I recently bumbped into a rather basic but interesting question on aeroelasticity. I've learned to derive the fluttering critical speed from Pines's theory but it involves some spring stiffness like for normal spring $K_h$ and torsion spring $K_t$.

However, I want to see if I can derive the same thing if the wing is now treated as a continuous beam-rod model (i.e. torsion and bending). The equations of motion are:

$ \frac{\partial^2}{\partial y^2} \left( EI \frac{\partial^2 h}{\partial y^2} \right) + m \frac{\partial^2 h}{\partial t^2} + m x_\alpha \frac{\partial^2 \alpha}{\partial t^2}+L=0$

$ -\frac{\partial}{\partial y} \left( GJ \frac{\partial \alpha}{\partial y} \right) + I_\alpha \frac{\partial^2 \alpha}{\partial t^2} + mx_\alpha \frac{\partial^2 h}{\partial t^2} - M = 0$

For simplicity, let $h(y,t)=0$ and $\alpha=s(y)e^{pt}$ so we can focus entirely on the torsion dynamic response. Let $L=qca_0(\alpha+\alpha_0)$ and $M=qcea_0(\alpha+\alpha_0)$. But now things get out of my control as I don't know how to solve for $p$ (to let its real part positive), given $s(y)$ unsolved too.

  • $\begingroup$ You can't "focus entirely on the torsion dynamic response". The essential feature of flutter is coupling between the torsion and flapping motions of the beam, and the fact that the two components of the motion can occur at the same frequency but 90 degrees out of phase with each other. $\endgroup$
    – alephzero
    May 28, 2017 at 14:59
  • $\begingroup$ You're correct, I was wrong when trying to decouple the system of equations. $\endgroup$ Jul 4, 2017 at 5:59

1 Answer 1


By writing $\alpha=s(y) e^{pt}$, you've assumed that the solution is separable, i.e. that it can be separated into a part that depends only on y and another part that depends only on t. This is pretty common for this type of equation and it should work. The typical next step here would be to separate the variables in the equation. i.e. move everything that depends on s(y) (and its derivatives) to one side, and everything that depends on $e^{pt}$ (and its derivatives) to the other side. Then, since the left depends only on y, and the right only on t, the only way this can work if is both sides are equal to the same constant. You then just need to find that constant. Typically you'll proceed by assuming something like $s(y) = A \cos(kpy)+B \sin(kpy)$, where A and B are unknown constants and k is the wavenumber. Then just plug in and solve. There will be an infinite number of solutions to the equation. Each solution is a combination of k, A, B that solve the equation (and the boundary conditions), and represents an eigenmode of vibration. Typically, you'll keep only the first few (or maybe only first one) eigenmodes, and then find the time response of the modes.

This page might help, http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf , or pick up any textbook on "vibrations of continuous systems".


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