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X(t) A motor system of mass M = 10 kg running at 2000 rpm was found to produce excess vibration due to rotating unbalanc

Posted: Tue Apr 26, 2022 4:54 pm
by answerhappygod
X T A Motor System Of Mass M 10 Kg Running At 2000 Rpm Was Found To Produce Excess Vibration Due To Rotating Unbalanc 1
X T A Motor System Of Mass M 10 Kg Running At 2000 Rpm Was Found To Produce Excess Vibration Due To Rotating Unbalanc 1 (230.92 KiB) Viewed 41 times
X(t) A motor system of mass M = 10 kg running at 2000 rpm was found to produce excess vibration due to rotating unbalance. The unbalance in the motor was estimated to be me = 0.1 kg.m. This is a 2-part question. You are to determine the design parameters for the isolator and absorber. OG се 1 Part (A) Consider the criterion F, Fmax = 90 N where F, is the force transmitted to the ground. Use an isolator characterized as a spring of spring isolator Part (A) constant k, see Fig Part (A). Assume zero damping (C = 0.) (A1) [5/40] What is the minimum reduction (%) to meet the criterion F, Fmax? (A2) [10/40] Assume an isolator of spring constant k = 20,000 N/m. Can F, Fmax be met? If not, use the table below and calculate the k value that can result in F, = Fmax- . Part (B) Let the primary system be the motor and the isolator determined in (A2). To further reduce the force transmission, consider adding an absorber of mass m, and ka, see Fig Part (B). (B1) [10/40] Find m, and ka, with the constraint u= m/m50.3, such that the force transmission to the ground can be reduced to the minimum. (B2) [5/40] With m and k, in (B1), what is the minimum force transmission to the ground. (B2) [10/40] Determine the natural frequencies of the absorber-motor system. 50.000 x.(t) m. k 20.000 10.000 x(t) 5.000 Excitation frequency (rpm) 2.0 Percentage redution 2.000 1.5 1.2 (1.1 WEEK H 1,000. 500 90 Amplification Part (B) 200 10 0.0051 0.0127 0.0254 0.051 0.127 0.254 0.51 1.27 2.54 0.00254 Static Deflection A (cm)