Finite element method

1. Sep 4, 2014

Chacabucogod

I was reading the finite element method in engineering by Rao and in the first example he ends up with a matrix that is singular.

The matrix is the following:

$$\begin{pmatrix} 2 &-2 & 0\\ -2 & 3&-1\\ 0&-1& 1 \end{pmatrix}$$

Which is a symmetric matrix as far as I can remember. Now you have 3 Nodes that will be the vector that will multiply this matrix the first node you already know is zero
so:

$$\begin{pmatrix} 0\\ f1\\ f2 \end{pmatrix}$$

The solution vector is

$$\begin{pmatrix} P1\\ 0\\ 1 \end{pmatrix}$$

Now he says that you can eliminate row 1 and column 1, but if you multiply the matrix by the first element of the vector only column one disappears. I tried solving this with the transpose method, but it seems there's no way to solve this matrix. Since there are only 2 unknowns can I completely forget ably the 3*3 matrix and do a 2*2 or do I have to take into account the third equation (the one that comes out of row 1)

Thank you

2. Sep 7, 2014

Stephen Tashi

You haven't stated a mathematical problem, so it isn't clear what you mean by "the solution".

I suggest you describe the problem completely or at least give a link to the passage in the book if Google books has the text.

3. Sep 7, 2014

the_wolfman

I not familiar with this particular book and I'm not sure where the book ends and where your interpretation begins.
But something is wrong.

That being said, in FE methods we often have an algebraic system of the form:

$A\vec x = \vec b$

For a simple system with 3 degrees of freedom we can write
$A=\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31}&a_{32}& a_{33} \end{pmatrix}$,

$\vec x=\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$

and
$\vec b=\begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}$

Now if we want to insert a Dirichlet boundary condition we have to modify the matrix A and the vector b, but the vector x is unchanged. (This is our unknown that we are solving for.) For instance if we want the displacement at $x_1=2$ then the matrix becomes

$A=\begin{pmatrix} 1 & 0 & 0\\ a_{21} & a_{22} & a_{23}\\ a_{31}&a_{32}& a_{33} \end{pmatrix}$

and the RHS vector becomes
$\vec b=\begin{pmatrix} 2\\ b_2\\ b_3 \end{pmatrix}$.

Note that if you multiple the first row of A with x you get the equation $x_1=2$.

Since we know $x_1$ we can use this information to simplify the equation. In this case we'd get a simplier 2x2 system with

$A=\begin{pmatrix} a_{22} & a_{23}\\ a_{32}& a_{33} \end{pmatrix}$,

$\vec x=\begin{pmatrix} x_2\\ x_3 \end{pmatrix}$

and
$\vec b=\begin{pmatrix} b_2-2a_{21}\\ b_3-2a_{31} \end{pmatrix}$.

4. Sep 8, 2014

Chacabucogod

Thank you. I guess That is what I was looking for