18.7 Boundary conditions

18.7.1 Elimination method

By matrix algebra: \[\begin{align} [K{\alpha\alpha}][Q_{\alpha}] + [K{\alpha\beta}][Q_{\beta}] =& [F_{\alpha}] \\ [K{\beta\alpha}][Q_{\alpha}] + [K{\beta\beta}][Q_{\beta}] =& [F_{\beta}] \\ \end{align}\]

Simple manipulation: \[\begin{align} K_{\alpha\alpha}\cdot q_{\alpha}=&\; F_{\alpha}-K_{\alpha\beta}\cdot q_{\beta} \\ F_{\beta}=&\; K_{\beta\alpha}\cdot q_{\alpha}+K_{\beta\beta}\cdot q_{\beta} \end{align}\]

@Chandrupatla2002

@Chandrupatla2002

18.7.2 Example: elimination method with symbols

\[\begin{bmatrix} k&-k&0\cr -k&2\,k&-k\cr 0&-k&k \end{bmatrix}\cdot \begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} F_{1}\cr F_{2}\cr F_{3} \end{bmatrix}\]

Impose a dof \[q_{1}=a_{1}\]

Store the stiffness row where the load is imposed

\[K_{1}=\begin{bmatrix} k&-k&0 \end{bmatrix}\]

Store the column for modification of the force vector

\[\begin{bmatrix} -k\cr 0 \end{bmatrix}\]

Examine the reduced matrix \[K=\begin{bmatrix} 2\,k&-k\cr -k&k \end{bmatrix}\]

The matrix is invertible

Reduce the length of the force vector \[F=\begin{bmatrix} F_{2}\cr F_{3} \end{bmatrix}\]

Create a modification vector from the column of the stiffness matrix times the imposed dislacement vector

\[F_{q}=\begin{bmatrix} -a_{1}\,k\cr 0 \end{bmatrix}\]

Reduce the length of the displacement vector.

\[Q=\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}\]

Modify the force vector.

\[F=F-F_q\]

\[F=\begin{bmatrix} a_{1}\,k+F_{2}\cr F_{3} \end{bmatrix}\]

The new problem is:

\[\begin{bmatrix} 2\,k&-k\cr -k&k \end{bmatrix}\cdot \begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} a_{1}\,k+F_{2}\cr F_{3} \end{bmatrix}\]

Invert and multiply.

\[\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} \frac{1}{k}&\frac{1}{k}\cr \frac{1}{k}&\frac{2}{k} \end{bmatrix}\cdot \begin{bmatrix} a_{1}\,k+F_{2}\cr F_{3} \end{bmatrix}\]

\[\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} \frac{a_{1}\,k+F_{2}}{k}+\frac{F_{3}}{k}\cr \frac{a_{1}\,k+F_{2}}{k}+\frac{2\,F_{3}}{k} \end{bmatrix}\]

Reassemble the displacement vector

\[Q=\begin{bmatrix} a_{1}\cr \frac{a_{1}\,k+F_{2}}{k}+\frac{F_{3}}{k}\cr \frac{a_{1}\,k+F_{2}}{k}+\frac{2\,F_{3}}{k} \end{bmatrix}\]

\[R_{1}=\begin{bmatrix} k&-k&0 \end{bmatrix}\cdot \begin{bmatrix} a_{1}\cr q_{2}\cr q_{3} \end{bmatrix}\]

\[R_{1}=a_{1}\,k-k\,\left(\frac{a_{1}\,k+F_{2}}{k}+\frac{F_{3}}{k}\right)\]

18.7.3 Example: elimination method with no applied force

\[\begin{bmatrix} 1&-1&0\cr -1&2&-1\cr 0&-1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 0\cr 0\cr 0 \end{bmatrix}\]

Impose a dof

\[q_{1}=1\]

Store the stiffness row where the load is imposed

\[K_{1}=\begin{bmatrix} 1&-1&0 \end{bmatrix}\]

Store the column for modification of the force vector

\[\begin{bmatrix} -1\cr 0 \end{bmatrix}\]

Examine the reduced matrix

\[K=\begin{bmatrix} 2&-1\cr -1&1 \end{bmatrix}\]

The matrix is invertible

Reduce the length of the force vector

\[F=\begin{bmatrix} 0\cr 0 \end{bmatrix}\]

Create a modification vector from the column of the stiffness matrix times the imposed dislacement vector

\[F_{q}=\begin{bmatrix} -1\cr 0 \end{bmatrix}\]

Reduce the length of the displacement vector.

\[Q=\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}\]

Modify the force vector.

\[F=\begin{bmatrix} 1\cr 0 \end{bmatrix}\]

The new problem is:

\[\begin{bmatrix} 2&-1\cr -1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 1\cr 0 \end{bmatrix}\]

Invert and multiply.

\[\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 1&1\cr 1&2 \end{bmatrix}\cdot \begin{bmatrix} 1\cr 0 \end{bmatrix}\]

\[\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 1\cr 1 \end{bmatrix}\]

Reassemble the displacement vector

\[Q=\begin{bmatrix} 1\cr 1\cr 1 \end{bmatrix}\]

\[R_{1}=\begin{bmatrix} 1&-1&0 \end{bmatrix}\cdot \begin{bmatrix} 1\cr q_{2}\cr q_{3} \end{bmatrix}\]

\[R_{1}=0\]

18.7.4 Example: elimination method with applied force

\[\begin{bmatrix} 1&-1&0\cr -1&2&-1\cr 0&-1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 0\cr 0\cr 1 \end{bmatrix}\]

Impose a dof \[q_{1}=1\]

Store the stiffness row where the load is imposed \[K_{1}=\begin{bmatrix} 1&-1&0 \end{bmatrix}\]

Store the column for modification of the force vector \[\begin{bmatrix} -1\cr 0 \end{bmatrix}\]

Examine the reduced matrix \[K=\begin{bmatrix} 2&-1\cr -1&1 \end{bmatrix}\]

The matrix is invertible

Reduce the length of the force vector \[F=\begin{bmatrix} 0\cr 1 \end{bmatrix}\]

Create a modification vector from the column of the stiffness matrix times the imposed dislacement vector \[F_{q}=\begin{bmatrix} -1\cr 0 \end{bmatrix}\]

Reduce the length of the displacement vector. \[Q=\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}\]

Modify the force vector. \[F=\begin{bmatrix} 1\cr 1 \end{bmatrix}\]

The new problem is: \[\begin{bmatrix} 2&-1\cr -1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 1\cr 1 \end{bmatrix}\]

Invert and multiply. \[\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 1&1\cr 1&2 \end{bmatrix}\cdot \begin{bmatrix} 1\cr 1 \end{bmatrix}\]

\[\begin{bmatrix} q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 2\cr 3 \end{bmatrix}\]

Reassemble the displacement vector \[Q=\begin{bmatrix} 1\cr 2\cr 3 \end{bmatrix}\]

\[R_{1}=\begin{bmatrix} 1&-1&0 \end{bmatrix}\cdot \begin{bmatrix} 1\cr q_{2}\cr q_{3} \end{bmatrix}\]

\[R_{1}=-1\]

18.7.5 Example: elimination method via partitioning with applied force

\[\begin{bmatrix} 1&-1&0\cr -1&2&-1\cr 0&-1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 0\cr 0\cr 1 \end{bmatrix}\]

Impose a dof

\[q_{1}=1\]

Restructure the matrix by reordering equations. \[K_{aa}=\begin{bmatrix} 2&-1\cr -1&1 \end{bmatrix}\]

\[K_{ba}=\begin{bmatrix} -1&0 \end{bmatrix}\]

\[K_{ab}=\begin{bmatrix} -1\cr 0 \end{bmatrix}\]

\[K_{bb}=\begin{bmatrix} 1 \end{bmatrix}\]

\[F_{a}=\begin{bmatrix} 0\cr 1 \end{bmatrix}\]

Compute the displacements which are not imposed: \[Q_{a}=\left(K_{aa}\right)^{-1}\cdot \left(F_{a}-K_{ab}\cdot Q_{b}\right)\]

Compute the reaction forces of the imposition: \[F_{b}=K_{ba}\cdot Q_{a}+K_{bb}\cdot Q_{b}\]

Thus: \[Q_{a}=\begin{bmatrix} 2\cr 3 \end{bmatrix}\]

\[F_{b}=-1\]

18.7.6 Example: Penalty approach

Penalty method

The matrix equations

\[\begin{bmatrix} 1&-1&0\cr -1&2&-1\cr 0&-1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 0\cr 0\cr 1 \end{bmatrix}\]

Gather all the imposed DOF into a vector \[A=\begin{bmatrix} 1\cr 0\cr 0 \end{bmatrix}\]

Although it may seem like the load is being artificially constrained by the non-imposed displacements, this isn’t the case because the no change in force is being applied and the stiffness matrix is not penalized for these DOF.

Modify the force vector

\[F_{a}=100\,A\]

\[F_{a}=\begin{bmatrix} 100\cr 0\cr 0 \end{bmatrix}\]

\[F=F+F_{a}\]

\[F=\begin{bmatrix} 100\cr 0\cr 1 \end{bmatrix}\]

Penalize the stiffness matrix

\[{K_{1}}_{1}={K_{1}}_{1}+C\]

\[{K_{1}}_{1}={K_{1}}_{1}+100\]

\[{K_{1}}_{1}=101\]

\[K=\begin{bmatrix} 101&-1&0\cr -1&2&-1\cr 0&-1&1 \end{bmatrix}\]

\[\begin{bmatrix} 101&-1&0\cr -1&2&-1\cr 0&-1&1 \end{bmatrix}\cdot \begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} 100\cr 0\cr 1 \end{bmatrix}\]

\[\begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}= \frac{1}{100} \begin{bmatrix} 1 & 1 & 1 \cr 1 & 101 & 101 \cr 1 & 101 & 201 \end{bmatrix}\cdot \begin{bmatrix} 100\cr 0\cr 1 \end{bmatrix}\]

\[\begin{bmatrix} q_{1}\cr q_{2}\cr q_{3} \end{bmatrix}=\begin{bmatrix} \frac{101}{100}\cr \frac{201}{100}\cr \frac{301}{100} \end{bmatrix}\]

Examine the displacement vector: \[Q=\begin{bmatrix} 1.01\cr 2.01\cr 3.01 \end{bmatrix}\]

\[R=-100\,\left(Q-A\right)\]

\[R=\begin{bmatrix} -1\cr -201\cr -301 \end{bmatrix}\]

18.7.7 Problems with the penalty approarch

@Chandrupatla2002