Lecture -Belt Drives

BELT DRIVES

INTRODUCTION:

The belts are used to transmit power from one shaft to another by means of pulleys

which rotate at the same speed or at different speeds. The amount of power

transmitted depends upon the following factors:

The velocity of the belt.

The tension under which the belt is placed on the pulleys.

The arc of contact between the belt and the smaller pulley.

The conditions under which the belt is used.

The belt drives are usually classified into the following three groups:

Light drives. These are used to transmit small powers at belt speeds up to about 10

m/s as in agricultural machines and small machine tools.

Medium drives. These are used to transmit medium powers at belt speeds over 10

m/s but up to 22 m/s, as in machine tools.

Heavy drives. These are used to transmit large powers at belt speeds above 22 m/s

as in compressors and generators.

TYPES OF BELTS:

There are three main types of belts:

Flat belt.

The flat belt as shown in Fig. .1 (a), is mostly used in the factories and workshops,

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

when the two pulleys are not more than 8 meters apart.

V- belt.

The V-belt as shown in Fig. .1 (b), is mostly used in the factories and workshops,

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.

Circular belt.

The circular belt or rope as shown in Fig. .1 (c) is mostly used in the factories and

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

another, when the two pulleys are more than 8 meters apart.

Fig. .1: Types of belts

Figure .2 shows non-reversing and reversing belt drives. (a) Non-reversing open

belt. (b) Reversing crossed belt. Crossed belts must be separated to prevent rubbing

if high-friction materials are used.

(c) Reversing open-belt drive.

Fig. .2: Non-reversing and reversing belt drives

Figure .3 shows belt geometry. (a) Open belt. (b) Crossed belt.

Where:

d = diameter of small pulley

D= diameter of large pulley

C = center distance

= angle of contact

Fig.3: Belt geometry

For open belt;

(

)

(

)

[ ( ) ]

( )

For crossed belt

(

)

[ ( ) ]

( )

Let;

n1: speed of small pulley in r.p.s (Driver pulley)

n1: speed of large pulley in r.p.s (Driven pulley)

The linear velocity of the belt on the driving pulley is,

& the linear velocity of the belt on the driven pulley is

When there is no slip, then ;

or,

POWER TRANSMITTED BY A BELT:

Fig. 4.4 shows the driving pulley (or driver) A and the driven pulley (or follower)

B. The driving pulley pulls the belt from one side and delivers it to the other side. It

is thus obvious that the tension on the former side (i.e. tight side) will be greater

than the latter side (i.e. slack side).

Let;

: Tensions in the tight side of the belt

: Tensions in the slack side of the belt

N: Number of belts ( N=1 for flat belt)

Then the power transmitted is;

( ) ( )

Fig. .4

The relation between the tension in tight side and in slack side is;

( )

Where;

( newton)

A: cross-sectional area of the belt ( m2)

: Allowable normal stress for belt material ( Pa)

: Coefficient of friction between belt and pulley

: Effective coefficient of friction between belt and pulley

: Groove angle (for flat belt )

: Centrifugal tension

( newton)

V: Linear velocity of belt in ( m/s)

m : Mass of belt per unit length in (kg/m)

Example 1:

A leather belt 9 mm 250 mm is used to drive a cast iron pulley 900 mm in

diameter at 336 r.p.m. If the active arc on the smaller pulley is 120 and the stress

in tight side is 2 MPa, find the power capacity of the belt. The density of leather

may be taken as 980 kg/m3, and the coefficient of friction of leather on cast iron is

0.35.

Solution:

Given;

t = 9 mm ; b = 250 mm ; D = 900 mm ; n2 = 336 r.p.m ;

d = 120 ; = 2 MPa ; = 980 kg/m3 ; = 0.35

Linear velocity of the belt;

Cross-sectional area of the belt;

Tensions in the tight side of the belt;

Centrifugal tension

( )

Contact angle;

Effective coefficient of friction between belt and pulley;

The tension in slack side can be found from;

( )

( )

Then the power capacity of the belt is;

( )

( )

Example 2:

Flat belt is required to transmit 30 kW from a pulley of 1.5 m effective diameter

running at 300 r.p.m. The angle of contact is spread over11/24 of the

circumference. The coefficient of friction between the belt and pulley surface is

0.3. Determine, taking centrifugal tension into account, width of the belt required.

It is given that the belt thickness is 9 mm, density of its material is

1100 kg / m3 and the related permissible working stress is 2.5 MPa.

Solution;

Given;

Power= 30KW; d = 1.5m ; n1 = 300 r.p.m ; t=9 mm

d = (11/24)*360= 165o ; = 2.5 MPa ; = 1100 kg/m3 ;

= 0.3




( )

Centrifugal tension

( ) ( )

Contact angle;



( )

( )

( )

The power capacity of the belt is;

( )

( )

Example 3:

Flat belt 100 mm wide and 10 mm thick is transmitting power at

1000 m/min. The tension on tight side is 4 times the tension on the slack side. If

the safe permissible stress on the belt section in 1.6 MPa, calculate the maximum

power that can be transmitted at this speed. Assume density of the leather as 1000

kg/m3.

Solution:

Given;

b= 100 mm; t = 10 mm ; V = 1000 m/min;

= 2.5 MPa ; = 1000 kg/m3 ;

( )

( )

Example 4:

A compressor, requiring 20 kW, is to run at about 250 r.p.m. The drive is by V-

belts from an electric motor running at 750 r.p.m. The diameter of the pulley on the

compressor is 1 metre while the centre distance between the pulleys is limited to

1.75 metre. Determine the number of V-belts required to transmit the power if

each belt has a cross-sectional area of 375 mm2, density 1000 kg / m

3 and an

allowable tensile stress of 2.5 MPa. The groove angle of the pulleys is 35. The

coefficient of friction between the belt and the pulley is 0.25.

Solution;

Given :

P = 20 kW; nD = 250 r.p.m. ; nd = 750 r.p.m. ; D = 1 m ;

C = 1.75 m ; A = 375 mm2; = 1000 kg /m3 = 2.5 MPa ;

2 = 35 or = 17.5 ; = 0.25




Centrifugal tension

( )

Contact angle;

(

) (

)



( )

( )

The power capacity of the belt is;

( )

( )

Documents

Lecture -Belt Drives