User:EML4500.f08.A-team.VandenBerg/HW7

 Comment: This is HW7 located in the user namespace of EML4500.f08.A-team.VandenBerg 22:09, 20 November 2008 (UTC)

Meeting 35: Monday, 17 November 2008. EML4500
35-1

Dimensions for Element # 1 of the 2 bar system: $$E_{1}^{(1)}=2, E_{2}^{(1)}=4$$

$$A_{1}^{(1)}=0.5, A_{2}^{(1)}=1.5$$

Dimensions for Element # 2 of the 2 bar system:

$$E_{1}^{(2)}=3, E_{2}^{(2)}=7$$

$$A_{1}^{(2)}=1, A_{2}^{(2)}=3$$

- Compute a solution for the 2 bar truss system with the tapered elements.

- Plot the deformed shape for this model using average E and average A.

- Plot the deformed shape for the previous model again using the average E and average A.

Frame element = truss (bar) element + beam element

- Truss bar element is the axial deformation

- Beam element is the transverse deformation

35-2

Model the frame with 2 elements.



Free Body Diagrams:



Circles on elements represent nodes, not pin connections.

35-3

In general, $$d_{i}^{(e)} \rightarrow f_{i}^{(e)}$$

where $$d_{i}^{(e)}$$ are the generalized displacements,

while $$f_{i}^{(e)}$$ are the generalized forces.

$$ e = 1, 2 $$

$$ i = 1,. . ., 6 $$

$$ \begin{Bmatrix} d_{3}^{(e)} \\ d_{6}^{(e)} \end{Bmatrix} $$ (rotational degrees of freedom) $$ \Rightarrow \begin{Bmatrix} f_{3}^{(e)} \\ f_{6}^{(e)} \end{Bmatrix} $$ (bending moments)

Two dimensional frame global degree of freedoms:



2 element stiffness matrix $$\underline{k}_{6x6}^{(e)}, e = 1 , 2 $$

Global stiffness matrix $$ \underline{K}_{9x9} = \overset{e=2}{\underset{e=1}{A}} \underline{k}_{6x6}^{(e)} $$

35-4



Meeting 36: Wednesday, 19 November 2008. EML4500
36-1



$$ \tilde {\mathbf{k}} _ {6 x 6} ^ {(e)}\ \tilde {\mathbf{d}} _ {6 x 1} ^ {(e)}\ =\ \tilde {\mathbf{f}} _ {6 x 1} ^ {(e)}$$

$$ \tilde {\mathbf{d}} ^ {(e)} = \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_6^{(e)} \end{Bmatrix} = \tilde {\mathbf{f}} ^ {(e)} = \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_6^{(e)} \end{Bmatrix} $$

Note: $$ \tilde {f} _{3}^ {(e)}\ =\ f_3^{(e)}\, \ \tilde {f} _{6}^ {(e)}\ =\ f_6^{(e)} $$

moments (about $$  \tilde {z}\ =\ z) $$

36-2

$$ \begin{matrix} \tilde {d} _1 & \tilde {d} _2 & \tilde {d} _3 & \tilde {d} _4 & \tilde {d} _5 & \tilde {d} _6 & \cdots\\ EA/L & 0 & 0 & -EA/L & 0 & 0 & \tilde {d}_1\\ \cdots & 12EI/L^3 & 6EI/L^2 & 0 & -12EI/L^3 & 6I/L^2 & \tilde {d}_2\\ \cdots & \cdots & 4EI/L & 0 & -6EI/L^2 & 2EI/L & \tilde {d}_3\\ \cdots & \cdots & \cdots & EA/L & 0 & 0 \tilde {d}_4\\ \cdots & \cdots & \cdots & \cdots & 12EI/L^3 & -6EA/L^2 & \tilde {d}_5\\ SYM & \cdots & \cdots & \cdots & \cdots & -EA/L & \tilde {d}_6 \end{matrix} $$

36-3

Aranuas, David

Dimensional Analysis
$$ \left [ \tilde {d}_1 \right ] = L = \left [ \tilde {d}_i \right ] \qquad i=1, 2, 4, 5$$

where,
 * $$ \left [ \tilde {d}_1 \right ] $$ is a dimension of length

and,
 * $$ \left [ \tilde {d}_i \right ] $$ is displacement

$$ \left [ \tilde {d}_3 \right ] = 1 = \left [ \tilde {d}_6 \right ] $$

where,
 * 1 has no dimensions

and,
 * $$ \left [ \tilde {d}_6 \right ] $$ is the rotational displacement



$$ \left [ \theta \right ] = \frac {[AB]}{[R]} = \frac {L}{L} = 1 $$

$$ \sigma = \Epsilon \varepsilon \rightarrow \left [ \sigma \right ] = \left [ \Epsilon \right ] \left [ \varepsilon \right ] = 1 $$


 * Why?

$$ \overline{AB} = R \theta $$

where,
 * $$ \overline{AB} $$ is the arclength

and,
 * $$ \theta $$ is in radians.

Verifying Dimensions and Element Force Displacement Relations
Using the definition shown above, the following dimensional analysis shows the rotational degrees of freedom are dimensionless. $$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

A dimensional analysis can also be done on other important parameters:

$$ \left[\epsilon\right] = \frac{[du]}{[dx]} = \frac{L}{L} = 1 $$

$$ \left[\sigma\right] = \left[E\right] = \frac{F}{L^2} $$

$$ \left[A\right] = L^2, [I] = L^4 $$

$$ \left[\frac{EA}{L}\right] = [\tilde{k}_{11}] = \frac{\frac{F}{L^2}{L^2}}{L} = \frac{F}{L} $$

$$ \left[\tilde{k}_{11}\tilde{d_1}\right] = [\tilde{k}_{11}][\tilde{d_1}] = F $$

$$ \left[\tilde{k}_{23}\tilde{d_3}\right] = [\tilde{k}_{23}][\tilde{d_3}] $$ where $$ \tilde{d_3} = 1 $$

$$ = \frac{\left[6\right][E][I]}{L^2} $$

$$ = \frac{\left(1\right)\left(\frac{F}{L^2}\right)(L^4)}{L^2} = F $$

Verifying Dimensions of $$ \tilde{k}_{ij} $$ and $$ \tilde{d}_{j} $$ for $$ i $$ and $$ j $$ = 1,...,6:
The matrix $$ \tilde{k_{ij}}\tilde{d_j} $$ is shown below

$$ \begin{bmatrix}\frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

The dimension of each term in the $$ \tilde{k_{ij}}\tilde{d_j} $$ matrix was calculated just as the above terms were in the previous notes and the resulting matrix is found to be:

$$ \begin{bmatrix}\frac{F}{L} & 0 & 0 & \frac{-F}{L} & 0 & 0 \\ 0 & \frac{F}{L} & F & 0 & \frac{-F}{L} & F \\ 0 & F & FL & 0 & -F & FL \\ \frac{-F}{L} & 0 & 0 & \frac{F}{L} & 0 & 0 \\ 0 & \frac{-F}{L} & -F & 0 & \frac{F}{L} & -F \\ 0 & F & FL & 0 & -F & FL \end{bmatrix} $$

The element force displacement relation in global coordinates from the element force displacement relation in local coordinates yields:

$$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} = \underline{f}^{(e)}_{6x1} $$

with $$ \underline{k}^{(e)}_{6x6} = {\underline{\tilde{T}}^{(e)}_{6x6}}^T\underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{T}}^{(e)}_{6x6} $$

from $$ \underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{d}}^{(e)}_{6x1} = \underline{\tilde{f}}^{(e)}_{6x1} $$

$$ \begin{bmatrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0 \\ 0 & 0 & 0 & -m^{(e)} & l^{(e)} & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{bmatrix} $$

The derivation of $$ \tilde{K}^{(e)} $$ from PVW, focusing only on bending effects:

$$ \frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial^{2}v}{\partial x^{2}}\right) - f_{t}(x) = m(x)\ddot{v} $$

Meeting 38: Monday, 24 November 2008. EML4500
PVW for beams:  $$ \int_{0}^{b} w(\underbrace{x}_{\tilde{x}}) [-\frac{\delta^2}{\delta x^2} ((EI) \frac{\delta^2 v}{\delta x^2}) +ft-m \ddot{v}] \  dx = 0$$ for all possible w(x)

Integration by parts of 1st term

$$ \alpha = \int_{0}^{L} \underbrace{w(x)}_{g(x)} \frac{\delta^2}{\delta x^2}{(EI) \frac{\delta^2 v}{\delta x^2}} \ dx $$

$$ \underbrace{\frac{\delta}{\delta x} (\underbrace{\frac{\delta}{\delta x}(EI) \frac{\delta^2 v}{\delta x^2}}_{r(x)})}_{r'(x)} $$

$$ \alpha = [\underbrace{w \frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{\beta_{1}}]^{2}_{0} - \int_{0}^{L} \underbrace{\frac{dw}{dx}}_{g'(x)} \underbrace{\frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{r(x)} \ dx$$

$$ \alpha = \beta_{1} - [\underbrace{w \frac{\delta}{\delta x} {(EI) \frac{\delta^2 v}{\delta x^2}}}_{\beta_{2}}]^{2}_{0} + \underbrace{\int_{0}^{L} \frac{\delta^2 w}{\delta x^2} (EI) \frac{\delta^2 v}{\delta x^2} \ dx}_{\gamma}$$

Note : symmetric

Eq. (1) becomes: $$ - \beta_{1} + \beta_{2} - \gamma + \int_{0}^{L} wft \ dx - \int_{0}^{L} wm \ddot{v} \ dx = 0 $$ for all possible w(x)

Let's focus on the stiffness term $$ \gamma $$ for now to derive the beam stiffness matrix and to identify the beam shape functions.



$$ v( \tilde{x}) = N_{2} (\tilde{x})\tilde{d}_{2} + N_{3} (\tilde{x})\tilde{d}_{3} + N_{5} (\tilde{x})\tilde{d}_{5} + N_{6} (\tilde{x})\tilde{d}_{6} $$

Recall : $$ u(\tilde{x}) = N_{1} (\tilde{x})\tilde{d}_{1} + N_{4} (\tilde{x})\tilde{d}_{4} $$



p.246:

$$ N_{2} (\tilde{x}) = 1 - \frac{3 \tilde{x}^2}{L^2} + \frac{2 \tilde{x}^3}{L^3} \ \tilde{d}_{2} $$

$$ N_{3} (\tilde{x}) = \tilde{x} - \frac{2 \tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^3} \ \tilde{d}_{3} $$

$$ N_{5} (\tilde{x}) = \frac{3 \tilde{x}^2}{L^2} - \frac{2 \tilde{x}^3}{L^3} \ \tilde{d}_{5} $$

$$ N_{6} (\tilde{x}) = - \frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2} $$

Meeting 40: Friday, 5 December 2008. EML4500
Note: Eq. (2) (39-2), dimension analysis:

$$ [u] {=} L $$

(31-3): $$ [N_1] {=} [N_4] {=} 1 $$

$$ Eq. (2) (39-2) \underbrace{[N_1]}_{1} \underbrace{[\tilde{d}_1]}_{L} + \underbrace{[N_4]}_{1} \underbrace{[\tilde{d}_4]}_{L} $$

$$ [V] {=} L $$

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ (displ. transv.)

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ (rot.)

$$ {[N_5]}{[\tilde{d}_5]} $$ and $$ {[N_6]}{[\tilde{d}_6]} $$ are to be done for homework

Derivation of the beam shape functions

$$ N_2, N_3, N_5, N_6 $$ (38-4)

p. (38-3) plots

Recall: Governing PDE for beams

p. (37-4), Eq. (1), without $$ f_t $$ (distributed transverse load) and without inertia force $$ m\ddot{v} $$ (static case):

$$ \frac{\delta^2}{\delta x^2} \left \{(EI) \frac{\delta^2v}{\delta x^2}\right \} {=} 0 $$

Further, consider constant EI:

$$ \frac{\delta^4}{\delta x^4} v {=} 0 $$ Integrate 4 times to get 4 constants.

$$ \Rightarrow V(x) {=} C_0 + C_1x^1 + C_2x^2 + C_3x^3 $$

To obtain $$ N_2(x) $$ ($$ \tilde{x} $$ {=} x for simplicity)

V(0) = 1,  V(L) = 0

V'(0)   =   V'(L) = 0

Use above boundary conditions to solve for $$ L_0, ... , L_3 $$

V(0) = 1 = C0

(1) $$ V(L) {=} 1 + C_1(L) + C_2(L)^2 + C_3(L)^3 {=} 0 $$

$$ V^{\prime}(x) {=} C_1 + 2C_2x + 3C_3x^2 $$

$$ V^{\prime}(0) {=} C_1 {=} 0 $$

(2) $$ V^{\prime}(L) {=} 2C_2(L) + 3C_3(L)^2 {=} 0 $$

(2) $$ \rightarrow C_3 {=} - \frac{2}{3} \frac{C_2}{L} $$

(1) $$ \rightarrow 0 {=} 1 + C_2L^2 + (\frac{-2}{3} \frac{C_2}{L})L^3 {=} 1 + C_2L^2[1 - \frac{2}{3}] $$

$$ C_2 {=} - \frac{3}{L^2} $$

$$ C_3 {=} - \frac{2}{3} \frac{1}{L}(- \frac{3}{L^2}) {=} \frac{2}{L^3} $$

Compare with expression for N2 on p. (38-4)

For N3: boundary conditions         $$ \tilde{d}_3 $$ rotation

V(0) = V(L) = 0

V'(0) = 1, V'(L) = 0

For N5: V(0) = 0, V(L) = 1    $$ \tilde{d}_5 $$ displacement

V'(0) = 0, V'(L) = 0

For N6: $$ d_6 $$ rotation

V(0) = V(L) = 0

V'(0) = 0, V'(L) = 1

See p. (39-1) for plots of N5, N6

Derive coefficient in $$ \underline{\tilde{k}} $$ (elem. stiffness matrix)

Coefficient for EA: Done

Coefficient for EI: To Be Done

$$ \tilde{k}_{22} {=} \frac{12EI}{L^3} \int^L_0 \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx $$

Vibrational Analysis Using FEA
$$\hat{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx}$$

In general,

$$\hat{k}_{ij} = \int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}dx}$$

With i,j = 2,3,5,6

Elastodynamics (trusses, 2-D frames, 3-D elasticity)

Using modal problem from lecture 31 using discrete Principle of Virtual Works:

$$ \mathbf{\bar{w}} = [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0$$

In the above equation, the boundary conditions are already applied

The following equation is true for all $$\mathbf{\bar{w}}$$:

$$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = \mathbf{\bar{F}}(t)$$

where $$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$

and $$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{V}}_0$$

The above will be referred to as (1)

These are the complete ordinary differential equations (ODEs), second order in time, and initial conditions governing the elastodynamics of the discretized continuous problem with multiple degrees of freedom.

Solving equation (1):

1) Consider the unforced vibrations problem:
$$\mathbf{\bar{M}}_{nxn}\mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn}\mathbf{v}_{nx1} = \mathbf{0}_{nx1}$$

Unforced meaning the equation is equal to zero. This will be referred to as equation (2)

Assume: $$\mathbf{v}(t)_{nx1} = (sinwt)\mathbf{\phi _{nx1}}$$

Where the phi matrix is not time dependent

Thus: $$\mathbf{\ddot{v}} = -\omega ^2 sin\omega t\mathbf{\phi }$$

This means equation (2) becomes:

$$-\omega ^2 sin\omega t\mathbf{\bar{M}}\mathbf{\phi } + \omega ^2 sin\omega t\mathbf{\bar{K}}\mathbf{\phi } = \mathbf{0}$$

Therefore:

$$\Rightarrow \mathbf{\bar{k}}\mathbf{\phi } = \omega ^2\mathbf{\bar{M}}\mathbf{\phi }$$

Which is the generalized eigenvalue problem, as it is of the form:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}$$

With lambda as the eigenvalue.

A standard eigenvalue problem: $$\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$$ is given when the B matrix is equal to the identity matrix.

This means that

$$\lambda =\omega ^2$$ is an eigenvalue.

$$(\lambda _i,\mathbf{\phi _i})$$ are eigenpairs.

with i = 1 through n

Now the mode i for the animation can be represented as:

$$\mathbf{v}_i(t) = (sinw_it)\mathbf{\phi _i}$$

for i = 1 through n

2) Modal superposition method:
Using orthogonal properties of the eigenpairs:

$$\mathbf{\phi _i}^T_{1xn}\mathbf{\bar{M}}_{nxn}\mathbf{\phi }_{nx1} = \delta _{ij} = \begin{cases} & \text{1 if } i=j \\ & \text{0 if } i\neq j \end{cases}$$

This delta is the Kronecker delta.

Mass orthogonality of the eigenvector.

Now, applying this to Eq (1) and (2) gives:

$$\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j \cdot \mathbf{\bar{k}}\mathbf{\phi }_j$$

$$ \mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j$$

Therefore:

$$\Rightarrow \phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j = \frac{1}{\lambda _j}\delta _{ij}$$

Equation (1) can be written as:

$$\mathbf{\bar{M}}(\sum_{j}^{}{\ddot{\zeta }_j\mathbf{\phi }_j}) + \mathbf{\bar{K}}(\sum_{j}^{}{\zeta _j\mathbf{\phi }_j}) = \mathbf{F}$$

In the above equation, the first term in parenthesis is equal to the second derivative of the d matrix, while the second term in parenthesis is equal to the d matrix.

We can also write:

$$\sum_{j}^{}{\ddot{\zeta }_j(\mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j} + \sum_{j}^{}{\zeta _j(\mathbf{\phi }_i^T\mathbf{\bar{K}}\mathbf{\phi }_j} = \mathbf{\phi }_i^T\mathbf{F}$$

The first term in parenthesis in this equation is equal to the Kronecker delta, while the second term in parenthesis is equal to an eigenvalue times the Kronecker delta.

Finally,

$$\Rightarrow \ddot{\zeta } + \lambda _i\zeta _i = \mathbf{\phi }_i^T\mathbf{F}$$

with i = 1 through n

MediaWiki VS. WebCT
It is the opinion of this group that neither of these options for completing homework are very appropriate. MediaWiki, while fancy looking and user friendly to viewers, can consume a large portion of time simply in writing the page. This consumption of time takes away from the learning experience in FEA, since FEA is not a web design course. MediaWiki has other problems also, such as having to deal with moderators, administrators, and others who may, intentionally or unintentionally, and with or without malice, change the content of a student's page. WebCT does not have this problem however, but WebCT is not much more viable, it is difficult to work on group projects, and those with older computers or those who devote too much memory to other applications on their computers can find the system slow and prone to crashing. Also, the proposed idea of the student's work lasting far into the future is somewhat ludicrous, since unless that student constantly checks the site, it will most likely become prone to vandals. If a student wishes to have his/her work last into the far future, it would be more logical for them to save their work on their computer, or a flash drive, or even in a filing cabinet. Those student's who would like to be recognized for their work or have their work published for all to see would be advised to contact the appropriate journal/magazine and submit their work to them. It is this groups' conclusion that more time has been spent writing code for MediaWiki than studying the material, which is an inappropriate use of time. Also, receiving/checking grades has become more of a hassle than it should be, and figuring out what was marked wrong and right is practically impossible, considering the only guide the student is given is a selection of other student's homework that is "better" than their own, but it is not known in what way it is "better" or whether or not it is entirely accurate. There lies the possibility that one group's homework may be academically more accurate, but misses the aesthetic quality of another group's work, and is therefor marked with a lower grade.

It is our opinion that neither MediaWiki nor WebCT is an appropriate way to submit homework for this class.

MATLAB PROBLEMS
Here is the Matlab code for the Tapered Two Bar Truss System:



Here is the code for the Two Bar Frame/Truss System:



Here is the code for the Electric Pylon System:

Results are as follows:

Highest Bending Moment was 6.2874 in Element 81 Highest Transverse Shear Stress was 5.4827 in Element 81 The first three eigens were 132.86, 2471.2, and 2942.6 occurring in elements 79, 81, and 82, respectively. The three lowest vibrational periods were 0.54511, 0.12639, and 0.11583

The frame model is not statically determinate since there are far too many unknowns and not enough equations.

The reactions for the frame model are the same as those for the truss model, with the exception of an additional two reactions that relate to the rotations at the nodes.









Below are the M-file functions used to complete the previous Matlab codes:

Contributions
--EML4500.f08.A-team.VandenBerg 22:12, 20 November 2008 (UTC)

--EML4500.f08.A-team.rieth 19:11, 3 December 2008 (UTC)

--EML4500.f08.A-team.melvin 01:26, 9 December 2008 (UTC)

--EML4500.f08.A-team.kirley 02:50, 9 December 2008 (UTC)

--Eml4500.f08.a-team.robinson 01:12, 9 December 2008 (UTC)

--EML4500.f08.A-team.morford 20:11, 9 December 2008 (UTC)