J3010 Unit 6

BELTING

J 3010/6/1

BELTING

OBJECTIVES

General Objective : To understand and apply the concept of belting

Specific Objectives : At the end of this unit you will be able to:

state the difference between open and close belt.

explain that power transmitted by the flat belt and V belt.

explain that ratio tension for flat and V belt.

calculate power transmitted by the belt with consider

centrifugal force.

UNIT 6

BELTING

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6.0 INTRODUCTION

The pulley rotating shaft is called driver. The pulley intended to rotate is

known, as follower or driven. When the driver rotates, it carries the belt that

grip between its surface and the belt. The belt, in turn, carries the driven

pulley which starts rotating. The grip between the pulley and the belt is

obtained by friction, which arises from pressure between the belt and the

pulley.

(a) Types of belts

The flat belt is mostly used in the factories and work shops,

where a moderate amount of power is to be transmitted, from one

pulley to another, when the two pulleys are not more than 10 m a part.

The V-belt is mostly used in the factories and work shops

where a great amount of power is to be transmitted, from one pulley

to another, when the two pulleys are very near to each other.

INPUT

In factories, the power has to be

transmitted from one shaft to another,

then belt driving between pulleys on

the shafts may be used.

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6.1 LENGTH OF AN OPEN BELT DRIVE

A

B

C

F

D

E

d

Fig.6.1 Open belt drive

O1 and O2 = Centers of two pulleys

r1 and r2 = radius of the larger and smaller pulleys

d = Distance between O1 and O2

L = Total length of the belt.

Angle AO1O2 = α1

Angle BO2C = α2

Angle AFE = θ1 (radian)

Angle BCD = θ2 (radian)

The types of belts:-

a. flat belt

b. V belt

O1 O2

K

r1 r2

α1 α2

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We know that the length of the belt,

L = Arc AFE + ED + Arc DCB + BA

Cos α1 = d

rr 21

α1 = cos-1

d

rr 21

θ1 = 2π - 2α1

= 2 (π - α1) (radian)

θ2 = α1 = 2 α2 (radian)

Arc AFE = r1θ1

Arc DCB = r2θ2

And ED = BA = KO2

Sin α1 = d

KO2

KO2 = d Sin α1

Finally the total of length of belt,


= r1θ1 + d Sin α1 + r2θ2 + d Sin α1

= r1θ1 + r2θ2 + 2d Sin α1

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Example 6.1

Find the length of belt necessary to drive a pulley of 480 cm diameter running

parallel at a distance of 12 meter from the driving pulley of diameter 80 cm. This

system is an open belt drive.

Solution 6.1

A

B

F C

D

E

d

Fig.6.2 Open belt drive





Angle AO1O2 = α1

Angle BO2C = α2



Radius of smaller pulley = 80/2 = 40 cm.

Radius of larger pulley = 480/2 = 240 cm.

Distance between the pulleys, d = 12m = 1 200 cm

We know that the length of belt is,


Cos α1 = d

rr 21 = 1200

40240

O1

O2

r1

r2

α1

K

α2

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α1 = cos-1

1200

40240

= cos-1

0.16667

= 80º

α1 = 1.396 radian

θ1 = 2π - 2α1

= 2 (π - 1.396 ) (radian)

= 3.491 radian

θ2 = 2α1 = 2 α2 (radian)

= 2 (1.396 )

= 2.792 radian

Arc AFE = r1θ1

= 240 x 3.491

= 837.84 cm

Arc DCB = r2θ2

= 40 x 2.792

= 111.68 cm

And ED = BA = KO2

Sin 80º = d

KO2

KO2 = d Sin α1 = 1 200 x 0.98481

= 1181.77 m

Finally the total of belt length is



= r1θ1 + r2θ2 + 2d Sin α1

= 240 x 3.491 + 40 x 2.792 +2 x 1181.77

= 837.84 + 111.68 + 2363.54

= 3313.06 cm

= 33.13 m

BELTING

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6.2 LENGTH OF CLOSE BELT DRIVE

K

A

D

C

F

B

E

d

Fig.6.3 Cross belt drive





Angle AO1O2 = α1

Angle DO2O1 = α2



We know that the length of the belt is


Cos α1 = d

rr 21

α1 = cos-1

d

rr 21

θ1 = 2π - 2α1

= 2 (π - α1) (radian)

θ1 = θ2

O1 O2

r1

r2

α1 α2

BELTING

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Arc AFE = r1θ1

Arc DCB = r2θ2

And ED = BA = KO2

Sin α1 = d

KO2

KO2 = d Sin α1

Finally the total of length belt,



= r1θ1 + r2θ2 + 2d Sin α1

Example 6.2

Find the length of belt for a cross belt drive system. The diameter of

the drive pulley is 480 cm which running parallel at a distance of 12

meter from the driving pulley which has a diameter of 80 cm.

Solution 6.2

K

A

D

C

F

B

E

d

Fig.6.4 Cross belt drive

O1 O2

r1 r2

α1 α2

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d = distance between O1 and O2


Angle AO1O2 = α1

Angle DO2O1 = α2



Radius of smaller pulley = 80/2 = 40 cm.

Radius of larger pulley = 480/2 = 240 cm.

Distance between the pulleys, d = 12m = 1 200 cm

We know that the length of the belt,


Cos α1 = d

rr 21 =1200

40240

α1 = cos-1

1200

40240

= cos-1

0.23333

= 76.5º

α1 = 1.335 radian

θ1 = 2π - 2α1

= 2 (π - α1) (radian)

= 2 (π - 1.335) (radian)

= 3.613 radian

θ1 = θ2

Arc AFE = r1θ1

= 240 x 3.613

= 867.12 cm

Arc DCB = r2θ2

= 40 x 3.613

= 144.52 cm

And

ED = BA = KO2

Sin α1 = d

KO2

KO2 = d Sin 76.5º =1 200 x 0.9724

ED = BA = KO2 = 1166.88 cm

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Finally the total of length belt,


= r1θ1 + d Sinα1 + r2θ2 + d Sin α1

= r1θ1 + r2θ2 + 2d Sin α1

= 240 x 3.613 + 240 x 3.613 + 2 x 1 200 Sin 76.5º

= 867.12 + 144.52 + 2333.76

= 3345.4 cm

L = 33.45 m

BELTING

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Activity 6A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT INPUT…!

6.1 Two pulleys, one with a 450 mm diameter and the other with a 200 mm

diameter are on parallel shaft of 1.95 m apart and are connected by a cross

belt. Find the length of the belt required and the angle of contact between

the belt and each pulley.

6.2 It is required to drive a shaft B at 620 rpm by means of a belt from a parallel

shaft A. A pulley of 30 cm diameter on shaft A runs at 240 rpm. Determine

the size of pulley on the shaft B.

6.3 Find the length of belt necessary to drive a pulley of 1.4 m diameter running

parallel at a distance of 1.7 meter from the driving pulley of diameter 0.5 m.

It is connected by an open belt.

1

2

2

1

d

d

N

N

N1 = speed diver in rpm

N2 = speed follower in rpm

d1 ,d2 = diameter pulley driver and

follower.

BELTING

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Feedback to Activity 6A

Have you tried the questions????? If “YES”, check your answers now

6.1 4.975 m; 199 or 3.473 radian.

6.2. 11.6 cm

6.3 6.51 m

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6.3 POWER TRANSMITTED BY A BELT

Fig. 6.5, shows the driving pulley (i.e., driver) A and the follower B.

The driving pulley pulls the belt from one side, and delivers the same to the

other.The maximum tension in the tight side will be greater than that

slack side.

Fig. 6.5

Torque exerted on the driving pulley

= ( T1 - T2) r1

Torque exerted on the driven or follower

= ( T1 - T2) r2

Power transmitted = Force x distance

= ( T1 - T2) v

INPUT

Power = (T1 –T2) v

Where, T1 =Tension in tight side

in Newton.

T2 = Tension in slack side.

V = velocity of belt

BELTING

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Example 6.3

The tension in the two sides of the belt are 100 N and 80 N respectively. If

the speed of the belt is 75 meters per second. Find the power transmitted by

the belt.

Solution 6.3

Given,

Tension in tight side,

T1 = 100 N

Tension in slack side,

T2 = 80 N

Velocity of belt, v = 75 m/s

Let P = Power transmitted by the belt

Using the relation,

Power = (T1 –T2) v

= ( 100 – 80) 75

= 1500 watt

= 1.5 kw.

BELTING

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Activity 6B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT INPUT…!

6.4. Find the tension in the tight side, if the tension in slack side 150 N. The speed

of the belt is 58 meters per second and the power transmitted by this

belt is 2 kw.

6.5 Diameter of driver is 50 mm. If the driver transmits 8 kw when it is rotating

at 300 rev/ min. Calculate velocity of driver.

6.6 The tension in the two sides of the belt are 300 N and 180 N respectively. If

the speed of the belt is 50 meters per second, find the initial tension and

power transmitted by the belt.

6.7 A 100 mm diameter pulley is belt-driven from a 400 mm diameter pulley.

The 400 mm pulley rotates at 480 rev/min. The number of rev/sec of the

100 mm diameter pulley is

A. 2 B. 32 C. 120 D. 1920

T0 = 2

21 TT

Where, T0 = Initial tension in the belt.

BELTING

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Feedback to Activity 6B


6.4 115.5 N

6.5 0.785 m/s

6.6 6 000 watt or 6 kw; 240 N

6.7 B. 32

BELTING

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6.4 RATIO OF TENSIONS.

Fig. 6.6

Consider a driven pulley rotating in the clockwise direction as in fig 6.6.

Let T1 = Tension in the belt on the tight side.

T2 = Tension in the belt on the slack side.

ө = An angle of contact in radians (i.e , angle subtended by the arc

AB, along which the belt touches the pulley, at the centre)

Now consider a small position of the belt PQ, subtending an angle at the

centre of the pulley as shown fig 6.6. The belt PQ is in equilibrium under the

following force:

i. Tension T in the belt at P.

ii. Tension T + T in the belt at Q.

iii. Normal reaction R, and

iv. Frictional force F = x R

Where is the coefficient of friction between the belt and pulley.

Resolving all the forces horizontally and equating the same,

INPUT

BELTING

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R = ( T + T) sin 2

+ T sin

2

(6.1)

Since the is very small, therefore, substituting,

sin 2

=

2

in equation 6.1

R = ( T + T) 2

+T

2

= 2

T +

2

T +

2

.T

= T. ( neglecting 2

T ) (6.2)

Now resolving the forces vertically,

x R = (T + T) cos 2

- T cos

2

(6.3)

Since the angle . is very small, therefore substituting cos 2

= 1 in

equation 6.3 or, x R = T + T – T = T

R =

T (6.4)

Equating the values of R from equation 6.2 and 6.4,

Or T. =

T

Or T

T = .

Integrating both sides from A to B,

0

1

2

T

TT

T

or loge 2

1

T

T =

2

1

T

T = e

ration of tension for flat belt

2

1

T

T = e

/sin ration of tension for V belt

where = half angle of groove

BELTING

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Example 6.4

Find the power transmitted by a belt running over a pulley of 60 cm diameter

at 200 rpm. The coefficient of friction between the pulley is 0.25, angle of lap

160 and maximum tension in the belt is 250 KN.

Solution 6.4

Given,

Diameter of pulley, d = 60 cm = 0.6 m

Speed of pulley N = 200 rpm

Speed of belt, v = 60

dN =

60

2006.0 xx = 2 = 6.28 m/ sec

Coefficient of friction,

μ = 0.25

Angle of contact, = 160 = 180

160 x = 2.7926 radian.

Maximum tension in the belt,

T1 = 250 kN

Let P = power transmitted by the belt.

Using the relation,

2

1

T

T = e

= e

0.25 x 2.7926

2

1

T

T = 2.01

T2 = 01.2

1T =

01.2

250 = 124.38 kN

Now using the relation,

P = (T1 –T2 ) v

= ( 250 - 124.38 ) 6.28

= 788.89 watt

P = 0.79 kw

BELTING

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6.5 CENTRIFUGAL TENSION.

Fig.6.7

The tension caused by centrifugal force is called centrifugal tension. At lower

speeds the centrifugal tension is very small, but at higher speeds its effect is

considerable, and thus should be taken into account.

Consider a small portion PQ of the belt subtending an angle d at the centre

of the pulley as show in fig 6.7.

Let M = mass of the belt per unit length,

V = linear velocity of the belt,

r = radius of the pulley over which the belt runs,

Tc = Centrifugal tension acting tangentially at P and Q.

Length of the belt PQ,

= r d

Mass of the belt PQ,

M = M r d

We know that centrifugal force,

= M v2/r

INPUT

BELTING

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Centrifugal force of the belt PQ

= r

vdrxM 2.

= 2vdxM

The centrifugal tension (Tc) acting tangentially at P and Q keeps the belt in

equilibrium.

Now resolving the force (i.e, centrifugal force and centrifugal tension)

horizontally and equating the same,

2 Tc sin 2

d = 2vdxM

since angle d is very small, therefore, substituting

sin 2

d =

2

d

2 Tc 2

d = 2vdxM

Tc = 2vdxM / d

Tc = M v2

eTT

TT

c

c

2

1 ratio tension for flat belt.

sin/

2

1 eTT

TT

c

c

ratio tension for V belt

6.5.1 Condition for the transmission of maximum power

The maximum power,

When T c = ⅓T1

It shows that the power transmitted is maximum ⅓ of the maximum tension

is absorbed as centrifugal tension.

The velocity of belt for maximum transmission of power may be

obtained from equation T1 = 3Tc = M v2.

BELTING

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v2 =

M

T13

v =

M

T

3

1

Example 6.5

An open belt drive connects two pulleys 1.2 m and 0.5 m diameter, on

parallel shafts 3.6 m apart. The belt has a mass of 0.9 kg/m length and the

maximum tension in it is not exceed 2 kN. The larger pulley runs at 200

rev/min. Calculate the torque on each of the two shafts and the power

transmitted. Coefficient of friction is 0.3 and angle of lap on the smaller

pulley is 168° ( 2.947 radian ).

Solution 6.5

Given

Diameter of larger pulley,

d1 = 1.2 m

Radius of the larger pulley,

r1 = 0.6

Diameter of smaller pulley,

d2 =0.5 m

Radius of the smaller pulley,

r2 = 0.25 m

Distance between two shaft,

D = 3.6 m

Mass of the belt per meter length,

M = 0.9 kg/m

Maximum Tension, T1 = 2 kN = 2 000 N

Speed of the larger pulley

= 200 rpm

Velocity of the belt,

V = 60

2002.1 xx = 12.57 m/s

T c = Mv2

= 0.9 x 12.572

= 142.2 N

BELTING

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e µθ

= e 0.3 x 2.947

= 2.421

Using the relation,

eTT

TT

c

c

2

1

2.142

2.1422000

2

T = 2.421

(T2 – 142.2) 2.421 = 1857.8

(T2 – 142.2) = 421.2

8.1857 = 767.37

T2 = 767.37 + 142.2

T2 = 909.57 N

We know that the torque on the larger pulley shaft (TL),

TL = ( T1 – T2) r1

= ( 2000 – 909.57) 0.6

= 654.26 Nm.

Torque on the smaller pulley shaft (Ts),

Ts = ( T1 – T2) r2

= (2000 – 909.57) 0.25

= 272.61 Nm.

Power transmitted, P = ( T1 – T2)v

= ( 2000 – 909.57) 12.57

= 13706.71 watt

= 13.71 kw

BELTING

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Activity 6C

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT

INPUT…!

6.8 Two pulleys, one with a 450 mm diameter and the other with a 200 mm

diameter are on parallel of shafts 1.95 m apart. What power can be

transmitted by the belt when the larger pulley rotates at 200 rev/min, if the

maximum permissible tension in the belt is 1 kN and the coefficient of

friction between the belt and pulley is 0.25. Angle of contact between the belt

and larger pulley is 3.477 radian.

6.9 Find the power transmitted by a V drive from the following data:

Angle of contact = 84º

Pulley groove angle = 45º

Coefficient of friction = 0.25

Mass of belt per meter length = 0.472 kg/m

Permissible tension = 139 N

Velocity of V belt = 12.57 m/s.

6.10 An open belt drive connects two pulleys 1.2 m and 0.6 m, on parallel shafts 3

m apart. The belt has a mass 0f 0.56 kg/m and maximum tension is 1.5 kN. The

driver pulley runs at 300 rev/min. Calculate the inertial tension, power

transmitted and maximum power. The coefficient of friction between the belt

and the pulley surface is 0.3.

BELTING

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Feedback to Activity 6C


6.8 2 740 watt or 2.74 kw

6.9 591 watt.

6.10 1074.5 N ; 8.0 kw ; 17.54 kw.

BELTING

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SELF-ASSESSMENT 6

You are approaching success. Try all the questions in this self-assessment section

and check your answers with those given in the Feedback on Self-Assessment 6

given on the next page. If you face any problems, discuss it with your lecturer.

Good luck.

1. A pulley is driven by a flat belt running at speed of 600 m/min. The

coefficient of friction between the pulley and the belt is 0.3 and the angle lap

is 160°. If the maximum tension in the belt is 700 N, find the power

transmitted by a belt.

2. A leather belt, 125 mm wide and 6 mm thick, transmits power from a pulley

of 750 mm diameter which run at 500 rpm. The angle of lap is 150° and

µ = 0.3. If the mass of 1 m3 of leather is 1 Mg and the stress in the belt does

not exceed 2.75 MN/m2, find the maximum power that can be transmitted.

3. A flat belt, 8 mm thick and 100 mm wide transmits power between two

pulleys, running at 1 600 m/min. The mass of the belt is 0.9 kg/m length. The

angle of lap in the smaller pulley is 165° and the coefficient of friction

between the belt and pulleys is 0.3. If the maximum permissible stress in the

belt is 2 MN/m2, find

(i) Maximum power transmitted; and

(ii) Initial tension in the belt.

4. The moment on a pulley, which produces the rotation of the pulley is called:

A. Momentum B. Torque C. Work D. Energy

5. If T1 and T2 are the tension in the tight and slack sides of a belt and θ is the

angle of contact between the belt and pulley. Coefficient of friction is μ, then

the ratio of driving tension will be:

A. eT

T

1

2 B. eT

T

2

1 C. 2

1

T

T D.

2

110log

T

T

BELTING

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Feedback to Self-Assessment 6

Have you tried the questions????? If “YES”, check your answers now.

1. 3.974 kw ( see example 6.4)

2. 18.97 kw

3. (i) 14.281 kw

(ii) 1.3221 kN

4. B. Torque

5. B. eT

T

2

1

CONGRATULATIONS!!!!…..

May success be with you

always….

J3010 Unit 6

Education

Transcript of J3010 Unit 6