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I have a fourth order partial differential equation of motion of a tube, with clamped boundary conditions, I don't know what would be the general solution for $W$:

$$EI \frac{d^4 w(x,t)}{dx^4} + MU^2 \frac{d^2 w(x,t)}{dx^2} + 2MU\frac{d^2 w(x,t)}{dx\,dt} +M \frac{d^2 w(x,t)}{dt^2}=0$$

I need to know the general solution (mode shape) for $w$ (displacement).

$M, E,I,U$ all are known and constant ($U$ is the velocity of a fluid inside the tube).

parham
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2 Answers2

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If you want a single Fourier mode, you're looking for a solution of the form $$W=e^{ikx+\omega t}$$ The only thing you're missing is the dispersion function $\omega(k)$. If you plug the above solution into your equation, you get the condition $$EI k^4-M (k U-i \omega )^2=0\ ,$$ which has the solution $$\omega = i \left(\pm\sqrt{\frac{EI}{M}} k^2 - U k\right)$$ The general solution is a (possibly infinite) superposition of modes which satisfy this relation. The clamped boundary conditions give you a constraint on the possible values of $k$.

yohBS
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if i use separation of variables [ w=S(x)*G(t) ] then i would have this Eq.:

A1 * (d^4 S/dx^4) + A2 * (d^2 S/dx^2) + lambda * S =0

in which A1 & A2 are known constants from E,I,M,...

in this case the term (d^2 w/dsdt) would be dropped down (eliminate)

then i have the roots of characteristic equation this way :

q1= (-A2/(2A1)) + [ sqrt(A2^2 - 4*lambda*A1) ]/(2A1)

q2= (-A2/(2A1)) - [ sqrt(A2^2 - 4*lambda*A1) ]/(2A1)

then my general solution would become:

S(x)=C1 sin [(sqrt(q1))*x] + C2 cos [(sqrt(q1))*x] + C3 sinh [(sqrt(q2))*x] + C4 cosh [(sqrt(q2))*x]

where C1,...,C4 are unknown constants to be determined by boundary conditions.

is my procedure correct ??

parham
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  • Why would you drop the cross derivative term $d^2w/dxdt$? – yohBS Oct 18 '13 at 06:17
  • ah , yes it is wrong! i can't use separation of variables for this equation, but i want a way to know the superposition equation for w(x,t) , i don't know what to do (i need for boundaries other than simply support) – parham Oct 18 '13 at 07:03
  • I don't know what you mean. Do you have the initial conditions, i.e. $W(x,t=0)?$ – yohBS Oct 18 '13 at 15:02
  • i know W(x,t=0)=W(x) because of no deflection, it's not zero because we may have an initial deflection W at any point x of the tube , i have 2 boundary conditions at 2 ends of tube (clamped) it means at x=0,L we have W(0,t)=W(L,t)=0 , dW(0,t)/dx = dw(L,t)/dx = 0 and omega is frequency because i am seeking the frequency – parham Oct 19 '13 at 07:01
  • in my problem ,for example for simply supported boundary condition , this is the general solution : w(x,t)=C exp(omegat) * sin(mpix/L) , but i'm looking for a general solution for clamped condition – parham Oct 19 '13 at 07:06
  • also i can suppose that w(x,t)=W(x)exp(omegat) , but i don't know how to deal with W(x) because we are looking for omega (unknown) and also this is a partial differential equation! please help me – parham Oct 19 '13 at 07:11