# Natural Vibration of Beam - PDE

1. Dec 5, 2012

### bugatti79

I am just wondering the author is doing in this calculation step.

Given $\displaystyle \rho A \frac {\partial^2 w}{\partial x^2} - \rho I \frac{\partial^4 w}{\partial t^2 \partial x^2} +\frac {\partial^2 }{\partial x^2}EI \frac {\partial^2 w}{\partial x^2}=q(x,t)$

where $w(x,t)=W(x)e^{-i \omega t}$

$\omega$ is the frequency of natural transverse motion and $W(x)$ is the mode shape of the transverse motion.

He substitutes the above into the PDE to get the following

$\displaystyle \frac {d^2 }{d x^2}EI \frac {d^2 W}{d x^2} - \lambda (\rho A W -\rho I \frac {d^2 W }{d x^2} ) =0$ where $\lambda=\omega^2$

However, I calculate the second derivative $w''(x)=e^{-i\omega t} W''(x)$ and $w''(t)=- \lambda e^{-i\omega t} W(x)$

What is incorrect on my part? Ie, where did the exponentials go?

thanks

2. Dec 5, 2012

### dextercioby

If the non-homogeneity function q goes away, so that the PDE hat 0 in the RHS, you can show that the exponential gets factored in the LHS, so that the resulting ODE will no longer contain it.

3. Dec 6, 2012

### bugatti79

1) How does the non homogeneity function just go away?

2) How is it shown that the exponential is absorbed in the LHS of PDE?

Thanks

Last edited: Dec 6, 2012
4. Dec 6, 2012

### bugatti79

On second thoughts is it something along these lines...

If $q(x,t)=0$then we can write the ODE in which each term on the LHS will have a $e^{-i \omega t}$ factor. Pull this out from each of the terms and thus we get

$e^{-i \omega t}[ODE]=0$ but $e^{-i \omega t}\ne 0$ therefore

$[ODE]=0$...?

5. Dec 6, 2012

### AlephZero

If "natural transverse motion" means the same as "free vibration", then q(x,t) can't depend on t by definition, otherwise you would have forced vibration not free vibration.

So this is the same idea as a vertical mass-on-a-spring, where the spring is stretched by the weight of the mass. You can find the equilibrium position as the statics problem
$$\frac{d^2}{dz^2}EI\frac{d^2}{dz^2}w_0(x) = q(x)$$
Then you measure w(x) from the equilibrum position. That makes the right hand side = 0.