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Ivan
3 years ago
5

An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir

ed that the amplitude of motion should not exceed 0.1 mm per 1 N of unbalance force. The mass of the motor is 120 kg and the operating speed is 720 rpm Use the concept of transfer function to determine the required stiffhess coefficient of each mount.
Engineering
1 answer:
Artyom0805 [142]3 years ago
8 0

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

\frac{A}{F} =  \frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}

Formula Used:

A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

\omega = \frac{2 \pi\times N}{60}

so,

\omega = \frac{2 \pi\times 720}{60} = 75.39 rad/s

Using the given formula:

Damping is negligible, so, \zeta = 0

\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

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Answer:

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Explanation:

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M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\  x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m

then we get  

M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}

We get the inertia as follows

I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}

We use the formula

σ = M*y/I

⇒ M = σ*I/y

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y=\frac{h}{2} =\frac{0.3m}{2}=0.15m

If M = Mmax, we have

(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4}   }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}

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