Answer to Question #238326 in Mechanical Engineering for Richard

Question

Question 2
A flat belt is required to transmit 22 kW from a 250 mm diameter pulley running at 1450
rpm to a 355 mm diameter pulley. The proposed belt is 3.5 mm thick.
The other system specifications are as follows:
 Coefficient of friction 0.7,
 Density of the belt is 1100 kg/m3
 Maximum permissible stress is 7 MPa.
 The distance between the shafts is 1.8 m.
2.1 Calculate the width required. [15]

Answer 1

Maximum tension T,

$P = \frac{2\pi N T}{1000}$

$T =\frac{22000\times1000}{2\pi\times1450} = 2414.76 N$

Width of belt b,

$b = \frac{T}{\sigma\times t}= \frac{2414.76}{7\times 3.5}=98.56$ mm