The factor of two belongs to one mode

The structure that survives losing a member and a determinate truss has no robustness at all asked what a truss does when one of its members is gone, and answered statically: remove the member, reanalyse, and read what the survivors carry. Both said what that leaves out. A member that fails does so suddenly, its force arrives at its neighbours dynamically, and the convention is to double the static change — “a factor with about as much derivation behind it as the demand ratio has”.

The factor has a derivation, and the essay on suddenly applied loads gave it: a load applied at once to a spring does twice as much work as the spring stores at equilibrium, so the spring overshoots to twice its static deflection and oscillates about it. That is exact, and it is a statement about one spring. This essay takes the member out of a truss that has many.

The truss, the member and the release #

The truss is a Pratt truss of eight 3 m panels, 3 m deep, with a counter-diagonal in each interior panel, carrying 100 kN at each bottom joint and sized so that every member is at 200 N/mm² — the counter-braced truss whose forces follow its sections, in real units. Its mass is its loads’ own, about ten tonnes at each loaded joint, and its members’ self-weight. The counters the sizing starved are left in the model at their tiny areas and out of every statistic.

The member lost is the first interior diagonal from the left support, joining the top of the first vertical to the second bottom joint, which carries 242 kN in tension. Losing it is equivalent to leaving it in place and applying, to the damaged truss, a pair of 242 kN forces pushing its two ends apart — the forces it was pulling them together with. Applied gradually, that pair produces the static redistribution the earlier essays computed. Applied at once, it sets the damaged truss vibrating. The response is followed through all of its modes, with 2 per cent damping, from the moment of release.

Three members, one release #

The three traces are three different answers to one question. The counter-diagonal in the same panel takes up most of the lost diagonal’s work: its compression rises from 112 to 354 kN, and dynamically it passes through 491 — a factor of 1.57 on its change, less than two, which is the one member the convention describes well.

The mid-span bottom chord’s static force does not change at all. The moment at mid-span is set by the loads and the reactions, and a diagonal near the support has nothing to do with it once the truss has settled. It swings anyway, to 1,115 kN, half as much again as it carries — a member the static check would not even list among the affected.

The third chord settles lower, at 556 kN from 666, and on the way goes the other way, up to 905. Its first move is in the opposite direction to its last, by more than twice the size of its change.

Why two, and why not two #

Write the damaged truss’s motion as a sum of its modes. The release is a force pattern applied at once, and each mode $n$ feels a share of it, $\Gamma_n$. Each mode then responds exactly like the spring of the suddenly loaded load essay: its coordinate rises to twice its static value, $2\Gamma_n/\omega_n^2$, and oscillates about $\Gamma_n/\omega_n^2$ — at its own frequency $\omega_n$.

A member’s change of force is a weighted sum of the modal coordinates, $\Delta N = \sum_n c_n q_n(t)$. Statically it is $\sum_n c_n \Gamma_n/\omega_n^2$ — call each term $s_n$. Dynamically each term reaches $2 s_n$, but at different times, and in the worst case they all arrive together, so the largest the member’s change can be is

$$|\Delta N|_{\max} \le 2 \sum_n |s_n|, \qquad \frac{|\Delta N|_{\max}}{|\Delta N_{\text{static}}|} \le \frac{2\sum_n |s_n|}{\left|\sum_n s_n\right|}.$$

The factor of two is the special case where every mode’s contribution to a member has the same sign — in particular, where there is only one mode. In a truss with thirty modes, a member’s static change is a sum of contributions of both signs, some of which cancel, and the factor can be anything from two to infinity. It is infinite in the limit of the mid-span chord, whose modal contributions cancel exactly in the static answer and do not cancel in motion, because the modes swing at different speeds. And a member can move the wrong way first when its largest modal contribution has the opposite sign to the sum, and belongs to a faster mode than the ones that eventually outweigh it.

Every member at once #

Plotted for every member, the convention is a line and the truss is a cloud. The members that carry most of the redistribution sit near the line, some below it and three above it; the worst, the bottom chord in the damaged panel, is at 2.97. The member that goes the wrong way is on the far side of the vertical axis from where its static change would put it. And the twenty-one members with no static change are a column of points on the axis itself, each swinging by up to 365 kN — nearly half the largest force anywhere in the intact truss.

Those twenty-one are the members a static study reports as unaffected and does not check. They are also not rare: the counter-braced truss’s redundancy is local, a diagonal’s loss is absorbed in its own panel and the ones beside it, and most of the truss’s members carry the same static force before and after. Dynamically, a local loss is a global event, because the release excites modes that involve the whole truss.

Where the energy goes #

The release does not choose a mode. It puts 44 per cent of its energy into the fourth mode, at 0.23 s, 30 per cent into the two lowest, near 0.56 s, and a tenth into modes shorter than a twentieth of a second. A single-degree-of-freedom model of the loss, which is what the factor of two is, has to choose one of those periods and one mode shape to put all the energy in, and whichever it chooses, it is wrong about most members. The spread of energy over modes of different periods is precisely what makes the members’ factors differ from two.

It is also why a structure has more than one period is not an academic statement here. The truss’s response to a sudden loss is a chord struck across many modes at once, and a member’s force is the sum of their motions at its own place.

How slowly a member has to go #

A real member does not vanish in an instant. A bolt group shears progressively, a strut buckles over a finite time, a tie heated in a fire loses its strength over minutes. Released over a time $t_r$, each mode’s overshoot is $1 + |\sin(\omega t_r/2)/(\omega t_r/2)|$ — the ramp result for one spring — so a release slower than a mode’s period barely excites it.

Because the energy is spread across modes of different periods, the factor falls in steps as the release slows past each. Released over a tenth of the first period, the short modes are already quiet and the factor has hardly moved. Released over half the first period, the largest factor is 2.09 and the unchanged members’ swing has fallen by two thirds; over a whole period, 1.14 and almost nothing. The time that decides whether a loss is sudden is the longest period that carries a substantial share of the release’s energy — for this truss, the fundamental — and it is a property of the damaged structure, not of the member that failed.

Damping, by comparison, does little. Five per cent instead of two lowers the instantaneous factor from 2.97 to 2.66, because the peaks arrive within the first swing, before damping has had time to act.

Every member it could lose #

The first loss was not a special case. Four of the sixteen members this truss can lose without becoming a mechanism are the verticals near mid-span, which the sizing left carrying almost nothing; losing them changes nothing, statically or dynamically. For the other twelve the largest factor on a member’s change exceeds two every time, and reaches 4.05 when a vertical near the support goes — and in ten of the twelve, somewhere in the truss a member with no static change swings by more than a tenth of the largest force the intact truss carries, by nearly half in the worst.

The chord that nothing touched, by a free body #

The mid-span chord’s behaviour is the least intuitive of the three, and a free body explains it without any modes at all. Cut the truss through the fourth panel and take the left-hand part. Statically the bottom chord’s force there is the moment about the top joint over the depth: the reaction times its lever arm less the loads times theirs, divided by 3 m. The lost diagonal is inside the free body, the loads and the reaction are unchanged, and so the chord’s static force is unchanged, 750 kN before and after.

In motion the free body has more forces on it. Every joint that is accelerating carries an inertia force, its mass times its acceleration, and by d’Alembert’s principle those forces enter the free body’s moment balance exactly as loads do. The release sets the joints near the lost member moving, the whole truss begins to sag and swing, and the moment of the inertia forces about the top joint of the fourth panel is no longer zero. The chord swings by that moment over the depth. Once the motion has died away the inertia forces are gone, and the chord is back to 750 kN, having passed through 1,115 on the way.

The same free body explains the member that moved the wrong way first. Near the lost diagonal the first inertia forces act on the joints the release pushed apart, and their moment about a nearby section can have the opposite sign to the eventual static redistribution, which is the net effect after the whole truss has found its new equilibrium. The fast local motion arrives first and the slow global one later, and a member between them sees the first before the second.

Two ways to add modes, and why neither is safe here #

Earthquake engineering met this problem long ago, from the other side. A structure’s modes are excited together, each has a peak, and the peaks do not coincide, so the codes combine them: the square root of the sum of the squares for well-separated modes, and a rule that accounts for correlation where modes are close in frequency, because two modes near one period are really one motion in a plane. Both rules assume the modal peaks are, in some statistical sense, unrelated in time.

A sudden release breaks that assumption. Every modal coordinate starts from rest at the same instant and rises together, so the early response is correlated by construction; and the truss here has two lowest modes 4 per cent apart in period, which beat against each other for several cycles. The two rules then disagree with the time history in both directions.

For the bottom chord in the damaged panel, twice each mode’s static share added absolutely gives a change of 1,202 kN — safe, and seven times the static change — while the root of the sum of their squares gives 527 kN, a little above the 508 the time history finds. For the counter-diagonal that takes up the lost member’s work, the same root-sum-square gives 223 kN against an actual 379: the combination rule that is close for the chord is unsafe by 40 per cent for the member carrying the redistribution. The absolute sum is safe everywhere and far too large nearly everywhere; the root-sum-square is sometimes close and sometimes short, and nothing in it says which.

What to do with the factor #

The factor of two is not wrong so much as addressed to the wrong object. It is exact for each mode, and a designer who wants a single number for each member has three honest options.

Use the bound. Twice the sum of the absolute modal contributions, $2\sum_n |s_n|$, is an upper bound on any member’s change, it needs only the modes and their static shares, and it is finite for every member — including the ones whose static change is zero, which it gives a swing for rather than nothing. It is also generous: for the chord in the damaged panel it is 2.4 times the swing the time history finds.

Run the dynamics. For a structure of this size a modal time history is a few seconds of computation and gives every member’s actual peak. The alternative-path procedures that progressive-collapse guidance sets out include a dynamic analysis alongside the static one with its factor, for exactly the structures whose response is not dominated by one mode.

Check the unaffected members. Whatever factor is used, the members whose static force does not change are the ones a static study never looks at, and they are where the largest surprises in these figures are. A robustness check that lists only the members whose static demand rose has checked the members the factor of two is closest to being right for.

The worst chord, by hand #

The bottom chord 1–2, in the damaged panel, carries 429 kN before the loss and 600 kN after — a static change of 171 kN. The convention gives it $429 + 2 \times 171 = 771$ kN. The modal analysis gives it 937 kN, a change of 508 kN, which is 2.97 times 171; undamped it would be 573. The bound $2\sum|s_n|$ for this member is 1,202 kN of change, because it assumes every mode peaks at the same instant with the same sign, and in the truss they do not.

The mid-span chord 3–4 has $\sum_n s_n = 0$ to the precision of the analysis, and $\sum_n |s_n|$ is not zero: its modal contributions cancel in the sum without being individually small, and the chord swings by 365 kN because they do not peak together.

A linear truss, a clean release and lumped masses #

Everything stays elastic. The overshoots here would yield the members that reach them, and a yielding member absorbs energy and caps its own force; the elastic peaks are what an elastic design must survive, and an upper bound on what a ductile one sees. Members in compression that overshoot are also candidates for buckling, which is not a capping mechanism.

The release is a clean ramp from the member’s force to nothing. A real failure may retain some force for a while — a bolt group that slips before it shears — or release in stages, and the time that matters is the time over which most of the force goes.

The mass is lumped at the joints, the loads’ mass at the bottom joints and the members’ own weight split to their ends. A deck carried by the truss would add mass and change the periods without changing the argument.

What the pictures cannot show #

That the numbers are for one truss. The factor for any member depends on how its modal contributions are signed, which depends on the geometry, the sizing and which member is lost, and another truss would give another cloud — but a cloud, not a line, whenever the release excites more than one mode.

Nor can they show the joint. A member that fails at a connection often takes the connection with it, and the joint that goes releases several members’ forces at once, each with its own pattern; the modal argument is the same, and the cancellations are different.

Still open: the frame that hangs its floor on its beams #

A truss that loses a diagonal redistributes by axial force, and its members stay nearly straight. A frame that loses a column carries the floor above it across the gap by bending and then, as the beams deflect, by catenary tension — a load path that develops only after large movement, stiffening as it goes. Whether the dynamic factor for that path, where the stiffness grows with the very deflection the release causes, is nearer one or nearer the factor of a linear system, and whether the beams’ connections can supply the rotation the catenary needs before it is reached, is the question this essay leaves for the structures that robustness rules were actually written for.