Question 2.7.2: An experiment is performed on a hysteretic system with known......
An experiment is performed on a hysteretic system with known spring stiffness of k = 4 × 10^4 N/m. The system is driven at resonance, the area of the hysteresis loop is measured to be ΔE = 30 N · m, and the amplitude, X, is measured to be X = 0.02 m. Calculate the magnitude of the driving force and the hysteretic damping constant.
At resonance, equation (2.125) yields
X=\frac{F_0 / k}{\sqrt{\left(1-r^2\right)^2+\beta^2}} (2.125)
X=\frac{F_0}{k \beta}
or k \beta=F_0 / X. The area enclosed by the hysteresis loop is equal to \pi k \beta X^2, so that
30\ N \cdot m =\pi k \beta X^2=\pi \frac{F_0}{X} X^2
and hence
F_0=\frac{30\ N \cdot m }{\pi X}=\frac{30\ N \cdot m }{\pi(0.02\ m )}=477.5\ N
From the resonance expression,
\beta=\frac{F_0}{X k}=\frac{477.5}{(0.02\ m )\left(4 \times 10^4\ N / m \right)}=0.60
which is the hysteretic damping constant calculated based on the principle of equivalent viscous damping.