The truss whose forces follow its sections

The calculus answer, rounded, is the worst one sized a truss from a catalogue and found that the problem, hard as it looks, is a knapsack: each member’s contribution to the deflection is independent of every other’s, so the members can be ranked and bought in order of deflection saved per unit of steel. It said why that was possible and where it stops. The truss was determinate, so its forces were fixed by statics before any section was chosen. In a redundant truss the forces follow the sections, a member made larger attracts force from its neighbours, and a section chosen for strength may stop being strong enough when the section beside it is enlarged for stiffness.

This essay puts one redundant member into every panel of the same truss and follows the three things that change: what “fully stressed” means, what stiffening does, and how much steel a deflection limit needs once the two are sized together.

A truss with a choice in every panel #

The truss is the eight-panel Pratt of the earlier essays, one panel deep, with a load of 10 at every bottom joint. Its diagonals run down toward mid-span, so under gravity they are in tension and its verticals in compression. Into each of the six interior panels goes a second diagonal, the other way — a counter — so that each panel’s shear can be carried by either diagonal or by both. The truss is now six times redundant: statics gives the shear in each panel and cannot say how it divides between the pair.

What divides it is their stiffness. A pair of diagonals in one panel is two paths for one load, and the stiffer path takes the larger share. So the forces in the truss depend on the areas given to its members, and the areas a designer wants to give them depend on the forces.

Throughout, the allowable stress and Young’s modulus are taken as one, so a volume is in units of force times length over stress, and a deflection in units of the allowable strain times a panel. For a truss of 3 m panels in S235 steel, a deflection of 24 units is 81 mm on a 24 m span.

Fully stressed, by iteration #

The obvious way to size such a truss for strength is the one every designer uses by hand: analyse it, give each member the area its force needs at the allowable stress, analyse again with the new areas, and repeat until nothing changes. That is fully stressed design, and for a determinate truss it takes one step, because the forces do not move.

For the redundant truss it takes many, and something happens that no single analysis shows. The counters near mid-span are in compression beside diagonals in tension, carrying a little of a small shear. Each resizing makes them a little smaller, so a little less stiff, so on the next analysis they attract a little less force — and are made smaller again. The iteration does not cut them. It starves them: after eight steps the middle counters are below a hundredth of a unit, and they go on shrinking toward nothing for as long as the iteration runs.

The counters nearer the supports behave differently. There the shear is large, and the counters settle at areas comparable to the diagonals beside them, each pair of members carrying the panel’s shear together, one in tension and one in compression, both at the allowable stress.

The result is lighter than the Pratt truss it was made from — 1,200 units of steel against 1,220 — and stiffer, 24.0 against 26.0. The redundant members that survived paid for themselves and more.

A family, not a truss #

Start the iteration somewhere else and it arrives somewhere else.

Started with counters a hundredth the size of everything else, the iteration ends with the first panel’s counter at 0.40; started with them at a hundred times, at 11.12. In between it lands at every value. The panels near the supports divide their shear differently in every design. Fully stressed design of a redundant truss is not an answer but a family of answers, and the iteration returns whichever member of the family is nearest where it started.

That would be alarming if the family differed in anything that mattered. It does not. Every design in it needs exactly 1,200 units of steel, and every one deflects exactly 24.00 at mid-span.

Why the split costs nothing #

The equal volume can be found by hand, in one panel. The first panel’s shear, beside the support, is the reaction less the first joint’s load: 35 − 10 = 25. Its two diagonals are at 45°, each $\sqrt 2$ long; if one carries a tension $T$ and the other a compression $C$, their vertical components must add to the shear, $(T + C)/\sqrt 2 = 25$, so $T + C = 35.4$. At full stress each needs an area equal to its force, so the pair’s volume is $(T + C)\sqrt 2 = 50$ — whatever the split. Giving the shear to one diagonal or dividing it between two costs exactly the same steel in the diagonals.

The split does change the verticals and chords, and the whole saving over the Pratt truss is in them. Summed over the truss, the diagonals of every fully stressed design carry 180 units of volume, the same as the Pratt’s diagonals; the chords carry 840, the same as the Pratt’s; and the verticals carry 40 against the Pratt’s 60. In the Pratt truss the verticals beside the supports are in compression, carrying the shear from one diagonal across to the next. With a counter in the panel, part of the shear crosses by the counter instead, and the vertical is relieved. The family is the set of splits that relieve the verticals by the full twenty — and there are many, because in these panels relieving them does not require a unique split.

This is the lower-bound argument of plasticity turned into a sizing rule. A fully stressed truss is one whose volume is $\sum |N| L / \sigma$ for a set of forces in equilibrium with the load, and the least such volume is a problem about equilibrium alone, in which stiffness has no part. Where that problem has many solutions, full stress is reached at all of them.

Why every fully stressed truss deflects the same #

The equal deflection is a different argument, and a shorter one. In a fully stressed truss every member that is present is strained by exactly the allowable stress over the modulus, stretched in the members in tension and shortened in those in compression. Its deflection is decided by those strains and the geometry, and the strains are decided by the signs of the forces — not by their sizes and not by the areas.

That is the drawing that is right except for a rotation put to a new use. Williot’s construction draws a truss’s deflected shape from its members’ extensions, one joint at a time, and needs nothing else. Every design in the family has the same members in tension and in compression, all at the same strain, so every one draws the same Williot diagram, and every one deflects 24.00. In a fully stressed truss the stiffness is fixed by the stress, and the areas are only how that stress was arranged.

The same argument says why the counter-braced truss is stiffer than the Pratt: its verticals near the supports went from compression into light tension or nearly nothing, so their strains changed sign or vanished, and the geometry of the deflection changed with them.

Stiffen one group, and another is overloaded #

The family’s stiffness is fixed. A designer who needs it stiffer has to add steel, and in a determinate truss adding steel to any member lowers its stress and nobody else’s. Here it does not.

Enlarging the chords is safe: the chords’ forces are nearly fixed by the bending moment, the stiffer chords take the deflection from 24.0 to 18.7, and nothing is overstressed. Enlarging every diagonal together is safe too, because the pairs keep their relative stiffness and so their split. Enlarging the verticals changes nothing at all — they were not doing much.

Enlarging only the Pratt diagonals is not safe. Each is now stiffer than its counter by more than before, so it draws shear out of the counter; the verticals, whose forces depended on that split, are handed a different one; and a vertical beside the support that nobody touched ends at 1.58 times its allowable stress, for a gain in stiffness of 5 per cent. Enlarging only the counters does the reverse, less violently: a vertical at 1.04.

The overstress arrives at once and grows faster than the stiffness does. A tenth more area in the diagonals buys 2 per cent of deflection and costs 16 per cent of overstress in a vertical; tripling them buys 11 per cent and costs 118. In a redundant truss a stiffening that is not proportional is a redistribution, and a strength check made before it says nothing about the truss after it.

The same happens in any structure with more than one path for a load. A continuous beam’s supports share its reactions by stiffness; a frame’s columns share its storey shear the same way; a composite section’s steel and concrete share its moment by their moduli. In each, stiffening one path unloads the others — and loads whatever sits between the paths and was sized for the old division.

Sized for both at once #

The remedy is to size for strength and stiffness together. Each member is given the larger of two areas — the one its force needs at the allowable stress, and the one the deflection limit asks of it, which for a single deflection is proportional to the square root of the product of its real and virtual forces, as the essay on the stiffest division of a truss found — and the truss is reanalysed with the new areas, until both the forces and the areas stop changing.

The counter-braced truss needs less steel at every limit, and its advantage grows as the limit tightens: 5 per cent at 24, 9 at half and at three tenths of it. The counters are not there for strength — at a loose limit the strength design already starved the middle ones — but they give the sizing a choice about how to route each panel’s shear, and a deflection limit is exactly the kind of requirement that rewards a choice. The Pratt truss has to buy stiffness member by member along a load path it cannot change; the counter-braced truss can change its load path as it buys.

Two load cases #

Everything so far has one load, the full one. A real truss also sees load on half its span, as snow drifts or a moving load crosses, and then the counters earn their name: the panels near mid-span, whose shear is small and changes sign between the two halves, need a diagonal each way. Sized fully stressed for the full load and both half loads together, the counter-braced truss needs 1,230 units of steel against the Pratt’s 1,265. The counters grow where the half loads make them useful — the second panel’s to 20.4, twice its size under the full load alone — and the middle ones are still starved: with members as good in compression as in tension, a single diagonal can carry a shear of either sign, and the half loads only change which sign.

The same truss, by hand, in the Pratt #

For the Pratt truss the numbers can be checked with the method of sections. The reactions are 35 each, and the shear falls by 10 at each loaded joint: 35, 25, 15 and 5 in panels one to four. Each diagonal carries its panel’s shear times $\sqrt 2$ and is $\sqrt 2$ long, so at full stress its volume is twice the shear: 70 for each end diagonal, and 50, 30 and 10 for the three interior diagonals on each side — 140 and 180. Each vertical carries the shear of the panel beyond it, except the first, which hangs the first joint’s load: 10, 15, 5 and nothing at mid-span, 60 in all. The chords carry the bending moment over the depth: 35, 35, 60 and 75 along the bottom from the support, 60, 75 and 80 along the top — 410 and 430, 840 together. The sum is 1,220.

For the counter-braced truss the same panel arithmetic gives the same 140 and 180 in the diagonals whatever the splits, and the analysis gives the same 840 in the chords; its verticals carry 40. The 20 units between the two trusses are the verticals’, and nowhere else.

The order of work, when the forces move #

A design office sizes a determinate truss in one pass: forces, then sections, then a deflection check, then, if the deflection is too large, larger sections where they do most good. Every step can be checked against the one before, because the forces never move. In a counter-braced truss each of those steps invalidates the one before it, and three habits follow from that.

Stiffen in proportion, or reanalyse. Scaling every member of the truss by one factor leaves every force where it was, because the ratios of stiffness that decide the splits are unchanged; so does scaling whole load paths together, as the diagonal pairs were. Anything else — the chords here and the diagonals there, the members the deflection calculation says are most effective — is a redistribution, and the strength check has to be run again on the truss as it now is.

Read the verticals. The members that were overloaded by stiffening were not the ones enlarged and not the ones that looked important. They were the verticals, which in a Pratt truss carry the shear from one diagonal to the next and in a counter-braced truss carry whatever part of it the counters do not. A member whose force is a difference between two load paths is the one a change in either path moves most.

Decide what the counters are for. Counters sized by an iteration are whatever the iteration left, and that can be almost nothing. If they are wanted for load reversal, or for the truss that has to survive losing a member, they need a size set by that requirement, and the fully stressed family then narrows to the designs that keep them.

The same three habits apply wherever a determinate structure is made redundant to stiffen it. A determinate truss has no robustness at all and no choice about its load paths; the redundancy that buys both also makes the forces a consequence of the sizing, and the sizing has to be done knowing it.

A pin-jointed, linear, single-material truss #

The truss is pin-jointed and elastic, every member with the same allowable stress in tension and compression. Real compression members are limited by buckling, so a counter in compression is worth less than the same counter in tension, and a design that relieves its verticals by putting its counters into compression is making a trade the allowable stress alone does not see. Real counter-diagonals are often flat bars or rods that can only carry tension, and then the pair in each panel is a switch rather than a share: only the diagonal in tension works.

The areas are continuous, as in the essay before the catalogue. With sections from a catalogue the family of equal designs becomes a set of neighbouring catalogue designs of slightly different weight, and the question of which the iteration lands on becomes a question about the catalogue’s steps.

What the pictures cannot show #

That the starved counters are still there. A member whose area has been iterated down to a millionth of its neighbours’ is, on the drawing, a dotted line; in the truss it is a member someone has to fabricate at a minimum practical size, and at that size it is not starved but carrying a little force, which moves the others a little. The family of equal designs exists for the continuous problem; the built truss is one member of it, displaced slightly by every section that could not be as small as the iteration asked.

Nor can they show where the iteration stops being safe to run. Each step here was a linear analysis of a truss that could carry its load; a truss whose resizing had removed a member it needed for stability would be a mechanism, and the iteration would stop only because the analysis failed.

Still open: the truss that must stay safe with a member gone #

Every design here, and every redundancy it kept, was chosen for strength and stiffness under loads that act on the whole truss. A counter-braced truss is also more robust than a Pratt: lose a diagonal and its counter can carry the panel’s shear, which in a Pratt truss nothing can. But a counter starved to nothing by the sizing is a counter that cannot do that. Whether a truss can be sized for least weight and still keep the counters it would need if one diagonal failed — and how much weight that alternate load path costs against what the redundancy saved — is the question where this essay’s arithmetic meets the one that asks what a structure does when part of it is gone.