User:Eml4500.f08.lulz.strack/hw3

Eigenvalues of the Global Stiffness Matrix
From Homework 2, the MATLAB code for the two bar truss system produced the global stiffness matrix (K) verified by manual calculations. This matrix was used to determine the eigenvalues. After running the M-file "TwoBarTrussSystem" and the command "format", the following output was produced.

>> K

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

>> eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4705   10.0295 The negative sign in front of the first two zeros implies that all four "zeros" are actually very small values. Due to the precision in MATLAB, it is assumed that the "zeros" are essentially zero.

Under this assumption, the result of four zeros as eigenvalues represents the four degrees of freedom of the two bar truss system that are fixed. Those four degrees of freedom are the x and y directional displacements at global nodes 1 and 3.

Comparison of Global F-d relationships and Elemental F-d relationships
From Homework 2, the elemental F-d relationships produced reactions at global nodes 1 and 3 as shown below.

$$ \begin{bmatrix} F_{1} \\ F_{2} \\ F_{5} \\ F_{6} \end{bmatrix}= \begin{bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ -4.4378 \end{bmatrix} $$

For the global F-d relationships method, the following calculations were performed to produce the reactions at all the nodes. Columns 1, 2, 5, and 6 were removed from the global stiffness matrix, and rows of the same rank were removed from the displacement matrix. This is because the four directional displacements of Nodes 1 and 3 are all zero because the nodes are fixed.

$$ d_1 = d_2 = d_5 = d_6 = 0 $$

$$ \left[ \begin{array}{cc} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right]\begin{bmatrix} d_{3} \\ d_{4} \end{bmatrix}= \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{bmatrix} $$

After running the M-file "TwoBarTrussSystem" and the command "format", the following output was produced. The global stiffness matrix (K) was the same one from the previous section.

>> d

d =

0        0    4.3520    6.1271         0         0

>> K34 = K(1:6,3:4)

K34 =

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

>> d34 = d(3:4)

d34 =

4.3520   6.1271

>> F = K34*d34

F =

-4.4378  -2.5622    0.0000    7.0000    4.4378   -4.4378

The reactions produced using the global F-d relationships method are identical to the results from the elemental F-d relationships method. The additional two values (rows 3 and 4) of the global method verify the load to the system.