(read the last sentence for the actual question).

Consider an elastically suspended solid body from a massless cantilever beam


The equations of motion of such a system are

$$\begin{bmatrix}m\\ & J_{c} \end{bmatrix}\begin{bmatrix}\ddot{\delta}\\ \ddot{\theta} \end{bmatrix}+\tfrac{3EI}{\ell^{3}}\begin{bmatrix}4 & -2\ell\\ -2\ell & \tfrac{4}{3}\ell^{2} \end{bmatrix}\begin{bmatrix}\delta\\ \theta \end{bmatrix}=0 \tag{1}$$

and with some parametrizations, such as $J_c = \eta\, m \ell^2$ and $\frac{3 E I}{\ell^3} = m \Omega^2$ to have the dimensionless parameter $\eta$ to represent the inertial properties of the body, and $\Omega$ to represent some reference frequency of the system.

The above is re-arranged into

$$\underbrace{\begin{bmatrix}\ddot{\delta}\\ \ddot{\theta} \end{bmatrix}}_{\ddot{x}}+\Omega^{2}\underbrace{\begin{bmatrix}4 & -2\ell\\ -\frac{2}{\eta\ell} & \frac{4}{3\eta} \end{bmatrix}}_{A}\underbrace{\begin{bmatrix}\delta\\ \theta \end{bmatrix}}_{x}=0 \tag{2}$$

where the problem $\ddot{x} + \Omega^2 A \,{x} = 0$ has known solutions using modal analysis.

The theory goes that if $u = {\rm X} x$ where ${\rm X}$ is a matrix of eigenvectors, then $A = {\rm X}^{-1} D {\rm X}$ where $D$ is a diagonal matrix. The system $ \ddot{u} + \Omega^2 D u = 0 $ has $n$ independent solution of the form $$ u_i = U_i \cos(\omega_i t) \tag{3}$$ with $\omega_i^2 = d_i \Omega^2$ and $d_i$ the diagonal elements of $D$.

All this is fine, but when it comes to calculating the matrix ${\rm X}$ I noticed a problem with the units. This is because the solution space $x=\begin{Bmatrix}\delta \\ \theta \end{Bmatrix}$ has the first element in units of length, and the second element with no units.

Specifically, the eigenvalues are $$w=2+\frac{2}{3\eta}\left(1\pm\sqrt{\left(3\eta+1\right)^{2}-3\eta}\right) \tag{4}$$ and the eigenvectors matrix $${\rm X}=\begin{bmatrix}6\ell\eta-6\eta+2-2\sqrt{\left(3\eta+1\right)^{2}-3\eta} & 6\ell\eta-6\eta+2+2\sqrt{\left(3\eta+1\right)^{2}-3\eta}\\ \frac{6}{\ell}+6\eta-2-2\sqrt{\left(3\eta+1\right)^{2}-3\eta} & \frac{6}{\ell}+6\eta-2+2\sqrt{\left(3\eta+1\right)^{2}-3\eta} \end{bmatrix} \tag{5}$$

But the units are non-sensical since $\ell$ has length units and $\eta$ is dimensionless. Terms like $\frac{6}{\ell}+6\eta$ are clearly erroneous here. This is not a problem with calculating ${\rm X}$ as it indeed diagonalizes $A$ and provides a numeric solution.

But how can the above eigenvectors make any physical sense?

To overcome this solution the DOFs should be chosen such that they have consistent units. If the modified solution was $x^\star = \begin{Bmatrix} \delta \\ \ell \theta \end{Bmatrix}$ then the eigenvectors have consistent units and the solution makes physical sense.

My question here is the numeric solution that is yielded by the inconsistent ${\rm X}$ a valid solution, and how so. Or is there a requirement for modal analysis to construct a solution vector $x$ with consistent units only.


1 Answer 1


I get the following eigenvectors:

$$\mathbf v_1= \left[ \begin {array}{c} 1\\ {l}^{-1}-1/3\,{\frac { 1}{l\eta}}-1/3\,{\frac {\sqrt {9\,{\eta}^{2}+3\,\eta+1}}{l\eta}} \end {array} \right] ; \\ \mathbf v_2=\left[ \begin {array}{c} 1\\ {l}^{-1}-1/3\,{\frac { 1}{l\eta}}+1/3\,{\frac {\sqrt {9\,{\eta}^{2}+3\,\eta+1}}{l\eta}} \end {array} \right], $$

where the units are consistent.

Those results are from my own program that I wrote with the symbolic program MAPLE. The corresponding eigenvalues are

$$\mathbf \lambda=\left[ \begin {array}{c} 2\,{\Omega}^{2}+2/3\,{\frac {{\Omega}^{2}}{ \eta}}+2/3\,{\frac {{\Omega}^{2}\sqrt {9\,{\eta}^{2}+3\,\eta+1}}{\eta} }\\ 2\,{\Omega}^{2}+2/3\,{\frac {{\Omega}^{2}}{\eta} }-2/3\,{\frac {{\Omega}^{2}\sqrt {9\,{\eta}^{2}+3\,\eta+1}}{\eta}} \end {array} \right] $$


$$X^{-1}\,A\,X= \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{bmatrix}.$$

  • 2
    $\begingroup$ I think I understand what you did. First we have the same eigenvalues, so that is good. Next you normalized the deflection component $\delta$ part of the eigenvector into 1,. Whereas I used my own procedure which solved $(A-\lambda 1) \pmatrix{v_1 \\ v_2} = \pmatrix{0 \\ 0}$ but treating the $0$ as a variable to be solve for together with $v_2$. Here is where the problem arises. Doing math with zeros is dangerous. Thank you. $\endgroup$ Commented May 2, 2022 at 18:13
  • $\begingroup$ Did you checked this equation $X^{-1}\,A\,X= \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{bmatrix}$ with your eigen vectors? $\endgroup$
    – Eli
    Commented May 2, 2022 at 18:53
  • $\begingroup$ yes I did and it gave the correct diagonal matrix. $\endgroup$ Commented May 2, 2022 at 18:55

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