Beam Buckling 1

The intuitive way I understand beam buckling:

Put force on a beam.  It compresses (linearly).  It takes a certain amount of energy to compress it linearly.  At some point, it might have so much energy stored in compression that it may be able to relieve some by returning to a longer length, but bent.  At that point, the beam has buckled.

The full elastic energy in the beam would be something like

\frac{1}{2}n \sum_{i=1}^{n} (k_1 \theta_i^2 + k_2 \Delta_i^2 + 2k_3 \theta_i \Delta_i )

Theta is the normal hinge angle, while delta is the distance each microspring has been distorted by.  k1, k2, and k3 can be figured out from the elasticity as before, but this time k3 can be negative (k1 and k2 are always positive) and the axis y from which you do the integrals might change.  Before, you just choose y to minimize I, which in turn minimized k1, which in turn minimized the elastic energy.  But now y influences both k1 and k3, so the y that minimizes total energy might not even be constant (eg., as a function of angle).

Maybe an example would clarify things.  Imagine an I beam bent normally.

The beam will bend on an axis through its middle, so the left side of the beam is compressed while the right side is expanded.  This takes less energy than if the left side was undistorted but the right was expanded twice as much.

However, now imagine that the entire beam already is compressed.

Now it may want to bend along an axis that is more towards one side than the other (since the two sides are ostensibly symmetric, there is an instability regarding which side the bending axis chooses — but imagine someone on the left side of the beam leans on it).  In this case, the left side is undistorted (beyond its original compression), but the expansion of the right side serves merely to relax the compression it already has.  So this bending doesn’t cost any energy.

Also, there are Lagrange multipliers.  Example, you might have the beam flat in the middle, and half the length to which it is compressed.

Euler buckling theory does not predict the phenomenon discussed above with the I Beams.  This might lead us to guess that our theory is wrong, but in fact Euler theory does predict that beams can withstand more compression before buckling than they actually can; using Euler theory in mechanical engineering requires adding safety factors to the predictions to account for this.  Allowing two modes of distortion simultaneously would seem to reduce the critical load for buckling, and perhaps better fit with experiment.

With that said, the math for two simultaneous modes is so much more complex that we will at most leave it for another day and for now make the simplifying assumptions that either all thetas are zero (pure compression) or all deltas are zero (pure bending

If all thetas are zero, then

\Delta = (a-L)/n

The total energy in the beam is

E = \frac{1}{2}k (a-L)^2



A is the cross-sectional area of the beam.

For pure bending, we want to take the extreme of

F=\frac{1}{2}nk \sum_{i=1}^{n/2} (\theta_i^2) +\lambda_1 (\theta_L + \sum_{i=1}^{n/2}\theta_i) + \lambda_2 \frac{L}{n} ((\cos \theta_L + \sum_{i=1}^{n/2} \cos (\theta_L + \sum_{j=1}^{i}\theta_j)) - a/2)

We are only looking at half the beam.  The first Lagrange multiplier reflects that, with symmetric bending, the slope in the middle of the beam must be zero.  The second reflects that the beam must go to x=a/2 in the middle.

theta_L is included if the beam is not constrained to be flat at the ends.

The gradient is

 \frac{dF}{d \theta_t} = nk\theta_i + \lambda_1 - \lambda_2 \frac{L}{n} ((\sum_{i=t}^{n/2} \sin (\theta_L + \sum_{j=1}^{i}\theta_j)))

(I’m using d rather than the partial symbol because I think this implementation of LaTeX has a bug in partial)

This can be converted to a differential equation (we are implicitly assuming small angles here):

 k y

(the primes on the Lambdas, for now, indicates they are different by being divided by n or dx, etc.)

The solution is

y(x) = A \sin(\sqrt{\frac{\lambda_2}{k}}x) +B \cos(\sqrt{\frac{\lambda_2}{k}}x) +\frac{\lambda_2 y(a/2) -\lambda_1}{\lambda_2}

Note that this is true whether lambda_L is zero or not — that only shows up in the boundary conditions.

Let’s begin by assuming the beam is held flat at the two ends.  Also, we’ll consider x=0 to be in the middle of the beam. Then the minimum value for lambda_2 to give this solution is

y(x) = B (\cos(2\pi x /a)+1)

The length of this curve (taking the Taylor series in small B) is

L = a + \frac{B^2 \pi^2}{a}


B = \frac{\sqrt{a(L-a)}}{\pi}

The energy in the beam (found as in this post) is


E = \frac{4EI(L-a)\pi^2}{a^2}

The energy of the bent beam is linear in (L-a), while the energy of the compressed beam is quadratic in (L-a).  This means that for L-a near zero, the compressed beam will have less energy, but there will come a point at which they have equal energy, and then for larger L-a the bent beam will have less energy.  This happens when

 L - a = \frac{8IL\pi^2}{Aa^2}

We find the force by multiplying by (the compression) k, giving

F_{critical}= \frac{8EI\pi^2}{a^2}

This is strange and interesting.  I think this method seems quite different from the normal derivation of the critical load, yet the form of the solution is similar … but not quite the same.  This expression is a factor of 2 greater than the normal critical load.

There are a few possible explanations:

1. Simple error at some point.

2. Although these methods give similar results, there is a fundamental physical difference between them that results in different answers (compare in normal beam bending how conservation of energy gives a factor of 2 difference from the minimum total potential energy)

3. The small angle approximation (implicitly used when we replaced theta in terms of the second derivative when we derived the differential equation) prevented correct low-order correction to the length of the beam, which we absolutely need in this method (normal Euler beam bending treats the x-coordinate as unchanging, so the length is not correctly estimated).

It might be interesting to resolve this discrepancy, but for now let’s be satisfied that the formulas agree to an order of magnitude, and move on.

Now let’s consider if the sides are free to bend.  The shape of the curve goes to

 B(\cos (d x) -\cos (\frac{da}{2}))

The boundary condition doesn’t give us enough information to solve for d yet.

We need to keep the length equal to L




The energy is

 \frac{EI B^2 d^3(ad + \sin(ad))}{4}


Doing math on that expression might prove irritating, but we note by graphing that it seems to have a maximum at ad=pi.

Though this might not count as a strict derivation, it can be used to strictly prove that is the minimum.

Plug in ad = pi and we get

 \frac{EI (L-a) \pi^2}{a^2}

Again, we can set this equal to the compressive energy in the beam, giving

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


F_{critical}= \frac{2EI\pi^2}{a^2}

Again, a factor of 2 difference from the normally accepted value.

We can try to resolve this later.

Leave a Reply