Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: Medium

Since the belt continuously runs over the pulleys, therefore some centrifugal force is caused, whose effect is to increase for tension on both, tight as well as the slack sides. The tension caused by centrifugal force os called centrifugal tension.at lower belt speeds (less than 10 m/s), the centrifugal tension is very small, but at higher belt speeds (more than 10 m/s), its effect is considerable and thus should be taken into account.

Consider a small portion PQ of the belt subtending an angle $d\theta$ the centre of the pulley as shown in fig.

Let m=mass of the belt per unit length in kg.

$v$ = Linear velocity of the belt in m/s. $r$ = Radius of the pulley over which the belt runs in metres, and $T_{C}$ = Centrifugal tension acting tangentially at P and Q in newtons. We know that length of the belt PQ = $r.d\theta$ And mass of the belt PQ = $mrd\theta$ Centrifugal force acting on the belt PQ $F_{C}=(mrd\theta)\times\frac{v^2}{r}=md\theta v^2$ The centrifugal tension TC acting tangentially at P and Q keeps the belt in equilibrium. Now resolving the forces (i.e. centrifugal force and centrifugal tension) horizontally and equating the same, we have $T_{C}\sin (\frac{d\theta}{2})+_{C}\sin (\frac{d\theta}{2})= F_{C}= md\theta v^2$ Since the angle $d\theta$ is very small, therefore putting $\sin (\frac{d\theta}{2})=\frac{d\theta}{2}$ , in the above expression, $2T_{C}(\frac{d\theta}{2})=md\theta v^2$ or $ T_{C}=mv^2$ 1.When the centrifugal tension is taken into account,then total tension in the tight side. $T_{t1} =T_{1}+ T_{C}$ And total tension in the slack side, $T_{t2} =T_{2}+ T_{C}$. Power transmitted, $P=(T_{t1} +T_{t2})v$ Thus we see that centrifugal tension has no effect n the power transmitted. 3.The ratio of driving tension may also be written as $2.3log(\frac{( T_{t1} - T_{C})}{ T_{t2}- T_{C}})=\mu\theta$ $T_{t1}$= Maximum or total tension in the belt.