The weight that bends a rib it cannot bend

The arch that leans instead of squashing found the load at which a steel rib buckles, and found that it buckles sideways at the crown: the first mode of a two-pinned parabola is antisymmetric, one half rising while the other falls. It found that number with a uniform load on the whole span, because that is what an eigenvalue needs, and it named what it had set aside: “asymmetric loading, and the second-order arch that never bifurcates because it was bending from the start.”

That is not a refinement of the buckling load. It replaces the question the buckling load answers, and the replacement matters most for exactly the ribs that essay was about, the slender ones.

A uniform load asks nothing of the mode #

A parabola is the funicular shape for a load spread uniformly over its span. The shape that carries itself is the general statement: an arch whose axis follows the thrust line of its load carries that load as pure compression, with no bending anywhere. So a parabolic rib under its own deck, whose weight is close to uniform over the span, has an axial force and nothing else. Loaded up, it stays exactly on its axis until the thrust reaches the value at which the straight-and-compressed state is no longer the only one available — the bifurcation load — and then it can leave it sideways.

That is the ideal the eigenvalue describes. On the rib here, 60 m span, 12 m rise, a steel box with a bending stiffness of $250{,}000\ \mathrm{kN\,m^2}$, the bifurcation load is 53.4 kN/m, which is the coefficient 46.2 from the earlier essay times $EI/L^3$.

The antisymmetric mode is the one that matters, and the reason is worth restating because it is the reason for everything below. That mode is inextensional: one half goes down, the other up, and the arc length hardly changes, so it costs only bending. A uniform load pushes the whole rib down, which is the symmetric mode’s direction, and does nothing to start the antisymmetric one.

Half a load is a uniform load and a twist #

Now put a live load on half the span. Six kilonewtons per metre on the left 30 m of the rib is, by superposition, three kilonewtons per metre on the whole span plus a load of +3 on the left half and −3 on the right. The first part is funicular and adds thrust. The second is antisymmetric — down on one side, up on the other — which is precisely the shape of load that pushes the rib into its first mode.

Solved, the rib’s displaced shape is its buckling mode to within less than one per cent. The loaded half sags, the unloaded half lifts, and the crown moves 106 mm sideways towards the unloaded side. There is no bifurcation for this rib to reach, because it is in the buckled shape already: the live load has done what an imperfection does to a column, and done it more thoroughly, since an imperfection is small and this is proportional to the load.

Why one mode takes the whole of the shape #

A 99.3 per cent fit sounds like a coincidence and is not. Any displaced shape of the rib can be written as a sum of its buckling modes, and under an axial load each mode’s share is amplified by its own factor, $1/(1 - w/w_i)$, with $w_i$ that mode’s bifurcation load. The half-span load’s antisymmetric part has no component along the symmetric modes at all, so the second mode, the symmetric one at 2.3 times the first, does not enter; its funicular part barely moves the rib. What is left is the first mode and the higher antisymmetric ones, and the next antisymmetric mode sits at 220 kN/m, about four times the first’s load, so its amplifier is 1.09 while the first mode’s is 1.51. Even before amplification most of the antisymmetric load’s displacement is first-mode shape, and the thrust then enlarges that share further.

That is also why the amplifier in the next sections works as well as it does. A single-mode prediction is only as good as the claim that one mode carries the response, and here the response is one mode almost exactly.

The moment the dead load does not cause #

The live load’s bending is easy to find on the undeformed rib. For a two-pinned parabola with a uniform load $q$ on half the span, the crown free body gives the thrust as half the full-span value, $qL^2/16f$, and the loaded quarter point carries

$$M = \frac{qL^2}{64} = \frac{6 \times 60^2}{64} = 337.5 \text{ kN·m}$$

sagging, with the unloaded quarter hogging by the same amount. The dead load adds nothing to that on a rigid rib, and on this one adds the few kilonewton-metres that its axial shortening leaves behind.

The second-order solve puts the thrust on the displaced rib, and the peak rises to 519 kN·m — half as much again. Nothing about the live load has changed. What changed is that the rib carries 676 kN of thrust, 563 kN of it from the dead load, and that thrust acts on the displacements the live load produced, at a lever arm of the rib’s sideways and vertical movement. It is the load that makes itself worse in a curved member: the axial force times the displacement it is acting across is a moment, which bends the rib further, which lengthens the lever arm.

That is why the second refutation above is a good one to have in mind. The dead load bends the rib by 9 kN·m and raises its bending moment by 173. Designing the rib for the live load’s moment, on the grounds that the dead load is funicular, is right about the dead load’s moment and wrong about the rib’s.

The amplifier the eigenvalue predicts #

Every second-order argument in structures ends in the same factor, and the arch is no exception. For a column with an initial bow in the shape of its first mode, the bow is amplified by $1/(1 - P/P_{cr})$; the half-loaded rib’s bending is in the shape of its first mode, so the same factor ought to apply with the load in place of $P$ and the bifurcation load in place of $P_{cr}$. Written in loads, with $w$ the dead load plus the live load spread over the whole span,

$$\frac{M_2}{M_1} \approx \frac{1}{1 - w/w_{cr}} = \frac{1}{1 - 18/53.4} = 1.51$$

against 1.50 solved.

Swept over the dead load, the prediction follows the solved curve within two per cent, from no dead load at all to 43 kN/m, where the moment has grown sevenfold. The first-order moment barely moves over that range: it rises from 339 to 359 kN·m. That flat line is what a designer would draw by hand, and it is the one line in the figure that has nothing to do with what the rib does.

So the eigenvalue from the earlier essay turns out to be exactly what the rib needs, but not as a limit. It is the denominator of an amplifier. The storey that cannot see the building lean made the same point about a building, where the sway amplifier is taken from an elastic critical load that the frame will never reach either.

There is a detail in how the amplifier is written. The same prediction can be stated in thrusts, $1/(1 - H/H_{cr})$, with the thrust the rib actually carries against the thrust at bifurcation, and on an ideal inextensible parabola the two forms are the same thing. On a real rib they differ slightly, and the thrust form always reads the higher — by under one per cent at this rise, and by up to seven over the range of rises in the figure further down. It errs on the safe side, which is the most that can be asked of an approximation, and the load form is the one to use when it matters.

The same numbers by hand #

Each number above can be recovered in a few lines. The thrust under the dead load on the whole span and the live load on half is

$$H = \left(\frac{g}{8} + \frac{q}{16}\right)\frac{L^2}{f} = (1.875 + 0.375)\times\frac{3600}{12} = 675 \text{ kN}$$

against 676 solved, the difference being the rib’s shortening. The bifurcation load is the earlier essay’s coefficient, $46.2\,EI/L^3 = 46.2 \times 250{,}000/216{,}000 = 53.4$ kN/m. The uniform load equivalent to this pattern, for the amplifier, is $15 + 6/2 = 18$ kN/m, a third of it. And the amplified moment is $337.5 \times 1.51 = 510$ kN·m against 519 solved, the gap being the few kilonewton-metres the dead load leaves by shortening the rib, amplified with the rest.

That is the case for doing the amplification by hand in a preliminary design and by a second-order solve in a final one. The hand version needs the eigenvalue, which is a table lookup for a parabolic rib, and gets the moment to within two per cent.

The whole span loaded bends the rib least #

For a beam, the worst live load is usually the most live load. For an arch, the case with the most load and the least bending is the same case.

The first-order worst case is familiar from the influence line for a quarter-point moment on a two-hinged arch: load the stretch that produces sagging there and leave the rest empty, which is a little under half the span. Loading the rest adds thrust, which relieves the moment. The second-order curve is the first-order one multiplied by an amplifier that grows with the total load, so its peak moves a step towards the longer loaded length, and it is higher by half. Loaded over the whole span, the rib is back to being funicular, and bends by 13 kN·m.

This is pattern loading in its arch form, and it is where the design moments of an arch rib come from. The load case that tests the rib’s bending is the half-loaded one, and its second-order moment is what the rib is checked for.

Where the rise stops mattering #

The amplification is least where the bifurcation load is highest, and the earlier essay’s best rise, 0.275 of the span, reappears here as a best rise of 0.29 for the amplification. The flatness reappears too: anywhere from a fifth to two fifths of the span, the rib’s moment is within a few per cent of its best. A shallow rib is the one to worry about. At 0.12 of the span its amplifier is 1.83; at a twentieth, these loads are the rib’s bifurcation load itself, and there is no second-order state to draw.

The rib yields long before it would buckle #

The last figure is the one that changes the design. Give the rib a section — 12,000 mm² of steel, 1,000 mm deep, a section modulus of 2.38 × 10⁶ mm³ — and factor the dead and live loads together until the most stressed section reaches yield.

First order, the rib yields at a load factor of 1.73: at service the worst section, the loaded quarter point, carries 60 N/mm² of axial stress and 145 of bending, and both rise in proportion to the load. That rib passes a check at the usual factors of 1.35 on dead load and 1.5 on live — 300 N/mm² at the worst section, against 355.

Second order, it yields at 1.18. At the same factored loads the moment is 962 kN·m rather than 518, the amplifier having risen to 1.86 because the factored load is a larger share of the bifurcation load, and the worst section’s stress is 488 N/mm². The rib that passed its first-order check fails its second-order one by more than a third, and it does so at a load factor of 1.18, while the buckling load that the eigenvalue reports sits at 2.97, two and a half times higher.

That gap is the reason for the first refutation. A rib at a third of its buckling load sounds safe, and in the sense of bifurcation it is. The margin that matters is the one between the amplified stress and the yield stress, and a third of the buckling load has already multiplied the bending by one and a half. Strong enough and still falls over drew the column that buckles while its stress is still low; this rib is the other way round, and more typical of real arches, which fail by yielding in bending that the thrust has amplified.

What a designer can change #

The amplifier has two inputs and a designer controls both. The load in its numerator is mostly dead load, so a lighter deck lowers it. The bifurcation load in its denominator scales with the rib’s bending stiffness, so a stiffer rib raises it. Doubling the rib’s stiffness takes the bifurcation load to 107 kN/m and the amplifier from 1.50 to 1.20, and the peak moment from 519 kN·m to 423 — less of a fall than the doubled stiffness suggests, because the first-order moment, the 346 kN·m the amplifier multiplies, does not depend on the rib’s stiffness at all. The stiffer rib is also a stronger one, so the stress falls further than the moment; but the lever it pulls is the denominator, not the numerator.

Bracing is the third input, and it is the one the earlier essay said to use: anything that resists the crown’s sideways movement raises the antisymmetric bifurcation load without adding weight. On a deck arch the deck does this through the spandrel columns; on a tied arch, through the hangers.

What happens between the two #

The stress curve is not the rib’s collapse curve. Once the worst section yields it loses stiffness, the amplifier grows because the rib’s effective stiffness has fallen, and the rib collapses by a combination of plasticity and instability somewhere above a factor of 1.18 and well below 2.97. That is the same interaction as the column that was never straight — the column curve is a curve rather than the Euler hyperbola cut off at yield because the imperfection’s amplified bending and the axial stress share the yield stress — and an arch rib’s design follows the same logic: either a second-order elastic analysis with the stresses checked against yield, or a column check with an effective length, which packages the amplification into a reduction factor.

Where the model stops #

The rib is in its plane. Out-of-plane buckling, where the rib twists and moves sideways out of its own plane, is not in this model at all, and for a rib braced only at its ends it can be the lower mode.

The rib is perfect. A real rib has an out-of-straightness as well as a half-span load, and the two add; at the loads here the half-span load’s displacement is much the larger, which is why the fabrication tolerance matters less for an arch carrying a live load than for one carrying only its weight.

The loads keep their direction. The earlier essay found that a load which turns with the structure is worth 7.6 per cent to the buckle; for a deck on hangers the load hangs vertically whatever the rib does, which is the case modelled.

What the pictures cannot show #

That the dead load is the cheapest thing to change and the hardest to argue with. A heavier deck slab, a thicker surfacing, a service duct added late: each is funicular on its own, and each lowers the rib’s margin against the live load it carries. A design review that checks the arch for its dead load by looking for bending in the rib will find none, and will conclude that the extra weight is carried by thrust, which is true and not the point.

Still open: the tied arch as a system #

Everything here has pinned springings on rigid ground. A tied arch takes its thrust in a tie, hangs its deck on hangers, and the deck’s own bending stiffness resists the antisymmetric mode directly — the role the deck is not there to carry the load gave a suspension bridge’s deck against the same kind of partial load — so the half-span live load is shared between the rib, which amplifies it, and the deck, which does not. How that share moves with the ratio of deck to rib stiffness, and whether a stiff deck under a slender rib is a rib amplifying almost nothing or a deck carrying almost everything, is the tied arch’s own second-order question.