The cable that pays the factor back

A member that fails suddenly hands its force to the structure around it all at once, and the convention is to double the static answer, because a load applied suddenly to a spring overshoots to twice its static deflection. The earlier essay found that doubling is exact for one mode and wrong for a truss with many. It ended on the structure the robustness rules were actually written for: a frame that loses a column, and carries the floor above across the gap — first by bending, and then, as the beams sag, by catenary tension.

That path has a property a truss member does not. It stiffens as it deflects. A beam in bending reaches its plastic collapse load and stops gaining strength; held against spreading at its far ends, it then develops an axial tension that grows with the sag, and the floor is held by a cable, stiff only because of its shape. The hope built into a great deal of robustness design is that a resistance which grows with the deflection the loss causes will need less than the factor of two. This essay computes it.

A beam that becomes a cable #

Take a beam over a column that is lost: two spans of 8 m, 600 mm deep, a plastic moment of 800 kN·m, its far ends held against rotation and against moving apart. Its plastic collapse load at the lost column, with hinges at both far ends and at the middle, is $4M_p/L$ = 400 kN. Below that it is elastic, and it reaches 400 kN at a deflection of 44 mm.

Past the collapse load a beam that is free to move apart at its ends would simply keep deflecting. One that is held cannot, because sagging stretches it, and the stretch is resisted by an axial force. The resistance of a fully clamped, axially restrained beam is a classical result: as a share of the plastic collapse load it is $1 + (w/h)^2$ while the extra sag $w$ is less than the section’s depth $h$, and $2w/h$ after it. The first is the bending plateau lifted by the growing axial force; the second is a cable whose tension is the section’s full axial capacity.

The solid curve is the static resistance. Its last part is a straight line through the origin, extended: past one depth of sag, the beam’s resistance is proportional to its deflection again, exactly as it was in the elastic range, only on a much softer spring.

The dashed curve is what a sudden loss can be held at. A column removed in an instant leaves the floor’s weight falling onto a beam that has no resistance yet; the beam deflects until the work its resistance has done equals the work the falling weight has done, and the load that reaches is the work done divided by the deflection — the area under the solid curve over the distance. That curve lags the static one everywhere past the elastic range, and at one depth of sag the beam that holds twice its collapse load slowly holds 1.28 times it suddenly.

The release, as it happens #

The same release, integrated in time with the floor’s weight as a mass at the lost column, makes the energy argument visible. Under 1.5 times the collapse load, 600 kN, the beam would settle at 468 mm if the load arrived slowly. Released at once, it swings down to 824 mm, reaching it 0.79 seconds after the loss, and oscillates about the static position.

The lowest point the time history reaches is the deflection the work balance predicts, to four figures. That is the check that the pseudo-static reading is not an approximation here but an identity: for a single degree of freedom with no damping, the turning point is where kinetic energy is zero, and that is where the work done by the load equals the work done on the beam.

Where the factor goes, and where it comes back #

The quantity the convention is about is the force the beam reaches at its lowest point over the load it is carrying — the dynamic factor, two for a linear spring.

In the elastic range it is exactly two, as for any linear spring. On the bending plateau it falls, to 1.18 at nine tenths of the collapse load, because a flat resistance does its work at full strength from the first millimetre past yield, and a beam whose resistance is flat absorbs energy faster per unit of deflection than one whose resistance is still rising.

Past the plateau the beam becomes a cable and the factor climbs straight back: 1.73 at one and a half times the collapse load, 1.88 at twice, 1.97 at three times. The cable’s resistance is a straight line through the origin, and a straight line under a sudden load overshoots by two, less only the small head start the plateau gave it. The stiffening that was supposed to rescue the dynamic factor does the opposite: it is what returns the beam to the behaviour of a spring.

So the factor is least where the beam is weakest relative to its load and greatest where it is strongest. A designer who takes 2 for the catenary regime is very nearly right; one who takes the plateau’s 1.2 for a catenary is wrong by half again.

How far it sags #

The deflection tells the same story the other way round. Where the force factor was least the deflection factor is largest: at 0.9 times the collapse load the beam settles at 40 mm and overshoots to 191 mm, nearly five times as far, because a flat resistance has to travel a long way to do a given amount of work. In the cable the overshoot is back near two — 468 mm slowly and 824 suddenly at 1.5 times.

That is the plateau’s bargain, and it is the important one. The bending plateau keeps the force factor low by spending deflection, and deflection is rotation at the beam’s ends.

The rotation the cable needs #

The beam turns at its ends through its deflection over its span. A sudden loss at the collapse load asks for 0.042 radians; at one and a half times it, 0.103. Those are not small numbers. A simple shear connection can supply a few hundredths of a radian before its bolts tear out or its plate fractures; a well-detailed moment connection a few hundredths more. Shaded on the figure as bands of that order, they are exhausted at 0.95 and 1.20 times the collapse load respectively — for the second, when the beam has sagged about 570 mm, a little under one depth, and the cable is only beginning.

A joint that delivers its resistance has to keep turning, and this is the case where the turning asked of it is largest. The catenary exists in the calculation from the moment the beam starts to sag; it exists in the structure only if every connection along the beam survives the rotation needed to reach it.

What a connection’s rotation is worth #

Turned round, the curve is a design chart: what load a frame can carry through a sudden column loss, given the rotation its connections can supply. Connections good for 0.03 radians carry a sudden loss only up to 0.94 of the beam’s plastic collapse load — less than the collapse load, so the catenary contributes nothing and the plateau is not even reached. Good for 0.07, 1.19. Only connections good for 0.15 radians — far beyond what an ordinary connection is detailed for — carry a sudden loss at twice the collapse load.

The gap between the dashed and solid lines is the dynamic penalty, and it widens with the rotation capacity: 1.11 against 0.94 at 0.03, 1.74 against 1.19 at 0.07, 3.85 against 2.04 at 0.15. A slow test of a frame’s catenary capacity therefore overstates its robustness against sudden loss by more the more ductile the frame is.

How much load an ordinary frame brings #

The load ratio along every axis above — the floor’s weight at the lost column over the beam’s plastic collapse load across the doubled span — is not a free choice. For a frame whose beams were designed for gravity on their own spans and nothing else, it is set by arithmetic.

A beam fixed at both ends and designed to collapse under its own span’s ultimate load $w_u$ has a plastic moment of $w_u L^2/16$. Lose the column at one end and the same beam spans $2L$, where a uniform load collapses it at a quarter of that intensity, $w_u/4$. The load it actually carries after the loss is the accidental combination — the permanent load and a fraction of the imposed one, about half of the ultimate load in an ordinary office. So the ratio is about $0.5\,w_u$ over $w_u/4$: about two. The beam is asked to carry roughly twice its collapse load across the gap, and on the figures above that is a dynamic factor of 1.88 and a rotation at the beam’s ends of 0.15 radians.

That is why the robustness rules do not trust bending alone, and why they ask for tying: a beam designed for its span cannot carry the gap in bending, by a factor of about two, and the only mechanism left is the cable. It is also why the rotation chart matters more than the factor chart. No ordinary connection supplies 0.15 radians, so a frame of ordinary connections that loses a column suddenly is carried, if at all, by something other than its beams’ catenary — the slab, the floors above acting as a Vierendeel, or ties detailed for exactly this.

The factor of two in that estimate is soft. A floor with a light imposed load, beams oversized for deflection, or composite action in the beams lowers it; a heavy floor or a corner column, with fewer beams to share the load, raises it. The conclusion is not: it sits on the right of the factor chart, in the catenary, where the dynamic penalty is nearly the linear spring’s.

How quick is sudden #

Every number here is for a column removed in an instant. The beam’s own elastic period, with the floor’s weight as its mass, is about half a second, and under 1.5 times its collapse load it takes 0.79 seconds to reach its lowest point. A column that is cut, crushed or blown out in a few hundredths of a second is gone long before the beam has begun to respond, and the release is sudden by any measure.

A release slower than the structure’s own period lets it follow the load down, and the factor falls toward one. A column that buckles under fire over minutes is not sudden at all, and a frame that survives a slowly failing column at its static capacity may not survive the same column removed by an impact. The design guides’ instantaneous removal is therefore the bounding case, not the typical one — but it is the case robustness is for, because the events that remove columns instantaneously are the ones nobody designed for.

The beam cut at its far ends, falling #

The free body is the beam, cut at its two far ends, with the falling floor’s weight at the middle. Statically, three things hold it: the bending moments at the hinges, the vertical shears at the ends, and — once it has sagged — the horizontal tension its ends are held against, whose vertical component at the middle is the tension times twice the slope.

Dynamically one more term enters: the floor’s inertia, a force equal to its mass times its acceleration, which is downward and growing while the beam is still giving way and reverses as the beam slows. The work balance is that free body integrated over the fall. The time history is the same free body solved step by step. They agree because they are the same equation.

The 1.5 times release by hand #

At 1.5 times the collapse load the static deflection is where $1 + (w/h)^2 = 1.5$: $w = 0.6\sqrt{0.5} = 0.424$ m beyond the 44 mm elastic limit, 468 mm in all.

The dynamic deflection is where the work under the resistance curve equals $1.5\,u$. Up to one depth of sag, 644 mm, the work is the elastic triangle, $0.044/2 = 0.022$, plus the plateau’s $0.6 + 0.6/3 = 0.8$, in units of the collapse load times metres: 0.822. The load’s work over the same 644 mm is $1.5 \times 0.644 = 0.966$, more than the beam has absorbed, so the beam is still falling. Past one depth the cable’s resistance $2w/h$ adds $(w^2 - h^2)/h$, and the balance $0.822 + (w^2 - 0.36)/0.6 = 1.5\,(0.044 + w)$ gives $w = 0.780$ m — 824 mm in all, where the beam’s resistance is $2 \times 0.780/0.6 = 2.60$ times the collapse load, and $2.60/1.5 = 1.73$.

What the model assumes #

A rectangular section held fully against spreading. The membrane curve is Jones’s for a rectangle whose ends cannot move apart at all. An I-section’s axial capacity is a smaller multiple of its plastic moment over its depth than a rectangle’s — about half — so its cable lifts the resistance more slowly per millimetre of sag; ends that can move a little — a column that bows inward, a floor plate that stretches — delay it further. Both change the numbers and neither changes the shape: a plateau, then a straight line.

Perfect plasticity, then a cable to any sag. Real steel hardens and real connections fracture. The curve here stops at nothing; the structure stops at the first connection that cannot turn further, which is the figure the rotation chart reads.

And one degree of freedom. The floor’s mass is lumped at the lost column. A real floor has its own modes, and a structure with many modes does not take the factor of two of one — but a beam whose response is dominated by one large sag is the case where one mode is closest to the truth.

What the curves cannot show #

They cannot show the rest of the frame. The beam’s ends are held against spreading by the columns and the floor beyond them, and those resist a tension that, at twice the collapse load, is the section’s full axial capacity. A force nobody put in the model arrives at the adjacent columns as a horizontal pull, and whether they can provide it is a separate check.

They cannot show damping. With no damping the beam oscillates for ever about its static position; with any, it settles. Damping does not change the first overshoot much, because it has had less than a cycle to act, and the first overshoot is what the factor is.

And they cannot show the floor slab. A composite floor adds its own membrane action, and the tension a slab can carry across a lost column changes what survives losing a member by more than any beam’s catenary.

What it comes to #

On the bending plateau the dynamic factor is low. 1.18 at nine tenths of the collapse load — paid for by a deflection nearly five times the static one.

In the catenary it climbs back toward two. 1.73 at 1.5 times the collapse load, 1.97 at three, because a cable’s resistance is a straight line and a straight line under a sudden load overshoots like a spring.

The cable needs rotation first. A sudden loss at 1.5 times the collapse load asks the connections for 0.10 radians.

And ordinary connections run out before the cable forms. Good for 0.03 radians, a sudden loss can be carried to 0.94 of the collapse load; good for 0.07, to 1.19.

Still open: the slab that hangs between the beams #

Every number here is one beam’s. A floor that loses a column has four beams meeting there and a slab spanning between them, and the slab, held at its edges by the beams that did not lose anything, can hang as a membrane in two directions at once. A membrane’s resistance grows with the square of its sag at first and linearly later, like the beam’s, but it reaches its straight line at a sag set by the bay rather than by a section depth, and its edges are held by beams whose own connections were never asked to carry tension. Whether the slab’s membrane arrives before or after the beams’ connections run out — and so whether a floor’s robustness belongs to its steelwork or to its concrete — is the question the single beam leaves.