Question 16.3: The 30-story building shown in Figure E16.3a has a uniform m...
The 30-story building shown in Figure E16.3a has a uniform mass and story stiffness throughout the height. The building mass can be assumed to be lumped at the floor levels and is 16.4 metric ton per floor. The floor to floor height is 3.66 \mathrm{~m} and the sum of the flexural rigidities of column in a story is 42 \times 10^{4} \mathrm{kN} \cdot \mathrm{m}^{2}. The building, which can be idealized as a shear beam, is subjected to a specified ground motion.
(a) If the ground motion consists of an acceleration pulse of duration t_{1}=0.60 \mathrm{~s}, as shown in Figure E16.3b, obtain the maximum base shear and the maximum top-story deflection in the first mode of vibration. Also determine the time at which these maximum values are attained. Then obtain the base shear and top displacements at this instant of time in the second and third modes of the building. By superposing the modal responses for the three modes, obtain an estimate of the total response.
An upper bound estimate of the total response in the first three modes can be obtained by taking the sum of the absolute maximum values in each of the three modes. Compare the upper bound estimate of the base shear with the estimate obtained earlier.
(b) Solve the example in part (a) with t_{1}=1.5 \mathrm{~s}.
The governing equation and the modal equations are given by Equations a and \mathrm{b} of Example 16.2, respectively. The effective modal force p_{n} is obtained from Equation \mathrm{f} of that example. When the ground acceleration a is in the form of a rectangular pulse, the solution to the modal equations are similar to those derived in Section 7.7.1:
The response spectrum for a rectangular pulse function was also derived in Section 7.7.1. For \omega_{n} t_{1} / 2 \pi \geq 1 / 2, the maximum displacement response occurs at t_{p}=\pi / \omega_{n} and its value is given by
\left(y_{n}\right)_{\max }=\frac{2 p_{n}}{\omega_{n}^{2}} \quad \text { (c) }
In this case the maximum response is the same as that for a rectangular pulse of infinite duration.
When \omega_{n} t_{1} / 2 \pi<1 / 2, the maximum response occurs at t_{p}=\pi / 2 \omega_{n}+t_{1} / 2 and its value is given by
\left(y_{n}\right)_{\max }=\frac{2 p_{n}}{\omega_{n}^{2}} \sin \frac{\omega_{n} t_{1}}{2}(\mathrm{~d})
For the given building, we have
Hence \omega_{1}=4.393, \omega_{2}=13.18, and \omega_{3}=21.97.
(a) For the first mode
\begin{aligned}\frac{\omega_{1} t_{1}}{2 \pi} &=\frac{4.393 \times 0.6}{2 \pi} \\&=0.419<0.5\end{aligned}
Hence the maximum occurs at
\begin{aligned}t_{p} &=\frac{\pi}{2 \omega_{1}}+\frac{t_{1}}{2} \\&=0.657 \mathrm{~s}\end{aligned}
and the value of the maximum modal response is
\left(y_{1}\right)_{\max }=\frac{4 m L a}{\pi \omega_{1}^{2}} \sqrt{\frac{2}{m L}} \sin \frac{\omega_{1} t_{1}}{2} \quad \text { (e) }
The corresponding response in the physical coordinate is
\left(u_{1}\right)_{\max }=\frac{8 a}{\pi \omega_{1}^{2}} \sin \frac{\omega_{1} t_{1}}{2} \sin \frac{\pi x}{2 L} \quad \text { (f) }
The top displacement is obtained by substituting x=L in Equation \mathrm{f}
\left\{u_{1}(L)\right\}_{\max }=\frac{32 a m L^{2}}{\pi^{3} k} \sin \frac{\omega_{1} t_{1}}{2} \qquad (\mathrm{g})
Equation g may be compared with Equation m of Example 16.2
Substitution of appropriate values in Equation g gives
\begin{aligned}\left\{u_{1}(L)\right\}_{\max } &=0.1319 a \sin \frac{\omega_{1} t_{1}}{2} \\&=0.1277 a \mathrm{~m}\end{aligned}
For the second mode, the response at t=0.657 \mathrm{~s} is obtained from Equation \mathrm{b}
The response in the physical coordinate is given by
The top displacement becomes
u_{2}(L, 0.657)=-0.00355 a \mathrm{~m}
Also, for the second mode
\frac{\omega_{2} t_{1}}{2 \pi}=1.258>0.5 \mathrm{~s}
Hence the maximum response in the second mode occurs at t_{p}=\pi / \omega_{2}=0.238 \mathrm{~s} and its value is given by
\left(u_{2}\right)_{\max }=\frac{8 a}{3 \pi \omega_{2}^{2}} \sin \frac{3 \pi x}{2 L}The top displacement is
\begin{aligned}\left\{u_{2}(L)\right\}_{\max } &=\frac{32 a m L^{2}}{27 \pi^{3} k} \\&=0.00489 a \mathrm{~m}\end{aligned}
For the third mode the response at t=0.657 \mathrm{~s} is obtained from
The top displacement is given by
u_{3}(L, 0.657)=0.00032 a \mathrm{~m}
The maximum response in the third mode occurs at t_{p}=\pi / \omega_{3}=0.143 \mathrm{~s} and its value is
\left(u_{3}\right)_{\max }=\frac{8 a}{5 \pi \omega_{3}^{2}} \sin \frac{5 \pi x}{2 L}
The top displacement is
\begin{aligned}\left\{u_{3}(L)\right\}_{\max } &=\frac{32 a m L^{2}}{125 \pi^{3} k} \\&=0.00106 a \mathrm{~m}\end{aligned}
The top-story deflection at t=0.657 \mathrm{~s} considering the first three modes is given by
\begin{aligned}u(L, 0.657) &=(0.1277-0.00355+0.00032) a \\&=0.1245 a \mathrm{~m}\end{aligned}The estimate obtained by taking the absolute sum of the modal maxima is
\begin{aligned}\{u(L)\}_{\max } &=(0.1277+0.00489+0.00106) a \\&=0.1338 a \mathrm{~m}\end{aligned}
The base shear is obtained from
V_{0}=\left(k \frac{\partial u}{\partial x}\right)_{x=0}
The base shear at t=0.657 \mathrm{~s} is therefore given by
while the estimate obtained by taking the absolute sum of the modal shears is
\begin{aligned}V_{\max } &=(0.1277+3 \times 0.00489+5 \times 0.00106) \frac{\pi k}{2 L} a \\&=795.4 a \mathrm{kN}\end{aligned}(b) In this case, \omega_{1} t_{1} / 2 \pi=1.048 is greater than 0.5. Hence the maximum response in the first mode occurs at t_{p}=\omega_{1} / \pi=1.398 \mathrm{~s} and is given by
\begin{aligned}\left\{u_{1}(L)\right\}_{\max } &=\frac{32 a m L^{2}}{\pi^{3} k} \\&=0.1319 a \mathrm{~m}\end{aligned}The modal response in the second mode at t=1.398 \mathrm{~s} is obtained from Equation a
\begin{aligned}y_{2}(t) &=\frac{p_{2}}{\omega_{2}^{2}}\left(1-\cos \omega_{2} t\right) \\&=\frac{2 L m a}{3 \pi} \sqrt{\frac{2}{m L}} \frac{1}{\omega_{2}^{2}}\left(1-\cos 1.398 \omega_{2}\right)\end{aligned}The response in the physical coordinate is
\begin{aligned}u_{2}(x, 1.398) &=\frac{4 a}{3 \pi \omega_{2}^{2}}\left(1-\cos 1.398 \omega_{2}\right) \sin \frac{3 \pi x}{2 L} \\&=0.000216 \sin \frac{3 \pi x}{2 L}\end{aligned}The top displacement becomes
u_{2}(L, 1.398)=-0.000216 a \mathrm{~m}
The modal response in the third mode at t=1.398 \mathrm{~s} is given by
y_{3}(t)=\frac{2 L m a}{5 \pi} \sqrt{\frac{2}{m L}} \frac{1}{\omega_{3}^{2}}\left(1-\cos 1.398 \omega_{3}\right)
The response in the physical coordinate is
\begin{aligned}u_{3}(x, 1.398) &=\frac{4 a}{5 \pi \omega_{3}^{2}}\left(1-\cos 1.398 \omega_{3}\right) \\&=0.000125 a \mathrm{~m}\end{aligned}The top-story displacement at t=1.398 \mathrm{~s} considering the first three modes is given by
\begin{aligned}u(L, 1.398) &=(0.1319-0.000216+0.000125) a \\&=0.1318 a \mathrm{~m}\end{aligned}
The estimate obtained by taking the absolute sum of the modal maxima is
\begin{aligned}\{u(L)\}_{\max } &=(0.1319+0.00489+0.00106) a \\&=0.1379 \mathrm{~m}\end{aligned}
The base shear at t=1.398 \mathrm{~s} is given by
The estimate obtained by taking the absolute sum of the modal shear is
When all modes of the system are included, the base shear is obtained from Equation q of Example 16.2
\begin{aligned}V(0)_{\max } &=2 a m L \\&=876.0 a \mathrm{kN}\end{aligned}
Thus the first three modes together account for 93.3 \% of the total base shear, while the first mode alone accounts for 81.0 \%