The rib that cannot lean on its own tie

The weight that bends a rib it cannot bend put a live load on half of a two-pinned parabolic rib and solved it to second order. Half a span of live load is a uniform load, which the parabola carries in pure compression, plus an antisymmetric load that bends the rib in very nearly the shape of its own antisymmetric buckling mode. So the rib never bifurcates; it bends from the first kilonewton, and the whole thrust, dead load included, amplifies the bending as an axial load amplifies a bowed column’s.

That essay ended on a tied arch, in which the deck hangs from the rib and closes its thrust, and asked how the half-span bending would be shared between the rib, which amplifies it, and the deck, which does not. The answer is that the rib does not amplify it either. The reason is a free body, and it is the most useful thing to know about a tied arch.

The same rib, free and hung #

The rib is the earlier essay’s: 60 m span, 12 m rise, a bending stiffness of $250{,}000\ \text{kN·m}^2$, under 25 kN/m of dead load over the span and 10 kN/m of live load on the left half. Standing free on pinned springings it carries the loads directly. Hung, it carries them through 23 vertical hangers at 2.5 m from a deck whose bending stiffness is a tenth of the rib’s, and the deck runs between the rib’s ends as its tie: pinned at one end, on a roller at the other, so the bearings carry only weight.

Both are solved as plane frames to second order — every member’s stiffness reduced by its own axial force, and the axial forces updated until they settle. The free rib’s numbers reproduce the earlier essay’s to the kilonewton-metre.

The free rib’s moment is 576 kN·m at first order and 1,303 at second: the thrust has amplified the half-span bending 2.26 times. The hung rib’s is 528 at first order — a little less than the free rib’s, because the deck takes a tenth of it — and 524 at second. The amplification has gone. It is not reduced; it is slightly below one.

Why the thrust has no lever arm to grow #

Cut the free rib at a section and take moments about the cut for the part to its left. The springing pushes in with the thrust $H$ along a line at the springing’s height, and the rib’s centreline at the cut is at height $y(x) + w(x)$, where $w$ is the rib’s deflection. So the moment in the rib is the moment a simple beam would carry, less $H$ times the rib’s height above the springing line — and that height includes $w$. Every millimetre the rib deflects is a millimetre of extra lever arm for the thrust, which is P-δ, and it is what the free rib’s 2.26 is.

Now cut the tied arch at the same section, through the rib and through the deck. The thrust is not reacted at a springing any more; it comes back through the deck, as the deck’s tension. The couple that the rib’s compression and the deck’s tension make has a lever arm equal to the distance between them — $y(x) + w_r(x) - w_d(x)$, with $w_r$ the rib’s deflection and $w_d$ the deck’s. The rib and the deck are joined by vertical hangers, so wherever a hanger is they deflect by almost the same amount, and the difference is the hanger’s stretch.

So the second-order term is $H(w_r - w_d)$, not $H w_r$. The thrust and the tie’s pull move together. The rib can deflect as much as it likes; if the deck goes with it, the couple’s lever arm does not change, and there is nothing for the thrust to amplify. What the free rib’s springings could not do — move down with the rib — the deck does.

The same statement can be made in energy, which is how the frame solution makes it. A member carrying a compression $N$ and bending into a shape $w$ loses stiffness by $\tfrac12 N\int w'^2\,dx$, which is what makes a column buckle; a member carrying a tension gains the same amount, which is what makes a stretched string hard to push sideways. The rib carries the thrust as compression and the deck carries it as tension, and the horizontal components of the two are equal. Where the hangers make the two deflect into the same shape, the rib’s loss and the deck’s gain are the same integral with opposite signs, and they cancel. All that is left to soften the system is the part of the rib’s deflection that the deck does not share.

The figure’s 0.99 is the same argument with the small terms kept. The deck in tension is a little stiffer against bending than its own stiffness says, and the rib’s slight extra lever arm, from the hangers’ stretch, is less than that; the net is a shade under one.

The critical load #

The same free body changes the buckling load, and by much more than the deck’s own stiffness could.

The free rib’s first mode is the sway the earlier essays drew: one half down, the other up, at 1.78 times the dead load plus half the live load. The hung rib’s first mode still has the rib swaying, but the shape drawn is a different object. To move sideways and down on one half, the rib has to drag every hanger out of the vertical, and each hanger is in tension, carrying its share of the deck: a tilted hanger in tension pulls the rib back towards the point above its foot, like a pendulum’s string. The deck, bending with the rib, is in tension too, which stiffens it. And where the rib and the deck move apart the hangers stretch. The hung rib buckles at 28.6 times the load.

The deck is not there to carry the load on a suspension bridge either, where its stiffness resists the cable’s antisymmetric mode directly. On a tied arch the deck is a tie first and a beam second, and its being a tie is worth far more than its being a beam: a deck a hundredth as stiff as the rib still gives a critical factor of 26.

How much of the bending the deck takes #

The deck’s stiffness still matters, for what it carries rather than for the rib’s stability.

The deck takes a share of the antisymmetric bending in proportion to its share of the combined stiffness, near enough: 6% with a deck a twentieth as stiff as the rib, 26% at a third, two-thirds at nearly twice the rib’s. It takes half at about 1.15 times the rib’s stiffness — not exactly at equal stiffness, because the rib is curved and longer than the deck, and its stiffness against the antisymmetric shape is a little more than its flat equivalent’s.

This settles the question the earlier essay put — whether a stiff deck under a slender rib is a rib amplifying almost nothing or a deck carrying almost everything. It is a deck carrying its share by stiffness, over a rib that amplifies nothing whatever the share. A designer can choose where the half-span bending goes — into a deep deck, as a Langer girder does, or into a deep rib with a light deck — and the choice is about where the steel is cheapest to put, not about stability. That is the same rule the stiffest path takes the load states for any two members in parallel; the tie has simply removed the one thing that would have made the rib a worse path than its stiffness says.

A deck as stiff as the rib #

The share is easiest to see with the two members equal.

With a deck as stiff as the rib, the antisymmetric bending divides almost equally — 289 kN·m in the rib, 318 in the deck — and neither is amplified. Together they carry 607 kN·m, a little more than the free rib’s first-order 576, because the deck’s bending is measured on a straight member and the rib’s on a curved one, and the two are not the same share of the same free body. The free rib, with all of the bending and all of the amplification, carries 1,303.

So the comparison a designer is really making is not between a rib and a deck. It is between a rib that has to be sized for 1,303 kN·m — more than twice its first-order moment, with the factor still growing as the load does — and a rib and a deck that between them are sized for 607, and whose sizes can be traded against each other freely. The deck was going to be there anyway; it has to span between the hangers and carry the traffic. Making it the tie, and hanging it vertically, buys the rest.

This is the arch’s version of what a column that was never straight does with a load that is held to its line: a bow amplified by an axial load is the free rib; the same bow with the axial load’s line of action forced to follow it is no amplification at all. A tie that moves with the rib is a line of action that follows the bow.

Soft hangers still hold #

The argument leans on the hangers, so the next question is how stiff they have to be.

Hangers a hundredth as stiff as a 50 mm rod — a rod of 5 mm, which no bridge would use — still give a critical factor of 7.5, more than four times the free rib’s, and still leave the half-span moment within 3% of its first-order value. A tenth as stiff gives 17.7, and stiffer hangers than the rod add little: the factor approaches the 32 or so that the pendulum action and the deck’s tension give with no stretch at all.

The reason the hangers matter so little to the amplification is that the half-span load pulls the rib and the deck in the same direction. The hangers carry the deck’s weight to the rib, so they are already in tension, and under the antisymmetric load the rib and deck deflect together by much more than the hangers stretch. The difference $w_r - w_d$ is a hanger’s elongation, a few millimetres, against the rib’s deflection of tens.

The load at which the free rib runs away #

The two ribs part company most clearly as the live load grows.

The free rib’s second-order moment curves away from its first-order line as the live load grows, because the live load adds to the thrust that amplifies it: at 40 kN/m on half the span the total is 45 kN/m averaged over the rib, close to the 53.5 at which the free rib bifurcates, and its moment is seven times the first-order one. The hung rib’s moment is a straight line in the load. A hung rib is a first-order structure in the plane of the arch, and its moments can be added from separate load cases as a beam’s can, which no free rib’s can.

The free body, by hand #

The lever-arm argument can be checked with the figure’s numbers. The thrust under the dead load and the half-span live load is about 1,150 kN. The free rib’s second-order moment exceeds its first-order one by $1{,}303 - 576 = 727$ kN·m, which is the thrust times an extra lever arm of $727/1{,}150 = 0.63$ m. The rib’s quarter point has moved 0.55 m down and 0.35 m sideways — the lever arm is the rib’s own displacement, and the displacement grows with the moment it causes, which is the amplification.

The hung rib deflects too — 239 mm at the quarter point under this load — and the deck beneath it 241 mm. Their difference, 1.8 mm, is a hanger’s stretch: a hanger 9 m long carrying about 90 kN, at 400,000 kN of axial stiffness, stretches by $90 \times 9/400{,}000 = 2$ mm. The thrust times 2 mm is 2 kN·m. That is the whole of the second-order term the hung rib can have, and it is smaller than the deck’s tension stiffening, which is why the figure’s ratio is 0.99 and not a little over one.

Where vertical hangers stop being enough #

The result is about the plane of the arch, with vertical hangers that stay in tension, and each of those limits it.

Out of plane, the hangers do nothing of the kind. A tied arch’s rib buckling sideways drags the hangers sideways too, and their tension pulls it back only as far as the deck is held laterally — which is the subject of the pendulum effect in out-of-plane design and is usually what governs the rib’s size.

Slack hangers lose it. A hanger that goes slack under a partial load stops tying the rib to the deck, and over that stretch the rib is free again. Vertical hangers under dead load rarely do; inclined hangers in a network arch can, and there the question becomes which of them are still working.

The rib is elastic. A rib near its strength yields at the quarter points, and its stiffness against the antisymmetric shape falls with the yield; the hung rib’s moments stay first-order, but the share between rib and deck moves towards the deck as the rib softens.

The rib’s own shortening. A rib that shortens elastically under its thrust loses some of that thrust to its own compression, and on a tied arch the deck’s stretch under the same force adds to it; the frame includes both, and on a flatter arch they are a larger part of the answer.

No snap. A shallow rib has a second way to lose its load, by snapping through symmetrically, and on a free rib below a rise of about a tenth of the span it is the one that governs. The rib here is at a fifth of its span and far from that; whether a deck hung beneath a flat rib suppresses the snap as it suppresses the sway — the deck would have to go down with the rib, which its own stiffness resists — is a separate calculation, not made here.

The rib between hangers. With the global mode suppressed, the next mode is the rib buckling locally between two hangers, as a column of the hanger spacing with the thrust in it. At 2.5 m it is far above anything here; for a rib with hangers at 10 m it is not, and it is the mode that the hangers’ spacing has to be chosen for.

The ends. The rib meets the deck at the bearings, and the deck’s end is where the tie force turns into the rib’s thrust. That joint carries the whole of $H$ through a knee that is neither a pin nor fixed, and its stiffness enters every number above — most of all the share of the bending the deck takes near the ends.

Still open: the hangers that are not vertical #

Vertical hangers make the rib and the deck deflect together at every hanger, and that is the whole of the argument. Inclined hangers — a network, or the Nielsen arrangement with a single slope — tie the rib to the deck along diagonals instead, so they also stop the rib and the deck sliding past each other along the span, and the structure becomes a truss in which the rib is a chord. That should suppress the second-order term more completely still, and raise the rib’s in-plane buckling load further; but a diagonal that goes slack under a partial load releases a panel of the truss at exactly the place the load is. Whether a network arch’s rib, with some of its hangers slack, is a first-order structure as a vertically hung rib is — or whether its second-order term comes back, panel by panel, wherever the hangers have stopped working — is the question of which hanger arrangement keeps the thrust’s lever arm fixed when the load is where it does the most.