Question 13.3: Design Analysis of a V-Belt Drive The capacity of a V-belt d...
Design Analysis of a V-Belt Drive
The capacity of a V-belt drive is to be 10 kW, based on a coefficient of friction of 0.3. Determine the required belt tensions and the maximum tension.
Given: A driver sheave has a radius of r_{1} = 100 mm, a speed of n_{1} = 1800 rpm, and a contact angle of ϕ = 153°. The belt weighs 2.25 N/m and the included angle is 36°.
Assumptions: The driver is a normal torque motor and driven machine involves light shock load.
We have ϕ = 153° = 2.76 rad and β = 18°. 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} (a)
where
F_c=\frac{w}{g} V^2=\frac{2.25}{9.81}\left\lgroup \frac{\pi \times 0.2 \times 1800}{60}\right\rgroup^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
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- or V-belt drives. The manufacturers provide detailed information on sizes and strengths. The following case study illustrates an application.
| Table 13.5 Service Factors K_{s} for V-Belt Drives |
||
| Driver (Motor or Engine) | ||
| Driven Machine | Normal Torque Characteristic | High or Nonuniform Torque |
| Uniform | 1.0–1.2 | 1.1–1.3 |
| Light shock | 1.1–1.3 | 1.2–1.4 |
| Medium shock | 1.2–1.2 | 1.4–1.6 |
| Heavy shock | 1.3–1.5 | 1.5–1.8 |