What is refuted here — page 10

A plate buckles into the shape that is cheapest to bend it into.

What decides it: The cheapest shape to bend a plate into has a half-wave 1.73 times its width, and the shape it buckles in has a half-wave equal to its width. The difference is the work term, which rewards a short wave, and the buckle is the compromise between the two rather than the minimum of either.

Tested in Four was never a fact about plates, at the figure it turns on · the essays about plate buckling.

A deeper web gives a class 4 girder more working material in compression.

What decides it: It gives it almost none. The effective width of the compression zone of a 12 mm web in S355 runs 593, 620, 633 and 646 mm as the web goes from 1200 to 1800 to 2400 to 3600 mm deep — nine per cent for a threefold change in depth — because the leading term of that width is 69.5εt and contains neither the depth nor the gradient.

Tested in Classified by a gradient it does not have, at the figure it turns on · the essays about plate buckling.

A bolted end plate's tension resistance is the sum of its bolt rows' resistances.

What decides it: Three rows at 90 mm pitch on a 12 mm plate carry 377 kN together and 657 kN added up — 57 per cent of the sum. The rows fold the plate on 429 mm of hinge line rather than on the 746 mm that three separate patterns would need, and 746 mm of fold does not exist in a plate whose rows are 90 mm apart.

Tested in The rows have to share one fold, at the figure it turns on · the essays about prying.

Adding another bolt row to a moment end plate adds another row's worth of tension.

What decides it: It adds the pitch. Each row after the second contributes 90 mm to the group's effective length where a row standing alone contributes 249 — 36 per cent of a row — so the third row of that group is worth 79 kN against the first row's 219.

Tested in The rows have to share one fold, at the figure it turns on · the essays about prying.

Preloading a bolt in tension makes the connection stronger.

What decides it: The two lines meet. An M20 preloaded to 171 kN in a 25 mm tee flange carries the same bolt force as an untightened one at every applied load above 138 kN, and the failure load is identical. What preload changes is the path, not the destination.

Tested in The bolt that was already stretched, at the figure it turns on · the essays about prying.

The bolt in a preloaded joint carries the preload plus the applied load.

What decides it: It carries the preload plus 18 per cent of the applied load. The clamped plates are a spring of 3,781 kN/mm in parallel with a bolt of 857, so the bolt takes 857/4,638 of any extension the joint is given, and the rest of the applied load goes into unloading the contact.

Tested in The bolt that was already stretched, at the figure it turns on · the essays about prying.

To compute a deflection of an indeterminate beam by the unit-load method, the unit load has to be applied to the indeterminate beam.

What decides it: The unit load can stand on any determinate structure the beam releases to. A propped cantilever of 8 m under 4 kN/m, asked for its movement at 4 m, gives 85.33/EI with the unit load on the cantilever with no prop, on the simple span with no fixity, on a girder with a hinge put in at 2 m, and on the propped cantilever itself.

Tested in Any structure will carry the unit load, at the figure it turns on · the essays about virtual work.

The principle of virtual displacements can find any force the principle of virtual forces can find.

What decides it: A virtual displacement statement is an equilibrium statement. Asked for the prop force of the same propped cantilever with three different admissible displacement fields, its residual is zero at every trial force from nothing to 24 kN, while the gap at the prop from a virtual force is a line through zero at exactly 12 kN.

Tested in Any structure will carry the unit load, at the figure it turns on · the essays about virtual work.

A response-spectrum analysis of a building whose stiffness is almost exactly centred will report almost no torsion.

What decides it: Not if it combines modes by the square root of the sum of squares. At an eccentricity of 0.20 m the two coupled modes are 2 per cent apart and correlated 0.96, the full combination gives a base torque of 2,401 kNm, and the root-sum-square gives 12,233 kNm, five times as much, from two modal torques that very nearly cancel.

Tested in The twist the combination rule invents, at the figure it turns on · the essays about mode shapes.

If the modal periods and damping ratios of an analysis are correct, its modal response is correct.

What decides it: For a 1,200 tonne isolated building with 20 per cent damping at the bearings and 2 per cent above, the classical analysis gives both modes' periods and damping ratios within 0.2 per cent of the exact complex eigenvalues, and a storey drift at the superstructure's frequency 6.1 times smaller than the exact response.

Tested in The damping that belongs to no mode, at the figure it turns on · the essays about mode shapes.

More damping in an isolation system always reduces the demand on the building above it.

What decides it: Raising the bearings' damping from 20 to 50 per cent takes the white-noise storey drift from 0.34 to 0.29 of its lightly damped value, where the classical analysis promises 0.32 to 0.20. The damping force at the bearings drives the superstructure's own mode: at 40 per cent the drift at that mode's frequency is 11.8 times what the classical analysis gives.

Tested in The damping that belongs to no mode, at the figure it turns on · the essays about mode shapes.

Checking a four-legged body along its two axes and its two diagonals finds its weakest direction.

What decides it: With one of the tank's four pads at 8,000 kN/m instead of 20,000, the weakest direction is 15 degrees off an axis, at 1,039 kN. Along the axes it is 1,066 and along the diagonal toward the soft pad 1,223; the rigid-ground hull's weakest is 1,333.

Tested in It tips inside its own hull, at the figure it turns on · the essays about overturning.

The diagonals of a lattice tower carry its shear, so they are heaviest at the base.

What decides it: Only if the legs are parallel. With the same loads, a face tapering from 6 m to 3 m has diagonals of 20.8 kN at the base and 23.8 at the top, and one tapering to 1.5 m has 15.0 at the base and 34.2 at the top, because converging legs carry part of the shear and the diagonals carry the moment about the point where the legs' lines meet.

Tested in The cut that needs a joint first, at the figure it turns on · the essays about method of sections.

Between first yield and collapse, a structure under repeated loading will settle into a stable residual state.

What decides it: It settles only below the shakedown load, which for the two-span beam is 126.3 kN against a first yield of 123.1 kN and a collapse load of 150. Above 126.3 kN no straight residual moment line keeps the midspans and the support inside the plastic moment for every load state, and the beam ratchets.

Tested in The load it can carry once, at the figure it turns on · the essays about shakedown.

For a given number of bolts, the layout with the larger polar moment of inertia is the stronger bolt group.

What decides it: A ring of six bolts with a polar moment 4 per cent smaller than a 150 by 75 mm rectangle of six has a weakest capacity of 150.3 kN against the rectangle's 135.8, and is the stronger of the two in 306 of 360 load directions, because its furthest bolt is 70.4 mm from the centroid rather than 83.9.

Tested in The bracket pushed from the wrong side, at the figure it turns on · the essays about bolt group.

The instantaneous centre method finds the same extra capacity in any bolt group, so a factor taken from one connection can be carried to another.

What decides it: Three rows of two bolts under a load of unknown direction gain 23.8 per cent at their weakest direction; a ring of six with a 4 per cent smaller polar moment gains 14.3. The factor is a property of how unevenly the elastic method loaded the group, not of the method that replaces it.

Tested in The better method flatters the worse layout, at the figure it turns on · the essays about bolt group.

A group of six bolts with one missing has five sixths of its capacity.

What decides it: Six bolts at 75 mm each way under a load through a point 150 mm across and 150 mm up carry 135.8 kN at their weakest direction. With the top far bolt missing they carry 90.2 kN — 66 per cent, not 83 — and with the middle near bolt missing, 131.1 kN, which is 97.

Tested in The bolt that was never fitted, at the figure it turns on · the essays about bolt group.

The column analogy is a method for rectangular frames, because the analogous column has to be a section and a section is made of straight pieces.

What decides it: The analogy needs an area, a centroid and two second moments along the member, and a curve has all four. A 30 m parabolic arch of 6 m rise returns a thrust of 375.0 kN and a moment diagram identically zero, which is the exact answer, and a circular arch of the same span and rise returns 383.1 kN and 80.4 kN·m at the springings — 0.03 per cent from a stiffness solution of the same arch.

Tested in The centre that hangs in the air, at the figure it turns on · the essays about Moment-area.

A frame has to be visibly lopsided before the coupling matters.

What decides it: The error rises smoothly from zero and passes 10 per cent of the free moment at a column-height ratio of 0.67 — a 5 m column beside a 3.35 m one — where the principal axes are tilted about six degrees and nothing on the drawing looks unusual.

Tested in The axes that have to be turned first, at the figure it turns on · the essays about Moment-area.

A support movement is a small perturbation that can be covered by the factors of safety on the loads.

What decides it: Ten millimetres on this frame adds 9.5 per cent to its largest moment and thirty adds 29.1, and neither figure contains any load. A factor applied to the load multiplies the load case and leaves the settlement case exactly where it was.

Tested in The load that is not a load, at the figure it turns on · the essays about Moment-area.

The column analogy is a method for rectangular frames and arches, and a frame of an awkward shape needs a different method.

What decides it: A portal, a pitched portal, a stepped frame and one with splayed legs, all at 10 kN/m over 8 m, agree with a stiffness solution of the same frame to 0.24, 0.15, 0.16 and 0.09 per cent of the free moment. Nothing in the construction asks what shape the centreline is.

Tested in Three, and what three is a property of, at the figure it turns on · the essays about Moment-area.

The analogy handles a two-bay frame if it is applied to the outer ring and the middle column is treated as a support.

What decides it: The outer ring of a two-bay portal gives −184.5 kN·m at the knees where the real frame carries −32.7. It is not an approximation: it is the exact answer to a single-bay frame of twice the span, and it is out by 190 per cent of one bay's free moment.

Tested in Three, and what three is a property of, at the figure it turns on · the essays about Moment-area.

Miner's rule is order-blind and the error that makes is small enough to ignore.

What decides it: Two sequences of identical cycles on the same detail fail at 12.36 and 6.36 million cycles. The Miner sum at the moment of failure is 0.99 for the first and 0.68 for the second, so the rule is exact for one arrangement and 48 per cent unconservative for the other.

Tested in The record played backwards, at the figure it turns on · the essays about fatigue.

A thicker plate is worse for fatigue because the crack has further to grow before it breaks through.

What decides it: That reasoning has the sign backwards. Remove the toe's magnification and integrate the same crack to the same fraction of the plate: a 100 mm plate then has a life of 2.05 million cycles against a 25 mm plate's 1.95 — very slightly better, not worse.

Tested in The rule that points sideways, at the figure it turns on · the essays about fatigue.

A force diagram is a device for finding forces, so it can only be drawn once the frame it describes has been fixed.

What decides it: The relation between a frame and its force diagram is symmetric: the reciprocal of the reciprocal is the frame again, every line parallel to the member it came from. A designer can therefore draw the force diagram first and construct the frame it belongs to. A bowstring whose web carries nothing is the frame belonging to the funicular's force diagram with the hangers laid beside it.

Tested in Every space a point, every joint a polygon, at the figure it turns on · the essays about graphic statics.

A truss whose web members carry no force uses less material than one whose web members work.

What decides it: On a 12 m span 2.4 m deep under five 10 kN hanger loads, the bowstring's force times length comes to 1,087 kN·m and the parallel-chord truss's to 1,021, although every diagonal of the bowstring is idle. Its chords are longer and carry the full 37.5 kN thrust from end to end.

Tested in Every space a point, every joint a polygon, at the figure it turns on · the essays about graphic statics.

An arch's reactions depend on the shape of its rib, which is why arches are shaped with care.

What decides it: For a three-hinged arch they depend on the three hinges alone. A parabolic arch, a circular arch and a portal frame through the same springings and crown hinge, under 100 kN at the quarter point of a 20 m span, all push on their abutments with 90.1 and 55.9 kN and a thrust of 50 kN. The rib's shape sets the bending — 187.5, 177.2 and 172.5 kN·m — and nothing else.

Tested in The arch that is only its three hinges, at the figure it turns on · the essays about graphic statics.

A funicular polygon can be drawn to pass through any points a designer chooses, by choosing the pole.

What decides it: Through three at most. A pole has two freedoms and the polygon's starting point one more. Put a polygon for four loads through both springings and a crown 5 m up, and at 4 m it passes 0.27 m above the parabolic rib through the same three points. Nothing but a change in the loads can bring it down onto the rib there.

Tested in The polygon that runs out of freedom, at the figure it turns on · the essays about graphic statics.

A two-hinged arch can be solved by drawing, like a three-hinged one, once its geometry and loads are known.

What decides it: The drawing gives a family of thrust lines through the two springings, one for every thrust, and nothing on the paper selects one. Compatibility does: for a parabolic rib under 100 kN at the quarter point the springings stay put only at a thrust of 55.7 kN, against 50.0 with a crown hinge, and finding it needs an integral of the rib's shape.

Tested in The polygon that runs out of freedom, at the figure it turns on · the essays about graphic statics.

A force diagram that closes has been drawn accurately.

What decides it: A joint with two unknown members is closed by the construction itself whatever angle its lines cross at, so the error a shallow crossing makes leaves no gap. A six-panel truss a fortieth of its span deep, drawn with every line up to 0.1° off, has members wrong by up to 1.84 per cent of the largest force and a closing gap of 0.08 per cent.

Tested in How wrong a drawing is, at the figure it turns on · the essays about graphic statics.

An isolated building's higher modes are coupled to the bearings less than its first, so the classical modal analysis is more trustworthy for them.

What decides it: The coupling does fall, from 0.74 of the diagonal terms in the first structural mode to 0.20 in the fourth for a six-storey building on bearings damped 20 per cent. The top storey's exact drift at each mode's own frequency is 3.3, 6.2, 8.6 and 10.9 times the classical, because each higher mode is driven by the ground roughly ten times less than the one below.

Tested in The top floor the bearings shake, at the figure it turns on · the essays about mode shapes.

More damping at the bearings always reduces the accelerations of the floors above them.

What decides it: For six storeys the roof's white-noise acceleration falls to 0.44 of its lightly damped value at about 28 per cent damping, then rises to 0.50 at 58 per cent, while the classical analysis goes on falling to 0.33. Past the minimum, extra damping shakes the top floor harder.

Tested in The top floor the bearings shake, at the figure it turns on · the essays about mode shapes.

The right amount of damping for an isolation system is a property of the bearings.

What decides it: It depends on the building above them. With the total mass and both periods held fixed, the bearing damping that minimises the roof's acceleration is 48 per cent for one storey, 32 for three and 28 for ten. The classical analysis puts it at about 57 per cent for all three.

Tested in The top floor the bearings shake, at the figure it turns on · the essays about mode shapes.

A modal analysis returns a structure's own mode shapes.

What decides it: For a square floor equally stiff both ways there are none to return: every direction is a mode. Four floors made stiffer by one part in a million along 0°, 20°, 45° and 70° return first modes along 90°, 110°, 135° and 160° — a right angle's worth of disagreement from a difference no instrument could detect — and the floor with no asymmetry returns whatever its arithmetic prefers.

Tested in Two modes that are really a plane, at the figure it turns on · the essays about mode shapes.

The square root of the sum of squares errs only for modes that are close but distinct, and otherwise gives the same answer however the modes were computed.

What decides it: For two modes of exactly equal period it depends on which pair the solver happened to return. Under ground motion along 30°, the exact response along x is 0.87 per unit spectral displacement, and the square-root combination gives anything from 0.35 to 0.94 depending on the pair; the complete quadratic combination gives 0.87 for every pair.

Tested in Two modes that are really a plane, at the figure it turns on · the essays about mode shapes.

A low modal assurance criterion between a model's mode and a measured one shows the model is wrong.

What decides it: For a doubly symmetric floor the model can be right to one part in a million and score anything from one to zero, depending on the direction of an asymmetry nobody specified. At 45° it scores 0.50, well below the 0.9 usually taken for a match.

Tested in Two modes that are really a plane, at the figure it turns on · the essays about mode shapes.

A loss of a tenth of the stiffness in part of a structure shows up as a comparable fall in its natural frequency.

What decides it: In a ten-storey shear building a 10 per cent loss in the first storey lowers the first frequency by 1.04 per cent, and the same loss in the top storey by 0.02 per cent. A frequency is weighted by each storey's share of the mode's strain energy, which is 18.9 per cent at the bottom and 0.4 at the top.

Tested in What a mode shape notices that a frequency does not, at the figure it turns on · the essays about mode shapes.

If a structure's measured mode shape still matches its model's, the structure is undamaged.

What decides it: A 10 per cent loss in the third storey leaves the first mode's shape within 0.92 per cent of the roof's movement at every floor, with a modal assurance criterion of 0.99997 against the intact shape. The same damage raises that storey's drift in the mode by 9.6 per cent.

Tested in What a mode shape notices that a frequency does not, at the figure it turns on · the essays about mode shapes.

A mode at zero frequency is a numerical artefact of a model with missing supports, to be filtered out of the results.

What decides it: It is the limit of a real mode. Weaken the spring holding a five-mass chain to the ground from as stiff as its own springs to a millionth of that, and the first frequency falls smoothly from 0.28 to 4.5 × 10⁻⁴ while its share of the mass rises from 88 to 100 per cent. With the spring gone, that mode carries every kilogram under a uniform acceleration and the other four carry none.

Tested in The modes at zero frequency, at the figure it turns on · the essays about mode shapes.

A suddenly applied load produces at most twice the effect of the same load applied slowly.

What decides it: Twice is the limit for one mass on one spring. In a free chain pushed suddenly at one end, the force in the middle spring swings about its average of 0.40 of the load to a peak of 1.22 — three times the average — because several flexible modes reach their peaks close together.

Tested in The modes at zero frequency, at the figure it turns on · the essays about mode shapes.

A bolt layout designed to survive a missing bolt has to spread its bolts evenly, and gives up a good deal of its complete capacity to do it.

What decides it: In a 150 by 150 mm field loaded 200 mm from its centre, the most robust layout found carries 231.6 kN complete and 174.7 with its worst bolt missing. The strongest layout found carries 234.3 and 149.7. The evenly spread ring carries 167.2 and 117.8, and loses the larger share of the two symmetric layouts.

Tested in The strongest layout leans on one bolt, at the figure it turns on · the essays about bolt group.

The layout that carries most with all its bolts is the safest one to have when a bolt turns out to be missing.

What decides it: Of four layouts in the same plate, the strongest carries the most complete, 234.3 kN, and loses the largest share to its worst omission, 36 per cent. The two-column layout loses 28 per cent and the most robust layout found 25.

Tested in The strongest layout leans on one bolt, at the figure it turns on · the essays about bolt group.

A tubular K joint can always be detailed so that the working lines meet on the chord's axis; the eccentricity is a matter of care in setting out.

What decides it: For 114.3 mm braces at 45 degrees on a 168.3 mm chord the lines meet at a gap of 6.7 mm. The smallest gap in which both toes can be welded is 12.6 mm and the smallest permitted overlap is 40.4 mm, so the concurrent joint lies in the band between them that no rule allows.

Tested in The joint whose lines may not meet, at the figure it turns on · the essays about noding eccentricity.

An end-plate joint in tension is as strong as the weaker of its two flanges, each checked as a tee stub on its own.

What decides it: An 18 mm end plate with its bolt 45 mm from the beam web carries 109.3 kN a bolt on a rigid base, and a 14 mm column flange with the bolt 30 mm from the column web carries 104.4. Bolted together they collapse at 92.3 kN, twelve per cent below the weaker, because the prying force at their tips has to be the same force for both.

Tested in The hinge that forms in the other flange, at the figure it turns on · the essays about prying.

A composite precast section can be analysed for long-term effects with one creep coefficient, as long as it is chosen for the right concrete.

What decides it: With the slab cast on a six-week-old beam, giving the slab its own creep but the beam's shrinkage takes 0.16 N/mm² of precompression from the soffit over thirty years; giving it its own shrinkage but the beam's creep takes 0.83. Nearly all of the effect is differential shrinkage, which no creep coefficient represents however it is chosen.

Tested in The slab that shrinks onto a finished beam, at the figure it turns on · the essays about creep.

U-frames can be replaced by an equivalent continuous restraint of stiffness K over a, whatever their spacing.

What decides it: On a 24 m chord with a smeared restraint of 0.35 N/mm per mm, the smeared answer is 1,897 kN at every spacing. Frames every 4 m give 1,548 kN, 18 per cent less; every 6 m, 691 kN, 64 per cent less. The smeared model is unconservative once the spacing passes about two thirds of the 5.1 m half-wavelength.

Tested in A row of frames is not a foundation, at the figure it turns on · the essays about continuous restraint.

A truss sized to be as stiff as possible at one joint is stiff everywhere.

What decides it: Made stiffest at the quarter point, the same truss deflects 11,918 there against the fully stressed 20,125 — and 43,716 at mid-span, more than the equal-area truss's 43,503.

Tested in The truss that is stiff by accident, at the figure it turns on · the essays about truss deflection.

The danger in analysing a nearly critical frame is numerical: the matrix is ill-conditioned and the computer loses accuracy.

What decides it: The two-bar frame's condition number costs the arithmetic three digits at no angle above half a degree. Its own sag reaches a tenth of its rise below 6.19 degrees, its bar force ten times the load below 2.87, and a millimetre of fabrication error changes that force by one per cent below 1.42.

Tested in Rigid by every test, and still folding, at the figure it turns on · the essays about determinacy.

A formula that uses the peak moment and the moments at the quarter points captures the shape of the diagram.

What decides it: The fixed-ended beam's diagram and a smooth wave through the same four readings both give the formula 1.923. Their exact factors are 1.723 and 1.409, twenty-two per cent apart.

Tested in The formula that reads four numbers, at the figure it turns on · the essays about moment gradient.

Adding weight to a body always makes it harder to overturn; the restoring moment is the weight times half the base.

What decides it: A block 6 m square on ground bearing 300 kPa at most, pushed 8 m up, tips at 958 kN when it weighs half of the ground's 10,800 kN capacity and at 561 kN when it weighs 81 per cent of it — against 3,260 kN on rigid ground.

Tested in The ballast that helps it over, at the figure it turns on · the essays about overturning.

Rounding every member of a truss up to the next section in a catalogue throws away the stiffness that careful proportioning found.

What decides it: With sections 25 per cent apart in area, the rounded Pratt truss's mid-span deflection times its steel is 1.050 to 1.066 times the least possible, against 1.047 to 1.053 for the same truss unrounded. The rounding adds steel, and across catalogues with steps from 10 to 50 per cent the steel it adds buys deflection at 0.998 to 1.030 times the rate any steel in the unrounded proportions would.

Tested in The calculus answer, rounded, is the worst one, at the figure it turns on · the essays about truss deflection.

When a deflection limit governs, the best a catalogue allows is the continuous optimum with each member rounded up to the next section.

What decides it: Made 1.5 times as stiff as its strength design, the truss needs 1,984 units of steel at the continuous bound, 1,997 at the exact discrete optimum and 2,009 stepping one member at a time; rounding the continuous optimum up needs 2,135. Across stiffening factors from 1.25 to 3 the rounded optimum is 7 to 13 per cent above the discrete one, the worst of the four catalogue methods.

Tested in The calculus answer, rounded, is the worst one, at the figure it turns on · the essays about truss deflection.

The rotation a bridge bearing must permit is governed by the live load and the temperature gradient, since those are the terms that change once the beam is in place.

What decides it: For a 12 m pretensioned beam with a 160 mm slab, release and the creep before erection turn each end 7.9 milliradians upward before the bearing exists. Every later term added regardless of sign — slab, surfacing, creep in service, imposed load and both temperature differences — is 4.8. With a level plate the bearing is asked for 9.2 milliradians plus its allowance.

Tested in The angle made in the casting yard, at the figure it turns on · the essays about end rotation.

In a single lap joint each plate carries half the eccentricity moment, P·e/2, at the end of the overlap, whatever its thickness.

What decides it: A 10 mm plate lapped on a 20 mm plate, 100 mm wide, over 60 mm, carrying 60 kN: where the thin plate enters the overlap it carries 0.49 of P·e/2 and the thick plate 1.34. Two equal 10 mm plates each carry 0.86, which is Goland and Reissner's factor.

Tested in The thicker plate takes the bend, at the figure it turns on · the essays about single lap.

A resilient pad under a forge hammer's anvil isolates the foundation block from the blow.

What decides it: For a 150 t foundation on soil with a 30 t anvil, the block's largest movement after one 18 kN·s blow is 0.86 mm on a rigid seat, 1.14 mm on a pad at four times the foundation's frequency and up to 1.36 mm near twice it. Only a pad softer than 0.6 of the foundation's frequency moves this block less than a rigid seat, and on it the anvil moves 11 mm per blow.

Tested in The pad that makes the blow worse, at the figure it turns on · the essays about vibration isolation.

In a K girder symmetric about mid-span, symmetry supplies the equation the central joint lacks, so the girder can be solved by sections under a symmetric load.

What decides it: The girder's one self-stress state is itself symmetric about mid-span: tension in the four diagonals at the central joint, compression in the four central chord members and in the half-verticals at the outer side of each middle panel. A symmetric load can call on it as freely as any other, and 100 kN at every top joint does, shifting the middle diagonals to 48.4 kN in tension and 63.4 kN in compression.

Tested in The redundancy only the sun can find, at the figure it turns on · the essays about method of sections.

A statically indeterminate member only matters when the load is large.

What decides it: With no load at all, one middle diagonal made 1 mm too long puts 23.6 kN into each of the four; the top chord 20 °C warmer than the rest, 30.4 kN; both chords 20 °C warmer than the web, 60.7 kN — against 63.4 kN from a uniform load of 100 kN a joint.

Tested in The redundancy only the sun can find, at the figure it turns on · the essays about method of sections.

A precast beam made continuous over a pier is a continuous beam from then on, so the moment at the pier is hogging.

What decides it: For two 12 m pretensioned beams joined with the slab at 28 days, the moment at the pier after thirty years is +214 kN·m, sagging — above the 159 kN·m that cracks the bottom of the joint. Joined at 90 days it is +59; at a year, −164, hogging.

Tested in The pier that bends the wrong way, at the figure it turns on · the essays about creep.

The deck slab's shrinkage relieves the joint, so a drier climate is the worse case for it.

What decides it: Drier air makes the prestress creep more and the slab shrink more, and at 28 days the thirty-year moment stays between +212 and +225 kN·m from 40 to 70 per cent humidity. In wet air the slab hardly shrinks, and the moment rises to +260 at 80 per cent and +328 at 90.

Tested in The pier that bends the wrong way, at the figure it turns on · the essays about creep.

U-frames carry the largest force when they are near the stiffness at which the chord switches from buckling on them to buckling between them, since that is where they are only just enough.

What decides it: For a 24 m chord at 930 kN with frames every 4 m, the largest frame force over every bow is least near that stiffness — 1.38 per cent of the chord force at 1,893 N/mm — and rises to 5.55 per cent for frames just stiff enough to hold the chord at 1.13 times its load. Very stiff frames approach 1.64 per cent.

Tested in What the frames hold is the bow, at the figure it turns on · the essays about continuous restraint.