What is refuted here — page 3

The bursting force behind an anchorage is a property of the end block, so a better analysis would pin it down.

What decides it: Three assumed spreads of the longitudinal stress, all satisfying equilibrium exactly — the transverse stress integrates to zero along the axis to one part in a hundred thousand, and the far free face closes to 10⁻¹¹ N/mm² — give bursting forces of 96, 174 and 198 kN on the same block. The grid is not the reason: refining it twice moves each answer by 0.3%.

Tested in The force that splits what it pushes on, at the figure it turns on · the anchorage zone ladder.

A section that cannot be prestressed will show it by failing a stress check.

What decides it: It shows it by having no feasible region at all. Raise the service moment on this section from 640 to 1,400 kNm and there is no combination of force and eccentricity anywhere that satisfies all four — not because any one of them is badly violated, but because the transfer pair and the service pair have stopped overlapping.

Tested in Four inequalities and a wedge, at the figure it turns on · the prestress limits ladder.

Moving the braces apart to make a link is a compromise that gives up most of the frame's stiffness in exchange for ductility.

What decides it: A link a tenth of the bay long keeps 79 per cent of the concentric frame's stiffness. The curve is steep only past about a third of the bay, and a moment frame — the far end of the same sweep — has 11 per cent. The first tenth of the bay costs a fifth of the stiffness and buys the whole mechanism.

Tested in The part that is meant to be weak, at the figure it turns on · the eccentric brace ladder.

A shear link and a long link do the same job with different proportions.

What decides it: They yield by different mechanisms and are credited with rotation capacities four times apart — 0.08 radians against 0.02. The boundary is a property of the section alone: 1.6Mp/Vp and 2.6Mp/Vp, which for the section here are 1,057 mm and 1,717 mm.

Tested in The part that is meant to be weak, at the figure it turns on · the eccentric brace ladder.

The joints of a branching column are complicated because several members meet there.

What decides it: A joint where the branch axes are concurrent and the loads are symmetric carries no moment whatever, however many members meet. What makes the joint hard is that the symmetry is a load-case assumption: take the load off one branch and the same joint carries 300 kNm over a lever arm the drawing shows and the design load case does not.

Tested in The tree that strength does not ask for, at the figure it turns on · the branching structure ladder.

A longer member restrained in the same way is a weaker member.

What decides it: Only until it has room for its preferred wave. The flange here carries 1,925 kN at 12 m, 1,880 at 21 m, 1,879 at 31 m and 1,881 at 40 m — flat to half a per cent over a length range of nearly four to one, because the buckle settles into a half-wavelength of 5,146 mm and simply repeats.

Tested in Held everywhere, and it forgets its length, at the figure it turns on · the continuous restraint ladder.

A section is at its most dangerous when its local and global buckling loads are equal.

What decides it: Measured on the direct-strength arithmetic and referred to the weaker of the two single-mode capacities, the loss at exact coincidence is 2.1 per cent. The worst point is a ratio of 0.45, at 23.2 per cent, and the curve is within a per cent of that for every ratio below about a half.

Tested in Two ways of buckling at once, at the figure it turns on · the mode interaction ladder.

A balanced structure is a structure whose supports are lightly loaded, since the moments cancel.

What decides it: The moments cancel and the forces add. Balancing this 900 kN leaf takes 2,700 kN at a third of the lever arm, and the trunnion then carries 3,600 kN — four times the leaf's own weight, at a bearing that would otherwise have carried 900.

Tested in Balanced, and four times as heavy, at the figure it turns on · the counterweight ladder.

Loading every span is the worst case, because it puts the most load on the beam.

What decides it: It is the worst case for the total load and for nothing else. On three equal spans, loading alternate spans raises the first span's sagging moment by 11 per cent over loading all three, and loading two adjacent spans raises the first interior support's hogging moment by 7 per cent. Neither maximum is produced by the arrangement with the most load on it.

Tested in The envelope is not a structure, at the figure it turns on · the load arrangement ladder.

The end rotation of a beam depends on what it is made of and how deep it is, like every other deflection quantity.

What decides it: The rotation does. The ratio θL/δ does not: integrating the curvature gives exactly 16/5 for a uniform load and exactly 3 for a load at mid-span, with E, I and L cancelling identically. A beam at any stated deflection limit has a known end rotation before anything else about it is decided.

Tested in The angle nobody limits, at the figure it turns on · the end rotation ladder.

A building that satisfies a roof-drift limit satisfies the inter-storey drift limit as well, since the second is the first divided by the number of storeys.

What decides it: The roof drift here is 1 in 328 of the height and the worst storey is 1 in 296 of its own — eleven per cent worse than the average the roof figure implies, and at storey 13 of 30, which no roof measurement locates.

Tested in Two motions with one name, at the figure it turns on · the drift components ladder.

Closing the hoops helps in proportion to the steel added, since the confining pressure is proportional to the hoop area per unit length.

What decides it: The pressure is, and the pressure that reaches the core is not. Between hoops the confinement arches, and the effectiveness coefficient falls from 0.79 at 100 mm centres to 0.46 at 250 — so halving the spacing buys the area twice over and the arching once more.

Tested in Squeezed sideways into a different material, at the figure it turns on · the confinement ladder.

Testing a scale model of a concrete member gives its strength, provided the model is geometrically similar and made of the same material.

What decides it: A 100 mm specimen of this material reads 3.10 N/mm² and a 1,500 mm member of it carries 1.14 — the test overestimates by a factor of 2.71. The specimen and the member are on opposite sides of a transition that happens at a size, and the model is on the flat side of it.

Tested in The bigger one is the weaker one, at the figure it turns on · the size effect ladder.

A load a few degrees out of plane is a few per cent of a problem.

What decides it: The two stress terms are M cos θ over Z_y and M sin θ over Z_z, and Z_y over Z_z is 10.3 for the section drawn here. At five degrees the corner stress is 1.90 times the straight-down answer, and the two terms are already equal at 5.5 degrees. The small angle multiplies the big ratio, and the product is not small.

Tested in Two moments and a neutral axis that obeys neither, at the figure it turns on · the biaxial bending ladder.

The neutral axis is perpendicular to the moment vector.

What decides it: It is perpendicular to the moment vector only when the two second moments are equal, which for a rolled I-section they are not — they differ by a factor of 31. The neutral axis follows tan α = (M_z/M_y)(I_y/I_z), so a moment five degrees off the web puts it at 69.7 degrees, and the corner that ends up furthest from it is the one that decides the section.

Tested in Two moments and a neutral axis that obeys neither, at the figure it turns on · the biaxial bending ladder.

A tie removes the thrust from an arch.

What decides it: It removes the thrust from the foundation and puts it in a member. The arch drawn here still pushes outwards with 2,168 kN at each springing; what changes is that an equal and opposite 2,168 kN runs back along the tie, so the bearings see nothing horizontal. The force is unchanged, its path is closed, and the tie is now the most heavily loaded member in the structure.

Tested in The thrust that never reaches the ground, at the figure it turns on · the tied arch ladder.

An arch with a slender rib is more vulnerable to a support that moves than a massive one.

What decides it: It is the other way round, and by a wide margin. What a spread costs is δ divided by the arch's own flexibility, so a stiff arch loses more thrust for the same movement: 25 mm takes 1.0 per cent off this steel rib's thrust and 100 per cent off a masonry arch of the same geometry a hundred times stiffer. The vulnerability belongs to the stiff one, which is why the buildings that lost arches to spreading foundations were made of stone.

Tested in The thrust that never reaches the ground, at the figure it turns on · the tied arch ladder.

The principle holds for any member, because it is a consequence of equilibrium.

What decides it: It is a consequence of a length scale, and the length scale is not always the section's. A bimoment on a thin-walled open section is self-equilibrating — no force, no moment, nothing a resultant can see — and it decays over √(EI_w/GJ), which for the 400 mm section here is 1.5 m, or 3.8 depths. On a longer, thinner section it is many more.

Tested in How far a wrong load reaches, at the figure it turns on · the Saint-Venant's principle ladder.

Shear reinforcement is shear reinforcement, so a load applied at the bottom of a beam needs the same links as one applied at the top.

What decides it: The two models have identical chord forces, identical struts and identical support reactions — and the vertical at the load carries nothing in one and the whole 600 kN in the other, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That is 1,200 mm² of hanger steel against the 96 mm² the shear calculation asks for over the same length, and it is additional to it rather than instead of it.

Tested in The load that has to be lifted, at the figure it turns on · the indirect support ladder.

A sectional analysis will find it, because the shear at that section goes up.

What decides it: The shear diagram is the same whichever face the load arrives on — it is fixed by the reactions and the load position, and neither has changed. A sectional model has no variable for which face the load came in on, which is exactly why the requirement is invisible to it and why the check is written into codes as a separate clause rather than as a term.

Tested in The load that has to be lifted, at the figure it turns on · the indirect support ladder.

The mode with the lowest elastic buckling stress is the one that governs the strength.

What decides it: Local buckling here is at 41 N/mm² and distortional at 287, a factor of seven apart, and it is the distortional mode that decides the member. Local buckling has an enormous post-buckling reserve — the plate sheds its middle and carries on — and the distortional mode has very little, because there is no stiffer neighbouring strip for the load to move into.

Tested in The mode between the two that get checked, at the figure it turns on · the distortional buckling ladder.

A column checked against its buckling load on the day it is loaded is safe thereafter, because nothing about it changes.

What decides it: Its modulus changes. Creep takes the effective modulus to E/(1 + φ), and the buckling load goes with it — from 11,580 kN on the day to 3,309 in the long term for a creep coefficient of 2.5, which is 29 per cent of it. A column at 46 per cent of its day-one critical load has no long-term equilibrium at all and diverges after 55 days.

Tested in The column that fails years later, at the figure it turns on · the creep buckling ladder.

A cracked reinforced concrete beam deflects according to its cracked second moment.

What decides it: It deflects according to an average of two curvatures, and the average is not either of them. The beam here has a gross second moment of 4.55 × 10⁹ mm⁴ and a cracked one of 1.50 × 10⁹ — a factor of three — and at service load it deflects 23.2 mm against 25.2 for the fully cracked section and 8.3 for the uncracked one. Part of the beam has never cracked, and the part that has is still carrying tension between the cracks.

Tested in Stiffer than its cracked section says, at the figure it turns on · the tension stiffening ladder.

A load at an angle to the grain has a strength between the two extremes, so interpolating between them is a reasonable first estimate.

What decides it: A straight line between 21 and 2.5 N/mm² gives 11.8 at forty-five degrees. Hankinson's formula gives 4.47 — 38 per cent of the linear guess, and 21 per cent of the strength along the grain. The weak direction starts governing as soon as it has any component at all, because the interpolation is in the reciprocals rather than in the strengths.

Tested in The material that has a direction, at the figure it turns on · the anisotropy ladder.

A collapse mechanism gives the collapse load.

What decides it: It gives an upper bound, which is the collapse load only if it happens to be the lowest of them. On the frame here the beam mechanism returns a load factor of 1.00 and the sway mechanism 1.25, against a true collapse factor of 0.714 — so using either as a capacity claims 40 or 75 per cent more than the frame has, on the unsafe side, from a work equation that is arithmetically correct.

Tested in Two ways of being wrong, at the figure it turns on · the bound theorems ladder.

A longer anchorage always anchors more force.

What decides it: Not if the bond is elastic and brittle. The anchorable force goes as tanh(αL), which has a ceiling: past about two decay lengths the extra bar transfers nothing, and for the bar drawn here the ceiling is 80 kN against the 137 kN the bar itself carries. What lets a real anchorage work is not its length but the bond's own ductility.

Tested in The force that arrives along a length, at the figure it turns on · the bond ladder.

The web of a beam carries the shear.

What decides it: Only if the chords are parallel. For a member whose depth is proportional to its bending moment — the triangular cantilever every crane jib and every stayed pylon is — the web shear is exactly zero along the whole length, because the two inclined chord forces resolve to the applied shear by themselves.

Tested in The shear the chords take, at the figure it turns on · the inclined chord ladder.

The elastic and plastic neutral axes of a section are the same line.

What decides it: They coincide only where the section is symmetric about the axis of bending. The elastic axis is the centroid, because stress is proportional to distance from it; the plastic axis is the equal-area axis, because every fibre carries the same stress and only areas matter. For the tee drawn here the two are 69 mm apart on a 400 mm section — so the section moves its own neutral axis as it yields, which no elastic calculation contains.

Tested in What is left after the first fibre yields, at the figure it turns on · the shape factor ladder.

A section's torsional stiffness depends on how its material is arranged, in the way its bending stiffness does.

What decides it: Only if it is closed. For an open section J is Σbt³/3, which contains the plate lengths and thicknesses and nothing whatever about the arrangement — so a channel, an angle, a tee and a flat strip rolled from the same plate have identical torsion constants. No other section property in this collection behaves that way; all four have completely different second moments, radii of gyration and shear centres.

Tested in The slit that costs a factor of six hundred, at the figure it turns on · the torsional constant ladder.

A cable-stayed bridge is a suspension bridge with straight cables.

What decides it: The two carry load by opposite mechanisms. A suspension cable is a funicular whose shape carries the load and whose deck exists to distribute it; a stay is an elastic support whose stiffness carries the load and whose deck is a continuous beam on springs. Nothing in this essay is a catenary, and the design quantity — EA sin³α/h — has no sag in it at all.

Tested in The cable that is a spring, at the figure it turns on · the Cable-stayed ladder.

An air-supported roof needs a lot of pressure, because it is holding up a building.

What decides it: It needs the load per unit plan area and nothing else. A fabric roof at 0.25 kN/m² under 0.6 of snow wants 850 pascals to lift it and about 1,200 to keep the membrane taut — 1.2 per cent of an atmosphere, 120 millimetres of water, and less than the difference across an ordinary internal door on a windy day.

Tested in Held up by the air inside, at the figure it turns on · the pneumatic structure ladder.

The worst case for the fabric is the heaviest load on it.

What decides it: The membrane carries the NET outward pressure, so the worst membrane force is not under the snow the pressure was raised for. It is the moment the snow melts and the pressure has not yet been turned down: the net pressure trebles and the fabric force trebles with it, under a load case nobody would write down.

Tested in Held up by the air inside, at the figure it turns on · the pneumatic structure ladder.

A plastic hinge in the rafter of a pitched portal forms at the apex.

What decides it: Only under a load symmetric enough for it to. Sweeping the hinge position along the rafter of the frame drawn here finds the lowest collapse load at 63 to 77 per cent of the way from eaves to apex, depending on how hard the frame is being pushed sideways — and assuming the apex overstates the collapse load by up to three per cent, on the unsafe side.

Tested in The point the mechanism turns about, at the figure it turns on · the instantaneous centre ladder.

A four-legged sling shares the load between four legs.

What decides it: The four leg forces all pass through the hook, so they contribute no moment about it and only the three force equations are available — four unknowns, three equations, one degree of static indeterminacy. What settles it is stiffness, and a millimetre of length difference in a four-metre wire rope leg moves about two kilonewtons between the diagonals. The industry designs a four-leg sling as though two legs carried everything, and the arithmetic says so.

Tested in The angle that doubles the force, at the figure it turns on · the rigging ladder.

Sling angle is a rule of thumb.

What decides it: It is a division. The leg force is W/(n sin β) exactly, which is 0.58 W at sixty degrees, 0.71 at forty-five, 1.00 at thirty and 1.93 at fifteen. The thirty-degree limit is where the curve passes one, and the reason it is a limit is that below it the leg force rises faster than any angle a rigger can judge by eye.

Tested in The angle that doubles the force, at the figure it turns on · the rigging ladder.

In a lap splice the bolts share the load equally.

What decides it: Only in the limit of a very short joint. Solving the compatibility of the two plates gives a hyperbolic cosine in the bolt forces with its minimum at the centre, so a joint of eight bolts has its end bolts at 1.09 of an equal share and its middle ones at 0.94. At twenty-four bolts the end bolt is at 1.89 and the elastic efficiency has fallen to 0.530.

Tested in The bolts that do not share, at the figure it turns on · the Long-joint ladder.

The code's long-joint reduction is the elastic non-uniformity, tabulated.

What decides it: It is milder — 0.75 at the floor against an elastic 0.59 for a twenty-four bolt joint — and it starts at fifteen diameters rather than where the distribution first becomes uneven. It is a statement about how far the end bolt can deform before the others catch up, not about how much it carries while everything is elastic.

Tested in The bolts that do not share, at the figure it turns on · the Long-joint ladder.

The worst speed for a bridge is the fastest one.

What decides it: The response against speed is a set of spikes, not a rising curve. They sit at v = d·f₁/k for integer k, which for an 18-metre coach on a 4 Hz span is 259, 130 and 86 km/h — all operating speeds — and between them the response is a fraction of the peak. Going faster than the first resonance is safer than sitting on it.

Tested in The train that arrives in time with itself, at the figure it turns on · the moving load resonance ladder.

If the dynamic response is too large, make the bridge stronger.

What decides it: Strength does not appear anywhere in the equation. What appears is mass, stiffness through f₁, and damping. Stiffening the span raises f₁ and moves the resonant speed up, which removes the problem only if it moves it out of the operating range; adding mass moves it down. Damping is the one change that reduces the peak without relocating it.

Tested in The train that arrives in time with itself, at the figure it turns on · the moving load resonance ladder.

Base isolation makes a building stronger against an earthquake.

What decides it: It makes it softer, and the superstructure above the plane is often designed for less strength than it would otherwise need. What falls is the DEMAND: shifting from half a second to two and a half moves the structure down the spectrum's falling branch and cuts the base shear by a factor of ten, and no part of that is a strength.

Tested in Made weaker on purpose, at the figure it turns on · the base isolation ladder.

Isolation is always beneficial — worst case, it does nothing.

What decides it: On soft ground the argument runs the other way. Compliant ground lengthens a 0.6 s building's period by 1.51 times at a shear wave velocity of 150 m/s, and what follows is not the benign case: the base shear falls, the displacement rises by 82%, and the effective damping falls from 5% to 1.5% because rocking radiates almost nothing back. A spectrum whose plateau extends past the period a designer is shifting towards removes the benefit as well. Mexico City in 1985 is the standing example, and it is why isolation is a site decision before it is a structural one.

Tested in Made weaker on purpose, at the figure it turns on · the base isolation ladder.

Restraining a member always raises its buckling load without changing its mode.

What decides it: The soil around a pipe adds a term that FALLS with the lobe count, because a long lobe has to push more soil out of the way than a short one. So the two-lobe mode stops being the cheapest and the pipe buckles into four or five — at eight times the bare pressure, in a shape a bare ring never takes.

Tested in The pressure that needs no direction, at the figure it turns on · the ring buckling ladder.

The stiffness method is a more accurate way of analysing a frame.

What decides it: It is the same equilibrium and the same compatibility a hand method writes, in an order a machine can follow. Assembled from the same element behaviour it returns the same answers to machine precision — the propped cantilever's prop comes out at 3wL/8 to a part in a billion — and where it differs from moment distribution the difference is that moment distribution was stopped early.

Tested in The matrix that replaced the hand methods, at the figure it turns on · the stiffness method ladder.

Bandwidth is a property of the structure.

What decides it: It is a property of how the nodes were numbered. The same one-bay, twenty-storey frame numbered column by column has a bandwidth of 65 and numbered storey by storey has 8, which is a factor of sixty-six in the work of solving it — for identical matrices containing identical numbers in different places, and identical answers.

Tested in The matrix that replaced the hand methods, at the figure it turns on · the stiffness method ladder.

A thicker wall carries more hoop force, because it is stronger.

What decides it: It carries exactly the same force. N = pR contains no thickness, so a 5 mm wall and a 20 mm wall under the same pressure at the same radius each carry 300 kN per metre — the thicker one simply carries it at a quarter of the stress. The thickness is in the stress and not in the force, and the two are not the same statement.

Tested in The force that is only a radius, at the figure it turns on · the hoop tension ladder.

A pressure vessel is equally likely to fail in either direction.

What decides it: The longitudinal force in a closed cylinder is exactly half the hoop force, for every pressure and every radius, because the end cap presents half the projected area per unit of cut. So the hoop stress is twice the longitudinal one and a vessel splits along its length, which is what every burst boiler in the historical record did.

Tested in The force that is only a radius, at the figure it turns on · the hoop tension ladder.

The coupling beam at the bottom is the one to design; the rest are the same or lighter.

What decides it: The bottom beam carries 87 kN and the worst carries 222, at level six of twenty — 30% of the way up. The axial force in the walls grows fastest where the two are trying hardest to bend differently from each other, which is not at the base, and detailing the ground-floor beam and repeating it upward under-reinforces the ones that decide the system.

Tested in Two walls that agreed to be one, at the figure it turns on · the wall coupling ladder.

Any reasonable smooth shape over the compression zone would do about as well.

What decides it: A triangle is 37% low and a full rectangle 20% high, on the same compression zone and the same concrete. Neither matches either integral. The equivalent rectangle works because of the two integrals and for no other reason, which is why it can be visibly unlike the curve and still be exact.

Tested in Deliberately the wrong shape, at the figure it turns on · the stress block ladder.

A stronger concrete needs different block factors.

What decides it: Not below C50. Alpha and beta are 0.8095 and 0.4160 at C20 and identical at C50, because they are properties of the SHAPE of the normalised curve and the design strength scales the whole curve without changing its shape. What changes them is the crushing strain or the curve exponent — the strain at 0.003 rather than 0.0035 gives 0.7778 and 0.4048.

Tested in Deliberately the wrong shape, at the figure it turns on · the stress block ladder.

The method is exact, because it makes no assumptions.

What decides it: It makes three, and they are the same three the closed forms make. Plane sections stays plane; each strip is at one stress; and the material's law is the same in the member as it was in the specimen. All three are visible in the transformed section, which is those same assumptions solved by hand. What the fibre model removes is the algebra, not the modelling.

Tested in The section calculation with no formula in it, at the figure it turns on · the fibre model ladder.

Adding load to a structure always makes it less safe.

What decides it: A pinnacle's weight has no lever arm about the pier's own centre, so it adds to the vertical force in the denominator of e = M/N and nothing to the moment in the numerator. The eccentricity at the base of this pier falls from 0.503 m to 0.447 m for 55 kN on top, which is what takes it back inside the middle third. Nothing about the masonry has changed.

Tested in The weight that makes it safer, at the figure it turns on · the buttress ladder.

The worst section in a pier is at its base, where the vertical force is greatest.

What decides it: Take the same pier with a thrust at five degrees rather than twenty-five and the base eccentricity falls to 0.594 m, comfortably inside a 2.68 m base — and the pier does not stand, because the line has left the stone near the top, where the pier is only 1.6 m wide and its own weight has not yet accumulated. The critical section is wherever the ratio of eccentricity to half-width peaks, and it moves with the angle of the thrust.

Tested in The weight that makes it safer, at the figure it turns on · the buttress ladder.

A grid of members following a shell's surface behaves like the shell.

What decides it: It carries two of the shell's three membrane resultants and none of the third. Four pin-jointed bars in a quadrilateral have zero in-plane shear stiffness, so N_xφ — which is how an asymmetric load reaches the end frames — has no path at all. The result is not a soft shell but a mechanism, and the grid is a row of arches that discovers this the first time the snow lies on one side.

Tested in A shell only if the grid takes shear, at the figure it turns on · the gridshell ladder.

A longer girder is proportionally harder to lift.

What decides it: It is a fourth power. z̄ goes as wL⁴/EI, so the same section at 30, 36, 40 and 44 m has factors of safety of 13.3, 6.9, 2.3 and none. The margin does not decline gently and then run out; it falls away and then stops existing, and there is no warning in the numbers a lift plan usually contains.

Tested in Hung from above and still unstable, at the figure it turns on · the lift stability ladder.

A slender masonry wall fails by buckling, so Euler's load is the thing to check.

What decides it: Euler's load for a 215 mm wall three metres high is 5,449 N per mm of run. The load the eccentricity rule allows is 565. A factor of 9.6 sits between them, and it does not close for any wall anybody builds — at a slenderness of 23 the ratio is still 10. The wall runs out of bearing width long before it runs out of stability, and no masonry code contains a critical stress.

Tested in It does not buckle, it runs out of width, at the figure it turns on · the wall slenderness ladder.

The largest pressure on a building is on the windward face.

What decides it: It depends on what is inside. With a dominant opening on the windward face the internal coefficient goes to +0.7 and the side faces reach a net −1.26 kN/m² against the windward face's +0.09. The cladding on the sides is then carrying fourteen times what the front is, and the front is nearly unloaded.

Tested in Most of it is suction, at the figure it turns on · the wind pressure ladder.

A different choice of redundants gives a different, less accurate answer.

What decides it: It gives the same answer. Releasing hinges over the supports of an 8, 10, 8 metre beam and releasing the supports themselves are two entirely different released structures with different shapes and different deflections, and the bending moment diagrams they produce agree to nine parts in 10^15 — machine precision. The choice is about arithmetic, not about the structure.

Tested in Choose what to take away, at the figure it turns on · the force method ladder.

The twisting in a plate is a Poisson's ratio effect and vanishes for a material with no Poisson coupling.

What decides it: Set nu to zero and the twist share goes UP, from 30.7% to 38.4%, while the deflection coefficient does not move at all. The twisting moment is proportional to (1 − nu), so removing the coupling increases it. It is a geometric consequence of a surface having two curvatures and a cross-derivative, not a material one.

Tested in A third of the load crosses sideways, at the figure it turns on · the plate torsion ladder.

A slab designed by the strip method is under-designed, since it ignores a mechanism.

What decides it: It is over-designed. Ignoring the twist and giving the whole load to two strips in bending is a lower-bound solution, so it cannot be unsafe — it is simply heavier than it needs to be, by exactly the fraction the twist would have carried. What it does get wrong is the corner, where the reinforcement it produces is in the wrong direction.

Tested in A third of the load crosses sideways, at the figure it turns on · the plate torsion ladder.

Steel is stiffer than timber, so a steel beam is lighter than a timber one of the same stiffness.

What decides it: Steel is nineteen times stiffer and nineteen times denser, and for a beam free to choose its depth the index is sqrt(E)/rho — where timber wins by 4.28 to one. The depth is the free variable and it is what a low modulus is spent recovering, at a mass penalty that is a square root rather than a first power.

Tested in The ranking belongs to the load case, at the figure it turns on · the material index ladder.