User:Eml4500.f08.bottle.barnes/HW3

Using the global FD relation, where displacements $$\displaystyle d_1, d_2, d_5, d_6 $$ all equal zero, the first two columns and the last two columns for the $$\displaystyle K $$ matrix can be ignored, yielding the following relation:


 * $$ \begin{bmatrix} K_{13} & K_{14}

\\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix} \begin{Bmatrix} d_3 \\d_4 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6  \end{Bmatrix}$$

which can also be expressed as


 * $$ \begin{bmatrix} k_{13}^{(1)} & k_{14}^{(1)}

\\ k_{23}^{(1)} & k_{24}^{(1)} \\ k_{33}^{(1)} + k_{11}^{(2)} & k_{34}^{(1)} + k_{12}^{(2)} \\ k_{43}^{(1)} + k_{21}^{(2)} & k_{44}^{(1)} + k_{22}^{(2)} \\ k_{31}^{(2)} & k_{32}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)} \end{bmatrix} \begin{Bmatrix} d_3 \\d_4 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} $$

Inputting the values for each $$\displaystyle K $$ and $$\displaystyle d $$ yields


 * $$ \begin{bmatrix} -{9 \over 16} & -{3 \sqrt 3 \over 16}

\\ -{3 \sqrt 3 \over 16} & -{3 \over 16} \\ {9 \over 16} + {5 \over 2} & {3 \sqrt 3 \over 16} + -{5 \over 2} \\ {3 \sqrt 3 \over 16} + -{5 \over 2} & {3 \sqrt 3 \over 16} + {5 \over 2} \\ -{5 \over 2} & {5 \over 2} \\ {5 \over 2} & -{5 \over 2} \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ 0 \\ 7 \\ F_5 \\ F_6 \end{Bmatrix} $$

Note that since $$\displaystyle F_3 $$ and $$\displaystyle F_4 $$ are known (because they are the applied loads, 0 and 7 respectively) rows three and four do not need to be computed.

Thus, the computed solution is


 * $$ F = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 0 \\ 7 \\ 4.4378 \\ -4.4378 \end{Bmatrix} $$

Notice that these results are the same as those determined from Method 1.