Structural Engineering & Statics
where the load goes
Deflection

The curvature nobody applied

Concrete shrinks as it dries, by about half a millimetre in every metre. In a symmetrically reinforced member that is a shortening and nothing else. In a member with more steel in one face than the other — which is every beam and every slab — the steel holds one side back and the section bends, with no load on it at all.

Assumes The deflection that arrives three years late and The movement nobody applied.

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

A deflection calculation is a division: a moment by a stiffness, integrated twice. Everything in it comes from the load.

A concrete member’s deflection contains several contributions that came from nowhere near the load, and one of them is a curvature the member acquired by drying out. It has no moment behind it, it satisfies equilibrium with nothing on the section, and it adds to the deflection exactly as though somebody had applied a bending moment along the whole span.

The least reliable number in the material decides the answer, brieflyWhat a twenty per cent error in the concrete's tensile strength does to a computed deflection, against how far past cracking the beam is. Well past the cracking moment it does almost nothing — at 1.9 times M_cr the spread is 56 per cent — because the section is nearly fully cracked and the interpolation has run out. Just above cracking it does everything: at 1.19 times M_cr the same twenty per cent moves the deflection by 3658 per cent. Tensile strength is the property with the widest scatter and the least direct test, and a beam designed to sit near its cracking moment has put the answer on it.0.40.60.811.21.41.61.8102030moment ÷ cracking momentdeflection spread for ±20% on f_ctm3658% at 1.19 M_crcracking
Fig. 1 What a twenty per cent error in the concrete’s tensile strength does to a computed deflection. Well past cracking it does almost nothing; just above cracking it does everything.

Which free body produced the number

Take a slice of a reinforced section and let the concrete shrink freely by a strain εcs\varepsilon_{cs}. The steel does not shrink. So the two cannot both do what they want, and the section settles at a compromise fixed by compatibility.

Write the compromise as a uniform strain plus a curvature — ε0+κy\varepsilon_0 + \kappa y, which is all a plane section can take. Two conditions close it: the axial forces sum to zero, because nothing is applied, and the moments sum to zero for the same reason.

If the steel is symmetric about the centroid, the second condition is satisfied at κ=0\kappa = 0 and the answer is a pure shortening. If it is not, the steel’s restraint has a lever arm about the centroid and the curvature is

κcs=εcs αe SI\kappa_{cs} = \varepsilon_{cs}\,\alpha_e\,\frac{S}{I}

where SS is the first moment of the reinforcement about the section’s own centroid and αe\alpha_e the modular ratio. A first moment of the steel — an asymmetry — is the entire driver, and where the steel is symmetric it is zero.

Two things follow immediately, and both are counter-intuitive.

Why the lightly reinforced member is worse

The first is that more steel does not mean more curvature.

Adding steel to the tension face increases SS and increases II, and past a point the second wins. Adding steel to the compression face reduces SS directly. So the worst case is a member with a modest amount of steel, all of it in one face — which is a lightly reinforced slab, the member nobody expects to have a problem.

The second is that cracking makes it worse rather than better. A cracked section has lost its tension zone, so its centroid has risen toward the compression face and the steel’s lever arm about that centroid has grown, while II has fallen by 40% or more. Both changes increase κcs\kappa_{cs}.

The beam is stiffer than its cracked section and softer than its gross oneMoment against mid-span deflection for a 350 × 550 mm beam spanning 8.0 m, with the two bounds it lies between. The uncracked line is what the gross transformed section gives; the cracked line is what the section at a crack gives; and the curve between them is the member, because between the cracks the concrete is still carrying tension and the average curvature is not either section's. At the service load the deflection is 24.8 mm — span over 323 — against 8.5 uncracked and 27.0 fully cracked, a factor of 3.18 between the bounds. The interpolation ζ = 1 − β(M_cr/M)² sits it 88 per cent of the way across, and β falls from one to a half under sustained or repeated load because the bond that does the dragging deteriorates.010203040050100150200250300mid-span deflection (mm)moment (kNm)uncrackedthe memberfully crackedM_cr = 58 kNm24.8 mm at service, span over 323
Fig. 2 A member between its two bounds. The uncracked line is what the gross section gives, the cracked line what the section at a crack gives, and the member sits between them because the concrete between the cracks is still carrying tension.

Which of those two bounds a member sits near is decided by how far past its cracking moment it is, and a beam is stiffer than its cracked section says by an amount that depends on the concrete’s tensile strength — the property with the widest scatter and the least direct test.

That sensitivity is the figure at the top of this essay, and it is worth reading as a warning about the whole subject: a member designed to sit just above its cracking moment has put its computed deflection on a number nobody can measure to better than twenty per cent, and a twenty per cent change there moves the answer by a factor of thirty-seven.

Creep, which multiplies a different thing

The other time-dependent effect is creep, and the two are frequently conflated because both take years and both make deflections worse.

The deflection that arrives years lateThe multiplier on a concrete member's deflection under a sustained load, against time. The elastic deflection arrives on the day the load does and is the 1.0 at the left. After a year it has been multiplied by 3.76, after five years by 4.15, and it approaches 4.27. Nothing has been added to the load and nothing about the strength has changed: this is a serviceability failure arriving on a structure that passed every strength check on the day it was built.1 d10 d100 d2.7 yr27 yr01234time under loaddeflection ÷ the deflection on day one1 year: ×3.765 years: ×4.15the deflection the calculation gives
Fig. 3 The multiplier on a member’s deflection under a sustained load. After a year it has been multiplied by 3.76 and it approaches 4.27, with nothing added to the load.

Creep is a multiplier on what the load did. The deflection that arrives three years late is the essay about it, and its arithmetic is a reduced effective modulus applied to the same moment.

Shrinkage is an addition that the load had no part in. A member carrying nothing at all still acquires its shrinkage curvature.

The two interact, because the shrinkage curvature itself develops slowly and is relieved by creep as it develops, so the usual treatment applies an age-adjusted modulus to both. But the distinction matters for where each is largest: creep is worst on a heavily and permanently loaded member, and shrinkage curvature is worst on a lightly loaded one. A lightly loaded slab can have more of its long-term deflection from shrinkage than from creep.

What it is worth on a real slab

The arithmetic is short enough to do once.

Take a 200 mm slab spanning 6 m, reinforced with 1,100 mm²/m in the bottom face only, at an effective depth of 165 mm. Cracked, the neutral axis sits about 45 mm from the top, so the steel’s lever arm about the cracked section’s centroid is around 120 mm. With a free shrinkage strain of 400 microstrain, a modular ratio of 15 and a cracked second moment of about 210×10⁶ mm⁴ per metre, the curvature comes to roughly 0.35 per kilometre.

A constant curvature over a simply supported span deflects it by κL2/8\kappa L^2/8, which for 6 m is 1.6 mm. On its own that is nothing — span over 3,800.

Now put the same slab at a span of 9 m and thin it to 175 mm, which is what a designer working to a span-to-depth table would do. The curvature roughly doubles, because the section is shallower and the steel more one-sided; the L2L^2 term more than doubles again; and the shrinkage deflection reaches something like 8 mm, against a total limit of 9 m over 250, which is 36. It has taken a fifth of the allowance before any load has been applied.

That is the shape of the whole effect. It is negligible on short, thick, symmetrically reinforced members and a substantial fraction of the allowance on long, thin, singly reinforced ones — which are exactly the members a span-to-depth rule produces, because the rule was calibrated on the total and not on its parts.

The cheapest fix is compression steel. Putting a modest area in the top face reduces SS directly, and unlike almost every other intervention in this collection it costs nothing structurally: the top bars are doing no work in a sagging region and are there purely to make the section shrink straight.

The general shape of the effect

Shrinkage curvature is one member of a family, and the family is easier to recognise than any of its members.

The strain it wants, the strain it is allowed, and the differenceA bridge deck 1.40 m deep with 14 °C at the top face falling away over 10% of the depth. The left curve is the free thermal strain αT(y); the straight line beside it is what a plane section will actually take, ε₀ + κy with ε₀ = 24.9 microstrain and κ = 0.049 per km. The right-hand block is E times the difference, and it reaches -3.10 N/mm² of compression at the surface and 1.43 of tension 140 mm below it. Its resultant force is 5.6e-13 kN and its resultant moment 2.1e-12 kNm, which is what self-equilibrating means: the field is invisible to every equilibrium check that could be made on the member.the sectionstrainfreeplanestress-3.10+1.43resultant force 5.6e-13 kN · resultant moment 2.1e-12 kNmno restraint anywhere, and the member is stressed from end to end
Fig. 4 The strain a section wants, the strain a plane section is allowed to take, and the difference between them. The difference is a stress field whose resultant force is 5.6×10⁻¹³ kN and whose resultant moment is of the same order.

Whenever a member is given a strain distribution it cannot take — because plane sections have to stay plane — the difference between what it wants and what it is allowed becomes a self-equilibrating stress field, invisible to every equilibrium check that could be made on it. A thermal gradient does it, differential shrinkage does it, a welding pass does it, and a section made of two materials with different time-dependent behaviour does it continuously.

Warmer on top: the same two limits, one field alongA 30 m deck 12 °C warmer on top than underneath, over a depth of 600 mm. Free to move, it takes a curvature of 0.240 per km and lifts 27.0 mm at midspan, carrying no stress at all. Held down, it carries 18 kNm and ±15.1 MPa at the extreme fibres and does not move. Every real deck is somewhere between, and where it sits is decided by the bearings rather than by the deck.free: 27.0 mm of camber, no stressheld: 18 kNm and ±15.1 MPa, no movementκ = α·ΔT/d = 0.240 × 10⁻⁶ per mm either way
Fig. 5 A deck warmer on top than underneath. Free to move it takes a curvature and lifts 27 mm at midspan with no stress at all; held down it carries 18 kNm and ±15.1 MPa and does not move.

The pair in that figure is the whole logic. A curvature and a moment are alternatives: a member free to bend takes the curvature and carries no stress, a member restrained carries the stress and does not move, and every real member is somewhere between with the position decided by its supports rather than by itself. The movement nobody applied is the essay about the restrained end of that spectrum; this one is about the free end, where the price is a deflection instead of a stress.

Where it accumulates

In a single member the effect is a deflection. In a building it is a difference between one part and another.

Two differences up the same building, peaking in different placesDifferential shortening between a perimeter column and the core of a 52-storey building, plotted up the height. The part driven by load peaks at level 26 — exactly half way up, because a floor near the top has almost nothing built above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it to shorten. The part driven by shrinkage does not care what is above it at all and accumulates all the way to the roof. Their sum is worst at level 52, at 56 mm, which across a 9 m bay is a floor out of level by one in 160.-60-40-20020020406080100120140160180column shorter than core (mm)height (m)from loadfrom shrinkagethe sumworst 56 mmat level 52one in 160
Fig. 6 Differential shortening up a tall building, split into the part driven by load and the part driven by shrinkage. The load-driven part peaks halfway up; the shrinkage-driven part accumulates all the way to the roof.

The two parts peak in different places for a reason worth keeping. The load-driven difference is largest halfway up, because a floor near the top has almost nothing above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it. The shrinkage-driven difference does not care what is above it at all — every metre of column shrinks whether or not it is loaded — so it accumulates monotonically and their sum is worst at the roof, at 56 mm across a 9 m bay, a floor out of level by one in 160.

That is the columns being shorter than the core with the shrinkage half separated out, and the separation matters because the two are corrected by different means: the load part by a construction-stage adjustment, the shrinkage part by making the two elements shrink alike.

The deflection as a sum

Putting it together, the deflection of a concrete member at the end of its life is a sum with at least five terms.

The elastic deflection under the load that is present. The creep multiplier on the permanent part of it. The shrinkage curvature, from no load at all. The cracking, which reduced the stiffness the first two terms divide by. And the shear deflection, which for most beams is small.

The shear term is a few per cent of a beam and most of a wallThe share of a beam's deflection carried by shear, against its span-to-depth ratio, for a 300 × 600 rectangle under a central point load. The ratio δs/δb is Q/λ² with Q = 3.111 for this section, so the share falls as the square of the slenderness: 2% at L/d = 12.35, 15% at L/d = 4.20, 60% at L/d = 1.44. At a span-to-depth ratio of two it is 43.75% — seven sixteenths exactly. The same member in concrete is solved beside it and the two curves cannot be told apart: E/G is 2.59 against 2.35, a 10.6% spread in the ratio, against a factor of four for every doubling of the depth. The material barely matters and the geometry decides.0510152025020406080100span-to-depth ratio L/dshear, as a percentage of the whole deflection2% at L/d = 12.3515% at L/d = 4.2060% at L/d = 1.447⁄16 exactly, at L/d = 2steel and concrete, solved separately: E/G 2.59 and 2.35,10.6% apart, against four times for every doubling of the depth
Fig. 7 The share of a deflection that is shear rather than bending, against the span-to-depth ratio. It is 2% for an ordinary beam and 60% for a member as deep as it is long — and the material barely matters.

Only the first of those five is what a deflection calculation classically computes, and stiffness is not strength is the essay about why the sum matters at all, and the others are between them frequently larger. That is why concrete deflection prediction is unreliable in a way steel deflection prediction is not: the steel beam’s answer is a division and the concrete beam’s is a sum of five terms, four of which depend on time, humidity, cracking history and construction sequence.

Cambered against the wet loadA 9 m composite beam whose flexural rigidity rises from 94 to 260 kN·m² when the slab sets, so the first two loads are carried by the bare steel and the rest by the composite section. Fabricated with 8.0 mm of camber, it moves through -6.5, 0.0, 2.5, 4.2 mm as the four stages arrive — 0.0 mm on the day the slab is poured, and 4.2 mm at the end, which is one part in 2160 of the span. The largest curvature it ever has is 8.0 mm of hog, and it has that with nothing on it. Every shape is drawn at the same exaggeration and the drawing is a diagram of a proportion: the vertical scale is 283221 times the horizontal.levelas fabricated: 8.0 mm of camber-6.50.02.54.2self-weight of the steel: 1.5 mm on EI = 94wet concrete: 6.5 mm on EI = 94finishes and services: 2.5 mm on EI = 260imposed load: 1.6 mm on EI = 260
Fig. 8 The same problem answered by building the member the wrong shape. Cambered at fabrication, it moves through four stages and finishes 4.2 mm down — one part in 2160 of the span.

Where the model stops

The free shrinkage strain is an estimate. It depends on the humidity, the member’s surface-to-volume ratio, the cement content and the aggregate, and it is quoted as a range of two to one. Everything computed from it inherits that range.

Shrinkage is not uniform across the section. The surfaces dry first and the core last, so a thick member has its own internal gradient — a self-equilibrating field within the section, on top of the curvature the reinforcement causes. The arithmetic here treats the shrinkage as a single number applied everywhere.

Cracking and shrinkage are not independent. The shrinkage curvature above was computed on a cracked section, and part of what cracked it may have been the restrained shrinkage itself — so a slab restrained at its ends cracks earlier than its applied moment predicts and then has a larger curvature for the same load. The sequence closes on itself and there is no closed form for it.

Restraint outside the member has been ignored. A slab cast between stiff walls cannot shorten, so it cracks instead, and a cracked slab has a different stiffness and a different shrinkage curvature. That coupling is not in any of the expressions above.

A beam of two stiffnesses has one diagram with a step in itA 9 m cantilever whose flexural rigidity is 0.4 times larger beyond 4.5 m — a deep root and a shallow tip — under a tip load of 10. The M/EI diagram therefore has a step in it at a station where nothing about the moment changes, and the method adds two areas where an integration would need two cases and two more constants. The shaded area is 556.72, which by the first theorem is the change of slope along the whole member. Its centroid is at 3.818 m, and the first moment about the tip is 2884.9 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 2430.0.10EI ÷ 0.4 beyond herecentroid at 3.82 mM/EIarea = 556.72 · first moment = 2884.9by double integration: 2430.0
Fig. 9 The method that handles a curvature diagram with a step in it. Shrinkage curvature is not constant along a member either, because the reinforcement is curtailed and the cracking is not uniform.

Why it is nearly always found late

There is a characteristic history to a shrinkage-curvature problem, and it is worth recognising because every stage of it looks like something else.

The slab is designed to a span-to-depth ratio and passes. It is built and it is flat. Over the first winter a line of cracking appears over the supports, which is attributed to the construction sequence. Over two or three years the mid-span sags visibly — visibly, because a person can see about one part in 250 of a span when there is a straight line such as a partition head or a window reveal to compare it against.

By then the investigation has three candidates and no way to separate them. The concrete may have been weaker than specified; the slab may have been loaded early; the props may have been struck before it had strength. All three are plausible, all three are unprovable years later, and none of them is what happened.

What settles it, when anything does, is the shape. A deflection caused by load is proportional to the load’s own moment diagram and is largest where the load is. A shrinkage deflection is proportional to a curvature that is nearly constant along the span, so it produces a circular arc — and a bay that has sagged uniformly, including the parts carrying nothing, has sagged for a reason the loading cannot explain.

The second tell is the neighbours. Shrinkage curvature depends on the reinforcement layout, so two bays with the same load and different bar arrangements deflect differently. A load-driven problem does not do that.

What the picture cannot show

The curvature has no moment diagram. There is nothing to draw for it in the place where a structural drawing would put the cause of a deflection, and no analysis output contains a member force corresponding to it.

That absence is the practical difficulty. A deflection that exceeds a limit is investigated by looking at the loads and the stiffnesses, and the term responsible may be neither — it may be a curvature that arrived because the top steel was curtailed and the bottom steel was not, in a member carrying nothing unusual at all.

Nor does any figure show that the effect is a rate. Half of the shrinkage happens in the first few months and the rest over years, so a slab that passed its deflection check at handover fails it three years later with nothing having changed on it.

Deflection goes as the fourth power of the spanDeflection against span for a constant load intensity and section, with two slower relationships drawn faintly behind it for comparison: the load itself, which grows in proportion to the span, and the bending moment, which grows as its square. Doubling the span multiplies the deflection by sixteen, while the moment only quadruples.11.522.533.54050100150200250300span, relative to the first1×5×16×81×moment: the squareload: the first powerdeflection: the fourth
Fig. 10 Why any of it matters more at long spans. A curvature integrates twice into a deflection, and the integration carries the span squared — so a curvature that is negligible at 5 m is not at 12.

The generalisation

The habit worth carrying is to treat a strain the member cannot take as a load case in its own right.

Every one of the effects in this essay begins the same way: some part of a section wants a strain that a plane section cannot accommodate, and the difference has to go somewhere. If the member is free, it goes into a curvature and produces a movement with no stress. If it is held, it goes into a self-equilibrating stress field with no movement. Real members are partly held and get some of each.

That is the same statement as a residual stress, as a thermal gradient, as differential creep between two materials, and as the locked-in stresses of a section built in stages. Five subjects, one mechanism, and the useful diagnostic is a question rather than a calculation: what strain does this member want, and what is it allowed?

Where this goes next

1 d10 d100 d2.7 yr27 yr0123time under loaddeflection ÷ the deflection on day one1 year: ×3.005 years: ×3.29the deflection the calculation gives The deflection that arrives three years late A concrete beam that passes every check on the day it is built goes on deflecting for a decade, and ends up three times where it started. Nothing about the load changed, and nothing about the strength was ever in question. free at one end10.8 mmno stressheld at both ends75.6 MPaheld by a spring of 100 kN/mm53.2 MPaE·α·ΔT = 210000 × 12×10⁻⁶ × 30 = 75.6 MPa, at every length The movement nobody applied A temperature change is the only load in this collection that a structure can decline. Let it move and it produces a movement with no stress; hold it and it produces a stress with no movement — and that stress contains no length, no area and no second moment, so a bracket and a bridge girder carry exactly the same one. 01020304050050100150200250mid-span deflection (mm)moment (kNm)uncrackedthe memberfully crackedM_cr = 47 kNm34.2 mm at service, span over 234 Stiffer than its cracked section says At a crack the concrete below the neutral axis has gone and the steel carries the tension alone. Between the cracks it has not gone — bond drags it back into tension, the steel strain drops, and the curvature averaged over a length of beam is neither section's. -40-30-20-10010020406080100120140column shorter than core (mm)height (m)from loadfrom shrinkagethe sumworst 43 mmat level 40one in 208 The columns are shorter than the core Every column in a tall building gets shorter as the building is built on top of it, and the core beside it gets shorter by a different amount. The floors between them tilt by the difference — and the difference is largest exactly half way up, because a floor near the top has almost nothing built above it and a floor near the bottom has almost nothing beneath it.

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.

CamberCompatibilityCracked sectionCreepCurvatureDeflectionDifferential shorteningModular ratioReinforcementRestraintSelf equilibratingServiceabilityShrinkageTension stiffeningThermal gradient