The truss sized to lose a diagonal

The truss whose forces follow its sections put a counter-diagonal into every interior panel of an eight-panel Pratt truss and sized it fully stressed. The iteration kept the counters near the supports, starved the two in the middle panels to nothing, and arrived at a family of designs that all needed 1,200 units of steel against the plain Pratt’s 1,220 and all deflected 24.0 at mid-span instead of 26.0. The counters paid for themselves, a little.

That essay ended on what else they might be for. A counter-braced truss is more robust than a Pratt: in a Pratt truss each panel’s shear crosses by one diagonal and nothing else, so losing that diagonal makes the truss a mechanism, not a weaker truss; in a counter-braced one, the panel has a second diagonal. But a counter starved to nothing by the sizing is not a second diagonal. The question was whether a truss can be sized for least weight and still keep the counters it would need if a diagonal failed, and what that alternative load path costs against the twenty units the redundancy saved.

What the truss can lose #

Before any sizing, the question of which members a truss can lose at all is a question about its geometry. Take the counter-braced truss with every member at the same area, remove each member in turn, and ask whether what is left can carry a load.

The interior panels can lose anything. A lost diagonal’s shear goes to its counter. A lost vertical’s job, carrying a joint’s load or turning the shear from one diagonal into the next, is taken by the diagonals either side, which meet at the same joints. And — less obviously — a lost chord can be survived too. A chord carries the panel’s share of the bending moment; with it gone, the two diagonals of that panel, which cross between the top and bottom joints, can carry the moment between them, one acting as the missing chord’s replacement and the other balancing it. That is the counter-braced panel’s own robustness, a second triangle in the same rectangle.

The end panels have nothing to offer. A Pratt truss’s end panels are triangles — the end diagonal running from the support up to the first top joint, the first vertical and the first length of bottom chord — and a triangle has no redundancy. Lose any of its three members and the joint it braced is free to move. The vertical is the exception only because a lost vertical’s load can still reach the end diagonal through the second panel’s diagonals; the end diagonal and the end bottom chord have no such alternative. Four members of this truss cannot be designed to be lost, at any weight. A design rule that asks for an alternative path round every member has to treat them as key elements, sized to resist the accident directly rather than to be lost.

The lightest truss, one diagonal short #

The fully stressed design starved its middle counters to nothing. A truss someone builds has no members of zero area, so take that design and give every counter at least one unit — a small bar, a twentieth of the largest diagonal — and ask what happens when a diagonal goes.

The load it has to carry when it does is not the full design load. A building or bridge losing a member is an accident, and the design rules for an accident use the loads expected to be present at the time — the permanent load and a fraction of the imposed — with partial factors of one. For a floor whose imposed load is about half its total, that comes to about two thirds of the factored design load, and two thirds is the share used here.

The lightest design is not robust. Every Pratt diagonal’s loss leaves some member at more than three and a half times its allowable stress, even at two thirds of the load, and the worst is a vertical at 5.47 times. Losing a counter is milder — the Pratt diagonals were carrying most of the shear anyway — but beside the supports, where the counters had settled at real sizes, it is still more than three times too much.

The member that fails is rarely the counter. The counter beside the lost diagonal is asked for its panel’s whole shear and is overloaded, but what is overloaded most is the vertical at the panel’s end. In the intact truss, where the stiffer path takes the load, that vertical carried only the difference between how the two diagonals either side of it split their shears, which the sizing had arranged to be small. With one diagonal gone the split is forced — all of the panel’s shear by the counter, which leans the other way — and the vertical has to turn the whole shear from the neighbouring panel’s diagonal into it. The stiffening that overloaded an untouched vertical was the same mechanism at a small scale: a vertical whose force is a difference between two load paths is the member a change in either path moves most.

Sized for the loss of any diagonal #

The remedy is to size the truss for every case at once: the intact truss under the full load, and the truss with each interior diagonal removed in turn under two thirds of it, each member given the area the worst of those cases asks of it, the forces recomputed with the new areas, and the whole repeated until nothing changes.

Three things happen, and the first is the one the counter-braced truss’s earlier essay had no way to get. The family of equal designs collapses to one. Started with counters at a hundredth of every other member, at one, or at a hundred times, the iteration lands on exactly the same areas every time. In the intact truss the split of each panel’s shear between its two diagonals was free, because every split cost the same; once each diagonal has to be able to carry the whole of its panel’s shear alone, the split is no longer a choice — each diagonal is sized by its partner’s loss.

The second is that every counter comes back, and the middle ones come back largest relative to where they were: from nothing to 4.71, which is exactly two thirds of the middle panels’ shear, 5, times $\sqrt 2$. The counters beside the supports grow from 11.19 to 23.57, two thirds of 25 times $\sqrt 2$. Each counter is now the area its panel needs with the Pratt diagonal gone, and each Pratt diagonal, in turn, the area its panel needs with the counter gone. In the intact truss they share, and neither is at its full stress: the robust design’s Pratt diagonals sit at 0.86 to 0.96 of the allowable under the full load and its counters at 0.43 to 0.64.

The third is the verticals. The first interior vertical, beside the end triangle, grows from 17.9 to 23.3, which is two thirds of the reaction: with the second panel’s Pratt diagonal gone, the whole of the end diagonal’s shear has to come down that vertical to the foot of the counter. The middle verticals, which carried almost nothing intact, grow from nothing or a fraction to between 3.9 and 9.2.

Where the extra steel goes #

At two thirds of the load the chords do not change at all, 840 units in both designs, and nor do the end diagonals at 140. Losing an interior diagonal changes how a panel’s shear crosses it; it does not change the bending moment at any section of the truss, and the moment is what the chords carry. Every one of the 104 extra units is in the web: 61 more in the interior diagonals and 43 more in the verticals.

The interior diagonals’ share has a closed form. Each diagonal is sized to carry its panel’s shear $V$ alone at the damaged load, a force of $\lambda V\sqrt 2$ over a length of $\sqrt 2$, so a volume of $2\lambda V$; a panel’s pair is $4\lambda V$; and the six interior panels’ shears, 25, 15, 5, 5, 15 and 25, add to 90. At $\lambda = 2/3$ that is 240 units against the intact design’s 180 — the 241 in the figure, to rounding. In the intact design the diagonal pairs cost $2V$ each panel whatever their split, as the earlier essay found; robustness costs $4\lambda V$, which is more as soon as $\lambda$ passes a half.

At four thirds of the load the chords grow too, to 1,240. The damaged truss is then carrying more than the intact one was sized for, and the chords, whose force depends only on the moment, are sized by the damaged case like everything else.

What it costs depends on the load it must carry #

The two-thirds load is the accidental combination’s, counted as though the member had been removed gently and the truss given time to redistribute. A member that fails suddenly — a diagonal cut by an impact, a connection that lets go — delivers its force to the rest of the truss in an instant, and the truss overshoots before it settles. The factor of two that is usually applied belongs to an undamped single mode suddenly loaded; counted that way, the damaged truss has to carry four thirds of its load.

Below about a third of the load, robustness costs almost nothing: the counters the intact design kept are big enough, and the middle ones need only small areas to carry a fraction of a small shear. By a third the robust truss weighs 1,221, what the plain Pratt truss weighs to within a unit, which is a fair way to read what the counters bought: everything they saved, spent on making the truss survive a gentle loss of a third of its load.

From about a half the cost climbs steadily, as each diagonal comes to be sized by its partner’s loss. At two thirds the truss needs 9 per cent more steel than the intact design and 7 per cent more than the Pratt; at the full load, 30 per cent more than intact. Counted as a sudden loss at four thirds, 2,074 — 73 per cent more than the intact design, because past the full load the chords join the web in growing.

The weight of robustness is decided less by the truss than by the load the rules ask it to carry damaged, and that load is a choice made outside the analysis. Two thirds is the accidental combination applied statically; four thirds is the same with a dynamic factor of two. Neither is a property of the truss.

A chord is the expensive loss #

The rules that ask for alternative load paths usually ask for them round any one member, not only the diagonals. The counter-braced truss can survive the loss of an interior chord; the question is at what price.

The web members are cheap to lose. A lost counter costs 13 units, because the Pratt diagonals were carrying most of the shear already; a lost vertical 36; a lost diagonal of either kind 104. In each case the lost member’s partner already spans the same panel or the same joint, and the alternative path is a short one.

A lost chord is a different order of thing. The top chords’ loss costs 2,458 units and the bottom chords’ 2,355 — about twice the intact truss. With a chord gone, the panel’s moment has to be carried by its two diagonals as a couple across the gap, and diagonals at 45° in a panel as deep as it is long carry a moment far less efficiently than a chord does: each is at most a panel’s depth from the other chord, but its force is inclined and its horizontal component is only $1/\sqrt 2$ of it. Every diagonal in the truss has to be able to do that in its own panel, so the whole web grows, and with it the verticals that balance the diagonals’ vertical components.

That is why the rules’ alternative-path checks so often end in tie forces and key elements rather than in a calculation round every member: some members are cheap to lose and some are not, and in a truss the chords are the members whose loss a web was never meant to carry.

The robust truss by hand, in one panel #

Take the second panel, next to the end triangle, where the shear is 25. Intact, its Pratt diagonal and its counter share the shear in tension and compression, and fully stressed their areas add to $25\sqrt 2 = 35.4$ whatever the split.

With the Pratt diagonal gone, at two thirds of the load, the counter carries the panel’s shear alone: $\tfrac23 \times 25 = 16.7$ vertically, so $16.7\sqrt 2 = 23.6$ along its length, in compression. Its area must be 23.6. With the counter gone instead, the Pratt diagonal carries the same 23.6 in tension. So the pair needs 47.1 of area against the intact pair’s 35.4 — a third more — and both members are then only partly used when both are present. The figure’s 23.57 for both members in the end panels is that number.

The vertical at the panel’s left end hangs the first bottom joint in the intact truss and carries the difference between the end diagonal’s shear and the second panel’s Pratt diagonal’s. With the Pratt diagonal gone, it is the only way down from the top of the end diagonal to the foot of the counter, and carries the whole damaged reaction, $\tfrac23 \times 35 = 23.3$. The figure’s first vertical at 23.33 is that.

The truss a design office would draw #

A design office given this truss would not reach the robust design by iterating an envelope; it would size for strength, check the alternative paths, and enlarge what failed. The figures say what it would find.

Size the counters for the lost diagonal, not by the iteration. The fully stressed iteration decides how much each counter carries intact; the lost-diagonal case decides how large it has to be. Once that is the rule, the counter in each panel is simply the panel’s damaged shear at the allowable stress, and so is the Pratt diagonal: the panel is sized as two independent single-diagonal panels, each carrying the whole shear at the accidental load, and the intact truss is checked afterwards.

Check the verticals in the damaged case. The verticals were the members that failed when a diagonal was lost, and the members that grew most in the robust design: from 40 units in all to 83. A vertical that turns shear from one diagonal into another is sized by the case in which its neighbours are lopsided, and losing a diagonal is the most lopsided case there is.

Treat the end triangles as key elements. The end diagonals and the first bottom chords cannot be bypassed. A truss whose robustness matters needs them designed for the accidental action directly — or a different end panel, with a second diagonal of its own, which a Pratt truss’s triangular ends do not have room for.

A pin-jointed truss, statically damaged #

The truss is pin-jointed and its members equally good in tension and compression. A real counter in compression is limited by buckling; a slender counter that carries 23.6 in compression needs much more than 23.6 of area, and a counter made as a flat bar that buckles at a small load is a diagonal that is absent whenever it would be in compression. A truss whose counters can only work in tension survives the loss of a Pratt diagonal only where the counter would be in tension — in a Pratt truss under gravity, nowhere.

The loss is counted statically. A member removed and the truss reanalysed is the state the truss settles in, not the overshoot on the way. The four-thirds load stands in for the overshoot with a factor that belongs to a single mode; a truss losing a diagonal redistributes through several, and its true dynamic demand may be lower than twice in some members and higher in others.

The truss stays elastic. Every damaged case is sized at the allowable stress. A truss allowed to yield in the accident would carry more than the elastic analysis says, and the rules for alternative paths usually allow that; a plastic count would lower every robust volume above, most where the damaged truss is most redundant.

What the counters were for #

In the intact truss the counters were a choice about how to route each panel’s shear, worth twenty units of steel and a little stiffness, and the fully stressed sizing used the choice by starving the counters it did not need. Asked to survive a lost diagonal, the truss has no choice left: each diagonal has to carry its panel’s shear alone, the family of designs becomes one design, and every counter comes back at the size its partner’s loss requires.

The price, at the accidental two thirds of the load, is 104 units — 9 per cent of the intact truss and five times what the counters saved. The Pratt truss cannot buy the same thing at any price, because it has nothing to buy it with. And the members that cost most to lose, the chords, are not the ones the counters were added for; the members that cannot be lost at all, in the end triangles, are not ones the counters can reach.

Still open: the truss that is allowed to yield when it is damaged #

Every case here was sized elastically at the allowable stress, as though the damaged truss had to be as good as new. An accident is the one state in which a structure is allowed to be damaged and still stand, and the damaged truss, allowed to yield, has a plastic capacity that the lower-bound theorem gives without any compatibility at all. Sized that way, the lost-diagonal case asks only that some distribution of forces carry the damaged load within the members’ strengths, the lost chord can borrow every diagonal in the truss instead of only its own panel’s, and the price of robustness becomes a question about equilibrium alone — which may cost far less, if the members that yield are ductile enough to let it happen.