Three equations at every joint

A triangle of pinned bars cannot fold and a square can, and the whole of plane trussing follows from it. The three-dimensional version of that sentence is shorter and less familiar: a tetrahedron cannot fold and a cube can, and almost nothing about the second half of that is obvious from looking at a cube.

The count, with a three in it #

A rigid body in space has six degrees of freedom, and a joint — a point that can move but not rotate — has three. So a pin-jointed frame of $j$ joints has $3j$ freedoms, and each requires an equation to be pinned down. Every member supplies one equation and every restraint component supplies one, which gives Maxwell’s rule in three dimensions:

$$m + r = 3j$$

against the plane version’s $2j$. The change is one digit and the consequences are out of proportion to it, because the shortfall it opens is not a small one.

The cube is the example worth carrying. Twelve edges, eight corners, six restraint components at the bottom: $12 + 6 = 18$ against $3 \times 8 = 24$. Six mechanisms, one for each face, and each of them is the square that folds, six times over. Every one of the six is invisible in a drawing and obvious the moment the frame is built and pushed.

Add a diagonal to each face and the count balances exactly. The result is a frame of eighteen bars for eight joints, which is a member-to-joint ratio of 2.25 against a plane truss’s typical 1.7 — and that ratio is the whole economic argument about three-dimensional frames. They need more bars per joint, and they get something for them.

The procedure is the one this collection first met in the plane, with the number of equations per joint changed from two to three and nothing else. Unknowns against equations, the sign of the difference naming the answer: fewer unknowns than equations is a mechanism, equal is solvable by statics alone, more needs stiffness. What changes in space is only how quickly the requirement grows, because every joint added asks for three equations rather than two.

The tetrahedron is the only unit #

In the plane the rigid unit is the triangle, and every rigid plane frame can be built by adding two members and one joint at a time. In space the rigid unit is the tetrahedron, and the corresponding rule adds three members and one joint at a time.

That is the operation every space frame in existence is built from, and it is worth stating as a rule about drawings: a new joint is stable if the three members reaching it are not coplanar. Three coplanar members leave the joint free to move perpendicular to their plane; two members leave it free to rotate about the line joining their far ends; one member leaves it on a sphere.

The tripod’s answer is worth checking by hand, because it is the one closed form in this essay. Three symmetric legs at an inclination whose vertical direction cosine is 0.8 carry $90 / (3 \times 0.8) = 37.5$ kN each. The solver returns 37.500, which is the whole of what a solver is for on a site where every figure is solved rather than sketched: to be trusted on the frames where nothing can be checked, having been checked on the ones where something can.

The count that does not see it, in three dimensions #

Every leg’s force has a line of action passing through the line joining the three feet. A rotation about that line lengthens no member, so no member resists it, and the structure is a mechanism whose count says it is determinate.

This is the count that does not see it in its three-dimensional form, and the diagnosis is identical. The count compares two integers; what actually decides the question is whether the equilibrium equations are independent, which is the rank of the equilibrium matrix. Here the rank is 2 where the count claims 3, and that shortfall of one is the mechanism.

The plane version of the same trap is easier to see and worth holding beside it: two frames can both satisfy m + r = 2j and only one of them stand, the other folding through a motion that costs no member any change of length at all. Nothing about the diagnosis changes with the dimension.

The three-dimensional cases are harder to spot, and that is the practical difference. A plane frame with a critical form usually looks wrong: three members meeting at a point, a panel with two parallel diagonals. A space frame with one can look entirely conventional, because the coplanarity that causes it is a relationship between three directions in space and a drawing shows two of them at a time.

The check is one line of arithmetic and it is never made by eye. Assemble the equilibrium matrix, take its rank, compare with the number of free freedoms. The solver behind every figure on this page does exactly that before it attempts anything else, and refuses to return forces when the rank is short.

What a space frame buys #

The counting is the cost. Here is the return.

Put the two numbers side by side. A single load on a plane truss produces a worst member force of 1.76 times the load; on the grid it is 0.36 times the load, a factor of five less. The plane truss’s worst member takes 10.9% of all the axial force in the frame; the grid’s takes 5.0%.

The mechanism of that improvement is not subtle: there are many routes from the load to the supports, and they share. A plane truss has exactly one route from a bottom node to each support, and every member on it is in series with every other. A grid has routes in two directions, four diagonals at every node, and thirty-three redundancies through which the sharing is negotiated.

Two consequences follow, and both are the reasons space frames are used where they are used.

A point load anywhere is tolerable. A plane truss is designed for loads at its panel points and suffers if one arrives elsewhere. A grid distributes any load over a region of itself, which is why plant, ductwork and lighting can be hung from a space deck almost anywhere without a member being changed.

Losing a member is survivable. With thirty-three redundancies the frame has thirty-three ways to be wrong and still stand.

The plane case is the stark one. Removing a single diagonal from a determinate truss does not weaken it; it turns it into a mechanism, which is why a structure that survives losing a member has to have had somewhere else for the load to go. A grid with thirty-three redundancies has thirty-three of those routes, and the arithmetic of the loss is a redistribution rather than a collapse.

Which free body produced the numbers #

Every force on this page comes from the same two-line construction, and it is the plane truss’s construction with a third row added.

At each free joint, sum the components of every member force meeting there, in all three directions, and set each sum equal to the applied load component. That is three equations per joint, and assembling them for the whole frame gives $\mathbf{A}\,\mathbf{t} = \mathbf{f}$ where $\mathbf{A}$ is the equilibrium matrix, $\mathbf{t}$ the member forces and $\mathbf{f}$ the loads.

For a determinate frame that system is square and is solved directly. For a redundant one it is not, and the extra information comes from the members’ stiffnesses in the usual way — the solve is $\mathbf{K}\mathbf{d} = \mathbf{f}$ with $\mathbf{K} = \mathbf{A}\,\text{diag}(EA/L)\,\mathbf{A}^{\!\top}$, and the member forces follow from the displacements, exactly as they do for a redundant plane frame.

The check that matters is made afterwards and independently: substitute the forces back into $\mathbf{A}\mathbf{t}$ and compare with $\mathbf{f}$. On every frame on this page the residual is of order $10^{-14}$ of the applied load, which is arithmetic rather than physics, and which would be of order one if the equilibrium matrix had been assembled with a sign error. The signs of an equilibrium matrix and of its transpose are the one thing a stiffness solve cannot tell apart on its own — both give plausible displacements — and the residual is what separates them.

A rectangular plan throws the argument away #

The sharing measured above is a property of a square grid, and it decays with the plan proportion far faster than anyone expects.

Two directions spanning the same point must deflect the same amount there, so the load divides in inverse proportion to each direction’s flexibility. With the same grid in both directions that flexibility goes as the fourth power of the span, and the share taken by the short direction is

$$\frac{w_{short}}{w} = \frac{L_{long}^4}{L_{short}^4 + L_{long}^4}$$

which is the slab’s own two-way rule arriving unchanged, because it depends on nothing but the two flexibilities.

At two to one the grid is a one-way structure carrying a two-way member count. Ninety-four per cent of the load goes the short way, the long-way members are carrying six per cent and are sized by their own slenderness rather than by anything applied, and every one of the four advantages listed above has quietly gone: the load has one route again, so a point load concentrates again, and a member loss on the short-way route is not shared out by members that are barely working.

The threshold is tighter than the table looks. By 1.25 to 1 the split is already 71/29, and the frame is buying a doubled connection count for a 29 per cent second load path. Roughly 1.2 to 1 is where a space grid stops being worth its node count on this argument alone, which is why space decks are built on square, round and near-square plans and why a long rectangular hall is roofed with plane trusses at a spacing.

The escape, where the plan cannot be changed, is to make the two directions unequal on purpose — a deeper or stiffer grid the long way, so the flexibilities come back toward each other. That works, and it is worth noticing what it costs: the material is being spent to restore a sharing that a square plan would have supplied for nothing.

The tolerance on a member length is a design value #

Thirty-three redundancies were counted above as an asset. They are also thirty-three self-stress states, and a self-stress state is what a member of the wrong length excites.

Put a bound on it. A 3 m tube of 1,000 mm² fabricated 3 mm long is asked for a strain of $10^{-3}$ if the frame refuses to move at all, and

$$N = \frac{\delta EA}{L} = \frac{0.003 \times 210{,}000 \times 1{,}000}{3{,}000} = 210\ \text{kN}$$

against a largest force from the applied load of 10.7 kN. The bound is twenty times the design force, from an error of one part in a thousand in a member length.

The frame is of course softer than fully rigid, so the realised force is some fraction of the bound — but the fraction is a property of the grid rather than of the error, and in a frame this redundant it is not small. That is the whole of the difference from a determinate structure, where the bound itself is zero: a determinate frame with a member 3 mm long is a frame with a joint 3 mm out of position and no force anywhere.

So three things that look like site practice are structural decisions:

Slotted holes or an adjustable node. Every proprietary node system has take-up in it somewhere, and the take-up is sized against this arithmetic rather than against convenience.

The erection sequence. A grid assembled from the middle outward accumulates its errors toward the perimeter, where the last members are the ones that will not fit; assembled from a measured datum, the error is shared.

And the last member in. Closing a highly redundant frame is the moment every accumulated tolerance is resolved at once, which is why that member is often deliberately made short and packed, or left until the roof is on and the dead load has taken the geometry where it is going to sit.

None of that appears in $\mathbf{A}\mathbf{t} = \mathbf{f}$, because the load vector is zero throughout.

The economics, which are not the statics #

Space frames are not used everywhere, and the reason is in the counting rather than in the performance.

The member-to-joint ratio is the cost. A determinate plane truss runs at about 1.7 members per joint; a determinate space frame needs 3 minus the restraint contribution, which for a large frame is essentially three members per joint. Nearly twice as many bars, and — more expensively — nearly twice as many ends, since every bar has two of them and every end is a connection.

The connection is the product. In a plane truss a joint is a gusset plate and a few bolts, and it is made by the same fabricator who cut the members. In a space frame a joint is a machined ball or a cast cluster receiving up to nine members from directions that vary from node to node, and it is a manufactured component with a catalogue number. That is why proprietary node systems exist at all, and why the geometry of a space deck is usually chosen from what a node system can make rather than from what the loads want.

So the argument is about spans. The load sharing measured above improves with the size of the grid, and the connection cost is per node whatever the span. Below roughly a fifteen-metre span, plane trusses or portal frames at a spacing win comfortably. Above thirty metres, with an irregular plan or with loads that can arrive anywhere, the grid wins — and it wins on all four of the properties above at once, which is unusual enough to be worth naming: it is lighter, it is more tolerant of a point load, it is more robust to a member loss, and it does not care about the plan shape.

And there is a fifth reason that is not structural. A grid built from one member length and one node is a kit, and a kit is assembled by semi-skilled labour at ground level and lifted. A one-off truss is fabricated in a shop and craned into position in pieces. That difference decides more space-frame roofs than any of the arithmetic above.

Where the model stops #

Every joint here is a pin. A real space-frame node is a machined ball or a welded cluster, and the members are bolted into it with an eccentricity of tens of millimetres. That eccentricity puts bending into members designed for axial force, and the secondary moments are usually assessed and usually small — which is the same accommodation a plane truss’s rigid joints get.

The redundancy that shares the load also traps forces. Thirty-three redundancies mean thirty-three self-stress states, and a member fabricated a few millimetres too long will develop force in every one of them with no load applied at all. Space frames are sensitive to fit-up in a way determinate frames are not, and the tolerance on a member length is a structural quantity.

Nothing here is a buckling calculation. Sixty members in compression is sixty stability problems, and in a grid whose members are short and stocky the answer is usually favourable — but the system can buckle too, as a shallow shell rather than as a member, and no count of members sees that at all.

And the support condition has been assumed. A grid supported round its whole edge is a very different structure from the same grid on four corner columns, and the difference is much larger than the equivalent difference for a beam. Edge supports let the load out in the shortest direction available; corner supports force it round the perimeter first.

What the pictures cannot show #

Every three-dimensional figure here is a projection, and a projection of a pin-jointed frame is a picture in which two members that cross on the page may or may not meet. The drawings are painted back to front and the members are drawn heavier where the force is larger, which helps and does not solve it.

The mechanism figure draws a frame that folds, and the folding cannot be drawn — the members are shown in their original positions with an arrow indicating a push. A mechanism is a motion, and a static drawing of one is a picture of a frame that looks perfectly serviceable.

And the spread figures report a count of members carrying more than a tenth of the maximum, which is a summary statistic standing in for a distribution. Two grids with the same count can have very different distributions, and the number is quoted because it is the honest short version rather than because it is the whole answer.

The ladder from here #

Later rungs on this anchor: the double-layer grid in detail — square-on-square, square-on-diagonal, and why the offset matters. Node systems as an engineering subject in their own right, where the connection cost dominates the member cost and decides the geometry. The single-layer grid shell, which is not a truss at all because its nodes must carry moment. System buckling of a shallow space deck, where the frame goes over as a surface. Tensegrity, where the count is satisfied by a state of self-stress rather than by members. Deployable and folding frames, which are mechanisms on purpose. And the historical case: Alexander Graham Bell built tetrahedral kites and towers in the 1900s from exactly this argument, twenty years before anybody wrote the counting rule down for three dimensions.