User:Eml4500.f08.delta 6.castrillon/HW 2

= Testing the Steps =

It is important to see how the steps to solve truss systems work.

Given


Take this truss example with the following data:

Global Picture
Releasing the truss system from its supports and showing the reactions, we have:



There are 6 unknowns for which we need to solve for, therefore we need to analyze each element. The matrix equation is as follows.


 * $$\begin{pmatrix}

f_1\\ ...\\ f_6\end{pmatrix}$$ = $$\begin{bmatrix} & &  \\  & k &  \\ & &  \end{bmatrix}$$ $$\begin{pmatrix} d_1\\ ...\\ d_6\end{pmatrix}$$

which corresponds to


 * $$\mathbf{F}$$=$$\mathbf{K}$$ $$\mathbf{d}$$.

To solve for the reactions, we have to now examine the whole system by breaking it down into elements.

Element Picture
Analyzing each element with its own force and dofs yields:





It is important to notice the chosen coordinate axis since they will allow for ease of analyzing each element. The tilda on the coordinate direction names designate a non-horizontal or non-vertical direction.

We now want to creat a matrix equation, just like for the global picture, for the element picture. The matrix equation consists on the basic force-distance relationship, however, this principle is applied at the element level. The notation is altered in that each term is designated to belong to the particular element with an upper script number in parenthesis so that we get, $$ F(e) = k(e) d(e) $$, named as the element force matrix, the element stiffness matrix, and the element distance matrix.

Force-Displacement Relationship
To build the element stiffness matrix, we have to follow a "recipe" for calculating each term. This stiffness matrix will end up measuring 4x4.

The recipe is as follows, and in the wise words of Prof. Vu-Quoc, "The notation is beautiful here. It tells you no more, no less."


 * k^{(e)} $$\begin{bmatrix}

l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2}\\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ = k $$^{(e)}$$

where the axial stiffness fo the bar "e" is


 * $$k^{(e)}=\frac{E^{(e)} A^{(e)}}{L^{(e)}}$$,

where $$E^{(e)}$$ is the element's material Young's Modulus, $$A^{(e)}$$ is the element cross-sectional area, and $$L^{(e)}$$ the element length. Also, $$l^{(e)}$$ and $$m^{(e)}$$ are te director cosines of the rotated, and convenient, coordinate axes with respect to the positive, and horizontal, x-direction (the global axis).

Director Cosines: The director cosines will yield the projections of vectors oriented with an inclination w.r.t. the positive x-axis on the horizontal or vertical directions. These horizontal and vertical directions are more precisely for this course the global x-y coordinates. From this diagram, we can derive our much needed $$l^{(e)}$$ and $$m^{(e)}$$ values. $$l^{(e)}$$ = $$\vec{\tilde{i}}$$ • $$\vec{\hat{i}}$$ = cosθ$$^{(e)}$$ $$m^{(e)}$$ = $$\vec{\tilde{i}}$$ • $$\vec{\hat{j}}$$ = sinθ$$^{(e)}$$ = cos(π/2-θ$$^{(e)})$$ Also, $$\vec{\tilde{i}}$$=cosθ$$^{(e)}$$•$$\vec{\hat{i}}$$+sinθ$$^{(e)}$$•$$\vec{\hat{j}}$$