**1 Answer**

written 5.0 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

$tan \ \phi \ = \ \frac{2 \xi r}{1 – r^2}$

$r = \frac{20.94}{75.73} = 0.276$

$tan \ 5 \ = \ \frac{2 \times \xi 0.276}{1 – (0.276)^2}$

$0.08748 \ = \ \frac{0.55 \xi}{0.9235}$

$\xi \ = \ 0.14690$