Buckling of a slender column - total energy I'm following Goldbart's Mathematics for Physics book, and I ran into a problem with exercise 1.4 (page 43). We have a formula for the energy stored in a slightly bent rod aligned on the $z$ axis:
$ U[y] = \int_0^L \frac{1}{2}YI (y'')^2 dz $
Then we apply this knowledge to the following situation:

Euler's problem: the buckling of a slender column. The rod is used as a column which supports a compressive load $Mg$ directed along the $z$ axis (which is vertical). Show that when the rod buckles slighly (i.e. deforms with both ends remaining on the z axis) the total energy , including the gravitational potential energy of the loading mass M, can be approximated by
$ U[y] = \int_0^L \frac{1}{2}YI (y'')^2 - \frac{1}{2}Mg(y')^2 dz $

I don't really know what to do, my total energy looks like:
$ U_{tot}[y] = \int_0^L \frac{1}{2}YI (y'')^2 dz + MgL $
I think that $L$ should actually vary depending on $y$, since it's the rod length that should be constant, not it's $z$ component, i.e.:
$  \int_0^L ds = \int_0^L \sqrt{1+y'^2} dz = const $
I tried including that as Lagrange multiplier in the energy term:
$ U_{tot}[y] = \int_0^L \frac{1}{2}YI (y'')^2 dz + MgL - \lambda (\int_0^L \sqrt{1+y'^2} dz - C) $
Two questions:


*

*First, which approach do I have to take to arrive at the desired equation? Variational calculus doesn't actually seem fruitful here, as I want to stay at the energy level (not drop down to differential equations)? Can I transform $MgL$ term into something that relates to the integral over $dz$? Nudges into the right direction would be welcome!

*Can I use variational calculus to find an $U$-minimizing $y$ when the integration boundaries depend on the function that's varied? ($L = L(y')$)

 A: I figured it out. It's a really horrid approximation that's only valid for small $y'$ and doesn't seem grounded in mathematical rigor. It's only reasonable because as soon as you get any appreciable buckling ($y' \neq 0$), the rod collapses anyway.
The arc length of the rod is:
$ \int_0^L ds = \int_0^L \sqrt{1+y'^2} dz \approx \int_0^L 1 + \frac{1}{2} y'^2 dz = L_v + \int_0^L \frac{1}{2} y'^2 dz $
where $L_v = L$ is the vertical length of the rod. The rod has constant arc-length ($L_{tot}$) tough, so $L_v$ has to depend on this integral:
$ L_v = L_{tot} - \int_0^L \frac{1}{2} y'^2 dz $
The potential energy of the mass $M$ at the top of the rod thus becomes:
$ U_M = MgL_v = MgL_{tot} - \int_0^L \frac{1}{2} Mg y'^2 dz $
We can ignore the constant part and plug the rest into our total energy to obtain the desired equation:
$ U_{tot} = \int_0^L \frac{1}{2} YI(y'')^2 - \frac{1}{2} Mgy'^2 dz $
I get the correct answer, but I'm curious how legitimate this approach is. It certainly doesn't seem very rigorous... Comments would be appreciated!
