The load it can carry once

The bound theorems of plastic collapse give a structure’s collapse load from two sides: any mechanism overestimates it, any moment field in equilibrium and inside the plastic moment underestimates it, and where they meet is the answer. That essay ended by naming the assumption both theorems rest on — that every load grows together in a fixed ratio — and saying that under loads that vary independently a structure can fail at a lower factor by a different mechanism, found by a different pair of theorems.

The essays on shakedown have so far computed that other failure for a section under a steady moment and a cycled temperature, and mapped it as three regions on two axes. Both are about a section. The theorems themselves — Melan’s for the lower bound and Koiter’s for the upper — are theorems about structures, and they are clearest on the simplest structure where loads can take turns: a beam continuous over two spans.

Two spans, two loads, four states #

Take a steel beam continuous over three supports, two spans of 8 m each, with a plastic moment of 200 kNm throughout. Each span carries a point load $W$ at its midspan. The loads are independent: either can be present or absent, so the beam sees four load states — span 1 loaded, span 2 loaded, both, and neither — in any order, as many times as its life delivers.

The elastic moments are standard results for a two-span beam. With span 1 loaded alone, the moment at its midspan is $\tfrac{13}{64}WL$ sagging, at the support $\tfrac{6}{64}WL$ hogging, and at the other midspan $\tfrac{3}{64}WL$ hogging. With both loaded, $\tfrac{10}{64}WL$ sagging at each midspan and $\tfrac{12}{64}WL$ hogging at the support. Span 2 alone is the mirror image of span 1.

Under proportional loading the numbers are the familiar ones. The largest elastic moment is the $\tfrac{13}{64}WL$ at a loaded midspan, so first yield is at $WL = \tfrac{64}{13}M_p$: $W = 123.1$ kN. Collapse is by the beam mechanism in one span — a hinge at the loaded midspan and one at the support — and the work equation $W \cdot \tfrac{L}{2}\theta = M_p(2\theta + \theta)$ gives $WL = 6M_p$: $W = 150$ kN. Redistribution of moment between first yield and collapse is worth 22 per cent of the load.

Under variable loading the answer is 126.3 kN, and the rest of this essay is where that number comes from and what happens above it.

Melan: find a straight line that fits #

Melan’s theorem says a structure shakes down — stops accumulating plastic deformation after some initial yielding — if there exists any time-independent residual moment field which, added to the elastic moments of every load state the structure can see, keeps the total inside the plastic moment everywhere.

For this beam the residual moment field has a simple form. A residual field is a set of internal forces in equilibrium with no load, and a beam with one redundant has exactly one of them: the moment diagram produced by a force at the central support with nothing else applied. It is a straight line from zero at each end to some value $m$ at the support, and $m/2$ at each midspan. The whole of Melan’s search is a search for one number.

For each critical section, write down the largest and smallest elastic moment over all four states — the envelope — and require the envelope plus the residual to stay inside $\pm M_p$:

• at a midspan: $\tfrac{13}{64}WL + \tfrac{m}{2} \le M_p$ and $-\tfrac{3}{64}WL + \tfrac{m}{2} \ge -M_p$;
• at the support: $0 + m \le M_p$ and $-\tfrac{12}{64}WL + m \ge -M_p$.

The first and last are the binding pair. The first wants $m$ negative enough to pull the sagging peak down; the last wants $m$ not so negative that the hogging peak at the support goes past $-M_p$. They are compatible while

$$-M_p + \tfrac{12}{64}WL \;\le\; m \;\le\; 2M_p - \tfrac{26}{64}WL,$$

and the interval closes when $\tfrac{38}{64}WL = 3M_p$, which is $WL = \tfrac{96}{19}M_p = 5.053\,M_p$. The shakedown load is $W = 126.3$ kN, and at that load the only residual that works is $m = -\tfrac{1}{19}M_p$, 10.5 kNm of hogging at the support.

The residual is small — five per cent of the plastic moment — and the elastic envelope only just crosses the plastic moment at the midspans. That smallness is the point. The beam has almost nothing to redistribute. Pushing the midspan peaks down means pushing the support peak up, and the support peak is already at 95 per cent of the plastic moment from the state with both spans loaded.

The shrinking wedge is Melan’s theorem drawn. Below first yield, the elastic moments alone fit, so the beam never yields and zero is an acceptable residual. Between first yield and 5.053 the elastic moments do not fit, but some negative residual makes them fit — and the first few cycles of loading will manufacture one, by yielding a little at the midspans. At 5.053 the wedge closes. Above it no residual field exists, and nothing the beam can do to itself will make every state elastic.

The gap between first yield and shakedown is 2.6 per cent. The gap between shakedown and collapse is 16 per cent. Under loads that vary, the reserve of strength that plastic design is built on has almost entirely gone.

Koiter: a mechanism nobody loads #

Koiter’s theorem comes at the same number from above. Choose a mechanism — a set of hinge rotations that is kinematically admissible, so the beam’s supports stay where they are — and credit each hinge with the elastic moment, over all the load states, that does the most work in that hinge’s direction of rotation. The load factor at which that credited work equals the plastic dissipation is an upper bound on the shakedown load, and the lowest over all mechanisms is the answer.

The difference from the collapse theorem is the crediting. Under proportional loading every hinge gets its moment from the same load state. Under Koiter each hinge may borrow from a different one.

Take the mechanism with a sagging hinge at each midspan and a hogging hinge at the support, each turning through $2\theta$, so the beam sags as a W. The dissipation is $6M_p\theta$. The credited work is $\tfrac{13}{64}WL \times 2\theta$ at each midspan and $\tfrac{12}{64}WL \times 2\theta$ at the support, $\tfrac{76}{64}WL\theta$ in all. So $WL = \tfrac{6 \times 64}{76}M_p = 5.053\,M_p$.

No load state produces those three moments together. The 13/64 at a midspan needs that span loaded and the other empty; the 12/64 at the support needs both loaded. A proportional analysis, which evaluates every state separately and takes the worst, never assembles this mechanism, because in no single state is it the mechanism the beam would form. Under variable loading it forms anyway, one hinge at a time, because each load state yields a little at the hinge it drives hardest.

The single-span beam mechanism, credited the same way, gives the same 5.053 — its midspan hinge borrows from one state and its support hinge from another — which is why the static and kinematic numbers meet at the first try. On a beam with more spans or more load positions the kinematic search has more mechanisms to try and the static search more envelope constraints, and the two still meet.

What happens above it #

The theorems say what the limit is. They say nothing about what the beam does past it, and the only way to see that is to load it.

Model the beam as elastic, with $EI = 42{,}000$ kNm², and with plastic hinges that can form at the three critical sections when their moment reaches $\pm M_p$. The residual support moment is then fixed by compatibility: every hinge rotation is a kink in a beam that must still meet its central support, and the force the support has to supply to close the resulting gap is the residual field. Apply the four load states in the order span 1, both, span 2, neither, and repeat.

The hero figure is that calculation at four loads. At 0.98 times the shakedown load, the first cycle yields the loaded midspan by 0.59 milliradians and then nothing moves again: the residual that yield created is one of the fields in the wedge, and every later state is elastic. At 1.02 times the shakedown load the beam gains 4.6 milliradians of rotation every cycle; at 1.05, 11.4; at 1.10, 22.9.

The rotation does not accumulate evenly. After the first cycle one midspan hinge is finished, and the growth alternates between the support, which yields hogging each time both spans are loaded, and the other midspan, which yields sagging each time its own span is. Each yield moves the residual field, and the next state finds it moved in exactly the wrong direction and yields back. The beam builds a mechanism out of two load states taking turns, and each turn adds to a permanent deflection.

One cycle, by hand #

The ratchet is short enough to follow state by state, at 1.05 times the shakedown load — $W = 132.6$ kN, $WL = 5.305\,M_p$ — and following it shows exactly which inequality each state breaks.

Two facts do all the work. The residual support moment $m$ changes only when a hinge rotates, and a hinge rotation $\theta$ at a midspan changes it by $-\tfrac{3EI}{4L}\theta$, while one at the support changes it by $-\tfrac{3EI}{2L}\theta$. For this beam $3EI/L$ is 15,750 kNm per radian.

Span 1 loaded. The elastic moment at its midspan is $\tfrac{13}{64} \times 5.305 \times 200 = 215.5$ kNm, above the plastic moment. The midspan hinge rotates until the residual has pulled it back to 200: that needs $m/2 = -15.5$, so $m = -31.1$ kNm of hogging, and a sagging rotation of $31.1 \times 4/15{,}750 = 7.9$ milliradians.

Both loaded. The elastic support moment is $-\tfrac{12}{64} \times 5.305 \times 200 = -198.9$ kNm, and with the $-31.1$ residual the total is $-230$ — past the plastic moment in hogging. The support hinge rotates until the total is $-200$: the residual has to come back to $-1.1$, a change of 30 kNm, which is a hogging rotation of $30 \times 2/15{,}750 = 3.8$ milliradians.

Span 2 loaded. Its midspan’s elastic moment is 215.5 again, and the residual now contributes only $-0.55$. The second midspan hinge rotates, sagging 7.6 milliradians, and drives the residual back to $-31.1$.

Neither. Nothing yields. The beam is unloaded carrying a residual of 31.1 kNm of hogging at the support and 15.5 at each midspan.

The second cycle begins with span 1 loaded, and its midspan now sees $215.5 - 15.5 = 200$ kNm: exactly the plastic moment, and no further rotation. That hinge is finished. But the next state, both spans loaded, finds the support at $-198.9 - 31.1 = -230$ again, and yields it another 3.8 milliradians; and span 2 loaded finds its midspan at 215.5 again and yields it another 7.6. The residual returns to where it was after every cycle, and the rotations do not.

That is the ratchet in one sentence: the state that loads the support and the state that loads the second midspan each need a residual the other one destroys. One pulls $m$ toward zero, the other pushes it back to $-31.1$, and each push costs plastic rotation. Below the shakedown load the two required residuals overlap — the wedge in the second figure — and a value exists that both accept. Above it they do not, and the beam pays for the disagreement every time the loads change places.

That deflection is the practical consequence. At 1.05 times the shakedown load the incremental mechanism turns its support hinge 3.8 milliradians a cycle, which lowers the ratcheting midspan by $3.8 \times 10^{-3} \times 4{,}000 = 15$ mm every cycle. A crane runway or a floor with its live load moving between bays sees thousands of cycles. At 132.6 kN — twelve per cent below the collapse load, and a load the collapse calculation says is safe — the beam would be a hundred and fifty millimetres lower after ten of them.

The rate is stiffness, the limit is not #

Double the beam’s stiffness and run the same calculation.

The shakedown load has not moved. It cannot: Melan’s inequalities contain the elastic moments, which for a beam of uniform stiffness depend only on geometry, and the plastic moment, and nothing else. What has halved is the rotation per cycle, because a stiffer beam needs a smaller kink to produce the same residual moment.

That separation is worth carrying. Whether a structure ratchets is a question of strength and geometry. How fast it ratchets is a question of stiffness. A design check that asks the first question needs no stiffness at all; a serviceability judgement about how many overloads a structure can tolerate needs the second, and gets a different answer for a beam twice as stiff.

Where shakedown governs, and where it does not #

Most buildings never approach this. The loads that vary independently on a floor are live loads, a fraction of the total, and a beam designed with load factors applied to the dead load as well is rarely loaded above its elastic limit in service, let alone its shakedown limit.

It governs where the variable part of the load is a large fraction of the whole and genuinely moves: crane girders, bridge decks under traffic, conveyor supports, and the continuous beams under moving plant. It governs where a design has relied on redistribution — a continuous beam sized to its plastic collapse load under the governing arrangement of loads — and the loads that produce that arrangement do not arrive together. And it governs for any structure whose worst states are different states: a load whose worst position depends on the member is exactly a load that will drive different hinges in turn.

The numbers here also suggest a reading of a rule that can look arbitrary. Design rules that permit moment redistribution in continuous members commonly cap it at a stated fraction of the elastic moment rather than allowing the full collapse mechanism. The 2.6 per cent between first yield and shakedown in this example is small because the support moment under both spans loaded is already near its limit. A limit on redistribution is a cruder statement of the same fact: the reserve that exists under one arrangement is spent differently under another.

Where the model stops #

Hinges at three sections only. The simulation lets plasticity form only at the midspans and the support, which is where the envelope peaks. A real beam spreads its yielding over a length, and a residual stress field inside each section — the kind a section keeps when the load is removed — accompanies the residual moment field between sections. The theorems apply to the stress field; the moment version is the beam idealisation of them.

Elastic–perfectly plastic. No strain hardening. Hardening raises the effective plastic moment with each excursion and can arrest ratcheting that the perfectly plastic model continues; it is also what makes low-cycle fatigue at a ratcheting hinge the failure that ends the process.

Quasi-static loading. The states are applied slowly, and the order matters to the details of the growth — which hinge yields first in each cycle — but not to whether it happens. Melan’s theorem is independent of the order, which is one of its strengths: it needs only the set of states, not their sequence.

Point loads at midspan. Uniformly distributed live loads span by span give a different set of envelope coefficients and a different shakedown factor; loads that move continuously along the beam need the whole influence-line envelope rather than four states. The method is unchanged: an envelope, one residual line, one inequality pair.

No instability. A beam gaining rotation every cycle eventually has hinges whose rotation capacity is exhausted, whose flanges have buckled locally, or whose lateral restraint has gone. None of that is here, and all of it arrives before the deflections become large.

Still open: the frame whose sway takes turns with its beams #

A continuous beam has one residual field, so Melan’s search is for one number. A portal frame under a vertical load that comes and goes and a horizontal load that reverses has two residual fields, one of them a sway, and its shakedown domain is a region in the plane of the two loads rather than a single factor. Whether that region is bounded by alternating plasticity at a column base — the reversing load yielding the same section in both directions — or by an incremental mechanism that combines a beam hinge driven by gravity with a sway hinge driven by wind, and which of the two a real frame reaches first, is the question the next structure asks.