From here, I read that if you have a straight beam, bend it with force at one end, and then create a beam with the resulting curvature, the equal and opposite force won't make a straight beam.
- What is the shape of a curved beam that will become straight given a point force on the end?
Assume the cross section is rectangular.
Either a closed solution or a way to approximate the shape will do.
Edit:
I'm trying to design a spring for a 3d printer. For technical reasons, printing a 3d helix is difficult. A design that has a constant cross section in one axis is best.
One design that I've seen is an zigzag-shaped spring. It seemed to be, however, that the stress isn't evenly distributed, with more strain on the corners and less on the straight parts.
My thought was to make an eye-shape with pointed corners that compresses to two flat lines. My intuition is that, if the shape compresses to a flat line then it's spreading the tensile forces evenly and won't break at a weak point.
From the link above, I read that I can't assume a constant $I$ in the Euler-Bernouli beam equation:
$${\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left(EI{\frac {\mathrm {d} ^{2}w}{\mathrm {d} x^{2}}}\right)=q $$
This PDF claims that the neutral axis of a curved beam isn't the same as in a straight beam (where it is just the centeroid). There is a formula for distance of the neutral axis from the inside of the curve:
$$ \text{Height of Neutral Axis} = {\frac h {\ln(r_o / r_i)}} - r_i$$
And the radius of a function $y$ at point $x$ is:
$$R =\left| \frac { \left(1 + y'^{\,2}\right)^{3/2}}{y''}\right|, \qquad\mbox{where}\quad y' = \frac{dy}{dx},\quad y'' = \frac{d^2y}{dx^2}$$
With those, I hoped to find the shape that, at rest, looks like an eye and when squeezed becomes two flat lines.
