Question 13.12: Find the response of the system of Example 13.9 using the Wi...
Find the response of the system of Example 13.9 using the Wilson- \theta method and a time step of (a) 0.1 \mathrm{~s}, and (b) 10.0 \mathrm{~s}. Select \theta=1.4.
(a) Again, the initial conditions are as in Example 13.9. The displacement at time point n+\theta is obtained from Equation 13.118.
For h=0.1 and \theta=1.4, that equation reduces to
\left\{306.12\left[\begin{array}{ll}2 & 0 \\0 & 1\end{array}\right]+\left[\begin{array}{rr}96 & -32 \\-32 & 32\end{array}\right]\right\} \mathbf{u}_{n+\theta}=\left[\begin{array}{c}0 \\100\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\0 & 1\end{array}\right]\left(306.12 \mathbf{u}_{n}+42.86 \dot{\mathbf{u}}_{n}+2 \ddot{\mathbf{u}}_{n}\right) \qquad (a)
Note that from Equation 13.119,
\mathbf{p}_{n+\theta}=(1-\theta) \mathbf{p}_n+\theta \mathbf{p}_{n+1} \qquad (13.119)
\mathbf{p}_{n+\theta}=(1-\theta)\left[\begin{array}{c}0 \\100\end{array}\right]+\theta\left[\begin{array}{c}0 \\100\end{array}\right]=\left[\begin{array}{c}0 \\100\end{array}\right] \qquad (b)
Table E13.12a Response of a two-degree-of-freedom system obtained by the wilson- \theta method, h=0. I s.
\begin{array}{lrc}\hline Time & {u_{1}} & u_{2} \\\hline 0.1 & 0.014 & 0.468 \\0.2 & 0.124 & 1.715 \\0.3 & 0.446 & 3.409 \\0.4 & 1.057 & 5.166 \\0.5 & 1.922 & 6.664 \\0.6 & 2.876 & 7.717 \\0.7 & 3.670 & 8.273 \\0.8 & 4.060 & 8.377 \\0.9 & 3.915 & 8.089 \\1.0 & 3.265 & 7.448 \\1.1 & 2.293 & 6.465 \\1.2 & 1.263 & 5.173 \\1.3 & 0.417 & 3.676 \\1.4 & -0.102 & 2.183 \\1.5 & -0.277 & 0.978 \\1.6 & -0.190 & 0.343 \\1.7 & 0.052 & 0.468 \\1.8 & 0.374 & 1.368 \\1.9 & 0.764 & 2.870 \\2.0 & 1.233 & 4.657 \\\hline\end{array}
Equation a is solved for u_{n+\theta}. Substitution of \mathbf{u}_{n+\theta} in Equation 13.116
\ddot{\mathbf{u}}_{n+\theta}=-2 \ddot{\mathbf{u}}_n+\frac{6}{(\theta h)^2}\left\{\mathbf{u}_{n+\theta}-\mathbf{u}_n-\theta h \dot{\mathbf{u}}_n\right\} \qquad (13.116)
gives \ddot{\mathbf{u}}_{n+\theta} :
\ddot{\mathbf{u}}_{n+\theta}=-2 \ddot{\mathbf{u}}_{n}+306.12\left(\mathbf{u}_{n+\theta}-\mathbf{u}_{n}\right)-42.86 \dot{\mathbf{u}}_{n} \qquad (c)
The accelerations at time point n+1 are now obtained from Equation 13.120
The displacements and velocities at t_{n+1} are calculated next from Equations 13.113 and 13.114,
respectively
This completes one cycle of iteration.
The displacement response obtained as above are presented in Table E13.12a for the first 2 \mathrm{~s}. The results are reasonably close to the exact values shown in Table E13.9a.
(b) The procedure is similar to that described in part (a). The response from 0 to 100 \mathrm{~s} and from 500 to 550 \mathrm{~s} at intervals of 10 \mathrm{~s} is shown in Table E13.12b. The response is stable and approaches the static response with the passage of time.
It is of interest to note that the displacement response for the first few time steps is very high. This is the result of using a very large time step in the integration and the presence of an initial acceleration. In fact, whenever the time step used in the integration is large in comparison to a certain mode period, the Wilson- \theta method will spuriously amplify the displacement response contribution from that mode during the first few time steps. In practice, this is not likely to
Table E13.12 b Response of a two-degree-of-freedom system obtained by the wilson- \theta method, h=10.0 \mathrm{~s}.
\begin{array}{rrr}\hline Time & {u_{1}} & {u_{2}} \\\hline 10.0 & 1.704 & 1434.000 \\20.0 & 4.404 & -1061.000 \\30.0 & -4.085 & 857.300 \\40.0 & 9.158 & -656.800 \\50.0 & -7.004 & 514.500 \\60.0 & 10.330 & -306.700 \\70.0 & -6.892 & 304.700 \\80.0 & 9.389 & -225.200 \\90.0 & -5.478 & 180.700 \\100.0 & 7.760 & -130.100 \\500.0 & 1.563 & 4.683 \\510.0 & 1.561 & 4.688 \\520.0 & 1.563 & 4.684 \\530.0 & 1.562 & 4.688 \\540.0 & 1.562 & 4.685 \\550.0 & 1.562 & 4.685 \\ \hline \end{array}
introduce serious errors in the total response, because the integration step will be large only in comparison to a higher mode period, and even the amplified response from such a mode will be small in comparison to the total response. However, in certain cases, this spurious amplification may be viewed as a disadvantage of the Wilson- \theta method.