0
4.6kviews
A machine 100kg mass has a 20kg rotor with 0.5mm eccentricity.
1 Answer
0
563views

The mounting springs have $k = 85 \times 10^3 N/m$ $\xi = 0.02$ (damping ratio)

The operating speed of machine is 600 rpm and the unit is constrained to move vertically.

Find:

1] Amplitude of machine

2] Force transmitted to the support.

m = 100 kg

$m_o$ = 20 kg

e = 0.5mm = 0.005m

K = $85 \times 10^3 N/m$

$\xi = 0.02$

N = 600 rpm.

Amplitude of M/c [X = ?]

$\frac{X}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{ (1- r^2)^2 + (2 \xi r)^2}}$

$\therefore$ $w = \frac{2 \pi N}{60} = \frac{2 \pi \times 600}{60} = 62.83$ rad/sec.

$wn = \sqrt{ \frac{k}{m}} = \sqrt{ \frac{85 \times 10^3}{100}}$

w_n = 29.15 rad/sec

$\therefore$ $r = \frac{62.03}{29.15} = 2.155$

$\frac{X}{\frac{20 \times 0.005}{100}} = \frac{(2.155)^2}{3.6452}$

$X = 1.2739 \times 20 \times \frac{0.005}{100}$

$X = 1.27 \times 10^{-4} m$

$\frac{F_{tr}}{F_o} = \frac{\sqrt{1 + (2 \xi r)^2}}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$\frac{F_{tr}}{F_o} = \frac{1.0037}{3.6452}$

$F_{tr} = 0.2753 \times F_o$

$= 0.2753 \times 20 \times 0.0005 \times (62.83)^2$

= 10.86 N

Please log in to add an answer.