What is refuted here — page 5

A model twice as flexible as the building predicts a frequency twice too low.

What decides it: A frequency is the square root of a stiffness, so a factor of two in stiffness is 1.41 in frequency. The same discrepancy looks enormous on a deflection and modest on a period, which is why it is noticed in one calculation and not in the other.

Tested in Stiffer than the model said, at the figure it turns on · the measured stiffness ladder.

Strength is a property of a material, so a member either carries a load or it does not.

What decides it: It carries it for a time that depends on the stress. Timber at 59% of its short-term strength lasts fifty years; at 72% it lasts about six weeks; at 80% about two days. One material, one load, and the answer is a duration rather than a yes or no.

Tested in The load that was left on too long, at the figure it turns on · the duration of load ladder.

The permanent load matters most, because it is applied for the whole life.

What decides it: Time to failure falls as roughly the twenty-first power of the margin over the curve's asymptote, so a stress a third higher is worth several thousand in time. On the history drawn, six hours of wind a year consumes more of the life than fifty years of dead load.

Tested in The load that was left on too long, at the figure it turns on · the duration of load ladder.

Specifying a stronger alloy gives a stronger member.

What decides it: Only where it is not welded. 6082-T6 at 260 N/mm² and 5083-H111 at 125 differ by a factor of two as delivered and by four per cent beside a weld, because the first loses half its proof stress in the heat-affected zone and the second, being annealed already, loses nothing.

Tested in The strength the welder gives back, at the figure it turns on · the heat affected zone ladder.

A sliding bearing releases the structure, so it delivers no horizontal force.

What decides it: It delivers μV, and on the bearing drawn that is 90 kN against a wind force of 120 — three quarters of the load the bearing was put there to release the structure from, arriving from the component that was supposed to do the releasing.

Tested in The roller that is not a roller, at the figure it turns on · the bearing friction ladder.

Halving the load on a bearing halves the friction force.

What decides it: The coefficient is 1.2/(10 + σ) in newtons per square millimetre, so it rises as the load falls. A fifth of the load gives two and a half times the coefficient and half the force, and a bearing at a tenth of its rating is at four times its quoted μ.

Tested in The roller that is not a roller, at the figure it turns on · the bearing friction ladder.

Friction is a single load case.

What decides it: It opposes whichever way the deck is moving, so it reverses with the season and has to be taken in both directions. And it is not present at all until the deck moves, so a case with friction and a case without it are both real and neither bounds the other.

Tested in The roller that is not a roller, at the figure it turns on · the bearing friction ladder.

A stronger steel gives a longer fatigue life.

What decides it: Nothing on an S–N plot depends on the yield stress. The same detail in S355 and S690 lies on the same line, so a member sized on strength in the stronger grade is smaller, carries a larger stress range, and has a shorter life — the stronger steel makes the fatigue worse.

Tested in The detail decides and the steel does not, at the figure it turns on · the detail category ladder.

The equivalent constant stress range is the average range.

What decides it: It is the cube root of the mean cube, which sits far nearer the heaviest band than the mean does. Using an arithmetic average of a traffic spectrum understates the damage by a large factor, and the factor grows as the spectrum widens.

Tested in The detail decides and the steel does not, at the figure it turns on · the detail category ladder.

A bigger object is more vulnerable in an earthquake.

What decides it: For rocking it is the reverse and by a large margin. Uplift happens at a/g = b/h with no size in it at all, and the pulse that overturns rises with the block's diagonal — a block three times the size of the one drawn needs four and a half times the ground motion to topple.

Tested in The block that is safer for being bigger, at the figure it turns on · the rocking ladder.

A rocking system needs damping to come to rest.

What decides it: There is no dashpot anywhere in the model. All the energy is lost at the instants the block lands on its other corner, at a velocity ratio of 1 − 1.5sin²α per landing — so the damping is a property of the block's shape and of nothing else.

Tested in The block that is safer for being bigger, at the figure it turns on · the rocking ladder.

An approximate analysis gives an approximate set of forces, so the error in it is unknown.

What decides it: The forces it returns are in equilibrium with the applied load exactly — the column shears add to the storey shear to machine precision, and every joint balances. That is the condition of the lower-bound theorem, so a frame proportioned for them is safe whatever the real distribution is. What is wrong is the compatibility, and that shows up as a distribution error rather than a capacity one.

Tested in The analysis that assumes the answer, at the figure it turns on · the portal method ladder.

The portal method and the cantilever method are two names for the same thing.

What decides it: They share the hinge positions and differ in the equation that fixes the last unknown. The portal method divides the storey shear in proportion to the bays a column serves; the cantilever method makes the column axial forces vary linearly across the plan, like fibre stresses in a section. On the same four-storey frame they give 0.167 and 0.150 of the storey shear to an outer column.

Tested in The analysis that assumes the answer, at the figure it turns on · the portal method ladder.

A continuous beam carries less moment than a simply supported one of the same span.

What decides it: It carries the same total. Sweep the stiffness of one span over a factor of twenty-five and the mid-span moment and the support moment both move a great deal; their combination does not move at all, and stays at wL²/8 to fifteen decimal places. What continuity buys is putting half of that total where the beam is deepest and best able to take it.

Tested in Two of these move and the third cannot, at the figure it turns on · the static moment ladder.

A support that settles changes the moments in a beam, so the total in a span changes with it.

What decides it: A settlement adds a complete field of bending moments in equilibrium with no applied load whatever. Adding it changes the support moment and the mid-span moment by equal and opposite amounts in the identity, so the sum is untouched — and the check holds to 1e-9 with the settlement in place.

Tested in Two of these move and the third cannot, at the figure it turns on · the static moment ladder.

A bigger edge disturbance reaches further into the shell.

What decides it: The reach contains neither the disturbance nor the load. It is 1/β = 0.78 root Rt, a property of the geometry alone, and the moment is under a twentieth of its edge value within about three of them whether the pressure is 0.1 N/mm² or 10. What the disturbance decides is the size of the moment, not how far it travels.

Tested in The length a structure was never given, at the figure it turns on · the edge disturbance ladder.

A lightly loaded member needs little reinforcement, because there is little to carry.

What decides it: The requirement has no load in it at all. It compares the cracked section's ultimate moment with the moment that cracks the gross section, and the second is f_ctm bh²/6 — width, depth and tensile strength, with nothing applied. A 400 mm slab needs 507 mm² per metre whether it carries a car park or nothing whatever.

Tested in The steel the concrete asks for, at the figure it turns on · the minimum reinforcement ladder.

The 0.26 in the minimum reinforcement rule is a fitted constant.

What decides it: It is a section modulus divided by a lever arm. Setting A_s f_yd z equal to f_ctm bh²/6, with z about 0.95d and h about 1.1d, gives 0.245·(f_ctm/f_yk)·bd — six per cent below the printed 0.26, and the six per cent is the margin rather than the rule.

Tested in The steel the concrete asks for, at the figure it turns on · the minimum reinforcement ladder.

Adding an axial force moves the neutral axis smoothly, so the N–M interaction is a smooth curve.

What decides it: The axis moves at a rate set by the width it happens to be crossing, and a tee's width changes by a factor of twenty-five at the flange. The axis crawls through the flange and shoots through the web — a rate ratio of 27 across the section — and the interaction curve has a kink exactly where the axis crosses the junction.

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

A calculated crack width of 0.28 mm means the cracks will be about 0.28 mm.

What decides it: It means the characteristic value of a scattered population is 0.28. Measured widths on nominally identical specimens scatter with a coefficient of variation near 40%, and the crack spacing itself varies by a factor of exactly two along the member by construction — the closest cracks can be is half the furthest. A member with a calculated 0.28 has cracks at 0.15 and at 0.4 and neither is a defect.

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

Analysing the finished excavation gives the prop forces.

What decides it: It gives one of them. The middle prop of this three-level scheme reaches 214 kN per metre while the dig is at 9.5 m and finishes at 88 — a factor of 2.44, and the larger number is nowhere in the final analysis. Only the lowest prop is worst on the last day, because it is the only one that has had no other.

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

The apparent earth pressure diagram is a pressure distribution.

What decides it: It is an envelope of prop loads measured on real excavations and back-figured into a rectangle. Nothing on a wall is loaded like that — the real pressure is neither rectangular nor constant with time — and the diagram's only claim is that dividing it up between the props reproduces their measured maxima, which here it does to two per cent.

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

Adding a pile at the middle of the group helps the corner piles.

What decides it: It adds to the axial term and to neither second moment, so the moment share of every other pile is unchanged. A pile on the axis of bending carries the mean load and nothing whatever of the moment — which is the same statement as a bolt at the centroid of a bolt group taking no share of the torque.

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

Joints in a segmental lining are a weakness that has to be compensated for.

What decides it: Eight joints in this ring drop the bending moment to 34% of the solid ring's and leave the hoop thrust exactly where it was — because the thrust comes from the mean ground stress and the moment from the difference. The joints are not tolerated; they are the reason the lining works.

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

Adding damping to a structure cannot make it less stable.

What decides it: Adding any internal damping at all, however small, drops this column's critical load from 2.0858 to 41/28 = 1.4643 — thirty per cent — and the limit as the damping goes to zero is not the undamped answer. Ziegler found it in 1952 and it was taken for an arithmetic error for a decade.

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

A column with two stiffnesses can be checked at the average of them.

What decides it: With the middle third at half the stiffness of the rest, the average is 0.83 of the uniform value and the true critical load is 0.61 — the average over-predicts by 36%, and it does so on the unsafe side. Using the weakest segment throughout gives 0.50 and is safe by about as much as the average is unsafe.

Tested in An average stiffness is not a safe stiffness, at the figure it turns on · the stepped column ladder.

A local weakening reduces a column's capacity in proportion to how much of it is weakened.

What decides it: It depends entirely on where. The same weak band of a sixth of the length costs 4.6% at the end of a pinned column and 32.6% in the middle — a factor of seven for the same amount of material removed from the same column. It is not proportional to the amount either: a middle third at 0.81, 0.50 and 0.18 of the surrounding stiffness leaves the average high by 7%, 36% and 184%, so the error grows faster than the weakening.

Tested in An average stiffness is not a safe stiffness, at the figure it turns on · the stepped column ladder.

A column of uniform section is the best use of a given amount of material.

What decides it: Clausen asked the question in 1851 and Keller answered it in 1960: for a pinned column of similar cross-sections the best distribution beats the uniform one by exactly four thirds, with a spindle profile that tapers to nothing at both pins. The result is reported here rather than drawn; what the figure beside it shows is the eigenvalue machinery any such search is run with, and that every assumed shape it is given returns a load ABOVE the true one — 9.870, 9.882, 10.000 and 12.000 against a reference of 9.8696 — because an assumed shape is a constraint and a constraint can only stiffen a column.

Tested in An average stiffness is not a safe stiffness, at the figure it turns on · the stepped column ladder.

A heavier vehicle is more likely to overturn on a curve.

What decides it: The overturning condition is v²/gR greater than b/2h, and there is no mass in it anywhere. A loaded lorry and an empty one of the same geometry go over at exactly the same speed — 160 km/h on the curve drawn — and only the height of the centre of gravity moves it. Doubling that height halves the ratio and divides the speed by root two.

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

The load on a buried pipe is the weight of the soil above it.

What decides it: It is 64% of it for a rigid pipe in a narrow trench and 172% of it for the same pipe under an embankment — a factor of 2.7 between two conduits with nothing different but which way the ground moved. Both curves start on the prism line at zero depth, because with no depth there is no shear to redistribute anything.

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

Arching is a special effect that needs a special theory.

What decides it: It is one differential equation and it is the silo's. A slice of the prism weighs gamma B dh, carries B d sigma, and sheds 2 K mu sigma dh to the sides — which integrates to Janssen's expression with a trench for a silo. Only the sign of the wall shear differs between the three cases in this essay.

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

The conjugate beam's supports have to be looked up in a table.

What decides it: Each line of the table is one boundary condition read twice. A real fixed end has zero slope and zero deflection; the conjugate's shear is the slope and its moment is the deflection, so it needs zero shear and zero moment — which is a free end. Every entry is derived in a sentence, and none of them has to be memorised.

Tested in The beam whose moment is a deflection, at the figure it turns on · the conjugate beam ladder.

The method is a trick, so it gives an approximate answer.

What decides it: It is exact and it is a different computation from the one it agrees with. Statics on the transformed beam and a double integration with boundary conditions agree here to five parts in a million, which is the trapezium rule at four hundred stations rather than the method — and both agree with PL³/3EI to the same order.

Tested in The beam whose moment is a deflection, at the figure it turns on · the conjugate beam ladder.

The conjugate beam works for any beam.

What decides it: A real beam with r redundants has a conjugate short of r restraints. A propped cantilever's conjugate is free at both ends and loaded — not a structure at all — and the missing equations in the real beam are exactly the missing restraints in the fictitious one. The method covers determinate beams exactly, and the reason is not a convention.

Tested in The beam whose moment is a deflection, at the figure it turns on · the conjugate beam ladder.

A beam's deflection check is a property of the beam.

What decides it: The increment on this 12 m beam is 32 mm if the partitions go up at a week and 14 mm if they go up at a year — over a factor of two, from a date that appears on no calculation. About 44% of the final creep has already happened by four weeks, so an early finish inherits nearly all of it.

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

A stronger concrete is stronger in tension in proportion.

What decides it: f_ctm goes as f_ck to the two-thirds, so the ratio between them falls from 11% at C20 to 6% at C80 — a factor of 1.8 over the range. Every rule that depends on the tensile strength therefore gets relatively harder as the concrete gets stronger, in members whose bending capacity has barely moved.

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

Restrained shrinkage cracking can be avoided by good design.

What decides it: The strain at which concrete cracks is f_ctm over E_cm, about 100 microstrain. Free drying shrinkage is 300 to 600. The ratio is four, so a fully restrained member cracks with certainty — and the only design decision available is how wide the cracks are and how many of them there are.

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

The amplification a soil layer produces is limited by its damping.

What decides it: It is limited by the impedance contrast with the rock beneath, which lets energy leave downward. The peak is one over alpha plus pi zeta over two, and with alpha at 0.055 and 5% damping the first term is a third of the second — assume rigid bedrock, alpha goes to zero, and the model predicts an amplification of 12.7 instead of 7.5.

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

A structure passing through a resonance sees the resonant amplification.

What decides it: It sees whatever it has time to build up. At the sweep rate drawn, a 2% damped oscillator with a steady-state amplification of 25 reaches 23.6; at a rate eight times faster it reaches 12.6, and at fifty times faster it reaches 10. The whole answer depends on one dimensionless group, the sweep rate divided by the square of the damping times the square of the frequency.

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

The largest response occurs at the resonant frequency.

What decides it: It occurs after it, by a lag that grows with the sweep rate — 1.2% of the frequency at the rate drawn and 20% at fifty times it. An instrument reading a run-up therefore sees its largest amplitude while the machine is already above its critical speed, and the response beats afterwards at the difference between the two frequencies.

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

A fabrication tolerance of a few millimetres is too small to matter structurally.

What decides it: Five millimetres in one diagonal of this six-by-four metre braced bay puts 100 kN into both diagonals and 83 into the chords, with nothing applied — a quarter of the force the diagonal was sized for. The erection tolerance on a whole frame, twenty-five millimetres, exceeds the design force on its own.

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

Any misfit produces a force.

What decides it: Only the component of it along the self-stress state does. A misfit orthogonal to that state — equal and opposite errors in the two diagonals here, in the right proportion — produces exactly zero, to machine precision. Which is why a member can be a centimetre out in a direction the structure does not care about.

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

The largest axial force in a pile is at its head, where the load is applied.

What decides it: For the 24 m pile drawn it is 1,282.87 kN at a depth of 13.77 m, against 800 kN at the head and 300 kN at the toe. With no head load and no base resistance at all the maximum sits at L/√2 = 0.7071 of the length, a number containing nothing of the soil, the pile or the load.

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

Restraint cracking is about the temperature drop, so a wall cast in winter is the problem.

What decides it: It is about the length-to-height ratio. At L/h = 2 the restraint at the top of the wall is exactly zero and the cracked height is 50 mm of 3,000; at L/h = 6.67 the top of the wall is still restrained at 0.304 and the whole height cracks. Same concrete, same drop, same season, thirty-two cracks against none.

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

Prestress makes the concrete stronger in shear.

What decides it: Nothing about the concrete changes: f_ct is 1.35 N/mm² with the prestress and without it. What changes is the state of stress the shear is added to. Pure shear has a principal tension equal to its own magnitude, so the limit with no prestress is exactly 1.35; adding 7.14 N/mm² of compression shrinks Mohr's circle and rotates it, and the limit becomes √(1.35² + 7.14×1.35) = 3.39. The strength is the same and the demand on it is not.

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

The worst shear stress in a twisted bar is at its corners, where the material is furthest from the axis.

What decides it: It is zero at a corner and largest at the middle of the longest side. The film comes down to the boundary from two directions at a re-entrant corner of the hole, so its slope — which is the stress — vanishes there. On the 100 mm square relaxed here the peak slope is 67.515 at the middle of a side and the corners carry nothing at all.

Tested in Two volumes, and both of them are torques, at the figure it turns on · the membrane analogy ladder.

Torsion has no plastic reserve worth having, because the outer fibres yield first and the section unravels.

What decides it: The sand heap gives the collapse torque exactly. For a circle it is 2πτ_y r³/3 against πτ_y r³/2 at first yield — a shape factor of exactly 4/3, which the relaxation reproduces as 1.3334. A square is better still at 1.6013, because more of its area is a long way from a boundary. Both are larger than the same sections give in bending.

Tested in Two volumes, and both of them are torques, at the figure it turns on · the membrane analogy ladder.

Adding an internal web to a box girder makes it stiffer in torsion, because there is more material resisting the twist.

What decides it: A centred web adds nothing measurable. For the 3.0 × 1.5 m box drawn, J is 2.36788 × 10¹² mm⁴ with the web and 2.36788 × 10¹² without it — a ratio of 1.000000000000, and the flow in the web comes out at 5.7 × 10⁻¹⁴ N/mm, which is the linear solver's zero. The two cells are identical, so they carry identical flows, so the difference across the wall they share is nothing.

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

The taller the building, the more there is to gain by hanging it.

What decides it: The gain falls with height. At two storeys the saving is 38.4 per cent and at forty it is 4.3, because a tall building's columns are heavily loaded and therefore stocky, and a stocky column has almost no reduction to give away. The argument is strongest for the buildings nobody would ever hang.

Tested in Hung from the top, and nine per cent lighter, at the figure it turns on · the hung structure ladder.

A hung floor and a propped floor end up in the same place.

What decides it: Every hanger is at its full design stress, because a tie has no reason not to be, so the elongation is 6.93 mm a storey all the way down and the bottom floor is 138.5 mm low. The column building's top floor is 119.9 mm low. The hung building moves 15.6 per cent more, and it moves where the doors are.

Tested in Hung from the top, and nine per cent lighter, at the figure it turns on · the hung structure ladder.

The sag is chosen to balance the cable force against the deck depth.

What decides it: It is chosen by the gradient at the ends. The end slope of a parabola is 4f/L, so an accessibility limit of 8 per cent caps the sag at 2 per cent of the span exactly — and that is the sag drawn here, giving a 4.57° approach and 1 in 12.5. The structural consequence, an abutment force of 21,875 kN, is what is left over after a ramp gradient has decided the shape.

Tested in The deck that is its own cable, at the figure it turns on · the stressed ribbon ladder.

The deck's bending stiffness carries the local effects at the abutment.

What decides it: A clamped end would carry θ√(EI·H) = 6,735 kNm, which is 4.21 times the deck's own capacity. No stressed ribbon is clamped: the end curves over a saddle of radius EI/M_cap = 203 m, turning through the end slope over an arc of 16.2 m. The detail exists because the moment does not fit, not because it is elegant.

Tested in The deck that is its own cable, at the figure it turns on · the stressed ribbon ladder.

Buckling needs compression, so a member in pure torsion cannot buckle.

What decides it: It buckles at T = 2πEI/L, into a helix of one full turn over the length for pinned ends. Nothing in the member is in compression before it deflects; the instability comes from the torque acquiring a component about the deflected bar's own transverse axis the moment there is any deflection at all, which drives more of the same.

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

A member stiffer in torsion resists torsional buckling better.

What decides it: The shear modulus does not appear in the critical torque, and neither does J. T_cr = 2πEI/L contains the *bending* stiffness only. A tube and the same tube slit along its length — differing by a factor of several hundred in torsional stiffness — have identical torsional buckling loads, because what buckles is a bending mode driven by a torque.

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

A bigger stiffener makes a stiffened plate stronger.

What decides it: Only up to γ*, and not at all past it. Below the threshold the plate buckles in a mode that carries the stiffener with it and the capacity rises with γ; at γ* = 31.5 for the plate drawn the stiffener becomes a node and the plate buckles between stiffeners at 74.1 N/mm²; above it the capacity is exactly flat, because the sub-panel mode does not deflect the stiffener and therefore cannot tell how stiff it is. A stiffener at 2γ* is worth precisely what one at γ* is worth.

Tested in The rib that is a boundary condition, at the figure it turns on · the stiffener rigidity ladder.

A stiffener works by carrying part of the compression.

What decides it: Carrying compression makes it worse. The stiffener's own area enters γ* through δ = A_s/bt, and it enters it in the numerator: a stiffener that carries load is a stiffener that has to be held up itself, so raising δ raises the rigidity demanded. The stiffener's job is geometric, and the load it takes is a cost of doing it.

Tested in The rib that is a boundary condition, at the figure it turns on · the stiffener rigidity ladder.

Instability needs an imperfection to trigger it.

What decides it: Not this one. A perfectly circular, perfectly straight, perfectly elastic tube reaches a maximum moment and then sheds it, because the ovalisation is *caused by* the bending rather than by any initial deviation. There is no bifurcation on the path at all — the moment simply turns over, at a flattening of exactly 2/9 of the radius, where the secant stiffness has fallen to 67 per cent of the undeformed value and the tangent stiffness is zero.

Tested in The tube that flattens itself, at the figure it turns on · the brazier buckling ladder.

Which mode a thin tube fails in depends on how thin it is.

What decides it: Not for the two buckling modes. M_brazier/M_local = 2√6/(9α), which contains no radius, no thickness, no length and no material — only the shell knockdown factor α. Ovalisation governs a tube made well enough that α exceeds 2√6/9 = 0.5443, and local buckling governs one that is not. Which mode ends a tube is a question about how it was manufactured.

Tested in The tube that flattens itself, at the figure it turns on · the brazier buckling ladder.

The two classical methods of slices agree closely enough to be interchangeable.

What decides it: Fellenius gives 1.081 and Bishop 1.192 at the critical circle — 11 per cent apart, and always in the same direction because Fellenius ignores the inter-slice forces and is statically inconsistent. The gap grows with the friction angle and with the pore pressure; at the trial circle here it is 22 per cent.

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

A big gate needs a big hoist, because the water pushing on it has to be overcome.

What decides it: Not on a radial gate. Every pressure on a circular surface is normal to it, so every one passes through the centre of curvature, so the moment of all 208 kN/m of resultant about a trunnion at that centre is 3.4 × 10⁻¹⁵ kNm — machine zero. A flat gate on the same hinge would deliver 353 kNm. The hoist sees the gate's own weight and 2.6 kNm of friction in the pin, and no water at all.

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