Paper buckling and 2nd Order NL ODE

[evan247]

Hi all,

I was looking at the buckling problem of a piece of paper with both ends clamped. When the two ends come closes they form a bulb-like shape and I was interested in deriving the shape numerically by solving NL ODE, which comes from energy methods (neglecting gravity).

The ODE I got is sth. like:

$$\frac{d^2\theta(s)}{ds^2}=cos(\theta(s))$$

where $$\theta(s)$$ is the angle the paper is inclined at, with respect to arc length s, and L is the total length of the paper.

Boundary conditions $$\theta(0)=\theta(L)=\pi/2, \theta'(0)=\theta'(L)$$.

It's a bit puzzling to me as there are 4 boundary conditions and the solution to the ODE doesn't seem to be able to satisfy all of them... Any thoughts? Or anything wrong with my energy approach? (Lagrangian = elastic energy - \lambda * constraint) Thanks a lot!

 

[JJacquelin]

Up to now, you have only two well defined boundary conditions :
theta(0)=pi/2
theta(L)=pi/2
The numerial values of the derivatives at s=0 and s=L are not specified. They can be equal just by symetry of the solution. But this equality iis not necessarily a well defined boundary condition.
By the way, the ODE can be analitically solved, thanks to elliptic integrals.

 

[coelho]

There is a method for solving ODEs of the form y''=f(y) http://en.wikipedia.org/wiki/Autonomous_system_(mathematics)#Special_case:_x.27.27_.3D_f.28x.29".

But in the case of your equation, the method leads indeed to elliptic integrals, if not worse things, like the inverse functions of them...

$$s=c_1\pm \int \frac{d\theta}{\sqrt{c_2+2sin(\theta)}}$$

 

[Studiot]

If your paper has deformed as much as a 'bulb shape' one or both of your 'supports' (the clamps) must have moved significantly from their original positions.
That is the supports do significant work.
You need to take this into account.

And welcome to Physics Forums.