0
799views
A machine weighing 1100 N has a rotor 220N with 5mm eccentricity. The operating speed is 600 rpm, the machine is mounted on springs with stiffness 88 KN/m.
1 Answer
0
11views

Damping is negligible and unit is constrained to move vertically.

Determine dynamic amplitude of the machine.

Redesign the mountings so that the dynamic amplitude is reduced to one half of the original value, but maintain same natural frequency.

W = 1100N, $m = \frac{1100}{9.81} = 112.13$ kg

$m_o = \frac{220}{9.81} kg = 22.43 kg$

$e = 5 \times 10^{-3} \ m$

N = 600 rpm

$K = 88 \times 10^3$

$\xi = 0$

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

Find - $X_1$ = ?

Redesign mountings when new amplitude, $X_2 = \frac{x1}{2}$

Case 1-

$\frac{X_1}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$\frac{X1}{\frac{22.43 \times 5 \times 10^-3}{112.13}} = \frac{r^2}{1-r^2}$

$Wn = \sqrt{ \frac{k}{m}}$

$Wn = \sqrt{ \frac{88 \times 10^3}{112.13}}$

$Wn = 28.01$ rad/sec

$r = \frac{w}{wn} = \frac{62.83}{28.01}$

R = 2.242

Since r > 1

$\therefore$ RHS will be

$\frac{X1}{1 \times 10^{-3}} = \frac{r^2}{r^2 – 1}$

$\frac{X1}{1 \times 10^{-3}} = \frac{r^2}{r^2 – 1}$

$X_1 = 1.25 \times 10^{-3} \ m$

NOTE: To maintain same frequency (Wn) let us introduce a damper with damping ratio.

$\frac{6.23 \times 10^{-4}}{1 \times 10^{-3}} = \frac{(2.242)^2}{\sqrt{ (1- 2.242)^2 + ( 2 \times \xi \times 2.242)^2}}$

$0.623 = \frac{5.02}{\sqrt{16.24 + 4 \xi^2 \times 5.02}}$

$0.623 = \frac{5.02}{\sqrt{16.24 + 20.08 \xi^2}}$

$16.24 + 20.08 \xi^2 = 64.93$

$\xi = 1.55$

Please log in to add an answer.