The polygon that runs out of freedom

The pole construction was presented as a drawing with a free choice in it. Pick any pole and the funicular polygon comes out a different shape each time, while the resultant, the reactions and the bending moment it reports stay the same. The freedom was the point: the pole decides the drawing, not the answer.

The three-hinged arch takes that freedom away. Its thrust line under any set of loads is a funicular polygon, and it must pass through all three hinges. The previous essay found it by adding up one triangle per load. This one finds it directly, as the funicular through three given points, and in doing so counts the freedoms the drawing actually has. There are exactly three of them. That count decides which arches a drawing can solve, and the first structure it cannot solve arrives the moment one hinge is taken away.

Three freedoms, counted #

Fix the loads — their sizes and the stations where they act — and ask how many different funicular polygons they have.

The pole has two coordinates. Its horizontal distance from the load line is the thrust, and it sets how deep the polygon is. Its position up and down the load line tilts every string by the same amount, which adds a straight line to the polygon’s shape. The polygon can also start anywhere on the first load’s line, which shifts it bodily up or down. Three numbers, and every polygon for these loads is one of them.

Written as a formula the three freedoms are clean. Measured from any straight line, every funicular for given vertical loads is a straight line plus the simply supported bending moment of those loads divided by the thrust. So its height at station $x$ is $a + bx + M_0(x)/H$, where $M_0$ is the moment a beam spanning the same stations would carry. The a and the b are the shift and the tilt; the H is the scale.

Three freedoms means three conditions can be met. Make the polygon pass through a point A, and one freedom is spent. Through B as well, and a second. Through C, and the third. Three points use all of the drawing’s freedom, and a fourth point cannot be met at all, except by the accident of lying on the polygon the first three already fixed.

The construction, with a trial pole #

The classical construction reaches the polygon through three points in two moves, and both are worth seeing because each spends one freedom.

Start with any pole at all, and draw its polygon from A. It will not pass through B; here it ends 6.39 m below it. Join its start to its end and call that its closing line. Now draw the ray through the trial pole parallel to the closing line, and see where it cuts the load line. That point, Q, is 95 kN from the top.

Q does not depend on the trial pole. It is the left reaction of a simple beam spanning from A to B under these loads, which the pole essay found in exactly this way: the ray parallel to a closing line divides the load line into the two reactions. So every pole whose polygon passes through A and B must lie on the line through Q parallel to A B — here horizontal, because A and B are level. The first move has used the freedom of tilt.

Why the trial pole drops out is a line of arithmetic worth having. A pole whose rays start V down from the top of the load line and H out from it gives a polygon whose first string has slope V/H. After the loads, the polygon has fallen by the simple beam’s left reaction times the span over H, less V times the span over H. So its closing line has slope $(V - R_1)/H$, where $R_1$ is that left reaction. The ray through the pole with that slope reaches the load line $V - (V - R_1) = R_1$ down from the top — whatever $V$ and $H$ were. A trial pole is a way of measuring $R_1$ with a straightedge, and any trial pole measures it.

The second move chooses how far along that line the pole sits. Every pole on it draws a polygon through A and B, deeper as the pole comes closer to the load line. The one that reaches C is found from the depth C must be at. The polygon’s height above the chord is the simple beam’s moment divided by the thrust, and C is 5 m above the chord at mid-span, where that beam’s moment is 490 kN·m. So the thrust is 490 divided by 5, which is 98 kN. That is the three-hinged arch’s thrust, and the four equilibrium equations give 98 kN too. The pole distance, the arch’s thrust and the beam’s mid-span moment over the rise are one number.

The family through two points #

Stop after the first move and look at what is left.

The polygons through A and B are one shape at many scales. Each is the chord raised by the beam’s moment divided by its own thrust, so halving the thrust doubles every height. The family is one-dimensional, a single number sliding from flat to tall, and every member of it is a perfectly good thrust line for an arch pinned at A and B under these loads. Each is in equilibrium with the loads, and each passes through both supports.

That is what the previous essay’s crown hinge was for. It is a third point, and it picks one member. Without it, the drawing has done all it can. It has narrowed an infinite variety of polygons to a single line of candidates, and it has no way to choose between them.

A masonry arch lives with that family rather than choosing from it. The line that must stay inside found that a stone arch stands if any one thrust line can be drawn within its thickness, and the candidates it searches among are exactly these, plus the ones that leave the springings above or below their centres. The safe theorem turns the drawing’s missing equation into a license. The arch will find some member of the family that fits, so it is enough to show that one exists. That works because masonry has no tension to lose and its stiffness is not worth computing. A steel two-hinged arch has no such license. Its rib carries bending and its designer needs the bending, so the member has to be the one the arch actually uses, and that is a question the family cannot answer.

A fourth point is one too many #

The same count limits the drawing in the other direction, and the limit is easy to underestimate.

D in the figure is not an arbitrary point. It lies on the parabola through A, B and C, which is the shape an arch rib through those three hinges would plausibly be given. The thrust line for these four loads, forced through the three hinges, passes 0.27 m above the rib at that station. So the rib carries a bending moment there of the thrust times 0.27 m, about 26 kN·m. The only way to bring the thrust line down onto the rib at D is to change the loads, because every other freedom has been used.

That is the whole difficulty of shaping an arch, stated as a count. A rib can be made to follow the thrust line of one set of loads — that is the shape that carries itself, and it is found by drawing the funicular of those loads. For any other set of loads the thrust line meets the rib at the three hinges and nowhere else guaranteed, and between them it strays by whatever the loads decide. A hinged arch is funicular for one loading and bends under every other, and the three hinges are the only three points of the rib that promise nothing will happen there.

The arch with one point missing #

Now remove the crown hinge. The arch is pinned at A and B and continuous over its crown, and its thrust line must pass through two points only.

The drawing’s answer is the family of the previous figure: every thrust gives a polygon through A and B. The structure’s answer is one member of it, and the member is chosen by something that is not a force. The rib bends under whatever moment the chosen thrust line leaves in it, and the bending spreads the springings apart or draws them together. The abutments do not move, so the thrust has to be the one that leaves the springings exactly where they were.

That is a condition of compatibility, and writing it down needs the rib’s stiffness. For a rib whose second moment grows as the secant of its slope — the classical assumption, under which the stiffness along the rib measured horizontally is constant — and whose shortening under thrust is neglected, the condition is that the bending moment weighted by the rib’s height, integrated along the span, vanishes. For the quarter-point load that gives 55.7 kN, where the hinged arch had 50.0.

The two-hinged arch’s thrust line is on the drawing, since it is one of the family, and nothing on the drawing says which one it is. It lies a little lower than the hinged one, closer to the rib. The rib therefore bends less: 166.3 kN·m against 187.5. And there is no construction with a straightedge that will find it, because the quantity that selects it is an integral of the rib’s shape against its own stiffness.

The missing number did not stop anyone. The integral is a ratio of two sums along the rib: the beam’s moment times the rib’s height above the chord, and the height squared. Both functions are already on the drawing, one as the polygon’s depth and the other as the rib’s own ordinates. So a table of twenty ordinates read off the sheet, multiplied and added, gives the one thrust that picks the member, and the polygon for that thrust is then drawn like any other. The drawing did everything except one division, and the division was the part that needed to know what the rib was made of.

This is where a drawing meets indeterminacy for the first time, and the meeting is exact. Counting the unknowns says the two-hinged arch has one redundant. The funicular count says the drawing has one freedom left over. They are the same missing equation, found from two directions.

When the rib is already the funicular #

The penalty for the crown hinge depends entirely on how far the loads are from the ones the rib was shaped for, and one figure shows the limit.

Under loads that the parabola is very nearly the funicular of, the two thrust lines coincide with each other and with the rib. The member of the family that compatibility picks is the member through C anyway, because the rib is already the thrust line, and a thrust line on the rib bends nothing and moves nothing. The hinge makes no difference. The residual bending of about 5 kN·m in both is the difference between nine point loads and a continuous one, which puts small kinks in the thrust line between stations.

So the hinge’s cost is not a fixed fraction. It is proportional to the departure of the loads from the rib’s own funicular. A bridge arch carrying mostly its own weight and a deck spread along it is close to this figure. The same arch with a single heavy vehicle at the quarter point is close to the previous one, 13 per cent more bending with the hinge than without it.

That is why the question is usually settled by the split between dead and live load rather than by any principle. The dead load is known before the arch is built and the rib is shaped to be its funicular, so under dead load the hinge is free. The live load moves, and the rib cannot be the funicular of a load that moves. A structure whose live load is a small fraction of its weight — a long masonry-faced bridge, a heavy roof — gives up little by having a crown hinge and gains its indifference to settlement and temperature. A light structure under a heavy moving load pays for the hinge on every crossing.

The thrust as a load crosses #

The influence line shows the difference between the two arches over every position of a load at once.

The hinged arch’s influence line is two straight lines. The previous essay drew them with a straightedge, because the reactions’ crossing point moves along two lines through the crown hinge. The two-hinged arch’s is a smooth curve, a quartic in the load’s position. At mid-span it gives 25L/128f, which is 0.78 kN per kN here, against the hinged arch’s L/4f. No straightedge draws a quartic.

The two curves enclose the same area. That is the influence-line statement of the previous figure: a load spread evenly across the span gives the same thrust in both, because under it the parabola is funicular and the hinge has nothing to do. Everything the crown hinge changes is in how a concentrated load is shared out along the span, and for a load near the crown the hinge makes the arch push harder on its abutments by more than a fifth.

The construction does not care where the springings are #

Nothing in the three-point construction assumed level springings, and one more figure is enough to show it.

With A and B at different levels the chord A B slopes, and the line through Q along which every correct pole must lie slopes with it. That is the only change. Q is still the left reaction of a simple beam on the inclined chord, 65 kN here, and the pole distance is still that beam’s moment at C divided by C’s height above the chord: 375 kN·m over 6.1 m, 61.6 kN. The equilibrium equations give the same thrust, with the vertical reactions coupled to it through the difference in level. The drawing never wrote those equations down.

What the drawing leaves out #

The figures in this essay report equilibrium exactly and compatibility only in one figure, and the one that does rests on assumptions stated in its caption.

The rib does not shorten. A thrust of 55.7 kN compresses the rib slightly, and a shorter rib draws the springings together and relieves some of the thrust. For a rib this steep it is a small correction. For a shallow one it is the whole of the bending, and the two-hinged thrust drawn here would overstate the true one.

The rib’s stiffness varies as the secant of its slope. Real ribs are usually made of one section throughout, so they are relatively stiffer near the crown than this assumption makes them, and the integral that picks the thrust shifts. The family of candidates does not change; which member is chosen does.

Temperature is invisible to the three-hinged arch and not to the two-hinged one. A rib that warms wants to lengthen. With a crown hinge it rises a little and nothing happens to the forces. Without one, the abutments hold its length fixed and the arch takes a thrust of its own. That is a second member of the family, selected by a strain rather than a load, and the drawing cannot see it any more than it could see the first.

An arch with no hinges at all #

Taking the hinges away one at a time runs the count down to its end. A three-hinged arch has three known points and the drawing three freedoms: determinate. A two-hinged arch has two known points and one freedom left over: one redundant, one integral of the rib needed. An arch fixed at both springings has no known point at all. Its thrust line may pass above or below either springing, because the ends carry moments, and all three of the drawing’s freedoms are left over.

The fixed arch is therefore three times indeterminate, and it needs three compatibility conditions, which the column analogy supplies by treating the rib as the cross-section of an imaginary column and its elastic centre as that section’s centroid. The graphic statics of the nineteenth century handled it the same way, with elastic weights hung on the rib in place of loads and a funicular drawn through them. It is still a drawing, but it is a drawing of stiffness, and it had to be taught separately because the drawing of forces had run out of freedom.

Still open: how wrong a drawing can be #

Every construction so far has been drawn as though a line had no width and a crossing were a point. The three-point construction ends by crossing lines, like every construction before it. In a shallow arch those lines — the reaction lines at the hinges, the chord and the ray parallel to it — cross at small angles, and the flatter the arch, the smaller the angle.

A crossing at a small angle moves a long way when either line moves a little. The previous essay found this through friction in the hinges. It arrives just as surely through the width of a pencil. Whether the error it causes is a flaw in the method or a fact about shallow arches that any method must face is a question with a surprisingly definite answer.