The column with no plateau

The column that yielded before it was loaded found why a hot-rolled steel column buckles below the lesser of its yield stress and Euler’s: residual stress from uneven cooling has put the flange tips past yield before any load arrives, the yielded steel stops contributing stiffness, and the elastic core that is left buckles early. It closed by noting that its whole argument assumed a plateau — a yielded fibre supplies nothing, which is true of mild steel — and named the other kind of material, whose curve bends over gradually: “the tangent modulus of a rounded stress–strain curve, and the Ramberg–Osgood parameters that make an aluminium column curve a different shape rather than a shifted one.”

This essay takes the plateau away. The materials it is about — aluminium alloys and stainless steels, and the high-strength steels whose curves are rounded too — have no stress at which anything in particular happens, and a column made of them loses its stiffness in a way a mild steel column never does.

A curve described by two numbers #

Ramberg and Osgood fitted the rounded curves of aircraft alloys with one expression, and it has been used for them ever since:

$$\varepsilon = \frac{\sigma}{E} + 0.002\left(\frac{\sigma}{\sigma_{0.2}}\right)^n.$$

The first term is the elastic strain. The second is the permanent strain, which is 0.2 per cent at the proof stress $\sigma_{0.2}$ by construction and grows as the $n$-th power of the stress. A large $n$ keeps the permanent strain negligible until close to the proof stress and then lets it run — a sharp knee, mild steel in the limit. A small $n$ lets it begin early and grow slowly. Heat-treated aluminium alloys have exponents of twenty to forty; annealed aluminium and austenitic stainless steels, five to ten.

A column does not care about the strain. It cares about the slope of the curve at the stress it is carrying — the tangent modulus, $E_t = d\sigma/d\varepsilon$ — because that is the stiffness with which the column resists a small extra bend at that stress.

With $n = 40$ the alloy keeps its whole modulus to four fifths of its proof stress and then loses it quickly, like a steel. With $n = 5$ it has lost 15 per cent of its stiffness at half the proof stress and more than half by four fifths. At the proof stress itself every curve has lost most of its modulus, and the sharper the knee, the more — which is the only place the sharp material is worse off, and it is the stress a slender column never reaches.

A column curve that sags and overshoots #

The tangent-modulus theory says a perfect pin-ended column buckles at the stress where Euler’s formula, with the tangent modulus in place of $E$, gives back that stress:

$$\sigma = \frac{\pi^2 E_t(\sigma)}{\lambda^2}.$$

For a sharp material $E_t$ is $E$ up to yield and nothing after, so the solution is Euler’s stress or the yield stress, whichever is lower — the familiar envelope with a corner where they meet, at a slenderness $\bar\lambda = \sqrt{\sigma_{0.2}/\sigma_E} = 1$. For a rounded material the equation has to be solved, stress by stress.

The curves have no corner and no plateau. At the slender end they follow Euler, because a slender column buckles at a stress low enough to keep its whole modulus. In the middle every curve sags below the envelope. And at the stocky end they rise above the proof stress: a short column of the $n = 5$ alloy buckles at 1.54 times it, because its curve goes on rising past the proof stress with a tangent modulus that is still a useful fraction of $E$. The proof stress is not a strength for such a column — it is a point on a curve the column walks straight through.

The picture is the same one the residual-stress essay drew for steel, arrived at by a different route. There the modulus was lost because some fibres had yielded and others had not; here every fibre loses modulus together, a little at a time, because the material itself softens. The effect on the column is the same — a stiffness that falls as the stress rises — and so is the shape it gives the curve through the middle.

Where the rounded material loses most #

Divided by the sharp envelope, every curve has its deepest point exactly at the corner. That is where the sharp material is at its best — carrying its full proof stress at its full Euler stress, with its full modulus to the last — and where the rounded material is carrying a stress high enough to have lost a good part of its modulus and not high enough to have gained anything from the rising curve. A rounded material is weakest, against the sharp idealisation, at precisely the slenderness where columns are most often designed, because λ̄ near one is where the material and the geometry are both being used.

The loss is large. With $n = 10$ the column at the corner carries a quarter less than the envelope; with $n = 5$, a third less. With $n = 40$, which is as sharp as heat-treated aluminium gets, still 12 per cent. None of it involves an imperfection, a residual stress or an eccentric load: these are perfect columns, and the whole shortfall is the material’s curvature.

The second number #

The exponent describes the knee’s sharpness, and it is tempting to think it describes the column curve. It does not, and the reason is in the offset.

The proof stress is defined by a permanent strain of 0.2 per cent, a fixed number, and how large that is depends on the material. For the aluminium, whose modulus is a third of steel’s, the elastic strain at the proof stress is 3.6 thousandths, and the offset is about half of it. For the stainless steel, with steel’s modulus, the elastic strain at the proof stress is 1.15 thousandths and the offset is one and three quarter times as large. The same exponent then describes a knee that is, relative to the elastic line, far more gradual in the stainless steel — and its column curve is correspondingly lower: 0.64 of the proof stress at the corner against the aluminium’s 0.72.

A column curve for a rounded material needs two numbers: the exponent $n$, and the ratio of the 0.2 per cent offset to the elastic strain at the proof stress, $0.002 E/\sigma_{0.2}$. Neither alone fixes the curve, and a curve fitted to one alloy is not a curve for another with the same exponent and a different modulus. This is why design rules for stainless steel carry column curves of their own rather than a carbon steel curve relabelled, and why aluminium rules sort alloys into classes by their temper.

For an austenitic stainless steel, whose exponent is typically five to seven, the curve at the corner is at 0.56 to about 0.6 of the proof stress — a perfect column of stainless steel at λ̄ = 1 carries four tenths less than the envelope says. The stocky end overshoots less than the aluminium’s, because the same offset is a larger strain in a stiffer material and the curve has flattened more by the time the column reaches it.

The room between two theories #

The tangent-modulus load is where a perfect column first can bend. Engesser, who proposed it, was told it was wrong: when a column bends, the fibres on its convex side unload, and unloading follows the elastic modulus, not the tangent one, so the column is stiffer than the tangent modulus says. The reduced modulus that follows, for a rectangular section $E_r = 4EE_t/(\sqrt E + \sqrt{E_t})^2$, gives a higher load. For half a century the two theories competed, and the tests landed between them, nearer the lower.

Shanley settled it in 1947, with a model column and with tests on aluminium alloy columns for aircraft. A real column starts to bend at the tangent-modulus load, while the load is still rising; the unloading that the reduced-modulus theory assumes happens only once it is bending, and the load it carries after that rises from the tangent-modulus value toward the reduced-modulus value, never beyond. So the tangent-modulus load is the one to design for, and the reduced-modulus load is an upper bound on what any real column can carry.

For a rounded material the room between the two is not a curiosity at the corner of a steel curve; it runs through the whole middle of the column curve, and for a soft knee it is large — 19 per cent at λ̄ = 0.5 and 11 at the corner for $n = 5$. That is why the question mattered in 1947 and why it was settled on aluminium. The debate about which theory was right was, in practice, a debate about materials with no plateau, and the answer came from the aircraft industry because that is where such materials were being loaded to their limits.

A stronger steel is a rounder steel #

Mild steel is the material with the plateau, and it is not the only steel. The high-strength steels that are replacing it in heavily loaded columns reach their strength by quenching, tempering or thermomechanical rolling, and most of them have no yield plateau at all: their curves bend over like an aluminium alloy’s, with exponents in the teens or twenties. What they do not change is the one number a stronger steel does not change, the modulus, so the second number — the offset over the elastic strain at proof, $0.002E/\sigma_{0.2}$ — falls as the strength rises: 1.5 for a steel of 275 N/mm², 0.6 for one of 690. A stronger steel is rounder in its exponent and sharper in its offset ratio, and its column curve lies between the mild steel’s and the aluminium’s.

It also moves the corner. The crossing of the squash and Euler lines is at a slenderness proportional to $\sqrt{E/\sigma_{0.2}}$, so a steel of 690 N/mm² reaches its corner at a column two thirds as slender as one of 275. The rounding bites at the corner, and the corner of a strong steel is at the proportions most of its columns have.

An imperfect column on a softening curve #

Every curve here is for a perfect column, and a real one is never straight. A bow amplifies under load, the bending it adds raises the stress on the concave side, and on a rounded material that side is where the tangent modulus falls first. The two effects compound: the bow makes one side of the column carry more stress than the mean, the extra stress costs that side more of its stiffness than the mean stress does, and the lost stiffness lets the bow grow faster.

The practical result is that a rounded material is more imperfection-sensitive than a sharp one near the corner, where both losses are largest. A mild steel column with a bow loses strength by the Perry mechanism alone; an aluminium or stainless column loses it by the Perry mechanism working on a modulus that is already falling. The design curves for these materials are lower than a Perry curve with the same imperfection would give for a sharp material, and the difference is the curvature of the material.

Welded, and softer again #

A welded aluminium column has a third source of softening. The heat of the arc anneals the alloy beside the weld, permanently, and a heat-treated alloy with a proof stress of 250 N/mm² and an exponent of thirty becomes, over a band thirty millimetres wide, a softer alloy with half the proof stress and an exponent nearer ten. A column with a weld along its length has a section made of two materials, each on its own curve, and the softened band loses its stiffness at a stress the parent metal would carry with its whole modulus.

Cold-formed stainless sections have the opposite effect at their corners, which are strain-hardened by the bending that made them and carry a higher proof stress with a sharper knee. A section’s column curve is then a weighted average over materials that differ across its width, and the stiffness it has at a given mean stress is not the stiffness of any one of them.

Which free body produced the number #

Take the perfect column at the moment it is about to bend, carrying a stress $\sigma$ everywhere. Cut it at mid-height and take the lower half as a free body. The axial force times a small lateral deflection $v$ is a moment that the section must resist by bending, and the section’s resistance to a small extra bend is its second moment times the stiffness its fibres have at that stress — the tangent modulus, if the load keeps rising so that no fibre unloads. Equilibrium of the bent free body is then the Euler equation with $E_t$ in it, $E_t I v'' + P v = 0$, and its first solution is the tangent-modulus load. Every number above solves that equation at one slenderness, with $E_t$ taken from the Ramberg–Osgood curve at the stress it returns.

The corner, by hand #

For the aluminium alloy with $n = 10$ at the corner, λ̄ = 1: Euler’s stress equals the proof stress, 250 N/mm², so the slenderness is $\lambda = \pi\sqrt{70{,}000/250} = 52.6$. Try a stress of 184 N/mm², 0.737 of the proof stress. The permanent-strain term’s slope is $0.002 \times 10 \times 0.737^9/250 = 5.13 \times 10^{-6}$ per N/mm², against the elastic compliance $1/70{,}000 = 1.43 \times 10^{-5}$, so $E_t = 1/(1.43 + 0.51) \times 10^5 = 51{,}500$ N/mm². Then $\pi^2 \times 51{,}500/52.6^2 = 184$ N/mm². The column at the corner buckles at 184 N/mm² with 74 per cent of its modulus, where the sharp envelope says 250.

Perfect columns and a single curve #

The columns are perfect. A real column has an initial bow, residual stresses from welding or cold forming, and an eccentricity at its ends, and every one of them lowers the curve further — most near the corner, where the tangent-modulus loss is also largest. The perfect-column curve is an upper bound on the material’s column curve, not a design curve.

The material has one curve in tension and compression, and the same one everywhere in the section. Cold-formed stainless steel sections have corners that were strain-hardened as they were bent, with a higher proof stress and a sharper knee than the flat faces; welded aluminium has a softened zone beside every weld. Both change the stiffness the section has at a given mean stress.

The load rises monotonically to the buckling load, which is what makes the tangent modulus the right stiffness. Under cyclic loads a rounded material’s unloading and reloading curves differ, and a column that has been loaded before is stiffer on reloading than its virgin curve says.

What the pictures cannot show #

That Ramberg and Osgood’s expression is a fit. It describes the curves of many alloys well up to a little beyond the proof stress and poorly after, and the stocky end of every column curve here, where the buckling stress passes the proof stress, is where the fit is least reliable. Two-stage fits exist for the strain range beyond the proof stress, and on them the overshoot is somewhat smaller.

Nor can they show the local buckling that thin-walled aluminium and stainless sections are prone to. A column curve for a perfect cross-section assumes the section keeps its shape, and the plates of a thin section have their own buckling stresses — the elastic ones are a geometric factor times the modulus, and the inelastic ones fall with the same softening — which can arrive first.

Still open: the plate with no plateau #

Every result here is for a column, which bends as a whole. A plate element of a thin-walled section buckles by bending across its width, and its elastic buckling stress, too, is the elastic modulus times a geometric factor. With a rounded material the plate’s inelastic buckling stress uses a combination of the tangent and secant moduli, because a plate bends in two directions at once, and the combination depends on which theory of plasticity is assumed. Whether a stainless or aluminium plate’s inelastic buckling stress is lowered by its rounded curve as much as a column’s is — and whether the theory that gives the lower answer is again the one the tests support — is the question that follows this one into the thin sections these materials are mostly made into.