Question #239623
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.
1
Expert's answer
2021-09-21T02:06:40-0400

m=100 kgs=800 kN/mLF=400 Nω=3000 rpm=314.2 rad/s(using 1 rev =  rad)x2=x1/4m = 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;

ωn=sm=800×103100=89.44 rad/s\omega_n= \sqrt\dfrac sm= \sqrt\dfrac{800×10³}{100}= 89.44 \textsf{ rad/s}


For the amplitude;

logex1x2=a×2π(ωn)2a2\log_e \dfrac{x_1}{x_2} = a× \dfrac{2π}{\sqrt{(\omega_n)²-a²}}


loge4=2πa(89.44)2a2\log_e4 = \dfrac{2πa}{\sqrt{(89.44)²-a²}}


a=2236log(2)(25(π2+log4)a =\dfrac{ 2236 log(2)}{(25 \sqrt{(π^2 + log4)}}


a=19.27a=19.27


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


xo=Fs=400800×103=0.0005mx_o = \dfrac Fs = \dfrac{400}{800× 10³}= 0.0005m


Amplitude of the forced vibration;


xmax=xoc2ω2s2+[1ω2ωn2]2x_{\textsf{max}} = \dfrac{x_o}{\sqrt{\dfrac{c²\omega²}{s²}+[1-\dfrac{\omega²}{\omega_n²}]²}}


=0.0005385.42×314.22(800×103)2+[1314.2289.442]2= \dfrac{0.0005}{\sqrt{\dfrac{385.4²×314.2²}{(800×10³)²}+[1-\dfrac{314.2²}{89.44²}]²}}


=0.00004408m=0.0044mm= 0.00004408m = 0.0044mm


[0.044mm; 35.2 N]


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