User:Eml4500.f08.Ateam.Slevinski/Homework4

HW 4 - 10/06/08


From the diagram of the two bar truss above, we can see the relationships between the global and local nodes of elements 1 and 2. From these relationships, we can derive the connectivity array. In Matlab, this is denoted as conn(e,j). In this array, the rows represent the elements (e) and the columns represent the local node numbers (j). The values inside the array are the global node numbers. Using the two bar truss diagram above, the connectivity array becomes:

conn = $$ \left[ \begin{array}{cc} {1} & {2} \\ {2}& {3} \end{array} \right]$$

For example, in this case the local node of element 1 corresponds to the global node 1. Therefore, the value of in the first row and column is 1.

Another array that we will need to derive is the location matrix master array or "LMM". This array, will represent the relationships between the local degrees of freedom (DOF's) and the global DOF's numbers. As in the connectivity array, the rows in the LMM represent the elements (e), but the columns represent the local DOF's (j). In Matlab this command is lmm(e,j).



The above diagrams show the elements of the two bar truss and their local DOF's. From these diagrams, we will populate our location matrix master array. The array becomes:

lmm = $$ \left[ \begin{array}{cccc} {1} & {2} & {3} & {4} \\ {3} & {4} & {5} & {6} \end{array} \right]$$

HW 4 - 10/13/08
When the eigenvalues of the 6x6 matrix $$ \underline{K}$$ are calculated, there are 4 zero values. Three of these values correspond to the three rigid body motions and the other corresponds to the mechanism.

Eigenvalue problem: For this problem we will use the following equation: $$ \underline{K}\underline{v}=\lambda\underline{v} $$

For this matrix, we will let {$$ \underline{u}_{1}, \underline{u}_{2}, \underline{u}_{3}, \underline{u}_{4} $$} be the pure eigenvectors corresponding to the four zero eigenvalues. The equation above now becomes:

$$ \underline{K}\underline{u}_{i}=0\cdot\underline{u}_{i}=\underline{0} $$

Here $$ \underline{K}$$ is a 6x6 matrix, $$ \underline{u}$$ is a 6x1 matrix, and i=1,...,4.

We can now compute a linear combination on the values of $$ \underline{u}_{i}, i=1,...,4$$ with a summation:

$$ \sum_{i=1}^4 \alpha_{i} \underline{u}_{i} =: \underline{W} $$.

In this summation, $$ \alpha_{i}$$ is a 1x1 matrix, $$ \underline{u}_{i}$$ is a 6x1 matrix, $$ \underline{W}$$ is a 6x1, and $$ \alpha_{i}$$ are real numbers. The $$ =$$ sign here means "equal by definition" and $$ \underline{W}$$ is a zero eigenvector corresponding to a zero eigenvalue. Our original equation for the eigenvalue problem now becomes:

$$ \underline{K}\underline{W}=\underline{K}(\sum_{i=1}^4\alpha_{i} \underline{u}_{i})=\sum_{i=1}^4 \alpha_{i}(\underline{K}\underline{u}_{i})=\underline{0}=0\cdot\underline{W} $$

Using the equation $$ \underline{K}\underline{v}=\lambda\underline{v} $$ we will now evaluate a three and four bar truss.



Given the values in the diagram, we can compute the stiffness matrix $$ \underline{K}$$.

HW: Plot the eigenvectors corresponding to the zero eigenvalues of the two-bar truss system.

For the two bar truss system, our matrix $$ \underline{K}$$ is defined as:

With this matrix, we can now plot the eigenvalues and eigenvectors using the Matlab code [V,D]=eig(K). This will display the eigenvectors (V) and eigenvalues (D):

HW2: Plot the eigenvectors corresponding to the zero eigenvalues of the three-bar truss system above. The matrix $$ \underline{K}$$ has been determined to be:

Using the same command from earlier ([V,D]=eig(A)) we can determine the eigenvalues and eigenvectors of this matrix: