What is refuted here — page 6

Superposition is a good approximation.

What decides it: It is exact or it is not available, and there is nothing in between. Under the three hypotheses the residual for the beam drawn is zero — not 10⁻⁶, zero, because the governing equation is linear in the load and a linear operator distributes over addition by definition. Break a hypothesis and the error is 37 to 62 per cent. There is no regime in which it is a few per cent out.

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

Adding load cases separately is conservative, because it ignores the way they help each other.

What decides it: The two failures point opposite ways. Let the section crack and the sum of the parts is 62 per cent SMALLER than the whole, because each case alone sees a stiffer beam than the pair does. Put a support a gap away and the sum is 37 per cent LARGER, because one case closes the gap and the other then finds a structure with an extra support in it. There is no direction to lean in.

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

Warping restraint is a refinement worth a few per cent.

What decides it: It is worth a factor of 2.67 on this member. The Saint-Venant answer alone gives 9.09° and the restrained answer 3.40°, because at kL = 4.50 the warping term carries most of the torque. Ignoring it is not conservative in a useful way — it is wrong by nearly three times, which is enough to make a member look unusable that is merely poor.

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

A mismatch of 80 microstrain is a stress of E times it, so the concrete sees 2.4 N/mm² and is close to cracking.

What decides it: That is the fully restrained answer, and nothing is fully restrained. The bar and the concrete strain together to a common value set by their relative axial stiffnesses, so at one and a half per cent steel the concrete carries 0.223 N/mm² — 1.9 per cent of the 12.0 N/mm² full restraint would have given, and 7.7 per cent of its tensile strength.

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

Any reinforcement that bonds to concrete will do, provided it is strong enough.

What decides it: Aluminium bars of the same area in the same concrete put in 0.528 N/mm², 2.36 times as much, because the mismatch is 13 × 10⁻⁶ instead of 2. Carbon fibre with a coefficient near zero is worse still per unit of mismatch. Strength is not the property that decides whether a reinforcement can be bonded into concrete; the third significant figure of its expansion coefficient is.

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

Elongation after fracture is a material property.

What decides it: The same steel — same uniform elongation, same localisation — tested at seven bar diameters over a fixed 50 mm gauge reports 22.9 per cent at 6 mm and 44.8 at 25 mm, a spread of 1.96 with nothing about the material changed. The localised extension in the neck is a length rather than a strain, so dividing it by a fixed gauge gives a bigger answer for a bigger bar.

Tested in The ductility that depends on the ruler, at the figure it turns on · the gauge length ladder.

A5 and A10 are close enough that either will do for a comparison.

What decides it: They differ by 26.4 per cent for the steel drawn — 27.5 per cent against 21.75 — because the second gauge length is twice the first and the localised part is divided by it. Comparing a bar certified at A5 with one certified at A10 is comparing two rulers, and the shorter ruler always flatters.

Tested in The ductility that depends on the ruler, at the figure it turns on · the gauge length ladder.

A single lap joint is a double lap joint with one shear plane instead of two, so it is worth half as much.

What decides it: It is worth less than half, and the difference is bending rather than shear. The two load paths are offset by a plate thickness, so the joint carries P·t/2 of moment and the peak stress is 3.57 times the mean against 1.0 for a symmetric joint. The code's answer is to reduce the bearing resistance to 1.5f_u·d·t from 2.5 — 60 per cent — and that factor is this eccentricity, tabulated.

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

A cope removes a small part of the section, so it removes a small part of the strength.

What decides it: It removes 33 per cent of the area and 73 per cent of the section modulus — more than twice as much resistance as material. The flange being cut away is the material furthest from the neutral axis, and what is being reduced is a second moment: the coped tee's Z is 3.84 × 10⁵ mm³ against the whole section's 1.44 × 10⁶, a ratio of 0.267 for a cope 11 per cent of the depth.

Tested in The section that is checked is not the one chosen, at the figure it turns on · the coped beam ladder.

The coped web is a plate like any other, so it buckles at the usual stress.

What decides it: The cope leaves it a free edge, and a plate free along one edge has a buckling coefficient near 0.425 against 4.0 for one held on both — a factor of 9.4. Treated as a bare outstand the web drawn buckles at 41 N/mm² against an applied 61, which fails; treated with the restraint the uncoped web beyond the cope supplies, it buckles at 5,740, which does not.

Tested in The section that is checked is not the one chosen, at the figure it turns on · the coped beam ladder.

The two available buckling models bracket the answer closely enough for design.

What decides it: They differ by a factor of 139 — 41 N/mm² against 5,740 — because one is a free outstand with nothing holding it and the other is a test-calibrated expression that keeps the restraint from the full-depth web beyond the cope. Neither is a refinement of the other, and which is used decides whether the detail passes.

Tested in The section that is checked is not the one chosen, at the figure it turns on · the coped beam ladder.

Two buildings with similar dynamic properties are the dangerous case, because they will resonate together.

What decides it: They are the safe case, and swaying together is exactly why. At a period ratio of one the cross-correlation is 1.000, the two structures move as one, and the gap needed collapses to the difference of the two drifts — 100 mm rather than 340. It is the mismatched pair that needs the room, which is the opposite of the intuition the word resonance supplies.

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

Measuring tension from frequency is a rough field estimate.

What decides it: It is the primary measurement and it is accurate to about a per cent. Inverting the string relation gives T = 4μL²f₁², and the two corrections are small and computable: bending stiffness raises the implied tension by 0.99 per cent at ξ = 410, and sag lowers it by essentially nothing at λ² = 0.13. There is no more accurate way to know a stay's force, which is why every cable-stayed bridge in the world is commissioned this way.

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

A reduction on the imposed load is safety margin being spent, so a designer who takes it is building closer to the edge than one who does not.

What decides it: The opposite. Summing the characteristic loads of n floors produces a total whose own probability of being exceeded is far below the fractile each floor was drawn at — at ten storeys and a coefficient of variation of 0.6 it is roughly the 99.9th percentile rather than the 95th. The reduction restores the intended fractile. Refusing it makes the design more conservative on the columns than on the beams, unevenly and by an amount nobody computed.

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

With enough floors the reduction should go to nothing, since the average of many independent loads converges on the mean.

What decides it: It converges on the mean and the mean is not zero. The factor is (1 + zv/√n)/(1 + zv), whose limit is 1/(1 + zv) — 0.503 for a coefficient of variation of 0.6 at the 95th percentile, with no n in it at all. Every bay really does carry its mean, and no amount of averaging reduces that.

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

The moment is small and shares itself out among the four members meeting at the joint, so no one of them sees much.

What decides it: It divides by rotational stiffness, 4EI/L or 3EI/L, and stiffness is dominated by the second moment of the largest member. The chords take 48 per cent each and the diagonal that caused the offset takes two. The member that produced the eccentricity is not the member that pays for it.

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

Steel is isotropic, so a plate is as good through its thickness as it is along it.

What decides it: Not in ductility, which is what decides this. Reduction of area at fracture runs about 60 per cent along the rolling direction, 45 across it and 15 through the thickness — and reduction of area converts to a true fracture strain exactly, as ln(1/(1 − Z)). Those three are 0.916, 0.598 and 0.163: a factor of four in the reported percentage is a factor of 5.6 in the strain the material can take.

Tested in The direction a plate was never tested in, at the figure it turns on · the lamellar tearing ladder.

A thicker plate is safer, because it is stronger.

What decides it: The demand is a transverse weld shrinkage divided into the thickness twice — once because the shrinkage scales as the deposit over the plate, and again because the strain is that movement over the thickness. It goes as A_w/t². But thickness is also what generates the restraint, and a thick plate is a stiff one. Thickness helps the arithmetic and hurts the mechanism, and the two are not weighed against each other by any check.

Tested in The direction a plate was never tested in, at the figure it turns on · the lamellar tearing ladder.

The problem is the weld metal, so a better electrode fixes it.

What decides it: The tear is in the parent plate, below the weld, along a plane of inclusions rolled flat before anybody thought about the joint. Nothing done to the deposited metal reaches it. What does work is geometry — a detail that turns the pull into the plane of the plate rather than through it — and buying steel with a specified through-thickness reduction of area.

Tested in The direction a plate was never tested in, at the figure it turns on · the lamellar tearing ladder.

Welding distortion is a workmanship matter, so it belongs to the fabricator and not to the calculation.

What decides it: The shrinkage force is computable from the heat input and the steel's own properties, and for an ordinary plate girder it is 312 kN — larger than most of the applied forces in the connection it is part of. Applied 210 mm off the centroid it is a curvature, and the girder comes out bowed 12.5 mm over 12 m, which is L/962 against a tolerance of L/1000.

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

A bow of ten millimetres in twelve metres is a cosmetic defect.

What decides it: It is an initial imperfection, and an imperfection is what the column curve is a fit to. At 120 N/mm² of applied stress the bow is amplified to 17.3 mm and adds 17 N/mm² of bending — which is not large, and is added to a member whose buckling resistance was derived assuming an imperfection the fabrication has now supplied on top of.

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

A pile that deflects too much sideways can be fixed by making it longer.

What decides it: Past about four characteristic lengths the head deflection stops moving to within a per cent, because the ground below that depth is never asked for anything. For the pile drawn that is 7.6 m of a 20 m pile; the other twelve are carrying axial load and nothing else. Length is the one remedy that does not work, and it is the first one reached for.

Tested in A pile has no length until the ground gives it one, at the figure it turns on · the lateral pile ladder.

Better ground would fix it, so the answer is sensitive to the site investigation.

What decides it: T goes as n_h to the minus one fifth. A hundredfold range of soil stiffness — from very soft to dense — moves the characteristic length by a factor of 2.5, and the head deflection by rather less than that. A lateral pile design is remarkably insensitive to the number the ground investigation argues about and remarkably sensitive to the section, which is the opposite of the usual expectation.

Tested in A pile has no length until the ground gives it one, at the figure it turns on · the lateral pile ladder.

A pile cap fixes the head, so it takes the moment out of the pile.

What decides it: It halves the head deflection and it moves the worst moment from 1.3 characteristic lengths down the pile to the head itself, where it is 30 per cent larger. The cap does not reduce the moment; it relocates it — into the connection, which is the one part of a pile anybody can inspect and the one part that is usually detailed as though it were a pin.

Tested in A pile has no length until the ground gives it one, at the figure it turns on · the lateral pile ladder.

A skew deck is a right deck of the same span, checked with a correction factor.

What decides it: At 45 degrees the longitudinal moment is 59 per cent of the right deck's and the reaction at the obtuse corner is 2.4 times the average — one number down and the other up, and both by large factors. A single correction cannot move two quantities in opposite directions, which is why skew is handled by analysis and not by a factor.

Tested in The deck that spans square, at the figure it turns on · the skew ladder.

The load is uniform, so the abutment reaction is uniform.

What decides it: Uniform at zero skew and only there — the grillage returns exactly 1.00 for a right deck, which is the check on the model rather than an input to it. By 30 degrees the obtuse corner is at 2.15 times the average and the acute one at 0.16, on a deck nobody loaded unevenly.

Tested in The deck that spans square, at the figure it turns on · the skew ladder.

Reinforcing a slab along the road is reinforcing it in the direction it spans.

What decides it: The principal moments rotate away from the carriageway as the skew increases, toward the perpendicular between the abutments. A deck reinforced along the road at high skew is reinforced across its own span direction, and the steel that is doing the work is the distribution steel.

Tested in The deck that spans square, at the figure it turns on · the skew ladder.

Halving the steel over the middle of a beam halves the length of bar required.

What decides it: A parabolic moment diagram is above rM at x/L = (1 ± √(1 − r))/2, so a bar covering that region is L√(1 − r) long. Dropping to half the capacity leaves a bar 0.707 of the span — a shortening of 29 per cent, not 50. The moment diagram is flattest exactly where it is largest, so the first steel curtailed is the steel that buys least.

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

A bar may stop where the moment diagram says it is no longer needed.

What decides it: It has to run a further a_l = z(cot θ − cot α)/2 because the beam is a truss and the diagonal compression delivering the shear arrives at an angle — 270 mm for the beam drawn — and then a full anchorage length beyond that. The tail is 1,420 mm at each end of a 9 m beam, which is 16 per cent of the span before any steel has been saved at all.

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

A deck plate is the flange of the main girder, and the local effects under a wheel are a separate check.

What decides it: They are not separate because they are in the same fibres of the same plate. At the rib-to-deck weld the girder contributes 120 N/mm², the crossbeam 34 and the trough 13, and the total of 167 is what the material sees. Treating the crossbeam as a support rather than as a structure understates it by 26 per cent.

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

The deck is designed for strength, with fatigue as a subsequent check.

What decides it: The stress range at the rib-to-deck weld is 47 N/mm² and that weld is a partial-penetration detail in a low fatigue category, running the entire length of the bridge. The strength check is never close and the fatigue check decides the plate thickness, the trough spacing and the crossbeam spacing together.

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

A beam buckles when its largest moment reaches the critical value, so only the peak matters.

What decides it: Four diagrams with identical peaks buckle at 1.00, 1.13, 1.36 and 2.72 times the same reference. The destabilising integral is ∫M(z)²φ²/EI_z — the SQUARE of the moment, weighted by where the beam wants to twist — so a diagram that is large over a short length is worth much less than a flat one of the same maximum.

Tested in The shape of the diagram, and not its peak, at the figure it turns on · the moment gradient ladder.

The factor is a code convenience, fitted to test data.

What decides it: It is an eigenvalue. Eliminating the lateral displacement from the coupled buckling equations leaves one functional in the twist, and C₁ is the ratio of its lowest eigenvalue to the uniform-moment case. Computed that way the uniform case returns 1.000, a uniformly distributed load 1.131 and a linear diagram in double curvature 2.719 — the table values, from mechanics.

Tested in The shape of the diagram, and not its peak, at the figure it turns on · the moment gradient ladder.

A moment reversal between restraints makes a beam worse, because part of it is in hogging.

What decides it: It makes it much better. A reversal means the moment passes through zero somewhere in the middle, which is exactly where the buckling mode is largest, so the destabilising integral loses its most heavily weighted region. Double curvature is the most favourable moment distribution there is, at 2.72 against 1.00.

Tested in The shape of the diagram, and not its peak, at the figure it turns on · the moment gradient ladder.

A thinner-walled tube of the same area is always a better column, because it has a larger second moment.

What decides it: Only until local buckling arrives. With the area fixed, I = A r²/2 so Euler's load rises as r², and the local buckling stress is αEt/(rβ√3) which falls as 1/r². The capacity is the lower of the two and it has a maximum at r/t = 129 for the member drawn — past that, making the tube larger makes it weaker.

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

Using a stronger steel makes the problem easier.

What decides it: It makes it worse. While the member is short enough there is a plateau of proportions that all reach the squash load, and inside it the design is free; the plateau closes at 20 m for S355, 10.3 m for S690 and 7.4 m for S960. A stronger steel closes the safe band sooner and pushes the optimum onto the coincident-mode point.

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

Rib shortening is a small correction, so it can be neglected.

What decides it: It is a 0.10 per cent correction to the thrust of the arch drawn and it is the whole of the crown moment. A parabolic arch under a uniform load is funicular, so the rigid solution gives exactly zero bending; the shortening reduces the thrust by 5 kN and leaves 27.9 kNm behind. A correction that is a thousandth of one quantity is all of another.

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

A young concrete has less of everything, in proportion.

What decides it: The modulus follows the strength to the power 0.3, not one to one. At seven days an ordinary mix has 77.9 per cent of its strength and 92.8 per cent of its stiffness; at three days, 59.8 and 85.7. A young structure is much nearer its final deflection than its final capacity, which is the right way round for the load it is carrying and the wrong way round for the load it might be dropped.

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

Confining pressure is free, so more of it is better.

What decides it: The tube is in a biaxial state and von Mises fixes the trade exactly: every newton of hoop tension is bought from the wall's own axial capacity. Optimising the sum without that constraint hands the whole wall to the hoop and lets a linear confinement law carry the column, for a claimed 32 per cent that is an extrapolation far outside anything it was fitted to.

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

A pushover is a cheap substitute for a dynamic analysis and answers the same question.

What decides it: It computes a property of the structure with no time in it. The demand still has to come from somewhere, and the whole apparatus of the method — the equivalent single-degree system, the participation factor, the equal-displacement rule — is the machinery for supplying one. It is an answer about capacity married to an assumption about demand.

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

The load pattern is a detail, since the structure's capacity is the structure's capacity.

What decides it: A uniform pattern loads the lower storeys harder and returns a different base shear from a triangular one on the same frame. The pattern is a guess at a mode shape that the structure stops having as soon as it yields — the analysis's central assumption failing at exactly the point it starts to matter.

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

A restrained brace is stronger, which is why it dissipates more energy.

What decides it: It is not stronger in tension at all — the core yields at the same force. What it has is a compression side equal to its tension side, and a loop that does not pinch. Over six cycles it dissipates 2.07 times as much energy as the ordinary brace, and the ratio grows from 1.46 on the first cycle to 2.69 on the last.

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

The casing has to be strong, since it is preventing the core from buckling.

What decides it: It has to be stiff. It carries no axial force by design — that is what the de-bonding layer is for — so its requirement is a single inequality on its Euler load against the fully hardened core force. It is a buckling check on a member carrying nothing, which is a shape of requirement almost nothing else in structures has.

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

Equipment on a floor can be checked against the ground spectrum, since that is the earthquake the site will have.

What decides it: The building is a filter between the two. On the roof of the eight-storey frame drawn, a rigid item feels 1.79 times what the ground spectrum gives, and one whose own period matches the building's feels 6.09 times. The ground spectrum describes a motion the equipment never experiences.

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

Removing the expansion joint removes the thermal movement problem.

What decides it: It relocates it. A 60 m deck expands 9 mm at each end and pushes the abutment into its backfill; granular soil under cyclic strain densifies, so the same movement costs more pressure every year. K climbs from 0.384 at rest toward 0.897, and the force on the abutment is 2.33 times what it started at.

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

Earth pressure is a load like any other, so it is computed once.

What decides it: This one grows with the number of summers. It reaches 97 per cent of its final value after about a hundred and twenty cycles, which is the structure's design life — so the governing load case is one that does not exist when the bridge opens and arrives gradually over a century.

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

Inclining the hangers is a refinement that saves a little steel in the chords.

What decides it: It changes the structural system. The same load on the same arch with crossing hangers gives 686 kNm in the tie and 831 in the arch — factors of 4.8 and 7.5. The chords stop being beams and become the chords of a truss, which is a different structure with the same geometry.

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

A bridge is designed for its finished condition and checked for construction afterwards.

What decides it: On a launched bridge the construction condition designs it. In service each section has one sign of moment; during the launch every section on the deck sees both, because each passes over every pier and through every span. The section has to take the worst of everything at every point, which no in-service arrangement produces anywhere.

Tested in Every section was somewhere else, at the figure it turns on · the launched bridge ladder.

The launching nose is a temporary works item.

What decides it: Without one the worst launch moment is 170,156 kNm against an in-service 40,500 — a factor of 4.2, and the deck would have to be four times as strong for a condition lasting weeks. With a nose of 0.66 spans it is 42,539, within four per cent of the service value. A structure that carries almost nothing decides the deck's section.

Tested in Every section was somewhere else, at the figure it turns on · the launched bridge ladder.

A longer nose is better, so it is a matter of how much can be afforded.

What decides it: It is a threshold. Below about half a span the nose has not reached the next pier before the cantilever gets long, and the worst moment is unchanged: 170,156 at no nose, 170,156 at 0.4 spans, and 44,648 at 0.6. The curve is a step, and the step is where the geometry starts working.

Tested in Every section was somewhere else, at the figure it turns on · the launched bridge ladder.

Breaking a floor into more levels saves material, because each member is smaller.

What decides it: The members are smaller and there are more of them, and the two cancel exactly. A level's weight per unit area of floor is 3ρqrL/8σ, which contains the span and not the spacing — so adding a level adds a whole system whose span is the longest yet. A second level costs 52 per cent more steel than one and a third costs 108.

Tested in Every level is a longer span, at the figure it turns on · the hierarchy ladder.

A deeper hierarchy is at least shallower overall, since each member is smaller.

What decides it: The structural zone grows with it: 730 mm for one level, 1,030 for two, 1,346 for three, because the levels stack and the last of them still spans the whole bay. Hierarchy costs weight and depth at the same time, which leaves it needing a different justification entirely.

Tested in Every level is a longer span, at the figure it turns on · the hierarchy ladder.

The steel is what a floor structure weighs.

What decides it: The slab is 88 per cent of it. A 130 mm deck weighs 3.25 kN/m² and the two levels of steel under it weigh 0.44 between them, so every decision about the steelwork is a decision about an eighth of the structure — and the deck's span, which sets the rest, is the variable nobody calls structural.

Tested in Every level is a longer span, at the figure it turns on · the hierarchy ladder.

A wider contact carries more friction, because friction is a shear stress acting over an area.

What decides it: Nothing in the bound contains an area. A block of 48 kN on a plane at 22° demands 18.0 kN of friction against a capacity of 26.7 kN whatever its footprint, and spreading it over twice the plan halves the contact pressure and doubles the area with no change to either number.

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

Overturning is a moment check and sliding is a force check, so they are separate calculations about separate things.

What decides it: They are two readings of one free body, and both are decided by the same two numbers. The wind force appears in both denominators and cancels; what is left is B/2 against μ times the lever H/2, which is a comparison of the body's shape with its contact.

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

Two dampers with the same slip load are interchangeable.

What decides it: Four bolts on two faces and eight bolts on one both slip at 280 kN, and they are different devices. The first has half the bolts to relax and twice the load in each; the second has half the sliding interfaces, so twice the wear per interface and twice the heat into one surface. The product is the same and none of its factors is.

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

A graphical solution is an approximation, and the algebraic one is exact.

What decides it: The construction is exact and the drawing of it is approximate — a different statement. Every polygon here closes identically because the forces in it satisfy the joint equations, and the error in a reading is the error of the paper, not of the method. The joint at the support closes on 30, 35.3 and 46.3 exactly as the solver reports them.

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

A drawing cannot check itself, so a graphical method needs an independent verification.

What decides it: Closure is the verification, and it is built into the construction. A joint drawn with one force wrong does not close, and the gap is the out-of-balance to scale — 19.1 for a diagonal guessed at 40.0 instead of 46.3. Nothing in the algebraic method reports an error that way.

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

Adding panels to a truss makes it stronger, because there is more material resisting the load.

What decides it: The chord force barely moves and moves the wrong way: 52.9 kN at four panels, 56.5 at six, 56.7 at eight and 57.7 at twelve. The couple carrying the moment has the same lever arm in every case, so nothing about the chords improves. What improves is the strut, and only because it is shorter.

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

Panel subdivision and truss depth are two ways of doing the same thing.

What decides it: Depth changes the chord force and subdivision does not. Doubling the depth of this truss halves every chord force, because the chords are a couple whose arm is the depth; doubling the panel count changes the largest chord force by 2 per cent and halves the buckling length instead. They act on different terms and neither substitutes for the other.

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