The deck is not there to carry the load

A cable under load takes the funicular shape of that load, and the fact is usually stated as a virtue. It is the founding result of the structural-form field: a shape that carries itself does so in pure tension, with no bending anywhere, and the arch is its reflection. Stated as a limitation instead, the same sentence reads: the cable’s shape is decided by the load, so change the load and the shape changes. Under a single point load the funicular shape is two straight lines meeting at a corner.

A kink is not a road. Whatever else the deck of a suspension bridge is for, it is for turning that corner into a curve.

The system, which is a beam on a string #

Take the deck as a beam of flexural rigidity $EI$ and the cable as a string under a horizontal force $H$ that the dead load has already established. A load applied to the deck is shared between them, and the governing equation is one line:

$$EI\,v'''' - H\,v'' = p(x)$$

The first term is the beam and the second is the string. A string resists a change of shape with $Hv''$ — its restoring force is proportional to its curvature rather than to its displacement, which is what distinguishes it from a beam on an elastic foundation, where the restoring force is $kv$ and the equation is $EIv'''' + kv = p$.

Both are fourth-order and both have a characteristic length, and the two lengths behave completely differently. A Winkler foundation gives $\ell = (4EI/k)^{1/4}$; a tensioned string gives

$$\ell = \sqrt{\frac{EI}{H}}$$

which for this deck is 183 m — a fifth of the span. That is the distance over which the girder spreads anything applied to it, and it is the single number the whole subject turns on.

Which free body produced the number #

Cut the deck just to one side of the load and take the piece to the left.

Three things cross the cut. The girder’s shear, $-EIv'''$; the vertical component the hangers have delivered, which is the integral of $Hv''$ over the length; and, at the far end, the reaction. The load is shared between the second of those and the first, and the split at every wavelength is the ratio of the two terms in the equation.

That last observation is what makes the problem exactly solvable. Expand the load as a sine series, $p = \sum p_n \sin(n\pi x/L)$; then every term is uncoupled, because $\sin$ is an eigenfunction of both operators:

$$v_n = \frac{p_n}{EI\,(n\pi/L)^4 + H\,(n\pi/L)^2}$$

and the cable’s share of the $n$-th harmonic is

$$\frac{H k^2}{EI k^4 + H k^2} = \frac{1}{1 + (k\ell)^2}, \qquad k = \frac{n\pi}{L}$$

Long-wavelength load goes to the cable; short-wavelength load goes to the girder. For this deck the first harmonic has $k\ell = 0.64$ and the cable takes 71% of it; the third has $k\ell = 1.91$ and the cable takes 21%; the ninth has $k\ell = 5.7$ and the cable takes 3%.

A point load is all harmonics at once, and its sharp features are exactly the short-wavelength content — so the girder takes the corner and the cable takes the rest, which is the whole mechanism in a sentence.

What the girder actually buys #

Read down that table and the girder’s contribution to carrying rises from nothing to almost everything. Read the last column and something else happens: the deflection falls by a factor of thirteen, and most of the fall has happened by the time the cable is still taking 83%.

So the girder is worth having long before it is carrying much. At $\mu = 4.93$ it has removed 40% of the deflection and 83% of the kink while accepting 17% of the load, which is the regime a real suspension bridge is built in — the Humber’s deck is a 4.5 m box on a 1,410 m span, and its $\mu$ is of this order.

The load case that sizes it #

A uniform load over the whole span is very nearly what the cable’s shape already suits, and it produces an almost uniform additional sag with little curvature change — the cable simply deepens, in the way a funicular polygon redraws itself. Half a span loaded is antisymmetric, its content is concentrated in harmonics the cable is poor at, and it twists the cable into an S — so the girder has to supply the difference over the whole span.

Pattern loading is not a refinement here, it is the design case. The same is true of a continuous beam, and for the same reason: a structure whose response depends on the shape of the load has a worst shape, and it is rarely the one that is easiest to imagine.

The generalisation worth carrying #

The deck’s job — turn a concentrated action into a distributed one before handing it on — is a job many structures do, and naming it makes several of them the same structure.

A railway sleeper distributes a wheel load along the rail’s own beam-on-elastic-foundation length before it reaches the ballast.

A pad footing does the same for a column onto soil, over the characteristic length of a beam on the ground.

A floor slab distributes a point load to several beams over a width set by its own stiffness against theirs, which is the stiffest path taking the load in its most literal form.

And a raft does it for a whole building.

In every case the distributing member is not the load-carrying member, its stiffness sets a length rather than a capacity, and the useful question about it is “over what distance?” rather than “how much?”. The suspension bridge is the clearest example because the two functions are carried by two visibly different objects.

The comparison the two shapes make #

Setting the three limiting systems side by side is the fastest way to see what the arrangement is.

The cable alone deflects 1.873 units, has a kink, and takes 100% of the load. It is perfectly efficient and unusable.

The girder alone deflects 3.797 — twice as far — and takes 100% of the load. It is a plain 900 m beam, which is not a structure.

Together they deflect 1.125, which is 60% of the better of the two on its own, and share the load 83 to 17. The system is stiffer than either component because they resist different things: the girder resists curvature and the cable resists displacement, and the load’s content is split between the two.

A thrust line has to stay inside the material it runs through, which is why masonry is thick: a change of load moves the line, and an arch has no way of taking a moment except by having the line still inside the stone.

That pairing generalises. A structure that carries load by shape needs some way of tolerating a change of shape, and there are only two of them: mass, which is what a masonry arch uses, or bending stiffness, which is what a stiffening girder is. A prestressed cable net uses a third — opposing curvature — and is worth naming because it is the one that does not need either.

Reading the parameter #

μ = L√(H/EI) is the only number in the problem, and it is worth translating into things a reader can picture.

$H$ is set by the dead load and the sag: $H = wL^2/8f$ for a parabolic cable, so a sag of a tenth of the span gives $H = 1.25\,wL$. Substituting,

$$\mu = L\sqrt{\frac{wL^2}{8f\,EI}}$$

which says three things at once. Longer spans give larger μ — the cable dominates more, the deck matters less for carrying and more for spreading. Shallower cables give larger μ, because a shallow cable pulls harder for the same load. And a stiffer deck gives smaller μ, in the obvious way.

For the deck drawn here, μ is 4.93 and the characteristic length is a fifth of the span. A 19th-century bridge with an enormous stiffening truss might have μ near 1; a modern box-girder deck on a 2 km span has μ of 10 or more, and its deck is spreading a load over a few per cent of the span rather than a fifth of it.

The cables supply the torsional stiffness too #

Everything above is about vertical stiffness, and the deck has a second one that is decided the same way and matters more.

A suspension bridge has two cables, one each side of the deck at a spacing $b$. Twist the deck about its own axis and one cable is pulled up while the other is pushed down — so both are being asked to change their sag, and both resist with the same gravity stiffness that resists a vertical deflection.

The lever arm is what makes it large. Each cable acts at $b/2$ from the axis, so the torsional stiffness the pair supplies goes as

$$K_\theta \;\sim\; 2 \times H \times \left(\frac{b}{2}\right)^2 = \frac{H b^2}{2}$$

— the square of the cable spacing. A deck twice as wide, on the same cables at the same tension, is four times as stiff in torsion and no stiffer at all in bending.

That is the arithmetic behind the most famous failure in the subject. Tacoma Narrows had a deck 11.9 m wide on an 853 m span — a ratio of 1 in 72 — with a shallow plate girder that contributed almost nothing torsionally and cables so close together that their own contribution was small. Its vertical stiffness was adequate; its torsional stiffness was a fraction of what a modern bridge of that span carries, and the mode that destroyed it was a torsional one.

Every response since has been an attack on the same term. Wider decks, which raise $b$. Trusses instead of plate girders, which are torsionally far stiffer as sections. Closed box decks, which are hundreds of times stiffer in torsion than an open section of the same material. And cross-bracing between the two cables, which ties them together so that neither can move without the other.

None of those does anything for the vertical behaviour this essay has been computing. They are answers to a second stiffness that the same cables happen to supply, in a mode the vertical calculation contains no term for.

The side spans are part of the same cable #

The model treats one span. A suspension bridge has three, and they are joined by a cable whose total length is very nearly fixed — which couples them in a way no single-span analysis can produce.

Load the main span and it sags. Sagging lengthens the main span’s cable, and the extra length has to come from somewhere: it is drawn out of the side spans, which therefore rise. So a live load on the middle of the bridge lifts its ends, and the amount depends on the side spans’ geometry rather than on anything about the main span.

What decides whether that happens is a detail at the top of the tower. If the cable sits in a sliding saddle, the horizontal force is equal on both sides by definition, the cable moves over the tower, and the coupling is complete. If the saddle is clamped, the cable cannot move, the two spans have different horizontal forces, and the difference is a horizontal force on the tower head that the tower has to carry in bending.

Both are built. A sliding saddle keeps the tower in pure compression — which is what a tower is good at — and hands the coupling to the cable. A fixed saddle keeps the spans independent and hands the difference to the tower. Modern practice usually fixes the saddle and jacks it into position during construction so that it starts balanced under dead load, which is a third answer: fixed in service, adjustable while the shape is being set.

The consequence for the arithmetic on this page is that the parameter $\mu$ is a property of the main span alone and the real behaviour is not. A short stiff side span raises the whole system’s stiffness and a long flexible one lowers it, and neither appears anywhere in the equation the deck’s own girder is sized from.

The same coupling explains a construction sequence that looks superstitious. A suspension bridge’s deck is erected in segments hung from the cable, and the order they are hung in decides the shape the cable takes as it goes — because every segment added to the main span draws cable out of the side spans and moves everything already placed. Erecting from the towers outward, from the centre outward, or symmetrically in pairs are three different geometries arriving at the same finished bridge, and the segment joints are welded or bolted at the moment when the shape happens to be right. It is a staged structure whose stages are a single continuous curve rather than a set of members.

Where the model stops #

$H$ is held constant, and it is not. In real deflection theory the cable’s horizontal force increases when live load is added, and that increase is what makes a full-span uniform load so cheap — the cable simply deepens. This model cannot represent it, so it under-credits the cable for full-span loading and the two load cases come out closer together here than deflection theory puts them.

Everything is linear. Suspension-bridge behaviour is geometrically nonlinear by nature: the stiffness depends on the tension and the tension depends on the deflection. The linearised equation is the first term of that and is excellent for small live-load ratios.

The hangers are treated as a continuous connection. They are discrete, they can only pull, and a bridge under a severe antisymmetric load can slacken the hangers on the unloaded half — at which point the deck is a plain beam over that length.

Nothing here is dynamic. A suspension bridge’s most famous problem is aeroelastic, and the stiffening girder’s torsional stiffness — not touched anywhere above — is what decides it. Tacoma Narrows had an adequate vertical stiffening girder and a catastrophic torsional one.

And the cable’s own weight is absent. It sets the dead-load shape, it sets $H$, and it is the reason a real bridge’s cable is a catenary rather than a parabola — a distinction that matters for the geometry and not for anything on this page.

What the pictures cannot show #

The deflected shapes are drawn at an exaggeration and the kink is drawn as a sharp corner, which is the model’s answer rather than a bridge’s. A real cable has bending stiffness of its own, small but not zero, and the corner is rounded over a length of a few metres — which is negligible structurally and is not negligible if the reader is being invited to believe a picture.

Nothing in the figures shows a hanger, and the hangers are what make the whole arrangement a single system. Drawn as a continuous connection, they suggest the deck is glued to the cable; drawn discretely, at 20 m centres on a 900 m span, they would show that the “distributed” transfer is 45 point loads.

And the sharing curve has $\mu$ on its horizontal axis, which is a number no reader has an intuition for. The useful translation is the last column of the table: a bridge whose girder spreads a load over a fifth of its span is a bridge with a real stiffening girder, and one that spreads it over a hundredth has a deck rather than a girder.

The ladder from here #

Later rungs on this anchor: the deflection theory proper, with $H$ as an unknown determined by the cable’s own extensibility, and the difference between it and the elastic theory that preceded it — Melan’s equation and the reason the Brooklyn Bridge was designed by a method that made it several times heavier than it needed to be. Torsional stiffness of the deck and the flutter speed, which is the criterion that actually governs a long-span bridge. The self-anchored suspension bridge, where $H$ is delivered into the deck rather than into the ground and the deck is in compression. Cable-stayed bridges, where the stays are inclined and the deck is a beam on a set of elastic springs rather than a string. Hanger slackening and the nonlinear analysis it requires. And the cable-stiffened roof, where the same equation governs and the load that matters is uplift.

The thing worth carrying away is the reframing rather than any number. A structure’s most important member is not always the one carrying the most force, and asking “what fraction does it carry?” of the stiffening girder gets an answer — 17% — that is close to useless. Asking “over what length does it spread?” gets 183 m, and that is the number the bridge was designed around.