User:Eml4500.f08.RAMROD.A/MyContributionHW2

Statically Indeterminate 3Truss System
A look at the system with global nodes identifited:



Element Picture
We want to look at the elemental picture with $$\displaystyle {\color{Green} Global }$$ and $$\displaystyle {\color{Blue} Local }$$ coordinate systems:



While the $$\displaystyle {\color{Blue} Local }$$ Coordinates are denoated by $$\displaystyle {\color{Blue} (\tilde{x}, \tilde{y}) } $$.

 * $$\displaystyle \tilde{x} $$. is defined as along the element from local node 1 to 2.

Vector Notation

 * A common occurrence when dealing with forces such as P or the lengths of elements is to use vector notation: $$\displaystyle ({i}, {j})$$ a long with the angle between $$\displaystyle x $$ and $$\displaystyle \tilde{x}$$; denoted as $$\displaystyle \theta $$.



Next Step of Finite Element Method
It is important to go back to the equation described in the recipe from [earlier articles] on Finite Element Method:

K(e) * d(e) = f(e)

Where:


 * K is the element stiffness matrix


 * d is the element displacement matrix


 * f is the element force matrix

It is now possible to determine ke using the aforementioned $$\displaystyle \theta $$ and what we call the direction cosines

Direction Cosines
$$\displaystyle l^e $$ and $$\displaystyle m^e $$.


 * $$\displaystyle l^e = \cos (\theta) $$


 * $$\displaystyle m^e = \cos (90-\theta) = \sin (\theta) $$

Element Stiffness Matrix, k
$$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e}\left[Direction Cosine Matrix \right] $$

Where: "$\displaystyle L = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $"
 * E = Young's Modulus of element
 * A = Cross sectional Area of element
 * L = Length of element

When the cosine matrix is added in, we get: $$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e} \begin{bmatrix} l^2 & (l\cdot m) & -l^2  & -(l\cdot m) \\ \\(l\cdot m) & m^2 & -(l\cdot m) & -m^2 \\ \\-l^2 & -(l\cdot m) & l^2 & (l\cdot m) \\ \\-(l\cdot m) & -m^2 & (l\cdot m) & m^2 \end{bmatrix}$$