The arch that gets shorter

An arch shaped to its load carries no bending. A parabola under a load uniform along the horizontal is exactly that: at every section the thrust line coincides with the axis, the eccentricity is zero, and the bending moment is zero along the whole span.

That is the ideal, and it is worth being precise about how ideal it is. For the 60 m arch drawn here, carrying 60 kN/m over a 6 m rise, the rigid solution gives a thrust of 4,500 kN and a crown moment of zero to within the arithmetic. Not small; zero, by construction, because the shape was chosen to make it so.

Then the arch carries that thrust, and a member carrying 4,500 kN of compression gets shorter.

The flexibility equation has two terms #

A two-hinged arch is indeterminate to the first degree, and the redundant to choose is the horizontal thrust. Release it — put the arch on a roller at one springing — and the released structure spreads. Apply a unit horizontal force and it pulls back. The thrust is whatever makes the two cancel:

$$H = \frac{\displaystyle\int \frac{M_0 y}{EI}\,ds}{\displaystyle\int \frac{y^2}{EI}\,ds \;+\; \int \frac{\cos^2\theta}{EA}\,ds}.$$

The numerator is the spreading the load causes. The first denominator term is the pulling-back a unit thrust achieves by bending the released arch, and it is the one every textbook writes.

The second term is the pulling-back it achieves by shortening the rib, and it is the one that gets dropped. It is there for a simple reason: a rib pushed at both ends gets shorter, its ends move toward each other, and the abutments therefore do not have to push as hard as they would on a rigid one.

The ratio is two lengths #

The relative size of the two denominator terms is worth doing in symbols, because the answer is a shape rather than a number.

The bending term is of order $L f^2 / EI$ — the arch’s own rise squared, integrated along it. The axial term is of order $L/EA$. Their ratio is

$$\frac{\text{axial}}{\text{bending}} \approx \frac{15}{8}\left(\frac{i}{f}\right)^2, \qquad i = \sqrt{I/A}.$$

The square of the radius of gyration over the rise. No load, no span, no material. For the arch drawn, $i = 143$ mm against a rise of 6,000, so $i/f = 0.0239$ and the axial term is 0.103 per cent of the bending one. The thrust lost is a little more than that, 0.109 per cent, because the rib with its thrust released — a curved beam on a roller — is itself shortened by the part of its shear that runs along it, and that strain is symmetric about the crown and does not cancel.

That is why the term is dropped, and it is a good enough reason for the thrust. It is not a good enough reason for anything else.

A tenth of a per cent is all of the moment #

The thrust falls from 4,500 kN to 4,495.1. That 4.9 kN does not disappear; it turns into bending.

The moment at any section is $M = M_0 - H y$. With the funicular thrust, $M_0$ and $Hy$ cancel exactly at every point. Reduce $H$ by 4.9 kN and the cancellation fails by $\Delta H \cdot y$, which at the crown is $4.9 \times 6$ — 29.3 kNm.

A 0.11 per cent change in the thrust is a 100 per cent change in the moment.

That is the whole page, and it is a general lesson about differences rather than about arches. When a quantity is computed as the difference of two nearly equal terms, a negligible error in either is not negligible in the result. The arch is the cleanest example in this collection because the difference is exactly zero in the ideal case, so any correction at all produces the entire answer.

Which free body produced the number #

Take the whole arch and release the horizontal restraint at one springing, so that it is a determinate curved beam on a pin and a roller.

Under the applied load that released structure spreads: the roller moves outward by an amount that is the first integral in the equation, computed by virtual work with a unit horizontal force at the roller. Now apply the real thrust $H$ inward. It closes the gap two ways at once — by bending the rib, which is the $y^2/EI$ term, and by compressing it, which is the $\cos^2\theta/EA$ term.

The compatibility condition is that the roller ends up where the abutment is. Both mechanisms contribute to closing the gap, so both belong in the denominator, and dropping one means the calculation believes the arch is stiffer than it is — which makes the computed thrust too large.

The free body is the released arch and the equation is compatibility rather than equilibrium, which is why the term is invisible to anybody checking the arch by statics. Equilibrium is satisfied by any thrust; only compatibility picks the right one.

The three lengths in the ratio #

It is worth being explicit about which lengths appear in $(i/f)^2$, because two of them are not the ones a reader expects.

The span does not appear. A 30 m arch and a 300 m arch of the same rise ratio and the same slenderness lose the same fraction of their thrust. That is a genuinely surprising result for a subject in which almost everything scales with span, and it comes from the fact that both integrals grow with the arc length in the same way.

The rise appears squared, and it is the rise rather than the rise ratio: the term is $(i/f)^2$, not $(i/L)^2$. So a long shallow arch and a short shallow arch of the same absolute rise behave the same, which is another way of saying the same thing.

And the radius of gyration appears squared, which is where the counter-intuitive design consequence comes from. Making a rib deeper increases $i$, which increases the ratio, which increases the sensitivity. A slender rib is less troubled by rib shortening than a stocky one — because the correction is a comparison between the rib’s axial and flexural stiffnesses, and a stocky rib’s flexural stiffness is relatively larger.

The word “buckle” is doing something specific there, and it is worth drawing, because the failure is not the one the arithmetic above suggests. A rib carrying 4,500 kN of compression does not squash.

On this rib the trade has already been lost in one direction. Its coefficient of 29.1 is, with a bending stiffness of $420{,}000\ \mathrm{kN\,m^2}$, a buckling load of $29.1 \times 420{,}000/60^3 = 56.6$ kN/m — less than the 60 kN/m it is drawn carrying. As a two-hinged arch this rib would not stand under its own design load. Built fixed at the springings, as concrete ribs of this proportion are, its buckling load doubles to 119 kN/m, and the correction this page computes changes shape with it.

The two modes matter here because they decide which stiffness the slenderness argument is really trading against. The rib is being asked to be flexurally soft, so that $(i/f)^2$ stays small and the shortening correction stays small; and it is being asked to be flexurally stiff, so that the antisymmetric mode stays out of reach. Both demands are on $I$, and they point opposite ways.

Shallow arches feel everything #

The ratio $(i/f)^2$ has the rise in the denominator, so a shallow arch is in a different regime.

At a two per cent rise the same arch loses 2.6 per cent of its thrust to shortening — twenty-five times as much — and the crown moment goes from 29.3 kNm to 705. That is not a correction any more; it is the design.

Two effects compound. The thrust itself is larger, because $H = wL^2/8f$ and the rise is in the denominator. And the fraction of it lost is larger, because $(i/f)^2$ has the rise squared in the denominator. A shallow arch is a structure carrying an enormous force with almost no lever arm, and every small thing that happens to that force becomes a large moment.

Temperature and shrinkage come through the same door #

Everything that changes the length of the rib enters the same equation with the same denominator, which means the arch that is sensitive to one is sensitive to all of them.

A temperature rise expands the rib and pushes the springings apart, so the abutments push back harder and the thrust rises:

$$H_t = \frac{\alpha\,\Delta T \displaystyle\int \cos\theta\,ds}{\displaystyle\int \frac{y^2}{EI}ds + \int \frac{\cos^2\theta}{EA}ds}.$$

For the arch drawn, 20 °C gives 4.3 kN — and a crown moment of 25.9 kNm, close to the 29.3 that rib shortening produced. Two entirely different causes, one calculation, comparable answers.

Shrinkage of a concrete arch does the reverse and is larger: 250 microstrain is equivalent to a temperature fall of 25 °C, and it gives −5.4 kN.

Creep does something subtler. It is a strain under sustained stress, so it multiplies the axial term rather than adding to the numerator: the effective $EA$ falls by $(1+\phi)$ and the arch loses a further 0.2 per cent of its thrust over its life. On a shallow arch that is another large moment arriving slowly.

Reading it as a stiffness ratio #

There is a way of stating the whole thing that makes it portable, and it is worth having because the same structure appears elsewhere in the collection.

The two denominator terms are the released structure’s flexibility against the redundant, computed two ways: how much it deflects because it bends, and how much because it stretches. Dropping the second is the assumption that members are axially rigid, which is the same assumption every hand truss analysis makes about its chords and every frame analysis makes by default.

For most structures that assumption is excellent, and the reason is exactly the ratio above: bending flexibility exceeds axial flexibility by $(L/i)^2$-ish factors, which are thousands. An arch is the case where the geometry makes the bending flexibility artificially small — because the redundant’s lever arm is the rise, which is deliberately a small fraction of the span — so the ratio between the two collapses.

The arch is not unusual in its physics; it is unusual in its geometry, and the same collapse happens in any structure whose redundant acts on a short lever. A tied arch’s tie, a shallow portal frame’s rafter, a truss with a very small depth: each has a redundant whose bending contribution has been made small by design, and each therefore feels its members’ axial flexibility. This is the general machinery of an indeterminate structure arriving in a geometry that makes one of its usually-ignored terms visible, and which terms a model admits is a decision that is almost always made by default.

The tied arch is the version worth drawing, because there the axial flexibility that gets dropped belongs to a member the designer chose the size of.

The tie makes the point harder to dismiss than rib shortening does, because nobody would claim a tie is rigid. It is a bar with a length and an area, its extension is the first thing anyone would compute about it, and it enters the denominator in exactly the position the $\cos^2\theta/EA$ term occupies for the rib. The two are the same term — an axial flexibility in series with a bending one — and only one of them is habitually written down.

What is actually built #

The response in practice is not to compute the correction more carefully. It is to remove the indeterminacy or to change the geometry, and both are visible in the built stock.

The three-hinged arch. Put a hinge at the crown and the structure is determinate: the thrust follows from statics, no compatibility equation exists, and rib shortening, temperature, shrinkage, creep and abutment movement all produce no force whatever. That is what the hinge is for, and it is why so many nineteenth-century arches have one.

Deeper ribs are not the answer. Increasing $I$ increases $i$, so $(i/f)^2$ rises and the sensitivity goes up. A deeper arch is more sensitive to shortening, not less — which is the opposite of the usual instinct and follows directly from the ratio.

Rise is the answer. The only geometric lever that reduces every one of these effects is a larger rise, and it reduces the thrust at the same time.

There is a fourth answer that is used more than the other three and appears in no textbook chapter on arches: build it and then adjust it. A large concrete arch is cast on falsework in segments with gaps left at the crown, and the gaps are closed by jacking against the two halves. The jacking force is chosen so that the arch arrives at the thrust the designer wanted rather than the thrust the geometry happened to produce — which sets the redundant directly instead of computing it. Shrinkage and creep that occur before the closure are then outside the structure’s history altogether, which removes the largest of the terms above rather than allowing for it.

The number is a real one, historically #

This is one of the few corrections in structural engineering that changed how things were built rather than merely how they were calculated.

Nineteenth-century masonry and iron arches were mostly built with three hinges or with construction joints that behaved as hinges, and the reason given at the time was settlement of the abutments. That is the same term in the same equation: an abutment that moves has released the redundant, and the whole family of unwanted forces disappears with it.

Twentieth-century concrete arches were built fixed, because a monolithic arch is stiffer and needs less material — and they are the ones that crack near the springings, from a combination of shrinkage, creep and temperature that the fixed-arch calculation predicts and the funicular argument does not. The hinges were not superstition and the fixity was not carelessness; each is a considered answer to the sensitivity this page computes, and they weigh it differently.

Where the moment actually lands #

The crown moment is the headline and it is not where the arch is designed.

The moment produced by a thrust deficiency is $\Delta H \cdot y$, and $y$ is the height of the axis above the springing line — largest at the crown and zero at the springings. So rib shortening produces a moment diagram shaped exactly like the arch itself: maximum at the crown, zero at the ends.

Temperature and shrinkage produce the same shape, because they enter through the same $\Delta H$. So do abutment movements. Every one of the effects on this page produces a moment proportional to the arch’s own ordinate, which means they all add or subtract cleanly and the arch has one shape of secondary moment rather than several.

That is convenient, and it is true only of the arch drawn here. It depends on the thrust acting along the springing line, which is where a two-hinged arch’s pins put it, so the secondary moments vanish at the springings and the crown is the section they govern.

A fixed arch is different, and the difference runs the opposite way to what the diagram above suggests. Its redundants act at the elastic centre, about two thirds of the way up the rise, so the same loss of thrust hogs the springings by twice what it sags the crown — and fixing the springings makes the rib lose several times as much thrust to shortening in the first place. The springings that make shortening worse works it through: on this rib, 112 kN·m at each springing and 58 at the crown, against the two-hinged 29 at the crown and nothing at the ends.

Which hinge arrangement is chosen therefore decides where this page’s moments land, and in a fixed arch they land at the section that is already designed for the largest live-load moment.

The measurement that would settle it #

An arch is one of the few structures whose secondary moments can be checked against a measurement that is easy to take, and the check is worth stating because the quantities involved are so small.

Measure the span. The rib shortening the equation predicts is a specific closure of the springings — for the arch drawn, the axial strain integrated along the rib, which is under a millimetre. That is at the edge of what a survey can resolve, and it is why nobody measures it directly.

Measure the crown deflection instead. It is much larger, because the geometry amplifies: a closure of the springings drops the crown by a factor of roughly $L/4f$ — two and a half here — and the drop is a first-order consequence of the same shortening. A crown that has dropped by more than the funicular calculation predicts is an arch whose thrust is lower than the rigid solution said, and the difference converts straight back into a moment.

That is a rare situation: a secondary moment that can be inferred from a deflection nobody had to instrument for, and it is why arch monitoring records the crown rather than the springings.

Where the model stops #

The arch is parabolic and the load is uniform. That is what makes the rigid crown moment exactly zero, which is what makes the correction exactly the whole answer. Any other combination has a non-zero base case and the correction is a correction again.

Second-order effects are absent. A shallow arch under high thrust is close to snapping through, and its real deflections amplify the moments this calculation produces. The linear flexibility equation does not know that.

The abutments are rigid. A real abutment yields, and its movement enters the compatibility equation exactly as the rib shortening does — usually with a larger effect than any of the terms here.

And creep is treated as a modulus reduction. That is the standard simplification and it is not a creep analysis; the real problem has a stress history in it, and the arch’s thrust is changing while it creeps.

Where the ladder goes #

Later rungs on this anchor: the fixed arch, which is indeterminate to the third degree and has three of these corrections rather than one. Abutment flexibility as a fourth term in the same denominator. Creep analysis with a changing stress, rather than a reduced modulus. The three-hinged arch as a deliberate escape, and what it costs in material. Shallow arches and their approach to snap-through. The tied arch, where the tie’s own extension is a fifth term with the same character. Temperature gradients through the rib, which produce a curvature rather than a length change. And the general lesson: what to do about a quantity computed as a difference of two nearly equal numbers.