# 30.4: Enforcing Boundary Conditions

By enforcing boundary conditions, such as those depicted in the system below, [K] becomes invertible (non-singular) and we can solve for the reaction force F1 and the unknown displacements u2 and u3, for known (applied) F2 and F3.

$[K] = \begin{bmatrix} k^1 + k^2 & -k^2\\ -k^2 & k^2 \end{bmatrix} = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$

$det[K] = ad - cb$

$det[K] = (k^1 + k^2)k^2 - {k^2}^2 = k^1k^2 \neq 0$

Unique solutions for $$F_1$$, $$\begin{Bmatrix}u_2\end{Bmatrix}$$ and $$\begin{Bmatrix}u_3\end{Bmatrix}$$ can now be found

$k^1u_2 = F_1$

$(k^1+k^2)u_2 - k^2u_3 = F_2 = k^1u_2 + k^2u_2 - k^2u_3$

$-k^2u_2 + k^2u_3 = F_3$

In this instance we solved three equations for three unknowns. In problems of practical interest the order of $$[K]$$ is often very large and we can have thousands of unknowns. It then becomes impractical to solve for $$\begin{Bmatrix}u\end{Bmatrix}$$ by direct inversion of the global stiffness matrix. We can instead use Gauss elimination which is much more suitable for solving systems of linear equations with thousands of unknowns.

## Gauss Elimination

We have a system of equations

$x - 3y + z = 4$

$2x - 8y + 8z = -2$

$-6x + 3y -15z = 9$

We wish to create a matrix of the following form

$\begin{bmatrix}1 & -3 & 1 & 4 \\ 2 & -8 & 8 & -2 \\ -6 & 3 & -15 & 9 \end{bmatrix}$

$\begin{bmatrix}11 & 12 & 13 & 1 \\ 0 & 22 & 23 & 2 \\ 0 & 0 & 33 & 3 \end{bmatrix}$

Where the terms below the direct terms are zero. We need to eliminate some of the unknowns by solving the system of simultaneous equations. To eliminate x from row 2 (where R denotes the row)

-2(R1) + R2

$-2(x -3y +z) + (2x - 8y + 8z) = -10$

$-2y + 6z = -10$

so that

$\begin{bmatrix}1 & -3 & 1 & 4 \\ 0 & -2 & 6 & -10 \\ -6 & 3 & -15 & 9 \end{bmatrix}$

To eliminate x from row 3

6(R1) + R3

$6(x -3y +z) + (-6x - 3y - 15z) = 33$

$-15y - 9z = 33$

$\begin{bmatrix}1 & -3 & 1 & 4 \\ 0 & -2 & 6 & -10 \\ 0 & -15 & -9 & 33 \end{bmatrix}$

To eliminate y from row 2

R2/2

$-y + 3z = -5$

$\begin{bmatrix}1 & -3 & 1 & 4 \\ 0 & -1 & 3 & -5 \\ 0 & -15 & -9 & 33 \end{bmatrix}$

To eliminate y from row 3

R3/3

$-5y - 3z = -11$

$\begin{bmatrix}1 & -3 & 1 & 4 \\ 0 & -1 & 3 & -5 \\ 0 & -5 & -3 & 11 \end{bmatrix}$

And then

-5(R2) + R3

$-5(-y +3z) + (-5y - 3z) = 36$

$-18z = 36$

$\begin{bmatrix}1 & -3 & 1 & 4 \\ 0 & -1 & 3 & -5 \\ 0 & 0 & -18 & 36 \end{bmatrix}$

$-2 = z$

Substituting z = -2 back in to R2 gives y = -1

Substituting y = -1 and z = -2 back in to R1 gives x = 3

This process of progressively solving for the unknowns is called back substitution.