Question

Figure 1 shows the layout of countershaft used to transmit power to a blower through a pulley drive (4-5). Pulley (driving sheave) 4 has a diameter of 125-mm and pulley (driven sheave) 5 has a diameter of 75 mm. Pulley 5 is mounted vertically below pulley 4 (as shown in the figure). Belt tension on the loose side is 20% of the tension on the tight side. A power of 7.5 kW is transmitted via the gear set (2-3) from a motor rotating at 1800 rpm. Gear 3 has a pitch diameter of 96 mm and gear 2 has a pitch diameter of 72 mm. Both gears have a pressure angle of 25o. The location of the pulley P4, gear 3 and bearings A and D are as shown in the figure. The shaft of pulley 4 is coupled with the shaft of gear 3 through a rigid coupling C mounted mid-span between pulley 4 and gear 3 (this coupling is assumed not to affect the strength and rigidity of the countershaft AD)

Answer 1

Given the Diameter of pulley (mu) = 125 mm Diameter of pulley (tau) = 75 mm Tension on loose side = 20% of tension of tension on tight - side power = 7.5 kW speed = 1800 rpm pitch diameter of gear(D) = 96 mm pitch diameter of gear (ohm) = 72 mm pressure angle = 25 degree F_n = 2T sin d theta/2 F_n - force on puller KN/m T = belt tension KN/m theta - Angle degrees \r\n total tension tau_T = tension in light side + tension in slack side Belt tension (T_1) = T_2 + T_epsilon(N) T_1/T_2 = e^mu_0 T_1 = tension in tight side T_2 = tension in loose side Tight side elongation proportionalto (T_1 - T_i) slack side contraction proportionalto (T_i - T_2) initial tension (T_i) = T_i + T_2/2 power transmission, p = (T_1 - T_2) V puller torque: - T_a = r_a(T_1 - T_2) T_b = r_b(T_1 - T_2) shear force diagram: - \r\n Bending moment: -