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