Example - Modal Analysis

Problem

The eigenfrequencies and eigenmodes of the following system with three degrees of freedom ( $\begin{array}{l}n=3\end{array}$ ) should be computed using Modal Analysis.

The following structural matrices are also given;

$\begin{array}{l}\mathbf{M}=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix},\quad \mathbf{K}= \begin{bmatrix} 2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 1 \end{bmatrix}\end{array}$

Solution

The system has the following characteristic equation:

$\begin{array}{l}\displaystyle \det ( \mathbf{K} - \lambda \mathbf{M} ) = - \lambda^3 + 5 \lambda^2 - 6 \lambda + 1 \stackrel{!}{=} 0\end{array}$

The characteristic polynomial is plotted in the following figure:

The roots of the characteristic polynomial are:

$\begin{array}{l}\displaystyle \lambda_1 = 0.198, \enspace \lambda_2 = 1.555, \enspace \lambda_3 = 3.247\end{array}$

which gives the following circular eigenfrequencies ( $\begin{array}{l}\omega_i = \sqrt{\lambda_i}\end{array}$ )

$\begin{array}{l}\displaystyle \omega_1 = 0.445, \enspace \omega_2 = 1.247, \enspace \omega_3 = 1.802\end{array}$

The eigenvectors are then found by solving the homogeneous system of equations for each eigenvalue. For the first eigenvalue $\begin{array}{l}\lambda_1 = 0.198\end{array}$ , the system of equations reads:

$\begin{gather} \left( \mathbf{K} - \lambda_1 \mathbf{M} \right) \pmb{\phi}_1 = \pmb{0} \\ % \left( \begin{bmatrix} 2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 1 \end{bmatrix} - 0.198 \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} \right) \pmb{\phi}_1 = \pmb{0} \end{gather}$

Thus, we have to solve the following homogeneous equation system:

$\begin{array}{l}\begin{bmatrix} 1.802 & - 1 & 0 \\ - 1 & 1.802 & - 1 \\ 0 & - 1 & 0.802 \end{bmatrix} \begin{bmatrix} \phi_{1,1} \\ \phi_{1,2} \\ \phi_{1,3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\end{array}$

Through Gauss-Jordan elimination we find the following row echelon form:

$\begin{array}{l}\begin{bmatrix} 1 & 0 & - 0.445 \\ 0 & 1 & - 0.802 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \phi_{1,1} \\ \phi_{1,2} \\ \phi_{1,3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\end{array}$

As is expected, due to its singularity, the system of equations has two non-zero rows and one zero row. Thus, $\begin{array}{l}\text{rank}(\mathbf{K} - \lambda_1 \mathbf{M}) = 2\end{array}$ . Through choosing one entry arbitrarily, we can solve for the eigenvector:

$\begin{array}{l}\displaystyle \pmb{\phi}_1 = \left[ 0.328, 0.591, 0.737 \right]^T\end{array}$

Note that the resulting eigenvector, given here, is already mass-normalized, i.e. $\begin{array}{l}\pmb{\phi}_1^T \mathbf{M} \pmb{\phi}_1 = 1\end{array}$ .

Seitenhierarchie

Example - Modal Analysis

Problem

Solution