User:Eml4500.f08.Lulz.Layton.Eric

3-bar Truss System


For element 1:

$$E^{(1)}=2$$ $$A^{(1}=3$$ $$L^{(1)}=5$$ $$\theta^{(1)}=30$$(deg)

For element 2:

$$E^{(2)}=4$$ $$A^{(2}=1$$ $$L^{(2)}=5$$ $$\theta^{(2)}=30$$(deg)

For element 3:

$$E^{(3)}=3$$ $$A^{(3}=2$$ $$L^{(3)}=10$$ $$\theta^{(3)}=45$$(deg)

If we break down each element into its free-body diagram we get the following 3 pictures:







If we put these 3 pictures together we get:



We cannot solve this problem by statics because we only have 2 useful equations.

$$\Sigma F_x=0$$

$$\Sigma F_y=0$$

The moment equation $$\Sigma M_A=0$$ is trivial and does not help us solve the problem.

But what about taking the moment about an arbitrary point B?



Following the math for this we get:

$$\Sigma\textbf{M}_B=\textbf{BA}\times\textbf{F}$$

$$\Sigma\textbf{M}_B=\textbf{BA}^'\times\textbf{F}$$

$$\textbf{BA}^'=\textbf{BA}\times\textbf{AA}^'$$

$$\Sigma\textbf{M}_B=(\textbf{BA}+\textbf{AA}^')\times\textbf{F}$$

$$\Sigma\textbf{M}_B=\textbf{BA}\times\textbf{F} + \textbf{AA}^'\times\textbf{F}$$

Where: $$\textbf{AA}^'\times\textbf{F}=0$$ which is trivial and doesn't help us.

Now if we look back at the three-bar truss example we see that node A is in equilibrium.

$$\Sigma_{i=0}^3\textbf{F}_i=\overrightarrow{0}$$

So, where A' is any point on the line of action of Fi:

$$\Sigma\textbf{M}_{Bi}=\Sigma\textbf{BA}_i^' x \textbf{F}_i$$

Where $$\Sigma_i\textbf{F}_i=0$$ and is trivial.

Now we need to find the global stiffness matrix. This can be found in the following manner:



So, for example we see:

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}$$

Axial displacement calculation for local node 2
During meeting 12, a calculation for the axial displacement of local node 1 for an arbitrary element e was considered. What follows is a similar calculation for local node 2 of the same element.

First, we define the displacement vector at local node 2 (Please note the [#] subscript was used in place of the "box" subscript used in class):

$$ \vec{\mathbf{d}^{(e)}_{[2]}} = d_{3}^{(e)} \vec{i} + d_{4}^{(e)} \vec{j} $$

Next, the axial displacement at node 2 is defined: $$ q_{2}^{(e)} = \vec{\mathbf{d}^{(e)}_{[2]}}\cdot\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = (d_{3}^{(e)}\vec{i} + d_{4}^{(e)}\vec{j})\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = d_{3}^{(e)}(\vec{i}\cdot\vec{\tilde{i}}) + d_{4}^{(e)}(\vec{j}\cdot\vec{\tilde{i}}) $$

From a previous homework question and the class notes, the following relations are valid:

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

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}\cos \theta^{(e)} + d_{4}^{(e)}\sin\theta^{(e)} $$

From the notes, the following relations are valid:

$$\cos\theta^{\left(e\right)} = l^{(e)}$$ $$\sin\theta^{\left(e\right)} = m^{(e)}$$

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}l^{(e)} + d_{4}^{(e)}m^{(e)} $$

Putting the above equation yields the following result, consistent with that of the axial displacement of local node 1:

$$ q_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{Bmatrix} $$

Proof of $$\mathbf{k}^{(e)} = \mathbf{T}^{(e)^{T}} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$
In this question we are asked to prove that the above matrix product generates the generic element stiffness matrix $$\mathbf{k}^{(e)}$$.

First, the three matrices used to generate the product:

$$\mathbf{T}^{(e)} = \begin{bmatrix} l^{(e)}&m^{(e)}&0&0 \\ 0&0&l^{(e)}&m^{(e)} \end{bmatrix}$$

$$\mathbf{T}^{(e)^{T}} = \begin{bmatrix} l^{(e)}&0 \\ m^{(e)}&0 \\ 0&l^{(e)} \\ 0&m^{(e)} \\ \end{bmatrix}$$

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

The product of the first two matrices:

$$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&0&0 \\ 0&0&l^{(e)}&m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix}$$

The entire product multiplied out:

$$\mathbf{T}^{(e)^{T}}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} l^{(e)}&0 \\ m^{(e)}&0 \\ 0&l^{(e)} \\ 0&m^{(e)} \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix} $$ $$ = \begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} \\ m^{(e)}l^{(e)} & m^{(e)^{2}} & -m^{(e)}l^{(e)} & -m^{(e)^{2}}\\ -l^{(e)^{2}} & -m^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)^{2}} & m^{(e)}l^{(e)} & m^{(e)^{2}} \\ \end{bmatrix} $$

As shown in previous homework question, the element stiffness matrix:

$$\mathbf{k}^{(e)} = \begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} \\ m^{(e)}l^{(e)} & m^{(e)^{2}} & -m^{(e)}l^{(e)} & -m^{(e)^{2}}\\ -l^{(e)^{2}} & -m^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)^{2}} & m^{(e)}l^{(e)} & m^{(e)^{2}} \\ \end{bmatrix}$$

Thus, it is easily shown that the following relation is indeed true:

$$\mathbf{k}^{(e)} = \mathbf{T}^{(e)^{T}} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$

John Saxon Eml4500.f08.Lulz.js 18:09, 8 October 2008 (UTC)