7793

20

2 4499

11 59

sin °

=

=

e .

.

Min number of belts required 8359 1560

=

= 5 36.

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Hence, for 5 belts

maximum power transmitted 5 4386

21930 W

21.

= ×

=

= 993 kW

∴ =

=

=

N

Belt speed 600 m min 1

F

v

2 480

11 59

41 4

.

.

–

==

=

= ( )

∴

600 60

10

1 2

v

P F F v

P

m s

Now power

1–

–

== ( ) ×

=

480 41 4 10

4386

– .

W for one belt

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. A single V-belt drive connects two pulleys each 240 mm diameter rotating

at 240 rev min–1. The coefficient of friction between belt and pulley is

0.36 and the groove angle is 40°. Determine the maximum power which

can be transmitted if the maximum belt tension is 300 N.

2. A single V-belt drive connects two pulleys each 360 mm diameter and

transmits 6 kW at 600 rev min–1. The groove angle is 40° and the

coefficient of friction is 0.35. Calculate the tight and slack side tensions

in the belt.

3. A multiple V-belt is to transmit 100 kW from a pulley rotating at

900 rev min–1. The groove angle is 40° and the coefficient of friction

between belt and pulley is 0.35. The angle of lap is 168° and the effective

pulley diameter is 220 mm. If the maximum permissible tension in each

belt is 900 N determine the minimum number of belts required.

4. A V-belt has a cross-sectional area of 160 mm2 and the maximum

permissible tensile stress in the material is 2.8 MN m–2. The belt is used

on a pulley of 200 mm effective diameter with a groove angle of 40° and

the angle of lap is 165°. The coefficient of friction between the belt and

pulley is 0.32. Determine the maximum power which may be transmitted

for a belt speed of 360 metres min–1.

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. F

F e

F

1

2

1 300

0 36

180

=

=

=

= °

µθ α

µ

θ

sin

.

N

(pulleys of eequal diameter)

radians

20

300

=

= °

∴ =

π

α

F e

2

0..

.

.

.

36

3 3067

2

27 3

300 27 3

11

π sin 20

N

°

=

=

∴ =

=

e

F

NNow

mm

0.12 m

P T

F F r

r

=

= ( )

=

=

ω

ω1 2

120

–

14

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2. Angular velocity

rad s

power

1

ω = ×

=

600 60

2

20

π

π –

PP

P T

T P

=

=

∴ =

=

6000

6000 20

W

Effective torq

ω

ω

π

uue N m

Now

T

T F F r

F F T

r

=

= ( )

∴ =

=

95 49

1 2

1 2

.

–

–

995 49 0 18

530 51 2

. .

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

ω = ×

=

= ( ) × ×

=

240 2 60

8

300 11 0 12 8

871 6

π

π

π

rads 1–

– .

.

P

WW

Maximum power W= 871 6.

15

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where µ = 0.35 θ = 180° = π radians

sin α = sin 20° = 0.342

Substitute in equation (1)

24 905 530 5

23 905 530 5

530

2 2

2

2

. – .

. .

.

F F

F

F

=

=

∴ = 55 23 905

22 2

22 2 530 5

5

1

1

.

.

– . .

=

=

∴ =

N

Now

F

F 330 5 22 2

552 7

. .

.

+

=

=

N

Tight side tension 552.7 N

Sllack side tension 22.2 N=

∴ =

=

=

∴

F

F e

e

F

1

2

0 35

0 342

3 2151

1

24 905

.

.

.

.

π

== 24 905 2. F

F

F e1

2

= µθ

αsin

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3. Power

kW

100 10 W3

P T

T P

P

=

∴ =

=

= ×

= ×

ω

ω

ω

100

900 2 6

π 00

30

100 10 30

3

=

∴ = ×

=

π rad s

Torque 1061 N m

N

1–

T π

oow

and m

T F F r

F F T

r

r

= ( )

∴ =

=

1 2

1 2

110

–

–

mm

0.11 m

N .....

=

∴ =

=

F F

F F

1 2

1 2

1061 0 11

9645

– .

– ........... ..................... 1( )

17

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Substituting this value in equation (1):

F F

F

F

1 1

1

1

0 04978 9645

0 95023 9645

9645 0 950

– .

.

.

=

=

= 223

101501F = N

Now

where 0.35

sin F

F e1

2

168

1

=

=

= °

=

µθ α

µ

θ

θ 668 2 360

2 932

20

1

2

0 35

×

=

= °

∴ = ×

π

.

.

radians

α

F

F e

22 932

3 0004

1

2

2 1

20 09

20 0

.

.

.

.

sin 20

°

=

=

∴ =

e

F

F

F F

99

0 049782 1F F= .

18

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Hence at least 12 belts are required.

4. Force stress area

maximum tension

= ×

∴ =F A1 σ

σ == =

=

∴ = ×

=

2 8 2 8

160

2 8 1601

. .

.

– –MN m N mm

mm

2 2

2A

F

4448

448

0 32

165

165 2

1

2

1

N

N

sin F

F e

F

=

=

=

= °

= ×

µθ α

µ

θ

.

π 3360

2 88=

= °

. radians

20α

Number of belts required maximum tension

max =

iimum tension per belt

=

=

10150 900

11 28.

19

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∴ =

=

=

× ° sin 20

448

448 14

2

0 32 2 88

2 6946

2

F e

e

F

. .

.

..

.

.

–

8

448 14 8

30 3

2

1 2

∴ =

=

=

= ( )

N

Power

a

F

T

F F r

ω

ω

nnd

where linear belt speed in m s

v r

v

=

=

ω

–11

1

1

m min

m s

∴ = ( )

=

=

∴ =

P F F v

v

P

1 2

360

6

44

–

–

–

88 30 3 6– .( ) ×

=The maximum power transmitted 25006 W

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

This lesson concludes our studies of belt drives.

You should now be able to calculate belt tensions and power transmitted by

belt drives given all the necessary information. You should also be able to

calculate the cross-sectional area of a belt and the number of belts required to

transmit a given power. This should enable you to carry out preliminary

design of flat and V-belt drives.

21

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