Question 9.10: The gravity dam and the impounded reservoir shown in Figure ...
The gravity dam and the impounded reservoir shown in Figure E9.10a are excited by an earthquake motion along the y axis, that is, in the vertical direction. Assuming that the ground motion does not vary along the length of the dam and the dam is long in comparison to its cross-sectional dimension, the system can be represented by the two-dimensional model shown in Figure E9.10a. It can be shown that when ground motion is e^{i \Omega t}, the total pressure on the face of the dam, expressed as a fraction of the hydrostatic force F_{s}=\frac{1}{2} w H^{2}, is given by
F(\Omega)=\frac{\tilde{F}(\Omega)}{F_{s}}=\frac{2 c^{2}}{g \Omega^{2} H^{2}} \frac{1-\cos (\Omega H / c)}{\cos (\Omega H / c)} \qquad (a)
where w is the unit weight of water, H the height of the reservoir =200 \mathrm{~m}, c the velocity of sound in water =1440 \mathrm{~m} / \mathrm{s}, and g is the acceleration due to gravity =9.81 \mathrm{~m} / \mathrm{s}^{2}. The variation of ground acceleration is also shown in Figure E9.9a. Calculate the variation of total hydrodynamic force on the dam face with time for a duration of 25 \mathrm{~s}.
Since the reservoir bottom is rigid and there is no loss of energy by wave radiation, the system is completely undamped. Also, it is the frequency response function rather than the unit impulse response function that is specified. The standard frequency domain analysis procedures using discrete Fourier transforms do not therefore work. However, the exponential window method is quite effective. In carrying out the analysis the following data is used: time step \Delta t=0.025 \mathrm{~s}; T_{0}=25 \mathrm{~s}; frequency step \Delta \Omega=2 \pi / T_{0}=0.2513 \mathrm{rad} / \mathrm{s}. The exponential decay factor a is obtained as follows
a=2 \frac{\ln 10}{T_{0}}=0.184 \qquad (b)
The exciting function g(t) is sampled at intervals of 0.025, giving N=1000 samples and scaled by the exponential decay function e^{-a k \Delta t}. Discrete Fourier transform of the scaled function \hat{g}(t) is obtained using a standard computer program. A discrete version of the frequency response function \hat{F}(\Omega) must be obtained next. Values of \hat{F}(\Omega) are obtained from Equation a at frequency intervals of \Delta \Omega=0.2513 \mathrm{rad} / \mathrm{s} by substituting \Omega=(n \Delta \Omega-i a) instead of n \Delta \Omega. Discretization is carried out upto a frequency of \pi / \Delta t=125.7 corresponding to n=500. The sample values of \hat{F}(\Omega) for \Omega>\pi / \Delta t are taken as being the complex conjugates of values for frequencies located symmetrically about \pi / \Delta t. Thus \hat{F}(N / 2+m)=\hat{F} *(N / 2-m), where * denotes a complex conjugate. The scaled response \hat{p}(t) is obtained by taking the inverse discrete Fourier transform of the product of \hat{G}(\Omega) and \hat{F}(\Omega). The desired response, that is the normalized total force on the dam face, is now calculated from p(k \Delta t)=e^{a k \Delta t} \hat{p}(k \Delta t). The results are presented for the first 500 sample points in Figure E9.10b.