Guessing the shape, and getting the load anyway

Euler’s load is the eigenvalue of a differential equation, and for a uniform pin-ended strut the equation has a closed-form solution that everybody knows. Almost no real column is uniform, and almost no real column is pin-ended, and the moment either of those is true the differential equation stops having a solution anybody can write down. What survives is a method that needs no equation at all: guess the shape the column will buckle into, write down the energy, and divide.

Every one of those numbers is above the reference and none is below it. That is not luck and it is the first of the two properties that make the method worth a page.

The criterion, which is a statement about energy #

A column at its critical load is in equilibrium in a bent shape. Bending it stores strain energy; the load, riding down as the column shortens along its own now-curved axis, does work. At the critical load the two are equal, and above it the work exceeds the energy and the column goes.

For an assumed shape $v(x)$ the two integrals are

$$U = \tfrac{1}{2}\int_0^L EI\,(v'')^2\,dx, \qquad W = \tfrac{1}{2}P\int_0^L (v')^2\,dx$$

— the second because a curve of slope $v'$ is longer than its chord by $\int \tfrac{1}{2}(v')^2 dx$, which is exactly how far the load descends — the same geometric shortening that drives a sway frame’s second-order moments. Setting $U = W$ and dividing gives Rayleigh’s quotient:

$$P = \frac{\displaystyle\int_0^L EI\,(v'')^2\,dx}{\displaystyle\int_0^L (v')^2\,dx}$$

Nothing in that derivation asks the shape to be right. It asks only that the shape satisfy the geometric boundary conditions — that it be zero where the column is held — because a shape that is not admissible is a shape of a different structure.

Why the answer is always too large #

Assuming a shape is not an approximation in the ordinary sense. It is a constraint: it forbids the column from taking any of the other shapes available to it, and it forbids them absolutely rather than penalising them. A column that has been forbidden something can only be stiffer, and a stiffer column carries more.

So the quotient is an upper bound, always, and the bound is tight only when the assumed shape happens to be the mode. The quartic in the hero figure is the extreme case: $x^2(L-x)^2$ is zero at both ends, which is admissible, but it also has zero slope at both ends, which a pinned column does not — it is the fixed-fixed mode, offered to a pinned column. The extra restraint is worth a factor of 4.26, and the figure prints it as 42 against π².

The three admissible guesses fail in a milder way, and their failure is instructive. A parabola $x(L-x)$ has constant curvature, so it has curvature at the ends where the true mode has none. That is a stiffness error concentrated where the moment should be zero, and it is worth 21.6%. The cubic $x(L^3 - 2Lx^2 + x^3)$ — the shape a uniformly loaded beam sags into — has zero curvature at both ends, matches the mode’s shape to within a per cent everywhere, and is 0.129% high.

Virtual work asks the same integral of two diagrams, and the parallel is exact: both methods take a product of curvatures over the member and both are insensitive to error for the same reason.

The best guess available for free is the deflected shape under a transverse load, because that shape already satisfies the moment boundary conditions the true mode satisfies. It is not an accident: an elastic curve is what the column would do if bent, and buckling is the column bending under its own axial force.

Which free body produced the number #

Cut the buckled column at a station $x$ and take the piece below the cut. Two things cross it: the axial force $P$, and a bending moment. Summing moments about the cut on that free body gives $M = P\,v(x)$, which is the whole of the buckling problem — the moment is caused by the deflection, and the deflection is caused by the moment.

The energy statement is that same free body integrated. $\int M \kappa\,dx / 2$ is the strain energy, and with $M = Pv$ and $\kappa = v''$ it is $\tfrac{P}{2}\int v v''\,dx$, which integrates by parts to $-\tfrac{P}{2}\int (v')^2 dx$ for any shape that vanishes at both ends. Equating that to $\tfrac{1}{2}\int EI (v'')^2 dx$ gives the quotient again, from equilibrium on a cut rather than from a work argument.

The two derivations are not independent — they are the same statement — but having both is what makes the geometric boundary condition’s role visible. The integration by parts throws away a term $[v v']$ at the ends, and it is legitimate only because the shape is zero there.

The property that makes a guess worth making #

This is the second property and it is the important one. The quotient is stationary at the true mode: its first variation vanishes there, so an error $\varepsilon$ in the shape produces an error of order $\varepsilon^2$ in the load.

The arithmetic is short enough to do. Take $v = \sin(\pi x/L) + \varepsilon \sin(2\pi x/L)$. The numerator picks up $(\pi/L)^4 + 16\varepsilon^2(\pi/L)^4$ and the denominator $(\pi/L)^2 + 4\varepsilon^2 (\pi/L)^2$, because the cross terms integrate to zero — the modes are orthogonal, and that is where the second-order property comes from. So

$$\frac{P}{P_{cr}} = \frac{1+16\varepsilon^2}{1+4\varepsilon^2} \approx 1 + 12\varepsilon^2$$

and the measured coefficient in the figure is 11.98 at $\varepsilon = 0.02$, drifting to 8.8 by $\varepsilon = 0.3$ as the quadratic stops being the whole story. Orthogonality of the modes is the reason a wrong shape gives a nearly right load, and the same orthogonality is what makes a building’s modes analysable one at a time.

The column that has no closed form #

This is what the method exists for. A tapered column, a stepped column, a column with a change of section at a splice, a column braced at one intermediate point: none of them has an eigenvalue anybody can write out, and every one of them has a Rayleigh quotient that takes three lines.

The tapered case also shows the guesses re-ranking. On a prismatic column the half sine is exact and the parabola 21.6% high. On the tapered one the half sine is 1.86% high — it is no longer the mode — and the parabola 9.1%, so the gap between a good guess and a poor one has narrowed considerably. That is because the tapered column’s true mode is itself closer to a parabola: with the ends soft, the curvature concentrates in the middle less than it does in a uniform member.

A guess that is good for one structure is not good for another, and the only reliable instruction is the one already given — use the static deflected shape of the structure being asked about, because that shape knows about the very stiffness variation the eigenvalue is sensitive to.

Where the method sits among the others #

The effective-length table is the closed-form method’s whole output, and it is a table of four numbers. Everything outside those four cases is either a chart fitted to computed results, a finite element eigenvalue, or a Rayleigh quotient — and the third of those is the only one a person can do on paper while thinking about the structure.

That last figure is the reason the direction of the bound has to be watched. A critical load that is too high produces an amplification factor that is too small, so the upper bound is on the unsafe side of every second-order calculation it feeds. A method whose error is reliably in one direction is more useful than one whose error is unbiased, but only if the direction is remembered.

The generalisation, and what it becomes #

Allowing more than one term in the assumed shape turns the quotient into the Rayleigh–Ritz method: write $v = \sum a_n \phi_n$, form the two integrals as matrices, and minimise the quotient over the coefficients. The minimisation is a generalised eigenvalue problem $\mathbf{K}\mathbf{a} = P\,\mathbf{K}_g\mathbf{a}$, and every entry of it is still an integral of the assumed functions.

That is the reference used in every figure on this page: ten sine terms, assembled into a pencil, solved by counting negative pivots. It is the same machine every other critical load on this site is found by, and it is also, with different shape functions, the finite element method — an element’s shape functions are an assumed shape over a short length, and its geometric stiffness matrix is the denominator of this page’s quotient discretised.

So the guess did not get replaced. It got made local and automated, which is a fair description of what happened to the whole of hand structural analysis between 1950 and 1980.

What the second term of the energy decides #

Stopping at the quadratic terms is what makes the quotient an eigenvalue problem, and it is also what makes it silent about everything after the critical load. Carry the potential energy to fourth order and the sign of the next coefficient decides the entire character of the failure.

A stable-symmetric structure — a plate, a flat panel — carries more after it buckles, in the way a rippling flange does, and that reserve is the whole of a thin web’s usefulness. An unstable-symmetric one — a shell, a shallow arch — sheds load the instant it goes, and its real strength is a fraction of the critical value. Both have the same quadratic form and therefore the same quotient. The number the energy criterion produces is the same number for a forgiving structure and a treacherous one, and imperfection sensitivity is the name for how much that matters.

That kink is worth carrying away from the method rather than from the figure. A Rayleigh quotient with one assumed shape tracks one mode; when the structure’s preferred mode changes, the guess that was excellent becomes a guess about the wrong thing, and the quotient goes on returning a confident upper bound on a mode the column has stopped being interested in. It is the failure mode of every energy method: the answer degrades quietly rather than visibly, and nothing in the calculation reports that the shape has stopped being appropriate.

Where the model stops #

The quotient finds a bifurcation, not a failure. It answers the question “at what load does a second equilibrium state appear?”, and for a column with an initial bow there is no second state — the deflection grows from zero and the critical load is an asymptote rather than an event.

Everything above is linear elastic. $EI$ is constant with load, which is exactly what stops being true once part of the section has yielded, and a hot-rolled column has yielded before it is loaded. The quotient can be run with a tangent modulus in place of $E$, but then it has to be solved iteratively, because the modulus depends on the answer.

Only the geometric boundary conditions are enforced. A shape that violates the natural conditions — zero moment at a pin, zero shear at a free end — is still admissible and still gives a bound, just a poorer one. The quartic in the hero figure violates a geometric condition, which is why it gives an answer to a different problem rather than a poor answer to this one.

The load must be conservative. The work integral assumes the load keeps pointing the same way as the column bends. A follower force — one that stays aligned with the member’s own axis — does not admit an energy criterion at all, and its stability has to be settled dynamically. The classic example, a cantilever under a tangential tip load, has no static critical load and flutters instead.

And the quotient is a single number about a whole member. It says nothing about where the member is weak, which is often what a designer actually wants to know.

What the pictures cannot show #

Every shape in the hero figure is drawn scaled to the same amplitude, because a buckling mode has no amplitude — it is an eigenvector, determined only up to a multiplier, and the drawing has to choose one. A reader comparing the widths of two curves is comparing two arbitrary decisions.

The figure also cannot show what makes the parabola bad. Its error is in the second derivative at the ends, and a second derivative is not visible in a line drawn at this scale: the parabola and the sine differ by less than 3% of the peak amplitude anywhere along the column, which is a difference no eye would call a fifth of anything.

And the taper is drawn as a set of short lines beside the column standing for $EI$. A column whose flexural rigidity varies does not look like that; it looks like a column that is deeper in the middle, and drawing it that way would have made the shape of the member compete with the shape being assumed for it.

The ladder from here #

Later rungs on this anchor: the Timoshenko quotient, which uses $\int M^2/EI$ instead of $\int EI (v'')^2$ and gives a better bound from the same shape because it weights the error where the moment is. Lower bounds on the critical load, which are far harder to obtain and are the reason nobody quotes them. The Rayleigh–Ritz method set out properly, with the convergence from above as terms are added. Southwell’s plot, which extracts the critical load from a real column’s measured deflections without ever taking it there. Dunkerley’s rule, the same style of argument for several loads acting together. Energy methods for plates and shells, where the shape has two arguments and the guess is much harder to make well. And the dynamic criterion, which is where non-conservative loads have to be sent.

Rayleigh wrote the quotient down in The Theory of Sound in 1877, for frequencies, and the buckling application is a transposition somebody else made — the two problems have the same shape because both ask when a quadratic form stops being positive definite. Which is a good description of what stability is, and a much better one than any picture of a column bending.