Hi

The eigenvalues of the Stiffness matrix [K] in a discrete elastic system: [K]*{U} = {F} represent the "spring constants" of this system when transformed into the "uncoupled configuration". In other words [\Lambda]*{U*} = {F*}, which can be solved in a one to one configuration (\Lambda is a diagonal matrix consisting of the eigenvalues).

The eigenvalues have nothing to do with directions. Eigenvectors actually represent the directions applied to produce this uncoupled configuration.

Hope that helps

Hello,

thank you for your answers. I apologize I realy thought the eigenvectors.

Dear Petr,

as Hady observed, when you compute the stiffness matrix [K] (square, symmetric and positive defined), you refer to a discrete elastic system.

Eigenvectors give indication, in the vector space wherein [K] is defined, on the reference system that allows the diagonalization of [K].

It is worth observing that such a reference system does not identify a Cartesian reference system. Therefore, its axes do not identify Cartesian axes and spatial directions (in an Euclidean sense), but they identify principal directions in the vector space wherein [K] is defined.

If you would like to associate eigenvectors and eigenvalues of [K] to a physical meaning, you may refer to a dynamical interpretation.

In detail, let the damping-free dynamical discrete problem [M]{U}"+[K]{U}=0 be considered ({U}: nodal displacements' vector, ": second time derivative,

[M] discrete mass matrix), and let the dynamical harmonic solution {U}={U_o}exp(i w t) be introduced (t: time variable, i: imaginary unit; w: circular frequency)

Then, it leads to the following eigenvalues problem:

(*)                  ([K]-w^2[M]){U}={0}.

Accordingly, if one assumes that [M]^(-1) is the identity matrix, scalar solutions (w_n)^2=\lambda_n of (*) identify eigenvalues of [K], and then they

are strictly related to the natural frequencies of the discrete system, whereas the vector solutions {U}_n of (*) identify eigenvectors of [K], and then they are

strictly related to the natural vibration modes of the system.

In this light, and under previous considerations, eigenvectors of [K] (i.e., system eigenmodes) can be retained as the "directions" (better, the modes) along which

you can find the highest (lowest) modal system stiffness.

Let me know 

Regards

Giuseppe

The previous answer is the only one to have barely commented on the other key aspect of your question: namely the link to the highest/lowest structural stiffness?  The answer is that, yes, since the structural matrix is a symmetric matrix then the maximum and minimum eigenvalues are indeed the highest and lowest structural stiffness.  The eigenvectors corresponding to these eigenvalues are the directions in which these extreme stiffnesses are observed.

As Giuseppe also touched upon, the easiest way to see that these eigenvalues give the extremes is to change to a coordinate system with coordinate axes aligned along the eigenvectors.  Then the stiffness matrix becomes diagonal with eigenvalues down the diagonal, and hence the extremes must be among the eigenvalues.

Thank you, that is exactly what i wanted to know!

Thank you very much.