Question: A leather belt transmits 30 KW from a pulley 750 mm diameter which runs at 500 rpm. The belt is in contact with the pullry over an arc of 160° and coefficient of friction, μ = 0.3. If the permissible stress in the belt is not to exceed 2 MN / m^2, find the maximum cross sectional area of the belt?
Power transmitted, P = 30 KW
Diameter of the pulley, d = 750 x 10^-3m
Speed of the pulley, N = 500 rpm
Angle of contact, Θ = 160° = 160 . π/180 = 2.7925 rad
Coefficient of friction, μ = 0.3
Permissible Stress, ft = Tmax/Areamax = 2Mn/m2
Here, Tmax = T1, tension on the tight side.
We have to find out the maximum cross-sectional area of the belt – Δmax
We have, T1/T2 = eμΘ, where T2 is tension on slack side.
= e0.3 x 2.7925 = 2.31118
or, T1 = T2 . 2.31118
Again, we have, Power transmitted, P = (T1-T2) . π . d . N /60.
Lets substitute the values of T1 and T2, thus solving for T1 and T2, we get –
(2.31118 . T2 – T2) . π . 750 . 10-3 . 500 / 600 = 30 . 103
T2 = 30 . 103 / 25.745 = 1165.277 N
T1 = T2 . 2.31118
= 2693.16 N = Tmax
Permissible Stress, ft = Tmax/Areamax = 2 x 106
∴ Maximum cross sectional area,
Δmax = Tmax / ft = 2693.16 / 2 x 106 = 0.0013465 m2