User:Eml4500.f08.echo.mott/HW3

= Homework Report 3 =
 * This report contains material for homework report 3 in EML4500.

Derivation of Element F-D with Respect to Global Coordinate System
$$\mathbf{k}_{4x4}^{(e)}\mathbf{d}_{4x1}^{(e)}=\mathbf{f}_{4x1}^{(e)}$$  (Equation 1)

This can equivalently be expressed in the following figure.



In matrix terms this is:

$$k^{(e)}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}=\begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix}$$  (Equation 2)

In short:

$$\hat{\mathbf{k}}_{2x2}^{(e)}\mathbf{q}_{2x1}^{(e)}=\mathbf{P}_{2x1}^{(e)}$$

$$q_i^{(e)}=$$ axial displacement of element e at local node i

$$P_i^{(e)}=$$ axial force of element e at local node i


 * Goal: Derive Equation 1 from Equation 2. This can be done with the help of previous notes.

The k matrix can be found using the theta's.

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

Where: $$l_{ }^{(e)}=cos( \theta^{(e)})$$

$$m_{ }^{(e)}=sin( \theta^{(e)})$$

Next, the 4x1 distance matrix can be found using trigonometry.

$$\mathbf{d}_{4x1}^{(e)}=\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}=\begin{Bmatrix} q_1^{(e)}\cdot cos\theta_1^{(e)} \\ q_1^{(e)}\cdot sin\theta_1^{(e)} \\ q_2^{(e)}\cdot cos\theta_1^{(e)} \\ q_2^{(e)}\cdot sin\theta_1^{(e)} \end{Bmatrix}$$

Similarly, the 4x1 force matrix can be derived.

$$\mathbf{f}_{4x1}^{(e)}=\begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}=\begin{Bmatrix} P_1^{(e)}\cdot cos\theta_1^{(e)} \\ P_1^{(e)}\cdot sin\theta_1^{(e)} \\ P_2^{(e)}\cdot cos\theta_1^{(e)} \\ P_2^{(e)}\cdot sin\theta_1^{(e)} \end{Bmatrix}$$

Therefore the result is:

$$k^{(e)}\begin{bmatrix} cos( \theta^{(e)})^2 & cos( \theta^{(e)})sin( \theta^{(e)}) & -cos( \theta^{(e)})^2 & -cos( \theta^{(e)})sin( \theta^{(e)}) \\cos( \theta^{(e)})sin( \theta^{(e)}) & sin( \theta^{(e)})^2 & -cos( \theta^{(e)})sin( \theta^{(e)}) & -sin( \theta^{(e)})^2 \\-cos( \theta^{(e)})^2 & -cos( \theta^{(e)})sin( \theta^{(e)}) & cos( \theta^{(e)})^2 & cos( \theta^{(e)})sin( \theta^{(e)}) \\-cos( \theta^{(e)})sin( \theta^{(e)}) & -sin( \theta^{(e)})^2 & cos( \theta^{(e)})sin( \theta^{(e)}) & sin( \theta^{(e)})^2\end{bmatrix}\begin{Bmatrix} q_1^{(e)}\cdot cos\theta_1^{(e)} \\ q_1^{(e)}\cdot sin\theta_1^{(e)} \\ q_2^{(e)}\cdot cos\theta_1^{(e)} \\ q_2^{(e)}\cdot sin\theta_1^{(e)} \end{Bmatrix}=\begin{Bmatrix} P_1^{(e)}\cdot cos\theta_1^{(e)} \\ P_1^{(e)}\cdot sin\theta_1^{(e)} \\ P_2^{(e)}\cdot cos\theta_1^{(e)} \\ P_2^{(e)}\cdot sin\theta_1^{(e)} \end{Bmatrix}$$

These relations can be expressed in the form:

$$\mathbf{q}_{2x1}^{(e)}=\mathbf{T}_{2x4}^{(e)}\mathbf{d}_{4x1}^{(e)}$$

Consider the displacement vector of local node 1, denoted by $$d_1^{(2)}$$:



$$\vec{d}_1^{(e)}=d_1^{(e)}\vec{i}+d_2^{(e)}\vec{j}$$

$$q_1^{(e)}=$$ axial displacement of node 1 is the orthogonal projection of the displacement

Vector $$\vec{d}_1^{(e)}$$ of node 1 on the axes $$\tilde{x}$$ of element e:

$$\Rightarrow q_1^{(e)}=\vec{d}_1^{(e)}\cdot\vec{i}=(d_1^{(e)}\vec{i}+d_2^{(e)}\vec{j})\cdot\vec{\tilde{i}}=d_1^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+d_2^{(e)}(\vec{j}\cdot\vec{\tilde{i}})=d_1^{(e)}\cdot cos\theta^{(e)}+d_2^{(e)}\cdot sin\theta^{(e)}=d_1^{(e)}\cdot l^{(e)}+d_2^{(e)}\cdot m^{(e)}$$

$$\mathbf{q}_1^{(e)}=l^{(e)}\cdot d_1^{(e)}+m^{(e)}\cdot d_2^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$

Similarly for node 2:

$$\Rightarrow q_2^{(e)}=\vec{d}_3^{(e)}\cdot\vec{i}=(d_3^{(e)}\vec{i}+d_4^{(e)}\vec{j})\cdot\vec{\tilde{i}}=d_3^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+d_4^{(e)}(\vec{j}\cdot\vec{\tilde{i}})=d_3^{(e)}\cdot cos\theta^{(e)}+d_4^{(e)}\cdot sin\theta^{(e)}=d_3^{(e)}\cdot l^{(e)}+d_4^{(e)}\cdot m^{(e)}$$

$$\mathbf{q}_2^{(e)}=l^{(e)}\cdot d_3^{(e)}+m^{(e)}\cdot d_4^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix} d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}$$

Combining:

$$\mathbf{q}_{2x1}^{(e)}=\mathbf{T}_{2x4}^{(e)}\mathbf{d}_{4x1}^{(e)}$$

$$\begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix}$$

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix}_{2x1} = \mathbf{T}_{2x4}^{(e)} \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}_{4x1}$$

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

Recall the element axial Force-Displacement relationship

$$\mathbf{\hat{k}}_{2x2}^{(e)} = \mathbf{q}_{2x1}^{(e)} \mathbf{P}_{2x1}^{(e)}$$

where:

$$\mathbf{q}_{2x2}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{d}_{2x1}^{(e)}$$

$$\mathbf{P}_{2x2}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{f}_{2x1}^{(e)}$$

therefore:

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


 *  Goal:  We want to have $$\mathbf{k}_{}^{(e)} \mathbf{d}_{}^{(e)} = \mathbf{f}_{}^{(e)}$$.

So we need to move $$\mathbf{T}_{}^{(e)}$$ from the right hand side to the left hand side by "pre-multiplying" the equation by $$\mathbf{T}_{}^{(e) -1}$$.

Unfortunately since $$\mathbf{T}_{}^{(e)}$$ is a rectangular matrix it can not be inverted.

However through usage of the Principle of Virtual Work we find $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

Which yields answer of $$[\mathbf{T}_{4x2}^{(e)T} \mathbf{\hat{k}}_{2x2}^{(e)}\mathbf{T}_{2x4}^{(e)}]_{4x4} \mathbf{d}_{4x1}^{(e)} = \mathbf{f}_{4x1}^{(e)}$$

and gives us $$\mathbf{k}_{}^{(e)} \mathbf{d}_{}^{(e)} = \mathbf{f}_{}^{(e)}$$

We could not solve as $$\mathbf{d}_{}^{} = \mathbf{K}_{}^{-1} = \mathbf{F}_{}^{}$$ because of the singularity associated with matrix $$\mathbf{K}_{}^{}$$. ($$det\mathbf{K}_{}^{} = 0$$)

For an unconstrained strut system there are 3 possible rigid body motions in 2-D.

These 3 motions are Axial Deformation, Rotation at Nodes and Translation at Nodes.

Closing the Loop between FEM and Statics

 * When using the Finite Element Method, the displacements of the nodes are found and the reactions at the nodes are computed using the displacements. In Statics, the reactions are found directly. In order to close the loop, the displacements must be found from statics - after finding the reactions.


 * The image below shows the virtual displacement of global node 2. The displacement vector for node 2 is the vector from point A to point D. The length from A to C is the deformation of element 1. The length from A to B is the deformation of element 2.




 * The figure presented above can also be drawn and the displacements computed using MATLAB. The code and output image are collapsed in the table below.




 * The true displacement of the open-end in the image above has both an x and a y component. For very small angles, the equations for the components can be approximated by the ones shown in the figure below.



Infinitesimal Displacement

 * Related to virtual displacement



The length of AC can be found by:

$$AC=\frac{|P_2^{(1)}|}{k^{(1)}}=\frac{|P_2^{(1)}|}{(E^{(1)}A^{(1)})/{L^{(1)}}}=\frac{5.1243}{(3\cdot1)/4}=6.8324$$


 * The position of point D, $$(x_D^{ },y_D^{ })$$, is unknown. Therefore, equations for lines AB & BC must be found.



$$\begin{align} \overrightarrow{PQ}&=(PQ)\tilde{\vec{i}} \\ &=(PQ)[cos\theta \vec{i}+sin\theta\vec{j}] \\ &=(x-x_p)\vec{i}+(y-y_p)\vec{j} \\ \end{align}$$

$$ x-x_p=(PQ) cos \theta$$

$$ y-y_p=(PQ) sin \theta$$

Therefore,

$$\frac{y-y_p}{x-x_p} = tan \theta$$

And so,

$$ y-y_p = (tan \theta)(x-x_p)$$

Shown below is the equation for a line perpendicular to the above line passing P:

$$y-y_p=tan(\theta+\frac{\pi}{2})(x-x_p)$$

3-bar Truss System


$$E^{(1)}=2$$              $$E^{(2)}=4$$                 $$E^{(3)}=3$$ $$A^{(1)}=3$$              $$A^{(1)}=1$$                 $$A^{(3)}=2$$ $$L^{(1)}=5$$              $$L^{(2)}=5$$                 $$L^{(3)}=10$$

$$\mathbf {P}=30 $$

$$\theta{(1)} = {30}=-\theta{(2)}$$

$$\theta{( 3)} = 45 $$

Local node numbering = convenience in assembly of k

Convenient local node numbering




 * Go from the local node 2 of one element and make that the local node 1 of the next element

$$\Sigma F_x=0$$; $$\Sigma F_y=0$$; $$\Sigma M_A=0$$, This step is trivial

What about $$\Sigma M_B$$ ?

3-D Explanation

$$\sum M_b = \overrightarrow {BA}x \overrightarrow{F} = \overrightarrow {BA'}   x \overrightarrow{F} $$

For all points $$A'$$ on the line of action of $$\overrightarrow{F}$$ ,

$$\overrightarrow{BA'} = \overrightarrow{BA} + \overrightarrow{AA'}$$

$$\begin{align} \sum M_b &= (\overrightarrow {BA} + \overrightarrow{AA'})x \overrightarrow{F} \\ &= \overrightarrow {BA} x \overrightarrow{F} + \overrightarrow{AA'} x \overrightarrow{F} \\ \end{align}$$

Where $$\overrightarrow{AA'}x \overrightarrow{F} = \overrightarrow{0}$$ because they are both on the same line of action.

Back to the 3-Bar Truss:

Node A in equilibrium: $$\sum_{i=0}^{3} \overrightarrow{F_i} = \overrightarrow{0}$$

$$\sum_{i} \overrightarrow{M_{b,i}} = \sum_{i} \overrightarrow{BA'_i}x\overrightarrow{F_i}$$, where $$\ A'$$ is any point on the line of action of $$\overrightarrow{F_i}$$

Therefore:

$$\sum_{i} \overrightarrow{M_{b,i}} = \sum_{i} \overrightarrow{BA}x\overrightarrow{F_i} = \overrightarrow{BA}x\sum_{i} \overrightarrow{F_i} = \overrightarrow{0}$$



$$\mathbf{K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 & 0 & 0\\k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 & 0 & 0\\k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)} & k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)} & k_{13}^{(2)} & k_{14}^{(2)} & k_{13}^{(3)} & k_{14}^{(3)}\\k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)} & k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)} & k_{23}^{(2)} & k_{24}^{(2)} & k_{23}^{(3)} & k_{24}^{(3)}\\0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} & 0 & 0 \\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} & 0 & 0 \\0 & 0 & k_{31}^{(3)} & k_{32}^{(3)} & 0 & 0 & k_{33}^{(3)} & k_{34}^{(3)}\\0 & 0 & k_{41}^{(3)} & k_{42}^{(3)} & 0 & 0 & k_{43}^{(3)} & k_{44}^{(3)} \end{bmatrix}= \begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16} & K_{17} & K_{18}\\K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26} & K_{27} & K_{28}\\K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36} & K_{37} & K_{38}\\K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46} & K_{47} & K_{48}\\K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56} & K_{57} & K_{58}\\K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66} & K_{67} & K_{68}\\K_{71} & K_{72} & K_{73} & K_{74} & K_{75} & K_{76} & K_{77} & K_{78}\\K_{81} & K_{82} & K_{83} & K_{84} & K_{85} & K_{86} & K_{87} & K_{88}\end{bmatrix}$$

$$K_{33} = k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}$$ and $$K_{34} = k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}$$

Contributing Members
Chris Mott 16:42, 8 October 2008 (UTC)

Brad LaCroix Eml4500.f08.echo.lacroix 16:55, 8 October 2008 (UTC)

Brandon Sell Eml4500.f08.echo.sell.br 17:04, 8 October 2008 (UTC)

Charles Marshall 17:13, 8 October 2008 (UTC)

Jason Bruce Eml4500.f08.echo.bruce.jp 17:41, 8 October 2008 (UTC)

Chris Sell Eml4500.f08.echo.sell.cm 17:59, 8 October 2008 (UTC)