User:Eml4500.f08.ateam.nobrega/Homework Report Two

A 2-Beam Truss System Solved Statically


$$\sum F_{x} = 0 = R_{1}cos(30) - R_{2}cos(-45)$$

$$\sum F_{y} = 0 = R_{1}sin(30) + R_{2}sin(-45) + P $$

$$ R_{1} = \frac{R_{2}cos(-45)}{cos(30)} = \frac{\sqrt{2}}{\sqrt{3}}R_{2} $$

$$ 0 = \frac{\sqrt{2}}{2\sqrt{3}}R_{2} - \frac{\sqrt{2}}{2}R_{2} + P $$

$$ 0 =\frac{\sqrt{2}(1 - \sqrt{3})}{2\sqrt{3}}R_{2} + P $$

$$ R_{2} = -\frac{2\sqrt{3}}{\sqrt{2}(1 - \sqrt{3})}P $$

$$ R_{1} = -\frac{2}{1 - \sqrt{3}}P $$

An Example of a Statically Indeterminate Problem
An example of a statically indeterminate problem is given below:



This system is statically indeterminate because the beam is not a two-force member, with more unknowns than equations relating to the system. The two equations available are the sum of forces in the x- and y-direction. An equation relating moments to the system is not applicable in this situation.

The Definition of a Moment
A force moment is a force applied to a system at a radial distance from the origin of that system. The SI units for a moment is the Newton-meter(N·m), the US Customary Units is the foot-pound(ft·lbf). The equation for a moment is:

$$ \overrightarrow{M_o} = \overrightarrow{F}\times\overrightarrow{r} $$

This is the definition of a moment about point O. A moment can also be referred to as the torque of a system.

Solving A Truss System Using Matrices
The Force Equation as seen in matrix form is as follows:



\begin{bmatrix} f_1\\ .\\     .\\      .\\      .\\      f_6 \end{bmatrix} =     \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} \cdot \begin{bmatrix} d_1\\ .\\     .\\      .\\      .\\      d_6 \end{bmatrix} $$

Where F is the global force column matrix, K is the global stiffness matrix, and d is the global displacement matrix.

It was pointed out that the previous example, a truss system with two elastic bars, was in fact a solvable system correcting that it was originally stated to be indeterminate. The class was then told that the system was to be solved and turned in with the next homework report. As a reminder it is important that when the system is broken up into it's individual elements to place any forces shared by two or more elements in the global force diagram on only one of the individual elements.



To begin solving the above problem one must compute the the k matrix for each element, a depiction of what is to go in to each k matrix is given below:



K^{(e)}= k^{(e)} \cdot \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} $$ Here l(e) and m(e) are director cosines for the x'-axis and the equation for k(e) is:



k^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$



The director cosines are based on the $${\hat{x}}$$ coordinate system of the individual elements where,

l(e) = î • i = cosθ(e)

m(e) = î • j = cos(π/2 - θ(e)) = sinθ(e)

Here,

î = cosθ(e) i + sin θ(e) j

î • i = (cosθ(e) i + sinθ(e) j) • i

î • i = cosθ(e) i • i + sinθ(e) j • i

î • i = cosθ(e)

Therefore, î • j also equals sinθ(e)

It is important to note that the matrix, k(1), is symmetric along the diagonal. This means that values with interchangeable row and column indices are equal; for example, k(13) is equal to k(31) and k(12) is equivalent to k(21). Thus, only three values need to be computed. The remaining coefficients have the same absolute value and only differ by a positive or negative sign. In general, the transpose of matrix k(ij)(e) is equal to the original matrix.

The yellow highlighted portion of matrix k(1) is referred to as the upper triangular portion; whereas the lower triangular portion is not highlighted. The blue highlighted portion provides the line of symmetry.

Element 1:
L(1) = 4 E(1) = 3 Ac(1) = 1 θ(1) = 30° - l(1) = cosθ(1) l(1) = cos30° = &radic; 3 /2 m(1) = sinθ(1) m(1) = sin30° = 1/2

Element 2:
L(2) = 2 E(2) = 5 Ac(2) = 2 θ(2) = -45° - l(2) = cosθ(1) l(2) = cos(-45°) = &radic; 2 /2 m(2) = sinθ(1) m(2) = sin(-45°) = &radic; 2 /2

The Stiffness Matrix:


K^{(e)}= k^{(e)} \cdot \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} $$



k^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$



k^{(1)} = \frac{(3)\cdot(1)}{4} = \frac{3}{4} $$



k^{(2)} = \frac{(5)\cdot(2)}{2} = 5 $$



k_{11}^{(1)} = (l^{(1)})^2 = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} $$



k_{12}^{(1)} = l^{(1)}m^{(1)} = (\frac{\sqrt{3}}{2})\cdot(\frac{1}{2}) = \frac{\sqrt{3}}{4} $$



k_{22}^{(1)} = (m^{(1)})^2 = (\frac{1}{2})^2 = \frac{1}{4} $$

Due to symmetry, where k12(1) = k21(1),



K^{(1)} = \frac{3}{4}\cdot \begin{bmatrix} \frac{3}{4} & \frac{\sqrt{3}}{4} & -\frac{3}{4} & -\frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} & -\frac{1}{4}\\ -\frac{3}{4} & -\frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4}\\ -\frac{\sqrt{3}}{4} & -\frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{1}{4} \end{bmatrix} =    \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix} $$

For Element 2,



k_{11}^{(2)} = (\frac{\sqrt{2}}{2})^2 = \frac{2}{4} = \frac{1}{2} $$



k_{12}^{(2)} = \frac{1}{2} $$



k_{22}^{(2)} = \frac{1}{2} $$



K^{(2)} = 5\cdot \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} =    \begin{bmatrix} \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2}\\ -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \end{bmatrix} $$

Observations:

1) The absolute values of all coefficients k(ij)(e), e=2 (I,j) =1,…,4 are the same =>  compute1 coefficient For other coefficients the corresponding sign (positive or negative).

2) (k(2))^T = k(2) . Thus k(2) is symmetric.

Elem. FD rel: [k(e)][d(e)]= [F(e)]

$$ d^{(e)} = \begin{bmatrix} d_{(1)}^{(e)}\\ d_{(2)}^{(e)}\\ d_{(3)}^{(e)}\\ d_{(4)}^{(e)}\\ \end{bmatrix}

F^{(e)} = \begin{bmatrix} F_{(1)}^{(e)}\\ F_{(2)}^{(e)}\\ F_{(3)}^{(e)}\\ F_{(4)}^{(e)}\\ \end{bmatrix} $$

The column matrices d and F are both 4x1.

The global FD relation is [K][d] = [F], where K is an nxn matrix, d is an nx1 matrix and F is a nx1. In the example provided n is equal to six. Professor Vu-Quoc also mentioned a book titled "The Dragons of Eden" by [| Carl Sagan].

Global Force Displacement relationship

Global Force Displacement Relationship
The FD Relation:

\begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66}\\ \end{bmatrix}

\begin{bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6}\\ \end{bmatrix} =

\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}

$$

In compact notation the above equation simplifies to:

$$[ K_{ij} ] \cdot { d_{j} } = { F_{i} }$$

The stiffness coefficient, K, is a 6x6 matrix; d and F are 6x1 matrices.


 * $$\sum_{j=1}^6 F_{i} = K_{ij} \cdot d_j $$


 * K = global stiffness matrix.
 * d = global displacement matrix.
 * F = global force matrix.
 * k = elemental stiffness matrix.
 * f = elemental force matrix.

To go from an elemental matrix to a global matrix the matrix has to go through an assembly process.

First identify the correspondence between elemental displacement dof's and global displacement dof's.

At the Global level:

{d1, d2, ............, d6}

At the Elemental level:

* Element 1:{d(1)1, d(1)2, d(1)3, d(1)4} * Element 2:{d(2)1, d(2)2, d(2)3, d(2)4}

Identification of the Global-Local Degrees of Freedom:



Node 2 d2 = d2(1) d3 = d3(1) = d1(2) d4 = d4(1) = d2(3)

Node 3 d5 = d3(2) d6 = d4(2)

Conceptual step of assembly: (topology of K)



The highlighted portion in the figure above represents the overlap between the two elemental matrices of K.

The Global Stiffness Matrix can be seen below,

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

Below are sample calculations when solving for the global stiffness matrix. One can see how both element stiffness matrices come together to form the global stiffness matrix,

$$ K_{11} = k_{11}^{(1)} = \frac{9}{16} = 0.5625 $$ $$ K_{22} = k_{22}^{(1)} = \frac{3\sqrt{3}}{16} = 0.3248 $$ $$ K_{33} = k_{33}^{(1)} + k_{11}^{(2)} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$ $$ K_{34} = k_{34}^{(1)} + k_{21}^{(2)} = \frac{3\sqrt{3}}{16} + -\frac{5}{2} = -2.1752 $$ $$ K_{44} = k_{44}^{(1)} + k_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$

Once all the values have been computed the following matrix will be the result of combining the element stiffness matrix 1 with the element stiffness matrix 2.

$$ \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0.00 & 0.00\\     0.3248 & 0.1875 & -0.3248 & -0.1875 & 0.00 & 0.00\\      -0.5625 & -0.3248 & 3.0625 & -2.1752 & 2.5 & -2.5\\      -0.3248 & -0.1875 & -1.9375 & 2.6875 & -2.5 & 2.5\\      0.00 & 0.00 & 2.5 & -2.5 & 2.5 & -2.5\\      0.00 & 0.00 & -2.5 & 2.5 & -2.5 & 2.5\\     \end{bmatrix} $$

Elimination of Known Degrees of Freedom (dofs)
In this step one will apply known degrees of freedom. This will allow us to reduce the global force displacement relation. For example in the problem being described it is known that global node 1 and global node 3 are both fixed. A fixed constraint is the equivalent of stating that displacements 1, 2, 5, 6 are all equal to 0. Knowing this information allows us to reduce the 6x1 global displacement column matrix to a 2x1. Which then allows one to reduce the global stiffness matrix from a 6x6 to a 6x2, as it is shown below.

First Apply Boundary Conditions:


 * Global Node 1: Fixed
 * Results in:
 * d1=0
 * d2=0


 * Global Node 3: Fixed
 * Results in:
 * d5=0
 * d6=0

By applying the fixed boundary conditions it becomes apparent that d1=d2=d5=d6=0. With elements d1=d2=d5=d6 of the global displacement matrix d =0 computation time can reduced by realizing that columns 1,2,5, and 6 can be removed from the global displacement matrix K. See the example below for further explanation.



\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}

$$

From the principle of virtual work the corresponding rows 1, 2, 5, and 6 of the global stiffness and global force matrix K and F may also be deleted. This allows the force displacement relation to reduced further. The resulting force displacement relation is shown below.

\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \\ \end{bmatrix}

\begin{bmatrix} d_{3}\\ d_{4}\\ \end{bmatrix} =

\begin{bmatrix} F_{3}\\ F_{4}\\ \end{bmatrix} $$

F3 =0 and F4=P by inspection of Global Truss System shown below.

Global displacements d₃ and d₄ can then solved for by inverting K and multiplying by F.

\begin{bmatrix} d_{3}=4.352\\ d_{4}=6.127\\ \end{bmatrix} =K^{-1}

\begin{bmatrix} F_{3}\\ F_{4}\\ \end{bmatrix} $$

Displacements d3 and d4 correspond to X and Y displacement of Node 2 in the global picture. See the following example for a review on inverting matrices.



K^{-1}=\frac{1}{\det{(K)}} \begin{bmatrix} K_{44} & K_{-34} \\ K_{-43} & K_{33} \\ \end{bmatrix} $$ With

\det{(K)}=(K_{33}\cdot K_{44})-(K_{34}\cdot K_{43}) $$

Please see the MATLAB solution at the end of this test for the full solution.

Computing the Element Forces from New Known Displacements
Since displacements of Element 1 are now known, the element forces can be determined.

Recall Element 1:

$$ k^{(1)} \cdot d^{(1)} = f^{(1)} $$

Where k and d have previously been defined:



k^{(1)} = \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix} $$



d^{(1)} =

\begin{bmatrix} 0\\    0\\     4.352\\     6.1271\\     \end{bmatrix} $$

To compute the element forces in Element 1, simply multiply the k and d matrix of Element 1. Note that the first two columns of matrix k will be multiplied to zero and can be ignored if a manual calculation is performed.



\begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}\cdot

\begin{bmatrix} 0\\    0\\     4.352\\     6.1271\\     \end{bmatrix}

= \begin{bmatrix} f_{1}^{(1)}\\ f_{2}^{(1)}\\ f_{3}^{(1)}\\ f_{4}^{(1)} \end{bmatrix}

$$

Executing matrix multiplication to solve for the element and reaction forces, F:



\begin{bmatrix} -4.4378\\ -2.5622\\ 4.4378\\ 2.5622\\ \end{bmatrix}= \begin{bmatrix} f_{1}^{(1)}\\ f_{2}^{(1)}\\ f_{3}^{(1)}\\ f_{4}^{(1)} \end{bmatrix}

$$

Observation: Note Element 1 is in equilibrium. If the sum of the forces in the x and y direction are computed, one will note that they sum to zero.


 * $$\sum f_{1}^{(1)} + f_{3}^{(1)}=0$$
 * $$\sum f_{2}^{(1)} + f_{4}^{(1)}=0$$

Verifying Equilibrium:



To prove that the magnitude and direction of the forces acting on Element 1 are equal and opposite, first compute the magnitude of the reaction. This is done by the root-sum-square method:


 * $$P_1^{(1)}=[(f_1^{(1)})^2 + (f_2^{(1)})^2]^{1/2}$$


 * $$P_1^{(1)}=[(-4.4374)^{2} + (-2.5622)^{2}]^{1/2} $$


 * $$P_1^{(1)}=5.124 $$

The magnitude of P2 is computed the same way:


 * $$P_2^{(1)}=[(f_3^{(1)})^2 + (f_4^{(1)})^2]^{1/2}$$


 * $$P_2^{(1)}=[(4.4374)^{2} + (2.5622)^{2}]^{1/2} $$


 * $$P_2^{(1)}=5.124 $$

Thus, the forces are equal in magnitude but opposite in direction. The element is in equilibrium.

Element 2 is also in equilibrium and this can be proven the same way as above.



Method 2: To solve for 2 bar truss system using statics method:



Two different cut methods can be used as demonstrated above. The red outlines the different applications of the Euler cut principle. Individual free body diagrams can be drawn for each element circled in red and the problem can be solved as statically determinate.

Verification of the equilibrium of node 2 statically:




 * $$\sum F_{x}=0=-P_{1}^{(1)}cos(30) + P_{2}^{(1)}cos(45)$$
 * $$\sum F_{y}=0=P - P_{1}^{(1)}sin(30) - P_{2}^{(1)}sin(45)$$

P, P1, and P2 are known from previous steps:


 * $$P_{applied}^{} = 7 $$
 * $$P_{1}^{(1)}=5.124$$
 * $$P_{2}^{(2)}=6.276$$

Plugging these known values into the force equations:


 * $$\sum F_{x}=-5.124 \cdot cos(30) + 6.276 \cdot cos(45)$$
 * $$\sum F_{y}=7 - 5.124 \cdot sin(30) - 6.276 \cdot sin(45)$$


 * $$\sum F_{x}=0$$
 * $$\sum F_{y}=0$$

Global node 2 is in equilibrium.

MATLAB


% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4;

nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)];

dof=2*length(nodes);

conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1);

%load vector R = zeros(dof,1); R(4) = P;

% Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ... nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results

k_local =

0.75       -0.75        -0.75         0.75

k =

0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local =

5   -5    -5     5

k =

2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K =

0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -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

R =

0    0     0     7     0     0

d =

0           0        4.352       6.1271            0            0

reactions =

-4.4378     -2.5622       4.4378      -4.4378

results =

1.7081      5.1244       8.5406       5.1244       17.081       0.6276       1.8828        3.138       1.8828        6.276

Team Members contributing to this report
Tatiana Nobrega Eml4500.f08.ateam.nobrega 14:50, 26 September 2008 (UTC) Taylor Boggs   Eml4500.f08.ateam.boggs.t 03:20, 26 September 2008 (UTC) Akash Shah     Eml4500.f08.ateam.shah 14:59, 26 September 2008 (UTC) Pedro Rivero   Eml4500.f08.ateam.rivero 21:42, 25 September 2008 (UTC) Josh McNally   Eml500.f08.ateam.mcnally 20:57, 25 September 2008 (UTC) Michael Carr   Eml4500.f08.ateam.carr 14:36, 25 September 2008 (UTC)