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? ($L = L(y')$)
  • $\begingroup$ This question is probably ok according to our homework policy. Generally we ask that you show your work (you did), and ask a conceptual question. You first question, "First, how do I actually solve the problem" might be considered foul of the homework policy, but if you include a little more of your thinking involving your second question "Can I use variational calculus..." you might be ok. $\endgroup$
    – Sean
    Mar 16, 2015 at 11:50
  • $\begingroup$ @Sean: Thanks for your pointer! I don't want a completed solution, but a hint (or a general method) on how to approach the problem would be great. I feel I'm attacking the problem from the wrong angles. As an aside, I'm self-studying, I want an answer because I'm interested, not because I have to present it in class tomorrow. $\endgroup$ Mar 16, 2015 at 12:17
  • $\begingroup$ @Christian: Hi I'm author of that particular problem in Gioldbart/Stone , and I agree that the analysis is a bit ugly - but it is correct as you can see by working out what quantities you need to keep to the required order. This is also how Euler did it in his original analysis. $\endgroup$
    – mike stone
    Nov 6, 2017 at 13:41

1 Answer 1


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!


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