Question

A 35-kg fan is bolted to a light horizontal catilever beam. The unbalance of its rotor is equivalent to a 40 g mass located 45-mm from the axis of rotation. Knowing that resonance occurs at (234) RPM, determine:
-the amplitude (in mm) of the steady state vibration at: a) (244) RPM b) (239) RPM, c) (235) RPM, d) (232) RPM
-also determine the imbalance force at each of these four speeds (in Newtons)

Answer 1

Guren m= 35 kg imbalance mo= 40g = Answee (mass of fan) 0.040kg e- 0.045 m = 45 mm Resonance oluus at 234 pm 24.1504 Hadis so Natural frequency on = 2x TX 234 60 Steady state is X- amplitude of vibration 71? 5 (1–942)2 moe 92 Wn ယ m and imbalance force F- moewe @ HE 1.04277 We 244 MPM 25.552 rad/s 13) 125.552 24.504 X2 0.640mm 0.040 X45 35 2 25.552 24.504 1.1752N F. 0.040 X 0.005x (25-55212, b W2 1.0214 239 rom or 25-028 radia 915 wn 1.241 mm 35 0.040x45 x((1+021412) SC- (1.52147275 (25-028)? 1.1215 W F. 0.040 X o.ous* c . با 235 rpm on 24.609 rad/s 913 om - 1.00429 X2 6-0314 mm 0.040x 45 *((1.0042972) Se -(1.0042924 35 0.040x 0.045x (24.60932. 1.09 N d. 232 rpm or 24.2949 gadis ME 0.991 x = 0-040 X 45 x 35 2.9733 mm (0.99172 S(1-10-990')? 0.000x 0.045*( 24.295) = Fa 1.0624 N