The beam the wind walks without pushing

The load it can carry once found the shakedown load of a beam continuous over two spans whose loads came and went one span at a time: 126.3 kN against a collapse load of 150, and a cycle-by-cycle calculation that walked above the limit exactly as Melan’s and Koiter’s theorems said it would. A two-span beam has one redundant, so Melan’s search there was for one number. Its last paragraph named the next structure, a portal frame “under a vertical load that comes and goes and a horizontal load that reverses”, which has three redundants and a shakedown domain that is a region in the plane of its two loads rather than a single factor. It asked whether that region is bounded by alternating plasticity at a column base or by an incremental mechanism that combines a beam hinge driven by gravity with a sway hinge driven by wind.

It is bounded by alternating plasticity at a base, on one side. On the other it is bounded by an incremental mechanism, but not the combined one. It is the beam’s own mechanism — the one the wind does no work in at all.

A portal and the plane of its two loads #

Take a portal frame 10 m wide and 5 m high, fixed at both bases, with a plastic moment of 200 kNm and a stiffness of 42,000 kNm² in the columns and the beam alike. It carries a gravity load $V$ at the middle of its beam, which is either there or not, and a wind $H$ at the level of the beam, which blows from the left or the right. So the frame sees four extreme states: gravity with wind from either side, and wind from either side alone.

Under loads applied once and together, the bound theorems give the familiar collapse boundary. Five sections can hinge: the two bases, the two knees and the midspan. The beam mechanism hinges at the knees and the midspan, $VL/2 = 4M_p$; the sway mechanism at the four corners, $Hh = 4M_p$; the combined mechanism at both bases, one knee and the midspan, $VL/2 + Hh = 6M_p$. In the plane of $v = VL/M_p$ and $\eta = Hh/M_p$ that is three straight lines: $v = 8$, $\eta = 4$, and $v/2 + \eta = 6$ between them.

The shakedown boundary sits inside it, and its shape is not what the collapse boundary would suggest. Along the gravity axis the two meet: a gravity load that only comes and goes, with no wind, can be carried for ever right up to its collapse load, $v = 8$. As soon as there is any reversing wind the shakedown boundary leans in, along a straight line from $v = 8$ down to about $v = 6.0$ at $\eta = 2.7$. Beyond that it is nearly level, at $\eta$ between 2.7 and 3.2, all the way back to the wind axis.

So the plane has two kinds of edge. Neither of them is the combined mechanism’s line, along which the frame actually collapses for most mixtures of the two loads.

Three ways to find the same number #

The shakedown boundary comes from three calculations that have to agree before it is drawn.

Koiter’s theorem works from above. A mechanism’s hinges are each credited with the elastic moment that does most work in that hinge’s direction of rotation — taken from whichever of the four load states gives it — and the factor at which the credited work equals the plastic dissipation is an upper bound on the shakedown load. Every mechanism of this frame is a mixture of the beam and sway mechanisms, and the credited work changes its form only where some hinge stops rotating, so eight mechanisms — the beam, the sway and their sums and differences, each in both senses — are a complete search.

Melan’s theorem works from below. The frame has three redundants, so its residual moment fields — moments in equilibrium with no load at all — form a three-dimensional space. The frame shakes down if some field in that space, added to every load state’s elastic moments, keeps all five sections inside $\pm M_p$. That is a question of whether ten linear inequalities in three unknowns have a solution, and eliminating the unknowns one at a time answers it exactly. Bisecting on the load factor finds the largest that passes.

A simulation works by doing it: plastic hinges at the five sections, the residual moments each hinge rotation produces from the frame’s own flexibility, and the four load states applied over and over in a fixed order.

The first two agree to six figures at every point in the load plane and for every frame tried. The third settles below their number and walks above it.

The mechanism the wind cannot move #

The surprising branch is the first one. Take the beam loaded with 160 kN, which is exactly its collapse load, $VL/M_p = 8$, and add 60 kN of wind from either side, $\eta = 1.5$. Applied once, the wind makes no difference: the beam mechanism’s knees do not move sideways, so the wind does no work in it, and the frame still collapses at a factor of 1.00. Applied as a sequence, the pair can only be carried for ever at 0.88 of its value.

The reason is in the crediting. Gravity alone puts 160 kNm of hogging into each knee. Wind blowing to the left adds 56 kNm of hogging to the left knee and takes 56 off the right; wind to the right does the opposite. So the left knee’s worst hogging, 216 kNm, comes with the wind from one side, and the right knee’s worst, also 216, comes with the wind from the other. The beam mechanism needs both knees hogging and the midspan sagging. Koiter credits it with both knees’ worst, although no single storm gives them together, and the credited work exceeds what gravity alone gives by enough to lower the factor by 12 per cent.

That is the same trick the two-span beam played, in a new form. There the midspan hinges borrowed their moments from one span loaded and the support hinge from both. Here the hinges borrow from winds in opposite directions. The wind does no net work in the beam mechanism, but it does work at each hinge, and the work at the two knees is collected from two different storms.

The combined mechanism cannot do this. Its hinges — the two bases, the leeward knee and the midspan — are all driven harder by the wind blowing the same way. Credited with each hinge’s worst, it is credited with one storm’s moments, and its shakedown factor is its collapse factor. Reversing the wind earns it nothing. That is why the line along which this frame collapses never bounds what it can carry for ever: in five frames tried, with the columns half as stiff as the beam or twice as stiff, and with the beam weaker or stronger than the columns, the combined mechanism never governed shakedown once.

The residual that moves three sections at once #

Melan’s side of the same number says what the frame is trying to do to itself, and why it cannot.

At 97 per cent of the shakedown load, the elastic envelope over the four states puts 204 kNm of sagging at the midspan, which is past the plastic moment, and 184 kNm of hogging at each knee, which is not. The first time gravity arrives, the midspan yields by 1.22 milliradians and stops. That rotation leaves a residual field behind it: 4.1 kNm of hogging at the midspan, 4.1 kNm of hogging at each knee, and 2.0 kNm of sagging at each base. The midspan’s peak comes down to 200. The knees’ peak goes up to 188. Every later load state fits inside the plastic moment, and the frame never yields again.

The residual is tiny — two per cent of the plastic moment — and its shape is the whole story. A hinge at the midspan cannot lower the midspan’s moment without raising the knees’ hogging by the same amount, because in a fixed-base portal the beam’s residual moment is a straight line between the knees and a kink at the midspan would need a load to make it. So the useful residual moves three sections together, and at the shakedown load the three are exactly balanced: the midspan’s envelope, 210 kNm, needs 10.5 kNm taken off; each knee’s, 189.5 kNm, can take exactly 10.5 kNm more. Any more load and there is no residual that fits.

That is why the wind’s knee moments matter. They never touch the midspan. They use up the room at the knees that the midspan’s residual needs, and they do it from opposite sides on alternate storms, so each knee’s room is used up by its own worst wind. It is the envelope that is not a structure in a sharp form: the envelope of moments over every load state is not something any one state produces, and under loads that come and go it is what the structure has to fit inside.

What the beam does above the limit #

The simulation applies gravity, then the wind from the right, then takes gravity off, then puts it on with the wind from the left, takes it off, and lets the wind drop. Just below the shakedown load, the first cycle yields the midspan by 1.22 milliradians and leaves a residual field that every later state fits inside. Just above it, every cycle adds 15.2 milliradians at the midspan, and at 8 per cent above, 40.6 — a hinge rotation that at this span is a permanent sag of tens of millimetres a cycle.

The frame does not sway any more each cycle. Its permanent lateral displacement is fixed after the first cycle and stays there. The part of the frame that walks is the beam, under a horizontal load, in a mechanism with no horizontal movement in it. Anyone watching the frame for drift would see nothing. Anyone measuring the beam’s sag after each storm would see it grow.

That makes it a different kind of failure from the one the structure that settles down found in a section under heat. There the ratchet was a growing curvature at one section; here it is a mechanism assembled across three sections, each yielding under a different load state, and the deformation it produces is the shape of the mechanism. The beam does not reach a limit and stop. It keeps going for as long as the loads keep coming, and a span sagging by tens of millimetres a storm reaches a serviceability limit, and then a ponding problem on a flat roof, long before it reaches anything a strength check would call failure.

A breath of wind costs the beam #

Hold gravity at the beam’s collapse load and turn the wind up from nothing. The collapse factor stays at exactly one until $\eta = 2$, where the combined mechanism starts to need less than the beam mechanism; the wind is irrelevant to collapse until then. The shakedown factor falls from the first breath of wind: 0.91 at $\eta = 1$, 0.84 at $\eta = 2$, a straight line, because the knee credits grow in proportion to the wind. Past about $\eta = 3.7$ the shakedown factor falls faster, and the base takes over.

The gap between shakedown and collapse is what a plastic design that checks only collapse is relying on without knowing it. With a modest wind it is 9 to 16 per cent — less than the partial factors on the loads, but not by much, and it arrives in a direction no collapse calculation looks in.

The other edge, where nothing walks #

The nearly level branch of the boundary is a different failure.

At a column base, the moment from the wind reverses with the wind, and gravity adds a part that comes and goes. The base’s moment therefore swings over a range, and no residual moment can shrink a range: a constant added to both ends moves the swing, it does not narrow it. When the range exceeds twice the plastic moment the base must yield one way at one extreme and back the other way at the other, every cycle. The simulation shows exactly that. After the first few cycles the base settles into a loop, yielding through 1.2 milliradians each way, and gains nothing from one cycle to the next. The midspan never yields.

That is alternating plasticity, and it does not walk. Nothing accumulates, and the frame looks no different after a thousand storms than after three. What accumulates is damage. Each cycle is a cycle of plastic strain at one section, and yielding one way, and then the other is the experience that ends in low-cycle fatigue after hundreds or thousands of cycles rather than millions. A designer would want to know which of the two kinds of edge a frame is near. On this frame the line between them is at about $\eta = 2.7$: below it a beam that sags, above it a base that cracks.

And at the far end, with wind alone and no gravity, the shakedown load is exactly the first-yield load, $\eta = 3.2$. A load that reverses completely is its own worst range, and the range is twice the elastic peak. Under a fully reversing wind a fixed-base portal has no plastic reserve at all: the 25 per cent between first yield and the sway mechanism’s collapse at $\eta = 4$ belongs to a wind that blows once.

Stiffness, which collapse never sees #

The collapse boundary contains no stiffness. It is plastic moments and geometry, and a frame with columns twice as stiff collapses under exactly the same loads. The shakedown boundary depends on the elastic moments, which depend on how the frame shares its loads, which depends on stiffness.

Stiffer columns take more of the wind’s moment at their bases and leave less at the knees. So the two branches move in opposite directions. The beam’s branch, which lives on the knees’ wind moments, moves out: with little wind the frame with stiff columns shakes down at $v = 7.23$ and the one with soft columns at 6.95. The base’s branch, which lives on the bases’ wind moments, moves in: under wind alone, 2.86 for stiff columns against 3.50 for soft. A designer who stiffens the columns of a portal to control its drift buys a beam that ratchets later and bases that alternate sooner. After the first yield made the point that plastic design is valuable precisely because it does not need the stiffnesses. Under loads that vary, that value is gone, and the stiffnesses are back in the answer.

What the calculation does not include #

The frame is the simplest one with these properties, and it leaves out several things a real portal has.

The hinges are at five fixed sections. A real gravity load is distributed, so the beam’s sagging hinge can form anywhere along it, and a pitched portal has hinges in its rafters near the apex and near the haunches. More hinge positions mean more mechanisms, which can only lower an upper bound. The numbers here are for the portal as idealised, not a bound on all portals.

The plastic moment is perfectly plastic. Strain hardening lets a section that has yielded carry a little more next time, which would let some frames shake down slightly above these lines; the cyclic softening of some steels does the opposite. What is left when the load comes off is about the unloading, and it is not elastic–perfectly plastic either.

The loads are four extremes. A real wind does not come in two values, and a real gravity load is several loads. The theorems need only the extremes, so this is generous to the frame only if the extremes are right, and the beam branch’s 12 per cent comes from the wind’s peak in each direction, however rarely each is reached.

Nothing is second-order. A frame swaying under wind carries its gravity load through a displacement, which adds to the base and knee moments, as the frame that leans found. On a stocky portal it is a few per cent. On a slender one it moves both branches inward.

Still open: the frame whose bases are not fixed #

Every base here is fixed, and the alternating branch belongs to the bases. Many portals stand on pinned bases, or on bases whose fixity is a bolted plate on a pad that rotates a little. A pinned base has no moment, so it cannot alternate, and the wind’s moment goes to the knees instead — which is exactly what the beam branch feeds on. Whether a pinned-base portal trades its alternating edge for a beam that ratchets at a much smaller wind, whether a partly fixed base that yields in its bolts rather than its column moves the alternating branch onto a component that can be replaced, and whether the second-order sway that a pinned frame suffers more makes either edge come sooner, is the question the same two loads put to the base that most portals actually have.