A determinate truss has no robustness at all

A structure that survives losing a member does so by having somewhere else for the load to go. The rung below this one establishes that. This one is about how much of that a truss actually has, and the answer for most trusses is none.

The hero is a member-removal study of an 8-panel Pratt truss with counters: every one of its 35 members taken out in turn, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism. The other 31 redistribute, and the worst of those asks a survivor for 2.04 times what it carried before.

Now take the counters away, and the picture stops being a picture.

Thirty-one of thirty-one #

Those two figures are the whole argument of the essay and they contain no numbers, which is unusual for this site and is the finding.

A statically determinate structure has exactly one set of member forces that satisfies equilibrium. That is what determinacy means: the equations of statics have a unique solution. Take a member away and the equations have no solution — not a different one, none — because the count of unknowns has fallen below the count of equations and nothing can be found to satisfy them.

So the member-removal question, asked of a determinate truss, does not return a small number. It returns no number, for every member, and a robustness assessment that reports a demand ratio has to report an empty set.

That is worth insisting on because it is the opposite of how robustness is usually discussed. The vocabulary — degree of robustness, reserve, residual capacity — is the vocabulary of a continuous quantity, and there is a discontinuity underneath it. Below indeterminacy there is no quantity to be small.

Which free body produced the empty set #

“There is no solution” is a strong claim, and it comes from counting rather than from any analysis, so it is worth doing the count.

A plane truss of $m$ members and $j$ joints on $r$ reaction components has $2j$ equations — two per joint — and $m + r$ unknowns. Determinate means $m + r = 2j$. The 8-panel Warren truss has 31 members, 17 joints and 3 reactions: $31 + 3 = 34 = 2 \times 17$.

Take one member away. Now $m + r = 33$ against 34 equations, and a system of 34 equations in 33 unknowns has no solution unless the load happens to lie in a very particular subspace — which, for an arbitrary set of nodal loads, it does not.

The physical reading of that arithmetic is a mechanism. One of the 34 equations cannot be satisfied, which means one joint cannot be held in equilibrium, which means it moves. The frame has one degree of freedom left and the load drives it.

Two things about the count are worth carrying, and both are traps.

It is necessary and not sufficient. A truss can satisfy $m + r = 2j$ and still be a mechanism, if the members are arranged so that a panel has no diagonal while another has two — the count does not see it, and neither does any spreadsheet built on it.

And it is global while the failure is local. Adding a redundant member at one end of a truss raises the count by one and does nothing whatever for a member removal at the other end. The Pratt truss’s four essential members are the demonstration: the frame is indeterminate to the eighth degree and still has four places where the arithmetic runs out.

What the counters buy #

Two figures, one member, and the difference between a structure and a heap is a pair of bars that were carrying nothing.

That is what a counter is. In a Pratt truss under gravity load the diagonals in one direction are in tension and the counter-diagonals are unstressed — they are there for load reversal, for wind, for a crane travelling the other way. Under the gravity case they are idle, and an efficiency-minded reviewer looking at a force table would delete them.

Deleting them removes the structure’s entire robustness and changes no force in the gravity case. The force table cannot see it, the deflection cannot see it, and the utilisation cannot see it. What sees it is a member-removal study, and the reason to run one is that nothing else asks the question.

And the counters wake up in the right place. The figure reports two previously idle members now working, and they are the ones adjacent to the loss — not because anything routed them there, but because the stiffest available path takes the load, and the stiffest path around a hole is the one closest to it.

The demand ratio is not a measure of how close it came #

Take the same truss the other way instead, and the statistic moves as far again in the opposite direction on a change that makes every member work harder.

That pair is a warning about the statistic rather than about the truss.

The demand ratio is a ratio, and a deep truss’s diagonals carry small forces because the lever arm is long. A redistribution that adds a fixed amount to a small force is a large multiple of it; the same addition to a large force is a small multiple. Neither says anything about whether the member has the capacity.

What decides survival is the absolute force against the member’s absolute capacity, and there is a further complication: a member sized for a small force is usually not sized for that force at all. It is sized by a slenderness limit, a minimum thickness, a handling rule or a connection — so it has enormous reserve on the quantity that matters and none on the quantity being reported.

Which means the deep truss may well be the more robust one in fact and the less robust one in the statistic. A robustness study that ranks arrangements by demand ratio is ranking them by how lightly loaded their members were, which is not the question.

The honest version of the check compares the redistributed force against the member’s design resistance, member by member, and reports a utilisation rather than a multiple. It is a longer output and it is the only one that decides anything.

What “spare capacity” actually means here #

A structure that redistributes has turned a question about arrangement into a question about margins, and the margins are not where a designer would expect them.

A member’s spare capacity is whatever was left over after the rule that sized it. Very few members in a truss are sized by their force. A tension chord might be; a slender internal member almost never is, because it is set by a slenderness limit, and a small compression member is set by whatever section the fabricator will handle. Those members have reserve factors of two, three or more against their design force — for reasons that have nothing to do with robustness and are entirely available to it.

And the heavily-utilised members are the ones with none. A chord at midspan designed to 98 per cent of its resistance has two per cent, and a redistribution asking it for 1.24 times its previous force will not be met.

So the members most likely to survive a redistribution are the ones nobody optimised, and a structure that has been optimised carefully across every member is one where the reserves have been removed everywhere at once. Robustness is, in part, an accidental by-product of the parts of design that are not calculations — minimum sizes, standard sections, repeated details, and the reluctance to specify forty different members where six will do.

That is an uncomfortable thing to rely on and it is worth naming rather than leaving implicit, because it is being spent. Every improvement in optimisation, in section availability and in analysis fidelity removes some of it, and nothing in the design process notices, because the quantity being removed was never in a calculation. A member sized by something other than its force is the general case, and this is one more consequence of it.

What indeterminacy is not #

There is a temptation to turn all of this into a count — degrees of indeterminacy as a score for robustness — and the temptation should be resisted for a reason that is visible in the hero.

The Pratt truss with counters has 35 members and four of them still leave a mechanism when removed. Indeterminate overall, and locally determinate in four places, because the redundancy is not distributed evenly through the frame: it lives in the panels that have counters and nowhere else.

So a structure’s degree of indeterminacy is a global count and the failure it is supposed to protect against is local. Counting members and joints does not see it, for exactly the reason it does not see a mechanism in a frame that has enough members in the wrong arrangement: the count is a necessary condition and the arrangement is the sufficient one.

The four essential members are the ones worth identifying and they cannot be identified by counting. Three of them are usually the chords at the supports and the end verticals — members through which the whole reaction passes, with nothing parallel to them anywhere. No amount of redundancy elsewhere reaches them.

That is what a key element is: a member whose removal leaves a mechanism, in a structure that is otherwise redundant. Codes require key elements to be designed for an accidental load rather than being allowed to be removed, and the reason is exactly the discontinuity this essay opened with — for those members there is no residual capacity to compute, so the only available strategy is to make them not fail.

Two ways to buy what a truss has not got #

If the removal study on a determinate truss returns nothing, there are only two responses, and codes use both.

Make the essential members not fail. Design them for an accidental action — an impact, a notional pressure, a vehicle strike — so that the scenario the removal study was standing in for does not occur. This is key element design, and its honesty is that it admits there is no alternative path and stops pretending to compute one.

Or add the path. Counters, a second load-bearing system, a floor plate that can act as a catenary, a tie that does nothing until a column goes. The cost is material that is idle for the life of the structure, and the benefit is a demand ratio to compute.

The second is much cheaper in a frame than in a truss, which is why the prescriptive rules are written the way they are. A building frame’s beams are continuous, its slabs are two-way, and its connections have some moment capacity whether anybody designed for it or not — so the alternative path is largely already there and the rules ask for ties to make sure of it. A truss has none of that: its members are pin-ended, its load paths are explicit, and there is nothing accidental to fall back on.

Which produces the practical ranking this essay ends on. A determinate truss holding up a roof over people is a structure with one line of defence, and the response is either redundancy bought deliberately or protection of the members that cannot be lost. What it must not be is a demand ratio, because there is not one — and a report that prints a number there has computed something else.

The one thing a determinate truss does have #

It is worth ending the negative case fairly, because determinacy buys something real and this essay has spent its length on what it costs.

A determinate truss cannot be locked up. Its member forces depend on the loads and on nothing else — not on the fabrication tolerances, not on the temperature, not on a support that settled. A member built to the wrong length in a determinate truss goes in with a struggle and then carries what the equations say; in an indeterminate one it carries a permanent force that no load case contains.

That is not a small property. It means the truss can be analysed exactly by hand, that its forces are checkable on site with no assumptions about stiffness, and that its behaviour does not drift as the structure ages, creeps or moves. One support too many is the general statement of what the alternative costs.

So the two are a trade rather than a ranking, and the sentence worth carrying is narrow. Determinacy buys predictability at the price of having no alternative path; indeterminacy buys the path at the price of forces that depend on things nobody measured. A truss is usually made determinate for the first reason and then relied on as though it had the second.

The counters in the Pratt truss are the interesting middle case, and they are why this essay is carried by that truss rather than the Warren. They make the frame indeterminate under gravity and determinate in effect, because a slack tension counter carries nothing and imposes nothing — so the truss has predictable forces under the load case it lives in, and a second path available for the case it does not.

That is a genuinely good arrangement and it happened for a different reason: the counters were put there for load reversal a century before anybody used the word robustness. Which is the ordinary way robustness has been achieved in practice — as a by-product of designing for a load case that turned out not to govern.

What to carry away #

Determinate means every member is essential. All 31 members of a Warren truss and all 29 of a Howe leave a mechanism, and the removal study returns an empty set rather than a small number.

The idle members are the robustness. A counter that carries nothing under gravity is the difference between a redistribution at 1.97 times and a mechanism.

The demand ratio measures how lightly loaded a member was. A deeper truss scores worse on it and is not worse; the check that decides anything is a utilisation.

And redundancy is local. A frame indeterminate overall still has members whose removal ends it, and no global count finds them.

Where the model stops #

Every member here is elastic and stays elastic. A real redistribution goes plastic in the most-demanded member and sheds further, so the elastic demand ratio is an upper bound on what any one member is asked for — provided the members are ductile, which a slender compression member is not.

The removal is instantaneous and the analysis is static. A member that fails does so suddenly and the load arrives at its neighbours dynamically, which is conventionally handled by doubling it — a factor with about as much derivation behind it as the demand ratio has.

Nothing here is a connection. Members are removed and joints are not, and a joint failure removes every member meeting at it. A robustness study that removes only members is asking the easier half of the question.

And the load is unchanged throughout. The accidental load case that removed the member is usually also applying something — a blast pressure, an impact, a fire — and the survivors are carrying that too.

Two neighbours in this field carry the same argument into structures that are not trusses. The chord is a continuous beam is what happens when a truss’s members are not pin-ended after all, which is a source of alternative path nobody designed for; and a structure that was never complete is the reminder that the arrangement being assessed existed in several other arrangements first, each with its own essential members.

The ladder from here #

Later rungs on this anchor: tie forces, which are the prescriptive alternative to a removal study and are a horizontal strength requirement with no analysis behind it. Key element design and the 34 kN/m² accidental pressure. Dynamic amplification on member loss, and where the factor of two comes from. Removal of a joint rather than a member. Robustness in a frame rather than a truss, where the alternative path is catenary action in the beams and requires a rotation nothing else in the design provides. And the economic question underneath all of it: redundancy costs material that is idle for the whole life of the structure, and how much to buy is a decision nobody has a method for.

The subject exists because of one building. Ronan Point, in 1968, lost a corner of every floor above and below the flat where a gas explosion blew out a load-bearing panel — a structure with no alternative path at all, in which each panel carried what was above it and nothing else could. The inquiry’s response was the tie-force rules, which are a prescription rather than an analysis, and they have outlasted several attempts to replace them with a calculation. This essay is one reason why: the calculation, run on an ordinary determinate structure, has nothing to report.