2. In Fig. 1 below, the temporal variation of the output displacement of a simpl

Question

2. In Fig. 1 below, the temporal variation of the output displacement of a simply supported cantilever beam undergoing a free vibration is shown. It is seen that the amplitude of the beam displacement is reduced to about half of the previous amplitude, after each half cycle, e.g., x1.5 = x1 / 2.5, x2 = x1.5 / 2.5, x2.5 = x2 / 2.5, etc. Determine the damping ratio of the beam system.

Fig. 1. Displacement-time curve measured on a vibrating cantilever beam

Explanation / Answer

Time Period,T = 1 sec

T = 2*Pi/wd

wd = 2*pi = 6.28

x2/x1 = e^(-zeta*wn)*(t2-t1) = e^-(zeta*wn) Since (t2-t1 = 1)

(1/2.5)^2 = e^-(zeta*wn)

zeta*wn = 1.833------(1)

wd = wn*Sqrt((1-zeta^2)) = 6.28 ------(2)

From 1 and 2

(1-zeta^2)/zeta^2 = (6.28/1.833)^2

1-zeta^2 = 11.744 zeta^2

zeta = 0.28

Damping ratio = 0.28

wn = 6.543 rad/s

wd = 6.281 rad/s

zeta = 0.28

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