User:Eml4500.f08.bottle.vitello/HW 3

Lecture Notes
This part of the lecture concerns the derivation of the element force diagram with respect to the global coordinate system.

Initially in the previous lectures, we derived an element force diagram with $$f \! $$ representing element force and $$ d \! $$ representing element displacement on a truss element. We replace these unknowns with $$ P \! $$ representing the element force and $$ q \! $$ representing the element displacement through the axis of the truss element.



With:

1.   $$q_i^{(e)}$$=axial displacement of element e at local node i   2.    $$P_i^{(e)}$$=axial force of element e at local node i

To derive $$ k \!$$ $$* \!$$$$d \!$$$$= \!$$$$f \!$$ from $$k \!$$$$* \!$$$$q \!$$$$=$$$$P \!$$, we need to find the relation between $$q\!$$ and $$d\!$$, and $$P\!$$ and $$f\!$$. For displacement, we derive that:

$$ q_{2X1}^{(e)}$$ $$ = \!$$ $$T_{2X4}^{(e)}$$ $$ * \!$$ $$d_{4X1}^{(e)}$$

In order to derive $$T \!$$, we must consider the displacement vector of the nodes in each truss element. Considering the previous truss system derived in the lecture notes, we derive vector $$ d \! $$ of node 1 on the axis x of element e, giving us:

$$q_1^{(e)}$$ $$=$$ $$d_1^{(e)}$$ $$ *$$ $$ \bar{i} $$ $$ = $$ $$d_1^{(e)}$$ $$ \tilde{i} $$ $$ + $$ $$ d_2^{(e)} \tilde {j} $$ $$ $$ = $$ $$d_1^{(e)}$$ $$ ($$ $$\tilde{i}$$$$*$$ $$\bar{i}$$$$)$$ $$ + $$ $$ d_2^{(e)}$$ $$ ( $$ $$ \tilde{j}$$ $$*$$$$ \bar{i}) $$ $$=$$$$ d_1^{(e)}$$$$*$$ $$\cos (\theta^{(e)}) \! $$ $$ + $$ $$ d_2^{(e)}$$$$ * $$ $$ \sin (\theta^{(e)}) \!$$ $$q_1^{(e)}$$$$= \!$$$$ d_1^{(e)}$$$$ * \!$$ $$ l^{(e)} \!$$ $$+ \!$$ $$d_2^{(e)} $$ $$ * \!$$ $$ m^{(e)} \!$$
 * $$ $$ \bar {i} $$

Using the same procedure as before for element 2 of the truss system gives us:

$$q^{(e)}_2= (d^{(e)}_3\bar{i}+d^{(e)}_4\bar{j}) \cdot \bar{\tilde{i}}$$

$$= d^{(e)}_3(\bar{i}\cdot \bar{\tilde{i}})+d^{(e)}_4(\bar{j} \cdot\bar{\tilde{i}}) $$

$$= d_3^{(e)}\cdot cos(\theta^{(e)})+d_4^{(e)}\cdot sin(\theta^{(e)})$$

$$q_2^{(e)}$$ $$ = \!$$ $$d_3^{(e)}$$$$ *\! $$$$ l^{(e)}\!$$ $$ + \!$$ $$d_4^{(e)}$$ $$ *\!$$ $$ m^{(e)}\!$$

Combining these two equations gives us the result for the $$ T \! $$ matrix:

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

The same argument can also be applied to the axial force on the truss system, using the same procedure used to find the axial displacement of the truss system:

$$P^{(e)} \!$$ $$ = \!$$ $$T^{(e)}\!$$ $$* \!$$ $$f^{(e)}\!$$

Gives:

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ T_{2X4}^{(e)} $$ $$ * \! $$ $$ \begin{Bmatrix} f_1^{(e)} \\f_2^{(e)} \\f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$

We can then replace the axial forces for the equations derive, representing the element forces:

$$k_{2X1}^{(e)}$$ $$*$$ $$ q_{2X1}^{(e)}$$$$=\!$$$$ P_{2x1}^{(e)}$$ and $$k^{(e)} \!$$$$ * \!$$ $$\underbrace{(T^{(e)}\! d^{(e)}) \!}_{q^{(e)}}=\!$$ $$ \underbrace{(T^{(e)} \!  f^{(e)})\!}_{P^{(e)}}$$ Using the inverse of $$ T \! $$, we can get the following solution for the element forces: $$ \begin{bmatrix} (T^{(e)})^{T}k^{(e)}I^{(e)} \end{bmatrix}*d^{(e)} = f^{(e)}$$ However, we cannot invert $$ T \!$$ with itself since it is a rectangular matrix with a two by two matrix $$ k \!$$; we instead use the Principal of Virtual Work (PVW) to find $$f_1 \! $$, $$ f_2 \!$$, $$ f_5 \!$$, and $$ f_6 \!$$. There is no need to find $$ f_3 \! $$ and $$ f_4 \!$$ since they are the applied load on the truss system, which is already known.

The following loop used for the two bar truss system to compare between the Finite Element Method (FEM) and the statics method went as follows, starting with the FEM method:

FEM Compute displacements Compute reactions

Then the loop goes back using the statics method:

Statics Compute reactions Compute displacements

By using the statics method, the reactions $$ f_1 \! $$, $$ f_2 \! $$, $$ f_5 \! $$, and $$ f_6 \! $$ are known and with it, the forces $$ P \! $$ are then known. This knowledge can then be used to find the axial displacement $$ q \! $$ at each node on the truss system:

$$q_2^{(1)} $$ $$ =\frac{P_{2}^{(1)}}{k^{(1)}} \! $$$$ = \!$$ $$AC \!$$

$$q_1^{(1)} $$$$ = \!$$ $$0 \!$$ ,fixed node $$1 \!$$

$$q_1^{(2)}$$ $$ =\frac{P_{2}^{(2)}}{k^{(2)}} $$ $$= \!$$ $$ -AB \! $$

$$q_2^{(2)} $$$$=\! $$ $$0 \! $$ ,fixed node $$ 2 \! $$

Another method used to find the displacement of global node 2 of the two body truss system is to find the deformed shape of the truss system itself.



The first part is to find the deformed lengths of $$ AC \! $$ and $$ AB \! $$:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

Then the x-y coordinates of points B and C are determined, leaving only two unknowns $$ x_D \! $$ and $$ y_D \! $$. To find this displacement, the formula used to determine said point is:

$$ x-x_{p}= (PQ)cos \theta \!$$ $$ y-y_{p}= (PQ)sin\theta \!$$

$$ y-y_{p}= tan(\theta + \frac{\pi }{2})(x-x_{p})$$

with $$ x_p \!$$ and $$ y_p \! $$ representing points B and C on the displacement diagram. However, we can simply use point A as our reference point since it is already on the point of origin on the x-y axis.

$$AD = (x_{d}-x_{a})\bar{i} + (y_{d}-y_{a})\hat{j} \!$$ $$AD = (x_{d} -0)\bar{i} + (y_{d} -0)\hat{j} \! $$

With the resulting definition becoming:

$$AD = d_{3}*\bar{i} + d_{4}*\hat{j} \! $$

This ultimately closes the loop for the two bar truss system. The same procedure can be used to find the displacement for a three bar truss system



The bars are broken up into their respected elements:



The overall momentum at point A is equal to zero and therefore, point A is in equilibrium, with the summation of forces of x and y equal to zero. Momentum along any other point near point A will also equal to zero. The sample approach to find the stiffness matrix for a two body system is also applied to a three body system.

Derivation of the Node 2 Axial to Global Displacement Matrix
For Node 1, it was found that

$$q^{(e)}_1 = (d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j}) \cdot \vec{\tilde{i}}$$

For Node 2

$$q^{(e)}_2 = (d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j}) \cdot \vec{\tilde{i}}$$

Given the definitions of

$$ l^{(e)} $$ and  $$ m^{(e)} $$

from the notes above, it follows that

$$q_2^{(e)}=l^{(e)}d_3^{(e)}+m^{(e)}d_4^{(e)}$$

and hence

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

Using the Global FD Relation
Using the global FD relation, where displacements $$\displaystyle d_1, d_2, d_5, d_6 $$ all equal zero, the first two columns and the last two columns for the $$\displaystyle K $$ matrix can be ignored, yielding the following relation:


 * $$ \begin{bmatrix} K_{13} & K_{14}

\\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix} \begin{Bmatrix} d_3 \\d_4 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6  \end{Bmatrix}$$

which can also be expressed in terms of the local $$\displaystyle k^{(e)} $$ as shown


 * $$ \begin{bmatrix} k_{13}^{(1)} & k_{14}^{(1)}

\\ k_{23}^{(1)} & k_{24}^{(1)} \\ k_{33}^{(1)} + k_{11}^{(2)} & k_{34}^{(1)} + k_{12}^{(2)} \\ k_{43}^{(1)} + k_{21}^{(2)} & k_{44}^{(1)} + k_{22}^{(2)} \\ k_{31}^{(2)} & k_{32}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)} \end{bmatrix} \begin{Bmatrix} d_3 \\d_4 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} $$

Inputting the values for each $$\displaystyle K $$ and $$\displaystyle d $$ yields


 * $$ \begin{bmatrix} -{9 \over 16} & -{3 \sqrt 3 \over 16}

\\ -{3 \sqrt 3 \over 16} & -{3 \over 16} \\ {9 \over 16} + {5 \over 2} & {3 \sqrt 3 \over 16} + -{5 \over 2} \\ {3 \sqrt 3 \over 16} + -{5 \over 2} & {3 \sqrt 3 \over 16} + {5 \over 2} \\ -{5 \over 2} & {5 \over 2} \\ {5 \over 2} & -{5 \over 2} \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ 0 \\ 7 \\ F_5 \\ F_6 \end{Bmatrix} $$

Note that since $$\displaystyle F_3 $$ and $$\displaystyle F_4 $$ are known (because they are the applied loads, 0 and 7 respectively) rows three and four do not need to be computed.

Thus, the computed solution is


 * $$ \textbf{F} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 0 \\ 7 \\ 4.4378 \\ -4.4378 \end{Bmatrix} $$

Notice that these results are the same as those determined from Method 1.

Multiplication with the Transpose
$$\ \mathbf{k}^{(e)} = \mathbf{T}^{(e)T} \mathbf{\hat{k}}^{(e)} \mathbf{T}^{(e)}  $$

Prove that the above is true by multiplying right hand side of the equation.

$$\ \mathbf{T}^{(e)} $$ is defined in class as $$\ \begin{bmatrix} l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)} \end{bmatrix}_{2x4} $$

$$\ \mathbf{\hat{k}}^{(e)}$$ is defined in class as $$\ \begin{Bmatrix} k&-k\\-k&k\end{Bmatrix}_{2x2}$$

$$\ \mathbf{T}^{(e)T} $$ is the transpose of $$\ \mathbf{T}^{(e)}_, $$ where $$\ \mathbf{T}^{(e)T}_{ij} = \mathbf{T}^{(e)}_{ji} $$ from $$\ 1 \le i \le n_, \ 1 \le j \le m $$ for matrix $$\ \mathbf{T}^{(e)}$$  of dimensions $$\ n * m $$

Therefore, $$\ \mathbf{T}^{(e)T} = \begin{bmatrix} l^{(e)}&0\\m^{(e)}&0\\0&l^{(e)}\\0&m^{(e)} \end{bmatrix}_{4x2} $$

$$\ \mathbf{T}^{(e)T}*\mathbf{\hat{k}}^{(e)}= \begin{bmatrix} l^{(e)}*k^{(e)}&-l^{(e)}*k^{(e)}\\m^{(e)}*k^{(e)}&-m^{(e)}*k^{(e)}\\-l^{(e)}*k^{(e)}&l^{(e)}*k^{(e)}\\-m^{(e)}*k^{(e)}&m^{(e)}*k^{(e)} \end{bmatrix}_{4x2} $$

$$\ \mathbf{T}^{(e)T}*\mathbf{\hat{k}}^{(e)}*\mathbf{T}^{(e)}= \begin{bmatrix} l^{(e)}*l^{(e)}*k^{(e)}&l^{(e)}*m^{(e)}*k^{(e)}&-l^{(e)}*l^{(e)}*k^{(e)}&-l^{(e)}*m^{(e)}*k^{(e)}\\l^{(e)}*m^{(e)}*k^{(e)}&m^{(e)}*m^{(e)}*k^{(e)}&-l^{(e)}*m^{(e)}*k^{(e)}&-m^{(e)}*m^{(e)}*k^{(e)}\\-l^{(e)}*l^{(e)}*k^{(e)}&-l^{(e)}*m^{(e)}*k^{(e)}&l^{(e)}*l^{(e)}*k^{(e)}&l^{(e)}*m^{(e)}*k^{(e)}\\-l^{(e)}*m^{(e)}*k^{(e)}&-m^{(e)}*m^{(e)}*k^{(e)}&l^{(e)}*m^{(e)}*k^{(e)}&m^{(e)}*m^{(e)}*k^{(e)} \end{bmatrix}_{4x4} $$

Bringing out the $$\ k^{(e)} $$ and combining common terms gives $$\ \mathbf{T}^{(e)T}*\mathbf{\hat{k}}^{(e)}*\mathbf{T}^{(e)}= k^{(e)}\begin{bmatrix} (l^{(e)})^2&l^{(e)}*m^{(e)}&-(l^{(e)})^2&-l^{(e)}*m^{(e)}\\l^{(e)}*m^{(e)}&(m^{(e)})^2&-l^{(e)}*m^{(e)}&-(m^{(e)})^2\\-(l^{(e)})^2&-l^{(e)}*m^{(e)}&(l^{(e)})^2&l^{(e)}*m^{(e)}\\-l^{(e)}*m^{(e)}&-(m^{(e)})^2&l^{(e)}*m^{(e)}&(m^{(e)})^2 \end{bmatrix}_{4x4} $$

This is equivalent to the $$\ \mathbf{k}^{(e)} $$ given in class.

Therefore, $$\ \mathbf{f}^{(e)}=\mathbf{k}^{(e)}\mathbf{d}^{(e)} $$ can be written as

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

Closing the Loop and Associated Problems
From above, the following was established:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

To find the coordinates of (xb, yb) and (xc, yc) of the two body truss system example, it is necessary to compute the equations for lines AB and BC.

This is found by computing the vector PQ as

$$ \vec{PQ} = (PQ)\vec{\tilde{i}} = (PQ)(cos\theta \vec{i} + sin\theta \vec{j})$$

$$ \vec{PQ} = (x - x_{p})\vec{i} + (y - y_{p})\vec{j}$$

The corresponding expressions thus become

$$ x - x_{p} = (PQ)cos\theta$$ $$ y - y_{p} = (PQ)sin\theta$$

or through rearrangement

$$ \frac{y - y_{p}}{x - x_{p}} = tan\theta$$

The expression for a line perpendicular to the above line and passing through P can be expressed as

$$ y - y_{p} = tan(\theta + \frac{\pi}{2})(x - x_{p})$$

From here, knowing that the angle for finding the b coordinates is 30 degrees and the angle for finding the c coordinates is -135 degrees, substitution provides

$$ (x_{b}, y_{b}) = (6.8324*cos(30), 6.8324*sin(30)) = (5.917, 3.416)$$

$$ (x_{c}, y_{c}) = (1.255*cos(-135), 1.255*sin(-135)) = (-0.887, -0.887)$$

Finally, to find (xD, yD) it necessary to understand the following expression

$$ \vec{AD} = (x_{D} - x_{A})\vec{i} + (y_{D} - y_{A})\vec{j}$$

Which can reduce to

$$ \vec{AD} = (x_{D})\vec{i} + (y_{D})\vec{j}$$

based on the choice of the origin.

By definition

$$ \vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

and knowing that d3=4.352 and d4=6.1271, vector AD becomes

$$ \vec{AD} = 4.352\vec{i} + 6.1271\vec{j}$$

such that

$$ (x_{D}, y_{D}) = (4.352, 6.1271)$$

Stiffness Matrix K and its Corresponding Eigenvalues
Given that the stiffness matrix K is defined as

$$\textbf{K}= \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & K_{33}^{(1)} + K_{11}^{(2)} & K_{34}^{(1)} + K_{12}^{(2)} & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & K_{43}^{(1)} + K_{21}^{(2)} & K_{44}^{(1)} + K_{22}^{(2)} & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}\,$$

with values determined in HW 2, K then becomes

$$\textbf{K}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5 \\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5 \\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}\,$$

Solving for the eigenvalues of K using MATLAB results in

K =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

EDU>> eig(K)

ans =

-0.0001  -0.0000    0.0000    0.0000    1.4706   10.0294

Due to round-off error with MATLAB, the solution to the eigenvalues of the K matrix are 1.4706, 10.0294, and four values of zero.

Two Bar Stress MATLAB Problem
% Matlab coding for HW 3 problem % For HW3, write a matlab code to plot the initial underformed % configuration of the two-bar truss system solved in class in % dotted line, and then superpose on the same figure the deformed % configuration in solid line. clear; close; % L(1) = 4 % theta(1) = 30 % k(1) = 3/4 % AC = 6.8324 % L(2) = 2 % theta(2) = -45 % k(2) = 5 % AB = 1.255 n_node = 8; n_elem = 6; P1_1 = 5.1243; P1_2 = 6.276; k_1 = .75; k_2 = 5; AC = P1_1/k_1; AB = P1_2/k_2; position(:,1) = [0; 0; 0]; position(:,2) = [4*cos(.5236); 4*sin(.5236); 0]; position(:,3) = [4*cos(.5236)+2*cos(.7854); 4*sin(.5236)-2*sin(.7854); 0]; position(:,4) = [4*cos(.5236)-1.255*cos(.7854); 4*sin(.5236)+1.255*sin(.7854); 0]; position(:,5) = [4*cos(.5236)+6.8324*cos(.5236); 4*sin(.5236)+6.8324*sin(.5236); 0]; position(:,6) = [4*cos(.5236)+6.8324*cos(.5236)-4*cos(1.047); 4*sin(.5236)+6.8324*sin(.5236)+4*sin(1.047); 0]; position(:,7) = [4*cos(.5236)-1.255*cos(.7854)+8*cos(.7854); 4*sin(.5236)+1.255*sin(.7854)+8*cos(.7854); 0]; position(:,8) = [7.8157; 8.1264; 0];   % intersection between AC and AB                                         % determined by zooming on graph % intersection for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end node_connect(1,1) = 1; node_connect(2,1) = 2; node_connect(1,2) = 2; node_connect(2,2) = 3; node_connect(1,3) = 2; node_connect(2,3) = 4; node_connect(1,4) = 2; node_connect(2,4) = 5; node_connect(1,5) = 4; node_connect(2,5) = 7; node_connect(1,6) = 5; node_connect(2,6) = 6; node_connect(1,7) = 1; node_connect(2,7) = 8; node_connect(1,8) = 3; node_connect(2,8) = 8; for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; axis([-1 10 -1 10]) plot3(xx,yy,zz,':') hold on end for i = 7 : 8; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,'-') hold on end title('Deformed and undeformed two-bar truss system') xlabel('x') ylabel('y') zlabel('z') view([0 0 1])     % xy plane view

The image below represents the two-bar truss system plotted in Matlab. The dotted line corresponds to the two-bar truss system before deformation, and the solid line includes deformation. The arrow pointing from Global Node 2 to Global Node 6 (which is the same as Global Node 2 for the deformed case) is the deformation vector.

Individual Contributions
Eml4500.f08.bottle.vitello 15:50, 5 October 2008 (UTC)

Eml4500.f08.bottle.hipps 18:38, 5 October 2008 (UTC)

Eml4500.f08.bottle.butler 23:23, 5 October 2008 (UTC)

Eml4500.f08.bottle.barnes 21:59, 7 October 2008 (UTC)

Eml4500.f08.bottle.brockmiller 00:30, 8 October 2008 (UTC)

Eml4500.f08.bottle.ranto 03:07, 8 October 2008 (UTC)

Eml4500.f08.bottle.loschak 04:38, 8 October 2008 (UTC)