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Can you please help with this equation,

I am still baffled by it. It is a beam with free ends supported by two rollers. The beam is under axial load (tension) and gravity is acting on it. (actually its a section of conveyor belt between two idlers)

$$EI\frac{\partial^4y}{\partial x^4}-F\frac{\partial^2y}{\partial x^2}=-q\frac{\partial^2y}{\partial t^2}--------------(1)$$

where q = total mass per unit length, F = Tension, a = roller spacing, g = acceleration due to gravity

This partial differential equation has both the beam and string equation in it so it is not simple to find similar examples to help with me with this.

Using separated variables and inserting $y(x,t)=X(x)Y(t)$ into the PDE where

$y_x=X'(x)Y(t)$ $and$ $y_t=X(x)Y'(t)$

$y_{xx}=X''(x)Y(t)$ $and$ $y_{tt}=X(x)Y''(t)$

$y_{xxx}=X'''(x)Y(t)$

$y_{xxxx}=X''''(x)Y(t)$

If I sub the above back into the PDE and separate for common terms, I get two solutions.

$EIX''''(x)-FX''(x)-γ^2X(x)=0$ --------------------------------------------------(2)

and

$qY''(t)-γ^2 Y(t)=0$ ---------------------------------------------------------------------(3)

I get the assumed general solutions for (3) to be as follows

$$Y(t)=Acosγt+Bsinγt$$

By substitution back into equation 3 and comparing like terms I was able to the value for γ i.e. $$γ^2=\frac{1}{q}$$

I have tried to solve equation 2 using the characteristic equation which looks like a quadratic but I will get two real and two imaginary roots

$$r^2_1,r^2_2=\frac{F}{2EI}±\sqrt{\frac{F^2}{4E^2I^2}+\frac{1}{qEI}}$$

I have also assumed the initial conditions $Y=0$ @ $t=0$ and $Y'=0$ @ $t=0$

My four boundary conditions are $X''(0)=0$, & $X''(a)=0$ and $X'''(0)=0$ & $X'''(a)=0$

Since it is from a fourth order equation, I will assume a general solution of four parts.

$X = Ccosh\frac{F}{2EI}-\sqrt{\frac{F^2}{4E^2I^2}+\frac{1}{qEI}}x+Dsinh\frac{F}{2EI}-\sqrt{\frac{F^2}{4E^2I^2}-\frac{1}{qEI}}x+Ecos\frac{F}{2EI}+\sqrt{\frac{F^2}{4E^2I^2}+\frac{1}{qEI}}x+Fsin\frac{F}{2EI}+\sqrt{\frac{F^2}{4E^2I^2}-\frac{1}{qEI}}x$

next step is use my boundary conditions, is the above general solution correct? If it is, it will give me the four equations for the four unknown?

Cheers,

Sergio

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  • $\begingroup$ What is the question actually??? $\endgroup$ – daw Sep 15 '14 at 12:56
  • $\begingroup$ sorry I am trying to show my working out, it is taking me time to write it on here, my question is how do I solve it? I think I am close just can't get the last bit and would like some input ( I am still not sure if my B.C's are correct for my problem either), I will post the final bit that I have tomorrow $\endgroup$ – Sergio Sep 15 '14 at 14:10
  • $\begingroup$ The boundary conditions should be $X''(0)=0$, etc. I think the next step is to find when the system of four boundary conditions (with four unknown constants in $X$) has a solution; this happens only for particular $\lambda$. It's likely to be complicated. $\endgroup$ – user147263 Sep 30 '14 at 12:59
  • $\begingroup$ Hi Care Bear, thanks I got a little excited with the boundary conditions, are you able to tell me a bit more about what you mean by 'only happens for a particular lambda? also I am not sure what the general solution is in the first place. I will update it with what I think it is, I have seen this show up in a text book I read. X=asinx + bcosx + csihx + dcoshx $\endgroup$ – Sergio Oct 6 '14 at 10:40
  • $\begingroup$ Will the solution be a numerical approximation or will I eventually get an exact solution? $\endgroup$ – Sergio Oct 6 '14 at 12:20

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