(a) The initial displacement and velocity vectors are zero. As in Example 13.9, the initial acceleration vector is obtained from the equation of motion at t=0, giving
\ddot{\mathbf{u}}_{0}=\left[\begin{array}{c}0 \\100\end{array}\right]
Iterations are now carried out with Equations 13.112,13.109, and 13.110.
\begin{aligned}&\left(\frac{4}{h^2} \mathbf{M}+\frac{2}{h} \mathbf{C}+\mathbf{K}\right) \mathbf{u}_{n+1}=\mathbf{p}_{n+1}+\mathbf{M}\left(\frac{4}{h^2} \mathbf{u}_n+\frac{4}{h} \dot{\mathbf{u}}_n+\ddot{\mathbf{u}}_n\right)+\mathbf{C}\left(\frac{2}{h} \mathbf{u}_n+\dot{\mathbf{u}}_n\right) \qquad (13.112)\\&\ddot{\mathbf{u}}_{n+1}=-\ddot{\mathbf{u}}_n+\frac{4}{h^2}\left(\mathbf{u}_{n+1}-\mathbf{u}_n-h \dot{\mathbf{u}}_n\right) \qquad (13.109)\\&\dot{\mathbf{u}}_{n+1}=-\dot{\mathbf{u}}_n+\frac{2}{h}\left(\mathbf{u}_{n+1}-\mathbf{u}_n\right) \qquad (13.110)\end{aligned}
For h=0.1 these equations reduce, respectively, to
\begin{aligned}& \left\{400\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+1}=\left[\begin{array}{c}0 \\100\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\0 & 1\end{array}\right]\left(400 \mathbf{u}_{n}+40 \dot{\mathbf{u}}_{n}+\ddot{\mathbf{u}}_{n}\right) \qquad (a)\\& \ddot{\mathbf{u}}_{n+1}=-\ddot{\mathbf{u}}_{n}+400\left(\mathbf{u}_{n+1}-\mathbf{u}_{n}-0.1 \dot{\mathbf{u}}_{n}\right) \qquad (b)\\& \dot{\mathbf{u}}_{n+1}=-\dot{\mathbf{u}}_{n}+20\left(\mathbf{u}_{n+1}-\mathbf{u}_{n}\right) \qquad (c)\end{aligned}
The response results obtained from Equations a, b, and c are shown in Table E13.10a for the first 2 \mathrm{~s}. Considering that the time step is rather coarse, the results are reasonably close to the exact values shown in Table E13.9a.
Table E13.10a Response of a two-degree-of-freedom system by the average acceleration method, h=0.1 \mathrm{~s}.
\begin{array}{lrc}\hline Time & {u_{1}} & u_{2} \\\hline 0.1 & 0.017 & 0.464 \\0.2 & 0.121 & 1.728 \\0.3 & 0.437 & 3.458 \\0.4 & 1.062 & 5.241 \\0.5 & 1.970 & 6.733 \\0.6 & 2.978 & 7.746 \\0.7 & 3.799 & 8.260 \\0.8 & 4.156 & 8.343 \\0.9 & 3.912 & 8.067 \\1.0 & 3.130 & 7.446 \\1.1 & 2.058 & 6.450 \\1.2 & 1.012 & 5.082 \\1.3 & 0.249 & 3.454 \\1.4 & -0.128 & 1.831 \\1.5 & -0.174 & 0.580 \\1.6 & -0.038 & 0.042 \\1.7 & 0.153 & 0.412 \\1.8 & 0.367 & 1.633 \\1.9 & 0.678 & 3.412 \\2.0 & 1.189 & 5.322 \\\hline\end{array}
(b) The initial conditions are identical to those in part (a). With h=10 \mathrm{~s}, the three equations governing the iterations become
\begin{aligned}&\left\{0.04\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+1}=\left[\begin{array}{c}0 \\100\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\0 & \end{array}\right]\left(0.04 \mathbf{u}_{n}+0.4 \dot{\mathbf{u}}_{n}+\ddot{\mathbf{u}}_{n}\right) \qquad (d)\\&\ddot{\mathbf{u}}_{n+1}=-\ddot{\mathbf{u}}_{n}+0.04\left(\mathbf{u}_{n+1}-\mathbf{u}_{n}-10 \dot{\mathbf{u}}_{n}\right) \qquad (e)\\&\dot{\mathbf{u}}_{n+1}=-\dot{\mathbf{u}}_{n}+0.2\left(\mathbf{u}_{n+1}-\mathbf{u}_{n}\right) \qquad (f)\end{aligned}
The response for the first 10 time steps is shown in Table E13.10b. Even with the large time step, the response is bounded and oscillates about the static response given by
\begin{aligned}\mathbf{u}_{s} &=\mathbf{K}^{-1} \mathbf{p} \\&=\left[\begin{array}{l}1.563 \\4.688\end{array}\right]\end{aligned} \qquad (g)