The calculus answer, rounded, is the worst one

The truss that is stiff by accident divided a fixed volume of steel among the members of a determinate truss and found the division that makes one joint as stiff as it can be: each member’s area in proportion to $\sqrt{|N n|}$, the geometric mean of the force the load puts in it and the force a unit load at the joint would. The strength designer’s division — area in proportion to force, every member at the same stress — came within four or five per cent of it at mid-span. The reason was the Cauchy–Schwarz inequality, and the gap closed exactly when the load and the unit load pull on the members in the same proportions.

Every area in that argument was a real number. Any member could be any size, and the optimum was a formula. Nobody builds a truss that way. The members come from a catalogue — angles, channels, hollow sections — in which the next size up is typically ten to thirty per cent larger than the last, and a designer takes, member by member, the smallest section that works. So the question that essay ended on is a practical one. Does the rounding eat the few per cent the proportioning found? And if a deflection limit is what governs, how should the extra steel be chosen when it can only come in steps?

The answers point in opposite directions. The first costs almost nothing. The second is where the money is, and the obvious method is the worst.

A staircase of areas, and the truss set on it #

The truss is the one the earlier essay used: an eight-panel Pratt truss, as deep as a panel is long, carrying ten kilonewtons at each top joint. Every force is fixed by statics, because the truss is determinate, and the forces decide everything else. A member’s area is measured here in kilonewtons at the allowable stress, so a member of area 75 working at the allowable stress carries 75 kN: the number is simply the force the member could carry.

The catalogue is idealised in one respect and only one. Its sections form a geometric progression, each a fixed factor larger than the one below — a quarter larger, to start with — so that its steps are the same size in proportion everywhere along it. Real tables are ragged, with steps of five per cent in some places and forty in others, and the argument below is built so that the raggedness can be dealt with afterwards: every result is computed for every possible position of the steps relative to the members’ forces, and reported as a band. The smallest section is a tenth of what the most heavily loaded member needs. That is the floor that kept the optimum from being a mechanism, now supplied by the catalogue itself rather than chosen: nothing smaller is on sale.

Rounding up has a characteristic shape — the staircase that a single member chosen from a list climbs as its load grows, seen here from the other side — and it is worth seeing before anything is measured. Take a member needing an area $x$. The section it gets is the first in the list at or above $x$, so the factor by which it is over-provided, section divided by need, starts at one when $x$ lands exactly on a section and rises toward the catalogue’s ratio as $x$ falls toward the section below. Plotted against $x$ on a logarithmic scale that is a sawtooth, and every member of the truss sits somewhere on it.

Where it sits is, for practical purposes, an accident of arithmetic. The two bottom-chord members at mid-span need 75.0 and get 75.7, one per cent over. The top-chord members beside them need 80.0 and get 94.6, eighteen per cent over, because 80 happens to fall just past a step. Nothing about their structural role decides it. The members needing less than the smallest section are the exception: a diagonal near mid-span needs 7.1 and gets 8.1, and the two verticals that carry nothing get 8.1 as well, so on these the overshoot is set by the floor rather than by the steps, and for the two that carry nothing it is infinite. Summed over the truss, the rounded members hold 9 per cent more steel than their forces need.

Rounding for strength leaves the stiffness where it was #

The number that matters for stiffness is not the deflection alone, since more steel always deflects less. It is the deflection for the steel used. In the unit-load sum,

$$\delta = \sum_i \frac{N_i\, n_i\, L_i}{E A_i},$$

which is the sum that names which member moved the roof, multiplying every area by the same factor divides $\delta$ by it, so the product of the deflection and the volume $V = \sum A_i L_i$ is a property of the proportions of the design and not its size. Its least possible value is the one the earlier essay found, $\delta^* V = (\sum L_i\sqrt{|N_i n_i|})^2/E$, and every design can be scored by how far above it it sits: the equal-area truss at 1.40, the fully stressed truss at about 1.05.

Round the fully stressed truss up to a catalogue whose sections are a quarter apart and it moves from 1.047–1.053 to 1.050–1.066, depending where the steps fall. At steps of a half, 1.046 to 1.078. Even a catalogue that doubles from one section to the next, which no real table does, leaves the truss between 1.075 and 1.112 — much nearer the fully stressed design than the equal-area one, which the same scale puts at 1.40. The lower edge of the band sometimes dips below the unrounded truss: the rounding has, by accident, moved steel a little toward the members the stiffest design wanted.

So the first answer is plain. The few per cent between strength and stiffness survive a catalogue. The rounding changes how much steel the truss has; it barely changes how well that steel is placed.

Why the rounding is second order #

The reason is the same inequality that made the strength design nearly the stiffest, used once more. Rounding multiplies each member’s area by a factor between one and the catalogue’s ratio. If every factor were the same, the design would simply be the fully stressed truss scaled up, and its deflection-times-steel would not move at all. What moves it is only the spread of the factors: some members got 1 per cent extra and others 22.

Write each factor as $1 + \varepsilon_i$ and expand. The volume grows by the steel-weighted mean of the $\varepsilon_i$; the deflection falls by the deflection-weighted mean. Where those two weightings agree, the two first-order changes cancel and what is left is second order — the size of the variance of the $\varepsilon_i$. Where they disagree, a first-order term survives, but its sign depends on which members happened to be rounded most, so it can go either way, and its size is the spread of the $\varepsilon_i$ multiplied by how far apart the two weightings are.

In the fully stressed truss the two weightings are close. A member holds steel in proportion to $|N_i| L_i$ and contributes deflection in proportion to $|n_i| L_i$, and those are the two patterns whose near-alikeness made the strength design nearly the stiffest in the first place. So the rounding’s cost is small for the same reason the strength design was good, and the part of it that is first order is as likely to help as to hurt, which is why the band straddles the line of the unrounded truss.

At steps of a quarter the factors are spread evenly, on a logarithmic scale, between 1.00 and 1.25; their standard deviation is about six per cent and its square 0.4 per cent. The band is 1.6 per cent wide, a few times that. A catalogue with steps of ten per cent gives a band too thin to see.

The extra steel is bought at the ordinary price #

The same fact seen the other way round: whatever steel the rounding adds, the truss gets the stiffness for it. Plot the rounded truss’s steel, as a multiple of what its members need, against its deflection, as a fraction of the unrounded truss’s, and the points lie along the curve on which deflection falls in inverse proportion to steel. A catalogue with steps of a half adds between 11 and 36 per cent to the steel and takes between 8 and 24 per cent off the deflection, almost exactly in step.

This has a consequence that a designer checking deflection last will recognise, and it is a reminder that stiffness is not strength even when one calculation supplies both. A truss sized from a catalogue is stiffer than the calculation that sized it, by roughly the steel the rounding added, and that is stiffness that can be counted: the rounded truss here deflects 23.8 units at mid-span against 25.7 for its members’ exact needs. In those units — the allowable stress times a panel’s length over the modulus — and with three-metre panels in ordinary steel, the difference is about six millimetres on a twenty-four-metre span, found in material the strength check had already paid for.

When the deflection limit governs #

The interesting case is the one the strength design does not settle. A long, lightly loaded truss is often strong enough with far less steel than it needs to be stiff enough, because deflection grows as a high power of span and stress does not: the deflection limit asks for more steel than the forces do. The designer must add steel, every member keeping at least its strength section, and the question is where. With continuous areas the earlier essay’s calculus answers it at once. The least steel that meets a deflection limit, every member at or above its strength area, is

$$A_i = \max\!\left(A_i^{\text{strength}},\; c\,\sqrt{|N_i n_i|}\right),$$

with $c$ raised until the limit is met: members whose stiffness share outgrows their strength area are enlarged along the square-root rule, and the rest are left alone. It is a lower bound. No catalogue can do better, because a catalogue only removes options.

With a catalogue there are four ordinary ways to proceed, and they are not equally good.

Scale every member alike. Multiply every strength section by one factor, round up, and raise the factor until the limit is met. On a geometric catalogue that is exactly moving every member up the same number of sections, which is what a designer does who “goes up a size throughout”.

Round the calculus. Compute the continuous optimum above, then give each member the next section at or above its area. This feels like the most careful method, since it starts from the right answer.

Step greedily. Starting from the strength design, repeatedly move one member up one section — always the member for which that step removes the most deflection per unit of steel it adds — until the limit is met.

Solve for the discrete optimum exactly. The deflection is a sum with one term per member, and so is the steel. Choosing one section per member to minimise the sum of the steel while the sum of the deflection terms stays under the limit is a multiple-choice knapsack, the problem of packing a bag to a weight limit when each item comes in several sizes, and it can be solved exactly by dynamic programming over the deflection budget.

The picture separates them into two groups with a gap between. The exact discrete optimum is between 0.5 and 0.8 per cent above the continuous bound at every stiffening factor from 1.25 to 3. The catalogue, chosen well, costs the stiffest truss less than one per cent. The greedy steps come within about one per cent of the exact optimum, and at some factors they find it.

The other two are 4 to 13 per cent worse. Scaling every member alike needs 4 to 10 per cent more steel than the optimum. And rounding the calculus up, the method that starts from the right answer, needs 7 to 13 per cent more — the worst of the four across nearly the whole range.

Why rounding the right answer up is wrong #

The continuous optimum puts every member exactly where its last unit of steel buys the same deflection as every other member’s. Round each member up to the next section and every member receives an overshoot of up to a quarter, taken without reference to the limit. The deflection then lands well below the limit — in this truss the rounded optimum at a stiffening factor of 1.5 deflects 14.8 units against a limit of 15.8, seven per cent stiffer than anyone asked for — and all of that surplus was paid for.

Rounding up is how a strength check must work, because a member one per cent too small for its force is unsafe whatever the rest of the truss does. It is not how a deflection limit works. The limit applies to a sum, and a sum can be met by rounding some members up and others down, or by leaving some at their strength sections and stepping others two sections at once. The discrete optimum does exactly that. Of the members the continuous bound would enlarge, the optimum gives some more than the bound and some less, and it balances them so the total lands just inside the limit — 15.83 against 15.85. An optimum spends every margin it is given, which is why the best design is the most sensitive one: a member found a size too small on site takes this truss over its limit, where the rounded calculus would have absorbed it.

The calculus answer is the right answer to a question with continuous areas. Rounded, it is an answer to no question at all: not the least steel, which it overshoots, and not the stiffest truss for its steel, which would need the steel re-proportioned.

Which members the good designs enlarge #

The greedy method has the advantage that its choices can be read one at a time, and the order is instructive. Its first six steps all go to the two diagonals that meet at mid-span, three sections each. Those are members the strength design made as thin as the catalogue allows, because the symmetric load puts only seven kilonewtons into each, and yet a unit load at mid-span sends 0.71 through each of them — the whole of its shear, carried at forty-five degrees. Their ratio of virtual to real force is the largest in the truss, and a thin member with a large virtual force is the cheapest place in the structure to buy deflection. Next come the verticals one panel either side of mid-span, and then the next pair of diagonals out, for the same reason. Only after that does the method begin on the chords, which hold most of the steel and most of the deflection, but already work hard for strength.

The optimum at a factor of 1.5 ends by enlarging twenty-four of the twenty-nine members, the whole of both chords among them, and the steel it adds is 44 per cent of the strength design’s. The five members it leaves alone are informative. The two verticals that carry nothing under the load, and the central vertical, which carries ten kilonewtons under the load and nothing under the unit load, have $N_i n_i = 0$: enlarging them changes no deflection, which is the member that is not worth stiffening arriving as a design result rather than a curiosity. The two end diagonals are the subtle case. The continuous bound would enlarge them, by nine per cent, but the next section up is a quarter larger than theirs, and at that price the same steel removes more deflection almost anywhere else; the discrete optimum leaves them at their strength sections. It is the catalogue, not the structure, that exempts them.

There is one detail that no continuous argument could produce. The discrete optimum is not symmetric, although the truss and its load are. One of the two bottom-chord members at the third panel is given 94.6 and its mirror image 75.7, a whole section apart. The asymmetric design is the cheapest that fits under the limit because moving both up would overshoot it, moving neither would fall short, and moving one is what the budget can afford. On a real drawing the designer would give both the larger section and pay the difference, for reasons that are about fabrication rather than steel, and it is worth knowing that the difference is the price of symmetry and not of stiffness.

What a smaller range of sections costs #

A designer rarely uses twenty-nine different sections. Fabrication and procurement push toward a few — one for the top chord, one for the bottom, one for the web — and the cost of that rationalisation is a second kind of rounding, much coarser than the catalogue’s, since every member of a group is given the section its most heavily loaded member needs.

It is a very different kind of rounding, and the stiffness efficiency says so. One section per group of members uses 1.50 times the steel of the member-by-member design and is 1.35 times as stiff; its deflection times its steel rises from 1.059 of the best to 1.173. Two sections — chords and web — are 1.76 times the steel, 1.50 times as stiff, and 1.240 of the best. One section throughout reproduces the equal-area truss exactly, 1.402 of the best, with 2.21 times the steel.

Unlike the catalogue’s rounding, this one is first order, because its overshoot is not scattered at random along a sawtooth. It is systematic: every chord member near a support is given the section the mid-span member needs, and those are the chord members that do least for the deflection at mid-span. The extra steel lands where both forces are small, which the square-root rule says is the worst place for it. A catalogue’s rounding is noise and costs its square; a group’s rounding is a bias and costs itself. That is worth knowing before a deflection limit is met by a rationalised design, because the rationalisation has already bought stiffness, but at a poor rate, and the steel the limit then asks for is added on top of steel that was spent badly.

The knapsack, and why the greedy steps nearly solve it #

A knapsack with a weight limit is, in general, a hard problem — hard in the technical sense that no known method solves every instance quickly — and a greedy method can be arbitrarily bad on it. Two things make this instance easy.

The first is that every member’s deflection term is convex in its area — the unit-load sum is a derivative of the stored energy, and each member’s share of that energy falls off smoothly as it is made larger: $N_i n_i L_i / E A_i$ falls less and less as the area grows, so each successive section up saves less deflection than the last. The greedy rule, which takes the best remaining rate of saving at each step, never passes a good step to take a worse one in the same member, and the only error it can make is at the very end, where the last step it takes may overshoot the limit by most of one section’s worth of deflection when a smaller step elsewhere would have landed exactly. That error is bounded by one section of one member.

The second is that the truss has many members and the steps are small against the whole. One section of one member is a percent or two of the truss’s steel, so the greedy method’s worst case is a percent or two, and the band it actually shows is 0 to 1.1 per cent. The exact method needs a budget cut into many thousands of pieces and a table over them, but on a truss of this size it takes a fraction of a second, and its answer is certified rather than hoped for.

Buckling, connections, and the bands that are not measured #

Buckling. Every section here is chosen by area. A compression member in a real truss is chosen by a buckling curve that allows for a member that was never straight, which makes its required area depend on its length and its section’s radius of gyration as well as its force, and moves the strength design away from the fully stressed one — the top chord gets heavier than its force alone asks. The stiffness arguments survive, since deflection still depends only on area, but the strength floor under every member moves up, and the continuous optimum has less room to work in.

Connections. A section is also its connections, and a truss with twenty-nine sections has more distinct joints to detail. The cost of rationalising is paid partly back in fabrication, and nothing on this page prices that.

The catalogue. The catalogue here is geometric and complete. A real table has gaps, sections of equal area and different shape, and sections that are cheaper per kilogram than their neighbours. The bands in the figures cover every position of a regular catalogue’s steps; they do not cover an irregular one, although the argument about why rounding is second order does not depend on regularity.

Every number, found twice #

Every force is from the method of joints on the determinate truss, once under the load and once under a unit load at mid-span, and every deflection is the unit-load sum over the members at their chosen sections. The continuous optimum is found by raising $c$ in the square-root rule until the limit is met. The exact discrete optimum is a dynamic programme over the deflection budget cut into twelve to twenty thousand pieces, each member’s deflection term rounded up to a whole piece so that the answer it returns is certain to meet the limit; any slack the rounding leaves is then spent by stepping members down one section while the limit still holds. The greedy method is a separate calculation that shares nothing with the programme but the sums, and it never needs less steel than the programme finds, which is the check that the programme is not reporting a design that does not exist.

A determinate truss, which is why this was easy #

The truss is determinate, so its forces are fixed by statics and do not change when its sections do. That is what makes the deflection a sum of independent terms, one per member, and what turns the discrete problem into a knapsack. In a redundant truss, changing one member’s section changes the forces in the others, the terms are no longer independent, and the problem becomes a genuinely hard discrete optimisation. The greedy method still works as a heuristic, and still reads as an order of priorities; the certificate that it is within one per cent is lost.

Still open: the truss whose forces change with its sections #

The knapsack held because each member’s force was decided before its area was. In a truss with one redundant member — a diagonal in both directions in one panel, say — the forces follow the stiffnesses, a member made larger attracts force from its neighbours, and a section chosen for strength may no longer be strong enough once the section beside it has been enlarged for stiffness. The fully stressed design becomes an iteration, the stiffest path takes the load, and the two may not converge to the same truss. Whether the discrete optimum of a redundant truss is still within a per cent of its continuous bound, and whether the strength check and the stiffness check can then disagree about which member is too small, is a question about the one kind of truss where proportioning and analysis cannot be done in either order.