1. 1 An open belt drive transmits power through a pulley having a diameter of 300 mm at a speed of 150 rpm. The belt is 10 mm thick and 150 mm wide and has a density of 970 kg/m 3. The driven pulley, which is 1200 mm diameter, has an angle of contact of 200°. Allowable stress of the belt is 0.15 MPa. Determine the power transmitted if the coefficient of friction 0.2 for both pulleys. EXAMPLE-1

2. 2 11 22    d1d1 d2d2 Motor  Pulley-1: n = 1500 rpm d 1 = 300 mm Pulley-2: d 2 = 1200 mm  2 = 200 o Coefficient of friction, f = 0.2 Leather flat belt: b t  = 200 kg/m 3 b = 1500 mm t = 10 mm  1 = 0.15 MPa

3. 3 EXAMPLE-1 Here, we have the 2 nd case, where the belt dimensions are known. So, we use the following formula: m = b t  = 0.15 x 0.01 x 970 = 1.455 kg/m. V =  dn /60 = (  x 0.300 x 1500)/60 = 23.56 m/s.  for flat-belt = 180 . where The pulley which governs the design is the one with the smaller. Thus, the smaller pulley governs the design in this problem since the smaller  gives smaller (1)

4. 4 EXAMPLE-1 Substituting all in (1) gives: (2) From the allowable stress of the belt Substituting T 1 into Eq.(2) gives: Power:

5. 5 EXAMPLE-2 A V-belt drive transmits power at speed of 1000 rpm of the smaller sheave. The sheaves pitch diameters are 175 mm and 350 mm. The center distance is 760 mm. The ratio of the belt tension is T 1 /T 2 = 4, the number of belts is 4, and the coefficient of friction is 0.15. The groove angle of the sheaves is 34° and the belt mass is 0.194 Kg/m. Determine the maximum power transmitted therough the belt system.

6. 6 EXAMPLE-2 11 22    d1d1 d2d2 Motor  V-Belt: Mass, m = 0.194 kg/m Groove angle,  = 34  Tension ratio, T 1 /T 2 = 3 Number of belt = 4 Pulley-1 (driver): Speed,  n = 1000 rpm Pitch diameter, d 1 = 175 mm Pulley-2 (driven): Pitch diameter, d 2 = 350 mm Center of distance, C = 760 mm Coefficient of friction, f = 0.2

7. 7 EXAMPLE-2 Here, we can use the 2 nd case. So, we use the following formula: where, T 1 / T 2 = 3, or T 1 = 3T 2 m = 0.194 kg/m. V = (  x  d x n)/60 = (  x  0.175 x 1000)/60 = 9.163 m/s.  for V-belt = 34 . (1) Angles of wrap of smaller and larger pulleys are respectively:

8. 8 EXAMPLE-2 The pulley which governs the design is the one with the smaller. Thus, the smaller pulley governs the design in this problem since the smaller  gives smaller Substituting all in (1) gives: Thus, the power transmitted by the belt: and