User:Eml4500.f08.Ateam.philbin/HW4

Homework 4

Mtg. 20 10-10-08

Changing global degrees of freedom to a local coordinate system aligned along the truss
The tilde vectors represent the axial and normal forces from the axis of the bar. The same equation is used to start the derivation.

$$ \tilde\mathbf {f}^{(e)}= \tilde\mathbf {k}^{(e)} \tilde\mathbf {d}^{(e)} = 0 $$

A transformation vector will be used to get $$ \tilde\mathbf {d} $$ as well as $$ \tilde {f} $$.

$$ \tilde\mathbf {d}^{(e)} = \tilde\mathbf {T}^{(e)} \mathbf{d}^{(e)} $$

$$ \tilde\mathbf {f}^{(e)} = \tilde\mathbf {T}^{(e)} \mathbf{f}^{(e)} $$

By substituting the values for $$ \tilde \mathbf{T} $$ it can be proven.

$$ \tilde \mathbf{T} = \begin{bmatrix} cos (\Theta) & sin (\Theta) & 0 & 0\\ -sin (\Theta) & cos (\Theta) & 0 & 0\\0 & 0 & cos (\Theta) & sin (\Theta)\\0 & 0 & -sin (\Theta) & cos (\Theta)\end{bmatrix}=\begin{bmatrix} \mathbf{R}^{(e)} & \mathbf{0}\\\mathbf{0} & \mathbf{R}^{(e)}\end{bmatrix}$$

$$ \tilde \mathbf{T}^{(1)} \mathbf{f}^{(1)}= \begin{bmatrix} cos (30) & sin (30) & 0 & 0\\ -sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & sin (30)\\0 & 0 & -sin (30) & cos (30)\end{bmatrix}\begin{bmatrix}-4.378\\-2.5622\\4.4378\\2.5622\end{bmatrix} = \tilde\mathbf{f}^{(1)}$$

$$\tilde \mathbf{f}^{(1)} = \begin{bmatrix} -5.1243\\0\\5.1243\\0\end{bmatrix} $$

This confirms the previous statics work for $$ \tilde \mathbf{f}^{(1)} $$.

By substituting the previous relationships into $$ \tilde \mathbf{f}^{(e)}= \tilde \mathbf{k}^{(e)} \tilde \mathbf{d}^{(e)}$$, this equation can be rewritten as follows.

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

Verify transformation matrix transpose theorem
If the transformation matrix, $$\tilde \mathbf{T}^{(e)}$$, is invertible then the equation can be changed into the following form.

$$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}]\ \mathbf{d}^{(e)}= \tilde \mathbf{f}^{(e)}$$

Now since only a square matrix can be inverted, the following process was investigated. Since $$\tilde \mathbf{T}^{(e)}$$ is composed of R matrices which are sin and cos functions, when the transpose of $$\tilde \mathbf{R}^{(e)}$$ is multiplied by $$\tilde \mathbf{R}^{(e)}$$, the result is the identity matrix.

$$ \mathbf {R}^{(e)T} \ \mathbf {R}^{(e)} = \begin{bmatrix} \sin^2 {\Theta} + \cos^2 {\Theta} & 0 \\ 0 & \sin^2 {\Theta} + \cos^2 {\Theta} \end{bmatrix}=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$

Therefore if both sides are divided by $$ \mathbf {R}^{(e)}$$, then the following relationship can be made.

$$\mathbf {R}^{(e)-1}=\mathbf {R}^{(e)T} $$

This simplifies the math involved significantly and allows for rectangular matrices to be used instead of solely square matrices. Since $$\tilde \mathbf{T}^{(e)}$$ is comprised of a diagonal $$\tilde \mathbf{R}^{(e)}$$ matrix, the previous relationship between the transpose and inverse can be applied to $$\tilde \mathbf{T}^{(e)}$$ as well.

To verify that $$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}] $$, the values from the 2-bar truss will be inserted. The first element's values will be used.

$$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}] = \begin{bmatrix} cos (30) & -sin (30) & 0 & 0\\ sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & -sin (30)\\0 & 0 & sin (30) & cos (30)\end{bmatrix}\begin{bmatrix}0\\0\\4.4352\\6.1271\end{bmatrix} \begin{bmatrix} cos (30) & sin (30) & 0 & 0\\ -sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & sin (30)\\0 & 0 & -sin (30) & cos (30)\end{bmatrix}$$

$$= \begin{bmatrix} 0.5625 & 0.325&-0.5625&-0.325\\0.325&0.1875&-0.325&-0.1875\\-0.5625&-0.325&0.5625&0.325\\-0.325&-0.1875&0.325&0.1875\end{bmatrix}= \mathbf{k}^{(1)}$$

This verifies the derivation made earlier.