Question 8.10: Repeat Example 8.9 with 10% damping in the system. Use the t...

The displacement response is now given by

u(t)=A \sin \omega_{d} t-B \cos \omega_{d} t \qquad (a)

where A and B are as in Equation 8.70, and

\begin{aligned}\omega_{d} &=\omega \sqrt{1-\xi^{2}} \\&=6.283 \sqrt{1-0.1^{2}} \\&=6.2515 \mathrm{rad} / \mathrm{s}\end{aligned}

Table E8. 10a Numerical evaluation of Duhamel’s equation, damped system.
\begin{matrix} \\ \hline \tau & p(\tau) &&& p(\tau) \mathrm{e}^{\xi \omega \tau} & \bar{A} & & p(\tau) \mathrm{e}^{\xi \omega \tau} & \bar{B}\\(s) & (\mathrm{kips}) & \mathrm{e}^{\xi \omega \tau} & \cos \omega_{\mathrm{d}} \tau & \cos \omega_{\mathrm{d}} \tau & \text{trapezoidal }& \sin \omega_{d} \tau & \sin \omega_{\mathrm{d}} \tau & \text{trapezoidal }\\\hline 0.0 & 0.0 & 1.0000 & 1.0000 & 0.000 & 0.000 & 0.0000 & 0.000 & 0.000 \\0.1 & 50.0 & 1.0648 & 0.8109 & 43.172 & 43.172 & 0.5852 & 31.156 & 31.156 \\0.2 & 86.6 & 1.1339 & 0.3150 & 30.392 & 117.276 & 0.9491 & 93.198 & 155.510 \\0.3 & 100.0 & 1.2074 & -0.3000 & -36.222 & 111.986 & 0.9540 & 115.186 & 363.894 \\0.4 & 86.6 & 1.2857 & -0.8015 & -89.240 & -13.476 & 0.5980 & 66.823 & 545.903 \\0.5 & 50.0 & 1.3691 & -0.9999 & -68.448 & -171.164 & 0.0158 & 1.084 & 613.810 \\0.6 & 0.0 & 1.4579 & -0.8200 & 0.000 & -239.612 & -0.5723 & 0.000 & 614.894 \\\hline\end{matrix}

Table E8. 10b Numerical evaluation of Duhamel’s equation, damped system.

The response calculations for the first 0.6 \mathrm{~s} are shown in Tables E8.10a and E8.10b. After 0.6 \mathrm{~s}, A and B remain unchanged and the response for subsequent intervals of time is given by Equation a with A and B equal to their values at 0.6 \mathrm{~s}.

The exact response to the sine-wave force for the first 0.6 \mathrm{~s} is given by

\begin{aligned}u(t)=& \frac{p_{0}}{k} \frac{1}{\left(1-\beta^{2}\right)^{2}+(2 \beta \xi)^{2}}\left\{\left(1-\beta^{2}\right) \sin \Omega t-2 \beta \xi \cos \Omega t\right\} \\&+\frac{p_{0}}{k} \frac{e^{-\omega \xi t}}{\left(1-\beta^{2}\right)^{2}+(2 \beta \xi)^{2}}\left\{2 \beta \xi \cos \omega_{d} t+\frac{\Omega}{\omega_{d}}\left(1+\beta^{2}-2 \frac{\omega_{d}^{2}}{\omega^{2}}\right) \sin \omega_{d} t\right\} \qquad (b)\end{aligned}

where p_{0}=100  \mathrm{kips}, \Omega=5.236  \mathrm{rad} / \mathrm{s}, and \beta=\Omega / \omega=0.833. The response values obtained from Equation b are also shown in Table E8.10b.