# Curve of a rod bent by force on both sides

Suppose we have a flexible rod (i.e. it can be bent without breaking apart) and we excert a force on both sides, like this:

If the force $F$ is not exactly horizontal, the rod will be bent and form a curve (like for example the dashed line). Is there an equation that describes the curve? How would I go about to determine it?

Let's assume that we had some initial non-parallel force that caused the deflection in the beginning, but afterward the force is horizontal. Let's also neglect gravity.

What you're describing here is actually a very interesting phenomenon known as Buckling. Now Buckling does not occur on all cases, so I could see how this sentence

Let's assume that we had some initial non-parallel force that caused the deflection in the beginning, but afterward the force is horizontal.

would make sense. But trust me, reality is much worse. All you need are two coaxial forces, such as the ones your diagram and you can have the dashed curve given that specific criteria are fulfilled.

The force needed to cause buckling, according to your diagram - that is two forces on the same bar is:

$F=\frac{\pi^2\,\text{E}\,I_{min}}{l^2}$

So this is, what is known as a critical force, that would cause buckling to start. Now, a procedure of solving a simple differential equation (the elastic line eq), found in most textbooks on Mechanics of Material you can find this equation:

$y(z)=f\sin\left(\frac{\pi\,z}{l}\right)$

You might notice the f term there. That is the amplitude, or the maximum deflection that the beam will experience. However, due to the nature of the harmonic solution of the differential equation, computing values for this term is impossible since once buckling starts, it's improbable that we can pinpoint the exact family (or shape of buckling). Hence, I would suggest reading a bit further on this topic, as there's alot of material that one could learn from.

• As an approximation of constant length, the amplitude is found to be $$f^2 = \frac{4 \ell \delta}{\pi^2 }$$ where $\delta$ is the axial displacement. – John Alexiou Jun 9 '14 at 1:11
• In terms of axial force, the amplitude is $$f^2 \approx \frac{4}{\pi^2} \frac{F \ell^2}{A E}$$. – John Alexiou Jun 9 '14 at 4:05