The cable that is a spring

A suspension bridge and a cable-stayed bridge look like variations on one idea and are not. They carry load by mechanisms that have almost nothing in common, and the difference shows up in every quantity either of them is designed by.

A suspension cable is a funicular. It has no bending stiffness, it takes the shape of whatever is on it, and its stiffness against a change of load comes entirely from having to change that shape — which is why the tension already in it decides how stiff it is and why the deck’s job is to spread a point load out until the cable’s shape is right for it.

A stay is none of that. It is straight, it reaches the deck at one point, and it is an elastic support. So the deck is a continuous beam on springs, and the whole of the design is what those springs are worth and what they cost.

Which free body produced the number #

Cut one stay and take the deck node it was holding. The stay pulls along its own line with force $T$; resolving vertically, the share of the deck load it lifts is $T \sin\alpha$.

Now let that node move down by $\delta$. The stay’s extension is not $\delta$ — it is $\delta \sin\alpha$, the component of the movement along the stay. So the force change is $(EA/L_s)\,\delta \sin\alpha$, and the vertical component of that is $(EA/L_s)\,\delta \sin^2\alpha$. The vertical stiffness the stay offers the deck is therefore

$$k = \frac{EA}{L_s}\sin^2\alpha = \frac{EA\sin^3\alpha}{h}$$

using $L_s = h/\sin\alpha$ for a stay from a tower anchorage a height $h$ above the deck. Two powers of the sine because the geometry appears twice — once in what the stay resists and once in how much it is stretched — and the third from converting the stay length into a tower height.

Written against how far the stay reaches along the deck, that is

$$k = \frac{EA\,h^2}{(h^2 + x^2)^{3/2}}$$

which is a cube, not a first power. A stay reaching twice the tower height along the deck is not half as stiff as one reaching the tower height; it is one part in 11.2.

That cube is why a stay curtain looks the way it does. The outer stays are far bigger than the inner ones, and they are still far softer, so the deck’s moments are largest at the outer end of the curtain and it is the outermost stays that decide the deck depth.

The bill the vertical component does not include #

Every stay pulls the deck towards the tower as well as up, and nothing cancels that until the far anchor pier. So the horizontal components accumulate:

$$N(x) = \sum_{\text{stays outboard of } x} V_i \cot\alpha_i$$

and at the tower the total is the entire deck load times a mean cotangent. For the fan drawn above — a 200 metre half-span, a tower 70 metres above the deck — that is 1.55 times the whole load the stays are lifting.

A stayed deck is a strut before it is a beam. That is the real cost of a flat stay, and it is the reason a cable-stayed deck is a closed box rather than a pair of plate girders: the box provides the area and the buckling stiffness for a compression that no beam calculation would have predicted.

It also explains something about the two classical arrangements. A fan anchors every stay at the tower top, so the outer stays are as steep as the geometry allows and the deck compression is as small as it can be — but every stay force has to be transferred through one crowded region of the tower head. A harp runs the stays parallel, spreading their anchorages down the tower, which makes the head a simple detail and the outer stays much flatter. It costs the deck about eighty per cent more compression for that convenience, and it costs the stays themselves more steel, because a flatter stay carrying the same vertical share carries more force over more length. Most real bridges are a semi-fan — anchorages spread over the top third of the tower — which is a compromise between an arithmetic that wants a fan and a fabrication that wants a harp.

The tower height that costs least, which is a derivative #

The geometry that decides all of this is one number: how tall the tower is above the deck. It is worth deriving rather than adopting.

Take a stay curtain over a half-span $L$ carrying $w$ per unit length, with stay steel at $\sigma_s$ and deck and tower material at $\sigma_d$. Each of the three bills integrates in closed form:

$$V_{stays} = \frac{w}{\sigma_s}\left(hL + \frac{L^3}{3h}\right), \qquad V_{deck} = \frac{w}{\sigma_d}\cdot\frac{L^3}{3h}, \qquad V_{tower} = \frac{2wLh}{\sigma_d}$$

The stay term and the deck term both fall as the tower rises — steeper stays are shorter for the load they carry and pull the deck less — and the tower term rises. The sum has a genuine minimum, and differentiating gives

$$h^* = \sqrt{\frac{(L^3/3)(1/\sigma_s + 1/\sigma_d)}{L/\sigma_s + 2L/\sigma_d}}$$

For ordinary stay and deck stresses that lands near $0.4 L$. Real cable-stayed bridges use 0.2 to 0.3, and the gap is the honest part of this arithmetic. There is no tower buckling in it, no fatigue limit on the stay stress, no erection, no navigation clearance and no wind. Every one of those pushes the tower down rather than up, and together they cost perhaps five per cent of the material the optimum would have saved. It is a good example of an optimum that is worth deriving precisely so that the reasons for not using it can be named.

The stiffness that depends on the tension already there #

There is one respect in which a stay is not a simple spring, and it is the respect it shares with a suspension cable.

A stay sags under its own weight, so part of any extension is taken up by straightening rather than by stretching, and the effective modulus is less than the steel’s. Ernst’s correction writes it as

$$E_t = \frac{E}{1 + \dfrac{\gamma^2 L_h^2 E}{12\sigma^3}}$$

with $\gamma$ the specific weight, $L_h$ the horizontal projection and $\sigma$ the stress already in the stay. The $\sigma^3$ is the point. At the working stress a 200 metre stay has lost about one per cent of its modulus; at a third of that stress it has lost a third of it.

So a stay is soft when it is slack and stiff when it is tight, and the live-load stiffness of the whole bridge depends on how much dead load the stays are carrying. That is a design variable rather than a fact: the stays are stressed during erection to a level chosen partly for the geometry and partly to keep this correction small. It is the same relationship a prestressed cable roof has with its pretension, arriving through a different mechanism.

What the springs cost the deck #

The most useful single number in this essay is a comparison. Analyse the deck as a continuous beam on rigid supports at the stay points, and then as a beam on the springs the stays actually are. The peak sagging moment goes up by a factor of about 2.7.

That factor is what makes a stayed deck a real structure rather than a slab hanging from cables. It is also why the stay spacing is what it is: closer stays give shorter deck spans, but also smaller and softer stays, and the two effects fight. Modern practice puts them at eight to twenty metres, close enough that the deck spans between them almost as a slab and nearly all of the deck’s depth is there for the compression rather than for the bending.

The trap in the arithmetic is the one every beam on flexible supports has: the support stiffnesses are not properties of the supports alone. A stay’s stiffness depends on the tower’s, and the tower deflects towards the loaded side, which softens every stay on that side at once. Analysing the deck against fixed spring constants is a first approximation that a real analysis has to leave behind.

The form that had to wait for a computer #

Cable-stayed bridges are old as an idea and new as a practice, and the gap between the two is a calculation.

The idea is at least as old as Löscher’s timber proposal of 1784 and appears in a patent by Poyet in 1821. Several were built in the nineteenth century and several fell down, including two of Navier’s own examples — which is why Navier, having studied them, recommended against the form in his 1823 report and effectively closed it for a century. The suspension bridge, which he recommended instead, went on to the Menai and the Brooklyn.

What was wrong was not the mechanism but the analysis. A stay curtain with $n$ stays is a structure with $n$ redundancies: cut every stay and a simply supported deck remains, so the forces in them cannot be found from equilibrium at all and have to come from compatibility. With three stays that is an afternoon’s work by the flexibility method; with twenty it is not work anybody was going to do by hand, and the nineteenth-century designs were built on the assumption that each stay carried its tributary load — which is the assumption that a beam on springs never satisfies.

Franz Dischinger’s Strömsund bridge of 1955 is the usual starting point for the modern form, and what had changed by then was not the steel. It was that the redundant analysis could be done. The form arrived in quantity in the 1960s alongside the matrix stiffness method, and it spread because a deck on twenty springs stopped being a research problem and became a run.

There is a second reason it spread, and it is about erection rather than analysis. A suspension bridge needs its main cable spun before any deck exists, which needs a catwalk, which needs the towers and the anchorages complete. A stayed bridge grows outward from each tower with each segment hung as it arrives, so it needs no falsework and no temporary works over whatever is being crossed. The form is easier to build than to calculate, which is exactly the wrong way round for the nineteenth century and the right way round for the twentieth.

The anchor span, which is the part nobody draws #

Every figure in this essay is half a bridge, and the half that is missing is where the load actually goes.

The stays on the main span pull the tower towards midspan. Something has to pull it the other way, and that is the back-stay curtain running to the anchor pier — which is why a cable-stayed bridge has a short, heavily loaded span on each side of its tower and why those spans are often ballasted or tied down. The anchor pier carries uplift under live load on the main span, and the tie-down at that pier is one of the more heavily worked details on the whole bridge.

The arrangement decides how the tower is loaded. If the back-stays are anchored at a pier, the tower sees almost pure compression and the horizontal components cancel at its head; if they instead run to intermediate points on the back span, the tower bends. Between the two lies most of the difference between a slender tower and a stocky one, and none of it appears in the material optimum derived above, which quietly assumed the tower carried axial load only.

Where the model stops #

The bridge is never in the state it is analysed in. A stayed deck is built cantilever by cantilever from the tower, each segment hung on its stay as it goes, and each stay stressed to a value chosen so that the finished geometry comes out level. The completed structure has a locked-in state of stress that no analysis of the completed structure can produce, and the stay forces in the finished bridge are a design choice rather than a result. That is the largest thing this essay leaves out and it is the largest thing the design actually consists of.

The deck’s compression makes it a second-order problem. A deck carrying axial force and bending together is a beam-column, and the axial force amplifies the moments. Everything above is first-order and the correction on a long span is not small.

And nothing here is dynamic. A stay is a taut string with a natural frequency of its own, low enough to be excited by wind and by traffic through the deck, and stay vibration is the single most common maintenance problem this form has. Rain running down a stay changes its section and can drive it unstably — the mechanism that puts dampers and helical ribs on stays everywhere. None of that is visible in a stiffness.

One more consequence deserves naming, because it is the reason a stayed bridge behaves so differently from a suspension one under an unbalanced load. Loading half the main span pulls the tower over towards the loaded side; the back-stays take up the pull, so the tower’s movement is resisted by their axial stiffness rather than by its own bending. The whole structure’s response to an unsymmetrical load is decided by the back-stays, and a bridge with generous back-stays is stiff in a way its main-span geometry does not show. That is the opposite of a suspension bridge, where half-span loading is the case that sizes the stiffening girder and the cable’s shape does the resisting.

The generalisation #

The idea worth carrying out of this essay is smaller than the bridge. An inclined tie is a spring whose stiffness goes as the cube of the sine of its angle, and whose horizontal component has to be carried by whatever it is tied to.

That statement is about a bridge here, and it is the same statement about a guyed mast, whose guys are the springs and whose mast carries the accumulated verticals as compression. It is the same statement about an outrigger truss, whose columns are the ties. It is the same statement about a bracing member in any frame, and about a hanger in a tied arch.

In every one of them the useful component and the awkward component are two resolutions of the same force, and the awkward one accumulates. Counting it is what turns a sketch of a load path into a design.