It is estimated that damping force exerted on the system is 20% of the critical and is of viscous nature. When the speed of rotation of machine is 2000 rpm, the amp of vertical motion of bed plate is 0.06mm, Calculate the total maximum force transmitted through each mounting to found.

No of springs = 4

K = $3.92 \times 10^6 \ N/m$

m = 1000 kg

$\xi $ = 0.20

$F \ \alpha \dot{x}$

F = c.$\dot{x}$

$\frac{F}{\dot{x}} = c$

Similarly $\frac{Fc}{\dot{x}} = C_c$

F = 0.20 fc

$c.\dot{x} = 0.20 Cc .\dot{x}$

$\xi = \frac{c}{Cc} = 0.20$

N = 2000 rpm

X = 0.06 mm

Solution:

$Tr = \frac{ft.r}{f_o} = \frac{\sqrt{ 1+ (2 \xi r)^2}}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

$r = \frac{w}{w_n}$

$w = \frac{2 \pi N}{60}$

$w = \frac{2 \pi \times 2000}{60} = 209.43$ r/sec

$w_n = \sqrt{ \frac{4k}{m}} = \sqrt{ \frac{4 \times 3.92 \times 10^6}{1000}}$

$w_n = 125.21 \ rad/sec$

$r = \frac{w}{w_n} = \frac{209.43}{125.21} = 1.67$

$T_r = \frac{1.20}{1.909}$

$T_r$ = 0.6284

$fo \ = \ \frac{mo \ . \ e. \ w^2}{un \ balanced}$

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

$\frac{0.06 \times 10^{-3}}{\frac{mo.e}{1000}} = \frac{2.7889}{1.909}$

$0.06 \times 10^{-3} = 1.46092 \times \frac{mo.e}{1000}$

mo.e = 0.041069 kg-m

$\therefore$ $Fo = 0.041069 \times (209.43)^2$

= 1801.366 N

$\frac{F_{tr}}{F_o} = 0.6284$

$F_{tr}$ = 1131.97N

($F_{tr}$) each = 282.99 N