The section that yielding cannot reach

The truss sized to lose a diagonal took an eight-panel Pratt truss with a counter-diagonal in every interior panel and asked what it costs to make it survive the loss of any one member. Sized elastically, with every damaged case reanalysed and every member held to its allowable stress, the answer at two thirds of the full load was 1,304 units of steel against the intact truss’s 1,200: the counters starved by the fully stressed design came back, the verticals doubled, and losing a chord cost about twice the truss.

It ended on what looked like a large saving left on the table. An accident is the one state in which a structure is allowed to be damaged and still stand. The rules for alternative load paths let a damaged structure yield, and the lower-bound theorem then says that a design is safe if any distribution of forces carries the load in equilibrium with no member past its strength — no compatibility, no stiffness, no question of which path the truss would choose elastically. The elastic sizing had asked each damaged truss to carry the forces it would pick; the plastic sizing need only show that some set of forces works. That freedom ought to be worth a great deal.

It is worth seven units.

The number first, and how it was found #

The least weight of a truss checked by the lower bound in several cases at once is a linear programme. In every case the forces in the members left are written as one equilibrium set — the elastic forces at unit areas will do — plus any combination of the truss’s self-stress states, which are the force patterns a redundant truss can carry with no load on it. The intact counter-braced truss has six, one for each interior panel; with a diagonal gone it has five. The unknowns are the members’ areas and, in each case, the five or six amounts of self-stress; every member in every case must carry its force within its strength, in tension or in compression; and the objective is the volume. Thirteen cases — the intact truss under the full load and each of the twelve interior diagonals removed under two thirds of it — make 886 inequalities in 101 unknowns, solved exactly, and the answer is checked twice: every case’s forces are rebuilt and tested against the areas, and the volume is compared with the programme’s dual.

The answer is 1,296.7. The elastic sizing found 1,304.1. Allowed to yield, the damaged truss is lighter by 7.4 units, out of the 104 that robustness costs — about a fourteenth of the price, and a little over half a per cent of the truss.

A cut through the damaged panel #

The reason is visible in one free body, and it is the oldest move in statics. Take the panel whose Pratt diagonal has been lost and cut the truss vertically through it.

In the intact truss that cut would meet four members — two chords and two diagonals — and three equations cannot fix four unknowns, which is exactly why the split of each panel’s shear between its two diagonals was free and why the fully stressed design found a whole family of equally light ones. With one diagonal gone the cut meets three. The part to the left carries a known reaction and a known joint load, and three equations answer three questions without reference to anything else in the structure: vertical equilibrium gives the counter’s force, and moments about the two chord joints give the chords’.

A force fixed by a section is fixed for every analysis that satisfies equilibrium, and the lower-bound theorem is nothing but a licence to choose among equilibrium solutions. It has nothing to choose here. The counter must carry 23.6 whatever the truss does, elastic or plastic, stiff or yielding, because no other member crosses the cut that could share it. The robust design’s interior diagonals were sized by exactly this — each by its partner’s loss — and so the plastic design’s interior diagonals are the elastic design’s, member for member, at $4\lambda$ times the sum of the panel shears: 240 units at two thirds of the load.

Where the seven units are #

The chords do not move: the moment at every section of the damaged truss is the moment at the same section of the intact one, because losing a diagonal changes how a panel’s shear crosses it and not the load. The end diagonals do not move: they sit in triangles and carry the reaction whatever happens inside. The interior diagonals do not move, for the reason the cut gives. Everything the lower bound saves is in the verticals, and there it saves 6.4 of the 43 units the elastic sizing had added.

The verticals are the one place compatibility had been doing work. A vertical in a counter-braced truss carries the difference between how the diagonals either side of it divide their shears, and in the damaged truss elastically analysed, that division in the undamaged panels is the one their stiffnesses produce — not the one that would suit the vertical. The lower bound is free to choose it. Beside a lost diagonal, the neighbouring panels can split their shears so as to hand the vertical as little turning as possible, and the vertical can be smaller. The first interior vertical cannot: with the second panel’s Pratt diagonal gone it is the only way from the top of the end diagonal down to the foot of the counter, a cut round its joint fixes it at the whole damaged reaction, 23.3, and it is 23.3 in both designs. The saving is in the middle verticals, which the elastic analysis had loaded more than equilibrium required.

A design office would not accept a truss that yields in service, so the fourth bar holds the intact truss elastic under the full load and lets only the damaged cases yield. It needs 1,297.8. The saving falls from 7.4 to 6.3: the verticals give back 7.7, and the diagonals take 0.8 of it back to keep the intact truss within its allowable stress.

How hard the plastic design has to yield #

A lower-bound design is safe only if the truss can actually reach the forces it assumes, and it reaches them by yielding: the members the elastic analysis would overload stretch or shorten plastically until the forces have moved to the distribution the design was sized for. That needs ductility, and how much it needs is a question the elastic analysis of the plastic design answers.

Two things in it are worth reading. After a counter’s loss the plastic design is exactly at its strength and no further, so no member has to yield at all: the Pratt diagonals, which take over, were already sized for it, and the verticals the elastic analysis loads are no larger than the ones the lower bound chose. After a Pratt diagonal’s loss the elastic forces overshoot by up to 31 per cent, in the vertical that the lower bound shrank. That vertical must yield until the neighbouring panels have redistributed their shears — a modest plastic strain in a short member in tension or compression, well within what a steel vertical can deliver if it is stocky enough to yield in compression rather than buckle.

That condition is the whole of the caveat. A lower-bound design is a promise that the structure can find the forces; a member that buckles instead of yielding sheds its load rather than holding it at its strength, and the promise then belongs to a different and much less forgiving theory. It returns, in a larger form, below.

The price against the load it must carry #

The two-thirds load is the accidental combination counted statically. The essay before this one followed the elastic design’s price from a tenth of the load to twice it; the lower bound can be followed over the same range.

The two lines lie on each other. The largest gap in relative terms is at the low end, where a third of the load costs 20.5 units elastically and 6.7 plastically: there the robust truss is barely different from the intact one, its counters already large enough, and the saving is the elastic analysis’s fussiness about how the shear divides in panels nowhere near the loss. Above about half the load the interior diagonals are sized by their partners’ losses, which no redistribution can touch, and the gap stays between seven and twenty-two units while the price climbs into the hundreds and then the thousands. At four thirds — the accidental load with a dynamic factor of two — the plastic count saves 14 of the 874 units the elastic sizing spends on robustness.

The price of robustness in this truss is a property of its statics, not of its stiffness. The elastic sizing had looked like a demand that the damaged truss be as good as new; it turns out to have been, almost exactly, the demand that each damaged panel be able to carry its own shear, which is what equilibrium alone asks.

The chord that cannot borrow #

The case where yielding looked most promising was the lost chord. Elastically, surviving the loss of any top chord cost 2,458 units, twice the intact truss, because the panel’s two diagonals had to carry its moment as a couple across the gap and every panel needed diagonals able to do it. The hope was that, allowed to yield, a lost chord’s force could be spread across every diagonal in the truss instead of only its own panel’s.

It cannot, and for the same reason. A cut through the panel whose top chord is gone meets the bottom chord and the two diagonals — three members — and the free body to its left fixes their forces: the panel’s moment has to be carried by its own diagonals, whatever any other panel does. Yielding takes 95 units off the top chord’s price, 4 per cent, and leaves it at nearly twice the truss. The members a design would most like to be able to lose are the ones whose alternative path a section fixes most completely.

The Pratt diagonals’ bar is the one that moved most, by 35 units. A truss asked to survive only a Pratt diagonal’s loss keeps its counters, which then carry the shear in compression, and the verticals around them; the plastic sizing has a little more room among the undamaged counters than when it must also survive a counter’s loss, and uses it.

What does move the price #

If yielding buys so little, the price must be set somewhere else, and the figure that shows where is the one about compression.

Every design so far has treated a diagonal as equally strong in tension and in compression. A counter is the member that carries its panel’s shear in compression when the Pratt diagonal beside it is lost, and a counter is typically a long, light member — a flat bar, an angle, a pair of channels — whose compression strength is set by buckling rather than by yield. Let the interior diagonals’ strength in compression be a fraction $\chi$ of their strength in tension, and keep the chords and verticals stocky.

The curve is exact and has a closed form, because the cut makes it one. With its Pratt diagonal gone a counter must carry its panel’s damaged shear $\lambda V$ in compression, as a force $\lambda V\sqrt 2$ along a length $\sqrt 2$, at a strength $\chi\sigma$ per unit area: a volume $2\lambda V/\chi$. With the counter gone, its partner carries the same shear in tension at the full strength: $2\lambda V$. A panel’s pair costs $2\lambda V(1 + 1/\chi)$, and over the six interior panels, whose shears add to 90, the diagonals need $120(1 + 1/\chi)$ at two thirds of the load. At $\chi = 1$ that is 240; at a half, 360.

A counter good for half its tension in compression adds 120 units, more than the 97 that robustness cost with stocky ones, and about sixteen times what yielding saved. At three tenths — a slender flat bar — the robust truss needs 1,576.7, nearly a third more than the intact one. The intact design itself does not care: with weak counters it simply routes every panel’s shear through the Pratt diagonals in tension and needs 1,220, the plain Pratt truss’s weight.

The drawing shows where the extra steel sits. The Pratt diagonals are unchanged, because whichever diagonal is lost they are in tension; the counters double, because they are in compression exactly when they are needed. The elastic essay’s closing worry — that a counter made as a slender bar is a diagonal that is absent whenever it would be in compression — is here priced. A counter-braced truss whose counters can only work in tension cannot survive the loss of a Pratt diagonal at all; one whose counters are good for half their tension can, at the price of doubling them.

A hand check on the second panel #

The second panel carries a shear of 25 under the full load, so $\tfrac23 \times 25 = 16.7$ damaged. With its Pratt diagonal gone, the cut in the first figure says the counter carries all of it: $16.7\sqrt 2 = 23.6$ along its length. At a strength of one in compression it needs 23.6 of area; at half, 47.1. With the counter gone the Pratt diagonal carries the same 23.6 in tension and needs 23.6 either way. The pair needs 47.1 with stocky counters and 70.7 with slender ones, against 35.4 in the intact truss, where the pair shared the shear at whatever split the sizing chose.

The vertical at the left of that panel, cut round its top joint with the Pratt diagonal gone, carries the whole damaged reaction: $\tfrac23 \times 35 = 23.3$. That number is in the elastic design, in the plastic one and in the cut, and nothing a sizing method does can change it.

What the comparison rests on #

The truss is pin-jointed, and its loss is counted statically. A real joint carries some moment, which gives a damaged panel a little more than three equations’ worth of help; and a member that fails suddenly overshoots, which the four-thirds load stands in for with a single-mode factor. Neither changes the conclusion that a section through a damaged panel fixes what crosses it.

The plastic design assumes the members can yield in both directions. The verticals it shrinks must be stocky enough to yield in compression, and the 31 per cent redistribution it needs after a Pratt diagonal’s loss must actually be available. For ordinary steel verticals that is a mild demand; for slender ones it is the same condition as the diagonals’, and their price rises the same way.

The lower bound is a statement about strength, not about deflection. The plastic design carries the damaged load; it says nothing about how far the truss sags while it does, and a damaged truss that has yielded its way to the forces it needs may hang lower than any limit the intact truss was checked against. Accidental design usually accepts that, and so does this comparison.

And the least weight is a least weight, not a design. Both robust designs have members of non-standard sizes, and the plastic one is not unique — several sets of vertical areas give the same volume. A real truss would be drawn from a catalogue of sections, and the rounding would cost more than the difference between the two methods.

What the yield was supposed to buy #

The idea that plasticity would make robustness cheap rests on a picture of redundancy as a pool: a damaged structure with spare members draws on all of them, and the more freedom it is given to choose its forces, the more of the pool it can use. For a counter-braced truss the picture is wrong in a specific way. The redundancy is not a pool; it is one degree of freedom per panel, and losing a diagonal spends that panel’s degree of freedom entirely. The panel becomes determinate, its forces become a matter of statics, and no theorem about choosing among equilibrium solutions has anything left in it to choose.

What remains free is the division of shear in the panels that were not damaged, and that touches only the verticals between them — six units here. The steel robustness asks for is the steel each panel needs to stand alone, and the one thing that changes how much that is, is whether the member left standing can carry its load in the direction the accident chose.

Still open: the truss whose panels share a cut #

Every panel of this truss is crossed by a section that meets two chords and two diagonals, so losing one diagonal makes its panel determinate. A K-truss, a double-intersection Warren, or a lattice girder with two crossing systems of diagonals offers sections that cut five or six members, and losing one of them leaves the section still redundant. In those trusses the lower bound has something to choose after a loss, and the plastic price of robustness could genuinely fall below the elastic one. Whether a multiple-lattice girder — the form nineteenth-century engineers chose for its redundancy and twentieth-century engineers abandoned because it could not be analysed by hand — is the truss in which allowing yield finally pays, and by how much, is a question about how many members a single cut through the damaged panel still meets.