A 500-r/min shaft is fitted with a 30-in (76.2-cm) diameter pulley weighing 250 lb (113.4 kg). This pulley delivers 35 hp (26.1 kW) to a load. The shaft is also fitted with a 24-in (61.0-cm) pitch-diameter gear weighing 200 lb (90.7 kg). This gear delivers 25 hp (18.6 kW) to a load. Determine the concentrated loads produced on the shaft by the pulley and the gear.
Calculation Procedure:
The largest concentrated load caused by the pulley occurs when the belt load acts vertically downward. Then the total pulley concentrated load is the sum of the belt load and pulley weight.
For a pulley in which the tension of the tight side of the belt is twice the tension in the slack side of the belt, the maximum belt load is Fp = 3 T/r, where Fp = tension force, lb, produced by the belt load; T = torque acting on the pulley, lb·in; r = pulley radius, in. The torque acting on a pulley is found from T = 63,000 hp/R, where hp = horsepower delivered by pulley; R = revolutions per minute (rpm) of shaft.
For this pulley, T = 63,000(35)/500 = 4410 lb·in (498.3 N·m). Hence, the total pulley concentrated load = 882 + 250 = 1132 lb (5035.1 N).
With a gear, the turning force acts only on the teeth engaged with the meshing gear. Hence, there is no slack force as in a belt. Therefore, Fg = T/r, where Fg = gear tooth-thrust force, lb; r = gear pitch radius, in: other symbols as before. The torque acting on the gear is found in the same way as for the pulley.
Thus, T = 63,000(25)/500 = 3145 lb·in (355.3 N·m). Then Fg = 3145·12 = 263 lb (1169.9 N). Hence, the total gear concentrated load is 263 + 200 = 463 lb (2059.5 N).
Related Calculations: Use this procedure to determine the concentrated load produced by any type of gear (spur, herringbone, worm, etc.), pulley (flat, V, or chain belt), sprocket, or their driving member. When the power transmission belt or chain leaves the belt or sprocket at an angle other than the vertical, take the vertical component of the pulley force and add it to the pulley weight to determine the concentrated load.