Find the response of the system of Example 13.9 using the linear acceleration method and a time step h of (a) 0.1 s, and (b) 0.5 s.

Question

Find the response of the system of Example 13.9 using the linear acceleration method and a time step [latex]h[/latex] of (a) [latex]0.1 \mathrm{~s}[/latex], and (b) [latex]0.5 \mathrm{~s}[/latex].

Accepted Answer

(a) The initial conditions are the same as in Examples 13.9 and 13.10. Equations 13.115, 13.113, and 13.114

[latex]\begin{aligned}&\left(\frac{6}{b^2} \mathbf{M}+\frac{3}{b} \mathbf{C}+\mathbf{K} ight) \mathbf{u}_{n+1}=\mathbf{p}_{n+1}+\mathbf{M}\left(\frac{6}{b^2} \mathbf{u}_n+\frac{6}{b} \dot{\mathbf{u}}_n+2 \ddot{\mathbf{u}}_n ight) \&+\mathbf{C}\left(\frac{3}{b} \mathbf{u}_n+2 \dot{\mathbf{u}}_n+\frac{h}{2} \ddot{\mathbf{u}}_n ight) \qquad \qquad \qquad \qquad \qquad \qquad (13.115)\&\ddot{\mathbf{u}}_{n+1}=-2 \ddot{\mathbf{u}}_n+\frac{6}{h^2}\left(\mathbf{u}_{n+1}-\mathbf{u}_n-h \dot{\mathbf{u}}_n ight) \qquad (13.113) \&\dot{\mathbf{u}}_{n+1}=-2 \dot{\mathbf{u}}_n-\frac{h}{2} \ddot{\mathbf{u}}_n+\frac{3}{b}\left(\mathbf{u}_{n+1}-\mathbf{u}_n ight) \qquad (13.114)\end{aligned}[/latex]

are used in the iterations. For [latex]b=0.1 \mathrm{~s}[/latex] these equations reduce to the following

[latex]\begin{aligned}&\left\{600\left[\begin{array}{ll}2 & 0 \0 & 1\end{array} ight]+\left[\begin{array}{rr}96 & -32 \-32 & 32\end{array} ight] ight\} \mathbf{u}_{n+1}=\left[\begin{array}{c}0 \100\end{array} ight]+\left[\begin{array}{ll}2 & 0 \0 & 1\end{array} ight]\left(600 \mathbf{u}_{n}+60 \dot{\mathbf{u}}_{n}+2 \ddot{\mathbf{u}}_{n} ight) \qquad (a)\&\ddot{\mathbf{u}}_{n+1}=-2 \ddot{\mathbf{u}}_{n}+600\left(\mathbf{u}_{n+1}-\mathbf{u}_{n}-0.1 \dot{\mathbf{u}}_{n} ight) \qquad (b) \&\dot{\mathbf{u}}_{n+1}=-2 \dot{\mathbf{u}}_{n}-0.05 \ddot{\mathbf{u}}_{n}+30\left(\mathbf{u}_{n+1}-\mathbf{u}_{n} ight) \qquad (c)\end{aligned}[/latex]

The response results obtained from Equations a, b, and c are presented in Table E13.11a for the first [latex]2 \mathrm{~s}[/latex]. Again, the results are reasonably close to the exact values shown in Table E13.9a.

(b) The stability criterion requires that [latex]h[/latex] be less than [latex]0.55[/latex] times the second mode period. The limiting value of [latex]h[/latex] is thus given by

[latex]\begin{aligned}h &<0.55 \frac{2 \pi}{8} \&=0.432 \mathrm{~s}\end{aligned}[/latex]

Table E13.11a Response of a two-degree-of-freedom system by the linear acceleration method, [latex]h=0.1 \mathrm{~s}[/latex].
[latex]\begin{array}{lcc}\hline Time & u_{1} & u_{2} \\hline 0.1 & 0.012 & 0.475 \0.2 & 0.109 & 1.763 \0.3 & 0.430 & 3.510 \0.4 & 1.081 & 5.286 \0.5 & 2.027 & 6.750 \0.6 & 3.060 & 7.732 \0.7 & 3.867 & 8.233 \0.8 & 4.165 & 8.330 \0.9 & 3.837 & 8.081 \1.0 & 2.993 & 7.465 \1.1 & 1.913 & 6.428 \1.2 & 0.926 & 4.973 \1.3 & 0.257 & 3.255 \1.4 & 0.426 & 1.593 \1.5 & 0.076 & 0.398 \1.6 & 0.004 & 0.014 \1.7 & 0.107 & 0.577 \1.8 & 0.268 & 1.950 \1.9 & 0.610 & 3.771 \2.0 & 1.232 & 5.595 \\hline\end{array}[/latex]

The selected value of [latex]h[/latex] does not satisfy this requirement. The response obtained with [latex]h=0.5 \mathrm{~s}[/latex] is therefore expected to be unstable. The response for the first [latex]10 \mathrm{~s}[/latex] shown in Table E13.11b. As expected, the results are unbounded.

Question 13.11: Find the response of the system of Example 13.9 using the li...

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