A rope pulley is supported by two deep groove ball bearings as shown in the figure. A force of FR = 7 kN acts on the pulley in the radial direction. If Fa = 0.5kN, axial force acts. The rolling diameter of the pulley is d = 300 mm and the rope speed is v = 4 m / s. The nominal life of the bearings is required as Lh = 27000 h. Since the diameter of the shaft where the bearing is bearing is 45 mm, choose the appropriate bearing. V = N.D.n

A rope pulley is supported by two deep groove ball bearings as shown in the figure. A force of FR = 7 kN acts on the pulley in the radial direction. If Fa = 0.5kN, axial force acts. The rolling diameter of the pulley is d = 300 mm and the rope speed is v = 4 m / s. The nominal life of the bearings is required as Lh = 27000 h. Since the diameter of the shaft where the bearing is bearing is 45 mm, choose the appropriate bearing. V = N.D.n

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter3: Torsion

Section: Chapter Questions

Problem 3.12.4P: The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm. The shaft...

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The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm. The shaft has a full quarter-circular fillet, and the smaller diameter D1= 100 mm. If the allowable shear stress at the stress concentration is 100 MPa, at what diameter will this stress be reached? Is this diameter an upper or a lower limit on the value of D2?
A stepped shaft ABC consisting of two solid, circular segments is subjected to uniformly distributed torque t1acting aver segment 1 and concentrated torque t2applied at C, as shown in the figure. Segment 1 of the shaft has a diameter of d1= 57 mm and length of L1= 0.75 m; segment 2 has a diameter d2— 44 mm and length L2= 0.5 m. Torque intensity /,"= 3100 N . m/m and T2= 1100 N. m. (a) Find reaction torque TAat support A. (b) Find the internal torque T(x) at two locations: .x = L1/2 and at .x = L1+ L2/2. Show these internal torques on properly drawn free-body diagrams.
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