Calculate the torque developed by a motor that delivers 200 hp rotation at 3600 rpm. Use the Torque value to determine the diameter of a solid steel shaft with an allowable shear stress of 20,520 psi.

Solution :

Given:

Power, P = 200 hp

              = 110000 lbf ft/s

Speed, N = 3600 rpm

Power ,

∴ Torque,

T = 291.93 lb -ft

---- allowable shear stress

Therefore, torque , T = 391.93 lb-ft

                              T   = 291.93 x 12 lb-in

∴ T 3503.16 lb-in

Now from the Torsion formula,

Here,   ---- polar moment of inertia.

∴

or

d = 0.9546 in

Therefore the diameter of the shaft is 0.9546 in.

sorry what a asnns

explanation:

answer:

uygtfjyftujtfj

explanation: