Answer to Question #239623 in Mechanical Engineering for Joe melwin Raja

Question

a machine of mass 100 kg is supported on openings of total stiffness 800 kn/m and has a rotating unbalanced element which results in a disturbing force of 400 n at a speed of 3000 r.p.m. assum-ing the damping ratio as 0.25, determine : 1. the amplitude of vibrations due to unbalance ; and 2. the transmitted force.

Answer 1

$m = 100\textsf{ kg}\\ s = 800\textsf{ kN/m}L\\ F = 400\textsf{ N}\\ \omega = 3000\textsf{ rpm}= 314.2\textsf{ rad/s}\\ \textsf{(using 1 rev = 2π rad)}\\ x_2 = x_1/4$

We know that angular speed or natural circular frequency of free vibrations;

$\omega_n= \sqrt\dfrac sm= \sqrt\dfrac{800×10³}{100}= 89.44 \textsf{ rad/s}$

For the amplitude;

$\log_e \dfrac{x_1}{x_2} = a× \dfrac{2π}{\sqrt{(\omega_n)²-a²}}$

$\log_e4 = \dfrac{2πa}{\sqrt{(89.44)²-a²}}$

$a =\dfrac{ 2236 log(2)}{(25 \sqrt{(π^2 + log4)}}$

$a=19.27$

$c= a × 2m = 19.27 × 2 × 10= 385.4\textsf{ N/m/s}$

$x_o = \dfrac Fs = \dfrac{400}{800× 10³}= 0.0005m$

Amplitude of the forced vibration;

$x_{\textsf{max}} = \dfrac{x_o}{\sqrt{\dfrac{c²\omega²}{s²}+[1-\dfrac{\omega²}{\omega_n²}]²}}$

$= \dfrac{0.0005}{\sqrt{\dfrac{385.4²×314.2²}{(800×10³)²}+[1-\dfrac{314.2²}{89.44²}]²}}$

$= 0.00004408m = 0.0044mm$

[0.044mm; 35.2 N]