Field

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.

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

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

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.

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

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

Every other member on this site 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.

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.

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.

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.

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.

All essays