Calculate Reaction Forces for Bearings in Mechanical System
What are the reaction forces (RF1x, RF1y, RF2x, and RF2y) that the bearings need to supply in a mechanical system transmitting 30 hp at 3,450 rpm?
The reaction forces that the bearings need to supply are RF1x = 127.84 lb, RF1y = 129.6 lb, RF2x = -127.84 lb, and RF2y = -129.6 lb.
Calculating Reaction Forces for Bearings
A gear is mounted between two bearings and a sheave in a mechanical system transmitting 30 hp at 3,450 rpm. The distances from the center of the left bearing to the centers of the gear, right bearing, and sheave are:
- Sheave: 8 inches
- Gear: 1.75 inches
- Right Bearing: 6 inches
The first step is to calculate the torque transmitted by the belt, given by:
Torque (T) = hp * rpm * 5252 / pi
Substitute the values:
T = 30 hp * 3450 rpm * 5252 / pi = 51480 lb-in
The tight side force (Ft) and slack side force (Fs) are calculated next:
- Ft = T / (tight/slack ratio) = 10296 lb
- Fs = Ft / (tight/slack ratio) = 2059.2 lb
Using the equations below, we can calculate the reaction forces:
- RF1x = Ft * dg / L
- RF1y = Ft * ds / L
- RF2x = Fs * dg / L
- RF2y = Fs * ds / L
Plugging in the values, we get:
- RF1x = 10296 lb * 1.75 in / 10 in = 127.84 lb
- RF1y = 10296 lb * 8 in / 10 in = 129.6 lb
- RF2x = 2059.2 lb * 1.75 in / 10 in = -127.84 lb
- RF2y = 2059.2 lb * 8 in / 10 in = -129.6 lb
Therefore, the reaction forces that the bearings need to supply are RF1x = 127.84 lb, RF1y = 129.6 lb, RF2x = -127.84 lb, and RF2y = -129.6 lb.