Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

# 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. $\left[ K \right] = \left[ {\begin{array}{*{20}{c}}{{k^1} + {k^2}}&{ - {k^2}}\\{ - {k^2}}&{{k^2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right] \;\;\;\;\;(27)$

$det\left[ K \right] = ad - cb \;\;\;\;\; (28)$

$det\left[ K \right] = \;\left( {{k^1} + {k^2}} \right){k^2} - {k^2}^2 = {k^1}{k^2} \ne 0 \;\;\;\;\; (29)$

Unique solutions for $${F_1}$$, $$\left\{ {{u_2}} \right\}$$ and $$\left\{ {{u_3}} \right\}$$ can now be found

$- {k^1}{u_2} = {F_1}$

$\left( {{k^1} + {k^2}} \right){u_2} - {k^2}{u_3} = {F_2} = {k^1}{u_2} + {k^2}{u_2} - {k^2}{u_3}$

$- {k^2}{u_2} + {k^2}{u_3} = {F_3}$ In this instance we solved three equations for three unknowns. In problems of practical interest the order of $$\left[ K \right]$$ is often very large and we can have thousands of unknowns. It then becomes impractical to solve for $$\left\{ u \right\}$$ 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

$x - 3y + z = 4 \;\;\;\;\; \rm{(30)}$

$2x - 8y + 8z = - 2 \;\;\;\;\; \rm{(31)}$

$- 6x + 3y - 15z = 9 \;\;\;\;\; \rm{(32)}$

when expressed in augmented matrix form

$\left[ {\left. {\begin{array}{*{20}{c}}1&{ - 3}&1\\2&{ - 8}&8\\{ - 6}&3&{ - 15}\end{array}} \right|\begin{array}{*{20}{c}}4\\{ - 2}\\9\end{array}} \right] \;\;\;\;\; \rm{(33)}$

We wish to create a matrix of the following form

$\left[ {\left. {\begin{array}{*{20}{c}}{11}&{12}&{13}\\0&{22}&{23}\\0&0&{33}\end{array}} \right|\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right] \;\;\;\;\; \rm{(34)}$

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       (35)

$- 2\left( {x - 3y + z} \right) + \left( {2x - 8y + 8z} \right) = - 10 \;\;\;\;\; \rm{(36)}$

$- 2y + 6z = - 10 \;\;\;\;\; \rm{(37)}$

So that

$\left[ {\left. {\begin{array}{*{20}{c}}1&{ - 3}&1\\0&{ - 2}&6\\{ - 6}&3&{ - 15}\end{array}} \right|\begin{array}{*{20}{c}}4\\{ - 10}\\9\end{array}} \right] \;\;\;\;\; \rm{(38)}$

To eliminate x from row 3

6(R1) + R3       (39)

$6\left( {x - 3y + z} \right) + \left( { - 6x + 3y - 15z} \right) = 33 \;\;\;\;\; \rm{(40)}$

$- 15y - 9z = 33 \;\;\;\;\; \rm{(41)}$

$\left[ {\left. {\begin{array}{*{20}{c}}1&{ - 3}&1\\0&{ - 2}&6\\0&{ - 15}&{ - 9}\end{array}} \right|\begin{array}{*{20}{c}}4\\{ - 10}\\{33}\end{array}} \right] \;\;\;\;\; \rm{(42)}$

To eliminate y from row 2

R2/2

$- y + 3z = - 5 \;\;\;\;\; \rm{(43)}$

$\left[ {\left. {\begin{array}{*{20}{c}}1&{ - 3}&1\\0&{ - 1}&3\\0&{ - 15}&{ - 9}\end{array}} \right|\begin{array}{*{20}{c}}4\\{ - 5}\\{33}\end{array}} \right] \;\;\;\;\; \rm{(44)}$

To eliminate y from row 3

R3/3

$- 5y - 3z = - 11 \;\;\;\;\; \rm{(45)}$

$\left[ {\left. {\begin{array}{*{20}{c}}1&{ - 3}&1\\0&{ - 1}&3\\0&{ - 5}&{ - 3}\end{array}} \right|\begin{array}{*{20}{c}}4\\{ - 5}\\{11}\end{array}} \right] \;\;\;\;\; \rm{(46)}$

And then

-5(R2) + R3       (47)

$- 5\left( { - y + 3z} \right) + \left( { - 5y - 3z} \right) = 36 \;\;\;\;\; \rm{(48)}$

$- 18z = 36 \;\;\;\;\; \rm{(49)}$

$\left[ {\left. {\begin{array}{*{20}{c}}1&{ - 3}&1\\0&{ - 1}&3\\0&0&{ - 18}\end{array}} \right|\begin{array}{*{20}{c}}4\\{ - 5}\\{36}\end{array}} \right] \;\;\;\;\; \rm{(50)}$

$- 2 = z \;\;\;\;\; \rm{(51)}$

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.