written 5.3 years ago by |
The table is bolted to the laboratory floor. Measurement indicate that the floor has a steady state vibration amp of 0.5 mm at a frequency of 30 Hz. What is the amp of acc of the flow monitoring device?
m = 35kg
k = $2 \times 10^5 N/m$
$\xi = 0.08$
Y = $0.5 \times 10^{-3} \ m$
f = 30 Hz
w = $2 \pi f = 2 \pi \times 30 = 188.49$ rad/s.
$\rightarrow$ Acceleration of flow monitoring device.
$= w^2. X$
$\frac{x}{y} = \frac{\sqrt{1+(2 \xi r)^2}}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$
$w_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{ 2 \times 10^5}{35}} = 75.59$ rad/sec
$r = \frac{188.49}{75.59} = 2.49$
$\frac{x}{0.5 \times 10^{-3}} = \frac{\sqrt{ 1+ (2 \times 0.08 \times 2.49)^2}}{\sqrt{(102.492^2)^2 + (2 \times 0.08 \times 2.49)^2}}$
= $\frac{1.0766}{5.215}$
= 0.2064
$X = 1.03 \times 10^{-4} \ m$
Acceleration of flow monitoring device = $w^2.X$
$= (188.49)^2 \times (1.03 \times 10^{-4})$
$= 3.67 \ m/sec^2$
Vibration Isolation and Transmissibility.
1] $T_r \ = \ \frac{FT_r}{F_o} \ = \ mo.e.w^2$
$T_r \ = \ \frac{ \sqrt{1 + (2 \xi r)^2}}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$
To find mo.e
$\frac{X}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$
2] $\eta_{iso} = 1 - Tr$
3] $tan \ \phi = \frac{2 \xi r}{1 – r^2}$ [phase angle]
4] $tan \ \alpha = \frac{cw}{k}$ [phase lag]
5] Angle between Fo and Frr
$\psi = \phi - \alpha$