What is refuted here — page 2

Moment-area is a shortcut, so it gives an approximate deflection.

What decides it: It is the same double integral with the constants fixed geometrically rather than algebraically, so it is exact wherever the integral is. Over twenty-five load positions on a simply supported span the two routes disagree by 7 × 10⁻¹³ of the answer, which is the trapezium rule the diagram is sampled with and not the method.

Tested in The area of a diagram is a rotation, at the figure it turns on · the Moment-area ladder.

The second theorem gives the deflection of a point.

What decides it: It gives the deviation of a point from a TANGENT drawn at another point, which is a deflection only when that tangent happens to be horizontal. For a cantilever it is; for the simply supported span in fig-2 the tangent at the left support is inclined at 213.333, and the mid-span deflection of 533.33 is the difference of two quantities neither of which is a deflection.

Tested in The area of a diagram is a rotation, at the figure it turns on · the Moment-area ladder.

Cambering a beam reduces the stresses in it.

What decides it: A camber is a change of unstressed shape, so it changes no stress at all — the same load on the same section gives the same moment and the same fibre stress whether the member was built straight or curved. What it changes is the ordinate a reader measures from, and the solver returns exactly zero for the stress difference because there is nothing to compute.

Tested in Built to the wrong shape on purpose, at the figure it turns on · the camber ladder.

A buckling load obtained from an assumed shape is an estimate, so it might be high or low.

What decides it: It is always high, and never low. Assuming a shape is imposing a constraint, a constraint can only stiffen, and a stiffer column carries more. Four assumed shapes here give 9.8696, 9.8824, 10.0000 and 12.0000 EI/L² against a ten-term reference of 9.8696, and one that violates the end conditions gives 42.

Tested in Guessing the shape, and getting the load anyway, at the figure it turns on · the stability energy ladder.

A shape that is visibly wrong will give a visibly wrong load.

What decides it: A parabola has the wrong curvature at both ends of a pin-ended strut — it is finite where the true mode's is zero — and it is 21.6% high. The static sag under the column's own weight is 0.129% high. The relation between the two errors is quadratic: blending in 5% of the second mode gives 2.97% in the load, and 10% gives 11.5%.

Tested in Guessing the shape, and getting the load anyway, at the figure it turns on · the stability energy ladder.

A column fails at the lower of its Euler load and its squash load.

What decides it: It sits under both over the whole range below λ = 91. With a flange-tip residual stress of 0.3fy, the buckling stress about the minor axis is 0.884fy at λ = 20, 0.766 at λ = 60 and 0.730 at λ = 77 — where the two straight answers say 1.00, 1.00 and 1.00. No imperfection appears anywhere in that calculation.

Tested in The column that had yielded before it was loaded, at the figure it turns on · the inelastic buckling ladder.

A structural element that has buckled has failed.

What decides it: A 1,000 × 1,000 panel of 6 mm web buckles in shear at 383 kN and carries 696 kN — 1.82 times as much — because the compression diagonal's job passes to the tension diagonal, which is straight and has a flange to pull against. At 4 mm the same panel carries 2.98 times its buckling load.

Tested in The panel that carries more after it has failed, at the figure it turns on · the tension field ladder.

A shear panel's tension field runs at 45°, along the principal direction of the shear.

What decides it: Only in the limit of an infinitely long panel does anything approach it, and then from below. Maximising the band's vertical component over its own width gives tan 2θ = d/a, so a square panel bands at 22.5°, one twice as long as it is deep at 13.3°, and one shorter than it is deep at 29.5°.

Tested in The panel that carries more after it has failed, at the figure it turns on · the tension field ladder.

A strut-and-tie model has to be the right one, so choosing it is a calculation.

What decides it: It has to be an equilibrium solution that nowhere exceeds strength, which is the lower-bound theorem's whole requirement. On the region drawn here, admissible lever arms from 40% to 95% of the depth ask the tie for 1,500 kN down to 632 kN — a factor of 2.4 — and every one of those structures stands up if it is built to carry what its own model says.

Tested in When there is no section to design, at the figure it turns on · the Strut-and-tie ladder.

A deep beam is just a beam, so the usual bending calculation applies with a larger depth.

What decides it: The bending calculation assumes strain is linear across the section, and within one depth of a discontinuity it is not. The map here is drawn on a member eight times as long as it is deep, and even at that slenderness the shaded D-regions take 38% of the span. A B-region survives only while the span exceeds three depths, so on a member as deep as half its span there is no station anywhere along it at which a section is legitimate.

Tested in When there is no section to design, at the figure it turns on · the Strut-and-tie ladder.

Shear connectors are needed where the bending stress is largest.

What decides it: The interface force per unit length follows the SHEAR diagram, not the moment diagram. It is largest at the supports and exactly zero at mid-span of a uniformly loaded beam, so the connection works hardest where the bending stress is smallest and does nothing at all at the point a designer checks.

Tested in Two beams, or one beam four times as stiff, at the figure it turns on · the composite action ladder.

A partly connected beam is somewhere between two loose beams and one solid one, in proportion to how much connection it has.

What decides it: The relation is strongly curved. Half of the available stiffness arrives at a connector stiffness of 26 per unit length and nine tenths of it at 250 — ten times as much connection for the second half of the benefit as for the first.

Tested in Two beams, or one beam four times as stiff, at the figure it turns on · the composite action ladder.

Compressive membrane action stops and catenary action starts when the arch runs out of depth, at a deflection of one thickness.

What decides it: The two overlap. The tie is already yielded at 0.6 of the thickness, which is where the catenary branch overtakes the arch, while the arch still has 40% of its lever arm left. Switching between them at w = t puts a step in the curve; the honest curve is where they cross.

Tested in The force nobody put in the model, at the figure it turns on · the membrane action ladder.

A beam is checked for bending at the extreme fibre and for shear at the neutral axis, and between them the section is covered.

What decides it: On a 533 mm I-section spanning 3,500 mm, the principal tension at the web-flange junction is 90.2 N/mm² against 85.9 at the extreme fibre — and at the junction the bending stress is 79.5 and the shear 30.8, so neither component is anywhere near its own maximum. The governing point is one that neither standard check evaluates.

Tested in The worst stress is not where the worst bending is, at the figure it turns on · the principal stress ladder.

A material carries a share of a moment in proportion to how much of the section it is.

What decides it: It carries it in proportion to its modulus times its second moment about the common neutral axis. A 6 mm steel plate beside a 150 × 300 timber joist is 3.85% of the area and carries 43.3% of the moment, at 96.2 N/mm² against the timber's 5.04 — a ratio of exactly 19.09, which is the ratio of the moduli and nothing else.

Tested in A section made of two materials, one of them pretended away, at the figure it turns on · the transformed section ladder.

The transformed section is a convenient approximation for a section made of two materials.

What decides it: It is exact for linear elasticity. Plane sections give equal strain at equal height, Hooke's law turns that into stress in the ratio of the moduli, and multiplying a width by that ratio reproduces both the force and its first moment exactly. Transforming to either material gives identical stresses, which is the check that there is no choice in it.

Tested in A section made of two materials, one of them pretended away, at the figure it turns on · the transformed section ladder.

A hole in a beam web is a loss of section, so the check is the reduced section against the applied moment.

What decides it: The reduced section costs 30.3% of the second moment and 0.75% of the deflection. What governs is local bending of the tees across the opening: 167.6 N/mm² of the 237.9 total at the corner — 70.5% of it — from a Vierendeel moment of 6.30 kNm that no section calculation contains.

Tested in The hole that costs nothing, and everything, at the figure it turns on · the web opening ladder.

Openings should be put near the supports, where the bending moment is small.

What decides it: The Vierendeel moment follows the shear, and the shear is largest at the supports. The total stress in the tee is 242 N/mm² at 5% of the span and 138 at mid-span, so the worst position for a hole is exactly the one the ceiling void makes convenient.

Tested in The hole that costs nothing, and everything, at the figure it turns on · the web opening ladder.

A Vierendeel girder is a deep truss, so making it deeper helps in the same way.

What decides it: The penalty against a triangulated frame of the same members GROWS with depth: 1.10 times at 8% of the span, 1.95 at 13%, and 4.98 at 25%. The truss improves as the reciprocal of the depth because a chord force is M/d; the Vierendeel's chord-bending term does not contain the depth at all, and by 25% it is 92% of the movement.

Tested in The truss with no diagonals, at the figure it turns on · the vierendeel ladder.

A folded sheet is stiffer because folding work-hardens the metal and stiffens the material.

What decides it: No material property appears anywhere in the gain. Both the flat and the folded second moment are B·t·(a length)²/12, so everything cancels and the ratio is (h/t)² — 10,000 here, for a fold 300 mm deep in a 3 mm sheet, with not a gram of material added and nothing about the steel involved.

Tested in Folded until it spans, at the figure it turns on · the folded plate ladder.

A folded profile should have many small folds, so that the sheet is well supported everywhere.

What decides it: The second moment goes as the square of the leg length, so halving the number of folds quadruples the stiffness of the same sheet. What limits it is the leg buckling as a plate, and the optimum is exactly where the leg's buckling stress reaches yield: 19 legs here, carrying 8.65 against 2.73 for the six-leg profile drawn.

Tested in Folded until it spans, at the figure it turns on · the folded plate ladder.

A beam curved on plan can be simply supported like any other, because the load is still vertical and the reactions are still two.

What decides it: Two vertical reactions leave the rotation about the chord joining them unrestrained, and the load has a moment about that chord — the arc bulges away from it. The structure is a mechanism. A curved beam needs its supports to hold a torque, which makes even the two-support case statically redundant.

Tested in Bending that arrives as twist, at the figure it turns on · the curved in plan ladder.

The bending moments in a redundant curved beam depend on the section, because a redundant structure always shares load in proportion to stiffness.

What decides it: The self-stress state of a circular arc on two torsionally restrained supports is uniform torsion with no bending anywhere in it, so both flexibilities cancel out of the compatibility equation. The bending moments come back statically determinate, and the same beam in an open section and a closed one carries identical moments — while deflecting 4.09 and 1.62 times a straight beam's respectively.

Tested in Bending that arrives as twist, at the figure it turns on · the curved in plan ladder.

A coefficient of friction above one is impossible, so the 1.4 used for a crack in monolithic concrete must be a fudge factor.

What decides it: It is the tangent of an angle, not a ratio of surface forces. Sliding along a rough crack cannot happen without lifting one face over the other, and the separation per unit slip is the slope of the asperities. A slope of 54.5° gives 1.4, a slope of 45° gives exactly 1.0, and a ground joint at 31° gives 0.6. Nothing above one requires anything unusual.

Tested in Shear across a crack that is already there, at the figure it turns on · the shear friction ladder.

More reinforcement across a joint always buys more shear resistance, in proportion.

What decides it: Only until the asperities crush. Clamping rises linearly with the steel area and the resistance with it, up to a cap of 5.5 N/mm² — reached here at a reinforcement ratio of 0.786%. Past that the line is flat and every further bar is decoration.

Tested in Shear across a crack that is already there, at the figure it turns on · the shear friction ladder.

A section with a larger second moment of area is the stronger section.

What decides it: The tee here has 11.23 × 10⁶ mm⁴ against the rectangle's 10.00 — twelve per cent more — and an elastic strength of 75.3 × 10³ mm³ against 100.0, which is a quarter less. Stiffness divides the second moment by nothing; strength divides it by the distance to the furthest fibre, and moving material toward one face increases both.

Tested in Two strengths, depending which way up, at the figure it turns on · the asymmetric section ladder.

The section that is stronger as a beam is the better column.

What decides it: The tee and the rectangle here have the same area and the same depth. The tee's elastic section modulus is 75.3 × 10³ mm³ against the rectangle's 100.0, so it is a quarter weaker in bending — and its radius of gyration is 61.2 mm against 57.7, so it is six per cent better in compression. The two rankings are computed from the same second moment divided by different things.

Tested in The one length a section takes into a column, at the figure it turns on · the radius of gyration ladder.

A tapered member is checked at the section carrying the largest moment.

What decides it: The capacity is climbing as fast as the demand. For the cantilever drawn here the root carries the largest moment and is at 70.4% of its capacity, while the section three metres out is at 79.2% of its own. The governing station is where the depth has grown to exactly twice the tip depth, and for this member that is the middle.

Tested in The section that changes along the span, at the figure it turns on · the tapered member ladder.

Tapering a beam saves material at the cost of stiffness, because material has been removed.

What decides it: The taper removes material where the moment is small, which is where the second moment was contributing least to the deflection. Against a prismatic member of the same mean depth — the same weight — the tapered cantilever deflects 101 mm rather than 161. It is lighter for a given strength and stiffer for a given weight.

Tested in The section that changes along the span, at the figure it turns on · the tapered member ladder.

A cube of twelve pinned bars is a rigid frame, because every joint is held by three members.

What decides it: Twelve members and six restraints against eight joints needs 3 × 8 = 24, and 12 + 6 is 18. The cube is six mechanisms short and every one of them is a face shearing into a rhombus. Adding one diagonal to each of the six faces makes it exactly determinate.

Tested in Three equations at every joint, at the figure it turns on · the space frame ladder.

A frame satisfying m + r = 3j is a determinate space frame.

What decides it: The count is necessary and never sufficient. Three legs to an apex satisfy it exactly, and moving the three feet into a straight line leaves the count untouched while the frame turns freely about that line. The rank of the equilibrium matrix is 2 where the count claims 3, and only the rank knows.

Tested in Three equations at every joint, at the figure it turns on · the space frame ladder.

A symmetric plan does not twist, so plan torsion is a problem for irregular buildings.

What decides it: The plan that twists worst here is perfectly symmetric: a central core with nothing else resisting anything. Its natural eccentricity is exactly zero and its torsional radius is 1.41 m against a plan radius of 10.10, so the accidental eccentricity that has to be assumed anyway amplifies the edge displacement by 12.25. Symmetry is not the criterion; distribution is.

Tested in The corner that moves most, at the figure it turns on · the plan torsion ladder.

A patch load is checked over the length it is applied on.

What decides it: The flange bends and yield lines form in it, spreading the load along the web. For the girder here a 100 mm bearing produces an effective loaded length of 559 mm — 5.6 times as long — and 82% of the yield resistance is that spread rather than the bearing. The length is not a decision anybody made.

Tested in The load that chooses its own length, at the figure it turns on · the patch loading ladder.

A thicker flange spreads the load further, so it is the cheap way to improve a patch-load resistance.

What decides it: It does spread the load further and it buys almost nothing. The effective length is built from two yield-line terms that scale oppositely in the flange thickness — one grows with it and the other is independent of it — so the resistance rises as the 0.19 power. Going from 20 to 40 mm of flange buys 18%; going from 8 to 12 mm of web buys 120%.

Tested in The load that chooses its own length, at the figure it turns on · the patch loading ladder.

A built-up column is as strong as its second moment of area says, because the chords are held at their spacing.

What decides it: The lattice holding them at that spacing is a shear connection with a finite stiffness, and the two flexibilities add. With battens at 1.2 m the column here reaches 312 kN against an Euler load of 1,341 — 23% of it, and an effective slenderness of 163 against a nominal 79.

Tested in The column made of two columns, at the figure it turns on · the Built-up column ladder.

Chords further apart always make a stronger column, because the second moment rises with the square of the spacing.

What decides it: It rises with the square and the shear stiffness of the lattice does not rise at all, so the harmonic sum runs into a ceiling — 31,500 kN here, whatever the spacing. And if the panel length is held rather than the lacing angle, the shear stiffness falls with spacing and the critical load has a maximum: 7,472 kN at 1,400 mm, past which a wider column is a weaker one.

Tested in The column made of two columns, at the figure it turns on · the Built-up column ladder.

A partly built structure carries less load, so it is further from failure than the finished one.

What decides it: The load falls and the capacity falls with it, and the capacity falls faster. A 457 mm beam landed on its supports with no deck has an elastic critical moment of 17.5% of its plastic capacity; once restrained at 3 m centres it has 154%. The load has fallen to self-weight, perhaps a fifth — the capacity has fallen by a factor of 8.8.

Tested in The most dangerous day is before it is finished, at the figure it turns on · the erection stability ladder.

The worst differential shortening is at the top of the building, where the most shortening has accumulated.

What decides it: The load-driven part is zero at the top. A floor at the roof has nothing built above it, so nothing shortens the columns beneath it after it has been set to level. The product of "how much is still to come" and "how many storeys can shorten" is a parabola, and its maximum is at exactly half the height — 20.3 mm at level 20 of 40, and 0.0 mm at level 40.

Tested in The columns are shorter than the core, at the figure it turns on · the differential shortening ladder.

A column that shortens by 127 mm over the height of a building leaves the top floor 127 mm low.

What decides it: Each floor is built to its correct level as construction proceeds, so what a finished building shows is only the movement that happened after each floor was set. The total is 127 mm and the largest post-installation difference between a column and the core beside it is 20 mm. Reporting the first as though it were the second overstates the problem sixfold.

Tested in The columns are shorter than the core, at the figure it turns on · the differential shortening ladder.

Support flexibility affects deflections and not forces, since a settlement is not a load.

What decides it: In a continuous beam it changes both. Three equally stiff supports under a uniform load give the middle one 62.5% of the load when they are rigid and 35.9% when they are soft, and the hogging moment over it falls from 45.0 to 73.9 in the other direction. The redistribution is the reaction that the softer support declines to provide.

Tested in The deflection that belongs to the support, at the figure it turns on · the support flexibility ladder.

A structure is safest when it is lightest, because every load has to be carried.

What decides it: For every other action in this collection, yes. For flotation the weight is the resistance and the load does not scale with it, so the factor rises as the building grows: 0.53 with nothing above the ground floor, 0.80 with two storeys, 1.07 with four. The empty substructure is the governing case and it exists before anything is built on it.

Tested in A basement is a boat, at the figure it turns on · the uplift ladder.

The parasitic moment opposes the primary one, so ignoring it is conservative.

What decides it: Over the middle support here the primary moment is +540 kNm and the secondary is +306 — the same sign, 57% as large, and adding to it. The total prestress moment there is 846 kNm rather than 540, and a design that dropped the secondary term would be out by more than half.

Tested in The prestress that pushes back, at the figure it turns on · the secondary prestress ladder.

A tension bar may be curtailed where the bending-moment diagram says it is no longer needed.

What decides it: The cut along the crack severs the tension chord as well as the stirrups, and it carries half the shear as chord tension: 563 kN at the support here, where the moment is zero. The whole chord force diagram is the bending one shifted 619 mm toward the support, and a bar stopped at the moment diagram's own station is 619 mm short.

Tested in The beam that becomes a truss, at the figure it turns on · the shear truss analogy ladder.

A member free to expand and bend carries no thermal stress.

What decides it: The deck here is on sliding bearings with nothing restraining it, and the bilinear temperature profile leaves 3.98 N/mm² of compression at the top face and 1.83 of tension 140 mm below it. The field has a resultant force of zero and a resultant moment of zero to machine precision, which is why nothing about the member's freedom removes it.

Tested in The stress nobody restrained, at the figure it turns on · the thermal gradient ladder.

Thermal effects can be described by a mean temperature and a gradient.

What decides it: Those two numbers fix the axial movement and the curvature and say nothing about the stress. A straight profile from +6.28 °C at the top to −2.57 at the bottom has exactly the same mean and exactly the same gradient as the real one here — the same 4.9 mm of camber over a 25 m span, the same 283 kNm if the deck is held — and leaves no stress at all.

Tested in The stress nobody restrained, at the figure it turns on · the thermal gradient ladder.

The worst-stressed point of a beam is where the bending stress is largest.

What decides it: On the short heavily loaded beam here the extreme fibre reaches a von Mises stress of 312 N/mm² and the web–flange junction reaches 367, because the junction has 94% of the bending stress and a shear stress of 129 as well. The governing point is neither the extreme fibre nor the neutral axis.

Tested in Both at once, and neither matters until it does, at the figure it turns on · the shear moment interaction ladder.

A plate that has buckled has reached its capacity.

What decides it: The web here buckles at a stress a quarter of yield and keeps 42.5% of its width working at full yield afterwards, because the corners are held by the flanges and the redistribution runs toward them. What is lost is the middle. A plate on both edges keeps going a long way past its own critical stress, which is why the effective width exists at all.

Tested in What is left after it ripples, at the figure it turns on · the Effective cross-section ladder.

A brace is a brace: make it stiff enough and the member is restrained.

What decides it: On the 8 m beam here, 447 kN/m at the compression flange reaches the fully braced answer of 447 kNm. The identical stiffness at the tension flange gives 153 — a 7% gain on the unbraced 143 — and a hundred times that stiffness at the same place gives 163. The curve has an asymptote well below the plateau, and no amount of stiffness reaches it.

Tested in The brace on the wrong flange, at the figure it turns on · the beam bracing ladder.

Bracing a beam at midspan doubles its capacity, because the unbraced length halves.

What decides it: The critical moment goes as √(EIz(GJ + EIw π²/L²))/L, so halving the length multiplies the first term by two and the term under the root by more than one. Here it is a factor of 3.12 rather than 2, and on a beam with negligible warping stiffness it would be exactly 2 — the factor is a property of the section, not of the arithmetic of halving.

Tested in The brace on the wrong flange, at the figure it turns on · the beam bracing ladder.

A stronger material lets a tower be taller.

What decides it: Strength appears nowhere in the answer. The height limit is (C·E·i²/ρg)^(1/3), and for a radius of gyration of 80 mm it comes out at 51.5 m in steel, 51.0 in aluminium and 50.0 in structural timber — because those three materials have specific stiffnesses of 27, 26 and 24 MN·m/kg. Three materials with nothing else in common give the same tower.

Tested in Too tall for nothing but itself, at the figure it turns on · the Self-weight buckling ladder.

A frame that satisfies Maxwell's count is statically determinate and rigid.

What decides it: Two collinear bars between two pins satisfy it exactly — b + r − 2j = 2 + 4 − 6 = 0 — and have one mechanism and one self-stress state at the same time. The count is the difference of the two dimensions and is zero whenever they are equal, which says nothing about whether either is zero.

Tested in The forces that are there with nothing applied, at the figure it turns on · the Self-stress ladder.

Prestressing a mechanism is a way of pretending it is not one.

What decides it: The geometric stiffness a self-stress state lends a mechanism is computable and it has a sign. Tension the two collinear bars and the quadratic form comes out at +1 per unit tension; put them in compression instead and it comes out at −1, which is a structure that snaps rather than one that stands.

Tested in The forces that are there with nothing applied, at the figure it turns on · the Self-stress ladder.

A member carrying only axial force is a member with no bending in it.

What decides it: The theorem fixes the line of action, not the route the material takes between the pins. A bar pinned at two points 10 m apart but bowed 1.2 m off the chord carries 240 kNm at its crown for a 200 kN force — the force times the offset, exactly — and its bending stress there is 24 times its axial stress.

Tested in The member with only one direction, at the figure it turns on · the two force member ladder.

Stored solids press on a silo wall in proportion to depth, like a liquid with a reduced coefficient.

What decides it: The vertical pressure saturates at γR/μK — 89 kN/m² for the 8 m silo here — and never passes it. At the base of a 30 m fill it has reached 84.6 against a liquid's 270, and doubling the height to 60 m would add four per cent. The relationship is exponential in depth and it has a ceiling.

Tested in The pressure that stops growing, at the figure it turns on · the silo pressure ladder.

The floor of a silo carries what is stored in it.

What decides it: The base of this silo carries 4,254 kN of a 13,572 kN fill. The other 9,318 kN is carried by friction on the walls, at 371 kN per metre of perimeter — which is a vertical load on the wall that a tank of the same fluid does not have, and it is why silo walls fail vertically.

Tested in The pressure that stops growing, at the figure it turns on · the silo pressure ladder.

A deflection can only be computed where a load acts.

What decides it: Put a load of zero there and differentiate anyway. The horizontal movement of a joint carrying no load at all comes out at 5.667 × 10⁻³ by the derivative and 5.667 × 10⁻³ by the unit load — the dummy load is a device for making the derivative exist, not a load the structure ever carries.

Tested in The deflection that is a derivative, at the figure it turns on · the strain energy ladder.

A stiffer building settles less.

What decides it: With uniform soil the mean settlement is the total load over the total spring stiffness and nothing the superstructure does can move it — computed here at 24.0 mm for a limp frame and 24.0 mm for one a million times stiffer, agreeing to 3 × 10⁻¹⁰. What stiffness changes is how unevenly the settlement is distributed, not how much of it there is.

Tested in The settlement that matters is the difference, at the figure it turns on · the differential settlement ladder.

The elastic bending moments are what the beam has, so a design that uses different ones is approximate.

What decides it: The elastic diagram is the one a beam has while every section is still elastic, which is a statement about a load that is not the design load. Any diagram in equilibrium with the load and everywhere inside the section's capacity is safe by the lower-bound theorem — and the redistributed one drawn here satisfies the free-moment identity to machine precision at every value of the redistribution.

Tested in The moment that was moved on purpose, at the figure it turns on · the moment redistribution ladder.

Providing half the shear connectors gives a beam halfway between two beams and one.

What decides it: The degree of interaction is not linear in the connection stiffness and is not linear in anything else convenient. Halving the connection on the 12 m beam here takes αL from 15.9 to 11.2 and the interaction from 96% to 93% — three points, not fifty. The curve is steep only at stiffnesses far below anything a real floor is built with.

Tested in Half the studs, and most of the beam, at the figure it turns on · the partial interaction ladder.

Shear connectors are spaced uniformly because the shear flow they carry is uniform.

What decides it: The elastic shear flow is largest at the supports and zero at mid-span, and the end connectors on this beam are asked for 1.8 times the mean. Uniform spacing is justified by ductility — a connector that deforms sheds what it cannot carry to its neighbours — which is the same argument that lets a cracked web assume every stirrup is at yield.

Tested in Half the studs, and most of the beam, at the figure it turns on · the partial interaction ladder.