Leather belting 75mm wide is to be used to transmit power between two parallel shafts which run at the same speed. The pulleys are 0.6 m diameter. At slow speeds the tension in the slack side is half the tension in the driving side and the angle of lap on the pulley is 180 degrees. If the maximum load on the belting is not to exceed 140 N/ cm of width, neglecting the effect of centrifugal force, estimate the maximum power which can be transmitted at 9.2 r/s.
Find the maximum torsional shear stress in a 75 mm diameter shaft transmitting 40 KW power at 1.33 r/s if the maximum twisting moment exceeds the minimum by 40 per cent. What is the greatest twist in degrees per metre length if G =87 GPa?
5) Width of the belt wd = 75 mm = 7.5 cm = 0.075 m
Diameter of the pulley D = 0.6 m
Therefore, radius of the pulley R = 0.3 m
Tslack = Tdrive/2
Angle of lap = 180 degree
Rotation speed w = 9.2 r/s = 9.2*2*pi rad/s = 57.8036 rad/s
Maximum tension Tmax = 140*wd = 140*7.5 = 1050 N
Tdrive = Tmax = 1050 N
Tslack = Tdrive/2 ...
This solution is provided in approximately 266 words. It uses tension and torque to find maximum power, as well as twisting moment and polar modulus to find greatest twist.