Question 8.11: Calculate the response of the tower in Example 8.9 to the lo...

The following properties have been obtained in Examples 8.9 and 8.10.

\begin{aligned}k &=100  \mathrm{kips} / \mathrm{in} . \\m &=2.533  \mathrm{kip} \cdot \mathrm{s}^{2} / \mathrm{in} . \\\omega &=6.283  \mathrm{rad} / \mathrm{s} \\\omega_{d} &=6.2515  \mathrm{rad} / \mathrm{s} \\u_{0} &=\dot{u}_{0}=\ddot{u}_{0}=0\end{aligned}

The coefficients in the recurrence formulas are determined from Equation 8.83.

\begin{aligned}A &=e^{-\xi \omega h}\left(\frac{\xi}{\sqrt{1-\xi^2}} \sin \omega_d h+\cos \omega_d b\right) \\B &=e^{-\xi \omega h}\left(\frac{1}{\omega_d} \sin \omega_d h\right) \\C &=\frac{1}{k}\left[\frac{2 \xi}{\omega h}+e^{-\xi \omega h}\left\{\left(\frac{1-2 \xi^2}{\omega_d h}-\frac{\xi}{\sqrt{1-\xi^2}}\right) \sin \omega_d b-\left(1+\frac{2 \xi}{\omega h}\right) \cos \omega_d h\right\}\right] \\D &=\frac{1}{k}\left\{1-\frac{2 \xi}{\omega h}+e^{-\xi \omega h}\left(\frac{2 \xi^2-1}{\omega_d h} \sin \omega_d h+\frac{2 \xi}{\omega h} \cos \omega_d h\right)\right\} \\A_1 &=-e^{-\xi \omega h}\left(\frac{\omega}{\sqrt{1-\xi^2}} \sin \omega_d h\right) \\B_1 &=e^{-\xi \omega h}\left(\cos \omega_d h-\frac{\xi}{\sqrt{1-\xi^2}} \sin \omega_d h\right) \\C_1 &=\frac{1}{k}\left[-\frac{1}{h}+e^{-\xi \omega h}\left\{\left(\frac{\omega}{\sqrt{1-\xi^2}}+\frac{\xi}{h \sqrt{1-\xi^2}}\right) \sin \omega_d h+\frac{1}{h} \cos \omega_d h\right\}\right] \\D_1 &=\frac{1}{k}\left\{\frac{1}{h}-\frac{e^{-\xi \omega h}}{h}\left(\frac{\xi}{\sqrt{1-\xi^2}} \sin \omega_d h+\cos \omega_d h\right)\right\}\end{aligned}  \qquad  (8.83)

\begin{gathered}A=0.81672, \quad B=0.08791, \quad C=0.0012073, \quad D=0.00062547, \\A_{1}=-3.4706, \quad B_{1}=0.70625, \quad C_{1}=0.016378, \quad D_{1}=0.018328\end{gathered}

The displacements and velocities obtained from Equations 8.78 and 8.80,

u_{n+1}=A u_n+B \dot{u}_n+C p_n+D p_{n+1}        (8.78)

\dot{u}_{n+1}=A_1 u_n+B_1 \dot{u}_n+C_1 p_n+D_1 p_{n+1}        (8.80)

respectively, are shown in Table E8.11. The theoretical response for a sine-wave loading is given by Equation \mathrm{b} of Example 8.10. The equation is used to obtain the theoretical response values for the first 0.6 \mathrm{~s} shown in Table E8.11. For finding the response at time intervals beyond the first 0.6 \mathrm{~s}, we need the displacement as well as the velocity at t=0.6 \mathrm{~s}. The displacement has already been calculated; the velocity is obtained by differentiating Equation b of Example 8.10 and substituting t=0.6 \mathrm{~s}. Displacement response beyond 0.6 \mathrm{~s} is now given by

u(t)=e^{-\xi \omega(t-0.6)}\left\{\frac{v_{0.6}+u_{0.6} \xi \omega}{\omega_{d}} \sin \omega_{d}(t-0.6)+u_{0.6} \cos \omega_{d}(t-0.6)\right\} \qquad (j)

where u_{0.6} and v_{0.6} are the displacement and velocity, respectively, at t=0.6 \mathrm{~s}.

Theoretical displacement values for t \geq 0.6 \mathrm{~s} are calculated from Equation \mathrm{j} and are entered in Table E8.11. The difference between the calculated and theoretical values of the response can be attributed to the difference between the actual forcing function and its piece-wise linear representation. To improve the accuracy of computations we need to use a smaller interval of time, so that the piece-wise linear representation more closely matches the actual function.

Table E8.11 Piece-wise linear representation of the excitation.
\begin{matrix} \hline & & & & & u_{n+1} & {\dot{u}_{n+1}} & {u_{n}} \\Time & {u_{n}} & {\dot{u}_{n}} & {p_{n}} & p_{n+1} & { (Eq. 8.73) } & { (Eq. 8.75) } & (\text{theoretical}) \\\hline 0.0 & 0.0000 & 0.0000 & 0.0 & 50.0 & 0.0313 & 0.9164 & 0.0000 \\0.1 & 0.0313 & 0.9164 & 50.0 & 86.6 & 0.2206 & 2.9448 & 0.0323 \\0.2 & 0.2206 & 2.9448 & 86.6 & 100.0 & 0.6062 & 4.5651 & 0.2254 \\0.3 & 0.6062 & 4.5651 & 100.0 & 86.6 & 1.0713 & 4.3453 & 0.6204 \\0.4 & 1.0713 & 4.3453 & 86.6 & 50.0 & 1.3928 & 1.6855 & 1.0961 \\0.5 & 1.3928 & 1.6855 & 50.0 & 0.0 & 1.3461 & -2.8245 & 1.4251 \\0.6 & 1.3461 & -2.8245 & 0.0 & 0.0 & 0.8510 & -6.6664 & 1.3772 \\0.7 & 0.8510 & -6.6664 & 0.0 & 0.0 & 0.1090 & -7.6618 & 0.8683 \\0.8 & 0.1090 & -7.6618 & 0.0 & 0.0 & -0.5845 & -5.7895 & 0.1105 \\0.9 & -0.5845 & -5.7895 & 0.0 & 0.0 & -0.9864 & -2.0601 & -0.5974 \\1.0 & -0.9864 & -2.0601 & 0.0 & & & & -1.0073 \\\hline \end{matrix}