What is refuted here — page 10

The plastic modulus is the elastic one times a shape factor of about 1.15.

What decides it: That is true of an I-section and of nothing else in this collection. A rectangle is exactly 1.5, a channel about 1.20, a tee 1.79 and a composite deck 1.80 — and the reason the range is so wide is that the two moduli are taken about two different axes, which for a tee are 23% of the depth apart.

Tested in The axis that moves when the section yields, at the figure it turns on · the Equal-area axis ladder.

A section has a neutral axis.

What decides it: It has one while it is elastic and a different one when it is fully plastic, and a continuous family of them in between. It also has two elastic section moduli, one to each extreme fibre, which for this tee differ by a factor of 2.78 — while the plastic section has one, because a fully yielded section does not care which fibre reached yield first.

Tested in The axis that moves when the section yields, at the figure it turns on · the Equal-area axis ladder.

Cracks are controlled by putting in more reinforcement.

What decides it: More steel helps by lowering its stress, which is one of two factors. The other is the crack spacing, and that is a bond length proportional to the bar diameter — so the same 1,340 mm² per metre gives 0.17 mm in 8 mm bars and 0.37 mm in 25 mm bars, a factor of 2.15 with the steel area held identical. Rearranging the same steel is free; adding steel is not.

Tested in The same steel, and a wider crack, at the figure it turns on · the crack width ladder.

More cover is better, because it protects the bar.

What decides it: It protects the bar and widens the crack that lets the environment reach it. The cover enters the crack spacing directly — about 3.4 times it — so raising cover from 15 mm to 70 nearly trebles the calculated crack width. Two durability requirements written in the same clause pull in opposite directions.

Tested in The same steel, and a wider crack, at the figure it turns on · the crack width ladder.

Stiffer props are safer props.

What decides it: A prop's load is an imposed-deformation effect: the wall moves toward the hole and the prop refuses. So the force is proportional to the prop's own stiffness, exactly as a lack-of-fit force is. A stiffer prop takes more load and gives the soil less chance to mobilise its own strength — and the wall's bending moment falls with the movement it is allowed.

Tested in Every prop has its own worst day, at the figure it turns on · the propped excavation ladder.

A group of n piles carries n times what one carries.

What decides it: For capacity, at ordinary spacings, very nearly — the block-failure check on this three-by-three group at five diameters gives a ratio of 4.5, so it does not govern at all, and it only begins to bite at two diameters with many piles or in soft clay with long piles. For stiffness it is nowhere near: the group settles 3.9 times what one pile settles at the same load per pile, and a hundred-pile group settles thirteen times.

Tested in Nine piles, and four times the settlement, at the figure it turns on · the pile group ladder.

The group efficiency factor is the important number.

What decides it: It is the number that is always computed and rarely governs. What governs nearly every pile group ever designed is settlement, and settlement is not in the efficiency formula at all — the interaction that softens the group is an elastic effect over the whole depth of the stress bulbs, not a strength effect at the pile shafts.

Tested in Nine piles, and four times the settlement, at the figure it turns on · the pile group ladder.

A stronger lining is a safer lining.

What decides it: A lining stiff enough to refuse the ground's distortion absolutely collects the whole free-ring moment, p₂R²/3. One flexible enough to go with it collects nothing, because there is no curvature change left to resist. The moment rises monotonically with the lining's own bending stiffness over six orders of magnitude of it, and thickening the lining moves it up the curve.

Tested in The lining that is stronger for being weaker, at the figure it turns on · the tunnel ring ladder.

The load on a tunnel lining is the weight of the ground above it.

What decides it: The thrust is the mean stress times the radius, and the mean is (σ_v + σ_h)/2 — so the horizontal stress is half of it, and at K₀ = 0.6 the "weight above" over-states the mean by a quarter. The bending, which is what usually sizes the section, comes from the *difference* between the two stresses and vanishes entirely at K₀ = 1 however deep the tunnel is.

Tested in The lining that is stronger for being weaker, at the figure it turns on · the tunnel ring ladder.

A structure is stable up to the load at which its stiffness matrix becomes singular.

What decides it: For a load with a potential, yes. Ziegler's two-bar column under a tangential end load has a stiffness determinant of exactly k² at every load — it varies over the whole axis by one part in 10¹⁶, which is round-off — and the column nevertheless loses stability at Pℓ/k = 2.0858. The static criterion is not a poor approximation there; it has no root at all.

Tested in The load it cannot buckle under, at the figure it turns on · the follower force ladder.

Energy methods give an upper bound on the critical load, so they are safe to use.

What decides it: They give a bound when a potential exists. Symmetrising this column's stiffness matrix — which is what treating the follower force as a dead weight amounts to — gives divergence at 2.000, which happens to be near 2.0858 and is near it by luck: it is the answer to a different question about a different structure.

Tested in The load it cannot buckle under, at the figure it turns on · the follower force ladder.

Centrifugal force is a real force acting outward on the body.

What decides it: It is the body's own mass times its own acceleration, moved across the equals sign. Nothing applies it and no reaction to it exists. The move is legitimate and it is exactly half done when the inertia term is added to the free body and the acceleration is left on the other side as well — which counts it twice.

Tested in The force that is really an acceleration, at the figure it turns on · the centrifugal load ladder.

A larger flywheel stores more energy and is therefore more dangerous.

What decides it: It stores more energy and bursts at exactly the same rim speed, because the hoop stress in a spinning ring is rho times the square of the rim speed with no radius in it. Energy per unit mass is half the same quantity, so the whole design is a competition between one material property and one speed, and size cancels out of both.

Tested in The force that is really an acceleration, at the figure it turns on · the centrifugal load ladder.

A stronger pipe is a safer pipe.

What decides it: A pipe stiffer than the fill beside it settles less than the fill, drags the interior prism down with it and receives more than the prism weight. A pipe softer than the fill does the opposite and sheds load into the ground either side. Making the conduit weaker moves the load down the page, which is why modern practice buries plastic rather than concrete.

Tested in The pipe decides what the soil weighs, at the figure it turns on · the soil arching ladder.

Cambering a beam solves its deflection problem.

What decides it: Camber is subtracted from the deflection before the finishes and from the deflection after them, so it cancels out of the difference exactly — to the last decimal place, not approximately. It moves the span/250 total check a great deal and moves the span/500 incremental check by nothing at all, and the second is usually the one that governs.

Tested in The limit that depends on a date, at the figure it turns on · the incremental deflection ladder.

Propping a composite beam during construction improves it.

What decides it: It improves the total and worsens the increment. An unpropped beam carries its wet concrete on the bare steel, so that deflection happens before any finish exists and belongs to neither check — the unpropped beam here is at 0.55 on total and 0.76 on increment against the propped beam's 0.72 and 0.96.

Tested in The limit that depends on a date, at the figure it turns on · the incremental deflection ladder.

The tensile strength is a material property, like the compressive strength.

What decides it: It depends on the test. A flexural test on a 100 mm prism reads half again what a direct tension test on the same concrete reads, and the flexural value falls with the depth of the specimen. A "strength" that changes with the size of the thing measured is a fracture property wearing a strength's clothes.

Tested in The strength that is never used, at the figure it turns on · the tensile strength ladder.

A steel that meets its specification is fit for its job.

What decides it: The Liberty ships, the Hasselt bridge and the King's Bridge in Melbourne all broke in a brittle manner in steel that met every requirement of its specification — because the specifications had no toughness requirement in them at all. What was specified was strength, and strength is not the property that was exceeded.

Tested in The same steel, brittle in January, at the figure it turns on · the transition temperature ladder.

A thicker plate of the same steel is stronger.

What decides it: It is more brittle. Constraint at a crack tip in a thick section suppresses the yielding that would have blunted it, so the transition temperature shifts upward with thickness — a 100 mm plate of this steel tolerates 27% of the flaw a 10 mm plate does, on the same day, with the same yield stress and the same certificate.

Tested in The same steel, brittle in January, at the figure it turns on · the transition temperature ladder.

The transition curve is a property of a particular steel and has to be measured for each one.

What decides it: Wallin's master curve has the same shape for every ferritic steel — 30 + 70 exp(0.019(T − T₀)) in MPa√m — with a single free parameter, the reference temperature. One test temperature fixes the entire curve, which makes toughness a material property that is a temperature rather than a strength.

Tested in The same steel, brittle in January, at the figure it turns on · the transition temperature ladder.

Soft ground is bad ground for a building in an earthquake.

What decides it: It is bad for a building whose period is near the site's and can be better than rock for one much stiffer. This lake-bed profile amplifies a two-second building by 7.5 and a 0.3 second building by 1.4 — a selectivity of 5.4 — so "bad ground" is not a property of the ground alone but of the pair.

Tested in The ground has a period of its own, at the figure it turns on · the site response ladder.

A response spectrum is a property of an earthquake.

What decides it: It is a property of an earthquake and a place. Five sites over one bedrock give five different spectra from the same rock motion, with fundamental periods from 0.30 to 2.03 seconds and amplifications from 1.9 to 7.5 — and a code's site classes are a two-parameter summary of exactly that scatter.

Tested in The ground has a period of its own, at the figure it turns on · the site response ladder.

Adding damping is the way to protect a machine passing through its critical speed.

What decides it: At a slow sweep, damping is the whole answer — four damping ratios spanning a factor of ten give peaks spanning a factor of 9.7. At a fast sweep the same four span a factor of 1.8, on steady-state values twenty times apart. Damping cannot limit a response that never had time to grow.

Tested in The resonance that ran out of time, at the figure it turns on · the transient resonance ladder.

Making the structure stiffer reduces the forces in it.

What decides it: Against a load, yes. Against a misfit the force is proportional to stiffness exactly — the exponent computes as 1.000 — because the compatibility equation divides the misfit by the structure's own flexibility. Doubling every member's area doubles the locked-in force and buys the member nothing, since its capacity has only doubled too.

Tested in Built to the wrong length, at the figure it turns on · the Fit-up ladder.

Taking the pile deeper is the safe response to a settlement problem.

What decides it: It is safe only if the extra length is below the settling ground. Inside it the drag is the area under a friction that grows linearly with depth, so it grows as the square: 482.87 kN at 24 m and 1,408.13 kN at 36.1 m, a factor of 2.92 for half as much length again, with the head load and base resistance held.

Tested in The ground that hangs on instead of holding up, at the figure it turns on · the downdrag ladder.

A restrained wall carries EαΔT, so a 32 degree drop puts 67 N/mm² into the concrete and it is hopeless.

What decides it: Both hypotheses fail. The restraint is 0.5 rather than 1, and early-age creep relieves about 35 per cent of what is left, so the stress that actually arrives is 3.83 N/mm². It is still more than the concrete has: the restrained strain is 190 microstrain against a strain capacity of 50, a factor of 3.78, and the wall cracks. But it cracks by a factor of four, not forty.

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

A shear crack forms at forty-five degrees.

What decides it: It does in pure shear, which is the case with no axial stress on the element. At the centroid of the web drawn, with 7.14 N/mm² of compression and 3.39 of shear, the principal tension acts at 68.3° to the axis and the crack runs across it at 21.7°. That flat angle is worth cot θ = 2.51 in the truss model afterwards, so the prestress is paid twice for one purchase.

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

Torsion of a closed section is statically determinate, since T = 2qA gives the flow directly.

What decides it: It is determinate for one cell and for no more than one. Two cells have two unknown flows and T = 2Σq_iA_i is still one equation; the section is torsionally redundant to the degree of the number of cells less one, and the missing statements are compatibility — every cell twists by the same amount because they share walls. A three-cell box is a simultaneous system before it is anything else.

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

Torsional buckling is a failure mode a designer has to check.

What decides it: Not for anything that stands still. The critical torque exceeds the torque that yields the same tube by a factor of 42.7 for the 168 × 5 section drawn, and the two meet only at a length of πE/τ_y = 3,142 radii — 256 m for that tube, and a number containing no thickness whatever. The mode is real, exact, and out of reach of every structure in this collection.

Tested in The buckling load with no compression in it, at the figure it turns on · the torque buckling ladder.

A slope analysis computes a factor of safety, so it is a check like any other.

What decides it: It is a check whose answer depends on which surface was tried. The circle a first guess puts through the toe from above the middle of the slope gives F = 1.650; searching 81 toe circles finds one at 1.192, which is 39 per cent lower. Nothing in the arithmetic identifies the critical surface — it has to be looked for, and an analysis reporting one circle has reported nothing.

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

The force on a curved gate has to be found by integrating the pressure round it.

What decides it: It can be, and it need not be. The horizontal component equals the pressure force on the surface's own vertical projection — γH²/2 = 176.6 kN/m for the gate drawn — and the vertical component equals the weight of the water standing above it, 110.5 kN/m. The integral round the arc gives 176.579 and 110.492, agreeing to four parts in a million. Two free bodies replace a calculus problem with two areas.

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

A hollow section is chosen over an open one for strength.

What decides it: For torsional movement the ratio is 80.2 to one. The same load at the same eccentricity on a closed section of the same depth moves the flange tip 0.148 mm against 11.9. Strength rarely decides between them; this does, and it appears in no check anybody performs.

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

The 5.65 in the proportional gauge length is an arbitrary convention.

What decides it: It is arbitrary in value and not in form. What matters is that the gauge is proportional to √S₀ rather than fixed, because the localised extension is proportional to √S₀ too, so the quotient is the same for every specimen size — every one of the seven diameters returns the identical number to six decimal places. 5.65 is a choice; being a multiple of √S₀ is not.

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

The joint rotates under load until the eccentricity disappears, so the bending is a transient.

What decides it: True for a bonded lap in thin sheet and false for a bolted one in plate. Goland and Reissner's factor is k = cosh(uc)/[cosh(uc) + 2√2 sinh(uc)] with uc ∝ (c/t)√(σ/E); for the 10 mm plate and 60 mm overlap drawn, uc is 0.06 and k is 0.855, so the rotation removes 11 per cent of the moment. A 1 mm sheet with a 20 mm lap at the same stress has k at 0.64, and k falls below 0.4 only where the stress is another order up as well.

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

A structural member's natural frequencies get further apart as the mode number rises.

What decides it: A beam's do, as n². A cable's do not: a taut string's are an arithmetic progression, every one exactly f₁ above the last, for ever. The 120 m stay drawn has modes at 1.006, 2.013, 3.019, 4.025 Hz and onward at 1.006 Hz intervals — so there is always a mode within half a spacing of any excitation frequency, and the second one sits in the middle of ordinary walking.

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

The coefficient of friction is a property of a material.

What decides it: It is a property of a pair of surfaces, and of their state. The same PTFE bearing sliding on the same steel returns 0.048 at 15.0 N/mm² of contact pressure and 0.091 at a fifth of the load — a factor of 1.9 on one number, with nothing about either material changed.

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

Making a body heavier improves both its stability checks equally, so weight is the answer to either failure.

What decides it: It improves both by the same factor and changes neither's standing against the other. The weight cancels out of the comparison entirely: overturning and sliding reach a factor of one at heights in the ratio B/μ whatever the body weighs, so a heavier body of the same shape fails the same way at the same slenderness, later.

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

A friction connection has an uncertain capacity, so it can never be the basis of a design.

What decides it: Uncertainty in a capacity matters when the capacity is a resistance to be exceeded. A slip load is a cap: what follows it is not failure but a second mechanism at a higher load, so the scatter moves where the device operates and not whether it works. Eight bolts at 100 kN on two faces at μ = 0.35 slip at 560 kN and bear at 900 kN, and neither number is a collapse.

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

Chord bending is the secondary stress that comes from the joints being welded rather than pinned.

What decides it: Those are two different moments with two different causes. Secondary bending comes from joint rotation and reaches 32.5 per cent of the axial stress in the worst member here. Chord bending comes from load applied between nodes and is present in a chord whose joints are genuine pins.

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

A truss with crossed diagonals in a panel is statically indeterminate, so it needs a stiffness analysis.

What decides it: Only if both diagonals can carry either sign. A tension-only pair has one member slack in every load case, so the count of active members is one per panel and the frame is determinate in every state it is ever in. The redundancy exists on the drawing and never in the structure.

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

Anchoring a tie is a detailing matter that follows the analysis.

What decides it: It is part of the equilibrium. A tie is only worth what it develops inside the node, and a 25 mm bar at 435 N/mm² needs 906 mm to develop by the code's own uniform-bond rule — several times the node's own size. The node's dimensions are therefore set by the anchorage, and the model has to be drawn to fit them.

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

The published coefficients for a laterally loaded pile are a property of the soil model, so they apply wherever that model does.

What decides it: They are a property of a long pile. The deflection coefficient comes back at 2.430 against Matlock and Reese's 2.435 for a pile 10.6 T long, at 2.617 for one 3.2 T long, and at 4.559 for a monopile 2.0 T long — an 87 per cent error, in the same ground, from the same equation, on a structure that is being built by the thousand.

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

Redistribution can be applied to the moment envelope, since the envelope is what the section is designed for.

What decides it: The envelope is not a state of the structure and satisfies no equilibrium identity — it misses the free-moment relation by up to 21 per cent, because it is assembled from seven different load arrangements at different stations. Redistribution is a move from one equilibrium state to another, so each arrangement has to be redistributed separately and the envelope rebuilt from the results.

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

The secondary moment is a parasitic effect, so a good profile is one that minimises it.

What decides it: Minimising it minimises nothing. A profile with no secondary moment at all is called concordant, and its total moment is the same as any linear transformation of it — so the concordant profile is not a better design, it is the same design with a different bookkeeping.

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

The effect is small for sections that are nearly symmetric.

What decides it: The relevant measure is the ratio of the two principal second moments, not the departure from symmetry. A zed purlin looks like a channel with its flanges rearranged and has I₁/I₂ = 18.5, against 3.90 for an angle whose asymmetry is far more obvious.

Tested in The axis a column buckles about, at the figure it turns on · the principal axes ladder.

Making the flanges narrower reduces the eccentricity, so a narrow channel is the answer.

What decides it: It reduces it, and it reduces the second moment with it. Going from 150 mm flanges to 60 takes the eccentricity from 63.0 mm to 19.1 — a factor of 3.3 — and takes the second moment from 44.6 × 10⁶ mm⁴ to 23.5, a factor of 1.9. What is bought in torsion is paid for in bending.

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

Shear is a secondary check on a beam, since bending governs at any sensible span.

What decides it: That is true of steel and false of timber. The two checks cross where the span-to-depth ratio equals f_m/f_v, which is 1.73 for steel — shorter than any beam — and 6.7 for softwood. A timber beam shallower than about six times its depth is a shear problem.

Tested in The shear that decides a timber beam, at the figure it turns on · the shear flow ladder.

A pin-ended column with no bracing has an effective length factor of 1.0, so it is checked at its own height.

What decides it: It has no lateral stiffness of its own, so it cannot buckle alone and cannot be checked alone. In the storey drawn it is at K = 2.60 — it sways with the frame it is standing in — while the column that does the bracing is at 3.61, which is off the end of every published alignment chart.

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

An influence line is found by re-solving the structure for every load position.

What decides it: That is one way and it needs 301 solves to draw the curve above. Reciprocity says the same curve is a single deflected shape: release the quantity, impose a unit displacement, and read the ordinates off. One analysis instead of three hundred, and the two agree because a load at A causing a deflection at B equals the same load at B causing the deflection at A — 153.334 both ways, to 3 parts in 10¹⁴.

Tested in An influence line is a deflected shape, at the figure it turns on · the reciprocity ladder.

Shear deflection is a small correction, worth a per cent or two.

What decides it: For a solid beam, yes. The sandwich panel drawn spends 25 per cent of its deflection on shearing its core at a span-to-depth ratio of 20, where a solid beam spends about one — and 58 per cent if the core's shear modulus is 6 N/mm² instead of 25. The ratio of face modulus to core shear modulus is 2,800 in the first case and 11,667 in the second.

Tested in The stiffness that belongs to the span, at the figure it turns on · the shear deflection ladder.

An inspection interval is a maintenance decision rather than a design one.

What decides it: It is the output of a crack-growth calculation and it is set at design. The interval is the time to grow from the smallest detectable crack to the critical one, divided by a factor — and both ends of that range come from this page's arithmetic rather than from any fatigue table.

Tested in Designed to be found in time, at the figure it turns on · the fatigue ladder.

The age-adjusted effective modulus is a good approximation to the creep integral.

What decides it: Not at the usually quoted ageing coefficient. The superposition integral leaves 1.72 N/mm² and the one-line shortcut at χ = 0.8 leaves 3.05 — a factor of 1.8. This creep function implies χ = 1.67, not 0.8, and the shortcut carries a number fitted to a different function.

Tested in The stress that leaks away, at the figure it turns on · the creep ladder.

Losing thirty-seven per cent of the slip resistance to a long slot is a serious penalty.

What decides it: It is smaller than the two variables around it. Going from one friction face to two takes the slotted joint from 216.72 kN to 433.44 — past what the standard-hole single-face joint carried — and dropping the surface from a blasted class to an untreated one costs more than any hole in the table.

Tested in The hole made bigger so the steel would fit, at the figure it turns on · the Slip-critical ladder.

If a punching check fails, make the column bigger.

What decides it: It is the weakest of the three geometric levers. Going from 400 × 400 to 600 × 600 — 2.25 times the concrete — takes the utilisation from 1.059 to 0.882. Reproportioning to 250 × 800 at 80 per cent less added area gets to 0.921, and 75 mm of extra slab depth gets to 0.670.

Tested in Turn the column, and the slab passes, at the figure it turns on · the punching shear ladder.

Shape the section for the curvature correction, since that is what the curvature adds.

What decides it: The correction is the smaller of the two effects and barely moves. Across the four sections drawn it runs from 1.378 to 1.450 — five per cent — while the straight-beam stress they are multiplying runs from 118.8 N/mm² to 220.3, which is 86 per cent. The section modulus is still doing most of the work.

Tested in The wide side goes inside, at the figure it turns on · the curved beam ladder.

A thinner web is the economical way to use post-buckling reserve, since the reserve is largest there.

What decides it: The reserve is a ratio and the capacity is not. Halving the web from 8 mm to 4 raises the reserve from 2.41 times to 6.77 and lowers the capacity from 1,187 kN to 416 — and the flange pull only halves, so the anchorage cost per kilonewton carried nearly doubles.

Tested in The tension has to pull on something, at the figure it turns on · the tension field ladder.

If m + r = 2j, a pin-jointed frame is statically determinate and stable.

What decides it: The count asks whether there are enough equations and never whether they are different from one another. Arrangements satisfying it exactly are drawn folding up, and the honest test is the rank of the equilibrium matrix — with the mechanism sitting, visibly, in its null space.

Tested in The count that does not see it, at the figure it turns on · the determinacy ladder.

A member is safe when the stress in it is below the yield stress of its material.

What decides it: Failure load against slenderness, with both modes on one plot. Squashing happens at σ_y·A and does not care about length; Euler buckling happens at π²EI/L² and cares about nothing else. The governing capacity is the lower, and the yield stress does not appear in the second at all — so a slender column of high-strength steel and one of ordinary steel buckle at exactly the same load, with no stress anywhere near a limit.

Tested in Strong enough and still falls over, at the figure it turns on · the buckling ladder.

An extra support is a free margin of safety — more supports, more capacity.

What decides it: Equilibrium cannot divide a load between two routes to the ground; the missing condition is compatibility, one equation per redundancy, and it brings EI into a calculation statics never needed. Load then goes where the stiffness is — and the same structure acquires a sensitivity it did not have, because a support that moves produces a full moment field with no load applied at all.

Tested in One support too many, and what it costs to know, at the figure it turns on · the indeterminacy ladder.

A connection is designed to carry the forces the frame analysis reports at that node.

What decides it: The frame analysis reports forces at a point, and the connection is a region across which those forces have to be transferred by a specific mechanism at specific offsets. A bolt group solved at an eccentricity of 150 mm has its worst bolt at 50.4 kN where the direct share is 16.7 — three times what the node forces suggest, and the extra is entirely a consequence of the connection having size.

Tested in The connection is not a point, and every diagram on this site says it is, at the figure it turns on · the connection design ladder.