# Design a beam that is straightened with force on one end

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.

• What is this for? A challenge?
– Wasabi
Jan 8, 2016 at 13:51
• I wouldn't be surprised if this is impossible. If you have a horizontal beam with a vertical load at the free end, it will bend such that the tangent at the free end is non-zero. An "equal and opposite" force will still be vertical and will therefore create two components: parallel and perpendicular to the free end, causing both axial and shear forces in the curved beam. Or can the second force be "equal and opposite but perpendicular to the free end"? Also, is this for small or large displacements? Are shear deformations considered?
– Wasabi
Jan 8, 2016 at 13:57
• To solve the first issue, should this be a non-linear analysis (in which case the first load will already have created axial stresses)?
– Wasabi
Jan 8, 2016 at 14:06
• The question seems pretty clear to me. The first paragraph is background, see the link. The question is 'What is the shape of a curved beam that will become straight given a point force on the end'. If you haven't seen the link this is referring to an initially curved cantilever loaded by a point load at the end. Jan 8, 2016 at 14:32
• To straighten a beam, apply two equal and opposite forces at the two ends axially. Maybe you need to include a sketch on how the force is applied. Jan 8, 2016 at 18:24

$y=\dfrac{Wx^2(3l-x)}{6EI}$ is the equation for deflection of a beam supported at one end and load at the other end (reference).

$E$ is modulus of elasticity. For steels it's 30x10^6 psi

$I$ (moment of inertia) is $\dfrac{bh^3}{12}$ for rectangular sections.

Just design the spring going up by the amount of calculated deflection.

• This is the equation for a straight beam. A curved beam will have different values, as the axial forces wedge against the supports. Jan 8, 2016 at 19:31
• Even without axial force, just perpendicular, the second moment of area isn't the same because the rod is already bent. The second moment of area is the integral of the square of the distance from the neutral axis and the neutral axis of a bent rod is not at its centeroid.
– Eyal
Jan 8, 2016 at 19:46
• ja72, the question included "or a way to approximate the shape will do." Modifying a straight beam by its deflection is a way to approximate the shape. Jan 8, 2016 at 22:00
• You should have mentioned that in the answer as a note before starting the answer Jan 10, 2016 at 5:06