
#1
Dec512, 04:57 PM

P: 652

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
Dec512, 05:28 PM

Sci Advisor
HW Helper
P: 11,863

If the nonhomogeneity 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
Dec612, 02:19 PM

P: 652

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 



#4
Dec612, 02:28 PM

P: 652

Natural Vibration of Beam  PDEIf ##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
Dec612, 03:03 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,344

So this is the same idea as a vertical massonaspring, 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). 



#6
Dec712, 01:03 PM

P: 652




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