What is refuted here — page 7

Making the chord continuous over the panel points reduces the load it has to carry.

What decides it: It redistributes and does not reduce. Sweeping the middle panel's stiffness over a factor of twenty-five moves the support moment and the span moment in opposite directions while their combination stays at wL²/8 = 28.1 kNm to within 3e-16. Continuity buys a distribution, and the free moment is a statics result that no stiffness can touch.

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

Load in the middle of a span is the worst case for every member, because that is where the structure works hardest.

What decides it: It is the worst case for the chords and among the best for the middle diagonals. The shear at the panel 9 to 12 m along a 24 m truss runs from −0.44 to +0.56 of a unit load depending on where that load stands, and a load at mid-span puts it near the middle of that range.

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

Concrete has one compressive strength, so every part of a strut-and-tie model is checked against the same number.

What decides it: The permitted stress falls as ties enter the node, because a bar crossing a compression field cracks it transversely and a cracked field carries less. The limits go 1.0, 0.85 and 0.75 of the same basis for nodes with no ties, one tie and two — one material, three numbers, decided by what is anchored there.

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

The angle of the shear cracks is a property of the concrete, so the truss angle used in design should match what is observed.

What decides it: The cracks form at whatever angle the elastic field had when they opened, and the truss is a lower-bound model of what the member does afterwards. Any angle between 21.8° and 45° is admissible, gives a different stirrup requirement — 315 kN at 45° against 787 kN from the same steel at cot θ = 2.5 — and is safe if the web can carry the strut stress it implies.

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

Shear and bending are separate checks, so the shear calculation does not change the flexural reinforcement.

What decides it: The same cut that severs the stirrups passes through the tension chord, and it carries V·cot θ/2 of chord tension: 750 kN at a support where the moment diagram says nothing at all. The bar has to be there, and a curtailment taken from the moment diagram stops it 788 mm too early.

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

A pile that deflects too much laterally can be made stiffer by driving it deeper.

What decides it: Past about four characteristic lengths the head deflection is flat to within a per cent, because the ground below that depth is never asked for anything. The 20 m pile drawn is 10.6 T long and would deflect 20.5 mm at 4 T, at 8 T and at 20 T. Length is the one remedy that does nothing.

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

A curved cantilever is a curved beam with one support taken away, so its torsion behaves the same way.

What decides it: The torsion of a supported arc is antisymmetric and passes through zero at mid-span; the cantilever's grows monotonically from zero at the free end to its largest value at the root. For the same 90° arc of 6 m radius under 16 kN/m the supported case peaks at 124 and the cantilever at 329, and the peak has moved from the quarter points to the support.

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

Torsion in a curved member is a secondary action, worth a check after the bending is settled.

What decides it: At a quarter circle the cantilever's torsion is 57 per cent of its bending and the ratio reaches 1.33 before a half circle. Nothing is applied eccentrically and no load acts anywhere but downward; the moment vector at one section is simply not parallel to the one at the next.

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

Shear friction and strut-and-tie are two ways of getting the same answer, so one of them is enough.

What decides it: They are models of different failures on the same member. Shear friction asks for steel crossing a plane and saturates once the asperities crush, at 0.79 per cent reinforcement and 5.50 N/mm²; the truss model asks for a tie along the top and a strut through the body, and the tie's 375 kN has nothing to do with the plane at all.

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

The bars across a shear plane carry the shear.

What decides it: They carry none of it. Sliding a rough crack cannot happen without separating it — the asperity slope is 54° here, whose tangent is the 1.40 written down as a coefficient — so the bars are stretched by the separation and clamp the faces together. They supply the normal force that lets the roughness carry the shear, which is why a bar not anchored on both sides supplies nothing.

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

Redistribution reduces the moment at a support, so the structure around that support is designed for less.

What decides it: It reduces the moment in the member that sheds it. At a beam-to-column joint the moments must still balance, so 91 kNm taken off the beam end is 91 kNm the column has to take instead — and the column was sized from the elastic analysis that has just been abandoned.

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

Raising a tendon over an interior support reduces the moment the prestress puts into the beam, because the eccentricity is smaller.

What decides it: It reduces the primary moment from 1,440 kNm to 480 and raises the secondary from 720 to 1,680. The total is 2,160 kNm in both cases, identical, because the drape was held constant — and the drape is the only property of the profile the total depends on.

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

The shear connection is worked hardest where the beam's bending stress is largest, so the studs should be concentrated at mid-span.

What decides it: The interface carries a shear flow, which follows the shear diagram and not the moment diagram. On a 12 m beam under a uniform load it is 282 N/mm at the supports and zero at mid-span, so the studs are needed exactly where the section is least stressed.

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

Halving the number of studs halves the composite action.

What decides it: Interaction depends on αL, and the curve is flat where a real design sits. At αL = 15.8 the beam is 96 per cent composite; the connection would have to fall a long way before the deflection moved appreciably, and dropping the stiffness from 400 to 60 takes a 9 m pair of planks from 98 per cent composite to 89.

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

A strut's capacity is computed from the second moment of area about its weaker drawn axis.

What decides it: The drawn axes of a 120 × 120 × 12 angle both give 3.733 × 10⁶ mm⁴, and neither is the minimum. The principal axes lie at 45° and the smaller of them gives 1.523 × 10⁶ — a factor of 2.45 below the value the drawing suggests, and the Euler load falls with it.

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

Using the radius of gyration rather than the second moment avoids the problem, since it is a single number for the section.

What decides it: It is a number per axis, exactly as the second moment is: r = √(I/A), and I is a function of direction. Four profiles of the same 2,760 mm² area carry between 28 and 1,801 kN as 3 m struts, in the ratio of the squares of their radii — and for an unsymmetric section the one that counts is about an axis nobody drew.

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

Applying the load over the web of a channel avoids the torsion, since the web is where the section is.

What decides it: The shear centre of a 100 × 250 × 8 channel is 38.0 mm outside the web, in the air. A load on the web centreline misses it by that whole distance, and there is no point on the material at which it could be applied to miss it by less.

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

A beam has an effective width, so one number describes its flange.

What decides it: The working fraction of the same 4 m flange on a 24 m span is 0.839 at mid-span and 0.603 at the support — 23.6 percentage points apart on one member under one load. The single-sinusoid answer of 0.815 is correct at mid-span and optimistic everywhere else.

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

Plane stress means the state is two-dimensional, so two principal stresses describe it.

What decides it: Plane stress means the third principal stress is zero, not that it is absent. Zero is a value and it enters every criterion that ranks the three — Tresca compares the largest with the smallest, and for a biaxially compressed point at −31.7 and −148.3 the smallest is −148.3 while the largest is the zero nobody wrote down.

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

A material's shear strength is one number.

What decides it: Timber has three, and they differ by a factor of three or four. Shear parallel to the grain is about 4.2 N/mm², rolling shear across the fibres about 1.2, and the crossover moves with it: at 4.2 the checks cross at span-to-depth 6.7 and at 1.2 they cross at 23.3, which is longer than most timber beams are.

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

The crossover depends on how heavily the beam is loaded, so a lightly loaded beam is safe from it.

What decides it: No load, no width and no span survives the cancellation. Bending utilisation goes as wL²/bd² and shear as wL/bd, so their ratio is L/d times f_v/f_m — the crossover is a property of the material alone and a beam either is or is not short enough, whatever it carries.

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

A plate's slenderness limit is a property of the steel, since it is quoted as a width-to- thickness ratio for each grade.

What decides it: It is a property of the edge conditions and the stress pattern, scaled by the steel. The same 10 mm plate of the same steel is slender beyond 152 mm as an outstand, 462 mm as an internal element in compression, and 1,130 mm as a panel in shear — a factor of 7.4 from the coefficient alone.

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

A stiffer longitudinal stiffener gives a stiffer plate.

What decides it: Up to a threshold and not past it. Below γ* the stiffener rides on the buckle; at γ* it stays straight and the plate buckles between stiffeners at 9.0 times the bare stress; above γ* nothing further happens at all, because the sub-panel mode does not know the stiffener is there. The curve is a ramp and then a horizontal line.

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

Second-order effects are a property of the member being checked.

What decides it: They are a property of the storey. The critical load is fixed at 8,203 kN by the bracing that exists, and adding columns that carry load and no stiffness moves the structure along the axis without changing it — the amplifier goes from 1.57 to 2.60 as the leaning load rises, with nothing about the braced column altered.

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

A base detailed with two holding-down bolts on the column's centreline is a pin, because the bolts cannot resist a moment about that line.

What decides it: The bolts are not what resists it. A 550 × 450 mm plate under 900 kN stays in full contact up to 82.5 kN·m of moment, and over that range it is a bearing surface resisting rotation with no tension in anything. The bolts do not carry a single newton until 247.5 kN·m.

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

The base's fixity is a property of the detail, so it is decided once on the drawing.

What decides it: It is a property of the axial load. The same plate under 900 kN lifts at 82.5 kN·m and under 300 kN lifts at 27.5 — the threshold is NL/6 and scales with the load — so a base that is stiff under gravity becomes flexible under an uplift combination, in the load case where the sway is largest.

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

Warping restraint is a refinement worth a few per cent.

What decides it: The stiffening is 1/[1 − tanh κ/κ] and it runs from 1.005 at κ = 200 to 1,201 at κ = 0.05. The 533 × 190 beam here sits at 1.32 over 9 m and at 1.56 over the segment length its restraints actually give it — worth more than a moment-gradient factor and available from a detail rather than from a load case.

Tested in The restraint that beats the gradient, at the figure it turns on · the moment gradient ladder.

A column buckles about its weaker principal axis, so the smaller of the two Euler loads is the answer.

What decides it: There is a third load and it is often the smallest. A 3.5 m tee has flexural loads of 5,642 and 534 kN and a torsional load of 478, and the coupled answer is 309 kN — 42 per cent below the Euler value a designer would have used.

Tested in The third root of the cubic, at the figure it turns on · the Flexural-torsional ladder.

Reciprocity is a property of the beam being simple, so it does not survive redundancy.

What decides it: It survives every redundancy. The theorem needs linearity, elasticity and small displacements and nothing else, and it is exactly what makes the flexibility matrix symmetric — which is why the same three hundred solves are unnecessary on a continuous beam too, where the influence line is a curve rather than a pair of straight lines.

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

The deflected shape gives the shape of the influence line but not its magnitude.

What decides it: It gives both, provided the imposed displacement is a unit. The ordinate of the deflected shape at any point is the value of the released quantity when a unit load stands at that point, in the units the quantity has — 2.400 for the moment at *x* = 4 on the span drawn.

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

Stiffening the most heavily loaded member is the most effective way to reduce a truss's deflection.

What decides it: The vertical at mid-span carries 15.0 kN — the whole of a panel load — and contributes 0.000 to the movement, because the unit load applied where the deflection is read puts no force in it. Its term F·f·L/EA is zero whatever its area, and stiffening it changes 2,019 units of movement by nothing whatever.

Tested in The member that is not worth stiffening, at the figure it turns on · the truss deflection ladder.

A truss's deflection goes as the fourth power of its span, like a beam's.

What decides it: At a fixed depth and a fixed panel load it goes as the 3.56 power, fitted over spans from four to fourteen panels. The exponent is not four because the load grows with the span as well and because the web's share falls away — the chords go from 55 to 92 per cent of the movement over the same range.

Tested in The member that is not worth stiffening, at the figure it turns on · the truss deflection ladder.

The effective stiffness of a member is a property of its cross-section.

What decides it: Bending deflection goes as the fourth power of the span and shear deflection as the second, so their ratio contains the span squared. The shear share falls from 58 per cent at a slenderness of 10 to a few per cent at 60, which means the same panel has a different effective stiffness at every length it is built at.

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

The distribution factors are part of the iteration.

What decides it: They are written down before any arithmetic happens. At a joint the out-of-balance is shared in proportion to 4EI/L, so on the 7-9-7 beam the factors are 0.563 and 0.438 and on the 6-14-6 beam 0.700 and 0.300 — properties of the geometry, fixed before the first cycle and unchanged by any of them.

Tested in Why it converges, and how fast, at the figure it turns on · the moment distribution ladder.

A hand method and a matrix method are different analyses, so agreement between them is a coincidence.

What decides it: They are the same equations solved two ways. Eight cycles of hand arithmetic reach −0.0, 156.9, 156.9, 0.0 kNm at the supports against a stiffness solution's 0.0, 156.9, 156.9, −0.0, with a largest disagreement of 2.7 × 10⁻³ kNm — which is convergence rather than coincidence.

Tested in Why it converges, and how fast, at the figure it turns on · the moment distribution ladder.

A stronger steel is safer against a crack, because it can carry more stress.

What decides it: It shortens the crack that takes the strength away. At a toughness of 100 MPa√m the fracture curve crosses the grade line at 33.5 mm for 275 N/mm² steel, 20.1 mm for 355 and 12.0 mm for 460 — so the higher grade reaches its own yield at a shorter flaw, and the flaw is what decides.

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

The critical crack length is a property of the detail, so it can be tabulated with the category.

What decides it: It is a property of the toughness and the stress, and the toughness is a function of temperature. The same 40 mm plate tolerates a 241 mm flaw at +20 °C and 97 mm at −10 — a factor of 2.5 for a change in the weather, on steel that met its specification on both days.

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

A member that does not collapse under a repeated load is safe under it.

What decides it: A ratcheting member never collapses. At 90 per cent of the plastic moment with a 260 °C profile it gains 51.0 per cent of the first-yield curvature every cycle, for ever — arriving somewhere further round each time, at a load that no collapse calculation would refuse.

Tested in The map with three regions, at the figure it turns on · the shakedown ladder.

Which regime a member is in depends on how large the total load is.

What decides it: It depends on the two loads separately. Holding the moment at 30 per cent of plastic and raising the thermal profile from 0 to 300 °C moves the member from elastic to shakedown and no further; holding the profile at 220 °C and raising the moment from 10 to 85 per cent moves it from shakedown to ratcheting. The map is two-dimensional and a total is one number.

Tested in The map with three regions, at the figure it turns on · the shakedown ladder.

A material that has yielded in tension is stronger in compression, because it has strain hardened.

What decides it: It is weaker. Mild steel taken to 1.20 per cent strain and pushed back yields again after 550 N/mm² of stress change against a yield stress of 275 — so the reverse yield happens at −275 rather than at the hardened value, and the elastic range is twice the yield stress and not once it.

Tested in Yielding one way, and then the other, at the figure it turns on · the ductility ladder.

The effect is small enough to ignore in a structure that reverses.

What decides it: It is what makes the closed loop possible at all. A material whose yield surface *grew* rather than translating would harden out of its own loop within a few cycles and stop dissipating; a real one settles into a stable loop whose enclosed area is the same on every circuit, and the area is the energy a structure sheds.

Tested in Yielding one way, and then the other, at the figure it turns on · the ductility ladder.

Testing fewer specimens gives a less precise estimate of the same strength.

What decides it: It gives a lower strength. The factor applied to the sample's own scatter is the Student quantile when the scatter is estimated from the same results, running from 7.73 at two specimens to 1.73 at thirty — so four specimens of a 38 N/mm² material report 23.8 N/mm² against 27.9 for a population known exactly.

Tested in The strength that belongs to the test programme, at the figure it turns on · the characteristic strength ladder.

A partial factor covers the scatter in the material.

What decides it: The scatter was spent getting to the fractile. Dividing the 5 per cent characteristic value of a 38 N/mm² population by 1.50 gives 18.6, below which one specimen in 22,161 would fall — which is not a statement about scatter but about a reliability index the factor was calibrated to.

Tested in The strength that belongs to the test programme, at the figure it turns on · the characteristic strength ladder.

A young concrete's stiffness can be scaled from its strength.

What decides it: They do not develop together. The modulus follows strength to the power 0.3, so at seven days the concrete has 78 per cent of its 28-day strength and 93 per cent of its stiffness, and at three days 60 and 86. A young structure is much nearer its final deflection than its final capacity.

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

A fatigue spectrum's average stress range is the number to design with.

What decides it: The reduction is cube-weighted, not arithmetic. On the traffic spectrum drawn, the band at 62 N/mm² is 2 per cent of the crossings and 58 per cent of the damage, and the equivalent constant range is 25.5 N/mm² — much nearer the heaviest band than the average of all five.

Tested in The cycles that do not count, at the figure it turns on · the fatigue ladder.

Every cycle a structure sees does some damage.

What decides it: Below the cut-off limit none does. The smallest band is 35 per cent of the crossings and contributes nothing at all, because its range sits under the constant-amplitude limit where the curve's slope changes from three to five and then stops.

Tested in The cycles that do not count, at the figure it turns on · the fatigue ladder.

Improving the detail category extends the life and changes nothing else.

What decides it: It changes which cycles count. On category 71 the two heaviest bands share the damage 58 to 42; on category 112 the heaviest band does 100 per cent of it, because the second band has dropped below the improved cut-off. A better detail does not merely last longer — it is damaged by a different part of the traffic.

Tested in The cycles that do not count, at the figure it turns on · the fatigue ladder.

Two points at the same radius from the centroid carry the same stress.

What decides it: On a box group they differ by a factor of 1.23. The direct stress is the same everywhere and the torsional stress has the same magnitude at equal radius — but the angle between them changes with position, so the vector sum does not.

Tested in The radius rule, and where it fails, at the figure it turns on · the weld group ladder.

The elastic method and the instantaneous centre differ by a constant factor.

What decides it: The gap is a curve. For a 3 × 2 bolt group it is 13.87 per cent at 150 mm of eccentricity and a different figure at every other — largest where the group is nearly concentric and vanishing where the load is far away and the group is effectively in pure torsion.

Tested in The radius rule, and where it fails, at the figure it turns on · the weld group ladder.

A bolt in tension carries the tension applied to it.

What decides it: It carries that plus the prying force. A tee stub with a 22 mm flange pulled at 140 kN per bolt develops 73.94 kN of prying at the flange tip, so the bolt sees 213.94 — 1.53 times what was applied — and nothing on the connection's drawing shows it.

Tested in The thickness that decides who fails, at the figure it turns on · the prying ladder.

Moving the bolts is a detailing preference with little structural effect.

What decides it: It moves the regime boundaries further than plate thickness does per millimetre spent. Taking the bolt's distance from the web from 50 mm to 80 mm on the same 22 mm flange drops the flange's own collapse load from 133.1 kN to 83.19, and pushes the thickness at which prying stops from 31.91 mm out to 40.36.

Tested in The thickness that decides who fails, at the figure it turns on · the prying ladder.

An oversize hole reduces a bolted joint's capacity because it removes more steel.

What decides it: The net section is barely touched and the slip resistance falls by fifteen per cent. Four M20 bolts at 172 kN on a blasted face carry 344 kN through standard holes and 292.4 kN through oversize ones — the same steel, the same bolts, the same preload, and a coefficient of 0.85.

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

A slotted hole in a preloaded joint is a movement joint.

What decides it: It is a fit-up device that the preload then locks. The joint still carries 216.72 kN with no movement at all; what the slot changes is what happens after slip, when the plates travel the length of the slot before bearing picks the load up — 22 mm rather than 2.

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

More bolts in an end connection means more capacity.

What decides it: Not for the mode that governs here. Three bolts at 70 mm pitch give a block shear capacity of 524.79 kN and two at 140 mm give 539.1 over the identical 180 mm tear-out length — the pair leaves more metal on the shear plane, so yield governs instead of rupture and the joint is stronger with a bolt taken out.

Tested in The end that is only a plate, at the figure it turns on · the coped beam ladder.

A double-coped end is a coped end with the second cut treated the same way.

What decides it: The first cut leaves a tee and the second leaves a plate, and a plate has no compression flange, two free edges instead of one, and a block that can tear out in either direction. The governing check moves from the section modulus of the remainder to the edge distance of the bolt group, which changes the capacity by 46 per cent between 55 mm and 110.

Tested in The end that is only a plate, at the figure it turns on · the coped beam ladder.

An oversize hole is irrelevant to a connection that fails by tearing rather than by bearing.

What decides it: It takes 4 mm off both planes at once. Going from a 22 mm hole to a 26 mm one drops the capacity from 524.79 kN to 489.51 — 6.7 per cent — because every hole on the shear plane and the half-hole on the tension plane are all subtracted from areas that are the whole of the calculation.

Tested in The end that is only a plate, at the figure it turns on · the coped beam ladder.

A ductility demand of 2.0 describes a structure in less trouble than one demanding 3.1.

What decides it: Not when the two are different structures. The regular frame demands 3.10 and spreads 1.33 per cent drift over eight storeys; the frame with a ground storey at half strength demands 1.96 and puts 5.58 per cent into one. The lower demand belongs to the building that fails.

Tested in Weaker in one place, and better on every average, at the figure it turns on · the pushover ladder.

A soft storey is dangerous wherever it is.

What decides it: Position decides it. The same 50 per cent weakening gives a drift concentration of 5.9 at the ground storey, 5.6 at the fourth, and 1.7 at the eighth — which is exactly the regular frame's own figure. A weak top storey is not an irregularity in any sense that matters.

Tested in Weaker in one place, and better on every average, at the figure it turns on · the pushover ladder.

An eccentricity of a few hundred millimetres is a detail beside a load of six hundred kilonewtons.

What decides it: It is the difference between passing and failing. The same 604 kN gives 0.921 of the resistance at no eccentricity, 0.976 at 120 mm, 1.059 at 300 and 1.196 at 600. The vertical load never changed and the check went from comfortable to twenty per cent over.

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

Weaker lacing means a smaller force in the lacing, since it attracts less.

What decides it: It attracts more. Reducing the diagonal area from 300 mm² to 80 takes the critical load from 3,832 kN to 2,601 and the shear at 1,500 kN axial from 17.4 kN to 55.2 — three times larger, on diagonals a quarter of the size. The flexibility the lacing supplies is what amplifies the force it must carry.

Tested in The lacing decides the force it has to carry, at the figure it turns on · the Built-up column ladder.