User:Eml4500.f08.wiki1.aguilar/hw3

Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System

 * $$ \textbf{k}_{4x4}^{(e)}\textbf{d}_{4x1}^{(e)}=\textbf{f}_{4x1}^{(e)}

$$

$$ {k}^{(e)} \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix} \begin{Bmatrix} {q}_{1}^{(e)}\\ {q}_{2}^{(e)} \end{Bmatrix} = \begin{Bmatrix} {P}_{1}^{(e)}\\ {P}_{2}^{(e)} \end{Bmatrix} $$

$${q}_{i}^{(e)}=$$ axial displacement of element 'e' at local node 'i' $${P}_{i}^{(e)}=$$ axial force of element 'e' at local node 'i'

We want to find the relationship between $$\textbf{q}^{(e)}$$ and $$\textbf{d}^{(e)}$$, and $$\textbf{P}^{(e)}$$ and $$\textbf{F}^{(e)}$$.

These relationships can be expressed in the form: $$ \textbf{q}_{2x1}^{(e)}\textbf{T}_{2x4}^{(e)}=\textbf{d}_{4x1}^{(e)} $$

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.



$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the prthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}+\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j})+\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+({d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\cos\theta^{(e)}=l^{(e)}$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\sin\theta^{(e)}=m^{(e)}$$

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}\end{bmatrix}_{1x2}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}  \end{Bmatrix}_{2x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}\end{bmatrix}_{1x2}\begin{Bmatrix}{d}_{3}^{(e)}\\{d}_{4}^{(e)}  \end{Bmatrix}_{2x1}$$

which leads us to:

$$\begin{Bmatrix} {q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}= \begin{bmatrix} l^{(e)}& m^{(e)}&0&0\\0&0& l^{(e)}& m^{(e)} \end{bmatrix} \begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)} \end{Bmatrix}$$

where this is the equation we set out to derive, $$ \textbf{q}_{2x1}^{(e)}\textbf{T}_{2x4}^{(e)}=\textbf{d}_{4x1}^{(e)} $$

Using the Global Free-body Diagram Relationship
Using d3, and d4 in $$ \textbf{K}\textbf{d}=\textbf{F} $$ and using boundary conditions to eliminate rows in the global displacement matrix we get:

$$ \begin{bmatrix} K_{1,3} & K_{1,4} \\ K_{2,3} & K_{2,4} \\ K_{3,3} & K_{3,4} \\ K_{4,3} & K_{4,4} \\ K_{5,3} & K_{5,4} \\ K_{6,3} & K_{6,4} \end{bmatrix}_{6x2} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}_{2x1}=\begin{Bmatrix}F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}_{6x1}$$

Note: We really only need to do the computations for rows 1,2,5, and 6 to get F1,F2,F5,and F6 because computation from rows 3 and 4 gives the applied load which is already known.

Closing the Loop between FEM and Statics: Virtual Displacement
Two-bar truss system:



Since our two-bar truss system is statically determinant because we can view element 1 and 2 as two force bodies. By statics, the reactions are known and therefore, so are the member forces $$P^{(1)}_1$$, and $$P^{(1)}_1$$.

We then compute axial displacement degrees of freedom. This is the amount of extension within the bars.

$${q}_{2}^{(1)} = \frac{{P}_{1}^{(1)}}{{k}^{(1)}}= \frac{{P}_{2}^{(1)}}{{k}^{(1)}}=AC$$

$${q}_{1}^{(1)} = 0$$    (Node 1 is fixed)

$${q}_{2}^{(1)} = \frac{{-P}_{2}^{(2)}}{{k}^{(2)}}=AB$$

$${q}_{2}^{(2)} = 0$$    (Node 2 is fixed)

How do we back out the displacement DOFs of node 2 from above results?

Note: The displacement of node 2 is in the direction of vector D.