# Structural dynamics: Why are modeshape-vectors referred to as orthogonal?

When dealing with an undamped MDOF dynamic system, one can find the natural frequency (eigenvalue) and corresponding modeshape-vector (eigenvector) for mode $$n$$ by solving the eigenvalue problem:

$$[K] \{ \phi \}_n = \omega_n^2 [M] \{ \phi \}_n$$

Furthermore, it is stated in a lot of litterature, that for any two modes, $$n$$ and $$r$$, their eigenvectors are orthogonal, because they can be shown to fulfill the following:

$$\{ \phi \}_n [M] \{ \phi \}_r = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \{ \phi \}_n [K] \{ \phi \}_r = 0$$

Two vectors are, however generally, defined to be orthogonal if: $$\;\;\; \{ \phi \}_n^T \{ \phi \}_r = 0 \;\;\;$$, and this is actually not true for the modeshape-vectors you find by solving the eigenvalue problem for an undamped dynamic system!

Just to be completely clear, i will give an example using a 2DOF system:

For such a system, the stiffness and mass matrices will be the following:

$$K = \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; M = \begin{bmatrix} m_1 & 0\\ 0 & m_2 \end{bmatrix}$$

I will just select some arbitrary values:

$$m_1 = 50 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; m_2 = 20 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; k_1 = 90 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; k_2 = 50$$

which gives:

$$K = \begin{bmatrix} 140 & -50 \\ -50 & 50 \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; M = \begin{bmatrix} 50 & 0\\ 0 & 20 \end{bmatrix}$$

By solving the eigenvalue problem using these matrices, i get the following eigenvectors:

$$\{ \phi \}_1 = \begin{Bmatrix} 0.575\\ 1.000 \end{Bmatrix} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \{ \phi \}_2 = \begin{Bmatrix} -0.695\\ 1.000 \end{Bmatrix}$$

which gives:

$$\{ \phi \}_1^T \{ \phi \}_2 = 0.6 \neq 0$$

By using the mass or stiffness matrix, sure enough we get:

$$\{ \phi \}_1 [M] \{ \phi \}_2 = 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \{ \phi \}_1 [K] \{ \phi \}_2 = 0$$

but still $$\{ \phi \}_1^T \{ \phi \}_2 \neq 0$$ so they are not orthogonal! So why do we call them orthogonal?

• They are mutualy orthogonal with repect to the inner product defined by the $M$ matrix. Commented Mar 26, 2021 at 12:37
• The eigenvalue problem is $[K] \{ \phi \}_n = \omega_n^2 [M] \{ \phi \}_n$. You forgot the ^2 in the post. Commented Mar 26, 2021 at 20:15
• JAlex. Sorry, i fixed it Commented Mar 27, 2021 at 3:02