5.4.1. Equations for Eigenvalue Analysis

5.4.1.1. Undamped system

To obtain Eigenvalue we reorganize matrices from linearization as (5.1). Damping matrix is ignored in this analysis option.

(5.1)\[\mathbf{M}\delta {{\mathbf{\ddot{q}}}_{\mathbf{I}}}+\mathbf{K}\delta {{\mathbf{q}}_{\mathbf{I}}}=\mathbf{0}\]

Where,

\(\delta {{\mathbf{q}}_{\mathbf{I}}}\) and \(\delta {{\mathbf{\ddot{q}}}_{\mathbf{I}}}\) are displacement, acceleration of the independent coordinate, respectively.

\(\mathbf{M}\) and \(\mathbf{K}\) are the mass matrix, stiffness matrix, respectively.

The equation of motion can be expressed as follows:

(5.2)\[\mathbf{M\ddot{x}}+\mathbf{Kx}=\mathbf{0}\]

Where, \(\mathbf{x}\) is independent coordinates.

In order to solve the (5.2), let’s assume the solution as follows:

\(\mathbf{x}=A{{e}^{\lambda t}}\)

Therefore, if we substitute \(\mathbf{x}\) to the equation of motion,

(5.3)\[{\lambda ^2}\mathbf{Mx}+\mathbf{Kx}=\mathbf{0}\]

(5.4)\[-\mathbf{Kx}={{\mathbf{\lambda }}^{2}}\mathbf{Mx}\]

If we multiply the inverse of modified mass matrix (\({{\mathbf{{M}'}}^{-1}}\)) to (5.4), then

(5.5)\[{{\mathbf{M}'}^{-1}}\mathbf{Kx}=-{{\lambda }^{2}}\mathbf{x}={{(i\lambda )}^{2}}\mathbf{x}\]

If we define the \(\mathbf{\hat{K}}\) as follows,

(5.6)\[\mathbf{\hat{K}}\equiv {{\mathbf{M}}^{-1}}\mathbf{K}\]

Then,

(5.7)\[\mathbf{\hat{K}x}={\lambda }'\mathbf{x}\quad \because {\lambda }'\equiv {{(i\lambda )}^{2}}\]

If we consider the standard form of Eigenvalue problem as follows:

(5.8)\[\mathbf{\hat{K}u}={\lambda }'\mathbf{u}\]

(5.9)\[(\mathbf{\hat{K}}-{\lambda }'\mathbf{I})\mathbf{u}=\mathbf{0}\]

In (5.8), the standard form of Eigenvalue problem is exactly same with our equation form of (5.7). Therefore we can get the Eigenvalue (\({\lambda }'\)) and Eigenvector (\(\mathbf{x}=\mathbf{u}\)) from the Eigensolver. Here, the real Eigenvalue (\({\lambda }\)) of the undamped system can be recalculated as follows:

(5.10)\[\lambda =\sqrt{-{\lambda }'}\,\,\ \because {\lambda }'=-{{\lambda }^{2}}={{(i\lambda )}^{2}},i=\sqrt{-1}\]

Note

The number of computed eigenvalues is the same the number of system DOFs.

5.4.1.2. Damped system

To obtain Eigenvalue we reorganize matrices from linearization as (5.11).

(5.11)\[\mathbf{M}\delta {{\mathbf{\ddot{q}}}_{\mathbf{I}}}+\mathbf{C}\delta {{\mathbf{\dot{q}}}_{\mathbf{I}}}+\mathbf{K}\delta {{\mathbf{q}}_{\mathbf{I}}}=\mathbf{0}\]

Where,

\(\delta {{\mathbf{q}}_{\mathbf{I}}}\) , \(\delta {{\mathbf{\dot{q}}}_{\mathbf{I}}}\) and \(\delta {{\mathbf{\ddot{q}}}_{\mathbf{I}}}\) are displacement, velocity and acceleration of the independent coordinate, respectively.

\(\mathbf{M}\), \(\mathbf{K}\) and \(\mathbf{C}\) are the mass matrix, stiffness matrix, and damping matrix respectively.

The equation of motion can be expressed as follows:

(5.12)\[\mathbf{M\ddot{x}}+\mathbf{C\dot{x}}+\mathbf{Kx}=\mathbf{0}\]

Where, \(\mathbf{x}\) is independent coordinates.

In order to make the (5.12) as the Eigenvalue problem, let’s modify the (5.12) as follows:

(5.13)\[\begin{split}\mathbf{{M}'}\equiv \left[ \begin{matrix} \mathbf{M} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \\ \end{matrix} \right], \mathbf{{K}'}\equiv \left[ \begin{matrix} -\mathbf{C} & -\mathbf{K} \\ \mathbf{I} & \mathbf{0} \\ \end{matrix} \right] , \mathbf{{x}'} \equiv \left\{ \begin{matrix} \mathbf{\dot{x}} \\ \mathbf{x} \end{matrix} \right\}\end{split}\]

Therefore, we can express the (5.12) as follows:

(5.14)\[\mathbf{{M}'{\dot{x}}'}-\mathbf{{K}'{x}'}=\mathbf{0}\]

(5.15)\[\begin{split}\left[ \begin{matrix} \mathbf{M} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \\ \end{matrix} \right]\left\{ \begin{matrix} {\mathbf{\ddot{x}}} \\ {\mathbf{\dot{x}}} \\ \end{matrix} \right\}-\left[ \begin{matrix} -\mathbf{C} & -\mathbf{K} \\ \mathbf{I} & \mathbf{0} \\ \end{matrix} \right]\left\{ \begin{matrix} {\mathbf{\dot{x}}} \\ \mathbf{x} \\ \end{matrix} \right\}=\mathbf{0}\end{split}\]

In order to solve the (5.14), let’s assume the solution as follows:

(5.16)\[\mathbf{{x}'}=A{{e}^{\lambda t}}\]

Therefore, if we substitute \(\mathbf{x}\) to the equation of motion,

(5.17)\[\lambda \mathbf{{M}'{x}'}-\mathbf{{K}'{x}'}=\mathbf{0}\]

(5.18)\[\mathbf{{K}'{x}'}=\mathbf{\lambda {M}'{x}'}\]

If we multiply the inverse of modified mass matrix (\({{\mathbf{{M}'}}^{-1}}\)) to (5.14), then

(5.19)\[{{\mathbf{{M}'}}^{-1}}\mathbf{{K}'{x}'}=\lambda \mathbf{{x}'}\]

If we define the \(\mathbf{\hat{K}}\) as follows,

(5.20)\[\mathbf{\hat{K}}\equiv {{\mathbf{{M}'}}^{-1}}\mathbf{{K}'}\]

Then,

(5.21)\[\mathbf{\hat{K}{x}'}=\lambda \mathbf{{x}'}\]

If we consider the standard form of Eigenvalue problem as follows:

(5.22)\[\mathbf{\hat{K}u}=\lambda \mathbf{u}\]

(5.23)\[(\mathbf{\hat{K}}-\lambda \mathbf{I})\mathbf{u}=\mathbf{0}\]

In (5.22), the standard form of Eigenvalue problem is exactly same with our equation form of (5.21). Therefore, we can get the Eigenvalue (\(\lambda\)) and Eigenvector (\(\mathbf{{x}'}=\mathbf{u}\)) from the Eigensolver. Here, the Eigienvalue (\(\lambda\)) can be defined as follows:

(5.24)\[\lambda ={{\lambda }_{r}}\pm i{{\lambda }_{i}}\,\,\,,\,i=\sqrt{-1}\]