User:Eml4500.f08.wiki1.ambrosio/hw4

Assignment (10/8): Derive $$\tilde{d}^{(e)}_2$$
(NOTE: $$\tilde{d}^{(e)}_1$$ was derived on Page 12-4.)

(the component of $$\overrightarrow{d}^{(e)}_{local node 1}$$ along $$\overrightarrow{\tilde{j}}$$, i.e. the $$\tilde{y}$$ axis)

$$ \begin{array}{lcl} \begin{align}

\tilde{d}^{(e)}_2 & = \overrightarrow{d}^{(e)}_{local node 1} \cdot \overrightarrow{\tilde{j}} \\

&=(d^{(e)}_1 \overrightarrow{i} + d^{(e)}_2 \overrightarrow{j}) \cdot \overrightarrow{\tilde{j}} \\

&= d^{(e)}_1 (\overrightarrow{i} \cdot \overrightarrow{\tilde{j}}) + d^{(e)}_2(\overrightarrow{j} \cdot \overrightarrow{\tilde{j}}) \\

&= d^{(e)}_1 (-\sin \theta ^{(e)}) + d^{(e)}_2(\cos \theta^{(e)}) \\

&= d^{(e)}_1 (-m^{(e)}) + d^{(e)}_2(l^{(e)}) \\

\end{align} \end{array} $$

$$ \tilde{d}^{(e)}_2 = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} d^{(e)}_1\\ d^{(e)}_2 \end{Bmatrix} $$

Where $$\tilde{d}^{(e)}_2 = q^{(e)}_2$$

Assignment (10/10): Interpretation of Transverse DOFs $$ \mathbf{\tilde{f}}^{(e)} = \mathbf{\tilde{T}}^{(e)} \mathbf{f}^{(e)} $$
First we note that $$\mathbf{\tilde{f}}^{(e)}_x$$ and $$\mathbf{\tilde{d}}^{(e)}_x$$ point in the same directions. Since $$\mathbf{\tilde{T}}^{(e)} $$ is only depenedent on $$\theta$$, we can also use it to transform from $$\mathbf{f}^{(e)}$$ to $$\mathbf{\tilde{f}}^{(e)}$$.

We can easily convert between $$\mathbf{\tilde{f}}^{(e)} $$ and $$\mathbf{f}^{(e)} $$ by using this relation.
 * $$ \mathbf{\tilde{f}}^{(e)} = \mathbf{\tilde{T}}^{(e)} \mathbf{f}^{(e)} $$

When expanded, the relation looks like:

\begin{pmatrix} \mathbf{\tilde{f}}^{(e)}_1\\ \mathbf{\tilde{f}}^{(e)}_2

\end{pmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{pmatrix} \mathbf{f}^{(e)}_1\\ \mathbf{f}^{(e)}_2 \end{pmatrix} $$

Assignment (10/10): Verify $$\mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{(e)\mathbf{T}} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}$$
So far we know that $$\mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$. If we multiply both sides by $$\mathbf{\tilde{T}}^{(e)}$$, we get
 * $$\mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)} \mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}^{(e)} \mathbf{f}^{(e)}$$

We can then obtain
 * $$\mathbf{\tilde{T}}^{(e)-1} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)} \mathbf{d}^{(e)} = \mathbf{f}^{(e)}$$

It has been proven that $$\mathbf{\tilde{T}}^{(e)-1} = \mathbf{\tilde{T}}^{(e)T}$$, so


 * $$[\mathbf{\tilde{T}}^{T} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}] \mathbf{d}^{(e)} = \mathbf{f}^{(e)}$$

Now notice the relation between the two equations.
 * $$[\mathbf{\tilde{T}}^{T} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}] \mathbf{d}^{(e)} = \mathbf{f}^{(e)} \Leftrightarrow  \mathbf{k}^{(e)}  \mathbf{d}^{(e)} =  \mathbf{f}^{(e)}  $$

We can now see,
 * $$ \mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{T} \mathbf{\tilde{k}}^{(e)} \mathbf{\tilde{T}}^{(e)}$$