User:Eml4500.f08.delta 6.lopez/HW2

***WEDNESDAY***
The following displays the F matrix calculated two different ways.

$$ \textbf{K}\begin{bmatrix} d_{1}=0\\ d_{2}=0\\ d_{3}\\ d_{4}\\ d_{5}=0\\ d_{6}=0 \end{bmatrix} =\begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$

Here, we delete the corresponding columns in the global stiffness matix, K. Due to the Principle of Virtual Work (PVW), we must also delete the corresponding rows (rows 1, 2, 5, 6).

Below, we have the Resulting Force Displacement relation:

$$ \begin{bmatrix} K_{33} &K_{34} \\ K_{43} & K_{44} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} $$ where the F matrix is found below.

$$ \textbf{F} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} = \begin{bmatrix} 0\\ P \end{bmatrix} $$ Before we continue, we will find the determinant and inverse of the K matrix. The determinant is simply equal to:

$$ det\; \textbf{K} = K_{33}K_{44}-K_{34}K_{43} $$

Now that we have the determinant, we can calculate the inverse using the formula below.

$$ \textbf{K}^{-1} = \frac{1}{det\; \textbf{K}} \begin{bmatrix} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{bmatrix} $$

*NOTE* The inverse of the K matrix cannot be found by using the transpose of the matrix, illustrated below.

$$ \textbf{K}^{T} = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} , where\; \textbf{K}^{-1}\neq \frac{1}{det\; \textbf{K}} \textbf{K}^{T} $$

With the inverse of the K matrix known, we can now find the unknown forces from the d matrix.

$$ \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{K}^{-1}\begin{bmatrix} 0\\ P \end{bmatrix} = \begin{bmatrix} 4.352\\ 6.127 \end{bmatrix} $$

Step 5: Compute Reactions
There are two main methods for solving this 2-bar truss system.

*Method 1*
For the first method, we will use the element force displacement relationship: $$ \textbf{k}^{e}\textbf{d}^{e}=\textbf{f}^{e}, where\; e=1,2 $$

We can now separate the truss into each individual bar, and compute the calculations for each bar alone.

Element 1:
$$where\; \textbf{d}^{(1)}= \begin{bmatrix} 0\\ 0\\ 4.352\\ 6.127 \end{bmatrix} = \begin{bmatrix} d_{1}^{(1)}\\ d_{2}^{(1)}\\ d_{3}^{(1)}\\ d_{4}^{(1)} \end{bmatrix} $$

Element 2:
$$ where\; \textbf{d}^{(2)}= \begin{bmatrix} 4.352\\ 6.127\\ 0\\ 0 \end{bmatrix}= \begin{bmatrix} d_{1}^{(2)}\\ d_{2}^{(2)}\\ d_{3}^{(2)}\\ d_{4}^{(2)} \end{bmatrix} since\; d_{3}=d_{1}^{(2)}, d_{4}=d_{2}^{(2)} $$