Today, I've tried whether I can derive the fluttering speed if an aeroplane wing is 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$

Let $L=qca_0(\alpha+\alpha_0)$ and $M=qcea_0(\alpha+\alpha_0)$. Here, apart from $h(y,t)$ and $\alpha(y,t)$ which are functions of time $t$ and coordinate $y$, the remaining variables are constants. Let $h(y,t)=s(y)e^{pt}$ and $\alpha(y,t)=k(y)e^{pt}$.

The goal is to solve for $q$ which $\alpha \rightarrow \infty$. How can I solve for $q$ to have $p>0$?

Thanks in advance. Your help is greatly appreciated.

  • $\begingroup$ But $h,\alpha $ are coupled, by setting $h=0$ pde becomes uncoupled ode? $\endgroup$
    – Narasimham
    May 28, 2017 at 5:51
  • $\begingroup$ Oh I see, it was a deadly mistake. But if I let h be the same form of alpha, would I get any luck in solving for p? $\endgroup$ May 28, 2017 at 7:04
  • $\begingroup$ Why should $h,\alpha$ have same frequency? $\endgroup$
    – Narasimham
    May 28, 2017 at 9:38
  • $\begingroup$ I'm not sure, but I've seen many textbooks assume the same thing. But you're right, that's a legit question to think about. $\endgroup$ May 28, 2017 at 10:23
  • $\begingroup$ It may be correct because they are oscillating and twisting together on the same beam. In popular literature just I remember that a figure of eight forms through a complex indep variable $ h \, e ^{i \alpha}. $ $\endgroup$
    – Narasimham
    May 28, 2017 at 11:43


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