# What is the physical interpretation of the eigenvalues of the stiffness matrix in the finite element method?

Given an equation of motion of an undamped system $\mathbf{M}\mathbf{\ddot{q}} + \mathbf{K}\mathbf{q} = \mathbf{f}$, $\mathbf{M}$ indicates the mass-matrix, $\mathbf{K}$ the stiffness matrix, $\mathbf{q}$ the time-dependent displacement, and $\mathbf{f}$ the applied force. The roots, $\omega^2$, of the equation $det(\mathbf{K} - \omega^2\mathbf{M}) = 0$ indicate the vibration frequencies.

My question is: what is the physical interpretation of the eigenvalues of the stiffness matrix $\mathbf{K}$?

Assume a 1d version of the equation. Then the K matrix becomes a k, spring constant.

In 1-D The equation reduces to : $m \ddot{q} + kq = f$

The q matrix also becomes x vector in one dimension. This is the differential equation for a forced mass-spring system. (@Jmac added the 1-d equation)

Similarly, The physical meaning of the matrix eigenvalues is how stiff the system is in the corresponding eigen vector direction. And as such, this determines how much elastic position/movement the force causes in the system.

The most interesting eigenvalues/vectors of K are the zero eigenvalues. These correspond to rigid body motions of the structure. These eigenvectors can be a useful check for structures that contain moving joints, etc, where the number of rigid body motions is not expected to be 6. They can also show problems with the FE model, for example "hourglassing" modes caused by inappropriate choice of elements. (Thanks to @alephzero for this paragraph)

• Great answer. I just want to expand a bit to say the entire equation reduced to 1-D gives you $m \ddot{q} + kq = F$ which is the differential equation for a forced mass-spring system. – JMac Feb 2 '17 at 20:35
• The most interesting eigenvalues/vectors of K are the zero eigenvalues, which correspond to rigid body motions of the structure. That can be a useful check for structures that contain moving joints, etc, where the number of rigid body motions is not expected to be 6. They can also show problems with the FE model, for example "hourglassing" modes caused by inappropriate choice of elements. – alephzero Feb 2 '17 at 22:27
• Thank you for the additions, I will incorporate these into the answer for future reference. @alephzero – Gürkan Çetin Feb 3 '17 at 17:01
• @JMac thank you for the expansion. I had little time so I didn't elaborate on the 1D version. Will add to the answer. – Gürkan Çetin Feb 3 '17 at 17:02
• So that is why the first mode of a structure is often (always?) a bending mode, it is a mode that contains less elastic energy than a torsional mode. Furthermore, it is natural that a first order mode of, for example, a beam has a lower eigenvalue than a higher order bending mode as it induces less strain on the beam and thus contains less elastic energy. – H. Vabri Nov 19 '18 at 13:39

The matrix $\mathbf{K}$ simply represents the force response to a unit displacement on each of the degrees of freedom of the system.

Consider a 2D cantilever beam of length $\ell$ with two degrees of freedom. The end displacement $\delta$ and the end slope $\theta$. You can assemble a stiffness matrix of the form $\mathbf{f} = \mathbf{K} \mathbf{x}$

$$\begin{vmatrix} F \\ M \end{vmatrix} = \begin{bmatrix} \frac{48 E I}{\ell^3} & -\frac{18 E I}{\ell^2} \\ -\frac{12 E I}{\ell^2} & \frac{6 E I}{\ell} \end{bmatrix} \begin{vmatrix} \delta \\ \theta \end{vmatrix}$$

The interpretation is the force/moment needed to achieve unit deflections $[1\;0]$ or $[0\;1]$ between the two columns of the matrix.