Stability

The tie that spends an afternoon as a strut

A tension member is chosen by its area and nothing else. A compression member is chosen by how that area is arranged. So a member whose force reverses under some load case is not merely being asked for the same number with the other sign — it is being designed against a different variable, and the same steel can carry seventy times more or less depending on a shape nobody chose for that purpose.

Assumes Strong enough and still falls over, The member with only one direction and Counting the unknowns, and finding out whether statics can answer.

Nothing in this collection is easier to design than a member in tension. Its capacity is its area times a stress; there is no length in the answer, no shape, no imperfection and no eigenvalue. A tie is the one structural element whose design is a division.

That easiness is exactly why reversal is dangerous. A member designed by dividing has never been asked the question a compression member is asked, and the answer to that question is not implied by the answer to the first one.

A Pratt truss of 8 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.54.454.493.3116.7116.793.354.454.4-93.3-116.7-124.4-124.4-116.7-93.3-35.0-21.0-14.0-21.0-35.0-73.2-73.252.331.410.510.531.452.3tensioncompression2 carrying nothing
Fig. 1 A Pratt truss with the joint equations assembled and solved. Fourteen members came back in tension, thirteen in compression, and two carrying nothing at all — for this arrangement of load.

Which free body produced the number

One member, cut out. In tension it is a two-force member carrying NN along its own line, and the free body says only that the stress is N/AN/A.

In compression the same free body is not enough. Take a cut somewhere along the member with the member very slightly bowed, and the axial force now has a lever arm about the cut equal to the bow. That moment bends the member, which increases the bow, which increases the moment. The equation that closes is an eigenvalue and its answer contains II rather than AA — or, written the way a designer uses it, the radius of gyration r=I/Ar = \sqrt{I/A}.

So the two designs read different columns of the same section table. Area for one; a length hidden inside the section for the other.

The one length a section carries into a columnFour profiles of equal area, with the radius of gyration r = √(I/A) drawn as the distance it is — a pair of lines either side of the centroid, at the depth the whole area would have to sit at to give the section the second moment it has. As a 4 m pin-ended column the same 6000 mm² of material carries between 58 and 4149 kN, in the ratio of the squares of those radii and of nothing else.the same, laid flatr = 8.7 mmλ = 46258 kNsquarer = 22.4 mmλ = 179389 kNtall rectangler = 57.7 mmλ = 692591 kNI-sectionr = 73.1 mmλ = 554149 kNthe dashed pair is ±r about the centroidthe bar is the Euler load at 4 m, to scale
Fig. 2 Four profiles of exactly equal area, with the radius of gyration drawn as the distance it is. As a 4 m pin-ended column, the same 6000 mm² carries between 58 and 4149 kN.

A factor of seventy-one, on the same quantity of the same steel, decided by nothing but arrangement. That is the one length a section carries into a column, and it is the whole content of the reversal problem: a tie chosen for 6,000 mm² was never chosen for a radius of gyration, so its compression capacity is an accident of the section list.

Where reversal comes from, which is not only the wind

Uplift is the obvious source and it is not the commonest.

Pattern loading. Any load that can be present or absent produces a set of arrangements, and the internal forces are different in each. In a truss with a light dead load and a heavy imposed one, a web member near the middle can be in tension under full load and in compression under load on half the span. Nothing about the weather is involved.

Wind uplift. A light roof under suction has its dead load reversed. The bottom chord of a truss, in tension all its life, goes into compression; the sag rods that hold the purlins, designed as ties, are asked to push.

Seismic and blast. Both are reversing by nature: every brace in a concentrically braced frame goes into compression on alternate half-cycles by construction, which is why the design of such a frame is dominated by what a brace does after it has buckled.

Erection and lifting. A structure hanging from two points has its moment diagram inverted along most of its length compared with the finished condition.

Removing each member in turnEvery member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.02 times what it carried before. A single number for robustness does not exist: it depends on which member goes.05101520253001234member removedworst force ÷ the force it was designed forat the top:nothing is left1.0 — carryingwhat it always did
Fig. 3 Every member of a truss removed in turn, with the worst demand on the survivors. Four of the thirty-five leave a mechanism; the rest redistribute, and by very different amounts.

The tension-only assumption, and what it costs

The commonest way of dealing with reversal is to declare that it does not have to be dealt with. A cross-braced bay is designed tension-only: both diagonals are provided, the compression one is assumed to buckle out of the way immediately, and the whole storey shear is carried by whichever diagonal is in tension.

The assumption is sound and it has two prices.

The first is stiffness. Only half the bracing is working at any instant, so the bay is half as stiff as its member sizes suggest — and a rod or a flat is slender enough that this is genuinely true rather than conservative.

The second is worse and is about slack.

Prestress buys a stiffness no change of material canFour cables of identical steel — 45 m, 1000 mm², E = 160000 MPa — differing only in the tension put into them before the load arrived. The initial stiffness is 8T₀/L exactly: 0.0, 40.0, 160.0, 640.0 kN/m at T₀ = 0, 225, 900, 3600 kN, and no property of the steel appears in that expression. The slack cable leaves the origin flat — it has no stiffness whatever at zero load, and its sag grows as the cube root of the load, reaching 1.818 m under the same 225 kN that puts 0.349 m into the tightest of them. Four curves of one cable: the tightest starts 16 times stiffer than the slackest that has any stiffness at all, and every other property they share.00.511.5050100150200midspan sag (m)total load on the cable (kN)T₀ = 0 kN · k₀ = 0.0T₀ = 225 kN · k₀ = 40.0T₀ = 900 kN · k₀ = 160.0T₀ = 3600 kN · k₀ = 640.0all at 5 kN/mat zero prestress the curveleaves the origin flat
Fig. 4 Four cables of identical steel differing only in the tension put into them beforehand. The initial stiffness is 8T₀/L exactly, and the slack one leaves the origin flat — no stiffness whatever at zero load.

A tension-only diagonal is a member that has to take up its own slack before it does anything. Until it does, the bay has no lateral stiffness at all, and it acquires that slack from erection tolerance, from the elongation of the diagonal that was in tension last time, and from any permanent set in the previous cycle. The initial stiffness of a sagging tie is 8T0/L8T_0/L — proportional to the pretension and containing no property of the steel — so a tie with no pretension has, exactly, none.

A slack guy is not a weak spring, it is barely a springErnst's tangent modulus — the stiffness a sagging cable actually offers, against the steel's own — plotted against tension as a fraction of breaking load, for a 90 m guy of 1200 mm². The correction goes as the cube of the tension, so the curve collapses rather than sloping: at 14 per cent of breaking load the guy has 92 per cent of its material stiffness, and at 3 per cent it has 10. That is why a leeward guy stops contributing long before it stops carrying load, and it is the whole reason a guyed mast is pretensioned at all.01020304000.20.40.60.81tension (% of breaking load)tangent modulus ÷ E92% at 14%10% at 3%the correction falls as the cube of the tension, so it is a cliff rather than a slope
Fig. 5 The tangent stiffness a sagging cable actually offers against its steel’s own. The correction goes as the cube of the tension, so the curve collapses rather than sloping.

That collapse is why a guy that goes soft stops helping long before it stops carrying load: at 14% of breaking load it retains 92% of its material stiffness, and at 3% it retains 10%. A leeward guy on a mast is in exactly the state a slackened tension-only brace is in.

The other way, which buys the stiffness back

The alternative is to design both diagonals for compression. It costs section — a rod becomes an angle, an angle becomes a hollow section — and it buys three things: the bay is twice as stiff, the slack question disappears, and the structure is no longer relying on a member’s failure as part of its intended behaviour.

Every path to the ground goes through the linkA braced bay 7 m by 4 m whose two diagonals stop 800 mm apart instead of meeting. The storey shear reaches the ground through the diagonals, and the vertical components they deliver to the beam have to pass through the segment between them: the link carries 47% of the applied shear as a shear force, at a lever arm short enough that its ends reach 0 kNm while the rest of the beam carries 0. The deflected shape drawn is the solved one, magnified — the real drift under this load is 0.008 mm. Everything outside the link is designed to stay elastic while the link is yielding, which is what makes the mechanism a choice rather than a hope.storey shearthe link — 800 mm0.68 kN in each diagonal per kN of shearlink shear 0.47 kN · storey drift 0.008 mm · deflection magnified 6057×
Fig. 6 A braced bay whose diagonals stop short of each other. Every path to the ground goes through the segment between them, which carries 47% of the applied shear at a very short lever arm.

Whichever route is chosen, the decision has to be made explicitly, because the analysis will not make it. A model with both diagonals present and both able to take compression is stiffer than the real bay if the real bay is tension-only, and the error is in the unsafe direction for drift.

The slenderness limit that is not about buckling

There is a rule in every steel standard that a member’s slenderness shall not exceed some number — 180, 250, 300 depending on what the member does — and it applies to ties as well as to struts.

The reason is not buckling, because a tie does not buckle. It is that a very slender tie sags under its own weight, vibrates in the wind, is damaged by anybody who leans on it during construction, and — the reason that matters here — is a member that cannot take any compression at all if the load ever reverses.

Length costs more than it looksThe same column section at four lengths, with the buckling capacity of each drawn as a bar. Capacity falls as the inverse square of the length, so a column three times as long carries a ninth as much.2× the length25% of the capacity3× the length11% of the capacity4× the length6% of the capacity6× the length3% of the capacityidentical section, identical material, identical end conditions
Fig. 7 The same section at four lengths. Capacity falls as the inverse square of the length, so a member three times as long carries a ninth as much and one six times as long carries a thirty-sixth.

A slenderness limit on a tie is therefore a reversal provision wearing other clothes. It quietly guarantees that a member designed by division has some compression capacity, without anybody having to work out how much.

When the reversal removes the structure rather than a force

A more serious version of the problem appears when the member that reverses is one the structure counts on for its geometry rather than for its force.

The count is necessary and not sufficientTwo pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.75 of the span.one panel braced twice, the next not at allm 9 + r 3 = 2j 12 · rank 11a mechanismthe same count, properly arrangedm 9 + r 3 = 2j 12 · rank 12stands up
Fig. 8 Two pin-jointed frames, each satisfying the member count exactly. One folds anyway, because the equations are not independent, and the ghosted outline is the motion that costs no member any change of length.

A truss whose diagonals are tension-only is, for each direction of loading, a different truss — half the diagonals are absent. If that reduced truss is still determinate and stable, nothing is wrong. If it is not, the bay is a mechanism in one direction and a structure in the other, and no member count run on the drawing will report it, because the drawing shows all the members.

Joint 3 of the truss, cut outOne joint of the truss with every force acting on it. Two equations — the horizontal and vertical sums — are enough for a joint with no more than two unknown member forces, which is the whole method.HV83.388.95.0-7.5ΣH = 0 and ΣV = 0, and nothing else is needed
Fig. 9 One joint with every force on it. Two equations settle a joint with no more than two unknowns, and the whole method assumes every member drawn at that joint is there in both directions.

The length that appears out of nowhere

There is a second quantity a tension design never produced, and it is easier to miss than the radius of gyration because it is not a property of the member at all.

A compression capacity needs an effective length, and an effective length is a property of what restrains the member rather than of the member itself. A tie running the full height of a bay is one member with two ends, and nothing in its tension design ever asked whether anything holds it anywhere in between. The moment it goes into compression, the answer decides everything: restrained at midpoint, its capacity is four times what it is unrestrained, because the ends decide the length and so does everything between them.

The awkward case is the crossing point of a pair of diagonals. Where two ties cross and are bolted together, the tension diagonal restrains the compression one at midlength — but only if it is in tension at that instant, which it is, and only if it is stiff, which a slack tie is not. So the effective length of the compression diagonal depends on the tension in the other one, which depends on the load, which is the case being checked. The design is circular and the standards resolve it with a rule of thumb rather than a calculation.

A tie that crosses nothing has no such help. A single diagonal in a bay, sized for 400 kN of tension at 5 m long, has an effective length of 5 m in compression and very probably a capacity in the tens of kilonewtons.

Four members that have done this

Sag rods. Provided to hold purlins in line during construction and against the down-slope component of the roof load, sized as ties, and put into compression the moment the roof goes into uplift. They are usually 12 mm rods at 2 m centres and their compression capacity is a rounding error.

Bottom chords of trusses. In tension for the whole of a building’s life under gravity and in compression under wind uplift on a light roof. The bottom chord is also the one member of a roof truss that is typically unrestrained along its length, because the bracing and the sheeting are both at the top. The unrestrained length is then the whole span.

Glazing and façade tie rods. Slender by intention, because they are meant to be invisible, and pretensioned for exactly the reason this essay describes — a rod with no pretension has no stiffness at all until it is straight.

Uplift anchors and holding-down bolts. Designed for the tension a wind case produces and asked for compression by nothing, until the base plate lifts and settles back and the bolt is the only thing in contact.

The pattern in all four is the same, and it is not carelessness. Each member was created to solve a problem in which it was in tension. The load case that reverses it belongs to a different part of the design, done by a different check, and the member’s presence in that check is not obvious from either side.

Where the model stops

A buckled brace is not gone. The tension-only idealisation says the compression diagonal carries nothing. It carries its post-buckling residual, which for a stocky brace is a substantial fraction of its capacity and which is what makes the beam and column of the bay see forces the idealisation does not contain. Under cyclic load that residual degrades pass by pass, so the structure changes as it is being loaded.

The reversal may not be the worst case for the member. A brace’s design is often set by the requirement that the members around it stay elastic while it yields — the capacity design idea behind the part that is meant to be weak — and that requirement is written on the brace’s actual strength rather than its required one. A brace made generously stronger for compression makes the beam and the connection worse.

Nothing here is dynamic. Under an earthquake the reversal happens many times, at a rate that changes the material’s behaviour, with the buckled shape accumulating. Everything above is a static comparison between two load cases.

The column curveFailure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.2040608010012014016018000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither
Fig. 10 The column curve, which is the second half of every reversal check. A stocky member crushes, a slender one buckles, and real members fall below both near the crossover.

What the picture cannot show

The truss at the top of this essay is drawn with its members coloured by sign, and the colouring is a property of the load case rather than of the truss. There is no drawing anywhere of “the members that reverse”, because that is a statement about a set of load cases and every drawing shows one.

Nor does any figure show the connection. A member that reverses reverses at its ends too, and a bolted connection designed for tension has its bolts in shear either way while a welded lap has a very different stress field in push than in pull. Cleats designed for a tie are frequently long, thin and eccentric — the arrangement with the least compression capacity available, chosen for reasons that had nothing to do with compression.

What a reversal does to the connection

The member is only half of it, and the other half is rarely checked at all.

A bolted lap in tension carries its load in bearing and shear, and reversing the sign changes nothing about either — the bolt bears on the other side of its hole, and it has to travel the clearance to get there. That travel is a slip, it is a millimetre or two per joint, and in a bracing system with several joints in series it is the largest single contribution to the drift. A joint that carries nothing until it slips is the essay about the mechanism; a reversing member does it twice per cycle.

A welded connection has no slip and a different problem. A lap weld in tension is loaded along the length of a stiff plate; in compression the same plate is a strut that has to be checked for buckling between its welds, and a long thin gusset carrying compression is a width nobody drew — the dispersion width in the plate rather than the width of the plate itself.

And a pinned connection in a tension-only system may simply come apart. A rod with a clevis at each end, tightened up, carries nothing in compression and rattles, which is fine mechanically and unacceptable for anything a person can hear.

The generalisation

The habit worth carrying is to ask, of every member, which variable decided it.

A member decided by area has no length in its design. A member decided by a radius of gyration has a length, an end condition and an imperfection in it. A member decided by a deflection limit has a second moment. When a load case changes which of those governs, the member is not being asked for more of the same thing; it is being asked a different question, and the previous answer contains no information about the new one.

This is the same structure of argument as which failure arrives first applied to a single member rather than to a structure, and it has the same practical form: the check that matters is not the one that is tightest, it is the one that was never made. A tie designed by division, in a structure where nothing was ever expected to reverse, is a member with no compression check in its history at all — and the day it acquires one, the answer has already been chosen for it by whoever picked the section.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

BracingBucklingCompressionCritical loadLoad arrangementLoad pathLoad reversalMechanismPrestressRadius of gyrationRobustnessSlendernessTensionTwo force memberZero force member