The pole decides the drawing, not the answer

Culmann’s line solves a body held by four forces by splitting them into two pairs, each of which has a crossing to hang a resultant on. It works once. Five forces have no second pairing left to make, and a floor beam carrying five columns is not an exotic object.

Karl Culmann’s book of 1866 answers that with a construction that has a free choice inside it, which is the first thing about it worth noticing and the last thing anybody expects of a method for getting numbers.

What the pole is a point on #

The pole is not a point on the beam. It is not a point on any structure. It sits on the force diagram — the sheet where the loads have been drawn end to end as lengths rather than placed as arrows — and it is put wherever the drafter likes.

That is the whole of the licence, and it is worth being precise about what it licenses. The force diagram here is one vertical line 155 kN long, divided at 30, 50, 95 and 120. Marking a point beside that line and joining it to each division produces six rays. Nothing in the loads decided where the point went; two numbers were invented — how far to the side, and how far down — and six directions came out.

Those six directions are then used as the directions of six lines drawn on the beam, one for each region between consecutive loads. That is the funicular polygon, and the name is a promise about what it is: funis is a rope, and the polygon is the shape a rope carrying these loads would hang in, if it were pulled at its ends with a horizontal force equal to the distance from the pole to the load line.

That horizontal force has a name of its own, the polar distance, and it is the invented number that matters. The other one — how far down the load line the pole sits — slides the whole polygon and tilts it, which the drawing below shows and which changes nothing that will be read off it.

Why the first and last strings cross where the resultant acts #

The construction’s first claim is that the resultant of all five loads passes through the point where the first and last segments, extended, meet.

The proof is the three-force theorem used repeatedly, which is what the construction for four forces also is underneath. Take the rope seriously. The tension in the first segment is a real force along that segment; so is the tension in the last. Between them the rope carries the five loads and nothing else. So the free body consisting of the whole rope has exactly three external forces on it: the tension at its left end, the tension at its right end, and the resultant of the five loads. Three forces in equilibrium meet at a point, and the two tensions meet where their lines cross — so the resultant’s line passes through that crossing, whatever the loads are and however many of them there are.

The crossing in the figure is at 5.24 m. The loads’ first moment about the left support, divided by their total, is $(30 \times 1.5 + 20 \times 3.5 + 45 \times 5 + 25 \times 7 + 35 \times 8.5)/155$, which is 5.2419 m. The two agree to four figures because they are the same statement: the resultant of a set of forces is the single force with the same total and the same moment, and the rope’s end tensions are a mechanism for reading the second one off a drawing.

Five loads took one construction. Fifty would take the same one, with fifty segments instead of five and no more thought. That is the difference from Culmann’s line, which pairs four forces and then has to be applied again, and again, on a sequence of partial resultants.

Three poles, three drawings, one answer #

The free choice is not a small one, and its effect on the drawing is not subtle.

Three polygons that look nothing alike, one answer. The construction is exact, and the arbitrariness is confined to a part of the drawing that no reading depends on — which is a property worth naming, because it is rarer than it sounds. An arbitrary choice inside a calculation is usually an approximation wearing a disguise.

Here it is not, and the reason is that the pole’s two coordinates enter the construction only through the directions of the rays, and the resultant’s position is a statement about where a line is rather than about which way anything points. Change the pole and every ray turns; the polygon changes shape; the crossing slides up and down a vertical line without leaving it.

A drafter uses that. Drawing a second construction with a different pole is a check that costs a minute and catches the only error a careful drawing is likely to contain, which is a slipped parallel. Two poles agreeing is strong evidence, because an error in one would have to repeat itself at a different angle to survive into the other — the same argument that makes a second Culmann pairing the check on the first, and a truss diagram that closes on itself the check on a whole frame at once.

What the polar distance buys #

If the pole changes nothing, the obvious question is why anybody would think about where to put it. The answer is that it changes one thing, and the thing it changes is the only reason the construction is worth more than the arithmetic.

A straight line of slope −1 on logarithmic axes is a product that does not move. Depth times polar distance is one number, 233.75 kN·m for this beam under these loads at this station, and the drafter’s two invented numbers have bought a choice of how many millimetres of paper a kilonewton-metre is drawn as.

That is a scale, and choosing a scale is what a drafter does all day. A short polar distance draws a deep polygon, which reads precisely and runs off the bottom of the sheet; a long one draws a shallow polygon that fits and cannot be measured. The good choice is the one that fills the paper, and it is decided by the size of the drawing rather than by anything about the structure — which is exactly what an arbitrary parameter ought to be decided by.

The number the product gives is a bending moment. That is the finding of this essay and it arrives before the section that explains it, because the sweep above is where it is visible: a quantity read off a drawing whose only purpose was to find a resultant turns out to have units of force times length and a value the beam’s own analysis produces.

The chord that gives the reactions #

Before the moment, the reactions, because the moment reading needs them and the construction gives them first.

The rope again. Draw the funicular anywhere, note where it crosses the two support verticals, and join those two crossings. That chord is the closing line, and it is the segment a rope would need if it were pinned at the two supports rather than pulled at its ends — the segment closing the polygon into a shape whose ends are held.

Every segment of the funicular is parallel to a ray, and the closing line is a segment of it, so the closing line is parallel to a ray too. That ray runs from the pole to a point on the load line, and the point it runs to divides the load line into two lengths. Those two lengths are the reactions.

The argument is the rope’s equilibrium taken in two halves. The tension in the first segment and the tension in the closing line meet at the left support; those two, plus whatever the support supplies, are three forces at a point. The triangle they make on the force diagram is the pole, the top of the load line, and the point where the closing ray cuts it — and the side of that triangle lying along the load line is the left reaction, 73.75 kN. The other reaction is what is left of the load line, 81.25.

Those two numbers are what moments about either support give, and the construction took no moments. It drew a line parallel to another line.

The polygon was the moment diagram the whole time #

The two shapes in that figure are the same shape.

The reason is short enough to give in full, and it is worth giving because it explains why a nineteenth-century drawing office computed bending moments without ever writing one down.

Take a station $x$ somewhere along the beam. The funicular’s depth below its closing line there, call it $y$, is the vertical distance between two lines whose slopes are both known: the closing line’s slope is $(c_R - V)/H$ where $c_R$ is the division the closing ray cuts, and every funicular segment’s slope is $(c_k - V)/H$ where $c_k$ is the total of the loads passed so far. Subtracting, the rate at which the gap opens is $(c_R - c_k)/H$. But $c_R$ is the left reaction and $c_k$ is the sum of the loads to the left of $x$, so $c_R - c_k$ is the shear force at $x$. The gap therefore satisfies

$$H\,\frac{dy}{dx} = V(x)$$

and the shear force is the derivative of the bending moment. Integrating from a support, where both $y$ and $M$ are zero, gives $H y = M$ everywhere. The polar distance is the constant of proportionality, and it is the number the drafter invented.

So the construction is not related to the bending moment diagram. It is the bending moment diagram, drawn at a scale of $1/H$, by a method that never mentioned bending. The drafter who wanted the resultant of five column loads has, without asking, produced the diagram the beam will be designed from, complete and exact, at every station rather than at the five the loads sit on.

This is the reason graphic statics was a profession rather than a curiosity. The method of sections answers one question per drawing. The funicular polygon answers four from one: the resultant’s position, the two reactions, and the moment everywhere — and the last of those is the one an engineer actually needs, arriving as a by-product of the first.

Which free body produced the number #

Three free bodies were used and they are easy to confuse, so it is worth separating them.

The whole rope, for the resultant. Its external forces are the two end tensions and the five loads, and the three-force theorem applied to the tensions and their resultant is what puts the crossing on the resultant’s line. The beam is not in this free body at all; the rope is a fiction invented to carry the same loads.

The rope pinned at both supports, for the reactions. Its external forces are the five loads and two support reactions along the first and last rays — and the closing line is what makes the shape consistent with being held at two points rather than pulled at two ends.

The part of the beam left of a cut, for the moment. This is the only free body that contains any beam. Its forces are the left reaction and whatever loads have been passed, and their moment about the cut is what $H y$ equals. The derivation above never draws that free body; it shows instead that the gap’s derivative is the shear, which is the same statement differentiated once.

The confusion worth avoiding is between the rope and the beam. The rope is not a model of the beam and does not resemble it. A rope carries these loads in tension with no bending anywhere; the beam carries them in bending with no significant axial force. The two share exactly one thing, which is the function $M(x)$ — for the rope it is $H$ times a shape, and for the beam it is a moment. Graphic statics exploits the coincidence and says nothing about why it holds beyond the algebra above.

A lopsided case, where nothing about the drawing changes #

The construction is indifferent to how unevenly the loads sit, which is worth demonstrating rather than claiming, because the symmetric example above hides the one place a reader might expect trouble.

Three-quarters of this load sits in the first four metres, the left support takes 118 kN against the right’s 42, and the largest moment is at 2.5 m. Every one of those facts is read off the same four operations — divide the load line, pick a pole, draw parallels, join the ends.

The pole in that figure was put at 117.75 kN down the load line deliberately, which is the second invented number being used rather than ignored. Setting it to the left reaction makes the closing ray horizontal, and therefore the closing line horizontal, and therefore the gaps measurable with a scale rule held the ordinary way. Nobody can know that value before drawing, which is why the practice was to draw once with any pole, read the reaction off the closing ray, and redraw with the pole at that division if the moments were wanted accurately. Two drawings, and the second one is the moment diagram with a horizontal baseline.

Where the model stops #

Parallel loads. Everything above takes the loads as vertical. The construction does not require it — a set of forces in any directions has a force diagram whose divisions are the tips of the vectors laid end to end, and the same pole and the same rays work — but the reading that the gap is a bending moment does, because it assumed the closing line and the segments could be compared by a vertical distance. For a general force system the funicular still finds the resultant and no longer draws a moment diagram.

A distributed load has to be chopped up. A load spread along a beam is replaced by a set of point loads, one per strip, and the funicular comes out as a polygon inscribed in the smooth curve the real load would give. The error is the difference between a chord and an arc over one strip, and it falls as the square of the strip width — so ten strips is close and twenty is closer, and neither is exact. The moment diagram of a uniformly loaded beam is a parabola and no polygon is a parabola.

Statically determinate only. The construction uses equilibrium and nothing else, so it reaches every force a determinate structure has and none of the forces in a redundant one. A propped cantilever’s funicular can be drawn for any prop force at all, and the drawing has no way to say which is right; deciding that needs a statement about stiffness, which is not a statement any of these lines can make.

And the drawing’s accuracy is the drafter’s. A parallel drawn with a set square is good to perhaps a quarter of a degree, and the depth of a shallow polygon is the difference of two nearly parallel lines. That is the practical argument for a short polar distance and the practical limit on how short: too deep and the polygon leaves the sheet, too shallow and the gap is the difference between two pencil widths.

What the picture cannot show #

The polygon is drawn at one pole, and the sweep two sections above is the only figure here that shows the choice as a continuum. Every other drawing has made the choice already, so the arbitrariness — which is the essay’s subject — is invisible in six of the seven figures and has to be taken on the evidence of the one.

Nothing here shows the rope. The funicular polygon is drawn as a line and is a shape a physical rope would take, and the physical rope would also stretch, sag under its own weight, and have a horizontal tension that varies with none of the drafter’s intentions. The polar distance is the tension in an imaginary rope, and imagining it is the whole of its existence. The rope that hangs in this shape for real is a different object with its own essay, and the number a drafter picks here to fill a sheet is the number a cable engineer cannot pick at all.

The assumption underneath the free choice #

The pole is free because the construction returns lines and ratios rather than lengths, and a drawing’s arbitrary parameters are safe exactly when nothing is read off it in absolute units.

That condition breaks the moment the depth is measured. The gap under the middle load is 3.90 m at one pole and 1.94 m at another, and neither of those is a distance on the beam — they are moments divided by a force nobody applied. A length on this drawing is not a length, and the only defence against reading one as a length is knowing which of the two diagrams a dimension was taken from. The space diagram’s metres are metres; the funicular’s are not.

That is an old and specific trap, and it is why every published funicular construction states its two scales — so many millimetres to the metre, so many to the kilonewton — and why the polar distance is written on the drawing beside the pole. A drawing that has lost its scales has lost its reactions and its moments and kept its resultant, which is the one reading that is a position rather than a size.

Still open: the force diagram drawn first #

Everything above starts from the loads and produces a shape. Reversing that is a design method rather than an analysis one, and it is where this construction went after the calculator took its analysis work away: the force diagram and the space diagram are each other’s duals, every closed polygon in one corresponding to a point in the other, so a designer who draws the force diagram first has chosen the forces and can ask what shape carries them. That reciprocity is Maxwell’s, from 1864, it has a theorem under it about which frames have a reciprocal figure at all, and it is the subject the next essay takes up.

After that, the pin that is not a point. Every line of action on every drawing here has been assumed to pass exactly through a pin’s centre, which is true only of a frictionless pin; a real one lets the force pass anywhere within a small circle, so each line becomes a band and each crossing becomes a patch. And the three-hinged arch, whose two reactions each pass through a hinge and are therefore found by pairing the load with the arch’s own geometry rather than with a support condition.