A 100lb motor is directly supported by a light horizontal beam which has a (vert

Question

A 100lb motor is directly supported by a light horizontal beam which has a (vertical static) deflection of 0.2 in

at the center of the beam due to the weight of the motor. The unbalace of the rotor is equivalent to a mass of 3.5 oz located 3inch from the axis of rotation.

knowing that the amplitude of the steady state vibration of the motor is 0.03 inch at a speed of 400 rpm determine

a) K in /ft , Wn inrad /sec

b) Fo in lb

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Explanation / Answer

Omega , w = 400*2pi/60 = 41.888 rad/sed

3.5 oz = 0.21875 lb ; 3 in = 1/4 ft

Rotating inertia = me* d*w*w = 0.21875 * 0.25 * 41.888*41.888 = 95.955 lb.ft/s2

z (dynamic) max = z static max * Rotating Inertia / K

0.03 = 0.2 * 95.955 / K

K = 639.7 = 640 lb/ft

b) Fo = 95.955 lb.ft/s2

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