Maximum permissible stress in belt:
Q: A belt 100mm wide and 8.0mm thick are transmitting power at belt speed of 160m/minute. The angle of lap for smaller pulley is 165º and the friction coefficient is 0.3. The maximum permissible stress in belt is 2MN/m2 and mass of belt is 0.9Kg/m. find power transmitted and initial tension in belt.
Sol.: Given
Width of belt(b) = 100mm
Thickness of belt(t) = 8mm
Velocity of belt(V) = 160m/min = 2.66m/sec
Angle of contact(?) = 165° = 165° X Π/180 = 2.88rad
Coefficient of friction(µ) = 0.3
Maximum permissible stress(f) = 2 X 10^{6} N/m^{2} = 2N/mm^{2}
Mass of the belt material(m) = 0.9 Kg/m
Power = ?
Initial tension (To) = ?
We know that, Tmax = σ.b.t = 2 X 100 X 8 = 1600N ...(i)
As m and velocity (V) is given, then
Using formula, T_{C }= mV^{2}, For finding T_{C}