# 30.4: Enforcing Boundary Conditions

- Page ID
- 32863

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 F_{1} and the unknown displacements u_{2} and u_{3}, for known (applied) F_{2} and F_{3}.

\[ [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**.