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Natural Vibration of Beam - PDE

  1. Dec 5, 2012 #1
    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. jcsd
  3. Dec 5, 2012 #2

    dextercioby

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    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.
     
  4. Dec 6, 2012 #3

    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
  5. Dec 6, 2012 #4
    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##...?
     
  6. Dec 6, 2012 #5

    AlephZero

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    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.

    You answered your own question 2).
     
  7. Dec 7, 2012 #6
    Thanks, that makes good sense. Now I can proceed :-)
     
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