Structural Engineering & Statics
where the load goes
Sections and stress

What is left after it ripples

A thin plate that buckles locally has not failed. It has stopped taking load in its middle and gone on taking it near its edges, so the member is now made of a different section from the one that was drawn — and the new one has its centroid somewhere else, which turns a concentric load into an eccentric one.

Assumes The plate that ripples, and the width that is left, The section that cannot reach its own strength and The material far from the middle does nearly all the work.

21 min read 11 figures Geometry beats materialWhich failure arrives firstScale changes everything

A stocky section reaches yield everywhere and then stops. That is what makes it stocky, and every plastic argument on this site quietly depends on it.

A thin one does something else. Long before the yield stress arrives anywhere, one of its plates goes out of plane — a web ripples, a flange waves — and the stress in the middle of that plate stops rising. It does not stop rising near the edges, where the plate is held by the plate next door, and the load goes on increasing.

So the member has not failed. It has become a different member: the one made of the parts that are still working.

A 2 mm plate, and the width it can beThe elastic critical stress of a plate in compression against its width, with the yield stress drawn across it. Below 92 mm the plate reaches yield before it buckles; above it the plate ripples first, and the fraction of the width still carrying load falls away — at 700 mm only 13 per cent of it is still working.1002003004005006007000200400600plate width (mm)slender beyond 92 mmyieldcritical stress — inverse square in the widthwhat the plate actually delivers, over its full width
Fig. 1 The event this page begins after. A plate held on both edges buckles at kπ2E/12(1−ν2)⋅(t/b)2k\pi^2E/12(1-\nu^2) \cdot (t/b)^2, which for the 196 by 2 web here is about a quarter of the yield stress. What happens next is not on that page.

Which free body produced the number

Take a strip of the buckled plate across its width.

Before buckling the stress is uniform. After it, the middle of the plate has gone out of plane and can carry no more; the strips near the two supported edges are still straight and go on being loaded. The stress distribution across the width becomes a saddle — high at the edges, low in the middle — and the total force is the area under it.

Winter’s device is to stop describing that distribution and describe the force instead. Replace the real plate carrying a non-uniform stress with a narrower plate carrying the edge stress uniformly, of width

beff=ρ b,ρ=λp−0.22λp2    for  λp>0.673b_{eff} = \rho\, b, \qquad \rho = \frac{\lambda_p - 0.22}{\lambda_p^2}\;\;\text{for}\;\lambda_p > 0.673

where λp=(b/t)/(28.4εk)\lambda_p = (b/t)/(28.4\varepsilon\sqrt{k}) is the plate’s slenderness written so that λp=1\lambda_p = 1 means it buckles exactly at yield. Below 0.673 nothing is lost at all.

The two constants are fitted rather than derived, and the shape is not: ρ\rho falls as 1/λp1/\lambda_p for large slenderness rather than as 1/λp21/\lambda_p^2, which is the statement that a buckled plate keeps a fixed width of working material rather than a fixed fraction of it.

The plateau, and the curve that leaves itWinter's reduction factor against plate slenderness, for a plate held on both edges and for an outstand held on one. Below λ_p = 0.673 the plate reaches yield before it buckles and nothing is lost at all — the flat part is not an approximation, it is the range in which the section is the section that was drawn. Above it the reduction is (λ_p − c)/λ_p², which falls as one over the slenderness for large λ_p rather than as one over its square: the plate does not lose everything past its buckling stress, it loses the middle and keeps the edges. This section's web sits at λ_p = 2.11 and keeps 43% of its width; its flange at 0.68 keeps 100%.00.511.522.530%20%40%60%80%100%plate slenderness λ_pwidth that still worksboth edges heldan outstand1 ÷ λ²webflangelip
Fig. 2 Winter’s curve and the plateau below it. The flat part is not an approximation — it is the range in which the section is the section that was drawn. The dashed line is 1/λp21/\lambda_p^2, which is what a plate would keep if buckling simply capped its stress, and the gap between the two is the post-buckling reserve.

The plates lose their material in different places

This is where it stops being a strength calculation and starts being a geometry one.

A plate supported on both edges — a web between two flanges — loses its middle. The two effective strips sit at the two ends, and the plate’s own centroid does not move.

A plate supported on one edge — an outstand, a flange with a free edge — loses its free end. The effective strip sits at the supported end, and the plate’s centroid moves toward the support.

So a section made of plates of both kinds does not shrink symmetrically, and its centroid is somewhere else.

The load did not move; the section didA lipped channel 200 by 65 mm at 2 mm thick, drawn twice on top of itself: the outline as fabricated, and the part of it still working once the plates have buckled. The web is held on both edges, so it loses its middle; the flanges are held at the web, so an unlipped one would lose its free edge. What survives is not symmetric with what was drawn, so the centroid moves 8.0 mm — and a load applied along the axis it was designed to arrives 8.0 mm off the section that has to carry it. At the 177 kN this section will take, that is 1.42 kNm of bending nobody applied.8.0 mmas drawnas it workswebλ 2.11 · keeps 43%flangeλ 0.68 · keeps 100%lipλ 0.61 · keeps 100%area left69%eccentricity8.0 mm256 kN of section, 177 kN of it effective, and 1.42 kNm that was not in the load case
Fig. 3 The lipped channel drawn twice on top of itself: the outline as fabricated, and the part of it still working. The web has lost its middle; the flanges are internal plates because the lips hold their far edges, so they have kept nearly everything. The result is a section whose centroid is 8.0 mm further from the web than the drawn one — and a load applied through the drawn centroid arrives 8.0 mm off.

The eccentricity is an output

That last sentence is worth saying slowly, because it inverts the usual order.

Everywhere else in this collection an eccentricity is an input. Somebody puts a load off the centroid; the moment follows. Here the load is applied exactly where it was designed to be applied, along the axis of the member as drawn, and the moment appears because the section moved.

For this channel: the effective area is 506 mm² of a gross 732, so the squash load falls from 256 kN to 177. At that 177 kN, the 8.0 mm shift is worth 1.42 kNm of bending that appears in no load case anywhere. On a 3 m column that is not a rounding error; it is a beam-column problem the member was not designed as.

And the direction is not fixed. Take the lips off the same channel and the flanges become outstands, lose their free edges, and drag the centroid the other way — a shift of −6.8 mm rather than +8.0. The same mechanism, on two sections that differ by 20 mm of lip, moves the load in opposite directions.

The middle third, computedThe kern of a 250 × 400 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±66.7 mm vertically and ±41.7 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows.rectangle±67 of 400 mm33.3% of the depth
Fig. 4 Where a load may sit before the far face goes into tension. An eccentricity that arrives from the section rather than from the load has to be checked against the same kern, and it consumes part of a budget the designer thought was untouched.
Loaded straight down, and moving sidewaysAn zed purlin with a moment applied about the horizontal axis. Its principal axes lie at -19.4° to the drawn ones, so the neutral axis runs at 59.0° rather than horizontally, and the section moves 166% as far sideways as it moves down. The product of inertia that causes it is 6.840 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now.the axis the drawing suggestsneutral axis, 59.0°it moves this wayprincipal axes at -19.4° · I₁/I₂ = 13.83
Fig. 5 And why a channel is the section this happens to. It has one axis of symmetry and not two, so it already had a centroid that was not in the middle of anything, and there is nothing to stop the effective one being somewhere else again.

Twice the metal is more than twice the section

The capacity of a fully effective section is proportional to its area, which is proportional to its thickness. The capacity of a thin one is not.

A thicker plate is doing two things at once: it is larger, and it is less slender. So ρ\rho rises as well as AA, and the two multiply. Fitted over a sweep from 0.6 mm to 8 mm on this channel, the squash load goes as

Nc∝t1.43N_c \propto t^{1.43}

which for the plain channel — whose flanges are outstands and buckle much earlier — is t1.75t^{1.75}.

The concrete form: going from 1.5 mm to 3.0 mm on the lipped channel takes the capacity from 111 kN to 302, a factor of 2.72 for twice the material. There are very few places in this subject where doubling the metal buys more than double, and they are all places where something was being lost that the extra metal stops losing.

Twice the metal is more than twice the sectionThe squash load of the effective section against its thickness, with the gross section's above it. The gross line is straight, because area is linear in thickness. The effective one is not: a thicker plate is both larger and less slender, so it keeps a greater fraction of itself as well as being bigger, and the capacity goes as t^1.43 fitted over the whole sweep. The gap between the two lines is what local buckling has taken — 31% of the section at 2 mm — and it closes only at a thickness at which nobody would be cold-forming anything.1.02.03.04.05.06.07.08.002004006008001000thickness (mm)squash load of what is left (kN)the section drawnthe section that works177 kNfitted power 1.43 · a section that never buckled gives exactly 1
Fig. 6 The two lines. The gross section’s is straight, because area is linear in thickness. The effective one is not, and the gap between them is what local buckling has taken — 31% at 2 mm — closing only at a thickness at which nobody would be cold-forming anything.

The thickness at which this channel’s plates are all fully effective is 6.15 mm. A 200 mm deep channel would have to be six millimetres thick before its web stopped buckling, which is not a cold-formed section at all. Every cold-formed member in service is on the curved part of that graph, which is why the effective-width calculation is not an occasional refinement in that world but the whole of the section calculation.

The lip is worth twenty millimetres of nothing much and all the difference

The two channels compared above differ by a 20 mm return on each flange, and the comparison is the clearest demonstration on the page of what an edge support is worth.

Without it, each flange is an outstand: 63 mm of free plate with k=0.43k = 0.43, giving λp=2.06\lambda_p = 2.06 and ρ=0.44\rho = 0.44. With it, the same 63 mm is a plate held at both ends with k=4k = 4, giving λp=0.68\lambda_p = 0.68 and ρ=0.997\rho = 0.997.

The steel is identical apart from the lips. The effective area goes from 44% to 69% of gross, and the squash load from 100 kN to 177 — a 77% increase for 6% more metal, all of it spent making an edge into a support rather than making anything thicker.

That is the same trade the whole subject keeps offering and rarely this starkly: geometry beating material by a factor no amount of material could buy.

The same material, three waysThree cross-sections of identical area, so identical weight and cost, with the second moment of area computed from each profile's own geometry. Only the arrangement differs, and the stiffest is many times the flattest.tall rectangleI = 10.00 × 10⁶1.0× the firstI-sectionI = 24.29 × 10⁶2.4× the firstteeI = 11.22 × 10⁶1.1× the firstevery section here has an area of 3000 — only the shape differsthe bar is the second moment of area, to scale
Fig. 7 The same steel in a different shape, which is the standing version of this argument for bending. The lipped-against-plain comparison is its local-buckling counterpart: the material has not moved far, and what has changed is which edges are held.

Classification is the same question, asked coarsely

The four-class system on section classification sorts sections by what they can reach before a plate buckles: a plastic moment, a yield moment, or something less. Class 4 is the class where something less applies, and this page is what “something less” means.

Read the other way round, the classes are a set of thresholds on λp\lambda_p and the effective width is what happens past the last of them. The web here at λp=2.11\lambda_p = 2.11 is deep into class 4; the lip at 0.61 is fully effective; and the flange at 0.677 is four thousandths of a slenderness unit past the plateau, keeping 99.7% of itself — which is a nice demonstration that the plateau is a real boundary and not a cliff.

Where the class limits come fromThe width-to-thickness ratio at which two kinds of plate reaches its own elastic critical stress at the yield stress, for three steel grades. A web, in compression (buckling coefficient 4) derives to 56.8, 46.2, 40.6 at 235, 355, 460 N/mm², against quoted limits of 42.0, 34.2, 30.0; A flange outstand (buckling coefficient 0.43) derives to 18.6, 15.2, 13.3 at 235, 355, 460 N/mm², against quoted limits of 14.0, 11.4, 10.0. The derived number is the larger every time, and by the same factor at every grade — web, in compression 1.35, flange outstand 1.33 — because both the derivation and the quoted limit go as one over the root of the yield stress. A constant ratio is what a fixed knockdown looks like: the derivation is for a perfect plate and the quoted limit is for a rolled one, carrying residual stress and not quite flat.web, in compressionk = 456.8 at 23546.2 at 35540.6 at 460quoted: 42ε1.35× the quoted limit, at every gradeflange outstandk = 0.4318.6 at 23515.2 at 35513.3 at 460quoted: 14ε1.33× the quoted limit, at every grade0102030405060width ÷ thickness
Fig. 8 The thresholds, drawn. Everything on this page lives past the last of them, and the effective width is the continuous function the class boundaries are a coarse sampling of.

Where the effective section is going next

The area is only half of what a member needs. A column also needs a second moment of area, and the effective section’s is not the drawn section’s either — the web has lost the material nearest its own centroid, which contributes least to II about the strong axis, so II falls much less than AA does.

That has an odd consequence worth stating: local buckling makes a thin section relatively more slender in area than in stiffness, so a member’s radius of gyration I/A\sqrt{I/A} actually goes up when its plates buckle. The section is worse in every absolute sense and better in that one ratio, which is why an effective-section column check has to use the effective area with the gross-section slenderness rather than mixing them.

Every strip counts by the square of its distanceA rectangular section divided into equal strips, with each strip's contribution to the second moment of area drawn beside it. The strips are identical in size; only their distance from the neutral axis differs.neutral axiscontribution of each striptotal I = 60.00 × 10⁶the outer strips do almost all of the work
Fig. 9 Why the second moment barely notices: II is weighted by the square of the distance from the axis, and what the web lost is the material sitting on the axis. Nearly a third of the area went and a much smaller share of the stiffness went with it.
How much of a flange works is decided by the span, not by the flangeThe working fraction of a flange overhang against its width as a fraction of the span, with the code rule and the exact elastic ceiling on the same axes. At b/L = 0.150 — the 3 m overhang on the 20 m span — the model gives 0.844 of the width and the code's min(L/8, b) gives 0.833, while the exact ceiling of L/2π per side is 3.183 m, wider than the flange itself, so it does not bind until b/L reaches 1/2π = 0.159. Doubling the flange at a fixed span moves the curve down, not the effective width up: at b/L = 0.050 the fraction is 0.979 and at 0.199 it is 0.759, so 4 times the flange buys 3.06 times the working width. Past b/L ≈ 0.22 the one-term shape rises above the exact ceiling and is optimistic; the ceiling governs there, and the curve is drawn no further than the model is good for.00.050.10.150.20.2500.20.40.60.81b ÷ L, the overhang as a fraction of the spanb_eff ÷ b3 m on 20 m: 0.844 of the flange worksthe modelenergy minimummin(L/8, b)the code ruleL / 2π per sidethe exact ceiling
Fig. 10 And the other family of effective widths on this site, which shares a name and not a mechanism. Shear lag reduces a wide flange because the stress has not had room to spread into it; local buckling reduces a plate because the stress has left the middle of it. Both end in a beffb_{eff} and neither derives from the other.

Where the model stops

The plate is assumed to redistribute toward its supported edges. It does when the edges are genuinely held. A flange whose lip is too small to act as a support does not — the lip buckles with it, the whole assembly goes as one, and the calculation has to be done for the flange-and-lip together as a distortional mode rather than plate by plate. That mode is not on this page and it governs a great many real cold-formed sections.

Each plate is treated alone. They are not alone: a buckle in the web and a buckle in the flange have to be compatible at the corner, and the true local mode is a coupled one whose half-wavelength is common to both. Treating them separately with fixed kk values is a decoupling that is convenient rather than true, and the error runs both ways.

And the material is elastic–perfectly plastic at the edges. The effective strip is assumed to reach fyf_y and stop, which for a cold-formed section is doubly awkward: the corners are cold-worked and stronger than the flats, and the flats retain residual stresses from the forming.

The scale at which all of this arrives

There is a reason this subject belongs to cold-formed steel and not to rolled steel, and it is a scaling argument rather than a manufacturing one.

Plate slenderness is b/tb/t, and buckling stress goes as (t/b)2(t/b)^2. Rolling produces sections whose thickness is set by what a mill can handle — 8 mm and up for anything structural — so a 200 mm rolled web has b/tb/t around 25 and is nowhere near the plateau. Cold forming starts from sheet, which arrives at 1 to 3 mm, so the same 200 mm web has b/tb/t around 100 and is four times as slender, which is sixteen times less buckling stress.

The consequence is that the two families of steel section fail in entirely different ways for entirely geometric reasons. Hot-rolled members yield, buckle overall, or fail at a connection. Cold-formed members lose part of their own section first, always, and every other check is made on what is left.

Length costs more than it looksThe same column section at four lengths, with the buckling capacity of each drawn as a bar. Capacity falls as the inverse square of the length, so a column three times as long carries a ninth as much.1× the length100% of the capacity1.5× the length44% of the capacity2× the length25% of the capacity3× the length11% of the capacityidentical section, identical material, identical end conditions
Fig. 11 What slenderness costs at the member scale, which is the same graph one level up. A plate is a column of its own, and b/tb/t is its slenderness — so the reason a thin plate buckles early is precisely the reason a long column does.

What the pictures cannot show

The effective section is a bookkeeping device and not a photograph. There is no part of the real plate that is doing nothing and no line across it where the material stops working. What is drawn as two strips of full-strength steel is really one plate carrying a saddle-shaped stress distribution whose integral happens to equal theirs.

Nor can the drawing show the buckle. The plate is out of plane by something of the order of its own thickness, in half-waves whose length is about the plate width, and every section in this page’s figures is drawn flat.

The order the checks have to be made in

One consequence of the section being an output rather than an input is that the usual sequence of a member check no longer works, and it is worth stating because it catches people.

Ordinarily: compute the section properties, compute the actions, compare. Here the section properties depend on the stress the section is carrying, the stress depends on the actions, and one of the actions — the moment from the centroid shift — depends on the section properties. The calculation is circular, and it is resolved by iterating: assume the section is fully effective, find the stress, reduce the plates, find the new centroid, recompute the stress at the new extreme fibre, reduce again.

For a member in pure compression the loop converges in one pass, because the stress is uniform and the reduction does not change it. For one in bending it does not: reducing the compression zone moves the neutral axis, which changes the stress the reduction was computed at, which changes the reduction. Two or three passes is usual and the answer moves by a few per cent between them.

A section property that depends on the load is not a section property, and everything a designer knows about the order of a calculation has to be set aside for this one family.

The assumption the figure rests on

kk — the buckling coefficient — is a number chosen for each plate from the support conditions assumed at its edges: 4 for a plate simply supported on both, 0.43 for an outstand simply supported at one. Both assume the neighbouring plate offers simple support, which is a lower bound: a thick flange partly clamps the web, and a real web’s kk is somewhere between 4 and 7. That single conservative choice runs through λp\lambda_p, ρ\rho, the areas, the shift and every number on this page.

The ladder from here

Later rungs on this anchor: the distortional mode, where a flange and its lip buckle as a unit and no plate-by-plate calculation applies. The iteration for a section in bending, where reducing the compression zone moves the neutral axis and changes the stress the reduction was computed at. Effective sections in the interaction with overall buckling, where the shift computed here becomes the eccentricity in a beam-column check. Cold-work of forming and the corner strength enhancement, which gives back some of what the flats lose. The direct strength method, which drops the plate-by-plate decomposition entirely and works from the whole section’s elastic buckling stresses. And the historical question of where the 0.22 came from, which is a curve fitted to tests on stiffened plates in the nineteen-forties and has been in every code since.

Where this goes next

1002003004005006007000200400600plate width (mm)slender beyond 370 mmyieldcritical stress — inverse square in the widthwhat the plate actually delivers, over its full width The plate that ripples, and the width that is left A wide thin plate in compression buckles at a stress that has nothing to do with the strength of the material. It then goes on carrying load — the middle drops out, and the edges work harder. flange outstandk = 0.4318.6 at 23515.2 at 35513.3 at 460quoted: 14ε1.33× the quoted limit, at every gradeweb, in bendingk = 456.8 at 23546.2 at 35540.6 at 460quoted: 42ε1.35× the quoted limit, at every grade0102030405060width ÷ thickness The section that cannot reach its own strength A section classification looks like a table of arbitrary numbers. Set a plate's buckling stress equal to the yield stress and the numbers fall out of a formula written three phases ago — larger than the quoted ones by a constant factor, at every grade. webb = 3.00 mb_eff = 2.531 mfree edgedrawn flange 2 × 3.00 m = 120000 mm² at 20 mm thickworking flange 2 × 2.531 m = 101238 mm², on a span of 20 m18762 mm² is there, and paid for, and carrying almost nothing The flange that is not all there Stress can only get into a wide flange through shear along its junction with the web, and shear takes distance to do it. So the width that is working is set by the span, and a flange 3 m wide on a 20 m span has 18762 mm² of steel that is there, and paid for, and hardly carrying anything. neutral axiscontribution of each striptotal I = 79.86 × 10⁶the outer strips do almost all of the work The material far from the middle does nearly all the work A strip of steel contributes to bending stiffness in proportion to the square of its distance from the centre. Move the same steel outward and the section gets stiffer for nothing.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

CentroidCold formedEccentricityEffective widthLipLocal bucklingPlate bucklingPost bucklingSection classificationSlendernessSquash loadStress redistributionThin walled