Chapter 14, Problem 27P

To determine

Calculate the global coefficient matrix of the two element region.

Answer to Problem 27P

The global coefficient matrix is $\underset{\_}{\left[\begin{array}{cccc}0.8802& -0.2083& 0& -0.6719\\ -0.2083& 1.533& -1.2& -0.125\\ 0& -1.2& 1.4083& -0.2083\\ -0.6719& -0.125& -0.2083& 1.005\end{array}\right]}$.

Explanation of Solution

Calculation:

Refer to Figure 14-60 in the textbook for the two-element region.

For element 1, local numbering 1-2-3 corresponds to global numbering 4-2-1.

From Figure 14.60, the given values are,

$x_1=8,x_2=3,x_3=0,y_1=0,y_2=12,\text{and}{y}_3=0$.

Calculate the value of $P_1$.

$\begin{array}{c}{P}_1=y_2-y_3\\ =12-0\\ =12\end{array}$

Calculate the value of $P_2$.

$\begin{array}{c}{P}_2=y_3-y_1\\ =0-0\\ =0\end{array}$

Calculate the value of $P_3$.

$\begin{array}{c}{P}_3=y_1-y_2\\ =0-12\\ =-12\end{array}$

Calculate the value of $Q_1$.

$\begin{array}{c}{Q}_1=x_3-x_2\\ =0-3\\ =-3\end{array}$

Calculate the value of $Q_2$.

$\begin{array}{c}{Q}_2=x_1-x_3\\ =8-0\\ =8\end{array}$

Calculate the value of $Q_3$.

$\begin{array}{c}{Q}_3=x_2-x_1\\ =3-8\\ =-5\end{array}$

Consider the expression for the area A.

$A=\frac{1}{2}\left(P_2{Q}_3-P_3{Q}_2\right)$        (1)

Substitute 0 for $P_2$, $-12$ for $P_3$, 8 for $Q_2$, and $-5$ for $Q_3$ in Equation (1).

$\begin{array}{c}A=\frac{1}{2}\left[\left(\left(0\right)\left(-5\right)-\left(-12\right)\left(8\right)\right)\right]\\ =48\end{array}$

Write the expression for co-efficient $C_{ij}$.

$C_{ij}=\frac{1}{4A}\left[P_i{P}_j+Q_i{Q}_j\right]$        (2)

By using Equation (2), the coefficient values $C_{11},C_{12},C_{13},C_{21},C_{22},C_{23},C_{31},C_{32},\text{and}{C}_{33}$ are calculated as 0.7969, $-0.125$, $-0.6719$, $-0.125$, 0.3333, $-0.2083$, $-0.6719$, $-0.2083$, and 0.8802 respectively.

Write the expression of the element coefficient matrix C.

$C^{\left(1\right)}=\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]$

Substitute 0.7969 for $C_{11}$, $-0.125$ for $C_{12}$, $-0.6719$ for $C_{13}$, $-0.125$ for $C_{21}$, 0.3333 for $C_{22}$, $-0.2083$ for $C_{23}$, $-0.6719$ for $C_{31}$, $-0.2083$ for $C_{32}$, and 0.8802 for $C_{33}$.

$C^{\left(1\right)}=\left[\begin{array}{ccc}0.7969& -0.125& -0.6719\\ -0.125& 0.3333& -0.2083\\ -0.6719& -0.2083& 0.8802\end{array}\right]$

For element 2, local numbering 1-2-3 corresponds to global numbering 2-4-3.

From Figure 14.60, the given values are,

$x_1=3,x_2=8,x_3=8,y_1=12,y_2=0,\text{and}{y}_3=12$.

Calculate the value of $P_1$.

$\begin{array}{c}{P}_1=y_2-y_3\\ =0-12\\ =-12\end{array}$

Calculate the value of $P_2$.

$\begin{array}{c}{P}_2=y_3-y_1\\ =12-12\\ =0\end{array}$

Calculate the value of $P_3$.

$\begin{array}{c}{P}_3=y_1-y_2\\ =12-0\\ =12\end{array}$

Calculate the value of $Q_1$.

$\begin{array}{c}{Q}_1=x_3-x_2\\ =8-8\\ =0\end{array}$

Calculate the value of $Q_2$.

$\begin{array}{c}{Q}_2=x_1-x_3\\ =3-8\\ =-5\end{array}$

Calculate the value of $Q_3$.

$\begin{array}{c}{Q}_3=x_2-x_1\\ =8-3\\ =5\end{array}$

Substitute 0 for $P_2$, $12$ for $P_3$, $-5$ for $Q_2$, and 5 for $Q_3$ in Equation (1) to find area A.

$\begin{array}{c}A=\frac{1}{2}\left[\left(\left(0\right)\left(5\right)-\left(12\right)\left(-5\right)\right)\right]\\ =30\end{array}$

By using Equation (2), the coefficient values $C_{11},C_{12},C_{13},C_{21},C_{22},C_{23},C_{31},C_{32},\text{and}{C}_{33}$ are calculated as 1.2, 0, $-1.2$, 0, 0.208, $-0.208$, $-1.2$, $-0.208$, and 1.408 respectively.

Write the expression of the element coefficient matrix C.

$C^{\left(2\right)}=\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]$

Substitute 1.2 for $C_{11}$, 0 for $C_{12}$, $-1.2$ for $C_{13}$, 0 for $C_{21}$, 0.208 for $C_{22}$, $-0.208$ for $C_{23}$, $-1.2$ for $C_{31}$, $-0.208$ for $C_{32}$, and 1.408 for $C_{33}$.

$C^{\left(2\right)}=\left[\begin{array}{ccc}1.2& 0& -1.2\\ 0& 0.208& -0.208\\ -1.2& -0.208& 1.408\end{array}\right]$

Calculate the global coefficient matrix.

$\begin{array}{c}C=\left[\begin{array}{cccc}{C}_{33}^{\left(1\right)}& C_{23}^{\left(1\right)}& 0& C_{31}^{\left(1\right)}\\ C_{23}^{\left(1\right)}& C_{22}^{\left(1\right)}+C_{11}^{\left(2\right)}& C_{13}^{\left(2\right)}& C_{21}^{\left(1\right)}+C_{12}^{\left(2\right)}\\ 0& C_{31}^{\left(2\right)}& C_{33}^{\left(2\right)}& C_{32}^{\left(2\right)}\\ C_{13}^{\left(1\right)}& C_{21}^{\left(1\right)}+C_{21}^{\left(2\right)}& C_{23}^{\left(2\right)}& C_{22}^{\left(2\right)}+C_{11}^{\left(1\right)}\end{array}\right]\\ =\left[\begin{array}{cccc}0.8802& -0.2083& 0& -0.6719\\ -0.2083& 0.3333+1.2& -1.2& -0.125+0\\ 0& -1.2& 1.4083& -0.2083\\ -0.6719& -0.125+0& -0.2083& 0.208+0.7969\end{array}\right]\\ =\left[\begin{array}{cccc}0.8802& -0.2083& 0& -0.6719\\ -0.2083& 1.533& -1.2& -0.125\\ 0& -1.2& 1.4083& -0.2083\\ -0.6719& -0.125& -0.2083& 1.005\end{array}\right]\end{array}$

Conclusion:

Thus, the global coefficient matrix is $\underset{\_}{\left[\begin{array}{cccc}0.8802& -0.2083& 0& -0.6719\\ -0.2083& 1.533& -1.2& -0.125\\ 0& -1.2& 1.4083& -0.2083\\ -0.6719& -0.125& -0.2083& 1.005\end{array}\right]}$.