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 10⁶ N/m²    = 2N/mm²

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  = mV², For finding T_C