The thrust that never reaches the ground

An arch works by pushing outwards. That is not a side effect of the shape, it is the shape: the funicular of a uniform load is a parabola, and the horizontal component of the force running along it is constant and equal to $wL^2/8f$ everywhere.

Which means the whole design of an arch is a question about what happens at the springings, and for most of the history of the form the answer was rock. A tied arch answers it differently. Run a bar between the two springings, let the thrust close on itself through that bar, and the bearings see nothing but weight — so the whole assembly reacts like a simply supported beam and can sit on two pads, or on a barge, or be lifted in overnight.

Which free body produced the number #

Cut the tie. What is left is an arch on a pin and a roller — a curved simply supported beam, statically determinate, carrying the load entirely in bending because nothing is stopping the springings from spreading.

So the tie force is the single redundant, and the force method gives it directly. The tie stretches by $X L/EA_t$; the arch, pushed apart by a unit pair, spreads by $\int y^2\,ds/EI$ plus a smaller axial term; and the load, acting on the cut structure, spreads it by $\int M_0 y\,ds/EI$. Setting the total movement to zero,

$$H = \frac{\displaystyle\int \frac{M_0 y}{EI}\,ds}{\displaystyle\int \frac{y^2}{EI}\,ds + \int \frac{\cos^2\phi}{EA}\,ds + \frac{L}{EA_t}}$$

With a rigid tie and an inextensible rib that expression returns $wL^2/8f$ exactly, which is worth pausing on: the funicular thrust, arrived at from compatibility rather than from statics, with no assumption anywhere that the arch is carrying no bending. It comes out that way because a parabolic arch under a uniform load has $M_0 \propto y$, and the integral in the numerator becomes the integral in the denominator times $wL^2/8f$.

Three flexibilities, and the tie is the smallest of them #

The denominator has three terms, and reading their sizes is the whole design.

For the arch here the bending term is $1.11\times10^{-3}$, the rib’s own axial shortening is $1.03\times10^{-5}$, and the tie is $3.16\times10^{-5}$. The tie contributes 2.7 per cent of the flexibility, and takes 2.7 per cent off the thrust.

That is the sense in which a tied arch is “nearly” a true arch, and it is also the trap. The tie is three times more flexible than the rib’s own shortening — the term nobody computes for a true arch either — so the ordinary shortcut of taking $H = wL^2/8f$ is out by three per cent here, and everything it is out by lands in one place.

Whatever thrust the arch does not get, it carries as a moment #

That is the sentence the whole form turns on, and it is exact rather than approximate. The moment in the rib at any station is

$$M(x) = M_0(x) - H\,y(x)$$

so if $H$ were the funicular value the moment would be zero everywhere. It is not, and the shortfall multiplies the arch’s own ordinate: 61 kN of missing thrust times a 9 m rise gives 550 kNm at the crown, and the computed peak is 739.

Against the free moment $wL^2/8$ of 20,250 kNm, that is 3.6 per cent — small, and not nothing. It is the moment the rib is sized for, and it exists only because the tie stretched 69 mm.

Make the tie ten times softer and the arithmetic runs away in exactly the way the expression predicts: the thrust falls to 1,738 kN and the peak moment rises to 4,608 kNm — a factor of six in the design moment for a factor of ten in a bar nobody looks at twice.

The rise ratio decides everything, including the tie #

$H = wL^2/8f$ is a division, and the thing being divided by is the rise. Halve the rise and the thrust doubles; and the tie, whose force is the thrust, doubles with it.

At the 0.15 rise ratio drawn — 9 m on a 60 m span — the tie carries 2,168 kN, which at 234 N/mm² is a bar of 9,300 mm². At a rise ratio of 0.075 the same load needs a tie of nearly twice that, and at 0.05 the tie is the largest member in the structure by a wide margin.

That is the trade a tied arch makes visible in a way a true arch does not. A shallow true arch delivers its enormous thrust to a foundation, where it is somebody else’s problem and is usually absorbed; a shallow tied arch delivers it to a bar, where it has to be paid for in steel at a rate proportional to $1/f$. The tie turns the rise ratio from an architectural decision into a priced one.

The rib is still a column #

Removing the foundation problem does not remove the other one. An arch rib carries its thrust as an axial compression along its whole length, and a slender member in compression along a curve is a column that happens to be curved.

For a tied arch the in-plane buckling question is nearly the same as for a two-hinged arch of the same geometry, because the tie holds the span in both. The out-of-plane question is not: a tied arch is often a pair of ribs leaning inwards with bracing between them, or a single rib with nothing at all, and out of plane it has the tie for company at deck level and free length above.

The governing check on most tied arches is therefore lateral rather than in-plane, which is not obvious from anything in the two-dimensional analysis above and is why the ribs on real ones lean towards each other.

The load case the tie cannot help with #

Under a uniform load the arch is close to its funicular and the tie is doing something simple. Put the load on half the span and neither is true.

$M_0$ is no longer proportional to $y$, so no value of $H$ makes the moment vanish, and the best the tie can do is remove the average. The computed peak moment goes from 739 kNm to 2,812, at the quarter point rather than the crown, and the thrust halves to 1,084 kN because half the load produces half the numerator.

A tied arch is therefore designed by its unsymmetric load case, not by its full one — which is the same discovery a continuous beam makes about pattern loading, reached from the other direction. The full load is the case the shape was chosen for, and the case the shape was chosen for is never the one that governs.

The structural response is the network arch: hangers that cross each other instead of hanging vertically, so that an unsymmetric load is carried by a truss action between rib and tie rather than by bending in either. That converts the problem from one about moment into one about which hangers go slack, which is a different and more tractable question.

The comparison that is the reason for the whole form #

The expression for $H$ prices something else, for free: a support that moves.

For a true two-hinged arch, a spread $\delta$ at the springings takes $\delta$ divided by the arch’s own flexibility off the thrust. For the steel rib here that is 22 kN for a 25 mm spread — one per cent, and no cause for concern.

Make the same arch a hundred times stiffer, which is what building it out of stone does, and the same 25 mm takes 100 per cent of the thrust away. The arch does not lose a per cent of its thrust; it stops being an arch.

That is the finding worth carrying furthest here, and it is the opposite of the intuition. Vulnerability to foundation movement is not a property of slenderness — it belongs to stiffness, because a stiff structure develops large forces for small movements, and the force it develops for a movement it did not want is the force it loses.

And a tied arch is immune to all of it. The tie holds the span, the bearings are free to slide, and a foundation that moves 25 mm — or 250 — changes nothing at all in the structure above it. That is the argument for the form on soft ground, and it is a different argument from the one about being able to lift it into place.

Where the model stops #

The analysis is linear and the geometry is the undeformed one. A tie stretching 69 mm on a 60 m span lowers the crown, which reduces the rise, which raises the thrust required, which stretches the tie further. The second-order correction is a per cent or two here and grows with the tie’s flexibility; a network arch with a slender tie deserves a geometrically nonlinear check.

The hangers are ignored entirely. They are treated as delivering the deck load to the rib as a smooth line load, which is right for many closely spaced hangers and wrong for the eight or ten a real bridge has. Discrete hangers put a sawtooth of local bending into both rib and tie, and a hanger that goes slack changes the structure rather than the loading.

The tie is treated as an axially loaded bar. On most tied arches it is the deck, which is also carrying traffic in bending and is far stiffer than a bar of the same area — a deck that stiffens the structure it is hung from is the whole of the stiffened-girder argument, and it applies here with the tension added.

And the rib is prismatic. Real ribs are deeper at the springings, which raises $\int y^2/EI$ where $y$ is smallest and changes the numbers by a few per cent in a direction the integral will take without complaint.

The member with no second chance #

The tie carries 2,168 kN in tension, permanently, and it is the only member in the structure whose loss is unsurvivable. Everything else in a tied arch has an alternative path; the tie has none, because it is the alternative path — it exists to close a force that would otherwise go to the ground.

Two consequences follow, and neither is in the analysis.

It is a fatigue detail rather than a strength one. A tie under permanent tension with traffic on the deck sees a stress range at every axle, at a connection to the arch springing that is one of the most heavily worked details on any bridge. The load that never came near failing anything is the governing case for it, and the calculation that decides its size is a count of cycles rather than a strength check.

And it is made of several ties. Real tied arches use multiple strands, or a box girder acting as the tie, or a deck in parallel with a bar — not because the arithmetic asks for it, but because a single-load-path member carrying two hundred tonnes for a hundred years is a category of thing that engineering does not build. That decision is invisible in every equation on this page.

What the pictures cannot show #

The thrust in the first figure is a number on an axis. What it is on site is a force of 2,168 kN — two hundred and twenty tonnes — running along a bar at the deck, in tension, permanently, for the life of the structure.

That is a member with no redundancy. Every other force in the arch has somewhere else to go if a member is lost; the tie does not, and a tied arch with a single tie is a structure with a single point of failure that no diagram on this page draws attention to. Real ones have multiple ties, or a tie made of many strands, or a deck acting as the tie in parallel with it — and the reason is not visible anywhere in the analysis, because the analysis has no term for what happens if the answer stops existing.

The assumption the figure rests on #

The tie is assumed to be straight, and to stay straight.

It is the one assumption here that gets broken routinely. A tie at deck level carries the deck’s own weight between hangers, so it sags; sag reduces its axial stiffness by the geometric mechanism a cable’s stiffness is made of; and reduced axial stiffness raises the flexibility term that this essay has shown is the whole design. On a bridge where the tie is the deck, the sag is negligible and the assumption is safe. On a roof truss where the tie is a slender rod spanning forty metres between two ends, it is not — and the honest model is not a bar with a lower modulus but a cable, whose stiffness depends on the tension the analysis is trying to compute.