Stability

The panel that carries more after it has failed

Everywhere else in this field a critical load is where the argument ends. A thin web is the exception — it buckles visibly, in waves anybody can see, and then goes on to carry nearly twice as much again by turning itself into a truss nobody drew.

Assumes The plate that ripples, and the width that is left, Strong enough and still falls over and The shear nobody draws.

A plate girder’s web is a large piece of very thin steel, and under shear it develops a diagonal compression and a diagonal tension at right angles to each other. The compression diagonal is a strut with almost no stiffness across it, and it does what a strut does: it goes. The web ripples, visibly, in a set of parallel folds running one way across the panel, and the girder does not fall down. It carries on, and it carries on carrying nearly twice as much again — which no other member in this collection does.

A buckled panel is a truss that nobody drewA 1000 × 1000 panel of 6 mm web, at d/t = 167. It buckles in shear at 63.8 N/mm², which is 383 kN — and it then carries 696 kN, 1.82 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221.3 N per millimetre of its length. A web that never buckled at all would have reached 953 kN, so the panel ends at 73% of a stocky web's capacity on a fraction of its steel.stiffeners at 1000 mmthe band at 22.5°the truss it has becomestiffener in compression,web in tensionbuckles at 383 kN · carries 696 kN · a stocky web would reach 953 kNσ in the band 252 N/mm² over 541 mm
Fig. 1 A 1,000 mm square panel of 6 mm web, at a depth-to-thickness ratio of 167. It buckles in shear at 63.8 N/mm², which is 383 kN, and it carries 696 kN. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221 N per millimetre of its length.

Failure and buckling have come apart, and that is unusual enough on this site to be worth stating carefully. Every other stability argument here treats the critical load as the answer. A plate is the exception, and the exception has a mechanism.

Why a plate is different from a strut

A column that buckles has nowhere to go: its whole cross-section is on the same buckle, and once it starts to bend the load’s own eccentricity makes things worse. A plate is not one strut. It is a great many parallel strips, and they are joined side by side.

A 8 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 370 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 47 per cent of it is still working.1002003004005006007000200400600plate width (mm)slender beyond 370 mmyieldcritical stress — inverse square in the widthwhat the plate actually delivers, over its full width
Fig. 2 The plate’s own version of the column curve, and the effective width that survives past it. A buckled plate sheds stress from the middle of its width toward the supported edges, so the parts of it that are still straight take over — which is a redistribution a column has no room for.

The strips near the middle of a buckled panel are the ones that have gone; the strips near a stiffened edge are held straight and can pick up more. So a plate has an internal load path available after buckling, and the effective-width idea is that path counted for a plate in direct compression.

Under shear the redistribution is not to the edges but across the diagonal, and it is more dramatic, because the two diagonals of a shear field carry opposite signs. Losing one leaves the other entirely intact.

Which free body produced the number

Take the panel, cut it on a vertical line, and take the piece to one side. What crosses the cut before buckling is a uniform shear stress on the full depth: V=τdtV = \tau\, d\, t, and the state of stress at every point is pure shear — equal tension and compression at 45°, which is what Mohr’s circle says about a point in pure shear.

The critical value of that shear comes from the plate’s own eigenvalue problem:

τcr=kτπ2E12(1ν2)(td)2,kτ=5.34+4(a/d)2\tau_{cr} = k_\tau\,\frac{\pi^2 E}{12(1-\nu^2)}\left(\frac{t}{d}\right)^2, \qquad k_\tau = 5.34 + \frac{4}{(a/d)^2}

For the panel above, kτ=9.34k_\tau = 9.34 and τcr=63.8\tau_{cr} = 63.8 N/mm². A web that never buckled would have gone at fy/3=158.8f_y/\sqrt3 = 158.8, so this one buckles at 40% of the shear its material could have carried.

Now cut the same panel after buckling, and the free body is different. The compression diagonal contributes nothing across the cut, so the shear it used to carry has gone somewhere. What remains is a band of tension running corner to corner, anchored on the flanges above and below, and its vertical component is the extra shear.

The angle, which is not assumed

The band is anchored between the flanges, so its width is limited by geometry: a band at angle θ\theta to the horizontal, running between the panel’s corners, has a clear width of

s=dcosθasinθs = d\cos\theta - a\sin\theta

and its vertical component per unit stress is ssinθs \sin\theta. Maximising that product is one line:

ddθ[d2sin2θa2(1cos2θ)]=dcos2θasin2θ=0tan2θ=da\frac{d}{d\theta}\left[\tfrac{d}{2}\sin 2\theta - \tfrac{a}{2}(1-\cos 2\theta)\right] = d\cos2\theta - a\sin2\theta = 0 \quad\Longrightarrow\quad \tan 2\theta = \frac{d}{a}

The band finds its own angle, and it is not 45°The shear the tension band contributes, swept over every angle it could take, for five panel proportions. Each curve is the band's width d·cos θ − a·sin θ times its own vertical component, so it is zero when the band is flat and zero again when it no longer fits between the flanges. The maxima are at 29.5°, 22.5°, 16.8°, 13.3°, 9.3°, which is ½·arctan(d/a) in every case — the optimum recovered from the geometry rather than quoted. A square panel bands at 22.5°, and a panel twice as long as it is deep at 13.3°.020406080050100150200250300350angle of the band (degrees)shear from the band (kN)a/d = 0.6 → 29.5°a/d = 1 → 22.5°a/d = 1.5 → 16.8°a/d = 2 → 13.3°a/d = 3 → 9.3°
Fig. 3 The shear the band contributes, swept over every angle it could take, for five panel proportions. Each curve is zero when the band is flat and zero again when it no longer fits between the flanges, and the maxima sit at 29.5°, 22.5°, 16.8°, 13.3° and 9.2° — which is ½·arctan(d/a) in every case.

A square panel bands at 22.5°, not 45°. The result is Basler’s θd/2\theta_d/2 recovered from the geometry rather than quoted, and it is worth pausing on because the 45° is such a natural thing to expect — the yield lines in a slab are not at 45° either, for a related reason: the principal tension in the pre-buckling field really is at 45°, and the band is not the principal direction of anything. It is the direction that gets the most vertical force out of a strip of finite width anchored on two flanges, and the flanges are what make the answer smaller.

The trend is the reverse of the intuition too. A longer panel bands at a flatter angle, because a flat band in a long panel is wide, and width is what the vertical component is being traded against.

What the material has left

The band cannot be stressed to yield, because the critical shear is still there underneath it: the panel does not unload when it buckles, it stops taking more shear in the buckled diagonal. Superposing a uniaxial tension on a state of pure shear at τcr\tau_{cr} and going to yield gives

σt=fy23τcr2\sigma_t = \sqrt{f_y^2 - 3\tau_{cr}^2}

which is 252 N/mm² here against a yield of 275 — so the buckling that has already happened has cost only 8% of the membrane stress available. Multiply out: σttssinθ=313\sigma_t\, t\, s \sin\theta = 313 kN of extra shear, on top of the 383 kN the panel buckled at, for a total of 696.

Where the reserve is largest

What a thin web is worth before and after it bucklesA 1000 mm panel at a/d = 1, with the web thickness varied. The lower curve is the load at which the panel buckles, which goes as the square of the thickness; the upper one is what it carries in the end, which is very nearly linear in it because the band's own force is a stress on an area. So the reserve is largest exactly where the buckling load is smallest: at d/t = 333 the panel carries 4.6 times the load it visibly failed at, and at d/t = 83 only 1.00 times. The flat line is the shear a web that never buckled would reach, which no panel here gets to.0501001502002503000500100015002000depth ÷ thicknessshear (kN)carried in the endbuckles herea web that neverbuckled at all
Fig. 4 Buckling load and ultimate load against depth-to-thickness ratio for the same panel. The lower curve goes as the square of the thickness; the upper one is very nearly linear in it, because the band’s contribution is a stress on an area. So the two diverge as the web gets thinner, and the reserve is largest exactly where the buckling load is smallest.

That divergence is the practical content of the whole subject.

d/td/t buckles at carries reserve of a stocky web
250 113 kN 338 kN 2.98× 53%
167 383 696 1.82× 73%
125 908 1,226 1.35× 97%
106 1,588 1.00× 100%

At d/t=106d/t = 106 the two limits coincide: τcr\tau_{cr} has climbed to fy/3f_y/\sqrt3 and the web yields before it buckles, so there is no post-buckling reserve to have because there is no buckling. Below that ratio a web is a stocky web and this page has nothing to say about it.

Above it, the thinner the web the larger the fraction of its strength that arrives after it has visibly failed. A 4 mm web reaches only 53% of what its material could carry, but it reaches three times what a linear buckling calculation would have allowed it — and it does so on 40% of the steel a stocky web would have needed for the same shear.

The shape of the load path afterwards

Three paths out of the same critical loadLoad against sideways movement past the critical load, for three systems whose critical loads are identical. The stable one climbs, so a real structure with a small crookedness reaches nearly the full load and keeps going. The unstable one falls symmetrically, so the imperfect structure has a maximum below the critical load and it matters not at all which way it leans. The asymmetric one falls one way and climbs the other, so the direction of the imperfection decides everything. All three are drawn at an imperfection of 0.02 radians.stable symmetric — a columnan imperfection is a nuisancecritical89%unstable symmetric — a shellan imperfection is a demolitioncritical77%asymmetric — a frameand it matters which waycritical
Fig. 5 Three post-buckling characters on one set of axes. A plate is the stable-symmetric one: the path rises after the bifurcation, so the structure stiffens rather than sheds, and the critical load is a lower bound on its strength instead of an upper one. A shell is the opposite and loses two thirds of its theoretical load to a defect nobody can see.

The whole of this page depends on which of those three curves a shear panel is on, and the answer is the top one. Stability is not a single quality: two structures with identical critical loads can be a bargain and a trap, and the difference is a fourth-order term in the energy that no eigenvalue calculation contains. A plate girder web is the most useful example of the bargain in the whole of structural engineering, and a cylindrical shell in compression is the standing example of the trap.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero
Fig. 6 And where the shear actually is. The tension field is worth most near the supports, where the shear is largest and the moment is smallest — which is the region a designer is least likely to be worrying about, and is the region a plate girder’s web is thinnest in for exactly that reason.

The truss it turns into

A Pratt truss of 6 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.29.429.447.147.129.429.4-47.1-52.9-52.9-47.1-15.0-10.0-15.0-38.6-38.623.27.77.723.2tensioncompression2 carrying nothing
Fig. 7 The structure the girder has become. The flanges are chords, the intermediate stiffeners are vertical posts in compression, and the buckled panels are diagonals in tension — which is a Pratt truss, and is exactly the arrangement a Pratt truss uses for exactly this reason.

The analogy is not decorative. After buckling, a plate girder has: two chords carrying the bending moment as a couple; a set of verticals carrying compression, because the tension band pulls the flanges together and the stiffeners hold them apart; and a set of tension diagonals at a shallow angle. A stiffened plate girder is a Pratt truss with the diagonals made of sheet.

Which explains the one detail that otherwise looks arbitrary: the intermediate stiffeners of a plate girder do almost nothing before buckling, and after buckling they are struts carrying real axial load. They exist to divide the web into panels — which raises kτk_\tau — and to anchor the band, in the way a brace need not be strong to be worth having, and the second job only begins on the day the first job stops working.

The count is necessary and not sufficientTwo pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.55 of the span.one panel braced twice, the next not at allm 9 + r 3 = 2j 12 · rank 11a mechanismthe same count, properly arrangedm 9 + r 3 = 2j 12 · rank 12stands up
Fig. 8 And the reason the analogy has a limit: a truss panel that loses its diagonal is a mechanism, while a web panel that loses its compression diagonal still has the other one. Triangulation in one direction is enough here because shear has a sign, and a girder is only ever asked to carry it in the direction the band was drawn for.

That last observation is the honest boundary of the truss picture. A tension field is one-directional. Reverse the shear and the band has to reform along the other diagonal, which it will — but the panel has to go through its unbuckled state to get there, and under repeated reversal the folds work back and forth and fatigue becomes the governing question rather than strength.

What it does to the flange

Shear stress across a sectionThe distribution of shear stress over an I-section, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.41 against a mean of 0.19 — a ratio of 2.15 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 2.15× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 9 Where the shear was supposed to go: a parabolic distribution over the web, with the flanges carrying almost none of it. That picture is the pre-buckling one, and the tension field replaces it with a uniform band that ends on the flange rather than running past it.

The band pulls, and it pulls on something. Its transverse component is delivered to the flange as a distributed pull of σttsin2θ\sigma_t\, t \sin^2\theta — 221 N per millimetre here, which over a 1 m panel is 221 kN trying to bend the flange inward between the stiffeners.

So a plate girder relying on tension field action has a flange doing a second job it was not sized for, and the amount of band that can be anchored is limited by the flange’s own bending capacity. That is the difference between the models: Basler assumed the flanges were infinitely flexible and anchored nothing, the Cardiff models compute an anchored length from the flange’s plastic moment, and the truth is in between and depends on a member nobody thinks of as a beam.

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 flange outstand (buckling coefficient 0.43) derives to 17.2, 15.2, 13.3 at 275, 355, 460 N/mm², against quoted limits of 12.9, 11.4, 10.0; A web, in bending (buckling coefficient 4) derives to 52.5, 46.2, 40.6 at 275, 355, 460 N/mm², against quoted limits of 38.8, 34.2, 30.0. The derived number is the larger every time, and by the same factor at every grade — flange outstand 1.33, web, in bending 1.35 — 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.flange outstandk = 0.4317.2 at 27515.2 at 35513.3 at 460quoted: 14ε1.33× the quoted limit, at every gradeweb, in bendingk = 452.5 at 27546.2 at 35540.6 at 460quoted: 42ε1.35× the quoted limit, at every grade0102030405060width ÷ thickness
Fig. 10 The same question asked of a section rather than a panel: is this piece of plate slender enough that its own local instability arrives before the material’s limit? A plate girder web is deliberately on the far side of that line, and the whole of this page is about what happens there.

Where the model stops

Nothing here sizes anything. The band model above is a Cardiff-type one with the interaction simplified to σt=fy23τcr2\sigma_t = \sqrt{f_y^2 - 3\tau_{cr}^2}, and design codes use rotated-stress-field or Basler formulations that differ from it and from each other by 10–20%. The argument is about where the reserve comes from.

The flange is assumed to anchor the whole band. It cannot. A real anchored width is set by the flange’s plastic moment and the stiffener spacing, and a girder with light flanges reaches considerably less than the numbers above.

The end panel has no neighbour to lean on. A tension field pushes horizontally on the panel next to it, and interior panels balance out — but the panel at the end of a girder has nothing beyond it, so either it is designed without tension field action or the end stiffener has to be a substantial vertical beam. The most common error in this subject is applying the interior-panel formula to the end one.

The web still has to carry the bending moment. Everything above treats a shear panel in isolation. In a real girder the same web carries direct stress from the flexure of the whole member, that stress interacts with the shear, and the interaction is a curve rather than a pair of separate checks.

And the deformations are large. A web at its ultimate shear has out-of-plane folds of the order of its own thickness or more. Nothing in the elastic buckling calculation applies to that geometry; the critical stress is used only as a marker for where the redistribution begins.

What the pictures cannot show

The hero draws the band as a clean parallelogram with a sharp edge, and the panel as flat. Neither is true. The membrane stress varies across the band and dies gradually rather than at a line, the folds are out of the plane of the page by several millimetres, and the drawing is a plan of a surface that is no longer plane.

The reserve figure plots two loads on one axis as though they were the same kind of quantity. They are not: the lower curve is a bifurcation load computed from linear elasticity, and the upper one is an ultimate load computed from a plastic mechanism. Nothing continuous connects them, and the vertical distance between the curves is a comparison of two theories rather than a path a panel travels along.

And no figure here shows the deflection. A girder working in tension field action has a shear stiffness far below the elastic value, so it sags more than a shear calculation predicts, and the extra movement arrives suddenly at the buckling load. The strength argument and the serviceability one point in opposite directions, and only the first is drawn.

The ladder from here

Later rungs on this anchor: Basler’s model and the rotated stress field set out side by side, with the assumptions that separate them named. The flange anchorage calculation, and the plastic hinges in the flange that decide the anchored length. End panels, bearing stiffeners and the horizontal thrust a tension field applies to whatever is beyond it. The interaction of shear and bending in a slender web, which is a surface rather than a curve. Corrugated webs, which do not buckle in shear at all and therefore have none of this — and pay for it in fabrication. Tension field action in aircraft structure, where Wagner published the complete diagonal-tension theory in 1929 for a skin far thinner than any girder’s and where the reserve is not an extra but the whole design basis. And repeated shear reversal, which turns a strength argument into a fatigue one.

The historical shape of the subject is worth noticing. Wagner had the complete theory in 1929 for thin aircraft skins, where webs are so slender that they were never expected to do anything else. Civil engineering, whose webs are thicker and whose factors of safety are larger, went on treating buckling as failure for another thirty years — and only started using the reserve when the arithmetic of a welded plate girder made a thinner web worth the trouble.

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Critical loadEfficiencyLoad pathLocal bucklingPlate bucklingPlate slendernessPost bucklingShear stressShear yieldStabilityStiffnessTriangulation