Copy of Belt and Rope Drive01

7/28/2019 Copy of Belt and Rope Drive01

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BELT

&

ROPE DRIVES


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The power or rotary motion from one

shaft to another at a considerable

distance is, usually transmitted by means

of flat belts, vee belts or ropes, running

over the pulleys.

Figure shows an open belt driveconsisting of pulleys A and B. The pulley

A is keyed to rotating shaft, and is known

as driver. The pulley B is keyed to a shaft,intended to be rotated, and is known as

follower.


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When the driver rotates , it carries the

belt due to grip between its surface andthe belt. The belt, in turn, carries the

driven pulley which starts rotating. The

grip between the pulley and the belt isobtained by friction, which arises from

the pressure between the belt and thepulleys. The friction is increased by

tightening the belt.


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d1 d2

Driver

Follower


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SPEEDS OF PULLEYS CONNECTED BY

BELT DRIVE :

Figure shows the two pulleys of diametersd1 and d2. Let the centers of the two

pulleys be l apart. Let the pulley of diameter

d1 drive the pulley of diameter d2.Former

pulley is called driver and the latter pulley is

called a follower, when the belts are

provided the pulleys are said to be

connected by open belt drive. In this case,

the direction of rotation of the follower is the

same as that of the driver.


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d1

d2

Driver

Follower


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en s es re a e rec on orotation of the follower should be opposite to

that of the driver the pulleys will be connecte

by belts as shown in figure.

In this case the pulleys are said to be

connected by cross belt drive.

Angular velocity of the driver = 1 rad/ secAngular velocity of the Follower = 2 rad/seThere is no slip between driver and belt,

linear velocity of the belt = v

V = 1 d1/2


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Similarly There is no slip between driver

and belt, linear velocity of the belt = v

V = 2 d2 /2Linear velocity of the belt ,1 d1 = 1 d2

2 21 d1 = 1 d2

1 = 2 N1 /60

2 = 2 N260

N1 d1 = N2 d2


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VELOCITY RATIO OF A COMPOUND

BELT DRIVE :

Sometimes the power transmitted fromone shaft to another through a number of

pulleys. This arrangement is known as

compound belt drive.

Let, d1 = diameter of pulley 1

N1 = Speed of pulley 1 in r.p.m.

d2,d3,d4,N2,N3,N4 = Corresponding

values for pulleys 2,3 and 4


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Velocity ratio of the pulleys 1 and2,

N2 = d1

N1 d2 (1)Similarly, velocity ratio of the pulleys 3 and 4

N4 = d3

N3 d4 (2)

Multiplying equation 1 with 2

N2 x N4 = d1 x d3

N1 N3 d2 d4

N4 = d1 x d3N1 d2 d4

[ as N2 = N3, being keyed to the same shaft]


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Problem 1 : In a workshop, an engine

drives a shaft by a belt. The diameter

of the engine pulley and the shaftpulley are 500mm and 250mm

respectively . Another pulley of 700mm

diameter on the same shaft drives apulley 280mm in diameter on a motor

shaft. If the engine runs at 180RPM,

Find the speed of the motor shaft.


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SOLUTION :

N4 = d1 x d3N1 d2 d4

N4 = 180 X 500X 700

250X280

= 900 R.P.M


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A B

Driver

Follower

POWER TRANSMITTED BY BELT :


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Driving pulley pulls the belt form one side, and

delivers the same to the other. The tension T1 in

the former side ( tight side) will be more than

tension T2 in the latter side i.e slack side.

V = velocity of the belt in m/s.

Effective driving force at circumference of thefollower = T1 T2

Work done = Force x Distance

= [T1 T2] V Nm/s

Power = [T1 T2] V J/s

= [T1 T2] V watt


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Problem 2 :Find the necessary

difference in tensions in kgf in the two

sides of a belt drive, when transmitting20h.p at 30m/sec.SOLUTION :

Power = [T1 T2] V watt

20 = [T1 T2] V

7520 = [T1 T2] 30

75

[T1 T2] = 20/0.4 = 50kgf


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RATIO OF TENSIONS :

Consider a follower pulley rotating in a clockwise

direction.

Let T1 = Tension in the belt on the tight side

T2 = Tension in the belt on the slack side = Angle of contact in radians

Now consider a small portion of the belt PQ,

subtending an angle at the centre of thepulley as shown in fig. Belt PQ is in equilibrium

under the following forces:


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T2T1


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Tension T in the belt at P

Tension T +T in the belt at Q,

Normal Reaction RAnd Frictional force F= RResolving all the forces horizontally ,

R = (T +T) sin + T sin 2 2

Since is very small, sin =

2 2R = (T +T) + T

2 2

= T


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Now resolving the forces vertically,

R = (T +T) cos /2 _ T cos/2 .eq.iii

Since angle is very small, substitutingcos/2 = 1 in eq. iii

R = (T +T) - T R = T / . (iv)

Equating the values of R from equations (ii)

and (iv)

T / = T

T / T =


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Integrating both sides from A to B,

T1

T / T = T2 0

Loge ( T1/T2) =

( T1/T2) = e

2.3log ( T1/T2) =


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Problem 3 :Find the power transmitted

by a belt running over a pulley of 600mm

diameter at 200r.p.m. The coefficient offriction between the pulley is 0.25, angle

of lap 160 and maximum tension in the

belt is 2.5kN.

SOLUTION :

Speed of the belt v = dN/60= x 0.6 x 200/60= 2 rad/sec


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2.3log ( T1/T2) =

2.3log ( T1/T2) = 0.25 x ( 160x /180)

log ( T1/T2) =0.6975 / 2.3 = 0.3033

2.5/T2 = 2.01

T2 = 2.5/2.01 = 1.24kN

Power transmitted by the bellt P

P = (T1 T2) v

=(2.5 1.24 ) 2 kW

= 7.92kW


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Copy of Belt and Rope Drive01