Structural form

The tree that strength does not ask for

A branching column carries a roof on many points and reaches the ground on one. Size every member by its stress and the optimum tree turns out to have no trunk at all — the best answer is a fan of straight struts from the base. Put buckling in and the trunk appears, at 57 per cent of the height.

Assumes The shape that carries itself, and the arch that is its reflection, Strong enough and still falls over and Three equations at every joint.

A roof needs supporting at many points. The ground is available at few. Between the two there is a family of structures that start as one member and end as many, and the tree is the obvious member of it — a trunk that divides, and divides again, until there are as many tips as the roof has points.

They are built because they look like trees, which is a perfectly good reason and not the one this essay is about — though it returns at the end, with better credentials than it started with. The question here is narrower and has a clean answer: given a load to deliver from a set of points down to one, how much material does a branching structure need, and is it less than the alternatives?

The answer arrives in two halves that point in opposite directions, and the disagreement between them is the whole finding.

The split is where buckling puts itA branching column carrying 400 kN to two points 3.0 m apart, over 9.0 m. Every member is drawn at the thickness it needs: the area is the larger of N/σ and what Euler asks of a strut of that length, and 3 of 3 members here are sized by buckling rather than by strength. The split sits at 57% of the height, which is where the total volume is least — 51% less than the fan of straight struts that carries the same load with the same stresses.the split, at 5.1 mvolume 0.100 × 10⁹ mm³ against 0.204 for a fan from the base3 of 3 members sized by buckling
Fig. 1 The structure, with every member drawn at the thickness it needs. The area of each is the larger of two requirements — the stress it carries and the load it must not buckle under — and on this tree the second governs for two of the three members. The split sits where the total volume is least.

Which free body produced the number

Cut just below a joint and take everything above it.

The loads at the tips are vertical and equal. The branches meeting at the joint carry axial forces along their own axes. For the sum of forces to vanish in both directions with no moment left over, three things have to be true at once: the branch axes have to pass through a common point, the resultant of the branch forces has to lie along the trunk, and the loads have to be symmetric about it.

When they are, the arithmetic is trivial. Each branch carries N=(P/n)/gN = (P/n)\,\ell/g, where \ell is its length and gg its rise — the vertical load divided by the cosine of its inclination — and the trunk carries the whole of PP. No member carries any bending at all, which is the property the form exists for and is the reason the same drawing appears as a funicular shape in another field.

Concurrent, symmetric, and free of momentThe joint of a branching column, with two branches at 38° to the trunk. With every branch carrying the same load the forces are concurrent and their resultant is the trunk force, so the joint carries no moment at all — a property of the geometry rather than of the connection. Every branch here is loaded, which is the case the geometry was found for.200 kN200 kNno moment400 kN down the trunkthe branch axes meet at a point — which is the whole of what keeps the joint free of bending
Fig. 2 The joint when the assumption holds. Two branch forces and a trunk force, concurrent, summing to nothing, with no moment anywhere. The concurrency is a geometric property that has to be built rather than assumed: a joint whose axes miss by 40 mm carries the resultant times 40 mm, and the members meeting there are large.

Strength alone wants no tree

Now ask for the volume of material, with every member sized so that its stress is exactly the working stress. Hold the height HH and the reach ss fixed and let the split happen at any level; call the branch rise gg, so the trunk is HgH - g long.

The trunk contributes P(Hg)/σP(H-g)/\sigma. Each branch is =s2+g2\ell = \sqrt{s^2+g^2} long and carries (P/n)/g(P/n)\ell/g, so the branches together contribute P2/(gσ)=P(s2+g2)/(gσ)P\ell^2/(g\sigma) = P(s^2+g^2)/(g\sigma). Add them:

V=Pσ(Hg+s2+g2g)=Pσ(H+s2g)V = \frac{P}{\sigma}\left(H - g + \frac{s^2 + g^2}{g}\right) = \frac{P}{\sigma}\left(H + \frac{s^2}{g}\right)

The gg terms cancel except for one, and what is left falls monotonically as gg grows. The volume is least when the branch rise is the whole height — that is, when the split is at the ground and the “tree” is a fan of straight struts from a single point to the tips.

The trunk is pure waste. Every millimetre of it carries the full load over a length that could have been part of a member also covering some of the horizontal distance.

Strength wants no tree at allThe volume of a fully stressed branching column against the height at which it splits. The lower curve counts strength only: it falls monotonically as the split moves down, reaching its minimum at the ground, because a fully stressed structure's volume is P(H + s²/g)/σ and g is the branch rise. The best tree by that measure is a fan of straight struts and the trunk is waste. The upper curve adds the buckling requirement — a strut of length ℓ needs a second moment of at least Nℓ²/π²E — and it has an interior minimum at 57% of the height, 51% below the fan. The tree is a stability structure, not a strength one.0.00.20.40.60.81.000.050.10.150.2height of the split ÷ total heightvolume (×10⁹ mm³, upper curve)least at 57% of the heightwith bucklingstrength only(scaled to fit)2 branches · one level · reach 3.0 m over 9.0 m
Fig. 3 The two curves, on one axis. The lower one counts strength only and falls all the way to the ground, exactly as the closed form says — the computation and the algebra agree to machine precision, which is what says neither is wrong. The upper one adds buckling and has an interior minimum, 51 per cent below the fan.

That result is a small instance of a much older one. Michell’s theorem, from 1904, says that the minimum-volume truss transferring given loads to given supports is made of members lying along the principal directions of a particular strain field, and it produces fans of straight members radiating from load points rather than hierarchies. Strength-optimal structures are not tree-shaped, and the fact has been known for a century.

Buckling is why trees exist

Nothing above mentions the length of a member except as something to multiply a force by. Buckling does.

A strut of length \ell carrying NN needs a second moment of at least N2/π2EN\ell^2/\pi^2E, and for a section of a given shape the second moment goes as the square of the area. So the area buckling demands goes as N\sqrt{N}\,\ell rather than as NN, and — this is the part that matters — it grows with the length while the strength requirement does not.

Put that requirement in and the fan of long struts becomes the expensive option. Each member of the fan is H2+s2\sqrt{H^2+s^2} long, carrying a share of the load over the full height; each branch of a tree is shorter, and the trunk carries the whole load over a length where it is stubby. The optimum moves up off the ground and settles, for the column drawn here, at 57 per cent of the height — with a volume 51 per cent below the fan’s.

The tree is a stability structure. It is not a way of carrying force efficiently; it is a way of carrying force over a distance without any single member being long.

Length costs more than it looksThe same column section at four lengths, with the buckling capacity of each drawn as a bar. Capacity falls as the inverse square of the length, so a column three times as long carries a ninth as much.1× the length100% of the capacity1.5× the length44% of the capacity2× the length25% of the capacity3× the length11% of the capacityidentical section, identical material, identical end conditions
Fig. 4 What length costs a compression member. A tie’s area depends only on its force; a strut’s depends on its length as well, and past a modest slenderness the length is the whole of it. Every branching structure, every space frame and every lattice is an answer to that asymmetry.
The column curveFailure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither
Fig. 5 The curve the constraint comes from. Below a slenderness of about 90 a member is limited by its strength and the tree buys nothing; above it the member is limited by its length and the tree buys a great deal. Where a given structure sits on this axis decides whether branching is worth the joints.

Every generation is worth less than the last

If one split helps, more should help further, and they do — with a diminishing return steep enough to explain why real branching structures have two or three levels and not eight.

Measured on the same column: one level of branching gives a volume of 0.100 units, two gives 0.074, three gives 0.064. The first split buys 26 per cent, the second 14, the third would buy less than eight — against a joint count that doubles each time and a fabrication cost that does not.

The reason for the diminishing return is that each generation halves the load and roughly halves the length, and the buckling requirement goes as N\sqrt{N}\ell — so each split reduces a member’s demanded area by a factor of about 2×2\sqrt{2}\times 2, while multiplying the number of members by two. The product improves, and less each time.

A frame with no plane to be drawn inThe tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15.30 kNm + r = 12 · 3j = 12 · residual 1.8e-15heavy is more force · one colour is tension and the other compressionlargest member force 12.25 kN
Fig. 6 The alternative answer to the same problem. A space frame delivers a distributed load to a few supports by making every member short, which is the same objective the tree pursues by a different route. The two compete directly, and the choice between them is usually made on the joints rather than on the members.

What the section does to the answer

The buckling requirement contains a shape factor, and it moves the answer more than anything else does.

A solid circular member has I=A2/4πI = A^2/4\pi. A circular hollow section of the same area, with a diameter-to-thickness ratio of twenty, has a second moment perhaps twenty times larger — so the area buckling demands falls by a factor of 20\sqrt{20}, and the volume of the whole tree falls with it. The column drawn here needs 0.100 units of material as a solid section and 0.022 as a tube: a factor of 4.5, from a decision about the section rather than about the form.

That is a general and slightly deflating point about form optimisation. The choice between a fan and a tree is worth 51 per cent; the choice between a solid bar and a tube is worth 350. Where the material sits within the section is a bigger lever than where the members sit within the structure, and it is available without any joints at all.

The joint that is only free of moment on one load case

Everything above assumes the tips are equally loaded. They are not, and the case that is not symmetric is the one that sizes the joint.

Take the load off one branch of a two-branch joint — a partially loaded roof, a snow drift on one side, a maintenance load — and the two branch forces no longer resolve into something along the trunk. The joint carries the difference, and the difference has a lever arm: the horizontal offset from the joint to the tip, which is metres rather than millimetres. On this column that is 300 kNm at a joint whose members were sized for axial force alone.

This is the standing hazard of every form-found structure, and it deserves stating in general terms. A shape found for one load case is only funicular for that load case. An arch is funicular for its own weight and carries bending under anything else; a cable takes the shape of the load it carries and changes shape when the load changes; a branching column is momentless under symmetric load and is a frame under any other.

One branch unloaded, and the joint has a momentThe joint of a branching column, with two branches at 38° to the trunk. With every branch carrying the same load the forces are concurrent and their resultant is the trunk force, so the joint carries no moment at all — a property of the geometry rather than of the connection. Take the load off one of them, which is what a partly loaded roof does, and the balance is gone: the joint carries 600 kNm, over a lever arm the drawing shows and the load case does not. This is the case that decides how a branching column is detailed, and it is not the one it is designed for.no load200 kN600 kNm200 kN down the trunkthe branch axes meet at a point — which is the whole of what keeps the joint free of bending
Fig. 7 The same joint with one branch unloaded. Nothing about the geometry has changed and the joint is now carrying a moment as large as anything else in the structure — over a lever arm that is a design dimension of the tree rather than a detail of the connection.
A line of thrust, and the masonry it has to stay insideAn arch ring of 9% of the span in thickness, rising 28% of the span, under its own weight as a uniform load. Any horizontal thrust between 3.85 and 5.23 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.thrust anywhere from 3.85 to 5.23 fitsH = 3.85, leastH = 5.23, most
Fig. 8 The same lesson in masonry. A shape found for one loading has a thrust line that leaves it under another, and what happens next is decided by whether there is enough material around the line to contain it. A branching column’s equivalent of “enough material” is a joint stiff enough to carry the moment the asymmetry produces.

The load path is chosen, and it is chosen visibly

There is one property of a branching column that no volume calculation captures and that is probably the real reason they are built.

Most structures hide their load path. A flat slab on columns gives no indication of how the load reaches the ground; a steel frame behind cladding gives none either; even a transfer structure, which is the most dramatic redirection of load in the subject, is usually a beam in a ceiling void that nobody sees. The route the force takes is a matter for the drawings.

A tree is the load path, at full size, in the room. Each split is a place where a force divides in a ratio that is visible from the geometry; each branch’s inclination is the direction the force is travelling; the trunk’s thickness against the branches’ is the accumulation. Somebody standing under one can read the whole structure without being told anything.

That is worth something, and it is worth saying plainly that it is an architectural value rather than a structural one — the volume argument above says the fan is nearly as good and the tube is four times better. But it is not nothing, and it is consistent with the rest of this collection’s habit: the drawing is the calculation, and a structure whose geometry is its own force diagram is a structure that can be understood by looking.

Where the model stops

The tree is planar and regular. Real branching columns are three-dimensional, branch into three or four rather than two, and are rarely symmetric — the roof grid they support decides where the tips must be, and the tips are not conveniently spaced.

The members are pin-ended. Every buckling length above is the member’s own length, which assumes the joints offer no rotational restraint. A welded tree joint offers a great deal, and taking credit for it moves the optimum down and reduces the volume — at the cost of the joint having to carry the moment that credit implies.

And the objective is material volume. That is the objective for which the mathematics is clean and it is almost never the objective a project has. Fabrication cost scales with joints, not with tonnage; erection cost scales with the number of pieces; and a tree with fourteen members and seven bespoke nodes can be more expensive than a fan with four straight ones twice the weight.

Two structures that are the same argument

The comparison the volume calculation sets up — a hierarchy against a fan — turns up twice more in this collection with the same arithmetic and different names.

A cable-stayed bridge against a suspension bridge. The stays are a fan from a tower to the deck; the suspension cable is a hierarchy, with the main cable carrying everything and hangers dividing off it. In tension there is no buckling constraint, so the fan is the efficient answer — and the cable-stayed form has displaced the suspension form for every span where the stays are short enough to be practical.

A column tree against a beam. A roof delivered to one point by a branching column is a roof whose spans are short; the same roof on a single column with a long beam is one long member in bending. Bending is the expensive way to carry anything, so the tree wins easily — and the comparison is not between two columns but between a column and a member whose depth would have to grow with its span.

What the three share is a question about whether load should be gathered before it is carried. Gathering costs joints and buys shortness; carrying costs length and buys simplicity. Every structure in this collection that looks like a hierarchy — a tree, a suspension cable, a secondary-and-primary floor framing grid — has answered that question one way, and every fan has answered it the other.

The same polygon, inverted into an archThe shape a string takes under 5 point loads, with a vertex at every load and a constant horizontal component of 27.9 throughout. Inverted, every tension becomes a compression of the same size and the shape carries the same loads as an arch.10148126H = 27.9, the same at every stationevery force reversed; the geometry untouched
Fig. 9 The form found for its own load case, inverted into compression. A branching column is a funicular structure for a symmetric set of tip loads, which is exactly what this polygon is for the loads it was drawn from — and it stops being one for the same reason.

What the pictures cannot show

The members are drawn at a thickness proportional to the area they need, which makes the branching visible and makes the drawing lie about the scale: a 9 m column carrying 400 kN has members of the order of 100 mm across, which at this scale would be hairlines.

Nor can the figures show the foundation. All of this is a comparison of superstructure volume, and the single most persuasive practical argument for a branching column has nothing to do with any of it: one foundation instead of eight. On a poor site that comparison can dominate everything computed above.

The assumption the figure rests on

The working stress is a single number applied to every member, in tension or compression, and the buckling requirement uses a fixed section shape. Both hide a real choice. The branches of a tree are shorter and less heavily loaded than the trunk, so the section that suits one does not suit the other, and a real tree is fabricated from a small number of sizes rather than from a continuum — which puts its volume above the optimum computed here by whatever the rounding costs.

The ladder from here

Later rungs on this anchor: Michell’s theorem itself, and why the minimum-volume structure for two loads and two supports is a fan rather than a hierarchy. The three-dimensional tree, where the branch directions have two angles and the joint has three equations. Joint stiffness and the buckling length credit it buys. The tapered branch, which is the fully stressed member when the axial force is constant and the buckling length is not. And the same argument run for tension structures, where the buckling constraint disappears entirely and the fan wins outright — which is why a cable-stayed mast has no branches and a tree does.

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.

Branching structureBucklingConcurrencyEquilibriumForm findingFree bodyFully stressed designFunicularJointLoad pathMaterial volumeOptimisationRadius of gyrationSlendernessSpace frame