Post on 03-Apr-2018
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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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