Deflection

The arch that gets shorter

A parabolic arch under a uniform load is funicular, so the perfect solution gives it no bending at all. Then the rib shortens under its own thrust by a tenth of a per cent, and every kilonewton-metre of moment the arch will ever carry comes from that.

Assumes The hinge put in on purpose, One support too many, and what it costs to know and Choose what to take away.

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.

A tenth of a per cent of the thrust is all of the moment. The thrust a two-hinged arch loses to its own axial shortening, against rise-to-span. The flexibility equation's denominator has two terms — ∫y²ds/EI for bending and ∫cos²θ ds/EA for shortening — and their ratio is about (15/8)(i/f)², the square of the radius of gyration over the RISE. At the 10 per cent rise drawn that is 0.10 per cent of the thrust, which sounds like a rounding error and is not: a parabolic arch under a uniform load is funicular, so the rigid solution has NO crown moment at all, and the 0.10 per cent that the rib shortening removes from the thrust leaves 28 kNm behind. The correction that is a tenth of a per cent of the thrust is a hundred per cent of the bending. At a two per cent rise the loss is 2.6 per cent, because a shallow arch's thrust is enormous and its lever arm is not.
Fig. 1 The thrust a two-hinged arch loses to its own axial shortening, against rise-to-span. At a tenth rise it is a tenth of a per cent, which sounds like a rounding error. The crown moment printed beside it says otherwise.

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=M0yEIdsy2EIds  +  cos2θEAds.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 Lf2/EIL f^2 / EI — the arch’s own rise squared, integrated along it. The axial term is of order L/EAL/EA. Their ratio is

axialbending158(if)2,i=I/A.\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=143i = 143 mm against a rise of 6,000, so i/f=0.0239i/f = 0.0239 and the axial term is 0.103 per cent of the bending one.

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. That 5 kN does not disappear; it turns into bending.

The moment at any section is M=M0HyM = M_0 - H y. With the funicular thrust, M0M_0 and HyHy cancel exactly at every point. Reduce HH by 5 kN and the cancellation fails by ΔHy\Delta H \cdot y, which at the crown is 5×6=305 \times 6 = 30 kNm — 27.9 kNm once the arithmetic is done properly.

A 0.10 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.

A line of thrust, and the masonry it has to stay inside. An arch ring of 6% of the span in thickness, rising 10% of the span, under its own weight as a uniform load. Any horizontal thrust between 57.69 and 107.15 puts a line of compression entirely inside the masonry, so the arch stands — and which of them it actually takes is not decided by statics. The two extremes are drawn: the minimum-thrust line, which rides high at the crown and low at the haunches, and the maximum-thrust line, which does the opposite.
Fig. 2 The same statement drawn as a line rather than as a number. The thrust line of the ideal arch lies on the axis; a slightly smaller thrust lies slightly outside it, and the eccentricity between the two IS the bending moment. A tenth of a per cent on the force is a few millimetres on the line, and a few millimetres of eccentricity on 4,500 kN is a moment.

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 HH inward. It closes the gap two ways at once — by bending the rib, which is the y2/EIy^2/EI term, and by compressing it, which is the cos2θ/EA\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(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(i/f)^2, not (i/L)2(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 ii, 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.

There is a best rise, and it is not the tallest arch. The dimensionless buckling load w_cr L³/EI of a two pinned parabolic rib, against its rise divided by its span. A flat arch buckles at almost nothing because the same load generates an enormous thrust in it; a very tall one buckles at less than its best because the rib has become long. The maximum is at f/L = 0.28, where the coefficient reaches 49.7, and the curve is flat enough on either side that anything from about 0.15 to 0.4 is within a tenth of it. At the rise drawn the coefficient is 49.7 and the mode is antisymmetric. Nothing here is read off a table: each point is the smallest eigenvalue of the rib's own stiffness against the geometric stiffness its own thrust produces.
Fig. 3 The constraint that stops a rib from getting slender. A rib thin enough to be indifferent to shortening is thin enough to buckle, and the two requirements pull in opposite directions on the same dimension. That is the whole of arch proportioning in one sentence, and neither half is about strength.

Shallow arches feel everything

The ratio (i/f)2(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 27.9 kNm to 704. That is not a correction any more; it is the design.

Two effects compound. The thrust itself is larger, because H=wL2/8fH = wL^2/8f and the rise is in the denominator. And the fraction of it lost is larger, because (i/f)2(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.

A three-pinned arch, rise 2.6 on span 9. A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 23.37, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch.
Fig. 4 The thrust against rise, which is the first half of the same story. Halving the rise doubles the thrust; it also quadruples the sensitivity of the moment to any change in it. A shallow arch is efficient in material and delicate in behaviour, and the two follow from the same ratio.

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:

Ht=αΔTcosθdsy2EIds+cos2θEAds.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, almost exactly the same as 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 EAEA falls by (1+ϕ)(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.

It moves, or it pushes. Never both, and never neither. A 30 m steel member 30 °C warmer than it was built, in three conditions. Free, it grows 10.8 mm and carries nothing. Held, it moves nothing and carries 75.6 MPa — which is E·α·ΔT and contains neither the length nor the area of the member, so the identical stress arises in a two-metre strut. Held by a spring it does some of each: 3.2 mm of movement and 53.2 MPa, and the split is decided by the spring rather than by the member.
Fig. 5 The same equation for a straight member, where a restrained length change becomes a force in proportion to how restrained it is. An arch is the curved case, and the curvature is what turns the resulting force into a moment rather than leaving it as an axial one.

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(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.

Where the peak moment goes when a restraint is added. The same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.
Fig. 6 The general machinery this is a special case of. Every indeterminate structure computes its redundant as a ratio of flexibilities, and every one of those flexibilities is a sum over the deformation mechanisms the model admits. Which mechanisms are admitted is a modelling decision, and it is usually made by default.

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 II increases ii, so (i/f)2(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 two-hinged 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 ΔHy\Delta H \cdot y, and yy 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 ΔH\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 has one awkward consequence. A fixed arch is designed for its springing moments, which are large, and the secondary moments are zero there — so the effects discussed here are invisible at the section that governs and dominate at the section that does not. On a two-hinged arch, where the springing moment is zero by definition, the crown is the governing section and these effects are the whole of what governs it.

Which hinge arrangement is chosen therefore decides whether this page matters at all, and it decides it by moving the location of the maximum rather than by changing any number.

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/4fL/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.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

ArchAxial shorteningCreepFlexibilityFunicularIndeterminacyTemperatureThrust