Question

A machine f mass 1000 kg is supported on springs which deflect 8 mm under static load with negligible damping of the machine vibrates with an amplitude of 5 mm when subjected to a vertical harmonic force at 80 percent of the resonant frequency. When the damper is fitted it is found that the resonant amplitude is 2 mm. Find ( I ) Amplitude of the damping force and ( ii ) damping co-efficient

Answer 1

Question #81302, Engineering / Mechanical Engineering

A machine f mass 1000 kg is supported on springs which deflect 8 mm under static load with negligible damping of the machine vibrates with an amplitude of 5 mm when subjected to a vertical harmonic force at 80 percent of the resonant frequency. When the damper is fitted it is found that the resonant amplitude is 2 mm. Find (I) Amplitude of the damping force and (ii) damping co-efficient

Solution

i) Amplitude of the damping force:

$$F = c \omega x _ {d}$$ $$\omega = \sqrt {\frac {k}{m}} = \sqrt {\frac {\overline {{W}}}{\frac {\delta}{\delta}}} = \sqrt {\frac {\overline {{m g}}}{\frac {\delta}{\delta}}} = \sqrt {\frac {g}{\delta}}$$ $$F = c \sqrt {\frac {g}{\delta}} x _ {d}$$ $$F = 31516 \sqrt {\frac {9.81}{0.008}} \cdot 0.002 = 2207 \cdot N.$$

ii) The damping co-efficient:

$$c = 2 \sqrt {m k} \zeta = 2 \sqrt {\frac {m W}{\delta}} \zeta = 2 \sqrt {\frac {m m g}{\delta}} \zeta$$ $$X = \frac {x _ {s t}}{1 - \beta^ {2}} \rightarrow 5 = \frac {x _ {s t}}{1 - 0.8 ^ {2}}$$ $$x _ {s t} = 1.8 \text{ mm}.$$ $$\frac {x _ {d}}{x _ {s t}} = \frac {1}{2 \zeta}$$ $$\frac {2}{1.8} = \frac {1}{2 \zeta} \rightarrow \zeta = 0.45$$ $$c = 2 \sqrt {\frac {(1000) ^ {2} (9.81)}{0.008}} \cdot 0.45 = 31516 \frac {N s}{m}$$