Question 9.8: Solve the problem of Example 9.6 by superposition of correct...
Solve the problem of Example 9.6 by superposition of corrective responses obtained by applying two force pulses of appropriate magnitude.
As in Example 9.6, response is obtained over a period of 2.4 \mathrm{~s}, so that N=24. A force pulse of magnitude R_{1} is applied at time point N-2, and a pulse of magnitude R_{2} is applied at N-1. The unknown magnitudes R_{1} and R_{2} are determined by solving Equations 9.131,
which can be expressed as
\left[\begin{array}{ll}\bar{h}(2 \Delta t) & \bar{h}(\Delta t) \\\dot{\bar{h}}(2 \Delta t) & \dot{\bar{h}}(\Delta t)\end{array}\right]\left[\begin{array}{l}R_{1} \\R_{2}\end{array}\right]=\left[\begin{array}{c}\Delta u(0) \\\Delta \dot{u}(0)\end{array}\right] \qquad (a)
Terms \bar{h}(2 \Delta t) and \bar{h}(\Delta t) are obtained from Equation 9.79
\bar{h}(t)=\frac{e^{-\xi \omega t}}{m \omega_d}\left\{\frac{\sin \omega_d t-e^{-\xi \omega T_0} \sin \omega_d\left(t-T_0\right)}{1-2 e^{-\xi \omega T_0} \cos \omega_d T_0+e^{-2 \xi \omega T_0}}\right\}(9.79)
with \Delta t=0.1 \mathrm{~s}. Derivatives \dot{\bar{h}}(2 \Delta t) and \dot{\bar{h}}(\Delta t) are obtained from Equation 9.132.
Also, as in Example 9.6, \Delta u(0)=-0.0793 in. and \Delta \dot{u}=-0.4105 in./s. Equation a thus reduces to
\left[\begin{array}{ll}0.3347 & 0.2708 \\0.0000 & 1.2361\end{array}\right]\left[\begin{array}{l}R_{1} \\R_{2}\end{array}\right]=-\left[\begin{array}{l}0.0793 \\0.4105\end{array}\right] \qquad (b)
On solving Equations b, we get R_{1}=0.03188 and R_{2}=-0.33212. The corrected response is now obtained by using the superposition equation
u(k \Delta t)=\bar{u}(k \Delta t)+R_{1} \bar{h}\{(k+2) \Delta t\}+R_{2} \bar{h}\{(k+1) \Delta t\} \quad k=1,2, \ldots, N-3 \qquad (c)
For k=1 to N-3, the corrected response should provide a good estimate of the true transient response. The results obtained are identical to those in Examples 9.6 and 9.7.