Structural Engineering & Statics
where the load goes
Equilibrium

The load that is really a lean

No frame is ever plumb. The columns are out of upright by something like a three-hundredth, and every tonne of gravity load standing on that lean has a horizontal component. The force that represents it is not a safety allowance — it is an exact statics substitution for a geometry nobody drew.

Assumes The free body is a choice, and choosing it well is the whole skill, Moving a force, and what it costs and Strong enough and still falls over.

19 min read 10 figures The load must go somewhereThe statics of things that do not moveWhich failure arrives first

A frame is drawn plumb and built leaning. The lean is small — a three-hundredth of the height is a typical erection tolerance, which on a 3.6 m storey is twelve millimetres — and it is not an error, because there is no such thing as a plumb building. Steel arrives with a mill tolerance on its straightness, columns are set with a plumb line against a rule, welds shrink, foundations settle differentially, and the finished frame is out of upright by an amount somebody wrote into a specification as acceptable.

The structural consequence of that lean is a horizontal force, and the size of it is fixed by nothing more than a free body.

The load that makes itself worseThe amplification of a deflection against the ratio of applied load to buckling load. A structure at half its buckling load deflects twice as far as first-order analysis predicts, and the curve runs away well before the load is reached.00.20.40.60.80246810applied load ÷ buckling load1.1×1.3×1.4×first-order analysis says the answer is always 1×one over one minus the ratio
Fig. 1 Why the lean matters at all. A structure standing off the vertical carries its own weight off the axis, and the deflection that follows is amplified — sharply, long before the buckling load is anywhere near.

Which free body produced the number

One storey, cut top and bottom. Its columns carry a vertical load NN down to the storey below, and they are out of plumb by an angle φ\varphi. Take moments about the base of the storey.

The vertical load’s line of action arrives at the top of the storey displaced sideways by φh\varphi h relative to where it leaves at the bottom. So it delivers a moment NφhN \varphi h about the storey base, which nothing in a plumb analysis contains. And NφhN\varphi h is exactly the moment of a horizontal force φN\varphi N applied at the top of the storey.

That is the whole derivation. It is a statics identity between a geometry and a force, and it holds regardless of what the frame is made of, how stiff it is, or whether it is braced. Two free bodies produce the same moment, so a designer may draw whichever is easier — and the force is easier, because every analysis program in existence accepts a horizontal load and none of them accepts “the building is not quite straight”.

The identity is the reason for the name. The force is notional: nothing pushes on the building. It is the leaning of the building, written down.

The size of it, which is proportional to gravity

Because the substitution is φN\varphi N, the notional force is proportional to the vertical load, and a column’s vertical load is a running total down the height.

A column is a running totalA column carrying 42 m² of floor at each of twelve levels, at 7 kN/m². Each floor adds 294 kN, so the load at the base is 3528 kN — the same tributary area counted twelve times. Nothing in the drawing changes down the height; only the number does.level 12294 kNlevel 11588 kNlevel 10882 kNlevel 91176 kNlevel 81470 kNlevel 71764 kNlevel 62058 kNlevel 52352 kNlevel 42646 kNlevel 32940 kNlevel 23234 kNlevel 13528 kN3528 kN into the foundation= 42 m² × 7 kN/m² × twelve floors
Fig. 2 Where the multiplier comes from. The notional force at a level is the out-of-plumb times the vertical load at that level, and the vertical load is the same tributary area counted once per storey.

Take a twelve-storey frame whose columns each carry 42 m² of floor at 7 kN/m². Each column collects 294 kN a floor and 3,528 kN at the base; with forty such columns the building weighs about 141,000 kN. At an out-of-plumb of 1/200 the equivalent horizontal force is 706 kN, applied as 59 kN at each floor.

That is a real load applied to a real bracing system, and it is worth comparing with the wind on the same building. A 48 m by 42 m elevation at a net 1.17 kN/m² gives about 2,400 kN — three times as much. Every instinct then says the wind governs and the lean is a detail.

The combination the lean governs, which contains nothing else

The instinct is wrong for a reason that has nothing to do with the size of the two numbers.

Wind appears in some load combinations. The notional force appears in all of them, because it is proportional to the gravity load and the gravity load is always there. In particular it appears in the combination where the imposed load is at its maximum and the wind is absent — the one every project treats as the vertical case, and the one in which a bracing system is not usually thought about at all.

In that combination the horizontal load is 706 kN and nothing else. The braced bay carries it, its diagonals carry it, its connections carry it, and the foundations under it carry an uplift they were not asked about. None of that appears in a model to which no horizontal load was applied.

The stiffness the ductility is bought withLateral stiffness against link length, as a fraction of the same bay braced concentrically. At a link of 800 mm — 13% of the bay — the frame keeps 73% of the concentric stiffness; at the far end of the range, where the diagonals meet the columns, it is a moment frame at 14%. The curve is steep at the left, which is the useful part of it: the first tenth of the bay costs a fifth of the stiffness and buys the whole of the yielding mechanism.0.00.20.40.60.81.000.20.40.60.81link length ÷ baystiffness ÷ the concentric brace's73% at e/L = 0.14a moment framea concentric bracethe link carries 55% of the storey shear at every length drawn
Fig. 3 What is being asked to carry it. A braced bay is stiff and its stiffness is a strong function of how it is arranged, which is the quantity the whole argument eventually lands on.

This is the same structural mistake as designing for the wind and forgetting the storey is free to sway, and it has the same tell: a set of results in which everything is checked and one load case was never run.

Amplification, which applies to it like anything else

A notional force is a horizontal load, so a second-order analysis amplifies it exactly as it amplifies the wind — and the amplification is not small in the frames where the notional force matters, because those are the flexible ones.

A column that was never straightLoad against lateral deflection at mid-height, for a column starting with an initial bow of 0.004. There is no critical value to reach: the deflection grows from the first increment, slowly at first and then without bound as the ratio approaches 1.00 — which is the Euler load, and which the column therefore never attains. The perfect column, drawn for comparison, sits on the vertical axis until it arrives there and then has no answer at all.00.010.020.030.040.0500.20.40.60.81lateral deflection at mid-heightload ÷ P꜀ᵣP ÷ P꜀ᵣ = 1.00, approached and never reachedinitial bow: δ₀ = 0.004
Fig. 4 The imperfect column has no critical load to reach. Its deflection grows from the first increment and runs away as the load approaches a value it therefore never attains.

The clean way to see why is the column that was never straight. A perfect column sits on the load axis with no deflection until it bifurcates; a bowed one has a deflection from the first increment, amplified by 1/(1−P/Pcr)1/(1 - P/P_{cr}), and never reaches the Euler load at all. A leaning frame is that column one scale up: the lean is the initial imperfection, the sway is the deflection, and the elastic critical load ratio αcr\alpha_{cr} plays the part of P/PcrP/P_{cr}.

At αcr=5\alpha_{cr} = 5 — a perfectly ordinary value for a braced multi-storey frame — the amplifier is 1.25, so the 706 kN is really 883. At αcr=3\alpha_{cr} = 3, which is where most standards insist on a full second-order analysis, it is 1.5.

A portal frame swaying under 20A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 40.0. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.20H 10.0H 10.0the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 5 A frame under a horizontal load, solved by the stiffness method because statics cannot divide the load between two columns. Whether the load came from wind or from a lean makes no difference to anything after this point.

Note what has and has not happened here. The amplification is a second-order effect and the notional force is a first-order substitution for a geometry. They are different things and they compound: the imperfection creates a sway, and the vertical load acting on that sway makes it larger.

The same substitution, one scale down

A frame is out of plumb; a member is out of straight. The two imperfections are the same idea at different scales, and they get combined rather than added, because a whole frame is unlikely to lean and have every member bowed the same way at once.

One section depth of bow costs six times the axial stressBending stress divided by axial stress in a two-force member, against how far the member wanders from the line between its pins — measured in its own section depths. The moment is the force times the offset and nothing else, so the ratio is y over Z/A, which for a rectangle is 6y/d exactly. A member bowed by one depth carries 6.00 times as much bending stress as axial, and one bowed by three carries 18. The line has no material in it, no length, and no load: it is a statement about a shape.0.01.02.03.0051015offset from the chord, in section depthsbending stress ÷ axial stress6.0 at one depth6y ÷ dfor a rectangle
Fig. 6 What a bow costs a member. A compression member that wanders from the line between its pins carries bending stress in proportion to the offset, and the ratio is a statement about the section’s shape and nothing else.

The member version is starker than the frame version because the ratio is so large. For a rectangular section, bending stress over axial stress is exactly 6y/d6y/d, so a bow of one section depth costs six times the axial stress. Compression members are slender and their bows are measured in fractions of a depth, but the multiplier is what makes an initial bow of L/1000 into a design case rather than a curiosity — and it is why the shape of the buckling curve is a statement about tolerances rather than about steel.

The two imperfections also do different jobs in a check. A sway imperfection loads the bracing system; a bow imperfection loads the member. A frame can be perfectly braced and contain a column that fails on its own bow.

Where the force is applied, and in which direction

Two decisions remain, and neither of them is a calculation.

Direction. The lean has no natural sign, so the force is applied in whichever direction is worst, in each orthogonal plan direction independently, and with the sign that adds to the wind rather than relieving it. That is a search over cases and not a load case, which is the second time this argument produces one.

Where it acts. The force is φ\varphi times the vertical load carried by that storey, applied at that storey’s floor and taken out at the one below. It is self-equilibrating over the whole building only in the sense that the ground eventually resists it — the base shear it produces is φ\varphi times the total weight, and it is a real base shear that a foundation resists like any other.

Weight is the only thing holding it downA body 8 m wide and 20 m tall weighing 4000 kN, under a wind pressure of 1 kN/m². The wind delivers 60 kN and an overturning moment of 600 kNm about the leeward toe; the weight restores 16000 kNm, a factor of 26.67. The resultant lands 0.15 m from the centre against a middle third of ±1.33 m, so the base is still wholly in bearing.60 kNW = 4000 kNmiddle third: ±1.33 mresultant at 0.15 mrestoring 16000 kNmoverturning 600 kNmfactor 26.67
Fig. 7 Where it ends up. A horizontal force at height is an overturning moment against a weight and a width, and the notional force is the one horizontal load that grows with the same weight that resists it.

That last observation is a small piece of good news. Because the notional force is proportional to the weight and overturning is resisted by the weight, the stability ratio against a notional force alone is 1/(2φ⋅H/B)1/(2\varphi \cdot H/B) — a number with no load in it, and one that a squat building passes without noticing.

Which frames it actually decides

The notional force governs a design when three things line up, and the combination is common enough to be worth naming.

A heavy structure with a small elevation. Racking, water tanks, plant platforms, the internal frames of a car park: high gravity load per square metre of façade, so φN\varphi N is large and the wind is small.

A sheltered or internal frame. A braced core inside a building whose façade delivers its wind to a different system carries almost no wind and all of the gravity, so its horizontal demand is entirely notional.

A very flexible frame. Low αcr\alpha_{cr} amplifies the notional force more than it amplifies anything else, because the notional force is itself proportional to the vertical load that is doing the amplifying.

The alignment chart, computed rather than looked upThe effective length factor against G = (EI/L) of the column ÷ (EI/L) of the beams, for a storey held against sway and for one free to sway. Every point on both curves is the lowest eigenvalue of the assembled frame, swept over 30 beam stiffnesses — not a nomogram, and nothing here is read off a chart. The non-sway curve runs from k = 0.505 at G = 0.01, where the beams are stiff enough to be built-in, to k = 0.981 at G = 20, where they are soft enough to be pins: the whole of it lies between a half and one. The sway curve starts at k = 1.003 and has no upper bound at all, reaching 4.16 at the same G — so the braced frame carries 18.0 times the load of the unbraced one at its worst point on this sweep.0.010.111001234G = (EI ÷ L) of the column ÷ (EI ÷ L) of the beamseffective length factor kfree to swayheld against swayk = 1.0 — a pin-ended columnk = 0.5 — beams rigid, sway preventedk = 0.505k = 0.981
Fig. 8 Why flexibility matters so much here. A storey free to sway has no upper bound on its effective length factor, and the braced and unbraced cases differ by a factor of eighteen in capacity at the soft end of this sweep.

The third of those is worth dwelling on because it is a feedback rather than a coincidence. A frame’s αcr\alpha_{cr} falls as its vertical load rises. The notional force rises with the same vertical load. So the two effects multiply, and the horizontal demand on a heavily loaded flexible frame grows faster than its load does.

What it looks like when it has been missed

The failure this argument protects against is quiet, and it has a recognisable shape in a set of results.

A bracing system sized for wind is usually generous, so the diagonals themselves are rarely the problem. What is not generous is everything the wind combination happened to make easy. A holding-down bolt group is sized on the worst combination that was run; if the gravity-only case was run with no horizontal load, the tension in the windward foundation of the braced bay is a number nobody computed. A transfer beam carrying a braced bay across an opening is checked for the horizontal shear it was given. A connection between the bracing and the floor plate is sized on the load path that was modelled.

None of those is wrong in the model. Each of them is wrong about the building, and the error is invisible because every check that was made passes. That is the characteristic signature of a missing load case rather than a mistaken one — nothing to find by reading the output, because the output is internally consistent and complete.

There is a second, subtler version on a structure with no bracing system at all. A frame designed as a set of gravity columns and simply supported beams, with the stability provided by a shear wall elsewhere, is drawn with pinned connections and it has none. The columns lean, the leaning delivers a horizontal force into the floor plate, and the floor plate delivers it to the wall — a diaphragm action that only exists if somebody drew it. Where the floor is precast planks with no topping, or a steel deck with a movement joint through it, that path is absent and the notional force has nowhere to go except into the connections that were called pins.

The check is one line long and worth making on any project: apply a horizontal force of a two-hundredth of the total weight in each direction, with no wind, and see what changes.

Where the model stops

The out-of-plumb is a specification, not a measurement. Nothing in this calculation knows how far the building actually leans. It uses the value the erection tolerance permits, which is a bound on a quantity nobody surveyed, and a building erected badly is outside the calculation entirely rather than merely closer to its limit.

Leans do not add up the way the arithmetic assumes. Applying φ\varphi to every storey in the same direction treats the whole building as one straight leaning stick. Real out-of-plumb alternates: a storey out one way, the next out the other, because each is corrected against the one below. Standards handle this with a reduction factor that falls with the number of storeys and the number of columns, which is a statistical statement dressed as a geometric one.

A settlement is not an imperfection. A frame tilted by differential foundation movement has a lean that arrived after erection, grows with time, and is not covered by an erection tolerance. It produces exactly the same φN\varphi N and is checked by nobody, because the settlement that matters is a difference and the tilt of a whole building is not usually measured at all.

The two mechanisms put their worst storey at opposite endsInter-storey drift, storey by storey, split into the two motions that make it. The racking component is largest at the bottom, where the storey shear is largest, and dies away at the top. The bending component is largest at the top, where the accumulated rotation is greatest, and is nothing at the base. Their sum has its worst storey at number 1 of 16 — neither where the shear puts it (storey 1) nor where the bending does (storey 16). This is why the roof drift is a poor guide: it is 1 in 1069 of the height here, and the worst storey is 1 in 722 of its own.0.00.20.40.60.81.01.21.40246810121416inter-storey drift (‰ of the storey height)storeyrackingbendingworst at storey 1αH = 1.16 · bending is 25% of the roof drift
Fig. 9 Where a sway ends up mattering. Inter-storey drift is what limits a frame in service, and it is not distributed the way the roof deflection suggests, so the storey that governs is not the one at either end.

What the picture cannot show

The amplification curve at the top of this essay is drawn for a structure with one deflected shape and one critical load. A real building has many, and the imperfection has to be applied in the shape of the mode it is amplifying for the substitution to be exact. Applying a uniform lean to a building whose critical mode is a soft storey near the base under-represents the imperfection where it matters and over-represents it everywhere else.

Nor does the picture show the thing the whole substitution is for. There is no drawing anywhere in a project of the building leaning, because the drawings are of the building as designed and the lean is a property of the building as built. The notional force exists precisely so that a quantity nobody will ever draw can be carried into an analysis that only accepts loads.

A brace is a stiffness requirement, not a strength oneCritical load against brace stiffness for a pinned column braced at 40% of its height. The curve climbs from the unbraced Euler load of 9.87EI/L² and flattens at 27.42EI/L², which is the Euler load of the braced segment — past that the column buckles in a shape the brace does not obstruct, and further stiffness buys nothing. The knee is at about 107EI/L³.050100150200250300350400010203040brace stiffness (units of EI/L³)critical load (units of EI/L²)ideal stiffness ≈ 107 EI/L³27.4 — braced9.87 — unbraced
Fig. 10 The same argument for a brace. What restrains an imperfection is a stiffness rather than a strength, and it reaches a knee past which more of it buys nothing.

The generalisation

The habit worth carrying is about the difference between a load and a load case.

Everything else in this field is a force somebody can point at. A notional horizontal force is a model of a geometry, converted into a force because the analysis will only take forces — and once it is in the model it is indistinguishable from a real one, which is exactly the property that makes it useful and exactly the property that makes it easy to forget.

The move generalises. A brace that need not be strong is doing the same job: what it restrains is not a load but a tendency to move, and the force in it is whatever the tendency produces. Prestress is the same conversion in reverse — a state of the structure, entered into the analysis as a set of applied forces. In every case the honest description is that a fact about the structure has been rewritten as a force, and that the arithmetic is exact while the fact is an assumption.

Which leaves one thing to carry into any check. When a horizontal load appears in a model, ask what it is a substitution for. If the answer is “the wind”, it belongs in the wind combination. If the answer is “the building is not straight”, it belongs in every one of them — and most importantly in the one that has no other horizontal load in it at all.

Where this goes next

00.20.40.60.80246810applied load ÷ buckling load1.1×1.3×1.4×2.0×3.3×10.0×first-order analysis says the answer is always 1×one over one minus the ratio The load that makes itself worse A structure that has leaned carries its weight off the axis, which makes it lean further. The amplification is one over one minus the load ratio, and it runs away long before the buckling load. 00.0050.010.0150.020.02500.20.40.60.81lateral deflection at mid-heightload ÷ P꜀ᵣP ÷ P꜀ᵣ = 1.00, approached and never reachedinitial bow: δ₀ = 0.002 The column that was never straight Euler's load is the load at which a perfectly straight column becomes indifferent to being bent. No column is perfectly straight, so no column ever reaches it — and the load it never reaches can still be measured. the storey that cannot driftk = 0.774N꜀ᵣ = 16.46 EI/L²the storey that cank = 1.317N꜀ᵣ = 5.69 EI/L²the same column, 2.9 times the load — the restraint is the whole of the difference Held, and not held One horizontal restraint at the head of a storey, carrying no vertical load whatever, moves the critical load of the columns beneath it by a factor of 2.89. Effective length is a property of the frame, not of the member. wall 146 mmframe 140 mmtogether 58 mm20 storeys at 3.5 m · 40 kN per floor How a tall building stands still A shear wall bends and a framed tube shears, and the two deflected shapes are the wrong way up for each other. Tie them together at every floor and the pair is stiffer than the sum of their stiffnesses — because near the base the wall holds the frame back and near the top the frame holds the wall.

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.

BracingCritical loadEccentricityEquivalent horizontal forceErection toleranceFree bodyImperfectionLateral systemLoad combinationLoad pathNotional loadOut of plumbP-deltaSecond orderSway stability