By Avraham Harnoy

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The shaft speed is 3600 RPM. The gearbox is designed to transmit maximum power of 20 kW. The diameter of the pitch circle of the gear is equal to 5 in. The right-hand-side bearing is supporting the total thrust load. Find the axial and radial loads on the right-hand-side bearing and the radial load on the left-side bearing. Classi¢cation and Selection of Bearings 29 Solution The angular velocity of the shaft, o, is: o¼ 2pN 2p3600 ¼ ¼ 377 rad=s 60 60 Torque produced by the gear is T ¼ Ft dp =2. Substituting this into the power equation, E_ ¼ T o, yields: E_ ¼ Ft dp o 2 Solving for the tangential force, Ft , results in Ft ¼ 2E_ 2 Â 20;000 N-m=s ¼ 836 N ¼ dp o 0:127 m Â 377 rad=s Once the tangential component of the force is solved, the resultant force, F, and the thrust load (axial force), Fa , can be calculated as follows: Fa ¼ Ft tan c Fa ¼ 836 N Â tan 30 ¼ 482 N and the radial force component is: Fr ¼ Ft tan f ¼ 836 N Â tan 20 ¼ 304 N The force components, Ft and Fr, are both in the direction normal to the shaft centerline.

Find the radial force on each of the two bearings supporting the shaft. The ratio of the two bearings’ length and diameter is L=D ¼ 0:5. The bearings are made of acetal resin material with the 32 Chapter 1 following limits: Surface velocity limit, V , is 5 m=s. Average surface-pressure limit, P, is 7 MPa. PV limit is 3000 psi-ft=min. 1-3 Find the diameter of the shaft in order not to exceed the stated limits. A bearing is made of Nylon sleeve. Nylon has the following limits as a bearing material: Surface velocity limit, V , is 5 m=s.

Find the diameter and length of each bearing that is required in order not to exceed the PV limit. Solution a. Reaction Forces Given: 26 Chapter 1 F IG. 1-7 Gear pressure angle. Rotational speed N ¼ 600 RPM Power E_ ¼ 5000 W Diameter of pitch circle dp ¼ 5 in. Pressure angle f ¼ 20 PVallowed ¼ 110;000 psi-ft=min L=D ¼ 0:5 (the bore diameter of the bearing, D, is very close to that of the journal, d) Conversion Factors: 1psi ¼ 6895 N=m2 1 ft=min ¼ 5:08 Â 10À3 m=s 1 psi-ft=min ¼ 35 N=m2 -m=s The angular velocity, o, of the journal is: o¼ 2pN 2p600 ¼ ¼ 52:83 rad=s 60 60 Converting the diameter of the pitch circle to SI units, dp ¼ 5 in: Â 0:0254 m=in: ¼ 0:127 m Classi¢cation and Selection of Bearings 27 The tangential force, Ft , acting on the gear can now be derived from the power, E_ : E_ ¼ T o where the torque is T¼ F t dp 2 Substituting into the power equation: E_ ¼ Ft dp o 2 and solving for Ft and substituting yields 2E_ 2 Â 5000 Nm=s ¼ 1253:2 N ¼ dp o 0:127 m Â 62:83 rad=s Ft ¼ In spur gears, the resultant force acting on the gear is F ¼ Ftr (Fig.