User:Eml4500.f08.ateam.rivero/w3

Homework Report 3: Monday, September 29
An alternative way of solving for the reactions is by using the Global Force Displacement relation. Since one knows the values for the displacement column matrix ( d : 6x1) and the global stiffness matrix ( K : 6x6). One can solve for the force column matrix ( F : 6x1).



\begin{bmatrix} &  &   &   &   &  & \\         &   &   &   &   &  & \\         &   &   &   K   &  & \\ &  &   &   &   &  & \\         &   &   &   &   &  & \\         &   &   &   &   &  & \\     \end{bmatrix} \cdot \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} $$

Example Calculation:



\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} \cdot \begin{bmatrix} 0\\      0\\      4.352\\      6.127\\       0\\       0\\     \end{bmatrix} =      \begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{bmatrix} $$

The form above can be reduced to the form below :



\begin{bmatrix} -0.5625 & -0.3248\\    -0.3248 & -0.1875\\       3.0625& -2.1752\\     -1.9375 &  2.6875\\      -2.5   &   2.5  \\      -2.5   &   2.5  \\ \end{bmatrix} \cdot \begin{bmatrix} 4.352\\     6.127\\     \end{bmatrix} =      \begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{bmatrix} $$

How to solve in MATLAB:

>> K = [-0.5625,-0.32476;-0.32476,-0.1875;3.0625,-2.1752;-2.1752,2.6875;-2.5,2.5;2.5,-2.5]

K =

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

>> d = [4.352;6.1271]

d =

4.3520   6.1271

>> F = K*d

F =

-4.4378  -2.5622    0.0003    7.0001    4.4377   -4.4377

Note: F3 and F4 return the given input values for P. Therefore, to make computations easier there is no      need to compute rows 3, 4 since row 3 and 4 will return the values for F3 and F4.

Results Comparison:

Method 1: Using the element force displacement relationship to calculate the force reactions.
 * F 1 = -4.4378
 * F 2 = -2.5622
 * F 5 = 4.4378
 * F 6 = -4.4378

Method 2: Using the global force displacement relationship to calculate the force reactions.
 * F 1 = -4.4378
 * F 2 = -2.5622
 * F 5 = 4.4377
 * F 6 = -4.4377

Method 3: Using statics to calculate the force reactions.
 * F 1 = -5.124cos(30o)= -4.4375
 * F 2 = -5.124sin(30o)= -2.5622
 * F 5 = 6.276cos(-45o)= 4.4378
 * F 6 = 6.276sin(-45o)= -4.4378



The following image to the left is a visual representation of closing the loop between the relationship of the finite element method and statics, which will be further described below.To be able to close the loop in the diagram above we need to know how to go from the computed force reaction to an unknown displacement. By the statics method described above, we know the reactions of each element. Therefore, we also know the axial force being applied to that member. If we know both the axial load Pe and the element stiffness ke. Then we can calculate the axial displacement of the element. See calculations below for further explanation.

Recall Equations for Axial Displacement :

$$ q_{2}^{(1)} = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$

$$ q_{1}^{(2)} = \frac{|P_{2}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552 $$



The image on the left is a geometric representation of the deformed shape of a two bar truss. Length AB is the geometric representation of the axial displacement of element 2. Length AC is the geometric representation of the axial displacement of element 1. The intersection of the perpendicular projection of line segment AB and the perpendicular projection of line segment AC, would be the new location of node 2 after the initial load is applied. If you connect a line from node 1 and node 3 to the new location of node 2 this would give you what is known as the deformed shape of the truss after being loaded.