What is refuted here — page 13

More reinforcement means less cracking.

What decides it: It means less crack *width* and exactly proportionately more cracks. The total opening across the restrained length is the crack-inducing strain times that length — 3.30 mm over the 20 m wall drawn — and no bar diameter, spacing or area appears in it. At 8 mm bars the wall makes 52.7 cracks of 0.063 mm; at 25 mm bars it makes 22.3 of 0.148. The product is 3.30 mm in both cases and at every point on the curve.

Tested in The steel decides how many, not how much, at the figure it turns on · the restraint cracking ladder.

Doubling the prestress doubles the shear the web can take.

What decides it: It is a square root of a product, so it flattens. Going from zero to 7.14 N/mm² multiplies the limit by 2.51; going from 7.14 to 14.29 multiplies it by only a further 1.36, reaching 4.59. The straight line a rule that added the two strengths would draw is above the curve everywhere past the origin.

Tested in The crack that never reached forty-five degrees, at the figure it turns on · the prestressed shear ladder.

A square is a good torsional section because it is compact.

What decides it: Compactness is the right variable and the square is not the best value of it. A circle of the same area gives J = 1.5914e7 mm⁴ against the square's 1.4053e7 — 13.2 per cent more stiffness for the same material, and it is the only solid section that does not warp at all. Every corner a section has is material sitting where the film is flat.

Tested in Two volumes, and both of them are torques, at the figure it turns on · the membrane analogy ladder.

If the web is not helping in the middle, put it somewhere it does help.

What decides it: It does help off-centre, and by very little. Moved to a tenth of the box width the web carries 141 N/mm and raises J by 3.3 per cent, which is the largest value anywhere in the range. The same walls slit open along one corner give J thirty-two times smaller, so closure is worth about a thousand times what the web's position is.

Tested in Two cells, one equation, and a web with nothing in it, at the figure it turns on · the multicell torsion ladder.

Hanging a building saves material, because tension is more efficient than compression.

What decides it: It saves exactly the average buckling reduction over the height and not one per cent more. For the 20-storey frame drawn, the hangers total 0.412 m³ of steel against 0.453 for the columns — a saving of 9.0 per cent, and the average χ is 0.9096. The two numbers are the same number, because the loads accumulate the same way in both structures and only χ differs.

Tested in Hung from the top, and nine per cent lighter, at the figure it turns on · the hung structure ladder.

A stressed ribbon is a shallow suspension bridge, so its forces are of the same kind.

What decides it: Of the same kind and not of the same size. H = wL²/8f is a reciprocal with no flat part, so the sag ratio is the whole answer: at a tenth of the span the horizontal force is 1.25 times the deck's weight and at a fiftieth it is 6.25. The ribbon is five times the anchorage for the same span and the same deck, and the difference is one design decision that was not made for structural reasons at all.

Tested in The deck that is its own cable, at the figure it turns on · the stressed ribbon ladder.

Stiffeners are added until the plate is strong enough, so more is straightforwardly better.

What decides it: The gain goes as (n+1)² — 4, 9, 16, 25 — and the rigidity each stiffener must have goes as roughly (n+1)⁴: γ* runs 32, 231, 924, 2,738. So the outstand needed runs 144, 280, 444, 638 mm and the steel runs 6, 23, 55, 106 per cent of the plate's own area. At four stiffeners there is more steel in the ribs than in the plate they are stiffening.

Tested in The rib that is a boundary condition, at the figure it turns on · the stiffener rigidity ladder.

A tube's bending capacity is its section modulus times a stress, so it is proportional to the wall thickness.

What decides it: Only while a stress governs. The ovalisation limit is M = (2√2/9)πErt²/√(1−ν²), which goes as t squared — because thinning the wall both weakens the section and makes it flatten more readily, and the two effects multiply. Over the r/t range drawn the yield moment falls by a factor of five and the Brazier moment by twenty-five.

Tested in The tube that flattens itself, at the figure it turns on · the brazier buckling ladder.

A factor of safety of 1.2 means the slope has 20 per cent in hand.

What decides it: In hand against what is not stated. The number is a ratio of moments, and it divides the strength rather than multiplying the load — so it is 20 per cent on the *strength*, of which 75 per cent is friction that will still be there and 25 per cent is cohesion that may not be. Raise the pore pressure ratio from 0.25 to 0.40 and F falls from 1.650 to 1.358, with no load applied by anybody.

Tested in The surface that has to be searched for, at the figure it turns on · the slope stability ladder.

The vertical component of pressure on a curved surface is the weight of the water actually resting on it.

What decides it: It is the weight of the water that would stand above it, real or imagined. For a surface curving the other way — the underside of an arch dam's face, or a sluice with water beneath it — the column above the surface contains no water and the vertical component points upward with exactly that magnitude. The free body is a column of fluid, and it does not require the fluid to be there.

Tested in Every pressure points at the pin, at the figure it turns on · the curved surface pressure ladder.

Second-order effects break superposition, which is why the amplification factor exists.

What decides it: The amplification factor itself is a linear operation, so a check made with it superposes perfectly and proves nothing. What actually breaks is that the exact amplification depends on the deflected shape and the two cases have different shapes — 2.5925 for the point load against 2.6208 for the uniform one, an error of 0.32 per cent. It is the smallest of the three failures and the only one anybody names.

Tested in The addition everything else rests on, at the figure it turns on · the superposition ladder.

Torsion in a beam is a strength problem, and the stresses are small.

What decides it: The stresses are usually small and the movement is not. A 400 mm open section on an 8 m span carrying 12 kN/m at 75 mm off its shear centre reaches 3.40° at mid-span, which moves the flange tip 11.9 mm sideways — 82 per cent of the member's own vertical sag of 14.5 mm, and 37 per cent of the span/250 that sag is checked against. Nothing in the calculation approaches a stress limit.

Tested in The movement with no limit against it, at the figure it turns on · the twist serviceability ladder.

Steel and concrete have the same coefficient of thermal expansion, which is why reinforced concrete works.

What decides it: They do not. Steel is about 12 × 10⁻⁶ and concrete about 10 — a mismatch of 17 per cent, and 80 microstrain over a 40 degree change. What is true is the comparison rather than the equality: glass against aluminium is 560 microstrain over the same range, aluminium against concrete 520, and steel against concrete is the smallest of seven pairs by a factor of five.

Tested in The coincidence reinforced concrete stands on, at the figure it turns on · the thermal compatibility ladder.

The bolt in a single lap is in shear.

What decides it: It is in shear and in bending, because its two shear planes are offset by the plate thickness. For the M20 drawn the shear stress is 191 N/mm² and the bending stress 382 — twice as much — and the combined utilisation is 1.03 rather than the 0.61 the shear alone gives.

Tested in The joint that is crooked by construction, at the figure it turns on · the single lap ladder.

The separation between two buildings must be the sum of their maximum displacements.

What decides it: Only if the two peaks arrive at the same instant, and they generally do not. The correct combination is √(u₁² + u₂² − 2ρu₁u₂), with ρ the cross-correlation coefficient of the two responses. For the pair drawn — 0.8 s against 1.4 s, 120 mm against 220 — the sum is 340 mm and the correct answer is 248, which is 27 per cent less.

Tested in The gap between two buildings, at the figure it turns on · the pounding ladder.

If they do collide, the heavier building wins.

What decides it: The heavier building takes the smaller velocity change and the collision is symmetric in impulse. For the 500 and 300 tonne pair drawn the impulse is 597 kN·s and the velocity changes are 1.19 and 1.99 m/s — in the ratio of the masses, reversed. The lighter structure absorbs the same impulse into less mass, which is why the damage in recorded poundings is overwhelmingly on the lighter and more flexible of the two.

Tested in The gap between two buildings, at the figure it turns on · the pounding ladder.

Cable sag makes the frequency calculation unreliable.

What decides it: For a stay it does not: λ² is 0.13 against the 4π² = 39.5 at which the first symmetric and antisymmetric modes cross, so the stay is far into the taut-string regime and the sag correction is below a thousandth. For a suspension bridge's main cable λ² is large, the mode order changes, and the same measurement returns whichever of two nearly-coincident modes it happened to excite.

Tested in The force read off a frequency, at the figure it turns on · the cable dynamics ladder.

The rule is about area, so a long-span floor gets more of a reduction than a short-span one.

What decides it: Only through the area it collects, which is the point. Two beams of different spans carrying the same tributary area get the same factor. What the area rule actually says is stranger: past the point where it bites, the design load is linear in area with a slope of exactly (5/7)ψ₀ — a half for an office — so every additional square metre of floor is worth half a square metre to the column, for ever.

Tested in The load that is never all there at once, at the figure it turns on · the live load reduction ladder.

A truss carries its load axially, so a joint transmits forces and not moments.

What decides it: True only if the working lines meet. Offset the diagonal's centroidal axis by 60 mm at a node carrying 260 kN and the resultant has a moment of 15.6 kNm about it, in equilibrium with nothing until the members take it. The chord picks up 34 N/mm² of bending against the 139 N/mm² of axial stress it already had — a quarter as much again, from geometry alone.

Tested in The joint that is not where it was drawn, at the figure it turns on · the noding eccentricity ladder.

The fix is to detail the joint so the lines meet, which costs nothing.

What decides it: Sometimes. Bolt gauge lines are fixed by the section, a diagonal too shallow to reach the chord's centroid cannot be moved there, and forcing the lines to meet often enlarges the gusset — which introduces a different eccentricity, out of plane. What is free is knowing which of those has been chosen.

Tested in The joint that is not where it was drawn, at the figure it turns on · the noding eccentricity ladder.

The way to control it is to weld more carefully.

What decides it: The force is set by the heat input and cannot be avoided; only its line of action can be moved. Welding symmetrically about the centroid puts the resultant on the neutral axis and the bow goes to zero exactly — same heat, same force, no moment. The remedy is geometry, and it is decided when the section is drawn rather than when it is welded.

Tested in The shape that came out of the shop, at the figure it turns on · the weld distortion ladder.

Curtailment is worth doing on any beam.

What decides it: On the 9 m beam drawn the geometry offers a 20 per cent saving in steel and the shift and the anchorage take it to 2.2. The saving scales with span and the tail does not, so curtailment is a decision that turns positive somewhere around ten to twelve metres and is a waste of drawing office time below it.

Tested in Where a bar may stop, at the figure it turns on · the curtailment ladder.

A stiffer crossbeam always helps, because it reduces the local deflection.

What decides it: It reduces System II's own stress and increases the share of the wheel that reaches it, because load goes to stiffness. The deck panel is a grillage rather than a hierarchy, and stiffening one member of a grillage moves the load onto it. Which of the two effects wins depends on the spacings and is not obvious from either member alone.

Tested in One plate and three structures, at the figure it turns on · the orthotropic deck ladder.

The optimum is the best place to design.

What decides it: It is the highest capacity and the worst configuration. Two modes at the same critical load is exactly the coincidence Koiter showed is most imperfection-sensitive, so the design with the largest theoretical capacity is the one whose real capacity depends most on how well it was made. Optimisation walks into it by construction, because the maximum of a minimum of a rising and a falling curve is always their intersection.

Tested in The best design is the most sensitive one, at the figure it turns on · the wall optimum ladder.

A shallower arch is a lighter structure, because the arch is doing more of the work.

What decides it: A shallower arch has a larger thrust — H = wL²/8f, so halving the rise doubles it — and it also feels every correction more, because the axial term is compared against the bending term through (i/f)². The same 60 m span drawn at a 1.2 m rise instead of 6 carries 22,500 kN against 4,500 — and at that two per cent rise the shortening removes 2.6 per cent of the thrust and leaves 704 kNm of crown moment where the funicular solution had none.

Tested in The arch that gets shorter, at the figure it turns on · the rib shortening ladder.

Temperature and shrinkage are separate load cases, checked afterwards.

What decides it: They enter the same flexibility equation with the same denominator, so the same arch that is sensitive to shortening is sensitive to both — and by comparable amounts. On the arch drawn, rib shortening leaves 27.9 kNm at the crown, a 20 °C rise leaves 25.9, and shrinkage leaves more than either. The three are one calculation.

Tested in The arch that gets shorter, at the figure it turns on · the rib shortening ladder.

Warm weather is good for concrete, because it gains strength faster.

What decides it: It gains faster and finishes lower. A day at 35 °C is worth 1.95 days at 20, so the hot pour is 43 per cent stronger at twelve hours — and 9.5 per cent weaker at twenty-eight days, because hydration products formed quickly are badly distributed and the pore structure never recovers. The two curves cross.

Tested in The strength it had on the day, at the figure it turns on · the maturity ladder.

The twenty-eight-day strength is what the structure is designed for, so it is the number that matters.

What decides it: It is the number the design uses and it is never the number a construction load meets. The strength at striking, at transfer and at the next pour are three different values, all lower, all decided by the programme rather than the mix — and the temperature history that produced them is the one variable nobody records.

Tested in The strength it had on the day, at the figure it turns on · the maturity ladder.

A filled tube is a composite member: the two materials add, and that is the whole of it.

What decides it: Adding them describes the ordinary case and misses what makes the type worth having. At d/t = 80 an empty tube buckles locally at 286 N/mm² — below its own yield of 355 — and a filled one reaches 508, because the wall can only go outward. The two components change each other's failure modes, which no summation contains.

Tested in Each one stops the other failing, at the figure it turns on · the filled tube ladder.

The steel tube confines the concrete, so the section is worth more than its parts.

What decides it: Only while it is stocky. Confinement needs the concrete to dilate, dilation needs strain, and a slender column buckles before it gets there — the bonus is 11 per cent at a relative slenderness of 0.18 and is gone by 0.5. The column drawn, at 0.42, has none of it.

Tested in Each one stops the other failing, at the figure it turns on · the filled tube ladder.

The ductility demand is the target displacement divided by the yield displacement.

What decides it: Which yield? First yield on the frame drawn is at 28 mm and the equal-area bilinear yield is at 70. The same target of 219 mm gives a ductility of 7.7 against the first and 3.1 against the second. The number quoted against first yield means nothing — first yield is one storey, early, on its own.

Tested in Pushed over until it will not stand, at the figure it turns on · the pushover ladder.

A brace's compression capacity is its buckling load, and that is the whole of the difference from its tension capacity.

What decides it: That is the first cycle. The buckled shape leaves a plastic hinge at midspan that does not straighten, so every subsequent cycle starts from a bent member and buckles lower: from 54 per cent of the tension capacity at the first cycle down to 12 per cent of that by the sixth. The asymmetry is not a property, it is a trend.

Tested in The brace that yields both ways, at the figure it turns on · the Buckling-restrained ladder.

A stiff item is safe, because a rigid body simply follows its support.

What decides it: It follows the FLOOR, whose peak acceleration is 8.01 m/s² against the ground's 3.50 — a factor of 2.29. Rigidity removes the resonance and not the amplification, and the amplification is the part that is present at every period.

Tested in The spectrum a floor hands on, at the figure it turns on · the floor spectrum ladder.

A flexible mounting protects equipment, so softer is always better.

What decides it: Only if it lands on the right side of the peak. Softening a component whose period is below the building's moves it TOWARD the building's period, and the ratio it sees can rise by a factor of three on the way. Detuning has a direction, and the safe direction is decided by which side of the tooth the component starts on.

Tested in The spectrum a floor hands on, at the figure it turns on · the floor spectrum ladder.

An integral bridge can be any length, since the pressure is bounded by the passive value.

What decides it: Bounded and not reached. The movement is proportional to length and the pressure to roughly its 0.6 power, so the force grows steadily: 201 N/mm of wall at 10 m, 332 at 65 m, 418 at 120 m. Nothing stops it before passive, and passive is 4.2 against the 0.9 a 60 m deck reaches — so the limit is set by what the abutment can carry, not by a ceiling.

Tested in The summer that is worse than the last, at the figure it turns on · the integral bridge ladder.

A tied arch carries its load by arch action, so its chords are axially loaded members.

What decides it: Under a load covering the whole span, yes — the thrust comes out at wL²/8f and the frame model reproduces it to 0.02 per cent. Under a load on half the span the vertical-hanger arrangement carries 3,316 kNm in the tie and 6,234 in the arch, because a two-chord system with vertical webs is a Vierendeel frame and has no truss action.

Tested in Cross the hangers and the bending goes, at the figure it turns on · the network arch ladder.

More hangers is always better, since the load is spread more finely.

What decides it: Only while they stay in tension. A hanger can only pull, and under the half-span load five of the sixteen inclined hangers come out in compression — which for a hanger means slack, and a slack hanger takes its triangle out of the truss. The design question is how many are still working, not how hard any one of them is pulling.

Tested in Cross the hangers and the bending goes, at the figure it turns on · the network arch ladder.

The independence of area is an empirical approximation with no mechanism behind it.

What decides it: It has a mechanism and the mechanism is what predicts it. Two rough surfaces touch only at their high points, those points flatten plastically until the true contact area reaches N over the hardness, and the shear strength of the junctions then gives F equal to N times the ratio of two material properties. The apparent area cancels because it was never carrying anything.

Tested in The area that is not in the equation, at the figure it turns on · the friction ladder.

A slender body is more at risk of overturning than a squat one, so slenderness is the thing to check.

What decides it: Slenderness decides which failure arrives, not how close it is. The 1.6 m wide hoarding overturns at a factor of 4.35 and slides at 6.11 when it is 4.5 m tall, and the two swap below 3.2 m — where the same body, squatter, slides before it tips.

Tested in Whether it tips or slides, at the figure it turns on · the overturning ladder.

Slip is a serviceability matter, so a slipping joint is a joint that has failed its purpose.

What decides it: That is true of a connection whose purpose is to not move, and false of a device whose purpose is to move. The loop enclosed by a slipping device — 52.83 kJ over the record drawn — is energy removed from the structure, and a device that never slipped would have removed none of it.

Tested in The force that is capped on purpose, at the figure it turns on · the friction ladder.

Drawing a polygon for every joint is the same work as writing two equations for every joint.

What decides it: It is the same work only if the polygons are drawn separately. Laid on one sheet they share their edges: a member separates the same two spaces wherever it is read, so its force is one line in the whole figure. A 21-member truss produces twenty-one lines rather than six polygons of four sides each.

Tested in One drawing solves the whole truss, at the figure it turns on · the truss ladder.

A subdivided truss has more members, so it weighs more.

What decides it: It has more members and each of them is smaller, and the second effect is the larger one over the range that matters. The governing compression member's required second moment of area — proportional to force times length squared — falls from 2,895 to 540 in the same units between four panels and twelve, a factor of 5.4, against a member count that rises from 13 to 45.

Tested in Halving the panel buys a shorter strut, at the figure it turns on · the truss ladder.

A truss carries load by axial forces alone, so its members carry no bending.

What decides it: True of the idealisation and false of the steel. The top chord of the eight-panel truss carries −605 kN axially and, under a deck delivering 25 kN/m between its nodes, a hogging moment of 22.5 kNm as well. Neither number appears in the other calculation, and the section has to carry both at once.

Tested in The chord is a continuous beam, at the figure it turns on · the truss ladder.

Reversal is a problem for slender members only, so a stocky diagonal removes the need for a counter.

What decides it: A stocky diagonal removes the need for a counter and pays for it in the whole panel. The reversed force is the same size as the forward one — ±0.56 against ±0.44 of a unit load here — so the member has to be designed as a strut over its full diagonal length, which is the cost the crossed pair exists to avoid.

Tested in Two diagonals, one of which is absent, at the figure it turns on · the truss ladder.

A strut-and-tie model is checked by verifying the struts, since concrete crushing is the failure the model is about.

What decides it: The struts are usually the easiest part. In the model drawn the strut carries 1921 kN spread over 1031 mm of width, which is 3.7 N/mm² against a limit of 18.1 — a fifth of the available strength. The node under the same bearing is at 16.0 N/mm² before the load has spread, which is where the model actually gets close to its limit.

Tested in The node is the part that is checked, at the figure it turns on · the Strut-and-tie ladder.

Flattening the truss is free, since it takes more shear from the same stirrups.

What decides it: It is free until the web crushes. The strut stress is V(cot θ + tan θ)/b_w z, least at 45° and rising either side, so at 900 kN the two curves cross at cot θ = 2.10 and 1,031 kN is the most the section will carry however it is reinforced. The angle is free inside a window the concrete sets.

Tested in The angle is a choice, not a property, at the figure it turns on · the concrete shear ladder.

Getting the soil modulus right is the most important part of a lateral pile design.

What decides it: The characteristic length is the fifth root of EI over the modulus gradient, so a hundredfold range of ground — 0.001 to 0.1 N/mm³, which is everything from soft clay to dense gravel — moves T by a factor of 2.51. The section's own stiffness enters through the same fifth root and the designer chooses it.

Tested in The pile that is too short to bend, at the figure it turns on · the lateral pile ladder.

The section for a curved member is chosen by strength, so an open section of adequate capacity is adequate.

What decides it: A 600 × 900 box of 16 mm wall has a torsion constant of 5.81 × 10⁹ mm⁴ closed and 4.01 × 10⁶ slit — a ratio of 1,449. Under the same 329 kNm the closed section twists 0.378° over the arc and the slit one 547°, which is not a stress check but an admission that an open section cannot be used here at all.

Tested in The torque that has nowhere to go, at the figure it turns on · the curved in plan ladder.

A shear-friction check is a strength check like any other, so a joint that passes it is fine in service.

What decides it: The mechanism does not exist until something has moved. At zero slip an already-cracked plane carries nothing at all; the bars reach yield at 0.55 mm of slip and the resistance reaches 5.50 N/mm² there. It is an ultimate-limit-state mechanism whose serviceability behaviour is a displacement, and no check anywhere limits it.

Tested in Two models of one bracket, at the figure it turns on · the shear friction ladder.

A 30 per cent redistribution is twice as demanding as a 15 per cent one.

What decides it: The rotation demanded is twice as large — 13.0 milliradians against 6.5 — and the section class needed is not a linear function of it. What is exactly linear is the saving: the design moment falls from 304 kNm to 258 at 15 per cent and to 213 at 30, in exact proportion to β, because statics rather than stiffness sets the redistributed diagram.

Tested in The moment that was shed has to land, at the figure it turns on · the moment redistribution ladder.

A larger eccentricity always gives a larger prestress effect.

What decides it: True of the drape and false of the support ordinate. The upward load a parabolic tendon applies is 8Pd/L² with d the drape measured from the chord joining the support ordinates — 17.63 kN/m here — and moving both ends of that chord together changes it by nothing at all.

Tested in The tendon that can be moved, at the figure it turns on · the secondary prestress ladder.

Uniform stud spacing is an approximation that a careful designer would improve on.

What decides it: It is a plastic redistribution rather than an approximation, and it is only available because the connector is ductile. The end studs are asked for 1.81 times the mean elastically; a stud that can slip several millimetres at constant load sheds the excess to its neighbours, and the uniform spacing then describes the state the beam reaches rather than the one it starts in.

Tested in The connection is busiest where the beam is not, at the figure it turns on · the composite action ladder.

The eccentricity is a small dimension and therefore a small effect.

What decides it: It is 38 mm against a 250 mm depth, and the section it is applied to has a torsion constant of 1.14 × 10⁵ mm⁴ against a second moment of 32.9 × 10⁶. Under 2.5 kNm the slit shape twists 93° over 6 m where the closed one twists 0.45 — a factor of 208 that turns a small eccentricity into the governing action.

Tested in The eccentricity a purlin cannot avoid, at the figure it turns on · the shear centre ladder.

Widening a flange widens the part of it that works.

What decides it: Not in proportion. Going from b/L = 0.050 to 0.199 — four times the flange — takes the working fraction from 0.979 to 0.759, so the working width grows by 3.06 rather than by 4. Past b/L = 1/2π the exact ceiling of L/2π per side binds and further flange buys nothing at all.

Tested in The flange works least where the shear is largest, at the figure it turns on · the effective width ladder.

The code's L/8 is a conservative approximation to the elastic answer.

What decides it: It is conservative and it is not an approximation to anything with 8 in it. The exact elastic ceiling is L/2π = 0.1592L; the code caps at L/8 = 0.125L, and the ratio is 2π/8 = 0.7854 exactly. A rule derived from the elasticity would carry 2π, and the rule in use carries a round number 21.5 per cent below it.

Tested in The flange works least where the shear is largest, at the figure it turns on · the effective width ladder.

The largest shear stress at a point is the radius of its Mohr's circle.

What decides it: Of that circle, yes. A point at 180 and 90 N/mm² with 40 of shear has principal stresses of 195.2 and 74.8 and a drawn radius of 60.2 — and a third principal stress of zero, so the largest shear on any plane through it is (195.2 − 0)/2 = 97.6, which is 1.62 times the radius that was drawn.

Tested in The circle nobody draws, at the figure it turns on · the principal stress ladder.

Von Mises and Tresca differ by at most 15 per cent, so the choice between them rarely matters.

What decides it: They differ by at most 15.5 per cent when both are evaluated correctly, and the gap is widest at pure shear where the two answers are 159 and 138 N/mm². The error from omitting the third circle is larger than the difference between the criteria, and it goes the unsafe way.

Tested in The circle nobody draws, at the figure it turns on · the principal stress ladder.

Adding longitudinal stiffeners is an efficient way to make a wide plate work.

What decides it: The gain goes as (n+1)² and the rigidity each stiffener needs goes as roughly (n+1)⁴. On a 3,000 × 14 mm plate the steel in the ribs runs 6, 22, 53 and 101 per cent of the plate's own area for one to four stiffeners — so at four there is more steel in the ribs than in the plate they stiffen.

Tested in The coefficient that is not four, at the figure it turns on · the plate buckling ladder.

A column stable under a load stays stable under that load.

What decides it: Not in concrete. Creep takes the effective modulus down and the buckling load with it — from 18,549 kN on the day to 5,300 in the long term, 29 per cent of it — so the amplifier grows with nothing added. At 8,480 kN, still only 46 per cent of the day-one critical load, the deflection diverges at 55 days for no new reason at all.

Tested in Counted, not checked, at the figure it turns on · the Second-order ladder.

Assuming a pinned base is conservative, since it gives the longer effective length.

What decides it: It is conservative for the column and unconservative for everything else. A stiffer base attracts moment: the base moment a pinned assumption reports is zero, and a real base in full contact delivers a moment to the foundation that the foundation was not designed for. The column check and the foundation check want opposite assumptions.

Tested in The pinned base that is not pinned, at the figure it turns on · the sway stability ladder.