Field

Sections and stress

How a cross-section resists a moment, and why where the material sits matters more than how much there is.

Every strip counts by the square of its distance. A rectangular section divided into equal strips, with each strip's contribution to the second moment of area drawn beside it. The strips are identical in size; only their distance from the neutral axis differs.

The material far from the middle does nearly all the work

A strip of steel contributes to bending stiffness in proportion to the square of its distance from the centre. Move the same steel outward and the section gets stiffer for nothing.

Bending is a push and a pull. A section carrying a bending moment, with the stress at every height computed as the moment times the distance from the neutral axis divided by the second moment of area. It is compression above and tension below, and zero exactly at the neutral axis.

Bending is a pair of forces, pushing and pulling

A bending moment is not a mysterious twisting. It is a push near the top of a section and a pull near the bottom, separated by a lever arm — a couple, made out of stress.

The same steel in a different shape, and a factor of forty

Four sections of identical area, identical weight and identical cost. The stiffest is dozens of times the stiffest of the flattest, and the only thing that changed was the arrangement.

Where plane sections stop staying plane. Strain across a cut face at four span-to-depth ratios, with the straight line the theory assumes drawn faintly behind. For a slender beam the two coincide; for a beam as deep as its span the real distribution is nothing like a straight line, and beam theory has no claim on it.

Plane sections stay plane, and what the assumption costs

Beam theory rests on one sentence about geometry. It is very nearly true for a slender member, wrong for a deep one, and everything in the subject that fails does so where it stops holding.

Shear stress across a section. The distribution of shear stress over an I-section, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.41 against a mean of 0.19 — a ratio of 2.15 — and it falls at the neutral axis, where the bending stress is zero.

The shear nobody draws

A stack of loose planks slides at its ends when it is loaded. Glue them and the sliding stops — and whatever the glue is now carrying is a stress that no bending calculation contains.

The shear centre of a channel. A channel of 80 by 200, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 31.7 outside the web to leave the section untwisted — a point in the air, outside the material entirely.

The point that is not in the section

A channel loaded down its web twists. To stop it, the load must be applied through a point outside the steel entirely — in the air beside the section, where nothing can be attached.

The neutral axis is wherever the first moment vanishes. A 300 by 500 section with 1200 mm² of steel at a depth of 450, carrying 150 kNm after the tension side has cracked. The neutral axis is no longer at mid-depth: it has risen to 137.0 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 18.0 N/mm² at the top fibre and the steel carries 309 N/mm²; the resulting couple is 371 kN on a lever arm of 404 mm, which multiplies back to the 150 kNm applied. The uncracked section would have had 3422×10⁶ mm⁴ against the cracked 1139×10⁶ — a loss of 67% of the stiffness.

When half the section has given up

Bending theory puts the neutral axis through the centroid. That is a consequence, not a rule — and when the tension side cracks, the same reasoning moves the axis somewhere else entirely.

Two plates bolted together are twice as strong as one plate. Two plates made to act as one section are eight times as stiff. The difference is entirely in the joint, and the joint has a calculable spacing.

Loaded straight down, and moving sideways. An equal angle with a moment applied about the horizontal axis. Its principal axes lie at 45.0° to the drawn ones, so the neutral axis runs at -30.6° rather than horizontally, and the section moves 59% as far sideways as it moves down. The product of inertia that causes it is -1.066 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now.

Loaded straight down, and it moves sideways

Every section this collection has drawn had an axis of symmetry, and that symmetry has been doing silent work. Take it away and a vertical load produces a neutral axis that is not horizontal, a deflection that is not vertical, and on one ordinary section a sideways movement larger than the downward one.

The middle third

A material that cannot be pulled imposes a condition on where the load may land, and the condition is a region rather than a point. For a rectangle it is the famous middle third; for every other section it is a shape nobody quotes, and one ordinary section's is nearly twice as generous as the rule allows.

The flanges go opposite ways, and the pair of them is the bimoment. A 305 by 165 mm I-section held against warping and twisted by 0.5 kN·m, with the section on the left and the two flanges seen in plan on the right. At the built-in end each flange bends in its own plane, one way at the top and the other at the bottom, through 29.3 mm at the free end — drawn 20 times its true size against the 6 m length. The pair of flange shears is 1.69 kN each, and 1.69 × 295 mm is 0.500 kN·m — the whole torque at that section, carried by two forces neither of which is a torque. The pair of flange moments is 3.04 kN·m each, and 3.04 × 295 mm is 0.897 kN·m², which is the bimoment. It puts 67.0 N/mm² into two diagonally opposite flange tips and takes the same out of the other two, so its net force and its net moment about every axis are zero — which is exactly why no member diagram has a place for it.

The section that cannot stay flat

Twist an I-section and its cross-section dishes out of its own plane. Stop that happening at one end and the member finds a second way to resist — the flanges bend in opposite directions — and the stress resultant that describes it has units nothing else in statics has.

The flange that is drawn, and the strip of it that is working. A plan of a flange 3 m each side of the web on a 20 m span, with the working strip shaded. The effective width is 2.531 m per side, 84.4% of what is drawn, so 101238 mm² of the 120000 mm² of flange is doing the work and 18762 mm² is not. Widening the flange does not move the shaded strip much, because its width is set by the span: the design code would allow 2.500 m per side here and no elastic flange of any width beats 3.183 m, which is L/2π. The lag exists because longitudinal stress can only enter the flange through shear along the junction with the web, so the free edges arrive late — 76.5% of the web's stress at the edge, in this case.

The flange that is not all there

Stress can only get into a wide flange through shear along its junction with the web, and shear takes distance to do it. So the width that is working is set by the span, and a flange 3 m wide on a 20 m span has 18762 mm² of steel that is there, and paid for, and hardly carrying anything.

Cut the throat, and the face carries a moment and a tension at once. A crane hook of trapezoid section, 50 to 120 mm radius and 40 to 15 mm wide, carrying 50 kN, drawn beside the section at the cut and the stress across it. The load hangs on a line through the centre of curvature, so cutting the throat and taking everything below the cut as the free body leaves a face carrying a direct tension of 50 kN and a moment of N·R = 3.985 kN·m about the section's own centroid, which sits a full R = 79.70 mm from the load line. The stress is a hyperbola, zero at r = 75.04 mm rather than at the centroid 4.65 mm outside it, reaching 248.7 N/mm² of tension at the inner fibre and 140.6 of compression at the outer. The straight-beam formula, drawn dashed, reports 161.7 N/mm² for the bending part against the true 222.8, and leaves the 26.0 N/mm² of direct tension out altogether — between them, 54% under the real peak, at the fibre where a hook actually breaks. This is why a hook is trapezoidal: both effects are worst inside, so the material goes there.

The bar that was bent before it was loaded

In a curved bar plane sections still stay plane, and the bending formula is wrong anyway. The fibres were different lengths before anything was applied, so an equal rotation of two plane faces produces unequal strain — the stress is a hyperbola, the neutral axis has moved inward, and a crane hook carries half as much again as My/I reports.

One point, every plane through it, one circle. A point carrying 140 N/mm² across one face, 0 across the other and 45 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 83.2 centred at 70.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 153.2 and -13.2, on planes 16.4° from the face the 140 acts on; the largest shear on any plane is 83.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 160.2.

The worst stress is not where the worst bending is

Every stress this collection has quoted is a stress on a particular plane, and neither the bending stress nor the shear stress is a property of the point. Turn the plane and both change; one pair of numbers does not, and on a short beam it peaks where neither of them does.

The same strain, two moduli, and a width multiplied to say so. A timber section with a steel plate in it, carrying 20.0 kNm. Plane sections stay plane, so the strain at a height is the same in both materials; Hooke's law then puts the stresses in the ratio of the moduli, which here is 19.09. Multiplying the stiffer material's WIDTH by that ratio gives a fictitious section of one material with the same neutral axis and the same forces — 595.2×10⁶ mm⁴ of it, against 351.0 for the same shape with the moduli ignored. The steel plate is 3.8% of the area and carries 43% of the moment, at 96 N/mm² against the timber's 5.0. The transform is not an approximation: it is compatibility and Hooke's law written down.

A section made of two materials, one of them pretended away

Multiplying a material's width by the ratio of the moduli produces a fictitious section of one material with the right neutral axis and the right forces. It is not a trick — it is compatibility and Hooke's law written down — and it says a stiff material takes what its modulus asks for.

A hole in a web is a Vierendeel panel. A 400 × 300 rectangular opening in a 533 deep beam, 15% along a 9 m span carrying 20 per metre — where the moment is 103 kNm and the shear 63 kN. The moment is a couple on the two tees, 301 kN on a lever arm of 343 mm, which is 70.3 N/mm² of uniform stress. The shear has nowhere to go but through the tees, so each carries 32 kN over the opening and bends in double curvature: a Vierendeel moment of 6.3 kNm and 167.6 N/mm² on top. So 70% of the stress at the corner exists because the hole has a LENGTH, and only 30% of the section's second moment has gone.

The hole that costs nothing, and everything

A service opening removes 30% of a beam's second moment and 0.7% of its deflection. What it costs is not that. Across the opening the shear has nowhere to go but through the two tees, and a tee carrying shear over a length bends.

A section modulus for each face, and only the smaller one is a strength. Four profiles of equal area with the second moment divided by BOTH distances to an extreme fibre rather than by the larger of them. A symmetric section has one section modulus and an asymmetric one has two, differing here by as much as 1.00 to one — so the same member has two bending strengths, and which of them applies is decided by the sign of the moment rather than by anything about the section. The bar is the smaller of the two, which is the one that governs when the moment can go either way.

Two strengths, depending which way up

A symmetric section has one section modulus. A tee has two, differing by a factor of three, so the same member has two bending strengths and which applies is decided by the sign of the moment. Turn it over and it is a different beam.

The one length a section carries into a column. Four profiles of equal area, with the radius of gyration r = √(I/A) drawn as the distance it is — a pair of lines either side of the centroid, at the depth the whole area would have to sit at to give the section the second moment it has. As a 4 m pin-ended column the same 3000 mm² of material carries between 7 and 3146 kN, in the ratio of the squares of those radii and of nothing else.

The one length a section takes into a column

A section has an area, a second moment, two section moduli, a shear centre and a torsion constant. A column has heard of exactly one of them, and it is none of those — it is the length √(I/A), which is where the whole area would have to sit to give the section the stiffness it has.

Two curves climbing together, and the one that catches up first. A 6 m member tapering from 200 to 600 mm, with the moment it carries and the moment it can carry drawn on the same scale below it. The demand rises linearly and the capacity as the square of the depth, so the gap between them closes and then opens again. It is narrowest at 3.00 m from the free end, where the member is 400 mm deep and 79% used, against 70% at the root where the moment is largest.

The section that changes along the span

A prismatic beam is checked where the moment is largest, and everyone knows where that is. A tapered one is not, because the capacity is moving too — and for a cantilever with a load at its tip the governing station is exactly where the depth has doubled, with no length, no load and no material in the answer.

The strain it wants, the strain it is allowed, and the difference. A bridge deck 1.40 m deep with 18 °C at the top face falling away over 10% of the depth. The left curve is the free thermal strain αT(y); the straight line beside it is what a plane section will actually take, ε₀ + κy with ε₀ = 32.0 microstrain and κ = 0.063 per km. The right-hand block is E times the difference, and it reaches -3.98 N/mm² of compression at the surface and 1.83 of tension 140 mm below it. Its resultant force is 8.3e-14 kN and its resultant moment 3.0e-12 kNm, which is what self-equilibrating means: the field is invisible to every equilibrium check that could be made on the member.

The stress nobody restrained

A bridge deck lying loose on its bearings, with nothing holding it anywhere, develops four newtons per square millimetre when the sun comes out. The stress is not caused by restraint. It is caused by plane sections, and it is invisible to every equilibrium check that could be made on the member.

$Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.$

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

The load did not move; the section did. A lipped channel 200 by 65 mm at 2 mm thick, drawn twice on top of itself: the outline as fabricated, and the part of it still working once the plates have buckled. The web is held on both edges, so it loses its middle; the flanges are held at the web, so an unlipped one would lose its free edge. What survives is not symmetric with what was drawn, so the centroid moves 8.0 mm — and a load applied along the axis it was designed to arrives 8.0 mm off the section that has to carry it. At the 177 kN this section will take, that is 1.42 kNm of bending nobody applied.

What is left after it ripples

A thin plate that buckles locally has not failed. It has stopped taking load in its middle and gone on taking it near its edges, so the member is now made of a different section from the one that was drawn — and the new one has its centroid somewhere else, which turns a concentric load into an eccentric one.

Two faces, a couple, and a core that does none of it. A sandwich section 61.4 mm deep: two 0.7 mm faces separated by 60 mm of core. The bending is carried as a couple between the faces — 41 N/mm² of tension in one and compression in the other, over a lever arm of 60.7 mm — and the core carries a shear stress of 0.047 N/mm² and nothing else. The parallel-axis term is 98.8% of the section's second moment; the faces' own bending about their own centroids is 0.004% of it, and the core's is 1.2%. Separated by nothing at all the same two faces would be 2.3e+4 times less stiff.

Two skins and the space between them

A sandwich panel is a section made of a material that carries the bending and a material that carries none of it. The parallel-axis term is not a correction here — it is 98.8 per cent of the second moment — and the shear deflection is not a correction either.

Four inequalities and a wedge

A prestressed section has to satisfy two stress limits when the force is largest and the load smallest, and two more when the force has relaxed and the load has arrived. Each is linear in one over the force — which turns a search for a prestress into a region on a page, and turns an impossible section into an empty one.

An eccentric load is three load cases, and only two of them are checked. A line load of 40 N/mm at 1.5 m from the axis of a 3.0 by 2.0 m box, replaced by the three cases it is equivalent to. Bending is the load on the axis. The torque 60 kNm per metre then splits into a set of edge forces that drives Bredt's shear flow and distorts nothing, and a set with the flange forces reversed — 10.0 kN/m up one web and down the other, 15.0 kN/m across the flanges — which carries no torque at all and squashes the rectangle into the rhombus drawn behind it. Its generalised load is exactly half the torque, so a box girder spends half of an eccentric load's torsion on changing its own shape, and no torsion calculation contains that half.

The section that will not keep its shape

A box girder is closed, so torsion costs it almost nothing. What an eccentric load actually does to it is something a torsion calculation contains no term for — the rectangle becomes a parallelogram, in its own plane, along the whole length of the span.

Pull it along the girder and it just unfolds. One period of a 30° trapezoidal corrugation, 300 mm of flat and 260 mm of incline, and the same period pulled along the girder's axis. The fold opens by bending the inclined panels out of the web's own plane, so the axial flexibility contains the plate's t³ where a flat web's would contain t — and the effective modulus that comes back from solving the cell as a frame is 222 N/mm², which is 10.6 parts in ten thousand of the steel's 210 GPa. A web with a thousandth of the stiffness carries a thousandth of the stress, which is why the flanges of a corrugated girder carry the whole moment and why the section has 9 per cent less second moment than the flat-webbed girder it replaces. The fold buys freedom from stiffeners and pays for it here.

The web that carries no bending

A corrugated web needs no stiffeners, because the folds give it in one direction a depth it does not have in its thickness. In the other direction the same folds make it an accordion — and a web that cannot be stretched cannot carry a bending stress at all.

Two moments and a neutral axis that obeys neither

Tilt the load on a rolled beam by five degrees and the neutral axis swings by seventy. The section is doubly symmetric, its product of inertia is exactly zero, and none of that helps — because what decides the axis is the moment ratio multiplied by a stiffness ratio of thirty.

What is left after the first fibre yields

The elastic section modulus stops at the moment the outermost fibre reaches yield. Nothing else in the section has, so it goes on taking load — and how much more it takes turns out to be a property of the shape alone, with no dimension, no stress and no material anywhere in the answer.

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 slit that costs a factor of six hundred

Bending stiffness cares where the material is, and changes by a factor of two or three between sensible sections of the same area. Torsional stiffness cares whether the material forms a closed loop, and the penalty for not doing so is an order of magnitude squared.

Deliberately the wrong shape

Concrete in compression follows a curve, and no design office has ever integrated it. Every code in the world replaces it with a rectangle of reduced depth and reduced intensity, and the answer is right to a fraction of a per cent — not because the shapes are similar, which they visibly are not, but because a bending calculation only ever asks a stress distribution two questions.

Eight slices is enough, and nobody would have guessed it. The error in a cracked section's moment capacity against the number of strips it was integrated with, for a 300 × 450 mm section with 1200 mm² of steel, measured against the same computation at 2048 strips. The point of the fibre method is that it contains no formula: slice the section, give every strip the strain the assumed curvature puts it at, move the neutral axis until the axial force balances, and sum. It handles a cracked section, a confined one, a prestressed one and a composite one with the same twenty lines. The discretisation costs 1.4% at 2 strips and 0.088% at 8 — and the convergence is not smooth, because what the error actually depends on is where the neutral axis falls relative to a strip boundary rather than on the strip count as such.

The section calculation with no formula in it

Every ordinary section result is a closed form, and each was derived once for one arrangement of material. Slice the section instead, give each strip the strain a curvature puts it at, and move the neutral axis until the axial force balances — and the same twenty lines answer for a cracked section, a confined one, a prestressed one and a composite one, having been told nothing about any of them.

The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 6 kN/m of wet concrete and 9 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 146 MPa; propped, the finished composite section takes everything and reaches 93 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two.

The section that changed while it was being loaded

A stress is computed from a moment and a section modulus. When part of the moment arrived while the section was a different shape, there is no single section modulus to divide by — the stresses add and the properties do not, and two identical finished beams can differ by half again in stress with nothing on the drawing to say which is which.

The neutral axis is wherever the first moment vanishes. A 300 by 600 section with 1800 mm² of steel at a depth of 540, carrying 250 kNm after the tension side has cracked. The neutral axis is no longer at mid-depth: it has risen to 234.5 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 15.4 N/mm² at the top fibre and the steel carries 301 N/mm²; the resulting couple is 541 kN on a lever arm of 462 mm, which multiplies back to the 250 kNm applied. The uncracked section would have had 6673×10⁶ mm⁴ against the cracked 3809×10⁶ — a loss of 43% of the stiffness.

Where the steel is, not how much of it

A section in bending resists a moment with a couple, and a couple is a force times a distance. The force is bought — it is an area of steel at a stress. The distance is free, decided by where the bars were put, and it is the variable almost nobody optimises because it does not appear on an order.

The dimension the capacity rides on is not a drawn dimension. Moment capacity of a 225 mm slab against the cover to the reinforcement. The line is very nearly straight, because capacity is A_s·f_yd·z and z is about 0.9d — so the capacity is proportional to a dimension that is not on the drawing. What is on the drawing is the overall depth, and the effective depth is what the cover, the link and half a bar diameter leave of it: 225 − 30 − 0 − 8 = 187 mm here. Every one of those three is a site tolerance rather than a design decision. Ten millimetres of bar position is 5.7% of this slab's capacity and 1.8% of a 600 mm beam's — the same workmanship costs 3.0 times as much in the shallow member, and the shallow member is the one whose steel is walked on before the pour.

The dimension nobody can measure

Every flexural capacity in reinforced concrete is proportional to the effective depth, and the effective depth is not on the drawing. It is what is left of the thickness after a cover, a link and half a bar diameter have been taken off it — and each of those is a site tolerance. In a slab, ten millimetres of workmanship is six per cent of the strength.

The answer is continuous and the catalogue is not

Every optimisation in this subject returns a number with three significant figures in it, and nothing with three significant figures can be bought. What can be bought is a rolled series whose steps are a quarter to a half apart, so the member that goes on the drawing is on average a tenth stronger than the one that was calculated and can be a third stronger for no reason at all.

A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel.

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

The steel that is sized by the concrete. Ultimate moment of a 1000 × 400 mm section against the area of tension steel in it, with the moment that cracks the section drawn across. The cracking moment is 77.2 kNm and contains no steel at all — it is f_ctm times the gross section modulus, 2.90 N/mm² times bh²/6 — so it is a horizontal line, and every section to the left of where the two meet is one whose first crack is its failure. The crossing is at 507 mm², and rearranging the two expressions gives 0.245·(f_ctm/f_yk)·bd against the 0.26 the codes print — the constant is a section modulus divided by a lever arm and not a fitted number. The rule as printed asks for 538 mm² here, which is 6% more than the derivation needs, and that margin is the whole of the safety in a check whose failure mode is sudden.

The steel the concrete asks for

Every other bar in a concrete member is there because of an action. This one is there because of the member itself — enough steel that the cracked section can carry more than the moment that cracked it, so that the first crack is not also the failure. The requirement contains no load, and both of its consequences run the wrong way round.

The axis moves when the section yields. Six sections, each drawn to its own scale, with their elastic neutral axis — the centroid, dashed — and their plastic neutral axis, the equal-area axis, solid. For the symmetric ones the two lines are the same line and the distinction never arises, which is why it is so easily missed. For the tee they are 23% of the depth apart, because the axis that makes the first moment of area vanish is not the axis that makes the two areas equal. The shape factors run from 1.144 to 1.800 across these six, and they are ratios of moduli taken about two DIFFERENT axes — which is also why an asymmetric section has two elastic section moduli, one to each extreme fibre, and only one plastic modulus. The tee's two elastic moduli differ by a factor of 2.78; a fully plastic section does not care which fibre reached yield first, so it has nothing to be two of.

The axis that moves when the section yields

An elastic section bends about its centroid. A fully plastic one bends about the axis that halves its area, and for anything symmetric those are the same line — which is why the distinction is almost never met. For a tee they are a fifth of the depth apart, and three things follow that the elastic calculation gives no warning of.

The same steel, and a crack three times as wide. Calculated crack width against bar diameter, with the area of steel held at 1340 mm² per metre throughout — so the spacing changes with the square of the diameter and the amount of reinforcement does not change at all. The width runs from 0.173 mm at 8 mm bars to 0.372 at 25, a factor of 2.15 for identical steel. The reason is in the crack spacing: after a crack forms the bar has to re-anchor the concrete's tensile force before the next one can, and the length that takes is proportional to the bar's diameter. Of the 337 mm spacing drawn, 35% is the cover term and 65% is the bar term — and the cover term is the one that puts crack control and durability in opposition, because cover protects the bar and widens the crack that reaches it.

The same steel, and a wider crack

A crack's width is the distance between cracks times the strain the steel carries over that distance. Neither of those is decided by how much reinforcement there is. The spacing is a bond length, so it goes as the bar diameter; the strain is set by the stress in the steel. Two arrangements of identical steel can differ by a factor of two in crack width, and the one that wins is the one with more, smaller bars.

A soap film over a hole, and its volume is the torsion constant. The stress function for Saint-Venant torsion of a square of 100 mm, relaxed on a 97 by 97 grid until it stopped moving — 385 sweeps. Its contours are the lines the shear stress runs along, its slope is the magnitude of that stress, and twice its volume is the torsion constant: 1.4053e+7 mm⁴ against a closed form of 1.4058e+7, an error of -0.035 per cent. The steepest slope is 67.515 at the boundary, at the point on it nearest the centre — which is why the peak shear in a solid section is at the middle of the longest side and never at a corner, where the film comes down to zero from two directions and its slope vanishes.

Two volumes, and both of them are torques

Torsion of a solid section that is not a circle has no elementary answer, and for thirty years after Saint-Venant posed it the only way to get one was to blow a soap film over a hole cut in a plate and measure it. The film is not an illustration of the solution. It is the solution, and so is a heap of sand poured on the same hole.

The web down the middle carries no torsion at all. A 2-cell box 3.0 m wide and 1.5 m deep under 4000 kNm of torque. Statics gives one equation, T = 2 Σ q_i A_i, and there are 2 unknown flows — so the section is torsionally redundant and the missing statements are that every cell twists by the same amount. Solving that system gives 444.4 and 444.4 N/mm in the cells, so the internal web carries 0.00 N/mm — 0.00 per cent of the outer wall's. J is 2.3679e+12 mm⁴ against 2.3679e+12 for the same outline with no internal web at all, a ratio of 1.000000. The same walls slit open would give 7.413e+10, so closure is worth 32 times and the internal web is worth what the ratio says.

Two cells, one equation, and a web with nothing in it

Bredt's formula answers a single closed cell because a single closed cell has one unknown and one equation. Put a web down the middle and there are two unknowns and still one equation — and the answer, when the missing statement is supplied, is that the new web carries exactly nothing.

Where a bar may stop

The moment diagram falls away from midspan, so the steel midspan needs is not needed everywhere, and curtailing it saves real money. Then two things get in the way, and between them they take nine tenths of what the moment diagram promised.

One plate, three systems, one stress. The stress at the rib-to-deck weld of an orthotropic deck, split by which system produced it. The main girder contributes 120 N/mm², the crossbeam 34 and the trough 13, and they add to 167 because the deck plate is the top flange of all three at once. A calculation that treats the crossbeam as a support rather than a structure understates the total by 26 per cent. The stress RANGE at that weld is 47 N/mm², which is the number the whole deck is designed by, because the weld is a low fatigue category running the entire length of the bridge.

One plate and three structures

An orthotropic deck is a single steel plate stiffened by troughs, sitting on crossbeams, sitting on main girders. Nothing about that is unusual until it is noticed that the plate is the top flange of all three, so a wheel standing on it loads every one of them at once.

The second moment of area is a function of direction. Second moment of area of an equal angle against the angle of the axis it is taken about, with the product of inertia beneath it. The maximum is 5.943 × 10⁶ mm⁴ and the minimum 1.523 × 10⁶, a ratio of 3.90, and they occur where the product of inertia passes through zero — at 45.0° from the drawn axis. The value the drawing suggests, 3.733 × 10⁶, is neither of them.

The axis a column buckles about

A strut buckles about the axis with the smallest second moment of area, and for a section with no axis of symmetry that axis is neither of the two on the drawing. An angle used as a strut is 2.45 times weaker than the number a designer reads off its own dimensions.

The shear centre of a channel. A channel of 100 by 250, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 38.0 outside the web to leave the section untwisted — a point in the air, outside the material entirely.

The eccentricity a purlin cannot avoid

A channel's shear centre is outside the material, so a load applied anywhere on the section misses it. The distance is fixed by the proportions rather than by the detailing, it is 38 mm on an ordinary purlin, and the torque it produces is not an error anybody made.

$The lag follows the shear, so it is worst at the supports. Effective width along the span of a simply supported beam under a uniform load, summed over 25 odd harmonics. Each harmonic has its own half-wavelength L/n and its own, smaller, effective width — 0.815 for the first, 0.400 for the third, 0.269 for the fifth — and near a support the short harmonics carry a bigger share of what little moment there is. So the working fraction is 0.603 at the support against 0.839 at mid-span, a difference of 23.6 percentage points on the same flange. The single sinusoid's answer, 0.815, is drawn as the flat line, and it is only right at mid-span. This is the behaviour a code reproduces by shortening L_e near a support, and the reason it does is that shear lag is driven by w‴, which is the shear force.$

The flange works least where the shear is largest

Shear lag is driven by the shear force rather than by the moment, so the effective width of a wide flange is not a property of the beam. It is a function of position along it, worst at the supports, and a single number quoted for a whole span is right at mid-span and nowhere else.

One point, every plane through it, one circle. A point carrying 180 N/mm² across one face, 90 across the other and 40 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 60.2 centred at 135.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 195.2 and 74.8, on planes 20.8° from the face the 180 acts on; the largest shear on any plane is 60.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 170.6.

The circle nobody draws

A plane stress state has three principal stresses and the third is zero. When the two on the drawing share a sign, the largest shear in the state involves the one that is not there — and the circle a designer has drawn is not the circle that governs.

A short timber beam is a shear problem, and a steel one never is. Utilisation of the bending and shear checks on one beam, against span-to-depth. The two cross where the ratio equals f_m ÷ f_v exactly — no load, no width and no span survives the cancellation — which for this timber is 6.7 and for steel is 1.73. So a timber beam shallower than about six times its depth is governed by shear parallel to the grain, and a steel beam would have to be shorter than twice its own depth before the same thing happened, which is not a beam. The third check is bearing across the grain, which does not move with the span at all: on the beam drawn it is at 0.40, and it is the one that governs.

The shear that decides a timber beam

A steel beam is never governed by shear, because its bending strength is only 1.73 times its shear strength and no beam is that short. Timber's ratio is 6.7 along the grain and 23 across it, so shear governs at proportions people build every day.

Cut the throat, and the face carries a moment and a tension at once. A crane hook of trapezoid section, 50 to 120 mm radius and 40 to 15 mm wide, carrying 100 kN, drawn beside the section at the cut and the stress across it. The load hangs on a line through the centre of curvature, so cutting the throat and taking everything below the cut as the free body leaves a face carrying a direct tension of 100 kN and a moment of N·R = 7.970 kN·m about the section's own centroid, which sits a full R = 79.70 mm from the load line. The stress is a hyperbola, zero at r = 75.04 mm rather than at the centroid 4.65 mm outside it, reaching 497.5 N/mm² of tension at the inner fibre and 281.3 of compression at the outer. The straight-beam formula, drawn dashed, reports 323.4 N/mm² for the bending part against the true 445.5, and leaves the 51.9 N/mm² of direct tension out altogether — between them, 54% under the real peak, at the fibre where a hook actually breaks. This is why a hook is trapezoidal: both effects are worst inside, so the material goes there.

The wide side goes inside

A crane hook's section is a trapezoid with its broad face towards the centre of curvature, and that is not a casting convenience. Turn the same section round — same area, same depth, same moment — and the stress at the fibre that breaks rises by thirty-nine per cent. The shape is doing two things at once, and only one of them is in a straight beam's arithmetic.

The distortion runs the length of the span, and a diaphragm stops it. Longitudinal stress at a corner of the box from distortional warping, along a 60 m span carrying 45 N/mm at 2.0 m off the axis. With no interior diaphragm it peaks at 73 N/mm², which is 83 per cent of the bending stress the girder was designed for. Two interior diaphragms take it to 18. The governing length is Winkler's: the distortion decays over 23.6 m, so a diaphragm helps its neighbours only if it is closer than that, and past it the spacing stops mattering.

One diaphragm is nearly none

A box girder's distortion decays over a length the section decides, and on a sixty-metre span that length is twenty-four metres. So a single diaphragm at midspan sits further from each end than the distortion can reach and removes a third of the problem; three diaphragms remove nine tenths. The spacing rule is not span over five — it is a property of the plates.

The four corners of an opening, on the tee's own interaction diagram. The plastic interaction of the tee left above and below a 400 × 300 mm opening — every point on it computed by sweeping the plastic neutral axis through the section rather than from an interaction formula. Its squash load is 1519 kN and its plastic moment 32.2 kNm. At the working load the global moment puts 301 kN into each tee — compression above the hole, tension below — and the shear puts the same Vierendeel moment of 6.3 kNm into all four corners. The line is the path the demand takes as the load rises, and it reaches the surface at a load factor of 3.89 against 1.49 for first yield at one corner: the elastic check is finding one corner and the mechanism needs all four.

Four corners and a mechanism

The check on a hole in a web adds an axial stress to a local bending stress and compares the sum with the yield stress at one corner. What ends the opening is four hinges arriving together — and the gap between the two is a factor that varies along the span from two and a half to exactly one.

Two webs of one box, carrying different amounts of the same load. The equivalent stress on each web of a 3.0 × 2.0 m box along the half span, under 40 N/mm at an eccentricity of 1500 mm, with the bending stress alone drawn for comparison. The two webs carry the same bending and opposite torsion, so one gets the sum and the other the difference: at the quarter point the loaded web is at 116 N/mm² and the far web at 11, a ratio of 10.9. A check made on the section rather than on the wall reports the average of two numbers that differ by that much.

Three actions on one web

A load applied off the centreline of a box girder is three actions at once, and every textbook decomposes it into them. What the decomposition does not say is that all three land on the same piece of plate — added on one web and subtracted on the other, so the two walls of one section differ by a factor of ten.

The worst section of a haunched rafter is inside the haunch. Utilisation along a 15.3 m portal rafter carrying 8 kN/m, with an eaves moment of 500 kNm and an apex moment of 150, haunched over 3 m from 906 mm deep down to the rafter's own 453. The moment is largest at the eaves and the depth is largest there too, so the eaves is at 0.42; the apex is at 0.31. The peak is 0.46 at 2.98 m — the haunch tip, where the section has just become the bare rafter and the moment is still 226 kNm. The dashed curve is the same rafter with no haunch, which reaches 1.02 and does not pass.

The section that governs is inside the haunch

A tapered cantilever has its worst section somewhere along it because the moment grows linearly and the modulus quadratically. A haunched rafter has the same competition with a step in it, and the step is where the check lands — at neither end of the member, at a station no formula names.

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