The truss that is stiff by accident

Which member moved the roof wrote a truss’s deflection as a sum of one term per member, $\delta = \sum N n L / EA$, where $N$ is the force the real load puts in the member and $n$ the force a unit load at the joint of interest would put there. The member that is not worth stiffening read the sum as a ranking: a member whose $N$ or $n$ is zero contributes nothing however slender it is. The drawing that is right except for a rotation drew the same sum as a construction. All three took the members’ areas as given.

The ranking invites the obvious next question, and it is a design question rather than a check. If some members contribute a great deal and others nothing, a given amount of steel ought to be divided among them according to what each contributes. What division of a fixed volume makes a truss as stiff as it can be? And how far from that is the truss a designer actually draws?

The stiffest division, by one line of calculus #

Fix the volume, $V = \sum A_i L_i$, and minimise $\delta = \sum N_i n_i L_i / E A_i$ — the sum that is also the derivative of the truss’s stored energy with respect to a load at the joint. Each term falls as its area grows, and each unit of area costs $L_i$ of volume. At the optimum every member gives the same return on its last unit of volume — the derivative of its term, $N_i n_i L_i / E A_i^2$, divided by its cost $L_i$, is the same for all of them — which fixes

$$A_i \propto \sqrt{|N_i\, n_i|}$$

and the deflection that results is

$$\delta^* = \frac{\left(\sum L_i \sqrt{|N_i n_i|}\right)^2}{E V}.$$

The stiffest truss gives each member area in proportion to the geometric mean of its two forces. Not in proportion to the force it carries, which is what a strength designer does, and not in proportion to how much it matters to the joint, which is what the ranking measured — but to the square root of their product.

The strength design gets almost all of it #

The truss drawn below is an eight-panel Pratt truss, as deep as a panel is long, carrying equal loads at its top joints, drawn three times with every member as wide as its area. The first has equal areas throughout. The second is fully stressed: each member’s area is in proportion to its force, so every member works at the same stress, which is the strength designer’s natural answer. The third is the stiffest at mid-span.

Their mid-span deflections, for the same total volume of steel, are 43,503, 32,702 and 31,318 in units of load times length over modulus times volume. The equal-area truss is 39 per cent short of the best. The fully stressed truss is 4 per cent short — and it was never asked about deflection at all.

Across the three standard forms the pattern holds. A Pratt truss sized for strength is 4.4 per cent short of its stiffest, a Howe 5.0, a Warren 2.6. The equal-area versions are 35 to 40 per cent short. Sizing for strength has already done nearly everything that sizing for stiffness could.

Why the accident happens: an inequality #

The near-coincidence is not luck and it has a one-line explanation. Scale the fully stressed areas, $A_i \propto |N_i|$, to the volume $V$ and put them into the deflection sum:

$$\delta_{\text{fs}} = \frac{\left(\sum L_i |N_i|\right)\left(\sum L_i |n_i|\right)}{E V}$$

That is a product of two sums, and $\delta^*$ is the square of a sum of geometric means of the same terms. The Cauchy–Schwarz inequality says the first is never smaller than the second, and that they are equal exactly when $|N_i|$ is proportional to $|n_i|$ in every member — when the real forces are a scaled copy of the virtual ones.

That condition has a physical reading. The virtual forces are those of a unit load at the joint being measured. The real forces are those of the load the truss carries. If the load is a point load at that joint, the two sets are proportional and the fully stressed truss is exactly the stiffest truss there, which is a classical result. Equal loads at every top joint are not a point load at mid-span, but they bend the truss in nearly the same way: the chords carry forces that rise toward mid-span under both, and only the web members differ much. So the two patterns are nearly proportional, and the inequality is nearly an equality.

The fully stressed truss is stiff by accident, and the accident is how alike two sets of forces are. Where the load and the unit load bend the truss the same way, strength and stiffness ask for the same steel.

The same truss, twice #

Count where the steel goes and the two good designs are nearly the same truss. The fully stressed truss puts 67 per cent of its steel in the chords and 33 in the web; the stiffest puts 65 in the chords and 35 in the web, and no group differs between them by more than 3 per cent of the volume. What separates them is inside the groups — a little more in the bottom chord’s end panels for strength, a little more in the diagonals near mid-span for stiffness — and not the proportions a designer would notice on a drawing.

The equal-area truss is the one that is different in kind. It puts 57 per cent of its steel in the web, because the web has more members and longer ones, and the web is where the forces are smallest under both loads. Every criterion — strength, stiffness at mid-span, anything a designer might ask — dislikes steel spent on members that do not carry much, and the equal-area truss is the design that spends most of its steel there.

It is worth being clear what the comparison is not saying. A truss is rarely designed with equal areas, and nobody needs an optimiser to be told not to. The point of the equal-area truss is that it sets the scale: a 39 per cent loss is what bad distribution costs, and against it, the 4 per cent between the strength design and the best possible is small enough to call an accident rather than a shortfall.

The arithmetic, once, on a truss small enough to do by hand #

The inequality is easy to see on a truss with two kinds of member. Suppose the chords and the web have equal total lengths, that in the chords the real and virtual forces are equal, 1 and 1, and that in the web they are 1 and 4. Then $\sum L|N| = 1 + 1 = 2$, $\sum L|n| = 1 + 4 = 5$ and $\sum L\sqrt{|Nn|} = 1 + 2 = 3$, so the fully stressed truss deflects $2 \times 5 / 3^2 = 1.11$ times the stiffest: 11 per cent short. Had the web’s forces been in the ratio 2 instead of 4, it would be 3 per cent short; at a ratio of 10, 27 per cent.

That is the size of the effect in one line: the fully stressed truss loses only as much as the ratio of virtual to real force varies from member to member. A truss whose members all see the load and the unit load in roughly the same proportion loses almost nothing, and one in which some members carry much more of the unit load’s force than of the real load’s loses a great deal.

Where the accident stops #

Move the joint of interest from mid-span to the quarter point and ask the same question. The unit load there bends the truss asymmetrically, and its forces are nothing like the symmetric load’s: in four members of the web the real and virtual forces now have opposite signs.

The stiffest design for the quarter point is very different from the fully stressed truss, and very much stiffer where it was asked to be: 11,918 against 20,125, the fully stressed truss 69 per cent short. The inequality has stopped being nearly an equality because the two force patterns have stopped being alike.

And that stiffest-at-the-quarter-point truss pays for it. At mid-span it deflects 43,716 — more than the equal-area truss. It has moved steel into the members that serve the quarter point and out of the ones that serve mid-span, and a truss is not stiff in general; it is stiff at a joint, for a load.

The four members whose forces reverse are the strangest part of the answer. For them $N n$ is negative, so their term reduces the deflection at the quarter point, and it reduces it more the thinner they are. The optimiser therefore sends them to the least area it is allowed to give them. Left unbounded, the problem has no answer at all: a member that makes a joint move less by stretching the wrong way would be made infinitely thin. A member whose force reverses makes the structure stiffer was a curiosity in the first essay on this ranking; here it is a member the optimum wants to delete.

The stiffest truss is a mechanism #

The formula $A \propto \sqrt{|N n|}$ gives no area to any member whose $N$ or $n$ is zero. In this Pratt truss three members carry nothing under the symmetric load, so the stiffest truss has none of them. So does the fully stressed truss, for the same reason: a member at zero force needs zero area to be at any stress.

A determinate truss has exactly as many members as it needs to be rigid, so a determinate truss with three members removed is not a truss. It is a mechanism, which would fold under any load but the one it was designed for. The deflection of 31,027 that the calculus found is the deflection of a structure that cannot stand. It is the determinate truss’s lack of robustness turned into an optimisation result: every member of a determinate truss is essential, and an optimiser that values members by their contribution to one load case will delete the ones that contribute nothing to it.

So a real design needs a floor — a least area every member must have, for handling, for buckling under other loads, for the connections to be made at all. The figure prices it. A floor of a tenth of the equal area costs the stiffest design less than a per cent; a quarter, 2 per cent; a half, 5 per cent. The optimum is not a truss; the truss is the optimum with its mechanism filled in, and the filling is cheap because the members that were deleted were contributing nothing.

Deeper trusses, and the web #

Make the truss deeper and the chords carry less, because the chord force is the bending moment over the depth, while the web carries the same shear over longer members. More of the deflection moves into the web, and the web is where the real and virtual forces are least alike — under the symmetric load the diagonals near mid-span carry little, under a unit load at mid-span they carry half of it all the way to the support. So the fully stressed truss drifts further from the stiffest as the truss deepens: 3 per cent short at a depth of half a panel, 7 per cent at two panels, 8 at the deepest drawn. The gap widens and it does not open. The equal-area truss stays 26 to 43 per cent short at every depth, because putting equal steel in members that carry very unequal forces is the one division every criterion dislikes.

Reassurance, and the quarter point #

The practical content is reassuring and has a sting in it. For a roof truss under its ordinary load, checked for its mid-span deflection, the strength design is the stiffness design to within a few per cent, and nothing is gained by optimising for stiffness separately. The effort is better spent on depth, which the ranking already showed moves the deflection far more than any redistribution of area can.

The sting is the quarter point, and it generalises. Wherever the deflection that matters is produced by a load that bends the structure differently from the load that sized it — a crane rail’s deflection under a wheel near one end, a floor’s under a partition near a support, a bridge’s under a lane load on one half — the strength design can be far from the stiffest, and far from where the check expects. The inequality says exactly when: when the force pattern of the load that sized the members is unlike the force pattern of a unit load at the joint where the deflection is wanted.

The same pair of forces, elsewhere #

The product $N n$ is not special to trusses. It is the integrand of every virtual-work deflection, and the same optimisation runs wherever it appears. A beam of fixed volume made stiffest at mid-span puts its depth where $M m$ is largest, which is why the section that changes along the span is deepest at mid-span under a load that is also largest there. The tree that strength does not ask for found that a fully stressed branching column wants no trunk, which is the same kind of result from the strength side: an optimiser for one criterion discards what the other criteria need.

And the general moral is the one the best design is the most sensitive one found for a tube. An optimum for one load at one point is a design that has given up everything that load at that point does not use — here, whole members — and its weakness is exactly the case it was not asked about.

Every deflection, found two ways #

Every force is from the method of joints on the determinate truss, once for the real load and once for a unit load at the joint of interest; every deflection is the unit-load sum over the members at their chosen areas. Each design’s deflection is checked a second way, by assembling the stiffness matrix of the truss with its members at those areas and solving for the joint’s movement, and the two agree to the precision of the arithmetic. The fully stressed and stiffest designs are scaled to exactly the same volume as the equal-area truss, and any member the design would make thinner than the floor is held at the floor with the remaining volume redistributed.

Buckling, self-weight and every other load case #

Buckling. A compression member sized by its area alone is sized as though it could not buckle. The fully stressed truss’s top chord is thick where its force is large and thin where it is small, and a thin compression member is a slender one; its real area is set by a buckling curve, not by a stress.

Self-weight. The steel is weightless here. In a long span the truss’s own weight is a large part of its load, and moving steel changes the load it was moved for.

More than one load case. A real truss is checked for many loads — snow on half the roof, wind from each side, a load at every joint in turn — and a stiffest design for one is not the stiffest for another. The fully stressed truss has the same limitation, which is why real trusses are sized for an envelope rather than a case.

Forces that statics fixes before stiffness can #

That the truss is determinate, so that its forces do not depend on its areas. In a redundant truss, moving steel moves the forces too — the stiffer members attract more load — and the stiffest division has to be found by iteration rather than from a formula. The fully stressed design becomes an iteration as well, and the two iterations need not converge to anything alike. The clean inequality of this page is a property of structures whose statics decides their forces before their stiffness does.

Still open: the truss whose members are chosen from a catalogue #

Every area here is continuous: any member can be any size. A real truss is built from sections in a catalogue — angles, tees, hollow sections in steps of a few per cent — and a designer picks the smallest that works, member by member. The steps put steel where no optimiser wanted it, and the rounding is largest in the members with the smallest forces, which are the ones the optimum was trying to delete. Whether a truss built from a catalogue is closer to the fully stressed design or to the equal-area one — and whether the few per cent this page found survive the rounding at all — is a question about a discrete optimum, and discrete optima do not come from a single line of calculus.