User:Eml4500.f08.bike.viale/homework report 2

The above figure from the previous lecture shows the combination of the two local 4x4 k matrices into a 6x6 global k matrix. The top right corner is the k matrix corresponding to the first element and the bottom left corner corresponds to the k matrix of element 2. The 4x4 section in the middle of this global matrix is a combination of the 2 local matrices. The global matrix in terms of the local matrix elements is as follows:

$$

\left[ \begin{array}{cccccc} K^{(1)}_{11} & K^{(1)}_{12} & K^{(1)}_{13} & K^{(1)}_{14} & 0 & 0 \\ K^{(1)}_{21} & K^{(1)}_{22} & K^{(1)}_{23} & K^{(1)}_{24} & 0 & 0 \\ K^{(1)}_{31} & K^{(1)}_{32} & (K^{(1)}_{33} + K^{(2)}_{11} & (K^{(1)}_{34} + K^{(2)}_{12})& K^{(2)}_{13} & K^{(2)}_{14} \\ K^{(1)}_{41} & K^{(1)}_{42} & (K^{(1)}_{43} + K^{(2)}_{21}) & (K^{(1)}_{44} + K^{(2)}_{22}) & K^{(2)}_{23} & K^{(2)}_{24} \\ 0 & 0 & K^{(2)}_{31} & K^{(2)}_{32} & K^{(2)}_{33} & K^{(2)}_{34} \\ 0 & 0 & K^{(2)}_{41} & K^{(2)}_{42} & K^{(2)}_{43} & K^{(2)}_{44} \end{array} \right]\begin{Bmatrix} d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \end{Bmatrix}=\begin{Bmatrix} F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{Bmatrix}

$$

Using the K values given earlier in class, some sample element calculations are as follows, with the rest able to be found using the symmetry of the matrix:

$$ K_{11} = \frac{9}{16}$$

$$ K_{12} = \frac{3\sqrt{3}}{16}$$

$$ K_{33} = \frac{49}{16}$$

$$ K_{34} = \textbf{-2.175}$$

$$ K_{43} = \textbf{-2.175}$$

$$ K_{44} = \textbf{2.6875}$$

The notation used above is for the global elements, rather than the local elements.

Also, due to the symmetry of the matrix, $$ \textbf{K}_{34} = \textbf{K}_{43}$$, therefore only one needs to be calculated when computing the values of the elements.

The green shaded top right and bottom left corners of the figure correspond to the $$ \textbf{0}$$ elements of the matrix.

After constructing this global matrix, the next step is to reduce the equation by utilizing the boundary conditions. In this particular problem, global nodes 1 and 2 are fixed, thus there is no displacement involved, and:

$$ \textbf{d} = \begin{Bmatrix} 0 \\ 0 \\ d_{3} \\ d_{4} \\ 0 \\ 0 \end{Bmatrix} $$

Utilizing the Principle of Virtural Work allows us to ignore the first, second, fifth and sixth column of the $$ \textbf{k}$$ matrix, as all $$ \textbf{F}$$ elements that it would produce would not be used in the calculation. Thus:

$$

\left[ \begin{array}{cc} 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{array} \right]\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}=\begin{Bmatrix} F_{3} \\ F_{4}\end{Bmatrix}

$$