User:Eml4500.f08.Ateam.paquette/Homework4

 Lecture notes 19 October 8, 2008 

The goal of this next step is to find $$ \tilde{T}_{4x4}^{(e)}$$ that transforms the set of local, or elemental, degrees of freedom, $${d_{4x1}^{(e)}}$$, to another set of local degrees of freedom, $$ \tilde{d}_{4x1}^{(e)}$$, so that the matrix $$ \tilde{T}_{4x4}^{(e)}$$ is invertible.



$$ \tilde{d}_{4x1}^{(e)} = \tilde{T}_{4x4}^{(e)} {d_{4x1}^{(e)}}$$

$$ \tilde{d}_1^{(e)} = [ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

The right side of the equation has been seen before in previous lectures when trying to find the axial displacement, or the orthogonal projection, of the node. That means we can also write the equation as:

$$ {q}_1^{(e)} = [ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

Derivation of $$ {d_1^{(e)}} $$:

$$\tilde {d}_1^{(e)} = {\vec d_1^{(e)}} \cdot \vec \tilde{i} $$

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

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

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

Where:

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

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

Therefore:

$$ \tilde{d}_1^{(e)} = [ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$ (1)

This process is similar when finding $$ \tilde{d}_2^{(e)}$$. The only difference is that the components of $$ \overrightarrow {d}_1^{(e)}$$ are found along the $$ \tilde {j}$$ or the $$ \tilde {y} $$ axis.

Derivation of $$ {d_2^{(e)}} $$:

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

$$\tilde {d}_2^{(e)} = \vec d_1^{(e)} \cdot \vec \tilde{j} $$

$$\tilde {d}_2^{(e)} = (d_1^{(e)} \tilde{i} + d_2^{(e)} \tilde{j}) \cdot \vec \tilde{j} $$

$$\tilde {d}_2^{(e)} = d_1^{(e)} (\vec i \cdot \vec \tilde{j}) + d_2^{(e)} (\vec j \cdot \vec \tilde{j}) $$

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

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

$$\tilde {d}_2^{(e)} = -m^{(e)} d_1^{(e)} +  l^{(e)} d_2^{(e)} $$

$$\tilde {d}_2^{(e)} = [ \begin{matrix} -m^{(e)} & l^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$ (2)

Equations (1) and (2) can now be put into a matrix form

$$ \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} \\  -m^{(e)}& l^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix} $$

where $$ \left[ \begin{matrix} l^{(e)} & m^{(e)} \\  -m^{(e)}& l^{(e)} \end{matrix} \right] $$ can be written as $$ {R_{2x2}^{(e)}} $$

Now the equation $$ \tilde{d}_{4x1}^{(e)} = \tilde{T}_{4x4}^{(e)} {d_{4x1}^{(e)}}$$ can be formed from these equations.

$$ \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)} \\\tilde{d}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$

The element above has been rotated for an easier understanding.

$$\tilde{f}^{(e)} = k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \tilde{d}^{(e)}$$

If you try to move this node transversely, nothing will happen. That is why columns 2 and 4 are all zeros, because they do not stretch the string.

$$k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} = \tilde k $$

This gives us the final equation of:

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