User:Eml4500.f08.Ateam.paquette/Homework3

 Lecture notes September 22, 2008 

The element force displacement relation can be derived with respect to the global coordinate system from the axial displacement and the axial forces. The internal forces and internal displacements of any element can be represented by the member forces in the local coordinate system instead of by two variables in the global coordinates. The elemental forces and elemental displacements can be shown with an axial displacement of element e at local node i, $$ {q_i^{(e)}}$$, and an axial force of element e at local node i, $$ {p_i^{(e)}}$$.



Our goal is to derive equation 1
 * $$ \mathbf{k^{(e)}} \mathbf{d^{(e)}} = \mathbf{f^{(e)}} $$

from equation 2
 * $$ \mathbf{k^{(e)}} \mathbf{q^{(e)}} = \mathbf{P^{(e)}} $$

The relationship between $$q^{(e)} $$ and $$d^{(e)} $$, and $$p^{(e)} $$ and $$f^{(e)} $$ also need to be found. We know that $$ {T^{(e)}} {d^{(e)}} = {q^{(e)}} $$. Now we consider the displacement vector of local node 1, $$ {d_1^{(e)}} $$

Node 1



The vector d1(e) can be broken down in to two unit vectors.


 * $$ {d_1^{(e)}} = d_i^{(e)} \vec i + d_2^{(e)} \vec j $$

The vector $$ {\vec d_1^{(e)}} $$ of node 1 is on axis $$ \tilde{x} $$. It can be shown that $$ q_1^{(e)} $$ is the axial displacement of node 1 and is the orthoganal projection of the displacement. $$ q_1^{(e)} $$ can now be related to $$ d_1^{(e)} $$.


 * $$\overrightarrow {q_1^{(e)}} = {\vec d_1^{(e)}} \cdot \vec \tilde{i} $$


 * $$ {d_1^{(e)}} = d_1^{(e)} \vec i + d_2^{(e)} \vec j $$.


 * $$\overrightarrow {q_1^{(e)}} = (d_1^{(e)} \tilde{i} + d_2^{(e)} \tilde{j}) \cdot \vec \tilde{i} $$


 * $$\overrightarrow {q_1^{(e)}} = d_1^{(e)} (\vec i \cdot \vec \tilde{i}) + d_2^{(e)} (\vec j \cdot \vec \tilde{i}) $$

These equations can be shortened because, as proven in Homework 1,


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


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

$$ q_1^{(e)} $$ can now be rewritten again:

$$ q_1^{(e)} = l^{(e)} d_1^{(e)} + m^{(e)} d_2^{(e)} $$


 * $$ = \left[ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} \right] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

The reactions and relationships are very similar for local node 2.

Node 2


 * $$ {d_2^{(e)}} = d_3^{(e)} \vec i + d_4^{(e)} \vec j $$.

Therefore $$ q_2^{(e)} = \vec d_3^{(e)} \cdot \vec \tilde{j} $$


 * $$ q_2^{(e)} = (d_2^{(e)} \tilde{i} + d_4^{(e)} \tilde{j}) \cdot \vec \tilde{j} $$


 * $$ q_2^{(e)} = d_3^{(e)} (\vec i \cdot \vec \tilde{j}) + d_4^{(e)} (\vec j \cdot \vec \tilde{j}) $$


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

and
 * $$(\vec j \cdot \vec \tilde{j}) = cos(\theta^{(e)})$$


 * $$ q_2^{(e)} = m^{(e)} d_3^{(e)} +  l^{(e)} d_4^{(e)} $$


 * $$ q_2^{(e)} = \left[ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} \right] \left[ \begin{matrix} d_3^{(e)} \\ d_4^{(e)} \end{matrix} \right] $$

After solving for the axial displacement in both nodes, an overall reaction can be formed.

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$