A 150-kg electric motor has a rotating unbalance of 0.5 kg, 0.2 m from the center of rotation. The motor is to be mounted at the end of a steel (E = 210 * 10^9 N/m^2) cantilever beam of length m. The operating range of the motor is from 500 to 1200 rpm. For what values of I, the beam’s cross-sectio

Question

A 150-kg electric motor has a rotating unbalance of 0.5 kg, 0.2 m from the center of rotation. The motor is to be mounted at the end of a steel [latex](E = 210 imes 10^{9} N/m^{2})[/latex] cantilever beam of length 1 m. The operating range of the motor is from 500 to 1200 rpm. For what values of I, the beam’s cross-sectional moment of inertia, will the steady-state amplitude of vibration be less than 1 mm? Assume the damping ratio is 0.1.

Accepted Answer

The maximum allowable value of [latex] \Lambda[/latex] is

[latex] \Lambda_{allow}=\frac{mX_{allow}}{m_{0}e}=\frac{(150 kg)(0.001 m)}{(0.5 kg) (0.2 m)}=1.5 [/latex] (a)

Since [latex] \Lambda_{allow}\gt 1[/latex] and [latex] \zeta \lt 1/\sqrt{2}[/latex] Figure 4.11 shows that two values of r correspond to [latex] \Lambda =\Lambda_{allow}[/latex]. These are determined using Equation (4.52)

[latex] 1.5=\frac{r^{2}}{\sqrt{(1-r^{2})+ (0.2r)^{2}} }[/latex] (b)

[latex] \Lambda (r,\zeta )=\frac{r^{2}}{\sqrt{(1-r^{2})+(2\zeta r)^{2}} }[/latex] (4.52)

Rearrangement leads to the following equation:

[latex]0.556r^{4}- 1.96r^{2}+1=0[/latex]

whose positive roots are

[latex] r=0.787,1.71[/latex]

However if r _ 0.787 corresponds to [latex] \omega =1200[/latex] rpm then [latex] \Lambda \lt \Lambda_{allow}[/latex] or all r in the operating range. Whereas if r = 0.787 corresponds to [latex] \omega =500[/latex] rpm then [latex] \Lambda \gt \Lambda_{allow}[/latex] for r over part of the operating range. Thus requiring [latex] r\lt 0.787[/latex] over the entire operating range yields.

[latex] \frac{(1200 rev/\min )(2\pi rad/rev)(1\min/60s)}{\omega _{n}} \lt 0.787[/latex] (e )

Or [latex] \omega _{n}\gt 159.7 rad/s[/latex] The one degree-of-freedom approximation for the natural frequency of the motor attached to the end of a cantilever beam of negligible mass is

[latex] \omega _{n}=\sqrt{\frac{3EI}{ml^{3}} }[/latex] (f)

Thus,

[latex] I\gt \frac{(159.7 rad/s)^{2}L^{3}m}{3E} =\frac{(159.7 rad/s)^{2}(1m)^{3}(150 kg)}{3(210 imes 10^{9}N/m^{2})}=6.07 imes 10^{-6}m^{4}[/latex] (g)

Using a similar reasoning r _ 1.71 should correspond to [latex] \omega =500[/latex]

[latex] \frac {(500 rev/\min)(2\pi rad/rev)(1\min/60s)}{\omega _{n}}\gt 1.71[/latex] (h)

or [latex] \omega _{n}\lt 30.6 rad/s[/latex]. This requirement leads to [latex] I \lt2.23 imes 10^{-7}m^{4}[/latex].

Thus the amplitude of vibration will be limited to 1 mm if [latex] I \gt6.08 imes 10^{-6}m^{4}[/latex] or [latex] I \lt2.23 imes 10^{-7}m^{4}[/latex]. However, other considerations limit the design of the beam. The smaller the moment of inertia, the larger the bending stress in the outer fibers of the beam at the support.

Question 4.5: A 150-kg electric motor has a rotating unbalance of 0.5 kg, ...

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