Structural Engineering & Statics
where the load goes
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Internal forces

What a cut reveals — shear, moment and axial force, and the diagrams that track them along a member.
The same beam, cut at x = 5. A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.

What a cut reveals, and why it was there all along

Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing.

 Load, shear and moment — a simple span. The applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.

The diagram is an integral, and that is why it can be drawn by eye

Load, shear and moment are one function and its two integrals. Once that is seen, the diagrams stop being things to calculate and become things to sketch.

 Where to put the supports. Peak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.

Where to put the supports, which is not at the ends

Moving the supports of a uniformly loaded beam inward by about a fifth of its length halves the worst bending moment. The load has not changed and nor has the beam.

 The collapse mechanism of a propped cantilever. A collapse mechanism, with the hinge position found by searching rather than quoted. Every position gives an upper bound on the collapse load; the lowest is 7.29, at a hinge 58.6 per cent along, which is a coefficient of 11.657 times Mp over the square of the span.

After the first yield, which is not the end

A steel beam whose extreme fibre has reached yield has not failed. It has started forming a hinge, and collapse waits until there are enough hinges to make a mechanism.

 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 30.6 to 19.6, and a hogging moment of 24.5 appears over the supports where there was none.

The moment over the support, and what it buys

Run a beam over its supports instead of stopping at each one, and the mid-span moment falls by a third while a new moment appears where there was none. Nothing was added but continuity.

 Influence line for the bending moment at x = 3. The bending moment at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 3.00, giving 2.100.

The worst place to stand

A bridge is not designed for a load. It is designed for a load that moves, and for every station along it there is a different position of that load that does the most damage.

 One slit, and the torsional stiffness falls by a factor of hundreds. A 200 by 200 box of 8 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 5.66×10⁷ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 1.31×10⁵ mm⁴. The ratio is 432 to one, so the same torque twists the slit section 432 times as far and raises a peak shear stress 36 times as high. Nothing about the material changed.

The moment that will not lie flat

A plane cut exposes three actions. A real cut exposes six, and the fourth of them behaves unlike the others — torsion is resisted by a loop of shear, and one slit down the length of a tube destroys it.

 The worst position is not the obvious one. Three axles totalling 320 units, marched across a span of 20 in steps of 0.02. The envelope is the largest moment each station ever sees; its peak is 1160.2 at 9.88 along the span, which is 0.12 off midspan and occurs under the axle nearest the resultant rather than under the heaviest one. Barré's construction, which places midspan halfway between that axle and the resultant, independently gives 1160.3 at 9.88. The dashed curve is the envelope the same total weight would produce as one load rather than three: its peak is 1600.0, which is 38% more — spreading a load out is worth something.

The train that is worse than its heaviest axle

An influence line says where to stand one load. A vehicle is several loads at fixed spacings, and the worst arrangement never puts the heaviest one at the peak.

 A torque diagram is a shear diagram about a different axis. A torque of 40 kNm applied 2 m along a member of 6 m held against twist at both ends. The two ends take 26.7 and 13.3 kNm, in inverse proportion to their distances, because the two halves are springs in parallel and torsional stiffness is GJ over length. The diagram steps at the applied torque and closes at the far end, exactly as a shear diagram does — the only difference is which axis the arrows turn about.

The internal force with no diagram

A cut through a member reveals four things, and this collection has drawn diagrams for three of them. The fourth is a torque, it obeys exactly the same rules, and whether it exists at all can depend on a decision the designer is free to make.

 Two triangles that cross zero, and a block that does not. Stress across a 300 × 700 mm section at each stage, compression positive. The prestress alone gives -6.33 MPa at the top and 20.61 at the bottom; at transfer, with only self-weight on it, the top is at -2.47 MPa and in service the section runs from 7.61 to 3.82 MPa — compression everywhere. The same beam with no prestress reaches -12.67 MPa at the bottom fibre, which is 4.2 times what the concrete can hold.

The load put on backwards

Every other structure in this collection waits for its load and then resists it. A prestressed one is given a load first — chosen, permanent, and pointing the wrong way — so that when the real one arrives the two nearly cancel and the material never has to do the thing it is bad at.

 The moment does not stop at the end of the beam. A portal frame of 8 m by 4 m with fixed bases, carrying 20 kN/m on the beam. The bending moment is drawn on the tension side of every member, and it runs round the corner without a break: 65.2 kNm arrives at the end of the beam and 65.2 kNm leaves down the column, which is the same number, since joint rotational equilibrium is one of the equations the frame solve satisfied. Midspan carries 94.8 kNm, and the two add to 160.0 — the 160.0 kNm of a simply supported span, to 0.0e+0 kNm. The corner takes 61% of the wL²/12 a fully built-in beam would have carried, because the columns are springs rather than walls: the beam-to-column stiffness ratio is 1.27. The beam's moment crosses zero 0.92 m from the corner and the column's 1.33 m above its base.

The moment that goes round the corner

At a rigid knee the bending moment does not stop at the end of the beam. It turns and runs down the column, and in the same instant the beam's shear becomes the column's axial force — while the block of steel that has to carry the turn appears on no member diagram anywhere.

 Two beams tied together, and the deeper one takes 89% of the load. Two simply supported beams of 6 m, one twice as deep as the other, tied together at midspan so that they have to move as one. A load of 100 kN stands on the tie. Point stiffness is 48EI/L³, so the deeper beam is 8 times as stiff — depth cubed, nothing else — and the load divides in that ratio: 11.1 kN into the shallow beam and 88.9 kN into the deep one, 11% against 89%. Both midspan points move 5.00 mm, which is the whole of the argument: the geometry of the load never entered it. The deflection is drawn 78 times full size — the real sag is 5.00 mm on a 6 m span, about 1 in 1200.

The stiffest path takes the load

When two members share a force the split can be argued about. When they share a displacement it cannot — stiffness settles it, and nothing about the load or the plan drawing gets a vote. The consequence is that stiffening a lightly loaded member raises its stress, and the way to unload something is to soften it.

 The ground pushes back hardest where the beam has gone down furthest. A strip 16.4 m long and 1 m wide on ground of subgrade modulus 50 × 10³ kN/m³, carrying 1000 kN at its centre. The beam settles 3.90 mm under the load and the ground pushes back in proportion — the arrows are k times the settlement above them, peaking at 195 kN per metre — so the pressure diagram is the settlement bowl and not an assumed distribution. The characteristic length 1/β is 2.56 m: the bowl crosses zero at 6.04 m, which is 3π/4 of it, and beyond that the arrows reverse because the beam has lifted off. By 8.05 m — one π/β — the disturbance is 4.3% of what it was, which is why the moment 641 kNm and the peak pressure 195 kN/m contain no length at all. The settlement is drawn 217 times full size — the real bowl is 3.90 mm deep over 16.4 m, about 1 in 4207 — and at true scale the beam would be a straight line.

The beam that sits on the ground

Every other beam in this collection is held at points. A footing is held everywhere, by something that pushes back in proportion to how far it is pushed — and that single change hands the structure a length it did not choose. Two or three of those lengths from the column, nothing knows the load happened.

 A truss drawn inside a solid, and solved as one. A deep member 4000 mm between bearings and 2000 mm deep, carrying 1200 kN at mid-span. The model is two struts and one tie, on a lever arm of 1600 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 750 kN and each strut at 960 kN, at 38.7° to the horizontal. Spread over a strut width of 812 mm the compression is 3.0 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 1724 mm² of steel. A beam calculation on the same member would have asked the tie for 702 kN, which is 7% less than the model does.

When there is no section to design

Beam theory needs a section, and a section needs the strain to be linear across it. Within about a depth of a support, a load, a corner or a hole it is not — and those are the regions structures actually fail in.

 Two beams, or one beam four times as stiff. Two 200 × 150 planks spanning 4 m under 6 per millimetre. Loose, they have 112.5×10⁶ mm⁴ between them and deflect 16.2 mm, with the two faces at the interface sliding past one another. Bonded, the pair has 450.0×10⁶ — exactly 4 times as much, because doubling a depth cubes — and deflects 4.0 mm at half the extreme-fibre stress. Nothing was added but a restraint on slip. With connectors of stiffness 200 the same beam deflects 5.4 mm, which is 89% of the way from one bound to the other.

Two beams, or one beam four times as stiff

Stack two planks and they bend as two beams whose faces slide past one another. Bond the faces and the pair has one neutral axis, four times the second moment and half the stress. Nothing was added but a restraint on slip.

 The same restraint, twice, with opposite signs. A 4 m strip of 200 mm slab whose ends cannot move apart, against deflection measured in its own thicknesses. The flat line is what a yield-line calculation gives, which is what the same strip would carry if its ends were free: 30.0 per unit width. The rising branch is compressive membrane action — the deflected strip is forced into an arch — and it peaks at 116.6, which is 3.89 times the yield-line load, at a deflection of 0.24 of the thickness. Past that the arch runs out of depth and the load falls back to the flexural one; past a deflection of one thickness there is no arch left and the reinforcement starts carrying the strip as a cable. It gets back to the arch's load at 2.17 thicknesses, which is one part in 9 of the span — a sag nobody would design for and exactly what a floor does instead of falling.

The force nobody put in the model

A slab strip whose ends cannot move apart is not the strip in the yield-line calculation. Deflecting shortens the chord between its ends, the ends do not come in, and the strip is forced into an arch — worth four times the load it was designed for, at a movement nobody would see.

 A check made on a perimeter, not on a section. One bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 604 kN across 4427 × 225 mm, a shear stress of 0.606 N/mm² against a resistance of 0.658.

A check made on a perimeter, not on a section

Every shear check in this collection is made on a plane cut through a member. A slab sitting on a column has no such plane, because the shear leaves in every direction at once — so the check is made on a closed line, and a line grows with the column while the load grows with the square of the bay.

 The plan a straight beam does not have. A beam of radius 12 m turning through 60°, seen from above, with bending drawn outward from the axis in one colour and torsion in the other. The load is vertical and uniform and nothing is applied off the axis. Bending reaches 446 and torsion 155; the two peaks are in different places, which is why the section has to be chosen for a combination rather than for either.

Bending that arrives as twist

A straight beam under a vertical load carries no torsion unless something applies one. A beam whose axis curves on plan carries torsion everywhere, from the same load, with nothing applied off the axis — and it cannot be simply supported at all.

 The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing.

Shear across a crack that is already there

Every shear calculation in this collection starts from an uncracked solid — a principal stress, a shear flow, a diagonal tension. This one starts after the crack, on a plane with no tensile strength at all, and the coefficient it uses is not a coefficient of friction. It is the slope of the roughness.

 One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

The torsion that goes away if you let it

A spandrel beam attracts a torque in proportion to its own torsional stiffness. Crack it and the stiffness falls by a factor of four, the torque falls with it, and nothing has failed — because the floor beam it was competing with picks up exactly what was shed. A canopy hung off the same spandrel is a different animal entirely.

 A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 540 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 306 kNm — 57% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 51.0 kN pressing down there and 25.5 kN lifting at each end, a reaction set that sums to -2e-13 because nothing external was applied. Its diagram is straight between supports to 2.0e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span.

The prestress that pushes back

On a simply supported beam a tendon is an internal matter and changes no reaction. Put the same beam on three supports and the tendon lifts it off the middle one, the support refuses, and the force it takes to hold the beam down is a reaction produced with no load applied at all.

 The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from.

The beam that becomes a truss

Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes.

 The same load, two diagrams, both in equilibrium. One span of a pair of 7 m spans under 5 kN/m, drawn twice. The elastic solution puts 31 kNm over the support and 17 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 21 and 21: the section the beam needs falls from 31 kNm to 21, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 31 kNm for either — and the second is legitimate for that reason alone. What it costs is 1.0 milliradians of rotation at the support, which the section has to be able to deliver.

The moment that was moved on purpose

The elastic analysis of a continuous beam gives one set of moments. It is not the only set the beam is allowed to have, and taking a smaller one at the support is legal, cheaper, and paid for in a rotation that has to be delivered before the design exists.

 Between two beams and one, and much nearer one. How composite a beam is, against the one dimensionless group that decides it: αL, where α² = K·EI∞/(EA*·EI₀). At αL = 0 the layers slide freely and the beam is two beams; past about 20 the connection is stiff enough that the last per cent is unbuyable. The beam drawn sits at αL = 15.8 and is 96% composite, deflecting 36.7 mm against 33.9 for full interaction and 110 for none. The curve is steep where a real design sits, which is why halving the number of studs does not halve anything.

Half the studs, and most of the beam

Bonding two layers together quadruples the stiffness of the pair. A real connection is a row of studs that deform, so the layers slide a little and the beam sits between the two answers — but not halfway between, and the number that decides where is a single dimensionless group.

 The force spreads, and the spreading needs a tie. The end block behind an anchorage of 1200 kN on a 200 mm plate, in a section 700 mm deep. Half the force enters at the quarter point of the plate and leaves at the quarter point of the section, so a strut between the two rises 125 mm and needs a transverse tie to turn it. Placing the tie 0.5 depths from the face makes that tie force 214 kN — and at exactly half a depth this reproduces Guyon's 0.25P(1 − a/h) to the digit, which makes that famous coefficient a lever arm somebody chose rather than a property of concrete. The bearing stress under the plate is 20.0 N/mm² against 5.7 once the force has spread.

The force that splits what it pushes on

A prestressing tendon delivers its whole force through a plate a fraction of the section deep. One depth further along the stress is uniform, and the spreading in between requires a transverse tension nobody applied — the force that splits end blocks, and the only number in the design that no equilibrium equation on the member can see.

 Three ways to apply the same force, and one depth to forget the difference. Three end loads on a member 400 mm deep, all with the same resultant and the same moment: a point load, the same force spread over a fifth of the depth, and the same force split in two. What is plotted is the difference between each of them and the beam-theory answer — the self-equilibrating remainder — as a fraction of the mean stress. The point load starts at 20 times it and is under a tenth of it by 0.77 depths; all three are under one per cent by about 1.18. That distance is the licence every figure in this collection is drawn under, and the exact strip eigenvalue agrees with it: 2.106 + 1.125i, whose real part puts one per cent at 1.09 depths and whose imaginary part means the remainder changes sign on the way out, which no statement of the principle mentions.

How far a wrong load reaches

Every figure in this collection applies a load as a point, a line or a uniform pressure, and no real load is any of those. The licence is Saint-Venant's, it is usually quoted as a principle, and it is really a statement about a wavelength.

 The same beam, the same load, and one of them has to lift it. Two lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical in both — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in the top-loaded model, and the whole 600 kN in the hung one, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That tie is 1200 mm² of steel against the 96 mm² the shear calculation asks for over the same length — a factor of 12.5, in the same place, and additional to it.

The load that has to be lifted

Every shear calculation on this site assumes the load arrives on the top of the beam and walks down a diagonal to the support. Hang the same load from the soffit and the diagonal cannot start — the load has to be carried up to the compression zone first, by a tie no sectional calculation contains.

 The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the design rule's length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength.

The force that arrives along a length

A bolt takes its force at a hole and a weld along a line. A reinforcing bar has no such place — it is a smooth cylinder in a hole of its own shape, and the only thing stopping it sliding out is a stress smeared over its surface. So the force in it is not a number, it is a function of position.

 The chords take the shear the web is credited with. A cantilever of 6 m tapering from 400 to 1200 mm, under a 120 kN tip load, with the shear divided between the web and the two inclined chords. The chord force is M/z and it is not horizontal, so its vertical component is real: V_web = V − (M/z)·dz/dx, which is Résal's result of 1899. At the root the web is left with 33 per cent of the applied shear — exactly d₀/d₁ for a straight taper, with no length, load or material in it. Turn the same member round and the figure becomes 300 per cent: the chords stop helping and start adding, at the section where the moment is largest as well. And the limit is worth having — a member whose depth is proportional to its moment leaves the web nothing to do at all, which is the triangular cantilever every crane jib is.

The shear the chords take

Every shear check in this collection has assumed the two chords of a beam are parallel, so that the whole of the shear crosses the web. Taper the member and that stops being true — and the sign of the correction is decided by which end the haunch is at.

 The free body that makes a hoop force a pressure times a radius. Half a ring cut along a diameter, with the pressure drawn normal to the wall wherever the wall is. Vertical equilibrium of the half ring is the whole derivation: the pressure acts over the projected width 2R whatever the shape of the arc, the two cut faces carry N each, so N = pR — 300 kN per metre here at 1 MPa on a 0.3 m radius. The result contains no wall thickness, no second moment, and no length along the pipe, which is why a hoop force is the one internal force in this collection that arrives with no lever arm attached to it. The stress does contain the thickness — 25 MPa at 12 mm — but the force does not, and a thicker wall carries exactly the same force at a lower stress.

The force that is only a radius

Every internal force in this collection arrives with a lever arm attached. A hoop force does not. Cut a cylinder along a diameter and the free body settles it in one line — pressure times radius, with no thickness, no second moment and no length in it — which is why a tank wall is thin and why its worst hoop force is not at the bottom.

 Two cantilevers, or one wall, and the beams decide which. The deflected shape of a coupled pair of 6 m walls, drawn against the two limits it lies between. Release the coupling beams entirely and the pair is two independent cantilevers, deflecting 111 mm. Make them rigid and it is one composite wall of the full width, deflecting 16 mm — 6.8 times stiffer, because the lever arm between the wall centroids is 8.40 m and everything inside either wall is smaller than that. Real beams of 600 × 350 mm over a 2.4 m opening land at 23 mm and carry 63% of the base overturning as an axial couple rather than as wall bending. The degree of coupling never reaches one, because a beam of finite depth cannot suppress the walls' curvature entirely.

Two walls that agreed to be one

A pair of shear walls with a row of doors between them is the commonest lateral system there is, and it has two readings that differ by a factor of seven. What decides which one applies is a beam 600 mm deep over a 2.4 m opening — and most of the overturning ends up as an axial couple that no bending diagram contains.

 The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 200 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 659 mm — 3.3 times the bearing, and 70% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to.

The support that is not a point

A reaction is drawn as a single arrow because the equilibrium equations only need its total. Underneath the arrow is a bearing of some width, delivering a pressure over that width, and almost everything a designer would like to know about the region near a support is a consequence of the width the arrow does not have.

 A force may be moved anywhere, at the price of a couple. A 80 kN force applied 250 mm off the centreline of a body, and the same force applied ON the centreline together with a couple of 20 kNm. The two systems are equivalent: they have the same resultant force and the same moment about every point in space, so no equilibrium equation written about the body can tell them apart. What they are not is the same loading — the stresses inside the body differ, and they differ over a distance of about the body's own depth. The offset is drawn to a scale that keeps the arrow on the body; the number beside it is the real one.

The moment the beam left behind

A beam reaction is drawn arriving on a column's centreline. It arrives on a cleat a hundred millimetres out from the face, and the difference is a couple that goes into the column and has to be shared between the lengths above and below it. Nothing about it appears in a frame model whose members meet at nodes.

 A strength with no mechanism in it, made of four. The shear a member carries with no links in it, split into the mechanisms that carry it, against the member's effective depth on a logarithmic axis. The three bands are calibrated to Taylor's measured shares at one 300 mm × 500 mm member and are then evaluated everywhere else, so the shape of the total is a prediction. Aggregate interlock is the band that dies: it depends on how tightly the crack faces are held together, crack width grows with member depth, and it falls from 62% of a shallow member's strength to 22% of a deep one's. That decay is the whole of the size effect, and the dashed line is the design code's fitted k = 1 + √(200/d), which knows nothing about interlock and falls by a factor of 1.52 where the model falls by 2.05 over the same twentyfold range. Dowel action is why the expression contains the flexural reinforcement ratio, which nothing in a truss analogy would predict.

The strength with no mechanism in it

A concrete member with no links in it carries shear, and the expression that says how much is three variables raised to fitted powers with a size term in front. There is no free body anywhere in it. What it is fitting is a competition between four things that carry shear across a crack, and only one of them explains why a deeper member is worse at it.

 An enhanced strength that is the strength of a tie. Bearing strength as a multiple of the design cylinder strength, against how far the load is allowed to spread, with the bursting tension the spread creates on the same axis. The enhancement is √(A₂/A₁) and it reaches 2.80 for the 250 mm pad on a 700 mm block drawn — 47.6 N/mm² against a design strength of 17.0. There is no material property in that statement beyond the one being enhanced, and the reason is on the second curve: a load that spreads does so along inclined struts, a pair of inclined struts has a horizontal component, and that component is 16.1% of the load. It has to be tied. 1099 mm² of steel is what the enhancement actually is, and the cap of three is not a property of concrete — it is the angle past which nobody believes the strut.

Three times as strong under a smaller pad

Press a small plate onto a large block of concrete and it will carry three times the stress a cylinder of the same concrete fails at. The enhancement is a ratio of areas with no material property in it, which should be a warning: what has actually been measured is not the concrete's strength but the strength of a tie holding it together.

 The tendon is a load, pointing the other way. A 14 m beam with a parabolic tendon dropping 260 mm to midspan, stressed to 1440 kN after losses. Its curvature pushes the beam up along its whole length with an intensity of 8Pe/L² = 15.28 kN/m, against an applied 17.63 kN/m — so 2.34 kN/m is left to bend anything, and the beam carries 57.4 kNm where an unstressed one carries 432 kNm. What the section then feels is 6.40 MPa of uniform compression and very little else.

The load that comes from changing direction

A force that travels in a straight line asks nothing of anything. Bend its path and it asks for a transverse load of F over R along every millimetre of the curve, and that load is real, is nowhere on the load schedule, and is the same statement behind a prestressing tendon, a hoop force, an arch thrust and a web that buckles with nothing applied to it.

 The hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure.

The analysis that assumes the answer

A rigid frame is indeterminate, so statics cannot finish it. The hand methods finish it anyway, by assuming where the bending moment is zero and treating those points as hinges. That is not a shortcut around the analysis — it is a different kind of answer, exact in equilibrium and wrong in compatibility, and knowing which half is which is what makes the error a bound rather than a mystery.

 Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so.

Two of these move and the third cannot

Cut one span of a continuous beam free and add up the forces on it. What comes out is that the mid-span moment plus the average of the two end moments equals the free bending moment of that span, with nothing else in it — no stiffness, no support settlement, no analysis at all. Continuity moves moment about. It does not reduce the total, and it never has.

 The edge, and the length over which it is forgotten. A cylinder of radius 4.00 m and wall 12 mm under 0.6 N/mm² of internal pressure, held at its base. Away from the base the wall carries the pressure as pure hoop tension and bends nowhere, which is why a pressure vessel is a cylinder. At the base the hoop force is zero, because the wall cannot grow there, and the difference is made up by a boundary layer of bending that dies out inward. The length it dies out over is 1/β = 170 mm — 0.778√(Rt), a geometric mean of the radius and the thickness — and the moment is under a twentieth of its edge value by 3.07 of them. Nothing in that length is the load. The base moment is p/2β², and the bending stress it produces is 1.82 times the membrane hoop stress the whole design is about, at every pressure, every radius and every thickness: the ratio is √3/√(1 − ν²) and contains none of them. The hoop force overshoots by 4.3% at 3.2 lengths in, which is the wall springing back past where it was going.

The length a structure was never given

A disturbance applied at one place dies out over a distance, and the distance is not something anybody chose. A beam forgets a badly applied load over its own depth. A beam on the ground forgets a point load over the fourth root of its stiffness against the soil's. A shell forgets a held edge over the square root of the radius times the thickness — a geometric mean of two lengths three orders of magnitude apart, which is neither of them and is not near either.

 The worst force in a pile is not at the top of it. A 0.6 m pile 24 m long through ground that is settling, carrying 800 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 300 kN the base takes. The maximum is 1282.87 kN at 13.78 m — 1.60 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length.

The ground that hangs on instead of holding up

A pile is driven through fill that has not finished settling. The fill goes down past the shaft, the friction along that length turns round, and the pile is now carrying the soil rather than the other way about. The worst force is not at the head and not at the toe, and nothing at that depth can be seen.

 Restraint is a fraction, and the length decides how far up it reaches. A 20 m wall 3.0 m high cast against a base that has already hardened — a length-to-height ratio of 6.67. The base holds the bottom of the wall at R = 0.50 and the top of it at 0.304, decaying as 0.609 to the power of the height in wall heights. The free contraction is 380 microstrain, of which 84 per cent is the wall cooling from its own hydration peak and the rest is drying; the concrete's own strain capacity is 50. Everything to the right of the dashed line cracks, which here is the bottom 3.00 m of it. Nothing has been loaded.

The steel decides how many, not how much

A wall cast on a base that has already set cools, tries to contract, and is not allowed to. What follows is not a stress problem with a strength on the other side of it. The movement is going to happen; the only question the reinforcement gets to answer is how many pieces it is divided into.

 Prestress buys shear as a square root, not as a sum. The shear stress an uncracked web can take before the principal tension reaches the concrete's tensile strength, against the axial compression the prestress put there. With no prestress it is 1.35 N/mm², the tensile strength itself, because pure shear has a principal tension of exactly its own magnitude at forty-five degrees. Adding compression gives √(f_ct² + σ_cp·f_ct), which is a square root and therefore flattens: the first newton of prestress is worth far more than the last. At the 7.14 N/mm² drawn the limit is 3.39 N/mm², a gain of 2.51, and doubling the prestress from there takes it only to 4.59. The straight line is what a rule that simply added the two strengths would have promised.

The crack that never reached forty-five degrees

Every shear expression for reinforced concrete is a curve fitted to tests, because a cracked section has no free body worth drawing. An uncracked prestressed web has one — a single point, a Mohr's circle and a principal tension — and it is the only shear check in the subject that is derived rather than measured.

 A pile has no length until the ground gives it one. Deflection, bending moment and soil reaction down a 0.6 m pile carrying 150 kN at a free head, in ground whose modulus grows by 0.005 N/mm³ per millimetre of depth. The one length in the problem is T, the fifth root of EI over n_h, which is 1.89 m here; the head moves 20.5 mm, the worst moment of 219 kNm is at 2.50 m — 1.32 T — and below about four T nothing happens at all. The classical coefficients come out of the finite differences rather than a table: 2.430 against Matlock and Reese's 2.435, and 0.772 against their 0.772.

A pile has no length until the ground gives it one

Almost every other structural member is handed a length by the drawing. A pile goes into the ground until it stops, and what decides how much of it is working is a fifth root of the ratio between its own stiffness and the soil's.

 A skew deck spans square, and the corner knows it. Plan of a 30-degree skew slab, 12 m along the road by 10 m wide, with the reaction per unit length of abutment drawn as a bar at each support point. The load takes the shortest route between the abutments, which is the square span of 10.4 m rather than the 12 m of carriageway — so the reaction runs to the two OBTUSE corners, where the abutments are closest, and drains away from the acute ones. Peak 2.15 times the average, least 0.16. Nothing about the loading is uneven; the geometry is.

The deck that spans square

A slab bridge crossing a road at an angle is loaded uniformly and does not carry uniformly. Load takes the shortest route between the abutments, which is not the direction the carriageway runs, and the reaction piles up in two corners.

 A truss drawn inside a solid, and solved as one. A deep member 5000 mm between bearings and 2500 mm deep, carrying 2400 kN at mid-span. The model is two struts and one tie, on a lever arm of 2000 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 1500 kN and each strut at 1921 kN, at 38.7° to the horizontal. Spread over a strut width of 1031 mm the compression is 3.7 N/mm² against a limit of 18.1 for concrete cracked across its own strut, and the tie needs 3448 mm² of steel. A beam calculation on the same member would have asked the tie for 1404 kN, which is 7% less than the model does.

The node is the part that is checked

A strut-and-tie model is safe if every part of it can carry what the model asks. The struts are wide and lightly stressed, the ties are steel and easy, and the whole of the difficulty collects at the points where they meet — in a volume of concrete a few hundred millimetres across that no drawing shows.

 Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice.

The angle is a choice, not a property

The truss inside a cracked concrete web has a strut angle, and nothing measures it. The designer picks it, the stirrup requirement falls as it flattens, the web stress rises, and every choice in between is a different structure that carries the same load.

 Most of a long pile is doing nothing. Head deflection against embedded length, both measured in the pile's own characteristic length T = 14.67 m. A pile shorter than about two T is a lever with nothing holding its foot and deflects several times as much; past four T the curve is flat to within a per cent, because the ground below that depth is never asked for anything. A lateral check is a check on the top four T of a pile, however deep it goes for its axial load — and adding length to fix a lateral deflection is the one remedy that does not work.

The pile that is too short to bend

A laterally loaded pile is long or short in its own characteristic length, and the two are different structures. The slender pile bends and the ground below four characteristic lengths never hears about it; the monopile is two lengths deep, rotates about a point, and every tabulated coefficient written for the first case is wrong for the second.

 Two diagrams for one load, and the second one has no straight-beam ancestor. Bending moment and torsion round a 90° arc of radius 6 m under a uniform load, built in at one end. The bending peaks at 576 and the torsion at 329, 57% of it. Both are zero at the free end and largest at the support, which is where a curved cantilever's bearing has to hold a torque it was probably not asked for.

The torque that has nowhere to go

A curved beam on two supports splits its torsion between them, and the two halves cancel at mid-span. A curved cantilever has one end, so every increment of torque accumulates toward it — and the largest action at the root of a curved balcony is one that a straight beam does not have at all.

 More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable.

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

 The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver.

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

 A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 1440 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 720 kNm — 50% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 102.9 kN pressing down there and 51.4 kN lifting at each end, a reaction set that sums to 0e+0 because nothing external was applied. Its diagram is straight between supports to 3.6e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span.

The tendon that can be moved

Lift a continuous beam's tendon at its interior support without changing its drape and nothing about the beam's total moment changes. The primary falls, the secondary rises by exactly as much, and the pressure line stays where it was — which turns a parasitic effect into a quantity a designer can place.

 The studs are evenly spaced and the demand is not. The force per unit length the shear connection carries along half of a 12 m composite beam, from Newmark's solution. It is largest at the support — 282 N/mm — falls to nothing at mid-span, and averages 156: the end studs are asked for 1.81 times the mean. Studs are nevertheless placed at a uniform spacing, and the justification is the one the variable-angle truss uses for its stirrups — a ductile connector sheds what it cannot carry to its neighbours, so the uniform distribution is a plastic redistribution and not a description of the elastic state.

The connection is busiest where the beam is not

A composite beam's studs are spaced evenly along it and the demand on them is not even at all. It peaks at the supports, where the bending stress is nothing, and falls to zero at mid-span, where the section is working hardest — so the connection is designed from a diagram nobody looks at.

 A check made on a perimeter, not on a section. One bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 604 kN across 4427 × 225 mm, a shear stress of 0.697 N/mm² against a resistance of 0.658.

Turn the column, and the slab passes

A flat slab that is comfortable under gravity fails its punching check the moment a moment arrives at the column, and nothing about the load has changed. The fix is not more concrete. It is the column's plan shape and, at equal area, which way round it is turned — worth more than adding half again as much column.

 The worst force in a pile is not at the top of it. A 0.75 m pile 30 m long through ground that is settling, carrying 1200 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 600 kN the base takes. The maximum is 2331.39 kN at 18.86 m — 1.94 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length.

The coating that takes the resistance with it

The cure for downdrag is to make the pile slippery, and it works — a bitumen slip layer takes the drag on this pile from 1,131 kN to 34. It also removes the shaft friction that was holding the pile up, in the same proportion and over the same length, and past a certain smoothness there is no neutral plane to find because there is no equilibrium.

 One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

The torque that should not be shed

A compatibility torque can be let go, because the load has somewhere else to go. What the rule does not say is what it costs the somewhere else — a twenty per cent rise in a floor beam's midspan moment, a crack width nobody limits, and a rotation the spandrel has to actually deliver. There is a size of torque past which shedding is the wrong answer, and no code states it.

 Two restraints, and only one of them cares how much was imposed. Crack width against the restrained strain, for the same 20 m wall restrained two ways. Edge restraint — a wall cast on a base — gives a width proportional to the strain, because the concrete has to accommodate the movement and the cracks are where it does. End restraint — a bay cast between two that have hardened — gives 0.32 mm at every strain on the axis, because the crack opens only until the steel can push the cracking force back into the concrete, and that force is a property of the section. At the 304 microstrain this wall is asked for, the two are 0.17 and 0.32 mm, and the movement is divided into 35 cracks and 19.

The bay that is cast last

A wall cast on a base cracks in a way the steel controls: more movement gives wider cracks, and reinforcement decides how many. A bay cast between two walls that have already hardened cracks in a way that has nothing to do with how much movement there was — and the threshold that separates a controlled crack from a single wide one is a quantity of steel rather than a limit on anything.

 A soft layer over stiff ground, and a moment neither has. The bending moment down a 20 m pile of bending stiffness 120 MN·m² with a free head, pushed sideways by 150 kN (its characteristic length in ground at 5.00 MN/m³ being 1.89 m), in three grounds: all of it at 5.00 MN/m³; all of it at 0.50 MN/m³; and soft ground 2.0 characteristic lengths (3.8 m) thick over the other (dotted: the interface). The largest moments are 219 kN·m, 346 kN·m and 427 kN·m; the head deflections 20.4 mm, 81.4 mm and 61.6 mm. The layered ground gives a larger moment than either ground on its own: the pile bends as a pile in the soft layer would, and then meets ground that holds it.

The layer a pile feels

A pile pushed sideways in uniform ground has one characteristic length, the fifth root of its stiffness over the ground's, and every answer is a multiple of it. Put a soft layer over stiff ground and that length stops existing. The head feels the top of the ground in proportion to the square of its own deflection, so an average weighted that way gets the head deflection within eight per cent. It gets the moment a quarter too low: with soft ground two characteristic lengths thick over stiff, the pile bends harder than it would in either ground alone, and the peak sits where the stiff ground takes hold.

 The interior panel is designed by the case that designs nothing else. The shear across the column-web panel at each kind of joint of a frame of two 8 m bays and two 4 m storeys with fixed bases, 457 mm beams and 254 mm columns with 8.6 mm webs, under five load cases, the worse direction of sway taken where there is one. Full gravity: roof corner 106, roof interior 0, floor exterior 143, floor interior 0 kN; One bay loaded: roof corner 115, roof interior 117, floor exterior 150, floor interior 154 kN; Sway: roof corner 25, roof interior 40, floor exterior 53, floor interior 89 kN; Gravity + sway: roof corner 131, roof interior 40, floor exterior 196, floor interior 89 kN; Pattern + sway: roof corner 141, roof interior 157, floor exterior 203, floor interior 242 kN. The floor interior panel is empty under full gravity and carries 89 kN under gravity with sway, the case that governs the members; under one bay loaded with sway it carries 242 kN — 2.7 times as much, more than any exterior panel's 203 kN, and 70 per cent of the web's shear resistance.

The panel that gravity leaves empty

A corner joint's panel carries the difference between a beam's flange couple and half a column's shear, and full gravity load is what makes that large. An interior joint has a beam on each side, and under full gravity their two couples cancel: the panel between them carries nothing at all, in the load case that sizes every member around it. It fills under the two loads that are not symmetric. A pattern load hands it one beam's moment, and sway hands it two beams' moments turning the same way. Together they put more shear through it than through any exterior joint, in a combination that governs no member.

 The perimeter the edge leaves, and where its middle is. A 400 × 400 mm column at the edge of a 260 mm slab (d = 225 mm) carrying 12 kN/m², on a 7.2 m end span. Heavy line: the control perimeter 2d out, cut by the slab's edge — 2,614 mm, against 4,427 for the same column inside the slab. Wide band beneath it: the reduced perimeter u1 of EN 1992-1-1, 2,214 mm, its legs cut back from the edge. Open circle: the perimeter's centroid, 363 mm in from the column's centre. Filled: the load's resultant, 631 mm in, set by the slab's end moment of 170 kN·m on a shear of 269 kN — 269 mm beyond it, into the slab. The general multiplier on the shear is 1.78; u1/u1 gives 1.18 and the shortcut 1.4.

The perimeter whose middle is not the column's

At an edge column the slab's own edge cuts the control perimeter, and what is left has its centroid 363 mm inside the column's centre. The slab's end moment moves the shear's resultant inward too, by a distance that grows with the end span. What twists the perimeter is the gap between the two: nothing at a 5 m end span, enough by 7.2 m to put the check at 1.23 of its resistance where the 1.4 shortcut reads 0.97. And a bigger column makes it worse.

 The studs reach out to where the concrete is enough. One column of a 260 mm flat slab (d = 225 mm) carrying 12 kN/m² on 400 × 400 mm internal columns, the moment it hands the column putting the shear 300 mm off centre, on a 10.0 m bay, to scale. The inner check, on the control perimeter 2d out (solid), needs reinforcement: 12 rails of studs, 6 perimeters of them from 0.5d at 0.75d spacing, 1,415 mm² on each perimeter. The outer perimeter (dashed), where the concrete alone carries the shear, stands 5.4d from the face — 9,173 mm long against the control perimeter's 4,427 — and the last studs must be within 1.5d inside it, 3.9d from the face.

The studs that send the check outward

Shear studs round a column fix a failing punching check, and they turn one check into three. The studs carry the control perimeter; past them the concrete alone must carry a perimeter long enough to need no help, and that outer perimeter stands further from the column the larger the load, so the studs follow it out. At a 10 m bay they reach four effective depths from the face, and the steel they need grows as the shear to the power 3.5. The third check, at the column face, is the one no stud reaches, and it ends the series at 11.8 m.

 The shear is read from the pressure line, not the tendon. The first 9.6 m of a two-span beam of 16 + 16 m, 400 × 900 mm, prestressed with 2,400 kN on a parabolic tendon and carrying 45 kN/m, with depths drawn from the top face (−450 mm) to the bottom (+450 mm): the tendon's centre (dashed) and the pressure line, where the resultant of the prestress actually acts once the supports have pushed back (solid). 0.8 m from the support the tendon falls 87.00 mm per metre, which is the slope the rule P·sin θ reads, worth 209 kN; the pressure line falls 76.39 mm per metre, worth 183 kN. The difference is the secondary shear, 26 kN, which the supports put back.

The relief that belongs to the pressure line

An inclined tendon carries part of a prestressed beam's shear, and the rule is to subtract its vertical component, P times the slope of the tendon. In a continuous beam the tendon also provokes reactions, and the reactions make a shear of their own — a constant in each span, too small to look at. It is not small where it matters. Beside the end support of a two-span beam it doubles the shear left for the links, and moving the tendon without changing the beam at all can make the tendon's slope report that there is none.

 Where the pile and the ground settle alike. For a pile 24 m long and 600 mm across, carrying 800 kN, in ground whose top 18 m settles — by 100 mm at the surface, less with depth — over a toe that resists by moving: on the left, how far the ground (dashed) and the pile (solid) settle down the pile's length; on the right, the force in the pile. Above 15.9 m the ground settles more than the pile and hangs on it; below, the pile settles more and the ground holds it up. The force peaks there, at 1,407 kN against the 800 kN applied, and the toe, settling 10.6 mm, mobilises 623 kN; the head settles 13.6 mm.

The plane where the pile and the ground agree

Downdrag is usually found by equilibrium: the ground hangs on the pile above a neutral plane, holds it up below, and the plane is where those balance the load and the toe's resistance. But the toe's resistance is not a number the designer can supply. It is whatever the toe's own settlement mobilises, and the toe settles because the ground drags the pile down. Solved by compatibility instead — the plane where the pile and the ground settle by the same amount — the drag, the neutral plane, the toe's force and the pile's own settlement all turn out to grow with how far the ground goes, and each kilonewton put on the head costs the pile three quarters of one, not half.

 What leaves the interior panel lands in the span. A frame of two 8 m bays and two 4 m storeys with fixed bases, 457 mm beams and 254 mm columns, with 20 kN/m on the left bay's beams only, against the stiffness of every beam-to-column joint as a multiple of the beam's EI/L, on a logarithmic scale. The shear across the floor interior joint's column-web panel (solid): 154 kN with rigid joints, 146 at EN 1993-1-8's rigid limit of 25 EI/L, 104 at 3.0 EI/L, 23 MN·m/rad, 63 at EI/L. The loaded beam's largest sagging moment (dashed), in kN·m: 71, 76, 102 and 126, against 160 for a simply supported span. The faint lines are EN 1993-1-8's boundaries for a frame that sways: nominally pinned below 0.5 EI/L, rigid above 25.

The panel a soft joint empties into the span

Under a load on one bay, an interior column's web panel carries the difference between the two beams' moments. Make the beam-to-column joints semi-rigid and the panel is relieved — by 5 per cent at the stiffness EN 1993-1-8 still calls rigid, by a third at three times the beam's EI/L — because a softer joint takes less moment. The moment it does not take lands at mid-span, where the beam was not designed for it, and in sway it lands in the columns. And soften only some of a frame's joints, and the sway goes looking for the ones still stiff: their panels carry up to a quarter more than they would in an all-rigid frame.

 Six millimetres of slip draw the Eurocode's line. The least degree of shear connection at which a 457 × 191 × 67 UB in S355 under a 130 mm slab of C30/37 2.5 m wide, with 19 mm studs of 80 kN can reach 95 per cent of its plastic moment without its end studs slipping more than 6 mm, against the span (solid), and EN 1994-1-1's minimum for ductile studs in a beam with equal steel flanges, 1 − (355/fy)(0.75 − 0.03L), not less than 0.4 (dashed). 6 m: 0.24 against 0.43; 8 m: 0.51 against 0.49; 10 m: 0.61 against 0.55; 12 m: 0.69 against 0.61; 14 m: 0.75 against 0.67; 16 m: 0.77 against 0.73; 18 m: 0.80 against 0.79; 20 m: 0.88 against 0.85. From 8 m up the two differ by at most 0.08: a stud model and a slip capacity reproduce the rule's slope as well as its level. At 6 m the slip alone would allow 0.24, and the rule asks for 0.43.

The slip the plastic moment asks of the studs

A composite beam with half the studs that full interaction needs still has 85 per cent of the full plastic moment, because the slab's compression block simply gets thinner and the steel finds its own neutral axis. That is the stress-block answer, and it assumes every stud slips as far as the beam asks. Follow the slip along the beam and the end studs of a 12 m beam need 6.6 mm to reach nine tenths of that moment and 9 mm to reach 95 per cent — past the 6 mm a stud is trusted to give. Require 6 mm and the Eurocode's minimum degree of shear connection comes out of the arithmetic, slope and all.

 The linear answer has the ground holding the ends down. The contact pressure along a strip 16 m long, EI 5.40 × 10⁵ kN·m², on a bed of 50 × 10³ kN/m² per metre of its length, under a column of 1,000 kN. With springs that pull as readily as they push (dashed) the bed presses 197 kN/m under the column and pulls down on the ends at up to 34 kN/m — tension the ground cannot supply. With the pulling springs released (solid), the strip touches the ground only from 3.97 m to 12.03 m, 8.05 m in all, and presses 213 kN/m under the column, 8 per cent more; beyond that it carries nothing.

The ends the ground was holding down

Winkler's springs pull as readily as they push, and on a long footing strip the linear answer quietly uses that: it has the ground holding the strip's ends down. Take the pull away and the correction under the column is modest — eight or nine per cent on the pressure and the moment — but the shape of the answer changes completely. A weightless strip longer than about eight metres keeps exactly eight metres of itself on the ground, whatever length was poured, and the rest rises off as straight cantilevers carrying nothing. The middle-third rule turns out to be a property of rigid footings: an eight-metre strip lifts at a fiftieth of the eccentricity the kern allows.

 The arch is only as strong as what holds its ends. Load against mid-span deflection, as a share of the thickness, for a slab strip 4,000 mm long and 200 mm thick, a metre wide, with its ends held by springs of 0.05, 0.5 and 5 times its own axial stiffness EA/L and rigidly; dashed is its flexural load, 30.0 kN/m. At 0.05 EA/L it peaks at 45.3 kN/m, 1.51 times the flexural load, at 0.67 of its thickness; at 0.50 EA/L it peaks at 74.8 kN/m, 2.49 times the flexural load, at 0.45 of its thickness; at 5.00 EA/L it peaks at 108.6 kN/m, 3.62 times the flexural load, at 0.28 of its thickness; held rigidly it peaks at 116.8 kN/m, 3.89 times the flexural load, at 0.24 of its thickness. A soft restraint lets the ends move out as the strip tries to arch, so the thrust builds slowly and the peak arrives later and lower.

The edge that cannot hold the arch

A slab strip whose ends cannot move apart arches inside its own depth and carries nearly four times its flexural load. "Cannot move apart" is a stiffness, not a fact, and the arch needs a lot of it: half the gain takes end springs about half as stiff as the strip itself in compression, and nine tenths takes springs five times as stiff. An interior panel is held by the floor around it and gets the arch. An edge panel is held on its outer side by whatever runs along the edge, and a band of slab two metres wide between columns eight metres apart supplies a twentieth of what the strip needs. The arch is a property of where the panel is.

 The shadow an opening casts on every perimeter. One column of a 260 mm flat slab (d = 225 mm) carrying 12 kN/m² on 400 × 400 mm columns, the shear 300 mm off centre, on a 10.0 m bay, to scale, with a 300 × 300 mm opening 225 mm (1.0d) from the column. The two tangents from the column's centre to the opening's corners (dashed) enclose 38.9 degrees, and every perimeter loses the part between them: 458 mm of the control perimeter 2d out (solid), 1,093 mm of the outer perimeter (dashed), the lost parts drawn faint. With the opening the outer perimeter stands 6.13d from the face instead of 5.36d, and the studs need 7 perimeters instead of 6; one of the 12 rails runs into the hole.

A hole takes a tenth of every perimeter

A 300 mm riser cut a slab's depth from a column throws a shadow from the column's centre, and EN 1992-1-1 deletes whatever part of a punching perimeter falls inside it. The outer perimeter, being longest, loses the most length — but it loses the same share as the control perimeter, a tenth of each, because the shadow is an angle. The cost arrives somewhere less obvious: a seventh ring of studs, 24 per cent more steel, and studs needed in a bay a third of a metre shorter than before.

 Heave puts the pile in tension where settlement put it in compression. For a 24 m pile 600 mm across carrying 200 kN, in clay whose top 18 m moves, by 100 mm at the surface: the axial force down the pile, compression positive, when the ground heaves (solid) and when it settles as far (dashed). Settling, the pile's compression peaks at 910 kN at 17.2 m and its toe carries 324 kN. Heaving, it is in tension below 8.9 m, peaking at 520 kN at 17.3 m, and its toe carries nothing: lifted, it parts from the soil beneath, so everything holding the pile down comes from the shaft below the plane.

The pile the ground pulls apart

Clay that swells after its trees are felled rises past the piles in it, and the same compatibility that put a pile into compression in settling ground puts it into tension in heaving ground, at nearly the same depth. Nearly everything else changes. A lifted toe parts from the soil and holds nothing, so the shaft alone anchors the pile; the head load that added to the downdrag now subtracts from the tension; and the tension peaks 17 metres down, far below the short cage a lightly loaded pile is given. A house's piles in swelling clay are tension members nobody reinforced.

 The same studs moved toward the supports slip less. Half of a 12.0 m composite beam, a 457 × 191 × 67 UB in S355 under a 130 mm C30/37 slab, with 50 per cent shear connection, 19 studs from the support (left) to mid-span (right): spaced evenly, and graded so that they crowd toward the support. Below, the slip along the half-span with the mid-span moment at 95 per cent of the beam's plastic moment: evenly spaced, the end studs must slip 8.8 mm; graded, 4.4 mm. The dashed line is the 6 mm a ductile stud is trusted to give. The plastic moment is the same for both, because the studs are the same in number and each reaches its capacity.

The same studs, moved to where the slip is

A composite beam with half the studs full interaction needs reaches 85 per cent of the full plastic moment, if its end studs can slip as far as the beam asks — nearly 9 mm on a 12 m span, past the 6 mm a stud is trusted to give. Keep the same nineteen studs and crowd them toward the supports, and the end studs need 4.4 mm. Nothing about the beam's strength changes; what changes is how soon the slab's force builds up, and the least connection the slip allows falls from 0.69 to 0.40.

 A core drilled into a slab the studs were laid out for without it. One column of a 260 mm flat slab (d = 225 mm) carrying 12 kN/m² on 400 × 400 mm columns, the shear 300 mm off centre, on a 10.0 m bay, to scale, its studs laid out for the slab without a hole — six rings of 14 mm studs on twelve rails, the last ring 4.25d from the face — and a 150 mm core 225 mm (1.00d) from the face. Its shadow from the column's centre takes 20.3 degrees of every perimeter. The studs' outer perimeter stands where the last ring leaves it, 5.75d out (solid); with the core the slab needs it at 5.74d (dashed), so the outer check is at 1.00 of its capacity, and the control check's studs at 0.79. No rail runs into it.

The core drilled after the slab was built

A building in use acquires new services, and a core drilled through a flat slab beside a column is a routine instruction. The studs round the column were laid out for a slab without the hole, and the drill finds whatever is there. Whether it cuts a stud matters less than it looks. What decides is the outer perimeter, which the hole pushes outward while the studs stay where they are, and whether the slab survives is set by a margin nobody chose: the room left when the studs were rounded up to a stock size and the rings to a whole number.

 Ground already loaded rocks stiffer than springs that are not. The moment on the column against its rotation through a storm that takes the moment to ±1,000 kN·m and back, for an 8 m strip of EI 540 MN·m² on a bed of 50 MN/m² under a 1,000 kN column: on elastic springs that cannot pull, and on ground that unloads and reloads 4 times as stiffly as it first loads, in the first storm and the third. On elastic springs the column turns 1.418 milliradians at the peak, a rocking stiffness of 0.70 GN·m/rad, the same every storm. On the remembering ground it turns 1.116 in the first storm and 0.868 in the third — 0.90 and 1.15 GN·m/rad — because rocking unloads half the strip, and ground that has carried the column unloads stiffly.

The footing the storms press down

A footing strip under a column that a storm rocks one way and then the other was expected to dig itself a bowl: its ends loaded half the time, compacted, and left hanging over the hollow. On ground that unloads and reloads four times as stiffly as it first loads, the bowl appears and the strip fills it. The column settles 26% more over three storms and then stops; the footing rocks 22% less than it did in its first storm and is 64% stiffer than on springs that forget, because rocking unloads half the strip and ground that has carried the column unloads stiffly. The storms change the settlement and the stiffness, and leave the moments almost where they were.

 Stressed bay by bay, the beam keeps half its secondary moment. The secondary moment along a 4-span beam of 16 + 16 + 16 + 16 m prestressed with 2,400 kN on a parabolic tendon, straight between supports: stressed in one operation, stressed one bay at a time, and bay by bay after creep with a coefficient of 2.0. At the first interior support it is 350 kN·m stressed at once, 164 bay by bay and 307 after creep; at the next, 234, 160 and 216.

The beam stressed one bay at a time

A long continuous beam is rarely stressed in one pull. It is cast and stressed a bay at a time, each new length of tendon coupled to the last, and at every stage the structure is a different beam. The secondary moments the finished beam carries are the sum of what each stage did to the beam that existed then — about half of what the one-stage analysis gives — and creep spends the following years taking most of the other half back.

 The loaded end stops; the unloaded one takes over. The moments at the four floor beam ends of the frame of two 8 m bays and two 4 m storeys, its beam ends joined to the columns by end plates of 3.0 EI/L initial stiffness and 120 kN·m resistance, under λ times 20 kN/m on the left bay and 20 kN at each floor, against λ: on the softening joints (solid) and on joints that keep their initial stiffness (dashed). At λ = 1 the loaded beam's inner end carries 78 kN·m and the unloaded beam's inner end, across the same column, 13. The loaded end reaches its resistance at λ = 2.69; at λ = 3 the four ends carry 89, 120, 81, 88 kN·m, against 97, 235, 39, 68 on stiff joints. The unloaded inner end has gone from 13 to 81, 6.2 times as much for three times the load.

The joint that softens under its own load

A bolted end plate is stiff until its moment passes two thirds of its resistance and then softens until, at its resistance, it is a third as stiff. Put such joints at every beam end of a frame under a pattern load and a push, and the most loaded joint softens first and hands its share across the column to the joint on the other side, which carries six times as much for three times the load. The redistribution grows as the frame approaches its strength. Every joint ends at its own stiffness, so the single secant stiffness an elastic analysis is allowed asks the loaded joint for half as much again as it can carry, and gives the exterior panel three quarters of what it gets.

 Whole spans miss the shear inside a span. The shear along three spans of 8.0, 8.0, 8.0 m, under 20 kN/m of dead load everywhere and 15 kN/m imposed: the largest and least when the imposed load stands wherever it raises or lowers the shear at each section (solid), and when it may only cover whole spans (dashed). At the supports the two agree. At the middle of the second span the exact envelope reaches 23.8 kN either way and the whole spans 10.0 kN; at a tenth of the way along it, 122.4 against 122.0.

The shear whole spans cannot find

A continuous beam is designed by loading its spans in patterns, whole spans at a time, and for the bending moment that is very nearly exact: the moment's influence line keeps one sign over each span, so a span is either worth loading or not. The shear's influence line changes sign at the section itself, inside the span, and no pattern of whole spans can load one side of a section and not the other. At the middle of the inner span of three equal spans the imposed shear is 0.198 qL; whole spans find 0.083 qL, and the straight line to qL/8 that is safe on a simple span finds 0.125.

 The order of building decides which supports keep the secondary moment. The secondary moment at each interior support of a five-span beam of 16 + 16 + 16 + 16 + 16 m prestressed with 2,400 kN on a parabolic tendon, built and stressed from one end, from both ends towards a closure bay, from the middle outwards, and all at once. From one end: 161, 175, 160, 219 kN·m; from both ends: 161, 172, 172, 161 kN·m; from the middle out: 215, 120, 120, 215 kN·m; all at once: 344, 258, 258, 344 kN·m.

The beam built towards its own middle

A continuous prestressed beam built one bay at a time from one end keeps about half the secondary moment a beam stressed at once would carry, and most of it at the end where it was finished. Built from both ends towards a closure bay, it is symmetric, and the closure's supports keep only a little more than the rest: 172 kN·m against 161. Built from the middle outwards, the first interior supports — the ones the load hogs hardest — keep 215, 63 per cent of the at-once value, for any length of beam. The order of construction is a design variable. Creep then takes back most of the choice.

 A prevailing wind leans the footing, and the lean stops. The 8 m strip of EI 540 MN·m² on a bed of 50 MN/m² under a 1,000 kN column on ground that unloads and reloads 4 times as stiffly as it first loads, rocked by storms that push to 1,000 kN·m one way and swing back all, three quarters, half, a quarter or none of that the other: the column's tilt at rest after each of eight storms, in milliradians, positive the way the storms push. After eight: 0.011, 0.064, 0.117, 0.171, 0.218. Each settles within three storms; the lean is set by how unequal the storms are, not by how many there have been.

The footing a prevailing wind leans

A footing rocked by storms that reverse evenly settles into a symmetric bowl and stops. A footing under a frame in a prevailing wind is pushed harder one way than the other, and its ground keeps the difference. It leans — and then it stops leaning, within three storms, at a tilt set by how unequal the storms are and how large the largest has been. Fifty years of storms lean it further only in the years when a storm beats every one before it, and between those years nothing that happens to the footing moves it at all.

 A jack set for the first day is wrong by the last. The secondary moment over the first interior support of a five-span beam of 16 + 16 + 16 + 16 + 16 m prestressed with 2,400 kN on a parabolic tendon, built from both ends towards a closure bay, against the creep coefficient: not jacked; jacked at closure by the whole of what the order took, 183 kN·m there; jacked by 0.60 of it, so as to reach the at-once moment at a creep coefficient of 2.0; and by 0.89, so that the shortfall on the day and the overshoot at the end are equal; dotted, stressed at once, 344. On the day and at 2.0: not jacked, 161 and 302; jacked by the whole, 344 and 372; for the end, 271 and 344; balanced, 324 and 365.

The jack that creep carries past the mark

A continuous beam built from both ends towards a closure bay keeps less secondary moment at its supports than the same beam stressed all at once, and creep gives back three-quarters of the difference over the years. Bridge builders do not always wait: before the closure is cast, the supports can be lowered on jacks so that the beam starts with the moments it would have had. Jacked by the whole difference, it does start there — and creep, which goes on returning the staged deficit while it relaxes the jacked correction, carries it past them, by a seventh of what was jacked. Jacked by 0.89 of the difference, it starts a ninth short and ends a ninth over. And it takes a jack at every interior support: with fewer, the moment goes to the wrong places.

 Pressed and pulled, the footing leans once and stops. The column's tilt at rest after each of eight storms reversing fully to 300 kN·m each way, on the 8 m strip of EI 540 MN·m², weighing 320 kN, on a bed of 50 MN/m² under a column carrying 1,000 kN, on ground that unloads and reloads 4 times as stiffly as it first loads, for loads swinging by 0, 50, 100, 150, 200 per cent. 0 per cent: −0.007 milliradians after the first storm and 0.003 after the eighth; 50 per cent: 0.112 milliradians after the first storm and 0.103 after the eighth; 100 per cent: 0.109 milliradians after the first storm and 0.091 after the eighth; 150 per cent: 0.113 milliradians after the first storm and 0.095 after the eighth; 200 per cent: 0.114 milliradians after the first storm and 0.102 after the eighth. The lean is towards the end the heavier half of each storm presses; it stops within three storms whether or not the strip ever hangs clear.

The footing that hangs from its column

A footing under a bracing bay's column is pressed hard when the wind pushes one way and pulled when it swings back, and if the pull is larger than the footing's weight it hangs from the column, clear of the ground, for part of every storm. It looks like a footing free to come down somewhere new each time, and walk. It does not walk. The ground under it keeps the memory of every press, and a footing that is not touching the ground cannot change that memory; it comes back down into its own bowl, leans once towards its harder press, and stops within three storms.

 The hogging region cannot give its studs away. For a 457 × 191 × 67 steel beam under a 130 mm slab, continuous over two 12.0 m spans under 30 kN/m, 40 studs a span: the largest slip (solid, left scale) and the force in the slab's bars over the middle support (dashed, its value in the legend), against the density of studs beyond the point of contraflexure as a share of the density elsewhere, the span's total held at 40. At the same density everywhere the largest slip is 0.68 mm and the bars carry 344 kN; at half the density beyond contraflexure, 1.04 mm and 251 kN; at a quarter, 1.69 mm and 118 kN. The support's hogging moment falls from 448 to 398 kN·m as the bars stop working.

The slip that peaks where the moment turns

A simply supported composite beam slips most at its supports, and its studs are crowded there. Make the same beam continuous over a middle support and the slip peaks twice: at the end support, and again, just as hard and the other way, at the point of contraflexure — not over the middle support, where the shear is largest but the slip falls to nothing. Studs graded toward the supports, as a simple span's would be, miss the second peak and buy a tenth; graded to the slip they buy more than a third. And the hogging region cannot give its studs away to the span, because its edge is where the second peak is.

 The first hinge comes latest where the beam's end and its span yield together. The load factor at which the first plastic hinge forms in the two-bay frame of 8 m bays and 4 m storeys, its beams of plastic moment 405 kN·m, under λ times 20 kN/m on the left bay and 20 kN at each floor, against the initial stiffness of its full-strength end plates (1.20 times the beam's plastic moment), softening as EN 1993-1-8 has it and kept linear. With softening, 3.39 at 100 EI/L, 3.78 at 25 and 3.91 at 3; kept linear, 3.32, 3.51 and 3.91. Stiff joints hinge first at the inner end and flexible ones in the span; the first hinge comes latest between, 4.52 at 10.0 EI/L with softening and 4.31 at 5.0 kept linear. Rigid joints give 3.25, and the beam collapses at 5.06 for every joint.

The joint stronger than its beam

A full-strength joint never reaches its own resistance: the beam beside it hinges first. But an end plate softens from two thirds of its resistance, and a joint only a little stronger than its beam is already on the soft part of its curve before the beam yields. The softening sheds moment from the beam's end into its span, so it delays the first hinge — by 8 per cent for a joint at 1.2 times the beam's strength, 18 for one at exactly the beam's, and not at all from 1.5 on. It never moves the collapse. And too much shedding hands the first hinge to the span, so the first hinge comes latest at the stiffness where the end and the span yield together.

 Where the shelter is decides whether the leading row works harder. The share of the cap's shear the leading row and the last row of three rows carry, as a multiple of the average pile's, against the thickness of soft ground over stiffer, in characteristic lengths, with the row factors applied at every depth (solid) or in the soft layer only (dashed). With the factors at every depth the leading row carries 1.24 and the last 0.83 at 2.0 lengths; in the soft layer only, 1.07 and 0.96. In uniform ground the usual reading gives 1.35 and 0.75.

The shelter that stops at the interface

Piles in a group shelter one another, and practice counts the shelter as a factor on each row's soil, calibrated on groups in uniform ground. Put the same group in soft ground over stiff and the calibration has to be read one of two ways — the shelter at every depth, or only where the ground is soft — and the two readings disagree about which row works hardest. The first gives the leading pile 17 per cent more moment at the cap; the second gives the last pile 24 per cent more in the ground.

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