Question 13.15: Find the response of the system of Example 13.9 using the ce...

As in Example 13.14, the iteration equation corresponding to degree-of-freedom 1 is

200 u_{1}^{n+1}=304 u_{1}^{n}+32 u_{2}^{n}-200 u_{1}^{n-1}

or

u_{1}^{n+1}=1.52 u_{1}^{n}+0.16 u_{2}^{n}-u_{1}^{n-1} \qquad (a)

The equation for degree-of-freedom 2 is obtained from Equation 13.128

with m=2.

400 u_{2}^{n+p / 2}=100+(800-32) u_{2}^{n+(p-1) / 2}+32 u_{1}^{n+(p-1) / 2}-400 u_{2}^{n+(p-2) / 2}

or

u_{2}^{n+p / 2}=0.25+1.92 u_{2}^{n+(p-1) / 2}+0.08 u_{1}^{n+(p-1) / 2}-u_{2}^{n+(p-2) / 2} \qquad (b)

The initial conditions are u_{1}^{0}=0, \dot{u}_{1}^{0}=0, \ddot{u}_{1}^{0}=0, u_{2}^{0}=0, \dot{u}_{2}^{0}=0, \ddot{u}_{2}^{0}=100. To start the iteration for degree-of-freedom 1 , we need u_{1}^{-1}. This is obtained from Equation 13.108:

\mathbf{u}_{-1}=\mathbf{u}_0+\frac{h^2}{2} \ddot{\mathbf{u}}_0-b \dot{\mathbf{u}}_0 \qquad (13.108)

\begin{aligned}u_{1}^{-1} &=u_{1}^{0}-b \dot{u}_{1}^{0}+\frac{h^{2}}{2} \ddot{u}_{1}^{0} \\&=0\end{aligned} \qquad (c)

The first step in the iteration corresponding to n=0 gives

\begin{aligned}u_{1}^{1} &=1.52 u_{1}^{0}+0.16 u_{2}^{0}-u_{1}^{-1} \\&=0\end{aligned} \qquad (d)

To start the iteration for degree-of-freedom 2 , we need u_{2}^{-1 / 2}. This is obtained from an equation similar to Equation c but with h replaced by h / m=h / 2 :

\begin{aligned}u_{2}^{-1 / 2} &=u_{2}^{0}-\frac{h}{2} \dot{u}_{2}^{0}+\frac{h^{2}}{8} \ddot{u}_{2}^{0} \\&=\frac{0.1^{2}}{8} \times 100=0.125 \qquad (e)\end{aligned}

The first substep in the iteration, obtained by substituting n=0 and p=1 in Equation \mathrm{b}, is

\begin{aligned}u_{2}^{1 / 2} &=0.25+1.92 u_{2}^{0}+0.08 u_{1}^{0}-u_{2}^{-1 / 2} \\&=0.25-0.125=0.125 \qquad (f) \end{aligned}

Before we carry out the second substep of iteration with n=0 and p=2, we will need u_{1}^{1 / 2}. This is obtained by linear interpolation between u_{1}^{0} and u_{1}^{1} :

\begin{aligned}u_{1}^{1 / 2} &=0.5\left(u_{1}^{0}+u_{1}^{1}\right) \\&=0\end{aligned} \qquad (g)

We now write Equation \mathrm{b} with n=0 and p=2 :

\begin{aligned}u_{2}^{1} &=0.25+1.92 u_{2}^{1 / 2}+0.08 u_{1}^{1 / 2}-u_{2}^{0} \\&=0.25+1.92 \times 0.125+0.08 \times 0 \\&=0.49\end{aligned} \qquad (h)

Table E13.15 Response of a two-degree-of-freedom system obtained by mixed explicit-explicit integration method, h=0.1 \mathrm{~s}.
\begin{array}{lrc}\hline Time & u_{1} & u_{2} \\\hline 0.1 & 0.000 & 0.490 \\0.2 & 0.078 & 1.809 \\0.3 & 0.409 & 3.576 \\0.4 & 1.115 & 5.344 \\0.5 & 2.141 & 6.779 \\0.6 & 3.224 & 7.742 \\0.7 & 3.999 & 8.255 \\0.8 & 4.174 & 8.390 \\0.9 & 3.689 & 8.156 \\1.0 & 2.738 & 7.483 \\1.1 & 1.670 & 6.296 \\1.2 & 0.808 & 4.647 \\1.3 & 0.307 & 2.789 \\1.4 & 0.010 & 1.142 \\1.5 & 0.003 & 0.153 \\1.6 & -0.029 & 0.105 \\1.7 & -0.056 & 0.997 \\1.8 & 0.104 & 2.545 \\1.9 & 0.621 & 4.323 \\2.0 & 1.532 & 5.934 \\ \hline\end{array}

The second step in the iteration for u_{1} is

\begin{aligned}u_{1}^{2} &=1.52 u_{1}^{1}+0.16 u_{2}^{1}-u_{1}^{0} \\&=1.52 \times 0+0.16 \times 0.49 \\&=0.078\end{aligned} \qquad (i)

Also,

\begin{aligned}u_{1}^{3 / 2} &=0.5\left(u_{1}^{1}+u_{1}^{2}\right) \\&=0.039\end{aligned}\qquad (j)

We can now carry out two substeps of iteration for u_{2}.