A 100-kg lathe is mounted at the midspan of a 1.8-m simply supported beam (E = 200×10^9 N/m, I = 4.3× 10^-6 m^4). The lathe has a rotating unbalance of 0.43 kg. m and operates at 2000 rpm. When a free vibrations test is performed on the system it is found that the ratio of amplitudes on successive

Question

A 100-kg lathe is mounted at the midspan of a 1.8-m simply supported beam (E = 200 imes 10^{9 }N/m, I = 4.3 imes 10^{-6} m^{4}). The lathe has a rotating unbalance of 0.43 kg . m and operates at 2000 rpm. When a free vibrations test is performed on the system it is found that the ratio of amplitudes on successive cycles is 1.8 to 1. Determine the steadystate amplitude of vibration induced by the rotating unbalance. Assume the damping is hysteretic.

Accepted Answer

The beam’s stiffeness is

[latex] k=\frac{48EI}{L^{3}}=\frac{48(200 imes 10^{09} N/m^{2})(4.3 imes 10^{-6}m^{4})}{(1.8 m)^{3}}= 7.08 imes 10^{6} N/m[/latex] (a)

The natural frequency and frequency ratio are

[latex] \omega _{n}=\sqrt{\frac{k}{m} }=\sqrt{\frac{7.08 imes 10^{6} N/m}{100kg} }= 266.1 rad/s[/latex] (b)

[latex] r=\frac{\omega }{\omega _{n}}= \frac{(2000 rev/\min)(2\pi rad/rev)(1\min/60 s)}{266.1 rad/s}[/latex] (c)

The logarithmic decrement and hysteretic damping coefficient are calculated as

[latex] \delta =\ln 1.8 = 0.588[/latex] [latex] h=\frac{\delta }{\pi} =0.187[/latex] (d)

The appropriate form of [latex] \Lambda [/latex] for hysteretic damping is

[latex] \Lambda _{h}(r,h)=\frac{r^{2}}{\sqrt{(1-r^{2})^{2}+h^{2}} }[/latex] (e)

[latex] \Lambda _{h}(0.787, 0.187)= \frac{(0.787)^{2}}{\sqrt{\left[1-(0.787)^{2} ight]^{2}+(0.187)^{2} } }=1.46[/latex] (f)

The lathe’s steady-state amplitude is

[latex] X=\frac{m_{0}e}{m}\Lambda _{h}(0.787,0.187)= \frac{0.43kg\cdot m}{100kg} (1.46)=6.3 mm[/latex] (g)

Question 4.20: A 100-kg lathe is mounted at the midspan of a 1.8-m simply s...

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