Question 6.8: A water tower with a total weight of 50 kips is represented ...

(a) For viscous damping, the energy loss per cycle is given by Equation 6.96

\begin{aligned} W_i &=c \Omega^2 \rho^2 \int_0^{2 \pi / \Omega} \cos ^2(\Omega t-\phi) d t \\ &=c \pi \Omega \rho^2 \end{aligned} \qquad (6.96)

W_{D}=c \pi \Omega \rho^{2}            (a)

With \Omega=10  \mathrm{rad} / \mathrm{s}, \rho=2 in., and W_{D}=65.2  \mathrm{kip} \cdot in., Equation a gives

c=0.519 \frac{\mathrm{kip} \cdot \mathrm{s}}{\text { in. }}

The natural frequency of the system is obtained from

\begin{aligned}\omega=\sqrt{\frac{k}{m}} &=\sqrt{\frac{52 \times 386.4}{50}} \\&=20.05  \mathrm{rad} / \mathrm{s}\end{aligned}

The damping ratio is given by

\begin{aligned}\xi=\frac{c}{2 m \omega} &=\frac{0.519 \times 386.4}{2 \times 50 \times 20.05} \\&=0.1\end{aligned}

(b) In this case the energy loss is given by Equation 6.104

W_i =\eta k \pi \rho_b^2 \qquad (6.104)

W=\eta k \pi \rho^{2}            (b)

Substituting for W, k, and \rho in Equation \mathrm{b} we get

\eta=0.1

(c) If the damping is viscous, the energy loss per cycle at a test frequency of 20  \mathrm{rad} / \mathrm{s} will be given by

\begin{aligned}W &=c \pi \Omega \rho^{2} \\&=0.519 \times \pi \times 20 \times 4 \\&=130.4  \mathrm{kip} \cdot \mathrm{in} .\end{aligned}

If damping were hysteretic, the energy loss per cycle will not change and will still be 65.2  \mathrm{kip} \cdot \mathrm{in}.