Chapter 33 : Transmission of Power by Belts and Ropes 685
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 and
dθ= Angle subtended by the belt AB at the centre of the pulley.
∴ Length of belt AB= r dθ
and total mass of the belt M =mr dθ
We know that centrifugal force of the belt AB,
PC=22
Mv ()mr d v md v 2
rrθ
==θ
Now resolving the forces (i.e., centrifugal force and centrifugal
tension) horizontally and equating the same,
2sin^2
C 2d
Tmdv
⎛⎞θ
⎜⎟=θ
⎝⎠Since dθ is very small, therefore substituting sin
22⎛⎞ddθθ
⎜⎟=
⎝⎠in the above equation,2 2
C 2d
Tmdv
⎛⎞θ
⎜⎟=θ
⎝⎠or TC = mv^2Notes: 1. When the centrifugal tension is taken into account, the total tension in the tight side
= T 1 + TC
and total tension in the slack side
= T 2 + TC- The centrifugal tension on the belt has no effect on the power transmitted by it. The
reason for the same is that while calculating the power transmitted, we have to use
the values :
= Total tension in tight side – Total tension in the slack side
=(T 1 + TC) – (T 2 + TC) = (T 1 – T 2 ).
33.16.MAXIMUM TENSION IN THE BELT
Consider a belt transmitting power from the driver to the follower.
Let σ = Maximum safe stress in the belt,
b = Width of the belt in mm, and
t = Thickness of the belt in mm.
We know that maximum tension in the belt,
T = Maximum stress × Cross-sectional area of belt
=σ bt
When centrifugal tension is neglected, then maximum tension,
T =T 1
and when centrifugal tension is considered, then maximum tension,
T =T 1 + TC
Example 33.10. A laminated belt 8 mm thick and 150 mm wide drives a pulley of 1·2 m
diameter at 180 r.p.m. The angle of lap is 190° and mass of the belt material is 1000 kg/m^3. If the
stress in the belt is not to exceed 1·5 N/mm^2 and the coefficient of friction between the belt and the
pulley is 0·3, determine the power transmitted when the centrifugal tension is (i) considered, and
(ii) neglected.
Fig. 33.11. Centrifugal tension.