A flat belt is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at 300 rpm. The angle of contact is spread over 11/24 of th e circumference and the coefficient of friction between belt and pulley surface is 0.3. Taking centrifugal tension into account, determine the

Question

A flat belt is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at 300 rpm. The angle of contact is spread over 11/24 of th e circumference and the coefficient of friction between belt and pulley surface is 0.3. Taking centrifugal tension into account, determine the width of the belt. Take belt thickness as 9.5 mm, density 1100 kg/m³ and permissible stress as 2.5 MN/m². [KU, June 2010]

Accepted Answer

Given : P = 35 kW, d = 1.5 m, n = 300 rpm, [latex] heta=\frac{11}{24} imes 360=165^{\circ}, \mu=0.3, t=9.5 \mathrm{~mm} ho=1100 \mathrm{~kg} / \mathrm{m}^3, \sigma_w=2.5 \mathrm{\ MN} / \mathrm{m}^2[/latex]

Belt speed, [latex]ν=\frac{\pi d n}{60}=\frac{\pi imes 1.5 imes 300}{60}=23.56 \mathrm{~m} / \mathrm{s}[/latex]

Let b = width of belt, mm
Mass of belt per metre length, [latex]m=b t imes 1 imes ho=b imes 9.5 imes 10^{-6} imes 1 imes 1100=0.0104 .5 imes b \mathrm{~kg} / \mathrm{m}[/latex]
Centrifugal tension, [latex]T_c=m ν^2=0.01045 imes 6 imes(23.56)^2=5.8 imes b \mathrm{~N}[/latex]
Maximum tension in belt, [latex]\mathrm{T}=\sigma_w imes b t=2.5 imes b imes 9.5=23.75 imes b \mathrm{~N}[/latex]
Tension on right side of belt, [latex]T_1=T-T_C=23.75 imes b-5.8 imes b=17.95 imes b \mathrm{~N}[/latex]

Now,

[latex]\begin{gathered}\frac{T_1}{T_2}=\exp (\mu heta)=\exp \left\lgroup 0.3 imes \frac{\pi}{180} imes 165 ight group=\exp (0.86394)=2.3725 \T_2=\frac{T_1}{2.3725}=\frac{17.95 imes b}{2.3725}=7.566 imes b\end{gathered}[/latex]

Power transmitted, [latex]P=\frac{\left(T_1-T_2 ight) ν}{10^3}[/latex]

[latex]35=\frac{(17.95-7.566) imes b imes 23.56}{10^3} b=143 \mathrm{~mm}[/latex]

Question A2.15: A flat belt is required to transmit 35 kW from a pulley of 1...

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