Question 13.3: Design Analysis of a V Belt Drive The capacity of a V belt d......

We have $\phi=153^{\circ}=2.76  rad \text { and } \beta=18^{\circ}$. The tight-side tension is estimated from Equation (13.20) as

$F_1=F_c+\left(\frac{\gamma}{\gamma-1}\right) \frac{T_1}{r_1}$      (13.20)

$F_1=F_c+\left(\frac{\gamma}{\gamma-1}\right) \frac{T_1}{r_1}$        (a)

where

$F_c=\frac{w}{g} V^2=\frac{2.25}{9.81}\left(\frac{\pi \times 0.2 \times 1800}{60}\right)^2=81.5  N$

$\gamma=e^{f \phi / \sin \beta}=e^{0.3(2.67) / \sin 18^{\circ}}=13.36$

$T_1=\frac{9549 kW }{n_1}=\frac{9549(10)}{1800}=53.05  N \cdot m$

Carrying the preceding values into Equation (a), we have

$F_1=81.5+\left(\frac{13.36}{13.36-1}\right) \frac{53.05}{0.1}=655  N$

Then, by Equation (13.19), the slack-side tension is

$F_2=F_1-\frac{T_1}{r_1}$      (13.19)

$F_2=655-\frac{53.05}{0.1}=124.5  N$

Based upon a service factor of 1.2 (Table 13.5) to $F_1$, Equation (13.22) gives a maximum tensile force:

$F_{\max }=k_s F_1$     (13.22)

$F_{\max }=1.2(655)=786  N$

applied to the belt.

The design of timing-belt drives is the same as that of flat belt or V belt drives. The manufacturers provide detailed information on sizes and strengths. Case Study 18.10 illustrates an application.