# Beam Buckling 2

I suggested several ways of resolving the factor of 2 discrepancy in the previous post, but one thing I did not suggest in that section was what I used to start the post:  Allow for the possibility of the beam being simultaneously compressed and bent.

Let’s look at that now.  We will still have the cross product term (k_3 in the first equation) equal to zero as a simplification.  Another simplifying assumption (which perhaps can be rigorously proved, though we will not bother) is that the beam is uniformly compressed along its neutral axis.

This means we don’t have to do much new math.  We can re-use the formulas for energy in a bent and compressed beam to get

$Energy = \frac{4EI(L-\delta-a)\pi^2}{a^2}+\frac{1}{2}\frac{EA}{L}\delta^2$

Differentiate with respect to delta, set equal to zero, and solve for delta.

(I’m using r for area there)

Substitute back in…

Now set equal to the compressed-only energy and solve

That is

$L - a = \frac{4IL\pi^2}{Aa^2}$

This is exactly half  of what it was last time, so the factor of 2 has been explained.