written 5.7 years ago by |
Cantilever beam of cross sectional moment of inertia $1.3 \times 10^{-6} m^4$ when the motor operates at 200 r.p.m., the phase difference between the operation of the motor and the response of the beam is 5˚ assuming the viscous damping, estimate the damping ratio of the beam.
I = $1.3 \times 10^{-6} m^4$
N = 200 r.p.m
$\phi$ = 5˚
M = 65 kg
$\xi$ = ?
$K \ beam = \frac{3EI}{L^3}$
$= \frac{3 \times 210 \times 10^9 \times 1.3 \times 10^-6}{(1.3)^3}$
$= 372.78 \times 10^3 N/m$
$w_n = \sqrt{ \frac{K \ beam}{m}} = \sqrt{ \frac{372.78 \times 10^3}{65}} = 75.73$ rad/sec
$w = \frac{2 \pi N }{60} = \frac{2 \times \pi \times 200}{60} = 20.94$ rad/sec
$tan \ \phi \ = \ \frac{2 \xi r}{1 – r^2}$
$r = \frac{20.94}{75.73} = 0.276$
$tan \ 5 \ = \ \frac{2 \times \xi 0.276}{1 – (0.276)^2}$
$0.08748 \ = \ \frac{0.55 \xi}{0.9235}$
$\xi \ = \ 0.14690$