Question 8.9: The water tower shown in Figure E8.9a is idealized as a sing...

The mass m of the tower and the frequency \omega are obtained first:

\begin{aligned}m &=\frac{978.8}{386.4}=2.533  \mathrm{kip} \cdot \mathrm{s}^{2} / \mathrm{in} . \\\omega &=\sqrt{\frac{k}{m}}=\sqrt{\frac{100}{2.533}}=6.283  \mathrm{rad} / \mathrm{s}\end{aligned}

The displacement response is obtained in terms of Duhamel’s integral as

u(t)=A \sin \omega t-B \cos \omega t \qquad (a)

where A and B have been defined in Equation 8.68. The response calculations for the first 0.6 \mathrm{~s} are shown in Tables E8.9a and E8.9b, where results are given for three alternative methods rectangular summation, trapezoidal method, and the Simpson’s method: 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}.

Because the forcing function is, in this case, a simple mathematical function, it is possible to obtain a closed-form solution. When the sine-wave function is expressed as

P(t)=p_{0} \sin \Omega t \qquad (b)

the response for the first 0.6 \mathrm{~s} is given by

u(t)=\frac{p_{0}}{k} \frac{1}{1-\beta^{2}}(\sin \Omega t-\beta \sin \omega t) \qquad (c)

where \beta=\Omega / \omega. We have

\begin{aligned}\Omega &=\frac{\pi}{0.6}=5.236  \mathrm{rad} / \mathrm{s} \\\beta &=\frac{\Omega}{\omega}=\frac{5.236}{6.283}=0.833 \\p_{0} &=100  \mathrm{kips}\end{aligned}

so that

u(t)=3.267(\sin 5.236 t-0.833 \sin 6.283 t) \qquad (d)

Table E8.9a Numerical evaluation of Duhamel’s equation, undamped system.
\begin{matrix} \hline \\ &&&& \bar{A} &&&&&\bar{B} \\\tau & P(\tau) & & p(\tau) \times & \text { Rectangular } & \text { Trapezoidal } & \text { Simpson's } && p(\tau) \times & \text { Rectangular } & \text { Trapezoidal } & \text { Simpson's } \\\text { (s) } & (\text { kips }) & \cos \omega \tau & \cos \omega \tau & \text { sum } & \text { method } & \text { method } & \sin\omega \tau & \sin\omega \tau & \text { sum } & \text { method } & \text { method }\\ \hline (1) &(2)&(3)&(4)&(5)&(6)&(7)&(8)&(9)&(10)&(11)&(12)\\\hline 0.0 & 0.0 & 1.000 & 0.00 & 0.00 & 0.0 & 0.0 & 0.000 & 0.00 & 0.0 & 0.0 &0.0 \\0.1 & 50.0 & 0.809 & 40.45 & 0.00 & 40.5 & & 0.588 & 29.39 & 0.0 & 29.4 \\0.2 & 86.6 & 0.309 & 26.76 & 40.45 & 107.7 & 188.6 & 0.951 & 82.36 & 29.4 & 141.1 & 199.9 \\0.3 & 100.0 & -0.309 & -30.90 & 67.21 & 103.5 & & 0.957 & 95.10 & 111.7 & 318.6 \\0.4 & 86.6 & -0.809 & -70.06 & 36.32 & 2.6 & 21.7 & 0.588 & 50.92 & 206.8 & 464.6 & 713.6\\0.5 & 50.0 & -1.000 & -50.00 & -33.74 & -117.5 & & 0.000 & 0.00 & 257.8 & 515.5 \\0.6 & 0.0 & -0.809 & 0.00 & -83.74 & -167.5 & -248.4 & -0.588 & 0.00 & 257.8 & 515.5 & 764.5\\ \hline\end{matrix}

\begin{aligned}\frac{\Delta \tau}{m \omega} &=\frac{0.1}{2.533 \times 6.283}=6.283 \times 10^{-3} \\\frac{\Delta \tau}{2 m \omega} &=3.142 \times 10^{-3} \\\frac{\Delta \tau}{3 m \omega} &=2.094 \times 10^{-3}\end{aligned}

Table E8.9b Numerical evaluation of Duhamel’s equation, undamped system.
\begin{matrix}\hline &&&\text{Rectangular }&\text{sum }&&\text{Trapezoidal }&\text{method }&&\text{Simpson’s }&\text{method }&\\ &&&&&A \sin ωt &&&A \sin ωt &&&A \sin ωt \\t & \sin \omega t & \cos \omega t & A=\bar{A} \frac{\Delta \tau}{m \omega} & B=\bar{B} \frac{\Delta \tau}{m \omega} & -B \cos \omega t & A=\bar{A} \frac{\Delta \tau}{2 m \omega} & B=\bar{B} \frac{\Delta \tau}{2 m \omega} & -B \cos \omega t & A=\bar{A} \frac{\Delta \tau}{3 m \omega} & B=\bar{B} \frac{\Delta \tau}{3 m \omega} & -B \cos \omega t \\\hline 0.1 & 0.588 & 0.809 & 0.0000 & 0.0000 & 0.0000 & 0.1271 & 0.0923 & 0.0000 & \\0.2 & 0.951 & 0.309 & 0.2542 & 0.1847 & 0.1847 & 0.3382 & 0.4434 & 0.1846 & 0.3949 & 0.4187 & 0.2458\\0.3 & 0.951 & -0.309 & 0.4223 & 0.7019 & 0.6182 & 0.3252 & 1.0010 & 0.6184 & \\0.4 & 0.588 & -0.809 & 0.2282 & 1.2994 & 1.1849 & 0.0080 & 1.4597 & 1.1856 & 0.0454 & 1.4946 & 1.2358 \\0.5 & 0.000 & -1.000 & -0.2434 & 1.6199 & 1.6199 & -0.3692 & 1.6197 & 1.6197 & \\0.6 & -0.588 & -0.809 & -0.5262 & 1.6199 & 1.6199 & -0.5262 & 1.6197 & 1.6197 & -0.5203 & 1.6013 & 1.6162 \\ \hline \end{matrix}

Table E8.9c Comparison of numerical solution with theoretical results.
\begin{array}{lcc}\hline & \text{Duhamel integral} & \text{Closed-form solution,}\\ t & A \sin \omega t-B \cos \omega t & \text{Equation d or e} \\\hline 0.1 & 0.0000 & 0.0333 \\0.2 & 0.1846 & 0.241 \mathrm{I} \\0.3 & 0.6184 & 0.6788 \\0.4 & 1.1856 & 1.2290 \\0.5 & 1.6197 & 1.6334 \\0.6 & 1.6197 & 1.6000 \\0.7 & 1.0000 & 0.9880 \\0.8 & 0.0000 & 0.0000 \\0.9 & -1.0000 & -0.9900 \\1.0 & -1.6197 & -1.6006 \\ \hline \end{array}

\begin{aligned}\text { At } t &=0.6 \mathrm{~s}, \\u(t) &=3.267(\sin 0.6 \times 5.236-0.833 \sin 0.6 \times 6.283) \\&=1.6 \text { in. } \\\dot{u}(t) &=3.267(5.236 \cos 0.6 \times 5.236-0.833 \times 6.283 \cos 6.283 \times 0.6) \\&=-3.272 \text { in. } / \mathrm{s}\end{aligned}

The exact response subsequent to 0.6 \mathrm{~s} is given by

u(t)=1.6 \cos \omega(t-0.6)-\frac{3.272}{6.283} \sin (t-0.6) \qquad (e)

The theoretical response values for the first 1 \mathrm{~s} obtained from Equations \mathrm{d} and e are compared with the results of the numerical evaluation of Duhamel’s integral using the trapezoidal method (Table E8.9c).