The arch that is only its three hinges

The essay on the hinge put in on purpose showed why an arch is given a third hinge at its crown. The hinge adds one equation to statics — the moment there is zero — and that is exactly enough to find the four reaction components of an arch pinned at both feet. It drew the answer once, for a symmetric arch under a symmetric load, as two lines and a triangle.

That drawing is worth much more than one answer. Asked what else it knows, it gives three results that the four equations hide. The rib between the hinges has nothing to do with the reactions. A load crossing the arch moves the construction along two straight lines, which is the arch’s influence line drawn without arithmetic. And the springings can be at any two levels without the drawing noticing. There is a fourth result that complicates the other three, and it comes from the friction circle of two essays back: a real hinge is not a point, and an arch built around three of them is determinate only to within a band.

The half that carries nothing is a strut #

Put a single load on the left half of the arch and look at the right half by itself.

The right half has forces on it at two points only: the reaction at its springing B and the force passed to it through the crown hinge C. Neither is a moment, because both are hinges. A body held by two forces must have them equal, opposite and on the line joining their points, however the body between those points is shaped. So the right half, whatever its curve, is a strut along the line B C, and the reaction at B lies along that line.

That fixes one direction. The whole arch is now held by three forces: the load, the reaction at B along the known line, and the reaction at A. Three forces in equilibrium meet at a point, so the reaction at A must pass through K, where the load’s line crosses the line B C. Two directions and one known force close a triangle, and the triangle, drawn to scale on the right, gives both reactions in full.

Here K is 7.5 m up, the reaction at A leans at 56° to the horizontal and carries 90.1 kN, and the one at B leans at 27° and carries 55.9. Both have the same horizontal component, 50 kN, the thrust — which is the triangle’s width. Four equilibrium equations give the same four numbers. They agree to the last figure the arithmetic carries, because they are the same statement written two ways.

The rib is not in the construction #

Nothing in that construction used the rib. The two-force argument needed the hinges at B and C and nothing between them. The three-force argument needed the load’s line, the hinge at A and the point K. The rib could be any shape that passes through the three hinges.

All three push on their abutments with the same forces. A parabolic arch, a circular one and a steel portal frame of the kind that roofs a warehouse are, as far as their foundations can tell, the same structure. That is a strong statement and a useful one. It means the foundation of a three-hinged structure can be designed before anybody has decided what the structure looks like, from the hinge positions alone.

What the rib does decide is its own bending, and the figure says how. The thrust line is where the resultant of everything to one side of a section passes, and for these loads it is the two straight lines from A to K and from K to B. The bending moment at any point of the rib is the thrust times the vertical distance between the rib and that line. The parabola strays furthest from the line, 3.75 m below K at the load, and carries 187.5 kN·m. The portal frame’s rafter strays 3.45 m at the same station and carries 172.5. The circle is between them.

At the three hinges every rib passes through the thrust line, because the thrust line passes through the hinges, and so the moment there is zero on every shape. That is what a hinge is from the rib’s point of view: a point where the rib is made to agree with the thrust line whatever else it does. Three such points fix the thrust line completely. A masonry arch without hinges has infinitely many thrust lines to choose from and survives if any one of them fits inside it. A three-hinged arch has exactly one, and it is drawn before the rib is.

Where K goes as the load moves #

Now move the load. Each position gives its own K, and the positions of K form a pattern that the equations would not suggest.

For any load on the left half, the right half is the strut, so K is on the line B C, wherever the load is. As the load moves from the crown out toward A, K slides up that line, beyond the crown, to 9.25 m for a load at 1.5 m. For any load on the right half, K slides along A C extended. So the locus of K is two straight lines crossing at the crown hinge, and the arch’s whole response to a moving load is written on them.

That is the sort of fact that is invisible in the algebra, where the thrust for a load at station x is a ratio of two expressions in x and the hinge coordinates, and nothing says the ratio is a line. On the drawing it cannot be anything else. A line through a fixed point crossed by a vertical at a moving point gives a crossing that moves linearly.

An influence line drawn with a straightedge #

Read the thrust off the construction at each position of the load and the result is the thrust’s influence line.

The thrust’s influence line is a triangle, zero at both springings and peaking under the crown hinge at L/4f — here one kilonewton of thrust for each kilonewton of load, because the span is four times the rise. It has straight sides because K moves along straight lines, and a corner at the crown because that is where K changes from one line to the other.

The triangle has a name in the theory of beams, and the surprising connection in this essay is that it is the same object. The thrust’s influence line is the influence line for the bending moment at mid-span of a simply supported beam of the same span, divided by the rise. A load at x puts a moment at the crown of the simple beam, and the arch carries exactly that moment as a couple: the thrust at the crown times the rise. The influence line for a beam’s moment is a triangle peaking at the section, and dividing it by five metres gives this one. The three-hinged arch is a simply supported beam that carries its mid-span moment as a thrust.

Everything a beam’s influence line does, this one does. A uniform load w over the whole span gives a thrust equal to w times the triangle’s area, ten metres here, which is wL²/8f. A train of wheel loads is placed to put the heaviest axles near the peak. The left vertical reaction’s influence line is a straight line from one to zero, identical to a simple beam’s — because for level springings the thrusts are horizontal and take no part in vertical equilibrium, so the arch shares its vertical reactions with the beam.

Several loads, one triangle each #

A real arch carries several loads at once, and the construction handles them one at a time. Draw each load’s triangle separately, with its own K on its own line, and add the reactions that come out. Superposition is exact here, because every step of the construction is linear in the load. A load twice as large gives the same K and a triangle twice the size.

Put the 100 kN at 5 m and add 60 kN at 14 m. The first load, as already drawn, gives a thrust of 50 kN and vertical reactions of 75 kN at A and 25 at B. The second is on the right half, so the left half is the strut. Its triangle gives vertical reactions of 18 and 42 kN, and a thrust of 36 kN — the left vertical reaction times the half-span over the rise. Added, the arch carries 86 kN of thrust with 93 and 67 kN vertically, which makes resultant reactions of 126.7 kN at A and 109.0 kN at B.

The two triangles can also be drawn as one figure. Lay the loads down one line, as in the funicular construction, and the two reactions close it from its ends. Their meeting point is the pole of the one funicular polygon that passes through all three hinges. That polygon is the thrust line of the arch under both loads together. The construction above found it by adding answers. Finding it directly, as the funicular through three given points, is a different construction, and it is where the next essay starts.

Springings at two levels #

That last property does not survive uneven springings, and it is where the drawing earns its keep over the equations.

With the springings at different levels, the thrusts at A and B are no longer on one horizontal line. Their moments about each other’s springing no longer vanish, and every vertical reaction depends on the thrust. The four equations are still four equations, but they are coupled, and solving them by hand means eliminating a thrust through two moment equations about points at different heights. That calculation is where sign errors live.

Written out for this arch, moments about A for the whole arch give the right vertical reaction times the 20 m span, plus the thrust times the 3 m by which B stands above A, equal to 80 kN times 14 m. Moments about C for the right half give the same vertical reaction times its 11 m lever arm, less the thrust times the 3 m by which C stands above B, equal to 80 kN times 5 m. Two equations, the thrust in both, with opposite signs, solved together: a right vertical reaction of 49.0 kN and a thrust of 46.5. Each sign depends on which support is higher than which. Take the first one the wrong way and the pair still solves, to a thrust of the right size that is plausible and false.

The drawing does not know the springings are at different levels. The load is on the right half now, so the left half is the strut, its line runs from A through C, and K is where that line meets the load’s line, 9.33 m up. The triangle gives 55.8 kN at A and 67.5 kN at B, with a thrust of 46.5 kN — the four coupled equations give the same. The crown hinge being off-centre changed nothing about the method either, and the circular rib swinging above its own crown hinge changed nothing at all.

The locus is still two straight lines through the crown hinge, of different slopes now, so the thrust’s influence line is still two straight lines with a corner under the crown. It peaks at 1.06 kN per kN rather than 1.00. The left vertical reaction’s influence line, though, is no longer the simple beam’s. It picks up a kink at the crown, because the thrust now takes part in vertical equilibrium and the thrust has a corner there. The drawing found that without being told to look for it.

The hinge that is not a point #

Everything above takes the three hinges to be points, and they are not. A real pin carries its force tangent to a friction circle of radius r sin φ round its centre. So the two-force half’s line is not the line through B and C. It is one of four common tangents to the circles there, and the left reaction is one of two tangents from K to the circle at A. That makes eight constructions, and eight answers.

The band is small on a well-proportioned arch with greased pins. At a rise of a quarter of the span and µ = 0.15 it is about one per cent either way, and no designer would size anything differently for it. It grows sharply as the arch flattens. At a rise of a twentieth, the same pins give ±4.4 per cent, and at µ = 0.4 — a steel pin that has been left to corrode for a few decades — ±11.

The reason is the geometry of the construction, not the pins. A friction circle moves each line sideways by a few centimetres. In a flat arch the two reaction lines cross at a shallow angle, and a few centimetres of sideways movement of either line moves their crossing a long way. That is the ill-conditioning the pin essay found for its bracket, arriving here through the arch’s rise. The hinges were added to make the arch determinate. Friction in them makes it determinate only to within a band, and the band is widest on exactly the flat arches whose thrust matters most, because a flat arch’s thrust is already large.

Of the eight constructions, a real arch uses one at a time, and which one depends on how each hinge is turning. As the load is applied, each half rotates about its springing, and the two halves rotate against each other at the crown. Friction resists each of those rotations, so it pushes every line to the side that opposes the motion. Under a load that only increases, the construction is definite: one tangent at each hinge, offset from the frictionless line in a known direction, and the thrust sits at one edge of the band. When the load moves, or reverses, or the temperature cycles, the hinges turn back and the construction jumps to the other edge. The band is the range the thrust can wander over in service. Two different arches it is not.

The practical answer is the one bridge engineers adopted: a hinge detailed to rotate on a curved bearing surface or a low-friction sliding face rather than on a steel pin in a hole, for the same reason that a sliding bearing needs a low-friction face to be a roller.

Which forces the drawing reads, and which it cannot #

The free bodies in this essay are the two halves of the arch and the arch as a whole, and every force on them is known in direction before it is known in size. That is the whole trick: two directions and one force always close a triangle.

What the construction cannot see is anything that is not a force on those free bodies. It cannot see the rib’s stiffness, because a three-hinged arch’s reactions do not depend on it. That is the arch’s advantage, and it is also why the construction says nothing about deflection. It cannot see the rib’s own weight except as loads lumped at stations. A rib that carries its weight continuously has a thrust line that is a curve between the stations, and the construction draws its chords. And it cannot see the arch deform. Every line is drawn on the unloaded geometry, and a shallow arch whose crown drops under load has a smaller rise and so a larger thrust than the drawing gives. How much larger depends on stiffness, and the arch that gets shorter measures the same effect in an arch without the crown hinge.

The assumption the three points rest on #

The construction rests on the three hinges carrying no moment and staying where they were drawn. The first is the friction question just measured. The second is a question of support. A three-hinged arch is determinate, so a settlement or a spread of its abutments is accommodated by rotation at the hinges without any change in the reactions. That is one of the reasons the crown hinge is put in, and it holds only while the geometry is close to the drawn geometry. A spread large enough to change the rise changes the thrust through the L/4f ratio, and a flat arch has a small f.

The same property made the three-hinged arch the long-span form of the nineteenth-century exhibition hall and train shed. The Galerie des Machines in Paris, built for 1889, spanned about 110 m with three-hinged steel arches. A structure of that size lengthens by several centimetres between a winter night and a summer afternoon. A two-hinged arch would resist that lengthening and take it as thrust and bending. A three-hinged arch simply lets its crown rise and fall, and its reactions do not change, because nothing in the construction depends on the rib’s length. The determinacy that the drawing exploits is the same determinacy that makes the arch indifferent to temperature.

It is also why the reactions are so robust to the rib and so sensitive to the hinges. A three-hinged arch whose crown hinge is built a hundred millimetres low has a different thrust. One whose rib is built a hundred millimetres out of shape between the hinges has the same thrust and different bending.

Still open: a polygon forced through three points #

Everything here used one load at a time, with superposition doing the rest. Under several loads at once, the thrust line is a funicular polygon, and it has to pass through all three hinges. The pole construction draws a funicular through nothing in particular: the pole is chosen freely and the polygon lands where it lands. Asking it to pass through three given points uses up all of that freedom. That is what makes a three-hinged arch solvable by drawing. It is also why an arch with only two hinges cannot be solved that way, and the gap between those two statements is where a drawing first meets a structure it cannot solve.