A wormgear is mounted at the end of the shaft as shown in the Figure. The gear h

Question

A wormgear is mounted at the end of the shaft as shown in the Figure. The gear has the same design as that discussed in Example Problem 10-7 and delivers 6.68 hp to the shaft at a speed of 101 rpm. The magnitudes and directions of the forces on the gear are given in the figure. Notice that there is a system of three orthogonal forces acting on the gear The power is transmitted by a chain sprocket at B to drive a conveyor removing cast iron chips from a machining system.. It is necessary to do the following: (a) Determine the magnitude of the torque in the shaft at all points (b) Compute the forces acting on the shaft at all power transmitting elements (c) Compute the reactions at the bearings (d) Draw the complete load, shear, and bending moment diagrams Neglect the weight of the elements on the shafts, unless otherwise noted. Forces on wormgear: W_tG = 962 lb tangential W_rG = 352 lb radial W-xG = 265 lb axial

Explanation / Answer

Power = 6.68 hp = 4.98 KW

Speed = 101 rpm

P = 2 * pi() *N * T/60

T = 4.98 * 103 * 60 / (3.14 * 101)

T = 942 N-m

Consider the vertical reaction at bearing A and bearing C as Fva and Fvc respectively

Radial force on worm gear will act in upward direction which is 352 lb

Fva + Fvc = 352 lb

Considering moment about Fva = 0

352 * 20 - Fvc *15 = 0

Bearing reactions are

Fva = 469.3 lb

Fvc = -117 lb

Force on sprocket = Fc = 962 lb which is force on the worm gear

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