Too tall for nothing but itself

Euler’s column has a force on top of it, constant all the way down. That single feature is what makes the governing equation $EIv'''' + Pv'' = 0$ with constant coefficients, what makes the buckled shape a sine, and what puts $\pi$ into every answer in the subject.

A mast, a chimney, a lighting column and a tree do not have that. Their load is their own weight: zero at the tip and largest at the base, varying continuously in between. The coefficient in the differential equation varies with height, the solution stops being harmonic, and the critical quantity is no longer a force at all — it is a load per unit length.

$$\frac{q_{cr}L^3}{EI} = 7.837$$

with no $\pi$ anywhere in it.

Which free body produced the number #

Cut the column at height $x$. The piece above it weighs $q(L-x)$, so that is the axial force at the cut — and it is a function of $x$ rather than a constant.

Feed that into the geometric stiffness. Each element of the column contributes a $K_g$ proportional to the axial force it carries, so the destabilising matrix is built from a varying force rather than a uniform one, and the eigenvalue problem

$$\left(K - \lambda K_g\right)\mathbf{v} = 0$$

is the same problem with a different second matrix. Solved on sixty elements it gives $\lambda L^3/EI = 7.8365$, against the value 7.837 that Greenhill obtained in 1881 by recognising the differential equation as Bessel’s of order one-third.

The same machinery, handed a constant axial force instead, reproduces $\pi^2EI/L^2$, $\pi^2EI/4L^2$ and $20.19EI/L^2$ to seven figures. That is what makes 7.8365 worth quoting rather than merely computed.

A column carries its own weight better than somebody else’s #

Put the total weight $qL$ at the tip instead and the column buckles at $\pi^2EI/4L^2$, which for the same column is $2.467\,EI/L^2$. Written the same way as the self-weight answer:

$$q_{cr}L \cdot \frac{L^2}{EI} = 7.837 \qquad\text{against}\qquad P_{cr}\frac{L^2}{EI} = 2.467$$

so the column will carry 3.18 times as much total weight if the weight is spread along it as it will if the same weight is put on top.

The reason is where the buckle is. A cantilever’s first mode has its largest curvature at the base and its largest displacement at the tip; the destabilising work a load does is its force times the shortening of the column beneath it, which is largest for a load at the top. Self-weight puts most of its force near the base, where the shortening beneath is small, so most of it is doing very little harm.

The load’s distribution matters as much as its size, which is a statement no column curve or effective length can accommodate.

The two together, and the interaction is a straight line #

Real masts have both — their own weight and something on top, an aerial, a tank, a lamp. Sweeping the reference load from all-tip to all-self-weight and reading the eigenvalue at each:

$$\frac{P}{P_{cr}} + \frac{qL}{(qL)_{cr}} = 1.00 \;\text{to}\; 1.01$$

across the whole range, with a maximum departure of 1.4%.

That is a linear interaction, and linear interactions are rare enough in stability to be worth noticing. Two loads that buckle a member in the same mode and differ only in where they are applied add their utilisations, and nothing more complicated is needed.

The practical form: a lighting column that is at 60% of its self-weight buckling load has 40% of its tip capacity left, and no correction factor is required to say so.

The height limit, and the cube root that flattens everything #

Write $q = \rho g A$ and $I = A i^2$, where $i$ is the radius of gyration. The area cancels:

$$L_{max} = \left(\frac{7.837\, E\, i^2}{\rho g}\right)^{1/3}$$

Three things are worth reading off that expression.

Strength does not appear. Not the yield stress, not the ultimate stress, not any material property except the modulus and the density. A column that falls over under its own weight has not been overstressed anywhere; it has run out of stiffness.

It is a cube root, so nothing moves it much. Quadrupling $EI$ buys 59%. Halving the density buys 26%. A material twice as stiff and half as dense — which does not exist — buys 59% again.

And the shape matters more than the material. $i$ enters squared inside a cube root, so $L_{max} \propto i^{2/3}$: doubling the radius of gyration buys $2^{2/3} = 1.587$, exactly. That is more than any material substitution available, and it is free — a hollow tube of the same area has several times the radius of gyration of a solid bar.

Three materials, one tower #

Now the result that makes the cube root worth having. Put a radius of gyration of 80 mm into the expression for four materials:

The first three agree to within three per cent, across a factor of nineteen in modulus and a factor of seventeen in density.

They agree because the answer depends on $E/\rho$ alone — the specific stiffness — and the specific stiffness of nearly every structural material in common use is within a small factor of 25 MN·m/kg. Steel is not stiffer than wood in the sense that matters here; it is stiffer and heavier in almost exactly the same proportion.

That is not a coincidence about materials; it is a fact about atoms, and it belongs to physics rather than to this site. What belongs here is the consequence: a self-weight height limit is not a material choice. It is a shape choice, and the only material that changes it much is concrete, whose specific stiffness is half everybody else’s.

The scaling argument underneath it #

There is a shorter way to reach the same conclusion, and it explains why the cube root had to be there.

Scale a column by a factor $\alpha$ in every dimension. Its weight per unit length goes as $\alpha^2$; its bending stiffness $EI$ goes as $\alpha^4$; its length goes as $\alpha$. So the group $qL^3/EI$ goes as $\alpha^2 \cdot \alpha^3/\alpha^4 = \alpha$.

The dimensionless measure of how close a column is to falling over under its own weight grows linearly with the scale. Double every dimension of a tower and it is twice as close to its own limit; there is no arrangement of material that avoids it, because the argument used no material property at all.

That is the same square-cube reasoning that runs through everything on this site about size, and it arrives here in its purest form: no stress, no strength, no factor of safety, just an exponent.

Taper, and why every mast has one #

A real mast is not prismatic. It tapers, because the base carries the whole weight and the tip carries nothing, and a section sized for the base is wasted everywhere else.

Taper cuts both ways for stability. It removes stiffness from the top of the column, where the buckling mode has its largest displacement; but it also removes weight from the top, where the weight is doing most of the destabilising. Computed on the same column with the stiffness falling linearly to a fraction of its base value:

A column that has given up 85% of its tip stiffness has given up 21% of its buckling load. That is a very good trade, and it is why masts, chimneys and trees are all tapered — the second-moment penalty is small because the top of the column contributes little bending resistance to a mode whose curvature is at the base.

Where the model stops #

The column is elastic and perfectly straight. At the slendernesses this matters at — a 50 m steel tube of 175 mm radius of gyration has a slenderness of nearly 300 — it genuinely is elastic, and it is emphatically not straight. A real mast has an out-of-straightness of the order of $L/500$, so it never bifurcates: it bends from the first increment of self-weight, and the eigenvalue is an asymptote rather than an event.

There is no wind. Every real mast is designed for a lateral load rather than for this, and the self-weight buckling load enters the calculation as the denominator in the amplification factor instead of as a capacity in its own right.

And the material is linear. For a tall timber pole it is not, and the modulus that matters for a load sustained for decades is the long-term one, which is perhaps 60% of the short-term value — worth 15% of the height limit.

What the pictures cannot show #

The mode is drawn at an amplitude that means nothing. An eigenvector has no scale, so what the figure gives is the shape the column would take and no information whatever about how far it goes.

Nor can it show the growth. A real column approaches the critical load by bending more and more, and the relationship between load and deflection is a hyperbola that never reaches the asymptote — so a photograph of a mast just below its critical load and one at half of it look very much the same, and the difference is a factor of two in the moment at the base.

The history, and the equation that was already solved #

Greenhill gave the answer in 1881, in a paper about how tall a tree or a mast could be. He recognised that the governing equation, with an axial force linear in the coordinate, becomes Bessel’s equation of order one-third under a change of variable — and Bessel’s equation had been solved decades earlier for an entirely different reason.

The number 7.837 is $\left(2j_{-1/3,1}/3\right)^2$ where $j_{-1/3,1}$ is the first zero of a Bessel function of order minus one-third. It is not a rounded value or a fitted one; it is exact, and the seven-thousandths agreement with the finite-element answer above is the finite element being checked rather than the constant.

What is worth carrying from that is the habit rather than the result. A varying coefficient turns a harmonic equation into a special-function one, and the special functions are almost always already tabulated — so a problem that looks intractable is often a problem in a different notation.

The assumption the figure rests on #

The column is fixed at its base and free at the top, which is the mast’s condition and gives 7.837. Change it and the number changes a great deal: pinned at both ends gives 18.57, and fixed at the base with the top held laterally gives 52.5. Those are the same three-to-twenty spread that the end conditions produce for a tip-loaded column, and a self-weight problem is no less sensitive to them.

Where the limit actually gets close #

It is fair to ask whether any of this ever governs, and the answer is that it governs in exactly one family of structures: the very slender ones that carry nothing.

A building is nowhere near it. A forty-storey core has a radius of gyration of several metres, so its self-weight height limit is measured in kilometres; what stops a building getting taller is drift, and after that lifts.

A chimney is closer. A 60 m reinforced concrete stack of 3 m diameter has a radius of gyration around 1.1 m and a limit near 250 m, so it is at a quarter of it — enough that the second-order amplification of its wind moment is a real term but not enough to threaten it.

A lighting column is closest of anything built. An 12 m aluminium column of 60 mm radius of gyration has a limit around 43 m: a factor of 3.6, which after imperfections and the amplification of the wind case is not the comfortable margin it sounds.

And a scaffold standard, a temporary prop, a falsework tower leg — those are the members where a designer meets this calculation for real, and they meet it because temporary works are the one place a structure is deliberately built as slender as it can possibly be.

The ladder from here #

Later rungs on this anchor: the Bessel solution itself, and why an axial force linear in the coordinate produces an equation of order one-third. The tapered column solved as a variational problem rather than by finite elements. Trees, whose height limit is set by water transport rather than by buckling, and which sit at a comfortable factor below this one anyway. The self-weight buckling of a shell — a chimney or a silo — where the mode is a local ripple rather than a global lean. Guyed masts, where the guys are springs along the height and the eigenvalue becomes a function of their stiffness. And the same question for a hanging column, where the weight puts the member in tension and there is no eigenvalue at all.