Find the speed lost by the driven pulley as a result of creep, if the stress on the right and slack side of the belt is 1.4 MPa and 0.5 MPa respectively. The Young’s modulus for the material of the belt is 100MPa.

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

Given :- $d_{1}$ = 1 m; $N_{1}$ = 200 rpm; $d_{2}$ = 2.25 m; $\sigma_{1}$ =1.4 $MP_{a}$=1.4 ×106 N/m$^2$ , $\sigma_{2}$ =0.5 $MP_{a}$=0.5 ×106 N/m$^2$ ; E= 100 $MP_{a}$=100×106 N/m$^2$,

Let $N_{2}$ = speed of the driven pulley.

Neglecting creep, we know that

$\frac{N_{2}}{N_{1}}$ = $\frac{d_{1}}{d_{2}}$ or $N_{2}$ = $N_{1}\frac{d_{1}}{d_{2}}$ = 200 × (1/2.25) = 88.9 rpm.

Consider creep, we know that

$\frac{N_{2}}{N_{1}}$ = $\frac{d_{1}}{d_{2}}\times \frac{(E+\sigma_{2})}{(E+\sigma_{1})}$

$N_{2}=200\times(\frac{1}{2.25})\times\frac{(100\times 10^{6}+\sqrt{0.5}\times 10^{6})}{(100\times 10^{6}+\sqrt{1.4}\times 10^{6})}$ = 88.7 rpm.

Speed lost by driven pulley due to creep,

= 88.9-88.7 = 0.2 rpm.